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Abstract

In this article, the double Laplace transform and Adomian decomposition method are used to solve the nonlinear singular one-dimensional parabolic equation. In addition, we studied the convergence analysis of our problem. Using two examples, our proposed method is illustrated and the obtained results are confirmed.

Keywords

Modified double Laplace transform inverse Laplace transform nonlinear singular parabolic decomposition method convergence

Introduction

One-dimensional linear and nonlinear parabolic equations arise in many different fields of Science and Engineering and are frequently modelled through linear and nonlinear partial differential equations. However, it is still difficult to get closed-form solutions for most models of real-life problems. In recent years, several researchers have paid attention to find the solution of these equations using various methods, such as Adomian decomposition method (ADM)^1–5 and Laplace decomposition method.⁶ The convergence of the decomposition method has been studied by several authors.^7–11 The aim of this article is to solve a nonlinear singular one-dimensional parabolic partial differential equation by combining the domain decomposition techniques and the double Laplace transform method. In addition, we investigate the convergence of our method. We first recall the following definitions which can be found in previous studies.^12–16 The double Laplace transform is defined as

F (p, s) = L_{x} L_{t} [f (x, t)] = \int_{0}^{\infty} e^{- p x} (\int_{0}^{\infty} e^{- s t} f (x, t) d t) d x

(1)

where x, t are real variables and p, s are complex values and further double Laplace transform of the first-order partial derivatives is given by

L_{x} L_{t} [\frac{\partial f (x, t)}{\partial t}] = s F (p, s) - F (p, 0)

(2)

It is easy to prove equation (2) as follows

L_{x} L_{t} [\frac{\partial f (x, t)}{\partial t}] = \int_{0}^{\infty} e^{- p x} (\int_{0}^{\infty} e^{- s t} \frac{\partial f (x, t)}{\partial t} d t) d x

(3)

using integral by parts for the right-hand side of equation (3), we get

L_{x} L_{t} [\frac{\partial f (x, t)}{\partial t}] = - F (p, 0) + s F (p, s)

Similarly, the double Laplace transforms for second partial derivatives with respect to x and t are defined by

L_{x} L_{t} [\frac{\partial^{2} f (x, t)}{\partial^{2} t}] = s^{2} F (p, s) - s F (p, 0) - \frac{\partial F (p, 0)}{\partial t}

(4)

where $L_{x} L_{t}$ denotes the double Laplace transform with respect to x, t and for more details see previous studies.^13,15,16 The following lemma is used in this article.

Lemma 1

The double Laplace transform of the functions $x^{n} (\partial v / \partial t)$ and $x^{n} f (x, t)$ are given by

L_{x} L_{t} (x^{n} \frac{\partial f (x, t)}{\partial t}) = {(- 1)}^{n} \frac{d^{n}}{d p^{n}} [s F (p, s) - F (p, 0)]

(5)

and

L_{x} L_{t} (x^{n} f (x, t)) = {(- 1)}^{n} \frac{d^{n} F (p, s)}{d p^{n}}

(6)

respectively. One can prove this lemma using the definition of double Laplace transform in equations (1), (2) and (4) as follows

L_{x} L_{t} (x^{n} \frac{\partial v}{\partial t}) = {(- 1)}^{n} \frac{d^{n}}{d p^{n}} [sV (p, s) - V (p, 0)]

on using equations (2) and (1)

\begin{matrix} \frac{d^{n}}{d p^{n}} (L_{x} L_{t} [\frac{\partial v (x, t)}{\partial t}]) = {(- 1)}^{n} \int_{0}^{\infty} \int_{0}^{\infty} e^{- px - st} (x^{n} \frac{\partial v (x, t)}{\partial t}) dt dx \\ {(- 1)}^{n} \frac{d^{n}}{d p^{n}} (L_{x} L_{t} [\frac{\partial v (x, t)}{\partial t}]) = L_{x} L_{t} (x^{n} \frac{\partial v}{\partial t}) \\ {(- 1)}^{n} \frac{d^{n}}{d p^{n}} [sV (p, s) - V (p, 0)] = L_{x} L_{t} (x^{n} \frac{\partial v}{\partial t}) \end{matrix}

Similarly, we can prove equation (6).

