I am trying to solve $$\frac{\partial u}{\partial t}=\nabla^2 u$$ in a disc of unit radius where $u=0$ at $r=1$ for all $t$ and $u(r,\theta, 0)=f(r).$
Using the substitution $u(r,\theta,t)=R(r)S(\theta)T(t)$, I separated my solution into three ODEs: \begin{align} T'+T\alpha&=0 \tag{1} \\ \frac{r}{R}\frac{d}{dr}\left(rR'\right)+\alpha r^2+\beta&=0 \\ S''-S\beta&=0, \end{align} where $\alpha, \beta\in\mathbb{R}$. Usually I know how to continue, but given the boundary conditions i'm unsure. For instance, if I consider $\alpha=0, \ \alpha>0$ and $\alpha<0$ for equation $(1)$, what boundary conditions do I use?
$$\frac{\partial u}{\partial t}=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r} \right)+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$$ $$u(r,\theta,t)=R(r)S(\theta)T(t)$$ I am afraid that your separation of variables is not correct. $$RS\frac{dT}{dt}=ST\frac{1}{r}\frac{d }{dr}\left(r\frac{dR}{dr} \right)+RT\frac{1}{r^2}\frac{d^2 S}{d \theta^2}$$ $$\frac{1}{T(t)}\frac{dT}{dt}=\frac{1}{R(r)}\frac{1}{r}\frac{d }{dr}\left(r\frac{dR}{dr} \right)+\frac{1}{S(\theta)}\frac{1}{r^2}\frac{d^2 S}{d \theta^2}$$ The term $\frac{1}{T(t)}\frac{dT}{dt}$ is function of $t$ only.
The term $\frac{1}{R(r)}\frac{1}{r}\frac{d }{dr}\left(r\frac{dR}{dr} \right)$ is function of $r$ only.
But the term $\frac{1}{S(\theta)}\frac{1}{r^2}\frac{d^2 S}{d \theta^2}$ isn't function of $\theta$ only, except if $\frac{d^2 S}{d \theta^2}=0$.
As a consequence if we want to separate the variables necessarily $S(\theta)$ is a linear function. Moreover for this linear function the boundary conditions implies $S(\theta)=$constant.
This is why your equation $S''-S\beta=0$ is not correct except for $\beta=0$.
It is simpler to forget $\theta$ and consider the simplified PDE where $u$ is no longer function of $\theta$ (This is consistent with the specified conditions) : $$\frac{\partial u}{\partial t}=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r} \right)$$ where $u=u(r,t)$ with conditions $$u(1,t)=0\quad\text{and}\quad u(r,0)=f(r)$$ The separation of variables leads to the equations : $$\frac{1}{T(t)}\frac{dT}{dt}=\frac{1}{R(r)}\frac{1}{r}\frac{d }{dr}\left(r\frac{dR}{dr} \right)=-\lambda=\text{constant}$$ $$T(t)=e^{-\lambda t}$$ $$R(r)=c_1J_0(\sqrt{\lambda}\:r)+c_2Y_0(\sqrt{\lambda}\:r)$$ Thus some particular solutions of the PDE are on the form : $$u(r,y)=e^{-\lambda t}\left(c_1J_0(\sqrt{\lambda}\:r)+c_2Y_0(\sqrt{\lambda}\:r)\right)$$
$J_0$ and $Y_0$ are the Bessel functions of order $0$ and of first and second kind respectively.
A more general solution is on the form of a discret linear combination of the above particular solution with various $\lambda$. $$u(r,t)=\sum_\lambda e^{-\lambda t}\left(c_{1,\lambda}J_0(\sqrt{\lambda}\:r)+c_{2,\lambda}Y_0(\sqrt{\lambda}\:r)\right)$$ Or on integral form : $$u(r,t)=\int e^{-\lambda t}\left(c_1(\lambda)J_0(\sqrt{\lambda}\:r)+c_2(\lambda)Y_0(\sqrt{\lambda}\:r)\right)d\lambda$$ $c_1(\lambda)$ and $c_2(\lambda)$ are arbitrary functions, to be determined according to the specified conditions. $$u(1,t)=0=\int e^{-\lambda t}\left(c_1(\lambda)J_0(\sqrt{\lambda})+c_2(\lambda)Y_0(\sqrt{\lambda})\right)d\lambda$$ Since $c_1(\lambda)$ and $c_2(\lambda)$ are arbitrary, on can chose $c_2(\lambda)=-c_1(\lambda)\frac{J_0(\lambda)}{Y_0(\lambda)}$ $$u(r,t)=\int e^{-\lambda t}\phi(\lambda)\left(Y_0(\sqrt{\lambda})\:J_0(\sqrt{\lambda}\:r)-J_0(\sqrt{\lambda})\:Y_0(\sqrt{\lambda}\:r)\right)d\lambda$$ where $\phi(\lambda)$ is an arbitrary function to be determined according to the condition $u(r,0)=f(r)$ : $$f(r)=\int \phi(\lambda)\left(Y_0(\sqrt{\lambda})\:J_0(\sqrt{\lambda}\:r)-J_0(\sqrt{\lambda})\:Y_0(\sqrt{\lambda}\:r)\right)d\lambda$$ The result will depend on the form of the function $f(r)$ which is not explicitly known. Analytically solving such an integral equation for $\phi(r)$ appears very difficult. This is the hitch with the method of separation of variable in this case of PDE and conditions.