5. Applications

unþ<sup>1</sup>ð Þ¼ x; t unð Þþ x; t

n ≥ 0,

164 Differential Equations - Theory and Current Research

variation, that is, .

transform is appropriate to use.

Therefore

Thus:

Then, Eq. (32) becomes

Eq. (33) can be written in the form:

þℓ ðt

0

where \* is a single convolution with respect to t.

δ δun

δ δun <sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup>

2 4 ðt

λð Þ x; ς ½ � Lunð Þþ x; ς Nu~nðx; ςÞ � hðx; ςÞ dς,

3

<sup>5</sup>, n <sup>≥</sup> <sup>0</sup>, (30)

(29)

(31)

(32)

ð33Þ

0

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � unð Þ <sup>x</sup>; <sup>t</sup>

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � unð Þ <sup>x</sup>; <sup>t</sup>

where is a general Lagrange multiplier, which can be identified optimally via the variational theory, the subscript denotes the nth approximation, is considered as a restricted

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

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

λð Þ x; ς ½ � Lunð Þþ x; ς Nu~nðx; ςÞ � hðx; ςÞ dς

<sup>¼</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> � �ℓ½ � Lunð Þþ <sup>x</sup>; <sup>t</sup> Nu~nðx; <sup>t</sup>Þ � <sup>h</sup>ðx; <sup>t</sup><sup>Þ</sup>

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

<sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup>

<sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> � �ℓ½ � Lunð Þþ <sup>x</sup>; <sup>t</sup> Nu~nðx; <sup>t</sup>Þ � <sup>h</sup>ðx; <sup>t</sup><sup>Þ</sup>

In this chapter, we assume that L is a linear partial differential operator given by , then,

δ δun

<sup>þ</sup><sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>∗</sup>½ � Lunð Þþ <sup>x</sup>; <sup>t</sup> Nu~nðx; <sup>t</sup>Þ � <sup>h</sup>ðx; <sup>t</sup><sup>Þ</sup> � �

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

$$u\_{\
u}(\mathbf{x},t) - u\_{xx}(\mathbf{x},t) + u(\mathbf{x},t) = 0 \quad , \quad u(\mathbf{x},0) = 0 \quad , \quad \frac{\partial u(\mathbf{x},0)}{\partial t} = \mathbf{x} \tag{37}$$

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

$$\begin{aligned} \ell[\boldsymbol{u}\_{n+1}(\mathbf{x},t)] &= \ell[\boldsymbol{u}\_n(\mathbf{x},t)] \\ + \ell \left[ \int\_0^t \overline{\lambda}(\mathbf{x}, t-\boldsymbol{\varsigma}) \left[ (\boldsymbol{u}\_n)\_{tt}(\mathbf{x}, \boldsymbol{\varsigma}) - (\boldsymbol{u}\_n)\_{\mathbf{x}\mathbf{x}}(\mathbf{x}, \boldsymbol{\varsigma}) + \boldsymbol{u}\_n(\mathbf{x}, \boldsymbol{\varsigma}) \right] d\boldsymbol{\varsigma} \right] \end{aligned} \tag{38}$$

or

$$\begin{aligned} \ell[\boldsymbol{u}\_{n+1}(\mathbf{x},t)] &= \ell[\boldsymbol{u}\_{n}(\mathbf{x},t)] \\ \ell + \ell \left[ \overline{\boldsymbol{\Lambda}}(\mathbf{x},t) \ast \left[ (\boldsymbol{u}\_{n})\_{\boldsymbol{\Lambda}}(\mathbf{x},t) - (\boldsymbol{u}\_{n})\_{\boldsymbol{\Lambda}\mathbf{x}}(\mathbf{x},t) + \boldsymbol{u}\_{n}(\mathbf{x},t) \right] \right] \\ &= \ell[\boldsymbol{u}\_{n}(\mathbf{x},t)] + \ell \left[ \overline{\boldsymbol{\Lambda}}(\mathbf{x},t) \right] \ell \left[ (\boldsymbol{u}\_{n})\_{\boldsymbol{\Lambda}}(\mathbf{x},t) - (\boldsymbol{u}\_{n})\_{\boldsymbol{\Lambda}\mathbf{x}}(\mathbf{x},t) + \boldsymbol{u}\_{n}(\mathbf{x},t) \right] \\ &= \ell[\boldsymbol{u}\_{n}(\mathbf{x},t)] + \ell \left[ \overline{\boldsymbol{\Lambda}}(\mathbf{x},t) \right] \left[ \boldsymbol{s}^{2} \ell \boldsymbol{u}\_{n}(\mathbf{x},t) - \boldsymbol{s} \boldsymbol{u}\_{n}(\mathbf{x},0) - \frac{\partial \boldsymbol{u}\_{n}}{\partial t}(\mathbf{x},0) \right] \\ &= \ell[\boldsymbol{u}\_{n}]\_{\boldsymbol{\Lambda}}(\mathbf{x},t) + \ell \boldsymbol{u}\_{n}(\mathbf{x},t) \end{aligned} \tag{39}$$

