Vibrating String — Question 2

Question 2 – Solution

Strategy. Fourier coefficients reveal the missing harmonics, while reflected traveling profiles preserve the corners exactly.

Step 1: Compute the pluck coefficients. Integrating separately on $[0,a]$ and $[a,L]$ gives $$b_n=\frac 2L\int_0^Lf(x)\sin\frac{n\pi x}{L}\,dx =\frac{2hL^2}{a(L-a)n^2\pi^2}\sin\frac{n\pi a}{L}.$$ Thus $$\boxed{u(x,t)=\sum_{n\ge 1}b_n\sin\frac{n\pi x}{L} \cos\frac{n\pi ct}{L}.}$$ The $O(n^{-2})$ coefficients give uniform convergence of displacement. The initial velocity is zero in $L^2$, and the solution has finite conserved energy. It solves the PDE weakly and classically away from the propagating slope corners; no globally $C^2$ solution is claimed.

Step 2: Identify the spectrum and energy. For $a=L/3$, $b_n=9h\sin(n\pi/3)/(n^2\pi^2)$, so exactly the positive multiples of three are absent. Initially all energy is elastic: $$E=\frac T2\int_0^Lf'^2\,dx =\frac{Th^2L}{2a(L-a)} =\boxed{\frac{9Th^2}{4L}\quad(a=L/3).}$$ The finite-energy wave evolution conserves this value.

Step 3: Verify the reflected and full returns. At $t=L/c$, each cosine is $(-1)^n$, and $\sin(n\pi(L-x)/L)=(-1)^{n+1}\sin(n\pi x/L)$. Hence $u(x,L/c)=-f(L-x)$. At $2L/c$ every cosine equals one, so $u=f$. The modal velocities vanish at both times, giving zero velocity in $L^2$.

Step 4: Construct exact corner-preserving snapshots. The reflected d’Alembert form is $u(x,t)=\tfrac 12[F(x-ct)+F(x+ct)]$. Oddness and periodicity enforce both endpoints and recover both initial data. For the stated numbers, the half-time profile is $$u(x,\tfrac 12)= \begin{cases}-3x/2,&0\le x\le 1/6,\\ 3x/4-3/8,&1/6\le x\le 5/6,\\ 3(1-x)/2,&5/6\le x\le 1. \end{cases}$$ The other two profiles are $f(x)$ and $-f(1-x)$; their corners are plotted at their exact positions.