Nonlinear singular one-dimensional parabolic partial differential equation

The main aim of this study is to know how to solve the nonlinear singular one-dimensional parabolic partial differential equation using modified double Laplace decomposition method (MDLDM).

First problem

Consider the following nonlinear singular one-dimensional parabolic partial differential equation

a (x) u_{t} - \frac{1}{x} {(x u_{x})}_{x} + b (x) u u_{x} + c (x) u^{2} = f (x, t)

(7)

subject to initial condition

u (x, 0) = f_{1} (x)

(8)

where $f (x, t)$ , $f_{1} (x), a (x), b (x)$ and $c (x)$ are known functions. Multiplying both sides of equation (7) by $1 / a$ we have

u_{t} - \frac{1}{a (x) x} {(x u_{x})}_{x} + \frac{b (x)}{a (x)} u (u_{x}) + \frac{c (x)}{a (x)} u^{2} = \frac{1}{a (x)} f (x, t)

(9)

The proposed approach includes the following steps.

The double Laplace transform is applied on both sides of equation (9) and single Laplace transform is adopted and applied to equation (8). The differential property is then used and gives

\begin{matrix} U (p, s) = \frac{F_{1} (p)}{s} \\ + \frac{1}{s} L_{x} L_{t} [\frac{1}{a (x)} \frac{1}{x} {(x u_{x})}_{x} - \frac{b (x)}{a (x)} (u u_{x}) - \frac{c (x)}{a (x)} u^{2}] \\ + \frac{1}{s} L_{x} L_{t} [\frac{1}{a (x)} f (x, t)] \end{matrix}

(10)

we rewrite equation (10) as follows

\begin{matrix} U (p, s) = \frac{F_{1} (p)}{s} \\ + \frac{1}{s} L_{x} L_{t} [\frac{1}{a (x)} \frac{1}{x} {(x u_{x})}_{x} - \frac{b (x)}{a (x)} N_{1} - \frac{c (x)}{a (x)} N_{2}] \\ + \frac{1}{s} (L_{x} L_{t} [\frac{1}{a (x)} f (x, t)]) \end{matrix}

(11)

where $F_{1} (p)$ is single Laplace transform of $f_{1} (x)$ and $N_{1} = u (u_{x}) and N_{2} = u^{2}$ .

The MDLDM gives the solution in the following form

\sum_{n = 0}^{\infty} u_{n} (x, t) = u (x, t)

(12)

where the term $u_{n}$ is to be recursively computed. The nonlinear operators can be defined as follows

N_{1} = \sum_{n = 0}^{\infty} A_{n}, N_{2} = \sum_{n = 0}^{\infty} B_{n}

(13)

where $A_{n}$ and $B_{n}$ are denoted by

\begin{matrix} A_{n} = \frac{1}{n!} {(\frac{d^{n}}{d λ^{n}} [N_{1} \sum_{i = 0}^{\infty} (λ^{n} u_{n})])}_{λ = 0} \\ B_{n} = \frac{1}{n!} {(\frac{d^{n}}{d λ^{n}} [N_{2} \sum_{i = 0}^{\infty} (λ^{n} u_{n})])}_{λ = 0} \end{matrix}

(14)

Using double inverse Laplace transform for equation (11) and using equations (12) and (13) we have

\begin{matrix} \sum_{n = 0}^{\infty} u_{n} (x, t) = f_{1} (x) \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [\frac{1}{a (x) x} {(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}]] \\ - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [\frac{b (x)}{a (x)} (x \sum_{n = 0}^{\infty} A_{n})]] \\ - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [(\frac{c (x)}{a (x)} \sum_{n = 0}^{\infty} B_{n})]] \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [\frac{1}{a (x)} f (x, t)]] \end{matrix}

(15)

we define the following recursive formula

u_{0} = f_{1} (x) + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [\frac{1}{a (x)} f (x, t)]]

