Toronto Math Forum
APM3462015F => APM346Home Assignments => HA5 => Topic started by: Rong Wei on October 17, 2015, 08:38:43 PM

http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter3/S3.2.P.html

The core is to invent a initial function that satisfies the boundary conditions as well as defined on the whole line.
\begin{equation} \end{equation}
a) need to think of an initial function, consider:
$$
f(x) = \left\{\begin{aligned}
&g(x) && L<x<0 \\
&0 && x=...2L, L, 0, L, 2L ... \\
&g(x) &&0<x<L \\
&extended\ to\ be\ 2Lperiodic \\
\end{aligned}
\right.$$
This function satisfy Dirichlet boundary conditions on the whole line.
Thus our solution is now:
\begin{equation} v(x,t) = \int_{\infty}^{\infty} G(x,y,t)f(y)dy \end{equation}
Depends on the relative position of (x,t), the integral can be valued piecewisely.
b) by similar fashion, consider:
$$
f(x) = \left\{\begin{aligned}
&g(x) && L<x\le 0 \\
&g(x) &&0\le x<L \\
&extended\ to\ be\ 2Lperiodic \\
\end{aligned}
\right.$$
This function is even, thus its derivative is an odd function, which satisfies Neumann condition.
\begin{equation} v(x,t) = \int_{\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} can be valued based on (x,t)'s position.
c) the condition ask for a function f(x) that has odd continuation at 0 + 2kL, but even continuation at L+2kL. Thus its period is 4L. consider:
$$
f(x) = \left\{\begin{aligned}
&g(x2L) && 2L<x\le L\\
&g(x) && L\le x<0 \\
&0 && x=...4L,2L , 0, 2L, 4L... \\
&g(x) &&0<x\le L \\
&g(2Lx) && L\le x<2L\\
&extended\ to\ be\ 4Lperiodic \\
\end{aligned}
\right.$$
Solution \begin{equation} v(x,t) = \int_{\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} then can be valued based on (x,t)'s position.

Professor, could you verify the answer to this problem? I couldn't figure it out myself but I am not sure about this solution.

Yes, it is correct but unfinished. Say, (a). What is $f(y)$ as defined?
\begin{equation*}
f(y)=\left\{\begin{aligned}
&f(y2\pi n) && 0< y2\pi n < \pi,\\
&f(2\pi ny) && \pi<y2\pi n <0
\end{aligned}\right.
\end{equation*}
and therefore after breaking $(\infty,\infty)$ into "elementary" intervals of the length $\pi$ described above, and plugging $z=y2\pi n$ and $z=2\pi ny$ we get
\begin{equation}
u(x,t) = \int _{0}^{\pi}\underbrace{\Bigl( \sum_{n=\infty}^\infty \bigl[G (xy 2\pi n,t) G(x+y+2\pi n,t)\bigr]\Bigl)}_{= G_D (x,y,t) }g(y)\,dy
\end{equation}
where $G_D (x,y,t) $ is a Green's function for our IBVP.
For (b) result is very similar (how?) and for (c ) it is a bit more complicated (not 2 but 4 types of intervals and respectively square brackets contain 4 different terms (and $4\pi n$ instead of $2\pi n$), but is derived in the same way