Liouville Theorem for Mhd System and Its Applications

COMMUNICATIONS ON doi:10.3934/cpaa.2018111 PURE AND APPLIED ANALYSIS Volume 17, Number 6, November 2018 pp. 2329–2350

LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS

Xian-gao Liu and Xiaotao Zhang School of Mathematic Sciences, Fudan University, Shanghai, China

(Communicated by Igor Kukavica)

Abstract. In this paper, we construct Liouville theorem for the MHD system and apply it to study the potential singularities of its weak solution. And we 3 mainly study weak axi-symmetric solutions of MHD system in R × (0,T ).

1. Introduction.

1.1. Model and related works. Let Ω ⊂ Rn be smooth bounded domain. The n- dimensional incompressible magnetohydrodynamics(MHD) system are the following coupled equations  1  u − µ ∆u + u · ∇u + ∇(p + |b|2) = b · ∇b,  t 1 2 b − µ ∆b + u · ∇b = b · ∇u, (1.1)  t 2  div u = 0, div b = 0, where u :Ω × (0,T ) 7→ Rn is fluid velocity, p :Ω × (0,T ) 7→ R is the press and n b :Ω × (0,T ) 7→ R is the magnetic field. µ1, µ2 are two positive constants. For simplicity, we denote Π = p+|b|2/2. And, along with (1.1), the initial and boundary values are: u(x, 0) = u0(x), b(x, 0) = b0, for all x ∈ Ω u = 0, b · n = 0, on ∂Ω n n for a given datum (u0, b0):Ω 7→ R × R , with div u0 = 0 and div b0 = 0 and n is the outward normal on ∂Ω. There are lots of works on the solution of the MHD equations (1.1). In particular, Duvaut and Lions [4] constructed a class of global weak solutions and the local strong solutions to the initial boundary value problem, and Sermange and Temam in [17] discussed some properties of such solutions. For the 2-dimensional case, the smoothness and uniqueness of solutions have been shown. But for n- dimensional(n ≥ 3), the problem is still open in general case like Navier-Stokes equations v − ∆v + v · ∇v + ∇q = 0, t (1.2) div v = 0, where v is fluid velocity and q is the press. For Navier-Stokes equations, many regularity criteria have been established(e.g. [18, 19,1, 24,9]), and some of these criteria can be extended to the 3-D MHD equations by assumptions only on u, see

2000 Mathematics Subject Classiﬁcation. 35Q35, 76W05, 35B65. Key words and phrases. Incompressible magnetohydrodynamics(MHD), Liouville theorem, singularties.

2329 2330 XIAN-GAO LIU AND XIAOTAO ZHANG

He and Xin [8]. We note that 3-D MHD system need assumptions both on u and b for Ladyzhenskaya-Prodi-Serrin class L3,∞, see Mahalov, Nicolaenko and Shilkin [14]. But there is still a gap between the case of existence and the case of regularity. Scheffer [16] began to study the partial regularity theory of Navier-Stokes equations. Scheffer’s works were improved by Caffarelli-Kohn-Nirenberg [2], Tian-Xin [21]. We mention that, according to [2, 21], the 1-D Hausdorff measure of the blow-up set must be zero. For self-similar singularities in the Navier-Stokes equations, the work of Neˇcas, R˚uˇziˇcka and Sverák[ˇ 15] and Tsai, Tai-Peng [22] showed the trivial solutions with some integration conditions. And then Koch, Nadirashvili, Seregin and Sverák[ˇ 11] directly studied the potential singularities of weak axi-symmetric solutions in R3 × (0,T ), by Liouville theorem for the Navier-Stokes equations. For MHD system, He and Xin [7] extend the result of [21] to it. And [14] extend the result of [15] and [22] to the MHD system. Moreover, Lei [13] constructed one kind of smooth axially symmetric solutions(uθ = br = bz = 0) of MHD in three dimensions. In this paper, we construct Liouville theorem for the MHD system and apply it to study the potential singularities of its weak solution. Because the term b · ∇b in the equations velocity u and equations about magnetic field b, the problem becomes more complicated.

1.2. Main result and outline. Under some conditions about u, b, we prove the Liouville Theorems for MHD system. And, we mainly study weak axi-symmetric solutions of the MHD system in R3 × (0,T ). For 2-D case, with the condition b2,1 −b1,2 = 0 or b1 = 0 or b2 = 0 or b1u2 = b2u1, we have the Liouville Theorems for the bounded weak solutions of MHD system. Meanwhile, with the integral condition s,r 2 |u|+|b| ∈ Lx,t(R ×(−∞, 0))(2/s+2/r ≥ 1, s ≥ 3, 3 ≤ r < ∞), we prove the Liouville 2,1 Theorem for the solutions in Cx,t . For 3-D case, with the conditions uθ = 0(no √ C √ C swirl), br = bz = 0 or |u(x, t)| ≤ 2 2 , bz = 0 or |u(x, t)| ≤ 2 2 , urbz = uzbr, x1+x2 x1+x2 we have the Liouville Theorems for the weak bounded axi-symmetric solutions of the MHD system. And, when µ1 = µ2, with the conditions uz,z = bz,z = 0, we get the Liouville Theorem for the weak bounded axi-symmetric solutions. Moreover, when u = 0, we obtain that b must be a constant vector, this means b is smoother than u, see section 6. As in [11], by scaling transformations and the Liouville Theorems which we have proved, we proved the regularity for axi-symmetric solutions of MHD √ C 3 systems with the conditions |u(x, t)| ≤ 2 2 ,”bz is bounded in R × (0,T )” or x1+x2 some other conditions. The paper is organized as follows. In section 2, we introduce the strong maximum principle which is very essential in this article. In section 3, we introduce the mild solution and bounded weak solution and their properties, for Stokes equations and heat equations. In section 4, we study the regularity of mild solution and bounded weak solution of MHD system, and the limit properties of bounded mild solution of MHD system. In section 5, we construct the Liouville theorem for MHD system, we mention that we need conditions of b to prove the Liouville theorem in R2 ×(−∞, 0). And for 3-dimensional case, we study the axi-symmetric solutions of MHD system. In section 6, we apply the Liouville theorem constructed in section 5, to study the potential singularities of the ﬁnite time weak solution of MHD system.

2. maximum principle. The strong maximum principle plays essential role in this article, and we mainly use the form in [11]. For reader’s convenient, we write LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2331 it as Lemma 2.1. Let Ω ⊂ Rn be a bounded domain and T > 0. We consider the parabolic equation

ut + a(x, t)∇u − ∆u = 0 (2.1) ∞ in Q = Ω × (0,T ). Besides a ∈ Lx,t(Ω × (0,T )). And we mention that u is a scalar valued function. Lemma 2.1. [11, Lemma 2.1] Assume that u is a bounded solution of the equation (2.1). Let K be a compact subset of Ω, and Ω0 ⊂ Ω0 ⊂ Ω, and τ > 0. Let M = 0 sup |u|. Then, for each ε > 0, there exists δ = δ(Ω, Ω ,K,T, ||a|| ∞ , τ, ε) > 0 Ω×(0,T ) Lx,t 0 such that supx∈K |u| ≥ M(1 − δ), then u(x, t) ≥ M(1 − ε) in Ω × (τ, T ).

