Title: Methods

Q: Homogeneous boundary conditions
A: A boundary condition is \emph{homogeneous} if whenever $f$ and $g$ satisfy the boundary conditions, then so does $\lambda f + \mu g$ for any $\lambda, \mu \in \C$ (or $\R$).

Q: Fourier Series
A: \textbf{Theorem} (Fourier). Any periodic function $f: S^1 \to \C$ can be expanded in the orthonormal basis $\{ e^{in\theta} \ | \ n\in\Z \}$ such that
\[
  f(\theta) = \sum_{n \in \Z}\hat{f}_n e^{in\theta},
\]
where the $n$th Fourier component is
\[
  \hat{f}_n = \frac{1}{2\pi} (e^{in\theta}, f) = \frac{1}{2\pi}\int_{-\pi}^\pi e^{-in\theta} f(\theta)\;\d \theta.
\]

Q: Real Fourier Series
A: If $f: S^1 \to \R$ is a real function, then
\[
  (\hat{f}_n)^* = \left(\frac{1}{2\pi}\int_{-\pi}^\pi e^{-in\theta}f(\theta)\;\d \theta\right)^* = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{in\theta}f(\theta)\;\d \theta = \hat{f}_{-n}.
\]
Hence the Fourier series becomes
\begin{align*}
  f(\theta) &= \hat{f}_0 + \sum_{n = 1}^\infty\left(\hat{f}_n e^{in\theta} + \hat{f}_{-n}e^{-in\theta}\right) \\
& = \hat{f}_0 + \sum_{n = 1}^\infty \left(\hat{f}_n e^{in\theta} + (\hat{f}_n e^{in\theta})^*\right) \\
 &= \hat{f}_0 + \sum_{n = 1}^\infty (a_n \cos n\theta + b_n \sin n\theta)\\
  &= \frac{a_0}{2} + \sum_{n = 1}^\infty (a_n \cos n\theta + b_n \sin n\theta)
\end{align*}
where $\displaystyle \hat{f}_n = \frac{a_n - ib_n}{2}$ such that
\begin{align*}
  a_n &= 2\Re \hat{f}_n= \frac{1}{\pi}\int_{-\pi}^\pi \cos n\theta f(\theta) \;\d \theta \\
 b_n &= -2\Im \hat{f}_n=-\frac{1}{\pi}\int_{-\pi}^\pi \sin n\theta f(\theta) \;\d \theta.
\end{align*}

Q: Parseval's theorem
A: \[
    (f, f) = \frac{1}{2\pi}\int_{-\pi}^\pi |f(\theta)|^2 \;\d \theta = \sum_{n\in \Z}|\hat{f}_n|^2.
  \]
\begin{proof}
  \begin{align*}
    (f, f) &= \frac{1}{2\pi}\int_{-\pi}^\pi |f(\theta)|^2 \;\d \theta\\
    &= \frac{1}{2\pi}\int_{-\pi}^{\pi} \left(\sum_{m \in \Z} \hat{f}_m^* e^{-im\theta}\right)\left(\sum_{n \in \Z} \hat{f}_n e^{in\theta}\right)\;\d \theta\\
    &= \frac{1}{2\pi}\sum_{m, n\in \Z}\hat{f}_m^* \hat{f}_n \int_{-\pi}^\pi e^{i(n - m)\theta}\;\d \theta\\
    &=  \sum_{m, n\in \Z}\hat{f}_m^* \hat{f}_n \delta_{mn}\\
    &= \sum_{n\in \Z}|\hat{f}_n|^2.
  \end{align*}
\end{proof}

Q: Convergence of Fourier Series with continuity
A: Suppose for some numbers $\{a_n\}_{n\in\Z}$, the partial sums $S_N=\sum_{n=-N}^N |a_n|$ converge as $N\to\infty$. Then $\sum_{n=-N}^N a_n e^{in\theta}$ converges uniformly to a function $f:S^1\to\C$, $f$ is continuous. and $a_n=\hat{f}_n$
\begin{proof}
\textbf{Uniform convergence}. Since $\sum_{n=-N}^N |a_n|$ converges, given $\varepsilon>0$, $\exists N_0$ such that $\forall N\geq M \geq N_0: \sum_{M \leq |n| \leq N} |a_n| < \varepsilon$. Hence for all $\theta\in S^1:$
\begin{align*}
\left| \sum_{M \leq |n| \leq N} a_n e^{in\theta} \right| &\leq \sum_{M \leq |n| \leq N} |a_n| |e^{in\theta}| \\
&= \sum_{M \leq |n| \leq N} |a_n| \\
&< \varepsilon.
\end{align*}
So $\sum_{n=-N}^N a_n e^{in\theta}$ converges uniformly to some $f$ as $N\to\infty$.
\textbf{Continuity}. The partial sums $\sum_{n=-N}^N a_n e^{in\theta}$ are all continuous functions and the uniform limit of a sequence of continuous functions is continuous.
\textbf{Fourier components}. The $m$th coefficient can be written as
\begin{align*}
a_m &= \lim_{N\to\infty} \sum_{n=-N}^N \left[ \frac{1}{2\pi}a_n (2\pi\delta_{mn}) \right] \\
&= \lim_{N\to\infty} \sum_{n=-N}^N \left[ \frac{1}{2\pi}a_n \int_{-\pi}^\pi e^{i(n-m)\theta} \d \theta \right] \\
&= \frac{1}{2\pi} \int_{-\pi}^\pi \left( \lim_{N\to\infty} \sum_{n=-N}^N a_ne^{in\theta}\right) e^{-im\theta} \d \theta \\
&= \frac{1}{2\pi} \int_{-\pi}^\pi f(\theta) e^{-im\theta} \d \theta \\
&=\hat{f}_m.
\end{align*}
\end{proof}

Q: Differentiating Fourier Series
A: Let $f:S^1\to\C$ be continuous. If $\sum_{n\in\Z} |n\hat{f}_n|$ converges, then $f$ is continuously differentiable and $\sum_{n=-N}^N n\hat{f}_n e^{in\theta}\to f'(\theta)$ uniformly as $N\to\infty$.
\begin{proof}
For $n\neq 0$, $\hat{f}_n\leq |n\hat{f}_n|$ so the CT gives that $\sum_{n=-N}^N |\hat{f}_n|$ converges. Hence we have the following uniform convergence as $N\to\infty$:
\begin{align*}
S_N f(\theta) &= \sum_{n=-N}^N \hat{f}_n e^{in\theta}\to f(\theta) \\
(S_N f)'(\theta)&=\sum_{n=-N}^Nin\hat{f}_n e^{in\theta}\to g(\theta)
\end{align*}
for some continuous $g(\theta)$. So $g=f'$ where $f$ is continuously differentiable.
\end{proof}

Q: Convergence of Fourier Series by differentiability
A: Let $f:S^1\to\C$ be $m$-times differentiable where $f^{(m)}$ is continuous except at $\{ \theta_1, \ldots \theta_r \}\in S_1$. If also $|f^{(m)}(\theta)|\leq M$ for $\theta\in S^1\setminus \{ \theta_1, \ldots \theta_r \}$, then
\[
|\hat{f}_n| \leq \frac{M}{|n|^m}.
\]
\begin{proof}
We can look at the $n$th Fourier coefficient:
\begin{align*}
  \hat{f}_n &= \frac{1}{2\pi}\int_{-\pi}^\pi e^{-in\theta}f(\theta) \;\d \theta\\
  &= \left[-\frac{1}{2\pi i n}e^{-in\theta}f(\theta)\right]^{\pi}_{-\pi} + \frac{1}{2\pi i n}\int_{-\pi}^\pi e^{-in\theta}f'(\theta)\;\d \theta\\
  &= \frac{1}{2\pi i n}\int_{-\pi}^\pi e^{-in\theta}f'(\theta)\;\d \theta\\
  &= \cdots\\
  &= \frac{1}{(in)^m}\frac{1}{2\pi} \int_{-\pi}^\pi e^{-in\theta} f^{(m)}(\theta)\;\d \theta \\
  |\hat{f}_n| &\leq \frac{1}{2\pi(|n|)^m} \int_{-\pi}^\pi |f^{(m)}(\theta)|\;\d \theta \\
&\leq \frac{1}{2\pi(|n|)^m} (2\pi M) \\
&= M/|n|^M.
\end{align*}
\end{proof}

Q: Sturm-Liouville form
A: We can rewrite a second order linear differential operator in \emph{Sturm-Liouville form}:
\begin{align*}
  \mathcal{L} y &= P\frac{\d^2 y}{\d x^2} + R \frac{\d y}{\d x} + Q y \\
  &= P\left[\frac{\d^2 y}{\d x^2} + \frac{R}{P}\frac{\d y}{\d x} + \frac{Q}{P}y\right] \\
  &= P\left[e^{-\int \frac{R}{P}\;\d x}\frac{\d}{\d x}\left(e^{\int \frac{R}{P}\;\d x}\frac{\d y}{\d x}\right)+ \frac{Q}{P}y\right]\\
  &= P p^{-1}\left[\frac{\d}{\d x}\left(p\frac{\d y}{\d x}\right) + \frac{Q}{P}p y\right] \\
 \mathcal{L} &= \frac{\d}{\d x}\left(p(x)\frac{\d}{\d x}\right) + q(x)
\end{align*}
where $p = \exp\left(\int \frac{R}{P}\right)$, $q = \frac{Q}{P}p$ and we drop a factor of $Pp^{-1}$.

