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### Messages - Roro Sihui Yap

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1
##### Chapter 8 / Re: Typo
« on: November 23, 2016, 05:48:49 PM »
Minor typo at the end of Theorem 2
It should be $m=-l,1-l,\ldots,l-1,l$ not $m=-l,1-l,\ldots,l-1,m$

2
##### Chapter 8 / Typo
« on: November 23, 2016, 12:36:48 AM »
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.15

Just below equation (15), I think that it should be
$$\frac{\rho^2 P''+2\rho P'+\lambda \rho^2 P}{P}+ \frac{\Lambda Y }{Y}=0$$

$$\frac{\rho^2 P''+2\rho P'+\lambda P}{\rho^2 P}+ \frac{\Lambda Y }{Y}=0$$

If not, equation (16) would not follow
$$\rho^2 P''+2\rho P'+(\lambda \rho^2 - l(l+1) )P=0.$$

3
##### Chapter 6 / 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.

4
##### Q5 / Q5
« on: November 03, 2016, 08:34:00 PM »
\begin{align}
\end{align}

Taking fourier transform,
\begin{align}
&-k^2\hat{u} +\hat{u}_{yy}=0 \\
\end{align}

From equation (3): $\hat{u}(k,y) = A(k)e^{-ky}+B(k)e^{ky}$
From equation (4):
$\hat{u}(k,0) = A(k) +B(k) = \hat{f}(k)$
$\hat{u_y}(k,1) = -kA(k)e^{-k} +kB(k)e^{k} = \hat{g}(k)$

Solving the two equations we get
$$A(k) = \frac{\hat{f}(k)ke^k - \hat{g}(k)}{2k\cosh(k)}$$
$$B(k) = \frac{\hat{f}(k)ke^{-k} + \hat{g}(k)}{2k\cosh(k)}$$

$$\hat{u}(k,y) = \frac{1}{2k\cosh(k)}\big[\hat{f}(k)ke^{k-ky} - \hat{g}(k)e^{-ky}+\hat{f}(k)ke^{-k+ky} + \hat{g}(k)e^{ky} \big]$$
$$\hat{u}(k,y) = \frac{1}{k\cosh(k)}\big[\hat{f}(k)k\cosh(k-ky) + \hat{g}(k)\sinh(ky) \big]$$

$$u(x,y) = \int_{-\infty}^{\infty} \big[\frac{\hat{f}(k)\cosh(k-ky)}{\cosh(k)} + \frac{\hat{g}(k)\sinh(ky)}{k\cosh(k)} \big] e^{ikx} dk$$

5
##### Chapter 5 / Small typos in section 5.2
« on: November 02, 2016, 01:59:12 PM »
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*}

\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

6
##### Q4 / Re: Q4
« on: October 28, 2016, 09:33:02 AM »
Since $f(x) = |x|$ is an even function, $b_n = 0 \ \ \forall n$
$a_0 = \frac{1}{\pi}\int_{-\pi}^\pi |x|\,dx = \pi$
$a_n = \frac{1}{\pi}\int_{-\pi}^\pi |x|\cos(nx) \,dx = \int_0^\pi \frac{2x}{\pi}\cos(nx) \,dx$
Integrating by parts
$a_n =\frac{2x}{n\pi}\sin(nx) \big|_{0}^{\pi}- \int_0^\pi \frac{2}{n\pi} \sin(nx) \,dx = \frac{2}{n^2\pi}\cos(nx)\big|_{0}^{\pi}$
$a_n = \begin{cases}\frac{-4}{n^2\pi} && n \ is \ odd \\0 && n \ is \ even\end{cases}$

$f(x) = \frac{\pi}{2} +\sum_{m=0}^\infty \frac{-4}{(2m+1)^2\pi}\cos((2m +1)x)$

7
##### TT1 / Re: TT1-P3
« on: October 20, 2016, 12:21:08 PM »
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 $x+3t > 0$,$x-3t > 0$
$$u(x,t) = \sin (x + 3t )$$

8
##### TT1 / Re: TT1-P3
« on: October 20, 2016, 09:17:09 AM »
In domain $0< x < t,$ $x - 3t < 0$

