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\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 77, pp. 1--7.\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 2004 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE-2004/77\hfil Blow-up of solutions]
{Blow-up of solutions to a nonlinear wave equation} 

\author[Svetlin G. Georgiev \hfil EJDE-2004/77\hfilneg]
{Svetlin Georgiev Georgiev} 

\address{University of Sofia, Faculty of Mathematics
and Informatics, Department of Differential Equations,
Bulgaria}
\email{sgg2000bg@yahoo.com}


\date{}
\thanks{Submitted March 16, 2004. Published May 26, 2004.}
\subjclass[2000]{35L10, 35L50}
\keywords{Wave equation, blow-up, hyperbolic space}


\begin{abstract}
 We study the solutions to the the radial 2-dimensional wave equation
 \begin{gather*}
 \chi_{tt}-{1\over r}\chi_r-\chi_{rr}+{{\sinh2\chi}\over {2r^2}}=g, \\
 \chi(1, r)=\chi_{\circ}\in {\dot H}^{\gamma}_{\rm rad},\quad
 \chi_t(1, r)=\chi_1 \in {\dot H}^{\gamma-1}_{\rm rad},
 \end{gather*}
 where $r=|x|$ and $x$ in $\mathbb{R}^2$.
 We show that this Cauchy problem, with values into a
 hyperbolic space, is ill posed in subcritical Sobolev spaces.
 In particular, we construct a function
 $g(t, r)$ in the space $L^p([0,1]L_{\rm rad}^q)$,
 with ${1\over p}+{2\over q}=3-\gamma$, $0<\gamma<1$,
 $p\geq 1$, and  $1<q\leq 2$, for which the solution satisfies
 $\lim_{t\to 0}\|{\bar \chi}\|_{{\dot H}^{\gamma}_{\rm rad}}=\infty$.
 In doing so, we provide a counterexample to estimates
 in \cite{r1}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}


\section{Introduction}

In this paper, we study the properties of the
solutions to the
Cauchy problem
\begin{gather}
\chi_{tt}-{1\over
r}\chi_r-\chi_{rr}+{{\sinh(2\chi)}\over
{2r^2}}=g(t, r), \label{e1} \\
\chi(1, r)=\chi_{\circ}\in {\dot H}^{\gamma}_{\rm
rad},\quad \chi_t(1,
r)=\chi_1 \in {\dot H}^{\gamma-1}_{\rm rad},
\label{e2}
\end{gather}
where $g\in L^p([0, 1]L_{\rm rad}^q)$, ${1\over
p}+{2\over
q}=3-\gamma$, $0<\gamma<1$, $p\geq 1$, $1<q\leq 2$,
$r=|x|$,
$x\in \mathbb{R}^2$, and $t\in [0, 1]$.

The homogeneous problem, i.e. \eqref{e1} with $g\equiv
0$, has been 
investigated
by several authors; see for example \cite[p.
89-141]{r1}, \cite{g1}, 
\cite{k3}, 
and references therein.
For the non-homogeneous problem, we construct function
$g$ in
$L^p([0, 1]L_{\rm rad}^q)$, and a solution ${\bar
\chi}$
to \eqref{e1}--\eqref{e2} such that
$\lim_{t\to 0}\|{\bar \chi}\|_{{\dot H}^{\gamma}_{\rm
rad}}=\infty$.
This implies that \eqref{e1}--\eqref{e2} is an ill
posed problem,
and provides a counter example to \cite[pages 3--4,
Eq. (9)]{r1}.

