Liouville type theorems for the steady axially symmetric Navier-Stokes and magnetohydrodynamic equations

In this paper we study Liouville properties of smooth steady axially symmetric solutions of the Navier-Stokes equations. First, we provide another version of the Liouville theorem of \cite{kpr15} in the case of zero swirl, where we replaced the Dirichlet integrability condition by mild decay conditions. Then we prove some Liouville theorems under the assumption $\|\f{u_r}{r}{\bf 1}_{\{u_r<-\f 1r\}}\|_{L^{3/2}(\mbR^3)}<C_{\sharp}$ where $C_{\sharp}$ is a universal constant to be specified. In particular, if $u_r(r,z)\geq -\f1r$ for $\forall (r,z)\in[0,\oo)\times\mbR$, then ${\bf u}\equiv 0$. Liouville theorems also hold if $\displaystyle\lim_{|x|\to \oo}\Ga =0$ or $\Ga\in L^q(\mbR^3)$ for some $q\in [2,\oo)$ where $\Ga= r u_{\th}$. We also established some interesting inequalities for $\Om\co \f{\p_z u_r-\p_r u_z}{r}$, showing that $\na\Om$ can be bounded by $\Om$ itself. All these results are extended to the axially symmetric MHD and Hall-MHD equations with ${\bf u}=u_r(r,z){\bf e}_r +u_{\th}(r,z) {\bf e}_{\th} + u_z(r,z){\bf e}_z, {\bf h}=h_{\th}(r,z){\bf e}_{\th}$, indicating that the swirl component of the magnetic field does not affect the triviality. Especially, we establish the maximum principle for the total head pressure $\Phi=\f {1}{2} (|{\bf u}|^2+|{\bf h}|^2)+p$ for this special solution class.

One of outstanding open problems for steady Navier-Stokes equations is: Is a smooth solution to (1.1) satisfying (1.3) identically zero? For the two dimensional case, the uniqueness result had been proved in [10]. For the three dimensional case, Galdi [9] first established the triviality, under an additional integrability condition u ∈ L 9/2 (R 3 ). Chae and Yoneda [7] also obtained a Liouville theorem if u ∈ X ∩ Y, where X, Y are two function spaces and X controls the high oscillation in large part of u and Y gives control on the decay rate of u for sufficiently large x. In [2], Chae showed that the condition ∆u ∈ L 6/5 (R 3 ) is enough to guarantee the triviality. Note that this condition is stronger than the finite Dirichlet condition ∇u ∈ L 2 (R 3 ), but both of them have the same scaling. In [14], the authors proved that any axially symmetric smooth solution to (1.1) without swirl u θ ≡ 0, but with finite Dirichelt integral must be zero.
A function f is said to be axially symmetric if it does not depend on θ. A vector-valued function u = (u r , u θ , u z ) is called axially symmetric if u r , u θ and u z do not depend on θ. A vector-valued function u = (u r , u θ , u z ) is called axially symmetric with no swirl if u θ = 0 while u r and u z do not depend on θ. For more information about smooth axially symmetric vector fields, one may refer to [17].
Assume that u(x) = u r (r, z)e r + u θ (r, z)e θ + u z (r, z)e z is a smooth solution to (1.1). The corresponding asymmetric steady Navier-Stokes equations read as follows. (1.5) In this paper, we want to investigate that what kinds of decay and integrability conditions we should prescribe on u r , u θ and u z to guarantee the triviality of axially symmetric smooth solution to (1.1). Some of our conditions are weaker than the finite Dirichlet integral condition.
To our purpose, we also need the vorticity ω(x) = curl u(x) = ω r e r + ω θ e θ + ω z e z , where The equations satisfied by ω r , ω θ and ω z are listed as follows.
This paper is structured as follows. In section 2, we first provide another version of the Liouville theorem of [14] for axially symmetric flows with no swirl, but replacing the Dirichlet integral condition by mild decay condition at infinity. Then we establish some Liouville theorems under where C ♯ is a universal constant to be specified later. In particular, if u r (r, z) ≥ − 1 r for any (r, z) ∈ [0, ∞) × R, then u ≡ 0. The Liouville theorem also holds under some decay or L q (R 3 ) (q ∈ [2, ∞)) integrability conditions on Γ = ru θ . We also establish some interesting inequalities for Ω, use the equation for u θ r and the relation between u r r and Ω. These indicate that ∇Ω can be bounded by Ω itself, which is not so clear at first sight due to the additional term − 1 r 2 ∂ z (u θ ) 2 . All these results will be extended to the axially symmetric MHD case in section 3, where we consider a special solution form u(x) = u r (r, z)e r + u θ (r, z)e θ + u z (r, z)e z and h(x) = h θ (r, z)e θ . Our results show that the swirl component of the magnetic field does not affect the triviality. Especially, we establish a maximum principle for the total head pressure Φ = 1 2 (|u| 2 + |h| 2 ) + p. In section 4, we investigate the corresponding problem for steady resisitve, viscous Hall-MHD equations and obtain similar results as the MHD case.

