This article was downloaded by: [Pennsylvania State University] On: 12 August 2014, At: 04:08 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK International Journal of Computer Mathematics Publication details, including instructions for authors and subscription information: http://www.tandfonline.com/loi/gcom20 Numerical Solution of Time-Fractional Fourth-Order Partial Differential Equations Shahid S. Siddiqi a & Saima Arshed b a Department of Mathematics, University of the Punjab, Lahore 54590, Pakistan. b Email: Accepted author version posted online: 11 Aug 2014. To cite this article: Shahid S. Siddiqi & Saima Arshed (2014): Numerical Solution of Time-Fractional Fourth-Order Partial Differential Equations, International Journal of Computer Mathematics, DOI: 10.1080/00207160.2014.948430 To link to this article: http://dx.doi.org/10.1080/00207160.2014.948430 Disclaimer: This is a version of an unedited manuscript that has been accepted for publication. As a service to authors and researchers we are providing this version of the accepted manuscript (AM). Copyediting, typesetting, and review of the resulting proof will be undertaken on this manuscript before final publication of the Version of Record (VoR). During production and pre-press, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal relate to this version also. PLEASE SCROLL DOWN FOR ARTICLE Taylor & Francis makes every effort to ensure the accuracy of all the information (the “Content”) contained in the publications on our platform. However, Taylor & Francis, our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content. Any opinions and views expressed in this publication are the opinions and views of the authors, and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information. Taylor and Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content. This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden. Terms & Conditions of access and use can be found at http:// www.tandfonline.com/page/terms-and-conditions
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This article was downloaded by: [Pennsylvania State University]On: 12 August 2014, At: 04:08Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registered office: MortimerHouse, 37-41 Mortimer Street, London W1T 3JH, UK
International Journal of Computer MathematicsPublication details, including instructions for authors and subscription information:http://www.tandfonline.com/loi/gcom20
Numerical Solution of Time-Fractional Fourth-OrderPartial Differential EquationsShahid S. Siddiqia & Saima Arshedb
a Department of Mathematics, University of the Punjab, Lahore 54590, Pakistan.b Email:Accepted author version posted online: 11 Aug 2014.
To cite this article: Shahid S. Siddiqi & Saima Arshed (2014): Numerical Solution of Time-Fractional Fourth-Order PartialDifferential Equations, International Journal of Computer Mathematics, DOI: 10.1080/00207160.2014.948430
To link to this article: http://dx.doi.org/10.1080/00207160.2014.948430
Disclaimer: This is a version of an unedited manuscript that has been accepted for publication. As a serviceto authors and researchers we are providing this version of the accepted manuscript (AM). Copyediting,typesetting, and review of the resulting proof will be undertaken on this manuscript before final publicationof the Version of Record (VoR). During production and pre-press, errors may be discovered which couldaffect the content, and all legal disclaimers that apply to the journal relate to this version also.
PLEASE SCROLL DOWN FOR ARTICLE
Taylor & Francis makes every effort to ensure the accuracy of all the information (the “Content”) containedin the publications on our platform. However, Taylor & Francis, our agents, and our licensors make norepresentations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose ofthe Content. Any opinions and views expressed in this publication are the opinions and views of the authors,and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be reliedupon and should be independently verified with primary sources of information. Taylor and Francis shallnot be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and otherliabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to orarising out of the use of the Content.
This article may be used for research, teaching, and private study purposes. Any substantial or systematicreproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in anyform to anyone is expressly forbidden. Terms & Conditions of access and use can be found at http://www.tandfonline.com/page/terms-and-conditions
Publisher: Taylor & FrancisJournal: International Journal of Computer MathematicsDOI: 10.1080/00207160.2014.948430
Numerical Solution of Time-Fractional
Fourth-Order Partial Differential Equations
Shahid S. Siddiqi ∗ and Saima Arshed †
Received 16 Jan 2014, revised 09 Jun 2014, accepted 12 Jul 2014
Abstract
A quintic B-spline collocation technique is employed for the numerical solution oftime-fractional fourth-order partial differential equations. These equations occur inmany applications in real life problems such as modeling of plates and thin beams, straingradient elasticity and phase separation in binary mixtures, which are basic elementsin engineering structures and are of great practical significance to civil, mechanicaland aerospace engineering. The time-fractional derivative is described in the Caputosense. Backward Euler scheme is used for time discretization and the quintic B-splinebased numerical method is used for space discretization. The stability and convergenceproperties related to the time discretization are discussed and theoretically proven. Thegiven problem is solved with three different boundary conditions, including clamped-type condition, simply supported-type condition, and a transversely supported-typecondition. Numerical results are considered to investigate the accuracy and efficiencyof the proposed method.
The subject fractional calculus is as old as the conventional calculus, but become not very popularfor a long time. However in current decade, fractional calculus attract enough attention, because ofits non-local property of fractional derivatives (and integrals). Thereby this considers the history andnon-local distributed effects. Oldham and Spanier [7], Podlubny [6] and Miller and Ross [8] providethe history and a comprehensive treatment of this subject. Fractional partial differential equations,as generalizations of classical integer order partial differential equations, are increasingly used tomodel problems in fluid flow, finance and other areas of application. Fractional derivatives providean excellent instrument for the description of memory and hereditary properties of various materialsand processes. Half-order derivatives and integrals prove to be more useful for the formulation of
∗Department of Mathematics, University of the Punjab, Lahore 54590, Pakistan.