3. mild solution and bounded weak solution for linear case. Let u = n n (u1, ··· , un), b = (b1, ··· , bn): R × (0, ∞) → R . We ﬁrst consider the following linear equations about cauchy problem:  ∂  ut − ∆u + ∇Π = fk,  ∂xk ∂ (3.1) b − ∆b = g ,  t k  ∂xk  div u = 0, div b = 0, in Rn × (0, ∞), and n u(·, 0) = u0, b(·, 0) = b0 in R (3.2)

Here fk = (f1k, ··· , fnk), gk = (g1k, ··· , gnk) for k = 1, ··· , n. Let P denote the Helmholtz projection of vector fields on divergence free fields and let S be solution operator of the heat equation. Then we get the formula Z t ∂ u(t) = S(t)u0 + S(t − s)P fk(s)ds, (3.3) 0 ∂xk Z t ∂ b(t) = S(t)b0 + S(t − s) gk(s)ds, (3.4) 0 ∂xk where u(t), b(t) denote the two functions u(·, t), b(·, t). These can be written as more clearly in terms of some kernels. We first deal with the formula about u. This is similar to the [11], and for reader’s convenient, we write it as follows. Let ∂2 Kij(x, t) = −δij∆ + Φ(x, t), (3.5) ∂xi∂xj where Φ is defined in terms of the fundamental solution of Laplace operator G and the the heat kernel Γ: Z Φ(x, t) = G(y)Γ(x − y, t)dy, n R then let ∂ Kijk = Kij, ∂xk then we can rewrite the equality (3.3) as Z Z t Z ui(x, t) = Γ(x − y, t)u0i(y)dy + Kijk(x − y, t − s)fjk(y, s)dyds (3.6) n n R 0 R 2332 XIAN-GAO LIU AND XIAOTAO ZHANG

Note that Kij has the following estimates: C |K (x, t)| ≤ (3.7) ij (|x|2 + t)n/2 C |∇ K (x, t)| ≤ (3.8) x ij (|x|2 + t)(n+1)/2 Now we deal with (3.4). From the theory of heat equation, one can easily rewrite it as Z Z t Z b(x, t) = Γ(x − y, t)b0(y)dy − Γk(x − y, t − s)gk(y, s)dyds (3.9) n n R 0 R where ∂ Γk = Γ. ∂xk Now we give the definition of mild solution and bounded weak solution of the linear system (3.1) and (3.2). Definition 3.1. If (u(t), b(t)) is continuous and satisfies the formulas (3.6) and (3.9), then we call it a mild solution of the Cauchy problem (3.1). Let n n VT = {φ : R × (0,T ) → R ; φ is smooth and div φ = 0} (3.10) n n ∞ Definition 3.2. Let u, b : R × (0,T ) → R be two vector fields in the space Lx,t. If u, b satisfy: (i) div u=0, div b=0 in Rn × (0,T )(in the sense of distribution) (ii) for each φ ∈ VT (see (3.10))

Z T Z Z T Z ∂ u(φt + ∆φ)dxdt = fk φdxdt. n n ∂x 0 R 0 R k ∞ n (iii) for each ψ ∈ C0 (R × (0,T )) Z T Z Z T Z ∂ b(ψt + ∆ψ)dxdt = gk ψdxdt. n n ∂x 0 R 0 R k Then we call (u, b) is a bounded weak solution of the Cauchy problem (3.1). ∞ Let f, g ∈ Lx,t, we have some standard estimates for u, b. In particular, if u0 = 0, b0 = 0, then for any α ∈ (0, 1) and p ∈ (0, ∞)

||u|| α ≤ C(α, R)||f|| ∞ n , (3.11) Cpar (Q(z0,R)) Lx,t(R ×(0,T ))

||∇ u|| p ≤ C(p, R)||f|| ∞ n , (3.12) x Lx,t(Q(z0,R)) Lx,t(R ×(0,T )) ||b|| α ≤ C(α, R)||g|| ∞ n , (3.13) Cpar (Q(z0,R)) Lx,t(R ×(0,T )) ||∇ b|| p ≤ C(p, R)||g|| ∞ n (3.14) x Lx,t(Q(z0,R)) Lx,t(R ×(0,T )) 2 where Q(z0,R) = Q((x0, t0),R) = B(x0,R) × (t0 − R , t0) is any parabolic ball ball n α contained in R × (0, ∞)). And the space Cpar is deﬁned by means of the parabolic distance p|x − x0|2 + |t − t0|. Taking diﬀerence on both sides of the equations, we obtain: k k ||∇ u|| α ≤ C(α, R)||∇ f|| ∞ n , (3.15) x Cpar (Q(z0,R)) x Lx,t(R ×(0,T )) k k ||∇ u|| p ≤ C(p, R)||∇ f|| ∞ n , (3.16) x Lx,t(Q(z0,R)) x Lx,t(R ×(0,T )) k k ||∇ b|| α ≤ C(α, R)||∇ g|| ∞ n , (3.17) x Cpar (Q(z0,R)) x Lx,t(R ×(0,T )) LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2333

k k ||∇ b|| p ≤ C(p, R)||∇ g|| ∞ n (3.18) x Lx,t(Q(z0,R)) x Lx,t(R ×(0,T )) for u0 = 0, b0 = 0. Then, by the equations (3.1), we have k k+2 ||∇ u || ∞ n ≤ C(T, k)||∇ f|| ∞ n , (3.19) x t Lx,t(R ×(0,T )) x Lx,t(R ×(0,T )) k k+2 ||∇ b || ∞ n ≤ C(T, k)||∇ g|| ∞ n , (3.20) x t Lx,t(R ×(0,T )) x Lx,t(R ×(0,T )) for u0 = 0, b0 = 0. Remark 3.3. The estimates (3.3), (3.5), (3.6), (3.7), (3.8), (3.11), (3.12), (3.15), (3.16), (3.19) are from the theory of linear Stokes equations, see [11, 20,6]. Esti- mates (3.4), (3.9), (3.13), (3.14), (3.17), (3.18), (3.20) are from the theory of heat kernel, see [12,6, 18]. Now, we consider the relation between mild solution and bound weak solution of the system (3.1). ∞ n ∞ n Lemma 3.4. For ﬁxed f, g ∈ Lx,t(R × (0,T )), let u, b ∈ Lx,t(R × (0,T )) be any weak solution of (3.1) in Rn ×(0,T ). And let v, e be the mild solution of the Cauchy problem (3.1) and (3.2) with u0 = 0, b0 = 0. Then

u(x, t) = v(x, t) + w1(x, t) + d1(t),

b(x, t) = e(x, t) + w2(x, t), n where w1, w2 satisfy the heat equations wt −∆w = 0 in R ×(0,T ) and d1 is bounded measurable Rn-valued functions on (0,T ). Moreover,

||w || ∞ n ≤ C(T )||u|| ∞ n , 1 Lx,t(R ×(0,T )) Lx,t(R ×(0,T ))

||d || ∞ ≤ C(T )||u|| ∞ n , 1 L (0,T ) Lx,t(R ×(0,T ))

||w || ∞ n ≤ C(T )||b|| ∞ n , 2 Lx,t(R ×(0,T )) Lx,t(R ×(0,T )) Proof. The proof is similar to the way of G.Koch, N.Nadirashvili, G.A.Seregin, V.S˘ver´ak(see [11], Lemma 3.1), so we omit it.

4. Bounded solutions of MHD. Now we consider the Cauchy problem for the MHD system:   ut − ∆u + u · ∇u + ∇Π = b · ∇b, bt − ∆b + u · ∇b = b · ∇u, (4.1)  div u = 0, div b = 0, n u(·, 0) = u0, b(·, 0) = b0, in R (4.2) The considerations of the section 3 can be repeated with fk = −uku + bkb, gk = ∞ n −ukb + bku. Moreover, if the solutions of (4.1) and (4.2) u, b are in Lx,t(R × (0,T )), then its definitions of mild solution and bounded weak solution follow as the definition 3.1 and the definition 3.2. We define two bilinear form as follows: ∞ n ∞ n ∞ n B1 : Lx,t(R × (0,T )) × Lx,t(R × (0,T )) → Lx,t(R × (0,T )), ∞ n ∞ n ∞ n B2 : Lx,t(R × (0,T )) × Lx,t(R × (0,T )) → Lx,t(R × (0,T )), where Z t Z B1(u, v)i(x, t) = Kijk(x − y, t − s)uk(y, s)vj(y, s)dyds, (4.3) n 0 R Z t Z B2(u, v)(x, t) = Γk(x − y, t − s)uk(y, s)v(y, s)dyds. (4.4) n 0 R 2334 XIAN-GAO LIU AND XIAOTAO ZHANG

