Toronto Math Forum
APM346-2012 => APM346 Math => Term Test 2 => Topic started by: Victor Ivrii on November 15, 2012, 08:20:35 PM
-
Consider the diffusion equation
\begin{equation*}
u_t -ku_{xx}=0 \qquad \text{for} \quad t>0,\ x \in (0,2\pi)
\end{equation*}
with the boundary conditions
\begin{equation*}
u_x(0,t)=u_x(2\pi,t)=0
\end{equation*}
and the initial condition
\begin{equation*}
u(x,0)=|\sin (x)|.
\end{equation*}
- (a) Write the associated eigenvalue problem.
- (b) Find all eigenvalues and corresponding eigenfunctions.
- (c) Show that the eigenfunctions associated to 2 different eigenvalues are orthogonal.
- (d) Write the solution in the form of a series expansion.
- (e) Write a formula for the coefficients of the series expansion.
Post after 22:30
-
Solutions for part (a) and part(b)
-
Hopeful solution to part c attached! :)
EDIT: Was not originally attached?
-
Part 1 of 3
-
Part 2 of 3
-
Part 3 of 3
-
Unfinished business:
According to formulae $A_n=\frac{1}{\pi}\int _0^{2\pi} |\sin(x)| \cos (\frac{1}{2}nx)\,dx$.
Chen: wrong coefficient $\frac{1}{2\pi}$ and wrong limits $-\pi$ to $\pi$ and two other errors in (d)--(e) I do not disclose.
I expect for forum: Correct solution for (d),(e) including calculation of coefficients and plugging them into series.
-
Solution to (d) and (e) is attached!
-
In forum I expect this integral to be taken.
Also coefficient $1$ in front of integral is wrong as $\int_0^{2\pi}\cos ^2(mx/2)\,dx$ is not $1$ but $\pi$ (see lectures)
-
Integration is done!
-
You are correct that $A_n=0$ for odd $n$. However your calculation for even $n=2m$ is wrong.
Hint: $A_{2m}=0$ for odd $m$. $A_{4k}=?$.
Also note that $\cos$-series start from $n=0$ but it carries coefficient $1/2$
-
Please check the attachment!
-
So, An is not zero when n=4k.
-
So, An is not zero when n=4k.
Yes, look $|\sin (x)|$ is $\pi$-periodic; $\cos(nx/2)$ is $\pi$-periodic iff $n=4k$. Now thinks go easier:
$$
A_{4k}=\frac{2}{\pi}\int_0^\pi \sin (x) \cos (2kx)\,dx=
$$
where we halved interval but doubled coefficient (due to $\pi$-periodicity), then follow your calculations
$$
\frac{1}{\pi} \int_0^\pi \Bigl[ \sin ((2k+1)x)-\sin ((2k-1)x)\Bigr]\,dx =
\frac{1}{\pi} \Bigl[ -\frac{1}{2k+1}\cos ((2k+1)x)+\frac{1}{2k-1}\cos ((2k-1)x)\Bigr]_0^\pi =
-\frac{4}{\pi(k^2-1)}$$
as in your calculations but we do not need to analyze cases. In particular, $A_0=\frac{4}{\pi}$.
So
$$
u(x,t)= \frac{2}{\pi} -\sum_{k=0}^\infty \frac{4}{\pi(k^2-1)}\cos (2kx)e^{-4k^2t}.
$$