Research ArticleA Singular Sturm-Liouville Problem with Limit Circle Endpointsand Eigenparameter Dependent Boundary Conditions
Jinming Cai and Zhaowen Zheng
School of Mathematical Sciences Qufu Normal University Qufu Shandong 273165 China
Correspondence should be addressed to Zhaowen Zheng zhwzheng126com
Received 5 December 2016 Accepted 23 January 2017 Published 21 March 2017
Academic Editor Chris Goodrich
Copyright copy 2017 Jinming Cai and Zhaowen Zheng This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited
In this paper we investigate a class of discontinuous singular Sturm-Liouville problems with limit circle endpoints andeigenparameter dependent boundary conditions Operator formulation is constructed and asymptotic formulas for eigenvaluesand fundamental solutions are given Moreover the completeness of eigenfunctions is discussed
1 Introduction
It is well known that many topics in mathematical physicsrequire the investigation of eigenvalues and eigenfunctions ofthe Sturm-Liouville problems The theory of regular Sturm-Liouville problems is well built since the foundation work ofWeyl on limit-pointlimit-circle classification [1] the singularSturm-Liouville problems (see [2ndash7] for real coefficients and[8] for complex coefficients) and more general Hamiltoniansystems (see [9 10]) are widely researchedMeanwhile a largenumber of researchers are interested in the discontinuousSturm-Liouville problem with inner discontinuous pointssince these problems are of wide applications in engineeringand mechanics (see [11ndash25]) Various physics applications ofthis kind of problems are found such as oscillation of linear ornonlinear equation (see [26ndash29]) and heat and mass transferproblems (see [30])
The regular Sturm-Liouville problems with transmissionconditions containing an eigenparameter on one of theboundary conditions have received a lot of attention inresearch (see [18ndash22]) Based on these results some research-ers studied the regular Sturm-Liouville problems with eigen-parameter on both of the boundary conditions (see [23ndash25]) In these papers Yang and Wang in [18] considered aSturm-Liouville problem with discontinuities at two pointsand eigenparameter dependent boundary condition at oneendpoint they obtained the fundamental solutions and gavethe asymptotic formulas of eigenvalues and fundamental
solutions Further they studied the discontinuous Sturm-Liouville problem with eigenparameter boundary conditionsat two endpoints in [24] and extended the results of [18]to finite discontinuities case In papers [19 22 23 25] theauthors obtained the estimations of eigenvalues and eigen-functions of the discontinuous Sturm-Liouville problemwithone inner point containing an eigenparameter in the bound-ary condition Sen et al considered the Sturm-Liouville prob-lem with two inner points containing an eigenparameter inthe boundary condition and got similar result respectively(see [20 21]) Besides the authors also discussed the com-pleteness of the eigenfunctions of a regular discontinuousSturm-Liouville problem in papers [18 25] All of themresearched the regular Sturm-Liouville problem Howeverlittle is known about the singular Sturm-Liouville problemswith limit-circle endpoints
We will consider the following singular discontinuousSturm-Liouville problem with two limit-circle endpoints andeigenparameter in the boundary conditions119871119910 fl minus (119901 (119909) 1199101015840 (119909))1015840 + 119902 (119909) 119910 (119909) = 120582119910 (119909) 119909 isin (119886 120585) cup (120585 119887) minusinfin lt 119886 lt 120585 lt 119887 lt infin (1)
with eigenparameter dependent conditions at the endpoints119886 and 119887 1198711119910 fl 1198861 [119910 119906] (119886) minus 1198862 [119910 V] (119886)= 120582 (1198871 [119910 119906] (119886) minus 1198872 [119910 V] (119886)) (2)
HindawiDiscrete Dynamics in Nature and SocietyVolume 2017 Article ID 9673846 12 pageshttpsdoiorg10115520179673846
the transmission conditions at 119909 = 120585 are1198713119910 fl 119910 (120585 + 0) minus 1205751119910 (120585 minus 0) minus 12057521199101015840 (120585 minus 0) = 01198714119910 fl 1199101015840 (120585 + 0) minus 1205753119910 (120585 minus 0) minus 12057541199101015840 (120585 minus 0) = 0 (4)
(we assume that not all 120585119895 (1 le 119895 le 4) are equal to zerosince in this case the boundary value problem (1)ndash(3) hasno transmission conditions) where both 119886 and 119887 are limit-circle points 1199061(119909) and V1(119909) are linearly independent real-valued solutions of equation minus(119901(119909)1199101015840(119909))1015840 + 119902(119909)119910(119909) = 0on (119886 120585) and 1199062(119909) and V2(119909) are linearly independent real-valued solutions of equation minus(119901(119909)1199101015840(119909))1015840 + 119902(119909)119910(119909) = 0on (120585 119887) and satisfy [1199061 V1](119886) = 0 [1199062 V2](119887) = 0 where[119906 V] = 119901(119906V1015840 minus 1199061015840V) is the sesquilinear form 119901(119909) = 111990121for 119909 isin (119886 120585) 119901(119909) = 111990122 for 119909 isin (120585 119887) and 120582 isin C is aspectral parameter 119902(119909) is a real-valued continuous functionon (119886 120585)cup(120585 119887) and has finite limits 119902(120585plusmn0) = lim119909rarr(120585plusmn0)119902(119909)119901119894 119886119894 119887119894 119888119894 119889119894 (119894 = 1 2) and 120575119894 (119894 = 1 2 3 4) are nonzero realnumbers
Here we research a singular Sturm-Liouville problemwith two limit-circle endpoints and the parameter 120582 is notonly in the equation but also in the boundary conditionsBased on the modified inner product we define a newself-adjoint operator 119860 such that the eigenvalues of such aproblem are coincided with those of 119860 We rebuild its funda-mental solutions get the asymptotic formulas for eigenvaluesand eigenfunctions and also discuss the completeness of itseigenfunctions
At first we introduce the following lemmas
Lemma 1 (the Lagrange identity see [5]) Let119863(119871) = 119910 | 119910 isin1198712[119886 119887] 1199101015840 isin 119860119862[119886 119887] 119871119910 isin 1198712[119886 119887] where for any interval
119868 sub R 1198712(119868) denotes the set of functions on 119868 which are squareintegrable on 119868 For any 119910 isin 119863(119871) 119911 isin 119863(119871lowast) one has119911119871119910 minus 119910119871lowast119911 = 119889119889119909 [119910 119911] (119909) (7)
where 119871lowast is the adjoint expression of 119871 which is given by 119871lowast119911 =minus(1199011199111015840)1015840 + 119902119911Lemma 2 (Greenrsquos formula see [5]) Let 119868 = [119886 119887] forarbitrary 119910 isin 119863(119871) 119911 isin 119863(119871lowast) 119863(119871) and 119863(119871lowast) are definedas above one has(119871119910 119911) minus (119910 119871lowast119911) = int119887
119886(119911119871119910 minus 119910119871lowast119911) 119889119909= [119910 119911] (119887) minus [119910 119911] (119886) (8)
Since 119871119910 is of limit-circle type at 119909 = 119886int1199091119886
119906119871119910 119889119909 int1199091119886
119910119871lowast119906 119889119909 and [119910 119906](1199091) exist forall119909 isin (119886 1199091)So [119910 119906](119886) = lim119909rarr119886+[119910 119906](119909) exists Similarly [119910 V](119886)[119910 119906](119887) and [119910 V](119887) existLemma 3 (see [5]) For any 119891 119892 120572 120573 isin 119863(119871) while 119863(119871) isdefined in Lemma 1 one has1003816100381610038161003816100381610038161003816100381610038161003816[119891 120572] [119891 120573][119892 120572] [119892 120573]1003816100381610038161003816100381610038161003816100381610038161003816 = [119891 119892] [120572 120573] (9)
2 Operator Formulation
In this section we introduce the Hilbert space 119867 = 1198712(119886 120585) oplus1198712(120585 119887) oplus C2 the inner product on 119867 is defined by⟨119865 119866⟩ = 12057411990121 int120585119886
with119865 = (119891 119879119886(119891) 119879119887(119891)) isin 119863(119860)Nowwe can rewrite prob-lem (1)ndash(4) in the operator form 119860119865 = 120582119865 for 119865 = (119891 119879119886(119891)119879119887(119891)) isin 119863(119860) Obviously we have the following lemmas
Lemma 4 The eigenvalues and eigenfunctions of problem(1)ndash(4) are corresponding to the eigenvalues and the firstcomponent of the corresponding eigenfunctions of operator 119860respectively
Lemma 5 The domain 119863(119860) is dense in 119867
Proof Let 119865 = (119891(119909) 1198911 1198912) isin 119867 119865 perp 119863(119860) and let 119862infin0 be afunctional set which has compact support and can be differ-ential infinitely such that 1205991(119909) isin 119862infin0 (119886 120585) 1205992(119909) isin 119862infin0 (120585 119887)Since 119862infin0 oplus 0 oplus 0 sub 119863(119860) for forall119866 = (119892(119909) 0 0) isin 119862infin0 oplus 0 oplus 0is orthogonal to 119865 namely
so 119891(119909) = 0 For all 119881 = (](119909) ]1 0) isin 119863(119860) ⟨119865 119881⟩ =(120574120588)1198911]1 = 0 1198911 = 0 since ]1 is arbitrary Similarly onegets 1198912 = 0 Above all one has 119865 = (0 0 0) which proves theassertion
Definition 6 Let 119882(119891 119892 119909) = 119891(119909)1198921015840(119909) minus 1198911015840(119909)119892(119909) denotethe Wronskian of functions 119891(119909) and 119892(119909) One has119882 (119891 119892 119886) = 11990121 [119891 119892] (119886) 119882 (119891 119892 119887) = 11990122 [119891 119892] (119887) (15)
Theorem 7 Operator 119860 is symmetric
Proof For each 119865 119866 isin 119863(119860) by the definition of inner pro-duct and operator 119860 we can get⟨119860119865 119866⟩ minus ⟨119865 119860119866⟩= 120574119882 (119891 119892 120585 minus 0) minus 119882 (119891 119892 120585 + 0) + 119882 (119891 119892 119887)minus 120574119882 (119891 119892 119886)+ 120574120588 [1198791015840119886 (119891) 119879119886 (119892) minus 119879119886 (119891) 1198791015840119886 (119892)]
then (4) implies that119882 (119891 119892 120585 + 0) = 120574119882 (119891 119892 120585 minus 0) (17)
using boundary condition (2) we get120574120588 [1198791015840119886 (119891) 119879119886 (119892) minus 119879119886 (119891) 1198791015840119886 (119892)] = 120574119882 (119891 119892 119886) (18)
and by boundary condition (3) we have1120579 [1198791015840119887 (119891) 119879119887 (119892) minus 119879119887 (119891) 1198791015840119887 (119892)] = 119882 (119891 119892 119887) (19)
Substituting (17)ndash(19) into (16) yields ⟨119860119865 119866⟩ minus ⟨119865 119860119866⟩ = 0This completes the proof
Moreover we have the following conclusion
Theorem 8 Operator 119860 is self-adjoint
Proof 119860 is self-adjoint if and only if for each 119865 =(119891(119909) 119879119886(119891) 119879119887(119891)) isin 119863(119860) ⟨119860119865 119882⟩ = ⟨119865 119880⟩ for some119880 isin 119867 implying that 119882 isin 119863(119860) and 119860119882 = 119880 where119882 = (119908(119909) ℎ 119903) 119880 = (120583(119909) 119896 119904) Concretely we shouldprove that the following properties hold
(i) 119908(119909) 1199081015840(119909) are absolutely continuous on (119886 120585) cup(120585 119887) 119871119908 isin 1198712((119886 120585) cup (120585 119887))(ii) ℎ = 1198871[119908 119906](119886) minus 1198872[119908 V](119886) 119903 = 1198891[119908 119906](119887) minus 1198892[119908
that is ⟨119871119891 119908⟩1 = ⟨119891 120583⟩1 According to the classical Sturm-Liouville theory (i) and (iv) hold So ⟨119860119865 119882⟩ = ⟨119865 119880⟩ forall119865 isin119863(119860) ⟨119871119891 119908⟩1 + 1205741205881198791015840119886 (119891) ℎ + 11205791198791015840119887 (119891) 119903
Substituting (24) into (23) we have ℎ = 1198871[119908 119906](119886) minus1198872[119908 V](119886) Further there exists 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = 119891 (120585 plusmn 0) = 1198911015840 (120585 plusmn 0) = 0[119891 119906] (119887) = 1198892[119891 V] (119887) = 1198891 (25)
Analogously we can get 119903 = 1198891[119908 119906](119887) minus 1198892[119908 V](119887) So (ii)holds Similarly one proves (v) Next let 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = [119891 119906] (119887) = [119891 V] (119887)= 119891 (120585 + 0) = 0119891 (120585 minus 0) = minus12057521198911015840 (120585 minus 0) = 12057511198911015840 (120585 + 0) = 120574
(26)
Then by (23) we have 1198713119908 = 0 Similarly we can get 1198714119908 =0 So 119860 is a self-adjoint operator
From the properties of self-adjoint operators we have thefollowing corollaries
Corollary 9 All eigenvalues of the singular Sturm-Liouvilleproblem (1)ndash(4) are real
Corollary 10 Let 1205821 and 1205822 be two different eigenvalues ofthe singular Sturm-Liouville problem (1)ndash(4) then the corre-sponding eigenfunctions 119891(119909) and 119892(119909) are orthogonal in thesense of12057411990121 int120585
In this section we construct the fundamental solutions ofproblem (1)ndash(4) and get the asymptotic approximation forfundamental solutions
Lemma 11 (see [5]) Let the real-valued function 119902(119909) becontinuous on 119868 = (119886 120585) cup (120585 119887) and let 119891(120582) 119892(120582) be givenentire functions Then for any 120582 isin C the equation119871119910 fl minus (119901 (119909) 1199101015840 (119909))1015840 + 119902 (119909) 119910 (119909) = 120582119910 (119909) 119909 isin (119886 120585) cup (120585 119887) (28)
Let 1205931(119909 120582) be the solution of (1) on the interval (119886 120585)which satisfies the initial conditions1205931 (119886 120582) fl [119910 119906] (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) fl [119910 V] (119886 120582) = 1198861 minus 1205821198871 (31)
by virtue of Lemma 11 we can define the solution 1205932(119909 120582) of(1) on (120585 119887) by the initial conditions
(1205932 (120585 + 0)12059310158402 (120585 + 0)) = (12057511205931 (120585 minus 0) + 120575212059310158401 (120585 minus 0)12057531205931 (120585 minus 0) + 120575412059310158401 (120585 minus 0)) (32)
Analogously we define the solutions 1205942(119909 120582) and 1205941(119909 120582) of(1) by the initial conditions1205942 (119887 120582) fl [119910 119906] (119887 120582) = 1198882 minus 120582119889212059410158402 (119887 120582) fl [119910 V] (119887 120582) = 1198881 minus 1205821198891
(1205941 (120585 minus 0)12059410158401 (120585 minus 0)) = (12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)120574 ) (33)
Now we consider Wronskian119882119894 (120582) fl 119882 (120593119894 120594119894 119909)= 120593119894 (119909 120582) 1205941015840119894 (119909 120582) minus 1205931015840119894 (119909 120582) 120594119894 (119909 120582)(119894 = 1 2) (34)