And let U1,U2 be the heat extension of the initial datum u0, b0. Then the solutions u, b become u = U1 + B1(u, u) − B1(b, b),

b = U2 + B2(u, b) − B2(b, u). Since estimate (3.8), one can easily obtain √ ||B (u, v)|| ∞ n ≤ C T ||u|| ∞ n ||v|| ∞ n , (4.5) 1 Lx,t(R ×(0,T )) Lx,t(R ×(0,T )) Lx,t(R ×(0,T )) and by heat kernel theory, we also get √ ||B (u, v)|| ∞ n ≤ C T ||u|| ∞ n ||v|| ∞ n . (4.6) 2 Lx,t(R ×(0,T )) Lx,t(R ×(0,T )) Lx,t(R ×(0,T )) ∞ n Now we give some regularity properties of mild solutions in Lx,t(R × (0,T )). ∞ n Lemma 4.1. Let (u, b) ∈ Lx,t(R × (0,T )) be a mild solution of (4.1) and (4.2) ∞ k/2+l k l k/2+l k l with u0, b0 ∈ L . Then for k, l = 0, 1 ··· the functions t ∇x∂tu, t ∇x∂tb 0 −1 −1 2 are bounded for T = ε(k, l)(||u0|| ∞ n + ||b0|| ∞ n ) (where ε(k, l) > 0 is a L (R ) L (R ) small constant), and we have k/2+l k l ||t ∇ ∂ u|| ∞ n 0 ≤ C(k, l)(||u || ∞ n + ||b || ∞ n ) (4.7) x t Lx,t(R ×(0,T )) 0 L (R ) 0 L (R ) k/2+l k l ||t ∇ ∂ b|| ∞ n 0 ≤ C(k, l)(||u || ∞ n + ||b || ∞ n ) (4.8) x t Lx,t(R ×(0,T )) 0 L (R ) 0 L (R ) Proof. We use the method of local existence and uniqueness of the solution to Navier-Stokes equations(see [6, 10]). For simplicity, we just give the estimate of the u, b, ∇u, ∇b, ∂tu, ∂tb. We deﬁne the approximation uj, bj (j = 1, 2, ··· ) by

u1(t) = S(t)u0,

b1(t) = S(t)b0,

uj+1(t) = S(t)u0 + B1(uj, uj)(·, t) − B1(bj, bj)(·, t),

bj+1(t) = S(t)b0 + B2(uj, bj)(·, t) − B2(bj, uj)(·, t), where u0, b0 are the initial data of the MHD system and the deﬁnition of operator B1 and B2 follows (4.3) and (4.4). Let

∞ n ∞ n Kj = Kj(T ) = sup (||uj||L (R ) + ||bj||L (R )), 0≤t≤T 0 0 1/2 1/2 ∞ n ∞ n Kj = Kj(T ) = sup (t ||∇uj||L (R ) + t ||∇bj||L (R )), 0≤t≤T and 00 00 ∞ n ∞ n Kj = Kj (T ) = sup (t||∂tuj||L (R ) + t||∂tbj||L (R )). 0≤t≤T

∞ n ∞ n Note that K0 = ||u0||L (R ) +||b0||L (R ). By heat kernel theory and the estimates (4.5) and (4.6), we have 1/2 2 Kj+1(T ) ≤ K0 + C1T Kj(T ) , 0 1/2 0 Kj+1(T ) ≤ CK0 + C2T Kj(T )Kj(T ), and 00 1/2 00 2 Kj+1(T ) ≤ CK0 + C3T Kj (T )Kj(T ) + C4TKj (T ). 1/2 1/2 We take T1 small so that maxi 4CiT1 K0 < 1 and T < 1/2, then it is easy to prove that 0 00 sup Kj(T ) ≤ 2K0, sup Kj(T ) ≤ 2CK0 and sup Kj (T ) ≤ 2CK0 (4.9) j j j LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2335 for any T ≤ T1(C ≥ 1). Let

∞ n ∞ n Lj(T ) = sup (||uj(t) − uj−1(t)||L (R ) + ||bj(t) − bj−1(t)||L (R )), 0≤t≤T 0 1/2 1/2 ∞ n ∞ n Lj(T ) = sup (t ||∇uj(t) − ∇uj−1(t)||L (R ) + t ||∇bj(t) − ∇bj−1(t)||L (R )), 0≤t≤T and 00 ∞ n ∞ n Lj (T ) = sup (t||∂tuj(t) − ∂tuj−1(t)||L (R ) + t||∂tbj(t) − ∂tbj−1(t)||L (R )), 0≤t≤T By direct calculation, we have 1/2 Lj+1(T ) ≤ C5T K0Lj(T ), 0 1/2 0 Lj+1(T ) ≤ C6T K0(Lj(T ) + Lj(T )), 00 1/2 00 Lj+1(T ) ≤ C7T K0(Lj(T ) + Lj (T )) + C8TK0Lj(T ) for T ≤ T1 with C5,C6,C7,C8 independent of K0,T . Take T2 ≤ T1 small such 1/2 1/2 −2 that (C5 + C6 + C7 + C8T )T2 K0 < 1/2(that means T2 < CK0 ), then 0 0 00 Lj+1(T )/Lj(T ) < 1/2, (Lj+1(T ) + Lj+1(T ))/(Lj(T ) + Lj(T )) < 1/2, (Lj+1(T ) + 00 Lj+1(T ))/(Lj (T )+Lj(T )) < 1/2 for any T ≤ T2 and j ≥ 1. Therefore, {Lj(T )}j≥1, 0 00 {Lj(T )}j≥1 and {Lj (T )}j≥1 converge to 0 as j → ∞, then we conclude that 1/2 1/2 the approximation {uj(t)}j≥1, {bj(t)}j≥1, {t ∇uj(t)}j≥1, {t ∇bj(t)}j≥1 and {t∂uj(t)}j≥1, {t∂bj(t)}j≥1 respectively has a unique limit function u(t), b(t), v1(t), d1(t), v2(t), d2(t). The uniqueness of solution can be proved by estimating the diﬀerence sup (||u (t) − u (t)|| ∞ n + ||b (t) − b (t)|| ∞ n ) of two solu- 0≤t≤T2 1 2 L (R ) 1 2 L (R ) tions (u1, b1) and (u2, b2), where T2 is very small(use the equation(3.6) and (3.9)). Finally, by the estimate (4.9), we obtain the result.

(k) (k) ∞ n Lemma 4.2. Let u , b ∈ Lx,t(R × (0,T )) be a sequence of mild solution of (k) (k) (k) (4.1) and (4.2) with initial conditions u , b . Assume that ||u || ∞ n + 0 0 Lx,t(R ×(0,T )) (k) ||b || ∞ n ≤ C, where C is independent of k. Then a subsequence of the Lx,t(R ×(0,T )) sequence u(k) converges locally uniformly in Rn × (0,T ) to a mild solution u ∈ ∞ n ∗ Lx,t(R × (0,T )) with initial datum u0, where u0 is the weak limit of a suitable (k) subsequence of the sequence u0 . Proof. It is easy to get this result by lemma 4.1 and the decay estimates (3.8) and heat kernel theory. We now consider the regularity of bounded weak solutions of (4.1) and (4.2). ∞ n n let u, b ∈ Lx,t(R × (0,T )) be the weak solutions of (4.1) in R × (0,T ), and let M = ||u|| ∞ n + ||b|| ∞ n . Then, like [11], we have the following Lx,t(R ×(0,T )) Lx,t(R ×(0,T )) estimates: k k ||∇ u|| p + ||∇ b|| p ≤ C(k, δ, R, M), (4.10) x Lx,t(Q(z0,R)) x Lx,t(Q(z0,R)) n for each Q(z0,R) ⊂ R × (δ, T ) and k = 0, 1, 2, ··· . And for each k = 0, 1, 2, ··· , we also have k k ||∇ u|| ∞ n + ||∇ b|| ∞ n ≤ C(k, δ, T, M) (4.11) x Lx,t(R ×(δ,T )) x Lx,t(R ×(δ,T )) The we obtain k k ||∇ ∂ (u − d )|| ∞ n + ||∇ ∂ b|| ∞ n ≤ C(k, δ, T, M) (4.12) x t 1 Lx,t(R ×(δ,T )) x t Lx,t(R ×(δ,T )) for each k = 0, 1, 2, ··· . 2336 XIAN-GAO LIU AND XIAOTAO ZHANG

