### Author Topic: Last year FinalExam  (Read 4038 times)

#### Aida Razi

• Sr. Member
• Posts: 62
• Karma: 15
##### Last year FinalExam
« on: December 13, 2012, 01:56:49 PM »
http://weyl.math.toronto.edu:8888/APM346-2011F-forum/index.php?topic=208.0

In last year final exam, problem 1, when you solved it by method of continuation, I was wondering if the first term will be:

u(x,t)=1/2(-Ï•(xâˆ’ct)+Ï•(x+ct))+ 1/2câˆ«ct+x0Ïˆ(s)ds +1/2câˆ«ctâˆ’x0Ïˆ(s)ds

"In our case correct formula is different

u(x,t)=1/2(Ï•(xâˆ’ct)âˆ’Ï•(x+ct))+ 1/2câˆ«ct+x0Ïˆ(s)ds +1/2câˆ«ctâˆ’x0Ïˆ(s)ds"

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### Re: Last year FinalExam
« Reply #1 on: December 13, 2012, 02:53:04 PM »
We are looking for solution to the Neumann problem, so by D'A formula
$$u(x,t)=\frac{1}{2}\bigl( \Phi(x+ct) +\Phi (x-ct)\bigr) +\frac{1}{2c}\int_{x-ct}^{x+ct} \Psi (y)\,dy$$
where right-hand expression was $0$ and $\Phi$, $\Psi$ are even continuations of $\phi$, $\psi$.

(a) As $x>ct$ ($t>0)$ we just plug $\phi,\psi$ instead of them, getting the same solution as of Cauchy problem.

(b)  As $0<x<ct$ we get $x+ct<0$ (so $\Phi(x+ct)= \phi(x+ct)$) and $x-ct<0$ so $\Phi (x-ct)=\phi(-x+ct)$; we get the first term as below. WQith $\Psi$ it is a bit more complicated: we integrate $\Psi(y)$ from $x-ct$ to $0$ and from $0$ to $x+ct$ and the latter is
$\int_0^{x+ct} \psi(y)\,dy$ while the former is $\int_0^{-x+ct} \psi(y)\,dy$  and we get the second and the third terms below. This is a correct formula generally
$$u(x,t)=\frac{1}{2}\bigl( \Phi(x+ct) +\Phi (-x+ct)\bigr) +\frac{1}{2c}\int_0^{x+ct} \psi (y)\,dy +\frac{1}{2c} \int_0^{x-ct}\psi(y)\dy.$$
Case $t<0$, $x+ct >0$  is like (a) and $t<0$, $x+ct >0$ is like (b).