gpt 6 astra ultra attempt on Math-491

Liouville problem for stationary Navier–Stokes D-solutions in R^3

Definitions and mathematical statement

Let $u:{\mathbb{R}}^3\to{\mathbb{R}}^3$ and $p:{\mathbb{R}}^3\to{\mathbb{R}}$ be smooth and satisfy the stationary unforced system with viscosity normalized to one: \[ -\Delta u+(u\cdot\nabla)u+\nabla p=0, \qquad\nabla\cdot u=0. \] Assume finite Dirichlet integral and uniform vanishing at infinity: \[ \int_{{\mathbb{R}}^3}|\nabla u|^2{\,\mathrm{d}} x<\infty, \qquad \lim_{R\to\infty}\sup_{|x|\ge R}|u(x)|=0. \] Must $u\equiv0$? The finite-energy condition here is on the gradient, not the kinetic energy $\int|u|^2$, which is not an extra assumption. If the conclusion holds, the equation also makes the pressure constant. The spatial domain has no obstacle or physical boundary supplying energy.

Formulation and concept references: [S1].

Short English statement

Must every smooth decaying stationary Navier--Stokes flow on all of three-dimensional space vanish when its gradient has finite square integral?

Research attempt

Scope, source audit, and attack (27 September 2026).

This GPT-6 Astra Ultra attempt concerns stationary solutions on all of $\mathbb R^3$, with the two hypotheses printed above. In particular, $\nabla u\in L^2$ is given, whereas $u\in L^2$ is not. There is no boundary on which one can prescribe a zero energy flux. The conclusion must hold without a quantitative rate in the uniform decay assumption.

The supplied primary source [S1], Introduction and Theorems 1.1--1.2, describes the $D$-solution problem and proves results with additional control on a velocity antiderivative or on line integrals. Those are conditional hypotheses, not consequences of the definition of a $D$-solution. Its introduction also identifies the classical sufficient condition $u\in L^{9/2}$, which is rederived below as a benchmark. Chae's 2026 paper likewise proves conditional results involving head pressure decay and records the remaining three-dimensional difficulty [E1]. No unconditional result is inferred from either title.

The route investigated here is the direct energy argument: normalize the pressure, retain every term in the cutoff identity, and determine exactly which terms are controlled by the stated assumptions. The proof track reaches a complete theorem with the extra $L^{9/2}$ hypothesis. The counterexample track constructs a smooth divergence-free field satisfying the functional hypotheses but violating that extra integrability. It is explicitly checked not to solve the equation, so it invalidates a proposed functional shortcut rather than the conjecture.

Step 1: what the Dirichlet hypothesis actually gives.

Lemma 487.1. Under the stated hypotheses, \[ u\in L^6(\mathbb R^3)\cap L^\infty(\mathbb R^3), \qquad \|u\|_6\leq C\|\nabla u\|_2. \]

Proof. Smoothness on compact sets and uniform decay outside large balls give boundedness. For $\varepsilon>0$, the Sobolev function $v_\varepsilon=(|u|-\varepsilon)_+$ has compact support, by uniform decay. The chain rule gives $|\nabla v_\varepsilon|\leq|\nabla u|$ almost everywhere. Apply the ordinary three-dimensional Sobolev inequality to its smooth approximations to obtain $\|v_\varepsilon\|_6\leq C\|\nabla u\|_2$. Let $\varepsilon\downarrow0$ and use monotone convergence. This also explains why a nonzero additive constant, permitted by a homogeneous gradient norm alone, is excluded here.$\square$

It follows by boundedness that $u\in L^q$ for every $q\geq6$, since $\int|u|^q\leq\|u\|_\infty^{q-6}\int|u|^6$. There is no reverse inclusion into $L^{9/2}$ on a space of infinite measure. The tempting interpolation with $L^2$ would require an assumption absent from the problem.

We use the standard $L^q$ boundedness of the Riesz transforms: $\|R_iR_j f\|_q\leq C_q\|f\|_q$ for $1<q<\infty$. The multiplier is $-\xi_i\xi_j/|\xi|^2$. This is the Calderón--Zygmund estimate used for pressure in [E1], equation $2.30$. Its boundedness is a standard analytic input, not a new assertion proved in this notebook. All exponents at which it is used are strictly between one and infinity.

Step 2: pressure normalization without a hidden decay assumption.

Define \begin{equation}\label{wsref-0-72-61-34-38-37-3a-70-72-65-73-73-75-72-65} p_0=\sum_{i,j=1}^3R_iR_j(u_i u_j). \end{equation} Since $u\in L^6$, this is a well-defined $L^3$ function and \[ \|p_0\|_3\leq C\|u\|_6^2. \] One must still check that the pressure appearing in the equation differs from $p_0$ by only a constant; simply solving the pressure Poisson equation leaves a possible harmonic term.