5. Liouville theorem in MHD. First, we consider the MHD system in two dimensions space. Theorem 5.1. Let (u, b) be a bounded weak solution of the MHD system in R2 × (−∞, 0). If one of the following conditions is satisﬁed: 2 1. b2,1 − b1,2 = 0 in R × (−∞, 0); 2 2. b1 = 0 in R × (−∞, 0); 2 3. b2 = 0 in R × (−∞, 0); 2 4. b1u2 = b2u1 in R × (−∞, 0); then u(x, t) = d1(t), b(x, t) = d0, where d1 is bounded measurable functions from 2 (−∞, 0) to R and d0 is a constant vector. Proof. In two dimensions space, the vorticity is a scalar, which is deﬁned by

ω = u2,1 − u1,2 (5.1) where uk,j = ∂uk/∂xj, that is say, the indices after comma mean derivatives. For magnetic field, we also use the definition of vorticity 0 ω = b2,1 − b1,2 (5.2) We first consider the first equation of system (4.1). Then the vorticity ω satisfies 0 ωt + u · ∇ω − ∆ω = b · ∇ω (5.3) Through conditions (1)-(4), we aim to prove the term b∇ω0 = 0 in R2 × (−∞, 0). k k According to section 4, we know that ∇xu, ∇xb are bounded. 1. If condition (1) holds, we have b∇ω0 = 0 in R2 × (−∞, 0), then equation (5.3) can be written as ωt + u · ∇ω − ∆ω = 0. (5.4)

This is similar to the Navier-Stokes equations, we know that u(x, t) = d1(t), where 2 d1 is bounded measurable functions from (−∞, 0) to R (see [11], Theorem 5.1). 0 With ω = b2,1 − b1,2 = 0 and div b = 0, we ﬁnd that b1, b2 are harmonic functions, then b is constant in x for each t by the classical Liouville theorem. Take it into the second equation of the system (4.1), we ﬁnd that b is constant in x and t. 2. If condition (2) holds, then b2,2 ≡ 0, thus b · ∇b ≡ 0. Therefore, the equation (5.3) can be written as (5.4), so u(x, t) = d1(t), where d1 is bounded measurable functions from (−∞, 0) to R2. Then, the second equation can be written as

bt − ∆b + u · ∇b = 0,

Then b1,1 satisfies (b1,1)t − ∆b1,1 + u · ∇b1,1 = 0 (5.5) Let M1 = sup b1,1 and M2 = inf b1,1, 2 2×(−∞,0) R ×(−∞,0) R 1 and assume that M1 > 0. Applying Lemma 2.1 to b1,1 − 2 (M1 +M2), we get that there exist arbitrarily large parabolic balls Q((¯x, t¯),R) = B(¯x, R) × (t¯− R2, t¯) ⊂ 2 1 ¯ R × (−∞, 0) such that b1,1 ≥ 2 M1 in Q((¯x, t),R). For such parabolic balls, we have Z 1 4 b1,1dxdt ≥ πM1R . (5.6) QR 2 But, on the other hand, we can obtain Z Z 3 b1,1dxdt = b1n1dxdt ≤ CR (5.7) QR QR LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2337 where n is the normal to the boundary of B(¯x, R). When R is big enough, we find that (5.6) contradicts to (5.7), unless M1 ≤ 0. By the same way, we conclude that 2 M2 ≥ 0. Therefore, b1,1 = 0 in R × (−∞, 0). In the same way, we conclude that 2 b1,2 = 0, b2,1 = 0, b2,2 = 0 in R × (−∞, 0). Therefore, b is constant in x for each t. Take it into the second equation of the system (4.1), we find that b is constant in x and t. 3. If the condition (3) holds, then the proof is similar to (2), we omit it. 4. If the condition (4) holds, then the second equation of (4.1) can be written as

bt − ∆b = 0

Then b1,1 satisﬁes (b1,1)t − ∆b1,1 = 0 Let

M1 = sup b1,1 and M2 = inf b1,1, 2 2×(−∞,0) R ×(−∞,0) R 1 and assume that M1 > 0. Applying Lemma 2.1 to b1,1 − 2 (M1 +M2), we get that there exist arbitrarily large parabolic balls Q((¯x, t¯),R) = B(¯x, R) × (t¯− R2, t¯) ⊂ 2 1 ¯ R × (−∞, 0) such that b1,1 ≥ 2 M1 in Q((¯x, t),R). For such parabolic balls, we 2 have u(x, t) = d1(t), where d1 is bounded measurable functions from (−∞, 0) to R . Z 1 4 b1,1dxdt ≥ πM1R . (5.8) QR 2 But, on the other hand, we can obtain Z Z 1 3 ω dxdt = b1n1dxdt ≤ CR (5.9) QR QR where n is the normal to the boundary of B(¯x, R). When R is big enough, we ﬁnd that (5.8) contradicts to (5.9), unless M1 ≤ 0. By the same way, we conclude that 2 M2 ≥ 0. Therefore, b1,1 = 0 in R × (−∞, 0). In the same way, we conclude that 2 b1,2 = 0, b2,1 = 0, b2,2 = 0 in R × (−∞, 0). Therefore, b is constant in x for each t. Take it into the second equation of the system (4.1), we ﬁnd that b is constant in x and t. Therefore, b · ∇b ≡ 0, then we have u(x, t) = d1(t), where d1 is bounded measurable functions from (−∞, 0) to R2. Then to next, we prove a Liouville theorem under integration condition. We get the idea from the way of dealing with Steady-state problems, see [5, 23,3].

Theorem 5.2. Let (u, b) be a weak solution of the MHD system in R2 × (−∞, 0). 2,1 2 2 Assume that u, b ∈ Cx,t (R × (−∞, 0) : R ) and satisﬁes s,r 2 |u| + |b| ∈ Lx,t(R × (−∞, 0)), (5.10) where 2/s + 2/r ≥ 1, s ≥ 3, 3 ≤ r < ∞. Then, u(x, t) = d1(t), b(x, t) = 0 where d1 is bounded measurable functions from (−∞, 0) to R2. ∞ Proof. We consider a standard cut-oﬀ function ψ ∈ Cc (R) such that 1, if |y| < 1, ψ(y) = 0, if |y| > 2,

2 and 0 ≤ ψ(y) ≤ 1 for 1 < |y| < 2. For each R, deﬁne φR(x, t) = ψ(|x|/R)ψ(t/R ), (x, t) ∈ R2 × (−∞, 0). 2338 XIAN-GAO LIU AND XIAOTAO ZHANG

2 2 Taking the inner product of (4.1)1 with uφR,(4.1)2 with bφR in L (R ×(−∞, 0)). Adding the two resulting integrations together, and integrating by parts, then we have Z Z 1 2 2 2 2 (|u| + |b| )(x, 0)φR(x, 0)dx + (|∇u| + |∇b| )φRdxdt 2 2 2 R R ×(−∞,0) Z Z 1 2 2 1 2 2 = (|u| + |b| )∂tφRdxdt + (|u| + |b| )∆φRdxdt 2 2 2 2 R ×(−∞,0) R ×(−∞,0) Z Z 1 2 1 2 + |u| (u · ∇)φRdxdt + |b| (u · ∇)φRdxdt 2 2 2 2 R ×(−∞,0) R ×(−∞,0) 6 Z Z X + Π(u · ∇)φRdxdt − (u · b)(b · ∇)φRdxdt ≡ Ii 2 2 R ×(−∞,0) R ×(−∞,0) i=1

We estimate Ii for i = 1, 2, ··· , 6 one by one. For I1, H¨olderinequality implies Z R Z 1 2 2 ∞ 2|I1| ≤ 2 ||∂tφ1||L (|u| + |b| )dxdt R 2 −2R R 2(1−2/s−2/r) ≤ C||∂tφ1||L∞ R (J1 + J2), Where 2/r 2/r Z R Z r/s ! Z R Z r/s ! s s J1 = |u| dx dt ,J2 = |b| dx dt . 2 2 −2R R −2R R

2(1−2/s−2/r) 2(1−2/s−2/r) It is easy to see that R J1 → 0 and R J2 → 0 as R → ∞ by the condition (5.10). In the same way, we can estimate I2. For I3, we have Z 0 Z 1 3 2|I3| ≤ ||∇φ1||L∞ |u| dxdt R −∞ B2R\BR  r/s 3/r Z 0 Z ! 3(1−2/s−2/r) s ≤ C||∇φ1||L∞ R  |u| dx dt → 0 as R → ∞. −∞ B2R\BR

