Theory and Modern Applications

Weak generalized and numerical solution for a quasilinear pseudo-parabolic equation with nonlocal boundary condition

Abstract

This paper investigates the one dimensional mixed problem with nonlocal boundary conditions, for the quasilinear parabolic equation. Under some natural regularity and consistency conditions on the input data, the existence, uniqueness, convergence of the weak generalized solution, and also continuous dependence upon the data of the solution are shown by using the generalized Fourier method. We construct an iteration algorithm for the numerical solution of this problem.

1 Introduction

D denotes the domain

$D:=\left\{0

Consider the equation

$\frac{\partial u}{\partial t}-\frac{{\partial }^{2}u}{\partial {x}^{2}}-\epsilon \frac{{\partial }^{3}u}{\partial {x}^{2}\phantom{\rule{0.2em}{0ex}}\partial t}=f\left(x,t,u\right),$
(1)

with the initial condition

$u\left(x,0\right)=\phi \left(x\right),\phantom{\rule{1em}{0ex}}x\in \left[0,1\right],$
(2)

the nonlocal boundary condition

$u\left(0,t\right)=u\left(1,t\right),\phantom{\rule{1em}{0ex}}t\in \left[0,T\right],$
(3)
${u}_{x}\left(1,t\right)=0,\phantom{\rule{1em}{0ex}}t\in \left[0,T\right],$
(4)

for a quasilinear pseudo-parabolic equation with the nonlinear source term $f=f\left(x,t,u\right)$.

The functions $\phi \left(x\right)$ and $f\left(x,t,u\right)$ are given functions on $\left[0,1\right]$ and $\overline{D}×\left(-\mathrm{\infty },\mathrm{\infty }\right)$, respectively. $\epsilon \in \left[0,{\epsilon }_{0}\right]$ is a small parameter, ${\epsilon }_{0}\ge 0$.

Denote the solution of problem (1)-(4) by $u=u\left(x,t,\epsilon \right)$.

This problem was investigated with different boundary conditions by various researchers by using Fourier methods [1].

In this study, we consider the initial-boundary value problem (1)-(4) with nonlocal boundary conditions (2)-(3). The periodic nature of (2)-(3) type boundary conditions is demonstrated in [2]. In this study, we prove the existence, uniqueness convergence of the weak generalized solution, continuous dependent upon the data of the solution; and we construct an iteration algorithm for the numerical solution of this problem. We analyze computationally convergence of the iteration algorithm, as well as treating a test example.

We will use the weak solution approach from [3] for the considered problem (1)-(4).

According to [4, 5] we assume the following definitions.

Definition 1 The function $v\left(x,t\right)\in {C}^{2}\left(\overline{D}\right)$ is called test function if it satisfies the following conditions:

Definition 2 The function $u\left(x,t,\epsilon \right)\in C\left(\overline{D}\right)$ satisfying the integral identity

$\begin{array}{r}{\int }_{0}^{T}{\int }_{0}^{1}\left[\left(\frac{\partial v}{\partial t}+\frac{{\partial }^{2}v}{\partial {x}^{2}}-\epsilon \frac{{\partial }^{3}v}{\partial {x}^{2}\phantom{\rule{0.2em}{0ex}}\partial t}\right)u+f\left(x,t,u\right)v\right]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}+{\int }_{0}^{1}\phi \left(x\right)\left[v\left(x,0\right)-\epsilon \frac{{\partial }^{3}v\left(x,0\right)}{\partial {x}^{2}}\right]\phantom{\rule{0.2em}{0ex}}dx=0,\end{array}$
(5)

for arbitrary test function $v=v\left(x,t\right)$, is called a generalized (weak) solution of problem (1)-(4).

2 Reducing the problem to countable system of integral equations

Consider the following system of functions on the interval $\left[0,1\right]$:

$\begin{array}{r}{X}_{0}\left(x\right)=2,\phantom{\rule{2em}{0ex}}{X}_{2k-1}\left(x\right)=4cos\left(2\pi kx\right),\\ {X}_{2k}\left(x\right)=4\left(1-x\right)sin\left(2\pi kx\right),\phantom{\rule{1em}{0ex}}k=1,2,\dots ,\end{array}$
(6)
${Y}_{0}\left(x\right)=x,\phantom{\rule{2em}{0ex}}{Y}_{2k-1}\left(x\right)=xcos\left(2\pi kx\right),\phantom{\rule{2em}{0ex}}{Y}_{2k}\left(x\right)=sin\left(2\pi kx\right),\phantom{\rule{2em}{0ex}}k=1,2,\dots .$
(7)

The system of functions (6) and (7) arise in [6] for the solution of a nonlocal boundary value problem in heat conduction.

It is easy to verify that the systems of function (6) and (7) are biorthonormal on $\left[0,1\right]$. They are also Riesz bases in ${L}_{2}\left[0,1\right]$ (see [7, 8]).

We will use the Fourier series representation of the weak solution to transform the initial-boundary value problem to the infinite set of nonlinear integral equations.

Any solution of (1)-(4) can be represented as

$u\left(x,t,\epsilon \right)=\sum _{k=1}^{\mathrm{\infty }}{u}_{k}\left(t,\epsilon \right){X}_{k}\left(x\right),$
(8)

where the functions ${u}_{k}\left(t,\epsilon \right)$, $k=0,1,2,\dots$ , satisfy the following system of equations:

$\begin{array}{r}{u}_{0}\left(t,\epsilon \right)={\phi }_{0}+{\int }_{0}^{t}{f}_{0}\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {u}_{2k}\left(t,\epsilon \right)={\phi }_{2k}{e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{f}_{2k}\left(\tau \right){e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {u}_{2k-1}\left(t,\epsilon \right)=\left({\phi }_{2k-1}-{\phi }_{2k}+\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}{\phi }_{2k}\right){e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{{u}_{2k-1}\left(t,\epsilon \right)=}+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}\left[{f}_{2k-1}\left(\tau \right)-\left(1-\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}\right)\left(t-\tau \right){f}_{2k}\left(\tau \right)\right]\\ \phantom{{u}_{2k-1}\left(t,\epsilon \right)=}\cdot {e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}$
(9)

where

$\begin{array}{c}{\phi }_{k}={\int }_{0}^{1}\phi \left(x\right){Y}_{k}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\hfill \\ {f}_{k}\left(x\right)={\int }_{0}^{1}f\left(x,t,u\right){Y}_{k}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\hfill \end{array}$

