Differential Equations and Dynamical Systems

, Volume 21, Issue 4, pp 309–321

# A Study of Delay Integrodifferential Parabolic Problems

## Authors

• Department of Mathematics and StatisticsIndian Institute of Technology Kanpur
• Dhirendra Bahuguna
• Department of Mathematics and StatisticsIndian Institute of Technology Kanpur
Original Research

DOI: 10.1007/s12591-013-0166-6

Raheem, A. & Bahuguna, D. Differ Equ Dyn Syst (2013) 21: 309. doi:10.1007/s12591-013-0166-6

## Abstract

In this paper, we study the existence and uniqueness of a weak solution for a delay integrodifferential parabaolic equation using the method of semidiscretization.

### Keywords

Method of semidiscretizationDelay integrodifferential parabolic problemsWeak solution

## Introduction

The main purpose of this paper is to prove the existence and uniqueness of a weak solution for the following delay integrodifferential parabolic problem using the method of semidiscretization in time,
\begin{aligned} \frac{\partial u}{\partial t}+Au&= f(x,\,t,\,u(x,\,t),\,u(x,\,t-\tau ),\,r(x,\,t)) \quad \text{ in } \quad Q=G\times I,\end{aligned}
(1.1)
\begin{aligned} r(x,\,t)&= \int \limits _{0}^{t}f(x,\,s,\,u(x,\,s),\,u(x,\,s-\tau ),\,r(x,\,s))ds,\end{aligned}
(1.2)
\begin{aligned} u(x,\,t)&= \phi (x,\,t), \quad (x,\,t) \in G\times [-\tau ,\,0],\end{aligned}
(1.3)
\begin{aligned} B_iu&= 0 \quad \text{ on } \quad \Gamma \times I, \quad i=1,\,2,\ldots ,\mu , \end{aligned}
(1.4)
\begin{aligned} C_iu&= 0 \quad \text{ on } \quad \Gamma \times I,\quad i=1,\,2,\ldots ,k-\mu , \end{aligned}
(1.5)
where $$G$$ is a bounded domain in the $$N$$-dimensional Euclidean space $$E_N$$ with the Lipschitz boundary $$\Gamma ,\,A$$ is a linear differential operator of order $$2k$$ of the form
\begin{aligned} A=\sum \limits _{|i|,|j|\le k}(-1)^iD^i\left( a_{ij}(x)D^j\right) \!, \quad \text{ and } \quad I=(0,\,T], \end{aligned}
(1.4) and (1.5) are respectively stable and unstable boundary conditions with respect to the operator $$A,$$ where $$B_i,\,C_i$$ are some linear operators.

Since 1930, various classical types of initial boundary value problem have been investigated by many authors using the method of semidiscretization; see for instance [4, 5, 7, 8] and references therein. The method of semidescretization in time is a very efficient tool in the study of an approximate solution and its convergence to the solution of the problem. This method has been used by many authors to study the solutions of abstract Cauchy problems with classical conditions. Rektorys [9] has used this method for the numerical studies.

We aim to use this method on a delay integrodifferential parabolic problems. Delay integrodifferential parabolic problems arise in the modeling of many physical problems, for instance, population dynamics, investigation of a pollution process in rivers, seas and other related problems.

In [3], Bouziani and Mechri have established the existence of a weak solution for an integrodifferential equation with a non-local boundary condition using the method of semidiscretization in time. Lakoud and Chaoni [6] have proved the existence and uniqueness of a weak solution for a hyperbolic integrodifferential equation with initial and integral conditions by using Rothe’s method. For the more applications of this method to integrodifferential equations, we refer the reader to [1, 2] and references cited therin. In the present work, we extend the application of Rothe’s method for a delay integrodifferential equation with an integral condition.

In this paper, first we consider the problem (1.1)–(1.5) on the interval $$[-\tau ,\,\tau ]$$ and prove the existence and uniqueness of a weak solution on the interval $$[-\tau ,\,\tau ]$$ using the method of semidiscretization. In second step, we consider this weak solution as a history function on the interval $$[-\tau ,\,\tau ]$$ and prove the existence and uniqueness of a weak solution on the interval $$[-\tau ,\,2\tau ].$$ Continuing this process, we obtain a unique weak solution of the given problem on the whole interval.

