# Solving an ODE to find the general solution as an infinite series for $u(r,t)$

I am trying to solve the following problem.

I have the two ODEs \begin{align} T'+\alpha\lambda T&=0 \\ \frac{d}{dr}\left(r\frac{dR}{dr}\right)+\lambda rR&=0, \end{align} with boundary conditions $$R(b)=0, \ R'(0)=0$$ and $$T(0)=f(r)$$. I am trying to solve the ODE for $$T$$ and hence write down the general solution of $$u(r,t)=R(r)T(t)$$ as an infinite series.

Transforming the ODE for $$R$$ into a Bessel equation of order $$0$$ when $$\lambda=k^2>0$$ and $$p=kr$$ gives the solution $$R_n(p)=A_nJ_0(p)+B_nY_0(p)\implies R_n(r)=A_nJ_0(kr)+B_nY_0(kr).$$ Solving the ODE for $$T$$ gives $$T(t)=Ce^{-\alpha k^2 t},$$ as the characteristic equation is $$\beta+\alpha k^2=0\implies\beta=-\alpha k^2$$. Does this then mean that the general solution for $$u$$ is,

\begin{align} u(r,t)&=\sum_{n=1}^{\infty}R_n(r)T_n(t) \\ u(r,t)&=\sum_{n=1}^{\infty}e^{-\alpha k^2 t}\left(A_nJ_0(kr)+B_nY_0(kr)\right)? \end{align} Is this methodology correct? Thank you kindly for you help.

• @Mattos Does the $Y_0$ term disappear? If a boundary condition was that $R$ is finite as $r\rightarrow 0^+$ I would understand that is does vanish, but these boundary conditions don't seem to indicate this. – user654924 Mar 17 at 2:49