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Messages - Jingxuan Zhang

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Term Test 1 / Re: P5
« on: February 16, 2018, 02:50:23 PM »
A direct computation yields:

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}}))

Term Test 1 / Re: P2
« on: February 16, 2018, 02:36:46 PM »
A direct computation yields:
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'\\

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

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 $x>ct$
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
&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
Provided, indeed, $k^2+\alpha^2\neq 0$.

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
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.

Quiz-2 / Re: Q2-T5102
« on: February 02, 2018, 09:23:39 PM »
Plug in D'Alembert's instantly we obtain:

            &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\\
\tag*{Part (a)}\\[5pt]
            &0 && x+ct< 0\\
            &x/2c+t/2 && x-ct <0 \leq x+ct\\
            &t && 0 \leq x-ct\\
\tag*{Part (b)}

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:

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!

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.

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

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.

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.

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

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 $(2)$.

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.

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)


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

\left( \begin{array}{c}
\psi(Y)\end{array} \right)
\left( \begin{array}{c}
\end{array} \right)

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
$$-t-1/2\leq x\leq t+1/2.$$

Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 07:18:57 PM »

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