## Assumptions and Preliminaries

($$\mathbf H _1$$)
The form $$((v,\,u))$$ corresponding to the operator $$A$$ and to the boundary operators $$B_i,\,C_i$$ (i.e. corresponding to the elliptic problem)
\begin{aligned} Au&= f \quad \text{ in } \quad G, \end{aligned}
(2.1)
\begin{aligned} B_iu&= 0 \quad \text{ on } \quad \Gamma , \quad i=1,\,2,\ldots ,\mu , \end{aligned}
(2.2)
\begin{aligned} C_iu&= 0 \quad \text{ on } \quad \Gamma , \quad i=1,\,2,\ldots ,k-\mu , \end{aligned}
(2.3)
is assumed to be bounded in $$W_2^{(k)}(G)$$ and $$V$$-elliptic, i.e. there exist positive constants $$K$$ and $$\alpha$$ (independent of $$u,\,v$$) such that
\begin{aligned} |((v,\,u))|&\le K \Vert v\Vert _{W_2^{(k)}(G)}\Vert u\Vert _{W_2^{(k)}(G)} \quad \forall v,\,u \in W_2^{(k)}(G), \nonumber \\ \text{ and } \quad ((v,\,v))&\ge \alpha \Vert v\Vert ^2_V \quad \forall v \in V.\nonumber \\ \text{ Here } \quad V&= \left\{ v| v \in W_2^{(k)}(G),\,B_iv=0 \quad \text{ on }\quad \Gamma ,\right. \nonumber \\&\left. \text{ in } \text{ the } \text{ sense } \text{ of } \text{ traces },\quad i=1,\,2,\ldots ,\mu \right\} \!, \nonumber \\ \text{ and } \quad ((v,\,u))&= \sum \limits _{|i|,\,|j|\le k}\int \limits _{G}a_{ij}(x)D^ivD^judx. \end{aligned}
(2.4)
($$\mathbf H _2$$)
Let $$\Phi :\,[-\tau ,\,0]\rightarrow L^2(G)$$ such that $$\Phi (t)(x)=\phi (x,\,t).$$ We assume that there exists $$k>0$$ such that
\begin{aligned} \left\| \Phi \left( t_1\right) -\Phi \left( t_2\right) \right\| \le k \left| t_1-t_2\right| , \quad \text{ for } \text{ all } \quad t_1,\,t_2 \in [-\tau ,\,0]. \end{aligned}
Moreover, there exist constants $$M_1,\,M_2$$ such that
\begin{aligned} \Vert \Phi (t)\Vert _V \le M_1, \quad \Vert \Phi ^{\prime }(t)\Vert _V \le M_2. \end{aligned}
Also we assume that $$\Phi (0)=0.$$
($$\mathbf H _3$$)
Suppose that there exists $$C>0$$ such that
\begin{aligned}&\left\| f\left( x,\,t_2,\,u_2,\,v_2,\,r_2\right) -f\left( x,t_1,u_1,v_1,r_1\right) \right\| \\&\le C \left( |t_2-t_1|+\left\| u_2-u_1\right\| +\left\| v_2-v_1\right\| +\left\| r_2-r_1\right\| \right) \!, \end{aligned}
and
\begin{aligned} t\in [0,\,T],\quad u\in V,\quad r \in L_2(G)\,\Longrightarrow \,f\in L_2(G). \end{aligned}

### Lemma 2.1

If the form $$((v,\,u))$$ be bounded in $$W_2^{(k)}(G)$$ and $$V$$-elliptic, then there exists precisely one weak solution of the problem (2.1)–(2.3).

### Proof

For a proof we refer p. 62 in [8].

### Lemma 2.2

If the form $$((v,\,u))$$ be bounded in $$W_2^{(k)}(G)$$ and $$V$$-elliptic, then same holds for
\begin{aligned} (((v,\,u)))=((v,\,u))+\frac{1}{h}(v,\,u)\quad \text{ where }\quad h>0. \end{aligned}

### Proof

For a proof we refer p. 78 in [8].

### Definition 2.3

The pair of functions $$u(t)=u(x,\,t),\,r(t)=r(x,\,t)$$ satisfying
1. (i)

$$u(t)=\Phi (t) \quad \text{ on } \quad [-\tau ,\,0],$$

2. (ii)

$$u \in L_2(I,\,V)\cap AC(I,\,L_2(G)),$$

3. (iii)

$$u^{\prime } \in L_2(I,\,L_2(G)),$$

4. (iv)

$$u(0)=\Phi (0) \in C(I,\,L_2(G)),$$

5. (v)

$$r \in C(I,\,L_2(G)),$$

6. (vi)

$$\int _{0}^{T}((v(t),\,u(t)))dt+\int _{0}^{T}(v(t),\,u^{\prime }(t))dt=\int _{0}^{T}(v(t),\,g(t))dt \quad \forall v\in L_2(I,\,V),$$

7. (vii)

$$r(t)=\int _{0}^{t}g(\tau )d\tau \quad \forall t \in [0,\,T],$$ where $$g(t)=g(x,\,t)=f(x,\,t,\,u(x,\,t),\,u(x,t-\tau ),\,r(x,\,t)),$$

is called a weak solution of problem (1.1)–(1.5) on the interval $$[-\tau ,\,T].$$

### Remark 2.4

The space $$L_2(I,\,H)$$ is the space of functions which are square integrable in the Bochner sense, i.e. Bochner integrable and satisfying
\begin{aligned} \int \limits _{I}\left\| y(t)\right\| ^2_H dt< +\infty , \end{aligned}
with the scalar product
\begin{aligned} \left( y_1,\,y_2\right) _{L_2(I,H)}=\int \limits _{I}\left( y_1(t),\,y_2(t)\right) _Hdt, \end{aligned}
where, for every fixed $$t,$$
\begin{aligned} \left( y_1(t),\,y_2(t)\right) _H \end{aligned}
is the scalar product in $$H.$$ The space $$H$$ being a Hilbert space, it can be shown that $$L_2(I,\,H)$$ is a Hilbert space as well.
We define a primitive function to the function $$y(t)$$ as
\begin{aligned} Y(t)=\int \limits _{0}^{t}y(\tau )d\tau . \end{aligned}
It can be shown that for $$y\in L_2(I,\,H)$$ the function $$Y(t)$$ is a continuous abstract function in the interval I, i.e.
\begin{aligned} Y \in C(I,\,H), \end{aligned}
even absolutely continuous we write
\begin{aligned} Y \in AC(I,\,H), \end{aligned}
and has a derivative almost everywhere equal to $$y(t).$$ We write
\begin{aligned} Y^{\prime }(t)=y(t)\quad \text{ in }\quad L_2(I,\,H). \end{aligned}

## Main Result

### Theorem 3.1

Assume that conditions $$\varvec{H}_1$$$$\varvec{H}_3$$ holds. Then there exists a unique weak solution of the problem (1.1)–(1.5) on the interval $$[-\tau ,\,T],$$ in the sense of Definition 2.3.