Lemma 487.2. The given smooth pressure satisfies $p=p_0+c$ as a distribution, for a constant $c$.

Proof. Taking divergence of the stationary equation and using $\nabla\cdot u=0$ gives \[ -\Delta p=\sum_{i,j}\partial_i\partial_j(u_i u_j) =-\Delta p_0. \] Let $w=\nabla(p-p_0)$. The equation also writes \[ w=\Delta u-\operatorname{div}(u\otimes u)-\nabla p_0. \] The right side is tempered because $u\in L^6$ and $u\otimes u,p_0\in L^3$. Moreover $\Delta w=0$. A tempered harmonic distribution is a harmonic polynomial: its Fourier transform is supported at the origin and is a finite linear combination of derivatives of the delta distribution. We use this basic distribution fact for each component.

For any $\phi\in C_c^\infty(\mathbb R^3)$, put $\phi_R(x)=R^{-3}\phi(x/R)$. Distributional differentiation and Hölder give \begin{align*} |\langle\Delta u,\phi_R\rangle| &\leq C_\phi R^{-5/2}\|u\|_6,\\ |\langle\operatorname{div}(u\otimes u),\phi_R\rangle| +|\langle\nabla p_0,\phi_R\rangle| &\leq C_\phi R^{-2}(\|u\|_6^2+\|p_0\|_3). \end{align*} For example, the $L^{6/5}$ norm of $\Delta\phi_R$ is $R^{-5+3(5/6)}\|\Delta\phi\|_{6/5}=R^{-5/2}\|\Delta\phi\|_{6/5}$; the $L^{3/2}$ norm of $\nabla\phi_R$ is $R^{-2}$ times a fixed constant. Thus $\langle w,\phi_R\rangle\to0$ for every such $\phi$.

If a component of the polynomial $w$ were nonzero, pairing it with $\phi_R$ would be a polynomial in $R$. The coefficients are its homogeneous polynomial pieces paired with $\phi$. The limit zero for every $\phi$ forces every coefficient to vanish, including the constant one. Hence $w=0$. A distribution with zero gradient on connected $\mathbb R^3$ is constant, proving the claim.$\square$

This argument does not assume that a general smooth $L^3$ function has a uniform pointwise limit at infinity. That implication would be false without further control of possible narrow peaks. The normalization needed below is the integral normalization $p_0\in L^3$.

Step 3: the exact cutoff identity and its unresolved term.

Fix $\phi\in C_c^\infty(\mathbb R^3)$ with $0\leq\phi\leq1$, $\phi=1$ on the unit ball and support in the ball of radius two. Write $\phi_R(x)=\phi(x/R)$ in this paragraph, and $A_R=\{R<|x|<2R\}$. This cutoff differs from the normalized test function in Lemma 487.2. Multiplying the equation by $u\phi_R$ and integrating gives \begin{equation}\label{wsref-0-72-61-34-38-37-3a-65-6e-65-72-67-79} \int |\nabla u|^2\phi_R =\frac12\int |u|^2\Delta\phi_R +\int\left(\frac{|u|^2}{2}+p_0\right)u\cdot\nabla\phi_R. \end{equation} All terms have compact support or a compactly supported derivative, so these integrations require no unproved global integrability of $|u|^2$. For the viscous term use $\nabla(|u|^2/2)=u\cdot\nabla u$. For convection use $(u\cdot\nabla)u\cdot u=u\cdot\nabla(|u|^2/2)$ and integrate the divergence. The pressure term integrates to $-\int p\,u\cdot\nabla\phi_R$; its constant part vanishes because $\int u\cdot\nabla\phi_R=0$.

Set $U_R=\|u\|_{L^6(A_R)}$ and $P_R=\|p_0\|_{L^3(A_R)}$. The first term is controlled by \begin{equation}\label{wsref-0-72-61-34-38-37-3a-64-69-66-66-75-73-69-6f-6e} R^{-2}\int_{A_R}|u|^2 \leq C R^{-2}|A_R|^{2/3}U_R^2 \leq C U_R^2\longrightarrow0. \end{equation} The two remaining absolute-value estimates are \begin{align} R^{-1}\int_{A_R}|u|^3 &\leq C R^{1/2}U_R^3,\label{wsref-0-72-61-34-38-37-3a-63-6f-6e-76-65-63-74-69-6f-6e}\\ R^{-1}\int_{A_R}|p_0||u| &\leq C R^{1/2}P_R U_R.\label{wsref-0-72-61-34-38-37-3a-70-72-65-73-73-75-72-65-66-6c-75-78} \end{align} Here the residual volume exponent is $1-3/6=1/2$ for the first integral and $1-1/3-1/6=1/2$ for the second. Both $U_R$ and $P_R$ tend to zero, but no rate has been proved that defeats the factor $R^{1/2}$.

