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:

$$\ell\left(\frac{\partial f(\mathbf{x},t)}{\partial t}\right) = sF(\mathbf{x},\mathbf{s}) - \mathbf{f}(\mathbf{x},0),\tag{14}$$

$$\ell\left(\frac{\partial^2 f(\mathbf{x},t)}{\partial t^2}\right) = s^2 F(\mathbf{x},\mathbf{s}) - s\mathbf{f}(\mathbf{x},0) - \frac{\partial f(\mathbf{x},0)}{\partial t}$$

Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method http://dx.doi.org/10.5772/intechopen.73291 159

$$\ell\left(\frac{\partial f(\mathbf{x},\mathbf{t})}{\partial t}\right) = \frac{d}{d\mathbf{x}}[F(\mathbf{x},\mathbf{s})].$$

$$\ell\left(\frac{\partial^2 f(\mathbf{x},\mathbf{t})}{\partial t^2}\right) = \frac{d^2}{d\mathbf{x}^2}[F(\mathbf{x},\mathbf{s})].$$

where f xð Þ ;s is the Laplace transform of ð Þ x; t .

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,

$$\ell[L\mathfrak{u}(\mathbf{x},\mathbf{t})] + \ell[\mathrm{Nu}(\mathbf{x},\mathbf{t})] = \ell[\mathfrak{g}(\mathbf{x},\mathbf{t})] \tag{15}$$

with the initial condition

<sup>y</sup>ð Þ¼ <sup>x</sup> <sup>1</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

ynþ<sup>1</sup>ð Þ¼ <sup>x</sup> ynð Þþ <sup>x</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

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

ynþ<sup>1</sup>ð Þ¼ <sup>x</sup> ynð Þþ <sup>x</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

or

Then, we have:

<sup>y</sup>1ð Þ¼ <sup>x</sup> <sup>1</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

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

158 Differential Equations - Theory and Current Research

nonlinear partial differential equations.

nonlinear partial differential equations.

have:

s

s

s

<sup>ℓ</sup> <sup>y</sup><sup>0</sup> � �<sup>2</sup> h i

s

<sup>y</sup>0ð Þ¼ <sup>x</sup> <sup>1</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>

<sup>y</sup>0ð Þ¼ <sup>x</sup> <sup>y</sup>1ð Þ¼ <sup>x</sup> y2ð Þ¼ <sup>x</sup> :…… <sup>¼</sup> <sup>y</sup>nð Þ¼ <sup>x</sup> <sup>1</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>

<sup>y</sup>ð Þ¼ <sup>x</sup> <sup>1</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>

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

In this section, we combined Laplace transform and variational iteration method to solve the

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

<sup>F</sup>ð Þ� <sup>x</sup>; <sup>s</sup> <sup>s</sup>f xð Þ� ; <sup>0</sup> <sup>∂</sup>fð Þ <sup>x</sup>; <sup>0</sup>

<sup>ℓ</sup> <sup>∂</sup>fð Þ <sup>x</sup>; <sup>t</sup> ∂t � �

> ¼ s 2

fð Þ x; t ∂t 2 � �

<sup>ℓ</sup> <sup>∂</sup><sup>2</sup>

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

<sup>ℓ</sup> <sup>4</sup>x<sup>2</sup> � � � <sup>ℓ</sup>ð Þ 2x <sup>ℓ</sup>ð Þ<sup>4</sup> � � <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

<sup>ℓ</sup> <sup>y</sup><sup>0</sup> � �<sup>2</sup> � <sup>y</sup>0<sup>∗</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> � � h i

<sup>ℓ</sup> <sup>y</sup><sup>0</sup> � �<sup>2</sup> � <sup>y</sup><sup>0</sup>

n <sup>∗</sup> y<sup>00</sup> n � �<sup>2</sup>

� � h i

� <sup>ℓ</sup> <sup>y</sup><sup>0</sup> n � �ℓ y<sup>00</sup> n � �<sup>2</sup>

> s 8 <sup>s</sup><sup>3</sup> � <sup>2</sup> s2 � � 4

s

<sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>

� � � �

¼ sFð Þ� x; s f xð Þ ; 0 , (14)