To estimate I4, we ﬁrst estimate the following term Z 0 Z 1 3 ||∇φ1||L∞ |b| dxdt R −∞ B2R\BR  r/s 3/r Z 0 Z ! 3(1−2/s−2/r) s ≤C||∇φ1||L∞ R  |b| dx dt → 0 as R → ∞. −∞ B2R\BR

Then Z 0 Z 1 2 2|I4| ≤ ||∇φ1||L∞ |b| |u|dxdt R −∞ B2R\BR 2/3 1/3 Z 0 Z ! Z 0 Z ! 1 3 3 ≤ ||∇φ1||L∞ |b| dxdt |u| dxdt R −2R B2R\BR −2R B2R\BR → 0 as R → ∞ LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2339

s/2,r/2 2 By the Calder´on-Zygmund theorem, we have Π ∈ Lx,t (R × (−∞, 0)), then

1 Z 0 Z |I5| ≤ ||∇φ1||L∞ |Π||u|dxdt R −∞ B2R\BR 2/3 1/3 Z 0 Z ! Z 0 Z ! 1 3/2 3 ≤ ||∇φ1||L∞ |Π| dxdt |b| dxdt R −2R B2R\BR −2R B2R\BR → 0 as R → ∞

Finally, the way of estimating I6 is similar to I4. From the above estimations, we conclude that |∇u| + |∇b| = 0 almost everywhere in R2 × (−∞, 0). With Theorem (5.1), we have the result.

Remark 5.3. For 2-D, R 0 (R (|u|s + |b|s)dx)r/sdt, where 2/s + 2/r = 1, s ≥ 3, 3 < −R BR r < ∞, is scale-invariant for natural scaling of MHD equations. Therefore, we can use the Liouville Theorem 5.2 to study its regularity. Unfortunately, for n-D case (3 ≤ n ≤ 4), the integration condition of the Liouville Theorem is not scale-invariant by the method of Theorem 5.2, we just obtain that Liouville Theorem holds when the condition (5.10) is replaced by

s,r n |u| + |b| ∈ Lx,t(R × (−∞, 0)), (5.11) where n/s + 2/r √≥ n/2, s ≥ 3, 3 ≤ r < ∞. That means more decay, like, if |u| + |b| ≤ C(|x| + t)−2 in R3 × (−∞, 0), then |u| + |b| ∈ L3(R3 × (−∞, 0)) for the bounded weak solutions, then the Liouville theorem holds.

Since the vorticity is no longer a scalar function in three dimensions space, the problem becomes very different. But one can obtain the similar result under the additional assumption when the solutions are axi-symmetric. A vector field u : R3 → R3 is called axi-symmetric if it is invariant under rotations about a suitable axis, and here we choose x3-coordinate as the ”suitable axis”. That is to say, the velocity field u is axi-symmetric if u(Rx) = Ru(x) for every rotation R of the form

 cos α − sin α 0  R =  sin α cos α 0  0 0 1

Let x x x x ~e = ( 1 , 2 , 0)T , ~e = (− 2 , 1 , 0)T , ~e = (0, 0, 1)T , r r r θ r r z then, in cylindrical coordimates (r, θ, z), the axi-symmetric ﬁelds are given by

u = ur~er + uθ~eθ + uz~ez where the coordinate functions ur, uθ and uz depend only on r, z and time t. The axi-symmetric magnetic ﬁelds also have the similar representation

b = br~er + bθ~eθ + bz~ez 2340 XIAN-GAO LIU AND XIAOTAO ZHANG where the coordinate functions br, bθ and bz depend only on r, z and time t. There- fore, in these coordinates, the MHD system (4.1) becomes u2 b2 u (u ) + u u + u u − θ − b b − b b + θ + Π =∆u − r , (5.12) r t r r,r z r,z r r r,r z r,z r ,r r r2 u u b b u (u ) + u u + u u + θ r − b b − b b − θ r =∆u − θ , (5.13) θ t r θ,r z θ,z r r θ,r z θ,z r θ r2 (uz)t + uruz,r + uzuz,z − brbz,r − bzbz,z + Π,z =∆uz, (5.14) (ru ) r ,r + u =0, (5.15) r z,z b (b ) + u b + u b − b u − b u =∆b − r , (5.16) r t r r,r z r,z r r,r z r,z r r2 u b b u b (b ) + u b + u b + θ r − b u − b u − θ r =∆b − θ , (5.17) θ t r θ,r z θ,z r r θ,r z θ,z r θ r2 (bz)t + urbz,r + uzbz,z − bruz,r − bzuz,z =∆bz, (5.18) (rb ) r ,r + b =0, (5.19) r z,z where ∆ = ∂rr + ∂r/r + ∂zz is the scalar Laplacian(expressed in the coordinates (r, θ, z)), and the indices after comma mean derivatives, i.e. ur,z = ∂ur/∂z. Then we consider the case of axi-symmetric ﬂows without swirl in the velocity ﬁelds(uθ = 0). As usual, ω = curl u, and in cylindrical coordinates we write

ω = ωr~er + ωθ~eθ + ωz~ez, (5.20)

For axi-symmetric ﬂows u without swirl we have ωr = ωz = 0 by direct calculation. Thus we can write

ω = ωθ~eθ. (5.21)

And in the magnetic ﬁelds, we assume that br = bz = 0. where ∆ = ∂rr +∂r/r+∂zz is the scalar Laplacian(expressed in the coordinates (r, θ, z)), and the indices after comma mean derivatives, i.e. ur,z = ∂ur/∂z.

Theorem 5.4. Let (u, b) be a bounded weak solution of the MHD system in R3 × (∞, 0). Assume that u, b are axi-symmetric, and uθ = 0(no swirl), br = bz = 0 in 3 T T R × (−∞, 0). Then, u = (0, 0, d3(t)) and b = (0, 0, 0) , where d3 :(−∞, 0) → R is a bounded measurable function.

Proof. With br = bz = 0, we can write (5.17) as b u b (b ) + u b + u b − θ r = ∆b − θ θ t r θ,r z θ,z r θ r2

Then bθ satisﬁes bθ bθ bθ bθ 2 bθ + ur + uz = ∆ + . (5.22) r t r ,r r ,z r r r ,r The term on the right side of equation (5.22) can be treated as the 5-D Laplacian 5 p 2 2 2 2 acting on SO(4)-invariant function in R . We write r = y1 + y2 + y3 + y4 and ˜ y5 = z, and let f(y1, ··· , y5) = f(r, z), then we have ∂2f 3∂f ∂2f ∆ f˜(y , ··· , y ) = + + (r, z) y 1 5 ∂r2 r∂r ∂z2 LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2341

Therefore, with a slight abuse of notation, we can write the equation (5.22) as bθ bθ bθ bθ + ur + uz = ∆5 . (5.23) r t r ,r r ,z r k 3 From section 4, we know that ∇xb are bounded in R × (−∞, 0), and by the 3 condition, b is axi-symmetric, thus bθ/r is bounded R × (−∞, 0). Let bθ bθ M1 = sup and M2 = inf , 3 r 3×(−∞,0) r R ×(−∞,0) R 1 and assume that M1 > 0. Applying Lemma 2.1 to the solution bθ/r − 2 (M1 + 5 M2) of equation (5.23), considered as an equation in R × (−∞, 0). With suitable 1 centers, we see that bθ/r ≥ 2 M1 in arbitrarily large parabolic balls, this means bθ is unbounded, a contradiction. Therefore, M1 ≤ 0. In the same way, we ﬁnd that M2 ≥ 0. Thus bθ ≡ 0. Therefore, we can write (5.12), (5.14), and (5.15) as u (u ) + u u + u u + Π = ∆u − r , r t r r,r z r,z ,r r r2 (uz)t + uruz,r + uzuz,z + Π,z = ∆uz, (ru ) r ,r + u = 0, r z,z This is similar to the Navier-Stokes equations, and ωθ satisﬁes ωθ ωθ ωθ ωθ 2 ωθ + ur + uz = ∆ + . r t r ,r r ,z r r r ,r T Therefore, u(x, t) = (0, 0, d3(t)) (see [11], Theorem 5.2). Next, we will prove other conditions that makes Lioullive theorem hold.

