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I have the following problem:

$w_t = \Delta w$ for $x \in \Omega$, $t>0$.
$w(x,0) = 0$ for $x \in \bar\Omega$.
$w(x,t) = 0$ for $x \in \partial \Omega$ and $t>0$.

We define the energy:

$\mathcal E(t):=\displaystyle\int_\Omega w^2(x,t) \,dx$

It is easy to show that ${\cal E}(t)$ is decreasing (because $\dfrac{d\mathcal E(t)}{dt}\leq 0$), but why $\mathcal E(0)=0$ and why this implies that the solution of the problem is $w=0$?? I can't see the relation.

Thanks for your answers :)

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$0 = \sqrt{ {\cal E}(t) } = \| w \|_{L^2(\Omega)}$ so $w$ has to be zero –  tom Sep 22 '13 at 15:43
    
Thanks. Obviously $\displaystyle\int_\Omega w^2=0\Rightarrow w=0$, but why $\displaystyle\int_\Omega w^2=0$ in $t=0$. –  yemino Sep 22 '13 at 15:50
    
Because energy is decreasing and nonegative. –  tom Sep 22 '13 at 15:59
    
Thanks! I get it –  yemino Sep 22 '13 at 16:06
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1 Answer

The intuition is obvious: It's the heat equation. The first boundary condition says that the region $\Omega$ starts with a temperature of $0$. And the boundary of $\Omega$ is kept at a temperature of $0$ for all time. So the temperature must be $0$ for all time.

You almost have the argument. $\mathcal E$ is an integral of an everywhere non-negative function, so it's always non-negative. It starts at $0$ and is non-increasing and non-negative, so the only thing it can do is stay at $0$. Now $w$ is something whose square integral is $0$ and it's at least continuous, so $w = 0$.

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First: Thanks! Obviously $\displaystyle\int_\Omega w^2=0\Rightarrow w=0$, but why $\displaystyle\int_\Omega w^2=0$ in $t=0$. The intuition is ok, but I don't understand the mathematical reasoning/argument. –  yemino Sep 22 '13 at 15:49
    
Because $w(x,t) = 0$ for $t=0$. That's the first boundary condition. –  kahen Sep 22 '13 at 15:52
    
I'm feeling so stupid!! Thanks man!! It was obviously. Thanks again :) –  yemino Sep 22 '13 at 15:56
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