By the dependence of solutions of initial value problemson the parameter one has that 119882119894(120582) (119894 = 1 2) are entirefunctions of 120582 and are independent of 119909Lemma 12 For every 120582 isin C 1198822(120582) = 1205741198821(120582)
Discrete Dynamics in Nature and Society 5
Proof By the definition of 119882119894(120582) we have1198821 (120582) = 1205931 (120585 minus 0 120582) 12059410158401 (120585 minus 0 120582)minus 12059310158401 (120585 minus 0 120582) 1205941 (120585 minus 0 120582)
using the transmission conditions (4) simple computationgives 119882 (1205932 1205942 120585 + 0) = 120574119882 (1205931 1205941 120585 minus 0) (36)
Thus for each 120582 isin C we have 1198822(120582) = 1205741198821(120582) This com-pletes the proof
Besides we set 119882(120582) fl 1198821(120582) = (1120574)1199082(120582)Theorem 13 The eigenvalues of problem (1)ndash(4) coincide withthe zeros of the function 119882(120582)Proof Let ]0(119909 1205820) be any eigenfunction corresponding toeigenvalue 1205820 then the function ]0(119909 1205820)may be representedin the form
where at least one of the constants 119898119894 (119894 = 1 2 3 4) is notzero We should show that 119882(1205820) = 0 Suppose to thecontrary that there exists 1205820 isin 119877 such that119882(1205820) = 1198821(1205820) =(1120574)1198822(1205820) = 0 Since eigenfunction ]0(119909 1205820) satisfiesboth boundary and transmission conditions (2)ndash(4) we have119871 119894]0(119909 1205820) = 0 (119894 = 1 2 3 4) while the determinant ofcoefficient matrix is not zero so we obtain 119898119894 = 0 (119894 =1 2 3 4) which is a contradiction then 119882(1205820) = 0Conversely let 120582 = 1205820 be a zero of function 119882(120582) then119882(1205820) = 1198821(1205820) = (1120574)1198822(1205820) = 0 therefore 120594119894(119909 1205820) =119896120593119894(119909 1205820) (119894 = 1 2) for some 119896 = 0 Since both 1205932(119909 1205820) and1205942(119909 1205820) satisfy the boundary condition (3)
satisfies problem (1)ndash(4) So function 120593(119909 1205820) is an eigen-function of problem (1)ndash(4) corresponding to eigenvalue 1205820This completes the proof
Theorem 14 The eigenvalues of problem (1)ndash(4) are analyti-cally single
Proof Let 120582 = 119904 + 119894119905 and we use the following notions forsimplicity 120593 = 120593(119909 120582) 1205931120582 = 1205971205931120597120582 and 12059310158401120582 = 12059712059310158401120597120582
We differentiate the equation 119860120594 = 120582120594 with respect to 120582 toobtain 119860120594120582 = 120594 + 120582120594120582 (39)
Using integration by parts we get⟨119860120594120582 120593⟩1 minus ⟨120594120582 119860120593⟩1= 120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585 (40)
Substituting (39) and 119860120593 = 120582120593 into the left side of (40) wehave ⟨120582120594120582 120593⟩1 minus ⟨120594120582 120582120593⟩1 = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1 (41)
Moreover
120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585= 11988911205932 (119887) minus 119889212059310158402 (119887)minus 120574 [(1198861 minus 1205821198871) 1205941120582 (119886) minus (1198862 minus 1205821198872) 12059410158401120582 (119886)]
(42)
By (31) we observe that1198821015840 (120582) = 1205931 (119886 120582) 12059410158401 (119886 120582) minus 12059310158401 (119886 120582) 1205941 (119886 120582)= 11988711205941 (119886) minus 119887212059410158401 (119886) + (1198862 minus 1205821198872) 12059410158401120582 (119886)minus (1198861 minus 1205821198871) 1205941120582 (119886)
(43)
then (40) becomes1205741198821015840 (120582) = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1+ 120574 (11988711205941 (119886) minus 119887212059410158401 (119886))minus (11988911205932 (119887) minus 119889212059310158402 (119887))
(44)
Next let 1205830 be an arbitrary zero of119882(120582) Since119882(1205830) = 0 weobtain 120593119894(119909 1205830) = 119896120594119894(119909 1205830) (119894 = 1 2) (119896 = 0) 119896 isin R Notingthat 1205830 is real a short calculation (44) becomes1205741198821015840 (1205830)
Since 120588 gt 0 120579 gt 0 120574 gt 0 and 119896 = 0 1198821015840(1205830) = 0 Hence theanalytic multiplicity of 1205830 is one which completes the proof
6 Discrete Dynamics in Nature and Society
Lemma 15 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205931 (119909 120582)
= (1198862 minus 11990421198872) 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591(46)
1198891198961198891199091198961205932 (119909 120582)= 1205932 (120585 + 0) 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
+ 1119901211990412059310158402 (120585 + 0) 119889119896119889119909119896 sin [1199012119904 (119909 minus 120585)]+ 1199012119904 int119909
120585
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205932 (120591) 119889120591(47)
Proof For the case of 119896 = 0 consider 1205931(119909 120582) as the solutionof the following nonhomogeneous problem
minus (119901 (119909) 119910 (119909)1015840)1015840 + 120582119910 (119909) = 119902 (119909) 119910 (119909)1205931 (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) = 1198861 minus 1205821198871 (48)
Using the method of constant variation 1205931(119909 120582) satisfies1205931 (119909 120582) = (1198862 minus 11990421198872) cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591
(49)
Then differentiating it with respect to 119909 we have (46) Theproof for (47) is similar so we omit the details
Similarly we have the following theorem
Lemma 16 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205941 (119909 120582) = 12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574 119889119896119889119909119896
sdot cos [1199011119904 (119909 minus 120585)] + 120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)1199011119904120574
sdot 119889119896119889119909119896 sin [1199011119904 (119909 minus 120585)] minus 1199011119904sdot int120585119909
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205941 (120591) 1198891205911198891198961198891199091198961205942 (119909 120582) = (1198882 minus 11990421198892) 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
+ 1198881 minus 119904211988911199012119904 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] minus 1199012119904sdot int119887119909
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205942 (120591) 119889120591(50)
Lemma 17 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120593(119909 120582)have the following estimations
Case 1 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11990421198872 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]
+ 119874 (|119904|119896+1 119890|119905|1199011(119909minus119886)) 1198891198961198891199091198961205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 119886)]
Case 2 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11988711199041199011 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)] + 119874 (|119904|119896 119890|119905|1199011(119909minus119886))
1198891198961198891199091198961205932 (119909 120582)= minus120575211988711199042 cos [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Proof The proof of formulas for 1205931(119909 120582) is identical to thoseof Titchmarshrsquos proof for 120593(119909 120582) (see [6]) so we only give theproof of formulas for 1205932(119909 120582) namely equality (51) the otherequalities are similar
Discrete Dynamics in Nature and Society 7
For 119896 = 0 by the estimations of 1205931(119909 120582) and 12059310158401(119909 120582) wehave 1205931 (120585 minus 0 120582) = minus11990421198872 cos [1199011119904 (120585 minus 119886)]+ 119874 (|119904| 119890|119905|1199011(120585minus119886)) 12059310158401 (120585 minus 0 120582) = 119901111990431198872 sin [1199011119904 (120585 minus 119886)]+ 119874 (|119904|2 119890|119905|1199011(120585minus119886))
(53)
then substituting (53) into (47) and observing (32) we have1205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119909 minus 119886)]+ 119874 (|119904|2 119890|119905|(1199011(120585minus119886)+1199012(119909minus120585))) (54)
differentiating (54) with respect to 119909 we have (51)Lemma 18 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120594(119909 120582)have the following estimations
Case 1 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus1199012119889212057521199043120574 sin [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+2 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988921199042 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
Case 2 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus119889112057521199042120574 cos [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+1 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988911199041199012 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] + 119874 (|119904|119896 119890|119905|1199012(119909minus119887))
(56)
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Theorem 19 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the function 1198822(120582)has the following asymptotic representations
Case 1 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= 120575211990111199012119887211988921199046 sin [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|5 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (57)
Case 2 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199011119887211988911199045 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (58)
Case 3 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199012119887111988921199045 cos [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (59)
Case 4 If 1198872 = 0 and 1198892 = 0 then1198822 (120582) = 1205752119887111988911199044 cos [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|3 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (60)
Proof By the definition of 1198822(120582) we have1198822 (120582) = 1205932 (119887 120582) 12059410158402 (119887 120582) minus 12059310158402 (119887 120582) 1205942 (119887 120582)= (1198881 minus 1205821198891) 1205932 (119887 120582) minus (1198882 minus 1205821198892) 12059310158402 (119887 120582) (61)
According to the equalities of 1205932(119909 120582) and 12059310158402(119909 120582) inLemma 17 we can obtain the formulas of 1198822(120582) in thistheorem
Corollary 20 The eigenvalues of the boundary value problem(1)ndash(4) are bounded below
Proof Putting 119904 = 119894119905 (119905 gt 0) in Theorem 19 we can obtain1198822(120582) = 1198822(minus1199052) rarr infin (119905 rarr infin) Hence 1198822(minus1199052) = 0 forsufficiently negative 120582 and sufficiently large 120582 This completesthe proof
4 Asymptotic Formulas forEigenvalues and Eigenfunctions
In this section we can get the asymptotic formulas forthe eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)ndash(4) Since the eigenvalues coincidewiththe zeros of the entire function119882(120582) it follows that they haveno finite limits
Theorem 21 The eigenvalues 120582119899 = 1199042119899 (119899 = 0 1 2 ) ofproblem (1)ndash(4) have the following asymptotic representationsas 119899 rarr infin
8 Discrete Dynamics in Nature and Society
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
the transmission conditions at 119909 = 120585 are1198713119910 fl 119910 (120585 + 0) minus 1205751119910 (120585 minus 0) minus 12057521199101015840 (120585 minus 0) = 01198714119910 fl 1199101015840 (120585 + 0) minus 1205753119910 (120585 minus 0) minus 12057541199101015840 (120585 minus 0) = 0 (4)
(we assume that not all 120585119895 (1 le 119895 le 4) are equal to zerosince in this case the boundary value problem (1)ndash(3) hasno transmission conditions) where both 119886 and 119887 are limit-circle points 1199061(119909) and V1(119909) are linearly independent real-valued solutions of equation minus(119901(119909)1199101015840(119909))1015840 + 119902(119909)119910(119909) = 0on (119886 120585) and 1199062(119909) and V2(119909) are linearly independent real-valued solutions of equation minus(119901(119909)1199101015840(119909))1015840 + 119902(119909)119910(119909) = 0on (120585 119887) and satisfy [1199061 V1](119886) = 0 [1199062 V2](119887) = 0 where[119906 V] = 119901(119906V1015840 minus 1199061015840V) is the sesquilinear form 119901(119909) = 111990121for 119909 isin (119886 120585) 119901(119909) = 111990122 for 119909 isin (120585 119887) and 120582 isin C is aspectral parameter 119902(119909) is a real-valued continuous functionon (119886 120585)cup(120585 119887) and has finite limits 119902(120585plusmn0) = lim119909rarr(120585plusmn0)119902(119909)119901119894 119886119894 119887119894 119888119894 119889119894 (119894 = 1 2) and 120575119894 (119894 = 1 2 3 4) are nonzero realnumbers
Here we research a singular Sturm-Liouville problemwith two limit-circle endpoints and the parameter 120582 is notonly in the equation but also in the boundary conditionsBased on the modified inner product we define a newself-adjoint operator 119860 such that the eigenvalues of such aproblem are coincided with those of 119860 We rebuild its funda-mental solutions get the asymptotic formulas for eigenvaluesand eigenfunctions and also discuss the completeness of itseigenfunctions
At first we introduce the following lemmas
Lemma 1 (the Lagrange identity see [5]) Let119863(119871) = 119910 | 119910 isin1198712[119886 119887] 1199101015840 isin 119860119862[119886 119887] 119871119910 isin 1198712[119886 119887] where for any interval
119868 sub R 1198712(119868) denotes the set of functions on 119868 which are squareintegrable on 119868 For any 119910 isin 119863(119871) 119911 isin 119863(119871lowast) one has119911119871119910 minus 119910119871lowast119911 = 119889119889119909 [119910 119911] (119909) (7)
where 119871lowast is the adjoint expression of 119871 which is given by 119871lowast119911 =minus(1199011199111015840)1015840 + 119902119911Lemma 2 (Greenrsquos formula see [5]) Let 119868 = [119886 119887] forarbitrary 119910 isin 119863(119871) 119911 isin 119863(119871lowast) 119863(119871) and 119863(119871lowast) are definedas above one has(119871119910 119911) minus (119910 119871lowast119911) = int119887
119886(119911119871119910 minus 119910119871lowast119911) 119889119909= [119910 119911] (119887) minus [119910 119911] (119886) (8)
Since 119871119910 is of limit-circle type at 119909 = 119886int1199091119886
119906119871119910 119889119909 int1199091119886