Q: Second order differential operators are self-adjoint
A: For a Sturm-Liouville operator $\mathcal{L}$, we integrate by parts numerous times to obtain
\begin{align*}
  (f, \mathcal{L}g) &= \int_a^b f^*\left(\frac{\d}{\d x}\left(p \frac{\d g}{\d x}\right) + qg\right)\;\d x \\
  &= [f^* pg']_a^b - \int_a^b \left(\frac{\d f^*}{\d x}p\frac{\d g}{\d x} - f^* qg\right)\;\d x\\
  &= [f^*pg' - f'^*pg]_a^b + \int_a^b \left(\frac{\d}{\d x}\left(p\frac{\d f^*}{\d x}\right) + qf^*\right)g\;\d x\\
  &= [(f^*g' - f'^*g)p]_a^b + (\mathcal{L}f, g),
\end{align*}
assuming that $p, q$ are real.
So 2nd order linear differential operators are self-adjoint with respect to this norm if $p, q$ are real and $p,f,g$ are all periodic.

Q: Weight functions
A: A \emph{weight function} $w$ is a function which is real, non-negative, and has only finitely many zeroes.
An \emph{inner product with weight} $w$, written $(\ph, \ph)_w$, is defined by
  \[
    (f, g)_w = \int_a^b f^*(x) g(x) w(x)\;\d x.
  \]
  An \emph{eigenfunction with weight} $w$ of $\mathcal{L}$ is a function $y: [a, b] \to \C$ obeying the differential equation
  \[
    \mathcal{L} y = \lambda w y,
  \]
  where $\lambda\in \C$ is the eigenvalue.

Q: Eigenvalues of a Sturm-Liouville operator
A: \begin{prop}
  The eigenvalues of a Sturm-Liouville operator are real.
\end{prop}
\begin{proof}
  Suppose $\mathcal{L} y_i = \lambda_i w y_i$. Then
  \begin{align*}
    \lambda_i (y_i, y_i)_w &= \lambda_i (y_i, w y_i) \\
                                     &= (y_i, \mathcal{L} y_i)\\
                                     &= (\mathcal{L} y_i, y_i) \\
                                     &= (\lambda_i w y_i, y_i) \\
                                     &= \lambda_i^* (y_i, y_i)_w.
  \end{align*}
  Since $(y_i, y_i)_w \not= 0$, we have $\lambda_i = \lambda_i^*$.
\end{proof}

Q: Eigenfunctions of a Sturm-Liouville operator
A: \begin{prop}
  Eigenfunctions with different eigenvalues (but same weight) are orthogonal.
\end{prop}
\begin{proof}
  Let $\mathcal{L} y_i = \lambda_i w y_i$ and $\mathcal{L} y_j = \lambda_j w y_j$. Then
  \begin{align*}
    \lambda_i (y_j, y_i)_w &= (y_j, \mathcal{L} y_i) \\
&= (\mathcal{L} y_j, y_i) \\
&= \lambda_j (y_j, y_i)_w.
  \end{align*}
  Since $\lambda_i \not= \lambda_j$, we must have $(y_j, y_i)_w = 0$.
\end{proof}

Q: Legendre Polynomials
A: Consider the Sturm-Liouville operator with
\[
  p(x) = 1 - x^2, q(x) = 0, \Omega = [-1,1].
\]
For self-adjointness we require the b.t. in
\[
  (g, \mathcal{L} f) = (\mathcal{L}g, f) + [p(x) (g^* f' - g'^* f)]^{1}_{-1}
\]
to vanish. Since $p(\pm 1) = 0$ this requires $f,g$ to be regular at $x = \pm 1$. We look for power series solutions $f(x)=\sum_{n = 0}^\infty a_n x^n$ of the eigenvalue problem (\emph{Legendre's equation})
\begin{align*}
0 &= \frac{\d}{\d x} \left( (1-x^2)\frac{\d f}{\d x}\right) +\lambda f \\
  0 &= (1 - x^2) \sum_{n = 0}^\infty a_n n(n - 1)x^{n - 2} - 2\sum_{n = 0}^\infty a_n nx^n + \lambda \sum_{n = 0}^\infty a_n x^n \\
0 &= a_n(\lambda - n(n + 1)) + a_{n + 2}(n + 2)(n + 1) \\
a_{n+2} &= \frac{n(n+1)-\lambda}{(n+1)(n+2)}a_{n}.
\end{align*}
This gives two LI solutions - one even and one odd. Since
\[
  \lim_{n \to \infty}\frac{a_{n + 2}}{a_n} = 1,
\]
to ensure convergence at $x=\pm 1$ we require $\lambda = \ell(\ell + 1)$ for some $\ell \in \N$ such that the series truncates. The resulting polynomial solutions $P_\ell(x)$, called \emph{Legendre polynomials}, when normalised such that $P_\ell(1) = 1$ are given by \emph{Rodrigue's Formula}:
\[
  P_\ell(x) = \frac{1}{2^\ell \ell!} \frac{\d^\ell}{\d x^\ell} (x^2 - 1)^\ell.
\]

Q: Roots of Legendre polynomials
A: $P_\ell(x)=0$ has all $\ell$ roots in $(-1, 1)$.
\begin{proof}
Suppose only $m < \ell$ roots $\{x_1, x_2, \cdots, x_m\}$ lie in $(-1, 1)$. Then let $Q(x) = \prod_{r = 1}^m (x - x_r)$.
Since $P_\ell(x)$ and $Q(x)$ change sign at the same places, we have
\[
  \int_{-1}^1 P_\ell(x) Q(x) \;\d x \neq 0. \tag{$*$}
\]
However, we can use a Legendre polynomial expansion
\[
  Q(x) = \sum_{r = 1}^m q_r P_r(x).
\]
Since $(P_\ell, P_r) = 0$ for all $r < \ell$, $(*)$ is $0$. Contradiction.
\end{proof}

Q: Uniqueness of solutions to Laplace's equation
A: \begin{prop}
  Let $\Omega$ be a compact domain, and $\partial \Omega$ be its boundary. Let $ f: \partial \Omega\to \R$. Then there is a unique solution to $\nabla^2 \phi = 0$ on $\Omega$ with $\phi|_{\partial \Omega} = f$.
\end{prop}
\begin{proof}
  Suppose $\phi_1$ and $\phi_2$ are both solutions such that $\phi_1|_{\partial \Omega} = \phi_2|_{\partial \Omega} = f$. Then let $\Phi = \phi_1 - \phi_2$. So $\Phi = 0$ on the boundary. Then
  \begin{align*}
    0 &= \int_\Omega \Phi \nabla^2 \Phi \;\d^n x \\
&= -\int_\Omega (\nabla \Phi) \cdot (\nabla \Phi)\;\d x + \int_{\partial \Omega} \Phi \nabla \Phi\cdot \mathbf{n} \;\d^{n -1 }x \\
&= -\int_\Omega |\nabla \Phi|^2 \;\d x.
  \end{align*}
So $\Phi$ is constant throughout $\Omega$. Since at the boundary, $\Phi = 0$, we have $\Phi = 0$ throughout, ie. $\phi_1 = \phi_2$.
\end{proof}
Asking that $\phi|_{\partial \Omega} = f(x)$ is a \emph{Dirichlet} boundary condition. We can instead ask $\mathbf{n}\cdot \nabla \phi|_{\partial \Omega} = g$, a Neumann condition. In this case, we have a unique solution up to a constant factor.

Q: Multipole expansion
A: One can quickly check that
\[
  \phi(\mathbf{r}) = \frac{1}{|\mathbf{r} - \mathbf{r}'|}
\]
solves Laplace's equation $\nabla^2 \phi = 0$ for all $\mathbf{r}\in \R^3 \setminus \mathbf{r}'$. For example, if $\mathbf{r}' = \hat{\mathbf{k}}$, where $\hat{\mathbf{k}}$ is the unit vector in the $z$ direction, then
\[
  \frac{1}{|\mathbf{r} - \hat{\mathbf{k}}|} = \frac{1}{\sqrt{r^2 + 1 - 2r \cos \theta}} = \sum_{\ell = 0}^\infty c_\ell r^\ell P_\ell(\cos \theta).
\]
In particular for $\theta=0$, since $P_\ell(1) = 0$, we have
\[
  \sum_{\ell = 0}^\infty c_\ell r^\ell = \frac{1}{1 - r} = \sum_{\ell = 0}^\infty r^\ell.
\]
So all the coefficients must be $1$. This is valid for $r < 1$.
More generally, we have
\[
  \frac{1}{|\mathbf{r} - \mathbf{r}'|} = \frac{1}{r'} \sum_{\ell = 0}^\infty \left(\frac{r}{r'}\right)^\ell P_\ell(\hat{\mathbf{r}}\cdot \hat{\mathbf{r}}').
\]
This is called the \emph{multiple expansion}, and is valid when $r < r'$. Thus
\[
  \frac{1}{|\mathbf{r} - \mathbf{r}'|} = \frac{1}{r'} + \frac{r}{r'^2}\hat{\mathbf{r}}\cdot \hat{\mathbf{r}}'.
\]
The first term $\frac{1}{r'}$ is known as the \emph{monopole}, and the second term is known as the \emph{dipole}, in analogy to charges in electromagnetism. The monopole term is what we get due to a single charge, and the second term is the result of the interaction of two charges. % expand - what does this mean?

