\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amsfonts} \usepackage{amsmath} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 115, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2006/115\hfil Solution to a semilinear pseudoparabolic problem]{Solution to a semilinear pseudoparabolic problem with integral conditions} \author[A. Bouziani, N. Merazga, \hfil EJDE-2006/115\hfilneg] {Abdelfatah Bouziani, Nabil Merazga} \address{Abdelfatah Bouziani \newline D\'{e}partement de Math\'{e}matiques, Centre Universitaire Larbi Ben M'hidi, Oum El Bouaghi 04000, Algeria} \email{aefbouziani@yahoo.fr} \address{Nabil Merazga \newline D\'{e}partement de Math\'{e}matiques, Centre Universitaire Larbi Ben M'hidi, Oum El Bouaghi 04000, Algeria} \email{nabilmerazga@yahoo.fr} \date{} \thanks{Submitted March 7, 2006. Published September 21, 2006.} \subjclass[2000]{35K70, 35A35, 35B30, 35B45, 35D05} \keywords{Semilinear pseudoparabolic equation; time-discretization method; \hfill\break\indent integral condition; a priori estimate; generalized solution} \begin{abstract} In this article, we use the Rothe time-discretization method to prove the well-posedness of a mixed problem with integral conditions for a third order semilinear pseudoparabolic equation. Also we establish the convergence of the method and an error estimate for a semi-discrete approximation. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \numberwithin{equation}{section} \allowdisplaybreaks \section{Statement of the problem} This paper concerns the problem of finding a function $v=v(x,t)$ satisfying, in a weak sense, the semilinear pseudoparabolic equation \begin{equation} \frac{\partial v}{\partial t}-\frac{\partial ^{2}v}{\partial x^{2}}-\eta \frac{\partial ^{3}v}{\partial x^{2}\partial t}=F(x,t,v)\,,\quad (x,t)\in (0,1)\times [0,T], \label{e1.1} \end{equation} subject to the initial condition \begin{equation} v(x,0)=V_{0}(x),\quad 0\leq x\leq 1, \label{e1.2} \end{equation} and to the integral conditions \begin{gather} \int_{0}^{1}v(x,t)dx=E(t),\quad 0 \leq t\leq T, \label{e1.3} \\ \int_{0}^{1}xv(x,t)dx=G(t),\quad 0 \leq t\leq T, \label{e1.4} \end{gather} where $F$, $V_{0}$, $E$ and $G$ are given functions which are sufficiently regular, and $T$ and $\eta $ are positive constants. This problem has a practical relevance, for instance in the context of soil thermophysics, \eqref{e1.1} describes the dynamics of moisture transfer in a subsoil layer $00$ and the system \eqref{e3.12} which is equivalent to \eqref{e3.11} admits a unique solution $(\lambda _{1},\mu _{1})\in \mathbb{R}^{2}$, hence problem \eqref{e3.7}-\eqref{e3.8}, with $j=1$ , is uniquely solvable. In the case where $\eta <\overline{s}$, then $D(h)$ vanishes only for $h= \overline{s}-\eta $, consequently the system \eqref{e3.12} which is equivalent to \eqref{e3.11} has a unique solution for every $h<\overline{s}-\eta $ and so is problem \eqref{e3.7}-\eqref{e3.8} with $j=1$. To summarize, let $n_{0}$ be the smallest positive integer satisfying $T/n_{0}\leq h_{0}$ where \begin{equation*} h_{0}:=\begin{cases} T, & \text{if }\eta \geq \overline{s}, \\ \min \{ \overline{s}-\eta ,T\} , & \text{if } \eta <\overline{s}. \end{cases} \end{equation*} Then we have shown that problem \eqref{e3.7}-\eqref{e3.8}, with $j=1$, admits a unique solution $w_{1}=w_{1}^{\prime}(\cdot ;\lambda _{1},\mu _{1})\in H^{2}(0,1)$ and consequently the solution $u_{1}\in H^{2}(0,1)$ of problem \eqref{e3.1}-\eqref{e3.3}, with $j=1$, is uniquely determined via the formula \eqref{e3.9} provided that $n>n_{0}$ holds. Replacing in \eqref{e3.7} with $j=2$, this latter is seen to admit a unique strong solution $w_{2}^{\prime}\in H^{2}(0,1)$ satisfying \eqref{e3.10} wiht $j=2$. Thanks to Lax-Milgram Lemma since $f_{2}+\frac{1}{\eta +h}u_{1}\in L^{2}(0,1) $. Similarly as above, the function $w_{2}^{\prime}(\cdot;\lambda ,\mu )$ is seen to be the unique solution of problem \eqref{e3.7}-\eqref{e3.8} with $j=2 $ for a suitable choice of $(\lambda ,\mu )$ if $n>n_{0}$ is true. Accordingly, the solution $u_{2}\in H^{2}(0,1)$ of problem \eqref{e3.1}-\eqref{e3.3} with $j=2$ is uniquely determined due to relation \eqref{e3.9}. Proceeding in this way step by step, we will be able to state the following result: \begin{theorem} \label{thm3.1} There exists $n_{0}\in \mathbb{N}$ such that, for all $n>n_{0}$ and for all $j=1,\dots ,n$, the problems \eqref{e3.7}-\eqref{e3.8} and \eqref{e3.1}-\eqref{e3.3} admit a unique solution $w_{j}\in H^{2}(0,1)$ and $u_{j}\in H^{2}(0,1)$ respectively. \end{theorem} So, for all $n>n_{0}$, we can define the Rothe approximation $u^{(n)}:I\to H^{2}(0,1)\cap V$ as a piecewise linear interpolation of the $u_{j}$ $(j=1,\dots ,n)$ with respect to