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

=== I will work on R3\mathbb{R}^3R3 (the proof on the torus T3\mathbb{T}^3T3 is the same but with Fourier series). State the incompressible Navier–Stokes system ===

{∂tu−νΔu+(u⋅∇)u+∇p=0,x∈R3, t>0,∇⋅u=0,x∈R3, t≥0,u(x,0)=u0(x),x∈R3,\begin{cases}
\partial_t u - \nu\Delta u + (u\cdot\nabla)u + \nabla p = 0, & x\in\mathbb{R}^3,\ t>0,\\[4pt]
\nabla\cdot u = 0, & x\in\mathbb{R}^3,\ t\ge0,\\[4pt]
u(x,0)=u_0(x), & x\in\mathbb{R}^3,
\end{cases}⎩⎨⎧∂tu−νΔu+(u⋅∇)u+∇p=0,∇⋅u=0,u(x,0)=u0(x),x∈R3, t>0,x∈R3, t≥0,x∈R3,
with viscosity ν>0\nu>0ν>0 and divergence-free initial datum ∇⋅u0=0\nabla\cdot u_0=0∇⋅u0=0. Assume sufficient regularity (classical solution or at least u∈C([0,T];X)∩C1((0,T];X′)u\in C([0,T];X)\cap C^1((0,T];X')u∈C([0,T];X)∩C1((0,T];X′) for suitable spaces XXX) so that the computations below are justified. I will prove:
# Any sufficiently smooth solution of the PDE satisfies the mild (integral) equation

u(t)=eνtΔu0−∫0teν(t−s)ΔP∇ ⁣⋅(u⊗u)(s) ds,(M)u(t)=e^{\nu t\Delta}u_0 - \int_0^t e^{\nu (t-s)\Delta} P\nabla\!\cdot (u\otimes u)(s)\,ds,
\tag{M}u(t)=eνtΔu0−∫0teν(t−s)ΔP∇⋅(u⊗u)(s)ds,(M)
where eνtΔe^{\nu t\Delta}eνtΔ is the heat semigroup and PPP is the Leray projection onto divergence-free vector fields.
# Conversely, any sufficiently regular function uuu solving (M) is a (classical) solution of the PDE; the pressure ppp is recovered up to an additive function of ttt by solving a Poisson equation.

Throughout I denote by GtG_tGt the heat semigroup: Gtf=eνtΔf=Kt∗fG_t f = e^{\nu t\Delta} f = K_t * fGtf=eνtΔf=Kt∗f where KtK_tKt is the heat kernel Kt(x)=(4πνt)−3/2e−∣x∣2/(4νt)K_t(x)=(4\pi\nu t)^{-3/2}e^{-|x|^2/(4\nu t)}Kt(x)=(4πνt)−3/2e−∣x∣2/(4νt).

==== 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)

==== Equation (1) is linear in uuu with a known inhomogeneity −P∇ ⁣⋅(u⊗u)-P\nabla\!\cdot(u\otimes u)−P∇⋅(u⊗u). Treat it as ====

∂tu−νΔu=−F(t),F(t):=P∇ ⁣⋅(u⊗u)(t).\partial_t u - \nu\Delta u = -F(t),\qquad F(t):=P\nabla\!\cdot(u\otimes u)(t).∂tu−νΔu=−F(t),F(t):=P∇⋅(u⊗u)(t).
For the linear homogeneous heat equation ∂tv−νΔv=0\partial_t v - \nu\Delta v = 0∂tv−νΔv=0 with initial data v(0)=v0v(0)=v_0v(0)=v0, the solution is v(t)=Gtv0v(t)=G_t v_0v(t)=Gtv0. For the inhomogeneous equation, the Duhamel principle (variation-of-constants formula) gives the unique solution (for sufficiently regular FFF) as

u(t)=Gtu0+∫0tGt−s(−F(s)) ds=Gtu0−∫0tGt−sF(s) ds.u(t) = G_t u_0 + \int_0^t G_{t-s}\big(-F(s)\big)\,ds
= G_t u_0 - \int_0^t G_{t-s} F(s)\,ds.u(t)=Gtu0+∫0tGt−s(−F(s))ds=Gtu0−∫0tGt−sF(s)ds.
Substituting F=P∇ ⁣⋅(u⊗u)F= P\nabla\!\cdot(u\otimes u)F=P∇⋅(u⊗u) yields exactly the mild formula

