Vibrating String — Question 5

Question 5 – Solution

Strategy. The same drag rate competes with different natural frequencies, so one string can exhibit all three damping regimes at once.

Step 1: Classify the modal roots. With $u=\sum q_n(t)\sin(n\pi x)$, each coefficient satisfies $q_n''+2\beta q_n'+(n\pi)^2q_n=0$. For $\beta=2\pi$, the discriminant sign is the sign of $4-n^2$. Thus mode one is overdamped, mode two is critical, and mode three is underdamped.

Step 2: Impose both initial conditions. The required coefficient functions, all with $q_n(0)=1$ and $q_n'(0)=0$, are $$\begin{aligned} q_1&=e^{-2\pi t}\left[\cosh(\sqrt 3\pi t)+\frac 2{\sqrt 3}\sinh(\sqrt 3\pi t)\right],\\ q_2&=e^{-2\pi t}(1+2\pi t),\\ q_3&=e^{-2\pi t}\left[\cos(\sqrt 5\pi t)+\frac 2{\sqrt 5}\sin(\sqrt 5\pi t)\right]. \end{aligned}$$ Then $\boxed{u=\sum_{n=1}^3q_n(t)\sin(n\pi x)}$ satisfies the damped PDE and both endpoints. The added sinh, linear and sine terms are necessary to make the initial velocities zero.

Step 3: Account for dissipated energy. Differentiate $E$ and integrate $u_xu_{xt}$ by parts: $$E'=[u_xu_t]_0^1+\int_0^1u_t(u_{tt}-u_{xx})\,dx =\boxed{-2\beta\int_0^1u_t^2\,dx\le 0}.$$ The boundary term is zero because fixed ends have zero velocity. Oscillation changes how energy is partitioned; drag still removes energy whenever the velocity is not identically zero.

Step 4: Identify the slow surviving mode. Expanding $q_1$ into two exponentials, its slower coefficient is $C=(1+2/\sqrt 3)/2$ and its slower decay rate is $\gamma=(2-\sqrt 3)\pi$. The other two modes decay as $e^{-2\pi t}$ times bounded or linear factors. Therefore $$\boxed{e^{\gamma t}u(x,t)\longrightarrow C\sin(\pi x) \quad\text{uniformly for }0\le x\le 1.}$$ For a nonnegative general drag parameter, mode $n$ is underdamped when $\beta<n\pi$; all positive modes are underdamped exactly when $0\le\beta<\pi$. The threshold compares drag with the lowest frequency, not with the highest one.