time, i.e. \begin{equation} \label{e3.13} u^{(n)}(t)=u_{j-1}+\delta u_{j}(t-t_{j-1}),\quad t\in [ t_{j-1},t_{j}] ,\quad j=1,\dots ,n, \end{equation} as well as the corresponding step function $\overline{u}^{(n)}:I\to H^{2}(0,1)\cap V$: \begin{equation} \overline{u}^{(n)}(t)=\begin{cases} u_{j} & \text{for } t\in (t_{j-1},t_{j}] ,\; j=1,\dots ,n. \\ U_{0} & \text{for }t\in [\frac{-T}{n},0]% \end{cases} \label{e3.14} \end{equation} \section{A priori estimates for the approximations} In this section, we will derive some a priori estimates which are the key points to establish Theorem \ref{thm2.2}. Note that, in the rest of the paper, positive constants are denoted by $C$, $C_{i}$ ($i=1,2,\dots $). \begin{lemma} \label{lem4.1} There exist $C>0$ such that, for all $n>n_{0}$, the solutions $u_{j}$ of the time-discretized problems \eqref{e3.1}-\eqref{e3.3}, $j=1,\dots ,n$, satisfy the estimates \begin{itemize} \item[(i)] $\| u_{j}\| \leq C$ \item[(ii)] $\| \delta u_{j}\| \leq C$. \end{itemize} \end{lemma} \begin{proof} First of all, we write problem \eqref{e3.7}-\eqref{e3.8} in a variational form. Suppose that $n>n_{0}$ and let $\phi $ be any function from the space $V$ defined in \eqref{e2.1}. A standard integration by parts yields \begin{equation} \label{e4.1} \int_{0}^{x}(x-\xi )\phi (\xi )d\xi =\Im _{x}^{2}\phi ,\quad \text{for all } x\in (0,1), \end{equation} where \begin{equation*} \Im _{x}^{2}\phi :=\Im _{x}(\Im _{\xi }\phi )=\int_{0}^{x}d\xi \int_{0}^{\xi }\phi (\eta )d\eta . \end{equation*} Hence, taking $x=1$ in \eqref{e4.1}, we get \begin{equation} \label{e4.2} \Im _{1}^{2}\phi =\int_{0}^{1}(1-\xi )\phi (\xi )d\xi =\int_{0}^{1}\phi (\xi )d\xi -\int_{0}^{1}\xi \phi (\xi )d\xi =0. \end{equation} Next, taking for all $j=1,\dots ,n$, the inner product in $L^{2}(0,1)$ of equation \eqref{e3.7} and $\Im _{x}^{2}\phi $, we get \begin{equation} -\int_{0}^{1}\frac{d^{2}w_{j}}{dx^{2}}(x)\Im _{x}^{2}\phi dx +\frac{1}{\eta +h}\int_{0}^{1}w_{j}(x)\Im _{x}^{2}\phi dx =\int_{0}^{1}(f_{j}(x)+\frac{1}{\eta +h}u_{j-1}(x))\Im _{x}^{2}\phi dx. \label{e4.3} \end{equation} Carrying out some integrations by parts and invoking \eqref{e4.2}, we obtain for each term in \eqref{e4.3}: \begin{align*} \int_{0}^{1}\frac{d^{2}w_{j}}{dx^{2}}(x)\Im _{x}^{2}\phi dx &= \frac{dw_{j}}{% dx}(x)\Im _{x}^{2}\phi \big\vert _{x=0}^{x=1} -\int_{0}^{1}\frac{dw_{j}}{dx} (x)\Im _{x}\phi dx \\ &= - w_{j}(x)\Im _{x}\phi \big\vert_{x=0}^{x=1} +\int_{0}^{1}w_{j}(x)\phi (x)dx \\ &= (w_{j},\phi ), \end{align*} \begin{align*} \int_{0}^{1}w_{j}(x)\Im _{x}^{2}\phi dx &= \int_{0}^{1}\frac{d}{dx}(\Im _{x}w_{j})\Im _{x}^{2}\phi dx \\ &= \Im _{x}w_{j}\Im _{x}^{2}\phi \big\vert _{x=0}^{x=1}-\int_{0}^{1}\Im _{x}w_{j}\Im _{x}\phi dx \\ &= -(w_{j},\phi )_{B_{2}^{1}}, \end{align*} and \begin{align*} &\int_{0}^{1}(f_{j}(x)+\frac{1}{\eta +h}u_{j-1}(x))\Im _{x}^{2}\phi dx \\ &= \int_{0}^{1}\frac{d}{dx}\big[ \Im _{x}\big(f_{j}+\frac{ 1}{\eta +h}u_{j-1}% \big)\big] \Im _{x}^{2}\phi dx \\ &= \Im _{x}\big(f_{j}+\frac{1}{\eta +h}u_{j-1}\big)\Im _{x}^{2}\phi % \big\vert _{x=0}^{x=1} -\int_{0}^{1}\Im _{x}\big(f_{j}+\frac{1}{\eta +h}% u_{j-1}\big) \Im_{x}\phi dx \\ &= -\big(f_{j}+\frac{1}{\eta +h}u_{j-1},\phi\big)_{B_{2}^{1}}\,. \end{align*} So that \eqref{e4.3} becomes \begin{equation} (w_{j},\phi )+\frac{1}{\eta +h}(w_{j},\phi )_{B_{2}^{1}} =\big(f_{j}+\frac{1 }{\eta +h}u_{j-1},\phi\big)_{B_{2}^{1}},\quad \forall j=1,\dots ,n. \label{e4.4} \end{equation} Now, testing this identity with $\phi =w_{j}$ which is in $V$ thanks to \eqref{e3.8}, with the help of the Cauchy-Schwarz inequality we obtain \begin{equation*} \| w_{j}\| ^{2}+\frac{1}{\eta +h}\| w_{j}\| _{B_{2}^{1}}^{2}\leq \big[ \| f_{j}\| _{B_{2}^{1}}+\frac{1}{\eta +h}\| u_{j-1}\| _{B_{2}^{1}}\big] \| w_{j}\| _{B_{2}^{1}}, \end{equation*} from where we deduce \begin{equation} \| w_{j}\| \leq \| f_{j}\| _{B_{2}^{1}}+\frac{1}{\eta +h}\| u_{j-1}\| _{B_{2}^{1}}, \label{e4.5} \end{equation} as well as \begin{equation} \| w_{j}\| _{B_{2}^{1}}\leq (\eta +h)\| f_{j}\| _{B_{2}^{1}}+\| u_{j-1}\| _{B_{2}^{1}}. \label{e4.6} \end{equation} Hence, \eqref{e3.9} gives for all $j=1,\dots ,n$, \begin{align*} \| u_{j}\| _{B_{2}^{1}} &\leq \frac{h}{\eta +h} \| w_{j}\| _{B_{2}^{1}} + \frac{\eta }{\eta +h}\| u_{j-1}\| _{B_{2}^{1}} \\ &\leq \frac{h}{\eta +h}((\eta +h)\| f_{j}\|_{B_{2}^{1}} +\| u_{j-1}\| _{B_{2}^{1}})+\frac{\eta }{ \eta +h}\| u_{j-1}\| _{B_{2}^{1}}, \end{align*} i.e., \begin{equation*} \| u_{j}\| _{B_{2}^{1}}\leq h\| f_{j}\| _{B_{2}^{1}} +\| u_{j-1}\| _{B_{2}^{1}}, \end{equation*} then, iterating this last inequality, we may arrive at \begin{equation} \| u_{j}\| _{B_{2}^{1}}\leq h\sum_{i=1}^{i=j}\| f_{i}\| _{B_{2}^{1}}+\| U_{0}\| _{B_{2}^{1}},\quad \forall j=1,\dots ,n. \label{e4.7} \end{equation} But, for all $i\geq 1$ we have in view of Assumption (H1): \begin{equation} \| f_{i}\| _{B_{2}^{1}} \leq \| f(t_{i},u_{i-1})-f(t_{i},0)\| _{B_{2}^{1}} +\| f(t_{i},0)\| _{B_{2}^{1}} \leq l\| u_{i-1}\| _{B_{2}^{1}}+M, \label{e4.8} \end{equation} where $M:=\max_{t\in I} \| f(t,0)\| _{B_{2}^{1}}$. So that substituting in \eqref{e4.7}, \begin{align*} \| u_{j}\| _{B_{2}^{1}} &\leq h\sum_{i=1}^{i=j}(l\| u_{i-1}\| _{B_{2}^{1}}+M)+\|U_{0}\| _{B_{2}^{1}} \\ &= jhM+(1+lh)\| U_{0}\|_{B_{2}^{1}}+lh\sum_{i=2}^{i=j} \| u_{i-1}\| _{B_{2}^{1}} \\ &\leq TM+(1+lh_{0})\| U_{0}\|_{B_{2}^{1}} +lh\sum_{i=1}^{i=j-1}\| u_{i}\| _{B_{2}^{1}}, \end{align*} from where it comes due to the discrete Gronwall's Lemma \begin{equation*} \| u_{j}\| _{B_{2}^{1}}\leq ( TM+(1+lh_{0})\| U_{0}\| _{B_{2}^{1}})e^{l(j-1)h}. \end{equation*} Then \begin{equation} \| u_{j}\| _{B_{2}^{1}}\leq C_{1},\quad j=1,\dots ,n, \label{e4.9} \end{equation} with $C_{1}:=(TM+(1+lh_{0})\| U_{0}\|_{B_{2}^{1}})e^{lT}$. Then, From \eqref{e3.6}, \eqref{e4.6} and \eqref{e4.8}, we have the estimate \begin{align*} \| \delta u_{j}\| _{B_{2}^{1}} &= \frac{1}{\eta +h} \| w_{j}-u_{j-1}\| _{B_{2}^{1}} \\ &\leq \frac{1}{\eta }(\| w_{j}\| _{B_{2}^{1}}+\| u_{j-1}\| _{B_{2}^{1}}) \\ &\leq \frac{1}{\eta } \Big((\eta +h)\| f_{j}\| _{B_{2}^{1}}+2\| u_{j-1}\| _{B_{2}^{1}}\Big) \\ &\leq \frac{1}{\eta } \Big(((\eta +h)l+2)\| u_{j-1}\| _{B_{2}^{1}}+(\eta +h)M \Big), \end{align*} finally, due to \eqref{e4.9}, \begin{equation} \| \delta u_{j}\| _{B_{2}^{1}}\leq C_{2},\quad j=1,\dots ,n, \label{e4.10} \end{equation} where $C_{2}:=\frac{1}{\eta }(\left[ (\eta +h_{0})l+2\right] C_{1}+(\eta +h_{0})M)$. Now, combining \eqref{e4.5} and \eqref{e4.8}, \begin{equation*} \| w_{j}\| \leq \big(l+\frac{1}{\eta +h}\big) \| u_{j-1}\| _{B_{2}^{1}}+M. \end{equation*} Consequently by \eqref{e4.9}, we get \begin{equation} \| w_{j}\| \leq C_{3},\quad j=1,\dots ,n, \label{e4.11} \end{equation} with $C_{3}:=(l+\frac{1}{\eta })C_{1}+M$. On the other hand, iterating \eqref{e3.9} we may obtain for $j=1,\dots ,n$ \begin{align*} u_{j} &=\frac{h}{\eta +h}w_{j}+\frac{\eta }{\eta +h}u_{j-1} \\ &=\frac{h}{\eta +h}w_{j}+\frac{\eta }{\eta +h}\Big(\frac{h}{\eta +h} w_{j-1}+ \frac{\eta }{\eta +h}u_{j-2}\Big) \\ &=\frac{h}{\eta +h}\big(w_{j}+\frac{\eta }{\eta +h}w_{j-1}\big) +\big(\frac{\eta }{\eta +h}\big)^{2}u_{j-2} \\ &=\dots \\ &= \frac{h}{\eta +h}\Big[ w_{j}+\frac{\eta }{\eta +h}w_{j-1} +(\frac{ \eta }{\eta +h})^{2}w_{j-2}+\dots +(\frac{\eta }{\eta +h}) ^{j-1}w_{1}\Big] +(\frac{\eta }{\eta +h})^{j}U_{0}. \end{align*} So that by \eqref{e4.11}, we have \begin{align*} \| u_{j}\| &\leq \frac{h}{\eta +h}\Big[ \|w_{j}\| +\frac{\eta }{\eta +h}\| w_{j-1}\| +(\frac{\eta }{\eta +h})^{2}\| w_{j-2}\| +\dots +(\frac{\eta }{\eta +h})^{j-1}\| w_{1}\| \Big] \\ &\quad +(\frac{\eta }{\eta +h})^{j}\| U_{0}\| \\ &\leq \frac{C_{3}h}{\eta +h}\Big[ 1+\frac{\eta }{\eta +h} +( \frac{\eta }{\eta +h})^{2}+\dots +(\frac{\eta }{\eta +h}) ^{j-1}\Big] +\| U_{0}\| , \end{align*} since $\frac{\eta }{\eta +h}<1$. But \begin{align*} 1+\frac{\eta }{\eta +h}+(\frac{\eta }{\eta +h})^{2}+\dots +( \frac{\eta }{\eta +h})^{j-1} &=\frac{1-(\frac{\eta }{\eta +h})^{j}}{1-\frac{\eta }{\eta +h% }} \\ &\leq \frac{1}{1-\frac{\eta }{\eta +h}} =\frac{\eta +h}{h}, \end{align*} then \begin{equation} \| u_{j}\| \leq \frac{C_{3}h}{\eta +h}\frac{\eta +h}{h} +\| U_{0}\| =C_{3}+\| U_{0}\| ,\quad \text{for } j=1,\dots ,n, \label{e4.12} \end{equation} which proves estimate (i) with $C:=C_{3}+\| U_{0}\| $. Finally, using % \eqref{e3.5}, \eqref{e4.11} and \eqref{e4.12}, we derive \begin{equation*} \| \delta u_{j}\| \leq \frac{1}{\eta }(\| w_{j}\| +\| u_{j}\|) \leq \frac{1}{% \eta }(2C_{3}+\| U_{0}\|),\quad \text{for } j=1,\dots ,n. \end{equation*} Thus, estimate (ii) is proved with $C:=\frac{1}{\eta }( 2C_{3}+\| U_{0}\|)$, and so the proof is complete. \end{proof} We deduce the following estimates that we shall use later. \begin{corollary} \label{coro4.2} For all $n>n_{0}$, the functions $u^{(n)}$ and $\overline{u}^{(n)}$ satisfies the inequalities \begin{itemize} \item[(i)] $\| u^{(n)}(t)\| \leq C$, $\| \overline{u}^{(n)}(t)\| \leq C$, for all $t\in I$, \item[(ii)] $\| \frac{du^{(n)}}{dt}(t)\| \leq C$, a.e. in $I$, \item[(iii)] $\| \overline{u}^{(n)}(t)-u^{(n)}(t)\| \leq \frac{C}{n}$, for all $t\in I$ \item[(iv)] $\| \overline{u}^{(n)}(t)-\overline{u}^{(n)}(t-\frac{T}{n}) \| \leq \frac{C}{n}$, for all $t\in I$. \end{itemize} \end{corollary} \begin{proof} Inequalities (i) follow immediately from Lemma \ref{lem4.1} (i) with the same constant, whereas inequality (ii) is an easy consequence of Lemma \ref% {lem4.1} (ii), also with the same constant, noting that we have \begin{equation*} \frac{du^{(n)}}{dt}(t)=\delta