The Liouville theorem for axially symmetric flows with no swirl
In [14], the authors have showed that the axially symmetric smooth solutions to (1.1) satisfying (1.3) must be zero in the absence of swirl. In the following, we provide another version of their Liouville theorem, replacing the Dirichlet integrability condition by mild decay conditions on u. Indeed this follows from Theorem 5.2 in [13] immediately. In Theorem 5.2 in [13], they proved that for any bounded weak solution u of the unsteady Navier-Stokes equations in R 3 × (−∞, 0), if u is axially symmetric with no swirl, then u = (0, 0, b 3 (t)) for some bounded measurable function b 3 : (−∞, 0) → R. By (1.2), u ≡ 0. Here we provide another proof of Theorem 2.1 by using the maximum principle of Ω.

Liouville type theorems conditioned on u r and u θ .
We look at the equation u θ directly: where C * is the optimal constant in the Sobolev inequality f L 6 (R 3 ) ≤ C * ∇ f L 2 (R 3 ) for any f ∈ C ∞ 0 (R 3 ) and 1 E is the characteristic function for a set E ⊂ R 3 . Then, we have u ≡ 0. In particular, if then (2.6) is automatically satisfied and u ≡ 0.
By the condition (1.2), we have (2.8) and . Now let R tends to infinity, we obtain Combining all these calculations, we conclude that , then rewrite (2.10) as Hence the problem reduces to the case of axially symmetric flows with no swirl, the theorem follows from Theorem 2.1.
In the case of smooth axially symmetric solutions to (1.1)-(1.2) with finite Dirichlet integral (1.3) one can have the optimal constant in the estimate (2.6).
Proof. Note that f (q) ≔ 4(q+1) (q+2) 2 is monotonic decreasing for q ∈ [1, ∞), and sup In the following, we will verify that (2.10) holds for q = 1. Since ∇u ∈ L 2 (R 3 ), then by Sobolev embedding theorem u ∈ L 6 (R 3 ) and since Letting R → ∞ in (2.9), we obtain Finally, we obtain (2.10) for q = 1. Finally we give a simple vanishing criteria based on the L 3 (R 3 ) integrability conditions for u r and u z . Note that we do not need any additional conditions on u θ .
Proof. Multiplying the first three equations in (1.5) by σ R (x)u r , σ R (x)u θ , σ R (x)u z and adding them together, and then integrating by parts, we obtain Since ∇u ∈ L 2 (R 3 ), then 1 2 |u| 2 + p ∈ L 3 (R 3 ) and the left hand side of (2.13) tends to Hence letting R → ∞ in (2.13), we can conclude that u ≡ 0.

Liouville type theorems conditioned on Γ = ru θ .
It follows from (1.5) that As is well known, for any harmonic function then h ≡ 0. Inspired by this, in the following we show that some decay or integrability conditions on Γ are also enough to ensure that Γ ≡ 0.
Theorem 2.6. Let u(x) be an axially symmetric smooth solution to (1.1)- (1.2). If one of the following conditions holds Proof. Assume (i), then the conclusion follows immediately from the maximum and minimum principle by Γ. Note that Γ(0+, z) = 0 for ∀z ∈ R, so r = 0 does not cause any trouble. Now we assume (ii). Multiplying the equation (2.14) by σ R |Γ| q−2 Γ with q ≥ 2, and then integrate over the whole space, we obtain that We estimate both sides as follows.
Combining these together, we see that for any 0 < R < R 1 < ∞: where we have used J 1 (R) is positive and monotonic increasing with respect to R. Assume that Γ ∈ L q (R 3 ) for some q ∈ [2, ∞), then by Hölder's inequality, In (2.16), we first fix R > 0 and let R 1 tends to ∞, we arrive at B R |∇|Γ| The problem is reduced to the case of axially symmetric flows with no swirl, then the theorem follows from Theorem 2.1. The proof is completed. (2.17) In [8], the authors have proved the following inequality by using ( u θ r , ∇u θ ) ∈ L 2 (R 3 ): Comparing with (2.17) and (2.18), heuristically, one may guess that the integrability of u θ in the zdirection is enough, the decay rate of u θ in the radial direction maybe a key issue. Unfortunately, the decay rates obtained in [8] seemed not good enough, this issue will be further investigated in [18].
From Theorem 2.2, it is natural to conject that a condition like u θ r X ≤ c ♯ , maybe enough to guarantee that u ≡ 0, where X is a function norm and c ♯ > 0 is a constant depending only on the dimension n = 3. Right now, we can not prove such a conjecture (see Corollary 2.12). However, we have some interesting conclusions.
Proof. It follows from (1.6), then Ω satisfies We first verify that Ω ∈ L 2 (R 3 ) and ∇Ω ∈ L 2 (R 3 ). Note that for any axially symmetric vector field f = ( f r , f θ , f z ), we have Hence by the definition of Ω, it suffices to verify that ∇ 2 u and ∇ 3 u belong to L 2 (R 3 ). These indeed follow from a standard bootstrap argument by regarding (u · ∇)u as a forcing term and using L p estimates for the Stokes system (see Theorem IV.2.1 in [9]). By using the cut-off function and integration by parts as in the proof of Theorem 2.6, we can conclude that Then Ω ≡ 0 if (2.19) holds. That is, ω θ = ∂ z u r − ∂ r u z ≡ 0. Together with ∂ r u r + u r r + ∂ z u z = 0, we can conclude that u r = u z ≡ 0. The equation for u θ in (1.5) reduces to Setting Λ = u θ r , then Same argument as in the proof of Theorem 2.2, we can show that Λ ≡ 0 and u θ ≡ 0.