119910119871lowast119906 119889119909 and [119910 119906](1199091) exist forall119909 isin (119886 1199091)So [119910 119906](119886) = lim119909rarr119886+[119910 119906](119909) exists Similarly [119910 V](119886)[119910 119906](119887) and [119910 V](119887) existLemma 3 (see [5]) For any 119891 119892 120572 120573 isin 119863(119871) while 119863(119871) isdefined in Lemma 1 one has1003816100381610038161003816100381610038161003816100381610038161003816[119891 120572] [119891 120573][119892 120572] [119892 120573]1003816100381610038161003816100381610038161003816100381610038161003816 = [119891 119892] [120572 120573] (9)
2 Operator Formulation
In this section we introduce the Hilbert space 119867 = 1198712(119886 120585) oplus1198712(120585 119887) oplus C2 the inner product on 119867 is defined by⟨119865 119866⟩ = 12057411990121 int120585119886
with119865 = (119891 119879119886(119891) 119879119887(119891)) isin 119863(119860)Nowwe can rewrite prob-lem (1)ndash(4) in the operator form 119860119865 = 120582119865 for 119865 = (119891 119879119886(119891)119879119887(119891)) isin 119863(119860) Obviously we have the following lemmas
Lemma 4 The eigenvalues and eigenfunctions of problem(1)ndash(4) are corresponding to the eigenvalues and the firstcomponent of the corresponding eigenfunctions of operator 119860respectively
Lemma 5 The domain 119863(119860) is dense in 119867
Proof Let 119865 = (119891(119909) 1198911 1198912) isin 119867 119865 perp 119863(119860) and let 119862infin0 be afunctional set which has compact support and can be differ-ential infinitely such that 1205991(119909) isin 119862infin0 (119886 120585) 1205992(119909) isin 119862infin0 (120585 119887)Since 119862infin0 oplus 0 oplus 0 sub 119863(119860) for forall119866 = (119892(119909) 0 0) isin 119862infin0 oplus 0 oplus 0is orthogonal to 119865 namely
so 119891(119909) = 0 For all 119881 = (](119909) ]1 0) isin 119863(119860) ⟨119865 119881⟩ =(120574120588)1198911]1 = 0 1198911 = 0 since ]1 is arbitrary Similarly onegets 1198912 = 0 Above all one has 119865 = (0 0 0) which proves theassertion
Definition 6 Let 119882(119891 119892 119909) = 119891(119909)1198921015840(119909) minus 1198911015840(119909)119892(119909) denotethe Wronskian of functions 119891(119909) and 119892(119909) One has119882 (119891 119892 119886) = 11990121 [119891 119892] (119886) 119882 (119891 119892 119887) = 11990122 [119891 119892] (119887) (15)
Theorem 7 Operator 119860 is symmetric
Proof For each 119865 119866 isin 119863(119860) by the definition of inner pro-duct and operator 119860 we can get⟨119860119865 119866⟩ minus ⟨119865 119860119866⟩= 120574119882 (119891 119892 120585 minus 0) minus 119882 (119891 119892 120585 + 0) + 119882 (119891 119892 119887)minus 120574119882 (119891 119892 119886)+ 120574120588 [1198791015840119886 (119891) 119879119886 (119892) minus 119879119886 (119891) 1198791015840119886 (119892)]
then (4) implies that119882 (119891 119892 120585 + 0) = 120574119882 (119891 119892 120585 minus 0) (17)
using boundary condition (2) we get120574120588 [1198791015840119886 (119891) 119879119886 (119892) minus 119879119886 (119891) 1198791015840119886 (119892)] = 120574119882 (119891 119892 119886) (18)
and by boundary condition (3) we have1120579 [1198791015840119887 (119891) 119879119887 (119892) minus 119879119887 (119891) 1198791015840119887 (119892)] = 119882 (119891 119892 119887) (19)
Substituting (17)ndash(19) into (16) yields ⟨119860119865 119866⟩ minus ⟨119865 119860119866⟩ = 0This completes the proof
Moreover we have the following conclusion
Theorem 8 Operator 119860 is self-adjoint
Proof 119860 is self-adjoint if and only if for each 119865 =(119891(119909) 119879119886(119891) 119879119887(119891)) isin 119863(119860) ⟨119860119865 119882⟩ = ⟨119865 119880⟩ for some119880 isin 119867 implying that 119882 isin 119863(119860) and 119860119882 = 119880 where119882 = (119908(119909) ℎ 119903) 119880 = (120583(119909) 119896 119904) Concretely we shouldprove that the following properties hold
(i) 119908(119909) 1199081015840(119909) are absolutely continuous on (119886 120585) cup(120585 119887) 119871119908 isin 1198712((119886 120585) cup (120585 119887))(ii) ℎ = 1198871[119908 119906](119886) minus 1198872[119908 V](119886) 119903 = 1198891[119908 119906](119887) minus 1198892[119908
that is ⟨119871119891 119908⟩1 = ⟨119891 120583⟩1 According to the classical Sturm-Liouville theory (i) and (iv) hold So ⟨119860119865 119882⟩ = ⟨119865 119880⟩ forall119865 isin119863(119860) ⟨119871119891 119908⟩1 + 1205741205881198791015840119886 (119891) ℎ + 11205791198791015840119887 (119891) 119903
Substituting (24) into (23) we have ℎ = 1198871[119908 119906](119886) minus1198872[119908 V](119886) Further there exists 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = 119891 (120585 plusmn 0) = 1198911015840 (120585 plusmn 0) = 0[119891 119906] (119887) = 1198892[119891 V] (119887) = 1198891 (25)
Analogously we can get 119903 = 1198891[119908 119906](119887) minus 1198892[119908 V](119887) So (ii)holds Similarly one proves (v) Next let 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = [119891 119906] (119887) = [119891 V] (119887)= 119891 (120585 + 0) = 0119891 (120585 minus 0) = minus12057521198911015840 (120585 minus 0) = 12057511198911015840 (120585 + 0) = 120574
(26)
Then by (23) we have 1198713119908 = 0 Similarly we can get 1198714119908 =0 So 119860 is a self-adjoint operator
From the properties of self-adjoint operators we have thefollowing corollaries
Corollary 9 All eigenvalues of the singular Sturm-Liouvilleproblem (1)ndash(4) are real
Corollary 10 Let 1205821 and 1205822 be two different eigenvalues ofthe singular Sturm-Liouville problem (1)ndash(4) then the corre-sponding eigenfunctions 119891(119909) and 119892(119909) are orthogonal in thesense of12057411990121 int120585
In this section we construct the fundamental solutions ofproblem (1)ndash(4) and get the asymptotic approximation forfundamental solutions
Lemma 11 (see [5]) Let the real-valued function 119902(119909) becontinuous on 119868 = (119886 120585) cup (120585 119887) and let 119891(120582) 119892(120582) be givenentire functions Then for any 120582 isin C the equation119871119910 fl minus (119901 (119909) 1199101015840 (119909))1015840 + 119902 (119909) 119910 (119909) = 120582119910 (119909) 119909 isin (119886 120585) cup (120585 119887) (28)
Let 1205931(119909 120582) be the solution of (1) on the interval (119886 120585)which satisfies the initial conditions1205931 (119886 120582) fl [119910 119906] (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) fl [119910 V] (119886 120582) = 1198861 minus 1205821198871 (31)
by virtue of Lemma 11 we can define the solution 1205932(119909 120582) of(1) on (120585 119887) by the initial conditions
(1205932 (120585 + 0)12059310158402 (120585 + 0)) = (12057511205931 (120585 minus 0) + 120575212059310158401 (120585 minus 0)12057531205931 (120585 minus 0) + 120575412059310158401 (120585 minus 0)) (32)
Analogously we define the solutions 1205942(119909 120582) and 1205941(119909 120582) of(1) by the initial conditions1205942 (119887 120582) fl [119910 119906] (119887 120582) = 1198882 minus 120582119889212059410158402 (119887 120582) fl [119910 V] (119887 120582) = 1198881 minus 1205821198891
(1205941 (120585 minus 0)12059410158401 (120585 minus 0)) = (12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)120574 ) (33)
Now we consider Wronskian119882119894 (120582) fl 119882 (120593119894 120594119894 119909)= 120593119894 (119909 120582) 1205941015840119894 (119909 120582) minus 1205931015840119894 (119909 120582) 120594119894 (119909 120582)(119894 = 1 2) (34)
By the dependence of solutions of initial value problemson the parameter one has that 119882119894(120582) (119894 = 1 2) are entirefunctions of 120582 and are independent of 119909Lemma 12 For every 120582 isin C 1198822(120582) = 1205741198821(120582)
Discrete Dynamics in Nature and Society 5
Proof By the definition of 119882119894(120582) we have1198821 (120582) = 1205931 (120585 minus 0 120582) 12059410158401 (120585 minus 0 120582)minus 12059310158401 (120585 minus 0 120582) 1205941 (120585 minus 0 120582)
using the transmission conditions (4) simple computationgives 119882 (1205932 1205942 120585 + 0) = 120574119882 (1205931 1205941 120585 minus 0) (36)
Thus for each 120582 isin C we have 1198822(120582) = 1205741198821(120582) This com-pletes the proof
Besides we set 119882(120582) fl 1198821(120582) = (1120574)1199082(120582)Theorem 13 The eigenvalues of problem (1)ndash(4) coincide withthe zeros of the function 119882(120582)Proof Let ]0(119909 1205820) be any eigenfunction corresponding toeigenvalue 1205820 then the function ]0(119909 1205820)may be representedin the form
where at least one of the constants 119898119894 (119894 = 1 2 3 4) is notzero We should show that 119882(1205820) = 0 Suppose to thecontrary that there exists 1205820 isin 119877 such that119882(1205820) = 1198821(1205820) =(1120574)1198822(1205820) = 0 Since eigenfunction ]0(119909 1205820) satisfiesboth boundary and transmission conditions (2)ndash(4) we have119871 119894]0(119909 1205820) = 0 (119894 = 1 2 3 4) while the determinant ofcoefficient matrix is not zero so we obtain 119898119894 = 0 (119894 =1 2 3 4) which is a contradiction then 119882(1205820) = 0Conversely let 120582 = 1205820 be a zero of function 119882(120582) then119882(1205820) = 1198821(1205820) = (1120574)1198822(1205820) = 0 therefore 120594119894(119909 1205820) =119896120593119894(119909 1205820) (119894 = 1 2) for some 119896 = 0 Since both 1205932(119909 1205820) and1205942(119909 1205820) satisfy the boundary condition (3)
satisfies problem (1)ndash(4) So function 120593(119909 1205820) is an eigen-function of problem (1)ndash(4) corresponding to eigenvalue 1205820This completes the proof
Theorem 14 The eigenvalues of problem (1)ndash(4) are analyti-cally single
Proof Let 120582 = 119904 + 119894119905 and we use the following notions forsimplicity 120593 = 120593(119909 120582) 1205931120582 = 1205971205931120597120582 and 12059310158401120582 = 12059712059310158401120597120582
We differentiate the equation 119860120594 = 120582120594 with respect to 120582 toobtain 119860120594120582 = 120594 + 120582120594120582 (39)
Using integration by parts we get⟨119860120594120582 120593⟩1 minus ⟨120594120582 119860120593⟩1= 120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585 (40)
Substituting (39) and 119860120593 = 120582120593 into the left side of (40) wehave ⟨120582120594120582 120593⟩1 minus ⟨120594120582 120582120593⟩1 = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1 (41)
Moreover
120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585= 11988911205932 (119887) minus 119889212059310158402 (119887)minus 120574 [(1198861 minus 1205821198871) 1205941120582 (119886) minus (1198862 minus 1205821198872) 12059410158401120582 (119886)]
(42)
By (31) we observe that1198821015840 (120582) = 1205931 (119886 120582) 12059410158401 (119886 120582) minus 12059310158401 (119886 120582) 1205941 (119886 120582)= 11988711205941 (119886) minus 119887212059410158401 (119886) + (1198862 minus 1205821198872) 12059410158401120582 (119886)minus (1198861 minus 1205821198871) 1205941120582 (119886)
(43)
then (40) becomes1205741198821015840 (120582) = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1+ 120574 (11988711205941 (119886) minus 119887212059410158401 (119886))minus (11988911205932 (119887) minus 119889212059310158402 (119887))
(44)
Next let 1205830 be an arbitrary zero of119882(120582) Since119882(1205830) = 0 weobtain 120593119894(119909 1205830) = 119896120594119894(119909 1205830) (119894 = 1 2) (119896 = 0) 119896 isin R Notingthat 1205830 is real a short calculation (44) becomes1205741198821015840 (1205830)
Since 120588 gt 0 120579 gt 0 120574 gt 0 and 119896 = 0 1198821015840(1205830) = 0 Hence theanalytic multiplicity of 1205830 is one which completes the proof
6 Discrete Dynamics in Nature and Society
Lemma 15 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205931 (119909 120582)
= (1198862 minus 11990421198872) 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591(46)
1198891198961198891199091198961205932 (119909 120582)= 1205932 (120585 + 0) 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
+ 1119901211990412059310158402 (120585 + 0) 119889119896119889119909119896 sin [1199012119904 (119909 minus 120585)]+ 1199012119904 int119909
120585
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205932 (120591) 119889120591(47)
Proof For the case of 119896 = 0 consider 1205931(119909 120582) as the solutionof the following nonhomogeneous problem
minus (119901 (119909) 119910 (119909)1015840)1015840 + 120582119910 (119909) = 119902 (119909) 119910 (119909)1205931 (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) = 1198861 minus 1205821198871 (48)
Using the method of constant variation 1205931(119909 120582) satisfies1205931 (119909 120582) = (1198862 minus 11990421198872) cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591
(49)
Then differentiating it with respect to 119909 we have (46) Theproof for (47) is similar so we omit the details
Similarly we have the following theorem
Lemma 16 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205941 (119909 120582) = 12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574 119889119896119889119909119896
sdot cos [1199011119904 (119909 minus 120585)] + 120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)1199011119904120574
sdot 119889119896119889119909119896 sin [1199011119904 (119909 minus 120585)] minus 1199011119904sdot int120585119909
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205941 (120591) 1198891205911198891198961198891199091198961205942 (119909 120582) = (1198882 minus 11990421198892) 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
+ 1198881 minus 119904211988911199012119904 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] minus 1199012119904sdot int119887119909
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205942 (120591) 119889120591(50)
Lemma 17 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120593(119909 120582)have the following estimations
Case 1 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11990421198872 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]
+ 119874 (|119904|119896+1 119890|119905|1199011(119909minus119886)) 1198891198961198891199091198961205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 119886)]
Case 2 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11988711199041199011 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)] + 119874 (|119904|119896 119890|119905|1199011(119909minus119886))
1198891198961198891199091198961205932 (119909 120582)= minus120575211988711199042 cos [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Proof The proof of formulas for 1205931(119909 120582) is identical to thoseof Titchmarshrsquos proof for 120593(119909 120582) (see [6]) so we only give theproof of formulas for 1205932(119909 120582) namely equality (51) the otherequalities are similar