certain electrochemical problems than the classical models. Fractional differentiation and integrationoperators are also used for extensions of the diffusion and wave equations. A great deal of effort hasbeen expended in attempting to find robust and stable numerical and analytical methods for solvingfractional partial differential equations of physical interest.The fourth-order problems play an important role in modern science and engineering. For example,bridge slabs, floor systems, window glasses and airplane wings can be modeled as plates with variousboundary supports which are governed by fourth-order partial differential equations.In this paper, the time-fractional fourth-order partial differential equation with a fractional derivativeof order α, (0 < α < 1) is considered as
∂αu
∂tα+ µ
∂4u
∂x4= f(x, t), x ∈ Ω = [0, L], 0 < t ≤ T, ( 1. 1 )
subject to the initial condition
u(x, 0) = g0(x), 0 ≤ x ≤ L,
where µ is the ratio of flexural rigidity of the beam to its mass per unit length, u is the transversedisplacement of the beam, t and x are the time and distance variables respectively. f(x, t) is thedynamic driving force per unit mass and g0(x) is continuous function.When α = 1, Eq.(1.1) reduces to fourth-order partial differential equation
∂u
∂t+ µ
∂4u
∂x4= f(x, t), x ∈ Ω = [0, L], 0 < t ≤ T. ( 1. 2 )
The following three types of boundary conditions are considered as
• For clamped boundary conditions:The clamped boundary condition is obtained by casting the end of a beam into concrete.The clamped boundary condition has zero displacement, but it also has zero slope. Thus, itsboundary conditions are
(I)
u(0, t) = u(L, t) = 0,ux(0, t) = ux(L, t) = 0, 0 ≤ t ≤ T.
• For simply-supported boundary conditions:Another simple boundary condition is the pinned end. The pinned end is free to rotate, soit cannot support a bending moment. Its displacement is zero, since it is pinned. Thus, thesimply-supported end has the following boundary conditions:
(II)
u(0, t) = u(L, t) = 0,uxx(0, t) = uxx(L, t) = 0, 0 ≤ t ≤ T.
• For a transversely supported boundary conditions:The boundary conditions are
The time-fractional derivative ∂αu∂tα in Eq.(1.1) is defined by the Caputo fractional derivative of
order α, given by
∂αu(x, t)
∂tα=
1Γ(1−α)
∫ t
0∂u(x,s)
∂sds
(t−s)α , 0 < α < 1,∂u(x,t)
∂t , α = 1.
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Yang et al. [4] developed a novel numerical scheme for solving fourth-order partial integro-differentialequation with a weakly singular kernel. In the time direction, a Crank-Nicolson time-stepping is usedto approximate the differential term and the product trapezoidal method is employed to treat theintegral term, and the quasi-wavelets numerical method for space discretization. Yang et al. [5] de-veloped a quasi-wavelet based numerical method for solving fourth-order partial integro-differentialequations with a weakly singular kernel. The forward Euler scheme is used for time discretiza-tion and the quasi-wavelet based numerical method for space discretization. The given problemis solved subject to three different boundary conditions, including clamped-type condition, simplysupported-type condition and a transversely supported-type condition. Khan et al. [10], proposeda numerical method for the solution of time-fractional fourth-order differential equations with vari-able coefficients using Adomian decomposition method (ADM) and He,s variational iteration method(HVIM). Zhang and Han [2] developed a quasi-wavelet based numerical method for time-dependentfractional partial differential equation. The time variable is discretized using the second order back-ward differentiation formula and the quasi-wavelet method is used for spatial discretization. Thestability and convergence properties related to the time discretization are discussed and theoreticallyproven. Zhang et al. [3] proposed quintic B-spline collocation method for the numerical solutionof fourth-order partial integro-differential equations with a weakly singular kernel. Khan et al. [1]used sextic spline solution for solving fourth-order parabolic partial differential equation. Dehghan[9] used finite difference techniques for the numerical solution of partial integro-differential equationarising from viscoelasticity. Li and Da [11] used finite central difference and finite element ap-proximations for the numerical solution of parabolic integro-differential equations. Lin and Xu [12]proposed finite difference and spectral approximations for the numerical solution of time-fractionaldiffusion equation.The aim of the paper, is to find the numerical solution of time-fractional fourth-order partial dif-ferential equations. The numerical solution is obtained using quintic B-spline collocation method.B-spline collocation method is widely used to solve differential and partial differential equations.The collocation method with B-spline basis functions represents an economical alternative, since itonly requires the evaluation of the unknown parameters at the grid points.The paper is organized into seven sections. Section 2, presents the detailed description about thequintic B-spline. In section 3, the backward Euler scheme is used for temporal discretization ofthe problem. The stability and error analysis related to time discretization are also discussed. Insection 4, quintic B-spline collocation method for the spatial discretization is discussed. In section 5,time-fractional semilinear fourth-order partial differential equation has been discussed. Numericalresults are presented in section 6, while the conclusion is presented in section 7.
2 The quintic B-spline
The interval [0, L] of domain has been subdivided as
0 = x0 < x1 < x2 < . . . < xn = L
To provide the support for the quintic B-spline near the end boundaries, ten additional knots havebeen introduced as
x−5 < x−4 < x−3 < x−2 < x−1 < x0
and
xn < xn+1 < xn+2 < xn+3 < xn+4 < xn+5.
The basis functions Bi(x), i = −2, . . . , N + 2 of quintic B-spline are defined as
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Bi(x) =1
h5
(x − xi + 3h)5, x ∈ [xi−3, xi−2],(x − xi + 3h)5 − 6(x − xi + 2h)5, x ∈ [xi−2, xi−1],(x − xi + 3h)5 − 6(x − xi + 2h)5 + 15(x − xi + h)5, x ∈ [xi−1, xi],(−x + xi + 3h)5 − 6(−x + xi + 2h)5 + 15(−x + xi + h)5, x ∈ [xi, xi+1],(−x + xi + 3h)5 − 6(−x + xi + 2h)5, x ∈ [xi+1, xi+2],(−x + xi + 3h)5, x ∈ [xi+2, xi+3],0, otherwise.
The values of successive derivatives B(v)i (x), i = −2, . . . , N + 2; v = 0, 1, 2, 3, 4 at nodes, are listed
in Table 1.
Table 1: Coefficients of quintic B-spline and its derivatives at nodes xi.xi−3 xi−2 xi−1 xi xi+1 xi+2 xi+3 else
Bi(x) 0 1 26 66 26 1 0 0B′
i(x) 0 5h
50h 0 −50
h−5h 0 0
B′′i (x) 0 20
h2
40h2
−120h2
40h2
20h2 0 0
B′′′i (x) 0 60
h3
−120h3 0 120
h3
−60h3 0 0
B(4)i (x) 0 120
h4
−480h4
720h4
−480h4
120h4 0 0
An approximation Un+1(x) to the exact solution un+1(x) at n + 1 time level, can be expressed interms of the quintic B-spline basis functions Bi(x) as
Un+1(x) =N+2∑
i=−2
ci(t)Bi(x) ( 2. 3 )
where ci are unknown time dependent quantities to be determined from the boundary conditionsand collocation form of the fractional differential equation.
