Author Topic: Problem 6  (Read 6670 times)

Bowei Xiao

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Problem 6
« 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?

Victor Ivrii

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Re: Problem 6
« Reply #1 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)

Bowei Xiao

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Re: Problem 6
« Reply #2 on: October 04, 2012, 10:30:41 PM »
so, the only condition for a is Ut = K Uxx?

Victor Ivrii

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Re: Problem 6
« Reply #3 on: October 05, 2012, 07:04:07 AM »
so, the only condition for a is Ut = K Uxx?

Obviously

Vitaly Shemet

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Re: Problem 6
« Reply #4 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). 

Victor Ivrii

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Re: Problem 6
« Reply #5 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
Code: [Select]
$u(0,t)=u(l,t)=0$ to produce
$u(0,t)=u(l,t)=0$

Peishan Wang

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Re: Problem 6
« Reply #6 on: October 08, 2012, 08:05:08 AM »
Professor can we assume that u is 0 at positive and negative infinity? Thanks!

Victor Ivrii

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Re: Problem 6
« Reply #7 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

Peishan Wang

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Re: Problem 6
« Reply #8 on: October 10, 2012, 11:09:21 PM »
Q6 part(a)

Peishan Wang

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Re: Problem 6
« Reply #9 on: October 10, 2012, 11:12:31 PM »
Q6 part(b)

Peishan Wang

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Re: Problem 6
« Reply #10 on: October 10, 2012, 11:12:51 PM »
Q6 part(c)

Victor Ivrii

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Re: Problem 6
« Reply #11 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)

Levon Avanesyan

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Re: Problem 6
« Reply #12 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 $c=0$.

Victor Ivrii

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Re: Problem 6
« Reply #13 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 $a_j>0$