Q: Conservation of heat
A: Let $\phi: \Omega \to \C$ satisfy the heat equation
  \[
    \frac{\partial \phi}{\partial t} = \kappa \nabla^2 \phi
  \]
  where $\kappa > 0$ is the \emph{diffusion constant} and $\nabla\phi \cdot \mathbf{n}=0$ on $\partial\Omega$.
Differentiating the total heat energy gives
\begin{align*}
  \frac{\d}{\d t}\int_{\Omega} \phi(\mathbf{x}, t)\;\d^n x &= \int_{\R^n} \frac{\partial \phi}{\partial t}\; \d^n x \\
&= \kappa \int_{\Omega} \nabla^2 \phi \;\d^n x \\
&= \kappa \int_{\partial\Omega} \nabla \phi \cdot \mathbf{n} \;\d^{n-1} x \\
&= 0.
\end{align*}

Q: Smoothing effect of heat flow
A: The \emph{heat equation} for a function $\psi: \Omega \to \C$ with a unit diffusion constant is
  \[
    \frac{\partial \psi}{\partial t} =\nabla^2 \phi.
  \]
Suppose we have an eigenfunction $\psi(\mathbf{x})$ of the Laplacian on $\Omega$, ie. $\nabla^2 \psi = -\lambda \psi$ on $\Omega$ and $\psi=0$ or $\nabla\psi\cdot\mathbf{n}=0$ on $\partial\Omega$. Then
\begin{align*}
  - \lambda \int_\Omega |\psi|^2 \d^n x &= \int_\Omega \psi^* \nabla^2 \psi\;\d^n x \\
&= \int_{\partial\Omega} \psi^* \mathbf{n}\cdot \nabla \psi \;\d^{n - 1}x - \int_\Omega |\nabla \psi|^2 \;\d^n x \\
&= - \int_\Omega |\nabla \psi|^2 \;\d^n x \\
\lambda &= \frac{\int_\Omega |\nabla \psi|^2 \;\d^n x}{\lambda \int_\Omega |\psi|^2 \d^n x} \\
& \geq 0.
\end{align*}
Now write the solution to the heat equation as
\[
  \psi(\mathbf{x}, t) = e^{t\nabla^2} \psi(\mathbf{x}, 0)
\]
and expand $\psi(\mathbf{x},0)$ in terms of the eigenvalues $\{ \psi_I(\mathbf{x}) \}$ of $\nabla^2$ to obtain
\begin{align*}
  \psi(\mathbf{x}, t) &= e^{t \nabla^2}\left(\sum_I c_I \psi_I(\mathbf{x})\right) \\
  &= \sum_I c_I e^{t \nabla^2 }\psi_I(\mathbf{x}) \\
  &= \sum_I c_I e^{-\lambda t}\psi_I(\mathbf{x}).
\end{align*}
Since $\lambda_I$ are all positive, we know that $c_I(t)$ decays exponentially with time. In particular, the coefficients corresponding to the largest eigenvalues $|\lambda_I|$ decays most rapidly. We also know that eigenfunctions with larger eigenvalues are in general more ``spiky''. So these smooth out rather quickly.

Q: Fundamental solution of the heat equation
A: If $\phi(\mathbf{x}, t)$ solves the heat equation, so too do $\phi_1(\mathbf{x}, t) = \phi(\mathbf{x} - \mathbf{x}_0, t - t_0)$, and $\phi_2(\mathbf{x}, t) = A \phi(\lambda \mathbf{x}, \lambda^2 t)$.
Let's try to choose $A$ such that
\[
  \int_{\R^n}\phi_2(\mathbf{x}, t) \;\d^n x = \int_{\R^n}\phi(\mathbf{x}, t) \;\d^n x.
\]
The total heat of the second solution is
\begin{align*}
  \int_{\R^n} \phi_2(\mathbf{x}, t)\;\d^n x &= A\int_{\R^n}\phi(\lambda \mathbf{x}, \lambda^2 t)\;\d^n x \\
&= A\lambda^{-n} \int_{\R^n}\phi(\mathbf{y}, \lambda^2 t)\;\d^n y
\end{align*}
where we substituted $\mathbf{y} = \lambda \mathbf{x}$. So if we set $A = \lambda^n$, then the total heat is the same.
Assuming the solution is independent of rescalings, we can pick $\lambda = \frac{1}{\sqrt{t \kappa}}$ and consider solutions of the form
\[
  \phi(\mathbf{x}, t) = \frac{1}{(\kappa t)^{n/2}} F\left(\frac{\mathbf{x}}{\sqrt{\kappa t}}\right) = \frac{1}{(\kappa t)^{n/2}}F(\boldsymbol\eta),
\]
where $\boldsymbol\eta = \frac{\mathbf{x}}{\sqrt{\kappa t}}$. For example, in $1 + 1$ dimensions
\[
  \frac{\partial\phi}{\partial t} = \frac{\partial}{\partial t}\left(\frac{1}{\sqrt{\kappa t}} F(\eta)\right) = \frac{-1}{2\sqrt{\kappa t^3}} F(\eta) + \frac{1}{\sqrt{\kappa t}} \frac{\d \eta}{\d t}F'(\eta) = \frac{-1}{2\sqrt{\kappa t^3}}[F + \eta F']
\]
On the other side of the equation, we have
\[
  \kappa\frac{\partial^2}{\partial x^2} \left(\frac{1}{\sqrt{\kappa t}} F(\eta)\right) = \frac{\kappa}{\sqrt{\kappa t}} \frac{\partial}{\partial x}\left(\frac{\partial \eta}{\partial x} F'\right) = \frac{1}{\sqrt{\kappa t^3}}F''.
\]
So the equation becomes
\begin{align*}
  0 = 2F'' + \eta F' + F &= (2F' + \eta F)' \\
2F' + \eta F &= \text{const}
\end{align*}
and the BCs require that the constant is zero. So we get
\[
  F' = \frac{-\eta}{2}F.
\]
We now see a solution
\[
  F(\eta) = a \exp\left(\frac{-\eta^2}{4}\right).
\]
By convention, we now normalize our solution $\phi(x, t)$ by requiring $ \int_{-\infty}^\infty \phi(x, t)\;\d x = 1$ giving $a = \frac{1}{\sqrt{4\pi}}$ and
\[
  \phi(x, t) = \frac{1}{\sqrt{4\pi \kappa t}}\exp\left(- \frac{x^2}{4\kappa t}\right).
\]

Q: Derivation of the wave equation
A: Consider a string $x\in [0, L]$ with mass per unit length $\mu$ undergoing small oscillations described by $\phi(x, t)$.
Consider two points $A$, $B$ separated by a small distance $\delta x$. Let $T_A$ ($T_B$) be the outward pointing tension tangent to the string at $A$ ($B$). Resolving forces gives
\begin{align*}
T_A\cos \theta_A &= T_B \cos \theta_B = T \\
 \mu \delta x \frac{\partial^2 \phi}{\partial t^2} &= T_B \sin \theta_B - T_A \sin \theta_A \\
  \implies \mu \frac{\delta x}{T}\frac{\partial^2 \phi}{\partial t^2} &= \frac{T_B \sin \theta_B}{T_B\cos \theta_B} - \frac{T_A \sin \theta_A}{T_A \cos \theta_A}\\
  &= \tan \theta_B - \tan \theta_A\\
  &= \left.\frac{\partial\phi}{\partial x}\right|_B - \left.\frac{\partial\phi}{\partial x}\right|_A\\
  &= \delta x\frac{\partial^2 \phi}{\partial x^2} + o(\delta x).
\end{align*}
Taking the limit $\delta x \to 0$ and setting $c^2 = T/\mu$, we have that $\phi(x, t)$ obeys the wave equation
\[
  \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} = \frac{\partial^2 \phi}{\partial x^2}.
\]

Q: Energy of an oscillating string
A: The \emph{kinetic energy} of a small element $\delta x$ of the string is
\[
  \frac{1}{2}\mu \delta x\left(\frac{\partial \phi}{\partial t}\right)^2.
\]
The \emph{total kinetic energy} of the string is hence the integral
\[
  K(t) = \frac{\mu}{2}\int_0^L \left(\frac{\partial \phi}{\partial t}\right)^2 \;\d x.
\]
The \emph{potential energy} of a small element $\delta x$ of the string is
\begin{align*}
  T\times \text{extension} &= T(\delta s - \delta x) \\
  &= T(\sqrt{\delta x^2 + \delta \phi^2} - \delta x)\\
  &=T\delta x\left(1 + \frac{1}{2}\left(\frac{\delta \phi}{\delta x}\right)^2 + \cdots\right) - T \delta x + o(\delta x^2)\\
  &= \frac{T}{2}\delta x \left(\frac{\delta \phi}{\delta x}\right)^2 + o(\delta x^2)
\end{align*}
Hence the total potential energy of the string is
\[
  V(t) = \frac{\mu}{2}\int_0^L c^2 \left(\frac{\partial \phi}{\partial x}\right)^2 \;\d x,
\]
using the definition of $c^2$.
Hence the total energy is
\[
E=\frac{\mu}{2}\int_0^L \left(\frac{\partial \phi}{\partial t}\right)^2 + c^2 \left(\frac{\partial \phi}{\partial x}\right)^2 \;\d x.
\]

