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### Messages - Jingxuan Zhang

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46
##### 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 $(2)$.

47
##### 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.

48
##### 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
$$-t-1/2\leq x\leq t+1/2.$$

49
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 07:18:57 PM »
Necessity:$$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}.$$

50
##### 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.)

51
##### 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
$$-2x^2=xf(y)+g(y).$$

52
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 10:23:04 AM »
I am trying to follow the quick way as in lecture below.
$$\text{ something wrong }.$$
Hence upon substitution,
$$\text{ some other thing wrong}.$$
Thus I have the unknown functions. I think the constants and explicit forms shows up as a result of not presume these functions beforehand.

I hope what follows is no longer nonsense. We indeed have the not-so-arbitrary functions as desired,
$$c_{1}xy, c_{3}x, c_{2}y, c_{4}.$$
They lost their arbitrariness in form by constrain $(2)$, since upon substitution
$$x f_{y}(y) + g_{y}(y)=h(x)$$
the particular form of $f,g$ is dictated to be polynomial in $y$ of degree less than one.

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