Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 .

share|improve this question
    
What do you mean by convergence? –  Mercy Jul 3 '12 at 19:15
    
i had missed to put as $t\to \infty $ –  Theorem Jul 3 '12 at 19:17
add comment

1 Answer

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.

share|improve this answer
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.