# PDE Problem - (really just an integral)

• May 6th 2006, 04:46 PM
PDE Problem - (really just an integral)
This problem isn't too bad in theory. I understand it, I'm just having great difficulty expressing the answer, so I'm wondering if anyone has some method, or insight how to "see" or express the answer.

Question:
{
Find $\displaystyle u(r,t)$ from the formula:
$\displaystyle ru = \frac{1}{2c}\int_{r-t}^{r+t}sv_0(s)ds$

If $\displaystyle v_0=1$ in the solid sphere $\displaystyle |r| \leq R$ and $\displaystyle v_0=0$ outside. ...
}

The "..." just represents the other part of the question that I'm not interested in.

As an example if we set:
$\displaystyle c=1$
$\displaystyle R=8$
$\displaystyle r=1$
$\displaystyle t=10$
$\displaystyle r -t =-9$
$\displaystyle r+t=11$

The integral would drop to:
$\displaystyle \frac{1}{2}\int_{r-t}^{r+t}sv_0(s)ds=\frac{1}{2}\int_{-9}^{8}sds$

This makes sense, that as long as we integrate below R we get our integral back. But when we get above R then the integral returns 0. My question is this I guess...
Is there an easy way to setup all these equalities and integrals? Because I am getting lost when I try to do it. The only time it really makes sense is when I plug in numbers and force my way through it. I'm just curious what a better way to accomplish this task of setting up cases is. Thanks in advance.
• May 6th 2006, 04:50 PM
Hmm... after looking at it again, maybe something like this:

If we drop the $\displaystyle \frac{1}{2c}$ for clarity:

$\displaystyle r+t>R$
$\displaystyle r-t>R$
$\displaystyle u(r,t)=0$

$\displaystyle r+t>R$
$\displaystyle r-t<R$
$\displaystyle u(r,t)=\int_{r-t}^{R}sds$

$\displaystyle r+t < R$
$\displaystyle r-t < R$
$\displaystyle u(r,t)=\int_{r-t}^{r+t}sds$

$\displaystyle r+t < R$
$\displaystyle r-t>R$
$\displaystyle u(r,t)=-\int_{R}^{r+t}sds$

grr... I don't know. I guess I'll just wait for a response :)
• May 7th 2006, 01:33 PM