(16)

and

\begin{matrix} u_{n + 1} (x, t) = L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [\frac{1}{a (x) x} {(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}]] \\ - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [\frac{b (x)}{a (x)} (x \sum_{n = 0}^{\infty} A_{n})]] \\ - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} L_{x} L_{t} [(\frac{c (x)}{a (x)} \sum_{n = 0}^{\infty} B_{n})]] \end{matrix}

(17)

where the double inverse Laplace transform with respect to p, s is denoted by $L_{p}^{- 1} L_{s}^{- 1}$ . We provided that the inverse exists for each term in the right-hand side of equation (15). For the purpose of illustration of MDLDM, the homogeneous form of the nonlinear singular one-dimensional parabolic partial differential equation is solved. We assume $a (x) = 4 / x^{2},$ $b (x) = - (1 / 2) x$ , $c (x) = 1$ and $f (x, t) = 0$ in equation (7), we have the following example.

Example 1

Consider the following nonlinear singular one-dimensional parabolic partial differential equation

\frac{4}{x^{2}} u_{t} - \frac{1}{x} {(x u_{x})}_{x} - \frac{1}{2} xu u_{x} + u^{2} = 0

(18)

subject to initial conditions

u (x, 0) = 0

(19)

By multiplying equation (18) by $x^{2} / 4$ and using equation (10), we obtain

U (p, s) = \frac{2}{p^{3} s} + \frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x u_{x})}_{x} + \frac{x^{3}}{8} u u_{x} - \frac{x^{2}}{4} u^{2}]

(20)

Applying the inverse double Laplace transform in equation (20) yields

u (x, t) = x^{2} + L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x u_{x})}_{x} + \frac{x^{3}}{8} u u_{x} - \frac{x^{2}}{4} u^{2}])

using the decomposition series for $u (x, t)$ , which is defined by equation (12), we have

\begin{matrix} \sum_{n = 0}^{\infty} u_{n} (x, t) = x^{2} + L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}]) \\ + L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x^{3}}{8} \sum_{n = 0}^{\infty} A_{n}]) \\ - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x^{2}}{4} \sum_{n = 0}^{\infty} B_{n}]) \end{matrix}

(21)

Using equations (16) and (17), we obtain

u_{0} = x^{2}

The other components are given by

\begin{matrix} u_{n + 1} = + L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}]) \\ + L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x^{3}}{8} \sum_{n = 0}^{\infty} A_{n}]) \\ - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x^{2}}{4} \sum_{n = 0}^{\infty} B_{n}]) \end{matrix}

(22)

The other components of the solution can be easily found using equation (22) as follows

\begin{matrix} u_{1} = L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x u_{x})}_{x} + \frac{x^{3}}{8} A_{0} - \frac{x^{2}}{4} B_{0}]) \\ = L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [x^{2}]) = x^{2} t \end{matrix}

where $u_{2}$ is given by

\begin{matrix} u_{2} = L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x u_{x})}_{x} + \frac{x^{3}}{8} A_{1} - \frac{x^{2}}{4} B_{1}]) \\ = L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x u_{x})}_{x} + \frac{x^{3}}{8} (u_{0} u_{1 x} + u_{1} u_{0 x}) - \frac{x^{2}}{2} u_{0} u_{1}]) \\ = L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [x^{2} t]) = \frac{x^{2} t^{2}}{2} \end{matrix}

and

\begin{matrix} u_{3} = L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x u_{x})}_{x} + \frac{x^{3}}{8} A_{2} - \frac{x^{2}}{4} B_{2}]) \\ = L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x}{4} {(x u_{x})}_{x} + \frac{x^{3}}{8} (u_{0} u_{2 x} + u_{1} u_{1 x} + u_{2} u_{0 x})]) \\ - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} L_{x} L_{t} [\frac{x^{2}}{4} (2 u_{0} u_{2} + u_{1}^{2})]) \\ = \frac{x^{2} t^{3}}{6} \end{matrix}

and so on. In the same manner, the rest of the components can be obtained by the same method and then the functions $u (x, t)$ in the closed form are readily found to be

u (x, t) = u_{0} + u_{1} + \dots

u (x, t) = x^{2} (1 + t + \frac{t^{2}}{2} + \frac{t^{3}}{6} + \dots) = x^{2} e^{t}