Discrete Dynamics in Nature and Society 7
For 119896 = 0 by the estimations of 1205931(119909 120582) and 12059310158401(119909 120582) wehave 1205931 (120585 minus 0 120582) = minus11990421198872 cos [1199011119904 (120585 minus 119886)]+ 119874 (|119904| 119890|119905|1199011(120585minus119886)) 12059310158401 (120585 minus 0 120582) = 119901111990431198872 sin [1199011119904 (120585 minus 119886)]+ 119874 (|119904|2 119890|119905|1199011(120585minus119886))
(53)
then substituting (53) into (47) and observing (32) we have1205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119909 minus 119886)]+ 119874 (|119904|2 119890|119905|(1199011(120585minus119886)+1199012(119909minus120585))) (54)
differentiating (54) with respect to 119909 we have (51)Lemma 18 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120594(119909 120582)have the following estimations
Case 1 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus1199012119889212057521199043120574 sin [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+2 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988921199042 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
Case 2 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus119889112057521199042120574 cos [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+1 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988911199041199012 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] + 119874 (|119904|119896 119890|119905|1199012(119909minus119887))
(56)
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Theorem 19 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the function 1198822(120582)has the following asymptotic representations
Case 1 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= 120575211990111199012119887211988921199046 sin [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|5 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (57)
Case 2 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199011119887211988911199045 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (58)
Case 3 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199012119887111988921199045 cos [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (59)
Case 4 If 1198872 = 0 and 1198892 = 0 then1198822 (120582) = 1205752119887111988911199044 cos [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|3 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (60)
Proof By the definition of 1198822(120582) we have1198822 (120582) = 1205932 (119887 120582) 12059410158402 (119887 120582) minus 12059310158402 (119887 120582) 1205942 (119887 120582)= (1198881 minus 1205821198891) 1205932 (119887 120582) minus (1198882 minus 1205821198892) 12059310158402 (119887 120582) (61)
According to the equalities of 1205932(119909 120582) and 12059310158402(119909 120582) inLemma 17 we can obtain the formulas of 1198822(120582) in thistheorem
Corollary 20 The eigenvalues of the boundary value problem(1)ndash(4) are bounded below
Proof Putting 119904 = 119894119905 (119905 gt 0) in Theorem 19 we can obtain1198822(120582) = 1198822(minus1199052) rarr infin (119905 rarr infin) Hence 1198822(minus1199052) = 0 forsufficiently negative 120582 and sufficiently large 120582 This completesthe proof
4 Asymptotic Formulas forEigenvalues and Eigenfunctions
In this section we can get the asymptotic formulas forthe eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)ndash(4) Since the eigenvalues coincidewiththe zeros of the entire function119882(120582) it follows that they haveno finite limits
Theorem 21 The eigenvalues 120582119899 = 1199042119899 (119899 = 0 1 2 ) ofproblem (1)ndash(4) have the following asymptotic representationsas 119899 rarr infin
8 Discrete Dynamics in Nature and Society
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
with119865 = (119891 119879119886(119891) 119879119887(119891)) isin 119863(119860)Nowwe can rewrite prob-lem (1)ndash(4) in the operator form 119860119865 = 120582119865 for 119865 = (119891 119879119886(119891)119879119887(119891)) isin 119863(119860) Obviously we have the following lemmas
Lemma 4 The eigenvalues and eigenfunctions of problem(1)ndash(4) are corresponding to the eigenvalues and the firstcomponent of the corresponding eigenfunctions of operator 119860respectively
Lemma 5 The domain 119863(119860) is dense in 119867
Proof Let 119865 = (119891(119909) 1198911 1198912) isin 119867 119865 perp 119863(119860) and let 119862infin0 be afunctional set which has compact support and can be differ-ential infinitely such that 1205991(119909) isin 119862infin0 (119886 120585) 1205992(119909) isin 119862infin0 (120585 119887)Since 119862infin0 oplus 0 oplus 0 sub 119863(119860) for forall119866 = (119892(119909) 0 0) isin 119862infin0 oplus 0 oplus 0is orthogonal to 119865 namely
so 119891(119909) = 0 For all 119881 = (](119909) ]1 0) isin 119863(119860) ⟨119865 119881⟩ =(120574120588)1198911]1 = 0 1198911 = 0 since ]1 is arbitrary Similarly onegets 1198912 = 0 Above all one has 119865 = (0 0 0) which proves theassertion
Definition 6 Let 119882(119891 119892 119909) = 119891(119909)1198921015840(119909) minus 1198911015840(119909)119892(119909) denotethe Wronskian of functions 119891(119909) and 119892(119909) One has119882 (119891 119892 119886) = 11990121 [119891 119892] (119886) 119882 (119891 119892 119887) = 11990122 [119891 119892] (119887) (15)
Theorem 7 Operator 119860 is symmetric
Proof For each 119865 119866 isin 119863(119860) by the definition of inner pro-duct and operator 119860 we can get⟨119860119865 119866⟩ minus ⟨119865 119860119866⟩= 120574119882 (119891 119892 120585 minus 0) minus 119882 (119891 119892 120585 + 0) + 119882 (119891 119892 119887)minus 120574119882 (119891 119892 119886)+ 120574120588 [1198791015840119886 (119891) 119879119886 (119892) minus 119879119886 (119891) 1198791015840119886 (119892)]
then (4) implies that119882 (119891 119892 120585 + 0) = 120574119882 (119891 119892 120585 minus 0) (17)
using boundary condition (2) we get120574120588 [1198791015840119886 (119891) 119879119886 (119892) minus 119879119886 (119891) 1198791015840119886 (119892)] = 120574119882 (119891 119892 119886) (18)
and by boundary condition (3) we have1120579 [1198791015840119887 (119891) 119879119887 (119892) minus 119879119887 (119891) 1198791015840119887 (119892)] = 119882 (119891 119892 119887) (19)
Substituting (17)ndash(19) into (16) yields ⟨119860119865 119866⟩ minus ⟨119865 119860119866⟩ = 0This completes the proof
Moreover we have the following conclusion
Theorem 8 Operator 119860 is self-adjoint
Proof 119860 is self-adjoint if and only if for each 119865 =(119891(119909) 119879119886(119891) 119879119887(119891)) isin 119863(119860) ⟨119860119865 119882⟩ = ⟨119865 119880⟩ for some119880 isin 119867 implying that 119882 isin 119863(119860) and 119860119882 = 119880 where119882 = (119908(119909) ℎ 119903) 119880 = (120583(119909) 119896 119904) Concretely we shouldprove that the following properties hold
(i) 119908(119909) 1199081015840(119909) are absolutely continuous on (119886 120585) cup(120585 119887) 119871119908 isin 1198712((119886 120585) cup (120585 119887))(ii) ℎ = 1198871[119908 119906](119886) minus 1198872[119908 V](119886) 119903 = 1198891[119908 119906](119887) minus 1198892[119908
that is ⟨119871119891 119908⟩1 = ⟨119891 120583⟩1 According to the classical Sturm-Liouville theory (i) and (iv) hold So ⟨119860119865 119882⟩ = ⟨119865 119880⟩ forall119865 isin119863(119860) ⟨119871119891 119908⟩1 + 1205741205881198791015840119886 (119891) ℎ + 11205791198791015840119887 (119891) 119903
Substituting (24) into (23) we have ℎ = 1198871[119908 119906](119886) minus1198872[119908 V](119886) Further there exists 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = 119891 (120585 plusmn 0) = 1198911015840 (120585 plusmn 0) = 0[119891 119906] (119887) = 1198892[119891 V] (119887) = 1198891 (25)
Analogously we can get 119903 = 1198891[119908 119906](119887) minus 1198892[119908 V](119887) So (ii)holds Similarly one proves (v) Next let 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = [119891 119906] (119887) = [119891 V] (119887)= 119891 (120585 + 0) = 0119891 (120585 minus 0) = minus12057521198911015840 (120585 minus 0) = 12057511198911015840 (120585 + 0) = 120574
(26)
Then by (23) we have 1198713119908 = 0 Similarly we can get 1198714119908 =0 So 119860 is a self-adjoint operator
From the properties of self-adjoint operators we have thefollowing corollaries
Corollary 9 All eigenvalues of the singular Sturm-Liouvilleproblem (1)ndash(4) are real
Corollary 10 Let 1205821 and 1205822 be two different eigenvalues ofthe singular Sturm-Liouville problem (1)ndash(4) then the corre-sponding eigenfunctions 119891(119909) and 119892(119909) are orthogonal in thesense of12057411990121 int120585
In this section we construct the fundamental solutions ofproblem (1)ndash(4) and get the asymptotic approximation forfundamental solutions
Lemma 11 (see [5]) Let the real-valued function 119902(119909) becontinuous on 119868 = (119886 120585) cup (120585 119887) and let 119891(120582) 119892(120582) be givenentire functions Then for any 120582 isin C the equation119871119910 fl minus (119901 (119909) 1199101015840 (119909))1015840 + 119902 (119909) 119910 (119909) = 120582119910 (119909) 119909 isin (119886 120585) cup (120585 119887) (28)
Let 1205931(119909 120582) be the solution of (1) on the interval (119886 120585)which satisfies the initial conditions1205931 (119886 120582) fl [119910 119906] (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) fl [119910 V] (119886 120582) = 1198861 minus 1205821198871 (31)
by virtue of Lemma 11 we can define the solution 1205932(119909 120582) of(1) on (120585 119887) by the initial conditions
(1205932 (120585 + 0)12059310158402 (120585 + 0)) = (12057511205931 (120585 minus 0) + 120575212059310158401 (120585 minus 0)12057531205931 (120585 minus 0) + 120575412059310158401 (120585 minus 0)) (32)
Analogously we define the solutions 1205942(119909 120582) and 1205941(119909 120582) of(1) by the initial conditions1205942 (119887 120582) fl [119910 119906] (119887 120582) = 1198882 minus 120582119889212059410158402 (119887 120582) fl [119910 V] (119887 120582) = 1198881 minus 1205821198891
(1205941 (120585 minus 0)12059410158401 (120585 minus 0)) = (12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)120574 ) (33)
Now we consider Wronskian119882119894 (120582) fl 119882 (120593119894 120594119894 119909)= 120593119894 (119909 120582) 1205941015840119894 (119909 120582) minus 1205931015840119894 (119909 120582) 120594119894 (119909 120582)(119894 = 1 2) (34)
By the dependence of solutions of initial value problemson the parameter one has that 119882119894(120582) (119894 = 1 2) are entirefunctions of 120582 and are independent of 119909Lemma 12 For every 120582 isin C 1198822(120582) = 1205741198821(120582)
Discrete Dynamics in Nature and Society 5
Proof By the definition of 119882119894(120582) we have1198821 (120582) = 1205931 (120585 minus 0 120582) 12059410158401 (120585 minus 0 120582)minus 12059310158401 (120585 minus 0 120582) 1205941 (120585 minus 0 120582)
using the transmission conditions (4) simple computationgives 119882 (1205932 1205942 120585 + 0) = 120574119882 (1205931 1205941 120585 minus 0) (36)
Thus for each 120582 isin C we have 1198822(120582) = 1205741198821(120582) This com-pletes the proof
Besides we set 119882(120582) fl 1198821(120582) = (1120574)1199082(120582)Theorem 13 The eigenvalues of problem (1)ndash(4) coincide withthe zeros of the function 119882(120582)Proof Let ]0(119909 1205820) be any eigenfunction corresponding toeigenvalue 1205820 then the function ]0(119909 1205820)may be representedin the form
where at least one of the constants 119898119894 (119894 = 1 2 3 4) is notzero We should show that 119882(1205820) = 0 Suppose to thecontrary that there exists 1205820 isin 119877 such that119882(1205820) = 1198821(1205820) =(1120574)1198822(1205820) = 0 Since eigenfunction ]0(119909 1205820) satisfiesboth boundary and transmission conditions (2)ndash(4) we have119871 119894]0(119909 1205820) = 0 (119894 = 1 2 3 4) while the determinant ofcoefficient matrix is not zero so we obtain 119898119894 = 0 (119894 =1 2 3 4) which is a contradiction then 119882(1205820) = 0Conversely let 120582 = 1205820 be a zero of function 119882(120582) then119882(1205820) = 1198821(1205820) = (1120574)1198822(1205820) = 0 therefore 120594119894(119909 1205820) =119896120593119894(119909 1205820) (119894 = 1 2) for some 119896 = 0 Since both 1205932(119909 1205820) and1205942(119909 1205820) satisfy the boundary condition (3)
satisfies problem (1)ndash(4) So function 120593(119909 1205820) is an eigen-function of problem (1)ndash(4) corresponding to eigenvalue 1205820This completes the proof
Theorem 14 The eigenvalues of problem (1)ndash(4) are analyti-cally single
Proof Let 120582 = 119904 + 119894119905 and we use the following notions forsimplicity 120593 = 120593(119909 120582) 1205931120582 = 1205971205931120597120582 and 12059310158401120582 = 12059712059310158401120597120582
We differentiate the equation 119860120594 = 120582120594 with respect to 120582 toobtain 119860120594120582 = 120594 + 120582120594120582 (39)
Using integration by parts we get⟨119860120594120582 120593⟩1 minus ⟨120594120582 119860120593⟩1= 120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585 (40)
Substituting (39) and 119860120593 = 120582120593 into the left side of (40) wehave ⟨120582120594120582 120593⟩1 minus ⟨120594120582 120582120593⟩1 = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1 (41)
Moreover
120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585= 11988911205932 (119887) minus 119889212059310158402 (119887)minus 120574 [(1198861 minus 1205821198871) 1205941120582 (119886) minus (1198862 minus 1205821198872) 12059410158401120582 (119886)]
(42)
By (31) we observe that1198821015840 (120582) = 1205931 (119886 120582) 12059410158401 (119886 120582) minus 12059310158401 (119886 120582) 1205941 (119886 120582)= 11988711205941 (119886) minus 119887212059410158401 (119886) + (1198862 minus 1205821198872) 12059410158401120582 (119886)minus (1198861 minus 1205821198871) 1205941120582 (119886)
(43)
then (40) becomes1205741198821015840 (120582) = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1+ 120574 (11988711205941 (119886) minus 119887212059410158401 (119886))minus (11988911205932 (119887) minus 119889212059310158402 (119887))
(44)
