Build up the characteristic polynomial (you should get a $4 x 4$ determinant) to get the eigenvalues and corresponding eigenfunctions for the following pair of PDEs (note we just ask you for constructing the determinant, not to calculate the eigenvalues; that can be done by a computer):
$$f_{tt} = v_1^2 f_{\theta \theta} \ \ \ for \ \ \ 0 < \theta < \pi \ \ \ \ \ \ \ (1)$$
$$f_{tt} = v_2^2 f_{\theta \theta} \ \ \ for \ \ \ \pi < \theta < 2\pi \ \ \ \ \ \ \ (2)$$
The periodic boundary conditions are (periodicity applies between $0$ and $2\pi$ only):
$$f(0, t) = f(2\pi, t) \ \ \ ; \ \ \ f_{\theta}(0, t) = f_{\theta}(2\pi, t) \ \ \ \ \ \ \ $$
As $\theta$ approaches $\pi$ value, the solutions for $(1)$ and $(2)$ must satisfy the following conditions:
$$\lim\limits_{\theta \to \uparrow \pi} f(\theta, t) = \lim\limits_{\theta \to \downarrow \pi} f(\theta, t) \ \ \ \ \ ; \ \ \ \ \lim\limits_{\theta \to \uparrow \pi} f_{\theta}(\theta, t) = \lim\limits_{\theta \to \downarrow \pi} f_{\theta}(\theta, t)$$
Note that the periodic boundary conditions apply for both PDEs.
NOTE: I've solved the wave equation with different kinds of periodic boundary conditions before. Based on that, I expect to get a solution of the form:
$$u(\theta, t) = C_0 + \sum_{n=1}^{\infty} \Big( A_n \cos(\lambda_n \theta) + B_n \sin(\lambda_n \theta)\Big) \Big( D_n \cos(v\lambda_n t) + E_n \sin(v\lambda_n t) \Big)$$
Besides, I am used to get the eigenvalues directly out of applying BCs. However here it doesn't seem to be possible...
My work out is below.
Any help is appreciated.
NOTE: Let me know if something else needs to be added. I won't hesitate to work out this question further. You can verify I do so here: Deriving the wave equation
Let's (for instance) work out the solution for $(1)$.
- Assume that $u(\theta, t) = \Theta(\theta)T(t)$
Plugging $u(\theta, t) = \Theta(\theta)T(t)$ into $u_{tt} = v_1^2 u_{\theta \theta}$ we get:
$$\frac{T''(t)}{v_1^2 T(t)} = \frac{\Theta''(\theta)}{\Theta(\theta)} = p$$
Where $p \in \Re$
We have three possible values for $p$ : positive, negative and zero.
Case I: $p = \mu^2$
We get the ODE:
$$\Theta''(\theta) = \Theta\mu^2$$
And the solution is:
$$\Theta(\theta) = c_1 \cosh(\mu \theta) + c_2 \sinh(\mu \theta)$$
Applying periodic BCs leads to get the following system:
$$c_1[1 - \cosh(2 \pi \mu)] = c_2 \sinh(2 \pi \mu)$$
$$c_2[1 - \cosh(2 \pi \mu)] = c_1 \sinh(2 \pi \mu)$$
Thus $c_1 = c_2$. But this is not enough information to get the (positive) eigenvalues. I must be doing something wrong...
Case II: $p = 0$
We get the ODE:
$$\Theta''(\theta) = 0$$
And the solution is:
$$\Theta(\theta) = C_3 \theta + C_4$$
Once we apply periodic boundary conditions we get that $C_3 = 0$ and $C_4$ is arbitrary.
Thus we get that:
$$\Theta(\theta) = C_4$$
We can conclude that zero is a valid eigenvalue and the eigenfunction associated with it is $\Theta(\theta) = C_4$.
Case III: $p = -\lambda^2$
We get the ODE:
$$\Theta''(\theta) = -\Theta\lambda^2$$
And the solution is:
$$\Theta(\theta) = C_5 \cos(\lambda \theta) + C_6 \sin(\lambda \theta)$$
Once we apply the periodic boundary condition we get the following system:
Once again I get meaningless equations:
$$c_7 = c_7$$
$$c_8 = c_8$$
But this is not enough information to get the (positive) eigenvalues. I must be doing something wrong...
I've also applied the last two conditions, but I end up with trivial equations... I must be missing something here.