Definition 3 Denote the set

$\left\{u\left(t,\epsilon \right)\right\}=\left\{{u}_{0}\left(t,\epsilon \right),{u}_{2k}\left(t,\epsilon \right),{u}_{2k-1}\left(t,\epsilon \right),k=1,2,\dots \right\},$

of functions continuous on $\left[0,T\right]$ satisfying the condition

$\underset{0\le t\le T}{max}|{u}_{0}\left(t,\epsilon \right)|+\sum _{k=1}^{\mathrm{\infty }}\left(\underset{0\le t\le T}{max}|{u}_{2k}\left(t,\epsilon \right)|+\underset{0\le t\le T}{max}|{u}_{2k-1}\left(t,\epsilon \right)|\right)<\mathrm{\infty },$

by B. Let

$\parallel u\left(t,\epsilon \right)\parallel =\underset{0\le t\le T}{max}|{u}_{0}\left(t,\epsilon \right)|+\sum _{k=1}^{\mathrm{\infty }}\left(\underset{0\le t\le T}{max}|{u}_{2k}\left(t,\epsilon \right)|+\underset{0\le t\le T}{max}|{u}_{2k-1}\left(t,\epsilon \right)|\right)$

be the norm in B. It can be shown that B is a Banach space [9].

We denote the solution of the nonlinear system (9) by $\left\{u\left(t,\epsilon \right)\right\}$.

Theorem 4

1. (1)

Assume the function $f\left(x,t,u\right)$ is continuous with respect to all arguments in $\overline{D}×\left(-\mathrm{\infty },\mathrm{\infty }\right)$ and satisfies the following condition:

$|f\left(t,x,u\right)-f\left(t,x,\stackrel{˜}{u}\right)|\le b\left(t,x\right)|u-\stackrel{˜}{u}|,$

where $b\left(x,t\right)\in {L}_{2}\left(D\right)$, $b\left(x,t\right)\ge 0$,

1. (2)

$f\left(x,t,0\right)\in {C}^{2}\left[0,1\right]$, $t\in \left[0,1\right]$,

2. (3)

$\phi \left(x\right)\in {C}^{2}\left[0,1\right]$.

Then the system (9) has a unique solution in D.

Proof For $N=0,1,\dots$ , let us define an iteration for the system (9) as follows:

$\begin{array}{rl}{u}_{0}^{\left(N+1\right)}\left(t,\epsilon \right)=& {u}_{0}^{\left(0\right)}\left(t,\epsilon \right)+{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,\left(2{u}_{0}^{\left(N\right)}\left(\tau ,\epsilon \right)\\ +4\sum _{k=1}^{\mathrm{\infty }}\left({u}_{2k}^{\left(N\right)}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}^{\left(N\right)}\left(\tau ,\epsilon \right)cos2\pi k\xi \right)\right)\right)\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\\ {u}_{2k}^{\left(N+1\right)}\left(t,\epsilon \right)=& {u}_{2k}^{\left(0\right)}\left(t,\epsilon \right)+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,\left(2{u}_{0}^{\left(N\right)}\left(\tau ,\epsilon \right)\\ +4\sum _{k=1}^{\mathrm{\infty }}\left({u}_{2k}^{\left(N\right)}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}^{\left(N\right)}\left(\tau ,\epsilon \right)cos2\pi k\xi \right)\right)\right)\\ \cdot sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\\ {u}_{2k-1}^{\left(N+1\right)}\left(t,\epsilon \right)=& {u}_{2k-1}^{\left(0\right)}\left(t,\epsilon \right)+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,\left(2{u}_{0}^{\left(N\right)}\left(\tau ,\epsilon \right)\\ +4\sum _{k=1}^{\mathrm{\infty }}\left({u}_{2k}^{\left(N\right)}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}^{\left(N\right)}\left(\tau ,\epsilon \right)cos2\pi k\xi \right)\right)\right)\\ \cdot \xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,\left(2{u}_{0}^{\left(N\right)}\left(\tau \right)\\ +4\sum _{k=1}^{\mathrm{\infty }}\left({u}_{2k}^{\left(N\right)}\left(\tau \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}^{\left(N\right)}\left(\tau \right)cos2\pi k\xi \right)\right)\right)\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ -\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}f\left(\xi ,\tau ,\left(2{u}_{0}^{\left(N\right)}\left(\tau ,\epsilon \right)\\ +4\sum _{k=1}^{\mathrm{\infty }}\left({u}_{2k}^{\left(N\right)}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}^{\left(N\right)}\left(\tau ,\epsilon \right)cos2\pi k\xi \right)\right)\right)\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}$
(10)

where, for simplicity, we let

$\begin{array}{c}\begin{array}{r}A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\\ \phantom{\rule{1em}{0ex}}=2{u}_{0}^{\left(N\right)}\left(\tau ,\epsilon \right)+4\sum _{k=1}^{\mathrm{\infty }}\left({u}_{2k}^{\left(N\right)}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}^{\left(N\right)}\left(\tau ,\epsilon \right)cos2\pi k\xi \right),\end{array}\hfill \\ {u}_{0}^{\left(N+1\right)}\left(t,\epsilon \right)={u}_{0}^{\left(0\right)}\left(t,\epsilon \right)+{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ \begin{array}{rl}{u}_{2k}^{\left(N+1\right)}\left(t,\epsilon \right)=& {u}_{2k}^{\left(0\right)}\left(t,\epsilon \right)+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \cdot sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}\hfill \\ \begin{array}{rl}{u}_{2k-1}^{\left(N+1\right)}\left(t,\epsilon \right)=& {u}_{2k-1}^{\left(0\right)}\left(t,\epsilon \right)\\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ -\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}f\left(\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}\hfill \end{array}$
(11)

where ${u}_{0}^{\left(0\right)}\left(t,\epsilon \right)={\phi }_{0}$, ${u}_{2k}^{\left(0\right)}\left(t,\epsilon \right)={\phi }_{2k}{e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}$, ${u}_{2k-1}^{\left(0\right)}\left(t,\epsilon \right)=\left({\phi }_{2k-1}-{\phi }_{2k}+\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}{\phi }_{2k}\right){e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}$.

