Toronto Math Forum

By separation of variables,
\begin{equation*}
u(x,t)= X(x) T(t)
\end{equation*}
\begin{align*}
& 4X(x) T''(t)= X''(x)T(t), \\[3pt]
& X(0)T(t)=X(2)T(t)=0,
\end{align*}
\begin{align*}
& 4\frac{ T''(t)}{T(t)}= \frac{X''(x)}{X(x)}, \\[3pt]
& X(0)=X(2)=0.
\end{align*}
\begin{equation}
\frac{T''(t)}{T(t)}=-\frac{1}{4}\lambda,\qquad \frac{ X''(x)}{X(x)}=-\lambda
\end{equation}
This is a Dirichlet-Dirichlet boundary eigenvalue problem for $X(x)$, we have
\begin{align*}
& \lambda_n = \frac{\pi^2n^2}{4}&& n=1,2,\ldots, \\[3pt]
&X_n(x)=\sin (\frac{\pi n x}{2}).
\end{align*}
Then plug $\lambda$ into (5) and solve for $T(t)$,
\begin{equation*}
T_n(t)=A_n \cos(\frac{\pi n}{4}t ) + B_n \sin (\frac{\pi n}{4}t )
\end{equation*}
Therefore,
\begin{equation*}
u_n(x,t)= \bigl( A_n \cos(\frac{\pi n}{4}t )+ B_n \sin (\frac{\pi n}{4}t )\bigr) \cdot
\sin (\frac{\pi n x}{2})
\,\,\,n=1,2,\ldots
\qquad\end{equation*}
\begin{equation}
u(x,t)= \sum_{n=1}^\infty \bigl( A_n \cos(\frac{\pi nt}{4} )+
B_n \sin (\frac{\pi nt}{4} )\bigr) \cdot
\sin (\frac{\pi n x}{2}).
\end{equation}
\begin{align}
&\sum_{n=1}^\infty A_n \sin (\frac{\pi n x}{2})=f(x),\\
&\sum_{n=1}^\infty \frac{\pi n}{4} B_n \sin (\frac{\pi n x}{2} )=0.
\end{align}
Calculate the coefficients $A_n$
\begin{multline*}
A_n = \int_0^2 \sin (\frac{\pi n x}{2}) f(x)\, dx = \int_0^1 x\sin (\frac{\pi n x}{2}) \,dx + \int_1^2 (2-x)\sin (\frac{\pi n x}{2})  \,dx  = \frac{2\sin \frac{\pi n x}{2}}{\bigr(\frac{\pi n}{2}\bigl)^2} = (-1)^{\frac{n-1}{2}} \frac{8}{(\pi n)^2}, \,\,\,\,0 \,\,\,\text{when n is even}
\end{multline*}
I also get $B_n = 0$,
The solution is given by
\begin{equation}
u(x,t)= \sum_{n=1 \,\text{odd} }^\infty (-1)^{\frac{n-1}{2}} \frac{8}{(\pi n)^2} \cos(\frac{\pi nt}{4})\cdot
\sin (\frac{\pi n x}{2}).
\end{equation}

