Toronto Math Forum

Problem 3

Part a. Using the proof of the maximum principle, prove the maximum principle for subharmonic functions, and the minimum principle for superharmonic functions.

Answer:
Maximum principle for subharmonic functions:
Let: $u\left(\mathbf{x}\right)$ be subharmonic, $ \Delta u\left(\mathbf{x}\right) \ge 0$, on a bounded domain $\Omega$ with boundary $\partial\Omega = \Sigma$, $\mathbf{x} \in \Omega$. Let: $v\left(\mathbf{x}\right) = u\left(\mathbf{x}\right) + \epsilon |\mathbf{x}|^2$ where $|\mathbf{x}|^2 = \sum_{i=1}^{n} x_i^2 $, $\epsilon > 0$. Let $\delta$ be the diameter of the set $\bar\Omega$, namely the largest value of $| \mathbf{x}|^2 = \delta^2$.
At any interior maximum point of a function $f$ we require $f_{x_i x_i} \le 0$, $\Delta f = \sum_{i=1}^{n}f_{x_i x_i} \le 0$ by the second derivative test. Notice that for $\mathbf{x} \in \Omega$:
$$ \Delta v\left(\mathbf{x}\right) = \Delta u\left(\mathbf{x}\right) + \Delta \epsilon |\mathbf{x}|^2 \ge 0 + 2 n \epsilon > 0  $$
So $v$ has no interior maximum on $\Omega$. But $\Omega$ is bounded so $v$ has a maximum somewhere on $\bar\Omega$. As there is no interior maximum of $v$ it must have its maximum on $\partial\Omega = \Sigma$. Let: $\mathbf{x_o}$ be the maximum of $v$ on $\bar\Omega$. Then, for $\mathbf{x} \in \bar\Omega$:
$$ u\left(\mathbf{x}\right) \le v\left(\mathbf{x}\right) \le v\left(\mathbf{x_0}\right) = u\left(\mathbf{x_0}\right) + \epsilon |\mathbf{x_0}|^2 \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) + \epsilon \delta^2 $$
As $\delta$ is some constant and we have this inequality for all $\epsilon > 0$, we take the limit as $\epsilon \rightarrow 0$ to yield:
$$ \lim_{\epsilon \rightarrow 0} \{u\left(\mathbf{x}\right) \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) + \epsilon \delta^2 \} \rightarrow \{ u\left(\mathbf{x}\right) \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) \} $$
We have the maximum of $u\left(\mathbf{x}\right)$ is attained on $\mathbf{x} \in \Sigma = \partial\Omega$, as needed. $\square$

Minimum principle for subharmonic functions:
Let: $u\left(\mathbf{x}\right)$ be superharmonic, $ \Delta u\left(\mathbf{x}\right) \le 0$, on a bounded domain $\Omega$ with boundary $\partial\Omega = \Sigma$, $\mathbf{x} \in \Omega$. Let: $v\left(\mathbf{x}\right) = - u\left(\mathbf{x}\right)$.
Then $v\left(\mathbf{x}\right)$ is subharmonic and attains its maximum on $\Sigma$, by the maximum principle for subharmonic functions.  This is equivalent to a minimum of $u$ and so $u\left(\mathbf{x}\right)$ attains its minimum on $\Sigma = \partial\Omega$. We are done. $\blacksquare$


Part b. Show that the minimum principle for subharmonic functions, and maximum principle for superharmonic functions do not hold.