From the condition of the theorem we have ${u}^{\left(0\right)}\left(t,\epsilon \right)\in B$. We will prove that the other sequential approximations satisfy this condition.

Let us write $N=0$ in (11).

${u}_{0}^{\left(1\right)}\left(t,\epsilon \right)={u}_{0}^{\left(0\right)}\left(t,\epsilon \right)+{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .$

Adding and subtracting ${\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau$ to both sides of the last equation, we obtain

$\begin{array}{rl}{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)=& {u}_{0}^{\left(0\right)}\left(t,\epsilon \right)+{\int }_{0}^{t}{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$

Applying the Cauchy Inequality to the last equation, we have

$\begin{array}{rcl}|{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)|& \le & |{\phi }_{0}|+{\left({\int }_{0}^{t}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}{\left({\int }_{0}^{t}{\left\{{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +{\left({\int }_{0}^{t}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}{\left({\int }_{0}^{t}{\left\{{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Applying the Lipschitzs Condition to the last equation, we have

$\begin{array}{rcl}|{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)|& \le & |{\phi }_{0}|+\sqrt{t}{\left({\int }_{0}^{t}{\left\{{\int }_{0}^{1}b\left(\xi ,\tau \right)|A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)|\phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\sqrt{t}{\left({\int }_{0}^{t}{\left\{{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Let us use

$|A{u}^{\left(0\right)}\left(\xi ,\tau \right)|\le |{u}^{\left(0\right)}\left(\tau ,\epsilon \right)|.$

Taking the maximum of both sides of the last inequality yields the following:

$\begin{array}{c}\underset{0\le t\le T}{max}|{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)|\le |{\phi }_{0}|+\sqrt{T}{\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}\parallel {u}^{\left(0\right)}\left(t,\epsilon \right)\parallel +\sqrt{T}{\parallel f\left(x,t,0\right)\parallel }_{{L}_{2}\left(D\right)},\hfill \\ \begin{array}{rl}{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)=& {\phi }_{2k}{e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \cdot sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .\end{array}\hfill \end{array}$

Adding and subtracting $\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,0\right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau$ to both sides of the last equation, we obtain

$\begin{array}{rcl}{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)& =& {\phi }_{2k}{e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\\ \cdot sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,0\right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$

Applying the Cauchy Inequality to the last equation, we have

$\begin{array}{r}|{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le |{\phi }_{2k}|+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}\cdot {\left({\int }_{0}^{t}{\left\{\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}{\left({\int }_{0}^{t}{\left\{\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Taking the summation of both sides with respect to k and using the Hölder Inequality yield the following:

$\begin{array}{rl}\sum _{k=1}^{\mathrm{\infty }}|{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)|\le & \sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k}|+\frac{1}{2\sqrt{2}\pi }{\left(\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}{\left(\sqrt{1+\epsilon {\left(2\pi k\right)}^{2}}\right)}^{2}}\right)}^{\frac{1}{2}}\\ \cdot {\left({\int }_{0}^{t}\sum _{k=1}^{\mathrm{\infty }}{\left\{{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\frac{1}{2\sqrt{2}\pi }{\left(\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}{\left(\sqrt{1+\epsilon {\left(2\pi k\right)}^{2}}\right)}^{2}}\right)}^{\frac{1}{2}}\\ \cdot {\left({\int }_{0}^{t}\sum _{k=1}^{\mathrm{\infty }}{\left\{{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Using the Bessel Inequality in the last inequality, we obtain

$\begin{array}{rl}\sum _{k=1}^{\mathrm{\infty }}|{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)|\le & \sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k}|+\frac{1}{4\sqrt{3}}\left({\int }_{0}^{t}\sum _{k=1}^{\mathrm{\infty }}\left\{{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\\ {{\cdot sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\frac{1}{4\sqrt{3}}{\left({\int }_{0}^{t}\sum _{k=1}^{\mathrm{\infty }}{\left\{{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Applying the Lipschitzs Condition to the last equation and taking the maximum of both sides of the last inequality yield the following:

$\begin{array}{c}\begin{array}{rl}\sum _{k=1}^{\mathrm{\infty }}\underset{0\le t\le T}{max}|{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)|\le & \sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k}|+\frac{1}{4\sqrt{3}}{\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}{\parallel {u}^{\left(0\right)}\left(t,\epsilon \right)\parallel }_{B}\\ +\frac{1}{4\sqrt{3}}{\parallel f\left(x,t,0\right)\parallel }_{{L}_{2}\left(D\right)},\end{array}\hfill \\ \begin{array}{rl}{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)=& \left({\phi }_{2k-1}-{\phi }_{2k}+\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}{\phi }_{2k}\right){e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ -\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .\end{array}\hfill \end{array}$

$\begin{array}{c}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,0\right)\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ \frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,0\right)\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ \frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}f\left(\xi ,\tau ,0\right)\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \hfill \end{array}$

to both sides of the last equation and applying the derivative to ${\phi }_{2k}$, we obtain

$\begin{array}{rcl}{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)& =& \left({\phi }_{2k-1}+\frac{1}{4{\pi }^{2}{k}^{2}}{\phi }_{2k}^{\prime \prime }+\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}{\phi }_{2k}^{\prime \prime }\right){e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\\ \cdot \xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,0\right)\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,0\right)\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ -\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}\\ \cdot \left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ +\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}\\ \cdot f\left(\xi ,\tau ,0\right)\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$

Applying the Cauchy Inequality to the last equation, we have the following:

$\begin{array}{r}|{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le |{\phi }_{2k-1}|+\frac{1}{4{\pi }^{2}{k}^{2}}|{\phi }_{2k}^{\prime \prime }|+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}|{\phi }_{2k}^{\prime \prime }|\\ \phantom{\rule{2em}{0ex}}+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}\cdot \left({\int }_{0}^{t}\left\{\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\\ {{\phantom{\rule{2em}{0ex}}\cdot \xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}{\left({\int }_{0}^{t}{\left\{{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}\cdot \left({\int }_{0}^{t}\left\{\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\\ {{\phantom{\rule{2em}{0ex}}\cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}\cdot {\left({\int }_{0}^{t}{\left\{\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}\cdot {\int }_{0}^{t}\left\{\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{1}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-f\left(\xi ,\tau ,0\right)\right]\\ {\phantom{\rule{2em}{0ex}}\cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d{\tau }^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}+{\left({\int }_{0}^{t}{e}^{-\frac{2{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{2em}{0ex}}\cdot {\left({\int }_{0}^{t}{\left\{\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{1}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}f\left(\xi ,\tau ,0\right)\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \right\}}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Taking the summation of both sides with respect to k and using the Hölder, Bessel, and Lipschitzs Inequalities yields the following:

