### Author Topic: Small typos in Chapter 6  (Read 2013 times)

#### Roro Sihui Yap

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##### Small typos in Chapter 6
« 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
\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.

#### Victor Ivrii

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##### Re: Small typos in Chapter 6
« Reply #1 on: November 09, 2016, 05:14:40 AM »
OK

#### XinYu Zheng

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##### Re: Small typos in Chapter 6
« Reply #2 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.
« Last Edit: November 09, 2016, 09:28:31 AM by XinYu Zheng »