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I would like to know if it is true that the solution of the equation $\partial_tu(x,t)=\triangle u(x,t)+f(x) ,t\ge0, u=0 $ for $x\in R^n , t=0$ converges to the solution of $\triangle u=-f, x\in R^n$ 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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What do you mean by convergence? – Mercy King Jul 3 '12 at 19:15
i had missed to put as $t\to \infty $ – Theorem Jul 3 '12 at 19:17

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 think if $f$ is $C^2$ in space and $C^1$ in time will be sufficient isn't it ? – Theorem Jul 3 '12 at 19:43
a) in your question $f$ is independent of time. b)What about $$f=e^e^x$$? – yohBS Jul 3 '12 at 19:45
will that be a problem ?? – Theorem Jul 3 '12 at 19:48
Yes. Neither the integral defining $F$ nor that defining the solution $w(x,t)$ will converge. – yohBS Jul 3 '12 at 19:52
I am getting confused now . What do i have to change to make your answer approperiate ? – Theorem Jul 3 '12 at 19:55

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