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### Messages - Victor Ivrii

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2476
##### Home Assignment 2 / Re: Problem 2
« on: September 29, 2012, 11:02:40 PM »
Sorry, I am still having trouble understanding how #2. b is asking to solve for v using v. Could someone help please?

We are looking for $u$, not $v$ -- but $v$ satisfies 1D wave equation and we know everything (well, almost everything) about it

2477
##### Home Assignment 2 / Re: Problem 2
« on: September 29, 2012, 07:14:54 PM »
In this question r is always positive right (since it's the distance to the origin)? Should u(r,0) and ut(r,0) be even functions of r? I guess we need additional information about u(r,0) and ut(r,0) so that v can be extended to negative values.

Thanks!

2478
##### Home Assignment 2 / Re: Problem 2
« on: September 29, 2012, 06:34:05 PM »
I'm even more confused after your response to Thomas' comment - why is part b) the only part that has x as a variable now?

Could you make it clear which version you change first, so I know if I should always check the html version instead of the pdf version for changes?

Usually changes come first to html.

2479
##### Misc Math / Re: Classification criteria for PDEs
« on: September 29, 2012, 06:31:37 PM »
Good question. However an answer is more complicated: among 2-nd order equations there are elliptic, hyperbolic, parabolic but also a lot of equations which are neither (and some of them are rather important). Ditto for higher order equations and the systems.

There is no complete classifications of PDEs and cannot be because any reasonable classification should not be based on how equation looks like but on the reasonable analytic properties it exhibits (which IVP or BVP are well-posed etc).

2D If we consider only 2-nd order equations with constant real coefficients then in appropriate coordinates they will look like either

u_{xx}+u_{yy}+\text{l.o.t} =f
\label{ell-2}

or

u_{xx}-u_{yy}+\text{l.o.t.} =f.
\label{hyp-2}

Here l.o.t. means "lower order terms". (\ref{ell-2}) are elliptic,   (\ref{hyp-2}) are hyperbolic.

What to do if one of the 2-nd derivatives is missing? We get

u_{xx}-cu_{y}+\text{l.o.t.} =f.
\label{par-2}

with $c\ne 0$ and  IVP $u|_{y=0}=g$ is well-posed in the direction of $y>0$ if $c>0$ and in direction $y<0$ if $c<0$. We can dismiss $c=0$ as not-interesting.

However this classification leaves out very important SchrÃ¶dinger equation

u_{xx} +i c u_y=0
\label{Schr-2}

with real $c\ne 0$. For it IVP $u|_{y=0}=g$ is well-posed in both directions $y>0$ and $y<0$ but it lacks many properties of parabolic equations (like maximum principle or mollification).

3D If we consider only 2-nd order equations with constant real coefficients then in appropriate coordinates they will look like either

u_{xx}+u_{yy}+u_{zz}\text{l.o.t} =f
\label{ell-3}

or

u_{xx}+u_{yy}-u_{zz}+\text{l.o.t.} =f.
\label{hyp-3}

(\ref{ell-3}) are elliptic,   (\ref{hyp-3}) are hyperbolic.

Also we get parabolic equations like

u_{xx}+u_{y}-cu_z+\text{l.o.t.} =f.
\label{par-3}

u_{xx}-u_{y}-cu_z+\text{l.o.t.} =f?
\label{crap-3}

Algebraist-formalist would call them parabolic-hyperbolic but since this equation exhibits no interesting analytic properties (unless one considers lack of such properties interesting) it would be a perversion.

Yes, there will be SchrÃ¶dinger equation

u_{xx} +u_{yy}+i c u_z=0
\label{Schr-3}

with real $c\ne 0$ but $u_{xx} -u_{yy}+i c u_z=0$ would also have IVP $u|_{z=0}=g$ well posed in both directions.

4D Here we would get also elliptic

u_{xx}+u_{yy}+u_{zz}+u_{tt}+\text{l.o.t.} =f,
\label{ell-4}

hyperbolic

u_{xx}+u_{yy}+u_{zz}-u_{tt}+\text{l.o.t.} =f,
\label{hyp-4}

but also ultrahyperbolic

u_{xx}+u_{yy}-u_{zz}-u_{tt}+\text{l.o.t.} =f
\label{uhyp-4}

which exhibits some interesting analytic properties but these equations are way less important than elliptic, hyperbolic or parabolic.

Parabolic and SchrÃ¶dinger will be here as well.

The notions of elliptic, hyperbolic or parabolic equations are generalized to higher-order equations but most of the randomly written equations do not belong to any of these types and there is no reason to classify them.

To make things even more complicated there are equations changing types from point to point, f.e. Tricomi equation

u_{xx}+xu_{yy}=0
\label{Tric}

which is elliptic as $x>0$ and hyperbolic as $x<0$ and at $x=0$ has a "parabolic degeneration". It is a toy-model describing stationary transsonic flow of gas.