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

$$\begin{split} \frac{\delta}{\delta u\_n} \mathfrak{E}[u\_{n+1}(\mathbf{x},t)] &= \frac{\delta}{\delta u\_n} \mathfrak{E}[u\_n(\mathbf{x},t)] \\ &+ \frac{\delta}{\delta u\_n} \mathfrak{E}\left[\overline{\lambda}(\mathbf{x},t)\right] \begin{bmatrix} s^2 \ell u\_n(\mathbf{x},t) - s u\_n(\mathbf{x},0) - \frac{\partial u\_n}{\partial t}(\mathbf{x},0) \\ -\mathfrak{E}(u\_n)\_{\text{xx}}(\mathbf{x},t) + \ell u\_n(\mathbf{x},t) \end{bmatrix} \end{split} \tag{40}$$

Example (4.2)

or

stationary we obtain:

This implies that:

or

Consider the nonlinear partial differential equation:

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>∗</sup>

<sup>¼</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> � �

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

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

<sup>¼</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> � �<sup>ℓ</sup> ð Þ un ttðx; <sup>t</sup>Þ � ð Þ un xxðx; <sup>t</sup>Þ þ <sup>u</sup><sup>2</sup>

2 6 4 ð Þ un ttð Þ� x; t ð Þ un xxð Þ x; t

2

<sup>∂</sup><sup>t</sup> ð Þ <sup>x</sup>; <sup>0</sup>

<sup>2</sup> � �

<sup>n</sup>ð Þ� <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>x</sup><sup>2</sup><sup>t</sup>

<sup>n</sup>ðx; <sup>t</sup>Þ � <sup>x</sup><sup>2</sup><sup>t</sup>

3 7 5

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

<sup>n</sup>ð Þ� <sup>x</sup>; <sup>t</sup> <sup>x</sup><sup>2</sup><sup>t</sup>

<sup>2</sup> � �

" # " #

<sup>þ</sup>u<sup>2</sup>

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

<sup>s</sup><sup>2</sup>ℓunð Þ� <sup>x</sup>; <sup>t</sup> sunð Þ� <sup>x</sup>; <sup>0</sup> <sup>∂</sup>un

�ℓð Þ un xxð Þþ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>u<sup>2</sup>

Taking the variation with respect to of Eq. (47) and making the correction functional

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � �<sup>t</sup> <sup>ℓ</sup> ð Þ un ttð Þ� <sup>x</sup>; <sup>t</sup> ð Þ un xxð Þ <sup>x</sup>; <sup>t</sup>

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

<sup>þ</sup>u<sup>2</sup>

<sup>n</sup>ð Þ� <sup>x</sup>; <sup>t</sup> <sup>x</sup><sup>2</sup><sup>t</sup>

2

� � (50)

ð45Þ

167

ð46Þ

(47)

ð48Þ

ð49Þ

Then, we have.

