The Wave Equation — Question 7

Question 7

A localized initial kick can leave a displaced plateau

An infinite string starts flat: $h(x)=0$. Its initial velocity is $g(x)=U$ for $|x|<a$ and zero elsewhere. Use $y(x,t)=(2v)^{-1}\int_{x-vt}^{x+vt}g(s)ds$. Ideal sharp edges may be smoothed.

1. Find $y(0,t)$ for $t\ge0$.

2. Explain how a stationary central plateau after the pulse fronts depart is compatible with wave propagation and finite energy.

Question 7 - Solution

A localized initial kick can leave a displaced plateau

Method and mathematical working

With zero initial displacement, only the integral of initial velocity remains. At the center, its integration interval overlaps a length that grows until it covers the entire nonzero interval.$$\begin{aligned}y(0,t)&=\frac{U}{2v}\,2\min(vt,a)\\&=\begin{cases}Ut,&vt<a,\\Ua/v,&vt\ge a.\end{cases}\\[4pt]vt>a&:\quad y(0,t)=Ua/v,\quad y_t(0,t)=0,\quad y_x(0,t)=0\Rightarrow u(0,t)=0.\end{aligned}$$

(a) Calculation and conclusion

For $vt<a$, the integration interval has length $2vt$ inside the kicked region, so $y(0,t)=Ut$. For $vt\ge a$, it contains all the initial kick, giving $y(0,t)=Ua/v$. The central displacement stops changing after $a/v$.

(b) Interpretation and limits

The advancing transition regions carry slope and velocity; a constant plateau has neither. Thus the ideal transverse-string energy density vanishes inside the plateau even though its displacement is nonzero. The disturbance propagates through moving fronts, not by every point oscillating forever.

Physical check

After the fronts pass, the central plateau has zero slope and zero velocity. Nonzero displacement alone does not require nonzero local elastic or kinetic energy.