Investigation of New Explicit of Solutions for Multi-Dimensional Nonlinear Differential
Equations Using Lie Symmetry Reduction with the Integrating Factors Method
Sadat R1* and Kassem M2
1Research student, Faculty of Engineering, Physics and Mathematics Department, Zagazig University, Egypt
2Professor of Mathematics, Faculty of Engineering, Physics and Mathematics Department, Zagazig University, Egypt
*Corresponding author:
Sadat R, Research student, Faculty of Engineering, Physics and Mathematics Department, Zagazig University, Egypt
Received: July 31, 2019
Published: August 23, 2019
Volume: 01; Issue: 03
To Cite This Article:Sadat R, Kassem M. Investigation of New Explicit of Solutions for Multi-Dimensional Nonlinear Differential Equations Using Lie Symmetry Reduction with the Integrating Factors Method. Peer Res Nest. 2019 - 1(3)PNEST.19.08.025.
This work is licensed under Creative Commons Attribution 4.0 License
IntroductionRecently, many methods are applied to reach analytical solutions of NLPDEs as Darboux
transformation [1, 2]. Tanh-Coth method [3], Hirota bilinear method [4, 5], HBM [6], the (G^’/G) expansion method [7], Exp-Function method [8] and Rational Function transformations [9]. Here, two equations namely, the (2+1)-dimensional Calogero-Bogoyavlenskii-Schiff (CBS) equation and the (3+1)-dimensional generalized BKP equation have been solved by Lie symmetry Reduction method [10-12]. Using the Lie Reduction method and (IF), we derive a novel combination of solu-tions for these equations.
Mathematical FormulationIn this paragraph, we reduce the nonlinear evolution equations to (ODEs) in three steps. For
each used Lie vector, we apply the following steps;
I. The independent variables (x; y; z; t) are reduced to a (PDE) in two variables (r, s).
II. EvaluateLieinfinitesimalsforthesePDEthenusetheevaluatedsymmetriesforareduc-tionofindependentvariablesfrom(r;s)toonevariable(η).
III. The reduced ODE is non-solvable equations, through their corresponding IF are reduced to new solvable ones.
where u(x, y,t) describes the interaction of Riemann wave propagation in the x-direction with thelong-wavepropagationinthey-direction[13].Najafietal.[3]deriveanalyticalsolutionsof(1) using the Tanh-Coth method. In (2016), San et al. [14] obtained the Noether-type operators associated with the partial Lagrangian for all possible arbitrary functions, followed by a double re-duction using symmetries. In (2017), Saleh et al. [15] used Lie symmetry with the SMM to present new solutions of CBS equation (1).
Here, we apply two stages of symmetry reduction method to reduce the (CBS) equation to ODEs. During the reduction process, some of the obtained ODEs had no quadrature. We thus solve them using their (IF). Equation (1) has 24 Lie vectors. From the adjoint table, three optimal vec-tors are deduced.
Abstract
Two stages of Lie symmetry reduction have been applied to (2+1) Calogero-Bogoyavlenskii-Schiff (CBS) and (3+1)-dimensional generalized BKP equations. These reductions lead in some cases, to (ODE’s) having no quadrature. Using the Integrating Factors (IF) method, we explore new closed form solutionsfortheseequations.Acomparisonwithpreviousworkinthefieldhasbeenpresented.
These vectors are utilized to reduce (CBS) (1) three independent vari-ables; (x; y; t) to two independent variables; (r; s)
Reduction of the independent variables in (CBS) equation using X2 Lie vector
Equation (1) is transformed through the vector 2X
t u∂ ∂
= +∂ ∂
to;
4 2 0r rs s rr rrsF F F F F+ + = (3)
where r = x, s = y and ( ) ( ), , ,F r s u x y t t= −. The above
equation has no exact solution but has six Lie vectors. We choose here to deal only with V1, V5 Lie vector as they lead to ODE’s having no closed form solution.
51 F, V rV
s rrF
s −
∂ ∂ ∂= + −
∂ ∂∂ ∂+
∂ ∂=
∂ (4)
Reduction using V1
Using V1 (CBS) transform to a nonlinear fourth degree ODE of the following form;
6 0ηη η ηηηθ θ θ− = (5)
a. Using Integrating Factor to obtain an exact solution
Wefirstdeduceequation(5)IFusingmaple.
1 2, 1ηµ θ µ= = (6)
The IF reduce equation (5) to;
3 22 0η ηηθ θ− = (7)
This equation has a closed form solution of the form;
( ) 21
2 cc
θ ηη
= − ++ (8)
where ( ) ( ) 1 2, , ,andF c cr s r sη θ η= − + = are the integration constants? Back substituting to (x, y, t) for
( ) ( ), , , , ,r x s y F r s u x y t t= = = − , we get;
( ) 21
2, ,u x y t c tx y c
= − + +− + + (9)
This solution in (9) is plotted in Figure 1 for two different times; t=0, t= 20 sec.
Sadat R*Research student, Faculty of Engineering, Physics and Mathematics
Department, Zagazig University, Egypt
Figure 1: u (x, y, t) at c1 = 1, c2 =1, t=0 sec.