Q: Energy of a wave
A: \begin{prop}
  Suppose $\phi: \Omega \times [0, \infty) \to \R$ obeys the wave equation $\frac{\partial^2 \phi}{\partial t^2} = c^2 \nabla^2 \phi$ inside $\Omega \times (0, \infty)$, and $\phi =$ constant on $\partial\Omega$. Then
\[
E=\frac{1}{2}\int_\Omega \left(\frac{\partial \phi}{\partial t}\right)^2 + c^2 \left(\frac{\partial \phi}{\partial x}\right)^2 \;\d x
\]
is constant.
\end{prop}
\begin{proof}
Differentiating gives
  \[
    \frac{\d E}{\d t} = \int_\Omega \frac{\partial^2 \psi}{\partial t^2}\frac{\partial \phi}{\partial t} + c^2 \nabla \left(\frac{\partial \phi}{\partial t}\right)\cdot \nabla \phi \;\d V.
  \]
  We integrate by parts in the second term to obtain
  \[
    \frac{\d E}{\d t} = \int_\Omega \frac{\partial \phi}{\partial t}\left(\frac{\partial^2 \phi}{\partial t^2} - c^2 \nabla^2 \phi\right)\;\d V + c^2 \int_{\partial \Omega} \frac{\partial \phi}{\partial t}\nabla \phi \cdot \mathbf{n} \ \d S.
  \]
  Since $\frac{\partial^2 \phi}{\partial t^2} = c^2 \nabla^2 \phi$ by the wave equation, and $\phi$ is constant on $\partial \Omega$, we know that
  \[
    \frac{\d E_\phi}{\d t} = 0.
  \]
\end{proof}

Q: Uniqueness of solutions to the wave equation
A: \begin{prop}
  Suppose $\phi: \Omega \times [0, \infty) \to \R$ obeys the wave equation $\frac{\partial^2 \phi}{\partial t^2} = c^2 \nabla^2 \phi$ inside $\Omega \times (0, \infty)$, and obeys, for some $f, g, h$,
  \begin{enumerate}
    \item $\phi(x, 0) = f(x)$
    \item $\frac{\partial \phi}{\partial t}(x, 0) = g(x)$
    \item $\phi|_{\partial \Omega\times [0, \infty)} = h(x,t)$.
  \end{enumerate}
  Then $\phi$ is unique.
\end{prop}
\begin{proof}
  Suppose $\phi_1$ and $\phi_2$ are two such solutions. Then $\psi = \phi_1 - \phi_2$ obeys the wave equation
  \[
    \frac{\partial^2 \psi}{\partial t^2} = c^2 \nabla^2 \psi,
  \]
  and
  \[
    \psi|_{\partial \Omega \times [0, \infty)} = \psi|_{\Omega \times \{0\}} = \left.\frac{\partial \psi}{\partial t}\right|_{\Omega \times \{0\}} = 0.
  \]
 Hence
  \[
    E_\psi(t) = \frac{1}{2}\int_\Omega \left[\left(\frac{\partial \psi}{\partial t}\right)^2 + c^2 |\nabla \psi|^2 \right] \;\d V
  \]
satisfies $E_\psi(0) = 0$. Since $E_\psi$ is constant, we have $E_\psi(t)=0$. Then we must have $\frac{\partial \psi}{\partial t} = 0$. Since $\psi(x,0)=0$, we have $\psi(x,t)=0$, i.e. $\phi_1=\phi_2$.
\end{proof}

Q: Bessel functions
A: Laplace's equation in cylindrical coordinates gives rise to Bessel's equation of order $n$:
\[
  x^2 \frac{\d^2 y}{\d x^2} + x\frac{\d y}{\d x} + (x^2 - n^2)y = 0.
\]
Since Bessel's equations are second-order, there are two LI solutions for each $n$, namely $J_n(x)$ and $Y_n(x)$. These are called Bessel functions of order $n$ of the first ($J_n$) or second ($Y_n$) kind.
The $J_n$'s are all regular at the origin. In particular, as $x \to 0$, $J_n (x) \sim x^n$. These look like decaying sine waves, but the zeroes are not regularly spaced. Also $J_{-n}(x)=(-1)^nJ_n(x)$.
On the other hand, the $Y_n$'s are similar but singular at the origin. As $x \to 0$, $Y_0(x) \sim \ln x$, while $Y_n(x) \sim x^{-n}$.
The orthogonality condition is
\[
  \int_0^1 J_n\left(k_{nj}\right)J_n\left(k_{ni}\right) r\;\d r = \frac{1}{2}\delta_{ij}[J_n'(k_{ni})]^2
\]
where $k_{ni}$ is the $i$th root of the $n$th Bessel function.

Q: Distribution
A: Pick a class of ``nice'' test functions, i.e. we can do whatever we want to them (eg. differentiate, integrate etc.) A main example is the set of infinitely smooth functions $\phi: \Omega\to\R$ with compact support on some set $K\subseteq \Omega$, written $C^{\infty}_{\mathrm{cpt}}(\Omega)$ or $D(\Omega)$.
We now define a distribution $T$ to be a linear map $T:D(\Omega) \to \R$.
The space of distributions is written $D'(\Omega)$

Q: Differentiating distrubutions
A: If $T$ is function which is also a distribution on a space $D(\Omega)$ of differentiable test functions with compact support, then integrating by parts
\begin{align*}
  T'(\phi) &= \int_\Omega T'(x)\phi(x) \; \d x \\
&=[T(x)\phi(x)]_{\partial\Omega}-\int_\Omega T(x)\phi'(x) \; \d x \\
&=-\int_\Omega T(x)\phi'(x) \; \d x \\
\end{align*}
since $\phi$ has compact support.
In general we define $T'[\phi]=-T[\phi']$ and with higher derivatives $T^{(n)}=(-1)^nT[\phi^{(n)}]$.

Q: Translation and scaling of Dirac delta
A: \begin{itemize}
  \item Translation:
    \[
      \int_{-\infty}^\infty \delta(x - a) \phi(x)\;\d x = \int_{-\infty}^\infty \delta(y) \phi(y + a)\;\d x.
    \]
  \item Scaling:
    \[
      \int_{-\infty}^\infty \delta(cx) \phi(x)\;\d x = \int_{-\infty}^\infty \delta(y) \phi\left(\frac{y}{c}\right) \frac{\d y}{|c|} = \frac{1}{|c|}\phi(0).
    \]
\end{itemize}
\begin{proof}
For any $a\in\R$
\begin{align*}
\delta_a[\phi] &= \phi(a) \\
&= \int_\R \delta(x-a)\phi(x) \; \d x \\
&= \int_\R \delta(y) \phi(y+a).
\end{align*}
and any $c\in\R$
\begin{align*}
\int_\R \delta(cx)\phi(x) \; \d x &= \int_\R \delta(y)\phi(y/c) \; \frac{\d y}{|c|} \\
&= \frac{1}{|c|} \phi(0).
\end{align*}
\end{proof}

Q: Dirac delta of a function with simple zeros
A: Suppose $f(x)$ is a continuously differentiable function with isolated simple zeros at $x_i$. Then near any of its zeros $x_i$, we have $f(x) \approx (x - x_i) \left.\frac{\partial f}{\partial x}\right|_{x_i}$. Then
    \begin{align*}
      \int_{-\infty}^\infty \delta(f(x)) \phi(x)\;\d x &= \lim_{\varepsilon\to 0} \sum_{i = 1}^n \int_{x_i-\varepsilon}^{x_i+\varepsilon} \delta\left((x - x_i)\left.\frac{\partial f}{\partial x}\right|_{x_i}\right) \phi(x)\;\d x \\
      &= \sum_{i = 1}^n \frac{1}{|f'(x_i)|} \phi(x_i).
    \end{align*}

Q: Fourier Series of Dirac delta
A: On an interval $[-L, L]$ we have
\[
  \delta(x) = \sum_{n \in \Z}\hat{\delta}_n e^{in\pi x/L}
\]
with
\[
  \hat{\delta}_n = \frac{1}{2L} \int_{-L}^L e^{in\pi x/L}\delta(x)\;\d x = \frac{1}{2L}
\]
So we have
\[
  \delta(x) = \frac{1}{2L} \sum_{n \in \Z} e^{in\pi x/L}.
\]
This \emph{does} make sense as a distribution. We let
\[
  S_N \delta(x) = \frac{1}{2L} \sum_{n = -N}^N e^{in \pi x/L}.
\]
Since $\phi$ is assumed smooth
\begin{align*}
  \lim_{N \to \infty} \int_{-L}^L S_N \delta(x) \phi(x)\;\d x &= \lim_{N \to \infty} \frac{1}{2L}\int_{L}^L \sum_{n = -N}^N e^{in\pi x/L}\phi(x) \;\d x\\
  &= \lim_{N \to \infty} \sum_{n = -N}^N \left[\frac{1}{2L}\int_{-L}^L e^{in\pi x/L} \phi(x) \;\d x\right]\\
  &= \lim_{N\to \infty} \sum_{n = -N}^N \hat{\phi}_{-n}\\
  &= \lim_{N \to \infty}\sum_{n = -N}^N \hat{\phi}_n e^{in\pi 0/L}\\
  &= \phi(0).
\end{align*}

