I am trying to separate and solve this PDE:
$$ \frac{\partial^2{S(u,v)}}{\partial{u}^2}+\frac{\partial^2{S(u,v)}}{\partial{v}^2}-\left[\frac{a^2}{2 \cosh(u) \cos(v)}\right] S(u,v) =0 $$
where $(u,v)$ are orthogonal curvilinear coordinates associated with a bipolar conformal mapping from Cartesian coordinates. Can someone suggest an analytic method to solve this in a separated closed-form that will allow me to satisfy boundary conditions on contours of constant $(u)$?

I have attempted a separation of variables approximation with $(v)$ held constant about some constant contour $(v=v_0)$:
$$ \frac{\partial^2{T(u)}}{\partial{u}^2}-\left[\frac{\gamma^2}{ \cosh(u)} +\beta^2 \right] T(u) =0 $$
where: $$ \gamma^2=\frac{a^2}{2 \cos(v_0)}$$
but am unable to find a solution for this ODE.


Perhaps transform under a Laplace transform?

\begin{align} \mathscr{L}\left(\frac{\partial S}{\partial v}\right) &= s U(u,v)-U(u,o) \\ \mathscr{L}\left(\frac{\partial^2 S}{\partial v^2}\right) &= s^2 U(u,v)-sU(u,o)-U_v(u,0)\\ \mathscr{L}\left(\frac{\partial S}{\partial u}\right) &= \frac{d U}{d u} \\ \mathscr{L}\left(\frac{\partial^2 S}{\partial u^2}\right) &= \frac{d^2 U}{d u^2} \end{align} The trickier part would be the transform of the term immediately to the left of the zero.

  • $\begingroup$ Thanks Bacon, that was helpful. The term immediately left of the zero is the problem. I found the Laplace transform for: $\endgroup$ – SwetDwiW Apr 13 '17 at 18:28
  • $\begingroup$ The Laplace transform for (1/cosh(u)) is 1/2[Digamma((3+s)/4)-Digamma((1+s)/4), but how do I handle the product with S(u,v)? $\endgroup$ – SwetDwiW Apr 13 '17 at 18:44

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