3 Discretization in time: backward Euler scheme
The time-fractional derivative ∂αu(x,t)∂tα is discretized by the first-order backward Euler scheme. Let
tn = n∆t, n = 0, 1, 2, . . . ,K, in which ∆t = TK is the time step size. un(x) is an approximation to
the value of u(x, t) at a time point t = tn, n = 0, 1, . . . ,K − 1.The time-fractional derivative in Eq.(1.1) at time point t = tn+1, can be approximated as
∂αu(x, tn+1)
∂tα=
1
Γ(1 − α)
∫ tn+1
0
∂u(x, s)
∂s
ds
(tn+1 − s)α
=1
Γ(1 − α)
n∑
j=0
∫ tj+1
tj
∂u(x, s)
∂s
ds
(tn+1 − s)α
=1
Γ(1 − α)
n∑
j=0
u(x, tj+1) − u(x, tj)
∆t
∫ tj+1
tj
ds
(tn+1 − s)α+ rn+1
∆t
=1
Γ(1 − α)
n∑
j=0
u(x, tj+1) − u(x, tj)
∆t
∫ tn+1−j
tn−j
dτ
τα+ rn+1
∆t
=1
Γ(1 − α)
n∑
j=0
u(x, tn+1−j) − u(x, tn−j)
∆t
∫ tj+1
tj
dτ
τα+ rn+1
∆t
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=1
Γ(2 − α)
n∑
j=0
u(x, tn+1−j) − u(x, tn−j)
∆tα(
(j + 1)1−α − j1−α)
+ rn+1∆t
=1
Γ(2 − α)
n∑
j=0
bju(x, tn+1−j) − u(x, tn−j)
∆tα+ rn+1
∆t , ( 3. 4 )
where bj = (j +1)1−α − j1−α and τ = (tn+1−s). The coefficients bj possess the following properties
• bj > 0, j = 0, 1, 2, . . . , n,
• 1 = b0 > b1 > b2 > . . . > bn, bn → 0 as n → ∞,
•∑n
j=0(bj − bj+1) + bn+1 = (1 − b1) +∑n−1
j=1 (bj − bj+1) + bn = 1.
The semi-discrete differential operator Lαt is defined as
Lαt u(x, tn+1) :=
1
Γ(2 − α)
n∑
j=0
bju(x, tn+1−j) − u(x, tn−j)
∆tα.
Then Eq.(3.4) takes the following form
∂αu(x, tn+1)
∂tα= Lα
t u(x, tn+1) + rn+1∆t . ( 3. 5 )
The truncation error between Lαt u(x, tn+1) and ∂αu(x,tn+1)
∂tα in [12] is denoted by rn+1∆t , and
rn+1∆t ≤ Cu∆t2−α, ( 3. 6 )
where Cu is a constant only related to u.
Using Lαt u(x, tn+1) as an approximation of 1
Γ(1−α)
∫ tn+1
0∂u(x,s)
∂sds
(tn+1−s)α , the finite difference scheme
corresponding to (1.1) takes the following form
Lαt u(x, tn+1) + µ
∂4u(x, tn+1)
∂x4= f(x, tn+1),
1
Γ(2 − α)
n∑
j=0
bju(x, tn+1−j) − u(x, tn−j)
∆tα+ µ
∂4u(x, tn+1)
∂x4= f(x, tn+1).
The above equation can be rewritten as
un+1(x) + α0 µ∂4un+1
∂x4= (1 − b1)u
n(x) +
n−1∑
j=1
(bj − bj+1)un−j(x) + bnu0(x) + α0f
n+1(x),
n = 1, 2, . . . ,K − 1, ( 3. 7 )
where un+1(x) = u(x, tn+1) and α0 = Γ(2 − α)∆tα,along with the boundary conditions (I, II, III) and the initial condition
u0(x) = g0(x), x ∈ [0, L]. ( 3. 8 )
The proposed scheme is a three time level scheme. In order to apply the proposed scheme, it isnecessary to find the values of u at the nodal points at the zeroth (u0) and first (u1) time levels.For n = 1, the scheme (3.5) takes the following form
u2(x) + α0µ∂4u2
∂x4= (1 − b1)u
1(x) + b1u0(x) + α0f
2(x).
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For the special case n = 0, that is the first time step, the scheme simply takes the following form
u1(x) + α0µ∂4u1
∂x4= u0(x) + α0f
1(x), ( 3. 9 )
where u0(x) = u(x, 0) = g0(x) is the value of u at the zeroth time level (the initial condition).Eqs.(3.7) and (3.9), together with boundary conditions (I, II, III) and initial condition (3.8) forma complete set of the semi-dicscrete problem of Eqs.(1.1)-(1.2).Following [12], it will be useful to define the error term rn+1 by
rn+1 := α0
(
∂αu(x, tn+1)
∂tα− Lα
t u(x, tn+1)
)
. ( 3. 10 )
Using Eq.(3.5) and Eq.(3.6) the error term rn+1 becomes∣
∣rn+1∣
∣ = Γ(2 − α)∆tα∣
∣rn+1∆t
∣
∣ ≤ Cu∆t2. ( 3. 11 )
Some functional spaces endowed with standard norms and inner products, that will be used hereafter,are defined as under
H2(Ω) =
v ∈ L2(Ω),dv
dx,d2v
dx2∈ L2(Ω)
,
H20 (Ω) =
v ∈ H2(Ω), v|∂Ω = 0,dv
dx|∂Ω = 0
,
Hm(Ω) =
v ∈ L2(Ω),dkv
dxkfor all positive integer k ≤ m
,
where L2 (Ω) is the space of measurable functions whose square is Lebesgue integrable in Ω. Theinner products of L2 (Ω) and H2(Ω) are defined, respectively, by
(u, v) =
∫
Ω
uvdx, (u, v)2 = (u, v) +
(
du
dx,dv
dx
)
+
(
d2u
dx2,d2v
dx2
)
,
and the corresponding norms by
‖v‖0 = (v, v)12 , ‖v‖2 = (v, v)
12
2 .