$\begin{array}{rcl}\sum _{k=1}^{\mathrm{\infty }}|{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)|& \le & \sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k-1}|+\left(\frac{1}{4\sqrt{6}\pi }+\frac{\pi }{\sqrt{6}}\right)\sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k}^{\prime \prime }|\\ +\frac{1}{4\sqrt{3}}{\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)|{u}^{\left(0\right)}\left(\tau ,\epsilon \right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\frac{1}{4\sqrt{3}}{\left({\int }_{0}^{t}{\int }_{0}^{1}{f}^{2}\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\frac{|t|}{4\sqrt{3}}{\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)|{u}^{\left(0\right)}\left(\tau ,\epsilon \right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\frac{|t|}{4\sqrt{3}}{\left({\int }_{0}^{t}{\int }_{0}^{1}{f}^{2}\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\frac{|t|\pi }{\sqrt{3}}{\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)|{u}^{\left(0\right)}\left(\tau ,\epsilon \right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ +\frac{|t|\pi }{2\sqrt{3}}{\left({\int }_{0}^{t}{\int }_{0}^{1}{f}^{2}\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Taking the maximum of both sides of the last inequality yields the following:

$\begin{array}{r}\sum _{k=1}^{\mathrm{\infty }}\underset{0\le t\le T}{max}|{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le \sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k-1}|+\left(\frac{1+4{\pi }^{2}}{4\sqrt{6}}\right)\sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k}^{\prime \prime }|\\ \phantom{\rule{2em}{0ex}}+\frac{1}{4\sqrt{3}}{\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}{\parallel {u}^{\left(0\right)}\left(t,\epsilon \right)\parallel }_{B}+\frac{1}{4\sqrt{3}}{\parallel f\left(x,t,0\right)\parallel }_{{L}_{2}\left(D\right)}\\ \phantom{\rule{2em}{0ex}}+\frac{|T|}{4\sqrt{3}}{\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}{\parallel {u}^{\left(0\right)}\left(t,\epsilon \right)\parallel }_{B}+\frac{|T|}{4\sqrt{3}}{\parallel f\left(x,t,0\right)\parallel }_{{L}_{2}\left(D\right)}\\ \phantom{\rule{2em}{0ex}}+\frac{|T|\pi }{2\sqrt{3}}{\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}{\parallel {u}^{\left(0\right)}\left(t,\epsilon \right)\parallel }_{B}+\frac{|T|\pi }{2\sqrt{3}}{\parallel f\left(x,t,0\right)\parallel }_{{L}_{2}\left(D\right)}.\end{array}$

Finally we have the following inequality:

$\begin{array}{rcl}{\parallel {u}^{\left(1\right)}\left(t,\epsilon \right)\parallel }_{B}& =& 2\underset{0\le t\le T}{max}|{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)|+4\sum _{k=1}^{\mathrm{\infty }}\left(\underset{0\le t\le T}{max}|{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)|+\underset{0\le t\le T}{max}|{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)|\right)\\ \le & 2|{\phi }_{0}|+4\sum _{k=1}^{\mathrm{\infty }}\left(|{\phi }_{2k}|+|{\phi }_{2k-1}|\right)+\left(\frac{1+4{\pi }^{2}}{\sqrt{6}}\right)\sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k}^{\prime \prime }|\\ +\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)\left({\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}{\parallel {u}^{\left(0\right)}\left(t,\epsilon \right)\parallel }_{B}\right)\\ +\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\parallel f\left(x,t,0\right)\parallel }_{{L}_{2}\left(D\right)}.\end{array}$

Hence $\phantom{\rule{0.25em}{0ex}}{u}^{\left(1\right)}\left(t,\epsilon \right)\in B$. In the same way, for a general value of N we have

$\begin{array}{rcl}{\parallel {u}^{\left(N\right)}\left(t,\epsilon \right)\parallel }_{B}& =& 2\underset{0\le t\le T}{max}|{u}_{0}^{\left(N\right)}\left(t,\epsilon \right)|+4\sum _{k=1}^{\mathrm{\infty }}\left(\underset{0\le t\le T}{max}|{u}_{2k}^{\left(N\right)}\left(t,\epsilon \right)|+\underset{0\le t\le T}{max}|{u}_{2k-1}^{\left(N\right)}\left(t,\epsilon \right)|\right)\\ \le & 2|{\phi }_{0}|+4\sum _{k=1}^{\mathrm{\infty }}\left(|{\phi }_{2k}|+|{\phi }_{2k-1}|\right)+\left(\frac{1+4{\pi }^{2}}{\sqrt{6}}\right)\sum _{k=1}^{\mathrm{\infty }}|{\phi }_{2k}^{\prime \prime }|\\ +\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)\left({\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}{\parallel {u}^{\left(N-1\right)}\left(t,\epsilon \right)\parallel }_{B}\right)\\ +\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\parallel f\left(x,t,0\right)\parallel }_{{L}_{2}\left(D\right)},\end{array}$

considering the induction hypothesis that ${u}^{\left(N-1\right)}\left(t,\epsilon \right)\in B$, we deduce that ${u}^{\left(N\right)}\left(t,\epsilon \right)\in B$, and by the principle of mathematical induction we obtain

$\left\{u\left(t,\epsilon \right)\right\}=\left\{{u}_{0}\left(t,\epsilon \right),{u}_{2k}\left(t,\epsilon \right),{u}_{2k-1}\left(t,\epsilon \right),k=1,2,\dots \right\}\in B.$

Now we prove that the iterations ${u}^{\left(N+1\right)}\left(t,\epsilon \right)$ converge in B, as $N\to \mathrm{\infty }$. We have

