\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 67, pp. 1--22.\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-2005/67\hfil Blow up of solutions for Klein-Gordon equations] {Blow up of solutions for Klein-Gordon equations in the Reissner-Nordstr\"{o}m metric} \author[S. G. Georgiev\hfil EJDE-2005/67\hfilneg] {Svetlin G. Georgiev} \address{Svetlin Georgiev Georgiev\hfill\break University of Sofia \\ Faculty of Mathematics and Informatics \\ Department of Differential Equations, Bulgaria} \email{sgg2000bg@yahoo.com} \date{} \thanks{Submitted March 14, 2005. Published June 27, 2005.} \subjclass[2000]{35L05, 35L15} \keywords{Partial differential equation; Klein-Gordon; blow up} \begin{abstract} In this paper, we study the solutions to the Cauchy problem \begin{gather*} (u_{tt}-\Delta u)_{g_s}+m^2u=f(u),\quad t\in (0, 1], x\in \mathbb{R}^3,\\ u(1, x)=u_0\in {\dot B}^{\gamma}_{p, p}(\mathbb{R}^3),\quad u_t(1, x)=u_1\in {\dot B}^{\gamma-1}_{p, p}(\mathbb{R}^3), \end{gather*} where $g_s$ is the Reissner-Nordstr\"o m metric; $p>1$, $\gamma\in (0, 1)$, $m\ne 0$ are constants, $f\in \mathcal{C}^2(\mathbb{R}^1)$, $f(0)=0$, $2m^2 |u|\leq f^{(l)}(u)\leq 3m^2 |u|$, $l=0, 1$. More precisely we prove that the Cauchy problem has unique nontrivial solution in $\mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+))$, $$u(t,r)= \begin{cases} v(t)\omega(r)&\mbox{for }t\in (0, 1],\; r\leq r_1\\ 0&\mbox{for } t\in (0, 1],\; r\geq r_1, \end{cases}$$ where $r=|x|$, and $\lim_{t\to 0}\|u\|_{{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+)}=\infty$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} In this paper, we study properties of the solutions to the Cauchy problem \begin{gather} (u_{tt}-\Delta u)_{g_s}+m^2u=f(u),\quad t\in (0, 1], x\in \mathbb{R}^3, \label{e1}\\ u(1, x)=u_0\in {\dot B}^{\gamma}_{p, p}(\mathbb{R}^3),\quad u_t(1, x)=u_1\in {\dot B}^{\gamma-1}_{p, p}(\mathbb{R}^3), \label{e2} \end{gather} where $g_s$ is the Reissner-Nordstr\"om \cite{t1}, $$g_s={{r^2-Kr+Q^2}\over {r^2}}dt^2-{{r^2}\over {r^2-Kr+Q^2}}dr^2- r^2d\phi^2-r^2\sin^2\phi d\theta^2,$$ the constants $K$ and $Q$ are positive, $m\ne 0$, $p\in (1, \infty)$ and $\gamma\in (0, 1)$ are fixed, $f\in \mathcal{C}^2(\mathbb{R}^1)$, $f(0)=0$, $2m^2 |u|\leq f^{(l)}(u)\leq 3m^2 |u|$, $l=0, 1$. More precisely we prove that the Cauchy problem \eqref{e1}-\eqref{e2} has a unique nontrivial solution $u$ in $\mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+))$ such that $\lim_{t\to 0}\|u\|_{{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+)}=\infty$. The Cauchy problem \eqref{e1}-\eqref{e2} may rewrite in the form \begin{gather} \begin{aligned} {{r^2}\over {r^2-Kr+Q^2}}u_{tt}-{1\over {r^2}} \partial_r((r^2-Kr+Q^2)u_r)&\\ -{1\over {r^2\sin\phi}}\partial_{\phi}(\sin\phi u_{\phi})- {1\over {r^2\sin^2\phi}}u_{\theta\theta}+m^2 u&=f(u), \end{aligned} \label{e1'}\\ \begin{gathered} u(1, r, \phi, \theta)=u_0\in {\dot B}^{\gamma}_{p, p}(\mathbb{R}^+ \times [0, 2\pi]\times [0, \pi]), \\ u_t(1, r, \phi, \theta)=u_1\in {\dot B}^{\gamma-1}_{p, p} (\mathbb{R}^+\times [0, 2\pi]\times [0, \pi]), \end{gathered}\label{e2'} \end{gather} where $x=r\cos \phi \cos \theta$, $y=r\sin\phi \cos \theta$, $z=r\sin\theta$, $\phi\in [0, 2\pi]$, $\theta\in [0, \pi]$. When $g_s$ is the Riemann metric, $m=0$, $f(u)=|u|^p$; $u_0, u_1\in \mathcal{C}_0^{\infty}(\mathbb{R}^3)$ in \cite[Section 6.3]{s1} is proved that there exists $T>0$ and a unique local solution $u\in \mathcal{C}^2([0, T)\times \mathbb{R}^3)$ of \eqref{e1}-\eqref{e2} such that $$\sup_{t4Q^2,\quad {1\over {1-K+Q^2}}> 1,\quad 1-K+Q^2>0, \] with 1-K+Q^2 is small enough such that$$ {{K-\sqrt{K^2-4Q^2}}\over 2}-2\sqrt{1-K+Q^2}>0. $$Also let f\in \mathcal{C}^2(\mathbb{R}^1), f(0)=0, 2m^2 |u|\leq f^{(l)}(u)\leq 3m^2 |u|, l=0, 1. Then the Cauchy problem \eqref{e1}-\eqref{e2} has a unique nontrivial solution u(t, r)=v(t)\omega(r)\in \mathcal{C}((0,1]{\dot B}^{\gamma}_{p, p} (\mathbb{R}^+)) for which$$ \lim_{t\to 0}\|u\|_{{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+)}=\infty. $$\end{theorem} This paper is organized as follows: In section 2 we prove that the Cauchy problem \eqref{e1}-\eqref{e2} has unique nontrivial solution {\tilde u}=v(t)\omega(r)\in \mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p} (\mathbb{R}^+)). In section 3 we prove that$$ \lim_{t\to 0}\|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+)}=\infty, $$where {\tilde u} is the solution, which is received in section 2. Let$$ C=\big({{p\gamma .2^{p\gamma}}\over {2^{p\gamma}-1}}\big)^{1/p}. $$Let A>0, Q>0, B>0, K>0, 1<\beta<\alpha be constants for which \begin{itemize} \item[(H1)] {8\over {1-K+Q^2}}\Bigl({8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A}}+4m^2\Bigr) \leq 1, {{\alpha A}\over m}>1 \item[(H2)] ${1\over{1-\alpha K+\alpha^2Q^2}} \Bigl({1\over {1-\alpha K+\alpha^2Q^2}}{{m^2}\over {\alpha^4 A^2}}- 2{{m^2}}r_1^2\Bigr) \Bigl(r_1-{1\over \beta}\Bigr)^2\geq 1$ and {{m^2}\over {\alpha^4(1-\alpha K+\alpha^2Q^2)A^2}}-2m^2r_1^2\geq 0, \item[(H3)] $C\Bigl({{2^{p(1-\gamma)}}\over {p(1-\gamma)}}\Bigr)^{1/p} {8\over {1-K+Q^2}}\Bigl({8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+m^2+{{6m^2}\over {AB}}\Bigr) < 1$ \item[(H4)] {1\over {\alpha^2}}{1\over {1-\alpha K+\alpha^2Q^2}} {{m^2}\over {\alpha^2 A^4}}- {1\over {\beta^2}}{{m^2}\over {A^2}}>0 \item[(H5)] \left( {{2^{p(1-\gamma)}}\over {p(1-\gamma)}}\right)^{1/p} {{16C}\over {1-K+Q^2}} \bigl({8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2A^2}}+4m^2\bigr)<1 \item[(H6)] K^2>4Q^2, A\geq {8\over {1-K+Q^2}}> 1, {6\over {AB}}<1, 1>{{2Q^2}\over K}>{{K-\sqrt{K^2-4Q^2}}\over 2}, 1-K+Q^2>0 is small enough such that \begin{gather*} 1>{{K-\sqrt{K^2-4Q^2}}\over 2}-3\sqrt{1-K+Q^2}>0,\\ {2\over {K-\sqrt{K^2-4Q^2}-2\sqrt{1-K+Q^2}}}\leq\beta<\alpha\leq 3, \end{gather*} \end{itemize} where$$ r_1={{K-\sqrt{K^2-4Q^2}}\over 2}-{{\sqrt 2}\over 4}\sqrt{1-K+Q^2}. $$\subsection*{Example} Let \begin{gather*} A={1\over {\epsilon^4}},\quad B={1\over {\epsilon}},\quad p={3\over 2}, \quad \gamma={1\over 3},\quad \alpha=3,\\ {1\over \beta}={{K-\sqrt{K^2-4Q^2}}\over 2}-{3\over 2}\sqrt{1-K+Q^2},\quad K={4\over 3}+{1\over 6}\epsilon^{20}-{3\over 2}\epsilon^2,\\ Q^2={1\over 3}+{1\over {6}}\epsilon^{20}-{1\over 2}\epsilon^2,\quad m^2=\epsilon^4, \end{gather*} where 0<\epsilon<<1 is enough small such that (H1)-(H6) hold. Then \begin{gather*} 1-\alpha K+\alpha^2 Q^2=1-3K+9Q^2=\epsilon^{20},\\ 1-K+Q^2=\epsilon^2. \end{gather*} \begin{remark} \label{rmk1} \rm