# Thread: PDE Problem - (really just an integral)

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

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

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

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

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

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.

2. Hmm... after looking at it again, maybe something like this:

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

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

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

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

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

grr... I don't know. I guess I'll just wait for a response

3. Nevermind y'all. I found a good method to understand all of these equalities.

Sorry if you read through all of that.