We will prove this result in several steps. First we consider the given problem on interval $$[-\tau ,\,\tau ].$$ In this interval problem (1.1)–(1.5) reduces to
\begin{aligned} \frac{\partial u}{\partial t}+ Au&= f(x,\,t,\,u(x,\,t),\,\phi (x,\,t-\tau ),\,r(x,\,t)) \quad \text{ in } \quad Q=G\times (0,\,\tau ],\end{aligned}
(3.1)
\begin{aligned} r(x,\,t)&= \int \limits _{0}^{t}f(x,\,s,\,u(x,\,s),\,\phi (x,\,s-\tau ),\,r(x,\,s))ds,\end{aligned}
(3.2)
\begin{aligned} u(x,\,t)&= \phi (x,\,t), \quad (x,\,t) \in G\times [-\tau ,\,0],\end{aligned}
(3.3)
\begin{aligned} B_iu&= 0 \quad \text{ on } \quad \Gamma \times (0,\,\tau ], \quad i=1,\,2,\ldots ,\mu , \end{aligned}
(3.4)
\begin{aligned} C_iu&= 0 \quad \text{ on } \quad \Gamma \times (0,\,\tau ],\quad i=1,\,2,\ldots ,k-\mu . \end{aligned}
(3.5)
By using the method of semidiscretization, we establish the existence and uniqueness of a weak solution of the problem (3.1)–(3.5) in the sense of Definition 2.3. In the next step, we use this weak as history function and by using the result of previous step, we obtain a unique weak solution of the given problem on the interval $$[-\tau ,\,2\tau ]$$ in the sense of Definition 2.3. Continuing this process we obtain a unique weak solution of the given problem on the interval $$[-\tau ,\,T]$$ in the sense of Definition 2.3.

## Discretization and A Priori Estimates

To apply the method of semidiscretization, we divide the interval $$I_1=[0,\,\tau ]$$ into the subintervals of length $$h_n=\frac{\tau }{n}$$ and replace Eqs. (3.1)–(3.5) by the following approximate equations.
\begin{aligned} Az_j^n+\frac{1}{h_n}z_j^n&= \frac{1}{h_n}z_{j-1}^n+f_{j-1}^n,\end{aligned}
(4.1)
\begin{aligned} B_iz_j^n&= 0 \quad \text{ on } \quad \Gamma , \quad i=1,\,2,\ldots ,\mu ,\end{aligned}
(4.2)
\begin{aligned} C_iz_j^n&= 0 \quad \text{ on } \quad \Gamma , \quad i=1,\,2,\ldots ,k-\mu , \end{aligned}
(4.3)
with
\begin{aligned} z^n_0&= \phi (x,\,0),\\ f^n_j&= f\left( x,\,jh_n,\,z_j^n,\,\phi \left( x,\,jh_n-\tau \right) ,\,s_j^n\right) \!,\\ s_j^n&= h_n\left( f_0^n+\cdots +f^n_{j-1}\right) ,\quad s_0^n=0. \end{aligned}
Now we find the weak solutions of the problem (4.1)–(4.3) starting with $$z_0^n=\Phi (0).$$

Existence and uniqueness of weak solutions $$z_j^n \in V$$ of the problem (4.1)–(4.3) is a consequence of the Lemmas 2.1 and 2.2.

For $$t \in [-\tau ,\,\tau ],$$ we define the abstract functions $$u_n(t),$$ given by
\begin{aligned} u_n(t)&= \left\{ \begin{array}{lll} \Phi (t), \quad \text{ if } \quad t\in [-\tau ,\,0] \\ z^n_{j-1}+\frac{z^n_j-z^n_{j-1}}{h_n}(t-t^n_{j-1}), \quad \text{ if } \quad t\in \tilde{I_j^n}=(t^n_{j-1},\,t_j^n]. \end{array}\right. \end{aligned}
(4.4)
Also for $$t \in I_1,$$ we define the abstract functions $$\tilde{r}_n(t),$$ as
\begin{aligned} \tilde{r}_n(t)&= \left\{ \begin{array}{lll} s_1^n \quad \text{ for } \quad t=0,\\ s_j^n \quad \text{ for } \quad t\in \tilde{I_j^n}=(t^n_{j-1},\,t_j^n]. \end{array}\right. \end{aligned}
(4.5)
Let
\begin{aligned} Z^n_j=\frac{z_j^n-z_{j-1}^n}{h_n}, \quad j=1,\,2,\ldots ,n. \end{aligned}