Q: Dirac delta wrt an arbitrary basis
A: Let $\{y_n(x)\}$ be a complete set of eigenfunctions on $[a, b]$ that are orthogonal with respect to a weight function $w(x)$. Then we can write
\[
  \delta (x - \xi) = \sum_{n} c_n y_n(x),
\]
with
\[
  c_n = \int_a^b y_n^*(x) \delta(x - \xi) w(x)\;\d x = y_n^*(\xi) w(\xi).
\]
So
\[
  \delta(x - \xi) = w(\xi) \sum_n y_n^* (\xi) y_n(x) = w(x) \sum_n y_n^*(\xi) y_n(x),
\]
using the fact that
\[
  \delta(x - \xi) = \frac{w(\xi)}{w(x)} \delta(x - \xi).
\]

Q: Orthogonality condition for Legendre Polynomials
A: For $\ell' < \ell$
\begin{align*}
  (P_\ell, P_{\ell'}) &= \int_{-1}^1 P_\ell(x) P_{\ell'}(x) \;\d x\\
  &= \frac{1}{2^\ell \ell!}\int_{-1}^1 \frac{\d^\ell}{\d x^\ell} (x^2 - 1)^\ell P_{\ell'}(x)\;\d x\\
  &= \frac{1}{2^\ell \ell!}\left[\frac{\d^{\ell - 1}}{\d x^{\ell - 1}} (x^2 - 1)^\ell P_{\ell'}(x)\right]_{-1}^1 - \frac{1}{2^\ell \ell!} \int_{-1}^1 \frac{\d^{\ell - 1}}{\d x^{\ell - 1}}(x^2 - 1)^\ell \frac{\d P_{\ell'}}{\d x}\;\d x\\
  &= - \frac{1}{2^\ell \ell!} \int_{-1}^1 \frac{\d^{\ell - 1}}{\d x^{\ell - 1}}(x^2 - 1)^\ell \frac{\d P_{\ell'}}{\d x}\;\d x.
\end{align*}
We can integrate by parts $\ell'$ times until we get the $\ell'$\textsuperscript{th} derivative of $P_{\ell'}$, which is zero.
In fact, we can show that
\[
  (P_\ell, P_{\ell'}) = \frac{2\delta_{\ell\ell'}}{2\ell + 1}.
\]

Q: Green's function solution of ODEs
A: Suppose $\mathcal{L}y = f$, where $f(x)$ is a bounded forcing term, and $\mathcal{L}$ is a bounded differential operator, say
\[
  \mathcal{L} = \alpha(x) \frac{\partial^2}{\partial x^2} + \beta(x) \frac{\partial}{\partial x} + \gamma(x).
\]
The Green's function $G(x, \xi)$ obeys $ \mathcal{L} G(x, \xi) = \delta(x - \xi)$ so we can see that $y(x) = \int_{a}^b G(x, \xi) f(\xi) \;\d \xi$ is a solution:
\[
  \mathcal{L}y = \int_a^b \mathcal{L} G(x, \xi) f(\xi)\;\d \xi = \int_a^b \delta(x - \xi) f(\xi) \;d\xi = f(x).
\]
The BCs $\phi(a)=\phi(b)=0$ imply $G(a, \xi) = G(b, \xi) = 0$. Also note $\mathcal{L} G = 0$ for all $x\neq\xi$. Let $\{y_1(x), y_2(x)\}$ be a basis of solutions to $\mathcal{L}G = 0$ on $[a, b]$ with BCs $y_1(a) = 0, y_2(b) = 0$. Then
\[
  G(x, \xi) =
  \begin{cases}
    A(\xi) y_1(x) & a \leq x < \xi\\
    B(\xi) y_2(x) & \xi < x \leq b.
  \end{cases}
\]
We require $G(x, \xi)$ continuous to avoid differentiating the $\delta-$function, i.e.
\[
  A(\xi) y_1 (\xi) = B(\xi) y_2(\xi).\tag{$*$}
\]
Now integrate over a small region $(\xi - \varepsilon, \xi + \varepsilon)$ surrounding $\xi$ using continuity of $\alpha,\beta,\gamma, G$ and boundedness of $G'$:
\begin{align*}
 \int_{\xi - \varepsilon}^{\xi + \varepsilon} \delta(x - \xi)\;\d x&= \int_{\xi - \varepsilon}^{\xi + \varepsilon} \left[\alpha(x) \frac{\d^2 G}{\d x^2} + \beta(x) \frac{\d G}{\d x} + \gamma(x) G\right]\;\d x \\
  1&=\lim_{\varepsilon \to 0} \int_{\xi - \varepsilon}^{\xi + \varepsilon} \alpha G'' \;\d x \\
&=\alpha(\xi)\left(\left.\frac{\partial G}{\partial x}\right|_{\xi^+} - \left.\frac{\partial G}{\partial x}\right|_{\xi^-}\right) \\
\frac{1}{\alpha(\xi)}&=B(\xi) y'_2 (\xi) - A(\xi) y_1'(\xi).
\end{align*}
Together with $(*)$, we know that
\[
  A(\xi) = \frac{y_2(\xi)}{\alpha(\xi) W(\xi)},\quad B(\xi) = \frac{y_1(\xi)}{\alpha(\xi)W(\xi)},
\]
and hence
\begin{align*}
  G(x, \xi) &= \frac{1}{\alpha(\xi)W(\xi)}
  \begin{cases}
    y_2(\xi) y_1(x) & a \leq x \leq \xi\\
    y_1(\xi) y_2(x) & \xi < x \leq b.
  \end{cases} \\
&= \frac{1}{\alpha(\xi)W(\xi)} [\Theta (\xi - x) y_2(\xi)y_1(x) + \Theta(x - \xi)y_1(\xi)y_2(x)].
\end{align*}
So our general solution is
\[
  y(x) =\int_a^x \frac{f(\xi)}{\alpha(\xi)W(\xi)}y_1(\xi) y_2(x)\;\d \xi+ \int_x^b \frac{f(\xi)}{\alpha(\xi)W(\xi)} y_2(\xi) y_1(x)\;\d \xi .
\]

Q: Necessary condition for existence of solutions to linear problem
A: If $\mathcal{L}\phi=f$ has a solution, then $f$ must be orthogonal to $\mathrm{coker} f= \ker (\mathcal{L}^\dagger)=\{ \psi : \mathcal{L}^\dagger\psi=0 \}$.
\begin{proof}
Let $\mathcal{L}\phi=f$ and $\psi\in\mathrm{coker} f$. Then
\begin{align*}
(\psi, f) &= (\psi, \mathcal{L} f) \\
&= (\mathcal{L}^\dagger \psi, \phi) \\
&= (0, \phi) \\
&= 0.
\end{align*}
\end{proof}

Q: Eigenfunction expansion of Green's function
A: If $\mathcal{L}$ is a Sturm-Liouville operator, then we expect to be able to expand
\[
G(x,\xi) = \sum_{n\in\Z} \hat{G}_n(\xi) Y_n(x)
\]
where $\{Y_n\}$ is an orthonormal basis of eigenfunctions of $\mathcal{L}$. Hence
\begin{align*}
\delta(x-\xi) &= \mathcal{L}G(x,\xi) \\
&=\sum_{n\in\Z} \hat{G}_n(\xi) \mathcal{L} Y_n(x) \\
&= w(x) \sum_{n\in\Z} \hat{G}_n(\xi) \lambda_n Y_n(x) \\
\int_a^b Y_m^*(x)\delta(x-\xi)&= \sum_{n\in\Z} \hat{G}_n(\xi) \lambda_n \int_a^b Y_m^*(x)Y_n(x) w(x)\; \d x \\
Y_m^*(\xi)&= \hat{G}_m(\xi) \lambda_m \\
\hat{G}_m(\xi) &= Y_m^*(\xi) / \lambda_m.
\end{align*}
Therefore
\[
G(x,\xi) = \sum_{n\in\Z} \frac{1}{\lambda_n} Y_n^*(\xi)Y_n(x).
\]

Q: Green's functions for initial value problems
A: Consider $\mathcal{L}y = f(t)$ subject to $y(0) = 0$ and $y'(0) = 0$.
Let $y_1(t), y_2(t)$ be a basis of solutions to $\mathcal{L} y = 0$. The Green's function obeys
\[
  \mathcal{L}G = \delta(t - \tau)
\]
subject to $G|_0=\partial_t G|_0=0$. The solution can be written
\[
  G(t, \tau) =
  \begin{cases}
    A(\tau) y_1(t) + B(\tau) y_2(t) & 0 \leq t < \tau\\
    C(\tau) y_1(t) + D(\tau) y_2(t) & t > \tau.
  \end{cases}
\]
Our initial conditions require that
\[
  \begin{pmatrix}
    y_1(0) & y_2(0)\\
    y_1'(0) & y_2'(0)
  \end{pmatrix}
  \begin{pmatrix}
    A\\
    B
  \end{pmatrix}
  =
  \begin{pmatrix}
    0\\0
  \end{pmatrix}
\]
Since the Wronskian is non-singular (by definition of $y_1, y_2$ being a basis), we must have $A, B = 0$.
Our continuity and jump conditions now require
\begin{align*}
  0 &= C(\tau) t_1(t) + D(\tau) y_2(t)\\
  \frac{1}{\alpha(\tau)} &= C(\tau) y_1'(\tau) + D(\tau) y_2'(\tau).
\end{align*}
We can solve this to get $C(\tau)$ and $D(\tau)$. Then the only contribution comes from $\tau < t$:
\[
  y(t) = \int_{0}^\infty G(t, \tau) f(\tau)\;\d \tau = \int_{0}^t G(t, \tau) f(\tau) \;\d \tau.
\]
So $y(t)$ depends only on the history of $f$. This expresses causality!