Equation \eqref{e1} is obtained from the wave map
equation with a 
source
term  when $x\in \mathbb{R}^2$ and the target is the
hyperboloid
$\mathcal{H}^2: u_1^2+u_2^2-u_3^2=-1$,
$\mathcal{H}^2\hookrightarrow \mathbb{R}^3$.
Let us consider the equation
$$
u_{tt}-\triangle u -(|u_t|^2-|\nabla_x u|^2)u=q(t, x),
\label{star}
$$
 when $x\in \mathbb{R}^2$ and the target is the
hyperboloid
$\mathcal{H}^2:$ $u_1^2+u_2^2-u_3^2=-1$,
$\mathcal{H}^2\hookrightarrow \mathbb{R}^3$.
Equation \eqref{star} is a wave map equation with a
source term,
where
\begin{gather*}
|u_t|^2=u_{1t}^2+u_{2t}^2-u_{3t}^2, \\
|\nabla_x u|^2=|\nabla_{x_1}u|^2+|\nabla_{x_2}u|^2, \\
|\nabla_{x_i}u|^2=u_{1x_i}^2+u_{2x_i}^2-u_{3x_i}^2,\quad
i=1, 2.
\end{gather*}
Let $q=(q_1, q_2, q_3)$, with
\begin{gather*}
q_1=\cosh\chi \cos\phi_1(\chi_{tt}-\Delta\chi)+{1\over
{r^2}}\sinh\chi 
\cos\phi_1
+{{\sinh^3\chi \cos\phi_1}\over {r^2}}, \\
q_2=\cosh\chi \sin\phi_1(\chi_{tt}-\Delta\chi)+{1\over
{r^2}}\sinh\chi 
\sin\phi_1+
{{\sinh^3\chi \sin\phi_1}\over {r^2}}, \\
q_3=\sinh\chi g.
\end{gather*}
For the hyperboloid $\mathcal{H}^2$, we have the
parametric
representation
\begin{gather}
u=(u_1, u_2, u_3),\\
u_1=\sinh\chi\cos\phi_1,\\
u_2=\sinh\chi\sin\phi_1,\\
u_3=\cosh\chi,\quad \chi\geq 0,\quad \phi_1\in
[0,2\pi].
\end{gather}
Let $x_1=r\cos\phi_1$, $x_2=r\sin\phi_1$. Then from
\eqref{star},
we get that $\chi$ satisfies \eqref{e1}.

Our main result is as follows.

\begin{theorem} \label{thm1}
Let ${1\over p}+{2\over q}=3-\gamma$, $p\geq 1$,
$1<q\leq 2$, 
$0<\gamma<1$.
Then there exist function
$g\in L^p([0, 1]L_{\rm rad}^q)$ and solution
${\bar\chi}$ to
\eqref{e1}--\eqref{e2} such that
$$
\lim_{t\to 0}\|{\bar \chi}\|_{{\dot H}^{\gamma}_{\rm
rad}}=\infty.
$$
\end{theorem}

Note that the case $p=1$, $q=2$ which is the energy
case can not be
reached by the setting in this theorem.

\section{Preliminary results}

Let $f$ be a real-valued  function satisfying
\begin{itemize}
\item[(H1)] $f\in \mathcal{C}^2[0, \infty)$,
\item[(H2)] $f(0)=0$, $f(1)=f'(1)=0$.
\end{itemize}
As an example of a function  satisfying (H1)-(H2), we
have
\begin{equation}
f(x)=(1-x)^2x. \label{stars}
\end{equation}
Certainly $f\in \mathcal{C}^2[0, \infty)$; therefore,
(H1) holds.
Note that $f(0)=0$ and $f'(x)=-2(1-x)x+(1-x)^2$ thus
$f(1)=f'(1)=0$;
therefore, (H2) holds.

In addition, assume that:
\begin{itemize}
\item[(H3)] ${1\over p}+{2\over q}=3-\gamma$,
$0<\gamma<1$,
$p\geq 1,\quad q> 1$

\item[(H4)] $0<\alpha\leq 2-q$,  $\beta>0$

\item[(H5)] Either $ \beta>\alpha$ with 
$\frac{\beta}{\alpha}<\frac{q}{2p(q-1)}$
or
$\beta<\alpha$ with $\frac{\beta}{\alpha} >
\frac{q(2p-1)}{2p}$.
\end{itemize}
Note that when (H3) holds,  $q\leq 2$; because if
$q>2$ then
$3-\gamma={1\over p}+{2\over q}<1+1=2$, from where
$\gamma>1$ which
contradicts $0<\gamma<1$.