Answer:
Let: $u\left(\mathbf{x}\right) = |\mathbf{x}|^2 = \sum_{i=1}^{n} x_i^2$ defined on some ball in $\mathbb{R}^n$ about the origin with radius larger than $0$.
Now: $\Delta u = \Delta |\mathbf{x}|^2 = 2 n > 0$ so $u$ is subharmonic. But clearly at $ \mathbf{x} = \mathbf{0}$, $u\left(\mathbf{x}\right) = |\mathbf{x}|^2 = 0$ but at any $\mathbf{x} \ne \mathbf{0}$, $ |\mathbf{x}|^2 > 0$. Then $u$ has an interior minimum and the minimum principle for subharmonic functions does not hold. $\square$
Let: $u\left(\mathbf{x}\right)$ be subharmonic, and $v\left(\mathbf{x}\right) = - u\left(\mathbf{x}\right)$. Then $v$ is superharmonic.
Because the minimum principle does not hold for subharmonic functions as shown, $u$ may have an interior minimum. This is equivalent to an interior maximum for $v$, so the maximum principle does not hold for superharmonic functions. We are done. $\blacksquare$


Part c. Prove that if $u, v, w$ are respectively harmonic, subharmonic, and superharmonic functions in the bounded domain $ \Omega$, coinciding on its boundary $ \{u \bigr|_\Sigma= v\bigr|_\Sigma= w\bigr|_\Sigma\}$, then $w \ge u \ge v$ in $\Omega$.

Answer:
Let: $\Delta u = 0$, $\Delta v \ge 0$, $ \Delta w \le 0$ on $ \Omega$, coinciding on its boundary $\partial\Omega = \Sigma$,  $ \{u \bigr|_\Sigma= v\bigr|_\Sigma, w\bigr|_\Sigma\}$.
Define: $f = v - u$, $g = w - u$.
Then: $\Delta f = \Delta v - \Delta u \ge  0$, $\Delta g = \Delta w - \Delta u \le 0$, and: $ \{u \bigr|_\Sigma= v\bigr|_\Sigma= w\bigr|_\Sigma\}$ so:
$$ \{ f \bigr|_\Sigma= \left(v-u\right)\bigr|_\Sigma = 0, \, g \bigr|_\Sigma= \left(w-u\right)\bigr|_\Sigma = 0 \} $$
$\Delta f \ge 0$, so $f$ is subharmonic and by the maximum principle attains its maximum on $\Sigma$. Then in $\Omega$, $f \le f \bigr|_\Sigma = 0$ so $f = v - u \le 0 \implies v \le u$ in $\Omega$.
Similarly, $\Delta g \le 0$ so $g$ is superharmonic and by the minimum principle attains its minimum on $\Sigma$. So in $\Omega$, $g \ge g \bigr|_\Sigma = 0$ so $g = w - u \ge 0 \implies w \ge u$ in $\Omega$.
We have in $\Omega$, $v \le u$, and $ w \ge u$. So $w \ge u \ge v$ inside of $\Omega$, as needed. $\blacksquare$

Problem 1

Part a. Find the solutions that depend only on $r$ of the equation
$$ \Delta u\left(x,y\right) := u_{xx} + u_{yy} = 0    $$

Answer:
We derived that transforming Laplacian $\Delta := \partial_x^2 + \partial_y^2$ into polar coordinates  $u\left(x,y\right) \rightarrow u\left(r,\theta\right)$ yields us the polar form of the Laplacian:
$$ \Delta := \partial_r^2 + \frac{1}{r} \partial_r + \frac{1}{r^2} \partial_\theta ^2 $$
$u\left(r\right)$ is constant with respect to $\theta$, $u_{\theta\theta} = 0 $, so we have:
$$ \Delta u\left(x,y\right) := u_{xx} + u_{yy} = 0 $$
$$ \implies \Delta u\left(r\right) := u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^2} u_{\theta\theta} = u_{rr} + \frac{1}{r} u_{r} = 0  $$
$$ \implies r u_{rr} + u_r = 0 \implies \partial_r \left(r u_r\right) = 0 \implies r u_r = A \implies u_r = \frac{A}{r} $$
$$ \implies u\left(r\right) =  A \log\left(r\right) + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$


Part b. Find the solutions that depend only on $\rho$ of the equation
$$ \Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = 0    $$