Next let 1205830 be an arbitrary zero of119882(120582) Since119882(1205830) = 0 weobtain 120593119894(119909 1205830) = 119896120594119894(119909 1205830) (119894 = 1 2) (119896 = 0) 119896 isin R Notingthat 1205830 is real a short calculation (44) becomes1205741198821015840 (1205830)
Since 120588 gt 0 120579 gt 0 120574 gt 0 and 119896 = 0 1198821015840(1205830) = 0 Hence theanalytic multiplicity of 1205830 is one which completes the proof
6 Discrete Dynamics in Nature and Society
Lemma 15 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205931 (119909 120582)
= (1198862 minus 11990421198872) 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591(46)
1198891198961198891199091198961205932 (119909 120582)= 1205932 (120585 + 0) 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
+ 1119901211990412059310158402 (120585 + 0) 119889119896119889119909119896 sin [1199012119904 (119909 minus 120585)]+ 1199012119904 int119909
120585
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205932 (120591) 119889120591(47)
Proof For the case of 119896 = 0 consider 1205931(119909 120582) as the solutionof the following nonhomogeneous problem
minus (119901 (119909) 119910 (119909)1015840)1015840 + 120582119910 (119909) = 119902 (119909) 119910 (119909)1205931 (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) = 1198861 minus 1205821198871 (48)
Using the method of constant variation 1205931(119909 120582) satisfies1205931 (119909 120582) = (1198862 minus 11990421198872) cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591
(49)
Then differentiating it with respect to 119909 we have (46) Theproof for (47) is similar so we omit the details
Similarly we have the following theorem
Lemma 16 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205941 (119909 120582) = 12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574 119889119896119889119909119896
sdot cos [1199011119904 (119909 minus 120585)] + 120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)1199011119904120574
sdot 119889119896119889119909119896 sin [1199011119904 (119909 minus 120585)] minus 1199011119904sdot int120585119909
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205941 (120591) 1198891205911198891198961198891199091198961205942 (119909 120582) = (1198882 minus 11990421198892) 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
+ 1198881 minus 119904211988911199012119904 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] minus 1199012119904sdot int119887119909
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205942 (120591) 119889120591(50)
Lemma 17 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120593(119909 120582)have the following estimations
Case 1 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11990421198872 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]
+ 119874 (|119904|119896+1 119890|119905|1199011(119909minus119886)) 1198891198961198891199091198961205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 119886)]
Case 2 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11988711199041199011 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)] + 119874 (|119904|119896 119890|119905|1199011(119909minus119886))
1198891198961198891199091198961205932 (119909 120582)= minus120575211988711199042 cos [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Proof The proof of formulas for 1205931(119909 120582) is identical to thoseof Titchmarshrsquos proof for 120593(119909 120582) (see [6]) so we only give theproof of formulas for 1205932(119909 120582) namely equality (51) the otherequalities are similar
Discrete Dynamics in Nature and Society 7
For 119896 = 0 by the estimations of 1205931(119909 120582) and 12059310158401(119909 120582) wehave 1205931 (120585 minus 0 120582) = minus11990421198872 cos [1199011119904 (120585 minus 119886)]+ 119874 (|119904| 119890|119905|1199011(120585minus119886)) 12059310158401 (120585 minus 0 120582) = 119901111990431198872 sin [1199011119904 (120585 minus 119886)]+ 119874 (|119904|2 119890|119905|1199011(120585minus119886))
(53)
then substituting (53) into (47) and observing (32) we have1205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119909 minus 119886)]+ 119874 (|119904|2 119890|119905|(1199011(120585minus119886)+1199012(119909minus120585))) (54)
differentiating (54) with respect to 119909 we have (51)Lemma 18 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120594(119909 120582)have the following estimations
Case 1 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus1199012119889212057521199043120574 sin [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+2 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988921199042 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
Case 2 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus119889112057521199042120574 cos [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+1 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988911199041199012 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] + 119874 (|119904|119896 119890|119905|1199012(119909minus119887))
(56)
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Theorem 19 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the function 1198822(120582)has the following asymptotic representations
Case 1 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= 120575211990111199012119887211988921199046 sin [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|5 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (57)
Case 2 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199011119887211988911199045 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (58)
Case 3 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199012119887111988921199045 cos [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (59)
Case 4 If 1198872 = 0 and 1198892 = 0 then1198822 (120582) = 1205752119887111988911199044 cos [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|3 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (60)
Proof By the definition of 1198822(120582) we have1198822 (120582) = 1205932 (119887 120582) 12059410158402 (119887 120582) minus 12059310158402 (119887 120582) 1205942 (119887 120582)= (1198881 minus 1205821198891) 1205932 (119887 120582) minus (1198882 minus 1205821198892) 12059310158402 (119887 120582) (61)
According to the equalities of 1205932(119909 120582) and 12059310158402(119909 120582) inLemma 17 we can obtain the formulas of 1198822(120582) in thistheorem
Corollary 20 The eigenvalues of the boundary value problem(1)ndash(4) are bounded below
Proof Putting 119904 = 119894119905 (119905 gt 0) in Theorem 19 we can obtain1198822(120582) = 1198822(minus1199052) rarr infin (119905 rarr infin) Hence 1198822(minus1199052) = 0 forsufficiently negative 120582 and sufficiently large 120582 This completesthe proof
4 Asymptotic Formulas forEigenvalues and Eigenfunctions
In this section we can get the asymptotic formulas forthe eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)ndash(4) Since the eigenvalues coincidewiththe zeros of the entire function119882(120582) it follows that they haveno finite limits
Theorem 21 The eigenvalues 120582119899 = 1199042119899 (119899 = 0 1 2 ) ofproblem (1)ndash(4) have the following asymptotic representationsas 119899 rarr infin
8 Discrete Dynamics in Nature and Society
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
Substituting (24) into (23) we have ℎ = 1198871[119908 119906](119886) minus1198872[119908 V](119886) Further there exists 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = 119891 (120585 plusmn 0) = 1198911015840 (120585 plusmn 0) = 0[119891 119906] (119887) = 1198892[119891 V] (119887) = 1198891 (25)
Analogously we can get 119903 = 1198891[119908 119906](119887) minus 1198892[119908 V](119887) So (ii)holds Similarly one proves (v) Next let 119865 isin 119863(119860) such that[119891 119906] (119886) = [119891 V] (119886) = [119891 119906] (119887) = [119891 V] (119887)= 119891 (120585 + 0) = 0119891 (120585 minus 0) = minus12057521198911015840 (120585 minus 0) = 12057511198911015840 (120585 + 0) = 120574
(26)
Then by (23) we have 1198713119908 = 0 Similarly we can get 1198714119908 =0 So 119860 is a self-adjoint operator
From the properties of self-adjoint operators we have thefollowing corollaries
Corollary 9 All eigenvalues of the singular Sturm-Liouvilleproblem (1)ndash(4) are real
Corollary 10 Let 1205821 and 1205822 be two different eigenvalues ofthe singular Sturm-Liouville problem (1)ndash(4) then the corre-sponding eigenfunctions 119891(119909) and 119892(119909) are orthogonal in thesense of12057411990121 int120585
In this section we construct the fundamental solutions ofproblem (1)ndash(4) and get the asymptotic approximation forfundamental solutions
Lemma 11 (see [5]) Let the real-valued function 119902(119909) becontinuous on 119868 = (119886 120585) cup (120585 119887) and let 119891(120582) 119892(120582) be givenentire functions Then for any 120582 isin C the equation119871119910 fl minus (119901 (119909) 1199101015840 (119909))1015840 + 119902 (119909) 119910 (119909) = 120582119910 (119909) 119909 isin (119886 120585) cup (120585 119887) (28)
Let 1205931(119909 120582) be the solution of (1) on the interval (119886 120585)which satisfies the initial conditions1205931 (119886 120582) fl [119910 119906] (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) fl [119910 V] (119886 120582) = 1198861 minus 1205821198871 (31)
by virtue of Lemma 11 we can define the solution 1205932(119909 120582) of(1) on (120585 119887) by the initial conditions
(1205932 (120585 + 0)12059310158402 (120585 + 0)) = (12057511205931 (120585 minus 0) + 120575212059310158401 (120585 minus 0)12057531205931 (120585 minus 0) + 120575412059310158401 (120585 minus 0)) (32)
Analogously we define the solutions 1205942(119909 120582) and 1205941(119909 120582) of(1) by the initial conditions1205942 (119887 120582) fl [119910 119906] (119887 120582) = 1198882 minus 120582119889212059410158402 (119887 120582) fl [119910 V] (119887 120582) = 1198881 minus 1205821198891
(1205941 (120585 minus 0)12059410158401 (120585 minus 0)) = (12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)120574 ) (33)
Now we consider Wronskian119882119894 (120582) fl 119882 (120593119894 120594119894 119909)= 120593119894 (119909 120582) 1205941015840119894 (119909 120582) minus 1205931015840119894 (119909 120582) 120594119894 (119909 120582)(119894 = 1 2) (34)
By the dependence of solutions of initial value problemson the parameter one has that 119882119894(120582) (119894 = 1 2) are entirefunctions of 120582 and are independent of 119909Lemma 12 For every 120582 isin C 1198822(120582) = 1205741198821(120582)
Discrete Dynamics in Nature and Society 5
Proof By the definition of 119882119894(120582) we have1198821 (120582) = 1205931 (120585 minus 0 120582) 12059410158401 (120585 minus 0 120582)minus 12059310158401 (120585 minus 0 120582) 1205941 (120585 minus 0 120582)
using the transmission conditions (4) simple computationgives 119882 (1205932 1205942 120585 + 0) = 120574119882 (1205931 1205941 120585 minus 0) (36)
Thus for each 120582 isin C we have 1198822(120582) = 1205741198821(120582) This com-pletes the proof
Besides we set 119882(120582) fl 1198821(120582) = (1120574)1199082(120582)Theorem 13 The eigenvalues of problem (1)ndash(4) coincide withthe zeros of the function 119882(120582)Proof Let ]0(119909 1205820) be any eigenfunction corresponding toeigenvalue 1205820 then the function ]0(119909 1205820)may be representedin the form
where at least one of the constants 119898119894 (119894 = 1 2 3 4) is notzero We should show that 119882(1205820) = 0 Suppose to thecontrary that there exists 1205820 isin 119877 such that119882(1205820) = 1198821(1205820) =(1120574)1198822(1205820) = 0 Since eigenfunction ]0(119909 1205820) satisfiesboth boundary and transmission conditions (2)ndash(4) we have119871 119894]0(119909 1205820) = 0 (119894 = 1 2 3 4) while the determinant ofcoefficient matrix is not zero so we obtain 119898119894 = 0 (119894 =1 2 3 4) which is a contradiction then 119882(1205820) = 0Conversely let 120582 = 1205820 be a zero of function 119882(120582) then119882(1205820) = 1198821(1205820) = (1120574)1198822(1205820) = 0 therefore 120594119894(119909 1205820) =119896120593119894(119909 1205820) (119894 = 1 2) for some 119896 = 0 Since both 1205932(119909 1205820) and1205942(119909 1205820) satisfy the boundary condition (3)
satisfies problem (1)ndash(4) So function 120593(119909 1205820) is an eigen-function of problem (1)ndash(4) corresponding to eigenvalue 1205820This completes the proof
Theorem 14 The eigenvalues of problem (1)ndash(4) are analyti-cally single
Proof Let 120582 = 119904 + 119894119905 and we use the following notions forsimplicity 120593 = 120593(119909 120582) 1205931120582 = 1205971205931120597120582 and 12059310158401120582 = 12059712059310158401120597120582
We differentiate the equation 119860120594 = 120582120594 with respect to 120582 toobtain 119860120594120582 = 120594 + 120582120594120582 (39)
Using integration by parts we get⟨119860120594120582 120593⟩1 minus ⟨120594120582 119860120593⟩1= 120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585 (40)
Substituting (39) and 119860120593 = 120582120593 into the left side of (40) wehave ⟨120582120594120582 120593⟩1 minus ⟨120594120582 120582120593⟩1 = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1 (41)
Moreover
120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585= 11988911205932 (119887) minus 119889212059310158402 (119887)minus 120574 [(1198861 minus 1205821198871) 1205941120582 (119886) minus (1198862 minus 1205821198872) 12059410158401120582 (119886)]
(42)
By (31) we observe that1198821015840 (120582) = 1205931 (119886 120582) 12059410158401 (119886 120582) minus 12059310158401 (119886 120582) 1205941 (119886 120582)= 11988711205941 (119886) minus 119887212059410158401 (119886) + (1198862 minus 1205821198872) 12059410158401120582 (119886)minus (1198861 minus 1205821198871) 1205941120582 (119886)
(43)
then (40) becomes1205741198821015840 (120582) = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1+ 120574 (11988711205941 (119886) minus 119887212059410158401 (119886))minus (11988911205932 (119887) minus 119889212059310158402 (119887))
(44)
Next let 1205830 be an arbitrary zero of119882(120582) Since119882(1205830) = 0 weobtain 120593119894(119909 1205830) = 119896120594119894(119909 1205830) (119894 = 1 2) (119896 = 0) 119896 isin R Notingthat 1205830 is real a short calculation (44) becomes1205741198821015840 (1205830)
Since 120588 gt 0 120579 gt 0 120574 gt 0 and 119896 = 0 1198821015840(1205830) = 0 Hence theanalytic multiplicity of 1205830 is one which completes the proof
6 Discrete Dynamics in Nature and Society
Lemma 15 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205931 (119909 120582)
