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### Messages - Jingxuan Zhang

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46
##### 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.)

47
##### 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
$$-2x^2=xf(y)+g(y).$$

48
##### 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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