The Wave Equation — Question 6

Question 6 – Solution

Strategy. Integrate the PDE against velocity so that acceleration and curvature become time derivatives of kinetic and elastic energies.

Step 1: Derive the work balance. Multiplying by $u_t$ and integrating by parts gives $$\boxed{E'(t)=\mathcal T[u_tu_x]_0^L+\int_0^L p\,u_t\,dx.}$$ At a fixed endpoint, $u=0$ for all time, hence $u_t=0$ there. Thus $E'=\int p\,u_t$. With no applied force, total energy is constant. The sign of the work rate depends on the force and velocity together.

Step 2: Prove uniqueness for the full initial state. The difference $w$ of two solutions with the same force, endpoints and initial data obeys the homogeneous equation and has zero initial energy. Its energy stays zero, so $w_t=w_x=0$ everywhere by continuity. The fixed endpoints then force the spatial constant to be zero. Initial displacement alone would not make the kinetic part of the initial difference energy vanish; both initial data are essential.

Step 3: Compute the standing-wave energy. Put $k=\pi/L$, so $\omega^2=(\mathcal T/\mu)k^2$. Then $u_{tt}=-\omega^2u$, $u_{xx}=-k^2u$, and both endpoints are zero. Using $\int_0^L\sin^2kx=\int_0^L\cos^2kx=L/2$ gives $$\boxed{K(t)=E_0\sin^2(\omega t),\qquad V(t)=E_0\cos^2(\omega t),\qquad E_0=\frac{\mathcal T A^2k^2L}{4}.}$$ Their sum is the constant $E_0$.

Step 4: Interpret the exchange of energy. For $A\ne 0$, at $\omega t=\pi/2$ the displacement and elastic energy are zero, but the velocity is nonzero and all energy is kinetic. At $\omega t=0$ the velocity is zero and all energy is elastic. The field exchanges these forms while oscillating; total-energy conservation does not require pointwise monotonic motion. The graph shows $K/E_0$, $V/E_0$ and their constant sum.