Editing Openai/6935580c-34e4-8001-9136-7bbdf1499790 (section)

==== Hypotheses. Let u(x,t)u(x,t)u(x,t) be a sufficiently smooth solution on [0,T]×R3[0,T]\times\mathbb{R}^3[0,T]×R3 (for example u∈C1([0,T];L2)∩C([0,T];H2)u\in C^1([0,T];L^2)\cap C([0,T];H^2)u∈C1([0,T];L2)∩C([0,T];H2); decays at infinity fast enough or use periodic/compact domain so boundary terms vanish). Let viscosity ν>0\nu>0ν>0. Assume ∇⋅u=0\nabla\cdot u=0∇⋅u=0. ====

Claim (energy equality).

12ddt∥u(t)∥L22+ν∥∇u(t)∥L22=0.\frac{1}{2}\frac{d}{dt}\|u(t)\|_{L^2}^2 + \nu\|\nabla u(t)\|_{L^2}^2 = 0.21dtd∥u(t)∥L22+ν∥∇u(t)∥L22=0.
Proof. Multiply the PDE by uuu and integrate over space:
# Start from the momentum equation

∂tu−νΔu+(u⋅∇)u+∇p=0.\partial_t u - \nu\Delta u + (u\cdot\nabla)u + \nabla p = 0.∂tu−νΔu+(u⋅∇)u+∇p=0.
# Take the L2L^2L2-inner product with uuu (i.e. integrate componentwise: ∫(⋯ )⋅u dx\int (\cdots)\cdot u\,dx∫(⋯)⋅udx). Each term:
* Time-derivative term: ∫(∂tu)⋅u dx=12ddt∫∣u∣2 dx=12ddt∥u∥L22,\int (\partial_t u)\cdot u\,dx = \tfrac12\frac{d}{dt}\int |u|^2\,dx = \tfrac12\frac{d}{dt}\|u\|_{L^2}^2,∫(∂tu)⋅udx=21dtd∫∣u∣2dx=21dtd∥u∥L22, because the map t↦∥u(t)∥L22t\mapsto \|u(t)\|^2_{L^2}t↦∥u(t)∥L22 is differentiable when u∈C1([0,T];L2)u\in C^1([0,T];L^2)u∈C1([0,T];L2) and the derivative equals 2⟨∂tu,u⟩2\langle \partial_t u,u\rangle2⟨∂tu,u⟩.
* Viscous term (integration by parts; boundary terms vanish by decay or periodicity): −ν∫(Δu)⋅u dx=ν∫∣∇u∣2 dx=ν∥∇u∥L22.-\nu\int (\Delta u)\cdot u\,dx = \nu\int |\nabla u|^2\,dx = \nu\|\nabla u\|_{L^2}^2.−ν∫(Δu)⋅udx=ν∫∣∇u∣2dx=ν∥∇u∥L22. (Use ∫(Δu)⋅u=−∫∇u:∇u\int (\Delta u)\cdot u = -\int \nabla u : \nabla u∫(Δu)⋅u=−∫∇u:∇u.)
* Pressure term: ∫(∇p)⋅u dx=−∫p (∇⋅u) dx=0,\int (\nabla p)\cdot u\,dx = -\int p\,(\nabla\cdot u)\,dx = 0,∫(∇p)⋅udx=−∫p(∇⋅u)dx=0, because ∇⋅u=0\nabla\cdot u=0∇⋅u=0.
* Nonlinear convection term: ∫(u⋅∇)u⋅u dx=∑i∑j∫uj∂jui ui dx=12∑j∫uj∂j(∣u∣2) dx=−12∑j∫(∂juj) ∣u∣2 dx=0,\int (u\cdot\nabla)u\cdot u\,dx = \sum_i\sum_j \int u_j \partial_j u_i \, u_i\,dx = \frac12\sum_j\int u_j \partial_j (|u|^2)\,dx = -\frac12\sum_j\int (\partial_j u_j)\,|u|^2\,dx = 0,∫(u⋅∇)u⋅udx=i∑j∑∫uj∂juiuidx=21j∑∫uj∂j(∣u∣2)dx=−21j∑∫(∂juj)∣u∣2dx=0, again because ∇⋅u=0\nabla\cdot u=0∇⋅u=0 (or integrate by parts, boundary terms vanish).
# Summing terms yields exactly

12ddt∥u∥L22+ν∥∇u∥L22=0.\frac12\frac{d}{dt}\|u\|_{L^2}^2 + \nu\|\nabla u\|_{L^2}^2 = 0.21dtd∥u∥L22+ν∥∇u∥L22=0.
This is rigorous whenever the above integrals and integrations by parts are justified — i.e. uuu is smooth enough and decays or the domain has suitable boundary conditions.