u(t)=eνtΔu0−∫0teν(t−s)ΔP∇ ⁣⋅(u⊗u)(s) ds.u(t)=e^{\nu t\Delta}u_0 - \int_0^t e^{\nu (t-s)\Delta} P\nabla\!\cdot (u\otimes u)(s)\,ds.u(t)=eνtΔu0−∫0teν(t−s)ΔP∇⋅(u⊗u)(s)ds.
This derivation is valid provided:
'' the convolution Gt−s∗F(s)G_{t-s} '' F(s)Gt−s∗F(s) is defined (e.g. F(s)F(s)F(s) lies in a space where Gt−sG_{t-s}Gt−s acts boundedly), and
* we may interchange integration in time with spatial convolution (standard if FFF is continuous into a Banach space).

Thus any sufficiently regular solution of the PDE satisfies (M). This completes the PDE ⇒\Rightarrow⇒ mild implication.

==== Now prove the converse: if uuu satisfies the mild equation (with enough regularity), then it satisfies the PDE and we can recover a pressure ppp. ====

Assume uuu is such that:
* u∈C([0,T];X)∩C1((0,T];X′)u\in C([0,T];X)\cap C^1((0,T];X')u∈C([0,T];X)∩C1((0,T];X′) for spaces where eνtΔe^{\nu t\Delta}eνtΔ and PPP act and the time integral is differentiable (standard choices: X=HsX=H^sX=Hs with large sss, or X=Cb2X=C^2_bX=Cb2, etc.),
* the integral in (M) is differentiable in ttt (this follows under usual smoothing and integrability hypotheses).

Differentiate (M) in ttt. Use facts:
* ∂t(eνtΔu0)=νΔeνtΔu0\partial_t (e^{\nu t\Delta}u_0)=\nu\Delta e^{\nu t\Delta}u_0∂t(eνtΔu0)=νΔeνtΔu0,
* for the integral term, differentiate under the integral sign and use the semigroup property: ddt∫0teν(t−s)ΔF(s) ds=F(t)+∫0tνΔeν(t−s)ΔF(s) ds.\frac{d}{dt}\int_0^t e^{\nu (t-s)\Delta} F(s)\,ds = F(t) + \int_0^t \nu\Delta e^{\nu (t-s)\Delta} F(s)\,ds.dtd∫0teν(t−s)ΔF(s)ds=F(t)+∫0tνΔeν(t−s)ΔF(s)ds.

Apply this with F(s)=P∇ ⁣⋅(u⊗u)(s)F(s)=P\nabla\!\cdot(u\otimes u)(s)F(s)=P∇⋅(u⊗u)(s). Differentiating (M) gives

∂tu(t)=νΔeνtΔu0−P∇ ⁣⋅(u⊗u)(t)−∫0tνΔeν(t−s)ΔP∇ ⁣⋅(u⊗u)(s) ds.\partial_t u(t)
= \nu\Delta e^{\nu t\Delta}u_0
* P\nabla\!\cdot(u\otimes u)(t) - \int_0^t \nu\Delta e^{\nu (t-s)\Delta} P\nabla\!\cdot(u\otimes u)(s)\,ds.∂tu(t)=νΔeνtΔu0−P∇⋅(u⊗u)(t)−∫0tνΔeν(t−s)ΔP∇⋅(u⊗u)(s)ds.
Group terms:

∂tu(t)−νΔu(t)=−P∇ ⁣⋅(u⊗u)(t).\partial_t u(t) - \nu\Delta u(t)
= -P\nabla\!\cdot(u\otimes u)(t).∂tu(t)−νΔu(t)=−P∇⋅(u⊗u)(t).
But this is exactly the projected PDE (1). Since PPP annihilates gradients, applying (I−P)(I-P)(I−P) to the PDE recovers the pressure gradient term: start from the original PDE form

∂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.
We already have projection PPP of this equation satisfied. To obtain ∇p\nabla p∇p explicitly, take divergence of the PDE (classically permitted for sufficiently smooth uuu):

∇⋅((u⋅∇)u+∇p)=0\nabla\cdot\big( (u\cdot\nabla)u + \nabla p \big) = 0∇⋅((u⋅∇)u+∇p)=0
because ∇⋅(∂tu−νΔu)=∂t(∇⋅u)−νΔ(∇⋅u)=0\nabla\cdot(\partial_t u - \nu\Delta u)=\partial_t(\nabla\cdot u)-\nu\Delta(\nabla\cdot u)=0∇⋅(∂tu−νΔu)=∂t(∇⋅u)−νΔ(∇⋅u)=0 when ∇⋅u=0\nabla\cdot u=0∇⋅u=0 initially and evolution preserves it (or check separately). Use ∇⋅((u⋅∇)u)=∑i,j∂i∂j(uiuj)\nabla\cdot((u\cdot\nabla)u)=\sum_{i,j}\partial_i\partial_j (u_i u_j)∇⋅((u⋅∇)u)=∑i,j∂i∂j(uiuj). Thus −Δp=∇⋅∇⋅(u⊗u)-\Delta p = \nabla\cdot\nabla\cdot(u\otimes u)−Δp=∇⋅∇⋅(u⊗u), i.e.

