An ODE from the Dirichlet problem

Question

While trying to solve the Dirichlet problem, I consider that Laplacian should be a radial function. Then an ODE is derived $v = v(r), r v^{(2)} + v^{(1)} + \lambda r v = 0$, while $\lambda$ is a constant. How to solve this equation?

Answer 1

Let $v(r)=g(\sqrt{\lambda}r)$, then $v'(r)=\sqrt{\lambda}g'(\sqrt{\lambda}r)$ and $v''(r)=\lambda g''(\sqrt{\lambda} r)$, so $$ 0=rv''(r)+v'(r)+\lambda rv(r)=\lambda r\left(g''(\sqrt{\lambda}r)+\frac{1}{\sqrt{\lambda}r}g'(\sqrt{\lambda}r)+g(\sqrt{\lambda}r)\right), $$ so $$ g''(x)+\frac{1}{x}g'(x)+g(x)=0. $$ This is the Bessel equation of order $0$, so the solution is $g(x)=AJ_0(x)+BY_0(x)$, see here.