Let \epsilon^2=1-K+Q^2. Note that from (H6) we have g(r)=r^2-Kr+Q^2>0 for r\in [0, r_1], g(r) is decrease function for r\in [0, r_1]. Also (for r\in [0, r_1]) we have$$ {{r^2}\over {r^2-Kr+Q^2}}\leq {1\over {1-K+Q^2}}. $$In deed, letting {\tilde r}={{K-\sqrt{K^2-4Q^2}}\over 2}, we have r_1={\tilde r}-{{\sqrt 2}\over 4}\epsilon. Note that function$$ {{r^2}\over {r^2-Kr+Q^2}} $$is increasing for r\in [0, r_1]. Therefore,$$ {{r^2}\over {r^2-Kr+Q^2}}\leq {{r_1^2}\over {r_1^2-Kr_1+Q^2}}\leq {8\over {\epsilon^2}}={8\over {1-K+Q^2}}. $$Note that the function$$ {1\over {r^2-Kr+Q^2}} $$is increasing for r\in [0, r_1]. Therefore, for r\in [0, r_1],$$ {1\over {r^2-Kr+Q^2}}\leq {8\over {1-K+Q^2}}. $$\end{remark} Here we will use the following definition of the {\dot B}^{\gamma}_{p, p}(M)-norm (\gamma\in (0, 1), p>1) (see \cite[p.94, def. 2]{r1}, \cite{s1})$$ \|u\|_{{\dot B}^{\gamma}_{p, p}(M)}=\Big( \int_0^2h^{-1-p\gamma}\|\Delta_h u\|_{L^p(M)}^pdh\Big)^{1/p}, $$where \Delta_h u=u(x+h)-u(x). \begin{lemma} \label{lem1} Let u(x)\in \mathcal{C}^2([0, r_1]), u(x)=0 for x\geq r_1, 01 we have$$ C\|u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}\geq \|u\|_{L^p([0, r_1])}. \end{lemma} \begin{proof} We have \begin{align*} \|u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}^p &=\int_0^2 h^{-1-p\gamma}\|\Delta_h u\|_{L^p([0, r_1])}^p dh\\ &=\int_0^2 h^{-1-p\gamma}\|u(x+h)-u(x)\|_{L^p([0, r_1])}^p dh\\ &\geq \int_1^2 h^{-1-p\gamma}\|u(x+h)-u(x)\|_{L^p([0, r_1])}^p dh\\ &=\int_1^2 h^{-1-p\gamma}\|u(x)\|_{L^p([0, r_1])}^p dh\\ &=\|u(x)\|_{L^p([0, r_1])}^p\int_1^2 h^{-1-p\gamma} dh\\ &=\|u(x)\|_{L^p([0, r_1])}^p {{2^{p\gamma}-1}\over {p\gamma 2^{p\gamma}}}; \end{align*} i.e., \|u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}^p\geq {{2^{p\gamma}-1}\over {p\gamma 2^{p\gamma}}}\|u(x)\|_{L^p([0, r_1])}^p. $$From this estimate, we have C\|u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}\geq \|u(x)\|_{L^p([0, r_1])} which completes the proof. \end{proof} \section{Existence of local solutions to the Cauchy problem \eqref{e1}-\eqref{e2}} Here and below we suppose that the positive constants A, K, Q, B, 1<\beta< \alpha satisfy (H1)-(H6). Let t\in (0, 1]. Let v(t) be function which satisfies the hypotheses: \begin{itemize} \item[(H7)] v(t)\in \mathcal{C}^3[0, \infty), v(t)>0 for all t\in [0, 1] \item[(H8)] v''(t)>0 for all t\in [0, 1], v'(1)=v'''(1)=0, v(1)\ne 0 \item[(H9)] \begin{gather*} \min_{t\in [0, 1]}v(t)\geq {1\over A},\quad \max_{t\in [0, 1]}v(t)\leq {{2}\over {A}},\\ \min_{t\in [0, 1]}{{v''(t)}\over {v(t)}}\geq {{m^2}\over {\alpha^2 A^2}},\quad \max_{t\in [0, 1]}{{v''(t)}\over {v(t)}}\leq {{2m^2}\over {\alpha^2 A^2}};\\ \lim_{t\to 0}[v''(t)-{{m^2}\over {\alpha^2 A^2}}v(t)]=+0,\quad v''(t)-{{m^2}\over {\alpha^2 A^2}}v(t)\geq 0\quad \mbox{for }t\in [0, 1] . \end{gather*} \end{itemize} Note that there exist a functions v(t) for which (H7)-(H9) hold. For example consider the function $$v(t)={{(t-1)^2+{{2\alpha^2 A^2}\over {m^2}}-1}\over {A^3{{\alpha^2}\over {m^2}}}}. \label{e3}$$ Then v(t)\in \mathcal{C}^3[0, \infty); v(t)> 0 for all t\in [0, 1] because (H1), we have {{\alpha A}\over m}>1; i.e., (H7) holds. Since \begin{gather*} v'(t)={{2(t-1)}\over {A^3{{\alpha^2}\over {m^2}}}},\quad v'(1)=0,\\ v''(t)={{2}\over {A^3{{\alpha^2}\over {m^2}}}}\geq 0\quad \forall\, t\in [0, 1],\\ v'''(t)=0,\quad v'''(1)=0, \end{gather*} it follows (H8). On the other hand$$ \min_{t\in [0, 1]}v(t)\geq {1\over A}, \quad \max_{t\in [0, 1]}v(t)\leq {{2}\over {A}},\quad {{v''(t)}\over {v(t)}}={{2}\over{(t-1)^2+{{2\alpha^2 A^2}\over {m^2}}-1}}, $$which implies \begin{gather*} \min_{t\in [0, 1]}{{v''(t)}\over {v(t)}}\geq {{m^2}\over {\alpha^2 A^2}},\quad \max_{t\in [0, 1]}{{v''(t)}\over {v(t)}}\leq {{2m^2}\over {\alpha^2A^2}},\\ v''(t)-{{m^2}\over {\alpha^2 A^2}}v(t)={{m^4}\over {\alpha^4 A^5}}(2-t)t,\quad \lim_{t\to 0}[v''(t)-{{m^2}\over {\alpha^2 A^2}}v(t)]=+0; \end{gather*} i.e., (H9) holds. Here and below we suppose that v(t) is a fixed function satisfying (H7)-(H9). In this section we will prove that the Cauchy problem \eqref{e1}-\eqref{e2} has unique nontrivial solution of the form$$ u(t, r)=\begin{cases} v(t)\omega(r)&\mbox{for } r\leq r_1,\\ 0 &\mbox{for } r\geq r_1, \end{cases} with t in (0, 1] and u\in \mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+)). Let us consider the integral equation $$u(t, r)=\begin{cases} \int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} u(t, s)\\ +s^2 m^2u(t, s)-f(u(t, s))s^2\Bigr)ds\,d\tau, &\mbox{for } 0\leq r\leq r_1,\\ 0&\mbox{for} r\geq r_1, \end{cases} \label{e*}$$ where u(t, r)=v(t)\omega(r) and t\in (0, 1]. \begin{theorem} \label{thm1} Let p\in (1, \infty), m\ne 0 and \gamma \in (0, 1) be fixed constants and the positive constants A, B, Q,K, \alpha>\beta>1 satisfy (H1)--(H6) and f\in \mathcal{C}^2(\mathbb{R}^1), f(0)=0, 2m^2|u|\leq f^{(l)}(u)\leq 3m^2 |u|, l= 0, 1. Let also v(t) is function for which (H7)--(H9) hold. Then the equation \eqref{e*} has unique nontrivial solution u(t, r)=v(t)\omega(r) for which w\in \mathcal{C}^2[0, r_1], u(t,r_1)=u_r(t, r_1)=u_{rr}(t, r_1)=0 for t\in (0, 1], u(t, r)\in \mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}[0, r_1]), for r\in [{1\over\alpha}, {1\over\beta}] and t\in (0, 1] u(t, r)\geq {1\over {A^2}}, for r\in [{1\over\alpha}, r_1] and t\in (0, 1] u(t, r)\geq 0, for r\in [0, r_1] and t\in (0, 1] |u(t, r)|\leq {2\over {AB}}, u(t, r)=0 for r\geq r_1, t\in (0, 1]. \end{theorem} \begin{proof} Let N=\{u(t, r)\in \mathcal{C}( [0, r_1]) : t\in (0, 1]\} with u(t, r)=u_r(t,r)=u_{rr}(t,r)=0 for t\in (0, 1], r\geq r_1, u(t, r)\in \mathcal{C}((0, 1] {\dot B}^{\gamma}_{p, p}[0, r_1]). For r\in [{1\over\alpha}, {1\over\beta}] and t\in (0, 1], we have u(t, r)\geq {1\over {A^2}}. For r\in [0,r_1] and t\in (0, 1], we have |u(t, r)|\leq {2\over {AB}}. For r\in[{1\over\alpha}, r_1] and t\in (0, 1], we have u(t, r)\geq 0\}. We remark that if u\in N is a solution of \eqref{e*}, u\in \mathcal{C}^2([0, r_1]). We define the operator R as follows \begin{align*} R(u) &=\int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} u(t, s)\\ &\quad +s^2m^2u(t, s)-s^2f(u)\Bigr)ds\,d\tau, \end{align*} for  0\leq r\leq r_1 and t\in (0, 1]. First we show that R:N\to N. For each u\in N, we have the following five statements: \\ (1) Since u\in \mathcal{C}([0, r_1]) and f\in \mathcal{C}^2(\mathbb{R}^1), from the definition of the operator R we have R(u)\in \mathcal{C}^2([0, r_1]), R(u)\vert_{r=r_1}=0, ${{\partial}\over {\partial r}}R(u)= {1\over {r^2-Kr+Q^2}}\int_{r_1}^r[ {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}u+s^2(m^2 u-f(u))]ds ,$ ${{\partial}\over {\partial r}}R(u)\big|_{r=r_1}=0,$ \begin{align*} {{\partial^2}\over {\partial r^2}}R(u) &= {{K-2r}\over {(r^2-Kr+Q^2)^2}}\int_{r_1}^r[ {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}u+s^2(m^2 u-f(u))]ds \\ &\quad +{{r^4}\over {(r^2-Kr+Q^2)^2}}{{v''(t)}\over {v(t)}}u(t, r)+ {{r^2}\over {r^2-Kr+Q^2}}(m^2 u(t, r)-f(u)). \end{align*} Since u(t, r_1)=0, f(u(t, r_1))=f(0)=0 we obtain {{\partial^2}\over {\partial r^2}}R(u)\big|_{r=r_1}=0. Note that R(u)=0 for r\geq r_1, t\in (0, 1] because u(t, r)=0 for r\geq r_1, t\in (0, 1] and f(u(t, r))=f(0)=0 for r\geq r_1, t\in (0, 1]. \noindent(2) For r\in [0, r_1], t\in (0, 1] we have |u(t, r)|\leq {2\over {AB}}. Then \begin{align*} &|R(u)|\\ &=\Bigl|\int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}u+s^2(m^2u-f(u))\Bigr) dsd\tau\Bigr|\\ &\leq \int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}|u|+s^2(m^2|u|+|f(u)|)\Bigr)ds d\tau\,. \end{align*} Since |f(u)|\leq 3 m^2 |u|), the above quantity is lees than or equal to \begin{align*} &\int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}|u|+4s^2m^2|u|\Bigr)ds d\tau\\ &=\int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+4s^2m^2\Bigr)|u|ds d\tau\\ &\leq {2\over {AB}}\int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+4s^2m^2\Bigr)ds d\tau\\ \end{align*} where we use {{r^2}\over {r^2-Kr+Q^2}}\leq {8\over {1-K+Q^2}}, {1\over {r^2-Kr+Q^2}}\leq {8\over {1-K+Q^2}} for r\in [0, r_1]. The above estimate is also less than or equal to \begin{align*} &{2\over {AB}}{8\over {1-K+Q^2}}\Bigl({8\over {1-K+Q^2}}\max_{t\in [0, 1]} {{v''(t)}\over {v(t)}}+4m^2\Bigr)\\ &= {2\over {AB}}{8\over {1-K+Q^2}}\Bigl({8\over {1-K+Q^2}} {{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr)\\ &\leq {2\over {AB}}. \end{align*} In the above inequality we use (H1). Consequently, |R(u)|\leq {2\over {AB}} \quad \text{for } r\in [0, r_1], t\in (0, 1]. (3)For r\in[{1\over\alpha}, r_1] and t\in (0, 1] we have u(t, r)\geq 0. Then \begin{align*} R(u) &= \int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}u(t, s)\\ &\quad +s^2m^2u(t, s)-s^2f(u)\Bigr)ds\,d\tau \end{align*} (where we use f(u)\leq 3m^2 u for r\in [{1\over\alpha}, r_1], t\in (0, 1]. The above quantity is greater than or equal to \begin{align*} &\int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} u(t, s)+s^2(m^2u(t, s)-3m^2u)\Bigr)ds\,d\tau \\ & \geq \int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} -2m^2s^2\Bigr)u(t, s)ds\,d\tau \\ &\geq \int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({1\over {\alpha^2(1-\alpha K+\alpha^2 Q^2)}} \min_{t\in [0, 1]}{{v''(t)}\over {v(t)}} -2m^2r_1^2\Bigr)u(t, s)dsd\tau \\ &= \int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{m^2}\over {\alpha^4(1-\alpha K+\alpha^2 Q^2)A^2}} -2m^2r_1^2\Bigr)u(t, s)ds\,d\tau \,. \end{align*} From (H2), we have {{m^2}\over {\alpha^4(1-\alpha K+\alpha^2 Q^2)A^2}} -2m^2r_1^2\geq 0. $$From this inequality and from u(t, r)\geq 0 for r\in[{1\over\alpha}, r_1], t\in (0, 1], r^2-Kr+Q^2>0, for r\in [0, r_1], we get$$ R(u)\geq 0\quad for\quad r\in[{1\over\alpha}, r_1],\quad t\in (0, 1]. (4) For r\in [{1\over\alpha}, {1\over \beta}] and t\in (0, 1] we have that u(t, r)\geq {1\over {A^2}}. Using f(u)\leq 3m^2 u for r\in [{1\over\alpha}, {1\over\beta}], t\in (0, 1], we have \begin{align*} R(u)&\geq \int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}u-2s^2m^2u\Bigr) dsd\tau\\ &\geq \int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} \Bigl({{s^4}\over {s^2-Ks+Q^2}} \min_{t\in [0, 1]}{{v''(t)}\over {v(t)}}u-2s^2m^2u\Bigr)ds\,d\tau\\ &= \int_{r_1}^r{1\over {\tau^2-K\tau +Q^2}}\int_{r_1}^{\tau} s^2\Bigl({{s^2}\over {s^2-Ks+Q^2}} \min_{t\in [0, 1]}{{v''(t)}\over {v(t)}}u-2m^2u\Bigr)ds\,d\tau\\ & \geq \int_{r}^{r_1}{1\over {\tau^2-K\tau +Q^2}}\int_{1\over\alpha}^{1\over\beta} s^2\Bigl({{s^2}\over {s^2-Ks+Q^2}} \min_{t\in [0, 1]}{{v''(t)}\over {v(t)}}u-2m^2u\Bigr)ds\,d\tau\\ &\geq \int_{1\over\beta}^{r_1}{1\over {\tau^2-K\tau +Q^2}} \int_{1\over\alpha}^{1\over\beta} s^2\Bigl({{s^2}\over {s^2-Ks+Q^2}} \min_{t\in [0, 1]}{{v''(t)}\over {v(t)}}u-2m^2u\Bigr)ds\,d\tau\\ &\geq {1\over {A^2}}\Bigl({1\over {1-\alpha K+\alpha^2 Q^2}}{{m^2}\over {\alpha^4 A^2}}- 2{{m^2}r_1^2}\Bigr)\Bigl(r_1-{1\over \beta}\Bigr)^2 {1\over {1-\alpha K+\alpha^2Q^2}}\geq {1\over {A^2}}, \end{align*} (see (H2)); i.e., for r\in [{1\over\alpha}, {1\over\beta}] and t\in (0, 1] we have R(u)\geq {1\over {A^2}}. \noindent(5) We have the estimate \begin{align*} \|\Delta_h R(u)\|_{L^p}^p &= \int_0^{r_1}\Bigl(\Bigl|\int_{r}^{r+h} {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1}\Bigl( {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} u(t, s)\\ &\quad +s^2(m^2u-f(u))\Bigr)ds\,d\tau \Bigr|\Bigr)^p dr\\ &\leq \int_0^{r_1}\Bigl(\int_{r}^{r+h} {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1}\Bigl( {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}|u|\\ &\quad + 4 m^2|u(t, s)| s^2\Bigr)ds,d\tau \Bigr)^pdr \end{align*} where we use that for s\in [0, r_1], {{s^4}\over {s^2-Ks+Q^2}}\leq {8\over {1-K+Q^2}}, {1\over {s^2-Ks+Q^2}}\leq {8\over {1-K+Q^2}} and u(t, r)=0 for r\geq r_1 and t\in (0, 1]. By (H9) the above estimate is less than or equal to \begin{align*} &\int_0^{r_1}\Bigl(\int_{r}^{r+h} {8\over {1-K+Q^2}}\int_{\tau}^{r_1}\Bigl( {{8}\over {1-K+Q^2}}\max_{t\in [0, 1]}{{v''(t)}\over {v(t)}}|u| + 4m^2|u| \Bigr)dsd\tau \Bigr)^p dr\\ &\leq \int_0^{r_1}\Bigl(\int_{r}^{r+h} {8\over {1-K+Q^2}}\int_{\tau}^{r_1}\Bigl( {{8}\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}|u|+ 4m^2|u|\Bigr)dsd\tau \Bigr)^pdr\\ &\leq \int_0^{r_1}\Bigl(\int_{r}^{r+h}\!