∂t

(13)

� � h i h i

$$
\mu(\mathbf{x}, \mathbf{0}) = h(\mathbf{x}) \tag{16}
$$

where L is a linear operator of the first-order, N is a nonlinear operator and g xð Þ ; t is inhomogeneous term. According to variational iteration method, we can construct a correction functional as follows:

$$u\_{n+1} = u\_n + \int\_0^t \lambda [Lu\_n(\mathbf{x}, \mathbf{s}) + N\tilde{u}\_n(\mathbf{x}, \mathbf{s}) - \mathbf{g}(\mathbf{x}, \mathbf{s})] ds \tag{17}$$

where λ is a Lagrange multiplier ð Þ λ ¼ �1 , the subscripts n denotes the nth approximation, u~<sup>n</sup> is considered as a restricted variation, that is, δu~<sup>n</sup> ¼ 0.

Eq. (17) is called a correction functional.

The successive approximation unþ<sup>1</sup> of the solution u will be readily obtained by using the determined Lagrange multiplier and any selective function u0; consequently, the solution is given by:

$$\mu = \lim\_{u \to \infty} u\_n$$

In this section, we assume that L is an operator of the first-order <sup>∂</sup> <sup>∂</sup><sup>t</sup> in Eq. (15).

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

$$\ell[L\mathfrak{u}(\mathbf{x},\mathbf{t})] + \ell[\mathcal{N}\mathfrak{u}(\mathbf{x},\mathbf{t})] = \ell[\mathcal{g}(\mathbf{x},\mathbf{t})] \tag{18}$$

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

$$s\ell[u(\mathbf{x},\mathbf{t})] - h(\mathbf{x}) = \ell[\mathbf{g}(\mathbf{x},\mathbf{t})] - \ell[\mathbf{Nu}(\mathbf{x},\mathbf{t})] \tag{19}$$

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

$$u(\mathbf{x}, \mathbf{t}) = G(\mathbf{x}, \mathbf{t}) - \ell^{-1} \left\{ \frac{1}{s} N u[\mathbf{x}, \mathbf{t}] \right\},\tag{20}$$

unþ<sup>1</sup>ð Þ¼� <sup>x</sup>; <sup>t</sup> <sup>x</sup> � <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

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

s ℓ½ � x � �

> 1 s<sup>2</sup> þ 2 s<sup>3</sup> þ 2 s4

uð Þ¼� x; t x 1 þ t þ t

þ u ∂2 u

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

x2 s þ 1 s

<sup>ℓ</sup>½ �¼ <sup>u</sup>ð Þ <sup>x</sup>; <sup>t</sup>

<sup>u</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

unþ<sup>1</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> x2 <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

This is the new modified variational iteration Laplace transform method.

Consider the following nonlinear partial differential equation:

∂u <sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>∂</sup><sup>u</sup> ∂x � �<sup>2</sup>

� � � �

˙˙˙ ˙˙˙ ˙˙˙

> <sup>2</sup> <sup>þ</sup> <sup>t</sup> <sup>3</sup> <sup>þ</sup> … � � <sup>¼</sup> <sup>x</sup>

> > <sup>ℓ</sup> <sup>∂</sup><sup>u</sup> ∂x � �<sup>2</sup>

<sup>ℓ</sup> <sup>∂</sup><sup>u</sup> ∂x � �<sup>2</sup>

<sup>ℓ</sup> <sup>∂</sup>un ∂x � �<sup>2</sup>

s

s

þ u ∂2 u ∂x<sup>2</sup>

þ u ∂2 u ∂x<sup>2</sup>

> þ un ∂2 un ∂x<sup>2</sup>

( ) " #

( ) " #

" #

<sup>u</sup>1ð Þ¼� <sup>x</sup>; <sup>t</sup> <sup>x</sup> � <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

<sup>u</sup>2ð Þ¼� <sup>x</sup>; <sup>t</sup> <sup>x</sup> � <sup>ℓ</sup>�<sup>1</sup> <sup>x</sup>

Therefore, we deduce the series solution to be:

Take the inverse Laplace transform to find that:

The new correction functional is given as

The solution in series form is given by:

which is the exact solution.

