 Solution

In the bilinear identity choose field $a$ to be the complex conjugate of the field at frequency $\omega$, equivalently its $-\omega$ partner, and field $b$ to be the field itself. The squared frequencies are equal, so the mass density terms cancel:Multiply by $i\omega$ and use ${\hat{v}}_i=-i\omega {\hat{u}}_i$, ${\hat{v}}_i^{\ast }=i\omega {\hat{u}}_i^{\ast }$ to obtainFor the real fields defined in this question as the sum of positive and negative frequency components, time averaging cancels the $e^{\pm 2i\omega t}$ terms. Their mean body-force work mass density is exactly ${\hat{f}}_i^{\ast }{\hat{v}}_i+{\hat{v}}_i^{\ast }{\hat{f}}_i$, and their outward mean elastic-wave energy flux is $-t_i^{\ast }{\hat{v}}_i-{\hat{v}}_i^{\ast }{t}_i$. Thus mean body-force power equals mean outward boundary power. Mean stored energy does not accumulate in a periodic steady field.

There is no extra factor $1/2$ here: the specified real field is $\hat{f}{e}^{-i\omega t}+{\hat{f}}^{\ast }{e}^{i\omega t}$, not $\mathrm{Re}(\hat{f}{e}^{-i\omega t})$. For the latter phasor convention, both averages instead have the usual factor $1/2$. At $\omega =0$ velocity and both powers vanish.