\! {{64}\over {(1-K+Q^2)^2}}{{2m^2}\over {\alpha^2A^2}}\int_{0}^{r_1}|u|ds +{{8}\over {1-K+Q^2 }}4m^2\int_0^{r_1}|u|ds\Bigr)d\tau \Bigr)^pdr\\ &\leq h^p\Bigl({{64}\over {(1-K+Q^2)^2}}{{2m^2}\over {\alpha^2A^2}} \|u\|_{L^p[0, r_1]}+ {8\over {1-K+Q^2}} 4m^2 \|u\|_{L^p[0, r_1]}\Bigr)^p\,; \end{align*} i.e,, \begin{align*} &\|\Delta_h R(u)\|_{L^p[0, r_1]}^p\\ &\leq h^p\Bigl({{64}\over {(1-K+Q^2)^2}}{{2m^2}\over {\alpha^2A^{2}}} \|u\|_{L^p[0, r_1]}+{8\over {1-K+Q^2}} 4m^2 \|u\|_{L^p[0, r_1]}\Bigr)^p. \end{align*} Consequently, \begin{align*} &\|R(u)\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}^p\\ &=\int_0^{2} h^{-1-p\gamma}\|\Delta_h R(u)\|_{L^p[0, r_1]}^pdh\\ &\leq \Bigl({{64}\over {(1-K+Q^2)^2}}{{2m^2}\over {\alpha^2A^2}} \|u\|_{L^p[0, r_1]}+ {{8.4m^2}\over {(1-K+Q^2)}} \|u\|_{L^p[0, r_1]}\Bigr)^p\int_0^2h^{-1+p(1-\gamma)}dh\\ &=\Bigl({{64}\over {(1-K+Q^2)^2}}{{2m^2}\over {\alpha^2A^{2}}}\|u\| _{L^p[0, r_1]}+ {{8.4m^2}\over {(1-K+Q^2)}} \|u\|_{L^p[0, r_1]}\Bigr)^p{{2^{p(1-\gamma)}}\over {p(1-\gamma)}}. \end{align*} Therefore, \begin{align*} &\|R(u)\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}\\ &\leq \Bigl({{64}\over {(1-K+Q^2)^2}}{{2m^2}\over {\alpha^2A^{2}}}\|u\| _{L^p[0, r_1]}+ {{8.4m^2}\over {(1-K+Q^2)}} \|u\|_{L^p[0, r_1]}\Bigr)\left({{2^{p(1-\gamma)}}\over {p(1-\gamma)}} \right)^{1/p}. \end{align*} From Lemma \ref{lem1}, we have \begin{align*} \|R(u)\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]} &\leq C\Bigl({{64}\over {(1-K+Q^2)^2}}{{2m^2}\over {\alpha^2A^{2}}} \|u\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}\\ &\quad +{{8.4m^2}\over {(1-K+Q^2)}} \|u\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}\Bigr) \left({{2^{p(1-\gamma)}}\over {p(1-\gamma)}}\right)^{1/p}. \end{align*} From the above inequality, if u\in {\dot B}^{\gamma}_{p, p}[0, r_1] we get R(u)\in {{\dot B}^{\gamma}_{p, p}[0, r_1]} for t\in (0, 1]. From statements (1)--(5) above, R:N\to N. \smallskip Now, let u, u_1\in N. Then \begin{align*} &\|\Delta_h (R(u)-R(u_1))\|_{L^p}^p\\ &=\int_0^{r_1}\Bigl(\Bigl|\int_{r}^{r+h} {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1}\Bigl( {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} (u(t, s)-u_1)\\ &\quad +s^2(m^2(u-u_1)-(f(u)-f(u_1)))\Bigr)ds\,d\tau \Bigr|\Bigr)^pdr. \end{align*} From the mean value theorem, |f(u)-f(u_1)|=|u-u_1\|f'(\xi)| where \xi\in (u, u_1) or \xi\in (u_1, u). Then |f(u)-f(u_1)|\leq 3m^2|\xi| |u-u_1|\leq 3m^2|u-u_1\|q|, $$where |q|=\max\{|u|, |u_1|\}. Since |u|\leq {2\over {AB}} for r\in [0, r_1], t\in (0, 1] we have$$ |f(u)-f(u_1)|\leq {{6m^2}\over {AB}}|u-u_1|. Now, we use that u(t, r)=0 for r\geq r_1 and t\in (0, 1], to obtain \newpage \begin{align*} &\|\Delta_h (R(u)-R(u_1))\|_{L^p}^p\\ &\leq \int_0^{r_1}\Bigl(\int_{r}^{r+h} {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1}\Bigl( {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} |u(t, s)-u_1|\\ &\quad +s^2m^2|u-u_1|+|f(u)-f(u_1)|\Bigr)ds\,d\tau \Bigr)^pdr\\ & \leq \int_0^{r_1}\Bigl(\int_{r}^{r+h} {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1}\Bigl( {{s^4}\over {s^2-Ks+Q^2}}\max_{t\in [0, 1]}{{v''(t)}\over {v(t)}} |u(t, s)-u_1|\\ &\quad +s^2m^2|u-u_1|+|f(u)-f(u_1)|\Bigr)ds\,d\tau\Bigr)^pdr\\ &\leq \int_0^{r_1}\Bigl(\int_{r}^{r+h} {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1}\Bigl( {{8}\over {1-K+Q^2}}\max_{t\in [0, 1]}{{v''(t)}\over {v(t)}} |u(t, s)-u_1|\\ &\quad +m^2|u-u_1|+{{6m^2}\over {AB}}|u-u_1|\Bigr)ds\,d\tau \Bigr)^p dr\\ &\leq \int_0^{r_1}\Bigl(\int_{r}^{r+h} {8\over {1-K+Q^2}}\int_{\tau}^{r_1}\Bigl( {{8}\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}} +m^2+{{6m^2}\over {AB}}\Bigr)\\ &\quad\times |u-u_1|dsd\tau \Bigr)^pdr\\ &\leq h^p\Bigl({8\over {1-K+Q^2}}\Bigr)^p\Bigl( {8\over {(1-K+Q^2)}}{{2m^2}\over {\alpha^2 A^{2}}}+m^2+ {{6m^2}\over {AB}}\Bigr)^p\|u-u_1\|_{L^p}^p; \end{align*} i.e., \begin{align*} &\|\Delta_h(R(u)-R(u_1))\|_{L^p[0, r_1]}^p\\ &\leq h^p\Bigl({8\over {1-K+Q^2}}\Bigr)^p\Bigl( {8\over {(1-K+Q^2)}}{{2m^2}\over {\alpha^2 A^{2}}}+m^2+ {{6m^2}\over {AB}}\Bigr)^p\|u-u_1\|_{L^p}^p. \end{align*} From the last inequality we get \begin{align*} \|R(u)-R(u_1)\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}^p &\leq \Bigl({8\over {1-K+Q^2}}\Bigr)^{p} \Bigl({8\over {(1-K+Q^2)}}{{2m^2}\over {\alpha^2 A^{2}}}+m^2+ {{6m^2}\over {AB}}\Bigr)^p\\ &\quad \times \|u-u_1\|_{L^p}^p \int_0^2 h^{-1+p(1-\gamma)}dh. \end{align*} From the above inequality and Lemma \ref{lem1}, \begin{align*} \|R(u)-R(u_1)\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]} &\leq \Big({{2^{p(1-\gamma)}}\over {p(1-\gamma)}}\Big)^{1/p} {8\over {1-K+Q^2}} \Bigl({8\over {(1-K+Q^2)}}{{2m^2}\over {\alpha^2 A^{2}}}\\ &\quad +m^2+{{6m^2}\over {AB}}\Bigr) \|u-u_1\|_{L^p[0, r_1]}\\ &\leq C \Big({{2^{p(1-\gamma)}}\over {(p(1-\gamma)}}\Big)^{1/p} {8\over {1-K+Q^2}} \Bigl({8\over {(1-K+Q^2)}}{{2m^2}\over {\alpha^2 A^{2}}}\\ &\quad +m^2+{{6m^2}\over {AB}}\Bigr) \|u-u_1\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}\\ &< \|u-u_1\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]} \end{align*} (see i3)). i.e., \|R(u)-R(u_1)\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}< \|u-u_1\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}. $$Consequently, the operator R:N\to N is contractive operator. \end{proof} \begin{lemma} \label{lem2} The set N is closed subset of \mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+)). \end{lemma} \begin{proof} Let t\in (0, 1] be fixed. Let \{u_n\} be a sequence of elements of the set N for which$$ \lim_{n\to \infty} \|u_n-{\tilde {\tilde u}}\|_{{\dot B}^{\gamma}_{p, p}(\mathbb{R}^+)} =0, $$where {\tilde {\tilde u}}\in {\dot B}^{\gamma}_{p, p}(\mathbb{R}^+). We have$$ \lim_{n\to \infty} \|u_n-{\tilde {\tilde u}}\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])} =0. $$We define$$ {\tilde u}=\begin{cases} {\tilde {\tilde u}} &\mbox{for } r\in [0, r_1],\\ 0& \mbox{for } r>r_1. \end{cases} $$We have$$ \lim_{n\to \infty} \|u_n-{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])} =0. $$First we note that for u\in N, R(u)  is continuous function of u and there exists R'(u) because f(u)\in \mathcal{C}^2(\mathbb{R}^1). In fact,$$ R'(u)=\int_r^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+s^2m^2-s^2f'(u)\Bigr)ds\, d\tau. From which, \begin{align*} &|R'(u)|\\ &\geq\int_r^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+s^2m^2-s^2|f'(u)|\Bigr)ds\, d\tau\\ &\geq \int_r^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+s^2m^2-3m^2s^2|u|\Bigr) ds,d\tau\\ &\geq \int_r^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+s^2m^2- {{6m^2}\over {AB}}s^2\Bigr)ds\,d\tau\\ &=\int_r^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+s^2m^2\Bigl(1- {{6}\over {AB}}\Bigr)\Bigr)ds\,d\tau\,. \end{align*} From (H6), 1> 6/(AB). Therefore, for s\in [0, r_1] we have {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}+s^2m^2\Bigl(1- {{6}\over {AB}}\Bigr)\geq 0. Then for r\in [0, r_1] we have \begin{align*} &|R'(u)|\\ &\geq \int_{1\over\alpha}^{r_1}{1\over {\tau^2-K\tau+Q^2}} \int_{1\over\alpha}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}} +s^2m^2\Bigl(1- {{6}\over {AB}}\Bigr)s^2\Bigr)ds\,d\tau\\ &\geq \Bigl(r_1-{1\over\alpha}\Bigr)^2{{m^2}\over {\alpha^2 A^2(1-\alpha K+\alpha^2 Q^2)^2}}>0. \end{align*} From this, for u\in N, there exists M:=\min_{x\in [0, r_1]}|R'(u)(x)|>0 $$because R'(u)(x) is continuous function of x\in [0, r_1]. Let$$ M_1=\max_{r\in [0, r_1]}\bigl|{{\partial}\over {\partial r}}(R'(u))(r)\bigr|. $$Now we prove that for each \epsilon>0 there exists \delta=\delta(\epsilon)>0 such that $|x-y|<\delta\quad\mbox{implies} \quad |u_m(x)-u_m(y)|<\epsilon\quad \forall m.