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### Messages - Calvin Arnott

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1
##### Home Assignment 8 / Re: Problem 4
« on: November 28, 2012, 10:13:06 PM »
Anyone knows full solution of Problem 4? One of my ideas is gravitational potential approaches zero at infinity to set B=0 as:
http://www.math.toronto.edu/courses/apm346h1/20129/L27.html

I'm curious to see how everybody else approached this one as well, I didn't have time this week to write an answer I was happy with. The idea I was going to use was essentially take equations (2)/(3) in lecture 27, and take a rotationally symmetric density function $f(x) = c$ for some constant $c$. Evaluate the integral to be some other constant $k$. Then use that the $r^{2-n}$ in $n = 3$ gives us a term proportional to $r$.

2
##### Home Assignment 8 / Re: Problem 3
« on: November 28, 2012, 09:55:02 PM »
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.

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.

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 $\Omega$.

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. $\blacksquare$

3
##### Home Assignment 8 / Re: Problem 2
« on: November 28, 2012, 09:51:31 PM »
Problem 2
Let: $\Delta u \ge 0$ in $B\left(y,r\right)$, $\omega_n$ be the volume of the $n$-dimensional unit ball, and $\sigma_n = n \omega_n$. Prove that:

Part a. $u\left(y\right)$ does not exceed the mean value of $u$ over the sphere $S\left(y,r\right)$ bounding the ball $B\left(y,r\right)$:

u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \label{2.a.0}

In $\Delta u \ge 0$, $u$ is twice differentiable, so the spherical mean is continuous and we have:

\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \rightarrow u\left(y\right) \label{2.a.1}

Thus, to prove $u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$, it is sufficient to show that $\frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$ increases with $r > 0$. By the divergence theorem we have for a $C^1$ function $F$ on a compact set $\Omega$ with boundary $\partial\Omega = \Sigma$:
$$\int_\Omega \left(\nabla \cdot F\right) \,dV = \int_\Sigma \left(F \cdot n\right) \, dS \text{, namely: }$$
$$\int_{B\left(y,r\right)} \left(\nabla \cdot \nabla \right)\left(x\right) \,dV = \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = \int_{S\left(y,r\right)} \left(\nabla u \cdot n\right)\left(x\right) \, dS = \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS$$
As $\Delta u \ge 0$, we then have:
$$0 \le \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS$$
Parametrizing by the normal $\nu = \frac{x-y}{|x-y|}$ on the boundary where $|x-y| = r$, we have: $\frac{\partial u}{\partial n} \left(x\right) = \partial_r u\left(y+ r \nu\right)$ on $T : \{ \nu : | \nu | = 1 \}$, and $dS = r^{n-1} d \nu$
What? See note below V.I..

Then:
$$\int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS = \int_{T} \partial_r\left( u\left(y+ r \nu\right)\right) r^{n-1} \, d \nu$$
By differentiating under the sign with $\partial_r$ and pulling the $r^{n-1}$ term out of the integral, we have:
$$= r^{n-1} \partial_r \int_{T} u\left(y+ r \nu\right) \, d \nu = r^{n-1} \partial_r \left(r^{1-n} \int_{S\left(y,r\right)} u\left(x\right) \, d S\right)$$
$$= r^{n-1} \partial_r\left(\sigma_n \frac{r^{1-n}}{\sigma_n } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
As $r>0$ and $\sigma_n>0$ by \eqref{2.a.1} we have:
$$0 \le \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) \implies$$

0 \le \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)  \label{2.a.2}

Then by \eqref{2.a.1} as $r \rightarrow 0$ we have equality in \eqref{2.a.0}, and as $r$ increases the spherical mean is non-decreasing by \eqref{2.a.2}, $u\left(y\right)$ constant. Thus we have \eqref{2.a.0} for $r \ge 0$:
$$u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \phantom{O} \blacksquare$$