The norm ‖.‖ of the space Hm (Ω) is defined as
‖v‖m =
(
m∑
k=0
∥
∥
∥
∥
dkv
dxk
∥
∥
∥
∥
2
0
)12
.
In this paper, instead of using the above standard H2-norm, it is prefer to define ‖.‖2 by
‖v‖2 =
(
‖v‖20 + α0µ
∥
∥
∥
∥
d2v
dx2
∥
∥
∥
∥
2
0
)1/2
, ( 3. 12 )
where α0 = Γ(2 − α)∆tα.In order to analize the stability and convergence, the following weak formulation of Eq.(3.7) andEq.(3.9) is required, i.e. finding un+1 ∈ H2
0 (Ω), such that for all v ∈ H20 (Ω),
(
un+1, v)
+ α0µ
(
∂4un+1
∂x4, v
)
= (1 − b1) (un, v) +
n−1∑
j=1
(bj − bj+1)(
un−j , v)
+ bn
(
u0, v)
+ α0
(
fn+1, v)
, ( 3. 13 )
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and
(
u1, v)
+ α0µ
(
∂4u1
∂x4, v
)
=(
u0, v)
+ α0
(
f1, v)
. ( 3. 14 )
The stability analysis for the semi-discrete problem is given in the following theorem.Theorem 1The semi-discrete problem is unconditionally stable in the sense that for all ∆t > 0, it holds
∥
∥un+1∥
∥
2≤
∥
∥u0∥
∥
0+ α0
n+1∑
j=1
∥
∥f j∥
∥
0
, n = 0, 1, 2, . . . ,K − 1, ( 3. 15 )
where ‖.‖2 is defined in (3.12).ProofMathematical induction is used to prove the theorem. When n = 0, let v = u1 in Eq.(3.14), it canbe written as
(
u1, u1)
+ α0µ
(
∂4u1
∂x4, u1
)
=(
u0, u1)
+ α0
(
f1, u1)
.
Using integration by parts two times, the above equation becomes
(
u1, u1)
+ α0µ
(
∂2u1
∂x2,∂2u1
∂x2
)
=(
u0, u1)
+ α0
(
f1, u1)
, ( 3. 16 )
where all the boundary contributions disappeared due to boundary conditions on v.Using the inequality ‖v‖0 ≤ ‖v‖2 and Schwarz inequality, Eq.(3.16) takes the following form
∥
∥u1∥
∥
2
2≤
∥
∥u0∥
∥
0
∥
∥u1∥
∥
0+ α0
∥
∥f1∥
∥
0
∥
∥u1∥
∥
0
≤∥
∥u0∥
∥
0
∥
∥u1∥
∥
2+ α0
∥
∥f1∥
∥
0
∥
∥u1∥
∥
2∥
∥u1∥
∥
2≤
(∥
∥u0∥
∥
0+ α0
∥
∥f1∥
∥
0
)
.
Suppose that the result hold for v = uji.e.
∥
∥uj∥
∥
2≤
(
∥
∥u0∥
∥
0+ α0
j∑
i=1
∥
∥f i∥
∥
0
)
, j = 2, 3, ..., n. ( 3. 17 )
Taking v = un+1 in Eq.(3.13), it can be written as
(
un+1, un+1)
+ α0µ
(
∂4un+1
∂x4, un+1
)
= (1 − b1)(
un, un+1)
+n−1∑
j=1
(bj − bj+1)(
un−j , un+1)
+ bn
(
u0, un+1)
+ α0
(
fn+1, un+1)
.
Using integration by parts two times, the above equation takes the following form
(
un+1, un+1)
+ α0µ
(
∂2un+1
∂x2,∂2un+1
∂x2
)
= (1 − b1)(
un, un+1)
+
n−1∑
j=1
(bj − bj+1)(
un−j , un+1)
+bn
(
u0, un+1)
+ α0
(
fn+1, un+1)
,
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where all the boundary contributions disappeared due to boundary conditions on v.Using the inequality ‖v‖0 ≤ ‖v‖2 and Schwarz inequality, the above equation becomes
∥
∥un+1∥
∥
2
2≤ (1 − b1) ‖u
n‖0
∥
∥un+1∥
∥
0+
n−1∑
j=1
(bj − bj+1)∥
∥un−j∥
∥
0
∥
∥un+1∥
∥
0+ bn
∥
∥u0∥
∥
0
∥
∥un+1∥
∥
0
+α0
∥
∥fn+1∥
∥
0
∥
∥un+1∥
∥
0,
∥
∥un+1∥
∥
2
2≤ (1 − b1) ‖u
n‖0
∥
∥un+1∥
∥
2+
n−1∑
j=1
(bj − bj+1)∥
∥un−j∥
∥
0
∥
∥un+1∥
∥
2+ bn
∥
∥u0∥
∥
0
∥
∥un+1∥
∥
2
+α0
∥
∥fn+1∥
∥
0
∥
∥un+1∥
∥
2,
∥
∥un+1∥
∥
2≤ (1 − b1) ‖u
n‖0 +
n−1∑
j=1
(bj − bj+1)∥
∥un−j∥
∥
0+ bn
∥
∥u0∥
∥
0+ α0
∥
∥fn+1∥
∥
0.
Using Eq.(3.17), the above equation becomes
∥
∥un+1∥
∥
2≤
∥
∥u0∥
∥
0+ α0
n∑
j=1
∥
∥f j∥
∥
0
(1 − b1) +
n−1∑
j=1
(bj − bj+1) + bn
+ α0
∥
∥fn+1∥
∥
0.
( 3. 18 )
Using properties of bj , it can be rewritten as
∥
∥un+1∥
∥
2≤
∥
∥u0∥
∥
0+ α0
n+1∑
j=1
∥
∥f j∥
∥
0
.
