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What steps should be taken in order to get a solution (that only depends on v) for the following?:

$\dfrac{\partial ^2f}{\partial v^2}+\dfrac{1}{v}\dfrac{\partial f}{\partial v}-\dfrac{1}{v^2}\dfrac{\partial^2f}{\partial u^2} = 0$

Sam
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  • are you sure that at the end is $-\frac{1}{v^2}\frac{\partial ^2f}{\partial^2 u^2}$ and not $+\frac{1}{v^2}\frac{\partial^2 f}{\partial u^2}$ ? – Surb Aug 02 '15 at 13:27
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    You've given the polar version of the Laplace Equation, off by a constant. You have to give boundary conditions for a solution. – Zach466920 Aug 02 '15 at 13:36
  • The end is definitely as written. There are no boundary conditions provided. – Sam Aug 02 '15 at 13:43

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You want a solution $f(u,v)=g(v)$? if that's the case you use that ansatz and find solutions of that form. You will get a ode given by $$ \frac{1}{v}\dfrac{d}{dv}v\dfrac{dg}{dv} = 0 $$ but are you sure you want this? if you want a full solution then use this $$ f(u,v) = g(v)h(u) $$ and you will get a similar equation for $v$ and a another equation for $u$.

Chinny84
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