Open access peer-reviewed chapter

Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method

By Tarig M. Elzaki

Submitted: October 25th 2017Reviewed: December 20th 2017Published: May 23rd 2018

DOI: 10.5772/intechopen.73291

Downloaded: 571

Abstract

Nonlinear equations are of great importance to our contemporary world. Nonlinear phenomena have important applications in applied mathematics, physics, and issues related to engineering. Despite the importance of obtaining the exact solution of nonlinear partial differential equations in physics and applied mathematics, there is still the daunting problem of finding new methods to discover new exact or approximate solutions. The purpose of this chapter is to impart a safe strategy for solving some linear and nonlinear partial differential equations in applied science and physics fields, by combining Laplace transform and the modified variational iteration method (VIM). This method is founded on the variational iteration method, Laplace transforms and convolution integral, such that, we put in an alternative Laplace correction functional and express the integral as a convolution. Some examples in physical engineering are provided to illustrate the simplicity and reliability of this method. The solutions of these examples are contingent only on the initial conditions.

Keywords

  • nonlinear partial differential equations
  • Laplace transform
  • modified variational iteration method

1. Introduction

In the recent years, many authors have devoted their attention to study solutions of nonlinear partial differential equations using various methods. Among these attempts are the Adomian decomposition method, homotopy perturbation method, variational iteration method (VIM) [1, 2, 3, 4, 5], Laplace variational iteration method [6, 7, 8], differential transform method and projected differential transform method.

Many analytical and numerical methods have been proposed to obtain solutions for nonlinear PDEs with fractional derivatives such as local fractional variational iteration method [9], local fractional Fourier method, Yang-Fourier transform and Yang-Laplace transform and other methods. Two Laplace variational iteration methods are currently suggested by Wu in [10, 11, 12, 13]. In this chapter, we use the new method termed He’s variational iteration method, and it is employed in a straightforward manner.

Also, the main aim of this chapter is to introduce an alternative Laplace correction functional and express the integral as a convolution. This approach can tackle functions with discontinuities as well as impulse functions effectively. The estimation of the VIM is to build an iteration method based on a correction functional that includes a generalized Lagrange multiplier. The value of the multiplier is chosen using variational theory so that each iteration improves the accuracy of the result.

In this chapter, we have applied the modified variational iteration method (VIM) and Laplace transform to solve convolution differential equations.

2. Combine Laplace transform and variational iteration method to solve convolution differential equations

In this section, we combine Laplace transform and modified variational iteration method to figure out a new case of differential equation called convolution differential equations; it is possible to obtain the exact solutions or better approximate solutions of these equivalences. This method is utilized for solving a convolution differential equation with given initial conditions. The results obtained by this method show the accuracy and efficiency of the method.

Definition (2.1)

Let fx,gxbe integrable functions, then the convolution of fx,gxis defined as:

fxgx=0xfxtgtdt

and the Laplace transform is defined as:

fx=Fs=0esxfxdx

where Re s > 0, where s is complex valued and is the Laplace operator.

Further, the Laplace transform of first and second derivatives are given by:

ifx=sℓfxf0iifx=s2fxsf0f0

More generally:

fnx=snfxsn1f0sn2f0sfn20fn10

and the one-sided inverse Laplace transform is defined by:

1Fs=fx=12πiαiα+iFsesxds

where the integration is within the regions of convergence. The region of convergence is half-plane α<Res.

Theorem (2.2) (Convolution Theorem)

If

fx=Fs,gx=Gs,

then:

fxgx=fxgx=fsgs

or equivalently,

1FsGs=fxgx

Consider the differential equation,

Lyx+Ryx+Nyx+Nyx=0E1

With the initial conditions

y0=hx,y0=kxE2

where Lis a linear second-order operator, Ris a linear first-order operator, Nis the nonlinear operator and Nyxis the nonlinear convolution term which is defined by:

Nyx=fyyy..yngyyy.yn

According to the variational iteration method, we can construct a correction functional as follows:

yn+1x=ynx+0xλξLynξ+Ry˜nξ+Ny˜nξ+Ny˜nξdξE3

Rynξ,Ny˜nξandNy˜nξare considered as restricted variations, that is,

δRy˜n=0,δNy˜n=0andδNy˜n=0,λ=1

Then, the variational iteration formula can be obtained as:

yn+1x=ynx0xLynξ+Rynξ+Nynξ+Ny˜nξdξE4

Eq. (4) can be solved iteratively using y0xas the initial approximation.

Then, the solution is yx=limnynx.

Now, we assume that L=d2dx2in Eq. (1).

Take Laplace transform () of both sides of Eq. (1) to find:

Lyx+Ryx+Nyx+Nyx=0E5
s2ysy0y0=Ryx+Nyx+Nyx=0E6

By using the initial conditions and taking the inverse Laplace transform, we have:

yx=px11s2Ryx+Nyx+Nyx=0E7

where pxrepresents the terms arising from the source term and the prescribed initial conditions.