This is the exact point where testing by $u$ without a cutoff would conceal a gap. Formally discarding the last integral in (ra487:energy) amounts to assuming that the energy flux through large scales vanishes. The given gradient energy makes the left side converge to the total Dirichlet integral by dominated convergence; it does not, by itself, justify discarding that flux.

There is a precise conditional criterion even at the present exponents: if a sequence $R_j\to\infty$ satisfies \begin{equation}\label{wsref-0-72-61-34-38-37-3a-74-61-69-6c-63-72-69-74-65-72-69-6f-6e} R_j^{1/2}\left(U_{R_j}^3+P_{R_j}U_{R_j}\right)\longrightarrow0, \end{equation} then (ra487:energy) implies $\nabla u=0$, and uniform decay implies $u=0$. This is a proved sufficient condition, not an established property of every $D$-solution. Absolute estimates are only one possible way to prove zero flux; a successful argument could instead exploit cancellation in the signed integral.

A complete benchmark: the extra $L^{9/2}$ hypothesis.

Proposition 487.3. If the solution in the problem also satisfies $u\in L^{9/2}(\mathbb R^3)$, then $u\equiv0$ and $p$ is constant.

Proof. The Riesz transform estimate now gives $p_0\in L^{9/4}$ with $\|p_0\|_{9/4}\leq C\|u\|_{9/2}^2$. The definitions of the Riesz transform on $L^3\cap L^{9/4}$ agree, by approximation with smooth compactly supported functions, so this is the same pressure as in Lemma 487.2. Writing $V_R=\|u\|_{L^{9/2}(A_R)}$ and $Q_R=\|p_0\|_{L^{9/4}(A_R)}$, Hölder gives \begin{align*} R^{-1}\int_{A_R}|u|^3 &\leq R^{-1}|A_R|^{1/3}V_R^3\leq C V_R^3,\\ R^{-1}\int_{A_R}|p_0||u| &\leq R^{-1}|A_R|^{1-4/9-2/9}Q_R V_R \leq C Q_R V_R. \end{align*} Both terms tend to zero because the global integrals defining these norms are finite. The diffusion term tends to zero by (ra487:diffusion). Let $R\to\infty$ in (ra487:energy); it follows that $\int|\nabla u|^2=0$. Smoothness and connectedness imply that $u$ is constant, and decay makes the constant zero. Substitution in the equation gives $\nabla p=0$. $\square$

The special exponent comes from the elementary expression \[ R^{-1}|A_R|^{1-3/q}\asymp R^{2-9/q}. \] It has exponent zero at $q=9/2$, positive exponent at $q=6$, and negative exponent for $3\leq q<9/2$. With the corresponding pressure space $L^{q/2}$, the same proof works in that latter range. This is criticality for this particular cutoff estimate; it should not be confused with the scaling-critical Navier--Stokes velocity space $L^3$. The $L^{9/2}$ theorem is known, as explicitly recorded in [S1]; the purpose of the reconstruction is to locate exactly the missing decay.

A functional counterexample to the proposed integrability upgrade.

For $1/2<\alpha\leq2/3$, put $\beta=(\alpha+1)/2$ and define \begin{equation}\label{wsref-0-72-61-34-38-37-3a-73-77-69-72-6c} U_\alpha(x)=(1+|x|^2)^{-\beta}(-x_2,x_1,0). \end{equation} This field is smooth everywhere. Its divergence vanishes: the radial gradient of $(1+|x|^2)^{-\beta}$ is orthogonal to $(-x_2,x_1,0)$, whose own divergence is zero. Also $|U_\alpha(x)|\leq C(1+|x|)^{-\alpha}$, so it vanishes uniformly. Direct differentiation gives $|\nabla U_\alpha(x)|\leq C(1+|x|)^{-\alpha-1}$ and consequently \[ \int_{|x|>1}|\nabla U_\alpha|^2 \leq C\int_1^\infty r^{-2\alpha}\,\mathrm dr<\infty. \] On a fixed angular belt away from the $x_3$ axis, $|U_\alpha|$ is bounded above and below by positive constants times $r^{-\alpha}$. In fact the angular factor is $\sin\theta$ and its $q$th power has a positive finite integral over the sphere. Hence \begin{equation}\label{wsref-0-72-61-34-38-37-3a-73-77-69-72-6c-69-6e-74-65-67-72-61-62-69-6c-69-74-79} U_\alpha\in L^q(\mathbb R^3) \quad\Longleftrightarrow\quad q\alpha>3 \qquad(q>0). \end{equation} For $\alpha=3/5$, this gives $U_\alpha\in L^6$ but $U_\alpha\notin L^{9/2}$ and $U_\alpha\notin L^2$. Moreover \[ R^{-1}\int_{A_R}|U_{3/5}|^3\asymp R^{1/5}. \] Thus even smoothness, divergence freedom, uniform decay, finite Dirichlet integral and the resulting $L^6$ bound together cannot force the absolute convection estimate to vanish.

