I am trying to find a general solution for the PDE $$\frac{\partial u}{\partial t}=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 u}{\partial\theta^2},$$ using the substitution $u(r,\theta, t)=R(r)S(\theta)T(t).$ The substitution gives the three ODES, \begin{align} S''-\beta S&=0 \tag{1}\\ \frac{d}{dr}\left(r\frac{dR}{dr}\right)+\left(\lambda r+\frac{\beta}{r}\right)R&=0 \tag{2} \\ T'+\lambda T&=0 \tag{3}. \end{align} The boundary conditions are $u(1,\theta,t)=0, \forall t>0, \ 0\leq \theta< 2\pi\ $ and $u(r,\theta,0)=f(r,\theta)$, for $0\leq r\leq 1,\ 0 \leq \theta< 2\pi$.
The question states that $\beta=-m^2$, which gives the following solution to $(1)$, $$S_m(\theta)=A_m\cos(m\theta)+B_m\sin(m\theta).$$ The question also states that $\lambda=k^2>0$. By transforming $(2)$ into a Bessel equation of order $m$ $$p\frac{d}{dp}\left(p\frac{dR}{dp}\right)+(p^2-m^2)R=0,$$ we see that this has solution $R_n(r)=J_m(\mu^n_m),$ where $\mu^n_m$ denotes the $n^{th}$ zero of the $m^{th}$ order Bessel function ($R$ is finite as $r\rightarrow 0^+$). Solving $(3)$ gives $$T(t)=C_1e^{-k^2t}.$$ But I am unsure if this is correct, as I do not know how to create an infinite series for $u(r,\theta, t)$. I would really appreciate some help, as I have been trying to solve the problem over the last couple of days. Thanks.