Answer:
We derived that transforming Laplacian $\Delta := \partial_x^2 + \partial_y^2 + \partial_z^2$ into spherical coordinates  $u\left(x,y,z\right) \rightarrow u\left(\rho,\theta,\phi\right)$ yields us the spherical form of the Laplacian:
$$ \Delta := \partial_\rho^2 + \frac{2}{\rho} \partial_\rho + \frac{1}{\rho^2}\left( \partial_\theta ^2 + \cot\left(\theta\right) \partial_\theta + \frac{1}{\sin\left(\theta\right)^2} \partial_{\phi \phi}\right) $$
$u\left(\rho\right)$ is constant with respect to $\theta$, $u_{\theta} = u_{\theta\theta} = u_{\phi \phi} = 0 $, and we have:
$$ \Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = 0 $$
$$ \implies \Delta u\left(\rho\right) := u_{\rho \rho} + \frac{2}{\rho} u_{\rho} = 0  $$
$$ \implies \rho^2 u_{\rho \rho} + 2 \rho u_{\rho} = 0 \implies \partial_\rho \left(\rho^2 u_{\rho}\right) = 0  $$
$$ \implies \rho^2 u_{\rho} = -A \implies u_{\rho} = -\frac{A}{\rho ^2} \implies u_{\rho} = - \frac{A}{\rho ^2} $$
$$ \implies u\left(\rho\right) =  A \rho^{-1} + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$


Part c. In the $n$-dimensional case, prove that for $u=u\left(r\right)$, with $r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}} $:
\begin{equation}
\Delta u = u_{rr} + \frac{n-1}{r}u_r = 0 \label{1.c.1}
\end{equation}

Answer:
$$ \text{Let: } r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right). $$
$$\implies \Delta u = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(r\right) = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right) = 0 $$
$$ \implies \sum_{i=1}^{n} [ \partial_{x_i}^2 u\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) ] = \sum_{i=1}^{n} [ \partial_{x_i}\frac{ x_i u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} ]$$
$$ = \sum_{i=1}^{n} [ \frac{ u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} - \frac{ x_{i}^2 u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} + \frac{ x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)}] $$
$$ = \sum_{i=1}^{n} [    \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}}          +     \frac{x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)}  ]$$
$$ = \sum_{i=1}^{n} [    \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} ]         +    \frac{\left(\sum_{j=1}^{n}  x_{j}^2\right) u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\sum_{j=1}^{n}  x_{j}^2}  $$
$$ = u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)         +   \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} \sum_{i=1}^{n} \left(    \frac{ \left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2 }{ \sum_{j=1}^{n} x_j^2} \right)   $$
$$ \text{Notice that: } \sum_{i=1}^{n} [    \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) }{ \left(\sum_{j=1}^{n} x_j^2\right)} ] = \sum_{i=1}^{n} \left(    \frac{ \sum_{j = 1}^{n} x_j^2 }{ \sum_{j=1}^{n} x_j^2} \right) - \sum_{i=1}^{n} \left(    \frac{ x_{i}^2}{ \sum_{j=1}^{n} x_j^2} \right)  $$
$$ = n -    \frac{ \sum_{i=1}^{n} x_{i}^2}{ \sum_{j=1}^{n} x_j^2} = n -1 \text{ so we have:}$$
$$ \implies \Delta u = u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)         +  \left(n-1\right) \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} $$
$$ =  u_{rr} + \frac{n-1}{r}u_r = 0 \text{, as needed. } \blacksquare $$


Part d. In the $n$-dimensional case, $\left(n \ne 2\right)$, prove that $u = u\left(r\right)$ satisfies the Laplace equation for $r \ne 0 \iff u = A r^{2-n} + B$, $\{ A, B \} \in \mathbb{R}$.