$$\begin{aligned} \ell \left[ \delta \boldsymbol{u}\_{n+1} (\mathbf{x}, t) \right] &= \ell \left[ \delta \boldsymbol{u}\_{n} (\mathbf{x}, t) \right] + \ell \left[ \overline{\mathcal{A}} (\mathbf{x}, t) \right] \left[ \mathbf{s}^{2} \ell \, \delta \boldsymbol{u}\_{n} (\mathbf{x}, t) + \ell \, \delta \boldsymbol{u}\_{n} (\mathbf{x}, t) \right] \\ &= \ell \left[ \delta \boldsymbol{u}\_{n} (\mathbf{x}, t) \right] \left\{ \mathbf{l} + \ell \left[ \overline{\mathcal{A}} (\mathbf{x}, t) \right] \left( \mathbf{s}^{2} + 1 \right) \right\} \end{aligned}$$

The extreme condition of requires that . Hence, we have:

$$(1 + (\mathbf{s}^2 + 1)\ell \overline{\mathcal{A}})(\mathbf{x}, t) = 0 \quad , \text{ and } \quad \overline{\mathcal{A}}(\mathbf{x}, t) = \ell^{-1} \left[ \frac{-1}{\mathbf{s}^2 + 1} \right] = -\sin t \tag{41}$$

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

$$\begin{aligned} \ell[u\_{n+1}(\mathbf{x},t)] &= \ell[u\_n(\mathbf{x},t)] \\ -\ell\left[\int\_0^t \sin\left(t-\varsigma\right)[(u\_n)\_{tt}(\mathbf{x},\varsigma)-(u\_n)\_{xx}(\mathbf{x},\varsigma)+u\_n(\mathbf{x},\varsigma)]d\varsigma\right] \\ &= \ell[u\_n(\mathbf{x},t)] - \ell[\sin t]\ell\left[(u\_n)\_{tt}(\mathbf{x},t)-(u\_n)\_{xx}(\mathbf{x},t)+u\_n(\mathbf{x},t)\right] \end{aligned} \tag{42}$$

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

$$\ell\left[u\_{\mathrm{l}}(\mathbf{x},t)\right] = \ell\left[\mathbf{x}t\right] - \ell\left[\sin t\right]\ell\left[\mathbf{x}t\right] = \frac{\varkappa}{s^2} - \frac{\varkappa}{s^2(s^2+1)}$$

The inverse Laplace transforms yields:

$$u\_1(\mathbf{x}, t) = \mathbf{x} \sin t \tag{43}$$

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

$$\ell\left[\mu\_2(\mathbf{x}, t)\right] = \ell\left[\mathbf{x}\sin t\right] - \ell\left[\sin t\right]\ell\left[0\right] \qquad \text{then} \qquad \mu\_2(\mathbf{x}, t) = \mathbf{x}\sin t$$

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

$$u\left(\mathbf{x},t\right) = \mathbf{x}\sin t\tag{44}$$

### Example (4.2)

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

δ δun

The extreme condition of requires that . Hence, we have:

sin ð Þ <sup>t</sup> � <sup>ς</sup> ð Þ un ttðx; <sup>ς</sup>Þ � ð Þ un xxðx; <sup>ς</sup>Þ þ unðx; <sup>ς</sup><sup>Þ</sup> � �d<sup>ς</sup>

<sup>¼</sup> <sup>ℓ</sup>½ �� unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � sin <sup>t</sup> <sup>ℓ</sup> ð Þ un ttðx; <sup>t</sup>Þ � ð Þ un xxðx; <sup>t</sup>Þ þ unðx; <sup>t</sup><sup>Þ</sup> � �

2 6 4 <sup>ℓ</sup>½ � unð Þ <sup>x</sup>; <sup>t</sup>

�ℓð Þ un xxð Þþ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>unð Þ <sup>x</sup>; <sup>t</sup>

<sup>∂</sup><sup>t</sup> ð Þ <sup>x</sup>; <sup>0</sup>

3 7 5

> 3 5

(40)

ð41Þ

(42)

ð43Þ

ð44Þ

<sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> � � <sup>s</sup><sup>2</sup>ℓunð Þ� <sup>x</sup>; <sup>t</sup> sunð Þ� <sup>x</sup>; <sup>0</sup> <sup>∂</sup>un

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup>

δ δun

166 Differential Equations - Theory and Current Research

þ δ δun

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

2 4

�ℓ ðt

The inverse Laplace transforms yields:

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

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

0

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � unð Þ <sup>x</sup>; <sup>t</sup>

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

Then, we have.