### Lemma 4.1

The sequence $$\{\mathrm{Z_j^n}\}$$ is uniformly bounded in $$L_2\mathrm{(G)}.$$

### Proof

As $$z_j^n$$ is the weak solution of the problem (4.1)–(4.3), therefore we have
\begin{aligned} \left( \left( v,\,z_j^n\right) \right) +\frac{1}{h_n}\left( v,\,z_j^n\right)&= \frac{1}{h_n}\left( v,\,z^n_{j-1}\right) +\left( v,\,f^n_{j-1}\right) \quad \forall v \in V. \end{aligned}
(4.6)
Putting $$j=1,\,v=Z_1^n$$ in Eq. (4.6), we get
\begin{aligned} \left( \left( Z_1^n,\,z_1^n\right) \right) +\frac{1}{h_n}\left( Z_1^n,\,z_1^n\right)&= \frac{1}{h_n} \left( Z_1^n,\,z_0^n\right) +\left( Z_1^n,\,f_0^n\right) \!,\\ \left( \left( Z_1^n,\,z_1^n\right) \right) +\left( Z_1^n,\,Z_1^n\right)&= \left( Z_1^n,\,f_0^n\right) \!,\\ \left\| Z_1^n\right\|&\le \left\| f_0^n\right\| \!. \end{aligned}
Subtracting (4.6) written for $$j=1,$$ from the same identity for $$j=2$$ and then putting $$v=Z_2^n,$$ we get
\begin{aligned} h_n\left( \left( Z_2^n,\,Z_2^n\right) \right) +\left( Z_2^n,\,Z_2^n\right)&= \left( Z_2^n,\,Z_1^n\right) +\left( Z_2^n,\,f^n_1-f^n_0\right) \!,\\ \left\| Z_2^n\right\|&\le \left\| Z^n_1\right\| +\left\| f_1^n-f_0^n\right\| \!. \end{aligned}
Similarly
\begin{aligned} \left\| Z_3^n\right\|&\le \left\| Z^n_2\right\| +\left\| f_2^n-f_1^n\right\| \!. \end{aligned}
Continuing this process, we get
\begin{aligned} \left\| Z_j^n\right\| \le \left\| f^n_0\right\| +\sum \limits _{i=1}^{j-1} \left\| f^n_i-f^n_{i-1}\right\| \!. \end{aligned}
(4.7)
Now using conditions $$\mathbf H _2$$ and $$\mathbf H _3,$$ we get
\begin{aligned} \left\| f_i^n-f_{i-1}^n\right\|&\le C\left[ h_n+\left\| z_i^n-z_{i-1}^n\right\| +kh_n+\left\| s_i^n-s_{i-1}^n\right\| \right] \\&\le Ch_n\left[ (1+k)+\left\| Z_i^n\right\| +\left\| f^n_{i-1}\right\| \right] \!. \end{aligned}
Hence from (4.7), we have
\begin{aligned} \left\| Z^n_j\right\|&\le \left\| f^n_0\right\| +\sum \limits _{i=1}^{j-1}Ch_n\left[ (1+k)+\left\| Z_i^n\right\| +\left\| f^n_{i-1}\right\| \right] \\&\le \left\| f^n_0\right\| +C\left[ (1+k)+c_2\right] \tau +Ch_n \sum \limits _{i=1}^{j-1} \left\| Z^n_i\right\| \!,\\ \left\| Z_j^n\right\|&\le c_3, \end{aligned}
where $$c_3=(\Vert f_0^n\Vert +(1+k)\tau +c_2\tau )e^{C\tau }.$$$$\square$$

### Lemma 4.2

The sequences $$\{u_n(t)\}$$ and $$\{\tilde{r}_n(t)\}$$ defined by (4.4) and (4.5) are uniformly bounded in the spaces $$L_2(I_1,\,V)$$ and $$L_2(I_1,\,L_2(G)),$$ respectively.