Part b. $u\left(y\right)$ does not exceed the mean value of $u$ over the ball $B\left(y,r\right)$:

u\left(y\right) \le \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \label{2.b.0}

$$u\left(y\right) = \frac{\sigma_n \left(n r^n\right)}{\left(n \sigma_n\right) r^n} u\left(y\right) = \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u\left(y\right)$$
By \eqref{2.a.0} we have that:
$$u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$$
So because $\{r, \omega_n, \sigma_n\} \ge 0$ we have that:
$$\implies \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u\left(y\right) \le \frac{\sigma_n \int_{0}^{r} t^{n-1}\, dt}{\omega_n r^n} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$$
$$= \frac{1}{\omega_n r^n} \sigma_n\int_{0}^{r}t^{n-1} \,d t \frac{1}{\sigma_n r^{n-1}} \int_T u\left(y +r \nu\right) r^{n-1} \, d \nu$$
Where we parameterized as in part a. with $\nu = \frac{x-y}{|x-y|}$ on the boundary $|x-y| = r$, and $\left(x\right) = u\left(y+ r \nu\right)$ on $T : \{ \nu : | \nu | = 1 \}$, $dS = r^{n-1} d \nu$. Canceling terms gives:
$$= \frac{1}{\omega_n r^n} \int_{0}^{r}t^{n-1} \,d t \int_T u\left(y +r \nu\right) \, d \nu$$
Which allows us to change the order of integration to yield our result:
$$= \frac{1}{\omega_n r^n} \int_{0}^{r} t^{n-1} \int_T u\left(y +r \nu\right) \, d \nu d t = \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V$$
$$\implies u\left(y\right) \le \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \phantom{O} \blacksquare$$

Part c. Formulate similar statements for functions satisfying $\Delta u \le 0$.

In part a. we used the fact that $\Delta u \ge 0$, and $\{ \sigma_n, r \} \ge 0$ to show that the spherical mean is non-decreasing in $r$:
$$0 \le \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
$$\implies 0 \le \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
Because we have equality at $r \rightarrow 0$:
$$\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \rightarrow u\left(y\right)$$
and $u\left(y\right)$ is constant with respect to $r$, we showed that for all $r > 0$
$$u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$$
If instead $\Delta u \le 0$, we would have the inequalities:
$$0 \ge \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
$$\implies 0 \ge \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
And $\frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S$ is non-increasing with $r$, which gives us by the same argument that for all $r>0$
$$u\left(y\right) \ge \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \phantom{O} \square$$
In part b. we merely integrated this result radially to extend the inequality for the interior of the sphere. Doing the same in the case of $\Delta u \le 0$ yields us:
$$u\left(y\right) \ge \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \phantom{O} \blacksquare$$

4
##### Home Assignment 8 / Re: Problem 1
« on: November 28, 2012, 09:47:14 PM »
Problem 1

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

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
$$\Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = 0$$

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 $r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}$:

\Delta u = u_{rr} + \frac{n-1}{r}u_r = 0 \label{1.c.1}

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

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

$$\text{Let: } n \in \mathbb{N} \setminus 2, \phantom{O} \{ x_1 \dots x_n \} \in \mathbb{R}^n, \phantom{O} r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right).$$
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 $r \ne 0$, $n \in \mathbb{N} \setminus 2$,
$$\Delta u\left(r\right) = 0 \iff u\left(r\right) = A r^{2-n} + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$

5
##### Home Assignment 7 / Re: Problem 4
« on: November 19, 2012, 09:38:35 PM »
Problem 4

Part a.

$$\Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r < a$$
$$u_r \bigr|_{r=a} = f\left(\theta\right)$$
\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.

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 Neumann problem on the disk of radius $a$, with BC $h\left(\theta\right) = f\left(\theta\right)$. It was derived that by changing to polar coordinates $\left(x,y\right) \mapsto \left(r,\theta\right)$ has a solution of the form:

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

Differentiating this equation for $u\left(r,\theta\right)$ with respect to $r$ gives us our boundary condition at $r=a$, $f\left(\theta\right)$.

$$u_{r}\left(r,\theta\right) = \sum_{n=1}^{\infty} n r^{n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$u_{r}\left(a,\theta\right) = \sum_{n=1}^{\infty} n a^{n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right) = f\left(\theta\right)$$

$$\text{Notice that: }\int_{0}^{2\pi}f\left(\theta\right) d \theta = \int_{0}^{\pi} d \theta - \int_{\pi}^{2\pi} d \theta = \pi - 2 \pi + \pi = 0$$

