Vibrating String — Question 10

Question 10 – Solution

Strategy. Orthogonality gives a conserved tail energy, while absolute Fourier tails give a different uniform displacement certificate.

Step 1: Construct the solution and normalize energy. Two integrations by parts give $$b_n=2\int_0^1x(1-x)\sin(n\pi x)\,dx =\frac{4(1-(-1)^n)}{(n\pi)^3}.$$ Thus $$\begin{gathered} \boxed{u(x,t)=\frac 8{\pi^3}\sum_{\substack{n\ge 1\\n\ {\mathrm{odd}}}} \frac{\sin(n\pi x)\cos(n\pi t)}{n^3},\qquad E=\frac 12\int_0^1(1-2x)^2\,dx=\frac 16.} \end{gathered}$$ Displacement and first-derivative series converge uniformly, and the PDE has a finite-energy interpretation. A globally $C^2$ corner solution is impossible: fixed endpoints would have $u_{tt}=0$ there, while the initial second spatial derivative is $-2$.

Step 2: Bound the conserved energy tail. Each omitted mode contributes $(n\pi)^2b_n^2/4$, independent of time. Orthogonality therefore gives $$\begin{gathered} E_{\mathrm{err}}=\frac{16}{\pi^4} \sum_{\substack{n>N\\n\ {\mathrm{odd}}}}\frac 1{n^4} \le\frac{16}{\pi^4}\sum_{n>N}\frac 1{n^4} \le\boxed{\frac{16}{3\pi^4N^3}}. \end{gathered}$$ The last inequality is the decreasing-function integral bound $\sum_{n>N}n^{-4}\le\int_N^\infty s^{-4}\,ds$.

Step 3: Certify displacement independently. Using $|\sin|,|\cos|\le 1$ and the same tail comparison, $$\boxed{\|u-u_N\|_\infty\le\frac 8{\pi^3}\sum_{n>N}\frac 1{n^3} \le\frac 4{\pi^3N^2}.}$$ This holds simultaneously for every $x$ and every $t$. For tolerance $\delta=10^{-3}$, it suffices to take $N\ge\sqrt{4/(\pi^3\delta)}\approx 11.36$, so $\boxed{N=12}$ suffices. The bound includes even indices for convenience; it is sufficient, not a claim of the smallest possible retained set.

Step 4: Choose an energy-based cutoff. Since $E=1/6$, $E_{\mathrm{err}}/E\le 32/(\pi^4N^3)$. For tolerance $\eta=10^{-4}$, it suffices that $$N\ge\left(\frac{32}{\pi^4\eta}\right)^{1/3}\approx 14.87, \qquad \boxed{N=15}.$$ Each wave mode preserves its energy and merely oscillates. There is no factor decaying with $n^2t$ as in diffusion, so waiting does not reduce the truncation’s energy error. These two cutoffs measure different errors and need not agree.