### Proof

Putting $$v=z_1^n$$ and $$j=1$$ in Eq. (4.6), we get
\begin{aligned} \left( \left( z_1^n,\,z_1^n\right) \right) +\frac{1}{h_n}\left( z_1^n,\,z_1^n\right)&= \frac{1}{h_n} \left( z_1^n,\,\Phi (0)\right) +\left( z_1^n,\,f_0^n\right) \!,\\ \frac{1}{h_n} \left\| z_1^n\right\| ^2&\le \frac{1}{h_n}\left\| z_1^n\right\| \left\| \Phi (0)\right\| +\left\| z_1^n\right\| \left\| f_0^n\right\| \!, \\ \left\| z_1^n\right\|&\le h_n\left\| f_0^n\right\| \!. \end{aligned}
As $$V$$-ellipticity of $$((z_1^n,\,z_1^n))$$ implies that $$((z_1^n,\,z_1^n))>0.$$
Again putting $$v=z_2^n$$ and $$j=2$$ in Eq. (4.6), we get
\begin{aligned} \left\| z_2^n\right\| \le h_n\left( \left\| f_0^n\right\| +\left\| f_1^n\right\| \right) \!. \end{aligned}
Continuing this process, we get
\begin{aligned} \left\| z_j^n\right\| \le h_n\left( \left\| f_0^n\right\| + \cdots +\left\| f_{j-1}^n\right\| \right) . \end{aligned}
(4.8)
Now using assumptions $$\mathbf H _2$$ and $$\mathbf H _3,$$ we get
\begin{aligned} \left\| f_1^n-f_0^n\right\|&= \left\| f\left( x,\,h_n,\,z_1,\,\Phi \left( h_n-\tau \right) ,\,s_1\right) -f(x,\,0,\,\Phi (0),\,\Phi (-\tau ),\,s_0)\right\| \\&\le C\left[ h_n+\left\| z_1-\Phi (0)\right\| +\left\| \Phi \left( h_n-\tau \right) -\Phi (-\tau )\right\| +\left\| s_1^n\right\| \right] \\&\le C\left[ (1+k)h_n+\left\| z_1^n\right\| +\left\| s_1^n\right\| \right] \!, \\ \left\| f^n_1\right\|&\le \left\| f_0^n\right\| +C\left[ (1+k)h_n+\left\| z_1^n\right\| +\left\| s_1^n\right\| \right] . \end{aligned}
Similarly we can show that
\begin{aligned} \left\| f^n_2\right\|&\le \left\| f_0^n\right\| +C\left[ 2(1+k)h_n+\left\| z_2^n\right\| +\left\| s_2^n\right\| \right] \!. \end{aligned}
Continuing this process, we get
\begin{aligned} \left\| f^n_j\right\| \le \left\| f_0^n\right\| +C\left[ j(1+k)h_n+\left\| z_j^n\right\| +\left\| s_j^n\right\| \right] \!. \end{aligned}
(4.9)
Using (4.9) in (4.8), we get
\begin{aligned} \left\| z_j^n\right\|&\le h_n\left[ j\left\| f_0^n\right\| +C(1+k)h_n(1+2+\cdots j-1)\right. \\&\left. +C\left( \left\| z_1^n\right\| + \cdots +\left\| z_{j-1}^n\right\| +\left\| s_1^n\right\| +\cdots +\left\| s_{j-1}^n\right\| \right) \right] \!. \end{aligned}
As $$jh_n \le \tau ,\,C(1+k)h_n^2(1+2+\cdots +j-1)\le \frac{C(1+k)\tau ^2}{2}.$$ So that we have
\begin{aligned} \left\| z_j^n\right\| \le a_1+Ch_n\sum \limits _{i=1}^{j-1}\left( \left\| z_i^n\right\| +\left\| s_i^n\right\| \right) \!, \end{aligned}
(4.10)
where
\begin{aligned} a_1=\tau \left\| f_0^n\right\| +\frac{C(1+k)\tau ^2}{2}. \end{aligned}
Similarly, we can show that
\begin{aligned} \left\| s_j^n\right\| \le a_2+Ch_n\sum \limits _{i=1}^{j-1}\left( \left\| z_i^n\right\| +\left\| s_i^n\right\| \right) \!, \end{aligned}
(4.11)
where
\begin{aligned} a_2=\tau \left\| f_0^n\right\| +\frac{C(1+k)\tau ^2}{2}. \end{aligned}
Let
\begin{aligned} \left\| z_j^n\right\| +\left\| s_j^n\right\| =b_j \quad \text{ and }\quad a_1+a_2=a. \end{aligned}
Adding (4.10) and (4.11), we get
\begin{aligned} \left\| z_j^n\right\| +\left\| s_j^n\right\|&\le \left( a_1+a_2\right) +2Ch_n\sum \limits _{i=1}^{j-1}\left( \left\| z_i^n\right\| +\left\| s_i\right\| \right) \!,\\ b_j&\le a+2Ch_n\sum \limits _{i=1}^{j-1}b_j, \\ b_j&\le a e^{2C(j-1)h_n}\le a e^{2C\tau }=c_1. \end{aligned}
This shows that
\begin{aligned} \left\| z_j^n\right\| \le c_1 \quad \text{ and } \quad \left\| s_j^n\right\| \le c_1. \end{aligned}
Now, we see that
\begin{aligned} \left\| f_j^n\right\|&\le \left\| f_0^n\right\| +C\left[ (1+k)\tau +2c_1\right] \!, \\ \left\| f_j^n\right\|&\le c_2, \end{aligned}
where
\begin{aligned} c_2&= \Vert f_0^n\Vert +C[(1+k)\tau +2c_1]. \end{aligned}
Now, putting $$v=z_j^n$$ in Eq. (4.6)
\begin{aligned} \left( \left( z_j^n,\,z_j^n\right) \right) +\frac{1}{h_n}\left( z_j^n,\,z_j^n\right)&= \frac{1}{h_n} \left( z_j^n,\,z_{j-1}^n\right) +\left( z_j^n,\,f_{j-1}^n\right) \!,\\ \left( \left( z_j^n,\,z_j^n\right) \right) +\left( z_j^n,\,Z_j^n\right)&\le \left( z_j^n,\,f_{j-1}\right) \!, \\ \left( \left( z_j^n,\,z_j^n\right) \right)&\le \left\| z^n_j\right\| \left\| Z^n_j\right\| +\left\| z^n_j\right\| \left\| f^n_{j-1}\right\| \!,\\ \alpha \left\| z_j^n\right\| _V^2 \le \left( \left( z_j^n,\,z_j^n\right) \right)&\le \left\| z^n_j\right\| _V\left\| Z^n_j\right\| +\left\| z^n_j\right\| _V\left\| f^n_{j-1}\right\| \!, \\ \left\| z_j^n\right\| _V&\le \frac{1}{\alpha }\left( c_3+c_2\right) =c_4. \end{aligned}
Now, for $$t\in [-\tau ,\,0],$$ by condition $$\mathbf H _2,$$ we have
\begin{aligned} \left\| u_n(t)\right\| _V=\Vert \Phi (t)\Vert _V \le M_1. \end{aligned}
Also, for $$t \in (0,\,\tau ],$$ we have
\begin{aligned} \left\| u_n(t)\right\| _V&= \left\| z_{j-1}^n\left( 1-\frac{t-t^n_{j-1}}{h_n}\right) +z_j^n \frac{t-t^n_{j-1}}{h_n}\right\| _V \\&\le \left\| z_{j-1}^n\left( 1-\frac{t-t^n_{j-1}}{h_n}\right) \right\| _V+\left\| z_j^n \frac{t-t^n_{j-1}}{h_n}\right\| _V \\&\le \left( 1-\frac{t-t^n_{j-1}}{h_n}\right) c_4+ \frac{t-t^n_{j-1}}{h_n}c_4=c_4. \end{aligned}
Now
\begin{aligned} \left\| u_n(t)\right\| ^2_{L_2(I_1,V)}=\int \limits _{-\tau }^{\tau }\left\| u_n(t)\right\| _V^2dt\le c_3^2\tau . \end{aligned}
Hence the sequence $$\{u_n(t)\}$$ is uniformly bounded in $$L_2(I_1,\,V).$$