u_{j},\quad \forall t\in (t_{j-1},t_{j}],\; 1\leq j\leq n. \end{equation*} As for inequalities (iii) and (iv), since we have \begin{equation*} \overline{u}^{(n)}(t)-u^{(n)}(t)=(t_{j}-t)\delta u_{j},\quad \forall t\in (t_{j-1},t_{j}],\; 1\leq j\leq n, \end{equation*} and \begin{equation*} \overline{u}^{(n)}(t)-\overline{u}^{(n)}(t-\frac{T}{n}) =u_{j}-u_{j-1},\quad \forall t\in (t_{j-1},t_{j}],\; 1\leq j\leq n, \end{equation*} it follows that \begin{equation*} \| \overline{u}^{(n)}(t)-u^{(n)}(t)\| \leq h \max_{1\leq j\leq n} \| \delta u_{j}\|, \quad \forall \,t\in I, \end{equation*} and \begin{equation*} \| \overline{u}^{(n)}(t)-\overline{u}^{(n)}(t-\frac{T}{n}) \| \leq h\max_{1\leq j\leq n} \| \delta u_{j}\| ,\quad \forall \,t\in I. \end{equation*} Hence, applying Lemma \ref{lem4.1} (ii), we get the desired inequalities (iii) and (iv) with $C:=\frac{T}{\eta }(2C_{3}+\| U_{0}\|)$. \end{proof} \section{Convergence and Existence result} Using relations \eqref{e3.5} and \eqref{e3.6}, we can rearrange the variational equations \eqref{e4.4} as follows \begin{equation*} (\delta u_{j},\phi )_{B_{2}^{1}}+(u_{j},\phi )+\eta (\delta u_{j},\phi )=(f_{j},\phi)_{B_{2}^{1}},\quad \forall \phi \in V,\; j=1,\dots ,n. \end{equation*} If we define, for all $n>n_{0}$, the abstract step function $\overline{f} ^{(n)}:I\times V\to L^{2}(0,1)$ by \begin{equation*} \overline{f}^{(n)}(t,v)=f(t_{j},v),\quad t\in (t_{j-1},t_{j}],\,\,j=1,\dots ,n, \end{equation*} the previous equations may be rewritten as \begin{equation} \big(\frac{du^{(n)}}{dt}(t),\phi\big)_{B_{2}^{1}} +\big({\overline{u} ^{(n)}(t),\phi }\big) +\eta\big(\frac{du^{(n)}}{dt}(t),\phi\big) \\ =\big(\overline{f}^{(n)}(t,\overline{u}^{(n)}(t-\frac{T}{n} )),\phi\big) _{B_{2}^{1}}, \label{e5.1} \end{equation} for all $\phi \in V$, $t\in (0,T]$. Before performing a limiting process in the approximation scheme \eqref{e5.1}, we have to prove some convergence assertions. \begin{theorem} \label{thm5.1} The sequence $\{ u^{(n)}\}_{n>n_{0}}$ converges under the the norm of $C(I,V)$ to some function $u\in C(I,V)$ and the error estimate \begin{equation} \| u^{(n)}-u\| _{C(I,V)}\leq \frac{C}{n^{1/2}}, \label{e5.2} \end{equation} takes place for all $n>n_{0}$. \end{theorem} \begin{proof} The main idea of the proof is to show that $\{ u^{(n)}\}_{n>n_{0}}$ is a Cauchy sequence in the Banach space $C(I,V)$. Let $u^{(n)}$ and $u^{(m)}$ be the Rothe functions \eqref{e3.13} corresponding to the step lengths $h_{n}=% \frac{T}{n}$ and $h_{m}=\frac{T}{m}$ respectively with $m>n>n_{0}$. Testing the difference of \eqref{e5.1} for $n$ and $m$, with $\phi =u^{(n)}(t)-u^{(m)}(t)\,(\in V)$, we get for all $t\in \,(0,T]$: \begin{align*} &\Big(\frac{d}{dt}\big(u^{(n)}(t)-u^{(m)}(t)\big) ,u^{(n)}(t)-u^{(m)}(t)\Big) _{B_{2}^{1}} \\ &+\big(\overline{u}^{(n)}(t)-\overline{u}^{(m)}(t),u^{(n)}(t)-u^{(m)}(t) \big) \\ &+\eta \Big(\frac{d}{dt}\big(u^{(n)}(t)-u^{(m)}(t)\big) ,u^{(n)}(t)-u^{(m)}(t)\Big) \\ & =\Big(\overline{f}^{(n)}(t,\overline{u}^{(n)}(t-\frac{T}{n }))-\overline{f} ^{(m)}(t,\overline{u}^{(m)}(t- \frac{T}{m})),u^{(n)}(t)-u^{(m)}(t)\Big) _{B_{2}^{1}}, \end{align*} or after some rearrangement \begin{equation} \begin{aligned} &\frac{1}{2}\frac{d}{dt}\| u^{(n)}(t)-u^{(m)}(t)\| _{B_{2}^{1}}^{2}+\frac{\eta }{2}\frac{d}{dt}\| u^{(n)}(t)-u^{(m)}(t)\| ^{2}+\| \overline{u}^{(n)}(t)- \overline{u}^{(m)}(t)\| ^{2} \\ &=\big({\overline{u}^{(n)}(t)-\overline{u}^{(m)}(t),( \overline{u}^{(n)}(t)-u^{(n)}(t))+(u^{(m)}(t)-\overline{u} ^{(m)}(t))}\big)\\ &\quad+\Big({\overline{f}^{(n)}(t,\overline{u}^{(n)}(t- \frac{T}{n}))-\overline{f}^{(m)}(t,\overline{u} ^{(m)}(t-\frac{T}{m})),u^{(n)}(t)-u^{(m)}(t)}\Big) _{B_{2}^{1}}. \end{aligned} \label{e5.3} \end{equation} But, from the fact that \begin{equation*} \overline{f}^{(n)}(t,\overline{u}^{(n)}(t-\frac{T}{n}) )=f(t_{j},u_{j-1}):=f_{j},\quad \forall t\in (t_{j-1},t_{j}],\,\,j=1,\dots ,n, \end{equation*} we deduce, in view of \eqref{e4.8}, that \begin{align*} \| \overline{f}^{(n)}\big(t,\overline{u}^{(n)}(t-\frac{T}{n} )\big)\| _{B_{2}^{1}} &\leq \max_{1\leq j\leq n} \| f_{j}\| _{B_{2}^{1}} \\ &\leq l \max_{1\leq j\leq n}\|u_{j-1}\| _{B_{2}^{1}}+M, \quad \forall t\in I. \end{align*} Therefore, due to \eqref{e4.9}, \begin{equation} \| \overline{f}^{(n)}\big(t,\overline{u}^{(n)}(t-\frac{T}{n} )\big)\| _{B_{2}^{1}}\leq lC_{1}+M,\quad \forall t\in I. \label{e5.4} \end{equation} Thus, from the identity \begin{equation*} (\overline{u}^{(n)}(t),\phi) = \Big(\overline{f}^{(n)}\big(t,\overline{u} ^{(n)} (t-\frac{T}{n})\big)-\frac{du^{(n)}}{dt}(t),\phi\Big)_{B_{2}^{1}} -\eta \big(\frac{du^{(n)}}{dt}(t),\phi\big), \end{equation*} for all $t\in I$, $\phi \in V$, which follows from \eqref{e5.1}, due