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

==== Define the Leray projector PPP as the orthogonal projection in L2(R3;R3)L^2(\mathbb{R}^3;\mathbb{R}^3)L2(R3;R3) onto divergence-free vector fields. In Fourier variables it is the multiplier ====

(Pf)^(ξ)=(I−ξ⊗ξ∣ξ∣2)f^(ξ).\widehat{(P f)}(\xi) = \Big(I - \frac{\xi\otimes\xi}{|\xi|^2}\Big)\hat f(\xi).(Pf)(ξ)=(I−∣ξ∣2ξ⊗ξ)f^(ξ).
Properties used:
* PPP is linear and bounded on many function spaces (e.g. L2L^2L2, LpL^pLp for 1<p<∞1<p<\infty1<p<∞, Sobolev spaces HsH^sHs).
* P∇q=0P\nabla q = 0P∇q=0 for any scalar qqq (because ∇q^(ξ)=iξq^(ξ)\widehat{\nabla q}(\xi)=i\xi\widehat q(\xi)∇q(ξ)=iξq(ξ) and the multiplier annihilates iξi\xiiξ direction).
* PPP commutes with Δ\DeltaΔ and with the heat semigroup: PΔ=ΔP,  PeνtΔ=eνtΔPP\Delta=\Delta P,\; P e^{\nu t\Delta}=e^{\nu t\Delta}PPΔ=ΔP,PeνtΔ=eνtΔP.

Apply PPP to the PDE. Since P∇p=0P\nabla p=0P∇p=0 we obtain

P∂tu−νPΔu+P((u⋅∇)u)=0.P\partial_t u - \nu P\Delta u + P\big((u\cdot\nabla)u\big) = 0.P∂tu−νPΔu+P((u⋅∇)u)=0.
Because uuu is divergence-free and PPP equals identity on divergence-free fields, Pu= ⁣uPu=\!uPu=u. Also PPP commutes with ∂t\partial_t∂t and Δ\DeltaΔ when applied to sufficiently regular uuu. Thus we have the projected PDE

∂tu−νΔu+P((u⋅∇)u)=0.(PDEP)\partial_t u - \nu\Delta u + P\big((u\cdot\nabla)u\big) = 0.
\tag{PDE\(_P\)}∂tu−νΔu+P((u⋅∇)u)=0.(PDEP)
Write the nonlinear term as a divergence of a tensor:

(u⋅∇)u=∇ ⁣⋅(u⊗u),(u\cdot\nabla)u = \nabla\!\cdot(u\otimes u),(u⋅∇)u=∇⋅(u⊗u),
componentwise ((u⊗u)ij=uiuj)((u\otimes u)_{ij}=u_i u_j)((u⊗u)ij=uiuj) and (∇⋅(u⊗u))i=∑j∂j(uiuj)(\nabla\cdot(u\otimes u))_i = \sum_j \partial_j (u_i u_j)(∇⋅(u⊗u))i=∑j∂j(uiuj). Thus (PDEP_PP) becomes

∂tu−νΔu+P∇ ⁣⋅(u⊗u)=0.(1)\partial_t u - \nu\Delta u + P\nabla\!\cdot(u\otimes u) = 0.
\tag{1}∂tu−νΔu+P∇⋅(u⊗u)=0.(1)