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Math Help - PDE Problem - (really just an integral)

  1. #1
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    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.
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  2. #2
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    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<R
     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
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  3. #3
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    Nevermind y'all. I found a good method to understand all of these equalities.

    Sorry if you read through all of that.
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