to \eqref{e4.10}, \eqref{e5.4} and Corollary \ref{coro4.2}(ii), we obtain \begin{equation} \label{e5.5} \begin{aligned} \vert (\overline{u}^{(n)}(t),\phi)\vert &\leq \Big[ \| \overline{f}^{(n)}\big(t,\overline{u} ^{(n)}(t-\frac{T}{n})\big)\|_{B_{2}^{1}} +\| \frac{du^{(n)}}{dt}(t)\| _{B_{2}^{1}} +\eta \| \frac{du^{(n)}}{dt}(t)\| \Big] \| \phi\| \\ &\leq C_{4}\| \phi \| ,\quad \forall t\in I,\; \forall \phi \in V, \end{aligned} \end{equation} with $C_{4}:=lC_{1}+M+C_{2}+2C_{3}+\| U_{0}\| $. Applying \eqref{e5.5} for \begin{equation*} \phi =(\overline{u}^{(n)}(t)-u^{(n)}(t))+( u^{(m)}(t)-\overline{u}^{(m)}(t)), \end{equation*} together with Corollary \ref{coro4.2} (iii), we can dominate the first term in the right-hand side of \eqref{e5.3} as follows \begin{equation} \begin{aligned} &\big({\overline{u}^{(n)}(t)-\overline{u}^{(m)}(t),( \overline{u}^{(n)}(t)-u^{(n)}(t))+(u^{(m)}(t)-\overline{u} ^{(m)}(t))}\big)\\ &\leq 2C_{4}\big(\| \overline{u}^{(n)}(t)-u^{(n)}(t) \| +\| \overline{u}^{(m)}(t)-u^{(m)}(t)\|\big)\\ &\leq C_{5}(\frac{1}{n}+\frac{1}{m}),\quad \forall t\in \,I, \end{aligned} \label{e5.6} \end{equation} with $C_{5}:=\frac{2C_{4}T}{\eta }(2C_{3}+\| U_{0}\|)$. It remains to majorize the second term in the right hand side in \eqref{e5.3}. To this end, we use the Cauchy inequality \begin{equation*} \alpha \beta \leq \frac{\varepsilon }{2}\alpha ^{2}+\frac{1}{ 2\varepsilon } \beta ^{2},\quad \forall \alpha ,\beta \in \mathbb{R},\quad \forall \varepsilon \in \mathbb{R}_{+}^{\ast }, \end{equation*} for every $\varepsilon >0$: \begin{equation} \begin{aligned} &\Big(\overline{f}^{(n)}(t,\overline{u}^{(n)}(t-\frac{T}{n} ))-\overline{f}^{(m)}(t,\overline{u}^{(m)}(t-\frac{ T}{m})),u^{(n)}(t)-u^{(m)}(t)\Big)_{B_{2}^{1}} \\ &\leq \| \overline{f}^{(n)}(t,\overline{u}^{(n)}(t- \frac{T}{n}))-\overline{f}^{(m)}(t,\overline{u} ^{(m)}(t-\frac{T}{m}))\| _{B_{2}^{1}}\| u^{(n)}(t)-u^{(m)}(t)\| _{B_{2}^{1}} \\ &\leq \frac{\varepsilon }{2}\| \overline{f}^{(n)}(t, \overline{u}^{(n)}(t-\frac{T}{n}))-\overline{f} ^{(m)}(t,\overline{u}^{(m)}(t-\frac{T}{m})) \| _{B_{2}^{1}}^{2} \\ &\quad +\frac{1}{2\varepsilon }\| u^{(n)}(t)-u^{(m)}(t)\| _{B_{2}^{1}}^{2},\quad \forall t\in \,I. \end{aligned} \label{e5.7} \end{equation} Now, let $t$ be arbitrary but fixed in $(0,T]$. Then there exist two integers $k$ and $i$ corresponding to the subdivision of $I$ into $n$ and $m$ subintervals respectively, such that $t\in (t_{k-1},t_{k}]\cap (t_{i-1},t_{i}]$. Hence thanks to the assumed Lipschitz continuity of $f$, \begin{align*} &\| \overline{f}^{(n)}\big(t,\overline{u}^{(n)}(t-\frac{T}{n })\big) - \overline{f}^{(m)}\big(t,\overline{u}^{(m)}(t- \frac{T}{m})\big)\| _{B_{2}^{1}}^{2} \\ &=\| f\big(t_{k},\overline{u}^{(n)}(t-\frac{T}{n}) \big)-f\big(t_{i}, \overline{u}^{(m)}(t-\frac{T}{m})\big) \| _{B_{2}^{1}}^{2} \\ &\leq l^{2}\Big[ \vert t_{k}-t_{i}\vert \big\{ 1+\| \overline{u}^{(n)}(t- \frac{T}{n})\| _{B_{2}^{1}}+\| \overline{u}^{(m)}(t-\frac{T}{m}) \| _{B_{2}^{1}}\big\} \\ &\quad +\| \overline{u}^{(n)}(t-\frac{T}{n})- \overline{u}^{(m)}(t -\frac{T}{m})\| _{B_{2}^{1}}\Big]^{2} \\ &\leq l^{2}\Big[ (h_{n}+h_{m})(1+\| u_{k-1}\| _{B_{2}^{1}}+\| u_{i-1}\| _{B_{2}^{1}})+\| \overline{u}^{(n)}(t-\frac{T}{n} )-\overline{u}^{(n)}(t)\| _{B_{2}^{1}} \\ &\quad +\| \overline{u}^{(n)}(t)-\overline{u} ^{(m)}(t)\| _{B_{2}^{1}}+\| \overline{u} ^{(m)}(t)-\overline{u}^{(m)}(t-\frac{T}{m}) \| _{B_{2}^{1}}% \Big] ^{2}\,. \end{align*} Then follows with consideration to \eqref{e4.9} and Corollary \ref{coro4.2} (iv) that \begin{align*} &\| \overline{f}^{(n)}\big(t,\overline{u}^{(n)}(t-\frac{T}{n })\big)- \overline{f}^{(m)}\big(t,\overline{u}^{(m)}(t- \frac{T}{m})\big)\| _{B_{2}^{1}}^{2} \\ &\leq l^{2}\big[ T(\frac{1}{n}+\frac{1}{m})( 1+2C_{1})+\frac{T}{\eta } (2C_{3}+\| U_{0}\| )(\frac{1}{n}+\frac{1}{m})+\| \overline{u} ^{(n)}(t)- \overline{u}^{(m)}(t)\| _{B_{2}^{1}}\big] ^{2} \\ &= l^{2}\big[ T(1+2C_{1}+\frac{1}{\eta }(2C_{3}+\| U_{0}\|))(\frac{1}{n}+ \frac{1}{m}) +\| \overline{u}^{(n)}(t)-\overline{u}^{(m)}( t)\| _{B_{2}^{1}}% \big] ^{2} \\ &\leq l^{2}\Big[ C_{6}^{2}(\frac{1}{n}+\frac{1}{m}) ^{2}+2C_{6}(\frac{1}{n}+ \frac{1}{m})(\| \overline{ u}^{(n)}(t)\| _{B_{2}^{1}}+\| \overline{u} ^{(m)}(t)\| _{B_{2}^{1}}) \\ &\quad +\| \overline{u}^{(n)}(t)-\overline{ u}^{(m)}(t)\| _{B_{2}^{1}}^{2} \Big] \\ &\leq (lC_{6})^{2}(\frac{1}{n}+\frac{1}{m}) ^{2}+4l^{2}C_{6}C_{1}(\frac{1}{n} +\frac{1}{m})+l^{2}\| \overline{u}^{(n)}(t)-\overline{u}^{(m)}(t) \| _{B_{2}^{1}}^{2},\quad \forall t\in I, \end{align*} with $C_{6}:=T\big(1+2C_{1}+\frac{1}{\eta }(2C_{3}+\| U_{0}\|)\big)$. Thus, setting $C_{7}:=(lC_{6})^{2}$ and $C_{8}:=4l^{2}C_{6}C_{1}$, we write \begin{equation} \begin{aligned} &\| \overline{f}^{(n)}(t,\overline{u}^{(n)}(t-\frac{T}{n} ))-\overline{f}^{(m)}(t,\overline{u}^{(m)}(t-\frac{ T}{m}))\| _{B_{2}^{1}}^{2} \\ &\leq C_{7}(\frac{1}{n}+\frac{1}{m})^{2}+C_{8}(\frac{1 }{n}+\frac{1}{m})+l^{2}\| \overline{u}^{(n)}(t)- \overline{u}^{(m)}(t)\| _{B_{2}^{1}}^{2},\quad \forall t\in I; \end{aligned} \label{e5.8} \end{equation} therefore, going back to \eqref{e5.7}, we have \begin{equation} \begin{aligned} &\Big(\overline{f}^{(n)}(t,\overline{u}^{(n)}(t-\frac{T}{n} ))-\overline{f}^{(m)}(t,\overline{u}^{(m)}(t-\frac{ T}{m})),u^{(n)}(t)-u^{(m)}(t)\Big)_{B_{2}^{1}} \\ &\leq \frac{\varepsilon }{2}C_{7}(\frac{1}{n}+\frac{1}{m}) ^{2}+\frac{\varepsilon }{2}C_{8}(\frac{1}{n}+\frac{1}{m})+\frac{ \varepsilon }{2}l^{2}\| \overline{u}^{(n)}(t)-\overline{ u}^{(m)}(t)\| _{B_{2}^{1}}^{2} \\ &\quad +\frac{1}{2\varepsilon }\| u^{(n)}(t)-u^{(m)}(t)\| _{B_{2}^{1}}^{2},\quad \forall t\in I. \end{aligned} \label{e5.9} \end{equation} Now, combining \eqref{e5.3}, \eqref{e5.6}, \eqref{e5.9} and \eqref{e2.3}, we get \begin{align*} &\frac{d}{dt}\Big(\| u^{(n)}(t)-u^{(m)}(t)\| _{B_{2}^{1}}^{2}+\eta \| u^{(n)}(t)-u^{(m)}(t)\| ^{2}\Big)+2\| \overline{u}^{(n)}(t)-\overline{u} ^{(m)}(t)\| ^{2} \\ &\leq \varepsilon C_{7}(\frac{1}{n}+\frac{1}{m}) ^{2}+(\varepsilon C_{8}+2C_{5})(\frac{1}{n}+\frac{1}{m} )+\frac{\varepsilon l^{2}}{2}\| \overline{u}^{(n)}( t)-\overline{u}^{(m)}(t)\| ^{2} \\ &\quad +\frac{1}{2\varepsilon }\| u^{(n)}(t)-u^{(m)}(t)\| ^{2},\quad \forall t\in I. \end{align*} Hence \begin{align*} &\eta \frac{d}{dt}\| u^{(n)}(t)-u^{(m)}(t)\| ^{2}+(2- \frac{\varepsilon l^{2}% }{2})\| \overline{u}^{(n)}(t)- \overline{u}^{(m)}(t)\| ^{2} \\ &\leq \varepsilon C_{7}(\frac{1}{n}+\frac{1}{m}) ^{2}+(\varepsilon C_{8}+2C_{5})(\frac{1}{n}+\frac{1}{m} )+\frac{1}{2\varepsilon }\| u^{(n)}(t)-u^{(m)}(t)\| ^{2}. \end{align*} Choosing $\varepsilon >0$ so that $2-\frac{\varepsilon l^{2}}{2}=0$, i.e. $\varepsilon =\frac{4}{l^{2}}$ and integrating the just obtained inequality between $0$ and $t$ taking into account the fact that $% u^{(n)}(0)=u^{(m)}(0)=U_{0}$, we get for all $t\in I$: \begin{align*} &\| u^{(n)}(t)-u^{(m)}(t)\| ^{2} \\ &\leq \frac{4C_{7}T}{\eta l^{2}}(\frac{1}{n}+\frac{1}{m})^{2} +\frac{2T}{\eta }(\frac{2C_{8}}{l^{2}}+C_{5})(\frac{1}{n}+\frac{1}{m}) +\frac{l^{2}}{8\eta }\int_{0}^{t}\| u^{(n)}(\tau )-u^{(m)}(\tau )\| ^{2}d\tau . \end{align*} Then, by Gronwall's Lemma, \begin{equation*} \| u^{(n)}(t)-u^{(m)}(t)\| ^{2}\leq \big[ C_{9}( \frac{1}{n}+\frac{1}{m} )^{2}+C_{10}(\frac{1}{n}+\frac{1}{m} )\big] e^{\frac{l^{2}}{8\eta }t}\quad \forall t\in I, \end{equation*} with $C_{9}:=\frac{4C_{7}T}{\eta l^{2}}$ and $C_{10}:=\frac{2T}{\eta }(\frac{2C_{8}}{l^{2}}+C_{5})$. Accordingly \begin{equation*} \| u^{(n)}(t)-u^{(m)}(t)\| \leq \big[ C_{9}( \frac{1}{n}+\frac{1}{m} )^{2}+C_{10}(\frac{1}{n}+\frac{1}{m} )\big] ^{1/2}e^{\frac{l^{2}T}{16\eta } },\quad \forall t\in I. \end{equation*} Then, taking the upper bound with respect to $t$ in the left-hand side of this inequality, \begin{equation} \label{e5.10} \| u^{(n)}-u^{(m)}\| _{C(I,V)}\leq \big[ C_{9}( \frac{1}{n}+\frac{1}{m} )^{2}+C_{10}(\frac{1}{n}+\frac{1}{m} )\big] ^{1/2}e^{\frac{l^{2}T}{16\eta }}, \end{equation} which shows that $\{ u^{(n)}\} _{n>n_{0}}$ is a Cauchy sequence in $C(I,V)$. This implies the existence of a function $u\in C(I,V)$ such that $u^{(n)}\to u$ in this space. Moreover, letting $m\to \infty $ in \eqref{e5.10} we obtain the error estimate \eqref{e5.2} with $C=\sqrt{C_{9}+C_{10}}e^{\frac{l^{2}T}{16\eta }}$, what completes the proof. \end{proof} We write down some results for the limit-function $u$. \begin{corollary} \label{coro5.2} The function $u$ possesses the following properties: \begin{itemize} \item[(i)] $u\in C^{0,1}(I,V)$; \item[(ii)] $u$ is strongly differentiable a.e. in $I$ and $\frac{du}{dt}\in L^{\infty }(I,L^{2}(0,1))$; \item[(iii)] $\overline{u}^{(n)}(t)\to u(t)$ in $V$ for all $t\in I$; \item[(iv)] $\frac{du^{(n)}}{dt}\rightharpoonup \frac{du}{dt}$ in $L^{2}(I,L^{2}(0,1))$. \end{itemize} \end{corollary} \begin{proof} On the basis of Corollary \ref{coro4.2} (i) and (ii), uniform convergence statement from Theorem \ref{thm5.1} and the continuous embedding $V\hookrightarrow Y:=L^{2}(0,1)$, \cite[Lemma 1.3.15]{Kac} is valid for our special situation yielding assertions (i), (ii) and (iv) of the present Corollary. The remaining assertion (iii) is an immediate consequence of the combination of Corollary \ref{coro4.2} (iii) with the convergence result stated in Theorem \ref{thm5.1}. \end{proof} Collecting all the obtained results, we can state our existence theorem. \begin{theorem} \label{thm5.3} The limit function $u$ from Theorem \ref{thm5.1} is the unique weak solution to problem \eqref{e1.5}-\eqref{e1.8} in the sense of