Messages

31

Quiz-5 / Re: Quiz 5, T5102

« on: March 08, 2018, 07:23:50 AM »

George:
The reason that the absolute value is missing is that the contour you choose must depend on the sign of $k$ If $k<0$ then your contour should instead be upper semicircle and if $k>0$ then you are right. This is done to satisfy the hypothesis of Jordan's lemma, which you implicitly used to control the integral over the arc as it increasese.

Also what property do you refer to? I think $\hat{xf}=i\hat{f}'$?

Beside: don't we always use $\frac{1}{2\pi}$ for the scaling? and this for one thing at least better suit Residual Theorem.
$$\hat{f}(\omega)=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{-i\omega x}}{(x-ai)(x+ai)}=\left\{\begin{align*}\left.i\frac{e^{-i\omega x}}{x+ai}\right|_{x=ai}&&\omega<0\\ \left.-i\frac{e^{-i\omega x}}{x-ai}\right|_{x=-ai}&&\omega>0\end{align*}\right.=\frac{e^{-|\omega|a}}{2a} $$
And so immediately
$$\hat{xf}(\omega)=i\hat{f}'(\omega)=-i \text{sgn}(\omega)\frac{e^{-|\omega|a}}{2}$$

32

Quiz-4 / Quiz 4 -- both sections

« on: March 01, 2018, 10:07:07 PM »

Note: Since problems for both sections are very similar I suggest to discuss them together, so certain parts of the solutions could be used without repetition
 (V.I.)

The only difference that in Wed section condition on the right end are $u_{xx}|_{x=l}=u_{xxx}|_{x=l}=0$,


This is problem 3 part 1,2 from http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter4/S4.2.P.html

The associated eigenvalue problem is
\begin{align}X^{iv}-\omega^4 X=0\label{1}\\X(0)=X'(0)=0\label{2}\\X(l)=X'(l)=0
\label{3}\end{align}
From (\ref{1}) we write
\begin{equation}X=A\cosh (\omega x) + B\sinh(\omega x)+C\cos(\omega x)+D\sin(\omega x)\label{4}\end{equation}
whence (\ref{2}) implies, if $\omega\neq 0 $,
\begin{equation}A+C=B+D=0\label{5}\end{equation}
and so (\ref{4}) becomes
\begin{equation}X=A(\cosh (\omega x) -\cos(\omega x)) +  B(\sinh(\omega x)-\sin(\omega x))\label{6}\end{equation}

Now the algebraic system in variable of $A,B$ obtained from (\ref{3}) has nontrivial solution if and only if the coefficient matrix is singular, that is:
\begin{equation}\left|\begin{array}{cc}\cosh (\omega l) -\cos(\omega l)&\sinh(\omega l)-\sin(\omega l)\\ \sinh (\omega l) +\sin(\omega l)&\cosh(\omega l)-\cos(\omega l)\end{array}\right|=2-2\cosh (\omega l)\cos(\omega l)=0\iff\cosh (\omega l)\cos(\omega l)=1\label{equation}.\end{equation}
The null space of this system is
\begin{equation}(A,B)'=t(-\sinh(\omega l)+\sin(\omega l),\cosh (\omega l) -\cos(\omega l))',t\in\mathbb{R}\label{8}\end{equation}
and so (\ref{6}) becomes
\begin{equation}X=(-\sinh(\omega l)+\sin(\omega l))(\cosh (\omega x) -\cos(\omega x))+(\cosh (\omega l) -\cos(\omega l))(\sinh(\omega x)-\sin(\omega x))\label{9}\end{equation}
up to a scalar multiple. This is the eigenspace.

The graphs are those of $1/\cosh$ and $\cos$, imagined to be in variable of $\omega l$. Their (infinitely many) intersections suffice (\ref{equation}).

33

Web Bonus Problems / Re: Web Bonus problem -- Reading Week

« on: February 17, 2018, 03:26:11 PM »

Part (b): normalize the integrand we have
$$I=\hat{u}(0)=\int e^{-\frac{1}{2}a x^2}\,dx=\sqrt{\frac{2\pi}{a}}\cdot\underbrace{\frac{1}{\sqrt{2\pi}}\int e^{-z^2/2}\,dz}_{1}=\sqrt{\frac{2\pi}{a}}$$
But here really $\sqrt{a}$ is ill-defined: $a\in\mathbb{C}$ might result in this root having multiple value. The sign therefore depends on the branch chosen for $a$. In particular the principal value of $I$ will have a positive sign if $\Im(a)\geq0$ and a negative sign otherwise.

EDIT: I consulted Ahlfors' text on complex variable, which I have completely forgotten now. Write $a=\alpha+i\beta$ then
$$\sqrt{\frac{2\pi}{\alpha}}=\pm\frac{\sqrt{2\pi}}{\alpha^2+\beta^2}\left(\sqrt{\frac{\alpha+\sqrt{\alpha^2+\beta^2}}{2}}-i\frac{\beta}{|\beta|}\sqrt{\frac{-\alpha+\sqrt{\alpha^2+\beta^2}}{2}}\right)$$

If $\alpha>0,\beta=0$ then of course this is just the simple formula on LS.