Consider the nonlinear partial differential equation:

$$u\_{\boldsymbol{u}}\left(\mathbf{x},t\right) - u\_{xx}\left(\mathbf{x},t\right) + u^{\boldsymbol{2}}\left(\mathbf{x},t\right) = \mathbf{x}^{\top}t^{\top} \qquad , \quad u\left(\mathbf{x},0\right) = 0 \quad , \quad \frac{\partial u\left(\mathbf{x},0\right)}{\partial t} = \mathbf{x} \tag{45}$$

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

$$\ell\left[u\_{n+1}(\mathbf{x},t)\right] = \ell\left[u\_n(\mathbf{x},t)\right] - \ell\left[\int\_0^t (t-\xi)\begin{bmatrix} (u\_n)\_{\boldsymbol{u}}(\mathbf{x},\boldsymbol{\xi}) - (u\_n)\_{\boldsymbol{\chi}\mathbf{x}}(\mathbf{x},\boldsymbol{\xi})\\ +u\_s^2(\mathbf{x},\boldsymbol{\xi}) - \boldsymbol{\chi^2}\boldsymbol{\xi^2} \end{bmatrix} d\boldsymbol{\xi}\right] \tag{46}$$

or

$$\begin{aligned} \ell[u\_{n+1}(\mathbf{x},t)] &= \ell[u\_n(\mathbf{x},t)] + \ell \left[ \overline{\lambda}(\mathbf{x},t) \* \begin{bmatrix} (u\_n)\_{\mathcal{U}}(\mathbf{x},t) - (u\_n)\_{\mathcal{X}}(\mathbf{x},t) \\ + u\_n^2(\mathbf{x},t) - \mathbf{x}^2 t^2 \end{bmatrix} \right] \\ &= \ell[u\_n(\mathbf{x},t)] + \ell \left[ \overline{\lambda}(\mathbf{x},t) \right] \ell \left[ (u\_n)\_{\mathcal{U}}(\mathbf{x},t) - (u\_n)\_{\mathcal{X}}(\mathbf{x},t) + u\_n^2(\mathbf{x},t) - \mathbf{x}^2 t^2 \right] \\ &= \ell[u\_n(\mathbf{x},t)] + \ell \left[ \overline{\lambda}(\mathbf{x},t) \right] \left[ \mathbf{s}^2 \ell u\_n(\mathbf{x},t) - s u\_n(\mathbf{x},0) - \frac{\partial u\_n}{\partial t}(\mathbf{x},0) \right] \\ &- \ell (u\_n)\_{\mathcal{X}}(\mathbf{x},t) + \ell u\_n^2(\mathbf{x},t) - \ell (\mathbf{x}^2 t^2) \end{aligned} \tag{47}$$

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

This implies that:

$$\ell + s\ell \overline{\mathcal{A}}(\mathbf{x}, t) = \mathbf{0} \quad , \quad \text{and} \quad \overline{\mathcal{A}}(\mathbf{x}, t) = \ell^{-1} \left[ \frac{-1}{s} \right] = -\mathbf{1} \tag{48}$$

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

$$\ell\left[\boldsymbol{\mu}\_{n+1}(\mathbf{x},t)\right] = \ell\left[\boldsymbol{\mu}\_{n}(\mathbf{x},t)\right] - \ell\left[\int\_{0}^{t} (t-\boldsymbol{\xi})\begin{bmatrix} (\boldsymbol{\mu}\_{n})\_{n}(\mathbf{x},\boldsymbol{\xi}) - (\boldsymbol{\mu}\_{n})\_{\infty}(\mathbf{x},\boldsymbol{\xi}) \\ + \boldsymbol{\mu}\_{n}^{2}(\mathbf{x},\boldsymbol{\xi}) - \boldsymbol{\lambda}^{2}\boldsymbol{\xi}^{2} \end{bmatrix} d\boldsymbol{\xi}\right] \tag{49}$$