$\begin{array}{r}{u}^{\left(1\right)}\left(t,\epsilon \right)-{u}^{\left(0\right)}\left(t,\epsilon \right)\\ \phantom{\rule{1em}{0ex}}=2\left({u}_{0}^{\left(1\right)}\left(t,\epsilon \right)-{u}_{0}^{\left(0\right)}\left(t,\epsilon \right)\right)+4\sum _{k=1}^{\mathrm{\infty }}\left[\left({u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)-{u}_{2k}^{\left(0\right)}\left(t,\epsilon \right)\right)+\left({u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(0\right)}\left(t,\epsilon \right)\right)\right]\\ \phantom{\rule{1em}{0ex}}=2{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{2em}{0ex}}+\frac{4}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{2em}{0ex}}+\frac{4}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{2em}{0ex}}-\frac{4}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \phantom{\rule{2em}{0ex}}\cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$

Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right side of (11), respectively, we obtain

$\begin{array}{r}|{u}^{\left(1\right)}\left(t,\epsilon \right)-{u}^{\left(0\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le 2|{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)-{u}_{0}^{\left(0\right)}\left(t,\epsilon \right)|+4\sum _{k=1}^{\mathrm{\infty }}\left(|{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)-{u}_{2k}^{\left(0\right)}\left(t,\epsilon \right)|+|{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(0\right)}\left(t,\epsilon \right)|\right)\\ \phantom{\rule{1em}{0ex}}\le \left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}|{u}^{\left(0\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{2em}{0ex}}+\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}{f}^{2}\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}.\end{array}$

Let

$\begin{array}{c}\begin{array}{rl}{A}_{T}=& \left[\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}|{u}^{\left(0\right)}\left(t,\epsilon \right)|\\ +\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}{f}^{2}\left(\xi ,\tau ,0\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\right],\end{array}\hfill \\ \begin{array}{rl}{u}^{\left(2\right)}\left(t,\epsilon \right)-{u}^{\left(1\right)}\left(t,\epsilon \right)=& 2\left({u}_{0}^{\left(2\right)}\left(t,\epsilon \right)-{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)\right)\\ +4\sum _{k=1}^{\mathrm{\infty }}\left[\left({u}_{2k}^{\left(2\right)}\left(t,\epsilon \right)-{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)\right)+\left({u}_{2k-1}^{\left(2\right)}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)\right)\right].\end{array}\hfill \end{array}$

Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right hand side of (9), respectively, we obtain

$\begin{array}{c}\begin{array}{r}|{u}^{\left(2\right)}\left(t,\epsilon \right)-{u}^{\left(1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le 2|{u}_{0}^{\left(2\right)}\left(t,\epsilon \right)-{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{2em}{0ex}}+4\sum _{k=1}^{\mathrm{\infty }}\left(|{u}_{2k}^{\left(2\right)}\left(t,\epsilon \right)-{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)|+|{u}_{2k-1}^{\left(2\right)}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)|\right)\\ \phantom{\rule{1em}{0ex}}\le \left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}{A}_{T},\end{array}\hfill \\ \begin{array}{rl}{u}^{\left(3\right)}\left(t,\epsilon \right)-{u}^{\left(2\right)}\left(t,\epsilon \right)=& 2\left({u}_{0}^{\left(3\right)}\left(t,\epsilon \right)-{u}_{0}^{\left(2\right)}\left(t,\epsilon \right)\right)\\ +4\sum _{k=1}^{\mathrm{\infty }}\left[\left({u}_{2k}^{\left(3\right)}\left(t,\epsilon \right)-{u}_{2k}^{\left(2\right)}\left(t,\epsilon \right)\right)+\left({u}_{2k-1}^{\left(3\right)}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(2\right)}\left(t,\epsilon \right)\right)\right].\end{array}\hfill \end{array}$

Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right hand side of (9), respectively, we obtain

$\begin{array}{r}|{u}^{\left(3\right)}\left(t,\epsilon \right)-{u}^{\left(2\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le 2|{u}_{0}^{\left(3\right)}\left(t,\epsilon \right)-{u}_{0}^{\left(2\right)}\left(t,\epsilon \right)|+4\sum _{k=1}^{\mathrm{\infty }}\left(|{u}_{2k}^{\left(3\right)}\left(t,\epsilon \right)-{u}_{2k}^{\left(2\right)}\left(t,\epsilon \right)|+|{u}_{2k-1}^{\left(3\right)}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(2\right)}\left(t,\epsilon \right)|\right)\\ \phantom{\rule{1em}{0ex}}\le \left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)|{u}^{\left(2\right)}\left(t,\epsilon \right)-{u}^{\left(1\right)}\left(t,\epsilon \right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}\\ \phantom{\rule{1em}{0ex}}\le {\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}}{3}\right)}^{2}{A}_{T}\\ \phantom{\rule{2em}{0ex}}\cdot {\left[{\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\left({\int }_{0}^{\tau }{\int }_{0}^{1}{b}^{2}\left({\xi }_{1},{\tau }_{1}\right)\phantom{\rule{0.2em}{0ex}}d{\xi }_{1}\phantom{\rule{0.2em}{0ex}}d{\tau }_{1}\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right]}^{\frac{1}{2}}\\ \phantom{\rule{1em}{0ex}}\le {\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}}{3}\right)}^{2}{A}_{T}\frac{1}{\sqrt{2}}{\left[{\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{2}\right]}^{\frac{1}{2}}.\end{array}$

In the same way, for a general value of N we have

$\begin{array}{r}|{u}^{\left(N+1\right)}\left(t,\epsilon \right)-{u}^{\left(N\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le 2|{u}_{0}^{\left(N+1\right)}\left(t,\epsilon \right)-{u}_{0}^{\left(N\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{2em}{0ex}}+4\sum _{k=1}^{\mathrm{\infty }}\left(|{u}_{2k}^{\left(N+1\right)}\left(t,\epsilon \right)-{u}_{2k}^{\left(N\right)}\left(t,\epsilon \right)|+|{u}_{2k-1}^{\left(N+1\right)}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(N\right)}\left(t,\epsilon \right)|\right)\\ \phantom{\rule{1em}{0ex}}\le {\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)}^{N}\frac{{A}_{T}}{\sqrt{N!}}{\left[{\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{2}\right]}^{\frac{N}{2}}\\ \phantom{\rule{1em}{0ex}}\le {\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)}^{N}{A}_{T}\frac{1}{\sqrt{N!}}{\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}^{N}.\end{array}$
(12)

Then the last inequality shows that the ${u}^{\left(N+1\right)}\left(t,\epsilon \right)$ converge in B.

