Wave problem $u_{tt}=u_{xx}-u_t, 0\le x\le 2\pi$ with $u(x,0)=\phi(x)$, $u_t(x,0)=\psi(x)$ and $u(0,t)=u(2\pi,t)=0$

Question

Solve the wave problem $$\begin{cases}u_{tt}=u_{xx}-u_t, 0\le x\le 2\pi\\ u(0,t)=u(2\pi,t)=0\\ u(x,0)=\phi(x)\\ u_t(x,0)=\psi(x)\end{cases}$$ Write the eigenvalues and eigenfunctions explicitly.

Are they orthogonal?

We propose the separated solution $u(x,t)=X(x)T(t)$.

Then $\frac{(T''+T')}{T}=\frac{X''}{X}=\lambda$.

If we define $\lambda=\beta^2,$ we have $X(x)=C\cos\beta x+D\sin\beta x$ and $T(t)=Ae^{r_1t}+Be^{r_2t}, r=-1/2\pm\frac{\sqrt{1+4\beta^2}}{2}$.

Applying boundary condition, we find that $X_n(x)=D_n\sin(\lambda_nx),\lambda=\sqrt{n/2}$

Applying initial conditions we find that $T(0)=A+B=\phi(x)$ and $T'(0)=Ar_1+Br_2=\psi(x)$

My question here is how will I find $A$ and $B$ if we don't know $\phi$ nor $\psi$ ?

How will I find the eigenvalues ?

Help me please

Answer 1

The boundary conditions give

$$ X(x) = \sin\left(\frac{n}{2}x\right) $$

As usual, we drop the remaining constant for now, since it'll absorbed into the constants of $T(t)$

This means $\lambda_n = X''/X = -n^2/4 = -\beta^2$ (you have a sign error). The other equation becomes

$$ T'' + T' + \frac{n^2}{4}T = 0 $$

The two roots of the characteristic are

$$ r = \frac{-1 \pm \sqrt{1-n^2}}{2} $$

Since $n \ge 1$, we rewrite

$$ r = \frac{-1 \pm \sqrt{-(n^2-1)}}{2} = \frac{-1 \pm i\sqrt{n^2-1}}{2} $$

This gives the solution

$$ T(t) = e^{-t/2}\left[A\cos\left(\frac{\sqrt{n^2-1}}{2}t\right) + B\sin\left(\frac{\sqrt{n^2-1}}{2}t\right) \right] $$

For ease of notation, let $\mu_n = \sqrt{n^2-1}$ to make things shorter, so we have

\begin{align} T(t) &= e^{-t/2}\left[A\cos\left(\frac{\mu}{2}t\right) + B\sin\left(\frac{\mu}{2}t\right) \right] \\ T'(t) &= e^{-t/2}\left[\left(-\frac{A}{2} + \frac{\mu B}{2}\right)\cos\left(\frac{\mu}{2}t\right) + \left(-\frac{B}{2} - \frac{\mu A}{2} \right)\sin \left(\frac{\mu}{2}t\right) \right] \end{align}

As usual, the full solution in series form is

$$ u(x,t) = \sum_n X_n(x)T_n(t) = \sum_{n=1}^\infty \sin\left(\frac{n}{2}x\right) e^{-t/2}\left[A\cos\left(\frac{\mu}{2}t\right) + B\sin\left(\frac{\mu}{2}t\right) \right] $$

We can now plug in the given initial conditions

\begin{align} u(x,0) &= \sum_n X_n(x)T_n(0) = \sum_{n=1}^\infty B_n \sin\left(\frac{n}{2}x\right) = \phi(x) \\ u_t(x,0) &= \sum_n X_n(x){T_n}'(0) = \sum_{n=1}^\infty \left(-\frac{A_n}{2} + \frac{\mu B_n}{2}\right) \sin\left(\frac{n}{2}x\right) = \psi(x) \end{align}

You'll need to find the Fourier series of these to functions to obtain the coefficients

\begin{align} B_n &= \frac{1}{\pi}\int_0^{2\pi} \phi(x) \sin\left(\frac{n}{2}x\right) dx \\ \frac{-A_n + \mu B_n}{2} &= \frac{1}{\pi} \int_0^{2\pi} \psi(x) \sin\left(\frac{n}{2}x\right) dx \end{align}

and solve the resulting system of equations. There's no specific form for the solution here, but you'll know what to do when you are given specific $\phi(x)$ and $\psi(x)$, i.e. just do the integrals.

Answer 2

You are wrong when you say $T(0)=\phi(x)$. We actually have $$\phi(x)=\sum_n(A_n+B_n)\sin(nx/2)$$ and $$\psi(x)=\sum_n(A_n r_1+B_nr_2)\sin(nx/2)$$ In principle, this can be inverted, using orthogonality of $\sin (nx)$ to find the constants $A_n$ and $B_n$. This is done, say, by multiplying both sides by $\sin(mx/2)$ and summing over $m$.