The error analysis for the solution of the semi-discrete problem is discussed in the followingtheorem.Theorem 2Let u be the exact solution of (1.1)-(1.2) and unK
n=0 be the time-discrete solution of Eqs.(3.13)and (3.14) with initial condition (3.8), then it holds
ProofLet en = u(x, tn) − un(x), for n = 1, by combining Eqs (1.1), (3.14) and (3.12), the error equationcan be written as
(
e1, v)
+ α0µ
(
∂2e1
∂x2,∂2v
∂x2
)
=(
e0, v)
+(
r1, v)
, ∀v ∈ H20 (Ω).
Let v = e1, noting e0 = 0 yields∥
∥e1∥
∥
2≤∥
∥r1∥
∥
0.
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This together with (3.11), gives
∥
∥u(t1) − u1∥
∥
2≤ Cub−1
0 ∆t2. ( 3. 21 )
Hence (3.20) is proved for the case n = 1.For inductive part, suppose (3.20) holds for n = 1, 2, 3, ..., s, i.e.
‖u(tn) − un‖2 ≤ Cub−1n−1∆t2. ( 3. 22 )
To prove the lemma for n = s + 1 the Eqs.(1.1), (3.13) and (3.11) are used and the error equationcan be written, for all v ∈ H2
0 (Ω), as
(
en+1, v)
+ α0µ
(
∂2en+1
∂x2,∂2v
∂x2
)
= (1 − b1) (en, v) +
n−1∑
j=1
(bj − bj+1)(
en−j , v)
+ bn
(
e0, v)
+(
rn+1, v)
.
The above equation, for v = en+1, can be written as
∥
∥en+1∥
∥
2
2≤ (1 − b1) ‖e
n‖0
∥
∥en+1∥
∥
0+
n−1∑
j=1
(bj − bj+1)∥
∥en−j∥
∥
0
∥
∥en+1∥
∥
0
+ bn
∥
∥e0∥
∥
0
∥
∥en+1∥
∥
0+∥
∥rn+1∥
∥
0
∥
∥en+1∥
∥
0.
Using the induction assumption and the fact thatb−1
j
b−1
j+1
< 1 for all non negative integer j, it can be
written as
∥
∥en+1∥
∥
2≤
(1 − b1) +
n−1∑
j=1
(bj − bj+1) + bn
Cub−1n ∆t2.
Using the properties of bj , the above equation becomes
∥
∥en+1∥
∥
2≤ Cub−1
n ∆t2.
ProofBy the definition of bn, it can be shown that
limn→∞
b−1n−1
nα= lim
n→∞
n−α
n1−α − (n − 1)1−α
= limn→∞
n−1
1 − (1 − 1n )1−α
=1
(1 − α)
Introduce a function Φ(x) := x−α
x1−α−(x−1)1−α .
Since Φ′
(x) ≥ 0 ∀x > 1 therefor Φ(x) is increasing on x for all x > 1. This means that n−αb−1n−1
increasingly tends to 1(1−α) as 1 < n → ∞. It is to be noted that n−αb−1
n−1 = 1 for n = 1, hence it
can be written as
n−αb−1n−1 ≤
1
(1 − α), n = 1, 2, . . . ,K.
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Consequently, for all n such that n∆t ≤ T ,
‖u(tn) − un‖2 ≤ Cub−1n−1∆t2
= Cun−αb−1n−1n
α∆t2−α+α
≤ Cu1
1 − α(n∆t)α∆t2−α
≤ Cu,αTα∆t2−α.
4 Discretization in space: quintic B-spline collocation method
Consider a uniform mesh with the grid points (xi, tn) to discretize the region [0, L] × [0, T ], wherexi = ih, i = 0, 1, 2, . . . , N , and tn = n∆t, n = 0, 1, 2, . . . ,K, K∆t = T . The quantities h and ∆t arethe grid sizes in the space and time directions, respectively.The space discretization of Eq.(3.7) is carried out using Eq.(2.3) and the collocation method isimplemented by identifying the collocation points as nodes. So, for i = 0, 1, 2, . . . , N the followingrelation can be obtained as
(
cn+1i−2 + 26cn+1
i−1 + 66cn+1i + 26cn+1
i+1 + cn+1i+2
)
+ α0µ120
h4
(
cn+1i−2 − 4cn+1
i−1 + 6cn+1i − 4cn+1
i+1 + cn+1i+2
)
= (1 − b1)(
cni−2 + 26cn
i−1 + 66cni + 26cn
i+1 + cni+2
)
+
n−1∑
j=1
(bj − bj+1)(
cn−ji−2 + 26cn−j
i−1 + 66cn−ji + 26cn−j
i+1 + cn−ji+2
)
+bn
(
c0i−2 + 26c0
i−1 + 66c0i + 26c0
i+1 + c0i+2
)
+ α0fn+1i , n = 1, 2, . . . ,K − 1. ( 4. 23 )
Simplifying the above relation leads to the following system of (N + 1) linear equations in (N + 5)unknowns cn+1
−2 , cn+1−1 , cn+1
0 , . . . , cn+1N+1, c
n+1N+2.
(
1 + α0µ120
h4
)
cn+1i−2 +
(
26 − 4α0µ120
h4
)
cn+1i−1 +
(
66 + 6α0µ120
h4
)
cn+1i
+
(
26 − 4α0µ120
h4
)
cn+1i+1 +
(
1 + α0µ120
h4
)
cn+1i+2 = Fi,
n = 1, 2, . . . ,K − 1, i = 0, 1, 2, . . . , N, ( 4. 24 )
where
Fi = (1 − b1)(
cni−2 + 26cn
i−1 + 66cni + 26cn
i+1 + cni+2
)
+
n−1∑
j=1
(bj − bj+1)(
cn−ji−2 + 26cn−j
i−1 + 66cn−ji + 26cn−j
i+1 + cn−ji+2
)
+bn
(
c0i−2 + 26c0
i−1 + 66c0i + 26c0
i+1 + c0i+2
)
+ α0fn+1i .
To obtain the unique solution of the system (4.24), four additional constraints are required. Theseconstraints are obtained from the boundary conditions (I, II, III). Imposition of the boundaryconditions enables us to eliminate the parameters c−2, c−1, cN+1 and cN+2 from the system (4.24).After eliminating c−2, c−1, cN+1, cN+2 the system is reduced to a penta-diagonal system of (N +1)linear equations in (N + 1) unknowns.
