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Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ with smooth boundary $\partial\Omega$. Consider the following initial-boundary value problem for the heat equation:

\begin{equation} \begin{cases} u_t=\Delta u\quad\quad\quad\;\; \text{in}\;\Omega\times(0,\infty) , \\ u=0 \quad\quad\quad\;\;\;\;\; \text{on}\;\partial\Omega\times(0,\infty), \\ u(x,0)=u_0(x),\quad x\in\Omega. \end{cases} \end{equation}

It is mentioned in one book that the solution $u(x,t)\rightarrow0$ as $t\rightarrow\infty$.

The method I can think of to prove this is the energy method by letting $H[u](t)=\frac{1}{2}\,\int_{\Omega}u^2(x,t)\,dx$. However, this method seems does not work. I can only show that $\int_{\Omega}u^2(x,t)\,dx\rightarrow0$ as $t\rightarrow\infty$ by using integration by parts and the Poincare inequality. How could I obtain the desired result? Some hints, please.

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  • $\begingroup$ I've never come across the notation $u_{xx}$ for $\Delta u$ before (except of course for $n=1$). $\endgroup$
    – joriki
    Oct 9, 2015 at 6:10
  • $\begingroup$ @joriki $ u_{xx}$ refers to $ \frac{\partial ^2 u }{ \partial x^2}$ $\endgroup$
    – Nizar
    Oct 9, 2015 at 6:45
  • $\begingroup$ sorry, a typing error. I've corrected it. $\endgroup$
    – LCH
    Oct 9, 2015 at 7:07
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    $\begingroup$ @Nizar: That doesn't make sense in $\mathbb R^n$ for $n\gt1$. $\endgroup$
    – joriki
    Oct 9, 2015 at 8:11

1 Answer 1

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In the present case $\Delta $ as an operator acting in $L^{2}(\Omega ,d% \mathbf{x})$ has non-positive discrete spectrum. Thus% \begin{equation*} \Delta =\sum_{n}\lambda _{n}|v_{n}><v_{n}|=\sum_{n}\lambda _{n}(.,v_{n})v_{n} \end{equation*} where the $\lambda _{n}$ are the non-positive eigenvalues and the $v_{n}$ the associated eigenfunctions. As far as I remember $0$ is not an eigenvalue in the Dirichlet case. But then all eigenvalues are negative. Now \begin{equation*} u(t)=\exp [\Delta t]u(0)=\sum_{n}\exp [\lambda _{n}t](u(0),v_{n})v_{n}% \overset{t\rightarrow \infty }{\rightarrow }0 \end{equation*}

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