My purpose was not to give exact definitions but to explain a situation.

2480
##### Home Assignment 2 / Re: problem 1 typo?
« on: September 29, 2012, 05:10:55 PM »
Consider this:
\begin{align*}
&u|_{t=0}=g(x),\\
&u_t|_{t=0}=h(x),\\
&u|_{x=0}=p(t)
\end{align*}
has a continuous solution if and only if $p(0)=g(0)$ (compatibility condition) but with the Neumann BC solution would be always $C$ (albeit not necessarily $C^1$.

BTW  heat equation
\begin{align*}
&u|_{t=0}=g(x),\\
&u|_{x=0}=p(t)
\end{align*}
also has a continuous solution if and only if $p(0)=g(0)$ (compatibility condition) but the discontinuity stays in $(0,0)$ rather than propagating along characteristics as for wave equation.

2481
##### Home Assignment 2 / Re: Problem 2
« on: September 29, 2012, 04:01:39 PM »
should the $\phi$ in eq. 6 read $\phi(r+ct)$ rather than $\phi(x+ct)$?

Yes! I fixed pdf (a bit of hassle to maintain two versions)

2482
##### Home Assignment 2 / Re: Problem 3
« on: September 29, 2012, 03:58:54 PM »
thanks, notes are more clear! However, my understanding for the problem is that I can define odd function to solve the Dirichlet boundary problem and define even function to solve Neumann boundary problem, isn't it?
You are right. However remember that method of continuation has certain preconditions (satisfied in this case)

2483
##### Home Assignment 2 / Re: Problem 3
« on: September 29, 2012, 03:13:13 PM »
What is meaning of "method of continuation"? I read the textbook and find it define the odd function to solve the problem, can I use that method?

See also Lecture 8  and note that it could be either odd or even

2484
##### Home Assignment 2 / Re: problem 1 typo?
« on: September 29, 2012, 03:10:20 PM »
Professor I have a question for part (c). Does the solution have to be continuous? For example I have f(x) on x>2t, g(x) on -2t<x<2t and h(x) on -3t<x<-2t. Should f(2t) = g(2t) and g(-2t)=h(-2t) (so the overall solution is continuous)?

My problem is that some of the f, g, h involve a constant K and I was wondering if I should use continuity to specify what K is.

Thanks a lot!

You should be able to find constants from initial and boundary conditions. Solutions may be discontinuous along lines you indicated

2485
##### APM346 Announcements / Re: Please register ASAP
« on: September 29, 2012, 02:42:53 PM »
Also again: select username whatever you want (make it short enough to save yourself a time while logging in) but change DisplayName (open Profile > Modify Profile > Account Settings and there will be Username and the next Name) matching to one on BlackBoard.

Credits for active and useful forum participation will be given only to those who are easily identified. Well, for identifying those who behaves badly I would go an extra mile

2486
##### Home Assignment 2 / Re: Problem 2
« on: September 29, 2012, 02:36:32 PM »
for the last part, can we just assume the same solution as c) but state a few assumptions instead? it's because I dont think the general solution of the equation would vary since no other IC were stated.

You should check when and if the general solution is continuous at $r=0$ and adjust it respectively

2487
##### Home Assignment 2 / Re: problem 1 typo?
« on: September 29, 2012, 09:04:56 AM »
Can we assume that for part (C) of Problem 1 that the Cauchy conditions are evenly reflected for x < 0?

Sure, you can but it will not be useful as your domain is $x>vt$ rather than $x>0$. Just use the general solution.

2488
##### Misc Math / Re: Example 8b
« on: September 28, 2012, 05:36:19 PM »
If we are looking for $t>0$ then we are interested in $0<x<ct$ and $x>ct$ because the original problem is dealing with $x>0$. Continuation is the method to reduce it to IVP but we need to come back

Transitions in (9), (12) from the middle to the r.h. expression is due to the trigonometric formulae. Everything is ok

PS. Please use \sin,  \cos etc instead of sin, cos in LaTeX code as those are predefined macros.

PPS. In Lecture 9 we use \erf but as this is not a predefined macro, we define it by ourselves

2489
##### Home Assignment 2 / Re: Problem 2
« on: September 28, 2012, 05:23:56 PM »
I think in the new variant (just posted) it will be more clear

2490
##### APM346 Misc / Re: Midterm conflicts with schedule!
« on: September 28, 2012, 02:33:17 PM »
Just to open up the headline a bit:
Both midterm exams conflict with my schedule. So, what am I supposed to do?

There is now a poll attached to the announcement. Vote in it

http://forum.math.toronto.edu/index.php?topic=27.0

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