# Fourier coefficients in oscillation problem with viscosity

On the oscillation problem of a rope with fixed extremities, $$\left\{\begin{matrix} \left.\begin{matrix}\left.\begin{matrix} u_{tt}(t,x) = a^2u_{xx}(t,x)\\ u(0,x) = \varphi(x)\\ u_t(0,x) = \psi(x) \end{matrix}\right\}\end{matrix}\right|\; \left\{\begin{matrix} a > 0\\ t > 0\\ 0 < x < L \end{matrix}\right.\\ u(t,0) = u(t,L) = 0 \;|\; t \geq 0 \end{matrix}\right.\qquad,$$

we can show, by assuming that $u$ is separable (id est, $u(t,x) \equiv X(x)T(t)$ for some pair of function $X$ and $T$) and writing

$$\frac{\ddot{T}(t)}{a^2T(t)} = \frac{\ddot{X}(x)}{X(x)} := -\lambda \in \mathbb{R}\;,$$

that $\lambda_n = (n\pi/L)^2$. Therefore,

$$\left.\begin{matrix} u(t,x) = \sum_{n=1}^{\infty} \left(A_n \cos\frac{n\pi a}{L}t + B_n \sin\frac{n\pi a}{L}t\right)\sin\frac{n\pi}{L}x \end{matrix}\;\right|\; A_n \wedge B_n \in \mathbb{R}\quad,$$

and setting $t = 0$ gives

$$\left.\begin{matrix}\left.\begin{matrix} \varphi(x) = \sum_{n=1}^{\infty} A_n \sin\frac{n\pi}{L}x \\ \psi(x) = \sum_{n=1}^{\infty} \frac{n\pi a}{L}B_n\sin\frac{n\pi}{L}x \\ \end{matrix}\right\}\end{matrix}\right| \;0 \leq x\leq L\quad,$$

which means

$$\left\{\begin{matrix} A_n = \frac{2}{L}\int_{0}^{L} \varphi(x)\left(\sin\frac{n\pi}{L}x\right)dx \\ B_n = \frac{2}{n\pi a}\int_{0}^{L} \psi(x)\left(\sin\frac{n\pi}{L}x\right)dx \\ \end{matrix}\right.\quad.$$

When we add viscosity — $f_v(t,x) = -\beta u_t(t,x)$ with $\beta > 0$ —, the first equation of the problem becomes

$$u_{tt}(t,x) + \beta u_t(t,x)= a^2u_{xx}(t,x)\;,$$

and

$$\frac{\ddot{T}(t)}{a^2\beta T(t)} + \frac{\dot{T}(t)}{a^2T(t)} = \frac{\ddot{X}(x)}{\beta X(x)} := -\lambda \in \mathbb{R}$$

gives $\lambda_n = (n\pi/L)^2/\beta$. This means

$$u(t,x) = \sum_{1 \leq n < \frac{\beta L}{2\pi a}} \left(A_n \exp\left\{\!\!\frac{\beta}{2} \left[\sqrt{1 - \left(\frac{2n \pi a}{\beta L}\right)^{\!\!2}} - 1\right] t\right\}\right. +$$ $$+\; \left.B_n \exp\left\{\!\!- \frac{\beta}{2} \left[\sqrt{1 - \left(\frac{2n \pi a}{\beta L}\right)^{\!\!2}} + 1\right]t\right\}\right) \sin\frac{n\pi}{L}x +$$

$$+\, \mathbb{I}_{n = \frac{\beta L}{2\pi a}} (C_n + D_nt) \exp\left(\!\!-\frac{\beta}{2}t\right) \sin\frac{\beta}{2a}x +$$

$$+ \sum_{n > \frac{\beta L}{2\pi a}} \left[E_n \cos\frac{\beta}{2}\sqrt{\left(\frac{2n \pi a}{\beta L}\right)^{\!\!2} \!- 1}\;t \,+\, F_n \sin\frac{\beta}{2}\sqrt{\left(\frac{2n \pi a}{\beta L}\right)^{\!\!2} \!- 1}\;t\right] \exp\left(\!\!-\frac{\beta}{2}t\right) \sin\frac{n\pi}{L}x\quad.$$

How do I calculate coefficients $A_n$ to $F_n$ in terms of $\varphi(x)$ and $\psi(x)$?

-
\left\{ \begin{align} a + b &= c \\ c + d &= e \\ example &= of \quad \TeX \end{align} \right\} – mixedmath Jun 6 '11 at 1:14
@Luke: no problem! I learned that here too, actually. – mixedmath Jun 6 '11 at 15:58
@Luke: good theory! That looks even better, you're right. Good catch. – mixedmath Jun 6 '11 at 21:53
I think your expression for $\psi(x)$ in the damped case is wrong. The separation leads to a damped oscillation equation, $\ddot T+\beta\dot T+(n\pi a/L)^2T=0$, with characteristic values $(\beta\pm\sqrt{\beta^2-(2n\pi a/L)^2})/2$. In any case, taking $\beta$ to $0$ should recover the undamped case, and there should be only one free coefficient per eigenmode; neither is the case in your expression. – joriki Jun 10 '11 at 2:07
@Luke: Yes, but $\exp(-\frac{\beta}{2}(1+\sqrt{\phantom{\cdot}})t)=\exp(-\frac{\beta}{2}t)\exp(-‌​\frac{\beta}{2}\sqrt{\phantom{\cdot}} t)$, not $\exp(-\frac{\beta}{2}t)\exp(\sqrt{\phantom{\cdot}}t)$. I see now that the non-oscillating solutions are also wrong, for a similar reason: $\exp(-\frac{\beta}{2}(1+\sqrt{\phantom{\cdot}})t)\neq\exp(-\frac{\beta}{2}t)\ex‌​p((1+\sqrt{\phantom{\cdot}})t)$. (By the way, I only get notified of your comments if you ping me ("@joriki").) – joriki Jun 11 '11 at 7:06

Basically, you can calculate the coefficients just like you calculated them in the undamped case: Substitute $t=0$ into both $u$ and $u_t$ and Fourier-transform. The only difference is that in the undamped case you'd already chosen the coefficients such that they were decoupled and each individually was given by a Fourier integral; now, linear combinations, each involving two coefficients, are given by Fourier integrals, and you need to solve the resulting $2\times2$ systems of linear equations to get the individual coefficients.