Answer:
$$ \text{Let: } n \in  \mathbb{N} \setminus 2, \phantom{O} \{ x_1 \dots x_n \} \in \mathbb{R}^n, \phantom{O} r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right). $$
By part c. we have that the Laplacian of $u\left(r\right)$ satisfies \eqref{1.c.1}:
$$ \Delta u = u_{rr} + \frac{n-1}{r}u_r = 0 $$
If $r \ne 0$, $u\left(r\right) = A r^{2-n} + B$, $u_{r} = A\left(2-n\right) r^{1-n}$, $u_{rr} = A\left(1-n\right)\left(2-n\right) r^{-n}$ and clearly:
$$ u_{rr} + \frac{n-1}{r}u_r = A \left(1-n\right) \left(2 - n\right) r^{-n} +  \frac{n-1}{r} A \left(2 - n\right) r^{1 - n} $$
$$ = A \left(1 - n\right) \left(2 - n\right) r^{-n} - A \left(1 - n\right) \left(2 - n\right) r^{-n} = 0 \phantom{O} \square$$
Thus $u$ satisfies Laplace's equation in $r$. Conversely, if $u\left(r\right)$ satisfies Laplace's equation in $r$ \eqref{1.c.1} for $r\ne 0$, then:
$$u_{rr} + \frac{n-1}{r}u_r = 0 \implies r^{n-1} u_{rr} + \left(n-1\right)r^{n-2}u_r = 0 \implies \partial_r\left(r^{n-1} u_{r}\right)= 0 $$
$$\implies r^{n-1} u_{r}= \left(2-n\right) A \implies u_{r}= \left(2-n\right)\frac{A}{r^{n-1}} $$
$$ \implies u\left(r\right) = A r^{2-n} + B \phantom{O} \square$$
Thus we have $u = u\left(r\right)$ satisfies Laplace's equation in $r \ne 0$, $n \in  \mathbb{N} \setminus 2$,
$$\Delta u\left(r\right) = 0 \iff u\left(r\right) = A r^{2-n} + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$

Problem 3

Part a. Solve:

$$ \Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r < a  $$
$$ u \bigr|_{r=a} = f\left(\theta\right) $$
$$ \text{in polar coordiantes } \left(r,\theta\right) \text{, with: }
f(x) = \left\{\begin{aligned}
&\phantom{-\ }1 &&: 0 < \theta < \pi\\
&-1 &&: \pi < \theta < 2\pi
\end{aligned}
\right.
$$


Answer:
If $a \le 0$, clearly the problem is not well-posed or yields only a singular point-solution, as for polar coordinates we take only $r \ge 0$. Let $a > 0$. Then this is the Dirichlet problem on the disk of radius $ a $, with BC $h\left(\theta\right) = f\left(\theta\right)$. Because we have $u = f$ on the circumference of the disk, we have that $|u| \le |f| = 1$ on the disk by the maximum principle, and $|u|$ is bounded. Then, it was derived that by changing to polar coordinates $\left(x,y\right) \mapsto \left(r,\theta\right)$ with Dirichlet BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a solution of the form:

$$ u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right) $$

$$ A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\cos\left(n\phi\right)d\phi $$

$$ B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\sin\left(n\phi\right)d\phi $$

Substituting in $f\left(\theta\right)$ and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$:

$$ \implies A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f\left(\phi\right)\cos\left(n\phi\right)d\phi = \frac{1}{\pi a^n}\left( \int_{0}^{\pi}\cos\left(n\phi\right)d\phi -  \int_{\pi}^{2\pi}\cos\left(n\phi\right)d\phi\right) $$

$$= \frac{1}{\pi a^n}\left( \frac{1}{n}\sin\left(n\phi\right) \bigr|_{0}^{\pi} -  \frac{1}{n}\sin\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = 0 \text{, as } n \in \mathbb{N}$$

$$ \implies B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f\left(\phi\right)\sin\left(n\phi\right)d\phi = \frac{1}{\pi a^n}\left( \int_{0}^{\pi}\sin\left(n\phi\right)d\phi -  \int_{\pi}^{2\pi}\sin\left(n\phi\right)d\phi\right) $$

$$= \frac{1}{\pi a^n}\left( -\frac{1}{n}\cos\left(n\phi\right) \bigr|_{0}^{\pi} + \frac{1}{n}\cos\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = \frac{1}{n \pi  a^n}2\left(1-\left(-1\right)^n\right) \text{, as } n \in \mathbb{N}$$

$$ \implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n \frac{2\left(1-\left(-1\right)^n\right)}{n \pi } \sin\left(n \theta\right) \phantom{\ } \square $$