Now let us show ${lim}_{N\to \mathrm{\infty }}{u}^{\left(N+1\right)}\left(t,\epsilon \right)=u\left(t,\epsilon \right)$. Noting that

$\begin{array}{r}2|{\int }_{0}^{t}{\int }_{0}^{1}\left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{\rule{2em}{0ex}}\cdot \left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{\rule{2em}{0ex}}\cdot \left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{\rule{2em}{0ex}}\cdot \left(t-\tau \right)\left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}\\ \phantom{\rule{2em}{0ex}}\cdot \left(t-\tau \right)\left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{1em}{0ex}}\le \left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}{\parallel u\left(\tau ,\epsilon \right)-{u}^{\left(N\right)}\left(\tau ,\epsilon \right)\parallel }_{B},\end{array}$

it follows that if we prove ${lim}_{N\to \mathrm{\infty }}{\parallel u\left(\tau ,\epsilon \right)-{u}^{\left(N\right)}\left(\tau ,\epsilon \right)\parallel }_{B}=0$, then we may deduce that $u\left(t,\epsilon \right)$ satisfies (9).

To this aim we estimate the difference ${\parallel u\left(t,\epsilon \right)-{u}^{\left(N+1\right)}\left(t,\epsilon \right)\parallel }_{B}$; after some transformation we obtain

$\begin{array}{r}|u\left(t,\epsilon \right)-{u}^{\left(N+1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}=2|{u}_{0}\left(t,\epsilon \right)-{u}_{0}^{\left(N+1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{2em}{0ex}}+4\sum _{k=1}^{\mathrm{\infty }}\left(|{u}_{2k}\left(t,\epsilon \right)-{u}_{2k}^{\left(N+1\right)}\left(t,\epsilon \right)|+|{u}_{2k-1}\left(t,\epsilon \right)-{u}_{2k-1}^{\left(N+1\right)}\left(t,\epsilon \right)|\right)\\ \phantom{\rule{1em}{0ex}}\le 2|{\int }_{0}^{t}{\int }_{0}^{1}\left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{\rule{2em}{0ex}}\cdot \left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left(\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{\rule{2em}{0ex}}\cdot \left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{\rule{2em}{0ex}}\cdot \left(t-\tau \right)\left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |\\ \phantom{\rule{2em}{0ex}}+4|\sum _{k=1}^{\mathrm{\infty }}\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}\\ \phantom{\rule{2em}{0ex}}\cdot \left(t-\tau \right)\left\{f\left[\xi ,\tau ,Au\left(\xi ,\tau ,\epsilon \right)\right]-f\left[\xi ,\tau ,A{u}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right]\right\}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau |.\end{array}$

Adding and subtracting $f\left(\xi ,\tau ,A{u}^{\left(N+1\right)}\left(\xi ,\tau ,\epsilon \right)\right)$ under the appropriate integrals to the right hand side of the inequality we obtain

$\begin{array}{rcl}|u\left(t,\epsilon \right)-{u}^{\left(N+1\right)}\left(t,\epsilon \right)|& \le & \left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)\\ \cdot {\left\{{\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)|u\left(\tau ,\epsilon \right)-{u}^{\left(N+1\right)}\left(\tau ,\epsilon \right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right\}}^{\frac{1}{2}}\\ +\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)\\ \cdot {\left\{{\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right\}}^{\frac{1}{2}}{\parallel {u}^{\left(N+1\right)}\left(t,\epsilon \right)-{u}^{\left(N\right)}\left(t,\epsilon \right)\parallel }_{B}.\end{array}$

Applying the Gronwall Inequality to the last inequality and using inequality (11), we have

$\begin{array}{r}{\parallel u\left(t,\epsilon \right)-{u}^{\left(N+1\right)}\left(t,\epsilon \right)\parallel }_{B}\\ \phantom{\rule{1em}{0ex}}\le \sqrt{\frac{2}{N!}}{A}_{T}{\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)}^{\left(N+1\right)}{\parallel b\left(x,t\right)\parallel }_{{L}_{2}\left(D\right)}^{\left(N+1\right)}\\ \phantom{\rule{2em}{0ex}}\cdot exp{\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)}^{2}{\parallel b\left(x,t\right)\right)\parallel }_{{L}_{2}\left(D\right)}^{2}.\end{array}$
(13)

For the uniqueness, we assume that problem (1)-(4) has two solutions u, v. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right hand side of $|u\left(t,\epsilon \right)-v\left(t,\epsilon \right)|$, respectively, we obtain

$\begin{array}{r}|u\left(t,\epsilon \right)-v\left(t,\epsilon \right){|}^{2}\\ \phantom{\rule{1em}{0ex}}\le {\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right)}^{2}{\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)|u\left(\tau ,\epsilon \right)-v\left(\tau ,\epsilon \right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}$

applying the Gronwall Inequality to the last inequality we have

$u\left(t,\epsilon \right)=v\left(t,\epsilon \right).$

The theorem is proved. □

3 Solution of problem (1)-(4)

Using the solution of the system (9) we compose the series

$2{u}_{0}\left(t,\epsilon \right)+4\sum _{k=1}^{\mathrm{\infty }}\left[{u}_{2k}\left(t,\epsilon \right)\left(1-x\right)sin2\pi kx+{u}_{2k-1}\left(t,\epsilon \right)cos2\pi kx\right].$

It is evident that the series converges uniformly on D. Therefore the sum

$u\left(\xi ,\tau ,\epsilon \right)=2{u}_{0}\left(\tau ,\epsilon \right)+4\sum _{k=1}^{\mathrm{\infty }}\left[{u}_{2k}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}\left(\tau ,\epsilon \right)cos2\pi k\xi \right],$

is continuous on D. We have

${u}_{l}\left(\xi ,\tau ,\epsilon \right)=2{u}_{0}\left(\tau ,\epsilon \right)+4\sum _{k=1}^{\mathrm{ł}}\left[{u}_{2k}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{u}_{2k-1}\left(\tau ,\epsilon \right)cos2\pi k\xi \right].$
(14)

