Problem 6

[Bowei Xiao]

In question 6 part a, Are we supposed we have condition like U(l,t)=U(0,t)=0?

[Victor Ivrii]

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]

so, the only condition for a is Ut = K Uxx?

[Victor Ivrii]

so, the only condition for a is Ut = K Uxx?

Obviously

[Vitaly Shemet]

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]

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).

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$u(0,t)=u(l,t)=0$ to produce
$u(0,t)=u(l,t)=0$

[Peishan Wang]

Professor can we assume that u is 0 at positive and negative infinity? Thanks!

[Victor Ivrii]

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]

Q6 part(a)

[Peishan Wang]

Q6 part(b)

[Peishan Wang]

Q6 part(c)

[Victor Ivrii]

(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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[Levon Avanesyan]

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]

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 $a_j>0$