34

Web Bonus Problems / Re: Web Bonus problem -- Reading Week

« on: February 17, 2018, 03:10:50 PM »

I find the second part perplexing: what are we to compare $I$ with? and shouldn't $I=\hat{u}(0)$ be immediate once the formula for $\hat{u}$ is determined by Adam?

35

Term Test 1 / Re: P5

« on: February 16, 2018, 02:50:23 PM »

A direct computation yields:
\begin{align*}

u&=\frac{1}{\sqrt{4t\pi}}\int_{-\infty}^{\infty} \exp{-\frac{(y-x)^2}{4t}-|y|}\,dx\\
&=\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{4t}-y)\,dx +\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{4t}-y)\,dx\\
&=\frac{\exp(x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+2t)^2}{4t})\,dx + \frac{\exp(-x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+2t)^2}{4t})\,dx\\
&=\frac{\exp(x+t)}{\sqrt{\pi}}\int_{\frac{x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz+ \frac{\exp(-x+t)}{\sqrt{\pi}}\int_{\frac{-x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz\\
&=\frac{\exp(x+t)}{2}(1-\text{erf}(\frac{x+2t}{\sqrt{4t}})) + \frac{\exp(-x+t)}{2}(1-\text{erf}(\frac{-x+2t}{\sqrt{4t}}))
\end{align*}

36

Term Test 1 / Re: P2

« on: February 16, 2018, 02:36:46 PM »

A direct computation yields:
\begin{align*}
u &=\frac{1}{4}\int_0^t\int_{x-2t+2t'}^{x+2t-2t'} \frac{8}{x'^2+1} \,dx'\,dt'\\
&=\frac{1}{4}\int_{x-2t}^x\int_{0}^{x'/2+x/2+t} \frac{8}{x'^2+1} \,dt'\,dx' +\frac{1}{4}\int_x^{x+2t}\int_{0}^{-x'/2+x/2+t} \frac{8}{x'^2+1} \,dt'\,dx'\\
&=\int_{x-2t}^x\frac{x'-x+2t}{x'^2+1} \,dx' + \int_x^{x+2t}\frac{x'+x+2t}{x'^2+1} \,dx' \\
&= \int_{x-2t}^x\frac{x'}{x'^2+1} \,dx + (x-2t) \int_x^{x-2t}\frac{1}{x'^2+1}\,dx'+
\int_x^{x+2t}\frac{x'}{x'^2+1} \,dx + (x+2t) \int_x^{x+2t}\frac{1}{x'^2+1}\,dx'\\
&=\ln(x^2+1)-2x\tan^{-1}(x)-\frac{1}{2}\ln((x-2t)^2+1)+(x-2t)\tan^{-1}(x-2t)-\frac{1}{2}\ln((x+2t)^2+1)+(x+2t)\tan^{-1}(x+2t)
\end{align*}

To my great shame, I failed to directly compute this during the actual sitting.

37

Quiz-3 / Re: Q3-T0101

« on: February 11, 2018, 02:56:12 PM »

Write $u=\varphi(x+ct)+\psi(x-ct)$ then Instantly by D'Alembert' s on $x>ct$
$$\varphi=\phi,\psi=0$$
\begin{equation}u=\phi(x+ct),x>ct\end{equation}
On $0<x<ct$ solution $\varphi=\phi$ still works, whereas from boundary condition
$$\phi(-s)'+\psi'(s)+\alpha\phi(-s)+\alpha\psi(s)=0 \implies (e^{\alpha s}\psi(s))'=(e^{\alpha s}\phi(-s))'-2\alpha\phi(-s)
\implies \psi(s)=\phi(-s)+2\alpha e^{-\alpha s}\int^s e^{-\alpha s'}\phi(s)\, ds'.$$
So that
\begin{equation}u=\phi(ct+x)+\phi(ct-x)+2\alpha e^{-\alpha (ct-x)}\int^{ct-x} e^{-\alpha s}\phi(s)\, ds,0<x<ct.\end{equation}
In particular, if $\phi(x)=e^{ikx}$, then
\begin{equation}u=\left\{\begin{aligned}
&e^{ik(x+ct)}&x>ct\\
&e^{ik(ct+x)}+e^{ik(ct-x)}-2\frac{ik\alpha+\alpha^2}{k^2+\alpha^2}e^{(ik-2\alpha) (ct-x)}&0<x<ct
\end{aligned}\right.
\end{equation}
Provided, indeed, $k^2+\alpha^2\neq 0$.

38

Technical Questions / Tag each equation in an IVP

« on: February 10, 2018, 04:14:10 PM »

I am trying to typeset in latex a system such as
$$\left\{\begin{array}{cc}f(x)&=k_1\\g(x)&=k_2\end{array}\right..$$
Is there anyway that allows me to label each individual equation such as with \tag? True, with align I can label but i need also the brace. True, with array I can brace but then cannot label. Thanks.

