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TheoremProved

Conservation of wave energy

Statement

Let u(x,t)u(x,t) solve the wave equation utt=c2uxxu_{tt}=c^2 u_{xx} on [0,L][0,L] with fixed-end boundary conditions u(0,t)=u(L,t)=0u(0,t)=u(L,t)=0. The total energy E(t)=12∫0L(ut2+c2ux2)dxE(t) = \tfrac{1}{2}\int_0^L \left(u_t^2 + c^2 u_x^2\right)dx is constant: E′(t)=0E'(t)=0 for all t>0t>0.

Why is it true?

Energy has two parts: kinetic 12∫ut2 dx\tfrac12\int u_t^2\,dx (mass moving) and elastic potential 12c2∫ux2 dx\tfrac12 c^2\int u_x^2\,dx (string stretched). The wave equation turns kinetic energy into elastic energy and back, at the same rate, keeping the total constant — exactly as a pendulum trades kinetic and potential energy without loss.

Proof sketch

Differentiate E(t)E(t) under the integral sign: E′(t)=∫0L(ututt+c2uxuxt)dxE'(t) = \int_0^L \left(u_t u_{tt} + c^2 u_x u_{xt}\right)dx.

Substitute utt=c2uxxu_{tt}=c^2 u_{xx} from the wave equation to get E′(t)=∫0L(c2utuxx+c2uxuxt)dx=c2∫0L(utuxx+uxuxt)dxE'(t) = \int_0^L \left(c^2 u_t u_{xx} + c^2 u_x u_{xt}\right)dx = c^2\int_0^L \left(u_t u_{xx} + u_x u_{xt}\right)dx. Recognize the integrand as the derivative ∂∂x(utux)\tfrac{\partial}{\partial x}(u_t u_x), so E′(t)=c2[utux]0LE'(t) = c^2\Big[u_t u_x\Big]_0^L.

Apply the fixed-end boundary conditions u(0,t)=u(L,t)=0u(0,t)=u(L,t)=0, which force ut(0,t)=ut(L,t)=0u_t(0,t)=u_t(L,t)=0 as well (differentiating the boundary conditions with respect to tt). Therefore E′(t)=c2(ut(L,t)ux(L,t)−ut(0,t)ux(0,t))=0E'(t) = c^2\big(u_t(L,t)u_x(L,t) - u_t(0,t)u_x(0,t)\big) = 0, so E(t)=E(0)E(t)=E(0) is constant for all t≥0t\ge 0.

Topics that use this theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. Lawrence C. Evans (2010). Partial Differential Equations · DOI:10.1090/gsm/019
  2. Walter A. Strauss (2008). Partial Differential Equations: An Introduction
  3. Jean le Rond d'Alembert (1747). Recherches sur la courbe que forme une corde tendue mise en vibration