Example (3.2)

u0ð Þ¼� x; t x

s

¼ �<sup>x</sup> � <sup>ℓ</sup>�<sup>1</sup> <sup>x</sup>

s2

¼ �<sup>x</sup> � xt � xt<sup>2</sup> � <sup>1</sup>

t � 1 ,

<sup>∂</sup>x<sup>2</sup> , u xð Þ¼ :<sup>0</sup> <sup>x</sup><sup>2</sup> (24)

h i ¼ �<sup>x</sup> � xt

3 xt3

http://dx.doi.org/10.5772/intechopen.73291

161

<sup>ℓ</sup> unð Þ un <sup>x</sup> � � � �

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

where G xð Þ ; t represents 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:

$$\frac{\partial}{\partial t}\mu(\mathbf{x},\mathbf{t}) - \frac{\partial}{\partial t}G(\mathbf{x},\mathbf{t}) + \frac{\partial}{\partial t}\ell^{-1} \left\{ \frac{1}{s}\ell[\mathbf{N}\mu(\mathbf{x},\mathbf{t})] \right\} \tag{21}$$

By the correction functional of the variational iteration method

$$u\_{n+1} = u\_n - \int\_0^t \left\{ (u\_n)\_{\xi}(\mathbf{x}, \xi) - \frac{\partial}{\partial \xi} G(\mathbf{x}, \xi) + \frac{\partial}{\partial \xi} \ell^{-1} \left\{ \frac{1}{\xi} \ell [\mathcal{N}u(\xi, \mathbf{t})] \right\} \right\} d\xi$$

or

$$u\_{n+1} = G(\mathbf{x}, \mathbf{t}) - \mathcal{C}^{-1} \left\{ \frac{1}{s} \ell[\mathbf{N} u\_n(\mathbf{x}, \mathbf{t})] \right\} \tag{22}$$

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

$$\mu(\mathbf{x}, \mathbf{t}) = \lim\_{\mathbf{u} \to \mathbf{s}} \mu\_n(\mathbf{x}, \mathbf{t})$$

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:

$$
\mu\_t + \mu u\_x = 0 \quad , \qquad \qquad \mathfrak{u}(\mathbf{x}, 0) = -\mathbf{x} \tag{23}
$$

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

$$\ell[\mu(\mathbf{x}, \mathbf{t})] = -\frac{\mathbf{x}}{\mathbf{s}} - \frac{1}{\mathbf{s}} \ell[\mu \mu\_{\mathbf{x}}]$$

The inverse Laplace transform implies that:

$$\mu(\mathbf{x}, \mathbf{t}) = -\mathbf{x} - \ell^{-1} \left\{ \frac{1}{\mathbf{s}} \ell[\mu \mu\_x] \right\},$$

By the new correction functional, we find:

Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method http://dx.doi.org/10.5772/intechopen.73291 161

$$
\mu\_{n+1}(\mathbf{x}, \mathbf{t}) = -\mathbf{x} - \ell^{-1} \left\{ \frac{1}{s} \ell \left[ \mu\_n(\mu\_n)\_x \right] \right\},
$$

