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Messages - Jingxuan Zhang

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76
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:
$$D(x)=\frac{\beta}{\sqrt{v^2+4\beta}}\exp(x(v/2-\sqrt{v^2/2+\beta}))$$

77
APM346––Home Assignments / Re: Section 2.3, Q 5
« on: February 02, 2018, 06:58:56 PM »
So if I understand your hint right, professor, should I get:
\left\{ \begin{align*} u(r,t)&=\frac{1}{r}(f(ct+r)-f(ct-r)) & r\neq0 \\ u(0,t)&=2f'(ct) \end{align*} \right.

78
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!

79
APM346––Home Assignments / Re: Section 2.3, Q 5
« on: February 01, 2018, 08:21:24 AM »
$$u=\frac{1}{r}(f(r+ct)+g(r-ct))$$
If $f(x)=-g(-x)$ then the solution will be continuous at zero.

80
APM346––Home Assignments / Re: Homework 1: Problem 3 (20)
« on: January 30, 2018, 08:04:03 PM »
$$u_{xy}=2u_{x} \implies u_{y}=2u+\varphi_{y}(y) \implies (e^{-2y}u)_{y}=e^{-2y}\varphi_{y}(y) \implies u=\varphi(y)+2e^{2y}(\int^{y}\varphi(s)e^{-2s}\,ds+\psi(x))$$
But it is really an ODE. If I have mistaken then please inform me. I have used intergration by part and standard ODE technique.

I made a lost of mistakes when I first post these, including: forget the arbitrary function after last integration wrt y; messed up the order of multiplication.

Edit:

Really such a mess is no better than something wrong. Sub in $v=u_{x}$ immediately $v=\varphi_{x}(x)e^{2y}\implies u=\psi(y)+\varphi(x)e^{2y}$.

81
APM346––Home Assignments / Re: Goursat Problem
« on: January 30, 2018, 01:13:04 PM »
Solved. Only till yesterday did I find out that domain of dependence really depends on the particular problem.

82
APM346––Home Assignments / Promblem 2.6 probelems 2,3
« on: January 27, 2018, 12:36:47 PM »
I am very lost in dealing with these systems. But isn'tthe extra information (about $u$ in left-half plane, and $v$)rather redundant, or at least only simplifies to a boundary value of $0$? since it seems to me
$$u(x,0)=u_{t}(x,0)\equiv 0 \implies u\equiv 0$$
But we have the follow up questions asking for discussion about reflection and refraction, whereas should this be the case there is really nothing very interesting to discuss. Also, pray what connexion is there between 2,3?

83
APM346––Home Assignments / Re: Goursat Problem
« on: January 27, 2018, 10:53:46 AM »
Indeed I drew that right-ward triangle but what should the domain of independence be? should that be a triangle with its tip point to the right (so that the two sides intersect the two information)?

84
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.

85
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
$$u_{t}+(x^2+1) u_{x}=0 \implies C=\arctan(x)-t \implies u=\varphi(\arctan (x)-t)$$

86
APM346--Lectures / Re: Uniqueness/ continuity of solutions
« on: January 24, 2018, 07:13:23 AM »
Look, at the picture. In the "yellow" domain initial condition $u(x,0)=f(x)$ does not define solution (why?).
In  the white domain initial condition could be impossible to satisfy (why?)
But you mentioned in another post that existence problem is at $(0,0)$. For the white domain indeed you have to have even initial function to meet the symmetrical curves, but $t-x^{2}=0$ does not require this, and so shouldn't it rather be no issue at origin?

87
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.

88
APM346––Home Assignments / Re: Homework 2: Problem 3
« on: January 24, 2018, 06:09:21 AM »
But plug in the operation you would have
$$\frac{2xy(x^{2}+1)}{y^{2}+1}\varphi'(\frac{x^{2}+1}{y^{2}+1})+x(x^{2}+1)(y^{2}+1)(\tan^{-1}y)\varphi'(\frac{x^{2}+1}{y^{2}+1})$$

Whereas if $u=\varphi\left(\sqrt{\frac{x^{2}+1}{y^{2}+1}} \right)$ then plugging in you will happily see
$$xy(x^{2}+1)^{1/2}(y^{2}+1)^{-1/2}\varphi'-xy(x^{2}+1)^{1/2}(y^{2}+1)^{-1/2}\varphi'=0.$$

Similarly the other problem you would have product of square root. As I mentioned, when you finish integrating the characteristic you cannot cancel that a half out from both side. That will rescale the constant term and is significant after you take exponential.

89
APM346--Lectures / Re: Uniqueness/ continuity of solutions
« on: January 24, 2018, 06:01:01 AM »
But should there be issue with existence? I think there at the origin we have this parabolic characteristic tangent to it and what's wrong with that?

90
APM346––Home Assignments / Goursat Problem
« on: January 23, 2018, 10:05:09 PM »
I am approaching homework 3 question 2. It is very similar to the question worked out in lecture and the one in text but I must admit I am still not very content with my understanding. In particular,

1. should we not expect the solution is only determined up to an unknown function in two region;
2. How exactly do we translate the region $x>c|t|$ to the characteristic? My current idea is it will be come the first quadrant ($\eta~\xi$), but this is merely empirical.

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