Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. Consider the following equation: $$u_t=\Delta u+f(u),\quad x\in\Omega, t>0$$ $$u(x,0)=\phi(x),\quad x\in\Omega$$ $$u(x,t)=0,\quad x\in\partial\Omega,t>0.$$ where

  • $\phi$ is nonnegative on $\Omega$ and $\phi=0$ on $\partial\Omega$
  • $f\in C^2[0,\infty)$ is convex and strictly positive on $(0,\infty)$ with $$\int_0^\infty\frac{1}{f(s)}ds<\infty.$$ Let $\lambda_1$ be the principal eigenvalue (the one with the largest magnitude) of the Laplace operator and $\psi$ be the corresponding eigenvector, i.e. $$\Delta\psi=\lambda_1\psi,$$ Furthermore assume that $\psi>0$ and $\int_\Omega\psi=1$. Define $$g(t)=\int_\Omega u(x,t)\psi(x)dx,$$ where $u$ is the solution of the original equation. Show that $$g'(t)\geq-\lambda_1g(t)+f(g(t)),\quad (*)$$ and if $g(0)$ is sufficiently large then $g$ will blow up in finite time.

The inequality $(*)$ is relatively easy to prove. We have $$\begin{aligned}g'(t)&=\int_\Omega u_t(x,t)\psi(x)dx\\ &=\int_\Omega\Delta u\psi+f(u)\psi dx\\ &=\int_\Omega -u\Delta\psi dx+\int_\Omega f(u)\psi dx\quad\text{ (Integration by parts)}\\ &=-\lambda_1\int_\Omega u\psi dx+\int_\Omega f(u)\psi dx\\ &\geq-\lambda_1g(t)+f(g(t))\quad\text{(Jensen's inequality)} \end{aligned}$$

However I have no idea how to show that $g$ must blow up in finite time, i.e. we need to prove that there exists $T*$ such that $$\limsup_{t\to T*}|g(t)|=\infty.$$ Any suggestions? Thanks.

  • $\begingroup$ One idea: try to figure out what $f$ being convex and $\int 1/f$ converging tells you about how fast $f$ grows as $x\to \infty$. I suspect we should have $\lambda g / f(g) \to 0$ as $g\to \infty$. If so then if $g_0$ is large enough then $g'>0$ and the inequality reads to $g'/f(g(t) > 1 + $"small" which can be integrated over $[0,t]$ to get something like $\int_{g(0)}^{g(t)} \frac{dg}{f(g)} \gtrsim t$. If $g$ does not blow up at finite time then you get a contradiction to $\int 1/f$ converging. $\endgroup$ – Winther Apr 10 '17 at 3:04

First, there is some $\bar{s} > 0$ such that $f(s) \ge 2 \lambda_1 s$ for all $s \ge s_0$. To see this, assume that $f(s_k) < 2 \lambda_1 s_k$ for some sequence $s_k \nearrow \infty$. By convexity, $f(s) < 2 \lambda_1 s$ for all $s \in [s_k,s_{k+1}]$ with $k=1,2,\ldots$ and hence all $s \ge s_1$. This yields a contradiction with $\int_{s_1}^\infty\frac{ds}{f(s)}$.

Assume that $g(0)$ is sufficiently large, precisely that $g(0) \ge \bar{s}$. Let $$ F(s) = \int_0^s \frac{ds}{f(s)}. $$ By assumption, $F \in C^2$ is a bounded increasing function. Taking into account $g' \ge -\lambda_1 + f(g)$, we can bound the derivative of $F(g)$ from below: $$ (F(g(t)))' = \frac{g'(t)}{f(g(t))} \ge 1 - \lambda_1 \frac{g(t)}{f(g(t))}. $$ If $g(t) \ge \bar{s}$, this is estimated by $1/2$, so $F(g)$ is increasing and in consequence also $g$ is increasing. Since $g(0) \ge \bar{s}$, all these inequalities hold for all $t \ge 0$.

In consequence, $(F(g(t)))' \ge 1/2$ and $F(g(t)) \ge t/2$, which yields a contradiction with boundedness of $F$.


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