Theorem 5.5. Let (u, b) be a bounded weak solution of the MHD system in R3 × (∞, 0). Assume that u, b are axi-symmetric. u satisﬁes C |u(x, t)| ≤ in 3 × (−∞, 0), (5.24) p 2 2 R x1 + x2 3 and bz = 0 in R × (−∞, 0). Then, u = b = 0 in R3 × (−∞, 0).

Proof. With the condition bz = 0, the equation (5.19) can be written as b r + b = 0 (5.25) r r,r For ﬁx z, t,(5.25) is an ordinary diﬀerential equation of br with respect to r, and −1 3 br = Cr or br = 0. Since br is bounded in R × (−∞, 0), thus br = 0. With 3 3 br = bz = 0 in R × (−∞, 0), we conclude that bθ = 0 in R × (−∞, 0)(see the proof of Theorem 5.4). Therefore, equation (5.13) can be rewrite as u u u (u ) + u u + u u + θ r = ∆u − θ , (5.26) θ t r θ,r z θ,z r θ r2 then we use the equations expressed in the cylindrical coordinates (r, θ, z) for the equation (5.26), and we set f = ruθ, we have 2 f + u f + u f = ∆f − f , t r ,r z ,z r ,r and f ≤ C. Then we have uθ = 0(this is the result of Theorem 5.3 in [11], we omit the proof). 2342 XIAN-GAO LIU AND XIAOTAO ZHANG

Theorem 5.6. Let (u, b) be a bounded weak solution of the MHD system in R3 × (∞, 0). Assume that u, b are axi-symmetric. u satisﬁes C |u(x, t)| ≤ in 3 × (−∞, 0), (5.27) p 2 2 R x1 + x2 3 T and urbz = uzbr in R × (−∞, 0). In addition, let µ1 = µ2. Then, u = (0, 0, 0) and b = (0, 0, 0)T .

Proof. With the condition urbz = uzbr, equations (5.16) and (5.18) can be written as b (b ) = ∆b − r , (5.28) r t r r2 (bz)t = ∆bz, (5.29) Then we rewrite (5.28) as b b 2 b r = ∆ r + r . (5.30) r t r r r ,r Like (5.23), the right term of (5.30) also can be treated as Laplacian acting on SO(4)-invariant functions in R5: br br = ∆5 . (5.31) r t r k By the condition, b is axi-symmetric, and from section 4, we know that ∇xb are 3 3 bounded in R × (−∞, 0), thus br/r is bounded R × (−∞, 0). Let br br M1 = sup and M2 = inf , 3 r 3×(−∞,0) r R ×(−∞,0) R 1 and assume that M1 > 0. Applying Lemma 2.1 to the solution br/r − 2 (M1 + 5 M2) of equation (5.23), considered as an equation in R × (−∞, 0). With suitable 1 centers, we see that br/r ≥ 2 M1 in arbitrarily large parabolic balls, this means br is unbounded, a contradiction. Therefore, M1 ≤ 0. In the same way, we ﬁnd that M2 ≥ 0. Thus br ≡ 0, and bzur ≡ 0. Then, by the equation (5.19), bz,z = 0. 2 Therefore, bz is a function in R × (−∞, 0).

From equation (5.29), we know that bz,x1 and bz,x2 satisfy

(bz,x1 )t = ∆bz,x1 , (5.32)

(bz,x2 )t = ∆bz,x2 . (5.33) k 2 From section 4, we know that ∇xb are bounded in R × (−∞, 0). Let

M1 = sup bz,x1 and M2 = inf bz,x1 , 2 2×(−∞,0) R ×(−∞,0) R 1 and assume that M1 > 0. Applying Lemma 2.1 to bz,x1 − 2 (M1 + M2), we get that there exist arbitrarily large parabolic balls Q((¯x, t¯),R) = B(¯x, R) × (t¯− R2, t¯) ⊂ 2 1 1 ¯ R × (−∞, 0) such that ω ≥ 2 M1 in Q((¯x, t),R). For such parabolic balls, we have Z 1 4 bz,x1 dxdt ≥ πM1R . (5.34) QR 2 But, on the other hand, we can obtain Z Z 3 bz,x1 dxdt = bzn1dxdt ≤ CR (5.35) QR QR LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2343 where n is the normal to the boundary of B(¯x, R). When R is big enough, we ﬁnd that (5.35) contradicts to (5.34), unless M1 ≤ 0. By the same way, we conclude that

M2 ≥ 0. Therefore, bz,x1 ≡ 0. Use the same method, we have bz,x2 ≡ 0. Therefore, bz is a constant. Then bz ≡ 0 or ur ≡ 0. 3 If bz ≡ 0, then from Theorem 5.5, we know that u = 0, b = 0 in R × (−∞, 0). 3 If ur ≡ 0, by the equation (5.15), we conclude that uz,z = 0 in R × (−∞, 0). Therefore, u = 0, b = 0 in R3 × (−∞, 0) by the Theorem 5.7. Theorem 5.7. Let (u, b) be a bounded weak solution of the MHD system in R3 × 3 (∞, 0). Assume that u, b are axi-symmetric and uz,z = bz,z = 0 in R × (−∞, 0). T T In addition, let µ1 = µ2. Then, u = (0, 0, d4(t)) and b = (0, 0, d0) , where d4 :(−∞, 0) → R is a bounded measurable function and d0 is a constent.

Proof. With the condition uz,z = 0, the equation (5.15) can be written as u r + u = 0 (5.36) r r,r

For ﬁx z, t,(5.36) is an ordinary diﬀerential equation of ur with respect to r, and −1 3 ur = Cr or ur = 0. Since ur is bounded in R × (−∞, 0), thus ur = 0. In the same way, we conclude that br = 0. Therefore, equations (5.13) and (5.17) can be written as u (u ) + u u − b b = ∆u − θ , (5.37) θ t z θ,z z θ,z θ r2 b (b ) + u b − b u = ∆b − θ , (5.38) θ t z θ,z z θ,z θ r2 Then we rewrite (5.37) and (5.38) as uθ uθ bθ uθ 2 uθ + uz − bz = ∆ + , (5.39) r t r ,z r ,z r r r ,r bθ bθ uθ bθ 2 bθ + uz − bz = ∆ + , (5.40) r t r ,z r ,z r r r ,r + − + − Let hθ = uθ + bθ, hθ = uθ − bθ, then by (5.39) and (5.40), hθ and hθ satisfy + + + + hθ hθ hθ 2 hθ + (uz − bz) = ∆ + , (5.41) r t r ,z r r r ,r − − − − hθ hθ hθ 2 hθ + (uz + bz) = ∆ + , (5.42) r t r ,z r r r ,r Like (5.23), the right term of (5.41) and (5.42) can be treated as Laplacian acting on SO(4)-invariant functions in R5: + + + hθ hθ hθ + (uz − bz) = ∆5 , (5.43) r t r ,z r − − − hθ hθ hθ + (uz + bz) = ∆5 , (5.44) r t r ,z r k k By the condition, b, u are axi-symmetric, and from section 4, we know that ∇xb, ∇xu 3 + − 3 are bounded in R × (−∞, 0), thus hθ /r, hθ /r are bounded R × (−∞, 0). Let + + hθ hθ M1 = sup and M2 = inf , 3 r 3×(−∞,0) r R ×(−∞,0) R 2344 XIAN-GAO LIU AND XIAOTAO ZHANG

+ 1 and assume that M1 > 0. Applying Lemma 2.1 to the solution hθ /r− 2 (M1+M2) of equation (5.43), considered as an equation in R5 × (−∞, 0). With suitable centers, + 1 + we see that hθ /r ≥ 2 M1 in arbitrarily large parabolic balls, this means hθ is unbounded, a contradiction. Therefore, M1 ≤ 0. In the same way, we ﬁnd that + − M2 ≥ 0. Thus hθ ≡ 0. By the same proof, we conclude hθ ≡ 0. Therefore, + − uθ = bθ ≡ 0 by the deﬁnition of hθ and hθ . Then,

Πr = 0, by the equation (5.12),

(uz)t + Πz = ∆uz, by the equation (5.14),

(bz)t = ∇bz, by the equation (5.18).

