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Solving 2nd order PDE of single variable

  1. May 12, 2014 #1
    I've been getting pretty rusty in terms of derivation in recent years. Encountered this problem which I can't derive the steps despite knowing the solution.

    [itex] \frac{\partial^2 u}{\partial r^2} + \frac{\partial u}{\partial r}\left(\beta + \frac{1}{r}\right)+\frac{\beta}{r}u=0[/itex]

    Known solution:

    [itex]u(r) = \beta \cdot \exp(-\beta r)[/itex]

    Thank you very much for your help!!
     
  2. jcsd
  3. May 12, 2014 #2
    Since the differential equation has the form of:
    ([itex]\frac{∂}{∂r}[/itex])2 u(r) + a(r) [itex]\frac{∂}{∂r}[/itex] u(r) + b(r) u(r) =0​
    With a(r) having a simple pole at r=0, and b(r) having no more than a pole of order 2 at r=0, it is sufficient to continue with the Frobenius Method:
    u(r) = rσ [itex]\sum[/itex] knrn
    You can then substitute this into the equation to get something like:
    [itex]\sum[/itex] kn ((n+σ)(n+σ-1)+(n+σ)+(n+σ+1)βr) rn+σ-1 = 0​
    This means that the indicial equation becomes:
    σ2-σ+σ=0
    σ=0, multiplicity 2.​
    Using σ=0 the next step is to determine the recursion relationship which will give you kn. This get's one independent solution that you can show is the same as the series for βe-βr.
    And then you're done.
     
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