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Dirichlet problem on unit disc in polar:

$u_{rr} + (1/r) + (1/r^2)u_{\theta\theta} = 0$

$u(1,\theta) = f$

Period in $\theta$ gives

$u(r,\theta) = \sum R_n(r) e^{in\theta}$

Inserted into our PDE gives

$R_n '' + (1/r)R_n'-(n^2/r^2)R_n = 0$ which is an Euler equation. Solutions are:

$R_n(r) = A ln(r) + B : n = 0$

$R_n(r) = Ar^n + Br^{-n} : n \neq 0$

Here's where I get confused. The book just jumps right into the following conclusion:

$R_n(r) = c_nr^{|n|} \rightarrow u = \sum c_n r^{|n|}e^{in\theta}$

I don't get why we're discarding all the negative n:s. Can anyone shed a light?

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1 Answer 1

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Take your $R_n(r) = Ar^n + Br^{-n}$. We want this to hold for all possible values on $r$, however we still want our solution to be physically plausible. This means that it can not run away to infinite, i.e. we need to care about when $r=0$. Setting $R_n(r) = c_nr^{|n|}$ fixes this problem, with $c_n = A,B$ depending on $n$.

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  • $\begingroup$ Thanks, but I'm still unsure how we can be sure just taking the absolute value produces the same solution. Are we saying $Ar^n + Br^{-n}$ CAN BE EXPRESSED AS $c_nr^{|n|}$, if we are free to let $c_n$ be arbitrary and vary with n? $\endgroup$ Jul 10, 2014 at 11:40
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    $\begingroup$ What we are saying with $c_nr^{|n|}$ is that when $n$ is negative, we will only regard $Br^{-n}$ as this part is the one that will be well-defined, i.e. non-singular at the origin. When $n$ is positive, $c_n = A$ and we will regard $Ar^n$ as in this case, this is the well-defined part. $\endgroup$
    – user128779
    Jul 10, 2014 at 11:43

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