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

=== Assistant: Short answer up front: the pressure is recovered by taking divergence of the Navier–Stokes equation and solving a Poiss… ===
Short answer up front: the pressure is recovered by taking divergence of the Navier–Stokes equation and solving a Poisson equation. Concretely, for a sufficiently regular, divergence-free velocity field uuu one gets

− Δp  =  ∑i,j∂i∂j(uiuj),-\,\Delta p \;=\; \sum_{i,j}\partial_i\partial_j\big(u_i u_j\big),−Δp=i,j∑∂i∂j(uiuj),
and a solution ppp is given (on R3\mathbb{R}^3R3) by convolution with the Newton kernel

p(x)  =  (Φ∗f)(x),Φ(x)=−14π∣x∣,f=∑i,j∂i∂j(uiuj).p(x)\;=\;(\Phi * f)(x),\qquad \Phi(x)=-\frac{1}{4\pi|x|},\quad f=\sum_{i,j}\partial_i\partial_j(u_i u_j).p(x)=(Φ∗f)(x),Φ(x)=−4π∣x∣1,f=i,j∑∂i∂j(uiuj).
This ppp satisfies −Δp=f-\Delta p=f−Δp=f (in the classical sense when uuu is smooth and decays, and in the distributional sense under weak hypotheses). Uniqueness is up to an additive constant.

Below is a short, step-by-step explanation.

==== Start from the incompressible Navier–Stokes 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,
and the incompressibility condition ∇⋅u=0\nabla\cdot u=0∇⋅u=0. Take divergence of the momentum equation:

∂t(∇⋅u)−νΔ(∇⋅u)+∇⋅((u⋅∇)u)+Δp=0.\partial_t(\nabla\cdot u)-\nu\Delta(\nabla\cdot u)+\nabla\cdot\big((u\cdot\nabla)u\big)+\Delta p = 0.∂t(∇⋅u)−νΔ(∇⋅u)+∇⋅((u⋅∇)u)+Δp=0.
Because ∇⋅u=0\nabla\cdot u=0∇⋅u=0 for all time, the first two terms vanish, leaving

Δp  =  − ∇⋅((u⋅∇)u).\Delta p \;=\; -\,\nabla\cdot\big((u\cdot\nabla)u\big).Δp=−∇⋅((u⋅∇)u).
Write the nonlinear term as a divergence of the tensor u⊗uu\otimes uu⊗u:

((u⋅∇)u)i=∑juj∂jui=∑j∂j(ujui),\big((u\cdot\nabla)u\big)_i=\sum_j u_j\partial_j u_i = \sum_j\partial_j(u_j u_i),((u⋅∇)u)i=j∑uj∂jui=j∑∂j(ujui),
so

∇⋅((u⋅∇)u)=∑i,j∂i∂j(uiuj).\nabla\cdot\big((u\cdot\nabla)u\big) = \sum_{i,j}\partial_i\partial_j(u_i u_j).∇⋅((u⋅∇)u)=i,j∑∂i∂j(uiuj).
Thus

−Δp=∑i,j∂i∂j(uiuj).-\Delta p = \sum_{i,j}\partial_i\partial_j\big(u_i u_j\big).−Δp=i,j∑∂i∂j(uiuj).
(That is the correct right-hand side — it involves the products uiuju_i u_juiuj, not a single uuu.)

==== Let ====

f(x):=∑i,j∂i∂j(uiuj)(x).f(x):=\sum_{i,j}\partial_i\partial_j\big(u_i u_j\big)(x).f(x):=i,j∑∂i∂j(uiuj)(x).
Assume uuu is smooth and decays sufficiently fast at infinity (for example u∈Cc∞(R3)u\in C^\infty_c(\mathbb{R}^3)u∈Cc∞(R3) or uuu Schwartz). The Newton potential Φ(x)=−14π∣x∣\Phi(x)=-\dfrac{1}{4\pi|x|}Φ(x)=−4π∣x∣1 satisfies −ΔΦ=δ-\Delta\Phi=\delta−ΔΦ=δ (Dirac). Define

p(x):=(Φ∗f)(x)=∫R3Φ(x−y) f(y) dy.p(x):=(\Phi*f)(x)=\int_{\mathbb{R}^3}\Phi(x-y)\,f(y)\,dy.p(x):=(Φ∗f)(x)=∫R3Φ(x−y)f(y)dy.
Standard properties of convolution with the Newton kernel give −Δp=f-\Delta p=f−Δp=f. In other words, convolution with Φ\PhiΦ inverts −Δ-\Delta−Δ on appropriate classes of right-hand sides fff. If uuu is smooth, then fff is smooth and the convolution yields a classical smooth ppp.

Uniqueness: if p1,p2p_1,p_2p1,p2 both solve −Δp=f-\Delta p=f−Δp=f on R3\mathbb{R}^3R3, then −Δ(p1−p2)=0-\Delta(p_1-p_2)=0−Δ(p1−p2)=0 and the only globally bounded harmonic functions are constants, so p1−p2p_1-p_2p1−p2 is constant. Hence ppp is unique up to an additive constant.

==== If uuu is not compactly supported but uiuj∈Lpu_i u_j\in L^puiuj∈Lp for some suitable ppp, you can interpret the equation in distributions and obtain ppp by either: ====
'' convolution p=Φ∗fp=\Phi''fp=Φ∗f in the sense of distributions (valid when fff is a tempered distribution), or
* the Riesz transform representation p=∑i,jRiRj(uiuj),p = \sum_{i,j} \mathcal{R}_i\mathcal{R}_j (u_i u_j),p=i,j∑RiRj(uiuj), where Rk\mathcal{R}_kRk are the Riesz transforms (Fourier multipliers with symbol iξk∣ξ∣\dfrac{i\xi_k}{|\xi|}∣ξ∣iξk). The Riesz transforms are bounded on LqL^qLq for 1<q<∞1<q<\infty1<q<∞, so if uiuj∈Lqu_i u_j\in L^quiuj∈Lq then p∈Lqp\in L^qp∈Lq too and the equality holds in LqL^qLq-sense.

These provide existence (and LpL^pLp-regularity) of ppp under standard hypotheses used in Navier–Stokes theory.

==== - On a bounded domain one solves a Poisson equation with appropriate boundary conditions (usually Neumann-type for pressure); existence then follows from elliptic theory (Fredholm alternative), again unique up to a constant. ====
* The pressure is always determined only up to an additive function of time; fixing a normalization (e.g. ∫Ωp=0\int_{\Omega} p = 0∫Ωp=0) removes that freedom.
* The key step is: divergence the PDE → get −Δp=-\Delta p =−Δp= known right-hand side → invert −Δ-\Delta−Δ (Newton potential or Riesz transforms).

If you want, I can (a) show the convolution step in full detail (verify −Δ(Φ∗f)=f-\Delta(\Phi*f)=f−Δ(Φ∗f)=f), (b) write the Riesz-transform proof with Fourier formulas, or (c) give the bounded-domain (Neumann problem) existence proof. Which would you prefer?