= (1198862 minus 11990421198872) 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591(46)
1198891198961198891199091198961205932 (119909 120582)= 1205932 (120585 + 0) 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
+ 1119901211990412059310158402 (120585 + 0) 119889119896119889119909119896 sin [1199012119904 (119909 minus 120585)]+ 1199012119904 int119909
120585
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205932 (120591) 119889120591(47)
Proof For the case of 119896 = 0 consider 1205931(119909 120582) as the solutionof the following nonhomogeneous problem
minus (119901 (119909) 119910 (119909)1015840)1015840 + 120582119910 (119909) = 119902 (119909) 119910 (119909)1205931 (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) = 1198861 minus 1205821198871 (48)
Using the method of constant variation 1205931(119909 120582) satisfies1205931 (119909 120582) = (1198862 minus 11990421198872) cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591
(49)
Then differentiating it with respect to 119909 we have (46) Theproof for (47) is similar so we omit the details
Similarly we have the following theorem
Lemma 16 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205941 (119909 120582) = 12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574 119889119896119889119909119896
sdot cos [1199011119904 (119909 minus 120585)] + 120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)1199011119904120574
sdot 119889119896119889119909119896 sin [1199011119904 (119909 minus 120585)] minus 1199011119904sdot int120585119909
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205941 (120591) 1198891205911198891198961198891199091198961205942 (119909 120582) = (1198882 minus 11990421198892) 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
+ 1198881 minus 119904211988911199012119904 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] minus 1199012119904sdot int119887119909
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205942 (120591) 119889120591(50)
Lemma 17 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120593(119909 120582)have the following estimations
Case 1 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11990421198872 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]
+ 119874 (|119904|119896+1 119890|119905|1199011(119909minus119886)) 1198891198961198891199091198961205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 119886)]
Case 2 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11988711199041199011 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)] + 119874 (|119904|119896 119890|119905|1199011(119909minus119886))
1198891198961198891199091198961205932 (119909 120582)= minus120575211988711199042 cos [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Proof The proof of formulas for 1205931(119909 120582) is identical to thoseof Titchmarshrsquos proof for 120593(119909 120582) (see [6]) so we only give theproof of formulas for 1205932(119909 120582) namely equality (51) the otherequalities are similar
Discrete Dynamics in Nature and Society 7
For 119896 = 0 by the estimations of 1205931(119909 120582) and 12059310158401(119909 120582) wehave 1205931 (120585 minus 0 120582) = minus11990421198872 cos [1199011119904 (120585 minus 119886)]+ 119874 (|119904| 119890|119905|1199011(120585minus119886)) 12059310158401 (120585 minus 0 120582) = 119901111990431198872 sin [1199011119904 (120585 minus 119886)]+ 119874 (|119904|2 119890|119905|1199011(120585minus119886))
(53)
then substituting (53) into (47) and observing (32) we have1205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119909 minus 119886)]+ 119874 (|119904|2 119890|119905|(1199011(120585minus119886)+1199012(119909minus120585))) (54)
differentiating (54) with respect to 119909 we have (51)Lemma 18 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120594(119909 120582)have the following estimations
Case 1 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus1199012119889212057521199043120574 sin [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+2 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988921199042 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
Case 2 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus119889112057521199042120574 cos [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+1 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988911199041199012 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] + 119874 (|119904|119896 119890|119905|1199012(119909minus119887))
(56)
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Theorem 19 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the function 1198822(120582)has the following asymptotic representations
Case 1 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= 120575211990111199012119887211988921199046 sin [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|5 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (57)
Case 2 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199011119887211988911199045 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (58)
Case 3 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199012119887111988921199045 cos [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (59)
Case 4 If 1198872 = 0 and 1198892 = 0 then1198822 (120582) = 1205752119887111988911199044 cos [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|3 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (60)
Proof By the definition of 1198822(120582) we have1198822 (120582) = 1205932 (119887 120582) 12059410158402 (119887 120582) minus 12059310158402 (119887 120582) 1205942 (119887 120582)= (1198881 minus 1205821198891) 1205932 (119887 120582) minus (1198882 minus 1205821198892) 12059310158402 (119887 120582) (61)
According to the equalities of 1205932(119909 120582) and 12059310158402(119909 120582) inLemma 17 we can obtain the formulas of 1198822(120582) in thistheorem
Corollary 20 The eigenvalues of the boundary value problem(1)ndash(4) are bounded below
Proof Putting 119904 = 119894119905 (119905 gt 0) in Theorem 19 we can obtain1198822(120582) = 1198822(minus1199052) rarr infin (119905 rarr infin) Hence 1198822(minus1199052) = 0 forsufficiently negative 120582 and sufficiently large 120582 This completesthe proof
4 Asymptotic Formulas forEigenvalues and Eigenfunctions
In this section we can get the asymptotic formulas forthe eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)ndash(4) Since the eigenvalues coincidewiththe zeros of the entire function119882(120582) it follows that they haveno finite limits
Theorem 21 The eigenvalues 120582119899 = 1199042119899 (119899 = 0 1 2 ) ofproblem (1)ndash(4) have the following asymptotic representationsas 119899 rarr infin
8 Discrete Dynamics in Nature and Society
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
Proof By the definition of 119882119894(120582) we have1198821 (120582) = 1205931 (120585 minus 0 120582) 12059410158401 (120585 minus 0 120582)minus 12059310158401 (120585 minus 0 120582) 1205941 (120585 minus 0 120582)
using the transmission conditions (4) simple computationgives 119882 (1205932 1205942 120585 + 0) = 120574119882 (1205931 1205941 120585 minus 0) (36)
Thus for each 120582 isin C we have 1198822(120582) = 1205741198821(120582) This com-pletes the proof
Besides we set 119882(120582) fl 1198821(120582) = (1120574)1199082(120582)Theorem 13 The eigenvalues of problem (1)ndash(4) coincide withthe zeros of the function 119882(120582)Proof Let ]0(119909 1205820) be any eigenfunction corresponding toeigenvalue 1205820 then the function ]0(119909 1205820)may be representedin the form
where at least one of the constants 119898119894 (119894 = 1 2 3 4) is notzero We should show that 119882(1205820) = 0 Suppose to thecontrary that there exists 1205820 isin 119877 such that119882(1205820) = 1198821(1205820) =(1120574)1198822(1205820) = 0 Since eigenfunction ]0(119909 1205820) satisfiesboth boundary and transmission conditions (2)ndash(4) we have119871 119894]0(119909 1205820) = 0 (119894 = 1 2 3 4) while the determinant ofcoefficient matrix is not zero so we obtain 119898119894 = 0 (119894 =1 2 3 4) which is a contradiction then 119882(1205820) = 0Conversely let 120582 = 1205820 be a zero of function 119882(120582) then119882(1205820) = 1198821(1205820) = (1120574)1198822(1205820) = 0 therefore 120594119894(119909 1205820) =119896120593119894(119909 1205820) (119894 = 1 2) for some 119896 = 0 Since both 1205932(119909 1205820) and1205942(119909 1205820) satisfy the boundary condition (3)
satisfies problem (1)ndash(4) So function 120593(119909 1205820) is an eigen-function of problem (1)ndash(4) corresponding to eigenvalue 1205820This completes the proof
Theorem 14 The eigenvalues of problem (1)ndash(4) are analyti-cally single
Proof Let 120582 = 119904 + 119894119905 and we use the following notions forsimplicity 120593 = 120593(119909 120582) 1205931120582 = 1205971205931120597120582 and 12059310158401120582 = 12059712059310158401120597120582
We differentiate the equation 119860120594 = 120582120594 with respect to 120582 toobtain 119860120594120582 = 120594 + 120582120594120582 (39)
Using integration by parts we get⟨119860120594120582 120593⟩1 minus ⟨120594120582 119860120593⟩1= 120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585 (40)
Substituting (39) and 119860120593 = 120582120593 into the left side of (40) wehave ⟨120582120594120582 120593⟩1 minus ⟨120594120582 120582120593⟩1 = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1 (41)
Moreover
120574 (120594112058212059310158401 minus 12059410158401120582120593110038161003816100381610038161003816120585119886) + (120594212058212059310158402 minus 120594101584021205821205932)100381610038161003816100381610038161003816119887120585= 11988911205932 (119887) minus 119889212059310158402 (119887)minus 120574 [(1198861 minus 1205821198871) 1205941120582 (119886) minus (1198862 minus 1205821198872) 12059410158401120582 (119886)]
(42)
By (31) we observe that1198821015840 (120582) = 1205931 (119886 120582) 12059410158401 (119886 120582) minus 12059310158401 (119886 120582) 1205941 (119886 120582)= 11988711205941 (119886) minus 119887212059410158401 (119886) + (1198862 minus 1205821198872) 12059410158401120582 (119886)minus (1198861 minus 1205821198871) 1205941120582 (119886)
(43)
then (40) becomes1205741198821015840 (120582) = ⟨120594 120593⟩1 + 2119894119905 ⟨120594120582 120593⟩1+ 120574 (11988711205941 (119886) minus 119887212059410158401 (119886))minus (11988911205932 (119887) minus 119889212059310158402 (119887))
(44)
Next let 1205830 be an arbitrary zero of119882(120582) Since119882(1205830) = 0 weobtain 120593119894(119909 1205830) = 119896120594119894(119909 1205830) (119894 = 1 2) (119896 = 0) 119896 isin R Notingthat 1205830 is real a short calculation (44) becomes1205741198821015840 (1205830)
Since 120588 gt 0 120579 gt 0 120574 gt 0 and 119896 = 0 1198821015840(1205830) = 0 Hence theanalytic multiplicity of 1205830 is one which completes the proof
6 Discrete Dynamics in Nature and Society
Lemma 15 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205931 (119909 120582)
= (1198862 minus 11990421198872) 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591(46)
1198891198961198891199091198961205932 (119909 120582)= 1205932 (120585 + 0) 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
+ 1119901211990412059310158402 (120585 + 0) 119889119896119889119909119896 sin [1199012119904 (119909 minus 120585)]+ 1199012119904 int119909
120585
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205932 (120591) 119889120591(47)
Proof For the case of 119896 = 0 consider 1205931(119909 120582) as the solutionof the following nonhomogeneous problem
minus (119901 (119909) 119910 (119909)1015840)1015840 + 120582119910 (119909) = 119902 (119909) 119910 (119909)1205931 (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) = 1198861 minus 1205821198871 (48)
Using the method of constant variation 1205931(119909 120582) satisfies1205931 (119909 120582) = (1198862 minus 11990421198872) cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591
(49)
Then differentiating it with respect to 119909 we have (46) Theproof for (47) is similar so we omit the details
Similarly we have the following theorem
Lemma 16 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205941 (119909 120582) = 12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574 119889119896119889119909119896
sdot cos [1199011119904 (119909 minus 120585)] + 120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)1199011119904120574
sdot 119889119896119889119909119896 sin [1199011119904 (119909 minus 120585)] minus 1199011119904sdot int120585119909
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205941 (120591) 1198891205911198891198961198891199091198961205942 (119909 120582) = (1198882 minus 11990421198892) 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
+ 1198881 minus 119904211988911199012119904 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] minus 1199012119904sdot int119887119909
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205942 (120591) 119889120591(50)
Lemma 17 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120593(119909 120582)have the following estimations
Case 1 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11990421198872 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]
+ 119874 (|119904|119896+1 119890|119905|1199011(119909minus119886)) 1198891198961198891199091198961205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 119886)]
Case 2 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11988711199041199011 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)] + 119874 (|119904|119896 119890|119905|1199011(119909minus119886))
1198891198961198891199091198961205932 (119909 120582)= minus120575211988711199042 cos [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Proof The proof of formulas for 1205931(119909 120582) is identical to thoseof Titchmarshrsquos proof for 120593(119909 120582) (see [6]) so we only give theproof of formulas for 1205932(119909 120582) namely equality (51) the otherequalities are similar
Discrete Dynamics in Nature and Society 7
For 119896 = 0 by the estimations of 1205931(119909 120582) and 12059310158401(119909 120582) wehave 1205931 (120585 minus 0 120582) = minus11990421198872 cos [1199011119904 (120585 minus 119886)]+ 119874 (|119904| 119890|119905|1199011(120585minus119886)) 12059310158401 (120585 minus 0 120582) = 119901111990431198872 sin [1199011119904 (120585 minus 119886)]+ 119874 (|119904|2 119890|119905|1199011(120585minus119886))
(53)
then substituting (53) into (47) and observing (32) we have1205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119909 minus 119886)]+ 119874 (|119904|2 119890|119905|(1199011(120585minus119886)+1199012(119909minus120585))) (54)