If instead we approached the Dirichlet problem on the disk from Poisson's formula, we have that $u\left(r,\theta\right)$ has a solution of the form:

$$ u\left(r,\theta\right) = \left(a^2 - r^2\right)\int_{0}^{2\pi}\frac{h\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} $$

 Substituting in $f\left(\theta\right)$ for $h\left(\theta\right)$, and splitting between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$ gives the irreducible integrals:

$$ = \left(a^2 - r^2\right)\left(\int_{0}^{\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} - \int_{\pi}^{2\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}\right) \phantom{\ } \blacksquare $$



Part b. Solve:

$$ \Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r > a  $$
$$ u \bigr|_{r=a} = f\left(\theta\right) $$
$$ \max |u| < \infty $$
$$ \text{in polar coordiantes } \left(r,\theta\right) \text{, with: }
f(x) = \left\{\begin{aligned}
&\phantom{-\ }1 &&: 0 < \theta < \pi\\
&-1 &&: \pi < \theta < 2\pi
\end{aligned}
\right.
$$

Answer:If $a \le 0$, clearly the problem is not well-posed. For $r < 0$ our BC $ u \bigr|_{r=a} = f\left(\theta\right) $ is undefined, as $r \ge 0$, and for $r = 0$ the singular point-BC does not specify a solution. Let $a > 0$. Then this is the Dirichlet problem on the exterior of the disk of radius $ a $, with BC $h\left(\theta\right) = f\left(\theta\right)$. We can view this as an equivalent problem on the interior of the disk by considering the rotationally invariant projection of the real line of the Riemann sphere to itself with the rule: $r\mapsto \frac{1}{r}$ : $\{ z : z \in \mathbb{R}_\infty , z \ge a \} \mapsto \{ \frac{1}{z} : \frac{1}{z} \in \mathbb{R}_{\infty} , \frac{1}{z} \le \frac{1}{a}\} $. This gives a mapping the Dirichlet problem on the exterior of the disk in $r$ with BC at $a$ to the Dirichlet problem on the interior of the disk in $\frac{1}{r}$ with BC at $\frac{1}{a}$. This map exists as $|u|$ bounded as $r \rightarrow \infty$ which implies that we have a removable
singularity of $u$ at our point at infinity, so $u$ can be analytically extended there, and so may the equivalent mapped point at $0$ when $\frac{1}{r} \
\rightarrow 0$ as $r \rightarrow \infty$.

We derived that the Dirichlet problem on the interior of the disk with BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a solution of the form:

$$ u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right) $$

$$ A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\cos\left(n\phi\right)d\phi $$

$$ B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\sin\left(n\phi\right)d\phi $$

Substituting in $f\left(\theta\right)$, mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$, gives us the solution on the exterior:

$$ \implies u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^{-n} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right) $$

$$ \implies A_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f\left(\phi\right)\cos\left(n\phi\right)d\phi = \frac{a^{n}}{\pi }\left( \int_{0}^{\pi}\cos\left(n\phi\right)d\phi -  \int_{\pi}^{2\pi}\cos\left(n\phi\right)d\phi\right) $$

$$= \frac{a^{n}}{\pi}\left( \frac{1}{n}\sin\left(n\phi\right) \bigr|_{0}^{\pi} -  \frac{1}{n}\sin\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = 0 \text{, as } n \in \mathbb{N}$$