Now, the first derivative of Eq. (7) is given by:

dyxdx=dpxdxddx11s2Ryx+Nyx+Nyx=0E8

By the correction functional of the irrational method, we have:

yn+1x=ynx0xynξξdpξd11s2Ryξ+Nyξ+Nyξ

Then, the new correction functional (new modified VIM) is given by:

yn+1x=ynx+11s2{Rynx+Nynx+Nynx],n0E9

Finally, we find the answer in the strain; if inverse Laplace transforms exist, Laplace transforms exist.

In particular, consider the nonlinear ordinary differential equations with convolution terms,

1yx2+2yyyy2=0,y0=y0=0E10

Take Laplace transform of Eq. (10), and making use of initial conditions, we have:

s2yx2s=yy22yy

The inverse Laplace transform of the above equation gives that:

yx=x2+11s2yy22yy

By using the new modified (Eq. (9)), we have the new correction functional,

yn+1x=ynx+11s2yy22yy

or

yn+1x=ynx+11s2yy22yyE11

Then, we have:

y0x=x2
y1x=x2+11s2[42x22x2=x2
y2x=x2,y3x=x2,.,ynx=x2

This means that:

y0x=y1x=y2x=.=ynx=x2

Then, the exact solution of Eq. (10) is yx=x2.

2yy22x+yy2=0,y0=1E12

Take Laplace transform of Eq. (12), and using the initial condition, we obtain:

sy12s2=y2yy2

Take the inverse Laplace transform to obtain

yx=1+x2+11sy2yy2

Using Eq. (9) to find the new correction functional in the form

yn+1x=ynx+11sy2ynyn2

or

yn+1x=ynx+11sy2ynyn2E13

Then, we have:

y0x=1+x2
y1x=1+x2+11s4x22x4=1+x2+11s8s32s24s=1+x2
y0x=y1x=y2x=.=ynx=1+x2

Then, the exact solution of Eq. (12) is:

yx=1+x2

3. Solution of nonlinear partial differential equations by the combined Laplace transform and the new modified variational iteration method

In this section, we present a reliable combined Laplace transform and the new modified variational iteration method to solve some nonlinear partial differential equations. The analytical results of these equations have been obtained in terms of convergent series with easily computable components. The nonlinear terms in these equations can be handled by using the new modified variational iteration method. This method is more efficient and easy to handle such nonlinear partial differential equations.

In this section, we combined Laplace transform and variational iteration method to solve the nonlinear partial differential equations.

To obtain the Laplace transform of partial derivative, we use integration by parts, and then, we have:

fxtt=sFxsfx0,E14
2fxtt2=s2Fxssfx0fx0t
fxtt=ddxFxs,
2fxtt2=d2dx2Fxs.

where fxsis the Laplace transform of xt.

We can easily extend this result to the nth partial derivative by using mathematical induction.

To illustrate the basic concept of He’s VIM, we consider the following general differential equations,

Luxt+Nuxt=gxtE15

with the initial condition

ux0=hxE16

where Lis a linear operator of the first-order, Nis a nonlinear operator and gxtis inhomogeneous term. According to variational iteration method, we can construct a correction functional as follows:

un+1=un+0tλLunxs+Nu˜nxsg(xs)dsE17

where λis a Lagrange multiplier λ=1, the subscripts ndenotes the nthapproximation, u˜nis considered as a restricted variation, that is, δu˜n=0.

Eq. (17) is called a correction functional.

The successive approximation un+1of the solution uwill be readily obtained by using the determined Lagrange multiplier and any selective function u0; consequently, the solution is given by:

u=limuun

In this section, we assume that Lis an operator of the first-order tin Eq. (15).

Taking Laplace transform on both sides of Eq. (15), we get:

Luxt+Nuxt=gxtE18

Using the differentiation property of Laplace transform and initial condition (16), we have:

suxthx=gxtNuxtE19

Applying the inverse Laplace transform on both sides of Eq. (19), we find:

uxt=Gxt11sNuxt,E20

where Gxtrepresents the terms arising from the source term and the prescribed initial condition.