differentiating (54) with respect to 119909 we have (51)Lemma 18 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120594(119909 120582)have the following estimations
Case 1 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus1199012119889212057521199043120574 sin [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+2 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988921199042 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
Case 2 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus119889112057521199042120574 cos [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+1 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988911199041199012 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] + 119874 (|119904|119896 119890|119905|1199012(119909minus119887))
(56)
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Theorem 19 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the function 1198822(120582)has the following asymptotic representations
Case 1 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= 120575211990111199012119887211988921199046 sin [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|5 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (57)
Case 2 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199011119887211988911199045 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (58)
Case 3 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199012119887111988921199045 cos [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (59)
Case 4 If 1198872 = 0 and 1198892 = 0 then1198822 (120582) = 1205752119887111988911199044 cos [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|3 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (60)
Proof By the definition of 1198822(120582) we have1198822 (120582) = 1205932 (119887 120582) 12059410158402 (119887 120582) minus 12059310158402 (119887 120582) 1205942 (119887 120582)= (1198881 minus 1205821198891) 1205932 (119887 120582) minus (1198882 minus 1205821198892) 12059310158402 (119887 120582) (61)
According to the equalities of 1205932(119909 120582) and 12059310158402(119909 120582) inLemma 17 we can obtain the formulas of 1198822(120582) in thistheorem
Corollary 20 The eigenvalues of the boundary value problem(1)ndash(4) are bounded below
Proof Putting 119904 = 119894119905 (119905 gt 0) in Theorem 19 we can obtain1198822(120582) = 1198822(minus1199052) rarr infin (119905 rarr infin) Hence 1198822(minus1199052) = 0 forsufficiently negative 120582 and sufficiently large 120582 This completesthe proof
4 Asymptotic Formulas forEigenvalues and Eigenfunctions
In this section we can get the asymptotic formulas forthe eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)ndash(4) Since the eigenvalues coincidewiththe zeros of the entire function119882(120582) it follows that they haveno finite limits
Theorem 21 The eigenvalues 120582119899 = 1199042119899 (119899 = 0 1 2 ) ofproblem (1)ndash(4) have the following asymptotic representationsas 119899 rarr infin
8 Discrete Dynamics in Nature and Society
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
Lemma 15 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205931 (119909 120582)
= (1198862 minus 11990421198872) 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591(46)
1198891198961198891199091198961205932 (119909 120582)= 1205932 (120585 + 0) 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
+ 1119901211990412059310158402 (120585 + 0) 119889119896119889119909119896 sin [1199012119904 (119909 minus 120585)]+ 1199012119904 int119909
120585
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205932 (120591) 119889120591(47)
Proof For the case of 119896 = 0 consider 1205931(119909 120582) as the solutionof the following nonhomogeneous problem
minus (119901 (119909) 119910 (119909)1015840)1015840 + 120582119910 (119909) = 119902 (119909) 119910 (119909)1205931 (119886 120582) = 1198862 minus 120582119887212059310158401 (119886 120582) = 1198861 minus 1205821198871 (48)
Using the method of constant variation 1205931(119909 120582) satisfies1205931 (119909 120582) = (1198862 minus 11990421198872) cos [1199011119904 (119909 minus 119886)]minus 1198861 minus 119904211988711199011119904 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)]+ 1199011119904 int119909
119886sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205931 (120591) 119889120591
(49)
Then differentiating it with respect to 119909 we have (46) Theproof for (47) is similar so we omit the details
Similarly we have the following theorem
Lemma 16 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the following integraland differential equations hold for 119896 = 0 and 119896 = 11198891198961198891199091198961205941 (119909 120582) = 12057541205942 (120585 + 0) minus 120575212059410158402 (120585 + 0)120574 119889119896119889119909119896
sdot cos [1199011119904 (119909 minus 120585)] + 120575112059410158402 (120585 + 0) minus 12057531205942 (120585 + 0)1199011119904120574
sdot 119889119896119889119909119896 sin [1199011119904 (119909 minus 120585)] minus 1199011119904sdot int120585119909
119889119896119889119909119896 sin [1199011119904 (119909 minus 120591)] 119902 (120591) 1205941 (120591) 1198891205911198891198961198891199091198961205942 (119909 120582) = (1198882 minus 11990421198892) 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
+ 1198881 minus 119904211988911199012119904 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] minus 1199012119904sdot int119887119909
119889119896119889119909119896 sin [1199012119904 (119909 minus 120591)] 119902 (120591) 1205942 (120591) 119889120591(50)
Lemma 17 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120593(119909 120582)have the following estimations
Case 1 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11990421198872 119889119896119889119909119896 cos [1199011119904 (119909 minus 119886)]
+ 119874 (|119904|119896+1 119890|119905|1199011(119909minus119886)) 1198891198961198891199091198961205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 119886)]
Case 2 If 1198872 = 0 then1198891198961198891199091198961205931 (119909 120582)= minus11988711199041199011 119889119896119889119909119896 sin [1199011119904 (119909 minus 119886)] + 119874 (|119904|119896 119890|119905|1199011(119909minus119886))
1198891198961198891199091198961205932 (119909 120582)= minus120575211988711199042 cos [1199011119904 (120585 minus 119886)] 119889119896119889119909119896 cos [1199012119904 (119909 minus 120585)]
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Proof The proof of formulas for 1205931(119909 120582) is identical to thoseof Titchmarshrsquos proof for 120593(119909 120582) (see [6]) so we only give theproof of formulas for 1205932(119909 120582) namely equality (51) the otherequalities are similar
Discrete Dynamics in Nature and Society 7
For 119896 = 0 by the estimations of 1205931(119909 120582) and 12059310158401(119909 120582) wehave 1205931 (120585 minus 0 120582) = minus11990421198872 cos [1199011119904 (120585 minus 119886)]+ 119874 (|119904| 119890|119905|1199011(120585minus119886)) 12059310158401 (120585 minus 0 120582) = 119901111990431198872 sin [1199011119904 (120585 minus 119886)]+ 119874 (|119904|2 119890|119905|1199011(120585minus119886))
(53)
then substituting (53) into (47) and observing (32) we have1205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119909 minus 119886)]+ 119874 (|119904|2 119890|119905|(1199011(120585minus119886)+1199012(119909minus120585))) (54)
differentiating (54) with respect to 119909 we have (51)Lemma 18 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120594(119909 120582)have the following estimations
Case 1 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus1199012119889212057521199043120574 sin [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+2 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988921199042 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
Case 2 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus119889112057521199042120574 cos [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+1 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988911199041199012 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] + 119874 (|119904|119896 119890|119905|1199012(119909minus119887))
(56)
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Theorem 19 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the function 1198822(120582)has the following asymptotic representations
Case 1 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= 120575211990111199012119887211988921199046 sin [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|5 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (57)
Case 2 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199011119887211988911199045 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (58)
Case 3 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199012119887111988921199045 cos [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (59)
Case 4 If 1198872 = 0 and 1198892 = 0 then1198822 (120582) = 1205752119887111988911199044 cos [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|3 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (60)
Proof By the definition of 1198822(120582) we have1198822 (120582) = 1205932 (119887 120582) 12059410158402 (119887 120582) minus 12059310158402 (119887 120582) 1205942 (119887 120582)= (1198881 minus 1205821198891) 1205932 (119887 120582) minus (1198882 minus 1205821198892) 12059310158402 (119887 120582) (61)
According to the equalities of 1205932(119909 120582) and 12059310158402(119909 120582) inLemma 17 we can obtain the formulas of 1198822(120582) in thistheorem
Corollary 20 The eigenvalues of the boundary value problem(1)ndash(4) are bounded below
Proof Putting 119904 = 119894119905 (119905 gt 0) in Theorem 19 we can obtain1198822(120582) = 1198822(minus1199052) rarr infin (119905 rarr infin) Hence 1198822(minus1199052) = 0 forsufficiently negative 120582 and sufficiently large 120582 This completesthe proof
4 Asymptotic Formulas forEigenvalues and Eigenfunctions
In this section we can get the asymptotic formulas forthe eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)ndash(4) Since the eigenvalues coincidewiththe zeros of the entire function119882(120582) it follows that they haveno finite limits
Theorem 21 The eigenvalues 120582119899 = 1199042119899 (119899 = 0 1 2 ) ofproblem (1)ndash(4) have the following asymptotic representationsas 119899 rarr infin
8 Discrete Dynamics in Nature and Society
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
For 119896 = 0 by the estimations of 1205931(119909 120582) and 12059310158401(119909 120582) wehave 1205931 (120585 minus 0 120582) = minus11990421198872 cos [1199011119904 (120585 minus 119886)]+ 119874 (|119904| 119890|119905|1199011(120585minus119886)) 12059310158401 (120585 minus 0 120582) = 119901111990431198872 sin [1199011119904 (120585 minus 119886)]+ 119874 (|119904|2 119890|119905|1199011(120585minus119886))
(53)
then substituting (53) into (47) and observing (32) we have1205932 (119909 120582)= 1199011120575211990431198872 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119909 minus 119886)]+ 119874 (|119904|2 119890|119905|(1199011(120585minus119886)+1199012(119909minus120585))) (54)
differentiating (54) with respect to 119909 we have (51)Lemma 18 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then for 119896 = 0 1 120594(119909 120582)have the following estimations
Case 1 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus1199012119889212057521199043120574 sin [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+2 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988921199042 119889119896119889119909119896 cos [1199012119904 (119909 minus 119887)]
Case 2 If 1198892 = 0 then1198891198961198891199091198961205941 (119909 120582)= minus119889112057521199042120574 cos [1199012119904 (120585 minus 119887)] 119889119896119889119909119896 cos [1199011119904 (119909 minus 120585)]
+ 119874 (|119904|119896+1 119890|119905|(1199011(119909minus120585)+1199012(120585minus119887))) 1198891198961198891199091198961205942 (119909 120582)= minus11988911199041199012 119889119896119889119909119896 sin [1199012119904 (119909 minus 119887)] + 119874 (|119904|119896 119890|119905|1199012(119909minus119887))
(56)
Each of these asymptotic equalities holds uniformly for 119909 as|120582| rarr infin
Theorem 19 Let 120582 = 1199042 119904 = 120590 + 119894119905 Then the function 1198822(120582)has the following asymptotic representations
Case 1 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= 120575211990111199012119887211988921199046 sin [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|5 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (57)
Case 2 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199011119887211988911199045 sin [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (58)
Case 3 If 1198872 = 0 and 1198892 = 0 then1198822 (120582)= minus12057521199012119887111988921199045 cos [1199011119904 (120585 minus 119886)] sin [1199012119904 (119887 minus 120585)]+ 119874 (|119904|4 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (59)
Case 4 If 1198872 = 0 and 1198892 = 0 then1198822 (120582) = 1205752119887111988911199044 cos [1199011119904 (120585 minus 119886)] cos [1199012119904 (119887 minus 120585)]+ 119874 (|119904|3 119890|119905|(1199011(120585minus119886)+1199012(119887minus120585))) (60)
Proof By the definition of 1198822(120582) we have1198822 (120582) = 1205932 (119887 120582) 12059410158402 (119887 120582) minus 12059310158402 (119887 120582) 1205942 (119887 120582)= (1198881 minus 1205821198891) 1205932 (119887 120582) minus (1198882 minus 1205821198892) 12059310158402 (119887 120582) (61)
According to the equalities of 1205932(119909 120582) and 12059310158402(119909 120582) inLemma 17 we can obtain the formulas of 1198822(120582) in thistheorem
Corollary 20 The eigenvalues of the boundary value problem(1)ndash(4) are bounded below
Proof Putting 119904 = 119894119905 (119905 gt 0) in Theorem 19 we can obtain1198822(120582) = 1198822(minus1199052) rarr infin (119905 rarr infin) Hence 1198822(minus1199052) = 0 forsufficiently negative 120582 and sufficiently large 120582 This completesthe proof
4 Asymptotic Formulas forEigenvalues and Eigenfunctions
In this section we can get the asymptotic formulas forthe eigenvalues and eigenfunctions of the singular Sturm-Liouville problem (1)ndash(4) Since the eigenvalues coincidewiththe zeros of the entire function119882(120582) it follows that they haveno finite limits
Theorem 21 The eigenvalues 120582119899 = 1199042119899 (119899 = 0 1 2 ) ofproblem (1)ndash(4) have the following asymptotic representationsas 119899 rarr infin
