Problem 2
Part a. Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation
$$ \Delta u := u_{xx} + u_{yy} = k^2 u $$
What ODE satisfies $u\left(r\right)$?
Answer:
We have that transforming Laplacian $\Delta_{\left(x,y\right)} := \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ into polar coordinates $u\left(x,y\right) \rightarrow u\left(r,\theta\right)$ yields us the polar form of the Laplacian:
$$ \Delta_{\left(r,\theta\right)} := \frac{\partial^2}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta ^2} $$
Since we assume a form of $u\left(r,\theta\right)$ which depends only on $r$, we have that $u\left(r\right)$ is constant with respect to $\theta$, so $u_{\theta\theta} = 0 $, and our PDE simplifies to the ODE:
$$ \Delta_{\left(r,\theta\right)} u\left(r,\theta\right) := u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^2} u_{\theta\theta} = u_{rr} + \frac{1}{r} u_{r} = k^2 u $$
We proceed by finding a solution for $u\left(r\right)$ in the form of Bessel's function, which solves the Bessel differential equation of the form:
$$ u_{zz} + \frac{1}{z} u_{z} +\left(1-\frac{s^2}{z^2}\right)u = 0 $$
$$ \text{Let: } u\left(r\right) = v\left(i k r\right) \implies u_{r} = i k v_{r}, \phantom{\ } u_{rr} = - k^2 v_{rr} $$
$$ \implies \Delta_{\left(r,\theta\right)} u\left(r\right) := u_{rr} + \frac{1}{r} u_{r} - k^2 u = 0 \rightarrow - k^2 v_{rr} + \frac{1}{r} i k v_{r} - k^2 v = 0 $$
Moreover, as $ k > 0$, $ k^2 \ne 0 $, so dividing through by $ -k^2 $ yields:
$$ v_{rr} - \frac{i}{k r} v_{r} + v = 0 $$
Notice that $ -\left(i * i\right) = -\left(-1\right) = 1$, so $ - \frac{i}{k r} = \frac{\left(-i\right)}{\left(1\right) k r} = \frac{\left(-i\right)}{\left(-i * i\right) k r} = \frac{\overline{i}}{\overline{i}} \frac{1}{i k r} = \frac{1}{i k r} $ so our ODE in $v\left(i k r\right)$, say $v\left(z\right)$ where $z = i k r$ is equivalent to:
$$ v_{rr} + \frac{1}{i k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$
It's clear then that for $ s = n = 0$:
$$ v_{zz} + \frac{1}{z} v_{z} + v = v_{zz} + \frac{1}{z} v_{z} +\left(1-\frac{s^2}{z^2}\right)v = 0$$
Because $n = 0$ is an integer, our solution for $v\left(z\right) $ is given in the form of: $ v\left(z\right) = A J_0\left(z\right) + B N_0\left(z\right) $ for some $ \{A,B\} \in \mathbb{R} $, where $J_0\left(z\right) $ is Bessel's function of the first kind of order $0$, and $N_0\left(z\right)$ is Bessel's function of the second kind, namely, the Neumann function.
$$ J_n\left(z\right) = \sum_{j=0}^{\infty}\frac{\left(-1\right)^j}{\Gamma\left(j+1\right)\Gamma\left(j+n+1\right)}\left(\frac{z}{2}\right)^{2j+n} $$
$$ N_n\left(z\right) = \lim_{s \to n} N_s\left(z\right) = \lim_{s \to n} \frac{J_{s}\left(z\right) \cos\left(\pi s\right) - J_{-s}\left(z\right)}{\sin\left(\pi s\right)}$$
$$ \implies v\left(z\right) = v\left(i k r\right) = u\left(r\right) = A J_0\left(i k r\right) + B N_0\left(i k r\right) \phantom{\ } \blacksquare $$
Part b. Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation
$$ \Delta u := u_{xx} + u_{yy} = -k^2 u $$
What ODE satisfies $u\left(r\right)$?
Answer:
As the Laplacian term of this PDE is identical to part a. we follow the same derivation by first changing to polar coordinates and assuming a radial solution. Then we have an ODE in $u\left(r\right)$:
$$ \Delta_{\left(r,\theta\right)} u\left(r,\theta\right) := u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^2} u_{\theta\theta} = u_{rr} + \frac{1}{r} u_{r} = - k^2 u $$
We wish to again factor this into a Bessel differential equation:
$$ u_{zz} + \frac{1}{z} u_{z} +\left(1-\frac{s^2}{z^2}\right)u = 0 $$
$$ \text{Let: } u\left(r\right) = v\left(k r\right) \implies u_{r} = k v_{r}, \phantom{\ } u_{rr} = k^2 v_{rr} $$
$$ \implies \Delta_{\left(r,\theta\right)} u\left(r\right) := u_{rr} + \frac{1}{r} u_{r} + k^2 u = 0 \rightarrow k^2 v_{rr} + \frac{1}{r} k v_{r} + k^2 v = 0 $$
Let $z = k r$. Again, $k > 0$ so $k^2 \ne 0$ and dividing through gives us:
$$ v_{rr} + \frac{1}{k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$
Which is clearly a Bessel's differential equation with $s = n = 0$:
$$ v_{zz} + \frac{1}{z} v_{z} + v = v_{zz} + \frac{1}{z} v_{z} +\left(1-\frac{s^2}{z^2}\right)v = 0 $$
So our solution for $v\left(z\right)$ is again $ v\left(z\right) = A J_0\left(z\right) + B N_0\left(z\right) $ for some $ \{A,B\} \in \mathbb{R} $.
$$ \implies v\left(z\right) = v\left(k r\right) = u\left(r\right) = A J_0\left(k r\right) + B N_0\left(k r\right) \phantom{\ } \blacksquare $$