Q: Fourier transform
A: The \emph{Fourier transform} of an (absolutely integrable) function $f: \R \to \C$ is defined as
  \[
    \tilde{f}(k) = \int_{-\infty}^\infty e^{-ikx}f(x)\;\d x
  \]
  for \emph{all} $k \in \R$. We will also write $\tilde{f} = \mathcal{F}[f]$.
Note that for any $k$, we have
\[
  |\tilde{f}(k)| = \left|\int_{-\infty}^\infty e^{-ikx}f(x)\;\d x\right| \leq \int_{-\infty}^\infty |e^{-ikx} f(x)|\;\d x = \int_{-\infty}^\infty |f(x)|\;\d x.
\]
Since we have assumed that our function is absolutely integrable, this definition makes sense.

Q: Linearity of Fourier transforms
A: If $f, g: \R \to \C$ are absolutely integrable and $c_1, c_2$ are constants, then
    \[
      \mathcal{F}[c_1 f(x) + c_2 g(x)] = c_1 \mathcal{F}[f(x)] + c_2 \mathcal{F}[g(x)].
    \]
\begin{proof}
\begin{align*}
\mathcal{F}[c_1 f(x) + c_2 g(x)] &=\int_\R e^{-ikx} [c_1 f(x) + c_2 g(x)] \; \d x \\
&=c_1\int_\R e^{-ikx} f(x) \; \d x +  c_2\int_\R e^{-ikx} g(x) \; \d x\\
 &=c_1 \mathcal{F}[f(x)] + c_2 \mathcal{F}[g(x)].
\end{align*}
\end{proof}

Q: Translation of Fourier Transforms
A: If $f:\R\to\C$ is absolutely integrable, then
\[
\mathcal{F}[f(x - a)](k) =  e^{-ika}\mathcal{F}[f(x)]
\]
So a translation in $x$-space becomes a re-phasing in $k$-space.
 \begin{proof}
\begin{align*}
      \mathcal{F}[f(x - a)](k) &= \int_{\R}e^{-ikx} f(x - a)\;\d x \\
      &= \int_{\R} e^{-ik(y + a)}f (y) \;\d y \\
      &= e^{-ika} \int_{\R} e^{-iky}f(y)\;\d y \\
      &= e^{-ika}\mathcal{F}[f(x)](k).
    \end{align*}
 \end{proof}

Q: Re-phasing of Fourier transforms
A: If $f:\R\to\C$ is an absolutely integrable function, then
\[
 \mathcal{F}[e^{-i\ell x}f(x)](k)=\mathcal{F}[f(k + \ell)].
\]
\begin{proof}
    \begin{align*}
      \mathcal{F}[e^{-i\ell x}f(x)](k) &= \int_{-\infty}^\infty e^{-ikx} e^{-i\ell x} f(x)\;\d x\\
      &= \int_{-\infty}^\infty e^{-i(k + \ell)x} f(x)\;\d x\\
      &= \mathcal{F}[f(x)](k + \ell).
    \end{align*}
\end{proof}

Q: Scaling of Fourier Transforms
A: If $f:\R\to\C$ is an absolutely integrable function, then
\[
 \mathcal{F}[f(cx)](k)= \frac{1}{|c|}\mathcal{F}[f(x)](k/c).
\]
\begin{proof}
    \begin{align*}
      \mathcal{F}[f(cx)](k) &= \int_{-\infty}^\infty e^{-ikx}f(cx)\;\d x\\
      &= \int_{-\infty}^\infty e^{-iky/c} f(y) \frac{\d y}{|c|}\\
      &= \frac{1}{|c|}\mathcal{F}[f(x)](k/c).
    \end{align*}
\end{proof}

Q: Convolution of Fourier Transforms
A: For functions $f, g: \R \to \C$, we define the \emph{convolution} as
    \[
      (f * g) (x) = \int_{-\infty}^\infty f(x - y) g(y)\;\d y.
    \]
Then
\[
\mathcal{F}[f * g]=\mathcal{F}[f] \mathcal{F}[g].
\]
\begin{proof}
Letting $u = x - y$:
 \begin{align*}
      \mathcal{F}[f * g(x)] &= \int_{-\infty}^\infty e^{-ikx} \left[\int_{-\infty}^{\infty} f(x - y)g(y) \;\d y\right] \;\d x\\
      &= \int_{\R^2} e^{-ik(x - y)} f(x - y) e^{-iky}g(y) \;\d y \;\d x\\
      &= \int_{\R} e^{-iku} f(u) \;\d u \int_{\R} e^{-iky} g(y)\;\d y\\
      &= \mathcal{F}[f] \mathcal{F}[g].
    \end{align*}
\end{proof}

Q: Fourier transform of a derivative
A: If $f:\R\to\C$ is a differentiable function with $f,f'$ absolutely integrable, then
\[
\mathcal{F}[f']=\mathcal{F}[f].
\]
\begin{proof}
Integrating by parts, we have
    \begin{align*}
      \mathcal{F}[f'(x)] &= \int_{-\infty}^\infty e^{-ikx} \frac{\d f}{\d x}\;\d x\\
      &= [e^{-ikx}f(x)]_{-\infty}^\infty -\int_{-\infty}^\infty \frac{\d}{\d x} (e^{-ikx}) f(x)\;\d x\\
      &= ik \int_{-\infty}^\infty e^{-ikx} f(x)\;\d x\\
      &= ik \mathcal{F}[f(x)]
    \end{align*}
where the boundary term vanishes since $f$ is absolutely integrable.
\end{proof}

Q: Derivative of a Fourier Transform
A: If $f:\R\to\C$ is absolutely integrable, then
\[
 \tilde{f}'=-i\mathcal{F}[xf(x)].
\]
\begin{proof}
    \begin{align*}
      \mathcal{F}[xf(x)] &= \int_{-\infty}^\infty e^{-ikx} xf(x)\;\d x \\
&= i \frac{\d}{\d k}\int_{-\infty}^{\infty} e^{-ikx} f(x)\;\d x \\
&= i\tilde{f}'(k).
    \end{align*}
\end{proof}

Q: Fourier Transform solution of linear ODEs
A: Suppose we have a differential equation
\[
  \mathcal{L}(\partial) y = f,
\]
where
\[
  \mathcal{L}(\partial) = \sum_{r = 0}^p c_r \frac{\d^r}{\d x^r}
\]
is a differential operator of $p$th order with constant coefficients. Then
\[
 y= \mathcal{F}^{-1}\left(\frac{\tilde{f}(k)}{\mathcal{L}(ik)}\right).
\]
\begin{proof}
Taking the Fourier transform of both sides of the equation
\begin{align*}
  \mathcal{F}[\mathcal{L}(\partial)y] &= \mathcal{F}[f(x)] \\
\mathcal{L}(ik) \tilde{y}(k) &= \tilde{f}(k) \\
\tilde{y}(k) &= \frac{\tilde{f}(k)}{\mathcal{L}(ik)}.
\end{align*}
\end{proof}

Q: The Fourier inversion theorem
A: Let $f:\R\to\C$. Then we have the duality
\[
   \mathcal{F}^{-1}[f(x)]=\mathcal{F}[f(-x)]/(2\pi)= \frac{1}{2\pi} \int_{-\infty}^\infty e^{ik x}f(x) \;\d x.
\]
\begin{proof}
For a period $L$ function we have
\[
  f(x) = \sum_{n \in \Z}\hat{f}_n e^{2inx\pi /L}.
\]
where
\[
  \hat{f}_n = \frac{1}{L} \int_{-L/2}^{L/2} e^{-2 in\pi x/L} f(x)\;\d x.
\]
Now define
\[
  \Delta k = \frac{2\pi }{L}.
\]
So we have
\[
  f(x) = \sum_{n \in \Z} e^{inx \Delta k} \frac{\Delta k}{2 \pi}\int_{-L/2}^{L/2} e^{-inx \Delta k} f(u) \;\d u
\]
As we take the limit $L \to \infty$, we can approximate
\[
  \int_{-L/2}^{L/2} e^{-ixn\Delta k} f(x) \;\d x = \int_{-\infty}^\infty e^{-ix (n \Delta k)} f(x)\;\d x = \tilde{f}(n \Delta k).
\]
Then for a non-periodic function, we have the definition of the Riemann integral
\begin{align*}
  f(x) &= \lim_{\Delta k \to 0} \sum_{n \in \Z} \frac{\Delta k}{2\pi} e^{in \Delta k x} \tilde{f} (n \Delta k) \\
&= \frac{1}{2\pi} \int_{-\infty}^\infty e^{ik x} \tilde{f}(k) \;\d k\\
&=\mathcal{F}^{-1}[\tilde{f}(k)].
\end{align*}
\end{proof}

