Toronto Math Forum

Minor typo at the end of Theorem 2
It should be $m=-l,1-l,\ldots,l-1,l$ not $m=-l,1-l,\ldots,l-1,m $

1. In Section 6.2, Just below Equation (4) it should be
$(\alpha_0 X'-\alpha X)(0)=(\beta_0 X'+\beta X)(l)=0$ instead of
$(\alpha_0 X'-\alpha X)=(\beta_0 X'+\beta X)(l)=0$


2. In Section 6.3, Equation (2) http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.3.html#mjx-eqn-eq-6.3.2
$\left\{\begin{aligned}
&\partial_x = \cos(\theta)\partial_r - r^{-1}\sin(\theta)\partial_\theta,\\ &\partial_y = \sin(\theta)\partial_r + r^{-1}\cos(\theta)\partial_\theta
\end{aligned}\right.\label{eq-6.3.2} $
The second equation should be $\partial_y$ not $\partial_x$


3.  In Section 6.3, Exercise 3
Since $ r \Delta u = \bigr(r u_r\bigl)_r +\bigr(\frac{1}{r}u_\theta\bigl)_\theta$, then
$ \Delta u = r^{-1}\bigr(r u_r\bigl)_r +r^{-1} \bigr(\frac{1}{r}u_\theta\bigl)_\theta $ not $ \Delta u = r^{-1}\bigr(r u_r\bigl)_r + \bigr(\frac{1}{r}u_\theta\bigl)_\theta $


4. In Section 6.4, just below Equation (13), ' was left out. It should be $\sin(n\theta')$
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.13

$G(r,\theta,\theta'):= \frac{1}{2\pi} \Bigl(1+2\sum_{n=1}^\infty r^n a^{-n}
\bigl(\cos(n\theta)\cos(n\theta')+\sin(n\theta)\sin(n\theta')\bigr) \Bigr)$


5. In Section 6.4, in the derivation of Equation (14)
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.14
Since $\frac{1}{2\pi} \Bigl(1+2Re \frac{ra^{-1}e^{i(\theta-\theta')}}{1-ra^{-1}e^{i(\theta-\theta')}} \Bigr) = \frac{1}{2\pi} \Bigl(1+2Re \frac{r\cos(\theta-\theta') + ir\sin(\theta-\theta')}{a-r\cos(\theta-\theta') - ir\sin(\theta-\theta')} \Bigr) = \frac{1}{2\pi} \Bigl(1+2\frac{ra\cos(\theta-\theta') - r^2 }{a^2-2ra\cos (\theta-\theta') +r^2} \Bigr)$

So instead of $ G(r,\theta,\theta')= \frac{1}{2\pi}
\Bigl(1+2 \frac{ra \cos(\theta-\theta')}{a^2-2ra\cos (\theta-\theta') +r^2} \Bigr)$
it should be $ G(r,\theta,\theta')= \frac{1}{2\pi} \Bigl(1+2\frac{ra\cos(\theta-\theta') - r^2 }{a^2-2ra\cos (\theta-\theta') +r^2} \Bigr)$
The final equation (14) is right. Just the step before it has typo.
 

1. I think it should be $\hat{h}(k)$ instead of $\hat{h}(x)$ in the proof of theorem 4a
\begin{equation*}
\hat{h}(x)=\frac{\kappa}{2\pi} \int e^{-ix k }h(x)\,dx =
\frac{\kappa}{2\pi} \iint e^{-ix k }f(x-y)g(y)\,dxdy;
\end{equation*}

2. In Example 2, Since $f(x)=e^{-\frac{\alpha}{2}x^2}$, it should be $f'=- \alpha x f$ and not $f'=\alpha x f$

3. In Example 2, last paragraph, If $\alpha=\pm i\beta$ then it should be $f=e^{\mp\frac{i\beta}{2 }x^2}$  and not $f=e^{\mp\frac{i}{2\beta }x^2}$

4. In Example 2, for the last equation, there shouldn't be an "x" variable in the exponent

\begin{equation*}
\hat{f}( k )=\frac{\kappa}{2\sqrt{\pi\beta}}
(1\mp i)e^{\pm\frac{i}{2\beta} k ^2}.
\end{equation*}

Instead of
\begin{equation*}
\hat{f}( k )=\frac{\kappa}{2\sqrt{\pi\beta}}
(1\mp i)e^{\pm\frac{i}{2\beta} k ^2x}.
\end{equation*}

