Large Time Existence of Special Strong Solutions to MHD Equations in Cylindrical Domains

. We investigate the problem of the existence of regular solutions to the three-dimensional MHD equations in cylindrical domains with perfectly conducting boundaries and under the Navier boundary conditions for the velocity ﬁeld. We show that if the initial and external data do not change too rapidly along the axis of the cylinder, then there exists a unique regular solution for any ﬁnite time.


Introduction
In this work we examine the existence and regularity of solutions to the magnetohydrodynamics equations (MHD) in 3d cylindrical domains. The governing system of equations reads where Ω = Ω × (−a, a), a > 0 and Ω ⊂ R 2 is a bounded subset with the C 2 -boundary. Here, the unknowns are: • the velocity field v : • the pressure p : Ω → R, p = p(x, t) = p(x 1 , x 2 , x 3 , t).
The external force f : Ω → R 3 , f = f (x, t), f = (f 1 (x 1 , x 2 , x 3 , t), f 2 (x 1 , x 2 , x 3 , t), f 3 (x 1 , x 2 , x 3 , t)), the viscosity coefficients ν, ν κ > 0 and the initial conditions v t0 , H t0 are given. System (1.1) describes the motion of a viscous, incompressible and resistive fluid filling a region Ω (see e.g. [1,2]). This motion under the presence of the magnetic fields generates electric field and electric currents, thereby evoking forces which alter the magnetic field and the fluid motion itself. Clearly, (1.1) is a combination of the Navier-Stokes equations (NSE) and Maxwell's equations. Since the problem of regularity of weak solutions to the 3d NSE for arbitrary smooth data is still open, we cannot get any better results for (1.1) than there are for the ordinary NSE. Therefore, to obtain a new result for (1.1) we need to make further assumptions. Two basic strategies are: either take the initial data small enough or require some conditional regularity of v (see e.g. [3]).
In the following paper we are interested in proving the existence of strong solutions (see Definition 2.1) to system (1.1) without any assumptions on the magnitude of the L 2 -norms of the initial and external data. Instead, we assume that the L 2 -norms of the derivatives of the initial and external data along the axis of the cylinder Ω are small. More precisely, we prove the following: Theorem 1. Suppose that v t0 , H t0 ∈ H 1 (Ω), div v t0 = div H t0 = 0, f ∈ L 2 (Ω T ), f ,x3 ∈ L 2 (t 0 , T ; L 6 5 (Ω)). Fix T > t 0 . If δ 2 (T ) is sufficiently small then there exists a unique strong solution to (1.1). This unique solution satisfies the following exponential growth estimate for all t

Let us introduce
where a generic constant c depends on ν, ν κ and Ω but not on T .
Before we briefly discuss the proof, we should clarify that all constants in the sequel are denoted by c and they may vary from line to line. They may depend on ν, ν κ and Ω but neither on t nor on the initial and external data. The definition of the space V k 2 (Ω T ), k ∈ N, is given in Sect. 2. The proof of the above theorem is based on the refined energy estimates and a fixed point principle. We derive these estimates by utilizing the multiplicative Sobolev inequality (see e.g. [4, Remark 2.1], [5,Remark (iii)]), which distinguishes the differentiation with respect to x 3 from the whole gradient (see Lemma 3.7). To exploit this inequality we introduce an auxiliary problem for (v ,x3 , H ,x3 ) (see Lemma 4.2). The first energy estimate for the solutions to this problem reads v ,x3 Next, we test (1.1) 1,2 with −Δv and −ΔH, respectively, and use the multiplicative Sobolev inequality to estimate the non-linear terms (see Lemma 6.1). First we obtain where ϕ is a positive, increasing function of exponential type and g(t) is a function such that T t0 g(t) dt is bounded only by the data. Using the continuity argument (see e.g. [6,Ch. 1,§1.3]) it is clear that for sufficiently small δ 2 (T ) we can eliminate ϕ from the right-hand side. Next, we use the Gronwall inequality, basic energy estimates and the continuity argument again. In the end we can conclude the estimate for v and H (see Lemma 6.1). Finally, the existence of solutions to (1.1) follows from the Leray-Schauder fixed point theorem (see Lemma 7.1).
There is a slight problem with the above idea, namely the multiplicative Sobolev inequality cannot be utilized directly for v and H because of the boundary conditions (1.1) 4 Now we easily observe that the functions v −v and H −H satisfy the assumptions of the multiplicative Sobolev inequality (see Remark 3.8), whereas the functionsv,H are two-dimensional, i.e. they depend merely on x 1 and x 2 . The mean value operator that we have just introduced, was successfully used in e.g. [4] or more recently in [7]. Let us shortly discuss the consequences of the smallness of δ 2 (T ). Let u ∈ {v, H}. By the definition In (3.2) we shall see that conditions (1.1) 4,5 imply that v 3 = H 3 = 0 on the top and the bottom of the cylinder. Thus, we have the Poincaré inequality for u 3 with respect to x 3 and we easily get . This means that the third components of the initial velocity and magnetic fields must be close to zero. Thus, the initial flow is close to two-dimensional flow and in light of Lemma 4.2 it remains so for any time T . Note that this limitation is a direct consequence of the boundary conditions for v and H on the top and the bottom of the cylinder. It would not hold if Ω was e.g. a cylinder periodic with respect to x 3 . Then, v 3 (t 0 ) L2(Ω) and H 3 (t 0 ) L2(Ω) could be arbitrarily large.
For further discussion and references of MHD equations in periodic, cylindrical domains we refer the reader to e.g. [8] and [9]. Cylindrical domains play an important role in studying plasma physics (see [10] and [11]) and to some extent are used as approximations of a torus (see [12]).
We end this Introduction remarking that to the best of our knowledge results like Theorem 1 have not appeared in the literature. However, in case of the whole space and slightly more general system (i.e. with the fractional Laplacian) a result similar to ours follows from [13][Theorems 1.2 and 1.4]. Furthermore, if we assume v = 0 on the boundary, then (1.1) was studied in [14][15][16] and [17], where the existence of regular solutions was proved either for short time or for any time but under the assumption of the smallness of the initial data. For a detailed summary of various results related to MHD equations we refer the reader to the Introduction in [18].