The wave in Figure 1 shows a row of peakons moving down-ward as time passes from zero to twenty.
Reduction using V5
Using the V5, (CBS) equation is reduced to a nonlinear ODE of the form;
210 6 8 4 6 11 6 0η η ηη η η η η ηηη ηηηηθ θ θ θθ θθ θ θ θ θ+ + + − − − − = (10)
The above equation has no exact solution. Using the integrat-ing factor, we can reduce it to
The IF reduce equation (10) to;3 2 2 22 4 3 4 0η ηη η η ηη ηηθ θθ θ θ θ θ+ − − − = (12)
This equation has an exact solution of the form;
(13)
where ( ) ( ) ( )ln , ,r s r F r sη θ η= − + = and c1, c2 are integration constants. Then back substituting to (x, y, t) where r=x, s=y, F (r, s) =u (x, y, t)-t, we obtain;
( ) ( ) ( ) ( ) ( )1 1 2 1 1, , 2 tanh 0.5 ln 5 4 0.5 5( ( ( ) 4 /4 5 ))u x y t c F x y c c c c t x= + − − + + + + + + (14)
This solution is plotted in Figure 2 in a complex domain. The peakon waves depicted in Figure 2 decay with time.
Figure 2(a): u (x, y, t) at c1 = 1, c2=1, t=0 sec.
Figure 2(b): u (x, y, t) at c1 = 1, c2=1, t=0 sec.
By comparison of the results obtained in [15], our solution is new
Reduction of the independent variables in (CBS) equation using X7 Lie vector
Equation (1) is transformed through the optimal vector X_7 to
( )16 8 8 0s rs r ss ssrF F F F F+ − = (1SS5)
where ( ) ( )2, 2 , , , , 0.5r y s x t F r s u x y t ty= = − + = − ,
This equation hasn’t an exact solution but has a six Lie vectors. We choose to work only with V1. This Lie vector leads to an ODE with no analytic solution. While the rest of the vector lead to solv-able ODEs.
1Vr s∂ ∂
= +∂ ∂
(16)
Using V1 transform (CBS) to a nonlinear fourth degree ODE of the following form;
3 0ηη η ηηηηθ θ θ− = (17)
b. Using Integrating Factor to obtain a closed form solution
WefirstinvestigatetheIFof(17)usingmaple.
1 2, 1ηµ θ µ= = (18)
We use the two-integrating factor to reduce equation (17) to;
3 2 0η ηηθ θ− = (19)
This equation has an explicit solution in the form;
( ) 21
4n cc
θη
= − ++ (20)
where ( ) ( ) ,, F r sr sη θ η= − + = and c1, c2 are integration constants.
Then back to (x, y, t) coordinates where ( ) ( )2, 2 , , , , 0.5r y s x t F r s u x y t ty= = − + = − , we obtain;
( ) 221
4, , 0.52
u x y t c tyy x t c
= − + +− − + + (21)
This solution is plotted in Figure 3. The wave peak position changes with time.
Figure 3: u (x, y, t) at c1 = 1, c2 =1, t=0, 1.5 sec.
Reduction of the independent variables in (CBS) equation using X21 Lie vector
Equation (1) is transformed through the vector X21 to;
28 2 4 2 4 0 r ss sss s rs r r sss sssF FF F rF F F F s F+ + + + + = (22)
The reduced equation (22) has no closed form solution but have six Lie vectors. We choose to work with V2, as follow;
In (2016), Yasar [11] use Lie symmetry analysis, Riccati equationandpowerseriesmethodtointroducespecificsolutionsfor (3+1) BKP equation. Also, Jadaun et al. [10] apply prorogation theorem to get the similarity variables and use these generators to solve the equation. In this paper, we consider the BKP equation in (3+1) dimensional space as [16, 17];
3 3 6 3 0yt xxxy y xx x xy xx zzu u u u u u u u− − − + − = (29)
This equation has 36 Lie vectors, and we will choose 13X to reduce the equation;
13 1 9 13,X tz X X X tzz u x y z u∂ ∂ ∂ ∂ ∂ ∂
= + + + = + + +∂ ∂ ∂ ∂ ∂ ∂
(30)
The PDE (29) functionoffourindependentvariables;(x;y;t;z)isfirstreducedto a PDE in three independent variables, (l; h; o), using its Lie vec-tors (30) then reduce to two independent variables (r; s) and then oneindependentη.
Using X13 Lie Vector to reduce the Independent Variables in (BKP) Equation
Equation (29) is transformed through the vector 13X tzz u∂ ∂
= +∂ ∂ to;
( ) 3 3 6 3 0oh lllh ll h lh l llK K K K K K K o− − − + − = (31)
This equation has no closed form solution, but possesses 12 Lie vectors; we choose V1;
1Vl h∂ ∂
= +∂ ∂ (32)
This vector transforms the equation (31) to;
( )6 6 3 0rs r rr rr rrrrF F F F F s− + + − = (33)
This equation has not closed-form solution, but have eight Lie vectors; we will choose here to work only with e5
3 25
1 2 18 1 43 3 3
e r s s rs r Fr s F∂ ∂ ∂ = − + + + + − ∂ ∂ ∂
(34)
Using e_5, transform (33) to the following form;
54 18 0ηη ηηηη ηη ηθ θ θ θ− − + = (35)
This is ODE has no analytical solution.
d. Using the Integrating Factor to obtain an explicit solution
Wefirstdeduceequation(35)IFusingmaple.
