TT2 problem 3

[Victor Ivrii]

 Consider the diffusion equation
\begin{equation*}
u_t -ku_{xx}=0 \qquad \text{for} \quad t>0,\ x \in (0,3\pi)
\end{equation*}
with the boundary conditions
\begin{equation*}
u(0,t)=0,\qquad u(3\pi,t)=0
\end{equation*}
and the initial condition
\begin{equation*}
u(x,0)=|\cos (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.

[Victor Ivrii]

a-b. Separation of variables brings $X''+\lambda X=0$, $X(0)=X(3\pi)=0$ and then $\lambda_n= -n^2/9$, $X_n=\sin (nx/3)$.

c. We know that $\sin (\pi nx/l)$ orthogonal on $[0,l]$.

d. From separation of variables we also have $T'/T=-k\lambda$ and then $T_n=A_n e^{-k n^2t/9}$. Then
\begin{equation}
u(x,t)= \sum_{n=1}^\infty A_n e^{-k n^2t/9} \sin (nx/3),
\label{K}
\end{equation}

e. Then $u(x,0)=\sum_{n=1}^\infty A_n \sin (nx/3)=|\cos (x)|$. We need to decompose $|\cos (x)|$ into series.  Then
\begin{equation}
 A_n = \frac{2}{3\pi} \int_0^{3\pi} |\cos (x)|\sin \bigl(\frac{nx}{3}\bigr)\,dx
\label{L}
\end{equation}
which does not look very easy since $\cos (x)$ changes sign several times on $(0,3\pi)$. Because those who did not make calculations got no reduction.

But there is still an easy way to calculate $A_n$. Since $|\cos (x)$ is $\pi$-periodic
\begin{gather*}
 A_{n} = \frac{2}{3\pi} \int_{\pi}^{2\pi} |\cos (x)|\bigl[ \sin \bigl(\frac{n(x-\pi)}{3}\bigr)+
\sin \bigl(\frac{nx}{3}\bigr) + \sin \bigl(\frac{n(x+\pi)}{3}\bigr)\bigr]\,dx=\\
\frac{2}{3\pi} \int_{\pi}^{2\pi} |\cos (x)| \sin \bigl(\frac{nx}{3}\bigr)
\bigl[ 2\cos \bigl(\frac{n\pi}{3}\bigr)+1\bigr) \bigr]\,dx
\end{gather*}
where we used formula for $\sin(\alpha)+\sin(\beta)$. Observe that $|\cos (x)|$ is even with respect to $\frac{3\pi}{2}$ and $\sin (n x/3)$ is even as $n$ is odd and odd as $n$ is even. Then $A_{2m}=0$ and 
\begin{equation*}
 A_{2m+1} =
\frac{2}{3\pi} \bigl[ 2\cos \bigl(\frac{(2m+1)\pi}{3}\bigr)+1\bigr) \bigr]
\int_{\frac{3\pi}{2}}^{2\pi} \bigl[  \sin\bigl( \frac{(m+2)x}{3} \bigr)   +  \sin\bigl( \frac{(m-1)x}{3} \bigr) \bigr]\,dx
\end{equation*}
where we used formula for $2\sin(\alpha)\cos (\beta)$. This integral is easy.