$$ \implies B_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f\left(\phi\right)\sin\left(n\phi\right)d\phi = \frac{a^{n}}{\pi }\left( \int_{0}^{\pi}\sin\left(n\phi\right)d\phi -  \int_{\pi}^{2\pi}\sin\left(n\phi\right)d\phi\right) $$

$$= \frac{a^{n}}{\pi }\left( -\frac{1}{n}\cos\left(n\phi\right) \bigr|_{0}^{\pi} + \frac{1}{n}\cos\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = \frac{a^n}{n \pi  }2\left(1-\left(-1\right)^n\right) \text{, as } n \in \mathbb{N}$$

$$ \implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{a}{r}\right)^n \frac{2\left(1-\left(-1\right)^n\right)}{n \pi } \sin\left(n \theta\right) \phantom{\ } \square $$

 If instead we approached the Dirichlet problem on the interior disk from Poisson's formula, we have that $u\left(r,\theta\right)$ has a solution of the form:

$$ u\left(r,\theta\right) = \left(a^2 - r^2\right)\int_{0}^{2\pi}\frac{h\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} $$

Substituting in $f\left(\theta\right)$ for $h\left(\theta\right)$, taking $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$, gives us the solution on the exterior in the form of two irreducible integrals:

$$ = \left(a^{-2} - r^{-2}\right) \left(\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^{-2}-2 a^{-1} r^{-1} \cos\left(\theta-\phi\right)+r^{-2}}\frac{d \phi}{2\pi}\right)$$

$$ = \frac{a^2 r^2}{a^2 r^2}\left(a^{-2} - r^{-2}\right)\left(\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^{-2}-2 a^{-1} r^{-1} \cos\left(\theta-\phi\right)+r^{-2}}\frac{d \phi}{2\pi}\right)$$

$$ =\left(r^2 - a^2\right)\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} $$

$$ = \left(r^2 - a^2\right)\left(\int_{0}^{\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} - \int_{\pi}^{2\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}\right) \phantom{\ } \blacksquare $$

Problem 1

Part a Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation

$$ \Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = k^2 u    $$

Answer:
As previously derived, transforming $u\left(x,y,z\right)$ into spherical coordinates $u\left(r,\theta,\phi\right)$ yields the PDE:

$$ \Delta u\left(r,\theta,\phi\right) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}\left(u_{\theta\theta} +\cot\left(\theta\right)u_\theta + \frac{1}{\sin\left(\theta\right)^2}u_{\phi\phi}\right) = k^2 u $$

Since we seek only $u\left(r,\theta,\phi\right)$ which depends $r$, we have that $u\left(r\right)$ is constant with respect to both of $\{\theta,\phi\}$. Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, and our equation simplifies to an ODE:

$$ \Delta u\left(r\right) := u_{rr} + \frac{2}{r} u_{r} = k^2 u $$

Substituting in $ u\left(r\right) = \frac{v\left(r\right)}{r} \implies u_{r} = \frac{v_r}{r} - \frac{v}{r^2} $, $u_{rr} = \frac{v_{rr}}{r} - \frac{v_{r}}{r^2} - \frac{v_{r}}{r^2}  + \frac{2 v}{r^3} $ yields:

$$  \Delta u := \left(\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}\right) + \frac{2}{r} \left( \frac{v_r}{r} - \frac{v}{r^2} \right) = k^2  \frac{v}{r}  $$

Multiplying through by $r$ and simplifying terms then gives us a nice expression:

$$ v_{rr} = k^2 v $$

Since we have that $ k > 0 $, $k^2 > 0$, our ODE has the solution, with $ \{A,B,C,D\} \in \mathbb{R} $:

$$ v\left(r\right) = A e^{k r} + B e^{-k r} = C \cosh\left(k r\right) + D \sinh\left(k r\right) $$

$$ \implies u\left(r\right) = \frac{1}{r} v\left(r\right) = \frac{1}{r}\left(A e^{k r} + B e^{-k r}\right) =\frac{1}{r} \left(C \cosh\left(k r\right) + D \sinh\left(k r\right)\right) \phantom{\ } \blacksquare$$