So our series representation for $f$ has a free coefficient and we may find a unique solution up to a constant $C \in \mathbb{R}$, despite losing a degree of freedom to differentiation. Our Fourier coefficients at $r = a$ are then given by:

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

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

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

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

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

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

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

$$\implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{r^n}{a^{n-1}}\right) \frac{2\left(1-\left(-1\right)^n\right)}{ \pi n^2 } \sin\left(n \theta\right) + C, \phantom{\ } C \in \mathbb{R} \phantom{\ } \blacksquare$$

Part b. Solve:

$$\Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r > a$$
$$u_r \bigr|_{r=a} = f\left(\theta\right)$$
$$\max |u| < \infty$$
\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.

If $a \le 0$, clearly the problem is not well-posed. For $r < 0$ our BC $u_r \bigr|_{r=a} = f\left(\theta\right)$ 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 Neumann 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 Neumann problem on the exterior of the disk in $r$ with BC at $a$ to the Neumann 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 $r \rightarrow \infty$.

Now- because for our BC again we have: $\int_{0}^{2\pi}f\left(\theta\right) d \theta$ so our BC has a free coefficient, and the Neumann problem on the interior of the disk with BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a unique solution up to a constant $C \in \mathbb{R}$ of the form:

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

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

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

Mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$ gives us the equivalent solution on the exterior:

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

Because we get a negative term when differentiating we have to be a bit more careful with the signs of our coefficients. Our BC at $\frac{1}{r} = \frac{1}{a}$, $u_r = f\left(\theta\right)$ gives us:

$$u_{r} = \sum_{n=1}^{\infty} \left(-n\right) r^{-n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$u_{r} \bigr|_{\frac{1}{a}} = \sum_{n=1}^{\infty} \left(-n\right) a^{-n-1} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)= f\left(\theta\right)$$

So our Fourier coefficients are given by:

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

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

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

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

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

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

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

$$\implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{a^{n+1}}{r^n}\right) \frac{2\left(\left(-1\right)^n - 1\right)}{ \pi n^2 } \sin\left(n \theta\right) + C, \phantom{\ } C \in \mathbb{R} \phantom{\ } \blacksquare$$

6
##### Home Assignment 7 / Re: Problem 3
« on: November 19, 2012, 09:38:03 PM »
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.

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:

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

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

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

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

7
##### Home Assignment 7 / Re: Problem 2
« on: November 19, 2012, 09:37:10 PM »
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 $u\left(r\right)$?

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 $u\left(r\right)$?

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 $\{A,B\} \in \mathbb{R}$.

$$\implies v\left(z\right) = v\left(k r\right) = u\left(r\right) = A J_0\left(k r\right) + B N_0\left(k r\right) \phantom{\ } \blacksquare$$

8
##### Home Assignment 7 / Re: Problem 1
« on: November 19, 2012, 09:35:54 PM »
Problem 1

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

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

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

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

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 $\{A,B,C,D\} \in \mathbb{R}$:

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

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

9
##### Technical Questions / Re: Typesetting piecewise functions
« on: November 18, 2012, 05:39:05 PM »
Thanks- that looks a lot nicer!

10
##### Technical Questions / Typesetting piecewise functions
« on: November 18, 2012, 05:09:34 PM »
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:

Code: [Select]
   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?

11
##### Home Assignment Y / Re: Problem 2
« on: November 14, 2012, 04:01:57 PM »
Nice! BTW, note where approximation is worse. Why here, what do you think?

Hmm, well, intuitively the Fourier series representation is quite smooth. Because it's so smooth, we would expect that whenever it tries to approximate a function that is jagged or discontinuous, it should have errors proportional somehow to how rough the approximated function is.

In this case $1-|y|$ is continuous at y=0, and at the points of periodic extension, and the derivative is piecewise continuous at these points, so we have convergence of the Fourier series. But the derivative has a jump discontinuity, so I would expect some kind of analogue to the Gibbs phenomenon, except in the estimation of the derivative of the function instead of the estimation of the function itself, where the derivative "overshoots" the estimated function's derivative, and converges slower here. I'm not sure how exactly to state that more mathematically, or to make that intuition rigorous if it is correct.

12
##### Home Assignment Y / Re: Problem 1
« on: November 12, 2012, 12:27:03 PM »
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.

