Periodic smooth incompressible energy identity

By [F1], $u$ is smooth, divergence-free, and periodic and satisfies the full equation in the Statement. Taking the $L^2$ inner product with $u$ gives $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }u{\mathrm{\parallel }}_2^2=-\int (u\cdot \mathrm{\nabla }u)\cdot ud\mu -\int \mathrm{\nabla }\pi \cdot ud\mu +\nu \int u\cdot \mathrm{\Delta }ud\mu$. The transport integral is $\frac{1}{2}\int u\cdot \mathrm{\nabla }\mathrm{\mid }u{\mathrm{\mid }}^{2}d\mu =0$, since it is the integral of a periodic divergence and $\mathrm{div}u=0$. Periodic integration by parts gives $\int \mathrm{\nabla }\pi \cdot ud\mu =-\int \pi \mathrm{div}ud\mu =0$ and $\int u\cdot \mathrm{\Delta }ud\mu =-\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2$. Hence $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }u{\mathrm{\parallel }}_2^2=-\nu \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2$.

Integrating the differential identity in step 1.1 from $0$ to $t$ and using $u(0)=u_0$ gives the asserted energy balance.

Integrating the full equation from step 1.1 over $Q_d$ shows that each component of ${\int }_{Q_d}ud\mu$ has derivative zero: the transport is $\mathrm{div}(u_{i}u)$, the pressure term is a periodic derivative, and the Laplacian term is a sum of periodic second derivatives. Since the initial mean is zero, $\bar{u}(t)=0$ throughout the lifespan.

If $\nu >0$, [F2] and step 1.1 imply $\frac{d}{dt}\mathrm{\parallel }u(t){\mathrm{\parallel }}_2^2\le -2\nu \mathrm{\parallel }u(t){\mathrm{\parallel }}_2^2$. Differentiating $e^{2\nu t}\mathrm{\parallel }u(t){\mathrm{\parallel }}_2^2$ therefore gives a nonpositive derivative, so $\mathrm{\parallel }u(t){\mathrm{\parallel }}_2^2\le e^{-2\nu t}\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_2^2$. Taking nonnegative square roots proves the stated decay.