Worked E0 check: constant-coefficient transport on the three-torus, explicit solution and conserved energy

(The explicit function solves the system.) Define $u(t,x):=u_0(x^1-2t,x^2,x^3)$. Then ${\mathrm{\partial }}_{t}u(t,x)=-2({\mathrm{\partial }}_1{u}_0)(x^1-2t,x^2,x^3)$ and ${\mathrm{\partial }}_{1}u(t,x)=({\mathrm{\partial }}_1{u}_0)(x^1-2t,x^2,x^3)$, so ${\mathrm{\partial }}_{t}u+2{\mathrm{\partial }}_{1}u=0$ pointwise; $u(0,\cdot )=u_0$, and $u$ is classical because integer $s\ge 5$ and inherits the $H^s$-bound of $u_0$ at every time, with higher smoothness whenever $u_0$ is smooth.

(Energy identity.) $e(t)=\int u(t{)}^{2}d\mu =\int u_0^{2}d\mu =e(0)$ by translation invariance [L3]; equivalently, differentiating under the integral and using the equation and [L3], $e^{\mathrm{\prime }}(t)=2\int u{\mathrm{\partial }}_{t}ud\mu =-4\int u{\mathrm{\partial }}_{1}ud\mu =-2\int {\mathrm{\partial }}_1(u^2)d\mu =0$.

(Identification with the E0 solution.) By [L2] there is a unique solution through $u_0$; the explicit function of step 1.1 is one, so the E0 solution is $u(t,x)=u_0(x^1-2t,x^2,x^3)$ and its energy is conserved. For the actual Galerkin energy, set $u_N(t)=P_{N}u(t)$, equivalently $u_N(t,x)=(P_N{u}_0)(x^1-2t,x^2,x^3)$. By [L4] this smooth finite Fourier polynomial solves ${\mathrm{\partial }}_t{u}_N+2{\mathrm{\partial }}_1{u}_N=0$ with datum $P_N{u}_0$, hence its projected equation. Each Fourier coefficient obeys ${\mathrm{\partial }}_t{\widehat{u}}_N(k)=-2i{k}_1{\widehat{u}}_N(k)$, so this is exactly the Galerkin trajectory. Every $D^{\alpha }{u}_N$ is the translate of $D^{\alpha }{P}_N{u}_0$; translation invariance gives $e_N(t)={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le s}\mathrm{\parallel }{D}^{\alpha }{P}_N{u}_0{\mathrm{\parallel }}_2^2=e_N(0)$ and $e_N^{\mathrm{\prime }}(t)=0$. Thus the actual energy inequality is verified with $C=0$, at every cutoff $N$, as well as the zero-order identity of step 2.1.