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

$$u\_0(\mathbf{x}, \mathbf{t}) = -\mathbf{x}$$

$$u\_1(\mathbf{x}, \mathbf{t}) = -\mathbf{x} - \ell^{-1} \left\{ \frac{1}{s} \ell[\mathbf{x}] \right\} = -\mathbf{x} - \ell^{-1} \left[ \frac{\mathbf{x}}{s^2} \right] = -\mathbf{x} - \mathbf{x}t$$

$$u\_2(\mathbf{x}, \mathbf{t}) = -\mathbf{x} - \ell^{-1} \left[ \mathbf{x} \left( \frac{1}{s^2} + \frac{2}{s^3} + \frac{2}{s^4} \right) \right] = -\mathbf{x} - \mathbf{x}t - \mathbf{x}t^2 - \frac{1}{3}\mathbf{x}t^3$$

$$\begin{array}{c c c} \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot \end{array}$$

˙˙˙

Therefore, we deduce the series solution to be:

$$\mu(\mathbf{x}, \mathbf{t}) = -\mathbf{x} \left( \mathbf{1} + \mathbf{t} + \mathbf{t}^2 + \mathbf{t}^3 + \dots \right) = \frac{\mathbf{x}}{t - 1},$$

which is the exact solution.

### Example (3.2)

<sup>u</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> <sup>G</sup>ð Þ� <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

Gð Þþ x; t

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

∂ ∂t

ð Þ un <sup>ξ</sup>ð Þ� <sup>x</sup>; <sup>ξ</sup> <sup>∂</sup>

ified variational iteration Laplace transform method; therefore, we have:

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

<sup>ℓ</sup>½ �¼� <sup>u</sup>ð Þ <sup>x</sup>; <sup>t</sup>

<sup>u</sup>ð Þ¼� <sup>x</sup>; <sup>t</sup> <sup>x</sup> � <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

Consider the following nonlinear partial differential equation:

unþ<sup>1</sup> <sup>¼</sup> <sup>G</sup>ð Þ� <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

uð Þ� x; t

By the correction functional of the variational iteration method

∂ ∂t

ðt

0

iteration method, and the solution u is given by:

The inverse Laplace transform implies that:

By the new correction functional, we find:

unþ<sup>1</sup> ¼ un �

160 Differential Equations - Theory and Current Research

condition.

or

Example (3.1)

where G xð Þ ; t represents the terms arising from the source term and the prescribed initial

∂ ∂t

<sup>∂</sup><sup>ξ</sup> <sup>G</sup>ð Þþ <sup>x</sup>; <sup>ξ</sup> <sup>∂</sup>

s

Eq. (22) is the new modified correction functional of Laplace transform and the variational

<sup>u</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> lim<sup>u</sup>!<sup>∞</sup> unð Þ <sup>x</sup>; <sup>t</sup>

In this section, we solve some nonlinear partial differential equations by using the new mod-

x s � 1 s <sup>ℓ</sup>½ � uux

> s <sup>ℓ</sup>½ � uux � �

s Nu½ � x; t � �

<sup>ℓ</sup>�<sup>1</sup> <sup>1</sup> s

� � � �

<sup>ℓ</sup>½ � Nuð Þ <sup>x</sup>; <sup>t</sup> � �

<sup>∂</sup><sup>ξ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

<sup>ℓ</sup>½ � Nunð Þ <sup>x</sup>; <sup>t</sup> � �

ut þ uux ¼ 0 , u xð Þ¼� ; 0 x (23)

<sup>ξ</sup> <sup>ℓ</sup>½ � Nuð Þ <sup>ξ</sup>; <sup>t</sup>

, (20)

dξ

(21)

(22)

Consider the following nonlinear partial differential equation:

$$\frac{\partial u}{\partial t} = \left(\frac{\partial u}{\partial x}\right)^2 + u \frac{\partial^2 u}{\partial x^2} \quad , \quad \mathbf{u}(\mathbf{x}.0) = \mathbf{x}^2 \tag{24}$$

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

$$\ell[\mu(\mathbf{x}, \mathbf{t})] = \frac{\mathbf{x}^2}{\mathbf{s}} + \frac{1}{\mathbf{s}} \ell \left[ \left( \frac{\partial \mu}{\partial \mathbf{x}} \right)^2 + \mu \frac{\partial^2 \mu}{\partial \mathbf{x}^2} \right]$$