From the conditions of Theorem 4 and from

$\underset{l\to \mathrm{\infty }}{lim}{u}_{l}\left(\xi ,\tau ,\epsilon \right)=u\left(\xi ,\tau ,\epsilon \right),$

it follows that

$\underset{l\to \mathrm{\infty }}{lim}f\left(\xi ,\tau ,{u}_{l}\left(\tau ,\xi ,\epsilon \right)\right)=f\left(\xi ,\tau ,u\left(\xi ,\tau ,\epsilon \right)\right).$

Using

${u}_{l}\left(\xi ,\tau ,\epsilon \right)$

and

${\phi }_{l}\left(x\right)=2{\phi }_{0}+4\sum _{k=1}^{l}\left[{\phi }_{2k}\left(1-x\right)sin2\pi kx+{\phi }_{2k-1}cos2\pi kx\right],\phantom{\rule{1em}{0ex}}x\in \left[0,1\right],$

on the left hand side of (5) we denote the obtained expression by ${J}_{l}$:

$\begin{array}{rcl}{J}_{l}& =& {\int }_{0}^{T}{\int }_{0}^{1}\left[\left(\frac{\partial v}{\partial t}+\frac{{\partial }^{2}v}{\partial {x}^{2}}\epsilon \frac{{\partial }^{3}v}{\partial {x}^{2}\phantom{\rule{0.2em}{0ex}}\partial t}\right){u}_{\left(l\right)}\left(x,t,\epsilon \right)+f\left(x,t,{u}_{\left(l\right)}\left(x,t,\epsilon \right)\right)v\left(x,t\right)\right]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ +{\int }_{0}^{1}{\phi }_{\left(l\right)}\left(x\right)\left[v\left(x,0\right)-\epsilon \frac{{\partial }^{3}v\left(x,0\right)}{\partial {x}^{2}}\right]\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(15)

Applying the integration by parts the formula on the right hand side, the last equation, and using the conditions of Theorem 4, we can show that

$\underset{l\to \mathrm{\infty }}{lim{J}_{l}}=0.$

This shows that the function $u\left(x,t,\epsilon \right)$ is a generalized (weak) solution of problem (1)-(4).

The following theorem shows the existence and uniqueness results for the generalized solutions of problem (1)-(4).

Theorem 5 Under the assumptions of Theorem  4, problem (1)-(4) possesses a unique generalized solution $u=u\left(x,t\right)\in C\left(\overline{D}\right)$ in the following form:

$u\left(x,t,\epsilon \right)=2{u}_{0}\left(t,\epsilon \right)+4\sum _{k=1}^{\mathrm{\infty }}\left[{u}_{2k}\left(t,\epsilon \right)\left(1-x\right)sin2\pi kx+{u}_{2k-1}\left(t,\epsilon \right)cos2\pi kx\right].$

4 Continuous dependence upon the data

In this section, we shall prove the continuous dependence of the solution $u=u\left(x,t,\epsilon \right)$ using an iteration method.

Theorem 6 Under the conditions of Theorem  4, the solution $u=u\left(x,t,\epsilon \right)$ depends continuously upon the data.

Proof Let $\varphi =\left\{\phi ,f\right\}$ and $\overline{\varphi }=\left\{\overline{\phi },\overline{f}\right\}$ be two sets of data which satisfy the conditions of Theorem 4. Let $u=u\left(x,t,\epsilon \right)$ and $v=v\left(x,t,\epsilon \right)$ be the solutions of problem (1)-(4) corresponding to the data ϕ and $\overline{\varphi }$, respectively, and

The solution $v=v\left(x,t,\epsilon \right)$ is defined by the following forms, respectively:

$\begin{array}{c}{v}_{0}\left(t,\epsilon \right)={\overline{\phi }}_{0}+{\int }_{0}^{t}{\overline{f}}_{0}\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ {v}_{2k}\left(t,\epsilon \right)={\overline{\phi }}_{2k}{e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\overline{f}}_{2k}\left(\tau \right){e}^{-\frac{{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ \begin{array}{rl}{v}_{2k-1}\left(t,\epsilon \right)=& \left({\overline{\phi }}_{2k-1}-{\overline{\phi }}_{2k}+\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}{\overline{\phi }}_{2k}\right){e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \cdot \left[{\overline{f}}_{2k-1}\left(\tau \right)-\left(1-\frac{{\left(2\pi k\right)}^{2}}{1+\epsilon {\left(2\pi k\right)}^{2}}\right)\left(t-\tau \right){\overline{f}}_{2k}\left(\tau \right)\right]\phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}\hfill \end{array}$

where

${\overline{\phi }}_{k}={\int }_{0}^{1}\overline{\phi }\left(x\right){Y}_{k}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\phantom{\rule{2em}{0ex}}{\overline{f}}_{k}\left(x\right)={\int }_{0}^{1}\overline{f}\left(x,t,u\right){Y}_{k}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$

For simplicity, let us write

$\begin{array}{c}A{v}^{\left(N\right)}\left(\xi ,\tau \right)=2{v}_{0}^{\left(N\right)}\left(\tau ,\epsilon \right)+4\sum _{k=1}^{\mathrm{\infty }}\left({v}_{2k}^{\left(N\right)}\left(\tau ,\epsilon \right)\left(1-\xi \right)sin2\pi k\xi +{v}_{2k-1}^{\left(N\right)}\left(\tau \right)cos2\pi k\xi \right),\hfill \\ {v}_{0}^{\left(N+1\right)}\left(t,\epsilon \right)={v}_{0}^{\left(0\right)}\left(t,\epsilon \right)+{\int }_{0}^{t}{\int }_{0}^{1}\overline{f}\left(\xi ,\tau ,A{v}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ \begin{array}{rl}{v}_{2k}^{\left(N+1\right)}\left(t,\epsilon \right)=& {v}_{2k}^{\left(0\right)}\left(t,\epsilon \right)+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \cdot \overline{f}\left(\xi ,\tau ,A{v}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}\hfill \\ \begin{array}{rl}{v}_{2k-1}^{\left(N+1\right)}\left(t,\epsilon \right)=& {v}_{2k-1}^{\left(0\right)}\left(t,\epsilon \right)+\frac{1}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \cdot \overline{f}\left(\xi ,\tau ,A{v}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ -\frac{4\pi k}{1+\epsilon {\left(2\pi k\right)}^{2}}{\int }_{0}^{t}{\int }_{0}^{1}{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\overline{f}\left(\xi ,\tau ,A{v}^{\left(N\right)}\left(\xi ,\tau ,\epsilon \right)\right)\\ \cdot \left(t-\tau \right)sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\end{array}\hfill \end{array}$

