\documentclass[reqno]{amsart} \usepackage{amsfonts,hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 138, 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 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/138\hfil Nonlinear Kirchhoff-Carrier wave equation] {Nonlinear Kirchhoff-Carrier wave equation in a unit membrane with mixed homogeneous boundary conditions} \author[N. T. Long\hfil EJDE-2005/138\hfilneg] {Nguyen Thanh Long} \address{Nguyen Thanh Long \hfill\break Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University HoChiMinh City, 227 Nguyen Van Cu Str., Dist. 5, Hochiminh City, Vietnam} \email{longnt@hcmc.netnam.vn \; longnt2@gmail.com} \date{} \thanks{Submitted August 3, 2004. Published December 1, 2005.} \subjclass[2000]{35L70, 35Q72} \keywords{Nonlinear wave equation; Galerkin method; quadratic convergence; \hfill\break\indent weighted Sobolev spaces} \begin{abstract} In this paper we consider the nonlinear wave equation problem \begin{gather*} u_{tt}-B\big(\|u\|_0^2,\|u_{r}\|_0^2\big)(u_{rr}+\frac{1}{r}u_{r}) =f(r,t,u,u_{r}),\quad 00$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} In this paper, we consider the initial and boundary value problem \label{1.1} \begin{gathered} u_{tt}-B\big(\|u\|_0^2,\|u_{r}\|_0^2\big)(u_{rr}+\frac{1}{r}u_{r}) =f(r,t,u,u_{r}),\quad 00$ is a given constant. Hosoya and Yamada \cite{6} considered problem \eqref{1.3}$_{3,4}$-\eqref{1.3} with $f=f(u)=-\delta|u|^{\alpha}u$ where $\delta>0$ and $\alpha\geq 0$ are given constants. In \cite{9} the authors studied the existence and uniqueness of the solution of the equation \begin{equation*} u_{tt}+\lambda \triangle^2u-B(\|\nabla u\|^2)\triangle u+\varepsilon|u_{t}|^{\alpha-1} u_{t}=F(x,t), \end{equation*} where $\lambda>0$, $\varepsilon>0$ and $0<\alpha<1$ are given constants. In the case of the term $\frac{1}{r}u_{r}$ appearing in equation \eqref{1.1}$_1$ we have to eliminate the coefficient $\frac{1}{r}$ by using Sobolev spaces with appropriate weight (see \cite{11}). On the other hand, problem \eqref{1.1} with general nonlinear right-hand side $f(r,t,u,u_{r},u_{t})$ given as a continuous function of five variables has not been studied completely yet. In the present paper, we study problem (\ref{1.1}) with some forms of the right-hand side $f$. In the first part, we study problem (\ref{1.1}) with the right-hand side $f(r,t,u,u_{r})$ where $f\in C^0([0,1]\times \mathbb{R}_+\times\mathbb{R}^2)$ satisfies the condition \begin{equation*} \frac{\partial f}{\partial r},\frac{\partial f}{\partial u}, \frac{\partial f}{\partial u_r}\quad\text{in}\quad C^0\big([0,1] \times \mathbb{R}_+\times \mathbb{R}^2\big). \end{equation*} It is not necessary that $f\in C^1\big([0,1]\times \mathbb{R}_+\times\mathbb{R}^2\big)$. First, we shall associate with equation \eqref{1.1}$_1$ a linear recurrent sequence which is bounded in a suitable function space. The existence of a local solution is proved by a standard compactness argument. Note that the linearization method in this paper and in papers (\cite{2,4,14,17,18,23} cannot be used in papers \cite{5,9,12,13,15,16,19}. In the second part, we consider problem (\ref{1.1}) corresponding to $f=f(r,u)$ and $B(\eta)=b_0+\eta$ with given constant $b_0>0$. We associate with equation \eqref{1.1}$_1$ a recurrent sequence ${u_{m}}$ (nonlinear) defined by \begin{align*} &\frac{\partial^2u_m}{\partial t^2}-\Big(b_0+\int_0^1\big|\frac{\partial u_m }{\partial r} (r,t)\big|^2rdr\Big)\big(\frac{\partial^2u_m}{\partial r^2}+ \frac{1}{r}\frac{\partial u_m} {\partial r}\big) \\ &=f(r,u_{m-1})+(u_{m}-u_{m-1})\frac{\partial f}{\partial u}(r,u_{m-1}) \quad \text{in }(0,1) \times (0,T), \end{align*} with $u_{m}$ satisfying $(1.1)_{2-3}$. The first term $u_0$ is chosen as $u_0=\widetilde{u}_0$. If $f\in C^2\big([0,1]\times\mathbb{R}\big)$, we prove that the sequence ${u_{m}}$ converges quadratically. The results obtained here relatively are in part generalizations of those in \cite% {2,4,14,17,18,23}. \section{Preliminary results, notation, function spaces} Put $\Omega =(0,1)$. We omit the definitions of the usual function spaces $L^{p}(\overline{\Omega})$, $H^{m}(\Omega )$, $W^{m,p}(\Omega )$. For any function $v\in C^0(\overline{\Omega})$ we define $\|v\|_0$ as \begin{equation*} \|v\|_0=\Big(\int_0^1rv^2(r)dr\Big)^{1/2} \end{equation*} and define the space $V_0$ as the completion of the space $C^0(\overline{\Omega})$ with respect to the norm $\|\cdot\|_0$. Similarly, for any function $v\in C^1(\overline{\Omega})$ we define $\|v\|_1$ as \begin{equation*} \|v\|_1=\Big(\int_0^1r[v^2(r)+|v'(r)|^2]dr\Big)^{1/2} \end{equation*} and define the space $V_1$ as completion of the space $C^1(\overline{\Omega})$ with respect to the norm $\|\cdot\|_1$. Note that the norms $\|\cdot\|_0$ and $\|\cdot\|_1$ can be defined, respectively, from the inner products \begin{gather*} \langle u,v\rangle=\int_0^1ru(r)v(r)dr, \\ \langle u,v\rangle+\langle u',v'\rangle=\int_0^1r[u(r)v(r)+u'(r)v^{% \prime}(r)]dr. \end{gather*} Identifying $V_0$ with its dual $V'_0$ we obtain the dense and continuous embedding $V_1\hookrightarrow V_0\equiv V_0'\hookrightarrow V_1'$. The inner product notation will be re-utilized to denote the duality pairing between $V_1$ and $V_1'$. We then have the following lemmas, the proofs of which can be found in \cite{2}: \begin{lemma} \label{lemma1} There exist constants $K_1>0$ and $K_2>0$ such that, for all $v\in C^1(\overline{\Omega})$ and $r\in \overline{\Omega}$, \begin{itemize} \item[(i)] $\|v'\|^2_0+v^2(1)\geq \|v\|_0^2$, \item[(ii)] $|v(1)|\leq K_1\|v\|_1$, \item[(iii)] $\sqrt{r}|v(r)|\leq K_2\|v\|_1$. \end{itemize} \end{lemma} \begin{lemma}\label{lemma2} The embedding $V_1\hookrightarrow V_0$ is compact. \end{lemma} \begin{remark}\label{remark1} \rm In Lemma \ref{lemma1}, the constants $K_1$ and $K_2$ can be given explicitly as $K_1=\sqrt{1+\sqrt{2}}$ and $K_2=\sqrt{1+\sqrt{5}}$. We also note that $lim_{r\to 0_+}\sqrt{r}v(r)=0$ for all $v\in V_1$ (see \cite[Lemma 5.40]{1}). On the other hand, from $H^1(\varepsilon,1)\hookrightarrow C^0([\varepsilon,1]), 0<\varepsilon<1$ and $\sqrt{\varepsilon}\|v\|_{H^1(\varepsilon,1)}\leq \|v\|_1$ for all $v\in V_1$, it follows that $v|_{[\varepsilon,1]}\in C^0([\varepsilon,1])$. From both relations we deduce that $\sqrt{r}v \in C^0(\overline{\Omega})$ for all $v\in V_1$. \end{remark} Now, we define the bilinear form $$\label{2.1} a(u,v)=hu(1)v(1)+\int_0^1ru'(r)v'(r)dr, \text{ for } u,v\in V_1,$$ where $h$ is a positive constant. Then for some uniquely defined bounded linear operator $A:V_1\to V_1'$ we have $a(u,v)=\langle Au,v\rangle$ for all $u,v\in V_1$. We then have the following lemma. \begin{lemma}\label{lemma3} The symmetric bilinear form $a(\cdot,\cdot)$ defined by \eqref{2.1} is continuous on $V_1\times V_1$ and coercive on $V_1$, i.e., \begin{itemize} \item[(i)] $|a(u,v)|\leq C_1\|u\|_1\|v\|_1$ \item[(ii)] $a(v,v)\geq C_0\|v\|_1^{2}$ \end{itemize} for all $u,v\in V_1$, where $C_0=\frac{1}{2}min\{1,h\}$ and $C_1=1+hK_1^2$. \end{lemma} The proof of Lemma \ref{lemma3} is straightforward and we omit it. \begin{lemma}\label{lemma4} There exists an orthonormal Hilbert basis $\{\widetilde{w}_{j}\}$ of the space $V_0$ consisting of eigenfunctions $\widetilde{w}_{j}$ corresponding to eigenvalues $\lambda_{j}$ such that \begin{itemize} \item[(i)] $0<\lambda_1\leq \lambda_{j}\uparrow +\infty$ as $j\to+\infty$, \item[(ii)] $a(\widetilde{w}_{j},v)=\lambda_{j}\langle \widetilde{w}_{j},v\rangle$ for all $v\in V_1$ and $j\in \mathbb{N}$. \end{itemize} Note that from $(ii)$ it follows that $\{\widetilde{w}_{j}/\sqrt{\lambda_j}\}$ is automatically an orthonormal set in $V_1$ with respect to $a(\cdot,\cdot)$ as inner product. The eigensolutions $\widetilde{w}_{j}$ are indeed eigensolutions for the boundary value problem \begin{gather*} A\widetilde{w}_{j}\equiv\frac{-1}{r}\frac{d}{dr} \big(r\frac{d\widetilde{w}_j}{dr}\big) =\lambda_{j}\widetilde{w}_{j}, \quad \text{in }\Omega,\\ \big|\lim_{r\to 0_+}\sqrt{r}\frac{d\widetilde{w}_j}{dr}(r)\big|<+\infty,\\ \frac{d\widetilde{w}_j}{dr}(1)+h\widetilde{w}_{j}(1)=0. \end{gather*} \end{lemma} The proof of the above Lemma can be found in \cite[Theorem 6.2.1]{24} with $V=V_1,H=V_0$ and $a(\cdot,\cdot)$ as defined by (\ref{2.1}). For functions $v$ in $C^2(\overline{\Omega})$, we define \begin{equation*} \|v\|_2=\Big(\int_0^1r[v^2(r)+|v'(r)|^2+|Av(r)|^2]dr\Big)^{1/2}, \end{equation*} and define the space $V_2$ as the completion of $C^2(\overline{\Omega})$ with respect to the norm $\|\cdot\|_2$. Note that $V_2$ is also a Hilbert space with respect to the scalar product \begin{equation*} \langle u,v\rangle+\langle u',v'\rangle+\langle Au,Av\rangle \end{equation*} and that $V_2$ can be defined also as $V_2=\{v\in V_1:Av\in V_0\}$. We then have the following two lemmas whose proof of which can be found in \cite{2}. \begin{lemma}\label{lemma5} The embedding $V_2\hookrightarrow V_1$ is compact. \end{lemma} \begin{lemma}\label{lemma6} For all $v\in V_2$ we have \begin{itemize} \item[(i)] $\|v'\|_{L^{\infty}(\Omega)}\leq \frac{1}{\sqrt{2}}\|Av\|_0$, \item[(ii)] $\|v''\|_{0}\leq \sqrt{\frac{3}{2}}\|Av\|_0$, \item[(iii)] $\|v\|^2_{L^{\infty}(\Omega)}\leq \big(2\|v\|_0 +\frac{1}{\sqrt{2}}\|Av\|_0\big)\|v\|_0$. \end{itemize} \end{lemma} Also the following lemma will be useful in Section 4. \begin{lemma}\label{lemma7} For all $u\in V_1$ and $v\in V_0$, $$\label{2.2} \langle u^2,|v|\rangle =\sqrt{2}(1+K_1^2)\|u\|_1^2\|v\|_0,$$ where the constant $K_1$ is given by Lemma \ref{lemma1}. \end{lemma} \begin{proof} It suffices to prove that inequality (\ref{2.2}) holds for $u\in C^1(\overline{\Omega})$ and $v\in C^0(\overline{\Omega})$. We have \begin{equation*} u(r)=u(1)-\int_r^1u'(s)ds. \end{equation*} Hence, it follows from Lemma \ref{lemma1} that \begin{equation*} u^2(r)\leq 2u^2(1)+2\Big(\int_r^1u'(s)ds\Big)^2 \leq 2K_1^2\|u\|_1^2+2(1-r)\int_r^1 |u'(s)|^2ds. \end{equation*} This implies $$\label{2.3} \begin{split} \langle u^2,|v|\rangle&=\int_0^1ru^2(r)|v(r)|dr \\ &\leq 