1 2, 1ηµ θ µ= = (36)
The IF reduce equation (35) to;
2 3 26 54 0ηη η ηθ θ θ− + = (17)
This equation has a closed-form solution of the form;
( ) ( ) ( ) ( ) 2, , , , , ; , , , , , , , , 0.5r l h s o F r s K l h o h y l x o t K l h o u x y t z z= =+ = = −= − = =
we obtain;( ) ( ) ( ) ( )
( )
3 1 3 31 1
2 5 22
3 6 1 3 6 3 6 1 3 6 1, , , 6 tan 6 tan tan 92 3 2 2 3 2 3
27 30.510 2
u x y z t x y t c x y t c x y t
c tz t t t x y
− = − + + + − − + + + + − + +
+ + + + + − +
(39)
This solution is plotted in Figure 5 for different values of times. It shows a change in the peak wave position as time increases and by increasing the time the surface of the wave follows a parabolic path whose amplitude decreases with time.
Figure 5(a): u (x, y, t, z) at c1 = 1, c2 = 1, y=0, t = 0.1 sec
Figure 5(b): u (x, y, t, z) at c1 = 1, c2 = 1, y=0, t = 0.3 sec.
Using 1 9 13X X X+ + Lie Vector to reduce the Independent
Variables in (BKP) Equation
Equation (29) is transformed through the vector 1 9 13X X X tz
x y z u∂ ∂ ∂ ∂
+ + = + + +∂ ∂ ∂ ∂ to;
( )3 3 6 9 3 3 3 3 123 3 6 3 3 0lh llll olll ooll oool l ll ol l l oo l ll o lo o lo
ll lo ll oo
K K K K K K K K K K K hK K K K K KohK ohK K K h+ + + + − − − + − − +
+ + + + − = (40)
This equation has not closed-form solution, but possesses 12 Lie vectors; we choose V1;
1 2V hol o K∂ ∂ ∂
= + +∂ ∂ ∂
(41)
Equation (40) is transformed to;
23 1 3 3 3 3 30.5 0.5 04 16 4 4 2 4 2ss ssss rs s s ss sss srF F F rF F F F sr r− − − − + − + − = (42)
This equation has not closed-form solution, but has eight Lie vectors; we choose to work only with e5;
2 2 35
3 91 0.5 62 2
e r s r s rsr s F∂ ∂ ∂ = + + + + − ∂ ∂ ∂ (43)
As this Lie vector leads to an ODE with no analytic solution. Using e5, transform (BKP) to a nonlinear fourth degree ODE of the form;
24 12 0ηη ηηηη ηη ηθ θ θ θ− − + = (44)
e. Using Integrating Factor to get an explicit solution
( ) ( ) ( )( )11 1 26 tan 6 6 6 tan tan 6 6 6c c cθ η η η η−= + − + + + (47)
where ( ) ( )3 5 2 23 27 1, , 3 12 10 2
r s F r s s r r s s rη θ η = − + = = − + + − + Then back to (l; h; o).
coordinates with
( ) ( ) 2, , , , , , , 0.5 l x y o x z K l h o u x y t z x t xtz= − + = − + = + −
That leads to;
(48)
This wave is plotted in Figure 6. A series of solitons inverse its flowatt=0.1sec.
The groups of soliton waves are decay with time and run to-wards left. Here, we compare our result in (48) with two solutions in[10,11],wefindthat;
i. Yasar [11], used some of the generators and obtained some traveling wave solution. His solutions in most cases, localized only in three dependent variables (x, y, t). We plot his result presented by X4 in Figure 7;
ii. In Figure 7, he presented travelling wave solution while we found a multi peaks solution as depicted in Figure 6.
iii. The solution in equation (23) presented in [10] in three variables (x, y, t) is different from our solution.
Conclusion
Here, we reduce the PDEs to ODEs through Lie vectors as previously done in [17] through two reduction stages. Some of
these ODEs have no solution. Some researchers in this step, use the SMM, power series method or Riccati equation method to solve non-solvable equations. We use the integrating factors as a tool to reduce the order and the nonlinearity in an ODE. This explore to
new solutions as it appears for the (2+1)-dimensional (CBS) and (3+1)-dimensional generalized BKP solutions compared with [14] and [10] results.
Figure 6(a): u (x, y, t, z) at c1 = 1, c2 = 1, z=0.1 at time t = 0.1 sec. Figure 6(b): u (x, y, t, z) at c1 = 1, c2 = 1, z=0.1 at time t = 1 sec.
Figure 7: u (x, y, t, z) for c1 = 1, c2 = c4 = c5 = at time t = 0.
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