5. At the start of theorem 5, I think it should be $(-\infty,\infty)$ instead of $(\infty,\infty)$

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter5/S5.2.html

Using the functions found above,

$\phi(x) = \sin(x)$ for $x > 0$
$\psi(x) = 0$ for $x > 0$

and

$\psi(x)=-\phi(-2x)=-sin(-2x)=sin(2x)$ for $x < 0$

In the region, $ 0<t<x<3t $, we have $x+3t > 0$,$x-3t < 0$
$$u(x,t) = \sin (x + 3t ) + \sin (2x - 6t)$$

In the region $ x > 3t > 0 $, we have $x+3t > 0$,$x-3t > 0$
$$u(x,t) = \sin (x + 3t ) $$

 Want to prove that $\partial_t E(t) = 0 $
$$\partial_t E(t) = \frac{1}{2}\int_0^L ( 2u_tu_{tt} + 2c^2u_{x}u_{xt} )\,dx + (c^2\alpha) u_t(0,t)u_{tt}(0,t)+  (c^2\beta) u_t(L,t)u_{tt}(L,t) $$

Use $u_{tt} = c^2u_{xx} $, $\alpha u_{tt}(0,t) = u_x(0,t) $ and $\beta u_{tt}(L,t) = -u_x(L,t)$
$$\partial_t E(t) = \int_0^L (c^2 u_tu_{xx} + c^2u_{x}u_{xt} )\,dx + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) $$
Since $\partial_x(u_tu_x) = u_{xt}u_x + u_tu_{xx}$
$$\partial_t E(t) = c^2(u_tu_x)\big|_{0}^{L} + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) $$
$$\partial_t E(t) = c^2u_t(L,t)u_x(L,t) - c^2u_t(0,t)u_x(0,t) + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) =0  $$

In an exam, should we always assume that U is continuous ?
or are we allowed to drop the constant term if it is not stated that U is continuous ?

\begin{align*}
& u_{tt}-9u_{xx}=0, &&&t>0, x>0,  \\
&u|_{t=0}= \phi (x),   &&u_t|_{t=0}= 3\phi'(x) &x>0, \\
&(u_x+2u_{t})|_{x=0}=0,  &&&t>0
\end{align*}

$ u = f(x+3t) + g(x-3t) $

From $u|_{t=0}= \phi (x)$, we get $f(x) + g(x) = \phi (x)$
From $u_t|_{t=0}= 3\phi'(x)$, we get $3f'(x) - 3g'(x) = 3\phi'(x)$, and thus $f(x) - g(x) = \phi (x) - \phi (0) $

Solving the equations, $ f(x) = \phi (x) - \frac{\phi (0)}{2}$ and $g(x) = \frac{\phi (0)}{2}$ for $x>0$ only

From $(u_x+2u_{t})|_{x=0}=0$, we get $f'(3t) + g'(-3t) + 6f'(3t) - 6g'(-3t)= 0 $
let $x = -3t$, since $t > 0$, we have $x < 0$
$7f'(-x) - 5g'(x) = 0 $
$-7f(-x) - 5g(x) = -k$ where k is some constant
$g(x) = \frac{k}{5} - \frac{7\phi(-x)}{5} +  \frac{7\phi (0)}{10} $ for $x < 0 $

when $ x > 3t$,
\begin{equation}u = \phi ( x + 3t ) \end{equation}

when $ 0 < x < 3t$,
\begin{equation}u = \phi ( x + 3t )  - \frac{7}{5} \phi (3t - x) + c\end{equation} where c is some constant

If we want the function to be continuous, as $ x \rightarrow 3t $, both of the above functions have to be equal.
when $x = 3t$,
(1) $u = \phi (6t)$                     
(2) $u = \phi (6t) - \frac{7}{5} \phi (0) + c $
In order for them to be equal $c = \frac{7}{5} \phi (0)  $

Thus, 
$u = \begin{cases}\phi ( x + 3t ) && x > 3t \\\phi ( x + 3t )  - \frac{7}{5} \phi (3t - x) + \frac{7}{5} \phi (0) && 0 < x < 3t \end{cases}$ 

\begin{equation}  \frac{dx}{1} = \frac{dy}{3} = \frac{du}{u} \end{equation}
From $\frac{dx}{1} = \frac{dy}{3} $, $3x - y = C_1$
From $\frac{dx}{1} = \frac{du}{u}$, $\ln u =  x + ln  C_2  $ 
So $u = C_2e^x = \phi(3x - y)e^x $

Since $u|_{x=0} = y$,
$\phi(-y) = y $ which means $ \phi(z) = -z $
Therefore, $u = (y - 3x)e^x$