9
##### TT1 / Re: TT1-P4
« on: October 19, 2016, 10:41:51 PM »
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 $\partial_x(u_tu_x) = u_{xt}u_x + u_tu_{xx}$
$$\partial_t E(t) = c^2(u_tu_x)\big|_{0}^{L} + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t)$$
$$\partial_t E(t) = c^2u_t(L,t)u_x(L,t) - c^2u_t(0,t)u_x(0,t) + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) =0$$

10
##### TT1 / Re: TT1-P1
« on: October 19, 2016, 10:35:16 PM »
a) Characteristic Equation

$$\frac{dt}{1} = \frac{dx}{xt} = \frac{du}{-u}$$
From $\frac{dt}{1} = \frac{dx}{xt}$, $\frac{t^2}{2} + \ln c = \ln x$, thus $x = ce^\frac{t^2}{2}$

b) General Solution
From $\frac{dt}{1} = \frac{du}{-u}$, $-t + \ln k = \ln u$
So $u = ke^{-t} = \phi(xe^{\frac{-t^2}{2}})e^{-t}$

c) Since $u|_{t=0} = \frac{1}{1+x^2}$,
$\phi(x) = \frac{1}{1+x^2}$
Therefore, $$u(x,t) = \frac{1}{1+x^2e^{-t^2}}e^{-t}$$

11
##### Q3 / Re: Q3
« on: October 14, 2016, 09:18:33 AM »
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 ?

12
##### Q3 / Re: Q3
« on: October 13, 2016, 10:35:12 PM »
Even if we do not consider u being continuous at the line $x = 3t$
Since $u|_{t=0}= \phi (x)$, then $u(0,0) = \phi (0)$
If we do not have the constant, $u(0,0) = \phi ( 0) - \frac{7}{5} \phi (0) \neq \phi (0)$
we need the constant $+\frac{7}{5} \phi (0)$

13
##### Q3 / Q3
« on: October 13, 2016, 08:48:28 PM »
\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 $x > 3t$,
$$u = \phi ( x + 3t )$$

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

14
##### Q2 / Q2
« on: October 06, 2016, 08:39:26 PM »
$u_{tt}-4u_{xx}=0$           $x > 0, t > 0$

$u|_{t=0}=0$
$u_t|_{t=0}= \begin{cases}\cos(x) && |x| < \pi/2 \\0 &&|x| \ge \pi/2\end{cases}$

General solution : $$u(x,t)=\frac{1}{2}\bigl[g(x+ct)+g(x-ct)\bigr]+ \frac{1}{2c}\int_{x-ct}^{x+ct} h(y)\,dy = \frac{1}{4}\int_{x-2t}^{x+2t} h(y)\,dy$$

when $(x-2t) \ge \frac{\pi}{2}$ and $(x+2t) \ge \frac{\pi}{2}$
$u = \frac{1}{4}\int_{x-2t}^{x+2t} 0\,dy = 0$

when $|x-2t| < \frac{\pi}{2}$ and $(x+2t) \ge \frac{\pi}{2}$
$u = \frac{1}{4}\int_{x-2t}^{x+2t} h(y)\,dy = \frac{1}{4}\int_{x-2t}^{\frac{\pi}{2}} \cos(y)\,dy + \frac{1}{4}\int_{\frac{\pi}{2}}^{x+2t} 0 \,dy = \frac{1}{4}(1 - \sin(x-2t))$

when $(x-2t) \le \frac{-\pi}{2}$ and $(x+2t) \ge \frac{\pi}{2}$
$u = \frac{1}{4}\int_{x-2t}^{x+2t} h(y)\,dy = \frac{1}{4}\int_{\frac{\pi}{2}}^{x+2t} 0\,dy + \frac{1}{4}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos(y)\,dy + \frac{1}{4}\int_{x-2t}^{\frac{-\pi}{2}} 0\,dy = \frac{1}{2}$

when $|x-2t| < \frac{\pi}{2}$ and $|x+2t| < \frac{\pi}{2}$
$u = \frac{1}{4}\int_{x-2t}^{x+2t} \cos(y)\,dy = \frac{1}{4}(\sin(x+2t) - \sin(x-2t))$

15
##### Q1 / Re: Q1-P3
« on: September 29, 2016, 09:32:27 PM »
$$\frac{dx}{1} = \frac{dy}{3} = \frac{du}{u}$$
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, $u = (y - 3x)e^x$

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