Messages

16

Web Bonus = Nov / Re: Web Bonus Problem to Week 8 (#3)

« on: November 09, 2015, 11:43:14 PM »

For the Robin boundary condition problem, we get transform $$\hat{u}(k,y) = A(k)e^{-|k|y} + B(k)e^{|k|y}$$
We discard $B(k)e^{|k|y}$, because it is unbounded.
The Robin boundary condition, $$(\hat{u}_y + \alpha\hat{u})|_{y=0} = \hat{f}(k)\\A(k) = -\frac{\hat{f}(k)}{(|k|-\alpha)}\\\hat{u}(k,y)= -\frac{\hat{f}(k)}{(|k| - \alpha)}e^{-|k|y}\\u(x,y) = \int_{-\infty}^{\infty}- \frac{\hat{f}(k)}{(|k| - \alpha)}e^{-|k|y+ikx} dk$$
For $\alpha<0$, the expression is a nice function (is correct from the physical point of view).
For $\alpha>0$, ?
For $\alpha = 0$, A(k) could be singular at $k = 0$, to avoid it we need $\int_{-\infty}^{\infty} f(x) dx = 0$.

17

HA7 / Re: HA7-P2

« on: November 05, 2015, 06:42:00 PM »

For the f^(ω) formula in (a), I think there should be a negative sign before iωx, however the final answer is right.

Yes, I miss the negative sign in my code.

18

HA7 / Re: HA7-P7

« on: November 05, 2015, 03:53:46 PM »

Yumeng, I think $e^{-|k|}+e^{|k|} = 2\cosh(|k|) $ not $ -2\sinh(|k|)$

19

HA7 / Re: HA7-P7

« on: November 05, 2015, 02:28:48 PM »

For part (a), similar to problem 6, we get $$\hat{u}(k,y) = A(k)e^{-|k|y}+B(k)e^{|k|y}$$
Because we have two boundary conditions $$\hat{u|}_{y=0} = \hat{f}(k)\\\hat{u|}_{y=1} = \hat{g}(k)$$
We cannot discard anything. Then,$$A(k) + B(k) = \hat{f}(k)\\A(k)e^{-|k|}+B(k)e^{|k|} = \hat{g}(k)$$
Then we get, $$A(k) = \frac{e^{|k|}\hat{f}(k)}{2\sinh(|k|)} - \frac{\hat{g}(k)}{2\sinh(|k|)}\\B(k)=-\frac{e^{-|k|}\hat{f}(k)}{2\sinh(|k|)}+\frac{\hat{g}(k)}{2\sinh(|k|)}$$
Hence, we get$$\hat{u}(k,y) = \frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}\\u(x,y) = \int_{-\infty}^{\infty} [\frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}]e^{ikx} dk$$

20

HA7 / Re: HA7-P6

« on: November 05, 2015, 12:31:17 PM »

For part (a), let make partial Fourier transform with respect to $x\mapsto k, u(x,t) \mapsto \hat{u}(k,t)$, then we get $$ \hat{u}_{yy} - k^2\hat{u} = 0\\\hat{u|}_{y=0} = \hat{f}(k)$$
Solving this ODE, then we have $$\hat{u}(k,y) = A(k)e^{-|k|y}+B(k)e^{|k|y}$$
And we discard the term $B(k)e^{|k|y}$, because it's unbounded.
So, we have $\hat{u} = A(k)e^{-|k|y}, A(k) = \hat{f}(k)$
Then we find the inverse Fourier transform of $e^{-|k|y}$, which is $\int_{-\infty}^{\infty} e^{-|k|y}e^{ikx} dk$. $$ = \int_{-\infty}^{0} e^{(y+ix)k} dk +\int_{0}^{\infty} e^{(-y+ix)k} dk\\=\frac{1}{y-ix}+\frac{1}{y+ix}\\=\frac{2y}{x^2+y^2}$$
Then$$ u(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(x')\frac{y}{(x-x')^2+y^2} dx'$$

21

HA7 / Re: HA7-P3

« on: November 03, 2015, 11:14:57 PM »

For part (a), by inverse Fourier transform we get $$ \int_{-\infty}^{\infty} \hat{f}(\omega)e^{i\omega x} d\omega = f(x)\\\int_{-\infty}^{\infty} e^{-\alpha |\omega|}e^{i\omega x } d\omega = \frac{2\alpha}{\alpha^2 + x^2}\\2\alpha[\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{\alpha^2 + x^2}e^{-i\omega x} dx] = e^{-\alpha |\omega|}\\\hat{f}(
\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{\alpha^2 + x^2}e^{-i\omega x} dx = \frac{1}{2\alpha}e^{-\alpha|\omega|}$$



