Topics

1

Chapter 8 / Typo

« on: November 23, 2016, 04:26:33 PM »

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.9

If we follow from equation(7) http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.7
(9) should be
\begin{equation}
\sin^2(\phi)
\Phi'' +\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
instead of
\begin{equation}
\sin^2(\phi)
\Phi'' +2\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
There will be a coefficient 2 if we substitute $\Phi(\phi)=L(\cos(\phi))$ into the equation and use the chain rule to get its derivatives but I think not now.

2

Chapter 9 / Typos?

« on: November 23, 2016, 01:36:19 PM »

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.

3

Chapter 6 / Change of Variable

« on: November 04, 2016, 01:20:01 AM »

\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*}

4

Chapter 2 / The definition of semilinear equation in 2.1 of textbook

« on: September 27, 2016, 05:22:03 PM »

In the textbook, it reads

Definition 2. If a=a(x,t) and b=b(x,t) equation is semilinear.

But in the previous section this should be linear equation with variable cofficients, right?
I think the next definition is consistent with what I know.

Definition 3. Furthermore if f is a linear function of u: f=c(x,t)u+g(x,t) original equation is linear.

In this case the last ODE is also linear.

But I don't understand this sentence. Is there an ODE?