Heat Equation on a disc Question: The temperature distribution $u(r, θ, t)$ in a circular disc of unit radius $(0 ≤ r < 1, 0 ≤ θ < 2π)$ evolves according to the heat equation with unit thermal diffusivity $$u_t = ∇^2u$$ where $∇^2$ is the Laplacian in Polar Coordinates. The outer edge of the disc is insulated so that the normal gradient of the temperature at the boundary satisfies $u_r(1, θ, t) = 0$. The initial temperature distribution is $u(r, θ, 0) = f(r)$. By seeking axisymmetric ($θ$-independent) solutions, show that the temperature at time $t$ is given by 
$$u(r, t) = A_0 + \sum_{k=1}^\infty A_k J_0 (\alpha_k r)exp(-\alpha_k^2 t)$$
Find the coefficients ${α_k}$ and obtain integral expressions involving $f(r)$ for the coefficients ${A_k}$. 
Can anybody help me with this question? I've used the seperated solution $u(r,t) = R(r)T(t)$, and found $T(t)=Aexp(-\lambda t)$. Using the fact that $R(0)=finite$ the conditon $u_r(1, θ, t) = 0$, I got $R'(1) = B\sqrt\lambda J_0'(\sqrt\lambda)=0$, where $J_0$ is the Bessel function of the first kind, index $0$. I dont really know how to find the eigenvalues and figure out what $\alpha_k$ is. Any help would be appreciated. Thanks.

They're determined by this condition "$R'(1) = B\sqrt\lambda J_0'(\sqrt\lambda)=0$"; that is "the eigenvalues are such that the square root is a zero of the derivative of $J_0$" (in particular finding the eigenvalues and finding $\alpha_k$ amounts to the same thing; though finding $A_k$ is done, as usual, by using your initial condition and orthogonality, and so will be defined in terms of integrals of $f$ against $J_0(\alpha_k r)$).
