Let $D =\{ (x,y) \in \mathbb{R}^2 \colon x^2 + y^2 < 1\}$ and consider the problem
\begin{cases} u_t = \alpha^2 (u_{xx} + u_{yy}), \hspace{0.5cm} (x,y) \in D, \hspace{0.4cm} t>0\\ u(x,y,0) = f(x,y) \hspace{0.8cm} (x,y) \in \overline{D}\\ u(x,y,t) = 0 \hspace{1.9cm} (x,t)\in \partial D \hspace{0.4cm} t >0\\ \end{cases}
I must solve this problem following the following conditions:
(i) Make a change of variable $\hat{x} = x/L, \hspace{0.2cm} \hat{y} = y/L$ to go to an analogous problem with $ \alpha= 1$. Then suppose $\alpha = 1$.
(ii) Separate the spatial variables from the temporal variable and obtain the problem: \begin{align} -\Delta v &= \lambda v\\ v|_{\partial D} &= 0 \end{align} It is possible to show that $\lambda> 0$. Then suppose $\lambda> 0$
(iii) Rewrite the problem in (ii) in polar coordinates.
(iv) Separate the variables $r$ and $\theta$ and show that the new separation constant $\eta$ is real and not negative.
(v) Solve the problem obtained in (iv) for the function $\theta$ and change the variable $\rho = \sqrt{\lambda}r$ to obtain the Bessel equation of order 2: $$\rho^2 R''(\rho) + \rho R'(\rho) + (\rho^2 - \eta^2)R(\rho) = 0$$
Remark: in the case where $\eta^2 = n \in \mathbb{Z}^{+}$, the general solution of the Bessel equation is of the form $R(\rho) = CJ_{n} + DY_{n}(\rho) = 0$, where $C$ and $D$ are arbitrary constants and $J_n$ and $Y_n$ are the Bessel functions of the first and second kind, respectively . The functions $Y_n$ are not bounded, so that the solutions of interest are the multiples of $J_n$.
(vi) Using the previous observations and the fact that the positive zeros of the Bessel function $J_n$ form a countable collection $\alpha_{n,m}$, $m \in \mathbb{N}$, show that the aigenvalues of the problem in (ii) are of the form $\lambda_{nm} = \alpha_{nm}^2$, $n \in \mathbb{Z}^{+}$, $m \in \mathbb{N}$. We therefore obtain a candidate for the solution of the original problem (in cylindrical coordinates and in the case $\alpha = 1$) of the form $$u(r,\theta,t) = \sum_{n=0}^{\infty}\sum_{m=1}^{\infty}{[A_{nm}\cos(n \theta) + B_{nm}\sin(n \theta)]J_{n}(\alpha_{nm}r)e^{-\alpha_{nm}^{2}t}} $$ Where the coefficients $A_{nm}$, $B_{nm}$ are determined by the initial condition (Do not calculate: it is necessary to use the orthogonality relations of the functions $J_n(\rho) = (1/2\pi)\int_{-\pi}^{\pi}{e^{(-in\theta + \rho \sin\theta)}}d \theta$).
Solve the original problem in the case that $f$ is a constant function.
I have spent hours trying to do this problem by following each of the stated steps, but I have not been able to solve it. In step (i) I obtain the problem in terms of a function $v$ with $\alpha = 1$. Now, applying the method of separation of variables, I obtain the problem in (ii). My difficulty starts from step (iii) when considering the problem in coordinates polar. How can I continue this? Any good help is appreciated.
My attempt:
(i) Assuming that $\alpha=1$, we get the problem
\begin{cases} u_t = u_{xx} + u_{yy}, \hspace{0.5cm} (x,y) \in D, \hspace{0.4cm} t>0\\ u(x,y,0) = f(x,y) \hspace{0.8cm} (x,y) \in \overline{D}\\ u(x,y,t) = 0 \hspace{1.9cm} (x,t)\in \partial D \hspace{0.4cm} t >0\\ \end{cases}
(ii) Let $u(x,y,t) = v(x,y)w(t)$. Then $u_t = vw'$, $u_{xx} = v_{xx}w$ and $u_{yy} = v_{yy}w$. Substituting in the problem we obtain
$vw' = v_{xx}w + v_{yy}w = (v_{xx} + v_{yy})w = \Delta v.w $ $\Rightarrow \frac{w'}{w} = \frac{\Delta v}{v} = - \lambda$, where $\lambda$ is a constant.
From this we get the ODE $w' = -\lambda w$ and the problem: \begin{equation} (1) \begin{cases} -\Delta v = \lambda v, \hspace{0.5cm} (x,y) \in D\\ v(x,y) = 0, \hspace{0.5cm} (x,y) \in \partial D \end{cases} \end{equation}
It is possible to show that $\lambda >0$. Suppose then $\lambda >0$.
(iii) Rewriting the problem $ (1) $ in polar coordinates we obtain: $$v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\theta \theta} = -\lambda v, \hspace{0.5cm } (r,\theta) \in D \hspace{0.9cm} (2)$$
(iv) By separation of variables, let $v(r, \theta) = R(r)S(\theta)$. Then
$v_r = R'S, \hspace{0.3cm} v_{rr} = R''S, \hspace{0.3cm} v_{\theta \theta} = RS''$. Substituting in (2) we have $$r^2 \frac{R''}{R} + r \frac{R'}{R} + \lambda r^2 = - \frac{S''}{S} = \eta$$
I must now show that $\eta$ is real and not negative, in (v) solve the problem obtained in (iv) for the function $\theta$ and obtain the Bessel equation from the change of variable $\rho = \sqrt{\lambda}r$ and in (vi) show that the eigenvalues in (ii) are of the form $\lambda_{nm} = \alpha_{nm}^2$, $n \in \mathbb{Z}^{+}$, $m \in \mathbb{N}$. Conclude by solving the initial problem in (i) when $f$ is constant. How can I do that? I have really tried to continue, but I have not been successful. I appreciate any help