# Ngô Quốc Anh

## October 29, 2010

### Blowup of semilinear heat equation with critical sobolev exponent via the concavity method

Filed under: PDEs — Ngô Quốc Anh @ 21:52

Let us consider the following semilinear heat equation

$u_t - \Delta u =u^{2^\star}, \quad (x,t) \in \Omega \times (0,T)$

together with the following conditions

$u(x,t)=0, \quad (x,t) \in \partial\Omega \times (0,T)$

and

$u(x,0)=u_0(x) \geqslant 0, \not\equiv 0$.

This equation corresponds formally to the $L^2$ gradient flow associated to the energy functional

$\displaystyle E(u)=\frac{1}{2}\int_\Omega |\nabla u|^2dx - \frac{1}{2^\star}\int_\Omega |u|^{2^\star}dx$.

We prove the following

If there exists some $t_0$ such that $E(u(t_0)) \leqslant 0$ then $u(u,t;u_0)$ blows up in finite time.

This result is adapted from a paper by Zhong Tan published in Commun. Partial Differential Equations in 2001 [here].

Proof. Suppose $t_{max}=\infty$ and denote

$\displaystyle f(t)=\frac{1}{2}\int_{t_0}^t \|u\|_2^2ds$.

We perform standard manipulations

$\displaystyle E(u({t_0})) = \int_{{t_0}}^t {\int_\Omega {u_t^2dxds} } + \frac{1}{2}\int_\Omega {|\nabla u{|^2}dx} - \frac{1}{{{2^ \star }}}\int_\Omega {|u{|^{{2^ \star }}}dx}$

and

$\displaystyle f'(t) = \frac{1}{2}\left\| {{u_0}} \right\|_2^2 + \int_{{t_0}}^t {\int_\Omega {\left( { - |\nabla u{|^2} + |u{|^{{2^ \star }}}} \right)dxds} }$

and

$\displaystyle f''(t) = - \int_\Omega {|\nabla u{|^2}dx} + \int_\Omega {|u{|^{{2^ \star }}}dx}$.

Therefore

$\displaystyle f''(t) \geqslant \left( {\frac{{{2^ \star }}}{2} - 1} \right)\int_\Omega {|\nabla u{|^2}dx} - {2^ \star }E(u({t_0})) + {2^ \star }\int_{{t_0}}^t {\int_\Omega {u_t^2dx} ds}$.

Since $E(u(t_0)) \leqslant 0$ we have

$\displaystyle\left( {\frac{{{2^ \star }}}{2} - 1} \right)\int_\Omega {|\nabla u{|^2}dx} - {2^ \star }E(u({t_0})) \geqslant 0$

for all $t \geqslant t_0$. If we had $t_{max}=\infty$, then there exists a constant $C>0$ such that for all $t>t_1>t_0$ there holds

$\displaystyle f''(t) \geqslant {2^ \star }\int_{{t_0}}^t {\int_\Omega {u_t^2dx} ds}$,

this inequality would yield

$\displaystyle\mathop {\lim }\limits_{t \to \infty } f'(t) = \mathop {\lim }\limits_{t \to \infty } f(t) = \infty$.

On the other hand, we know that

$\displaystyle f''(t) \geqslant {2^ \star }\int_{0}^t {\int_\Omega {u_t^2dx} ds}$

and

$\displaystyle\begin{gathered} f(t)f''(t) \geqslant \frac{{{2^ \star }}}{2}\left( {\int_0^t {\left\| {u(s)} \right\|_2^2ds} } \right)\left( {\int_0^t {\left\| {{u_s}(s)} \right\|_2^2ds} } \right) \hfill \\ \qquad\qquad= \frac{{{2^ \star }}}{2}{\left( {\int_0^t {\int_\Omega {u{u_t}dxds} } } \right)^2} \hfill \\ \qquad\qquad= \frac{{{2^ \star }}}{2}{\left( {f'(t) - f'(0)} \right)^2} \hfill \\ \end{gathered}$

and as $t \to \infty$ we have for some $\alpha>0$ and for all $t \geqslant t_0$ such that

$\displaystyle f(t)f''(t) \geqslant (1 + \alpha ){(f'(t))^2}$.

Hence $f(t)^{-\alpha}$ is concave on $[t_0,\infty]$, $f(t)^{-\alpha} >0$ and $\lim_{t \to \infty} f(t)^{-\alpha}=0$. This contradiction proves that $t_{max}<\infty$.