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I have the following problem:

Let $\Omega \subset \mathbb{R}^n$ be an open and bounded subset with piecewise smooth boundary $\partial\Omega$.

$a:\Omega\to]0,\infty[$ is a smooth function.

$f:\Omega\to\mathbb{R}$ is a smooth function.

Show that the following cauchy problem has at most one classical solution.

$\cases{-\nabla\cdot(a\nabla u)=f\quad in\quad \Omega\\u=0\quad on \quad \partial\Omega}$

I think that the energy method could be useful, but i'm not sure how to use it in this case.

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Follows directly from standard trick. Let $u,v$ be two classical solutions. Their difference $w=u-v$ solves the homogeneous problem $$-\nabla \cdot (a\nabla w ) = 0, \quad w = 0 \text{ on }\partial \Omega .$$

Multiply by $w$ and integrate over $\Omega$: $$ 0 = \int_\Omega -w\nabla \cdot (a\nabla w ) \overset{\tiny \substack{integration \\by\ parts} }{=} \int_\Omega \nabla w \cdot (a\nabla w) + \int_{\partial \Omega} w a\nabla w\cdot \nu \ dS \\= \int_\Omega a |\nabla w|^2 + 0 $$ so $\nabla w=0$ so $w$ is (locally) constant, and $w|_{\partial \Omega } = 0$ so $w=0$. Therefore, any two solutions coincide.

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  • $\begingroup$ Why do you multiply by w and integrate over Ω? $\endgroup$ – jm2501 Feb 7 at 15:15
  • $\begingroup$ @jm2501 Because it works. I pulled it out of a hat; it is a trick, but a standard one. $\endgroup$ – Calvin Khor Feb 7 at 15:31
  • $\begingroup$ Can you please show me that hat ^_^ I feel like it has something to do with drichlet's energy function? $\endgroup$ – jm2501 Feb 7 at 15:34
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    $\begingroup$ @jm2501 This particular hat is called integration by parts. You can call it that if you want $\endgroup$ – Calvin Khor Feb 7 at 15:36

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