For part (b), let $g(x) = xf(x) = \frac{x}{(x^2+\alpha^2)}$, where $f(x)$ is defined in part (a).
Then, $$\hat{g}(\omega) = i\hat{f'}(\omega) = \frac{-i\omega e^{-\alpha |\omega|}}{2|\omega|}\ ( |\omega|' = \frac{\omega}{|\omega|})$$



For part (c), is similar to problem 2, for $g(x) = f(x)\cos(\beta x)$, where $f(x)$ is defined in part (a). $\hat{g}(\omega) = \frac{1}{2}[\hat{f}(\omega - \beta) + \hat{f}(\omega + \beta)]$ $$=\frac{1}{4\alpha}[e^{-\alpha |\omega - \beta|}+e^{-\alpha |\omega + \beta|}]$$
for $g(x) = f(x)\sin(\beta x), \hat{g}(\omega) = \frac{1}{2i}[\hat{f}(\omega - \beta) - \hat{f}(\omega + \beta)]$
$$=\frac{1}{4i\alpha}[e^{-\alpha |\omega - \beta|}-e^{-\alpha |\omega + \beta|}]$$


For part (d), let $f(x) = \frac{\cos(\beta x)}{(\alpha^2 + x^2)}, g(x) = xf(x), \hat{g}(\omega) = i\hat{f'}(\omega)$ $$\hat{f}(\omega) = \frac{1}{4\alpha}[e^{-\alpha |\omega - \beta|}+e^{-\alpha |\omega + \beta|}]\\\hat{f'}(\omega) = -\frac{1}{4}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}+\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]\\\hat{g}(\omega) = -\frac{i}{4}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}+\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]$$
Similarly, let $f(x) = \frac{\sin(\beta x )}{(\alpha^2 + x^2)}, g(x) = xf(x), \hat{g}(\omega) = i\hat{f'}(\omega)$ $$\hat{f}(\omega) = \frac{1}{4i\alpha}[e^{-\alpha |\omega - \beta|}-e^{-\alpha |\omega + \beta|}]\\\hat{f'}(\omega) = -\frac{1}{4i}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}-\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]\\\hat{g}(\omega) = -\frac{1}{4}[\frac{\omega - \beta}{|\omega-\beta|}e^{-\alpha|\omega-\beta|}-\frac{\omega + \beta}{|\omega+\beta|}e^{-\alpha|\omega+\beta|}]$$

22

Web Bonus = Oct / Re: Web bonus problem : Week 4 (#1)

« on: November 03, 2015, 09:36:09 PM »

For part (a), first we separate variables, let $u(x,t) = X(x)T(t)$
Then, $$\frac{T''(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda$$
Take $\lambda = \omega^2$, where $\omega > 0$
Then, we have $$X(x) = A\cos(\omega x) + B\sin(\omega x)\\ T(t) = C\cos(\omega x) + D\sin(\omega x)$$
Plug in Boundary Conditions, $ u(0,t) = u(a,t) = u(x,0) = u(x,b) = 0$
$$X(0) = A\cos(0) + B\sin(0) = A = 0\\X(x) = B\sin(\omega x)\\X(x)=Bsin(\omega a) = 0$$
To get non-trivial solutions, we must have $sin(\omega a) = 0$, hence $\omega a = n\pi, a=\frac{n\pi}{\omega}$, where $n\in Z$
Similarly, we get $$T(t) = D\sin(\omega b) = 0$$
To get non-trivial solutions, we must have $sin(\omega b) = 0$, hence $\omega b = n'\pi, b=\frac{n'\pi}{\omega}$, where $n'\in Z$
Hence, $\frac{a}{b}\in Q$. If $\frac{a}{b}\notin Q$, we can only get trivial solution such that $B=D=0$.


For part (b), if $\frac{a}{b}\in Q$, take $p=n\pi, q=n'\pi$,  then $\omega = \frac{p}{a} = \frac{q}{b} $
Then $$X(x) = B\sin(\frac{px}{a})\\T(t) = D\sin(\frac{qt}{b})$$
Take $B = D = 1$, then there exists a nontrivial solution such that  $$u(x,t)=\sin(\frac{px}{a})\sin(\frac{qt}{b})$$

23

HA7 / Re: HA7-P5

« on: November 03, 2015, 08:11:49 PM »

For part (a), given that $|x| \leq a$, we get $$ \hat{f}(\omega) = \frac{1}{2\pi}\int_{-a}^{a} e^{-i\omega x } dx\\= \frac{1}{2\pi}\frac {e^{-i\omega a}-e^{i\omega a}}{(-i\omega)}\\=\frac{\sin(\omega a)}{\pi\omega}$$


