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Messages - Jingxuan Zhang

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91
APM346––Home Assignments / Re: Homework 2: Problem 3
« on: January 23, 2018, 09:50:05 PM »
When you solve the integral curve the one half cannot be cancelled lightly
$$\int\frac{x}{x^{2}+1} \,dx = \ln \sqrt{x^2+1}+C$$
or otherwise what you get is distorted in scale.

92
APM346––Home Assignments / Re: Homework 2: Problem 3
« on: January 23, 2018, 07:20:20 PM »
Keisen you missed the square root. That cannot be cancelled lightly. Continuity I think is automatic since the characteristic is continuous.

93
APM346––Home Assignments / Re: Homework 2: Problem 2 (a)
« on: January 23, 2018, 08:45:10 AM »
I think what the question really means is that how $\varphi$ should be defined, that is, defined throughout the plane, so that the continuity condition would be met.
Remember there used to be question like this in second year calculus. As for the reference in Folland, see pp. 14-16.

The function $\varphi(xy)$ is a composite funciton, just as $\varphi(x/y)$. They really differ only in that the former is a composite of two continuous funciton, and therefore itself is automatically continuous, and the latter not so, and therefore the continuity can only be met by restrict the unknown function $\varphi$ to a certain type: constant.

94
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

95
APM346––Home Assignments / Re: Homework 2: Problem 2 (a)
« on: January 22, 2018, 06:57:34 PM »
Jaisen

I appreciate your effort. Here is an argument I came up with just now, after refering to second year calculus text, the beloved Folland book.

Suppose $\varphi \in C$, then in particular $\forall k \in \mathbb{R}, \lim_{t \to 0} \varphi(kt/t) = \text{ const.}$, since it is necessary for continuity that when the argument closes up to a limit point in the domain from arbitrary path the limit of function value remains the same. This gives $\varphi \equiv \text{const}$. Conversely if $\varphi$ is constant then indeed the equation is fulfilled and continuity is achieved.

Whereas the second continuity is achieved whenever $\varphi \in C$, since then $u$ will be a composition of two continuous functions $\varphi$ and $xy$.

96
APM346--Lectures / Continuity of solution
« on: January 20, 2018, 02:49:57 PM »
This I found on homework 2: to determine the condition for the solution of
$$yu_{x}+xu_{y}=0$$
to be continuous at origin. Similar problems were covered in one previously lecture and several pictures were drawn. But even then I only vaguely felt those diagrams familiar (phase portrait covered in ODE), not their concrete connection. Please inform me how should I infer from the characteristic the condition for solution continuity, or even better, how to handle intersection of characteristics (as in another part of the question, where the circular characteristics shrink to a point).

97
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 $(2)$.


98
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.

99
APM346--Lectures / Chain rule
« on: January 15, 2018, 04:41:28 PM »
Pray what kind of chain rule gives us the following result, which I found in the PDF version (p.35, strangely not online) with little rephrasing:
Quote
$x=\frac{1}{2}(\xi+\eta), t=\frac{1}{2}(\xi-\eta)$ and therefore due to chain rule $v_{\xi}=\frac{1}{2c}(cv_{x}+v_{t})$ and $v_{\eta}=\frac{1}{2c}(cv_{x}-v_{t}).$
And also what is $v$? is it the same one as in the previous chapter, viz., a factor of the operator?

100
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
$$-t-1/2\leq x\leq t+1/2.$$

101
APM346--Lectures / Re: Integral curves
« on: January 09, 2018, 10:05:37 AM »
Please confirm with me if the following is correct. Suppose we have
$$a(x,y)u_{x}+b(x,y)u_{y}=f.$$
Then our LS expression is the directional derivative of $u$ on
$$\frac{dx}{a(x,y)}=\frac{dy}{b(x,y)}.$$
But how are we to read the RS expression under this context? Is it something related to this field? Suppose $f=0$, then it means $u$ is constant along each integral curve of the above field. But what if $f$ is some other function?

102
The term $uu_{x}$ determines $(2),(4)$ to be both Quasilinear.

103
Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 07:18:57 PM »
Necessity:$$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}.$$

104
Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 06:00:24 AM »
Quote from: Jaisen Kuhle
Suppose x not equal to 0

When we solve the quadratic we arrive at imaginary values for x. I'm not sure if I ought to continue.

I agree with your first part. Now suppose x is not identically $0$, what function of $f_{yy}(y)$ multiply to $x$ will give you such a quadratic in $x$?
Also the matter seems not to be with possible complex values, but that in this case the quadratic formula gives a relation $x=F(y).$

Quote from: Victor Ivrii
Is it ever possible?
No. Hence there is no common solution if $x$ is not identically zero. But if it is then upon substituting $u=u(y)=g(y)$ and so
$$u_{xx}=0=y^2 \implies y=0.$$
Thus $u$ can only be defined on the origin, and takes any constant value. (Though I doubt if derivatives are well defined then.)

105
Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 07:55:36 PM »


As for $(\ref{Q})$,
$$u_{xx}=y^2 \implies u=\frac{x^2y^2}{2} + xf(y)+g(y); u_{yy}=x^2 \implies f_{yy}(y)=g_{yy}(y)=0 \implies u = \frac{x^2y^2}{2} + x(ay+b) + (cy+d).$$
Soppose $u$ solves $(\ref{R})$, we would quickly arrive at what seems to me a contradiction
$$-2x^2=xf(y)+g(y).$$

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