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### Messages - Victor Ivrii

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31
##### APM346--Lectures / Re: Classification of PDEs (Chapter Problems 1) clarification
« on: January 11, 2018, 07:12:19 PM »
Quote
What does it exactly mean the right-hand expression?
Quote
It is only for semi-linear. That you can move some terms containing lower order derivatives only (including function) to the right, so on the left only linear part remains.

32
##### APM346--Lectures / Re: Overdetermined systems
« on: January 11, 2018, 04:59:19 AM »
1) THis is problem not from the homework: it was added just few days ago and homework description from "all problems" was changed to "problems 1--5" at http://www.math.toronto.edu/courses/apm346h1/20181/lectures.html (Week 2). We do not study such systems in the course.

2) It contained an error (fixed now) which did not make it senseless.

3) For overdetermined systems the usual case is as in
http://forum.math.toronto.edu/index.php?topic=882.30
a) Solution does not exist unless right hand expression satisfies certain compatibility conditions (depending on the system).
b) If solutions exist then there are less of them than in the case of determined system.

The simplest example (which you actually studied in Calculus II)
\left\{ \begin{aligned} &u_x=f,\\ &u_y=g. \end{aligned}\right.
Compatibility condition $f_y=g_x$ and under this condition $u$ is defined up to a constant (so the general solution does not contain an arbitrary function of one variable).

4) Another example is the Maxwell system. There are 6 unknown (components of electric $\mathbf{E}$ and magnetic $\mathbf{H}$ fields) and 8 equations. The right-hand expressions may contain densities of the charges $\rho$ and currents $\mathbf{j}$.

33
##### APM346--Lectures / Re: Integral curves
« on: January 09, 2018, 10:29:36 AM »
$$\frac{dx}{a(x,y)}=\frac{dy}{b(x,y)}=\frac{du}{f(x,y,u)}$$
so if integral curve is parametrized by $\tau$: $dx= a\,d\tau$, $dy=b\,d\tau$ then along this curve $du=f\,d\tau$, we get ODE along integral curve.

Examples of variable coefficients will be considered during tutorial and  the next lecture.

34
##### APM346--Lectures / Integral curves
« on: January 09, 2018, 03:38:46 AM »
Someone claimed "I took MAT244 but never heard about integral curves". LOL, on the bottom of page 12 of Boyce-DiPrima 10E
Quote
The geometrical representation of the general solution (17) is an infinite family of curves called integral curves. Each integral curve is associated with a particular value of c and is the graph...
after which integral curve or integral curves is repeated more than 30 times.

35
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 09, 2018, 03:32:30 AM »
Explanation: 1) $u_{xxyy}=u_{yyxx}$ due to the property of partial derivatives; in other words operators $\partial_x$ and $\partial_y$ (of partial differentiations) commute.

2) Proving that $f_{yy}=g_{xx}$ (*) is sufficient (at least in convex domains): we denote $v:= (u_{yy}-g)$ and if $u_{xx}=f$ we conclude that $v_{xx}=0$ due to (*).

Therefore if $v=v_x=0$ for $y=0$, then $v=0$ identically. But $u=\phi(y) +\psi(y)x$ and $v=\phi '' (y)+\psi'' (y)x-g(x,y)$ which indeed implies $v(0,y)=\phi''(y)-g(0,y)$ $v_x(0,y)=\psi''(y)-g_x(0,y)$.

___
This topic is closed now; I will give points over coming weekends

36
##### APM346--Lectures / Re: Classification of PDEs (Chapter Problems 1) clarification
« on: January 07, 2018, 11:08:26 PM »
Quasinear are subclass of Nonlinear , and Semilinear are subclass of Quasilinear

It will not be an error to call $u_{xx}+u_{yy}=u^2$ nonlinear or quasilinear, but it is in fact semilinear (linear expression, plus non-linear expression, depending only on lower order terms).

It will not be an error to call $u_{x}+u^2u_{y}=0$ nonlinear, but it is in fact quasilinear.

These problems ask for the most precise classification

37
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 09:19:50 PM »
Jingxuan
Correct and minimalist proof of necessity. Obviously $u_{xx}=y^2$, $u_{yy}=-x^2$ does not satisfy and solution does not exist.

To prove that this compatibility condition $f_{yy}=g_{xx}$ is sufficient (under additional assumption that domain is convex, you can figure out generalization) solve $u_{xx}=f$, $u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$ where $\phi$, $\psi$ so far are unknown and will be chosen later. Such $u$ obviously exist for any $\phi$, $\psi$.

What about $v:=u_{yy}-g=0$? Obviously $v_{xx}=0$ and we need to satisfy $v(0,y)=0$, $v_x(0,y)=0$, which are, in fact, $\phi '' -g(0,y)=0$, $\psi ''- g_x(0,y)=0$, and we can satisfy these two equations.  Four arbitrary constants $C_1=\phi(0)$, $C_2=\phi'(0)$, $C_3=\psi(0)$ and $C_4=\psi'(0)$ appear.

38
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 12:35:25 PM »
Adam nailed it: solution does not exist. It is the general situation for overdetermined systems: solution does not exist unless compatibility conditions are satisfied. If compatibility conditions are satisfied, then there exist fewer solutions.

You studied in Calculus II and ODE such system

\left\{\begin{aligned}
&u_x = f,\\
&u_y=g
\end{aligned}\right.

In other words, when $f\,dx+g\,dy$ is an exact differential. And the necessary (and for simple-connected domains, sufficient) condition was $f_y=g_x$.

For the system

\left\{\begin{aligned}
&u_{xx} = f,\\
&u_{yy}=g
\end{aligned}\right.

such condition is $f_{yy}=g_{xx}$. Prove the necessity!

39
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 10:03:05 AM »
So $f=f(y)$, $g=g(y)$ and
$$2x^2 + xg(y)+ h(y)=0$$
is an identity. Is it ever possible?

40
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 09:43:14 AM »
Wrong again. What are $f$ and $g$?

41
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 08:44:15 AM »
Jaisen, you arrived to the equality I maked by red. And it is not just an equality, it must be an identity, that is, fulfilled for all $x$ and $y$.

What are $f$ and $g$ here by  your definition? And which of these functions make that equation to be an identity?

42
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 06:15:05 AM »
The answer to the question "why solution does not include arbitrary functions off one variable" the answer is simple: because it is overdetermined system. One can arrive to this conclusion by the following reasoning: solution to equation $u_{xx}=0$ includes arbitrary functions of $y$ but does not include arbitrary functions of $x$; solution to $u_{yy}=0$ does not include arbitrary functions of $y$. Then the common solution to both equations contains neither.

Pending: what is the general solution to (14) from the previous page

43
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 11:13:31 PM »
Quote
Perhaps a hint? Are we supposed to know this from the solution or prior to arriving at the solution?
In this particular case we can arrive to the correct conclusion in either way: we can get it in the process of solving or we can get it without solving.

44
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 08:32:37 PM »
Jingxuan, I removed excessive quotation.

With (13)  you found the general solution. Great!

45
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 06:40:52 PM »
Guys, there is already a perfect solution by Jaisen, and Tristan made the first part correctly. There is no point to post more solutions! But the interesting question: Why this solution includes just 4 arbitrary constants rather than arbitrary functions of one variable?

And please find the general solutions to two other overdetermined systems

\left\{\begin{aligned}
&u_{xx}=y^2,\\
&u_{yy}=x^2
\end{aligned}\right.
\label{Q}

and

\left\{\begin{aligned}
&u_{xx}=y^2,\\
&u_{yy}=-x^2
\end{aligned}\right.
\label{R}

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