In order to find the value of C2 =[
c20, c
21, . . . , c
2N
]T, it is first needed to find the value of C1 =
[
c10, c
11, . . . , c
1N
]T. The value of C1 is obtained, solving Eq.(3.9) using quintic B-spline collocation
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method, as
(
1 + α0µ120
h4
)
c1i−2 +
(
26 − 4α0µ120
h4
)
c1i−1 +
(
66 + 6α0µ120
h4
)
c1i
+
(
26 − 4α0µ120
h4
)
c1i+1 +
(
1 + α0µ120
h4
)
c1i+2 =
(
c0i−2 + 26c0
i−1 + 66c0i + 26c0
i+1 + c0i+2
)
+ α0f1i ,
i = 0, 1, 2, . . . , N. ( 4. 25 )
The above Eq.(4.25) is a system of (N + 1) linear equations in (N + 5) unknowns c1−2, c
1−1, c1
0, . . . ,c1N , c1
N+1, c1N+2. To obtain the unique solution of this system, eliminate c−2, c−1, cN+1 and cN+2
using boundary conditions (I, II, III).The time evolution of the approximate solution Un+1 is determined by the time evolution of thevector Cn+1. This is found by repeatedly solving the recurrence relationship, once the initial vector
C0 =[
c00, c
01, . . . , c
0N
]T, has been computed from the initial condition.
The initial state vector C0 can be determined from the initial condition u(x, 0) = g0(x) which gives(N + 1) equations in (N + 5) unknowns. For determining these unknowns, the following relationsat the knots, are used
ux(x0, 0) = u′
(x0), ux(xN , 0) = u′
(xN ),
uxx(x0, 0) = u′′
(x0), uxx(xN , 0) = u′′
(xN ),
which give penta-diagonal system of equations written in the following matrix form
The time-fractional semilinear fourth-order partial differential equation with a fractional derivativeof order α, (0 < α < 1) is considered as
∂αu
∂tα+ µ
∂4u
∂x4+ h (x, t, u, ux, uxx, uxxx) = f(x, t), x ∈ Ω = [0, L], 0 < t ≤ T,
subject to the initial condition
u(x, 0) = g1(x), 0 ≤ x ≤ L.
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Using section 3, the required numerical scheme for time-fractional semilinear fourth-order partialdifferential equation is obtained as
un+1(x) + α0 µ∂4un+1
∂x4+ h
(
x, tn+1, un+1, un+1
x , un+1xx , un+1
xxx
)
= (1 − b1)un(x) +
n−1∑
j=1
(bj − bj+1)un−j(x) + bnu0(x) + α0f
n+1(x), ( 5. 27 )
n = 1, 2, . . . ,K − 1,
where un+1(x) = u(x, tn+1) and α0 = Γ(2 − α)∆tα,along with the boundary conditions (I, II, III) and the initial condition
u0(x) = g1(x), x ∈ [0, L].
The space discretization of Eq.(5.27) is carried out using Eq.(2.3) and the collocation method isimplemented by identifying the collocation points as nodes as discussed in section 4.
6 Numerical Results
In this section, the proposed method is tested on the five problems. Let tn = n∆t, n = 0, 1, 2, . . . ,K,h = 1
N , where K denotes the final time level tK and N + 1 is the number of nodes. In order tocheck the accuracy of the proposed method, the maximum norm errors and L2 norm errors betweennumerical and exact solutions are gr iven by the following definitions
Maximum norm error : ‖eK‖∞ = max0≤i≤N
∣
∣u (xi, tK) − UKi
∣
∣ ,
L2 norm error : ‖eK‖2 =1
N
(
N∑
i=0
∣
∣u (xi, tK) − UKi
∣
∣
2
)
12
.
Example 1Following is the time-fractional fourth-order partial differential equation
∂0.75u
∂t0.75+ 0.01
∂4u
∂x4= f(x, t), x ∈ [0, 4π], 0 < t ≤ T, ( 6. 28 )
with the initial condition
u(x, 0) = 1 − cos 2x, x ∈ [0, 4π],
and clamped boundary condition
u(0, t) = u(4π, t) = 0,
ux(0, t) = ux(4π, t) = 0, 0 ≤ t ≤ T.
The exact solution of the problem is
u(x, t) = 2 (t + 1) sin2 x.
The maximum norm error and L2 norm error for two different grid sizes N = 50 and N = 100 with∆t=0.00001 are presented in Table 2. In Table 2, ∆t is chosen small enough to avoid contaminationof temporal error and attribute most of the errors to the spatial discretization. The time level K are
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Table 2: The errors ‖eK‖∞ and ‖eK‖2 for different N taken ∆t=0.00001.N K ‖eM‖∞ ‖eM‖2
50 100 2.0964 × 10−5 1.7489 × 10−6
500 6.9909 × 10−5 5.9425 × 10−6
1000 1.17692 × 10−4 1.0453 × 10−5
1500 1.5975 × 10−4 1.4443 × 10−5
100 100 1.4771 × 10−5 1.0535 × 10−6
500 1.7313 × 10−5 1.0858 × 10−6
1000 2.9146 × 10−5 1.5952 × 10−6
1500 3.9562 × 10−5 2.1897 × 10−6
Table 3: The errors ‖eK‖∞ and ‖eK‖2 of different time steps with N = 100∆t ‖eK‖∞ Rate ‖eK‖2 Rate
varied to test the accuracy of the proposed method, which indicates that the proposed method issubstantially efficient. It can be observed from the Table 2 that the proposed method approximatesthe exact solution very efficiently.The maximum norm error and L2 norm error for different time steps ∆t with N = 100 and thecorresponding rates of convergence at T = 0.1 are presented in Table 3. From Table 3, it is easy tosee that the temporal convergence rate of the numerical results obtained by the proposed method isin good agreement with the theoretical estimation.In order to indicate the effects of the proposed method for larger time level K, the exact solutionand the numerical solution are plotted using N = 40, K = 1000 and ∆t = 0.00001 as shown inFig. 1. It is clear from the Fig. 1 that the numerical solution is highly consistent with the exactsolution, which indicates that the proposed method is very effective. In Fig. 2, the exact solution isrepresented by solid line and the numerical solution is represented by dotted line at K = 1000 timelevel.The temporal rate of convergence of L2 norm error as a function of the time step ∆t for α = 0.75 isshown in Fig. 3.
