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For $$ \begin{cases}u_t=\Delta u & \text { in } \Omega \times(0, \infty), \\ \frac{\partial u}{\partial \nu}=0 & \text { on } \partial \Omega \times(0, \infty), \\ u(x, 0)=\phi(x), & x \in \Omega\end{cases} $$ it's easy to prove that $$\int_\Omega u(x,t) dx$$ is conserved in time and the energy $$ E(t)=\int_\Omega | \nabla u(x,t) |^2$$ is decreasing in time t by Green's formula.

My question:

How to show that the limit of $E(t)$ is $0$ instead of a nonzero constant, so that the solution will tend to a constant for large $t$.

Attempt:1.Multiply $u$ on both side of the equation then we get $$ \frac{1}{2} \frac{d}{d t} \int_{\Omega} u^2+\int_{\Omega}|\nabla u|^2=0 $$ if $E(+\infty) \neq 0$, then we will have $\frac{d}{d t} \int_{\Omega} u^2$ is a negative constant when $t$ tends to $+\infty$, which is a contradiction.

2.Suppose this constant is $C$, then $$\int_\Omega \phi(x) dx = \int_\Omega C dx = C|\Omega|,$$ so $C=\frac{\int_\Omega \phi(x)}{|\Omega|}$

Am I correct?

Elio Li
  • 721
  • This is actually not as easy as you'd expect. https://math.stackexchange.com/questions/4887510/uniqueness-and-continuous-dependence-on-the-data-of-heat-equation/4902188#4902188 – K.defaoite May 15 '24 at 21:13
  • Sorry, I just realized you are using Neumann instead of Dirichlet boundary conditions. Disregard my prior comment. – K.defaoite May 15 '24 at 21:14
  • Your solution in (1) looks correct to me but I don't understand what you are doing in (2). – K.defaoite May 15 '24 at 21:18
  • I mean that $\int_\Omega u(x,t) dx$ doesn't change w.r.t time (Green formula), so $\int_\Omega u(x,0) dx= \int_\Omega u(x,+\infty) dx$. – Elio Li May 15 '24 at 22:46
  • This is probably easy to do via eigenfunctions of $\Delta$ with respect to neumann boundary data. The case for Dirichlet boundary data is explained in Partial Differential Equations I by Michael Taylor. – Mason May 16 '24 at 05:01

0 Answers0