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Messages - Roro Sihui Yap

Pages: [1] 2
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
$$\frac{\rho^2 P''+2\rho P'+\lambda \rho^2 P}{P}+
\frac{\Lambda Y }{Y}=0$$

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

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}
&u_{xx}+u_{yy}=0,\qquad -\infty<x<\infty, 0<y<1,  \\
&u|_{y=0}=f(x), \quad u_y|_{y=1}=g(x).
\end{align}

Taking fourier transform,
\begin{align}
&-k^2\hat{u} +\hat{u}_{yy}=0 \\
&\hat{u}|_{y=0}=\hat{f}(k) \quad \hat{u_y}|_{y=1}=\hat{g}(k)
\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*}

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

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
APM346--Misc / Re: Halloween Challenge 2
« on: October 21, 2016, 05:47:44 PM »
10a)
$$u(x,t) = f(x+t) + g(x-t) $$

From $u|_{t=0} = 0$, $\ \ f(x) + g(x) = 0$ $\rightarrow f(x)= -g(x)$
From $u|_{x=0} = 0$, $\ \ f(t) + g(-t) = 0 \ \ $
Using $f(x)= -g(x), \ $we get $\ -g(t) + g(-t) =0 $ $\rightarrow g(-t)= g(t)$

From $u|_{x=a} = 0$, $\ \ f(a+t) + g(a -t) = 0$ $\rightarrow  g(t-a) = g(t+a)$ $\rightarrow  g(x) = g(x +2a)\label{A}$
From $u|_{t=b} = 0$, $\ \ f(x+b) + g(x-b) = 0$ $\rightarrow g(x-b) = g(x+b)$ $\rightarrow g(x) = g(x+2b)\label{B}$

$g$ is a periodic function, suppose g(x) is not constant function, let the period of $g$ be $p$, the smallest number such that $g(x) = g(x+np)$ where $n$ is any integer and $p \ne 0$,

Then, $2a = k_1p$ and $2b = k_2p$ where $k_1,k_2$ are integers
$$\frac{2a}{k_1} = \frac{2b}{k_2} \implies \frac{a}{b} = \frac{k_1}{k_2}$$

Contradiction, we assumed a and b are not commensurable. Thus $g(x)$ must be a constant function, suppose $g(x) = k$, then $f(x) = -k$ where k is any number

$$u(x,t) = f(x+t) + g(x-t) = k - k = 0$$
The only solution is the trivial solution

8
APM346--Misc / Re: Halloween Challenge 3
« on: October 21, 2016, 04:40:13 PM »
Problem 3j)
 \begin{equation}
u(x,t)=\int_0^\infty G_D(x,y,t)g(y)\,dy+k
\int_0^{t} G_{D\,y}(x,0,t-\tau)h(\tau) \,d\tau; 
\end{equation}

Since $g(y) = 0 $ and $h(t)=\left\{\begin{aligned} &1 &&t<1, \\&0 &&t\ge 1.\end{aligned}\right.$
 
$$u(x,t) =\left\{\begin{aligned} &k\int_0^{t} G_{D\,y}(x,0,t-\tau) \,d\tau  &&t<1, \\&k\int_0^{1} G_{D\,y}(x,0,t-\tau) \,d\tau  &&t\ge 1.\end{aligned}\right.$$
$$G_{D\,y}(x,y,t) = \partial_y\big[ \frac{1}{2\sqrt{k\pi t}}\big(e^{-\frac{(x-y)^2}{4kt}} - e^{-\frac{(x+y)^2}{4kt}}\big)\big]$$
$$G_{D\,y}(x,y,t) =  \big[ \frac{1}{2\sqrt{k\pi t}}\big(e^{-\frac{(x-y)^2}{4kt}}(\frac{x-y}{2kt}) + e^{-\frac{(x+y)^2}{4kt}}(\frac{x+y}{2kt})\big)\big]$$
$$G_{D\,y}(x,0,t-\tau) =  \big[ \frac{1}{2\sqrt{k\pi (t-\tau)}}\big(e^{-\frac{x^2}{4k(t-\tau)}}(\frac{x}{k(t-\tau)})\big)  \big] =  \big[ \frac{x}{2k^{3/2}\sqrt{\pi }}\big(e^{-\frac{x^2}{4k(t-\tau)}}(\frac{1}{(t-\tau)^{3/2}})\big)  \big]$$

Let $z = \frac{x}{2\sqrt {k(t-\tau)}}$
$\frac{dz}{d\tau} = \frac{x}{4\sqrt {k}(t-\tau)^{3/2} }$

$$u(x,t) =\left\{\begin{aligned} &\frac{2}{\sqrt\pi}\int_\frac{x}{2\sqrt {tk}}^{\infty}  e^{-z^2}\,dz &&t<1, \\&\frac{2}{\sqrt\pi}\int_\frac{x}{2\sqrt {tk}}^{\frac{x}{2\sqrt {(t-1)k}}}  e^{-z^2}\,dz  &&t\ge 1.\end{aligned}\right.$$
$$u(x,t) =\left\{\begin{aligned} &1 - erf (\frac{x}{2\sqrt {tk}}) &&t<1, \\& erf (\frac{x}{2\sqrt {(t-1)k}}) - erf (\frac{x}{2\sqrt {tk}})  &&t\ge 1.\end{aligned}\right.$$