This is not a stationary Navier--Stokes counterexample. To verify that point rather than leave it implicit, write $f(r)=(1+r^2)^{-\beta}$. Then \[ \Delta U_\alpha=g(r)(-x_2,x_1,0),\qquad g(r)=f''(r)+\frac4r f'(r) =-2\beta(1+r^2)^{-\beta-2} \left(5+(2-\alpha)r^2\right), \] where the expression at zero is understood by continuity. Also $(U_\alpha\cdot\nabla)U_\alpha=-f(r)^2(x_1,x_2,0)$. If some smooth pressure made the stationary equation hold, its gradient would equal $g(r)(-x_2,x_1,0)+f(r)^2(x_1,x_2,0)$. The integral of this vector field around a circle of radius $r>0$ in the plane $x_3=0$ is $2\pi r^2g(r)$, which is nonzero. A gradient has zero integral around a closed curve, a contradiction.

For radial cutoffs the signed convection term of this example is actually zero because $U_\alpha\cdot\nabla\phi_R=0$. Thus the example has a second lesson: absolute estimates may lose essential cancellation. It disproves a functional implication used by a naive proof; it does not show that the true stationary energy flux is nonzero for an actual $D$-solution.

Attempt through head pressure: what the sign does and does not give.

Let $Q=p_0+|u|^2/2$ and $\omega=\nabla\times u$. A calculation using the equation and the pressure Poisson identity yields \begin{equation}\label{wsref-0-72-61-34-38-37-3a-62-65-72-6e-6f-75-6c-6c-69} -\Delta Q+u\cdot\nabla Q=-|\omega|^2. \end{equation} Indeed $\Delta(|u|^2/2)=|\nabla u|^2+u\cdot\nabla Q$, while $\Delta p_0=-\sum_{i,j}\partial_i u_j\partial_j u_i$; subtracting gives $\Delta Q-u\cdot\nabla Q=|\omega|^2$. This identity motivates head-pressure methods such as [E1], but does not by itself eliminate the right side.

Even if one has additionally justified the pointwise normalization $Q(x)\to0$, the maximum principle gives only $Q\leq0$. For example, the nonzero scalar function $q(x)=-(1+|x|^2)^{-1/2}$ satisfies \[ q(x)\to0,\qquad -\Delta q=-3(1+|x|^2)^{-5/2}<0. \] It is therefore false that a decaying nonpositive solution of an equation of the form $-\Delta q=-f$, $f\geq0$, must vanish. This scalar test is not asserted to arise from a Navier--Stokes pair; it pinpoints the missing information in using the maximum principle alone. A proof must couple (ra487:bernoulli) to the velocity equation and obtain an additional integral or level-set estimate. No such unconditional estimate is established here.

Verification and outcome.

The companion exact-arithmetic script records the exponents $2-9/q$, the integrability thresholds $q\alpha>3$, and the coefficients in $g(r)$ for rational test parameters, including $\alpha=3/5$. Those finite algebra checks help catch a wrong sign or scale; they are not numerical evidence for an infinite-domain Liouville theorem. The mathematical verification above separately checks pressure constants, compact support of each test, the zero mode in the harmonic pressure argument, and the difference between a manufactured field and a solution.

Outcome: UNRESOLVED. The proved results are the $L^6$ consequence of the hypotheses, the canonical pressure normalization, the exact energy identity, the tail criterion (ra487:tailcriterion), and the known $L^{9/2}$ sufficient condition. The functional counterexample explains why an integrability upgrade cannot be made without using further content of the stationary equation.

The first unresolved step for this route is proving, from the original hypotheses alone, that the signed flux $\int (|u|^2/2+p_0)u\cdot\nabla\phi_R$ tends to zero along some sequence of radii. A stronger sufficient goal is (ra487:tailcriterion); another is $u\in L^{9/2}$. Neither is proved for arbitrary $D$-solutions. Concrete continuations are to exploit cancellation between pressure and convection, to control low-frequency velocity tails using the equation, or to obtain a head-pressure level-set estimate strong enough to close (ra487:bernoulli). None has been completed here, and no counterexample satisfying the stationary system has been produced.

Sources

• S1 Coiculescu and Jincheng Yang, Conditional Liouville Theorems for the Navier–Stokes Equations, arXiv v1 $2025$. https://arxiv.org/html/2506.14533v1. Formulation source.
• E1 D. Chae, Liouville Type Theorems for the Stationary Navier--Stokes Equations in $\mathbb R^3$, Communications in Mathematical Physics 407 $2026$, article 53, Introduction and equation $2.30$. https://link.springer.com/article/10.1007/s00220-026-05555-y. Primary published source, 5 February 2026; conditional results and pressure estimates verified 27 September 2026.