Toronto Math Forum
APM3462012 => APM346 Math => Misc Math => Topic started by: Aida Razi on December 13, 2012, 01:56:49 PM

http://weyl.math.toronto.edu:8888/APM3462011Fforum/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
instead of what you wrote:
"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"

We are looking for solution to the Neumann problem, so by D'A formula
$$
u(x,t)=\frac{1}{2}\bigl( \Phi(x+ct) +\Phi (xct)\bigr) +\frac{1}{2c}\int_{xct}^{x+ct} \Psi (y)\,dy
$$
where righthand 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 $xct<0$ so $\Phi (xct)=\phi(x+ct)$; we get the first term as below. WQith $\Psi$ it is a bit more complicated: we integrate $\Psi(y)$ from $xct$ 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
$$
u(x,t)=\frac{1}{2}\bigl( \Phi(x+ct) +\Phi (x+ct)\bigr) +\frac{1}{2c}\int_0^{x+ct} \psi (y)\,dy +\frac{1}{2c} \int_0^{xct}\psi(y)\dy.
$$
Case $t<0$, $x+ct >0$ is like (a) and $t<0$, $x+ct >0$ is like (b).