Therefore, the exact solution is given by

u (x, t) = x^{2} e^{t}

Second problem

Consider the following nonlinear singular one-dimensional parabolic partial differential equation

u_{t} - \frac{a}{x} {(x u_{x})}_{x} + h (x) u u_{x} + k (x) u_{x} = f (x, t)

(23)

subject to

u (x, 0) = f_{1} (x)

(24)

where the term $(1 / x) (x u_{x})_{x}$ is called Bessel’s operator; $f_{1} (x)$ , $h (x)$ and $k (x)$ are known functions; and a is constant. In order to obtain the solution of equation (23), we use MDLDMs as follows:

Step 1. Multiply both sides of equation (7) by x.

Step 2. Using Lemma 1 and definition of the Laplace transform of partial derivatives for equations in step 1 and applying single Laplace transform for initial condition, we have

\begin{matrix} \frac{dU (p, s)}{dp} = \frac{1}{s} \frac{d F_{1} (p)}{dp} + \frac{1}{s} L_{x} L_{t} (\frac{dF (p, s)}{dp}) \\ - \frac{1}{s} L_{x} L_{t} (a \frac{\partial}{\partial x} [x u_{x}] - xk (x) u_{x} - xh (x) u u_{x}) \end{matrix}

(25)

where $F (p, s)$ and $F_{1} (p)$ are Laplace transform of the functions $f (x, t)$ and $f_{1} (x)$ , respectively.

Step 3. Applying the integral for both sides of equation (12) from 0 to p with respect to p, we have

\begin{matrix} U (p, s) = \frac{F_{1} (p)}{s} + \frac{1}{s} \int_{0}^{p} dF (p, s) - \int_{0}^{p} \frac{1}{s} L_{x} L_{t} [a {(x u_{x})}_{x}] dp \\ + \frac{1}{s} \int_{0}^{p} L_{x} L_{t} [xh (x) u u_{x} + xk (x) u_{x}] dp \end{matrix}

(26)

Step 4. The double Laplace decomposition method is representing the solution of nonlinear singular one-dimensional parabolic partial differential equation as $u (x, t)$ in equation (12).

Step 5. By applying double inverse Laplace transform for equation (26) and using equation (14), we obtain

\begin{matrix} u (x, t) = L_{p}^{- 1} L_{s}^{- 1} (\frac{F_{1} (p)}{s}) + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} dF (p, s)] \\ - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [a {(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}] dp] \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L [xh (x) \sum_{n = 0}^{\infty} A_{n}] dp] \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [xk (x) \sum_{n = 0}^{\infty} C_{n}] dp] \end{matrix}

(27)

where the definition of $C_{n}$ is similar to $A_{n}$ , in particular, we have

u_{0} = L_{p}^{- 1} L_{s}^{- 1} (\frac{F_{1} (p)}{s}) + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} dF (p, s)]

(28)

and

\begin{matrix} u_{n + 1} (x, t) = - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [a {(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}] dp] \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [xh (x) \sum_{n = 0}^{\infty} A_{n}] dp] \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [xk (x) \sum_{n = 0}^{\infty} C_{n}] dp] \end{matrix}

(29)

In order to illustrate the accuracy and reliability of the scheme, we assume $a = 1$ , $h (x) = 2 x$ , $k (x) = - x^{2}$ and $f (x, t) = x^{2} - 4 t$ in equation (23), and we have the following example.