Notation
The Boundary of the Domain The boundary S := ∂Ω is a union of three sets: (the lateral surface), where ϕ(x 1 , x 2 ) = c ϕ is a closed C 2 -curve in the plane x 3 = const.

Shorthand Notation
Some formulas appear often in the sequel. To shorten the notation we use d 2 (T ) -a constant from the energy estimate, see Theorem 1, δ 2 (T ) -the smallness assumption from Theorem 1, E 2 (T ) -a sum of some norms of the initial and the external data from Theorem 1,

Function Spaces
Throughout the paper we use the standard Lebesgue L p (Ω) and Sobolev W k 2 (Ω) = H k (Ω) spaces. We also need V k 2 (Ω T ), which is defined as follows:

Weak and Strong Solutions
Definition 2.1. By a weak solution to problem (1.1) we mean a pair of functions (v, H)

Auxiliary Remarks and Tools Boundary Conditions
Since Ω is a cylinder, we can easily determine the unit outward normal vector n and the unit tangent vectors τ i , i = 1, 2. We have: Using the above formulas in (1.1) 4,5 we immediately get the following identities on S T and Let us observe that the discussed boundary conditions suggest the following integration by parts formula: Let u and w belong to H 1 (Ω) and satisfy rot u × n = rot w × n = 0 and u · n = w · n = 0. Then Finally, let us note that there are other possible choices for the boundary conditions for the magnetic field. For an insightful discussions we refer the interested reader into [19][20][21] and [22].

Divergence-Free Magnetic Field
At first sight it might seem that system (1.1) is overdetermined since it contains four equations for H while H has only three unknown components. This is perfectly fine and consistent with the MHD theory (see e.g. [23,Ch.2 ]). However, it may appear that condition (1.1) 5 violates the equation div H = 0. This issue was partly addressed in [15, Lemma 2.1]. In Lemma 3.12 we show that if div H t0 = 0 and H satisfies (1.1) 5 , then div H = 0 for all t ≥ t 0 .

Mean Value Operator
The proof of Theorem 1 is based on the multiplicative Sobolev inequality, which requires that a given function u satisfies: u 3 | ST = u 3 | SB = 0 and Ω u j (x 1 , x 2 , x 3 ) dx 3 = 0 for j = 1, 2. Since we do not know whether Ω v j dx 3 and Ω H j dx 3 are equal to zero for j = 1, 2, so we define the mean value for v = (v 1 , v 2 ) and H = (H 1 , H 2 ) with respect to the x 3 -variable as follows: if z : R 3 → R then we writē Such defined functions have several properties, which we summarize in lemmas below.
Proof. Consider the first integral. We see that in light of (3.6) 4 we only need to examine From (3.6) 1 it follows that the above integral vanishes.
To investigate the second integral we use (3.6) 4 twice. Since divb = 0 (see Lemma 3.1) we conclude that the third component of Δb is zero and the first two components do not depend on x 3 . Thus, we need to examine The first integral on the right-hand side vanishes due to (3.6) 1 . The second integral is equal to The boundary conditions for a imply a ,x3 = 0 (we saw it in (3.2) 2 ). This concludes the proof.
Proof. We have The subsequent lemma will be very useful for estimating the integrals originating from the non-linear terms.
Multiplying the above formula by c and integrating over Ω results in four integrals on the right-hand side. We denote these integrals by J k . If c =w then which follows from Lemma 3.1. Similarly J 3 = 0 and (3.7a) is proved. We now set c = w −w. Then, Lemma 3.1 implies that J 4 = 0 which justifies (3.7b). This ends the proof.
Remark 3.6. From the above Lemma it follows that In light of the above Remark we see that every integral containing the non-linear term generates four new integrals. As we shall see these new integrals will be much easier to handle.