Let $z^{1/\alpha}={r\over {t^{\beta/\alpha}}}$. Then
$z^{2/\alpha}={r^2\over t^{2\beta/\alpha}}$,
\begin{gather*}
{{\partial z^{2/\alpha}}\over {\partial
t}}=-{{2\beta}\over\alpha}
{1\over t}z^{2/\alpha},\quad
{{\partial^2 z^{2/\alpha}}\over {\partial t^2}}=
{{2\beta(\alpha+2\beta)}\over{\alpha^2}}
{1\over {t^2}}z^{2/\alpha},
\\
{{\partial z^{2/\alpha}}\over {\partial r}}=
{2\over
{t^{{\beta}\over\alpha}}}z^{1\over\alpha},\quad
{{\partial^2 z^{2/\alpha}}\over {\partial r^2}}=
{2\over {t^{2\beta\over\alpha}}}.
\end{gather*}
Let $f$ be a function  satisfying (H1)--(H2) and let
\begin{gather}
\chi_{\circ}=
\begin{cases}
f(r^2)&\mbox{for } r\leq 1,\\
0& \mbox{for } r\geq 1,
\end{cases} \label{e3} \\
\chi_1= \begin{cases}
-{{2\beta}\over \alpha}r^2f'(r^2)&\mbox{for } r\leq
1,\\
0 &\mbox{for } r\geq 1,
\end{cases} \label{e4}\\
B_1={{4\beta^2}\over
{\alpha^2}}z^{4/\alpha}f''\bigl(z^{2/\alpha}\bigr)
+{{2\beta(\alpha+2\beta)}\over {\alpha^2}}z^{2/\alpha}
f'\bigl(z^{2/\alpha}\bigr), \label{e5} \\
B_2=z^{2/\alpha}f''(z^{2/\alpha})+f'(z^{2/\alpha}),
\label{e6} \\
g=\begin{cases}
{{B_1}\over {t^2}}-{4\over {t^{{2\beta}/\alpha}}}B_2+
{{\sinh(2{\bar\chi})}\over {2r^2}}&\mbox{for } r\leq 
t^{\beta/\alpha},\\
0 &\mbox{for } r\geq t^{\beta/\alpha},
\end{cases} \label{e7}
\end{gather}
and let
\begin{equation}
{\bar\chi}=\begin{cases}
f(z^{2/\alpha})&\mbox{for } r\leq t^{\beta/\alpha}\\
0&\mbox{for } r\geq t^{\beta/\alpha}.
\end{cases} \label{e8}
\end{equation}
Note that ${\bar \chi}$  is a solution
of \eqref{e1}--\eqref{e2}. Indeed, for $z\leq 1$ we
have
\begin{gather*}
{\bar \chi}_t=-{{2\beta}\over
{t\alpha}}z^{2/\alpha}f'(z^{2/\alpha}),
\\
{\bar \chi}_{tt}={{4\beta^2}\over {\alpha^2}}{1\over 
{t^2}}z^{4\over\alpha}
f''(z^{2/\alpha})+{{2\beta(\alpha+2\beta)}\over
{\alpha^2}}
{{z^{2/\alpha}}\over {t^2}}f'(z^{2/\alpha}),
\\
{\bar\chi}_r={2\over 
{t^{\beta/\alpha}}}z^{1\over\alpha}f'(z^{2/\alpha}),
\\
{\bar\chi}_{rr}={4\over {t^{{2\beta}\over
{\alpha}}}}z^{2/\alpha}
f''(z^{2/\alpha})+{2\over
{t^{{2\beta}\over\alpha}}}f'(z^{2/\alpha}).
\end{gather*}
Then, for $z\leq 1$, we have
$$
{\bar\chi}_{tt}-{1\over
r}{\bar\chi}_r-{\bar\chi}_{rr}+
{{\sinh{2{\bar\chi}}}\over {2r^2}}=
{{B_1}\over {t^2}}-{4\over
{t^{{2\beta}\over\alpha}}}B_2+
{{\sinh(2{\bar\chi})}\over {2r^2}}\equiv g,
$$


\begin{lemma} \label{lm1}
Under Assumptions (H1)-(H5), the function $g$ defined
by \eqref{e7}
is in the space $L^p([0, 1]L^q_{\rm rad})$.
\end{lemma}

