Equipartition of energy in one dimensional wave equation  $u_{tt}-u_{xx}=0 $

Suppose $g$ and $h$ have compact support and $u\in C^2(\mathbb{R}\times [0,\infty))$ solves the wave equation 
\[u_{tt}-u_{xx}=0     in   \mathbb{R}\times (0,\infty)\]
and $u=g$, $u_t=h$ on $\mathbb{R}\times \{t=0\}$. The kinetic energy of the wave is 
\[k(t)=\int_{-\infty}^{+\infty}u_t^2(x,t)dx\]
and the potential energy is 
\[p(t)=\int_{-\infty}^{+\infty}u_x^2(x,t)dx.\]
Prove that 
\[k(t)=p(t)\]
for $t$ large enough. 

Wave Equation Partial Differential Equations Hyperbolic PDEs
Nicolasp Nicolasp
163
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Answer

By D'Alembert's formula, we have the explicit solution $$u(x,t)=\frac{1}{2} \left[ g(x+t) +g(x-t) \right] +\frac{1}{2} \int_{x-t}^{x+t} h(y) \, dy.$$ We compute $$u_t=\frac{1}{2} \left[ g'(x+t) -g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) +h(x-t)\right],$$ and $$u_x=\frac{1}{2} \left[ g'(x+t) +g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) -h(x-t)\right].$$ This implies 
\[k(t)-p(t) = \frac{1}{2} \int_{-\infty}^{\infty} \left( \frac{1}{2} \left[ g'(x+t) -g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) +h(x-t)\right] \right)^2 \, dx\\ - \frac{1}{2} \int_{-\infty}^{\infty} \left( \frac{1}{2} \left[ g'(x+t) +g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) -h(x-t)\right] \right) ^2 \, dx\]
There exists $L>0$ such that $\text{supp}(g'), \text{supp}(h)\subseteq [-L,L]$. We proceed to show this integral is zero by showing $u_t^2=u_x^2$ in three different cases: 

Case 1: If $x+t > L$ $(x>L-t)$, then $g'(x+t)=h(x+t)=0 \Rightarrow u_t^2=u_x^2$.

Case 2: If $x-t<-L$ $(x<t-L)$, then $g'(x-t)=h(x-t)=0 \Rightarrow u_t^2=u_x^2$. 

Case 3: For the last interval $L-t \geq x \geq t-L$, there is no solution of x when $t>L$.

Therefore, $k(t)=p(t)$ for large enough $t$.

Erdos Erdos
4.7K
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