As $$\Vert s_j^n\Vert \le c_1,$$ we have $$\Vert \tilde{r}_n(t)\Vert \le c_1.$$

Now,
\begin{aligned} \left\| \tilde{r}_n(t)\right\| ^2_{L_2(I_1,L_2(G))}=\int \limits _{0}^{\tau }\left\| \tilde{r}_n(t)\right\| ^2dt\le c^2_1\tau . \end{aligned}
Hence the sequence $$\{\tilde{r}_n(t)\}$$ is bounded in $$L_2(I_1,\,L_2(G)).$$$$\square$$

### Remark 4.3

By Lemma 4.2 the sequences $$\{u_n(t)\}$$ and $$\{\tilde{r}_n(t)\}$$ are bounded in the spaces $$L_2(I_1,\,V)$$ and $$L_2(I_1,\,L_2(G)),$$ respectively, so that there exist subsequences $$\{u_{n_k}(t)\}$$ and $$\{\tilde{r}_{n_k}(t)\}$$ such that
\begin{aligned} u_{n_k}\rightharpoonup u \quad \text{ in } \quad L_2\left( I_1,\,V\right) \!, \end{aligned}
and
\begin{aligned} {\tilde{r}_{n_k}}\rightharpoonup r \quad \text{ in } \quad L_2\left( I_1,\,L_2(G)\right) \!. \end{aligned}
We define the sequence
\begin{aligned} {\tilde{u}}_n(t)&= \left\{ \begin{array}{lll} \Phi (t) \quad \text{ if } \quad t\in [-\tau ,\,0],\\ z_j^n \quad \text{ for } \quad t\in \tilde{I_j^n}=(t^n_{j-1},\,t_j^n]. \end{array}\right. \end{aligned}
(4.12)

### Lemma 4.4

We have that $$\tilde{u}_{n_k} \rightharpoonup u \quad \text{ in } \quad L_2(I_1,\,V).$$

### Proof

For a proof we refer to p. 209 in [8].

Let us define $$\{U_n(t)\}$$ as
\begin{aligned} {U_n}(t)&= \left\{ \begin{array}{lll} \Phi ^{\prime }(t) \quad \text{ if } \quad t\in [-\tau ,\,0],\\ Z_j^n \quad \text{ if } \quad t\in \tilde{I_j^n}=(t^n_{j-1},\,t_j^n]. \end{array}\right. \end{aligned}
(4.13)

### Remark 4.5

By Lemma 4.1, we can show that the sequence $$\{U_n(t)\}$$ is bounded in $$L_2(I_1,\,L_2(G)).$$ As sequence $$\{U_n(t)\}$$ is bounded in $$L_2(I_1,\,L_2(G)),$$ we have a subsequence $$\{U_{n_k}(t)\}$$ in $$L_2(I_1,\,L_2(G))$$ converging weakly to a function $$U(t),$$ i.e.,
\begin{aligned} U_{n_k} \rightharpoonup U\quad \text{ in } \quad L_2\left( I_1,\,L_2(G)\right) . \end{aligned}
Let
\begin{aligned} w(t)=\left\{ \begin{array}{lll} \Phi (t) \quad \text{ if } \quad t\in [-\tau ,\,0],\\ \int _{0}^{t}U(s)ds \quad \text{ if } \quad t\in (0,\,\tau ]. \end{array}\right. \end{aligned}
From (4.4) and (4.13), it follows that
\begin{aligned} u_{n_k}(t)=\left\{ \begin{array}{lll} \Phi (t) \quad \text{ if } \quad t\in [-\tau ,\,0],\\ \int _{0}^{t}U_{n_k}(s)ds \quad \text{ if } \quad t\in (0,\,\tau ]. \end{array}\right. \end{aligned}
Now by Remarks 4.3 and 4.5 and the uniqueness of the weak limit, we have
\begin{aligned} w=u \quad \text{ in } \quad L_2\left( I_1,\,L_2(G)\right) \!. \end{aligned}
Again by the Remark 2.4, we have
\begin{aligned} u\in AC\left( I_1,\,L_2(G)\right) \!, \end{aligned}
and
\begin{aligned} u^{\prime }(t)=U(t)\quad \text{ in } \quad L_2(G) \quad \text{ for } \text{ almost } \text{ all } \quad t\in I_1. \end{aligned}
Now from
\begin{aligned} u(t)=\left\{ \begin{array}{lll} \Phi (t), \quad \text{ if } \quad t\in [-\tau ,\,0],\\ \int _{0}^{t}U(s)ds, \quad \text{ if } \quad t\in (0,\,\tau ], \end{array}\right. \end{aligned}
we have
\begin{aligned} u(0)=\Phi (0) \quad \text{ in }\quad C\left( I_1,\,L_2(G)\right) \!. \end{aligned}
$$\square$$

### Lemma 4.6

We have that $$\tilde{r}_{n_k} \rightarrow r(t)$$ in $$L_2(G)$$ uniformly in $$I_1.$$

### Proof

For a proof we refer to p. 366 in [8].

Let
\begin{aligned} g(t)=g(x,\,t)=f(x,\,t,\,u(x,\,t),\,\phi (x,\,t-\tau ),\,r(x,\,t)), \end{aligned}
where $$u,\,r$$ are as in the Remark 4.3.
Since $$u \in V,\,r\in L_2(G),$$ we have
\begin{aligned} g \in L_2(G). \end{aligned}
$$\square$$

### Lemma 4.7

We have that
\begin{aligned} r(t)=\int \limits _{0}^{t}g(s)ds. \end{aligned}

### Proof

For a proof we refer p. 372 in [8].