Example 2

Consider the following nonlinear singular one-dimensional parabolic partial differential equation

u_{t} - \frac{1}{x} {(x u_{x})}_{x} + 2 xu u_{x} - x^{2} u_{x} = x^{2} - 4 t

(30)

subject to

u (x, 0) = 0

(31)

Using the above steps, we have

\begin{matrix} \sum_{n = 0}^{\infty} u_{n} (x, t) = x^{2} t - 2 t^{2} - L_{p}^{- 1} L_{s}^{- 1} \\ [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [{(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}] dp] \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [2 x^{2} \sum_{n = 0}^{\infty} A_{n}] dp] \\ - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [x^{3} \sum_{n = 0}^{\infty} C_{n}] dp] \end{matrix}

Using equations (28) and (29), we obtain

u_{0} = x^{2} t - 2 t^{2}

and

\begin{matrix} u_{n + 1} = - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [{(x {(\sum_{n = 0}^{\infty} u_{n})}_{x})}_{x}] dp] \\ + L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [2 x^{2} \sum_{n = 0}^{\infty} A_{n}] dp] \\ - L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [x^{3} \sum_{n = 0}^{\infty} C_{n}] dp] \end{matrix}

(32)

The other components of the solution can be easily found using equation (32) as follows

\begin{matrix} u_{1} = - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [{(x u_{x})}_{x} - 2 x^{2} A_{0} + x^{3} C_{0}] dp) \\ = - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [4 xt + 8 x^{3} t^{3}] dp) = 2 t^{2} + 2 x^{2} t^{4} \end{matrix}

where $u_{2}$ is defined by

\begin{matrix} u_{2} = - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [{(x u_{x})}_{x} - 2 x^{2} A_{1} + x^{3} C_{1}] dp) \\ = - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [{(x u_{x})}_{x} - 2 x^{2} (u_{0} u_{1 x} + u_{1} u_{0 x}) + x^{3} u_{0} u_{1}] dp) \\ = - L_{p}^{- 1} L_{s}^{- 1} (\frac{1}{s} \int_{0}^{p} L_{x} L_{t} [8 x t^{4} - 8 x^{3} t^{3} + 16 x^{3} t^{6}] dp) \\ = \frac{8}{5} t^{5} - + 2 x^{2} t^{4} + \frac{16}{7} x^{2} t^{7} \end{matrix}

so the solution u(x, t) is given by

u (x, t) = u_{0} + u_{1} + \dots

therefore, the exact solution is given by

u (x, t) = x^{2} t

Now, we study the convergence of the used method by considering a more general parabolic partial differential equation, more precisely, we investigate the convergence of the method for the one-dimensional parabolic partial differential equation with Bessel operator.

Convergence analysis

The aim of this section is to discuss our method for the nonlinear singular one-dimensional parabolic partial differential equation given by

u_{t} - \frac{1}{x} {(x u_{x})}_{x} + u u_{x} - u_{x} = f (u)

(33)

for all $u, v \in H$ . We define H as $H = L_{μ}^{2} ((a, b) \times [0, T])$ and

\begin{matrix} u : (a, b) \times [0, T] \to R \times R, with ∥ u {∥^{2}}_{H} = \int_{Q} x u^{2} (x, t) dxdt \\ (u, v) = \int_{Q} xu (x, t) v (x, t) dxdt \end{matrix}

where $Q = (a, b) \times [0, T]$ and

H = {\begin{cases} (u, v) : (a, b) \times [0, T], with \\ L_{p}^{- 1} L_{s}^{- 1} [\frac{1}{s^{2}} \int_{0}^{p} L_{x} L_{t} [u (x, t)] (p, s) d p] (x, t) < \infty \end{cases}}

For more details, see Atangana and Noutchie.¹⁰ We can rewrite equation (33) in the following form

L (u) = x u_{t} = u_{x} + x u_{x} - xu u_{x} + x u_{x} + xf (u)

(34)

For L hemicontinuous operator, consider the following hypothesis:

(H1) $k ∥ u - v ∥^{2} \leq (L (u) - L (v), u - v); k > 0, \forall u, v \in H$

(H2) whatever $M^{*} > 0$ , $\exists C (M^{*}) > 0$ such that for $u, v \in H$ with $M^{*} \geq ∥ u ∥$ , $M^{*} \geq ∥ v ∥$ and $∥ u_{x} ∥ \leq M_{1}^{*}, ∥ v_{x} ∥ \leq M_{2}^{*}$ , we have

(L (u) - L (v), w) \leq C (M^{*}) ∥ u - v ∥ ∥ w ∥

for every $w \in H$ .^11,17,18

Theorem 1 (Sufficient condition of convergence)

The MDLDMs applied to the nonlinear singular one-dimensional pseudohyperbolic equation (34) without initial and boundary conditions converge towards a particular solution.