or

$$\ell[u\_{n+1}(\mathbf{x},t)] = \ell[u\_n(\mathbf{x},t)] + \ell[-t]\ell\begin{bmatrix} (u\_n)\_{tt}(\mathbf{x},t) - (u\_n)\_{xx}(\mathbf{x},t) \\ + u\_n^2(\mathbf{x},t) - \mathbf{x}^2 t^2 \end{bmatrix} \tag{50}$$

$$\text{Let } u\_0(\mathbf{x}, t) = u(\mathbf{x}, 0) + t \frac{\partial u}{\partial t}(\mathbf{x}, 0) = \mathbf{x}t \text{ then, from Eq. (50), we have: }$$

$$\ell \left[ u\_1(\mathbf{x}, t) \right] = \ell \left[ \mathbf{x}t \right] + \ell \left[ -t \right] \ell \left[ \mathbf{0} - \mathbf{0} + \mathbf{x}^2 t^2 - \mathbf{x}^2 t^2 \right]$$

$$u\_1(\mathbf{x}, t) = \mathbf{x}t$$

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,

$$
\mu\_t - 6uu\_x + u\_{xxx} = 0 \qquad , \quad u(x,0) = \frac{6}{x^2} \quad , \quad x \neq 0
$$

The Laplace variational iteration correction functional is

$$\begin{aligned} \ell[\boldsymbol{u}\_{n+1}(\mathbf{x},t)] &= \ell[\boldsymbol{u}\_{n}(\mathbf{x},t)] \\ + \ell \left[ \begin{matrix} \boldsymbol{t} \\ \overline{\lambda}(\mathbf{x},t-\boldsymbol{\varepsilon}) \\ + (\boldsymbol{u}\_{n})\_{\mathrm{xxx}}(\mathbf{x},\boldsymbol{\varepsilon}) \end{matrix} \begin{matrix} (\boldsymbol{u}\_{n})\_{t}(\mathbf{x},\boldsymbol{\varepsilon}) - 6\boldsymbol{u}\_{n}(\mathbf{x},\boldsymbol{\varepsilon})(\boldsymbol{u}\_{n})\_{\mathrm{x}}(\mathbf{x},\boldsymbol{\varepsilon}) \\ + (\boldsymbol{u}\_{n})\_{\mathrm{xxx}}(\mathbf{x},\boldsymbol{\varepsilon}) \end{matrix} \right] d\boldsymbol{\varepsilon} \right] \end{aligned} \tag{52}$$

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

Solve the following nonlinear partial differential equations by new Laplace variational itera-

<sup>3</sup> <sup>þ</sup> xt<sup>2</sup>, u xð Þ¼ ; <sup>0</sup> <sup>0</sup>

<sup>4</sup>, u xð Þ¼ ; <sup>0</sup> <sup>1</sup>

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

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

169

<sup>2</sup> sin 2x, u xð Þ¼ ; <sup>0</sup> sin <sup>x</sup>

<sup>2</sup>, u xð Þ¼ ; <sup>0</sup> <sup>1</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>x</sup>

, u xð Þ¼ ; <sup>0</sup> <sup>x</sup><sup>2</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>0</sup>

<sup>2</sup>, u xð Þ¼ ; <sup>0</sup> <sup>1</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>x</sup>

<sup>6</sup>, u xð Þ¼ ; <sup>0</sup> <sup>0</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>0</sup>

<sup>1</sup>Þut <sup>þ</sup> uux <sup>¼</sup> <sup>1</sup> � <sup>e</sup>�<sup>x</sup> <sup>t</sup> <sup>þ</sup> <sup>e</sup>�<sup>x</sup> ð Þ, u xð Þ¼ ; <sup>0</sup> <sup>e</sup>�<sup>x</sup>

1

, u xð Þ¼ ; 0 1 þ x

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

<sup>11</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup> <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> <sup>x</sup><sup>2</sup> cos <sup>2</sup>t, u xð Þ¼ ; <sup>0</sup> x, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>0</sup>

