A closed magnetic loop (see picture below) is partially submerged in the solar interior and partly in the corona. When observed over time, it is clear that it behaves like an elastic band that can carry waves.
Find the general solution for the wave equation for the waves in the closed loop. Assume periodicity and apply interface conditions at the photosphere. For this case you need not worry about the curvature of the loop and you can stretch it as a straight line, but you need to apply periodicity at the ends. Solve means finding the eigenfrequencies and also the eigenfunctions.
HINTS:
1) You have to deal with 2 wave equations. Note that the wave travels through different mediums; in the upper half through the corona and in the lower half through the solar interior.
2) You have to end up with a $[4 \times 4]$ determinant. You do not have to work out the determinant; you only have to show it. The eigenfrequencies (eigenvalues) can easily be done by a computer by equating the solution of the determinant to zero and solving that equation.
I could set up the problem correctly:
We have two different Wave Equations:
$$f_{tt} = v_1^2 f_{\theta \theta} \ \ \ for \ \ \ 0 < \theta < \pi \ \ \ \ \ \ \ (1)$$
$$f_{tt} = v_2^2 f_{\theta \theta} \ \ \ for \ \ \ \pi < \theta < 2\pi \ \ \ \ \ \ \ (2)$$
The periodic boundary conditions are as follows (periodicity applies between $0$ and $2\pi$ only):
$$f(0, t) = f(2\pi, t) \ \ \ ; \ \ \ f_{\theta}(0, t) = f_{\theta}(2\pi, t) \ \ \ \ \ \ \ $$
The interface conditions at the photosphere are:
$$\lim\limits_{\theta \to \uparrow \pi} f(\theta, t) = \lim\limits_{\theta \to \downarrow \pi} f(\theta, t) \ \ \ \ \ ; \ \ \ \ \lim\limits_{\theta \to \uparrow \pi} f_{\theta}(\theta, t) = \lim\limits_{\theta \to \downarrow \pi} f_{\theta}(\theta, t)$$
But I am stuck in how to solve it. The method I normally apply is as follows:
$1)$ Assume separation of variables AKA $u(\theta, t) = \Theta(\theta)T(t)$.
$2)$ Study what are the solutions you get for the spatial part when the constant is: positive, zero and negative.
But I am not getting the eigenfunctions.
We expect solutions of the form:
$$(\theta, t) \rightarrow \Theta (\theta) e^{\pm i \omega t}$$
My in-detail work is here: Constructing a determinant to get the eigenvalues associated to a pair of wave equations
Any help is appreciated.
EDIT $0$:
These are my solutions:
- After applying the periodic boundary conditions one gets:
$$\Theta (\theta) = A_1 \sin \Big( n \frac{\omega}{v_1} \theta \Big) \ \ \ \ \ (SOL.1)$$
$$\Theta (\theta) = A_2 \sin \Big( n \frac{\omega}{v_2} (2\pi - \theta) \Big) \ \ \ \ \ (SOL.2)$$
- After applying interface conditions one gets:
$$A_1 \sin \Big( n \frac{\omega}{v_1} \pi \Big) = A_2 \sin \Big( n \frac{\omega}{v_2} \pi \Big) \ \ \ \ \ \ \ \ \ (5)$$
$$n\frac{A_1}{v_1} \cos \Big( n \frac{\omega}{v_1} \pi \Big) = -n\frac{A_2}{v_2} \cos \Big( n \frac{\omega}{v_2} \pi \Big) \ \ \ \ \ \ \ \ \ (6)$$
My questions now are:
1) Do these solutions make sense to you all?
2) how can I construct a $[4 \times 4]$ determinant out of $(5)$ and $(6)$?
EDIT $1$:
I missed the cosine term. Now it should be OK.
- After applying the periodic boundary conditions one gets:
$$\Theta (\theta) = A_1 \sin \Big( n \frac{\omega}{v_1} \theta \Big) + B_1 \cos \Big( n \frac{\omega}{v_1} \theta \Big) \ \ \ \ \ (S.SOL.1)$$
$$\Theta (\theta) = A_2 \sin \Big( n \frac{\omega}{v_2} (2\pi - \theta) \Big) + B_2 \cos \Big( n \frac{\omega}{v_2} (2\pi - \theta) \Big) \ \ \ \ \ (S.SOL.2)$$
- After applying interface conditions one gets:
$$A_1 \sin \Big( n \frac{\omega}{v_1} \pi \Big) + B_1 \cos \Big( n \frac{\omega}{v_1} \pi \Big) = A_2 \sin \Big( n \frac{\omega}{v_2} \pi \Big) + B_2 \cos \Big( n \frac{\omega}{v_2} \pi \Big) \ \ \ \ \ \ \ \ \ (5)$$
$$n\frac{A_1}{v_1} \cos \Big( n \frac{\omega}{v_1} \pi \Big) - n\frac{B_1}{v_1} \sin \Big( n \frac{\omega}{v_1} \pi \Big) = -n\frac{A_2}{v_2} \cos \Big( n \frac{\omega}{v_2} \pi \Big) + n\frac{B_2}{v_2} \sin \Big( n \frac{\omega}{v_2} \pi \Big) \ \ \ \ \ \ \ \ \ (6)$$
Now thinking how to construct the determinant...
EDIT $2$:
OK so let's finish it: the determinant is:
$$ \begin{vmatrix} \sin \Big( n \frac{\omega}{v_1} \theta \Big) & \cos \Big( n \frac{\omega}{v_1} \theta \Big) & 0 & 0 \\ 0 & 0 & \sin \Big( n \frac{\omega}{v_2} (2 \pi - \theta) \Big) & \cos \Big( n \frac{\omega}{v_2} (2 \pi - \theta) \Big) \\ \sin \Big( n \frac{\omega}{v_1} \pi \Big) & \cos \Big( n \frac{\omega}{v_1} \pi \Big) & -\sin \Big( n \frac{\omega}{v_2} \pi \Big) & -\cos \Big( n \frac{\omega}{v_2} \pi \Big) \\ \frac{n}{v_1} \cos \Big( n \frac{\omega}{v_1} \pi \Big) & -\frac{n}{v_1} \sin \Big( n \frac{\omega}{v_1} \pi \Big) & \frac{n}{v_2} \cos \Big( n \frac{\omega}{v_2} \pi \Big) & -n\frac{B_2}{v_2} \sin \Big( n \frac{\omega}{v_2} \pi \Big) \\ \end{vmatrix} $$
Note we are given the values of $R$ and $L$ and thus we can compute the determinant (L and R are related by $L = R \theta$).
EDIT $3$:
OK the title asks for finding the general solution for the wave equation for waves in a closed loop.
Above I just gave the spatial solutions.
Thus, let me know provide you with the complete general solutions I get (which, of course, have the form $u(\theta, t) = \Theta(\theta)T(t) \ $)
$$(G.S.1)$$
$$u(\theta, t) = A_0 + B_0t + \sum_{n=1}^{\infty} \Big( A_n \cos(n\omega t) + B_n \sin(n\omega t)\Big) \Big( C_n \cos(n\frac{\omega}{v_1} \theta) + D_n \sin(n\frac{\omega}{v_1} \theta) \Big)$$
$$(G.S.2)$$
$$u(\theta, t) = E_0 + F_0t + \sum_{n=1}^{\infty} \Big( E_n \cos(\omega t) + F_n \sin(\omega t)\Big) \Big( G_n \cos(\frac{\omega}{v_2} (2 \pi - \theta)) + D_n \sin(\frac{\omega}{v_2} (2 \pi - \theta))$$