# Toronto Math Forum

## APM346-2016F => APM346--Lectures => Chapter 6 => Topic started by: Roro Sihui Yap on November 08, 2016, 05:02:57 PM

Title: Small typos in Chapter 6
Post by: Roro Sihui Yap on November 08, 2016, 05:02:57 PM
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 (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 (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 (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.

Title: Re: Small typos in Chapter 6
Post by: Victor Ivrii on November 09, 2016, 05:14:40 AM
OK
Title: Re: Small typos in Chapter 6
Post by: XinYu Zheng on November 09, 2016, 09:17:20 AM
Some more typos in Section 6.5:

1. Equation (6) should contain $h(\theta ')$, not $g(\theta ')$.
2. In the derivation immediately following equation (6), we have a line that says $-\frac{a}{\pi}\mathrm{Re}\log (1-ra^{-1}e^{i(\theta-\theta')})=-\frac{a}{2\pi}\log(a^{-2}(1-ra^{-1}e^{i(\theta-\theta')})(1-ra^{-1}e^{-i(\theta-\theta')}))$. The $a^{-2}$ in the logarithm should not be there at this stage. It should be there in the line right after, so the final expression is correct.
3. At the very bottom of the page where it says "...where for sector $\{r<a, 0<\theta<\alpha\}$ we should set...", the $B_n$ is missing.