Now by Remark 2.4, we can say $$r \in C(I_1,\,L_2(G)).$$

As in Eq. (4.6), the presence of the term $$f^n_{j-1},$$ instead of $$f^n_j,$$ suggests to define
\begin{aligned} \hat{u}_n(t)&= \left\{ \begin{array}{lll} \Phi (t) \quad \text{ for } \quad t\in [-\tau ,\,0],\\ z_{j-1}^n \quad \text{ for } \quad t\in \tilde{I_j^n}=(t^n_{j-1},\,t_j^n], \end{array}\right. \\ \hat{r}_n(t)&= \left\{ \begin{array}{lll} 0 \quad \text{ for } \quad t=0,\\ s_{j-1}^n \quad \text{ for } \quad t\in \tilde{I_j^n}=(t^n_{j-1},\,t_j^n], \end{array}\right. \end{aligned}
and
\begin{aligned} \bar{f}_n(t)=\left\{ \begin{array}{lll} f^n_0 \quad \text{ for } \quad t=0,\\ f_{j-1}^n \quad \text{ for } \quad t\in \tilde{I_j^n}=(t^n_{j-1},\,t_j^n]. \end{array}\right. \end{aligned}
It is clear that
\begin{aligned} \hat{u}_{n_k}\rightarrow u, \quad \hat{r}_{n_k}\rightarrow r \quad \text{ in }\quad L_2(G)\quad \text{ uniformly } \text{ in }\quad I_1. \end{aligned}
Now for $$t \in (0,\,\tau ),$$ we have
\begin{aligned} \bar{f}_n=f\left( x,\,(j-1)h_n,\,\hat{u}_n(t),\,\hat{r}_n(t)\right) . \end{aligned}
Using conditions $$\mathbf H _2$$ and $$\mathbf H _3,$$ it immediately follows, that
\begin{aligned} \bar{f}_{n_k} \rightarrow g \quad \text{ in } \quad L_2(G) \quad \text{ uniformly } \text{ in } \quad I_1. \end{aligned}
We write (4.6) for $$n_k,$$ as
\begin{aligned} \left( \left( v,\,z_j^{n_k}\right) \right) +\left( v,\,Z_j^{n_k}\right) =\left( v,\,f_{j-1}^{n_k}\right) \quad \forall v \in V. \end{aligned}
Let $$v(t)$$ be an arbitrary function in $$L(I_1,\,V).$$
Using the definition $$\tilde{u}_n(t),\,U_n(t)$$ and $$\bar{f}_n(t),$$ we have for almost all $$t \in I_1,$$
\begin{aligned} \left( \left( v(t),\,\tilde{u}_{n_k}(t)\right) \right) +\left( v(t),\,U_{n_k}(t)\right) =\left( v(t),\,\bar{f}_{n_k}(t)\right) \!. \end{aligned}
Integrating between the limits $$0$$ to $$\tau ,$$ we obtain
\begin{aligned} \int \limits _{0}^{\tau }\left( \left( v(t),\,\hat{u}_{n_k}(t)\right) \right) dt+\int \limits _{0}^{\tau } \left( v(t),\,U_{n_k}(t)\right) dt=\int \limits _{0}^{\tau }\left( v(t),\,\bar{f}_{n_k}(t)\right) dt. \end{aligned}
Taking the limit as $$n_k \rightarrow \infty ,$$ we have
\begin{aligned} \int \limits _{0}^{\tau }((v(t),\,u(t)))dt+\int \limits _{0}^{\tau } (v(t),\,u^{\prime }(t))dt=\int \limits _{0}^{\tau }(v(t),\,g(t))dt. \end{aligned}
(4.14)
Thus there exists a pair $$u(t)=u(x,\,t),\,r(t)=r(x,\,t)$$ on the interval $$[-\tau ,\,\tau ],$$ satisfying all the properties of Definition 2.3. Thus we have proved the existence of a weak solution of the problem (3.1)–(3.5) in the sense of Definition 2.3.

To show the uniqueness of this weak solution, we use an idea of [8]. For this we assume that $$u_1(t)=u_1(x,\,t),\,r_1(t)=r_1(x,\,t)$$ and $$u_2(t)=u_2(x,\,t),\,r_2(t)=r_2(x,\,t)$$ be two weak solution of the given problem on the interval $$[-\tau ,\,\tau ].$$ Let $$u(t)=u_1(t)-u_2(t),\,r(t)=r_1(t)-r_2(t).$$

