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.

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