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 differ-

<sup>2</sup> <sup>2</sup>

2Þut þ uux ¼ 2t þ x þ t

4Þut þ uux ¼ 1 þ t cos x þ

<sup>6</sup>Þut <sup>þ</sup> uux � <sup>u</sup> <sup>¼</sup> <sup>e</sup><sup>t</sup>

<sup>3</sup>Þut <sup>þ</sup> uux <sup>¼</sup> <sup>2</sup>x<sup>2</sup><sup>t</sup> <sup>þ</sup> <sup>2</sup>xt<sup>2</sup> <sup>þ</sup> <sup>2</sup>x<sup>3</sup><sup>t</sup>

5Þut þ uux ¼ 0, u xð Þ¼� ; 0 x

<sup>7</sup>Þutt � uxx � <sup>u</sup> <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> xt <sup>þ</sup> <sup>x</sup><sup>2</sup><sup>t</sup>

<sup>8</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>2</sup>xt <sup>þ</sup> <sup>x</sup><sup>2</sup><sup>t</sup>

Þut þ uux ¼ 0, u xð Þ¼ ; 0 x Þut þ uux ¼ 0, u xð Þ¼� ; 0 x Þut þ uux ¼ 0, u xð Þ¼ ; 0 2x Þut þ uux ¼ uxx, u xð Þ¼� ; 0 x Þut þ uux ¼ uxx, u xð Þ¼ ; 0 2x Þut þ uux ¼ uxx, u xð Þ¼ ; 0 4 tan 2x

<sup>9</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> <sup>6</sup>xt <sup>x</sup><sup>2</sup> � <sup>t</sup>

<sup>10</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> x2 <sup>þ</sup> <sup>t</sup>

Exercises

tion method:

6. Conclusions

ential equations.

or

$$\begin{split} \ell[\boldsymbol{u}\_{n+1}(\mathbf{x},t)] &= \ell[\boldsymbol{u}\_{n}(\mathbf{x},t)] + \ell\left[\overline{\boldsymbol{\lambda}}(\mathbf{x},t) \ast \left[ (\boldsymbol{u}\_{n})\_{t}(\mathbf{x},t) - \boldsymbol{\delta}(\boldsymbol{u}\_{n})(\mathbf{x},t)(\boldsymbol{u}\_{n})\_{\mathbf{x}}(\mathbf{x},t) + (\boldsymbol{u}\_{n})\_{\mathbf{xxx}}(\mathbf{x},t) \right] \right] \\ &= \ell[\boldsymbol{u}\_{n}(\mathbf{x},t)] + \ell\left[\overline{\boldsymbol{\lambda}}(\mathbf{x},t)\right] \ell\left[ (\boldsymbol{u}\_{n})\_{t}(\mathbf{x},t) - \boldsymbol{\delta}(\boldsymbol{u}\_{n})(\mathbf{x},t)(\boldsymbol{u}\_{n})\_{\mathbf{x}}(\mathbf{x},t) + (\boldsymbol{u}\_{n})\_{\mathbf{xxx}}(\mathbf{x},t) \right] \\ &= \ell[\boldsymbol{u}\_{n}(\mathbf{x},t)] + \ell\left[\overline{\boldsymbol{\lambda}}(\mathbf{x},t)\right] \left[ \boldsymbol{s}\boldsymbol{\ell}\boldsymbol{u}\_{n}(\mathbf{x},t) - \boldsymbol{u}\_{n}(\mathbf{x},0) - \ell\left[\boldsymbol{\delta}(\boldsymbol{u}\_{n})(\mathbf{x},t)(\boldsymbol{u}\_{n})\_{\mathbf{x}}(\mathbf{x},t) - (\boldsymbol{u}\_{n})\_{\mathbf{xxx}}(\mathbf{x},t)\right] \right] \end{split}$$

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

$$\begin{aligned} \ell \left[ \delta \mathfrak{U}\_{n+1}(\mathbf{x}, t) \right] &= \ell \left[ \delta \mathfrak{u}\_n(\mathbf{x}, t) \right] + \ell \left[ \overline{\mathcal{A}}(\mathbf{x}, t) \right] \left[ s \ell \operatorname{\mathfrak{G}} \mathfrak{u}\_n(\mathbf{x}, t) \right] \\ &= \ell \left[ \delta \mathfrak{u}\_n(\mathbf{x}, t) \right] \left[ \ell + s \ell \left[ \overline{\mathcal{A}}(\mathbf{x}, t) \right] \right] \end{aligned}$$

