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I would like to know if it is true that the solution of the equation $\partial_tu(x,t)=\Delta u(x,t)+f(x) ,t\ge0, u=0 $ for $x\in R^n , t=0$ converges to the solution of $\Delta u=-f, x\in R^n$ as $t\to \infty$ ? How can i show if its true ?

What i am thinking is to use duhamels principle and find the solution and what should i do next ?

Thank you for your help .

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    i had missed to put as $t\to \infty $2012-07-03

1 Answers 1

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If $f$ is well behaved, the solution $u(t)$ can be obtained analytically. First, define $w(t,x) \equiv u(t,x)+F(x)$ where $\Delta F(x)=f$. Assuming $f$ is not pathological, such $F$ exists and can be calculated directly, because we know the kernel of Laplace's equation:

$F(x)=\int_{\mathbb{R}^n} \frac{f(y)}{|x-y|}dy $

This $w$ obeys a homogeneous equation:

$\partial_t w= \partial_t (u+F)=\partial_t u=\Delta u+f=\Delta (u+F)= \Delta w$

The solution for $w$ can also be obtained because we know the kernel of the heat equation:

$w(t,x)=\frac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} w(0,y)e^{-\frac{|x-y|^2}{4t}}dy$

It is seen from this solution that if $w(x,0)$ decays fast enough with x, then $w\to 0$ for $t\to\infty$ and for all $x$. Therefore, $\Delta u \to -f$ as needed.

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    I'm not sure, ut I think that if $f$ does not vanish fast enough at infinity then the statement that $\Delta u\to f$ will simply not hold.2012-07-03