Author Topic: HA5-P2  (Read 705 times)

Rong Wei

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Zaihao Zhou

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Re: HA5-P2
« Reply #1 on: October 18, 2015, 11:40:19 AM »
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:
$$
f(x) = \left\{\begin{aligned}
&-g(-x) && -L<x<0 \\
&0 && x=...-2L, -L, 0, L, 2L ... \\
&g(x) &&0<x<L \\
&extended\ to\ be\ 2L-periodic \\
\end{aligned}
\right.$$
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:
$$
f(x) = \left\{\begin{aligned}
&g(-x) && -L<x\le 0 \\
&g(x) &&0\le x<L \\
&extended\ to\ be\ 2L-periodic \\
\end{aligned}
\right.$$
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:
$$
f(x) = \left\{\begin{aligned}
&-g(x-2L) && -2L<x\le -L\\
&-g(-x) && -L\le x<0 \\
&0 && x=...-4L,-2L , 0, 2L, 4L... \\
&g(x) &&0<x\le L \\
&g(2L-x) && L\le x<2L\\
&extended\ to\ be\ 4L-periodic \\
\end{aligned}
\right.$$

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.
« Last Edit: October 18, 2015, 01:12:54 PM by Zaihao Zhou »

Emily Deibert

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Re: HA5-P2
« Reply #2 on: October 20, 2015, 08:05:11 PM »
Professor, could you verify the answer to this problem? I couldn't figure it out myself but I am not sure about this solution.

Victor Ivrii

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Re: HA5-P2
« Reply #3 on: October 21, 2015, 05:39:37 AM »
Yes, it is correct but unfinished. Say, (a). What is $f(y)$ as defined?
\begin{equation*}
f(y)=\left\{\begin{aligned}
&f(y-2\pi n) && 0< y-2\pi n < \pi,\\
-&f(2\pi n-y) && -\pi<y-2\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 $z=y-2\pi n$ and $z=2\pi n-y$ we get
\begin{equation}
u(x,t) = \int _{0}^{\pi}\underbrace{\Bigl( \sum_{n=-\infty}^\infty \bigl[G (x-y -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