Remark 2.10. This theorem shows that if the swirl component is smaller than the other two components in the sense of (2.19), then u ≡ 0.
In the following, we will use the equation for Λ ≔ u θ r and the relation between u r r and Ω, to derive some interesting inequalities for Ω, showing that ∇Ω can be bounded by Ω itself in some senses.
Although Ω satisfies an elliptic equation (2.20), but due to the extra term − 1 r 2 ∂ z (u 2 θ ) in (2.20), this property is not so clear at first sight.

Theorem 2.11. Let u(x) be an axially symmetric smooth solution to (1.1)-(1.2) with finite Dirichlet integral (1.3). Then
Furthermore, if we assume u θ r ∈ L 3/2 (R 3 ), then (2.23) Proof. It follows from (2.21) that It follows from (1.5) that Λ satisfies the following equation Since Λ ∈ L 2 and ∇Λ ∈ L 2 (R 3 ), then same as before, we can conclude where we have used the classical Gagliardo-Nirenberg inequality on R n : with p, q, s ≥ 1 and α ∈ [0, 1] satisfy the identity Then we obtain We still need to explore the relation between u r r and Ω by introducing the stream function, this relation was already known in the unsteady case [5,16]. By the divergence free condition, ∂ r (ru r ) + ∂ z (ru z ) = 0, one can introduce a stream function ψ θ such that Since ω θ = ∂ z u r − ∂ r u z , we have Setting ϕ = ψ θ r , then it is easy to see that The second order operator (∂ 2 r + 3 r + ∂ 2 z ) can be interpreted as the Laplace operator in R 5 , see [12,13]. Introduce y = (y 1 , y 2 , y 3 , y 4 , z), r = y 2 1 + y 2 2 + y 2 3 + y 2 4 , ∆ y = (∂ 2 r + Hence we have ϕ = (−∆ y ) −1 Ω and u r r = −∂ z ϕ. By simple calculations, one has where w(r) = r −2 and in the last step we have used the boundedness of Riesz operators in weighted Sobolev spaces (Lemma 2 in [12]). See also Corollary 2 in [4] for a similar weighted estimate for a singular integral operator. Similarly, we also have Then by (2.28), we have To estimate ∇∂ z Λ, we first derive the equation for ∂ z Λ: Then by integrate by parts, we have
Remark 2.13. We remark here all the quantities in (2.22) and (2.23) have same scaling properties.

Liouville type theorem for steady MHD equations
In this section, we will investigate the steady Magnetohydrodynamics equations (MHD). The steady MHD equations are listed as follows.
We also consider the weak solution to (3.1) with finite Dirichlet integral: Then following the argument developed in [9], one can show that any weak solution to (3.1) satisfying (3.2) is smooth. A natural problem is whether this solution is zero or not. We first present the following simple Liouville theorem for MHD equations, by prescribing the L 3 integrability only on u. Similar regularity criteria has been developed in [11] and [20] for unsteady MHD equations.  Let (u, h) be a smooth solution to (3.1) in R 3 with finite Dirichlet integral. If u ∈ L 3 (R 3 ), then u = h ≡ 0.
Proof. Multiplying the first and third equation in (3.1) by σ R u and σ R h respectively, and integrating by parts, we finally obtain It is well-known that u ∈ L 6 (R 3 ), h ∈ L 6 (R 3 ) and p ∈ L 3 (R 3 ). Since we assume that u ∈ L q (R 3 ) for q ∈ [1,3], then u ∈ L 3 (R 3 ). It is easy to see that H 1 → ∇u 2 L 2 + ∇h 2 L 2 as R → ∞, and also Hence letting R → ∞ in (3.3), we obtain In the following, we derive the equation for the total head pressure Φ ≔ 1 2 (|u| 2 + |h| 2 ) + p. Assume that (u, h) is a smooth solution to steady MHD equations (3.1), then by simple calculations, we obtain This yields Consider the special case, where (u, h) are axi-symmetric, and of the following special form then (h · ∇)(u · h) = h θ r ∂ θ (u θ h θ ) ≡ 0. Then we obtain the following important maximum principle.