This implies that:

$$1 + s \,\, \ell \overline{\lambda} (\mathbf{x}, t) = 0 \quad , \text{ and } \; \overline{\lambda} (\mathbf{x}, t) = \ell^{-1} \left[ \frac{-1}{s} \right] = -t \tag{53}$$

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

$$\ell\left[u\_{n+1}(\mathbf{x},t)\right] = \ell\left[u\_n(\mathbf{x},t)\right] + \ell\left[\int\_0^t (-1)\begin{bmatrix} (u\_n)\_\iota(\mathbf{x},\boldsymbol{\xi}) - 6(u\_n)(\mathbf{x},\boldsymbol{\xi})(u\_n)\_\iota(\mathbf{x},\boldsymbol{\xi})\\ + (u\_n)\_{\scriptscriptstyle x\boldsymbol{x}}(\mathbf{x},\boldsymbol{\xi}) \end{bmatrix} d\boldsymbol{\xi}\right]$$

or

$$\ell\left[\boldsymbol{\mu}\_{n+1}(\mathbf{x},t)\right] = \ell\left[\boldsymbol{\mu}\_{n}\right] + \ell\left[-1\right]\ell\left[(\boldsymbol{\mu}\_{n})\_{\boldsymbol{\iota}} - (\boldsymbol{\mu}\_{n})(\boldsymbol{\mu}\_{n})\_{\boldsymbol{\iota}} + (\boldsymbol{\mu}\_{n})\_{\boldsymbol{\iota}\boldsymbol{\iota}\boldsymbol{\iota}}\right] \tag{54}$$

Let then, from Eq. (54), we have:

Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method http://dx.doi.org/10.5772/intechopen.73291 169

$$\begin{aligned} \ell\left[u\_1(\mathbf{x},t)\right] &= \ell\left[\frac{6}{\mathbf{x}^2}\right] + \ell\left[-1\right]\ell\left[\frac{288}{\mathbf{x}^5}\right] = \frac{6}{\mathbf{x}^2} - \frac{288}{\mathbf{x}^3}\ell \\\ u\_2(\mathbf{x},t) &= \frac{6}{\mathbf{x}^2} - \frac{288}{\mathbf{x}^3}t - \frac{6048}{\mathbf{x}^8}t^2 \quad , \text{---} \dots \end{aligned}$$

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

#### Exercises

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

168 Differential Equations - Theory and Current Research

Consider the physics nonlinear boundary value problem,

The Laplace variational iteration correction functional is

þℓ ðt

functional stationary we obtain:

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

Let then, from Eq. (54), we have:

This implies that:

or

0

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>∗</sup> ð Þ un <sup>t</sup>

2 4

<sup>ℓ</sup>½ �¼ unþ<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � unð Þ <sup>x</sup>; <sup>t</sup>

λð Þ x; t � ς

Example (4.3)

or

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

þð Þ un xxxð Þ x; ς

<sup>¼</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> � �<sup>ℓ</sup> ð Þ un <sup>t</sup>ðx; <sup>t</sup>Þ � <sup>6</sup>ð Þð un <sup>x</sup>; <sup>t</sup>Þð Þ un <sup>x</sup>ðx; <sup>t</sup>Þ þ ð Þ un xxxðx; <sup>t</sup><sup>Þ</sup> � � <sup>¼</sup> <sup>ℓ</sup>½ �þ unð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup> <sup>λ</sup>ð Þ <sup>x</sup>; <sup>t</sup> � � <sup>s</sup>ℓunð Þ� <sup>x</sup>; <sup>t</sup> unðx; <sup>0</sup>Þ � <sup>ℓ</sup> <sup>6</sup>ð Þð un <sup>x</sup>; <sup>t</sup>Þð Þ un <sup>x</sup>ðx; <sup>t</sup>Þ � ð Þ un xxxðx; <sup>t</sup><sup>Þ</sup> � � � �

