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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 $\psi(x)$?

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    $\left\{ \begin{align} a + b &= c \\ c + d &= e \\ example &= of \quad \TeX \end{align} \right\}$2011-06-06
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    @Luke: no problem! I learned that here too, actually.2011-06-06
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    @Luke: good theory! That looks even better, you're right. Good catch.2011-06-06
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    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.2011-06-10
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    Much better, but I think there's a factor of $\beta/2$ missing in the frequencies of the non-critically damped oscillations.2011-06-10
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    What? But there is "$\exp\left(\!\!-\frac{\beta}{2}t\right)$" in all three.2011-06-10
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    @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: $\exp(-\frac{\beta}{2}(1+\sqrt{\phantom{\cdot}})t)\neq\exp(-\frac{\beta}{2}t)\exp((1+\sqrt{\phantom{\cdot}})t)$. (By the way, I only get notified of your comments if you ping me ("@joriki").)2011-06-11
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    @joriki Oh, geez, I know both those things… I managed to make two very silly mistakes at once… I guess I was sleepier than I thought. [facepalm] Thanks, I'll fix it ASAP.2011-06-11
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    I think the non-oscillating parts are OK now (though it seems preferable to keep the same sign in front of the $1$ and let the sign in front of the square root change), but you're still missing the factor of $\beta/2$ in the frequencies of the non-critically damped oscillations.2011-06-11
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    @joriki I really can't see where, can you edit it? (BTW, I changed the phrasing in the overdamped part of the formula because once the values are given, the exponent with $A_n$ is probably going to be positive and the $B_n$ one surely negative.)2011-06-11
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    OK, I've added the missing factors -- don't quite see why you understood them in the non-oscillating part but not in the oscillating part -- what's the difference?2011-06-11
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    @joriki Wow, I guess my brain would only perceive the lack of $\beta/2$ in an exponential function. I wrote down the underdamped case on paper before editing and felt comfortable with the root alone multiplying $t$. Go figure… Now I finally see what you meant.2011-06-12

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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.

Is that enough, or do you want more details?