Yes thanks. I modify my solution to solve for the sector instead of the half disk
\begin{equation}
\Delta =\partial_r^2 + \frac{1}{r}\partial_r +
\frac{1}{r^2}\partial_\theta^2
\end{equation}
Let's plug in $u = R(r)P(\theta) $ and I get
\begin{equation*}
R''(r)P(\theta) + \frac{1}{r}R' (r)P(\theta) + \frac{1}{r^2} R(r) P''(\theta)=0
\end{equation*}
which could be rewritten as
\begin{equation*}
\frac{r^2 R''(\rho) + r R' (r)}{R(r)}+
\frac{P''(\theta)}{P(\theta)}=0
\end{equation*}
The two terms must be constant
\begin{align}
&r^2 R'' +r R' = \lambda R\\[3pt]
&P(\theta)=-\lambda P(\theta).
\end{align}
The initial conditions can be translated as
\begin{align}
& u =1\qquad &&\text{for  }  r = 4,\\
& u=0 &&\text{for  }  \theta = -\frac{5\pi}{6}, \frac{5\pi}{6} \end{align}
Therefore the boundary condition for P is $P(-\frac{5\pi}{6}) = P(\frac{5\pi}{6}) = 0$ and this is a Dirichlet-Dirichlet boundary problem with $l = \frac{5\pi}{3}$.
Thus we have
\begin{align}
&\lambda_{n} = (\frac{3n}{5})^2
&P_{n} = \sin(\frac{3n}{5}(\theta + \frac{5\pi}{6}))
&& n = 1, 2, ...
\end{align}
\begin{equation}
r^2 R'' +r R' - (\frac{3n}{5})^2 R = 0
\end{equation}
The Euler equation for this is $k(k-1) + k - (\frac{3n}{5})^2 = 0$ and we have $ k = \frac{3n}{5}, -\frac{3n}{5}$
Since we are looking for continuous solutions,
\begin{equation}
R_n = A_n r^{\frac{3n}{5}}\\
u = \sum_{n = 1}^{\infty}A_n r^\frac{3n}{5} \sin(\frac{3n}{5}(\theta + \frac{5\pi}{6}))
\end{equation}
Apply the initial condition
\begin{equation}
\sum_{n = 1}^{\infty}A_n 4^\frac{3n}{5} \sin(n(\theta + \frac{5\pi}{6})) = 1
\end{equation}
Then we calculate the coefficients by
\begin{equation}
A_n=\frac{2}{4^\frac{3n}{5}\frac{5\pi}{3}}\int_{-\frac{5\pi}{6}}^{\frac{5\pi}{6}}\sin(\frac{3n}{5}(\theta+\frac{5\pi}{6}))\,d\theta=\frac{2}{4^\frac{3n}{5} \frac{5\pi}{3}}\int_0^{\frac{5\pi}{3}}\sin(\frac{3n}{5}x) \,dx=\frac{2}{4^\frac{3n}{5}\frac{5\pi}{3}\frac{3n}{5}}(\cos(\frac{3n}{5}x)|_{\frac{5\pi}{3}}^0)=\frac{1}{4^{\frac{3n}{5}-1}\frac{3\pi}{5} \frac{3n}{5}}\,\,\,\text{n odd, 0 otherwise}
\end{equation}
Therefore
\begin{equation}
u = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{1}{4^{\frac{3n}{5}-1}\frac{3\pi}{5} \frac{3n}{5}}r^{\frac{3n}{5}} \sin(n(\theta + \frac{5\pi}{6})) = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{100}{9n\pi4^{\frac{3n}{5}}} r^{\frac{3n}{5}} \sin(n(\theta + \frac{5\pi}{6}))
\end{equation}

The general solution for u is
\begin{equation*}u = \phi(x+3t) + \psi(x-3t)\end{equation*}
Now impose the initial conditions (2) and (3)
\begin{align}
&\phi(x) + \psi(x) = f(x)\\
&\phi'(x) + \psi'(x) = g(x)\end{align}
Solve the above to give
\begin{align}
&\phi(x) = \frac{1}{2}f(x) + \frac{1}{6}\int_{0}^{x} g(x') dx'  \\
&\psi(x)  = \frac{1}{2}f(x) - \frac{1}{6}\int_{0}^{x} g(x') dx'\end{align}
Therefore, when x > 3t,
\begin{equation}u(x) = \phi(x+3t) + \psi(x-3t) =  \frac{1}{2}[f(x+3t) + f(x-3t)] + \frac{1}{6}\int_{x-3t}^{x+3t} g(x') dx'\end{equation}
To find what u is when x<3t, impose initial condition (4)
\begin{equation*}\phi'(3t) + \psi'(-3t) = h(t)\end{equation*}
Which implies
\begin{equation*}\phi(3t) - \psi(-3t) = 3\int_{0}^{t} h(t') dt' + C\end{equation*}
Let $x =-3t, x<0, t = -\frac{x}{3}$
\begin{equation*}\phi(-x) - \psi(x) = 3\int_{0}^{-x/3} h(t')dt' + C\\
\psi(x) = \phi(-x) - 3\int_{0}^{-x/3} h(t')dt' + C\end{equation*}
Then for u to be continuous, $\psi(x)$ must be continuous at 0,
\begin{equation} \psi(0_{+}) = \frac{1}{2}f(0) = \phi(0) + C =\frac{1}{2}f(0)+C=\psi(0_{-})\end{equation}
We get C = 0
Therefore when x< 3t,
\begin{equation} u = \frac{1}{2}[f(x+3t)+f(3t-x)] + \frac{1}{6}[\int_{0}^{x+3t} g(x') dx' + \int_{0}^{3t-x} g(x') dx' ]- 3\int_{0}^{t-x/3} h(t')dt'\end{equation}
Now plug in $f, g, h$, we get
\begin{align}
&u = \cos (x+3t) + 3\cos (x-3t), x>3t\\[2pt]
&u = \cos (x+3t) + 2\cos (3t-x) + 1, 0<x<3t
\end{align}

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.8
The coefficient should be $\frac{1}{4\pi c^2 t}$ instead of $\frac{1}{4\pi c t}$

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.18
$v^\pm_\omega$ should be$-\frac{1}{4\pi}\iiint |x-y|^{-1}e^{\mp i\omega c^{-1}|x-y|}f(y)\, dy $ instead of
$v^\pm_\omega= -\frac{1}{4\pi}\iiint |x-y|^{-1}e^{\mp i\omega c^{-1}|x-y|}\, dy.$

