# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment 3 => Topic started by: Bowei Xiao on October 04, 2012, 06:26:59 PM

Title: Problem 6
Post by: Bowei Xiao on October 04, 2012, 06:26:59 PM
In question 6 part a, Are we supposed we have condition like U(l,t)=U(0,t)=0?
Title: Re: Problem 6
Post by: Victor Ivrii on October 04, 2012, 06:49:38 PM
In question 6 part a, Are we supposed we have condition like U(l,t)=U(0,t)=0?

No, it was at (a) to be integral on $(-\infty,\infty)$. $u(l,t)=u(0,t)=0$ (or Neumann) would be in (b)
Title: Re: Problem 6
Post by: Bowei Xiao on October 04, 2012, 10:30:41 PM
so, the only condition for a is Ut = K Uxx?
Title: Re: Problem 6
Post by: Victor Ivrii on October 05, 2012, 07:04:07 AM
so, the only condition for a is Ut = K Uxx?

Obviously
Title: Re: Problem 6
Post by: Vitaly Shemet on October 07, 2012, 10:47:13 AM
For 6b boundary conditions include u(l,t) or no? So that u(0,t)=u(l,t)=0 (Dirichlet) or u_x(0,t)=u_x(l,t)=0 (Newman).
Title: Re: Problem 6
Post by: Victor Ivrii on October 07, 2012, 11:19:17 AM
For 6b boundary conditions include u(l,t) or no? So that u(0,t)=u(l,t)=0 (Dirichlet) or u_x(0,t)=u_x(l,t)=0 (Newman).

Or Dirichlet on one end and Neumann on another (but one can notice that ends are independent so it would be just a footnote).

Please surround your math snippets by dollar signs for MathJax to kick out. Like
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$u(0,t)=u(l,t)=0$ to produce
$u(0,t)=u(l,t)=0$
Title: Re: Problem 6
Post by: Peishan Wang on October 08, 2012, 08:05:08 AM
Professor can we assume that u is 0 at positive and negative infinity? Thanks!
Title: Re: Problem 6
Post by: Victor Ivrii on October 08, 2012, 08:10:53 AM
Professor can we assume that u is 0 at positive and negative infinity? Thanks!

If you mean that $u=0$ as $|x|\ge C$, then the answer is no
If you mean that $u$ fast decays as $|x|\to \infty$, then the answer is yes
Title: Re: Problem 6
Post by: Peishan Wang on October 10, 2012, 11:09:21 PM
Q6 part(a)
Title: Re: Problem 6
Post by: Peishan Wang on October 10, 2012, 11:12:31 PM
Q6 part(b)
Title: Re: Problem 6
Post by: Peishan Wang on October 10, 2012, 11:12:51 PM
Q6 part(c)
Title: Re: Problem 6
Post by: Victor Ivrii on October 11, 2012, 04:41:10 AM
(a) and (b) are done completely, (c) is not. Question: is in (c) $u(x,t)=c$ is a solution for any constant $c$?

Remark. Your scanning is good but not great: try to scan to black-white not colour; then under correct threshold setting blue lines will disappear but text will remain dark. Size would be smaller. BTW, one can attach up to 4 images (< 192kb) to a single post (in our settings)
Title: Re: Problem 6
Post by: Levon Avanesyan on October 11, 2012, 01:21:49 PM
Question: is in (c) $u(x,t)=c$ is a solution for any constant $c$?
No. From Robin conditions we can see that  $u(x,t)=c$ is a solution only when $c=0$.
Title: Re: Problem 6
Post by: Victor Ivrii on October 11, 2012, 03:27:56 PM
Question: is in (c) $u(x,t)=c$ is a solution for any constant $c$?
No. From Robin conditions we can see that  $u(x,t)=c$ is a solution only when $c=0$.

This is a complete answer to (c) as $a_j>0$