Then, uz,r and bz satisfy u (u ) = ∆u − z,r , z,r t z,r r2 b (b ) = ∆b − z,r . z,r t z,r r2 Therefore, applying Lemma 2.1 to uz,r/r and bz,r/r, we conclude that uz,r = bz,r ≡ 0. Thus uz and bz are constant functions in x for each t. Take bz into the equation (5.18), we have the result.

Now we introduce a special case for MHD system in Rn × (−∞, 0). Theorem 5.8. Let (u, b) be a bounded weak solution of the MHD system in Rn × n (−∞, 0). If u = 0 in R × (−∞, 0), then b(x, t) = d0, where d0 is a constant vector. k n Proof. By section 4, with the condition, we know that ∇xb is bounded in R × (−∞, 0). Because u = 0 in Rn × (−∞, 0), the second equation of (4.1) can be written as

bt − ∆b = 0.

Then b1,1 satisﬁes (b1,1)t − ∆b1,1 = 0. Let

M1 = sup b1,1 and M2 = inf b1,1, n n×(−∞,0) R ×(−∞,0) R 1 and assume that M1 > 0. Applying Lemma 2.1 to b1,1 − 2 (M1 +M2), we get that there exist arbitrarily large parabolic balls Q((¯x, t¯),R) = B(¯x, R) × (t¯− R2, t¯) ⊂ n 1 ¯ R × (−∞, 0) such that b1,1 ≥ 2 M1 in Q((¯x, t),R). For such parabolic balls, we have Z 1 n+2 b1,1dxdt ≥ πM1R . (5.45) QR 2 But, on the other hand, we can obtain Z Z n+1 b1,1dxdt = b1n1dxdt ≤ CR , (5.46) QR QR where n is the normal to the boundary of B(¯x, R). When R is big enough, we ﬁnd that (5.45) contradicts to (5.46), unless M1 ≤ 0. By the same way, we conclude that n M2 ≥ 0. Therefore, b1,1 = 0 in R × (−∞, 0). In the same way, we conclude that n bi,j = 0(i, j = 1, 2, ··· , n) in R × (−∞, 0). Therefore, b is constant in x for each t. Take it into the second equation of the system (4.1), we obtain the result. LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2345

6. Singularities. In this section we will consider the potential singularity in the solutions of the Cauchy problem for the MHD system (4.1) and (4.2). We aim to show that singularities generate bounded ancient solution, which are the solutions defined in Rn × (−∞, 0). More precisely, an ancient weak solution of the MHD system is a weak solution defined in Rn × (−∞, 0), and (u, b) is an ancient mild solution if there is a sequence Tl → −∞ such that (u(·,Tl), b(·,Tl)) is well defined n and (u, b) is a mild solution of the Cauchy problem in R × (Tl, 0) with initial data (u(·,Tl), b(·,Tl)).

Lemma 6.1. Let (ul, bl) be a sequence of bounded mild solution of the MHD system n deﬁned in R × (−∞, 0)(for some initial data) with a uniform bound |ul| + |bl| ≤ C, and Tl → −∞. Then, there a subsequence along which (ul, bl) converges locally uniformly in Rn × (−∞) to an ancient mild solution (u, b) satisfying |u| ≤ C in Rn × (−∞). Proof. It is easy to prove with the regularity results in section 4. Now assume that the mild solution develops a singularity in ﬁnite time, and that (0,T ) is its maximal time interval of the existence. There are two situations:

1. there exists a positive number C0, such that

sup n |b(x, τ)| (x,τ)∈R ×(0,t) lim ≤ C0, (6.1) t→T − sup n |u(x, τ)| (x,τ)∈R ×(0,t) − 2. there exists a positive sequence {tk}k≥1 such that tk → T as k → ∞ and

sup n |u(x, τ)| lim (x,τ)∈R ×(0,tk) = 0. (6.2) k→∞ sup n |b(x, τ)| (x,τ)∈R ×(0,tk)

First, we assume that case (1) holds((6.1) holds). Let h(t) = sup n |u(x, t)| x∈R and H(t) = sup0≤s≤t h(s). It is easy to see that there exist a sequence tk ↑ T such that h(tk) = H(tk). Now we choose a sequence of numbers γk ↓ 1. For all k, let n Nk = H(tk) and choose xk ∈ R such that Mk = |u(xk, tk)| ≥ Nk/γk. Then we set

(k) 1 y s v (y, s) = u(xk + , tk + 2 ), (6.3) Mk Mk Mk

(k) 1 y s e (y, s) = b(xk + , tk + 2 ). (6.4) Mk Mk Mk (k) (k) n The functions v and e are deﬁned in R × (Ak,Bk), where 2 2 Ak = −Mk tk, and Bk = Mk (T − tk) > 0 (6.5) and satisfy (k) (k) n (k) |v | ≤ γk, |e | ≤ γkC0 in R × (Ak, 0) and |v (0, 0)| = 1. (6.6) (k) (k) n Also, v , e are mild solution of the MHD system in R × (Ak, 0) with initial (k) (k) data v0 (y) = (1/Mk)u0(xk + y/Mk), e0 (y) = (1/Mk)b0(xk + y/Mk). By Lemma 6.1, there is a subsequence of v(k), e(k) converging to an ancient mild solution (v, e) of the MHD system. Note that, by the construction, we have |v| ≤ 1, |e| ≤ C0 in Rn × (−∞, 0) and v(0, 0) = 1. Then we consider the situation (2)((6.2) holds). Now we let h(t)=sup n |b(x, t)| x∈R and H(t) = sup0≤s≤t h(s), and do the same scaling like (6.3) and (6.4). Therefore, by the construction, we have |e| ≤ 1 in Rn × (−∞, 0) and e(0, 0) = 1, particularly, n v = 0 in × (−∞, 0). Note that lim sup n |e(x, t)| = 0 with the bounded R t→−∞ x∈R 2346 XIAN-GAO LIU AND XIAOTAO ZHANG initial data of (u, b), then by Theorem 5.8, we have b = 0 in Rn × (−∞, 0), a contradiction. Therefore, we just need consider the case (1) in the following content.

Theorem 6.2. Let (u, v) be a weak axi-symmetric solution in R3 × (0,T ) which ∞ 3 0 0 belongs to Lx,t(R × (0,T )) for each T < T . Assume that u satisfies C |u(x, t)| ≤ in 3 × (0,T ), (6.7) p 2 2 R x1 + x2 3 and bz is bounded in R × (0,T ). Then, |u| + |b| ≤ M = M(C) in R3 × (0,T ). Moreover, (u, b) is a mild solution of the MHD system (for a suitable initial data). Proof. We first prove the statement assuming that u is a mild solution (for a suitable initial data). Arguing by contradiction, assume that (u, b) is a mild solution which is bounded in R3×(0,T 0) for each T 0 < T and develops a singularity at time T and case (1) is true, that means (6.1) holds. Let v(k) and b(k) be as in the construction (6.3) 0 0 and (6.4). We write xk = (xk, x3k), where xk = (x1k, x2k). With the assumption 0 (k) (k) (6.7), we find that |xk| ≤ C/Mk. This implies that the functions v , e are axi- symmetric with respect to an axis parallel to the y3-axis and at distance at most C from it. Therefore we can assume (by passing to suitable subsequence) that the limit function v is axi-symmetric with respect to a suitable axis. Moreover, since assumption (6.7) is scale-invariant, it will again be satisfied (in suitable coordinates) 3 by v, and in addition ez = 0 by the assumption of ”bz is bounded in R × (0,T )”. Then applying Theorem 5.5, we see that v = 0. On the other hand, |v(0, 0)| = 1, this is a contradiction. Recall that Lemma 3.4, weak solution u can be decomposed as u = v + ω1 + d1. Thus applying the condition (6.7), we can obtain that d1 = 0. Therefore, u, b are mild solutions of the MHD system.

3 Remark 6.3. If the condition ”bz is bounded in R × (0,T )” in Theorem 6.2 is re- 3 placed by ”bzur = urbz in R ×(0,T )”, the result also holds in the same way(applying Theorem 5.6). If the condition in Theorem 6.2 is replaced by ”u, b are axi-symmetric and uz,z = 3 bz,z = 0 in R × (0,T ), and µ1 = µ2”, then, the result still holds in the same way(applying Theorem 5.7).