Take the inverse Laplace transform to find that:

$$
\mu(\mathbf{x}, \mathbf{t}) = \mathbf{x}^2 + \mathcal{E}^{-1} \left\{ \frac{1}{s} \ell \left[ \left( \frac{\partial \mu}{\partial \mathbf{x}} \right)^2 + \mu \frac{\partial^2 \mu}{\partial \mathbf{x}^2} \right] \right\},
$$

The new correction functional is given as

$$u\_{n+1}(\mathbf{x}, \mathbf{t}) = \mathbf{x}^2 + \ell^{-1} \left\{ \frac{1}{s} \ell \left[ \left( \frac{\partial u\_n}{\partial \mathbf{x}} \right)^2 + u\_n \frac{\partial^2 u\_n}{\partial \mathbf{x}^2} \right] \right\}^2$$

This is the new modified variational iteration Laplace transform method.

The solution in series form is given by:

$$\mu\_0(\mathbf{x}, \mathbf{t}) = \mathbf{x}^2$$

$$\mu\_1(\mathbf{x}, \mathbf{t}) = \mathbf{x}^2 + \mathcal{C}^{-1} \left\{ \frac{6\mathbf{x}^2}{\mathbf{s}^2} \right\} = \mathbf{x}^2 + 6\mathbf{x}^2 \mathbf{t}$$

$$\mu\_2(\mathbf{x}, \mathbf{t}) = \mathbf{x}^2 \left( 1 + 6\mathbf{t} + 36\mathbf{t}^2 + 72\mathbf{t}^3 \right)$$

$$\begin{array}{c c c} \cdot & \cdot\\ \cdot & \cdot \end{array}$$

The series solution is given by:

$$\mu(\mathbf{x}, \mathbf{t}) = \mathbf{x}^2 \left( \mathbf{1} + 6\mathbf{t} + 36\mathbf{t}^2 + 72\mathbf{t}^3 + \dots \right) = \frac{\mathbf{x}^2}{1 - 6t}.$$

˙˙˙

### Example (3.3)

Consider the following nonlinear partial differential equation:

$$\frac{\partial u}{\partial t} = 2u \left( \frac{\partial u}{\partial x} \right)^2 + u^2 \frac{\partial^2 u}{\partial x^2} \quad , \quad \text{u(x.0)} = \frac{x+1}{2} \tag{25}$$

Example (3.4)

Consider the following nonlinear partial differential equation:

<sup>ℓ</sup>½ �� <sup>u</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>e</sup>

<sup>u</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> te�<sup>x</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

unþ<sup>1</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> te�<sup>x</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

4. New Laplace Variational iteration method

This is the new modified variational iteration Laplace transform method.

˙ ˙

<sup>þ</sup> <sup>u</sup> � <sup>u</sup><sup>2</sup> <sup>¼</sup> te�<sup>x</sup> , u xð Þ¼ :<sup>0</sup> <sup>0</sup>,

�<sup>x</sup> <sup>¼</sup> <sup>ℓ</sup> te�<sup>x</sup> <sup>þ</sup> <sup>u</sup><sup>2</sup> � <sup>∂</sup><sup>u</sup>

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> te�<sup>x</sup> <sup>þ</sup> <sup>u</sup><sup>2</sup> � <sup>∂</sup><sup>u</sup>

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> te�<sup>x</sup> <sup>þ</sup> un

<sup>u</sup>0ð Þ¼ <sup>x</sup>; <sup>t</sup> te�<sup>x</sup> <sup>u</sup>1ð Þ¼ <sup>x</sup>; <sup>t</sup> te�<sup>x</sup> <sup>u</sup>2ð Þ¼ <sup>x</sup>; <sup>t</sup> te�<sup>x</sup>