Exact solution
1.0
1.5
2.0t
020
40x
0.0
0.5
1.0
1.5
2.0
u
Numerical solution
1.0
1.5
2.0t
020
40x
0.0
0.5
1.0
1.5
2.0
u
Figure 1: The results at N=40, K=1000 and ∆t = 0.00001 for Example 1
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10 20 30 40 50x
0.5
1.0
1.5
2.0
u
Figure 2: The exact and numerical solution at K=1000. Dotted line: numerical solution,Solid line: exact solution
Figure 3: Errors as a function of the time ∆t for α = 0.75
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Example 2Following is the time-fractional fourth-order partial differential equation taking α = 0.5
∂0.50u
∂t0.50+ 0.05
∂4u
∂x4= f(x, t), x ∈ [0, 1], 0 < t ≤ T, ( 6. 29 )
For α = 1, the above equation becomes
∂u
∂t+ 0.05
∂4u
∂x4= f(x, t), x ∈ [0, 1], 0 < t ≤ T, ( 6. 30 )
with the initial condition
u(x, 0) = sinπx, x ∈ [0, 1],
and simply-supported boundary condition
u(0, t) = u(1, t) = 0,
uxx(0, t) = uxx(1, t) = 0, 0 ≤ t ≤ T.
The exact solution of the problem is
u(x, t) = (t + 1) sinπx.
The maximum norm error and L2 norm error for two different grid sizes N = 40 and N = 80 with∆t=0.00001 are presented in Table 4 for both the cases α = 0.5 and α = 1. In Table 4, ∆t ischosen small enough to avoid contamination of temporal error and attribute most of the errors tothe spatial discretization. The time level K are varied to test the accuracy of the proposed method,which indicates that the proposed method is substantially efficient. It can be observed from theTable 4 that the proposed method approximates the exact solution very efficiently.The maximum norm error and L2 norm error for different time steps ∆t with N = 100 and thecorresponding rates of convergence at T = 0.1 are presented in Table 5 for both the cases α = 0.5and α = 1. From Table 5, it is easy to see that the temporal convergence rate of the numericalresults obtained by the proposed method for time-fractional fourth-order partial differential equationis of O(∆t2−α) and for fourth-order partial differential equation is of O(∆t).In order to indicate the effects of time-fractional fourth-order PDE approach for larger time level K,the exact solution and the numerical solution are plotted using N = 40, K = 500 and ∆t =0.00001as shown in Fig. 4. It is clear from the Fig. 4 that the numerical solution is highly consistentwith the exact solution, which indicates that the proposed method is very effective. In Fig. 5, theexact solution is represented by solid line and the numerical solution is represented by dotted lineat K = 1000 time level.The temporal rate of convergence of L2 norm error as a function of the time step ∆t for α = 0.50 isshown in Fig. 6.
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Table 4: The errors ‖eK‖∞ and ‖eK‖2 for different N taken ∆t=0.00001.α = 0.5 N K ‖eM‖∞ ‖eM‖2
40 100 1.2964 × 10−4 1.4494 × 10−5
500 2.3241 × 10−4 2.5985 × 10−5
1000 2.8285 × 10−4 3.1624 × 10−5
1500 3.1247 × 10−4 3.4935 × 10−5
80 100 2.4905 × 10−5 1.9689 × 10−6
500 5.0599 × 10−5 4.0002 × 10−6
1000 6.3211 × 10−5 4.9972 × 10−6
1500 7.0617 × 10−5 5.5828 × 10−6
α = 1 N K ‖eM‖∞ ‖eM‖2
40 100 2.4509 × 10−6 2.7402 × 10−7
500 1.2161 × 10−5 1.3596 × 10−6
1000 2.4091 × 10−5 2.6934 × 10−6
1500 3.5796 × 10−5 4.0021 × 10−6
80 100 5.7615 × 10−7 4.5548 × 10−8
500 2.8591 × 10−6 2.2603 × 10−7
1000 5.6648 × 10−6 4.4785 × 10−7
1500 8.4186 × 10−6 6.6555 × 10−7
Exact solution
1.0 1.5 2.0
t
1020 30 40
x
0.0
0.5
1.0
u
Numerical solution
1.0 1.5 2.0
t
1020 30 40
x
0.0
0.5
1.0
u
Figure 4: The results at N=40, K=500 and ∆t = 0.00001 for Example 2
0.2 0.4 0.6 0.8 1.0x
0.05
0.10
0.15
0.20
0.25
0.30
u
Figure 5: The exact and numerical solution at K=1000. Dotted line: numerical solution,Solid line: exact solution
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Table 5: The errors ‖eK‖∞ and ‖eK‖2 of different time steps with N = 100α = 0.5 ∆t ‖eK‖∞ Rate ‖eK‖2 Rate
Figure 6: Errors as a function of the time ∆t for α = 0.5
Example 3Following is the time-fractional fourth-order partial differential equation
∂0.90u
∂t0.90+ 0.01
∂4u
∂x4= f(x, t), x ∈ [0, 1], 0 < t ≤ T, ( 6. 31 )
with the initial condition
u(x, 0) =
(
π5 sinπx +1
π5cos πx −
1
π5cos 3πx
)
, x ∈ [0, 1],
and transversely supported boundary condition
u(0, t) = u(1, t) = 0,
uxx(0, t) −8
π9ux(0, t) = uxx(1, t) −
8
π9ux(1, t) = 0, 0 ≤ t ≤ T.
The exact solution of the problem is
u(x, t) = (t + 1)
(
π5 sinπx +1
π5cos πx −
1
π5cos 3πx
)
.