Theorem 6.4. Let (u, v) be a weak axi-symmetric solution in R3 × (0,T ) which ∞ 3 0 0 belongs to Lx,t(R × (0,T )) for each T < T . Assume that u satisﬁes

C 3 |u| ≤ √ in R × (0,T ), (6.8) T − t and there exists some R0 > 0 such that C q |b(x, t)| + |u(x, t)| ≤ for x2 + x2 ≥ R and 0 ≤ t ≤ T . (6.9) p 2 2 1 2 0 x1 + x2 3 In addition, assume that bz is bounded in R × (0,T ). Then, |u| + |b| ≤ M = M(C) in R3 × (0,T ). Moreover, (u, b) is a mild solution of the MHD system (for a suitable initial data). Proof. Condition (6.9) implies that (u, b) is a mild solution of the MHD system for a suitable initial data. Therefore, it is smooth in open subsets of R3 × (0,T ). By LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2347

(6.1) and the condition (6.8), we have

CC1 3 |b| ≤ √ in R × (0,T ), (6.10) T − t for some C1 ≥ C0. Let q 0 2 2 f(x, t) = |x ||u(x, t)| = x1 + x2 |u(x, t)|, (6.11) q 0 2 2 g(x, t) = |x ||u(x, t)| = x1 + x2 |b(x, t)|. (6.12) 0 where x = (x1, x2). According to Theorem 6.2, it is enough to prove that f is bounded in R3 × (0,T ). We prove it by contradiction. There are two situations: 1. there exists a positive number C2, such that

sup n |g(x, τ)| (x,τ)∈R ×(0,t) lim ≤ C2, (6.13) t→T − sup n |f(x, τ)| (x,τ)∈R ×(0,t) − 2. there exists a positive sequence {tk}k≥1 such that tk → T as k → ∞ and

sup n |f(x, τ)| lim (x,τ)∈R ×(0,tk) = 0. (6.14) k→∞ sup n |g(x, τ)| (x,τ)∈R ×(0,tk)

Let F (t) = sup 3 f(x, t) and G(t) = sup 3 g(x, t). Assume that f is not R ×(0,t) R ×(0,t) 3 bounded and (6.13) holds. Choose tk ↑ T and xk ∈ R such that Mk = f(xk, tk) = 0 3 −2 F (tk) ↑ ∞. Let λk = |xk|, then for y ∈ R and s ∈ (−T λk , 0), deﬁne (k) (k) 0 0 2 v (y, s) = v (y , y3, s) = λku(λky , λky3 + x3k,T + λks),

(k) (k) 0 0 2 e (y, s) = e (y , y3, s) = λkb(λky , λky3 + x3k.T + λks),

We mention that the sequence λk is bounded because of (6.9). According to the deﬁnition, v(k), e(k) satisfy C |v(k)| ≤ √ in 3 × (−T λ , 0) (6.15) −s R k and CC |e(k)| ≤ √ 1 in 3 × (−T λ , 0). (6.16) −s R k −2 Let sk = −(T − tk)λk . From the construction, we have M |v(k)(y, s)| ≤ k in 3 × (−T λ , s ) (6.17) |y0| R k k and C M |e(k)(y, s)| ≤ 2 k in 3 × (−T λ , s ). (6.18) |y0| R k k By the inequality min{1/a, 1/b} ≤ 2/(a + b)(a, b > 0), the above estimates can lead to 2CM |v(k)(y, s)| ≤ √ k in 3 × (−T λ , s ), (6.19) 0 R k k Mk −s + C|y | and 2CC C M |e(k)(y, s)| ≤ √ 1 2 k in 3 × (−T λ , s ). (6.20) 0 R k k Mk −s + C|y | 2348 XIAN-GAO LIU AND XIAOTAO ZHANG

3 (k) Let γ = {y ∈ R : |(y1, y2)| = 1, y3 = 0}, then, by the construction, (|v (·, sk)|)|γ = 3 2 −2 Mk. Now let e1 = (1, 0, 0). For x ∈ R and τ ∈ (Ak, 0], where Ak = Mk (−T λk − sk), we deﬁne (k) 1 (k) x τ w (x, τ) = v (e1 + , sk + 2 ) Mk Mk Mk and (k) 1 (k) x τ h (x, τ) = e (e1 + , sk + 2 ). Mk Mk Mk 3 p 2 2 Let Λk = {x ∈ R : (x1 + Mk) + x2 ≤ Mk/2}, then we have (k) (k) 3 |w (0, 0)| = 1 and |w (x, τ)| ≤ 2 in (R \ Λk) × (Ak, 0), (6.21) and (k) 3 |h (x, τ)| ≤ 2C2 in (R \ Λk) × (Ak, 0). (6.22) Note that (6.19), (6.20), (6.15), (6.16) imply

(k) 2CMk |w (x, τ)| ≤ √ in Λk × (Ak, 0), (6.23) p 2 2 Mk −τ + C (x1 + Mk) + x2

(k) 2CC1C2Mk |h (x, τ)| ≤ √ in Λk × (Ak, 0), (6.24) p 2 2 Mk −τ + C (x1 + Mk) + x2 C |w(k)(x, τ)| ≤ √ in 3 × (−T λ , 0) (6.25) −τ R k and CC |h(k)(x, τ)| ≤ √ 1 in 3 × (−T λ , 0). (6.26) −τ R k Note that (w(k), h(k)) are the mild solution of the MHD system. Therefore, from the estimate (6.25) and (6.26), we can choose a subsequence of the sequence (w(k), h(k)), (k) (k) (k) (k) which we denote by (w1 , h1 ), such that the sequence (w1 , h1 ) converge uniformly on compact subsets of R3 × (−∞, 0) to an ancient mild solution (w, h). (k) (k) Since the solutions (w , h ) are axi-symmetric and Mk ↑ +∞, it is easy to prove that w is independent of the x2-variable. In addition, λk is bounded, thus w is also independent of the x2-variable. Moreover, because bz is bounded in 3 3 ¯ R × (0,T ), we have hz = 0 in R × (−∞, 0). Applying Theorem 5.1 to (w, ¯ h), T ¯ T ¯ wherew ¯ = (w1, w3) , h = (h1, h3) . We conclude that (w, ¯ h) must vanish iden- tically, that means w = h = 0 in R3 × (−∞, 0). This would give a contradiction with |w(k)(0, 0)| = 1, if we could prove that w(k)(0, 0) → w(0, 0), which is (k) not obvious since supx |w (x, t)| may not be uniformly bounded as τ → 0. To this aim, we use the representation formula (3.6) and (3.9) with the initial data (k) (k) (k) (k) (k) (k) w1 (x, −1), h1 (x, −1). Let fjl = −w1l w1j + h1l h1j , then Z Z t Z (k) (k) w1i (x, t) = Γ(x − y, t)w1i (y, −1)dy + Kijl(x − y, t − s)fjl(y, s)dyds n n R −1 R Z t Z =I1 + Kijl(x − y, t − s)fjl(y, s)dyds 3 −1 R \Λk Z t Z + Kijl(x − y, t − s)fjl(y, s)dyds −1 Λk (k) (k) (k) =I1 + I2 + I3 , LIOUVILLE THEOREM FOR MHD SYSTEM AND ITS APPLICATIONS 2349 where (x, t) ∈ B¯(0, 1) × [−1, 0]. By (6.21), (6.22), the estimate (3.11) and heat (k) (k) kernel theory, it is easy to see the sequence I1 ,I2 have subsequence converge uniformly. By (6.23), (6.24) and the kernel decay (3.8), we have Z 0 Z +∞ Z 1 1 (k) √ 0 |I3 | ≤ C 2 2 2 dx dx3dτ 0 0 2 −1 −∞ |x |≤Mk/2 ( −τ + |x |/Mk) (Mk /4 + x3) (k) By direct calculation, I3 → 0 as k → ∞. Therefore, we can choose a subsequence (k) (k) (k) of the sequence w1 , which we again denote by w , such that the sequence w converge uniformly in B¯(0, 1) × [−1, 0]. For the case (2)((6.14) holds), we take Mk = g(xk, tk) = G(tk), then in the same (k) (k) way, we get the result(When we estimate I3 for h1 , we use (3.13) and heat kernel theory).

Acknowledgments. The ﬁrst author’s research is partially supported by NSFC No.11131005.

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