<sup>u</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> te�<sup>x</sup>

To illustrate the idea of new Laplace variational iteration method, we consider the following

is a known analytical function. According to the variational iteration method, we can construct

where L is a linear partial differential operator given by , N is nonlinear operator and

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

∂u <sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>e</sup>

� u

� u

� un

∂x � �<sup>2</sup>

> ∂x � �<sup>2</sup>

<sup>2</sup> � <sup>∂</sup>un ∂x � �<sup>2</sup>

( ) " #

" #

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

( ) " #

�<sup>x</sup> (26)

163

http://dx.doi.org/10.5772/intechopen.73291

(27)

ð28Þ

∂u ∂x � �<sup>2</sup>

∂2 u ∂t <sup>2</sup> þ

> s 2

Take the inverse Laplace transform to find that:

The new correct functional is given as:

The solution in series form is given by:

The series solution is given by:

general differential equations in physics.

a correction functional for Eq. (28) as follows:

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

$$u\_{n+1}(\mathbf{x}, \mathbf{t}) = \frac{\mathbf{x} + \mathbf{1}}{2} + \mathcal{E}^{-1} \left\{ \frac{1}{\mathbf{s}} \, \ell \left[ 2u\_n \left( \frac{\partial u\_n}{\partial \mathbf{x}} \right)^2 + u\_n^2 \, \frac{\partial^2 u\_n}{\partial \mathbf{x}^2} \right] \right\}$$

Then, we have:

$$u\_0(\mathbf{x}, \mathbf{t}) = \frac{\mathbf{x} + \mathbf{1}}{2}$$

$$u\_1(\mathbf{x}, \mathbf{t}) = \frac{\mathbf{x} + \mathbf{1}}{2} + \mathbf{t}^{-1} \left\{ \frac{\mathbf{x} + \mathbf{1}}{4} \frac{\mathbf{1}}{s^2} \right\} = \frac{\mathbf{x} + \mathbf{1}}{2} \left[ \mathbf{1} + \frac{\mathbf{t}}{2} \right]$$

$$u\_2(\mathbf{x}, \mathbf{t}) = \frac{\mathbf{x} + \mathbf{1}}{2} \left( \mathbf{1} + \frac{\mathbf{t}}{2} + \frac{\mathbf{3}}{8} \mathbf{t}^2 + \frac{\mathbf{1}}{8} \mathbf{t}^3 + \frac{\mathbf{1}}{64} \mathbf{t}^4 \right)$$

$$\vdots$$

$$\vdots \qquad \vdots \qquad \vdots \qquad \vdots$$

The series solution is given by:

$$\mu(\mathbf{x}, \mathbf{t}) = \frac{\mathbf{x} + 1}{2} \left( 1 + \frac{\mathbf{t}}{2} + \frac{\frac{1}{2} \cdot \frac{3}{2}}{2!} \mathbf{t}^2 + \dots \right) = \frac{\mathbf{x} + 1}{2} (1 - t)^{-\frac{1}{2}} \mathbf{t}$$

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

### Example (3.4)

<sup>u</sup>0ð Þ¼ <sup>x</sup>; <sup>t</sup> x2

<sup>u</sup>2ð Þ¼ <sup>x</sup>; <sup>t</sup> x2 <sup>1</sup> <sup>þ</sup> 6t <sup>þ</sup> 36t2 <sup>þ</sup> 72t<sup>3</sup> � � ˙˙˙ ˙˙˙ ˙˙˙

<sup>u</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> x2 <sup>1</sup> <sup>þ</sup> 6t <sup>þ</sup> 36t2 <sup>þ</sup> 72t3 <sup>þ</sup> … � � <sup>¼</sup> <sup>x</sup><sup>2</sup>

<sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>∂</sup><sup>2</sup> u

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

s ℓ 2un

u0ð Þ¼ x; t

<sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>x</sup> <sup>þ</sup> <sup>1</sup>