\begin{proof} Note that
$$
\|g\|_{L^q_{\rm
rad}}^q=\int_0^{t^{\beta/\alpha}}|g|^qr\,dr
=\int_0^{t^{\beta/\alpha}}\Bigr|{{B_1}\over {t^2}}
-{4\over {t^{{2\beta}\over\alpha}}}B_2+
{{\sinh(2{\bar\chi})}\over {2r^2}}\Bigl|^qr\,dr.
$$
By making the change of variable
$r=z^{1\over\alpha}t^{\beta/\alpha}$,
$$
dr=t^{\beta/\alpha}{1\over\alpha}z^{{1\over\alpha}-1}dz,\quad
r\,dr=t^{{2\beta}\over\alpha}{1\over\alpha}z^{{2\over\alpha}-1}dz
$$
(with $t$ fixed)
and
\begin{align*}
\|g\|_{L^q_{\rm rad}}^q
&={{t^{{2\beta}\over\alpha}}\over\alpha}
\int_0^1\Bigl|B_1{1\over {t^2}}-{4\over
{t^{{2\beta}\over\alpha}}}B_2+
{{sinh(2{\bar\chi})}\over
{2t^{{2\beta}\over\alpha}z^{2/\alpha}}}
\Bigl|^qz^{{2\over\alpha}-1}dz\\
&\leq c_1
{{t^{{{2\beta}\over\alpha}-2q}}\over\alpha}\int_0^1|B_1|^q
z^{{2\over\alpha}-1}dz+c_2 
{{t^{{{2\beta}\over\alpha}(1-q)}}\over\alpha}
\int_0^1|B_2|^qz^{{2\over\alpha}-1}dz \\
&\quad + c_3
{{t^{{{2\beta}\over\alpha}(1-q)}}\over\alpha}
\int_0^1\Bigl|{{\sinh(2{\bar\chi})}\over
{2z^{2/\alpha}}}\Bigl|^q
z^{{2\over\alpha}-1}dz.
\end{align*}
Let
\[
I_1=\int_0^1|B_1|^qz^{{2\over\alpha}-1}dz,\quad
I_2=\int_0^1|B_2|^qz^{{2\over\alpha}-1}dz,\quad
I_3=\int_0^1\Bigl|{{\sinh(2{\bar\chi})}\over
{2z^{2/\alpha}}}\Bigl|^q
z^{{2\over\alpha}-1}dz.
\]
 From the definition of $B_1$  and $B_2$
 and since (H1) and (H4) hold (we note that $\alpha\in
(0, 1)$
because $1<q\leq 2$),
we have
$|B_1|^qz^{{2\over\alpha}-1}\in \mathcal{C}[0, 1]$,
$|B_2|^qz^{{2\over\alpha}-1}\in \mathcal{C}[0, 1]$.
Then $I_1<\infty$, $I_2<\infty$. Now we consider
$I_3$. Let $B$ be 
constant
for which
$$
B-2\cosh(2{\bar\chi}){\bar\chi}'\geq 0.
$$
Such a constant $B$ exists because
${\bar\chi}(z^{2/\alpha})$
and ${\bar \chi}'(z^{2/\alpha})$
are bounded functions for $z\in [0, 1]$ and
$\mathop{\rm supp} {\bar \chi}\subset [0, 1]$,
$\mathop{\rm supp} {\bar \chi}'\subset [0, 1]$. Let
$$
p(z)=Bz^{2/\alpha}-\sinh (2{\bar\chi}).
$$
Then
$$
p'(z)={2\over\alpha}z^{{2\over\alpha}-1}(B-2\cosh(2{\bar\chi}){\bar\chi}')\geq
0
\quad \forall z\in [0, 1].
$$
Consequently, $p(z)$ is an increasing function for
$z\in[0, 1]$.
Therefore, $p(z)\geq p(0)=0$ for all $z\in [0, 1]$ or
$Bz^{2/\alpha}\geq sinh(2{\bar\chi})$
 for $z\in [0, 1]$. Then
$$
I_3\leq \int_0^1\Bigl|{{Bz^{2/\alpha}}\over
{2z^{2/\alpha}}}\Bigr|^q
z^{{2\over\alpha}-1}dz\equiv \mbox{const}<\infty.
$$
Consequently,
$$
\|g\|_{L^q_{\rm rad}}^q\leq {\bar
c}_1t^{{{2\beta}\over\alpha}-2q}+
{\bar c}_2t^{{{2\beta}\over {\alpha}}(1-q)}
$$
and from here
\begin{gather*}
\|g\|_{L^q_{\rm rad}}\leq {\tilde
c_1}t^{{{2\beta}\over{\alpha q}}-2}+
{\tilde c_2}t^{{{2\beta}\over {\alpha
q}}-{{2\beta}\over \alpha}},
\\
\|g\|_{L^p([0, 1]L^q_{\rm rad})}^p\leq
{\overline{\overline c_1}}
\int_0^1t^{{{2\beta p}\over{\alpha q}}-2p}dt+
{\overline{\overline c_2}}
\int_0^1t^{{{2\beta p}\over {\alpha q}}-{{2\beta
p}\over 
\alpha}}dt<\infty
\end{gather*}
because (H5) holds. Here $c_1$, $c_2$, $c_3$, ${\bar
c}_1$, ${\bar 
c}_2$,
${\tilde c}_1$, ${\tilde c}_2$, ${\overline{{\bar
c}_1}}$,
${\overline{{\bar c}_2}}$ are positive constants.
\end{proof}