39

Quiz-2 / Re: Q2-T5102

« on: February 02, 2018, 09:23:39 PM »

Plug in D'Alembert's instantly we obtain:

\begin{align}     
&u=\left\{\begin{aligned}
            &0 && x+ct\leq -\pi/2 \text{ or } x-ct \geq \pi/2\\
            &\cos(x-ct)/2 && x-ct <\pi/2 \leq x+ct\\
            &\cos(x)\cos(ct) && -\pi/2 < x-ct < x+ct < \pi/2\\
            &\cos(x+ct)/2 && x-ct \leq -\pi/2 < x+ct\\
            \end{aligned}\right.
\tag*{Part (a)}\\[5pt]
&u=\left\{\begin{aligned}
            &0 && x+ct< 0\\
            &x/2c+t/2 && x-ct <0 \leq x+ct\\
            &t && 0 \leq x-ct\\
            \end{aligned}\right.
\tag*{Part (b)}
\end{align}

40

Web Bonus Problems / Re: Web bonus problem--Week 5

« on: February 02, 2018, 08:17:28 PM »

Okay let me confess: it was really due to shame of not being able to compute that integral by hand that I did not attempt to Wolfram it. From there I have:
$$D(x)=\frac{\beta}{\sqrt{v^2+4\beta}}\exp(x(v/2-\sqrt{v^2/2+\beta}))$$

41

Web Bonus Problems / Re: Web bonus problem--Week 5

« on: February 01, 2018, 03:47:24 PM »

a. Following the hint let $\tilde{u}(y,t):=u(y+vt,t)= v(y+vt,t)e^{-\beta t}=\tilde{v}(y,t)e^{-\beta t}$ then
$$(4) \implies \tilde{u}_t -\tilde{u}_{yy}+\beta \tilde{u} = 0 \implies \tilde{v}_t -\tilde{v}_{yy} = 0 \implies \tilde{v} = \frac{1}{\sqrt{4\pi t}}e^{-y^2/4t} \implies u = \frac{1}{\sqrt{4\pi t}}e^{-((x-vt)^2+4\beta t^2)/4t}$$

b. But I really don't know how to use that program!

42

Web Bonus Problems / Re: Web bonus problem -- Week 4

« on: January 26, 2018, 05:50:25 AM »

1. All functions $f,g,h,r,s$ are assumed to be very good functions. "Bad things" can happen only at $(0,0)$ and propagate along $x=ct$.

Hence I again advocate my answer in reply #2. Ioana this is what I meant, and especially that imposing the function to coincide at one point does not require it to be equal everywhere.

43

Quiz-1 / Q1-T0101-P1,2

« on: January 25, 2018, 01:10:48 PM »

1. General solution of
$$u_{xy}=e^{x+y}\implies u_{x}=e^{x+y}+\varphi_{x}(x)\implies e^{x+y}+\varphi(x)+\psi(y)$$
2. General solution of
$$u_{t}+(x^2+1) u_{x}=0 \implies C=\arctan(x)-t \implies u=\varphi(\arctan (x)-t)$$

44

Web Bonus Problems / Re: Web bonus problem -- Week 4

« on: January 24, 2018, 06:18:10 AM »

Ioana would you explain how to you get this? I thought the condition should be only at a point, such as origin as in my edited post. For I don't see the connexion from $$u(x,0)=g(x)$$ to $$u_{t}(x,0)=g'(x)$$, and how this helps to ensure continuity.

45

Web Bonus Problems / Re: Web bonus problem -- Week 4

« on: January 22, 2018, 09:30:08 PM »

Typo found: the boundary condition has the wrong argument. Indeed--fixed, V.I.
1. On the region $x>ct$, $u$ is automatically $C^{n}$ since the conditions are good. On $0<x<ct$:
a. intersection at origin requires $$r(0)=g(0)$$
b. plugging in the formula and match to $(3)$ gives $$r'(t)=\frac{1}{2}(h(ct)+h(-ct))+\frac{c}{2}(g'(ct)-g'(-ct))$$
c. d. I think the idea is to equate mixed partials but I am still trying to figure out what exactly is needed to infer the condition. Save for morrow.

Edit:
b. Was wrong. The approach was wrong, I should not plug in the general solution formula but rather just examine the initial data. To be $C^{1}$ it only needs
the various directional derivative agree to each other when close up to origin, viz.,
$$\lim_{x\to 0}u_{x}(x,0)=\lim_{t\to 0} u_{t}(0,t) \implies g'(0)=r'(0). $$

Wrong, since you equalize limits $u_t$ and $u_x$. Hint: equalize limits $u_t$ and $u_t$

c. This I think my idea was right. To have the mixed partial agree in particular we must have
$$\lim_{x\to 0} u_{tx}(x,0)=\lim_{x\to 0}u_{xt}(x,0) \implies h'(0)=(\frac{d}{dt}g')(0)=0. $$
Completely wrong. Use equation
d. Similarly
$$\lim_{x\to 0} u_{txx}(x,0)= \lim_{x\to 0}u_{xxt}(x,0)\implies h''(0)=(\frac{d}{dt}g'')(0)=0. $$
There are other equation arised by permuting partials but those are fulfilled automatically. Please correct me but anyway someone else please attempt the other part.
The same