Part b. Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation

$$ \Delta u := u_{xx} + u_{yy} + u_{zz} = -k^2 u    $$


Answer:
Following the derivation of part a., we substitute in spherical coordinates: $u\left(x,y,z\right) \rightarrow u\left(r,\theta,\phi\right)$, assume a form of $u\left(r,\theta,\phi\right)$ which depends only on $r$, $u\left(r\right)$. Then make the substitution $ u\left(r\right) = \frac{v\left(r\right)}{r} $ and simplify the resulting ODE. Because the left hand side of our equation is unchanged between part a. and b., this derivation yields identical equations on the left hand side. The difference in the two problems appears only in the right hand side of the expression, where $ \Delta u = -k^2 u $ in place of $ \Delta u = k^2 u $. After simplification this yields the ODE:

$$ v_{rr} = - k^2 v $$

Which, as $k > 0$, $-k^2 < 0$ has a solution in the form, for some $ \{A,B,C,D\} \in \mathbb{R} $:

$$ v\left(r\right) = A e^{i k r} + B e^{-i k r} = C \cos\left(k r\right) + D \sin\left(k r\right) $$

$$ \implies u\left(r\right) = \frac{1}{r} v\left(r\right) = \frac{1}{r}\left(A e^{i k r} + B e^{-i k r}\right) =\frac{1}{r} \left(C \cos\left(k r\right) + D \sin\left(k r\right)\right) \phantom{\ } \blacksquare$$

I ran into a problem typesetting the piecewise functions of problem 4 when converting my solutions for homework 6 to mathjax here: http://forum.math.toronto.edu/index.php?topic=118.msg626#msg626

Namely, I can't seem to get everything to line up nicely without cheating and using \phantom{}, which is just a really clumsy solution. Here's the code I'm using:

   f\left(x\right) = \left\{
     \begin{array}{lr}
       1 & : |x| \le 1\\
       0 & : |x| > 1
     \end{array}
   \right.\\
Which looks like this:

$$
   f\left(x\right) = \left\{
     \begin{array}{lr}
       1 & : |x| \le 1\\
       0 & : |x| > 1
     \end{array}
   \right.\\$$

Is there a better way of typesetting this?

Yeah, this method works by temporarily transforming the initial function $u(x,y)$ by $y \mapsto k$ and using the properties of the Fourier transform to get an ODE in the other variable $x$ of the function $U(x,k)$. Using that the Fourier transform is unique and invertible, after solving the ODE in $x$ we transform our ODE solution $U(x,k)$ back from $k \mapsto y$ to get a solution for the PDE in terms of $u(x,y)$.

If we had $e^{i k x}$ in the last line, our final transform would be an inverse Fourier transform mapping $k \mapsto x$ on the $U(x,k)$ solution, so our answer would be a single variable function $u(x,x)$. Since we wanted a solution to the initial PDE in both $(x,y)$ we instead apply the inverse transformation $k \mapsto y$, as is written in the initial solution.