13
##### Home Assignment Y / Problem 2
« on: November 12, 2012, 12:17:59 PM »
Problem
Consider the Laplace equation in the half-strip:
$$\Delta u = u_{xx} + u_{yy} = 0 : \{x > 0, -1 < y < 1\}$$
with the boundary conditions:
$$u(x,-1) = u(x,1) = 0$$
$$u(0,y) = 1 - |y|, \phantom{\ } \max_{\{x,y\}} |u| < \infty$$

Part  a. Write the associated eigenvalue problem
First we separate variables in $\{x,y\}$. Let: $(x,y) = X(x)Y(y)$.
$$\implies u_{xx} = X''(x)Y(y), \phantom{\ } u_{yy} = X(x)Y''(y)$$
$$u_{xx} + u_{yy} = 0 \rightarrow X''(x)Y(y) + X(x)Y''(y) = 0$$
$$\implies \frac{X''(x)}{X(x)} = - \frac{Y''(y)}{Y(y)} = \lambda$$
For some constant $\lambda$, as each side depends only on one variable and so is constant with repsect to the other. Then we have two ODEs with BCs, and our eigenvalue problem is:
$$X''(x) - \lambda X(x) = 0, \phantom{\ } Y''(y) + \lambda Y(y) = 0$$
$$Y(-1) = Y(1) = 0, \phantom{\ } \max_{\{x,y\}} |X(x)Y(y)| < \infty, \phantom{\ } u(0,y) = 1 - |y| \phantom{\ } \square$$

Part b. Find all eigenvalues and corresponding eigenfunctions
Answer As previously derived, we make the assumption that all eigenvalues $\lambda_n = \beta_n^2 > 0$. Solving first for $Y(y)$ in:
$$\{Y(y) \phantom{\ } | Y''(y) + \lambda Y(y) = 0, \phantom{\ } Y(-1) = Y(1) = 0\}$$
Let: $\lambda = \beta^2$. Then the ODE $Y''(y) + \lambda Y(y) = Y''(y) + \beta^2 Y(y) = 0$ has solution $Y(y) = A \cos(\beta y) + B \sin(\beta y)$ for some constants $\{A,B\} \in \mathbb{R}$. Plugging in our BCs $\{Y(-1) = Y(1) = 0\}$ yields:
$$Y(-1) = (A \cos(\beta y) + B \sin(\beta y))\bigr|_{y=-1} = A \cos(-\beta) + B \sin(-\beta) = 0$$
$$Y(1) = (A \cos(\beta y) + B \sin(\beta y))\bigr|_{y=1} = A \cos(\beta) + B \sin(\beta) = 0$$
Using that $\sin$ is odd, and $\cos$ is even, we have the equations:
$$A \cos(\beta) - B \sin(\beta) = A \cos(\beta) + B \sin(\beta) = 0$$
Adding and subtracting these equations, then dividing by $2$, gives us:
$$A \cos(\beta) = B \sin(\beta) = 0$$
Now, as $\nexists \beta \in \mathbb{R}$ where $\cos(\beta) = \sin(\beta) = 0$, and we discard the trivial solution $X(x) \equiv 0$ where $A = B = 0$, we must have either $A = 0, B \ne 0, \sin(\beta) = 0 \implies \beta = n \pi$, or $A \ne 0, B = 0, \cos(\beta) = 0 \implies \beta = (n \pi - \frac{\pi}{2})$, for $n \in \mathbb{N}$. Because our BC $u(0,y) = 1 - |y|$ is clearly an even function in $y$, we choose the case where $A \ne 0, B = 0, \cos(\beta) = 0 \implies \beta = (n \pi - \frac{\pi}{2})$. Then our eigenfunction for $Y_n(y)$, and eigenvalues $\lambda_n = \beta_n^2 > 0$, are given by:
$$n \in \mathbb{N} : \lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2, \phantom{\ } Y_n(y) = \cos(\beta_n y) = \cos((n \pi - \frac{\pi}{2}) y)$$
Next, using that $\lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2$, we solve for $X(x)$ in:
$$\{X(x) \phantom{\ } | X''(x) - (n \pi - \frac{\pi}{2})^2 X(x) = 0, \phantom{\ } \max_{\{x\}} |X(x)| < \infty \}$$
This ODE has solutions in the form $X(x) = C e^{(n \pi - \frac{\pi}{2}) x} + D e^{-(n \pi - \frac{\pi}{2}) x}$. But, because $x > 0$, and $(n \pi - \frac{\pi}{2}) > 0$, we must have that $C = 0$ since we require $|X(x)|$ to be bounded as $x \rightarrow \infty$. Otherwise: $C \ne 0 \implies \lim_{x \to \infty} |X(x)| \ge |C e^{(n \pi - \frac{\pi}{2}) x}| \rightarrow \infty$. So each $X_n(x)$ is given by:
$$n \in \mathbb{N} : X_n(x) = e^{-(n \pi - \frac{\pi}{2}) x}$$

