nickthequick
Dec5-11, 09:40 PM
Hi,
If I have a forced wave equation
u_{tt}-c^2u_{xx}= f(x,t)
what is my associated energy law?
For instance, in the homogeneous case
\Box u=0
I know that
E(t)=\frac{1}{2}\int u_t^2 +c^2|u_x|^2 \ dx
which implies that \frac{d E(t)}{d t} is equal to zero. (just use integration by parts)
I'm not sure if I should just conclude that
\frac{d E}{dt}= \int u_t f(x,t) dx
by following the same logic as was used to show that this quantity is zero in the homogeneous case. Or we can more generally see that the forced wave equation has a Lagrangian given by
L = \frac{1}{2}\iint -u_t^2 +c^2|u_x|^2 - 2u f(x,t) \ dx \ dt
which means that the associated Hamiltonian, or energy, would be
E(t) = \frac{1}{2}\int u_t^2 +c^2|u_x|^2 + 2u f(x,t) \ dx
which leads to a different result than what I quoted above.
Any help is appreciated!
Thanks,
Nick
If I have a forced wave equation
u_{tt}-c^2u_{xx}= f(x,t)
what is my associated energy law?
For instance, in the homogeneous case
\Box u=0
I know that
E(t)=\frac{1}{2}\int u_t^2 +c^2|u_x|^2 \ dx
which implies that \frac{d E(t)}{d t} is equal to zero. (just use integration by parts)
I'm not sure if I should just conclude that
\frac{d E}{dt}= \int u_t f(x,t) dx
by following the same logic as was used to show that this quantity is zero in the homogeneous case. Or we can more generally see that the forced wave equation has a Lagrangian given by
L = \frac{1}{2}\iint -u_t^2 +c^2|u_x|^2 - 2u f(x,t) \ dx \ dt
which means that the associated Hamiltonian, or energy, would be
E(t) = \frac{1}{2}\int u_t^2 +c^2|u_x|^2 + 2u f(x,t) \ dx
which leads to a different result than what I quoted above.
Any help is appreciated!
Thanks,
Nick