Consider the nonlinear heat equation in smooth bounded domain $\Omega$ \begin{align*} u_t-\Delta u+f(u)&=0\\ u&=0 \ \text{on} \ \partial \Omega\\ u(0,x)&=g(x) \end{align*} where $f\in C^1$ denote $F$ by the primitive of $f$ such that $F(0)=0$.Assume that $u$ is smooth. Show that the function $E(u(t))$ defined as $$E(u)=\frac{1}{2}\|\nabla u\|_{L^2(\Omega)}^2+\int_{\Omega}F(u)\;dx$$ is decreasing.

So following the hint, we have that $$\int_{\Omega}uu_t-u\Delta u +uf(u)\;dx=0\ \tag{1}$$ $$\int_\Omega u_t^2-u_t\Delta u+u_tf(u)\;dx=0 \tag{2}$$ Therefore, using integration by parts and the boundary value condition, we have $\int uu_t=0$ and $\int_\Omega |\nabla u|^2+uf(u)=0$. Next to show that $E(u)$ is decreasing, taking the derivative would yield $|\nabla u|\Delta u+F(u)$. What to continue after having done all this (I m not sure all the calculation I did is correct)? Thanks for all the hint.


Notice that $$\begin{aligned} E(u)&=\frac{1}{2}\int_\Omega\nabla u\cdot\nabla u\;dx+\int_\Omega F(u)\;dx\\\\ &=-\frac{1}{2}\int_\Omega u\Delta u\;dx+\int_\Omega F(u)\;dx \end{aligned}$$ and thus $$\begin{aligned} \frac{d}{dt}E(u)&=-\frac{1}{2}\int_\Omega (u_t\Delta u+u\Delta u_t)\;dx+\int_\Omega F'(u)u_t\;dx\\\\ &=-\int_\Omega u_t\Delta u\;dx+\int_\Omega f(u)u_t\;dx \end{aligned}$$ So, by $(2)$, $$\frac{d}{dt}E(u)=-\int_\Omega u_t^2\;dx\leq 0$$ which implies that $E(u(t))$ is decreasing.

  • $\begingroup$ @Pedro: Can we do the other way around starting from amathnerd's eq(1)? I tried but got stuck too. $\endgroup$ – Patrick Windance Dec 22 '15 at 8:48
  • $\begingroup$ @PatrickWindance When we calculate $\frac{d}{dt}E$, the terms of $(2)$ appear naturally. Thus, I don't think so we can do it from $(1)$ $\endgroup$ – Pedro Dec 23 '15 at 14:19

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