where ${v}_{0}^{\left(0\right)}\left(t\right)={\overline{\phi }}_{0}$, ${v}_{2k}^{\left(0\right)}\left(t\right)={\overline{\phi }}_{2k}{e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}$, ${v}_{2k-1}^{\left(0\right)}\left(t\right)=\left({\overline{\phi }}_{2k-1}-4\pi k{\overline{\phi }}_{2k}\right){e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}$. From the condition of the theorem we have ${v}^{\left(0\right)}\left(t,\epsilon \right)\in B$. We will prove that the other sequential approximations satisfy this condition. First of all, we consider ${u}^{\left(1\right)}\left(t,\epsilon \right)-{v}^{\left(1\right)}\left(t,\epsilon \right)$:

$\begin{array}{r}{u}^{\left(1\right)}\left(t,\epsilon \right)-{v}^{\left(1\right)}\left(t,\epsilon \right)\\ \phantom{\rule{1em}{0ex}}=2\left({u}_{0}^{\left(1\right)}\left(t,\epsilon \right)-{v}_{0}^{\left(1\right)}\left(t,\epsilon \right)\right)+4\sum _{k=1}^{\mathrm{\infty }}\left[\left({u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)-{v}_{2k}^{\left(1\right)}\left(t,\epsilon \right)\right)+\left({u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)-{v}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)\right)\right]\\ \phantom{\rule{1em}{0ex}}=\left({\phi }_{0}-{\overline{\phi }}_{0}\right)+2{\int }_{0}^{t}{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-\overline{f}\left(\xi ,\tau ,A{v}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\right]\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{2em}{0ex}}+4\left({\phi }_{2k}-{\overline{\phi }}_{2k}\right){e}^{-{\left(2\pi k\right)}^{2}t}\\ \phantom{\rule{2em}{0ex}}+4{\int }_{0}^{t}{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-\overline{f}\left(\xi ,\tau ,A{v}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\right]{e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{2em}{0ex}}+\left(4\left({\phi }_{2k-1}-{\overline{\phi }}_{2k-1}\right)-16\pi k\left({\phi }_{2k}-{\overline{\phi }}_{2k}\right)\right){e}^{-\frac{{\left(2\pi k\right)}^{2}t}{1+\epsilon {\left(2\pi k\right)}^{2}}}\\ \phantom{\rule{2em}{0ex}}+4{\int }_{0}^{t}{\int }_{0}^{1}\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-\overline{f}\left(\xi ,\tau ,A{v}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\right]\\ \phantom{\rule{2em}{0ex}}\cdot {e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{2em}{0ex}}-16\pi k{\int }_{0}^{t}{\int }_{0}^{1}\left(t-\tau \right)\left[f\left(\xi ,\tau ,A{u}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)-\overline{f}\left(\xi ,\tau ,A{v}^{\left(0\right)}\left(\xi ,\tau ,\epsilon \right)\right)\right]\\ \phantom{\rule{2em}{0ex}}\cdot {e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$

$\begin{array}{c}2{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right)\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\phantom{\rule{2em}{0ex}}2{\int }_{0}^{t}{\int }_{0}^{1}\overline{f}\left(\xi ,\tau ,0\right)\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ 4{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right){e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ 4{\int }_{0}^{t}{\int }_{0}^{1}\overline{f}\left(\xi ,\tau ,0\right){e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ 4{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right){e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ 4{\int }_{0}^{t}{\int }_{0}^{1}\overline{f}\left(\xi ,\tau ,0\right){e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}\xi cos2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ -16\pi k{\int }_{0}^{t}{\int }_{0}^{1}f\left(\xi ,\tau ,0\right){e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ -16\pi k{\int }_{0}^{t}{\int }_{0}^{1}\overline{f}\left(\xi ,\tau ,0\right){e}^{\frac{-{\left(2\pi k\right)}^{2}\left(t-\tau \right)}{1+\epsilon {\left(2\pi k\right)}^{2}}}sin2\pi k\xi \phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \hfill \end{array}$
$\begin{array}{r}|{u}^{\left(1\right)}\left(t,\epsilon \right)-{v}^{\left(1\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{1em}{0ex}}\le 2|{u}_{0}^{\left(1\right)}\left(t,\epsilon \right)-{v}_{0}^{\left(1\right)}\left(t,\epsilon \right)|+4\sum _{k=1}^{\mathrm{\infty }}\left(|{u}_{2k}^{\left(1\right)}\left(t,\epsilon \right)-{v}_{2k}^{\left(1\right)}\left(t,\epsilon \right)|+|{u}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)-{v}_{2k-1}^{\left(1\right)}\left(t,\epsilon \right)|\right)\\ \phantom{\rule{1em}{0ex}}\le 2max|{\phi }_{0}-{\overline{\phi }}_{0}|+4\sum _{k=1}^{\mathrm{\infty }}max|{\phi }_{2k}-{\overline{\phi }}_{2k}|+max|{\phi }_{2k-1}-{\overline{\phi }}_{2k-1}|\\ \phantom{\rule{2em}{0ex}}+\frac{2\sqrt{6}|T|}{3}\sum _{k=1}^{\mathrm{\infty }}max|{\phi }_{2k}^{\prime \prime }-{\overline{\phi }}_{2k}^{\prime \prime }|\\ \phantom{\rule{2em}{0ex}}\cdot \left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}{b}^{2}\left(\xi ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\tau \right)}^{\frac{1}{2}}|{u}^{\left(0\right)}\left(t,\epsilon \right)|\\ \phantom{\rule{2em}{0ex}}+\left(2\sqrt{T}+\frac{\sqrt{3}}{3}+\frac{\left(1+2\sqrt{2}\pi \right)\sqrt{3}|T|}{3}\right){\left({\int }_{0}^{t}{\int }_{0}^{1}}^{}\end{array}$