The Wave Equation — Question 2

Question 2 – Solution

Strategy. Initial velocity contributes an integral over a growing interval; it is not simply two displaced copies of the initial displacement.

Step 1: Verify the integral solution. Leibniz differentiation gives $$u_t=\frac{g(x+ct)+g(x-ct)}2,\quad u_{tt}=\frac c2\bigl(g'(x+ct)-g'(x-ct)\bigr),\quad u_{xx}=\frac 1{2c}\bigl(g'(x+ct)-g'(x-ct)\bigr).$$ Thus $u_{tt}=c^2u_{xx}$. At $t=0$, the integral is zero and $u_t(x,0)=g(x)$. The $C^1$ regularity of $g$ makes these second derivatives continuous.

Step 2: Integrate at the center. Symmetry gives $u(0,t)=c^{-1}\int_0^{\min(ct,1)}(1-s^2)^2\,ds$. Consequently $$\boxed{u(0,t)= \begin{cases} t-\dfrac{2c^2t^3}{3}+\dfrac{c^4t^5}{5},&0\leq ct\leq 1,\\[3pt] \dfrac 8{15c},&ct\geq 1. \end{cases}}$$ The two expressions agree at $ct=1$.

Step 3: Locate the moving fronts and plateau. For $t>0$, the integral is positive exactly when $[x-ct,x+ct]$ overlaps $(-1,1)$ in positive length. It is therefore zero for $|x|\geq 1+ct$. When $ct\geq|x|+1$, the integration interval contains the entire support of $g$, so $$\boxed{u(x,t)=\frac 1{2c}\int_{-1}^1g(s)\,ds=\frac 8{15c}.}$$ The same value holds at equality because the endpoints of $g$ contribute nothing. Behind the fronts, the string is displaced but locally stationary and flat. Displacement itself need not return to zero after a velocity pulse.

Step 4: Interpret the plotted profiles. Since $g\geq 0$, every interval integral lies between zero and the full integral. Thus $0\leq u\leq 8/(15c)$ for every $x,t$. For $c=1$, the fronts are at $x=\pm(1+t)$; a plateau appears once $t\geq 1$ and occupies $|x|\leq t-1$. Its height stays $8/15$ as its width grows. Zero local velocity on that plateau does not mean zero local displacement.