From (4.14), we have
\begin{aligned}&\int \limits _{0}^{\tau }((v(t),\,u(t)))dt+\int \limits _{0}^{\tau }(v(t),\,u^{\prime }(t))dt\nonumber \\&\quad =\int \limits _{0}^{\tau }\left( v(t),\,f\left( t,\,u_1(t),\,\Phi (t-\tau ),\,r_1(t)\right) -f\left( t,\,u_2(t),\,\Phi (t-\tau ),\,r_2(t)\right) \right) dt \nonumber \\&\qquad \forall v\in L_2(I,\,V), \end{aligned}
(4.15)
Also, we have
\begin{aligned} r(t)\!=\!\int \limits _{0}^{t}\big [f\left( s,\,u_1(s),\,\Phi (s\!-\!\tau ),\,r_1(s)\right) \!-\!f\left( s,\,u_2(s),\,\Phi (s\!-\!\tau ),\,r_2(s)\right) \big ]ds. \end{aligned}
(4.16)
Now we divide the interval $$[0,\,\tau ]$$ into subintervals of equal lengths $$l$$ such that $$4lC<1.$$ As functions $$u(t)$$ and $$r(t)$$ are continuous on $$I_1,$$ the norms $$\Vert u(t)\Vert$$ and $$\Vert r(t)\Vert$$ will attain their maximum value on the interval $$[0,\,l],$$
\begin{aligned} \max \limits _{t\in [0,\,l]}\Vert u(t)\Vert =u_0, \quad \max \limits _{t\in [0,\,l]}\Vert r(t)\Vert =r_0. \end{aligned}
We assume that $$\Vert u(t)\Vert$$ attains its maximum value at $$t=l_1.$$
Now we construct the abstract function
\begin{aligned} v(t)=\left\{ \begin{array}{lll} u(t) \quad \text{ for } \quad t\in [0,\,l_1],\\ 0 \quad \text{ for } \quad t\in (l_1,\,\tau ]. \end{array}\right. \end{aligned}
As $$v\in L_2(I_1,\,V),$$ substituting this $$v$$ in (4.15) and after some simplification, we will obtain $$u_0=0,\,r_0=0.$$ This implies that $$u(t)=0,\,r(t)=0 \;\text{ in } \;[0,\,l].$$
Repeating above process finite times, we obtain
\begin{aligned} u(t)=0, \quad r(t)=0 \quad \text{ in } \quad I_1. \end{aligned}
In this way uniqueness of weak solution is proved .

Thus there exists a unique weak solution of the (1.1)–(1.5) on the interval $$[-\tau ,\,\tau ].$$

In the next step, we prove the existence and uniqueness of a weak solution on the interval $$[-\tau ,\,2\tau ].$$ In this step we assume the weak solution obtained in previous step as history function on the interval $$[-\tau ,\,\tau ].$$
\begin{aligned} \text{ Let } \quad u(x,\,t)&= u_1(x,\,t) \quad \text{ if } \quad t\in [0,\,\tau ] \quad \text{ and }\\ w(x,\,t)&= u(x,\,t+\tau ) \quad 0\le t \le \tau . \end{aligned}
Replacing $$t$$ by $$t+\tau$$ in Eqs. (1.1)–(1.5) on the interval $$[0,\,\tau ]$$ and putting $$u(t+\tau ,\,x)=w(t,\,x)$$ and $$r(t+\tau ,\,x)=r_1(t,\,x),$$ we get the following problem
\begin{aligned} \frac{\partial }{\partial t}w(x,\,t)+ A w(x,\,t)&= f\left( x,\,t+\tau ,\,w(x,\,t),\,w(x,\,t-\tau ),\,r_1(x,\,t)\right) \!, \\&\text{ in } \quad Q=G\times (0,\,\tau ), \\ r_1(x,\,t)&= \int \limits _{0}^{t}f\left( x,\,s+\tau ,\,w(x,\,s),\,w(x,\,s-\tau ),\,r_1(x,\,s)\right) ds,\\ w(x,\,t)&= u_1 (x,\,t), \quad (x,\,t) \in G\times [-\tau ,\,0],\\ B_iw&= 0, \quad \text{ on } \quad \Gamma \times (0,\,\tau ) \quad i=1,\,2,\ldots ,\mu , \\ C_iw&= 0, \quad \text{ on } \quad \Gamma \times (0,\,\tau )\quad i=1,\,2,\ldots ,k-\mu . \end{aligned}
Above problem is same as the problem (3.1)–(3.5). So by the first step, we get a unique weak solution $$w=w(x,\,t),\; r_1=r_1(x,\,t)$$ on the interval $$[0,\,\tau ]$$ of the above problem.
We have that
\begin{aligned} w(x,\,t)&= u(x,\,t+\tau ) \\&= u^{\prime }(x,\,t) \quad (\text{ say }) \quad \text{ for }\quad t\in [\tau ,\,2\tau ], \\ \text{ and } \quad r_1(x,\,t)&= r(x,\,t+\tau ) \nonumber \\&= r^{\prime }(x,\,t) \quad (\text{ say }) \quad \text{ for }\quad t\in [\tau ,\,2\tau ]. \end{aligned}
We define
\begin{aligned} u_2(x,\,t)&= \left\{ \begin{array}{lll} u(x,\,t) \quad \text{ if } \quad t \in (0,\,\tau ], \\ u^{\prime }(x,\,t) \quad \text{ if } \quad t \in (\tau ,\,2\tau ], \end{array}\right. \\ r_2(x,\,t)&= \left\{ \begin{array}{lll} r(x,\,t) \quad \text{ if } \quad t \in (0,\,\tau ],\\ r^{\prime }(x,\,t) \quad \text{ if } \quad t \in (\tau ,\,2\tau ]. \end{array}\right. \nonumber \end{aligned}
It can be shown that, pair $$u(x,\,t)=u_2(x,\,t),\,r(x,\,t)=r_2(x,\,t),$$ is a unique weak solution of the problem (1.1)–(1.5) on the interval $$[-\tau ,\,2\tau ].$$ Continuing this process, we can obtained a unique weak solution of the problem (1.1)–(1.5) on the whole interval $$[-\tau ,\,T]$$.$$\square$$

## Acknowledgments

The authors thank the referee for his valuable suggestions. The first author acknowledges the sponsorship from CSIR, India, under its Research Grant 09/092(0652)/2008-EMR-1. The second author acknowledges the financial help from the Department of Science and Technology, New Delhi, under its research project SR/S4/MS:796/12.