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.2.html#mjx-eqn-eq-9.2.3
This one I am not sure but I think the coefficient $\frac{1}{2}$ is missing if we follow equation (2)
Otherwise we could not use the quadratic form in (4) which leads to theorem 1 in this section.
Therefore, it should be
\begin{equation}
\iint_{\Sigma} \Bigl(
\frac{1}{2}\bigl(u_{t}^2+|\nabla u|^2\bigr)\nu_t -c^2 u_t \nabla u \cdot \nu_x\Bigr) \,d\sigma=0
\end{equation}
instead of
\begin{equation}
\iint_{\Sigma} \Bigl(
\bigl(u_{t}^2+|\nabla u|^2\bigr)\nu_t -c^2 u_t \nabla u \cdot \nu_x\Bigr) \,d\sigma=0
\end{equation}
And the equation after (5) http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.2.html#mjx-eqn-eq-9.2.5 also needs to be changed.

\begin{equation*}x=r\cos (\theta)\end{equation*}, \begin{equation*}y=r\sin(\theta)\end{equation*}
\begin{equation*}
\left\{
\begin{aligned}
&r=\sqrt{x^2+y^2},\\
&\theta = \arctan \bigl(\frac{y}{x}\bigr);
\end{aligned}\right.
\end{equation*}

\begin{equation}
\left\{\begin{aligned}
&r_x=\frac{x}{\sqrt{x^2+y^2}}=\frac{r\cos(\theta)}{r}=\cos(\theta),\\ 
&r_y=\frac{y}{\sqrt{x^2+y^2}}=\frac{r\sin(\theta)}{r}=\sin(\theta),\\ 
&\theta_x = \frac{1}{(\frac{y}{x})^2+1} \cdot -\frac{y}{x^2}=-\frac{y}{y^2+x^2}=-\frac{r\sin(\theta)}{r^2}=-r^{-1}\sin(\theta),\\ 
&\theta_y = \frac{1}{(\frac{y}{x})^2+1} \cdot \frac{1}{x}=\frac{x}{y^2+x^2}=\frac{r\cos(\theta)}{r^2}=r^{-1}\cos(\theta).
\end{aligned}\right.
\end{equation}
By the chain rule
\begin{equation}
\left\{
\begin{aligned}
&\partial_x = r_x\partial_r  + \theta_x\partial_\theta  =\cos(\theta)\partial_r - r^{-1}\sin(\theta)\partial_\theta,\\
&\partial_y = r_y\partial_r  + \theta_y\partial_\theta  = \sin(\theta)\partial_r + r^{-1}\cos(\theta)\partial_\theta
\end{aligned}\right.
\end{equation}
Plug the above into the laplacian and expand the squares
\begin{multline*}
\Delta = \partial_x^2+ \partial_y^2=
\bigl(\cos(\theta)\partial_r - r^{-1}\sin(\theta)\partial_\theta\bigr)^2 +
\bigl(\sin(\theta)\partial_r +
r^{-1}\cos(\theta)\partial_\theta\bigr)^2 =\\
\bigl(\cos(\theta)\partial_r\cos(\theta)\partial_r -\cos(\theta)\partial_r\cdot r^{-1}\sin(\theta)\partial_\theta -r^{-1}\sin(\theta)\partial_\theta \cos(\theta)\partial_r+r^{-1}\sin(\theta)\partial_\theta r^{-1}\sin(\theta)\partial_\theta\bigr) + \\
\bigl(\sin(\theta)\partial_r\sin(\theta)\partial_r+ \sin(\theta)\partial_r r^{-1}\cos(\theta)\partial_\theta + r^{-1} \cos(\theta) \partial_\theta \sin(\theta)\partial_r+r^{-1}\cos(\theta)\partial_\theta r^{-1}\cos(\theta)\partial_\theta\bigr).
\end{multline*}
By the product rule, we have
\begin{multline*}
\Delta=
\bigl(\cos^2(\theta)\partial_r^2 + \frac{1}{r^2}\cos(\theta)\sin(\theta)\partial_\theta +r^{-1}\sin^2(\theta)\partial_r-2r^{-1}\sin{\theta}\cos{\theta}\partial_r\partial_\theta+ r^{-2}\cos^2(\theta)\partial_\theta^2 - r^{-2}\cos(\theta)\sin(\theta)\partial_\theta\bigr) +\\
\bigl(\sin^2(\theta)\partial_r^2 - \frac{1}{r^2}\sin(\theta)cos(\theta)\partial_\theta + r^{-1}\cos^2(\theta)\partial_r + 2r^{-1}\sin{\theta}\cos{\theta}\partial_r\partial_\theta +
r^{-2}\sin^2(\theta)\partial_\theta^2+r^{-2}\sin(\theta)\cos(\theta)\partial_\theta\bigr)=\\
\partial_r^2 +\frac{1}{r}\partial_r +\frac{1}{r^2}\partial_\theta^2.
\end{multline*}