Taking the variation with respect to of the last equation and making the correction

ð Þ un <sup>t</sup>ð Þ� x; ς 6unð Þ x; ς ð Þ un <sup>x</sup>ð Þ x; ς

<sup>ð</sup>x; <sup>t</sup>Þ � <sup>6</sup>ð Þð un <sup>x</sup>; <sup>t</sup>Þð Þ un <sup>x</sup>ðx; <sup>t</sup>Þ þ ð Þ un xxxðx; <sup>t</sup><sup>Þ</sup> � � � �

dς

3 5

(52)

ð53Þ

ð54Þ

" #

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

<sup>1</sup>Þut <sup>þ</sup> uux <sup>¼</sup> <sup>1</sup> � <sup>e</sup>�<sup>x</sup> <sup>t</sup> <sup>þ</sup> <sup>e</sup>�<sup>x</sup> ð Þ, u xð Þ¼ ; <sup>0</sup> <sup>e</sup>�<sup>x</sup> 2Þut þ uux ¼ 2t þ x þ t <sup>3</sup> <sup>þ</sup> xt<sup>2</sup>, u xð Þ¼ ; <sup>0</sup> <sup>0</sup> <sup>3</sup>Þut <sup>þ</sup> uux <sup>¼</sup> <sup>2</sup>x<sup>2</sup><sup>t</sup> <sup>þ</sup> <sup>2</sup>xt<sup>2</sup> <sup>þ</sup> <sup>2</sup>x<sup>3</sup><sup>t</sup> <sup>4</sup>, u xð Þ¼ ; <sup>0</sup> <sup>1</sup> 4Þut þ uux ¼ 1 þ t cos x þ 1 <sup>2</sup> sin 2x, u xð Þ¼ ; <sup>0</sup> sin <sup>x</sup> 5Þut þ uux ¼ 0, u xð Þ¼� ; 0 x <sup>6</sup>Þut <sup>þ</sup> uux � <sup>u</sup> <sup>¼</sup> <sup>e</sup><sup>t</sup> , u xð Þ¼ ; 0 1 þ x <sup>7</sup>Þutt � uxx � <sup>u</sup> <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> xt <sup>þ</sup> <sup>x</sup><sup>2</sup><sup>t</sup> <sup>2</sup>, u xð Þ¼ ; <sup>0</sup> <sup>1</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>x</sup> <sup>8</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>2</sup>xt <sup>þ</sup> <sup>x</sup><sup>2</sup><sup>t</sup> <sup>2</sup>, u xð Þ¼ ; <sup>0</sup> <sup>1</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>x</sup> <sup>9</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> <sup>6</sup>xt <sup>x</sup><sup>2</sup> � <sup>t</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>6</sup><sup>t</sup> <sup>6</sup>, u xð Þ¼ ; <sup>0</sup> <sup>0</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>0</sup> <sup>10</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> x2 <sup>þ</sup> <sup>t</sup> <sup>2</sup> <sup>2</sup> , u xð Þ¼ ; <sup>0</sup> <sup>x</sup><sup>2</sup>, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>0</sup> <sup>11</sup>Þutt � uxx <sup>þ</sup> <sup>u</sup> <sup>þ</sup> <sup>u</sup><sup>2</sup> <sup>¼</sup> <sup>x</sup><sup>2</sup> cos <sup>2</sup>t, u xð Þ¼ ; <sup>0</sup> x, <sup>u</sup>tð Þ¼ <sup>x</sup>; <sup>0</sup> <sup>0</sup> 12Þut þ uux ¼ 0, u xð Þ¼ ; 0 x 13Þut þ uux ¼ 0, u xð Þ¼� ; 0 x 14Þut þ uux ¼ 0, u xð Þ¼ ; 0 2x 15Þut þ uux ¼ uxx, u xð Þ¼� ; 0 x 16Þut þ uux ¼ uxx, u xð Þ¼ ; 0 2x 17Þut þ uux ¼ uxx, u xð Þ¼ ; 0 4 tan 2x