Problem 4j)
 \begin{equation}
u(x,t)=\int_0^\infty G_N(x,y,t)g(y)\,dy-
k \int_0^{t} G_{N}(x,0,t-\tau)h(\tau) \,d\tau
\end{equation}

Since $g(y) = 0 $ and $h(t)=\left\{\begin{aligned} &1 &&t<1, \\&0 &&t\ge 1.\end{aligned}\right.$
 
$$u(x,t) =\left\{\begin{aligned} &-
k \int_0^{t} G_{N}(x,0,t-\tau) \,d\tau  &&t<1, \\&-
k \int_0^{1} G_{N}(x,0,t-\tau)\,d\tau  &&t\ge 1.\end{aligned}\right.$$
 
$$G_{N}(x,0,t-\tau) =  \big[ \frac{1}{\sqrt{k\pi (t-\tau)}} \ e^{-\frac{x^2}{4k(t-\tau)}}  \big] $$

Let $z = \frac{x}{2\sqrt {k(t-\tau)}}$
$\frac{dz}{d\tau} = \frac{x}{4\sqrt {k}(t-\tau)^{3/2} }$

$$u(x,t) =\left\{\begin{aligned} &\int_\frac{x}{2\sqrt {tk}}^{\infty}  \frac{-x}{z^2\sqrt \pi}e^{-z^2}\,dz &&t<1, \\&\int_\frac{x}{2\sqrt {tk}}^{\frac{x}{2\sqrt {(t-1)k}}}  \frac{-x}{z^2\sqrt \pi}e^{-z^2}\,dz  &&t\ge 1.\end{aligned}\right.$$
 
Integrate by parts, let $w =\frac{x}{\sqrt \pi}e^{-z^2} $, $\frac{dw}{dz} =\frac{-2xz}{\sqrt \pi}e^{-z^2} $
let $\frac{dv}{dz} = \frac{-1}{z^2}$, $v = \frac{1}{z}$

$$u(x,t) =\left\{\begin{aligned} &\frac{x}{z\sqrt \pi}e^{-z^2}\big|_{\frac{x}{2\sqrt {tk}}}^{\infty} +  \int_\frac{x}{2\sqrt {tk}}^{\infty}  \frac{2x}{\sqrt \pi}e^{-z^2}\,dz &&t<1, \\&\frac{x}{z\sqrt \pi}e^{-z^2}\big|_{\frac{x}{2\sqrt {tk}}}^{\frac{x}{2\sqrt {(t-1)k}}} + \int_\frac{x}{2\sqrt {tk}}^{\frac{x}{2\sqrt {(t-1)k}}}  \frac{2x}{\sqrt \pi}e^{-z^2}\,dz  &&t\ge 1.\end{aligned}\right.$$

$$u(x,t) =\left\{\begin{aligned} &-\frac{2\sqrt{tk}}{\sqrt\pi}e^\frac{-x^2}{4tk}+  x (1- erf(\frac{x}{2\sqrt{tk}})) &&t<1, \\&\frac{2\sqrt{(t-1)k}}{\sqrt\pi}e^\frac{-x^2}{4(t-1)k}-\frac{2\sqrt{tk}}{\sqrt\pi}e^\frac{-x^2}{4tk}+  x (erf(\frac{x}{2\sqrt{(t-1)k}}- erf(\frac{x}{2\sqrt{tk}}))  &&t\ge 1.\end{aligned}\right.$$
 

9
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 $x+3t > 0$,$x-3t > 0$
$$u(x,t) = \sin (x + 3t ) $$

10
TT1 / Re: TT1-P3
« on: October 20, 2016, 09:17:09 AM »
In domain $ 0< x < t,$ $x - 3t < 0 $

11
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 $\partial_x(u_tu_x) = u_{xt}u_x + u_tu_{xx}$
$$\partial_t E(t) = c^2(u_tu_x)\big|_{0}^{L} + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) $$
$$\partial_t E(t) = c^2u_t(L,t)u_x(L,t) - c^2u_t(0,t)u_x(0,t) + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) =0  $$

12
TT1 / Re: TT1-P1
« on: October 19, 2016, 10:35:16 PM »
a) Characteristic Equation

\begin{equation}  \frac{dt}{1} = \frac{dx}{xt} = \frac{du}{-u} \end{equation}
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, \begin{equation} u(x,t) = \frac{1}{1+x^2e^{-t^2}}e^{-t}\end{equation}
 

 

13
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 ?

14
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) $ 

15
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$,
\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, 
$u = \begin{cases}\phi ( x + 3t ) && x > 3t \\\phi ( x + 3t )  - \frac{7}{5} \phi (3t - x) + \frac{7}{5} \phi (0) && 0 < x < 3t \end{cases}$ 

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