$ We suppose that there exists {\tilde \epsilon}>0 such that for every \delta>0 there exist natural m and x, y\in [0, r_1], |x-y|<\delta for which |u_m(x)-u_m(y)|\geq {\tilde \epsilon}. We choose {\tilde {\tilde\epsilon}}>0 such that {\tilde{\tilde\epsilon}}0 such that for every u\in N$$ |R(u)(x)-R(u)(y)|<{\tilde {\tilde\epsilon}}\quad \mbox{for for all $x, y\in [0, r_1]$: $|x-y|<\delta_1$}. Then we may choose 0<\delta<\min\big\{\delta_1,{{(M{\tilde \epsilon}-{\tilde {\tilde \epsilon}}) AB}\over {2M_1}}\big\} such that there exist natural m and x_1\in [0, r_1], x_2\in [0, r_1] for which |x_1-x_2|<\delta and |u_m(x_1)-u_m(x_2)|\geq {\tilde\epsilon}. In particular $$|R(u_m)(x_1)-R(u_m)(x_2)|<{\tilde {\tilde\epsilon}}. \label{e4*}$$ Then by the mean value theorem, R(0)=0, R(u_m)(x_1)=R'(\xi)(x_1)u_m(x_1), R(u_m)(x_2)= R'(\xi)(x_2)u_m(x_2), \begin{align*} &|R(u_m)(x_1)-R(u_m)(x_2)|\\ &=|R'(\xi)(x_1)u_m(x_1)-R'(\xi)(x_2)u_m(x_2)|\\ &=|R'(\xi)(x_1)u_m(x_1)-R'(\xi)(x_1)u_m(x_2)+R'(\xi)(x_1)u_m(x_2)- R'(\xi)(x_2)u_m(x_2)|\\ & \geq |R'(\xi)(x_1)u_m(x_1)-R'(\xi)(x_1)u_m(x_2)|-|R'(\xi)(x_1) -R'(\xi)(x_2)\|u_m(x_2)|\\ &= |R'(\xi)(x_1)\|u_m(x_1)-u_m(x_2)| -\Bigr|{{\partial}\over {\partial r}}(R'(\xi))(\eta)\Bigl| |x_1-x_2\|u_m(x_2)|\\ &\geq M{\tilde \epsilon}-M_1\delta{2\over {AB}} \geq {\tilde {\tilde \epsilon}}, \end{align*} which is a contradiction to \eqref{e4*}. Consequently, for each \epsilon>0 there exists \delta=\delta(\epsilon)>0 such that $$|x-y|<\delta \quad\mbox{implies} |u_m(x)-u_m(y)|<\epsilon\quad \forall m. \label{e4'}$$ On the other hand, from the definition of the set N we have $$|u_m|\leq {2\over {AB}}\quad \forall m. \label{e4''}$$ From \eqref{e4'} and \eqref{e4''}, we conclude that the set \{u_n\} is compact subset of \mathcal{C}([0, r_1]). Then there exists subsequence \{u_{n_{k}}\} and function u\in \mathcal{C}([0, r_1]) for which: for every \epsilon>0 there exists M=M(\epsilon)>0 such that for every n_k>M we have |u_{n_{k}}(x)-u(x)|<\epsilon for every x\in [0, r_1]; u(x)=0 for x>r_1. From this and from \lim_{k\to \infty}\|u_{n_{k}}-{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}=0 we have: For every \epsilon>0 \exists M=M(\epsilon)>0 such that for every n_k>M we have \max_{x\in [0, r_1]}|u_{n_{k}}-u|<\epsilon,\quad \|u_{n_{k}}-{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<\epsilon. Then for every n_k>M we have \begin{gather*} |u-{\tilde u}|\leq |u-u_{n_k}|+|u_{n_k}-{\tilde u}|<\epsilon+ |{\tilde u}-u_{n_k}|, \\ \int_0^{r_1}|u-{\tilde u}|dx<\epsilon r_1+\int_0^{r_1} |{\tilde u}-u_{n_k}|dx, \end{gather*} Using the H\"older's inequality, $\|u-{\tilde u}\|_{L^1[0, r_1]} < \epsilon r_1+r_1^{1/q}\Bigl(\int_0^{r_1} |{\tilde u}-u_{n_k}|^pdx\Bigr)^{1/p}, \quad {1\over p}+{1\over q}=1,$ for h>0, we have $h^{-1-p\gamma}\|u-{\tilde u}\|_{L^1[0, r_1]}< h^{-1-p\gamma}\epsilon r_1+r_1^{1/q}h^{-1-p\gamma}\Bigl(\int_0^{r_1} |{\tilde u}-u_{n_k}|^pdx\Bigr)^{1/p},$ \begin{align*} &\int_1^2h^{-1-p\gamma}dh\|u-{\tilde u}\|_{L^1[0, r_1]}\\ &< \int_1^2 h^{-1-p\gamma}dh \epsilon r_1+r_1^{1/q}\int_1^2 h^{-1-p\gamma} \Bigl(\int_0^{r_1} |{\tilde u}-u_{n_k}|^pdx\Bigr)^{1/p}dh, \end{align*} Using H\"older's inequality and that for h>1 we have h^{(-1-p\gamma)p}\leq h^{-1-p\gamma}, \begin{align*} &{{1-2^{-p\gamma}}\over {p\gamma}}\|u-{\tilde u}\|_{L^1[0, r_1]}\\ &<{{1-2^{-p\gamma}}\over {p\gamma}} \epsilon r_1+r_1^{1/q}\int_1^2 h^{-1-p\gamma}\Bigl(\int_0^{r_1} |{\tilde u}-u_{n_k}|^pdx\Bigr)^{1/p}dh \\ &\leq {{1-2^{-p\gamma}}\over {p\gamma}} \epsilon r_1+r_1^{1/q} \Bigl(\int_1^2 h^{(-1-p\gamma)p}\int_0^{r_1} |{\tilde u}-u_{n_k}|^pdxdh\Bigr)^{1/p}\\ &\leq {{1-2^{-p\gamma}}\over {p\gamma}} \epsilon r_1+r_1^{1/q} \Bigl(\int_1^2 h^{-1-p\gamma}\int_0^{r_1} |{\tilde u}-u_{n_k}|^pdxdh\Bigr)^{1/p}\,. \end{align*} Using that for x> r_1, u_{n_k}(x)={\tilde u}(x)=0, the above expression equals \begin{align*} &{{1-2^{-p\gamma}}\over {p\gamma}} \epsilon r_1+r_1^{1/q} \Bigl(\int_1^2 h^{-1-p\gamma}\int_0^{r_1} |\Delta_h({\tilde u}-u_{n_k})|^pdxdh\Bigr)^{1/p}\\ &\leq {{1-2^{-p\gamma}}\over {p\gamma}} \epsilon r_1+r_1^{1/q} \Bigl(\int_0^2 h^{-1-p\gamma}\int_0^{r_1} |\Delta_h({\tilde u}-u_{n_k})|^pdxdh\Bigr)^{1/p}\\ &={{1-2^{-p\gamma}}\over {p\gamma}} \epsilon r_1+r_1^{1/q} \|{\tilde u}-u_{n_k}\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}\\ &<\epsilon({{1-2^{-p\gamma}}\over {p\gamma}} r_1+r_1^{1/q})\; \end{align*} i.e., for every \epsilon>0, {{1-2^{-p\gamma}}\over {p\gamma}}\|u-{\tilde u}\|_{L^1[0, r_1]}< \epsilon\Bigl({{1-2^{-p\gamma}}\over {p\gamma}} r_1+r_1^{1/q}\Bigr)\,. $$Consequently u={\tilde u} a.e.(almost everywhere) in [0, r_1], |u|^p=|{\tilde u}|^p a.e. in [0, r_1]. >From here |u_n-u|=|u_n-{\tilde u}| a.e., |u_n-u|^p=|u_n-{\tilde u}|^p a.e. in [0, r_1]. Since u_n(x)=u(x)=0 for x>r_1 we have |\Delta_h(u_n-u)|^p=|\Delta_h(u_n-{\tilde u})|^p, |\Delta_h u|^p=|\Delta_h {\tilde u}|^p a.e. in [0, r_1], for h>0. Therefore, u\in {\dot B}^{\gamma}_{p, p}([0, r_1]) and$$ \int_0^{r_1}|u_n-u|^pdx=\int_0^{r_1}|u_n-{\tilde u}|^p dx. Now, we show that \lim_{n\to\infty} \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}=0. Note that \begin{align*} &\|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}\\ &= \Bigl(\int_0^2 h^{-1-p\gamma}\int_0^{r_1}|\Delta_h(u_n-u)|^p dx \,dh\Bigr)^{1/p}\\ &= \Bigl(\int_0^2 h^{-1-p\gamma}\int_0^{r_1}|\Delta_h(u_n-{\tilde u})|^p dx\, dh \Bigr)^{1/p}\\ &=\|u_n-{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])} \to_{n\to\infty} 0. \end{align*} Consequently, for every sequence \{u_n\} with elements from N, which converges in {\dot B}^{\gamma}_{p, p}([0, r_1]) there exists function u\in \mathcal{C}([0, r_1]), u\in {\dot B}^{\gamma}_{p, p}([0, r_1]), for which \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}\to_{n\to\infty} 0. Below we suppose that the sequence \{u_n\} is a sequence of elements of the set N which converges in {{\dot B}^{\gamma}_{p, p}([0, r_1])}. Then there exists u\in \mathcal{C}([0, r_1]), u(x)=0 for x>r_1, u\in {\dot