2K_1^2\|u\|_1^2\int_0^1r|v(r)|dr+2\int_0^1r(1-r)|v(r)|dr\int_r^1|u'(s)|^2ds. \end{split}%$$ Note that the first integral herein can be estimated as \begin{equation*} \int_0^1r|v(r)|dr\leq \Big(\int_0^1rdr\Big)^{1/2}\Big(\int_0^1r|v(r)|^2dr \Big)^{1/2} =\frac{1}{\sqrt{2}}\|v\|_0. \end{equation*} Reversing the order of integration in the last integral of (\ref{2.3}), we estimate that integral as \begin{equation*} \begin{split} &\int_0^1r(1-r)|v(r)|dr\int_0^1|u'(s)|^2ds \\ &=\int_0^1|u'(s)|^2ds\int_0^sr(1-r)|v(r)|dr \\ &\leq \int_0^1|u'(s)|^2ds\Big(\int_0^sr(1-r)^2dr\Big)^{1/2}\Big( \int_0^sr|v(r)|^2dr\Big)^{1/2} \\ &\leq \frac{1}{\sqrt{2}}\|u'\|_0^2\|v\|_0\leq \frac{1}{\sqrt{2}} \|u\|_1^2\|v\|_0. \end{split} \end{equation*} From the two estimates above, we obtain (\ref{2.2}) and the lemma is proved. \end{proof} For a Banach space $X$, we denote by $\|\cdot \|_{X}$ its norm, by ${X}'$ its dual space and by ${L}^{p}(0,T;{X})$,$1\leq p\leq \infty$ the Banach space of all real measurable functions $u:(0,T)\to {X}$ such that \begin{gather*} \|u\|_{{L}^{p}(0,T;{X})}=\Big(\int_{0}^{T}\|u(t)\|_{{X}}^{p}dt\Big)^{1/p} <\infty\quad \mbox{for } 1\leq{p}<\infty, \\ \|u\|_{{L}^{\infty }(0,T;{X})}=\mathop{\rm ess\, sup}_{00$for all$\xi,\eta\geq 0$, \item[(H3)]$f\in C^0(\overline{\Omega}\times\mathbb{R}_+\times\mathbb{R}^2)$and$\partial f/\partial r, \partial f/\partial u, \partial f/\partial u_{r} \in C^0(\overline{\Omega}\times\mathbb{R}_+\times\mathbb{R}^2)$. \end{itemize} With$B$and$f$satisfying assumptions (H2) and (H3), respectively, we introduce the following constants, for any$M>0$and$T>0$: $$\label{3.1} \begin{gathered} \widetilde{K}_0=\widetilde{K}_0(M,B)=\sup\{B(\xi,\eta):0\leq\xi,\eta \leq M^2\},\\ \widetilde{K}_1=\widetilde{K}_1(M,B)=\sup\big\{\big(\big|\frac{\partial B}{\partial \xi}\big| +\big|\frac{\partial B}{\partial \eta}|\big)(\xi,\eta):0\leq \xi,\eta \leq M^2\big\},\\ \overline{K}_0=\overline{K}_0(M,T,f)=\sup_{(r,t,u,v)\in A_*}|f(r,t,u,v)|,\\ \overline{K}_1=\overline{K}_1(M,T,f)=\sup_{(r,t,u,v)\in A_*}\big(\big| \frac{\partial f} {\partial r}\big|+\big|\frac{\partial f}{\partial u}\big| +\big|\frac{\partial f} {\partial v}\big|\big)(r,t,u,v), \end{gathered}$$ where \begin{equation*} \begin{split} A_{*}&=A_{*}(M,T) \\ &=\{(r,t,u,v):0\leq r\leq 1, 0\leq t\leq T, |u| \leq M\sqrt{2+1/\sqrt{2}}, |v|\leq M/\sqrt{2}\}. \end{split}% \end{equation*} For each$M>0$and$T>0we put \begin{gather*} \begin{aligned} W(M,T)=&\Big\{v\in L^{\infty}(0,T;V_2): \dot{v}\in L^{\infty}(0,T;V_1) \text{ and } \ddot{v} \in L^2(0,T;V_0),\\ &\text{with } \|v\|_{L^{\infty}(0,T;V_2)}, \ |\dot{v}\|_{L^{\infty}(0,T;V_1)}, \|\ddot{v}\|_{L^{2}(0,T;V_0)} \leq M\Big\}, \end{aligned} \\ W_1(M,T)=\big\{v\in W(M,T):\ddot{v} \in L^{\infty}(0,T;V_0)\big\}. \end{gather*} We shall choose as first initial termu_0=\widetilde{u}_0$, suppose that $$\label{3.2} u_{m-1}\in W_1(M,T),$$ and associate with problem (\ref{1.1}) the following variational problem: Find$u_{m}$in$W_1(M,T)(m\geq 1)so that $$\label{3.3} \begin{gathered} \langle \ddot{u}_{m}(t),v\rangle+b_{m}(t)a(u_{m}(t),v)=\langle F_{m}(t),v\rangle\quad \forall v\in V_1,\\ u_{m}(0)=\widetilde{u}_0,\quad \dot{u}_{m}(0)=\widetilde{u}_1, \end{gathered}$$ where \label{3.4} \begin{gathered} \begin{aligned} b_{m}(t)&=B\big(\|u_{m-1}(t)\|_0^2, \|\nabla u_{m-1}(t)\|_0^2\big)\\ &=B\Big(\int_0^1u_{m-1}^2(r,t)rdr, \int_0^1\big|\frac{\partial u_{m-1}}{\partial r}(r,t) \big|^2rdr\Big), \end{aligned}\\ F_{m}(r,t)=f\big( r,t,u_{m-1}(t),\nabla u_{m-1}(t)\big). \end{gathered} Then, we have the following result. \begin{theorem}\label{theorem1} Let assumptions (H1)--(H3) hold. Then there exist a constantM>0$depending on$\widetilde{u}_0$,$\widetilde{u}_1$,$B$,$h$and a constant$T>0$depending on$\widetilde{u}_0$,$\widetilde{u}_1$,$B$,$h$,$f$such that, for$u_0=\widetilde{u}_0$, there exists a linear recurrent sequence$\{u_{m}\}\subset W_1(M,T)$defined by \eqref{3.3}-\eqref{3.4}. \end{theorem} \begin{proof} The proof consists of several steps.\newline \textbf{Step 1:} The Galerkin approximation (introduced by Lions [10]). Consider as in Lemma \ref{lemma4} the basis$w_{j}=\widetilde{w}_{j}/\sqrt{\lambda}_{j}$for$V_1$and put $$\label{3.5} u_{m}^{(k)}(t)=\sum_{j=1}^k c_{mj}^{(k)}(t)w_{j},$$ where the coefficients$c_{mj}^{(k)}$satisfy the system of linear differential equations $$\label{3.6} \begin{gathered} \langle \ddot{u}_{m}^{(k)}(t),w_{j}\rangle +b_{m}(t)a \big(u_{m}^{(k)}(t),w_{j}\big) =\langle F_{m}(t),w_{j}\rangle, \quad 1\leq j\leq k,\\ u_{m}^{(k)}(0)=\widetilde{u}_{0k},\quad \dot{u}_{m}^{(k)}(0)=\widetilde{u}_{1k}, \end{gathered}$$ where $$\label{3.7} \begin{gathered} \widetilde{u}_{0k} \to \widetilde{u}_0 \quad \text{strongly in } V_2,\\ \widetilde{u}_{1k}\to \widetilde{u}_1 \quad \text{strongly in } V_1. \end{gathered}$$ Suppose that$u_{m-1}$satisfies (\ref{3.2}). Then it is clear that system (\ref{3.6}) has a unique solution$u_{m}^{(k)}$on an interval$0\leq t\leq T_{m}^{(k)}\leq T$. The following estimates allows us to the take constant$T_{m}^{(k)}=T$for