Q: Parseval's theorem for Fourier Transforms
A: \begin{thm}
  Suppose $f, g: \R \to \C$ are sufficiently well-behaved that $\tilde{f}$ and $\tilde{g}$ exist and we indeed have $\mathcal{F}^{-1}[\tilde{f}] = f, \mathcal{F}^{-1}[\tilde{g}] = g$. Then
  \[
    (f, g) = \int_\R f^*(x) g(x)\;\d x = \frac{1}{2\pi}(\tilde{f}, \tilde{g}).
  \]
  In particular, if $f = g$, then
  \[
    \|f\|^2 = \frac{1}{2\pi} \|\tilde{f}\|^2.
  \]
\end{thm}
\begin{proof}
  \begin{align*}
    (f, g) &= \int_\R f^*(x) g(x)\;\d x\\
    &= \int_{-\infty}^\infty f^*(x) \left[\frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx} \tilde{g}(x) \;\d k\right]\;\d x\\
    &= \frac{1}{2\pi} \int_{-\infty}^\infty \left[\int_{-\infty}^\infty f^*(x) e^{ikx} \;\d x\right] \tilde{g}(k)\;\d k\\
    &= \frac{1}{2\pi} \int_{-\infty}^\infty \left[\int_{-\infty}^\infty f(x) e^{-ikx} \;\d x\right]^* \tilde{g}(k)\;\d k\\
    &= \frac{1}{2\pi} \int_{-\infty}^\infty \tilde{f}^*(k) \tilde{g}(k) \;\d k\\
    &= \frac{1}{2\pi}(\tilde{f}, \tilde{g}).
  \end{align*}
\end{proof}

Q: Fourier Transform of distributions
A: For a distribution $T$, its Fourier Transform must be given by
\[
\mathcal{F}T[\phi]=2\pi T[\mathcal{F}^{-1}[\phi]].
\]
\begin{proof}
From Parseval's theorem for Fourier Transforms we have
\[
(f,g)=\frac{1}{2\pi} (\mathcal{F}[f],\mathcal{F}[g])
\]
or equivalently
\[
(\mathcal{F}[f],\tilde{g})=2\pi(\tilde{f},\mathcal{F}^{-1}(g)).
\]
Now suppose $T_g$ is a distribution given by $T_g[\phi]=(g,\phi)$ where $g$ is a function. Then
\begin{align*}
(FT_g)[\phi]&=(\mathcal{F}[g],\phi) \\
&=2\pi (g,\mathcal{F}^{-1}[\phi]) \\
&=2\pi T_g[\mathcal{F}^{-1}[\phi]].
\end{align*}
Hence for consistency we must define
\[
\mathcal{F}T[\phi]=2\pi T[\mathcal{F}^{-1}[\phi]].
\]
\end{proof}

Q: Fourier Transform of the Dirac delta
A: Write $T_1=(1,\ph)$, i.e. $1$ as a distribution. Then
\[
  \mathcal{F}[\delta(x)] = \int_{-\infty}^\infty e^{ikx} \delta (x)\;\d x = 1 = T_1
\]
and conversely
\[
  \mathcal{F}^{-1}[T_1] = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx}\;\d k = \delta (x).
\]
\begin{proof}
\begin{align*}
\mathcal{F}\delta[\phi]&=2\pi\delta[\mathcal{F}^{-1}[\phi]] \\
&=2\pi\delta\left[\frac{1}{2\pi}\int_\R e^{ikx}\phi(k) \; \d k\right] \\
&=\int_\R \phi(k) \; \d k \\
&=(1,\phi) \\
&=T_1
\end{align*}
and conversely using the above
\begin{align*}
\mathcal{F}T_1[\phi] &= 2\pi T_1[\mathcal{F}^{-1}\phi] \\
&= 2\pi(1,\mathcal{F}^{-1}\phi) \\
&= 2\pi\int_\R \mathcal{F}^{-1}\phi(k) \; \d k \\
&= 2\pi\delta\left[\int_\R e^{ikx}\mathcal{F}^{-1}\phi(k) \; \d k\right]  \\
&= 2\pi\delta[\phi].
\end{align*}
\end{proof}

Q: Fourier transform of Heaviside Step Function
A: Let $\theta_\varepsilon(x)=\theta(x)e^{-\varepsilon x}$ for $\varepsilon>0$. Then
\begin{align*}
[\mathcal{F}\theta_\varepsilon](k) &= \int_\R e^{-ikx} \theta(x) e^{-\varepsilon x} \; \d x\\
&= \int_0^\infty e^{-(ik+\varepsilon)x} \; \d x \\
&=\frac{1}{ik+\varepsilon}.
\end{align*}
This suggests $[\mathcal{F}\theta](k)=\frac{1}{ik}$.
\begin{proof}
\begin{align*}
\mathcal{F}\theta_\varepsilon[\phi] &= \int_{-\infty}^\infty \frac{\phi(k)}{ik+\varepsilon} \; \d k\\
&= \int_{|k|>\delta} \frac{\phi(k)}{ik+\varepsilon} \; \d k + \int_{-\delta}^{+\delta}  \frac{\phi(k)-\phi(0)}{ik+\varepsilon} +  \frac{\phi(0)}{ik+\varepsilon}\; \d k  \\
&= \int_{|k|>\delta} \frac{\phi(k)}{ik+\varepsilon} \; \d k + \int_{-\delta}^{+\delta}  \frac{1}{i} \frac{\partial\phi}{\partial k} \; \d k -i\phi(0)[\log(ik+\varepsilon)]_{-\delta}^{+\delta}+ O(\varepsilon) \\
&\to \int_{|k|>\delta} \frac{\phi(k)}{ik} \; \d k + \int_{-\delta}^{+\delta}  \frac{1}{i} \frac{\partial\phi}{\partial k} \; \d k +\pi\phi(0)+ O(\varepsilon) \text{ as } \varepsilon\to 0 \\
&\to \fint \frac{\phi(k)}{ik} +  \pi\phi(0) \text{ as } \delta\to 0
\end{align*}
where the integral excludes $k=0$. Hence
\[
\mathcal{F}\theta(k)=\text{p.v.}\frac{1}{ik} + \pi\delta(k)
\]
where principal value means the behaviour at $k=0$ is determined by the delta function.
\end{proof}

Q: Response Theory
A: Consider an amplifier $\mathcal{L}:I(t)\to O(t)$. By Fourier's inversion theorem, we have the \emph{synthesis} of the signal
\[
  I(t) = \frac{1}{2\pi} \int e^{i\omega t} \tilde{I}(\omega) \;\d \omega
\]
where $\tilde{I}(\omega)$ is the \emph{resolution} of $I(t)$ and
\[
  O(t) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{i\omega t} \tilde{R}(\omega) \tilde{I}(\omega)\;\d \omega.
\]
where $\tilde{R}(\omega)$ is the \emph{transfer function} which modifies the amplitudes/phases of $\tilde{I}(\omega)$.
The product $O(t) = \mathcal{F}^{-1}[\tilde{R}(\omega) \tilde{I}(\omega)]$ is a \emph{convolution}
\[
  O(t) = \int_{-\infty}^\infty R(t - u) I(u)\;\d u,
\]
where
\[
  R(t) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{i\omega t} \tilde{R}(\omega) \;\d \omega
\]
is the \emph{response function}.
By causality, if $\forall t<t_0: I(t)=0$, then $\forall t<t_0: R(t)=0$.