For part (b), we take $g(x) = xf(x)$, $\hat{g}(\omega) = i\hat{f'}(\omega)$, where $f(x)$ is defined in part (a).$$\hat{f'}(\omega) = \frac{a\cos(\omega a)}{\pi\omega} - \frac{\sin(\omega a)}{\pi\omega^2}\\\hat{g}(\omega) = i( \frac{a\cos(\omega a)}{\pi\omega} - \frac{\sin(\omega a)}{\pi\omega^2})$$

For part (c), by Inverse Fourier Transformation, $$\int_{-\infty}^{\infty} \hat{f}(\omega)e^{i\omega x} d\omega=\int_{-\infty}^{\infty} \frac{\sin(\omega a)}{\pi\omega}e^{i\omega x} d\omega= f(x) = 1$$ where $|x|\leq a$
We take $x=0, a=1$, then we get $$\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin(\omega)}{\omega} d\omega = 1 $$
Hence, $$\int_{-\infty}^{\infty} \frac{\sin(\omega)}{\omega} d\omega = \pi$$

24

HA7 / Re: HA7-P2

« on: November 03, 2015, 06:24:13 PM »

For part (d), given $g(x) = xe^{-\alpha |x|}\cos(\beta x)$, where $f(x) =  e^{-\alpha |x|}\cos(\beta x)$
By theorem $\hat{g}(\omega) = i \hat{f'}(\omega)$
By part (b), we have $\hat{f}(\omega) = \frac{1}{2}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)}+\frac{\alpha}{\pi(\alpha^2+(\omega + \beta)^2)}]$
Then, $$ \hat{f'}(\omega) = -\frac{\alpha}{\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} + \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$
Hence, $$ \hat{g}(\omega) = -\frac{i\alpha}{\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} + \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$
Similarly, for $g(x) =  xe^{-\alpha |x|}\sin(\beta x)$, where $f(x) =  e^{-\alpha |x|}\sin(\beta x)$
By part (b), we have $\hat{f}(\omega) = \frac{1}{2i}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)} - \frac{\alpha}{\pi(\alpha^2+(\omega + \beta)^2)}]$
Then, $$ \hat{f'}(\omega) = -\frac{\alpha}{i\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} - \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$
Hence, $$ \hat{g}(\omega) = -\frac{\alpha}{\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} - \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$

25

HA7 / Re: HA7-P2

« on: November 03, 2015, 05:33:18 PM »

For part (c),
      Given $g(x) = xe^{-\alpha |x|}$ where $f(x) = e^{-\alpha |x|}$. Hence, $g(x) = xf(x)\ then, \hat{g}(\omega) = i\hat{f'}(\omega)$
We have $$\hat{f}(\omega) = \frac{\alpha}{\pi(\alpha^2 +\omega^2)}\\ \hat{f'}(\omega) = -\frac{2\alpha\omega}{\pi(\alpha^2+\omega^2)^2}$$
Then we get $$\hat{g}(\omega) = -\frac{2i\alpha\omega}{\pi(\alpha^2+\omega^2)^2}$$

26

HA7 / Re: HA7-P2

« on: November 03, 2015, 01:01:08 PM »

For part (b), let $f(x)=e^{-\alpha|x|}$, for $g(x) = e^{-\alpha |x|}\cos(\beta x)$, where $\beta >0$.
$$\cos(\beta x) = \frac{1}{2}[e^{i\beta x} + e^{-i\beta x}]\\ g(x) = \frac{1}{2}[f(x)e^{i\beta x }+f(x)e^{-i\beta x}]\\\hat{g}(\omega) = \frac{1}{2}[\hat{f}(\omega - \beta)+\hat{f}(\omega +\beta)]$$
Combine the solution that we found in part (a), we have $$\\\hat{g}(\omega) = \frac{1}{2}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)}+\frac{\alpha}{\pi(\alpha^2 + (\omega +\beta)^2)}]$$
Similarly, for $g(x) = e^{-\alpha |x|}\sin(\beta x)$, where $\beta >0$.$$\sin(\beta x) = \frac{1}{2i}[e^{i\beta x} - e^{-i\beta x}]\\\hat{g}(\omega) = \frac{1}{2i}[\hat{f}(\omega - \beta)-\hat{f}(\omega +\beta)]\\\hat{g}(\omega) = \frac{1}{2i}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)}-\frac{\alpha}{\pi(\alpha^2 + (\omega +\beta)^2)}]$$

27

HA7 / Re: HA7-P2

« on: November 03, 2015, 12:30:16 PM »