As special notation we have
$$
{\dot H}^s(\mathbb{R}^n)\equiv {\dot F}^s_{2,
2}(\mathbb{R}^n),\quad
0 <s<\infty
$$
where (see \cite[p. 94, def. 2]{r1})
$${\dot F}^s_{2, 2}(\mathbb{R}^n)={\dot B}^s_{2,
2}(\mathbb{R}^n), 
\quad
-\infty<s<\infty.
$$
 As in  \cite[p. 30-31]{s1},  when  $f(r)$ is a
function with compact
support in $[0, 1]$, we have
$$
\|f\|_{\dot H^{\gamma}_{\rm rad}}:=
\Bigl(\int_0^1 h^{-1-2\gamma}\|\Delta_h f\|_{L^2_{\rm 
rad}}^2dh\Bigr)^{1/2},
$$
where
$\Delta_hf=f(r+h)-f(r)$, $0<\gamma<1$.

\begin{lemma} \label{lm2}
Under assumptions (H1)--(H5), the function
$\chi_{\circ}$
defined by \eqref{e3} is in the space ${\dot
H}^{\gamma}_{\rm rad}$.
\end{lemma}

\begin{proof} By definition of the norm,
$$
\|\chi_{\circ}\|_{{\dot H}^{\gamma}_{\rm 
rad}}^2=\int_0^1h^{-(1+2\gamma)}
\int_0^1
|f((r+h)^2)-f(r^2)|^2r\,dr\,dh
$$
Using the Mean Value Theorem,
\begin{align*}
\|\chi_{\circ}\|_{{\dot H}^{\gamma}_{\rm rad}}^2
&= 
\int_0^1h^{-(1+2\gamma)}\int_0^{1}|f'(\xi)|^2[(r+h)^2-r^2]^2r\,dr\,dh\\
&=
\int_0^1h^{-(1+2\gamma)}\int_0^{1}|f'(\xi)|^2(h^2+2rh)^2r\,dr\,dh\\
&\leq 
Q_1\int_0^1h^{-(1+2\gamma)}\int_0^{1}(h^4r+4r^2h^3+4r^3h^2)\,dr\,dh
\\
&= Q_1\int_0^1h^{-(1+2\gamma)}\Bigl({{h^4}\over
2}+{4\over 3}h^3+ 
h^2)dh \\
&= Q_1{1\over {2(4-2\gamma)}}+{4\over 3}Q_1{{1}\over
{3-2\gamma}}
+Q_1{{1}\over {2-2\gamma}},
\end{align*}
where $Q_1$ is positive constant. Therefore,
$\|\chi_{\circ}\|_{{\dot H}^{\gamma}_{\rm rad}}\leq
Q_2$ for some
constant $Q_2$ and $\chi_{\circ}\in {\dot
H}^{\gamma}_{\rm rad}$.
\end{proof}

\begin{lemma} \label{lm3}
Under Assumptions (H1)-(H5), the function ${\chi_1}$
defined by 
\eqref{e4}
is in the space ${\dot H}^{\gamma-1}_{\rm rad}$.
\end{lemma}

\begin{proof}  We have that
$L^2_{\rm rad}\hookrightarrow {\dot H}^{\gamma-1}_{\rm
rad}$. On the
other hand
$$
\|\chi_1\|^2_{L^2_{\rm rad}}={{4\beta^2}\over
{\alpha^2}}
\int_0^1r^5|f'(r^2)|^2dr={{2\beta^2}\over
{\alpha^2}}\int_0^1r^4
|f'(r^2)|^2dr^2<\infty
$$
because $f'(r^2)\in \mathcal{C}[0, 1]$. Consequently,
$\chi_1\in {\dot H}^{\gamma-1}_{\rm rad}$.
\end{proof}