1 þ t 2 þ 3 8 t 2 þ 1 8 t 3 þ 1 64 t 4

˙˙˙ ˙˙˙ ˙˙˙

Consider the following nonlinear partial differential equation:

∂u ∂x � �<sup>2</sup>

x þ 1

x þ 1

x þ 1 2

> t 2 þ 1 2 : 3 2 2! t <sup>2</sup> <sup>þ</sup> …

� �

<sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

∂u <sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>2</sup><sup>u</sup>

unþ<sup>1</sup>ð Þ¼ x; t

u1ð Þ¼ x; t

uð Þ¼ x; t

u2ð Þ¼ x; t

x þ 1 <sup>2</sup> <sup>1</sup> <sup>þ</sup>

s2 � �

<sup>¼</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>6</sup>x<sup>2</sup>

<sup>∂</sup>x<sup>2</sup> , u xð Þ¼ :<sup>0</sup> <sup>x</sup> <sup>þ</sup> <sup>1</sup>

∂un ∂x � �<sup>2</sup>

x þ 1 2

1 s2 � �

� �

4

( ) " #

<sup>¼</sup> <sup>x</sup> <sup>þ</sup> <sup>1</sup> 2

<sup>¼</sup> <sup>x</sup> <sup>þ</sup> <sup>1</sup>

<sup>2</sup> ð Þ <sup>1</sup> � <sup>t</sup> �<sup>1</sup>

2,

1 þ t 2 � �

<sup>þ</sup> <sup>u</sup><sup>2</sup> n ∂2 un ∂x<sup>2</sup>

t

1 � 6t

<sup>2</sup> (25)

<sup>u</sup>1ð Þ¼ <sup>x</sup>; <sup>t</sup> x2 <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>6</sup>x<sup>2</sup>

The series solution is given by:

162 Differential Equations - Theory and Current Research

Example (3.3)

Then, we have:

The series solution is given by:

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

Consider the following nonlinear partial differential equation:

$$\frac{\partial^2 u}{\partial t^2} + \left(\frac{\partial u}{\partial x}\right)^2 + u - u^2 = \text{te}^{-x}, \quad \text{u(x,0)} = 0, \frac{\partial u}{\partial t} = e^{-x} \tag{26}$$

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

$$s^2 \ell[\mu(\mathbf{x}, \mathbf{t})] - e^{-\mathbf{x}} = \ell \left[ t e^{-\mathbf{x}} + \mu^2 - \left( \frac{\partial \mu}{\partial \mathbf{x}} \right)^2 - \mu \right]$$

Take the inverse Laplace transform to find that:

$$u(\mathbf{x}, \mathbf{t}) = te^{-\mathbf{x}} + \ell^{-1} \left\{ \frac{1}{s^2} \ell \left[ te^{-\mathbf{x}} + u^2 - \left(\frac{\partial u}{\partial \mathbf{x}}\right)^2 - u \right] \right\}.$$

The new correct functional is given as:

$$u\_{n+1}(\mathbf{x}, \mathbf{t}) = te^{-\mathbf{x}} + \ell^{-1} \left\{ \frac{1}{s^2} \ell \left[ te^{-\mathbf{x}} + \mu\_n{}^2 - \left(\frac{\partial u\_n}{\partial \mathbf{x}}\right)^2 - \mu\_n \right] \right\}$$

This is the new modified variational iteration Laplace transform method.

˙ ˙

The solution in series form is given by:

$$\begin{aligned} u\_0(\mathbf{x}, \mathbf{t}) &= t e^{-\mathbf{x}} \\ u\_1(\mathbf{x}, \mathbf{t}) &= t e^{-\mathbf{x}} \\ u\_2(\mathbf{x}, \mathbf{t}) &= t e^{-\mathbf{x}} \\ \vdots \end{aligned} \tag{27}$$

The series solution is given by:

$$u(\mathbf{x}, \mathbf{t}) = t e^{-\mathbf{x}}$$