Lemma 3.2. Let (u, h) be a axially symmetric smooth solution to (3.1) with the form (3.7). Then we have the following important inequality
Hence we have the following maximum principle for Φ in any bounded domains Ω: In our cases, as shown in [9], by adding a constant if necessary, one has lim |x|→∞ p(x) = 0, so lim |x|→∞ Φ(x) = 0. Hence by maximum principle, we have If (u, h) are axi-symmetric with the form (3.7), then (3.1) can be rewritten as Proof. Define Π(r, z) = h θ (r,z) r , then it follows from (3.11) that Since lim |x|→∞ Π(x) = 0, by the equation (3.12), we have the maximum and minimum principle, which implies that Π(x) ≡ 0, i.e. h ≡ 0.
Now we consider the non-resistive, inviscid MHD equations: For smooth axially symmetric solution (u, h, p) to (3.13) with the form (3.7), we can derive the Bernoulli's law for the total head pressure: (3.14) One can use this Bernoulli's law to establish the existence of weak solutions to steady axially symmetric MHD equations with nonhomogeneous boundary conditions by following the approach developed in [15], for the details please refer to [19]. By slightly modifying the proof in [14], we can derive the following Liouville theorem for (3.13).
Theorem 3.4. Let (u, h, p) be an axially symmetric solution to (3.13) with the form (3.7) and finite Dirichlet integral. If u has no swirl and the corresponding head pressure Φ ∈ L 3 (R 3 ) satisfies the condition Φ(x) ≤ 0 for any x ∈ R 3 , then u = h ≡ 0.
(ii) Let g(r) = 1 r . Then Since we have an essential singularity at r = 0, we need to integrate this identity over the cylin- where P t 0 t = {(r, z) ∈ P + : r ∈ (t 0 , t), z ∈ R}.
If u is axially symmetric with no swirl, then the formulas (3.18) and (3.20) imply u ≡ 0. Indeed, from the last two inequalities it follows that R [p(t, z) + u 2 r (t, z)]dz ≡ 0 for all t > 0, and thus, by virtue of (3.20) and u θ (x) = 0, we obtain u r ≡ h θ ≡ 0. Therefore, by (3.13) we conclude that u z ≡ 0.

Liouville theorem for steady viscous resistive Hall-MHD equations
The steady Hall-MHD equations read as follows.
(u · ∇)u + ∇p = (h · ∇)h + ∆u, ∀x ∈ R 3 , div u = 0, (u · ∇)h − (h · ∇)u + ∇ × ((∇ × h) × h) = ∆h, ∀x ∈ R 3 , div h = 0, lim  In this section, we consider the smooth solution to (4.1) with finite Dirichlet integral (3.2). Comparing with the well-known MHD system, the Hall term ∇ × ((∇ × h) × h) is included due to the Ohm's law, which is believed to be a key issue for understanding magnetic reconnection. Note that the Hall term is quadratic in the magnetic field and involves the second order derivatives. A derivation of Hall-MHD system from a two-fluids Euler-Maxwell system for electrons and ions was presented in [1], through a set of scaling limits. They also provided a kinetic formulation for the Hall-MHD, and proved the existence of global weak solutions for the incompressible viscous resistive Hall-MHD system. The authors in [5] have showed unsteady Hall-MHD system without resistivity may develop finite time singularity for a special class of axially symmetric datum. Different from the steady MHD case, a weak solution to (4.1) with finite Dirichlet integral may not be smooth. Chae and Wolf [6] have investigated the partial regularity of suitable weak solutions to steady Hall-MHD equations, showing the set of possible singularities has Hausdorff dimension at most one. The authors in [3] Let (u, h) be a smooth solution to (4.1) in R 3 with finite Dirichlet integral. If u ∈ L 3 (R 3 ), then u = h ≡ 0.
Proof. Similar to the proof in Theorem 3.1, we only need to check the effect of the Hall-term H = Then we finish the proof.
If we also consider the axially symmetric smooth solution (u, h) to (4.1) with the form u(x) = u r (r, z)e r + u θ (r, z)e θ + u z (r, z)e z , h(x) = h θ (r, z)e θ , (4.2) then (4.1) can be rewritten as