Proof

To verify the convergence of equation (34), we use the definition of our operator L, and we have the following form

\begin{matrix} L (u) - L (v) = (u_{x} - v_{x}) + (x u_{x} - x v_{x}) \\ - (\frac{1}{2} x u_{x} - \frac{1}{2} x v_{x}) + (x {(u_{x})}^{2} - x {(v_{x})}^{2}) \\ + x (f (u) - f (v)) \\ = {(u - v)}_{x} + x {(u - v)}_{xx} - \frac{1}{2} x {(u^{2} - v^{2})}_{x} \\ + (x (u_{x} - v_{x}) (u_{x} + v_{x})) \\ + x (f (u) - f (v)) \end{matrix}

therefore

\begin{array}{l} (L (u) - L (v), u - v) = ({(u - v)}_{x}, u - v) \\ + (x {(u - v)}_{x x}, u - v) % \\ - (\frac{1}{2} x {(u^{2} - v^{2})}_{x}, u - v) % \\ + (x (u_{x} + v_{x}) (u_{x} - v_{x}), u - v) % \\ + (x (f (u) - f (v)), u - v) % \end{array}

(35)

According to the properties of the differential operator $\partial / \partial x$ in H. There exist $α, β, θ, ζ$ and $η$ such that

({(u - v)}_{x}, u - v) \geq α ∥ u - v ∥^{2}

(36)

using Cauchy–Schwarz inequality

\begin{matrix} - (x {(u - v)}_{xx}, u - v) \leq | x | ∥ {(u - v)}_{xx} ∥ ∥ u - v ∥ \\ \leq b β ∥ u - v ∥^{2} \Leftrightarrow \\ (x {(u - v)}_{xx}, u - v) \geq - b β ∥ u - v ∥^{2} \end{matrix}

(37)

\begin{matrix} (\frac{1}{2} x {(u^{2} - v^{2})}_{x}, u - v) \\ \leq \frac{1}{2} | x | η ∥ {(u - v)}_{x} ∥ ∥ u + v ∥ ∥ u - v ∥ \\ \leq b M^{*} η ∥ u - v ∥^{2} \\ \Leftrightarrow \\ - (\frac{1}{2} x {(u^{2} - v^{2})}_{x}, u - v) \geq - b M^{*} η ∥ u - v ∥^{2} \end{matrix}

(38)

and

\begin{matrix} - ((x (u_{x} - v_{x}) (u_{x} + v_{x})), u - v) \\ \leq | x | ∥ {(u + v)}_{x} ∥ ∥ {(u - v)}_{x} ∥ ∥ u - v ∥ \\ \leq M_{1}^{*} M_{2}^{*} b ζ ∥ u - v ∥^{2} \Leftrightarrow \\ ((x (u_{x} - v_{x}) (u_{x} + v_{x})), u - v) \geq - M_{1}^{*} M_{2}^{*} b ζ ∥ u - v ∥^{2} \end{matrix}

(39)

According to Cauchy–Schwarz inequality, where $σ > 0$ as f is Lipschitzian, we obtain

\begin{matrix} (- xf (u) + xf (v), u - v) \leq | x | ∥ f (u) - f (v) ∥ ∥ u - v ∥ \\ \leq b ∥ f (u) - f (v) ∥ ∥ u - v ∥ \\ \leq b σ ∥ u - v ∥^{2} \Leftrightarrow (x (f (u) - f (v)), u - v) \\ \geq - b σ ∥ u - v ∥^{2} \end{matrix}