all$m$and$k$. \noindent\textbf{Step 2:} \textit{A priori estimates}. Put $$\label{3.8} S_{m}^{(k)}(t)=X_{m}^{(k)}(t)+Y_{m}^{(k)}(t)+ \int_0^t\|\ddot{u}% _{m}^{(k)}(s)\|_0^2ds,$$ where $$\label{3.9} \begin{gathered} X_{m}^{(k)}(t)=\|\dot{u}_{m}^{(k)}(t)\|_0^2 +b_{m}(t)a\big(u_{m}^{(k)}(t),u_{m}^{(k)}(t)\big),\\ Y_{m}^{(k)}(t)=a\big(\dot{u}_{m}^{(k)}(t),\dot{u}_{m}^{(k)}(t)\big) +b_{m}(t)\|Au_{m}^{(k)}(t) \|_0^2, \end{gathered}$$ where$A$is defined by (\ref{2.1}). Then it follows that \begin{equation*} \begin{split} S_{m}^{(k)}(t)=&S_{m}^{(k)}(0)+\int_0^tb_{m}'(s)\big[a\big(% u_{m}^{(k)}(s),u_{m}^{(k)}(s)\big) +\|Au_{m}^{(k)}(s)\|_0^2\big]ds \\ & +2\int_0^t\langle F_{m}(s),\dot{u}_{m}^{(k)}(s)\rangle ds+2\int_0^ta\big( F_{m}(s), \dot{u}_{m}^{(k)}(s)\big)ds \\ &- \int_0^tb_{m}(s)\langle Au_{m}^{(k)}(s),\ddot{u}_{m}^{(k)}(s)\rangle ds +\int_0^t \langle F_{m}(s),\ddot{u}_{m}^{(k)}(s)\rangle ds \\ =&S_{m}^{(k)}(0)+I_1+\dots+I_5. \end{split}% \end{equation*} We shall estimate step by step all integrals$I_1,\dots,I_5$. \noindent\textit{Integral$I_1$}: Using assumption (H2), we obtain from \eqref{3.1}$_2$and \eqref{3.4}$_1$that \begin{equation*} \begin{split} |b_{m}'(t)|&\leq 2\big|\frac{\partial B}{\partial \xi}\big(% \|u_{m-1}(t)\|_0^2, \|\nabla u_{m-1}(t)\|_0^2\big)\langle u_{m-1}(t),\dot{u} _{m-1}(t)\rangle\big| \\ &\quad+2\big|\frac{\partial B}{\partial \eta}\big(\|u_{m-1}(t)\|_0^2, \|\nabla u_{m-1}(t)\|_0^2\big)\langle \nabla u_{m-1}(t),\nabla \dot{u} _{m-1}(t)\rangle\big| \\ &\leq 4M^2\widetilde{K}_1. \end{split} \end{equation*} Combining (\ref{3.8})-(\ref{3.9}) we obtain \begin{equation*} I_1\leq \frac{4M^2\widetilde{K}_1}{b_0}\int_0^tS_{m}^{(k)}(s)ds. \end{equation*} \noindent\textit{Integral$I_2$}: Since$u_{m-1}\in W_1(M,T)$, it follows from Lemma \ref{lemma6} that $$\label{3.10} |u_{m-1}(r,t)|\leq M\sqrt{2+1/\sqrt{2}}, |\nabla u_{m-1}(r,t)| \leq M/\sqrt{2 },\quad \text{a.e. on } \Omega\times(0,T).$$ By the Cauchy-Schwarz inequality, it follows from (\ref{3.1})$_3$that \begin{equation*} I_2\leq 2\int_0^t\|F_{m}(s)\|_0\big\|\dot{u}_{m}^{(k)}(s)\big\|_0ds\leq 2 \overline{K}_0 \int_0^t\sqrt{X_{m}^{(k)}(s)}ds. \end{equation*} \textit{Integral$I_3$}: Using Lemma \ref{lemma3}, we have \begin{equation*} I_3\leq 2C_1\int_0^t\|F_{m}(s)\|_1\big\|\dot{u}_{m}^{(k)}(s)\big\|_1ds. \end{equation*} On the other hand, from (\ref{3.1})$_{3-4}$and (\ref{3.10}) we obtain \begin{equation*} \|F_{m}(s)\|_1^2\leq \frac{1}{2}\overline{K}_0^2+\frac{1}{2}\overline{K}_1^2 \big[1+(1+\sqrt{3}M\big]^2. \end{equation*} Then we deduce, from (\ref{3.9})$_2$that \begin{equation*} I_3\leq \frac{\sqrt{2}C_1}{\sqrt{C_0}}\big[\overline{K}_0^2 +\big(1+(1+\sqrt{ 3})M\big)^2 \overline{K}_1^2\big]^{1/2}\int_0^t\sqrt{Y_{m}^{(k)}(s)}ds. \end{equation*} \noindent\textit{Integral$I_4$}: Using the inequality$|ab|\leq\frac{3}{4} a^2+\frac{1}{3}b^2\quad\forall a,b \in \mathbb{R}$, we get from \eqref{3.1}$_1$and (\ref{3.8})-(\ref{3.9}) that \begin{equation*} \begin{split} I_4&\leq \widetilde{K}_0\int_0^t\|Au_{m}^{(k)}(s)\|_0\big\|\ddot{u} _{m}^{(k)}(s)\big\|_0ds \\ &\leq \frac{3}{4}\widetilde{K}_0^2\int_0^t\|Au_{m}^{(k)}(s)\|_0^2ds+\frac{1}{ 3} \int_0^t\big\|\ddot{u}_{m}^{(k)}(s)\|_0^2ds \\ &\leq \frac{3\widetilde{K}_0^2}{4b_0}\int_0^tS_{m}^{(k)}(s)ds+\frac{1}{3} S_{m}^{(k)}(t). \end{split}% \end{equation*} \noindent\textit{Integral$I_5$}: We use again inequality$|ab|\leq \frac{3}{ 4}a^2+\frac{1}{3}b^2\quad\forall a,b \in \mathbb{R}$, we get from \eqref{3.1}$_3$and \eqref{3.8} that \begin{equation*} I_5\leq \int_0^t\|F_{m}(s)\|_0\big\|\ddot{u}_{m}^{(k)}(s)\|_0ds \leq \frac{3 }{4}T\overline{K}_0^2+\frac{1}{3}S_{m}^{(k)}(t). \end{equation*} Combining the above estimates for$I_1,\dots,I_5$, we get $$\label{3.11} S_{m}^{(k)}(t)\leq 3S_{m}^{(k)}(0)+\overline{C}_1(M,T)+\overline{C}_2(M) \int_0^tS_{m}^{(k)}(s)ds,$$ where $$\label{3.12} \begin{gathered} \overline{C}_1(M,T)=\frac{45}{4}T\overline{K}_0^2+\frac{9}{2C_0}TC_1^2 \big[\overline{K}_0^2 +\big(1+(1+\sqrt{3})M\big)^2\overline{K}_1^2\big],\\ \overline{C}_2(M)=1+\frac{3}{4b_0}\big(3\widetilde{K}_0^2+16M^2 \widetilde{K}_1\big). \end{gathered}$$ Now, we need an estimate on the term$S_{m}^{(k)}(0)$. We have \begin{equation*} \begin{split} S_{m}^{(k)}(0)&=X_{m}^{(k)}(0)+Y_{m}^{(k)}(0) \\ & =\|\widetilde{u}_{1k}\|_0^2+a(\widetilde{u}_{1k},\widetilde{u}_{1k}) +B% \big(\|\nabla \widetilde{u}_0\|_0^2\big)\big(a(\widetilde{u}_{0k}, \widetilde{u}_{0k}\big)+\|A\widetilde{u}_{0k}\|_0^2\big). \end{split}% \end{equation*} By means of the convergence (\ref{3.7}), we can deduce the existence of a constant$M>0$independent of$k$and$m$such that $$\label{3.13} S_{m}^{(k)}(0)\leq M^2/6.