Take the first partial derivative with respect to t of Eq. (20) to obtain:

tuxttGxt+t11sNuxtE21

By the correction functional of the variational iteration method

un+1=un0tunξxξξG(xξ)+ξ11ξNuξt

or

un+1=Gxt11sNunxtE22

Eq. (22) is the new modified correction functional of Laplace transform and the variational iteration method, and the solution uis given by:

uxt=limuunxt

In this section, we solve some nonlinear partial differential equations by using the new modified variational iteration Laplace transform method; therefore, we have:

Example (3.1)

Consider the following nonlinear partial differential equation:

ut+uux=0,ux0=xE23

Taking Laplace transform of Eq. (23), subject to the initial condition, we have:

uxt=xs1suux

The inverse Laplace transform implies that:

uxt=x11suux

By the new correction functional, we find:

un+1xt=x11sununx

Now, we apply the new modified variational iteration Laplace transform method:

u0xt=x
u1xt=x11sx=x1xs2=xxt
u2xt=x1x1s2+2s3+2s4=xxtxt213xt3
˙˙˙˙˙˙˙˙˙

Therefore, we deduce the series solution to be:

uxt=x1+t+t2+t3+=xt1,

which is the exact solution.

Example (3.2)

Consider the following nonlinear partial differential equation:

ut=ux2+u2ux2,ux.0=x2E24

Taking Laplace transform of Eq. (24), subject to the initial condition, we have:

uxt=x2s+1sux2+u2ux2

Take the inverse Laplace transform to find that:

uxt=x2+11sux2+u2ux2

The new correction functional is given as

un+1xt=x2+11sunx2+un2unx2

This is the new modified variational iteration Laplace transform method.

The solution in series form is given by:

u0xt=x2
u1xt=x2+16x2s2=x2+6x2t
u2xt=x21+6t+36t2+72t3
˙˙˙˙˙˙˙˙˙

The series solution is given by:

uxt=x21+6t+36t2+72t3+=x216t

Example (3.3)

Consider the following nonlinear partial differential equation:

ut=2uux2+u22ux2,ux.0=x+12E25

Using the same method in the above examples to find the new correction functional in the form:

un+1xt=x+12+11s2ununx2+u2n2unx2

Then, we have:

u0xt=x+12
u1xt=x+12+1x+141s2=x+121+t2
u2xt=x+121+t2+38t2+18t3+164t4
˙˙˙˙˙˙˙˙˙

The series solution is given by:

uxt=x+121+t2+12.322!t2+=x+121t12,

which is the exact solution of Eq. (25).

Example (3.4)

Consider the following nonlinear partial differential equation:

2ut2+ux2+uu2=tex,ux.0=0,ut=exE26

Taking the Laplace transform of the Eq. (26), subject to the initial conditions, we have:

s2uxtex=tex+u2ux2u

Take the inverse Laplace transform to find that:

uxt=tex+11s2tex+u2ux2u

The new correct functional is given as:

un+1xt=tex+11s2tex+un2unx2un

This is the new modified variational iteration Laplace transform method.

The solution in series form is given by:

u0xt=texu1xt=texu2xt=texE27
˙˙

The series solution is given by:

uxt=tex

4. New Laplace Variational iteration method

To illustrate the idea of new Laplace variational iteration method, we consider the following general differential equations in physics.

E28

where L is a linear partial differential operator given by, N is nonlinear operator andis a known analytical function. According to the variational iteration method, we can construct a correction functional for Eq. (28) as follows:

un+1xt=unxt+0tλ¯xςLunxς+Nu˜n(xς)h(xς),n0,E29

whereis a general Lagrange multiplier, which can be identified optimally via the variational theory, the subscriptdenotes the nth approximation,is considered as a restricted variation, that is,.

Also, we can find the Lagrange multipliers, by using integration by parts of Eq. (28), but in this chapter, the Lagrange multipliers are found to be of the form, and in such a case, the integration is basically the single convolution with respect to t, and hence, Laplace transform is appropriate to use.

Take Laplace transform of Eq. (29); then the correction functional will be constructed in the form:

un+1xt=unxt+0tλ¯xςLunxς+Nu˜n(xς)h(xς),n0,E30

Therefore

un+1xt=unxt+λ¯xtLunxt+Nu˜n(xt)h(xt)=unxt+λ¯xtLunxt+Nu˜n(xt)h(xt)E31

where * is a single convolution with respect to t.

To find the optimal value of, we first take the variation with respect to. Thus:

δδunun+1xt=δδununxt+δδunλ¯xtLunxt+Nu˜n(xt)h(xt)E32

Then, Eq. (32) becomes

E33

In this chapter, we assume that L is a linear partial differential operator given by, then, Eq. (33) can be written in the form:

E34

The extreme condition ofrequires that. This means that the right hand side of Eq. (34) should be set to zero; then, we have the following condition:

E35

Then, we have the following iteration formula

un+1xt=unxt0ttςLunxς+Nu˜n(xς)h(xς),n0,E36

5. Applications

In this section, we apply the Laplace variational iteration method to solve some linear and nonlinear partial differential equations in physics.

