Toronto Math Forum
APM346-2012 => APM346 Math => Home Assignment 3 => Topic started by: Bowei Xiao on October 04, 2012, 06:26:59 PM
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In question 6 part a, Are we supposed we have condition like U(l,t)=U(0,t)=0?
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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)
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so, the only condition for a is Ut = K Uxx?
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so, the only condition for a is Ut = K Uxx?
Obviously
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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).
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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
$u(0,t)=u(l,t)=0$ to produce
$u(0,t)=u(l,t)=0$
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Professor can we assume that u is 0 at positive and negative infinity? Thanks!
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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
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Q6 part(a)
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Q6 part(b)
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Q6 part(c)
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(a) and (b) are done completely, (c) is not. Question: is in (c) $u(x,t)=c$ is a solution for any constant $c$?
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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 $c=0$.
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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 $a_j>0$