The maximum norm error and L2 norm error for two different grid sizes N = 20 and N = 40 with∆t=0.00001 are presented in Table 6. In Table 6, ∆t is chosen small enough to avoid contamination
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Table 6: The errors ‖eK‖∞ and ‖eK‖2 for different N taken ∆t=0.00001.N K ‖eM‖∞ ‖eM‖2
20 100 1.7601 × 10−4 2.7996 × 10−5
500 3.0843 × 10−4 4.8934 × 10−5
1000 3.7339 × 10−4 5.9205 × 10−5
1500 4.1152 × 10−4 6.5235 × 10−5
40 100 4.8103 × 10−5 5.4639 × 10−6
500 8.6046 × 10−5 9.7178 × 10−6
1000 1.0469 × 10−4 1.1807 × 10−5
1500 1.1564 × 10−4 1.3034 × 10−5
of temporal error and attribute most of the errors to the spatial discretization. The time level K arevaried to test the accuracy of the proposed method, which indicates that the proposed method issubstantially efficient. It can be observed from the Table 6 that the proposed method approximatesthe exact solution very efficiently.The maximum norm error and L2 norm error for different time steps ∆t with N = 100 and thecorresponding rates of convergence at T = 0.1 are presented in Table 7. From Table 7, it is easy tosee that the temporal convergence rate of the numerical results obtained by the proposed method isin good agreement with the theoretical estimation.In order to indicate the effects of the proposed method for larger time level K, the exact solutionand the numerical solution are plotted using N = 80, K = 1000 and ∆t = 0.00001 as shown inFig. 7. It is clear from the Fig. 7 that the numerical solution is highly consistent with the exactsolution, which indicates that the proposed method is very effective. In Fig. 8, the exact solution isrepresented by solid line and the numerical solution is represented by dotted line at K = 1000 timelevel.The temporal rate of convergence of L2 norm error as a function of the time step ∆t for α = 0.90 isshown in Fig. 9.
Table 7: The errors ‖eK‖∞ and ‖eK‖2 of different time steps with N = 100∆t ‖eK‖∞ Rate ‖eK‖2 Rate
Example 4Following is the time-fractional semilinear fourth-order partial differential equation
∂0.5u
∂t0.5+ 0.1
∂4u
∂x4+ u(1 + u) = f(x, t), x ∈ [0, 1], 0 < t ≤ T, ( 6. 32 )
with the initial condition
u(x, 0) = sinπx, x ∈ [0, 1],
and simply-supported boundary condition
u(0, t) = u(1, t) = 0, t ≥ 0
uxx(0, t) = uxx(1, t) = 0, t ≥ 0.
The exact solution of the problem is
u(x, t) = (t + 1) sinπx.
The Maximum norm error and L2 norm error for two different grid sizes N = 40 and N = 80 with∆t = 0.00001 are presented in Table 8. In Table 8, ∆t is chosen small enough to avoid contaminationof temporal error and attribute most of the errors to the spatial discretization. The Time level Kare varied to test the accuracy of the proposed method, which indicates that the proposed method issubstantially efficient. It can be observed from the Table 8 that the proposed method approximatesthe exact solution very efficiently.The Maximum norm error and L2 norm error for different time steps ∆t with N = 80 at T = 0.1are presented in Table 9. From Table 9, it is easy to see that the temporal convergence rate ofthe numerical results obtained by the proposed method is in good agreement with the theoreticalestimation.
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Table 10: The errors ‖eK‖∞ and ‖eK‖2 for different N taken ∆t = 0.00001N K ‖eM‖∞ ‖eM‖2
50 100 4.9710 × 10−4 3.7272 × 10−5
500 4.5161 × 10−4 3.7668 × 10−5
1000 4.5545 × 10−4 3.7977 × 10−5
1500 4.5897 × 10−4 3.8253 × 10−5
100 100 2.5416 × 10−4 1.5143 × 10−5
500 2.5594 × 10−4 1.5263 × 10−5
1000 2.5845 × 10−4 1.5413 × 10−5
1500 2.6085 × 10−4 1.5561 × 10−5
Example 5Following is the time-fractional semilinear fourth-order partial differential equation
∂0.99u
∂t0.99+ 10
∂4u
∂x4+ uux = f(x, t), x ∈ [0, 1], 0 < t ≤ T, ( 6. 33 )
with the initial condition
u(x, 0) = sinπx, x ∈ [0, 1],
and simply-supported boundary condition
u(0, t) = u(1, t) = 0, t ≥ 0
uxx(0, t) = uxx(1, t) = 0, t ≥ 0.
The exact solution of the problem is
u(x, t) = (t + 1) sinπx.
The Maximum norm error and L2 norm error for two different grid sizes N = 50 and N = 100with ∆t = 0.00001 are presented in Table 10. In Table 10, ∆t is chosen small enough to avoidcontamination of temporal error and attribute most of the errors to the spatial discretization. TheTime level K are varied to test the accuracy of the proposed method, which indicates that theproposed method is substantially efficient. It can be observed from the Table 10 that the proposedmethod approximates the exact solution very efficiently.The Maximum norm error and L2 norm error for different time steps ∆t with N = 60 at T = 0.1are presented in Table 11. From Table 11, it is easy to see that the temporal convergence rate ofthe numerical results obtained by the proposed method is in good agreement with the theoreticalestimation.
Table 11: The errors ‖eK‖∞ and ‖eK‖2 of different time steps with N = 60∆t ‖eK‖∞ Rate ‖eK‖2 Rate
In this paper, an efficient numerical method has been developed for the solution of time-fractionalfourth-order partial differential equations. The backward Euler scheme is used in time direction andquintic B-spline collocation method for spatial derivatives. It has been shown that the discretizationin time is unconditionally stable and the numerical solution converges to exact solution with orderO(∆t2−α). The quintic B-spline collocation method exhibits high degree of accuracy in dealingwith the given problems. The performance of the proposed method for the considered problemswas measured by calculating the maximum norm error and L2-norm error, presented in Tables 2-11.The parameters N , ∆t and K are varied in order to test the accuracy of the proposed method. Theproposed method is also valid and efficient for different values of α, (0 < α < 1) and provide efficientnumerical results for the three boundary conditions (I, II, III). It is observed from the numericalexamples that the proposed method possesses high degree of efficiency and accuracy.
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