By means of the change of variables η=x−t and ξ=x+t, show that the wave equation for u=u(x,t)
∂x2∂2u−∂t2∂2u=0
is equivalent to the equation
∂η∂ξ∂2U=0
where U(η,ξ)=u(x,t). Hence show that the solution to (∗) on x∈R and t>0, subject to the initial conditions
u(x,0)=f(x),∂t∂u(x,0)=g(x)
u(x,t)=21[f(x−t)+f(x+t)]+21∫x−tx+tg(y)dy
Deduce that if f(x)=0 and g(x)=0 on the interval ∣x−x0∣>r then u(x,t)=0 on ∣x−x0∣>r+t.
Suppose now that y=y(x,t) is a solution to the wave equation (∗) on the finite interval 0<x<L and obeys the boundary conditions
y(0,t)=y(L,t)=0
for all t. The energy is defined by
E(t)=21∫0L[(∂x∂y)2+(∂t∂y)2]dx
By considering dE/dt, or otherwise, show that the energy remains constant in time.