\section*{Proof of main result}

Let (H1)-(H5) hold. By \eqref{stars},
$f(z^{2/\alpha})=(1-z^{2/\alpha})^2 z^{2/\alpha}$.
Let  $g$ be defined by \eqref{e7} and ${\bar\chi}$ by
\eqref{e8}.
 From Lemma \ref{lm1}, $g\in L^p([0, 1]L^q_{\rm
rad})$.
Also, we have ${\bar\chi}(1, r)=\chi_{\circ}$,
${\bar\chi}_t(1, 
r)=\chi_1$.
 From Lemma \ref{lm2},  ${\bar\chi}(1, r)\in {\dot
H}^{\gamma}_{\rm 
rad}$,
  and from Lemma \ref{lm3}, ${\bar\chi}_t(1, r)\in
{\dot 
H}^{\gamma-1}_{\rm rad}$.
 Therefore, ${\bar\chi}$ is solution of
\eqref{e1}--\eqref{e2}.
Let $\theta=t^{\beta/\alpha}$ which is in $(0, 1]$.
Then
\begin{equation}
\|{\bar\chi}\|_{{\dot H}^{\gamma}_{\rm rad}}^2=
\int_0^{1}h^{-(1+2\gamma)} \int_0^{\theta}
\Bigl|f\Bigl(\Bigl({{r+h}\over\theta}\Bigr)^2\Bigr)-f\Bigl(
{{r^2}\over {\theta^2}}\Bigr)\Bigr|^2
r\,dr\,dh\Bigr)=:I.
\label{e9}
\end{equation}
Then
$$
I\geq \int_{({\sqrt 2}-{1\over
2})\theta}^{1}h^{-(1+2\gamma)}
\int_{\theta\over 2}^{{\theta\over {\sqrt 3}}}
\Bigl|f\Bigl(\Bigl({{r+h}\over\theta}\Bigr)^2\Bigr)-f\Bigl(
{{r^2}\over {\theta^2}}\Bigr)\Bigr|^2 r\,dr\,dh.
$$
and
\begin{align*}
f\Bigl(\Bigl({{r+h}\over\theta}\Bigr)^2\Bigr)-f\Bigl(
{{r^2}\over {\theta^2}}\Bigr)
&= \Bigl[1-{{(r+h)^2}\over {\theta^2}}\Bigr]^2
{{(r+h)^2}\over
{\theta^2}}-\Bigl(1-{{r^2}\over{\theta^2}}\Bigr)^2
{{r^2}\over{\theta^2}}\\
&=
{{[\theta^2-(r+h)^2]^2(r+h)^2-(\theta^2-r^2)^2r^2}\over
{\theta^6}}.
\end{align*}
Note that the numerator of the above expression is
\begin{align*}
L&:=[\theta^2-(r+h)^2]^2(r+h)^2-(\theta^2-r^2)^2r^2\\
&=(\theta^2-r^2)(2rh+h^2)(\theta^2-3r^2)+(3r^2-2\theta^2)(2rh+h^2)^2+
(2rh+h^2)^3.
\end{align*}
For $r\in ({\theta\over 2}, {{\theta}\over {\sqrt
3}})$ we have
\begin{align*}
L&\geq (3r^2-2\theta^2)(2rh+h^2)^2+ (2rh+h^2)^3\\
&=(2rh+h^2)^2(3r^2-2\theta^2+2rh+h^2)\\
&=(2rh+h^2)^2[2r^2+(r+h)^2-2\theta^2].
\end{align*}
For $h\in (({\sqrt 2}-{1\over 2})\theta, 1)$ and
$r\in ({\theta\over 2}, {{\theta}\over {\sqrt 3}})$,
\begin{gather*}
r+h\geq {\theta\over 2}+{\sqrt 2}\theta-{1\over
2}\theta={\sqrt 
2}\theta,\\
(r+h)^2\geq 2\theta^2.
\end{gather*}
Therefore, for $h\in (({\sqrt 2}-{1\over 2})\theta,
1)$ and
$r\in ({\theta\over 2}, {{\theta}\over {\sqrt 3}})$,
\begin{gather*}
L\geq (2rh+h^2)^2 2r^2\geq h^4r^2, \\
\Bigl|f\bigl(\bigl({{r+h}\over\theta}\bigr)^2\bigr)-f\bigl(
{{r^2}\over {\theta^2}}\bigr)\Bigr|^2\geq
{{h^8r^4}\over 
{\theta^{12}}}.
\end{gather*}
Consequently
\begin{align*}
I&\geq \int_{({\sqrt 2}-{1\over
2})\theta}^{1}h^{-(1+2\gamma)}
\int_{\theta\over 2}^{{\theta\over {\sqrt
3}}}{{h^8r^5}\over 
{\theta^{12}}}\,dr\,dh\\
&={1\over {6\theta^{12}}}
\int_{({\sqrt 2}-{1\over 2})\theta}^{1}h^{7-2\gamma}
\Bigl({{\theta^6}\over {27}}-{{\theta^6}\over
{64}}\Bigr)dh\\
&={{37}\over {6\times 27\times 64\theta^6
.(8-2\gamma)}}
\Bigl[1-({\sqrt 2}-{1\over
2})^{8-2\gamma}\theta^{8-2\gamma}\Bigr]\\
&=
{{37}\over {6\times 27\times 64\theta^6
.(8-2\gamma)}}-
{{37}\over {6\times 27\times 64 (8-2\gamma)}({\sqrt
2}-{1\over 
2})^{8-2\gamma}}
\theta^{2-2\gamma}
\end{align*}
which approcahes zero as $\theta\to 0$.
From this statement and \eqref{e9}, we obtain
$$
\lim_{t\to 0}\|{\bar\chi}\|_{{\dot H}^{\gamma}_{\rm
rad}}=\infty.
$$