$$ Note that, from the assumption (H3), we have$\lim_{T\to 0_+}\sqrt{T}% \overline{K} _{i}(M,T,f)=0,i=0,1$. Then, from (\ref{3.12}) we can always choose the constant$T>0$such that $$\label{3.14} \begin{gathered} \big(M^2/2+\overline{C}_1(M,T)\big)exp\big[T\overline{C}_2(M)\big]\leq M^2,\\ \Big(1+\frac{1}{\sqrt{b_0C_0}}\Big) \sqrt{8M^2T\widetilde{K}_1 +\sqrt{2}T\overline{K}_1} \exp\big[\frac{1}{\sqrt{2}}T\overline{K}_1+\big(4+\frac{2C_1}{b_0C_0}\big) M^2T\widetilde{K}_1\big]<1. \end{gathered}$$ It follows from (\ref{3.11}) and (\ref{3.13})-(\ref{3.14}) that \begin{equation*} S_{m}^{(k)}(t)\leq M^2 \exp[-T\overline{C}_2(M)]+\overline{C}_2(M) \int_0^tS_{m}^{(k)}(s)ds \end{equation*} for$0\leq t\leq T_{m}^{(k)}\leq T$. By using Gronwall's lemma we deduce from here that \begin{equation*} S_{m}^{(k)}(t)\leq M^2 \exp[-T\overline{C}_2(M)] \exp[\overline{C}_2(M)t]% \leq M^2 \end{equation*} for all$t\in [0,T_{m}^{(k)}]$. So we can take constant$T_{m}^{(k)}=T$for all$k$and$m$. Therefore, we have$u_{m}^{(k)}\in W_1(M,T)$for all$m$and$k$. We can extract from$\{u_{m}^{(k)}\}$a subsequence$\{u_{m}^{(k_{i})}\}$such that \begin{gather*} u_{m}^{(k_{i})}\; \to \; u_{m}\quad\text{in } {L}^{\infty}(0,T;V_2) \text{ weak$^\star$}, \\ \dot{u}_{m}^{(k_{i})}\; \to \;\dot{u}_{m}\quad \text{in } {L}% ^{\infty}(0,T;V_1)\text{ weak$^\star$}, \\ \ddot{u}_{m}^{(k_{i})} \;\to \;\ddot{u}_{m}\quad \text{in } {L}^2(0,T;V_0) \text{ weak}, \\ \end{gather*} where$u_{m}\in W(M,T)$. Passing to the limit in (\ref{3.6}), we have$u_{m} $satisfying (\ref{3.3}) in$L^2(0,T)$, weak. On the other hand, it follows from (\ref{3.2})-\eqref{3.3}$_1$and$u_{m}\in W(M,T)$that$\ddot{u}_{m}=-b_{m}(t) Au_{m}+F_{m}\in L^{\infty}(0,T;V_0)$, hence$u_{m}\in W_1(M,T)$and the proof of Theorem \ref{theorem1} is complete. \end{proof} \begin{theorem}\label{theorem2} Let assumptions (H1)-(H3) hold. Then: \begin{itemize} \item[(i)] There exist constants$M>0$and$T>0$satisfying \eqref{3.13}-\eqref{3.14} such that problem \eqref{1.1} has a unique weak solution$u_{m}\in W_1(M,T)$. \item[(ii)] On the other hand, the linear recurrent sequence${u_{m}}$defined by \eqref{3.2}-\eqref{3.4} converges to the solution$u$of problem \eqref{1.1} strongly in the space $$W_1(T)=\big\{v\in L^{\infty}(0,T;V_1):\dot{v}\in L^{\infty}(0,T;V_0)\big\}.$$ \end{itemize} Furthermore, we have the estimate $$\|u_{m}-u\|_{L^{\infty}(0,T;V_1)}+\|\dot{u}_{m}-\dot{u}\|_{L^{\infty}(0,T;V_0)} \leq Ck_{T}^{m}\quad \forall m\geq 1,$$ where \begin{equation*} \begin{split} k_{T}=&\Big(1+\frac{1}{\sqrt{b_0C_0}}\Big)\sqrt{8M^2T\widetilde{K}_1 +\sqrt{2}T\widetilde{K}_1}\\ &\times \exp\Big[\frac{1}{\sqrt{2}}T\overline{K}_1 +\big(4+\frac{2C_1}{b_0C_0}\big) M^2T\widetilde{K}_1\Big]<1, \end{split} \end{equation*} and$C$is a constant depending only on$T, u_0, u_1$and$k_{T}$. \end{theorem} \begin{proof} \textit{Existence of the solution.} First, we note that$W_1(T)$is a Banach space with respect to the norm (see \cite{10}): \begin{equation*} \|v\|_{W_1(T)}=\|v\|_{L^{\infty}(0,T;V_1)}+\|\dot{v}\|_{L^{% \infty}(0,T;V_0)}. \end{equation*} We shall prove that$\{u_{m}\}$is a Cauchy sequence in$W_1(T)$. For this, set$v_{m}=u_{m+1}-u_{m}$. Then$v_{m}satisfies the variational problem \begin{gather*} \begin{aligned} &\langle \ddot{v}_{m}(t),w\rangle+b_{m+1}(t)a\big(v_{m}(t),w\big) +\big(b_{m+1}(t)-b_{m}(t)\big) \langle Au_{m}(t),w\rangle\\ &=\langle F_{m+1}(t)-F_{m}(t),w\rangle\quad \forall\in w\in V_1, \end{aligned} \\ v_{m}(0)=\dot{v}_{m}(0)=0. \end{gather*} Takingw=\dot{v}_{m}$herein, after integrating in$t$, we get \begin{equation*} \begin{split} X_{m}(t)=&\int_0^tb_{m+1}'(s)a\big(v_{m}(s),v_{m}(s)\big)ds \\ &-2\int_0^t\big(b_{m+1}(s)-b_{m}(s)\big)\langle Au_{m}(s),\dot{v} _{m}(s)\rangle ds \\ &+2\int_0^t\langle F_{m+1}(s)-F_{m}(s),\dot{v}_{m}(s)\rangle ds, \end{split} \end{equation*} where \begin{equation*} X_{m}(t)=\|\dot{v}_{m}(t)\|_0^2+b_{m+1}(t)a\big(v_{m}(t),v_{m}(t)\big). \end{equation*} On the other hand, from$(3.1)_{2,4}$and (\ref{3.2}) we obtain \begin{gather*} |b_{m+1}'(t)|\leq 4M^2\widetilde{K}_1, \\ \begin{split} |b_{m+1}(t)-b_{m}(t)|&\leq 2\widetilde{K}_1M\|v_{m-1}(t)\|_0+2\widetilde{K} _1M\|\nabla v_{m-1}(t)\|_0 \\ &\leq 4\widetilde{K}_1M\|v_{m-1}(t)\|_1, \end{split} \\ \|F_{m+1}(t)-F_{m}(t)\|_0\leq \sqrt{2}\overline{K}_1\|v_{m-1}(t)\|_1. \end{gather*} It follows that \begin{equation*} \begin{split} &\|\dot{v}_{m}(t)\|_0^2+b_0C_0\|v_{m}(t)\|_1^2 \\ &\leq 4M^2\widetilde{K}_1C_1\int_0^t\|v_{m}(s)\|_1^2ds+8M\widetilde{K} _1\int_0^t \|v_{m-1}(s)\|_1\|Av_{m}(s)\|_0\|\dot{v}_{m}(s)\|_0ds \\ &\quad+2\sqrt{2}\overline{K}_1\|v_{m-1}(s)\|_1\|\dot{v}_{m}(s)\|_0ds \\ &\leq 4M^2\widetilde{K}_1C_1\int_0^t\|v_{m}(s)\|_1^2ds \\ &\quad+\big(16M^2\widetilde{K}_1+2\sqrt{2}\overline{K}_1\big) \int_0^t\|v_{m-1}(s)\|_1 \|\dot{v}_{m}(s)\|_0ds \\ &\leq \big(8M^2\widetilde{K}_1+\sqrt{2}\overline{K}_1\big) \|v_{m-1}\|_{W_1(T)}^2 \\ &\quad+2\Big[\frac{1}{\sqrt{2}}\overline{K}_1+\Big(4+\frac{2C_1}{b_0C_0}\Big) M^2 \widetilde{K}_1\Big]\int_0^t\big(\|\dot{v}_{m}(s)\|_0^2+b_0C_0\|v_{m}(s) \|_1^2\big)ds. \end{split}% \end{equation*} Using Gronwall's lemma we deduce that \begin{equation*} \begin{split} &\|\dot{v}_{m}(t)\|_0^2+b_0C_0\|v_{m}(t)\|_1^2 \\ &\leq \big(8M^2\widetilde{K}_1+\sqrt{2}\overline{K}_1\big) \|v_{m-1}\|_{W_1(T)}^2 \exp\Big\{2T\Big[\frac{1}{\sqrt{2}}\overline{K}_1 + \Big(4+\frac{2C_1}{b_0C_0}\Big)M^2\widetilde{K}_1\Big]\Big\}, \end{split}% \end{equation*} for$0\leq t\leq T$. Hence \begin{equation*} \|v_{m}\|_{W_1(T)}\leq k_{T}\|v_{m-1}\|_{W_1(T)} \quad\forall m\geq 1, \end{equation*} where \begin{equation*} k_{T}=\big(1+\frac{1}{\sqrt{b_0C_0}}\big)\sqrt{8M^2T\widetilde{K}_1 +\sqrt{2} T\overline{K}_1} \exp\Big[\frac{1}{\sqrt{2}}T\overline{K}_1+\big(4+\frac{2C_1 }{b_0C_0}\big) M^2T\widetilde{K}_1\Big]<1. \end{equation*} Hence \begin{equation*} \|u_{m+p}-u_{m}\|_{W_1(T)}\leq \|u_1-u_0\|_{W_1(T)}\frac{k_{T}^{m}}{1-k_{T}} \end{equation*} for all$m$and$p$. It follows that$\{u_{m}\}$is a Cauchy sequence in$W_1(T)$. Therefore, there exists$u\in W_1(T)$such that $$\label{3.15} u_{m}\to u\quad \text{strongly in } W_1(T).