8 Discrete Dynamics in Nature and Society
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
Case 1 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (62)
Case 2 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 1) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (63)
Case 3 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 1) 1205871199012 (119887 minus 120585) + 119874 (1119899) (64)
Case 4 If 1198872 = 0 and 1198892 = 0 then1199041015840119899 = (119899 minus 12) 1205871199011 (120585 minus 119886) + 119874 (1119899) 11990410158401015840119899 = (119899 minus 12) 1205871199012 (119887 minus 120585) + 119874 (1119899) (65)
Proof By applying the well-known Rouche theorem we canobtain these conclusions (see [3] Theorem 23)
According to Theorem 21 and Lemmas 17 and 18 wecan obtain the following asymptotic representations of theeigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899)Theorem 22 The eigenfunctions 120593(119909 120582119899) and 120594(119909 120582119899) (119899 =0 1 2 ) of problem (1)ndash(4) have the following asymptoticrepresentations as 119899 rarr infin
Case 1 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
120593 (119909 12058210158401015840119899) = minus1198872 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]3 sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]3 sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(66)
Case 2 If 1198872 = 0 and 1198892 = 0 then
120593 (119909 1205821015840119899) = minus1198872 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119874 (1198992) 119909 isin (120585 119887)
Discrete Dynamics in Nature and Society 9
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
120593 (119909 12058210158401015840119899) = minus1198872 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (119899) 119909 isin (119886 120585) 119901111988721205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]3 sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119886)] + 119874 (1198992) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = 11988911205752120574 [ (119899 minus 1) 1205871199011 (120585 minus 119886)]2 cos [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 1) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [ (119899 minus 1) 1205871199011 (120585 minus 119886)] sin [1199012 (119899 minus 1) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(67)
Case 3 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [ (119899 minus 1) 1205871199012 (119887 minus 120585)] sin [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [1199011 (119899 minus 1) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 1205821015840119899) = minus119901211988921205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]3 sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (1198992) 119909 isin (119886 120585) minus1198892 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (1198992) 119909 isin (119886 120585) minus1198892 [ (119899 minus 1) 1205871199012 (119887 minus 120585)]2 cos [(119899 minus 1) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (119899) 119909 isin (120585 119887)
(68)
Case 4 If 1198872 = 0 and 1198892 = 0 then120593 (119909 1205821015840119899) = minus1198871 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [(119899 minus 12) 120587120585 minus 119886 (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) 119874 (119899) 119909 isin (120585 119887) 120593 (119909 12058210158401015840119899) =
minus1198871 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (119909 minus 119886)] + 119874 (1) 119909 isin (119886 120585) minus11988711205752 [(119899 minus 12) 1205871199012 (119887 minus 120585) ]2 cos [1199011 (119899 minus 12) 1205871199012 (119887 minus 120585) (120585 minus 119886)] cos [(119899 minus 12) 120587119887 minus 120585 (119909 minus 120585)] + 119874 (119899) 119909 isin (120585 119887)
10 Discrete Dynamics in Nature and Society
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
120594 (119909 1205821015840119899) = 11988911205752120574 [(119899 minus 12) 1205871199011 (120585 minus 119886) ]2 cos [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) ] cos [(119899 minus 12) 120587120585 minus 119886 (119909 minus 120585)] + 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199011 (120585 minus 119886) ] sin [1199012 (119899 minus 12) 1205871199011 (120585 minus 119886) (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
120594 (119909 12058210158401015840119899) = 119874 (119899) 119909 isin (119886 120585) minus11988911199012 [(119899 minus 12) 1205871199012 (119887 minus 120585) ] sin [(119899 minus 12) 120587119887 minus 120585 (119909 minus 119887)] + 119874 (1) 119909 isin (120585 119887)
(69)
5 Completeness of Eigenfunctions
In this section we get the property of spectrum for the oper-ator 119860 and discuss the completeness of the eigenfunctions ofproblem (1)ndash(4)
Theorem 23 The operator 119860 has only point spectrum that is120590(119860) = 120590119901(119860)Proof We only need to prove that if 120574 is not an eigenvalueof 119860 then 120574 isin 120588(119860) Here we investigate the equation (119860 minus120574)119885 = 119865 isin 119867 where 120574 isin R 119885 = (119911(119909) 119879119886(119911) 119879119887(119911)) and119865 = (119871119891 119879119886(119891) 119879119887(119891)) Consider the initial value problemminus (119901 (119909) 1199111015840 (119909))1015840 + 119902 (119909) 119911 (119909) minus 120574119911 (119909) = 119891 (119909) 119909 isin (119886 120585) cup (120585 119887)119911 (120585 + 0) minus 1205751119911 (120585 minus 0) minus 12057521199111015840 (120585 minus 0) = 01199111015840 (120585 + 0) minus 1205753119911 (120585 minus 0) minus 12057541199111015840 (120585 minus 0) = 0
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 0 satisfying the transmission conditions (4) Let
be the solution of the equation minus(119901(119909)1199111015840(119909))1015840 + 119902(119909)119911(119909) minus120574119911(119909) = 119891(119909) satisfying120601 (120585 + 0) minus 1205751120601 (120585 minus 0) minus 12057521206011015840 (120585 minus 0) = 01206011015840 (120585 + 0) minus 1205753120601 (120585 minus 0) minus 12057541206011015840 (120585 minus 0) = 0 (73)
As 120574 is not an eigenvalue of problem (1)ndash(4) we have120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886))minus (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886)) = 0120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887))minus (1198881 [119911 119888] (119887) minus 1198882 [119911 V] (119887)) = 0(75)
The second and third components of (119860minus120574)119885 = 119865 isin 119867meanthat (1198861 [119911 119906] (119886) minus 1198862 [119911 V] (119886))minus 120574 (1198871 [119911 119906] (119886) minus 1198872 [119911 V] (119886)) = 119879119886 (119891) (1198881 [119911 119906] (119887) minus 1198882 [119911 V] (119887))minus 120574 (1198891 [119911 119906] (119887) minus 1198892 [119911 V] (119887)) = 119879119887 (119891)
(76)
Substituting (74) into (76) we can get that 119899 is uniquelysolvable So 119911(119909) is uniquely determined Observing that (119860minus120574119868)minus1 is defined on all of 119867 we get that (119860 minus 120574119868)minus1 is boundedbyTheorem 8 and the closed graph theoremThus 120574 isin 120588(119860)Hence 120590(119860) = 120590119901(119860)Lemma 24 The eigenvalues of problem (1)ndash(4) are countablyinfinite and can cluster only at infin
Lemma 25 The operator 119860 has compact resolvent that is foreach 120575 isin R120590119901(119860) (119860 minus 120575119868)minus1 is compact on 119867 (see [14]Theorem 633)
By the above lemmas and the spectral theorem forcompact operator we obtain the following theorem
Theorem26 The eigenfunctions of problem (1)ndash(4) expandedto become eigenfunctions of 119860 are complete in 119867 that is letΦ119899 = (120593119899(119909) 119879119886(120593119899) 119879119887(120593119899)) 119899 isin N be a maximum set oforthogonal eigenfunctions of119860 where 120593119899(119909) 119899 isin N are eigen-functions of problem (1)ndash(4) Then for all 119865 isin 119867 119865 =suminfin119899=1⟨119865 Φ119899⟩Φ119899Remark 27 In this paper the spectral properties of singu-lar Sturm-Liouville problems with one inner discontinuouspoint are considered if there are two or even multi-innerdiscontinuous points 1205851 1205852 120585119899 in the interval (119886 119887) we
Discrete Dynamics in Nature and Society 11
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
can obtain similar results by defining a more complicatedHilbert space
Conflicts of Interest
The authors declare that there are no conflicts of interest
Authorsrsquo Contributions
Jinming Cai prepared the manuscript and corrected the mainresults and Zhaowen Zheng gave the main thought andrevised the manuscript
Acknowledgments
This research was partially supported by the NSF of China(Grants 11271225 and 11671227)
References
[1] H Weyl ldquoUber gewohnliche Differentialgleichungen mit Sin-gularitaten und die zugehorigen Entwicklungenrdquo Mathematis-che Annalen vol 68 no 2 pp 220ndash269 1910
[2] JWeidmann SpectralTheory ofOrdinaryDifferential Operatorsvol 1258 of Lecture Notes in Mathematics Springer new YorkNY USA 1987
[3] Z Cao Ordinary Differential Operator Shanhai Science andTechnology Press Shanghai China 1986 (Chinese)
[4] M A Naimark Linear Differential Operators Part 2 HarrapLondon UK 1968
[5] A Zettl Sturm-Liouville Theory vol 121 of Mathematical Sur-veys and Monographs American Mathematics Society Provi-dence RI USA 2005
[6] E C TitchmarshEigenfunction Expansions Associatedwith Sec-ond-Order Differential Equations Part 1 Oxford UniversityPress London UK 1962
[7] Z Zheng and W Zhang ldquoCharacterization of eigenvalues inspectral gap for singular differential operatorsrdquo Abstract andApplied Analysis vol 2012 Article ID 271657 10 pages 2012
[8] J Qi Z Zheng and H Sun ldquoClassification of Sturm-Liouvilledifferential equations with complex coefficients and operatorrealizationsrdquo Proceedings of The Royal Society of London SeriesAMathematical Physical and Engineering Sciences vol 467 no2131 pp 1835ndash1850 2011
[9] Z Zheng J Qi and S Chen ldquoEigenvalues below the lowerbound of minimal operators of singular Hamiltonian expres-sionsrdquo Computers ampMathematics with Applications vol 56 no11 pp 2825ndash2833 2008
[10] Z Zheng ldquoInvariance of deficiency indices under perturbationfor discrete Hamiltonian systemsrdquo Journal of Difference Equa-tions and Applications vol 19 no 8 pp 1243ndash1250 2013
[11] M Demirci Z Akdogan and O S H Mukhtarov ldquoAsymptoticbehavior of eigenvalues and eigenfunctions of one discontin-uous boundary value problemrdquo International Journal of Com-putational Cognition vol 2 article 3 pp 101ndash113 2004
[12] D Buschmann G Stolz and J Weidmann ldquoOne-dimensionalschrodinger operators with local point interactionsrdquo Journal furdie Reine und Angewandte Mathematik vol 467 pp 169ndash1861995
[13] M Kadakal O S Mukhtarov and F S Muhtarov ldquoSome spec-tral properties of Sturm-Liouville problem with transmissionconditionsrdquo Iranian Journal of Science and Technology Trans-action A Science vol 29 no 2 pp 229ndash245 2005
[14] A P Wang Research on weimann conjecture and differentialoperators with transmission conditions [PhD thesis] InnerMongolia University 2006 (Chinese)
[15] O S H Mukhtarov and M Kadakal ldquoSome spectral proper-ties of one Sturm-Liouville type problem with discontinuousweightrdquo Siberian Mathematical Journal vol 46 no 4 pp 681ndash694 2005
[16] C T Fulton ldquoTwo-point boundary value problems with eigen-value parameter contained in the boundary conditionsrdquo Pro-ceedings of the Royal Society of Edinburgh Section A Mathemat-ics vol 77 no 3-4 pp 293ndash308 1977
[17] E Sen andA Bayramov ldquoAsymptotic formulations of the eigen-values and eigenfunctions for a boundary value problemrdquoMathematical Methods in the Applied Sciences vol 36 no 12pp 1512ndash1519 2013
[18] Q Yang and W Wang ldquoAsymptotic behavior of a differentialoperator with discontinuities at two pointsrdquo MathematicalMethods in the Applied Sciences vol 34 no 4 pp 373ndash383 2011
[19] O S Mukhtarov M Kadakal and F S Muhtarov ldquoEigenvaluesand normalized eigenfunctions of discontinuous Sturm-Liou-ville problem with transmission conditionsrdquo Reports on Mathe-matical Physics vol 54 no 1 pp 41ndash56 2004
[20] E Sen ldquoA Sturm-Liouville problemwith a discontinuous coeffi-cient and containing an eigenparameter in the boundary con-ditionrdquo Physics Research International vol 2013 Article ID159243 9 pages 2013
[21] M Kadakal and O S Mukhtarov ldquoSturm-Liouville problemswith discontinuities at two pointsrdquo Computers amp Mathematicswith Applications vol 54 no 11-12 pp 1367ndash1379 2007
[22] M Kadakal and O S Mukhtarov ldquoDiscontinuous Sturm-Liou-ville problems containing eigenparameter in the boundary con-ditionsrdquo Acta Mathematica Sinica vol 22 no 5 pp 1519ndash15282006
[23] Z Akdogan M Demirci and O S H Mukhtarov ldquoSturm-Liouville problems with eigendependent boundary and trans-missions conditionsrdquo Acta Mathematica Scientia vol 25 no 4pp 731ndash740 2005
[24] Q Yang and W Wang ldquoSpectral properties of Sturm-Liouvilleoperators with discontinuities at finite pointsrdquo MathematicalSciences vol 6 no 1 pp 1ndash9 2012
[25] K Li and Z Zheng ldquoSpectral properties for Sturm-Liouvilleequations with transmission conditionsrdquo Acta MathematicaScientia vol 35 no 5 pp 910ndash926 2015 (Chinese)
[26] F Meng and Y Huang ldquoInterval oscillation criteria for a forcedsecond-order nonlinear differential equations with dampingrdquoAppliedMathematics and Computation vol 218 no 5 pp 1857ndash1861 2011
[27] Z Zheng X Wang and H Han ldquoOscillation criteria for forcedsecond order differential equations with mixed nonlinearitiesrdquoApplied Mathematics Letters vol 22 no 7 pp 1096ndash1101 2009
[28] J Shao F Meng and X Pang ldquoGeneralized variational oscil-lation principles for second-order differential equations withmixed-nonlinearitiesrdquoDiscrete Dynamics in Nature and Societyvol 2012 Article ID 539213 10 pages 2012
[29] J Shao and F Meng ldquoGeneralized variational principleson oscillation for nonlinear nonhomogeneous differential
![Fractional singular Sturm-Liouville problems on the half-line · PDF fileKittipoomAdvancesinDifferenceEquations20172017:310 Page2of22 inganappropriatekernel.Recently,Ansari[ ]introducedthefractionalSturm-Liouville](https://static.documents.pub/doc/80x56/5a7435ac7f8b9a63638babb6/fractional-singular-sturm-liouville-problems-on-the-half-line-kittipoomadvancesindierenceequations20172017310.jpg)
