The Wave Equation — Question 9

Question 9 – Solution

Strategy. Wave energy controls velocity and slope, but a spatially uniform motion can carry the mean displacement indefinitely.

Step 1: Track the constant spatial mode. Integrating the PDE and using matching periodic slopes gives $$m''(t)=\frac{c^2}{2\pi}[u_x]_{-\pi}^{\pi}=0.$$ Hence $$\boxed{m(t)=m(0)+m'(0)t,\qquad P(t)=2\pi\mu m'(0).}$$ The mean velocity, rather than the mean displacement, is conserved.

Step 2: Derive the periodic energy balance. Multiplication by $u_t$ and integration by parts gives $E'=\mathcal T[u_tu_x]_{-\pi}^{\pi}=0$ by periodicity. The energy contains $u_t$ and $u_x$, but not $u$ itself. A nonzero constant mean velocity is therefore compatible with a linearly growing mean displacement and a fixed total energy.

Step 3: Verify a drifting example. For $u=Vt+A\cos x\cos(ct)$, the term $Vt$ and the oscillatory term each solve the equation. Orthogonality gives $$\boxed{m(t)=Vt,\quad P=2\pi\mu V,\quad E=\pi\mu V^2+\frac{\pi\mathcal T A^2}{2}.}$$ The cosine’s kinetic and elastic contributions sum to the second constant. For $V\ne 0$, displacement is unbounded in time even though the energy is finite and constant.

Step 4: Bound displacement when the mean cannot drift. If $m'(0)=0$, then $m(t)=m(0)$. At each time the continuous function $w=u-m(0)$ has zero spatial mean, so it has a zero at some $x_*$ (or is identically zero). Integrate its derivative from that point over a path of length at most $2\pi$. Cauchy–Schwarz gives $$|w(x,t)|\leq\sqrt{2\pi}\,\|u_x(\cdot,t)\|_2 \leq\sqrt{\frac{4\pi E(0)}{\mathcal T}}.$$ Thus $\boxed{|u(x,t)|\leq|m(0)|+\sqrt{4\pi E(0)/\mathcal T}}$ for all time. The estimate need not be sharp; it identifies the missing mean-velocity condition.