Definition \ref{def2.1}. Moreover, $u$ depends continuously upon data $f$ and $U_{0}$, namely the inequality \begin{equation} \label{e5.11} \max_{0\leq s\leq t} \| u^{\ast }(s)-u^{\ast\ast }(s)\| \leq C\Big(\| U_{0}^{\ast }-U_{0}^{\ast \ast }\| +\int_{0}^{t}\| f^{\ast }(s,u^{\ast }(s))-f^{\ast \ast }(s,u^{\ast \ast }(s))\| _{B_{2}^{1}}ds\Big), \end{equation} holds for all $t\in I$, with some positive constant $C$ depending only on $\eta $. \end{theorem} \begin{proof} \textbf{Existence. }It suffices to check all the points (i)-(iv) of Definition \ref{def2.1}. Obviously, in light of Corollary \ref{coro5.2}, the first two points of Definition \ref{def2.1} are already fulfilled. Moreover, since $u^{(n)}\to u$ in $C(I,V)$ as $n\to \infty $ and, by definition, $u^{(n)}(0)=U_{0}$, it follows that $u(0)=U_{0}$ holds in $V$ so the initial condition \eqref{e1.6} is also fulfilled, that is point (iii) of Definition \ref{def2.1} takes place. To show that $u$ obeys the integral equation \eqref{e2.4}, we investigate the behaviour as $n\to \infty $ of the integral relation \begin{equation} \begin{aligned} &\big({u^{(n)}(t)-U_{0},\phi }\big)_{B_{2}^{1}} +\int_{0}^{t}\big({\overline{u}^{(n)}(\tau ),\phi } \big)d\tau +\eta \big({u^{(n)}(t)-U_{0},\phi }\big)\\ &=\int_{0}^{t}\Big(\overline{f}^{(n)}\big(\tau ,\overline{u}^{(n)}( \tau -\frac{T}{n})\big),\phi\Big)_{B_{2}^{1}}d\tau ,\quad \forall \phi \in V,\; \forall t\in I, \end{aligned} \label{e5.12} \end{equation} which results from \eqref{e5.1} by integration over $(0,t)\subset I$ noting that $u^{(n)}(0)=U_{0}$. This requires some additional convergence statements. Firstly, since $u^{(n)}\rightarrow u$ in $C(I,V)$ and since for all fixed $\phi \in V$, the linear functional $v\mapsto (v,\phi )_{B_{2}^{1}}$ is continuous on $V$, we deduce that \begin{gather} \big({u^{(n)}(t),\phi }\big)\underset{n\rightarrow \infty }{\longrightarrow } \big({u(t),\phi }\big),\quad \forall \phi \in V,\;\forall t\in I, \label{e5.13} \\ \big({u^{(n)}(t),\phi }\big)_{B_{2}^{1}}\underset{n\rightarrow \infty }{ \longrightarrow }\big({u(t),\phi }\big)_{B_{2}^{1}},\quad \forall \phi \in V,\;\forall t\in I. \label{e5.14} \end{gather}% Secondly, by virtue of \eqref{e5.5} the Lebesgue Theorem of dominated convergence may be applied to the convergence statement (iii) from Corollary \ref{coro5.2} giving \begin{equation} \int_{0}^{t}\big({\overline{u}^{(n)}(\tau ),\phi }\big)d\tau \underset{n\rightarrow \infty }{\longrightarrow }\int_{0}^{t} \big(\overset{}{u(\tau ),\phi }\big)d\tau ,\quad \forall \phi \in V,\;\forall t\in I. \label{e5.15} \end{equation}% Thirdly, in view of Assumption (H1), we have \begin{align*} & \Vert \overline{f}^{(n)}\big(\tau ,\overline{u}^{(n)}(\tau -\frac{T}{n}) \big)-f(\tau ,u(\tau ))\Vert _{B_{2}^{1}} \\ & =\Vert f\big(t_{j},\overline{u}^{(n)}(\tau -\frac{T}{n})\big)-f(\tau ,u(\tau ))\Vert _{B_{2}^{1}} \\ & \leq l\big[|t_{j}-\tau |(1+\Vert u_{j-1}\Vert _{B_{2}^{1}}+\Vert u(\tau )\Vert _{B_{2}^{1}})+\Vert \overline{u}^{(n)}(\tau -\frac{T}{n})-u(\tau )\Vert _{B_{2}^{1}}\big], \end{align*}% for all $\tau \in (t_{j-1},t_{j}]$, $1\leq j\leq n$; therefore \begin{equation*} \Vert \overline{f}^{(n)}\big(\tau ,\overline{u}^{(n)}(\tau -\frac{T}{n})\big)% -f(\tau ,u(\tau ))\Vert _{B_{2}^{1}}\leq \frac{C}{n}+l\Vert \overline{u}% ^{(n)}(\tau -\frac{T}{n})-u(\tau )\Vert _{B_{2}^{1}}, \end{equation*}% for all $\tau \in I$, where $C:=lT(1+C_{1}+\Vert u\Vert _{C(I,V)})$. However, with consideration to estimates (iii)-(iv) from Corollary \ref{coro4.2} and inequality \eqref{e5.2}, we can write \begin{align*} \Vert \overline{u}^{(n)}(\tau -\frac{T}{n})-u(\tau )\Vert _{B_{2}^{1}}& \leq \Vert \overline{u}^{(n)}(\tau -\frac{T}{n})-\overline{u}^{(n)}(\tau )\Vert \\ & \quad +\Vert \overline{u}^{(n)}(\tau )-u^{(n)}(\tau )\Vert +\Vert u^{(n)}(\tau )-u(\tau )\Vert \\ & \leq C(\frac{1}{n}+\frac{1}{n^{1/2}}),\quad \forall \tau \in I, \end{align*}% whence \begin{equation*} \Vert \overline{f}^{(n)}\big(\tau ,\overline{u}^{(n)}(\tau -\frac{T}{n})\big) -f(\tau ,u(\tau ))\Vert _{B_{2}^{1}}\leq \frac{C}{n^{1/2}},\quad \forall \tau \in I, \end{equation*}% and then \begin{equation} \overline{f}^{(n)}\big(\tau ,\overline{u}^{(n)}(\tau -\frac{T}{n})\big) \underset{n\rightarrow \infty }{\longrightarrow }f(\tau ,u(\tau ))\quad \text{in }B_{2}^{1}(0,1),\quad \forall \tau \in I. \label{e5.16} \end{equation}% Now, due to \eqref{e5.4} the function $|(\overline{f}^{(n)}\big(\tau , \overline{u}^{(n)}(\tau -\frac{T}{n})\big),\phi )_{B_{2}^{1}}|$ is uniformly bounded with respect to both $\tau $ and $n$. So the Lebesgue Theorem of dominated convergence may be applied again to \eqref{e5.16} yielding \begin{equation} \int_{0}^{t}(\Big(\overline{f}^{(n)}\big(\tau ,\overline{u}^{(n)}(\tau - \frac{T}{n})\big),\phi \Big)_{B_{2}^{1}}d\tau \underset{n\rightarrow \infty } {\longrightarrow }\int_{0}^{t}(f(\tau ,u(\tau )),\phi )_{B_{2}^{1}}d\tau , \label{e5.17} \end{equation}% for all $\phi \in V$ and all $t\in I$. Then, by \eqref{e5.13}, \eqref{e5.14} , \eqref{e5.15} and \eqref{e5.17}, a limiting process $n\rightarrow \infty $ in \eqref{e5.12} leads to \begin{equation*} \big({u(t)-U_{0},\phi }\big)_{B_{2}^{1}}+\int_{0}^{t}\big({u(\tau ),\phi } \big)d\tau +\eta \big({u(t)-U_{0},\phi }\big)=\int_{0}^{t}(f(\tau ,u(\tau )),\phi )_{B_{2}^{1}}d\tau , \end{equation*}% for all $\phi \in V$ and $t\in I$. Finally, the differentiation of this last identity with respect to $t$ recalling that $u:I\rightarrow L^{2}(0,1)$ is strongly differentiable for $a.e$. $t\in I$, leads to the required identity \eqref{e2.4} by the aid of the equalities $\frac{d}{dt}({u(t),\phi } )_{B_{2}^{1}}=(\frac{du}{dt}(t),\phi )_{B_{2}^{1}}$ and $\frac{d}{dt}({u(t),\phi })=(\frac{du}{dt}(t),\phi )$ which hold for all $t\in I$ and all $\phi \in V$. Thus, $u$ weakly solves problem \eqref{e1.5}-\eqref{e1.8}. \newline \textbf{Uniqueness and continuous dependence upon data. }Let $u^{\ast }$ and $u^{\ast \ast }$ be two weak solutions of problem \eqref{e1.5}-\eqref{e1.8} corresponding respectively to $(U_{0}^{\ast },f^{\ast })$ and $(U_{0}^{\ast \ast },f^{\ast \ast })$ instead of $(U_{0},f)$. Subtracting the identity \eqref{e2.4} written for $u^{\ast \ast }$ from the same identity written for $u^{\ast }$ and inserting $\phi =u^{\ast }(t)-u^{\ast \ast }(t)$ in the resulting relation, we get by integration over $(0,\tau )$, with $\tau \in I: $ \begin{align*} & \frac{1}{2}\Vert u^{\ast }(\tau )-u^{\ast \ast }(\tau )\Vert _{B_{2}^{1}}^{2}-\frac{1}{2}\Vert u^{\ast }(0)-u^{\ast \ast }(0)\Vert _{B_{2}^{1}}^{2}+\int_{0}^{\tau }\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert ^{2}dt \\ & +\frac{\eta }{2}\Vert u^{\ast }(\tau )-u^{\ast \ast }(\tau )\Vert ^{2}- \frac{\eta }{2}\Vert u^{\ast }(0)-u^{\ast \ast }(0)\Vert ^{2} \\ & =\int_{0}^{\tau }\Big(f^{\ast }(t,u^{\ast }(t))-f^{\ast \ast }(t,u^{\ast \ast }(t)),u^{\ast }(t)-u^{\ast \ast }(t)\Big)_{B_{2}^{1}}dt, \end{align*}% hence, ignoring the first and the third terms in the left hand side, we obtain \begin{align*} & \Vert u^{\ast }(\tau )-u^{\ast \ast }(\tau )\Vert ^{2} \\ & \leq \frac{1}{\eta }\Vert u^{\ast }(0)-u^{\ast \ast }(0)\Vert _{B_{2}^{1}}^{2}+\Vert u^{\ast }(0)-u^{\ast \ast }(0)\Vert ^{2} \\ & \quad +\frac{2}{\eta }\int_{0}^{\tau }\Vert f^{\ast }(t,u^{\ast }(t))-f^{\ast \ast }(t,u^{\ast \ast }(t))\Vert _{B_{2}^{1}}\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert _{B_{2}^{1}}dt \\ & \leq (\frac{1}{2\eta }+1)\Vert u^{\ast }(0)-u^{\ast \ast }(0)\Vert ^{2}+ \frac{\sqrt{2}}{\eta }\max_{0\leq t\leq \tau }\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert \\ & \times \int_{0}^{\tau }\Vert f^{\ast }(t,u^{\ast }(t))-f^{\ast \ast }(t,u^{\ast \ast }(t))\Vert _{B_{2}^{1}}dt \\ & \leq (\frac{1}{2\eta }+1)\Vert U_{0}^{\ast }-U_{0}^{\ast \ast }\Vert \max_{0\leq t\leq \tau }\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert +\frac{ \sqrt{2}}{\eta }\max_{0\leq t\leq \tau }\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert \\ & \quad \times \int_{0}^{\tau }\Vert f^{\ast }(t,u^{\ast }(t))-f^{\ast \ast }(t,u^{\ast \ast }(t))\Vert _{B_{2}^{1}}dt \\ & \leq \Big[(\frac{1}{2\eta }+1)\Vert U_{0}^{\ast }-U_{0}^{\ast \ast }\Vert + \frac{\sqrt{2}}{\eta }\int_{0}^{\tau }\Vert f^{\ast }(t,u^{\ast }(t))-f^{\ast \ast }(t,u^{\ast \ast }(t))\Vert _{B_{2}^{1}}dt\Big] \\ & \quad \times \max_{0\leq t\leq \tau }\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert , \end{align*}% where \eqref{e2.3} has been used. Consequently for all $s\in \lbrack 0,\tau ] $, we have \begin{align*} & \Vert u^{\ast }(s)-u^{\ast \ast }(s)\Vert ^{2} \\ & \leq \Big[(\frac{1}{2\eta }+1)\Vert U_{0}^{\ast }-U_{0}^{\ast \ast }\Vert + \frac{\sqrt{2}}{\eta }\int_{0}^{\tau }\Vert f^{\ast }(t,u^{\ast }(t))-f^{\ast \ast }(t,u^{\ast \ast }(t))\Vert _{B_{2}^{1}}dt\Big] \\ & \quad \times \max {0\leq t\leq \tau }\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert , \end{align*} whence \begin{align*} & \max_{0\leq s\leq \tau }\Vert u^{\ast }(s)-u^{\ast \ast }(s)\Vert ^{2} \\ & \leq \Big[(\frac{1}{2\eta }+1)\Vert U_{0}^{\ast }-U_{0}^{\ast \ast }\Vert + \frac{\sqrt{2}}{\eta }\int_{0}^{\tau }\Vert f^{\ast }(t,u^{\ast }(t))-f^{\ast \ast }(t,u^{\ast \ast }(t))\Vert _{B_{2}^{1}}dt\Big] \\ & \quad \times \max_{0\leq t\leq \tau }\Vert u^{\ast }(t)-u^{\ast \ast }(t)\Vert , \end{align*} from which the estimate \eqref{e5.11} follows with $C:=\max (\frac{1}{2\eta }+1,\frac{\sqrt{2}}{\eta })$. 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