−Δp=∑i,j∂i∂j(uiuj).(Poisson)-\Delta p = \sum_{i,j}\partial_i\partial_j(u_i u_j).
\tag{Poisson}−Δp=i,j∑∂i∂j(uiuj).(Poisson)
Solve the Poisson equation for p(⋅,t)p(\cdot,t)p(⋅,t) (up to additive function of ttt) by convolution with the Newton potential: for each fixed ttt,

p(⋅,t)=(−Δ)−1∇⋅∇⋅(u⊗u)(⋅,t)=RiRj(uiuj),p(\cdot,t) = (-\Delta)^{-1}\nabla\cdot\nabla\cdot (u\otimes u)(\cdot,t)
= \mathcal{R}_i\mathcal{R}_j (u_i u_j),p(⋅,t)=(−Δ)−1∇⋅∇⋅(u⊗u)(⋅,t)=RiRj(uiuj),
where Rk\mathcal{R}_kRk are Riesz transforms. This recovers ppp (up to function of time) and shows that

∇p=−(I−P)∇ ⁣⋅(u⊗u),\nabla p = -(I-P)\nabla\!\cdot(u\otimes u),∇p=−(I−P)∇⋅(u⊗u),
so the full PDE

∂tu−νΔu+∇p+∇ ⁣⋅(u⊗u)=0\partial_t u - \nu\Delta u + \nabla p + \nabla\!\cdot(u\otimes u)=0∂tu−νΔu+∇p+∇⋅(u⊗u)=0
holds. Since P∇p=0P\nabla p=0P∇p=0, the splitting P∇ ⁣⋅(u⊗u)+(I−P)∇ ⁣⋅(u⊗u)=∇ ⁣⋅(u⊗u)P\nabla\!\cdot(u\otimes u) + (I-P)\nabla\!\cdot(u\otimes u)=\nabla\!\cdot(u\otimes u)P∇⋅(u⊗u)+(I−P)∇⋅(u⊗u)=∇⋅(u⊗u) recovers the original nonlinear term plus the pressure.

Hence a sufficiently regular mild solution satisfies the PDE with pressure given by the solution of the Poisson equation above.

==== Summarize the equivalence: ====
* If uuu is a sufficiently smooth (classical) solution of the incompressible Navier–Stokes PDE with divergence-free initial data, then uuu satisfies the mild equation (M).
* Conversely, if uuu satisfies (M) and is regular enough to differentiate in time and to apply spatial differential operators and Riesz transforms, then uuu is a classical solution of the PDE; the pressure is recovered by solving −Δp=∂i∂j(uiuj)-\Delta p = \partial_i\partial_j (u_i u_j)−Δp=∂i∂j(uiuj).

Remarks on hypotheses: the equivalence requires assumptions that permit the manipulations above (application of PPP, differentiation under the integral sign, convolution with the heat kernel, use of Riesz transforms). Typical functional settings where everything is justified:
* u∈C([0,T];Hs(R3))∩C1((0,T);Hs−2(R3))u\in C([0,T];H^s(\mathbb{R}^3))\cap C^1((0,T);H^{s-2}(\mathbb{R}^3))u∈C([0,T];Hs(R3))∩C1((0,T);Hs−2(R3)) with s>52s>\tfrac{5}{2}s>25, or
* uuu smooth and rapidly decaying (Schwartz) in space for each t>0t>0t>0.

Under such hypotheses the derivation is rigorous.

==== The mild formulation is extremely useful because it converts the PDE into an integral fixed-point equation (Picard iteration / contraction mapping), and because the heat semigroup eνtΔe^{\nu t\Delta}eνtΔ provides explicit smoothing estimates used to control the nonlinear term. The steps above show precisely how the pressure is eliminated (via projection), how Duhamel’s formula is derived, and how to reconstruct the pressure from a mild solution. ====