ODE:
Let $u = u(t, x):\ [0, \infty)\times [0, L] \to \mathbb{R}$ be unknown and let $u_0 = u_0(x):\ [0,L] \to \mathbb{R}, u_1 = u_1(x):\ [0,L] \to \mathbb{R}$ be given. Let $L > 0, c > 0, 0 \leq \mu < \frac{\pi c}{L}\ $be constants.
At this point, I would like to solve the following ODE: $$ \frac{1}{c^2} \frac{d^2b_n}{dt^2}(t) + \frac{2\mu}{c^2} \frac{db_n}{dt}(t) = -\left(\frac{n\pi}{L} \right)^2 b_n(t),\\ b_n(0) = b_{0,n},\\ \frac{db_n}{dt}(0) = b_{1,n} $$ for all $n \in \mathbb{N}$.
What I know:
Consider the characteristic equation $$\lambda^2 + 2\mu\lambda + c^2\left(\frac{n\pi}{L}\right)^2 = 0.\\ \therefore \lambda = -\mu \pm\sqrt{\mu^2-\left(\frac{n\pi c}{L}\right)^2} $$ so the ODE solution is $$ b_n(t) = e^{-\mu t}\left\{ b_{0,n}\cos{\left(\sqrt{\left(\frac{n\pi c}{L}\right)^2 -\mu^2}t\right)} + \frac{b_{1,n}+\mu b_{0,n}}{\sqrt{\left(\frac{n\pi c}{L}\right)^2 -\mu^2}} \sin{\left(\sqrt{\left(\frac{n\pi c}{L}\right)^2 -\mu^2}t\right)}\right\}. $$
If $\mu=0$, then $$ b_n(t) = b_{0,n}\cos{\left( \frac{n\pi c}{L}t \right)} + \frac{L}{n\pi c}b_{1,n}\sin{\left( \frac{n\pi c}{L}t \right)} $$.
If $\mu= \frac{\pi c}{2L}$, then $$ b_n(t) = e^{-\frac{\pi c}{2L}t}\left\{b_{0,n}\cos{\left( \sqrt{n^2-\frac{1}{4}}\frac{\pi c}{L}t \right)} + \frac{2Lb_{1,n}+\pi cb_{0,n}}{2\pi c \sqrt{n^2-\frac{1}{4}}}b_{1,n}\sin{\left( \sqrt{n^2-\frac{1}{4}}\frac{\pi c}{L}t \right)}\right\}. $$
- When $\mu=0$, the PDE is the wave equation? How do the Fourier coefficients $b_n$ behave depending on these two values of $\mu$? Could you please consider the difference?