$$ We also note that$u\in W_1(M,T)$. Then from the sequence$\{u_{m}\}$we can deduce a subsequence$\{u_{m_{j}}\}$such that \begin{gather*} u_{m_{j}} \; \to\; u\quad \text{in }L^{\infty}(0,T;V_2) \text{ weak$^\star$}, \\ \dot{u}_{m_{j}}\; \to \; \dot{u}\quad\text{in }L^{\infty} (0,T;V_1)\text{ weak$^\star$}, \\ \ddot{u}_{m_{j}}\; \to \;\ddot{u}\quad \text{in } L^2(0,T;V_0)\text{ weak}, \end{gather*} with$u\in W(M,T)$. Noticing \eqref{3.1}$_{1-2}$we have $$\label{3.16} \begin{split} &\Big|\int_0^T\langle b_{m}(t)Au_{m}(t)-B\big(\|u(t)\|_0^2,\|\nabla u(t)\|_0^2\big) Au(t),w(t)\rangle dt\Big| \\ &\leq C_1\widetilde{K}_0\|u_{m}-u\|_{L^{\infty}(0,T;V_1)}\|w\|_{L^1(0,T;V_1)} \\ &\quad+4C_1M\widetilde{K}_1\|u_{m-1}-u\|_{L^{\infty}(0,T;V_1)}\|u\|_{L^{ \infty}(0,T;V_1)} \|w\|_{L^1(0,T;V_1)} \end{split}%$$ for all$w\in L^1(0,T;V_1)$. It follows from (\ref{3.15})-(\ref{3.16}) that $$\label{3.17} b_{m}(t)Au_{m}\to B\big(\|u(t)\|_0^2,\|\nabla u(t)\|_0^2\big)Au \quad \text{in } L^{\infty}(0,T;V_1') \text{ weak^\star}.$$ Similarly $$\label{3.18} \|F_{m}-f(r,t,u,u_{r})\|_{L^{\infty}(0,T;V_0)}\leq \sqrt{2}\overline{K}_1 \|u_{m-1}-u\|_{L^{\infty}(0,T;V_1)}.$$ Hence, from (\ref{3.15}) and (\ref{3.18}), we obtain $$\label{3.19} F_{m} \to f(r,t,u,u_{r}) \quad\text{strongly in } L^{\infty}(0,T;V_0).$$ Then, taking limits in (\ref{3.3}) with$m=m_{j}\to +\infty$, there exists$u\in W(M,T)$satisfying $$\label{3.20} \begin{gathered} \langle \ddot{u}(t),w\rangle +B\big(\|u(t)\|_0^2,\|\nabla u(t)\|_0^2\big) a\big(u(t),w\big)=\langle f(r,t,u,u_{r}),w\rangle\quad w\in V_1,\\ u(0)=\widetilde{u}_0,\quad \dot{u}(0)=\widetilde{u}_1. \end{gathered}$$ On the other hand, from (\ref{3.17}) and (\ref{3.19})-(\ref{3.20}) we have \begin{equation*} \ddot{u} =-B\big(\|u\|_0^2,\|\nabla u\|_0^2\big)Au+f(r,t,u,u_{r})\in L^{\infty}(0,T;V_0). \end{equation*} Hence,$u\in W_1(M,T)$and the proof of existence complete. \noindent \textit{Uniqueness of the solution.} Let$u_1, u_2$, be weak solutions of problem \eqref{1.1}$_{1-3}$such that$u_1$and$u_2$are in$W_1(M,T)$. Then$w=u_1-u_2$satisfies the variational problem \begin{gather*} \langle \ddot{w}(t),v\rangle +\widetilde{b}_1(t)a\big(w(t),v\big) +\big(\widetilde{b}_1(t)-\widetilde{b}_2(t)\big) \langle Au_2(t),v\rangle =\langle \widetilde{f}_1(t)-\widetilde{f}_2(t),v\rangle \;\forall v\in V_1, \\ w(0)=\dot{w}(0)=0, \end{gather*} where \begin{equation*} \widetilde{b}_{i}(t)=B\big(\|u_{i}(t)\|_0^2,\|\nabla u_{i}(t)\|_0^2\big), \widetilde{f}_{i}(t)=f(r,t,u_{i},\nabla u_{i}),\quad i=1,2\,. \end{equation*} Taking$v=\dot{w}and integrating by parts, we obtain \begin{align*} \|\dot{w}(t)\|_0^2+\widetilde{b}_1(t)a\big(w(t),w(t)\big) &=\int_0^t \widetilde{b}_1'(s) a\big(w(s),w(s)\big)ds \\ &\quad -2\int_0^t\big(\widetilde{b}_1(s)-\widetilde{b}_2(s)\big)\langle Au_2(s), \dot{w}(s)\rangle ds \\ &\quad +2\int_0^t\langle \widetilde{f}_1(s)-\widetilde{f}_2(s),\dot{w} (s)\rangle ds. \end{align*} Put \begin{equation*} X(t)=\|\dot{w}(t)\|_0^2+b_0C_0\|w(t)\|_1^2. \end{equation*} Then \begin{equation*} X(t)=\frac{1}{\sqrt{b_0C_0}}\Big[4\Big(1+\frac{C_1}{\sqrt{b_0C_0}}\Big)M^2 \widetilde{K}_1 +\sqrt{2}\overline{K}_1\Big]\int_0^tX(s)ds \end{equation*} for allt\in [0,T]$follows. Using Gronwall's lemma we deduce$X(t)=0$, i.e.,$u_1=u_2$and the proof of Theorem \ref{theorem2} is complete. \end{proof} \begin{remark}\label{remark3} \rm In the case of$B\equiv 1$and$f=f(t,u,u_{t})$with$f\in C^1(\mathbb{R}_+\times\mathbb{R}^2)$and$f(t,0,0)=0$for all$t\geq 0$, and with the homogeneous Dirichlet boundary condition instead of$(1.1)_2$, some results have been obtained in \cite{4}. In the case of$f$being in$C^1(\overline{\Omega}\times\mathbb{R}_+\times\mathbb{R}^2)$and$B\equiv 1$we have previously obtained some results in \cite{2}. We emphasize here that in the above, however, we do not need to assume that$f$is in$C^1(\overline{\Omega}\times\mathbb{R}_+\times \mathbb{R}^2)$. \end{remark} \section{A special case} In this section, we consider initial boundary value problem (\ref{1.1}) with an autonomous right-hand side independent of$u_{r}$and an affine coefficient function$B$. Under these assumptions, we obtain stronger conclusion on the approach results in a quadratic convergence of the approximation (Theorem \ref{theorem4}). We make the following assumptions: \begin{itemize} \item[(H4)]$B(\eta)=b_0+\eta$with$b_0>0$a given constant. \item[(H5)]$f\in C^2(\overline{\Omega}\times\mathbb{R})$. \end{itemize} With$f$satisfying assumption (H5), for any$M>0$we put \begin{gather*} \overline{K}_0=\overline{K}_0(M,f)=\sup_{(r,u)\in \overline{A}_*}|f(r,u)|, \\ \overline{K}_1=\overline{K}_1(M,f)=\sup_{(r,u)\in \overline{A}_*} \Big(\Big| \frac{\partial f}{\partial r}\Big|+\Big|\frac{\partial f}{\partial u} \Big| \Big)(r,u), \\ \overline{K}_2=\overline{K}_2(M,f)=\sup_{(r,u)\in \overline{A}_*} \Big(\Big| \frac{\partial^2 f}{\partial r\partial u}\Big|+\Big|\frac{\partial^2 f} { \partial u^2}\Big|\Big)(r,u), \end{gather*} where \begin{equation*} \overline{A}_{*}=\overline{A}_{*}(M)=\big\{(r,u):0\leq r\leq 1, |u|\leq M \sqrt{2+1/\sqrt{2}}\big\}. \end{equation*} We shall choose as a (constant in time) starting point$u_0$the initial data$\widetilde{u}_0$. Assume$u_{m-1}\in W_1(M,T)$and consider the variational problem (\ref{3.3}), where $$\label{4.1} \begin{gathered} b_{m}(t)=b_0+\|\nabla u_{m}(t)\|_0^2,\\ F_{m}(r,t)=f_{m}(r,t,u_{m})=f(r,u_{m-1})+(u_{m}-u_{m-1})\frac{\partial f}{\partial u} (r,u_{m-1}), \end{gathered}$$ with \begin{equation*} f_{m}(r,t,u)=f(r,u_{m-1})+(u-u_{m-1})\frac{\partial f}{\partial u} (r,u_{m-1}). \end{equation*} Then we have the following theorem. \begin{theorem}\label{theorem3} Let (H1), (H4), and (H5) hold. Then there exist constants$M>0$and$T>0$and the recurrent sequence$\{u_{m}\}\in W_1(M,T)$defined by (\ref{3.3}) and (\ref{4.1}). \end{theorem} \begin{proof} The idea is the same as in the proof of Theorem \ref{theorem1}. As there we define$u_{m}^{(k)}$by (\ref{3.5})-(\ref{3.7}), where the functions$b_{m}$and$F_{m}$appearing in (\ref{3.5}) are replaced by \begin{gather*} b_{m}^{(k)}(t)=b_0+\|\nabla u_{m}^{(k)}(t)\|_0^2, \\ F_{m}^{(k)}(r,t)=f_{m}\big(r,t,u_{m}^{(k)}\big)=f(r,u_{m-1})+\big( u_{m}^{(k)}-u_{m-1}\big) \frac{\partial f}{\partial u}(r,u_{m-1}), \end{gather*} respectively. With$S_{m}^{(k)}, X_{m}^{(k)}, Y_{m}^{(k)}$defined by (\ref{3.8})-(\ref{3.9}), where the function$b_{m}$appearing in$X_{m}^{(k)}$and$Y_{m}^{(k)}$are replaced by$b_{m}^{(k)}, it follows that \begin{align*} S_{m}^{(k)}(t)=&S_{m}^{(k)}(0)+ \int_0^t {b_m^{(k)}}'(s)\Big[a\big( u_{m}^{(k)}(s), u_{m}^{(k)}(s)\big)+\|Au_{m}^{(k)}(s)\|_0^2\Big]ds \\ & +2\int_0^t\big\langle F_{m}^{(k)}(s),\dot{u}_{m}^{(k)}(s)\big\rangle ds +2\int_0^t a\big(F_{m}^{(k)}(s), \dot{u}_{m}^{(k)}(s)\big)ds \\ &- \int_0^t b_{m}^{(k)}(s)\big\langle Au_{m}^{(k)}(s),\ddot{u}_{m}^{(k)}(s) \big\rangle ds +\int_0^t\big\langle F_{m}^{(k)}(s),\ddot{u}_{m}^{(k)}(s) \big\rangle ds. \end{align*} We can estimateS_{m}^{(k)}in a manner similar to (\ref{3.11}) as \begin{equation*} S_{m}^{(k)}(t)\leq 3S_{m}^{(k)}(0)+\widetilde{D}_0(M,T)+D_1(M) \int_0^tS_{m}^{(k)}(s)ds +D_2\int_0^t\big(S_{m}^{(k)}(s)\big)^2ds, \end{equation*} where \begin{align*} \widetilde{D}_0 =&\widetilde{D}_0(M,T)=\frac{21}{2}(\overline{K}_0+M \overline{K}_1)^2 \\ &+6C_1T\Big(\overline{K}_0+(1+M+\sqrt{1+3M^2})\overline{K}_1+\overline{K}_0 +M\overline{K}_2\sqrt{3+3M^2/2}\Big)^2, \\ D_1=&D_1(M) \\ =&\frac{9}{4}+3(1+C_1)/C_0+21\overline{K}_1^2/2b_0C_0+\frac{6C_1}{b_0C_0} \big(4\overline{K}_1^2+(3+3M^2/2)\overline{K}_2^2), \\ D_2=&\frac{3}{b_0^2}\big(\sqrt{b_0}+\frac{3}{4C_0}\big). \end{align*} From convergence (\ref{3.7}) we can deduce the existence of a constantM>0$independent of$k$and$m$such that$S_{m}^{(k)}(0)\leq M^2/6$. Next, we can always choose a constant$T>0$, so that $$\label{4.2} \widetilde{D}_0(M,T)\leq M^2/2 \quad\text{and}\quad \Big(1+\frac{D_1}{M^2D_2} \Big) exp(TD_1)\leq 1+\frac{4D_1}{3M^2D_2}.$$ Then $$\label{4.3} S_{m}^{(k)}(t)\leq \frac{3}{4}M^2+D_1(M)\int_0^tS_{m}^{(k)}(s)ds+D_2\int_0^t \big(S_{m}^{(k)}(s)\big)^2ds.$$ On the other hand, the function $$\label{4.4} S(t)=\frac{D_1\exp(D_1t)}{\frac{4D_1}{3M^2}-D_2[\exp(D_1t)-1]}, \quad 0\leq t\leq T,$$ is the maximal solution of the Volterra integral equation with non-decreasing kernel \cite{8} \begin{equation*} S(t)=\frac{3}{4}M^2+D_1(M)\int_0^tS(s)ds+D_2\int_0^tS^2(s)ds,\quad 0\leq t\leq T. \end{equation*} From (\ref{4.2})-(\ref{4.4}) $$\label{4.5} S_{m}^{(k)}(t)\leq S(t)\leq M^2,\quad 0\leq t\leq T$$ follows for all$k$and$m$. Hence$u_{m}^{(k)}\in W_1(M,T)$for all$k$and$m$. Then, in a manner similar to the proof of Theorem \ref{theorem1}, we can prove that the limit$u_{m}\in W_1(M,T)$of the sequence$\{u_{m}^{(k)}\} $when$k\to +\infty$is the unique solution of variational problem (\ref{3.3}) and (\ref{4.1}). The proof of Theorem \ref{theorem3} is complete. \end{proof} The following result gives a quadratic convergence of the sequence$\{u_{m}\} $to a weak solution of problem (\ref{1.1}) corresponding to$f=f(r,u)$and$B(\eta)=b_0+\eta$. \begin{theorem}\label{theorem4} Let assumptions (H1) and (H4)-(H5), hold. Then \begin{itemize} \item[(i)] There exist constants$M>0$and$T>0$such that problem (\ref{1.1}) corresponding to$f=f(r,u)$and$B(\eta)=b_0+\eta$has a unique weak solution$u\in W_1(M,T)$. \item [(ii)] On the other hand, the recurrent