Noted that the Fourier transform of $f(x)$ is given as
correct formula
$$\hat{f}(\omega) = \frac{k}{2\pi}\int_{-\infty}^{\infty} f(x)e^{-i\omega x}dx$$
 where
$$k = 1, \omega = \frac{\pi n}{l}$$ WTH? There is no $l$ or $n$, $\omega\in \mathbb{R}$ is continuous
For part (a), given $\alpha > 0$
$$f(x) = e^{-\alpha |x|}\\
 \hat{f}(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-\alpha |x|}e^{-i\omega x} dx \\
= \frac{1}{2\pi}[\int_{-\infty}^{0} e^{\alpha x - i\omega x} dx + \int_{0}^{\infty} e^{-\alpha x - i\omega x} dx]\\
 = \frac{1}{2\pi}[\int_{0}^{\infty} e^{-\alpha x + i\omega x} dx + \int_{0}^{\infty} e^{-\alpha x - i\omega x} dx]\\
 = \frac{1}{2\pi}[\frac{1}{\alpha - i\omega}+\frac{1}{\alpha + i\omega}]\\ = \frac{\alpha}{\pi(\alpha^2 + \omega^2)}$$

28

Test 1 / Re: TT1-P2

« on: October 22, 2015, 12:39:17 AM »

For part a) we have $$ c = 1,\ g(x) = 0 = h(x),\ f(x,t) = \frac{x}{(x^2 + 1)^2}\\Then\ u(x,t) = \frac{1}{2}\int_{0}^{t} \int_{x-(t-t')}^{x+(t-t')} \frac{x'}{(x'^2+1)^2} dx'dt'\\ = \frac{1}{-4}\int_{0}^{t} \frac{1}{x'^2+1}\Big|_{x-(t-t')}^{x+(t-t')} dt'\\= \frac{1}{-4}\int_{0}^{t} \frac{1}{(x+(t-t'))^2+1} - \frac{1}{(x-(t-t'))^2+1)} dt'\\ for\ \frac{1}{4}\int_{0}^{t} \frac{1}{(x-(t-t'))^2+1)} dt'\\ take\ z = x - t + t'\ and\ dz = dt'\\ = \frac{1}{4}\int_{x-t}^{x} \frac{1}{z^2+1} dz\\= \frac{1}{4}[\arctan(x) - \arctan(x-t)]\\similar\ way\ to\ get\ \frac{1}{-4}\int_{0}^{t} \frac{1}{(x+(t-t'))^2+1}\\Finally\ u(x,t) = \frac{1}{4}[\arctan(x+t) - \arctan(x-t)]$$
For par b) $$\lim_{t\to +\infty} u(x,t) = \lim_{t\to +\infty} \frac{1}{4}[\arctan(x+t) - \arctan(x-t)]\\as\ t\to +\infty, x+t\to +\infty, x-t\to -\infty\\ hence, \arctan(x+t)\to \frac{\pi}{2} + 2k\pi, \arctan(x-t)\to -\frac{\pi}{2} + 2k\pi\\ \lim_{t\to +\infty} u(x,t) = \lim_{t\to +\infty} \frac{1}{4}[\arctan(x+t) - \arctan(x-t)]\\= \frac{1}{4}[\frac{\pi}{2} + 2k\pi + \frac{\pi}{2} - 2k\pi] = \frac{\pi}{4}$$

I am not sure this is correct.

29

Test 1 / Re: TT1-P1

« on: October 21, 2015, 11:35:38 PM »

For part a) b) I get,  $$\frac{dt}{t^2+1} = \frac{dx}{1}\\Let\ t = \tan{\theta}\\So\  \theta = \arctan{\ t}\\So\  1 + t^2 = 1 + \tan^2{\theta} = \sec^2{\theta}\\and\ \frac{dt}{d\theta} = \sec^2{\theta}\\So\ \int \frac{1}{1+t^2} dt = \int \frac{sec^2{\theta}}{\sec^2{\theta}} d\theta\\= \int 1 d\theta\\= \theta\\= \arctan{t}\\For \int 1 dx = x\\ Hence\ x = \arctan{t}\\the\ characteristic\ curves\ are\ t = \tan{x}\\then\ we\ have\ c = x - \arctan{t}\\ u = \phi{(c)} = \phi{(x - \arctan{t})}$$

30

Textbook errors / Misprint for 2.4

« on: October 01, 2015, 03:59:03 PM »

Hi Professor,
    I think in "2.4 1D Wave equation reloaded:characteristic coordinates", for the last graph, the equation on the right side should be   
    x' = x + c(t - t').