Author Topic: Problem 3  (Read 6253 times)

Qitan Cui

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Re: Problem 3
« Reply #15 on: October 01, 2012, 09:14:09 PM »
part 5    complete

Rouhollah Ramezani

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Re: Problem 3
« Reply #16 on: October 01, 2012, 09:50:25 PM »
Should we consider solutions on different intervals (i.e. x>2t and 0<x<2t) since they differ in part (a) and part (d)?

To be impeccable, Yes. (but I guess solution where $x>2t$ is too trivial/standard to be of any particular interest here.)

Victor Ivrii

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Re: Problem 3
« Reply #17 on: October 02, 2012, 07:50:45 AM »
Again MZ solutions are perfect and it is a cut-off. Only one remark: in (b) we have even initial functions and continuation must be even due to Neumann, in (c) we have odd initial functions and continuation must be odd due to Dirichlet. So in fact in these parts the answer is obviously the same for $x>t$ and $x<t$.


RR: you obviously put the wrong limits in some of the integrals. Not that it matters for credits -- MZ got them

The rest is a flood, sorry. BTW, one can attach several docs to the post (OK, as configured currently it is up to 4 of the total size 192K). Also pdf is not the best possible format for a forum as forum does not display it: most of the browsers either cannot display pdf files or can display pdf files as standalone but not embedded into html page. *** png or jpg are better suited for a forum

Chiara Moraglia

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Re: Problem 3
« Reply #18 on: October 05, 2012, 06:58:15 PM »
Hi, I understand how to derive the solutions for Problem 3, but now I have a question. For example, in 3.a, the solution is
u=x/2 for 0<x<2t, but u(0,x)=x/2 which does not equal 0, the given initial condition. Also, du/dt(0,x)=0 which does not equal 1, the given initial condition. I know however that these initial conditions are satisfied for u=t as x>2t. Could you please explain this?

Victor Ivrii

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Re: Problem 3
« Reply #19 on: October 05, 2012, 08:56:36 PM »
Hi, I understand how to derive the solutions for Problem 3, but now I have a question. For example, in 3.a, the solution is
u=x/2 for 0<x<2t, but u(0,x)=x/2 which does not equal 0, the given initial condition. Also, du/dt(0,x)=0 which does not equal 1, the given initial condition. I know however that these initial conditions are satisfied for u=t as x>2t. Could you please explain this?

Domain $ 0<x<2t$ does not approach line $t=0$ (except a single point) where initial condition must be satisfied.