sequence$\{u_{m}\}$defined by (\ref{3.3}) and (\ref{4.1}) converges quadratically to the solution$u$strongly in the space$W_1(T)$in the sense $$\|u_{m}-u\|_{W_1(T)}\leq C\|u_{m-1}-u\|_{W_1(T)}^2,$$ where$C$is a suitable constant. Furthermore, we have also the estimation $$\|u_{m}-u\|_{W_1(T)}\leq \frac{\beta^{2^m}}{\mu_T(1-\beta)} \quad \text{for all}\quad m,$$ where $$\mu_{T}=\Big(1+\frac{1}{\sqrt{b_0C_0}}\Big)(1+b_0C_0)\sqrt{TK_{T}^{(2)}exp(TK_{T}^{(1)})}$$ and$\beta=4M\mu_T<1$. \end{itemize} \end{theorem} Note that the last condition is always satisfied by taking a suitable$T>0$. \begin{proof} First, we shall prove that$\{u_{m}\}$is a Cauchy sequence in$W_1(T)$. For this, set$v_{m}=u_{m+1}-u_{m}$. Then$v_{m}satisfies the variational problem \label{4.6} \begin{gathered} \begin{aligned} &\langle \ddot{v}_{m}(t),w\rangle +b_{m+1}(t)a\big(v_{m}(t),w\big) +\big(b_{m+1}(t)-b_{m}(t)\big)\langle Au_{m}(t),w\rangle\\ &=\langle F_{m+1}(t)-F_{m}(t), w\rangle,\quad \forall w\in V_1, \end{aligned}\\ v_{m}(0)=\dot{v}_{m}(0)=0, \end{gathered} with \begin{gather*} \begin{aligned} F_{m+1}-F_{m}&=f(r,u_{m})-f(r,u_{m-1}) +(u_{m+1}-u_{m})\frac{\partial f}{\partial u}(r,u_{m})\\ &\quad -(u_{m}-u_{m-1})\frac{\partial f}{\partial u}(r,u_{m-1})\\ &=v_{m}\frac{\partial f}{\partial u}(r,u_{m})+\frac{1}{2} v_{m-1}^2\frac{\partial^2 f}{\partial u^2}(r,\lambda_{m}), \end{aligned} \\ \lambda_{m}=u_{m-1}+\theta v_{m-1},\quad (0<\theta <1), \\ b_{m+1}(t)-b_{m}(t)=\|\nabla u_{m+1}(t)\|_0^2-\|\nabla u_{m}(t)\|_0^2. \end{gather*} Takingw=\dot{v}_{m}$in (\ref{4.6}) and integrating in$t$, we get \begin{equation*} \begin{split} &\|\dot{v}_{m}(t)\|_0^2+b_{m+1}(t)a\big(v_{m}(t),v_{m}(t)\big) \\ & =2\int_0^t\langle \nabla u_{m+1}(s), \nabla \dot{u}_{m+1}(s)\rangle a\big( v_{m}(s),v_{m}(s)\big)ds \\ &\quad-2\int_0^t\big(\|\nabla u_{m+1}(s)\|_0^2-\|\nabla u_{m}(s)\|_0^2\big) \langle Au_{m}(s),\dot{v}_{m}(s)\rangle ds \\ & \quad +2\int_0^t\big\langle v_{m}\frac{\partial f}{\partial u}(r,u_{m}), \dot{v}_{m}(s) \big\rangle ds +\int_0^t\big\langle v_{m-1}^2\frac{\partial^2 f}{\partial u^2} (r,\lambda_{m}), \dot{v}_{m}(s)\big\rangle ds \\ &=J_1+\dots+J_4. \end{split} \end{equation*} We can estimate herein the integrals$J_1,\dots,J_4$step by step as \begin{gather*} J_1\leq 2M^2\int_0^ta\big(v_{m}(s),v_{m}(s)\big)ds\leq 2C_1M^2\int_0^t\|v_{m}(s)\|_1^2ds, \\ J_2\leq 4M^2\int_0^t\|v_{m}(s)\|_1\|\dot{v}_{m}(s)\|_0ds,\quad \text{(by \eqref{4.5})}, \\ J_3\leq 4\overline{K}_1\int_0^t\|v_{m}(s)\|_0\|\dot{v}_{m}(s)\|_0ds \leq 4 \overline{K}_1\int_0^t\|v_{m}(s)\|_1\|\dot{v}_{m}(s)\|_0ds, \\ J_4\leq 4\overline{K}_2\int_0^t\langle v_{m-1}^2(s),|\dot{v}_{m}(s)|\rangle ds \leq \overline{K}_2\sqrt{2}(1+K_1^2)\int_0^t\|v_{m-1}(s)\|_1^2 \|\dot{v} _{m}(s)\|_0 ds, \end{gather*} where the last inequality follows from \eqref{4.5} and Lemma \ref{lemma7}. Combining the above estimates, we obtain \begin{equation*} \begin{split} &\|\dot{v}_{m}(t)\|_0^2+b_0C_0\|v_{m}(t)\|_1^2 \\ &\leq 2C_1M^2 \int_0^t\|v_{m}(s)\|_1^2ds +2(2M^2+\overline{K} _1)\int_0^t\|v_{m}(s)\|_1\|\dot{v}_{m}(s)\|_0ds \\ &\quad+\overline{K}_2\sqrt{2}(1+K_1^2)\int_0^t\|v_{m-1}(s)\|_1^2\|\dot{v} _{m}(s)\|_0ds. \end{split}% \end{equation*} Letting$Z_{m}(t)=\|\dot{v}_{m}(s)\|_0^2+b_0C_0\|v_{m}(s)\|_1^2$, the above inequality can be written as $$\label{4.7} Z_{m}(t)\leq K_{T}^{(1)}\int_0^tZ_{m}(s)ds+K_{T}^{(2)}\int_0^tZ_{m-1}^2(s)ds,$$ where \begin{equation*} K_{T}^{(1)}=\frac{2C_1M^2}{b_0C_0}+\frac{2M^2+\overline{K}_1}{\sqrt{b_0C_0}} +\frac{(1+K_1^2)\overline{K}_2}{\sqrt{2}b_0C_0}, \quad K_{T}^{(2)} =\frac{(1+K_1^2)\overline{K}_2}{\sqrt{2}b_0C_0}. \end{equation*} Hence, we deduce from (\ref{4.7}) that $$\label{4.8} \|v_{m}\|_{W_1(T)}\leq \mu_{T}\|v_{m-1}\|_{W_1(T)}^2,$$ where$\mu_{T}$is the constant \begin{equation*} \mu_{T}=\big(1+\frac{1}{\sqrt{b_0C_0}}\big) \big(1+b_0C_0\big)\sqrt{TK_{T}^{(2)}\exp(TK_{T}^{(1)})}. \end{equation*} From (\ref{4.8}), we obtain $$\label{4.9} \|u_{m}-u_{m+p}\|_{W_1(T)}\leq \frac{{\beta}^{2^m}}{\mu_T(1-\beta)}$$ for all$m$and$p$where$\beta=4M_{\mu_T}<1$. It follows that$\{u_{m}\}$is a Cauchy sequence in$W_1(T)$. Then there exists$u\in W_1(T)$such that$u_{m}\to u$strongly in$W_1(T)$. Thus, and by applying a similar argument as used in the proof of Theorem \ref{theorem2},$u\in W_1(M,T)$is the unique weak solution of problem (\ref{1.1}) corresponding to$f=f(r,u)$and$B(\eta)=b_0+\eta$. Passing to the limit as$p\to +\infty$for$m$fixed, we obtain estimate (\ref{4.5}) from (\ref{4.9}). This completes the proof of Theorem \ref{theorem4}. \end{proof} \subsection*{Acknowledgments} The author wants to thank Professor Dung Le for this help, and the anonymous referees for their valuable suggestions. \begin{thebibliography}{99} \bibitem{1} Adams, R. A.; \textit{Sobolev Spaces}, Academic Press, NewYork, 1975. \bibitem{2} Binh, D. T. T.; Dinh, A. P. N; Long, N. T.; \textit{Linear recursive schemes associated with the nonlinear wave equation involving Bessel's operator}, Math. Comp. 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