B}^{\gamma}_{p, p}([0, r_1]), \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}\to_{n\to\infty} 0. Now we suppose that u(r_1)\ne 0. Since u\in \mathcal{C}([0, r_1]), u_n\in \mathcal{C}([0, r_1]), u_n(r_1)=0 for every natural n, there exist \epsilon_2>0 and \Delta_1\subset [0, r_1], r_1\in \Delta_1, such that |u_n|<{{\epsilon_2}\over 2}, \quad |u|>\epsilon_2 $$for every natural n and every x\in \Delta_1. Then for every natural n and for every x\in \Delta_1,$$ |u_n(x)-u(x)|>{{\epsilon_2}\over 2}. $$Let \epsilon_3>0 be such that $$\epsilon_3<{{\epsilon_2}\over 2} {{1-2^{-p\gamma}}\over {p\gamma}}\mu(\Delta_1)r_1^{-{1\over q}}, \quad {1\over p}+{1\over q}=1, \label{e4}$$ where \mu(\Delta_1) is the measure of the set \Delta_1. There exists M>0 such that for every n>M we have \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<{\epsilon_3}. Consequently for every n>M and for every x\in \Delta_1 we have$$ |u_n(x)-u(x)|>{{\epsilon_2}\over 2},\quad \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<{\epsilon_3}. Also using the H\"older's inequality, we have \begin{align*} {{\epsilon_2}\over 2}\mu(\Delta_1) &< \int_{\Delta_1}|u_n(x)-u(x)|dx\\ &\leq \int_0^{r_1}|u_n(x)-u(x)|dx\\ &\leq \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}. \end{align*} For h>0, we have \begin{gather*} h^{-1-p\gamma}{{\epsilon_2}\over 2}\mu(\Delta_1)\leq h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}, \\ \int_1^2h^{-1-p\gamma}{{\epsilon_2}\over 2}\mu(\Delta_1)dh\leq \int_1^2 h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}dh, \\ {{1-2^{-p\gamma}}\over {p\gamma}}{{\epsilon_2}\over 2}\mu(\Delta_1)\leq \Bigl(\int_1^2 h^{(-1-p\gamma)p} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}\,. \end{gather*} Since h>1, h^{(-1-p\gamma)p}\leq h^{-1-p\gamma} and the above expression is less than or equal to \Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}\leq Now using that u_n=u=0 for x>r_1, the above expression is less than or equal to \begin{align*} &\Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &\leq \Bigl(\int_0^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &= \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}r_1^{1/q} < \epsilon_3 r_1^{1/q}. \end{align*} Therefore, {{1-2^{-p\gamma}}\over {p\gamma}}{{\epsilon_2}\over 2}\mu(\Delta_1)< \epsilon_3 r_1^{1/q}, $$which is a contradiction with \eqref{e4}. Consequently, u(r_1)=0. From this, u(t, r)=0 for r\geq r_1. Then u_r(t, r)=u_{rr}(t, r)=0 for every r\geq r_1. Now we suppose that the inequality$$ |u(t, r)|\leq {2\over {AB}} $$is not hold for every r\in [0, r_1]. Since u\in \mathcal{C}([0, r_1]) we may take \epsilon_4>0 and \Delta_2\subset [0, r_1] such that$$ |u|\geq {2\over {AB}}+{\epsilon_4}\quad \mbox{for} \quad r\in \Delta_2. $$Then for every natural n and for every r\in \Delta_2, we have$$ |u_n-u|\geq |u|-|u_n|\geq {2\over {AB}}+\epsilon_4-{2\over {AB}}=\epsilon_4. $$Let \epsilon_5>0 be such that $${{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_4\mu(\Delta_2)> \epsilon_5 r_1^{1/q}. \label{e5}$$ There exist M>0 such that for every n>M we have \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<\epsilon_5. Consequently for every n>M and for every x\in \Delta_2 we have$$ |u_n(x)-u(x)|\geq\epsilon_4,\quad \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<\epsilon_5. $$Also using the H\"older's inequality, we have$$ \epsilon_4\mu(\Delta_2)< \int_{\Delta_2}|u_n(x)-u(x)|dx \leq \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}. For h>0 we have \begin{gather*} h^{-1-p\gamma}\epsilon_4\mu(\Delta_2)\leq h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}, \\ \int_1^2h^{-1-p\gamma}\epsilon_4\mu(\Delta_2)dh\leq \int_1^2 h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}dh, \end{gather*} Using H\"older's inequality and that for h>1, h^{(-1-p\gamma)p}\leq h^{-1-p\gamma}, we have \begin{align*} &{{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_4\mu(\Delta_2)\\ &\leq \Bigl(\int_1^2 h^{(-1-p\gamma)p} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &\leq \Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}\,. \end{align*} Using that u_n=u=0 for x>r_1, the above expression is less than or equal to \begin{align*} &\Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ & \leq \Bigl(\int_0^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &=\|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}r_1^{1/q} < \epsilon_5 r_1^{1/q}. \end{align*} Therefore, {{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_4\mu(\Delta_2)< \epsilon_5 r_1^{1/q}, $$which is a contradiction with \eqref{e5}. Therefore, |u|\leq {2\over {AB}} for every r\in [0, r_1]. Now suppose that the inequality$$ |u(t, r)|\geq {1\over {A^2}} $$is not true for every r\in [{1\over\alpha}, {1\over\beta}]. Since u\in \mathcal{C}([0, r_1]) we may take \epsilon_6>0 and \Delta_3\subset [{1\over\alpha}, {1\over\beta}] such that$$ |u|\leq {1\over {A^2}}-{\epsilon_6}\quad for \quad r\in \Delta_3. $$Then for every natural n and for every r\in \Delta_3 we have$$ |u_n-u|\geq |u_n|-|u|\geq {1\over {A^2}}+\epsilon_6-{1\over {A^2}}=\epsilon_6. $$Let \epsilon_7>0 be such that $${{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_6\mu(\Delta_3)> \epsilon_7 r_1^{1/q}. \label{e6}$$ There exist M>0 such that for every n>M we have \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<\epsilon_7. Consequently, for every n>M and for every x\in \Delta_3, we have$$ |u_n(x)-u(x)|>\epsilon_6,\quad \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<\epsilon_7. $$Also using the H\"older's inequality, we have$$ \epsilon_6\mu(\Delta_3)< \int_{\Delta_3}|u_n(x)-u(x)|dx\leq \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}. For h>0 we have \begin{gather*} h^{-1-p\gamma}\epsilon_6\mu(\Delta_3)\leq h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}, \\ \int_1^2h^{-1-p\gamma}\epsilon_6\mu(\Delta_3)dh\leq \int_1^2 h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}dh, \end{gather*} Using the H\"older's inequality and that for h>1, h^{(-1-p\gamma)p}\leq h^{-1-p\gamma}, we have \begin{align*} {{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_6\mu(\Delta_3) &\leq \Bigl(\int_1^2 h^{(-1-p\gamma)p} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &\leq \Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}\,. \end{align*} Using that u_n=u=0 for x>r_1, the above expression is less than or equal to \begin{align*} &\Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &\leq \Bigl(\int_0^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &= \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}r_1^{1/q} < \epsilon_7 r_1^{1/q}. \end{align*} Therefore, {{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_6\mu(\Delta_2)< \epsilon_7 r_1^{1/q}, $$which is a contradiction with \eqref{e6}. Therefore, |u|\geq {1\over {A^2}} for every r\in [{1\over\alpha}, {1\over\beta}]. Now suppose that the inequality$$ u(t, r)\geq 0 $$is not true for every r\in [{1\over\alpha}, r_1]. Then from u\in \mathcal{C}([0, r_1]) and from u_n\geq 0 for every natural n and for every r\in [{1\over\alpha}, r_1], we may take \epsilon_8>0 and \Delta_4\subset [{1\over\alpha},r_1] such that for every natural n and for every r\in \Delta_4 we have$$ |u_n-u|\geq \epsilon_8. $$Let \epsilon_9>0 be such that $${{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_8\mu(\Delta_3)> \epsilon_9 r_1^{1/q}. \label{e7}$$ There exist M>0 such that for every n>M we have \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<\epsilon_9. Consequently, for every n>M and for every x\in \Delta_4 we have$$ |u_n(x)-u(x)|>\epsilon_8,\quad \|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}<{\epsilon_9}. Also using the H\"older's inequality, $\epsilon_8\mu(\Delta_4)< \int_{\Delta_4}|u_n(x)-u(x)|dx \leq \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}.$ For h>0 , \begin{gather*} h^{-1-p\gamma}\epsilon_8\mu(\Delta_4) \leq h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}, \\ \int_1^2h^{-1-p\gamma}\epsilon_8\mu(\Delta_4)dh\leq \int_1^2 h^{-1-p\gamma} \Bigl(\int_0^{r_1}|u_n(x)-u(x)|^pdx\Bigr)^{1/p} r_1^{1/q}dh \end{gather*} Using H\"older's inequality and that for h>1, h^{(-1-p\gamma)p}\leq h^{-1-p\gamma} we have \begin{align*} {{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_8\mu(\Delta_4) &\leq \Bigl(\int_1^2 h^{(-1-p\gamma)p} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &\leq \Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|u_n(x)-u(x)|^pdxdh\Bigr)^{1/p} r_1^{1/q}. \end{align*} Using that u_n=u=0 for x>r_1, the above expression is less than or equal to \begin{align*} &\Bigl(\int_1^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &\leq \Bigl(\int_0^2 h^{-1-p\gamma} \int_0^{r_1}|\Delta_h(u_n(x)-u(x))|^pdxdh\Bigr)^{1/p} r_1^{1/q}\\ &=\|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}r_1^{1/q} < \epsilon_9 r_1^{1/q}. \end{align*} Therefore, {{1-2^{-p\gamma}}\over {p\gamma}}\epsilon_8\mu(\Delta_4)< \epsilon_9 r_1^{1/q}, $$which is a contradiction with \eqref{e7}. Therefore, |u|\geq 0 for every r\in [{1\over\alpha}, r_1]. Consequently u\in N. Then for every sequence \{u_n\}\subset N, which converges in {\dot B}^{\gamma}_{p, p}([0, r_1]) there exists u\in N for which$$ \lim_{n\to \infty}\|u_n-u\|_{{\dot B}^{\gamma}_{p, p}([0, r_1])}=0. $$\end{proof} From lemma \ref{lem2} we have that the set N is closed subset of \mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}([0, r_1])). Since \mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}([0, r_1])) is complete metric space and R:N\to N is contractive operator the equation \eqref{e*} has unique nontrivial solution {\tilde u}\in N. From \eqref{e*} we have that {\tilde u}(r)\in \mathcal{C}^2[0, r_1] and {\tilde u}(t, r_1)= {\tilde u}_r(t, r_1)={\tilde u}_{rr}(t, r_1)=0. Let {\tilde u} is the solution from the Theorem \ref{thm1}, i.e {\tilde u} is the solution to the equation \eqref{e*}. Then {\tilde u} is solution to the Cauchy problem \eqref{e1}-\eqref{e2} with initial data$$ u_0=\begin{cases} \int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(1)}}\omega(s)\\ +s^2(m^2v(1)\omega(s)-f(v(1)\omega(s))\Bigr)ds\,d\tau &\mbox{for }r\leq r_1,\\ 0 &\mbox{for } r\geq r_1, \end{cases} $$and$$ u_1=\begin{cases} \int_{r}^{r_1}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v'''(1)}} \omega(s)\\ +s^2(m^2v'(1)\omega(s)-f'(u)v'(1)\omega(s)\Bigr)dsd\tau=0 &\mbox{for } r\leq r_1,\\ 0\mbox{for } r\geq r_1. \end{cases} $$From the proof of the Theorem \ref{thm1}, we have u_0\in {\dot B}^{\gamma}_{p, p}(\mathbb{R}^+), u_1\in {\dot B}^{\gamma-1}_{p, p}(\mathbb{R}^+), {\tilde u}\in \mathcal{C}((0, 1]{\dot B}^{\gamma}_{p, p}[0, r_1]). \section{Blow-up of solutions to the Cauchy problem \eqref{e1}-\eqref{e2}} Let v(t) be the same function as in Theorem \ref{thm1}. \begin{theorem} \label{thm2} Let m^2\ne 0, \gamma\in (0, 1), p>1 be fixed and the constants A, B, Q, K, 1<\beta<\alpha satisfy the conditions (H1)-(H6). Let f\in \mathcal{C}^2(\mathbb{R}^1), f(0)=0, 2m^2 |u|\leq f^{(l)}(u)\leq 3m^2 |u|, l=0, 1. Then the solution {\tilde u} of the Cauchy problem \eqref{e1}-\eqref{e2} satisfies$$ \lim_{t\to {0}}\|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}[0, r_1]}=\infty. \end{theorem} \begin{proof} For t\in (0,1], we have \begin{align*} &\|\Delta_h R({\tilde u})\|_{L^p}^p\\ &= \int_0^{r_1}\Bigl| \int_{r}^{r+h} \! {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\Bigr|^pdr\\ &=\int_0^{1\over \alpha}\Bigl| \int_{r}^{r+h} \! {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\Bigr|^pdr\\ &\quad +\int_{1\over\alpha}^{r_1}\Bigl| \int_{r}^{r+h} \hskip -14pt {1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\Bigr|^pdr. \end{align*} Let \begin{align*} I_1&=\int_0^{1\over\alpha}\Bigl| \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}\\ &\quad +s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\Bigr|^pdr, \\ I_2&=\int_{1\over\alpha}^{r_1}\Bigl| \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{\tau}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}\\ &\quad + s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\Bigr|^pdr. \end{align*} As in proof of Theorem \ref{thm1}, for I_1 we have the estimate I_1\leq C^p {{2^{p(1-\gamma)}}\over {p(1-\gamma)}} {{8^p}\over {(1-K+Q^2)^p}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr)^p \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}^ph^p. Using that u(t, r)=0, f(u(t, r))=0 for r\geq r_1), for I_2, we have \begin{align*} I_2&=\int_{1\over\alpha}^{r_1}\Bigl| \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\alpha}^{1\over\beta} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\\ &\quad + \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\beta}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau \Bigr|^pdr\\ &\leq\int_{1\over\alpha}^{r_1}\Bigl( \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\alpha}^{1\over\beta} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\\ &\quad + \Bigl|\int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}} \int_{1\over\beta}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}\\ &\quad + s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\Bigr| \Bigr)^pdr. \end{align*} Let \begin{gather*} I_{21}= \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\alpha}^{1\over\beta} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau, \\ I_{22}=\Bigl|\int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}} \int_{1\over\beta}^{r_1} [{{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}+ s^2(m^2{\tilde u}-f({\tilde u}))]dsd\tau\Bigr|. \end{gather*} Then I_2\leq\int_{1\over\alpha}^{r_1}\Bigl(I_{21}+I_{22}\Bigr)^pdr. For I_{21} we have the following estimate, we use that for r\in [{1\over\alpha}, {1\over\beta}] u\geq 0, f(u)\geq 2m^2u, therefore -f(u)\leq -2m^2 u, \begin{align*} I_{21} &\leq\int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\alpha}^{1\over\beta} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{\tilde u}- s^2m^2{\tilde u}\Bigr)dsd\tau\\ &= \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\alpha}^{1\over\beta} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{{A^2{\tilde u}}\over {A^2}}- s^2m^2{{A^2{\tilde u}}\over {A^2}}\Bigr)dsd\tau\\ &=\int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\alpha}^{1\over\beta} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{{1}\over {A^2}}- s^2m^2{{1}\over {A^2}}\Bigr)A^2 {\tilde u}dsd\tau. \end{align*} From (H4) we have that for s\in [{1\over\alpha}, {1\over\beta}], {{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{1\over {A^2}}- {{s^2m^2}\over {A^2}} \geq {1\over {\alpha^2}}{1\over {1-\alpha K+\alpha^2Q^2}}{{m^2}\over {\alpha^2 A^4}} -{1\over {\beta^2}}{{m^2}\over {A^2}}>0. $$On the other hand we have {\tilde u}\geq {1\over {A^2}} for every t\in (0, 1] and every r\in [{1\over\alpha}, {1\over \beta}]; therefore, A^2 {\tilde u}\geq 1 and A^2 {\tilde u}\leq A^{2p}{\tilde u}^p. Consequently$$ I_{21}\leq \int_{r}^{r+h}{1\over {\tau^2-K\tau+Q^2}}\int_{1\over\alpha}^{1\over\beta} \Bigl({{s^4}\over {s^2-Ks+Q^2}}{{v''(t)}\over {v(t)}}{{1}\over {A^2}}- s^2m^2{{1}\over {A^2}}\Bigr)A^{2p} {\tilde u}^pdsd\tau. $$From (H6),$$ {8\over {1-K+Q^2}}\leq A\leq A^2. From this inequality, we have \begin{align*} I_{21}&\leq \int_{r}^{r+h}{8\over {1-K+Q^2}}\int_{1\over\alpha}^{1\over\beta} [{{v''(t)}\over {v(t)}}-{{m^2}\over {\alpha^2 A^2}}]A^{2p} {\tilde u}^p dsd\tau\\ &\leq {8\over {1-K+Q^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \int_0^1 {\tilde u}^p ds h\\ &= {8\over {1-K+Q^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \|{\tilde u}\|^p_{L^p}h\\ &\leq {8\over {(1-K+Q^2)^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \|{\tilde u}\|^p_{L^p}h. \end{align*} Now we use Lemma \ref{lem1} to get I_{21}\leq C^p{{8^2}\over {(1-K+Q^2)^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}h. $$As in proof of Theorem \ref{thm1} for I_{22} we have$$ I_{22}\leq C {8\over {(1-K+Q^2)}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr) \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}h. Consequently, \begin{align*} I_2&\leq [C{8\over {(1-K+Q^2)}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr) \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}h\\ &\quad +C^p {{8^2}\over {(1-K+Q^2)^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}h]^p, \end{align*} \begin{align*} \|\Delta_h R({\tilde u})\|_{L^p}^p &\leq [C{8\over {(1-K+Q^2)}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr) \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}h\\ &\quad + C^p{{8^2}\over {(1-K+Q^2)^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}h]^p\\ &\quad + C^p{{8^p}\over {(1-K+Q^2)^p}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr)^p \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}h^p. \end{align*} Then \begin{align*} &\|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}^p\\ &\leq \Big[\Big[C{8\over {(1-K+Q^2)}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr) \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}\\ &\quad + C^p{{8^2}\over {(1-K+Q^2)^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}\Big]^p\\ &\quad +C^p{{8^p}\over {(1-K+Q^2)^p}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr)^p \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}\Big]\int_0^2h^{-1+p(1-\gamma)}dh\\ &={{2^{p(1-\gamma)}}\over {p(1-\gamma)}}\Big[\Big[ C{8\over {(1-K+Q^2)}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr) \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}\\ &\quad +C^p{{8^2}\over {(1-K+Q^2)^2}} {{v''(t)-{{m^2}\over {\alpha^2 A^2}}v}\over {v}}A^{2p} \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}\Big]^p\\ &\quad+ C^p{{8^p}\over {(1-K+Q^2)^p}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr)^p \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}\Big], \end{align*} and \begin{align*} \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}} &\leq 2C{{8.2^{1-\gamma}}\over {(1-K+Q^2)(p(1-\gamma))^{1/p}}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr) \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}\\ &\quad + C^p{{8^2 .2^{1-\gamma}}\over {(1-K+Q^2)^2(p(1-\gamma))^{1/p}}} {{v''(t)-{{m^2}\over {\alpha^2A^2 }}v}\over {v }}A^{2p} \|{\tilde u}\|^p_{{\dot B}^{\gamma}_{p, p}}. \end{align*} Let \begin{gather*} D=2C{{8. 2^{1-\gamma}}\over {(1-K+Q^2)(p(1-\gamma))^{1/p}}}\Bigl( {8\over {1-K+Q^2}}{{2m^2}\over {\alpha^2 A^2}}+4m^2\Bigr), \\ F=C^p{{8^2.2^{1-\gamma}}\over {(1-K+Q^2)^2(p(1-\gamma))^{1/p}}} {{v''(t)-{{m^2}\over {\alpha^2A^2 }}v}\over {v }}A^{2p}. \end{gather*} From (H5) we have that D<1. Then \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}\leq D \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}+ F\|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}^p. $$from this inequality,$$ (1-D)\|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}\leq \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}^{p},\quad {{1-D}\over F}\leq \|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}^{p-1}\,. $$Since \lim_{t\to 0}F=+0, we have$$ \lim_{t\to 0}\|{\tilde u}\|_{{\dot B}^{\gamma}_{p, p}}=\infty.  \end{proof} \begin{thebibliography}{00} \bibitem{s1} Shatah, J., M. Struwe; \emph{Geometric wave equation}. Courant lecture notes in mathematics 2(1998). \bibitem{t1} Taylor, M.; \emph{Partial differential equations III}. Springer-Verlag, 1996. \bibitem{r1} Runst, T., W. Sickel.; \emph{Sobolev spaces of fractional order, Nemytskij operators and nonlinear partial differential equations}, 1996, New York. \end{thebibliography} \end{document}