Solve the eigenproblem for −∆ in the semicircle $x^2 + y^2 \le R^2$ , y ≥ 0 at homogeneous Dirichlet condition as boundary condition. Use the result to find solve the heat equation: An end-to-end massive rod initially has a temperature of 100◦ C after being in boiling water for a long time. Cross section of the rod is semicircular (as $x^2 + y^2 \le R^2$ , y ≥ 0) > with radius R. The cylinder is lowered at time t = 0 into a mixture of ice and water. Mantle surface is then assumed to have a temperature of 0◦ C.
First part:
Solve the eigenvalueproblem (where we use the radial Laplacian since the domain is circular): $$\Delta u=0 \ \ \ \ \ \ \, x^2+y^2\le R^2, t>0\\ \Delta u= u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}\\ u=R\Theta T\\ u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=u_t$$
By separation of variables:
$$R_{rr}\Theta T+\frac{1}{r}R_{r}\Theta T+\frac{1}{r^2}R\Theta_{\theta\theta} T= T_t R\Theta$$
divide by $R\Theta T$:
$$\frac{R_{rr}}{R}+\frac{1}{r}\frac{R_{r}}{R}+\frac{1}{r^2}\frac{\Theta_{\theta\theta}}{\Theta} =\frac{T_t}{T}$$
Separation of variables gives two ODE:
$$\frac{R_{rr}}{R}+\frac{1}{r}\frac{R_{r}}{R}+\frac{1}{r^2}\frac{\Theta_{\theta\theta}}{\Theta}=-\lambda$$
and
$$\frac{T_t}{T}=-\lambda$$
The latter gives: $T(t)=e^{-\lambda t}$
The former we solve by separation of variables a second time:
$$r^2\frac{R_{rr}}{R}+r\frac{R_{r}}{R}+\frac{\Theta_{\theta\theta}}{\Theta}=-\lambda r^2$$
Separation of variables yields two ODE:
$$r^2\frac{R_{rr}}{R}+r\frac{R_{r}}{R}+\lambda r^2=\mu$$
$$\frac{\Theta_{\theta\theta}}{\Theta}=-\mu$$
With Dirichlet conditions, the latter gives $\Theta(\theta)=A\sin \frac{n\pi}{\alpha} \theta$, and since the domain is a half-circle it follows that $\alpha=\pi$. This gives : $\Theta(\theta)=A\sin n \theta$.
Solve the radial part, which is the Bessel equation on a bounded domain, $\lambda>0$ and we have the general solution:
$R(r)=AJ_n\big(\sqrt{\lambda} r\big)+BJ_n\big(\sqrt{\lambda}$
Since the domain is bounded, B=0, so we get:
$R(r)=AJ_n\big(\sqrt{\lambda} r\big)$
Now we find $\lambda$, we use Dirichlet on R, $u(R)=0$. However, 0 is not an option, so we use the Bessel zeros, $u(R)=\alpha_{nk}$ and get:
$$\alpha_{nk}=J_n\big(\sqrt{\lambda} R\big)$$
which gives $\lambda=\bigg(\frac{\alpha_{nk}}{J_n(R)}\bigg)^2$.
So $R(r)=J_n\big(\frac{\alpha_{nk}}{J_n(R)} r\big)$
This gives finally:
$$u(r,\theta,t)_n=A_nJ_n\big(\frac{\alpha_{nk}}{J_n(R)} r\big)\sin n \theta e^{\frac{\alpha_{nk}}{J_n(R)}t}$$
But how do I find A using the part of the original question where at t=0 the temperature is 0?
Thanks