\subsection{Comments}
Let (H1)-(H5) hold. By \eqref{stars},
$f(z^{2/\alpha})=(1-z^{2/\alpha})^2z^{2/\alpha}$.
Then the function ${\bar\chi}$ defined by \eqref{e8}
is a solution to 
the Cauchy
problem
\begin{gather}
\chi_{tt}-{1\over r}\chi_r-\chi_{rr}=g_1(t, r),
\label{e1'} \\
\chi(1, r)=\chi_{\circ}\in {\dot H}^{\gamma}_{\rm
rad},\quad
\chi_t(1,r)=\chi_1\in {\dot H}^{\gamma-1}_{\rm rad},
\label{e2'}
\end{gather}
where $\chi_{\circ}$ and $\chi_1$ are the functions
defined
with \eqref{e3} and \eqref{e4},
$$
g_1= \begin{cases}
{B_1\over t^2}- {4\over t^{2\beta/\alpha}} B_2
&\mbox{for } r\leq t^{\beta/\alpha}\\
0 &\mbox{for } r\geq t^{\beta/\alpha}.
\end{cases}
$$
Then because of (H5),
$$
\|g_1\|_{L^p([0, 1]L^q_{\rm rad})}^p\leq l_1
\int_0^1t^{{{2\beta p}\over{\alpha q}}-2p}dt+
l_2
\int_0^1t^{{{2\beta p}\over {\alpha q}}-{{2\beta
p}\over 
\alpha}}dt<\infty,
$$
where $l_1$, $l_2$ are positive constants. In
\cite[Corollary 1.3]{k1} it is proved that
$$
\|{\bar\chi}\|_{\mathcal{C}([0, 1]{\dot
H}^{\gamma}_{\rm rad})}\leq
\|\chi_{\circ}\|_{{\dot H}^{\gamma}_{\rm rad}}
+\|\chi_1\|_{{\dot H}^{\gamma-1}_{\rm
rad}}+\|g_1\|_{L^p([0, 1]L^q_{\rm 
rad})}.
$$
Therefore,
${\bar \chi}$ is in $\mathcal{C}([0, 1]{\dot
H}^{\gamma}_{\rm rad})$.

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\end{document}