Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Messages - Jingxuan Zhang

Pages: 1 2 [3] 4
31
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*}

32
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.



33
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, $k^2+\alpha^2\neq 0$.

34
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.

35
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}



36
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:
$$D(x)=\frac{\beta}{\sqrt{v^2+4\beta}}\exp(x(v/2-\sqrt{v^2/2+\beta}))$$

37
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!

38
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.

39
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
$$u_{t}+(x^2+1) u_{x}=0 \implies C=\arctan(x)-t \implies u=\varphi(\arctan (x)-t)$$

40
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.

41
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

42
Web Bonus Problems / Re: Web bonus problem--Week 3
« on: January 19, 2018, 02:56:50 PM »
Let me admit the middle of $(5)$ was not only mystery but also wrong. The reason is I incorrectly thought the integrand is dependent on $t$. I hope this should correct it:
$$(\int _0^t u(x,t')\,dt')_{t}=\frac{d}{dt}\int _0^t u(x,t')\,dt' =u(x,t) \tag{8}$$
simply by FTC.

Thus $(6)$ is also wrong, since really from above we have
$$(\int _0^t u(x,t')\,dt')_{tt}=u_{t}(x,t)=\int _0^t u(x,t')_{tt}\,dt'.\tag{9}$$

It is only after that do we have, if $Lu:=u_{tt}-c^{2}u_{xx}$,

$$ L(\int _0^t u(x,t')\,dt') = \int _0^t Lu(x,t')\,dt'= 0\tag{10}$$

by equation in $(2)$.


43
Web Bonus Problems / Re: Web bonus problem--Week 3
« on: January 18, 2018, 10:19:21 PM »
Evaluated at $t=0$
$$\int _0^t u(x,t')\,dt'=0$$
whereas by assumption
$$(\int _0^t u(x,t')\,dt')_{t}=\int _0^t u_{t}(x,t')\,dt'+u(x,t)=g(x)\tag{5}$$
Also for each real $x$
$$ L(\int _0^t u(x,t')\,dt') = \int _0^t Lu(x,t')\,dt' + 2u_{t}(x,t) = 0\tag{6}$$
by formula $(1)$ and the equation in $(2)$. We thus recovered $(4)$ and uniqueness forces $v=\int _0^t u(x,t')\,dt'$. But then
$$\int _0^t u(x,t')\,dt' =\int _0^t \frac{1}{2}\bigl( g(x+ct') +g(x-ct')\bigr) dt' =
\frac{1}{2c} \bigl(\int _0^{x+ct}g(x')dx' -(-\int _{x-ct}^0 g(x')dx')\bigr)=\frac{1}{2c}\int_{x-ct}^{x+ct}g(x')\,dx'.\tag{7}$$

Someone else please do the Also.

44
Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 15, 2018, 06:50:44 AM »
Differentiating Ioana's $(A)$ and equating it with his(her?) $(B)$, we have the symbolic system
$$\left( \begin{array}{cc|c}
1 & -1 & 2x\\
1+x& 1-x&3x^{2}\\
\end{array} \right)

\implies

\left( \begin{array}{c}
\varphi'(X)\\
\psi'(Y)
\end{array} \right)
=
\left( \begin{array}{c}
X\\
-Y
\end{array} \right)
\text{ where X, Y are the arguments of $\varphi, \psi$ resp. }
\implies

\left( \begin{array}{c}
\varphi(X)\\
\psi(Y)\end{array} \right)
=
\left( \begin{array}{c}
X^{2}/2+C_{1}\\
-Y^{2}/2+C_{2}
\end{array} \right)

\implies
u=(x+t)^{2}/2 - (x-t)^{2}/2 + const.
$$

Fixed now. For uniqueness we impose that the characteristics intersect the initial data, that is, precisely when
$$x^{2}-2x+2C, x^{2}+2x-2C$$
both have solution. This happens whenever
$$-t-1/2\leq x\leq t+1/2.$$

45
Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 07:18:57 PM »
Necessity:$$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}.$$

Pages: 1 2 [3] 4