Q: General response for linear systems
A: Let $I(t) = \mathcal{L}_m O(t)$ where
\[
  \mathcal{L}_m = \sum_{i = 0}^m a_i \frac{\d^i}{\d t^i}
\]
with $a_i \in \C$. Taking the Fourier transform
\begin{align*}
  \tilde{I}(\omega) &= \sum_{i = 0}^m a_j (i\omega)^j \tilde{O}(\omega) \\
\tilde{O}(\omega) &= \frac{\tilde{I}(\omega)}{a_0 + i\omega a_1 + \cdots + (i\omega)^m a_m}\\
 \tilde{R}(\omega) &= \frac{1}{a_0 + i\omega a_1 + \cdots + (i\omega)^m a_m}
\end{align*}
The denominator of $R(\omega)$ has $m$ roots $c_j \in \C$ with multiplicity $c_j$ ($1\leq j \leq J$) by FTA s.t.
\[
  \tilde{R}(\omega) = \frac{1}{a_m} \prod_{j = 1}^J \frac{1}{(i\omega - c_j)^{k_j}}.
\]
By repeated partial fractions
\[
  \tilde{R}(\omega) = \sum_{j = 1}^J \sum_{r = 1}^k \frac{\Gamma_{rj}}{(i\omega - c_j)^r}
\]
for some $\Gamma_{rj} \in \C$. By linearity, it is enough to find $\mathcal{F}^{-1}$ for
\[
  \frac{1}{(i\omega - \alpha)^p}.
\]
To compute this, consider the function
\[
  h_0(t) =
  \begin{cases}
    e^{\alpha t} & t > 0\\
    0 & \text{otherwise}.
  \end{cases}
\]
The Fourier transform is then
\[
  \tilde{h}_o(\omega) = \int_{-\infty}^\infty e^{-i\omega t} h_0(t) \;\d t = \int_0^\infty e^{(\alpha - i \omega)t} \;\d t = \frac{1}{i\omega - \alpha}
\]
provided $\Re(\alpha) < 0$ (so that $e^{(\alpha - i\omega) t} \to 0$ as $t \to \infty$). Similarly, we can let
\[
  h_1(t) =
  \begin{cases}
    te^{\alpha t} & t > 0\\
    0 & \text{otherwise}
  \end{cases}
\]
Then we get
\[
  \tilde{h}_1(\omega) = \mathcal{F}[t h_0(t)] = i\frac{\d}{\d \omega} \mathcal{F}[h_0(t)] = \frac{1}{(i \omega - \alpha)^2}.
\]
Proceeding inductively we find the solution
\[
  h_p(t) =
  \begin{cases}
    \frac{t^p}{p!} e^{\alpha t} & t > 0\\
    0 & \text{otherwise}
  \end{cases}
\]
provided $\Re(\alpha) < 0$.

Q: Discrete Fourier Transform
A: If $|f(x)|\ll 1$ for $x \not\in [-R, S]$ for some $R, S > 0$. Then we can approximate the FT as
\[
  \tilde{f}(k) = \int_{-R}^S e^{-ikx} f(x)\;\d x.
\]
If we `sample' $f(x)$ at $ x_j = -R + j\frac{R + S}{N}$ for $N\in\N$ and $0\leq j < N$, then
\[
  \tilde{f}(k) \approx \frac{R + S}{N} \sum_{j = 0}^{N - 1} f(x_j) e^{-ik x_j}.
\]
Lets cache $\tilde{f}(k)$ for $k = k_m = \frac{2\pi m}{R + S}$. Then
\begin{align*}
  \tilde{f}(k_m) &\approx \frac{R + S}{N} e^{\frac{2\pi i m R}{R + S}} \sum_{j = 0}^{N - 1} f(x_j) e^{-\frac{2\pi ijm}{N}}\\
  &= (R+S) e^{\frac{2\pi i m R}{R + S}} \left[ \frac{1}{N}\sum_{j = 0}^{N - 1} f(x_j) \omega^{-jm}\right],
\end{align*}
where $\omega = e^{\frac{2\pi i}{N}}$ is an $N$th root of unity and in brackets we have
\[
  F(m) = \frac{1}{N}\sum_{j = 0}^{N - 1} g(\omega^j) \omega^{-jm}= \frac{1}{N}\sum_{j = 0}^{N - 1} e_m^* (\omega^j) g(\omega^j) = (e_m, g)
\]
where $g: G \to \C$ is defined by $g(\omega^j) = f(x_j)$.
We can expand our $g$ as
\[
  g = \sum_{m = 0}^{N - 1} (e_m, g) e_m = \sum_{m = 0}^{N - 1} F(m) e_m.
\]
Writing $f$ instead of $g$, we recover the formula
\[
  f(x_j) = g(\omega^j) = \sum_{m = 0}^{N - 1} F(m) e_m(\omega^j).
\]

Q: Basis for functions on roots of unity
A: Consider the space of \emph{all} functions $g: G \to \C$ is a finite-dimensional vector space, isomorphic to $\C^N$. This has an inner product
\[
  (f, g) = \frac{1}{N} \sum_{j = 0}^{N - 1} f^*(\omega^j) g(\omega^j).
\]
Now let $e_n: G \to \C$ be the function
\[
  e_m(\omega^j) = \omega^{jm}.
\]
Then the set of functions $\{e_m\}$ for $m = 0, \cdots, N - 1$ is an orthonormal basis.
\begin{proof}
We can compute
\begin{align*}
  (e_m, e_m) &= \frac{1}{N} \sum_{j = 0}^{N - 1} e_m^*(\omega^j) e_m(\omega^j)\\
  &= \frac{1}{N} \sum_{j = 0}^{N - 1} \omega^{-jm} \omega^{jm}\\
  &= \frac{1}{N} \sum_{j = 0}^{N - 1} 1\\
  &= 1.
\end{align*}
For $n \not =m$, we have
\begin{align*}
  (e_n, e_m) &= \frac{1}{N} \sum_{j = 0}^{N - 1} e_n^* (\omega^j) e^m(\omega^j)\\
  &= \frac{1}{N} \sum_{j = 0}^{N - 1} \omega^{j(m - n)}\\
  &= \frac{1}{N} \frac{1 - \omega^{(m - n)N}}{1 - \omega^{m - n}}\\
&=0.
\end{align*}
\end{proof}

Q: Parseval's Theorem for functions on roots of unity
A: $(f,f)=\sum_{m=0}^{N-1} |\hat{f}_m|^2$.
\begin{proof}
\begin{align*}
(f,f)&=\frac{1}{N} \sum_{j=0}^{N-1} f(\omega^j)^* f(\omega^j) \\
&=\frac{1}{N} \sum_{j=0}^{N-1}\sum_{m,k=0}^{N-1} \hat{f}_m^*e_m^*(\omega^j) \hat{f}_k e_k(\omega^j) \\
&=\sum_{m,k=0}^{N-1} \hat{f}_m^*\hat{f}_k \left[\frac{1}{N} \sum_{j=0}^{N-1} e_m^*(\omega^j) e_k(\omega^j)\right] \\
&=\sum_{m,k=0}^{N-1} \hat{f}_m^*\hat{f}_k \delta_{mk} \\
&=\sum_{m=0}^{N-1} |\hat{f}_m|^2.
\end{align*}
\end{proof}

Q: Cauchy data
A: Recall that to find a unique solution of Laplace's equation $\nabla^2 \phi = 0$ on a bounded domain $\Omega \subseteq \R^n$, we imposed the Dirichlet BC $\phi|_{\partial \Omega} = f$ or the Neumann BC $\mathbf{n}\cdot \nabla \phi|_{\partial \Omega} = g$.
All the BCs and ICs restrict the value of $\phi$ on some co-dimension 1 surface, ie. a surface whose dimension is one less than the domain. This is known as \emph{Cauchy data} for our PDE.

Q: Well-posed problem
A: A PDE problem is said to be well-posed if its Cauchy data implies
  \begin{enumerate}
    \item A solution exists;
    \item The solution is unique;
    \item A ``small change'' in the Cauchy data leads to a ``small change'' in the solution.
  \end{enumerate}

Q: Integral curves, characteristics and transverse curves
A: Let $\mathbf{V}(x, y): \R^2 \to \R^2$ be a vector field. The \emph{integral curves} associated to $\mathbf{V}$ are curves whose tangent $\left(\frac{\d x}{\d s}, \frac{\d y}{\d s}\right)$ is just $\mathbf{V}(x, y)$.
The integral curves of a PDE are called characteristics.
Transverse curves are nowhere parallel to a characteristic at intersection. These can be used to parameterise the characteristics and are a natural curve for Cauchy data.

Q: The method of characteristics for 1st order PDEs
A: Suppose $\phi: \R^2 \to \R$ obeys
\[
  a(x, y) \frac{\partial \phi}{\partial x} + b(x, y) \frac{\partial \phi}{\partial y} = f(x,y), \quad \phi |_T = g(t)
\]
for some curve $T$ parameterised by $t$.
We can define a vector field $\mathbf{V}$ as
\[
  \mathbf{V}(x, y) =
  \begin{pmatrix}
    a(x, y)\\
    b(x, y)
  \end{pmatrix}
\]
such that $\mathbf{V}\cdot \nabla \phi = 0$. The integral curves of $\mathbf{V}$ with $s=0$ on $T$ for each $t\in\R$ satisfy
\[
  \left.\frac{\partial x}{\partial s}\right|_t = a(x, y),\quad \left.\frac{\partial y}{\partial s}\right|_t = b(x, y).
\]
Hence, along an integral curve of $\mathbf{V}$, we obtain the characteristic equation
\[
  \frac{\partial \phi}{\partial s} = \frac{\d x(s)}{\d s} \frac{\partial \phi}{\partial x} + \frac{\d y(s)}{\d s} \frac{\partial \phi}{\partial y}.
\]
Hence our PDE is reduced to the ODE
\[
  \left.\frac{\partial \phi}{\partial s}\right|_t = f(x(s),y(s)), \quad\phi(0,t)=g(t).
\]