I have a difficult PDE to solve here:
An infinitely long massive rod has been lying in 0 degrees water over a long period of time. The cross section of the rod is quarter-circular with radius $R$. The massive rod i sunken in boiling water at $t=0$. The surface is then assumed to attain a temperature of 100 C. Solve the heat conduction problem
So it is clear that the PDE is
$$\Delta u=\frac{1}{\alpha}\frac{\partial T}{\partial t}, \ \ \ \ \ \ \ u=R\Theta T\\ u(r,\theta,0)=100 \ \ \ \ \ \ \ \ \ t>0\\ u(R,\theta,t)=0 \ \ \ \ \ \ \ \ \ \ 0\le r\le R\\ u(r,0,t)=0 \ \ \ \ \ \ \ \ \ \ \ 0\le\theta \le \frac{\pi}{2}$$
So we have the PDE:
$$R_{rr}\Theta T+\frac{1}{r}R_r\Theta T+\frac{1}{r^2}R\Theta_{\theta\theta}T=\frac{1}{\alpha}R\Theta T_t$$
Separation of variables gives the three following ODEs and respective solutions:
$$T_t=-\lambda\alpha T \longrightarrow T(t)=C_1e^{-\alpha\lambda t}\\ R_{rr}+\frac{1}{r}R_r+\bigg(\lambda-\frac{\mu}{r^2}\bigg)R=0 \longrightarrow J(r)=C_2J_{n^2}\bigg(\sqrt{\lambda}r\bigg)\\ \Theta_{\theta\theta}+\mu\Theta=0 \longrightarrow \Theta(\theta)=C_3\cos n\theta+C_4\sin n\theta$$
Using ICs, assuming they are correctly deduced from the problem we get:
$$100=C_1e^{-\alpha\lambda 0}\longrightarrow C_1=100 \\ \alpha_{n,k}=C_2J_{n^2}\bigg(\sqrt{\lambda}R\bigg)\longrightarrow \lambda=\bigg(\frac{\alpha_{n,k}}{C_2J_{n^2}(R)}\bigg)^2\\ IC(3)\longrightarrow \Theta(\theta)=C_4\sin n\theta $$
So the total function is:
$$u_n(r,\theta,t)=100\cdot C_4 J_{n^2}\bigg[\bigg(\frac{\alpha_{n,k}}{C_2J_{n^2}(R)}\bigg)r\bigg]\sin n\theta e^{-\alpha\lambda t}$$
Where $\alpha_{n,k}$ are the Bessel zeros, and $\alpha$ is the thermal diffusivity constant of the material.
But this should be developed into a Fourier series and the constants are still elusive.
What is wrong here and what is the right approach?
Thanks