Example (5.1)

Consider the initial linear partial differential equation

E37

The Laplace variational iteration correction functional will be constructed in the following manner:

un+1xt=unxt+0tλ¯xtςuntt(xς)unxx(xς)+un(xς)E38

or

un+1xt=unxt+λ¯xtuntt(xt)unxx(xt)+un(xt)=unxt+λ¯xtuntt(xt)unxx(xt)+un(xt)=unxt+λ¯xts2unxtsunx0untx0unxxxt+unxtE39

Taking the variation with respect toof Eq. (39), we obtain:

δδunun+1xt=δδununxt+δδunλ¯xts2unxtsunx0untx0unxxxt+unxtE40

Then, we have.

The extreme condition ofrequires that. Hence, we have:

E41

Substituting Eq. (41) into Eq. (38), we obtain:

un+1xt=unxt0tsintςuntt(xς)unxx(xς)+un(xς)=unxtsintuntt(xt)unxx(xt)+un(xt)E42

Let, then, from Eq. (42), we have:

The inverse Laplace transforms yields:

E43

Substituting Eq. (43) into Eq. (38), we obtain:

Then, the exact solution of Eq. (37) is:

E44

Example (4.2)

Consider the nonlinear partial differential equation:

E45

The Laplace variational iteration correction functional will be constructed as follows:

E46

or

un+1xt=unxt+λ¯xtunttxtunxxxt+un2xtx2t2=unxt+λ¯xtuntt(xt)unxx(xt)+un2(xt)x2t2=unxt+λ¯xts2unxtsunx0untx0unxxxt+un2xtx2t2E47

Taking the variation with respect toof Eq. (47) and making the correction functional stationary we obtain:

This implies that:

E48

Substituting Eq. (21) into Eq. (19), we obtain:

E49

or

un+1xt=unxt+tunttxtunxxxt+un2xtx2t2E50

Let, then, from Eq. (50), we have:

Then, the exact solution of Eq. (45) is:

Again, the exact solution is obtained by using only few steps of the iterative scheme.

Example (4.3)

Consider the physics nonlinear boundary value problem,

E51

The Laplace variational iteration correction functional is

un+1xt=unxt+0tλ¯xtςuntxς6unxςunxxς+unxxxxςE52

or

un+1xt=unxt+λ¯xtunt(xt)6un(xt)unx(xt)+unxxx(xt)=unxt+λ¯xtunt(xt)6un(xt)unx(xt)+unxxx(xt)=unxt+λ¯xtsunxtun(x0)6un(xt)unx(xt)unxxx(xt)

Taking the variation with respect toof the last equation and making the correction functional stationary we obtain:

This implies that:

E53

Substituting Eq. (53) into Eq. (52), we obtain:

or

E54

Letthen, from Eq. (54), we have:

Then, the exact solution of Eq. (51) is:

Exercises

Solve the following nonlinear partial differential equations by new Laplace variational iteration method:

1)ut+uux=1ext+ex,ux0=ex2)ut+uux=2t+x+t3+xt2,ux0=03)ut+uux=2x2t+2xt2+2x3t4,ux0=14)ut+uux=1+tcosx+12sin2x,ux0=sinx5)ut+uux=0,ux0=x6)ut+uuxu=et,ux0=1+x7)uttuxxu+u2=xt+x2t2,ux0=1,utx0=x8)uttuxx+u2=1+2xt+x2t2,ux0=1,utx0=x9)uttuxx+u2=6xtx2t2+x6t6,ux0=0,utx0=010)uttuxx+u2=x2+t22,ux0=x2,utx0=011)uttuxx+u+u2=x2cos2t,ux0=x,utx0=012)ut+uux=0,ux0=x13)ut+uux=0,ux0=x14)ut+uux=0,ux0=2x15)ut+uux=uxx,ux0=x16)ut+uux=uxx,ux0=2x17)ut+uux=uxx,ux0=4tan2x

6. Conclusions

The method of combining Laplace transforms and variational iteration method is proposed for the solution of linear and nonlinear partial differential equations. This method is applied in a direct way without employing linearization and is successfully implemented by using the initial conditions and convolution integral. But this method failed to solve the singular differential equations.

Conflict of interest

The author declares that there is no conflict of interest regarding the publication of this chapter.

Answers

1)uxt=t+ex,2)uxt=t2+xt,3)uxt=1+x2t2,4)uxt=t+sinx5)uxt=xt1,6)uxt=x+et,7)uxt=1+xt,8)uxt=1+xt9)uxt=x3t3,10)uxt=t2+x2,11)uxt=xcost,12)uxt=x1+t13)uxt=xt1,14)uxt=2x1+2t,15)uxt=xt1,16)uxt=2x1+2t17)uxt=4tan2x

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Tarig M. Elzaki (May 23rd 2018). Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method, Differential Equations - Theory and Current Research, Terry E. Moschandreou, IntechOpen, DOI: 10.5772/intechopen.73291. Available from:

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