(40)

substituting equations (36)–(41) into equation (35) gives

\begin{matrix} (L (u) - L (v), u - v) \\ \geq (α - b β - b M^{*} η - M_{1}^{*} M_{2}^{*} b ζ - b σ) ∥ u - v ∥^{2} \\ k ∥ u - v ∥^{2} \leq (L (u) - L (v), u - v) \end{matrix}

So the hypothesis (H1) holds, where

k = α - b β - b M^{*} η - M_{1}^{*} M_{2}^{*} b ζ - b σ > 0

Now we check the hypotheses (H2) for the operator $L (u)$ , for every $M^{*} > 0$ , there exists a constant $C (M^{*}) > 0$ such that for $\forall u, v \in H$ with $∥ u ∥ \leq M^{*}, ∥ v ∥ \leq M^{*}$

(L (u) - L (v), u - v) \leq C (M^{*}) ∥ u - v ∥ ∥ w ∥

$\forall w \in H$ . For that we have

\begin{matrix} (L (u) - L (v), u - v) = ({(u - v)}_{x}, w) + (x {(u - v)}_{xx}, w) \\ - (\frac{1}{2} x {(u^{2} - v^{2})}_{x}, w) \\ + (x (u_{x} + v_{x}) (u_{x} - v_{x}), w) \\ + (x (f (u) - f (v)), w) \end{matrix}

(41)

There exist $α_{1}, α_{2}, α_{3}, β_{1}$ and $σ_{1}$ such that, using the Schwartz inequality and the fact that u and v are bounded, we obtain the following

\begin{matrix} ({(u - v)}_{x}, w) \leq α_{1} ∥ u - v ∥ ∥ w ∥ \\ (x {(u - v)}_{xx}, w) \leq b α_{2} ∥ u - v ∥ ∥ w ∥ \\ (x {(u - v)}_{x}, w) \leq b β_{2} ∥ u - v ∥ ∥ w ∥ \\ - (\frac{1}{2} x {(u^{2} - v^{2})}_{x}, w) \\ \leq \frac{1}{2} α_{3} | x | ∥ u + v ∥ ∥ u - v ∥ ∥ w ∥ \\ \leq b α_{3} M^{*} ∥ u - v ∥ ∥ w ∥, \\ (x (u_{x} + v_{x}) (u_{x} - v_{x}), w) \leq | x | ‖ \frac{\partial}{\partial x} (u + v) ‖ \\ ∥ {(u - v)}_{x} ∥ ∥ w ∥ \\ \leq b β_{1} M_{1}^{*} M_{2}^{*} ∥ u - v ∥ ∥ w ∥ \\ (x (f (u) - f (v)), w) \leq b σ_{1} ∥ u - v ∥ ∥ w ∥ \end{matrix}

where $| x | < b$ we have

\begin{matrix} (L (u) - L (v), w) \\ \leq (α_{1} + b α_{2} + b α_{3} M^{*} + b β_{1} M_{1}^{*} M_{2}^{*} + b σ_{1}) \\ ∥ u - v ∥ ∥ w ∥ \\ = C (M^{*}) ∥ u - v ∥ ∥ w ∥ \end{matrix}

where $C (M^{*}) = α_{1} + b α_{2} + b α_{3} M^{*} + b β_{1} M_{1}^{*} M_{2}^{*} + b σ_{1}$ and therefore (H2) holds.

Conclusion

Modified double Laplace Adomian decomposition method (MDLADM) has been known to be an efficient and accurate method for solving nonlinear singular one-dimensional parabolic partial differential equation. In this work, we have applied the MDLADM for solving a nonlinear parabolic equation. The method was tested to obtain exact solutions in two examples.

Footnotes

Academic Editor: Ramoshweu Lebelo

Declaration of conflicting interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for funding this research (group no. RG-1435-043).

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On a nonlinear singular one-dimensional parabolic equation and double Laplace decomposition method

Abstract

Keywords

Introduction

Lemma 1

Nonlinear singular one-dimensional parabolic partial differential equation

First problem

Example 1

Second problem

Example 2

Convergence analysis

Theorem 1 (Sufficient condition of convergence)

Proof

Conclusion

Footnotes

Declaration of conflicting interests

Funding

References