Thus, our eigenvalues are given by:
$$n \in \mathbb{N} : \lambda_n = (n \pi - \frac{\pi}{2})^2$$
With corresponding eigenfunctions:
$$u_n(x,y) = A_n X_n(x) Y_n(y) = A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y), \phantom{\ } A_n \in \mathbb{R} \phantom{\ } \square$$

Part c. Write the solution in the form of a series expansion

We have for $n \in \mathbb{N}, A_n \in \mathbb{R}$ the eigenvalues $\lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2$ and eigenfunctions $u_n(x,y) = A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y)$. Then the series expansion of our solution $u(x,y)$ is given by:
$$u(x,y) = \sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y)$$
Applying our final BC: $u(0,y) = 1 - |y|$:
$$u(0,y) = (\sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y))\bigr|_{x=0}$$
$$= \sum_{n=1}^{\infty} A_n \cos(\pi(n - \frac{1}{2}) y) = 1 - |y| : \{ -1 < y < 1 \}$$
Proceeding; recall that the set of half-integer cosines:
$$\{ X_n(x) = \cos(\pi(n - \frac{1}{2}) x) : n \in \mathbb{N} \}$$
$$\text{With inner product: } \langle f,g \rangle = \int_{-1}^{1} f(x)\overline{g(x)} dx$$
forms an orthonormal basis for the space of even functions which converge on $[-1,1]$ in the $L^2$ sense: $\| f \| = \langle f,f \rangle ^{\frac{1}{2}} < \infty$. Then the projection of $f(y) = 1 - |y|$ onto the function space is given by:
$$f(y) = 1 - |y| = \sum_{n=1}^{\infty} A_n X_n(y) = \sum_{n=1}^{\infty} \langle f,X_n \rangle \frac{X_n(y)}{\| X_n \| ^2}$$
$$\implies A_n = \langle f,X_n \rangle \cdot \frac{1}{\| X_n \|^2} = \int_{-1}^{1} f(x)\overline{X_n(x)}dx \cdot 1$$
$$= \int_{-1}^{1} (1-|x|)\cos(\pi(n - \frac{1}{2}) x)dx = 2 \int_{0}^{1} (1-x)\cos(\pi(n - \frac{1}{2}) x)dx$$
$$= \frac{8 - \sin(n \pi)}{(2 n \pi - \pi)^2} = \frac{8}{\pi^2(2 n - 1)^2} = A_n$$
Finally then, our solution $u(x,y)$ is given by the series:
$$u(x,y) = \sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) = \sum_{n=1}^{\infty} \frac{8}{\pi^2(2 n - 1)^2} e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) \phantom{\ } \blacksquare$$ \\ \\

14
##### Home Assignment Y / Problem 1
« on: November 11, 2012, 12:49:08 PM »
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 $u(x,y)$:
$$\mathcal{F}^{-1}_k (U(x,k)) = \mathcal{F}_k^{-1}(\mathcal{F}_y(u(x,y))) \mapsto u(x,y)$$
$$\implies u(x,y) = \mathcal{F}^{-1}_k (\frac{2}{1+k^2} e^{- |k| x}) = \int_{-\infty}^{\infty}(\frac{2}{1+k^2} e^{- |k| x}) e^{i k y} \frac{dk}{2 \pi} \phantom{\ } \blacksquare$$ \\ \\

15
##### Term Test 2 / Re: Scope of Term Test 2
« on: November 10, 2012, 06:00:39 PM »