I take the definition of the Fourier transform $ \hat{f} $ for a function $ f $ to be: $ F\left(k\right) = \hat{f}\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx, $ with inverse Fourier transform $ f\left(x\right) = \check{F}\left(x\right) = \int_{-\infty}^{\infty}F\left(k\right) e^{i k x}\frac{dk}{2 \pi} $.
Problem: Find a solution using the Fourier transform for the Laplace equation in the half-plane (it was a misprint in the problem. V.I.):
$$ \Delta u = u_{xx} + u_{yy} = 0 : \{x > 0, -\infty < y < \infty\} $$
with the boundary conditions:
$$ u(0,y) = e^{-|y|}, \phantom{\ } \max_{\{x,y\}} |u| < \infty $$
Answer: Now, $x > 0$ so we cannot perform a Fourier transform over $x$. We proceed instead by partially transforming with respect to $y \mapsto k$, $u(x,y) \mapsto \mathcal{F}_y (u)(x,k)$:
$$ \text{Let: } U(x,k) = \mathcal{F}_y (u)(x,k) = \int_{-\infty}^{\infty} u(x,y) e^{-i k y} d y $$
We know that this transformation converges and exists as we have that $|u|$ is bounded by some constant $c < \infty $. Recall that the Fourier transform $\mathcal{F} $ is an operator, and that for a function $f(x)$ with transform $\mathcal{F}(f)(k)$, a property of the transform $\mathcal{F}$ is that for $ g(x) = \partial_x f(x) $, $ \mathcal{F}(g)(k) = i k \mathcal{F}(f)(k)$. Applying the partial transformation $ \mathcal{F}_y $ to our PDE and BC then yields us:
$$ \mathcal{F}_y((u_{xx} + u_{yy})(x,y) = (0)(x,y)) \mapsto \mathcal{F}_y(u_{xx})(x,k) +  \mathcal{F}_y(u_{yy})(x,k) =  \mathcal{F}_y(0)(x,k) = (0)(x,k) $$
$$ = \mathcal{F}(\partial_{x}^2 u)(x,k) + \mathcal{F}(\partial_{y}^2 u)(x,k) = \partial_{x}^2 \mathcal{F}(u)(x,k) + (i k)^2 \mathcal{F}(u)(x,k) = 0 $$
$$ \text{And BC: } \mathcal{F}_y(u(0,y) = e^{-|y|}) \mapsto \mathcal{F}_y(u)(0,k) = \mathcal{F}_y (e^{-|y|})(k) = \frac{2}{1+k^2} $$
Where we differentiated $ \mathcal{F}_y(\partial_{x}^2 u)(x,k) $ under the sign, and used that because $ \mathcal{F} $ is an operator, we must have $ \mathcal{F}(0) = 0 $. Using $ U(x,k) = \mathcal{F}_y (u)(x,k) $, we then have an ODE in $x$ with BC:
$$ U_{xx} + (i k)^2 U = U_{xx} -k^2 U = 0  $$
$$ U(0,k) = \mathcal{F}_y (e^{-|y|})(k) = \frac{2}{1+k^2} $$
Using our previously derived transformation for $ e^{-|y|} $. This has solution $ U(x,k) = \Psi(k) e^{- |k| x} + \Phi(k) e^{+ |k| x}$, with $\{ \Psi(k),\Phi(k)\}$ arbitrary functions of $k$, where we take $|k|$ in place of $k$ to control the asymptotic behavior as $\{|k|,x\} \rightarrow \infty$. Since $x > 0$ we must exclude $ e^{|k| x} $ in our solution since it grows without bound as $ |k| \rightarrow \infty$ and so doesn't allow for the convergence of the inverse Fourier transformation. Then $U(x,k) = \Psi(k) e^{- |k| x}$. Plugging in our BC:
$$  U(0,k) = \frac{2}{1+k^2} = (\Psi(k) e^{- |k| x})\bigr|_{x=0} = \Psi(k) $$
$$ \implies \Psi(k) = \frac{2}{1+k^2} \implies U(x,k) = \frac{2}{1+k^2} e^{- |k| x} $$
Finally, we apply the inverse Fourier Transformation $\mathcal{F}^{-1}$ with respect to $k$ on $U(x,k)$ to solve in terms of $u(x,y)$:
$$\mathcal{F}^{-1}_k (U(x,k)) = \mathcal{F}_k^{-1}(\mathcal{F}_y(u(x,y))) \mapsto u(x,y) $$
$$ \implies u(x,y) = \mathcal{F}^{-1}_k (\frac{2}{1+k^2} e^{- |k| x}) = \int_{-\infty}^{\infty}(\frac{2}{1+k^2} e^{- |k| x}) e^{i k y} \frac{dk}{2 \pi} \phantom{\ } \blacksquare$$ \\ \\