2

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 :)

Daniel Robert-Nicoud
  • 30,663
yemino
  • 21

1 Answers1

2

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 $w = 0$.

kahen
  • 16,130
  • 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