Auxiliary Tools
Proof. In cylindrical domains we have the following Sobolev inequality (see [4,Remark 2 Using the Poincaré inequality with respect to x 3 we conclude the proof.
Moreover, if w satisfies rot w × n = 0, w · n = 0 on the boundary, then we also have where the first equality follows from (3.6) 1 . From (3.6) 2 we infer that a −a (g −ḡ) dx 3 = 0.
Since g 3 −ḡ 3 = u 3,x −ū 3,x = 0 (see (3.2) and Lemma 3.1), all assumption from Lemma 3.7 are satisfied and the Remark is proved. Lemma 3.9. (see Theorem 1.1 in [24]) Suppose that z is a solution to the following problem rot z = α, div z = β, z × n = 0 or z · n = 0. Then The direct consequence of the above Lemma is the following useful inequality: Lemma 3.10. Suppose that (w −w) ∈ H 2 (Ω) and rot w × n = 0, w · n = 0. Then Proof. We set z = w −w in Lemma 3.9. Clearly z · n = 0 on S (see Lemma 3.1), thus Next we set z = rot (w −w). Then z × n = 0 (see Lemma 3.1). Then, by Lemma 3.9 and the above inequality we get and since div (w −w) = 0 we conclude the proof. Lemma 3.11. (Gagliardo-Nirenberg interpolation inequality, see [25]) Suppose that u : Ω → R, Ω ⊂ R n . Fix 1 ≤ q, r ≤ ∞ and m ∈ N. Suppose that a real number α ≥ 0 and j ∈ N are such that , where s > 0 is arbitrary and the constants c 1 and c 2 do not depend on u.
Finally, we justify why (1.1) 5 does not violate the condition div H = 0. Proof. Let h := div H. We take div of both sides in (1.1) 2 and we get
We only need to find the boundary condition for h on S L . Using the vector identities −Δ = rot rot −∇ div,

Multiplying this equation by n and projecting the result onto S L yields
Since e 3 = [0, 0, 1] is tangent to S L we add to the above equality ∂ x3 (rot H × n) = 0
Summarizing, we get Multiplying the first equation by h, integrating with respect to Ω and t we easily see that h ≡ 0 for all t ∈ (t 0 , T ) a.e. This concludes the proof.

Existence and Estimates of Solutions to Evolutionary Systems Lemma 3.13. Let us consider the Stokes problem
. Sketch of the proof. An identical problem but with slip boundary conditions with friction was considered in [26]. In cylindrical domains there is a tight relation between the Navier and slip boundary conditions, see e.g. [27,Lemma 6.5]. If we were to follow the details from [26], we would have to make sure that the problem for p has proper boundary conditions. Applying div to both sides of (3.9) 1 yields Δp = div F.
To derive the boundary conditions we multiply (3.9) 1 by n and project the result onto S. From (3.9) 3,4 and (3.2) we immediately obtain −νΔv · n + ∇p · n = F · n on S. Vol. 20 (2018) Strong Solutions to MHD Equations in Cylindrical Domains 1023 The first term on the left-hand side is zero. Indeed, we have −Δv · n = rot rot v · n = 0 because the tangential component of rot v is zero and therefore by the Stokes' theorem the normal component of rot rot v is zero, too. Summarizing, we have Δp = div F in Ω, ∇p · n = F · n on S, (3.10) which is a simpler version of Problem (2.29) from [26] (in [26] condition (3.10) reads ∇p · n = F · n + b). Therefore, the whole proof of the main Theorem in [26] can be safely repeated.
Lemma 3.14. Consider the following initial-boundary value problem Then, there exist a unique solution H such that H ∈ W 2,1 p (Ω T ) and Sketch of the proof. Since the boundary is not smooth we proceed as follows. First we reflect H outside Ω with respect to x 3 . This is possible due to (3.11) 2,3 , which imply (3.2). In that way we obtain (3.11) inΩ := Ω × (−3a, 3a).
In the second step we introduce a partition of unity

Basic Energy Estimates
In this Section we establish energy estimates for (v, H) and (v ,x3 , H ,x3 ).
Proof. We multiply (1.1) 1,2 by v and H, respectively, integrate over Ω and add to each other The sum of the two first integrals on the right-hand side is equal to zero. To the last integral we apply the Hölder and Cauchy inequalities. To get H 1 -norms we use Lemma 3.9. We have d dt v Integration with respect to time concludes the proof.

(4.3)
Proof of Lemma 4.2. We multiply (4.2) 1 and (4.3) 1 by w and K, respectively, integrate over Ω and add to each other 1 2 On the right-hand side we have nine terms under the integral sign. The second and the ninth terms vanish due to (4.2) 2 and (4.3) 2 . The fourth and the seventh terms have opposite sign. Hence Next, we apply Lemma 3.9 on the left-hand side and the Hölder, Young and Sobolev inequalities on the right-hand side 1 2 . Vol. 20 (2018) Strong Solutions to MHD Equations in Cylindrical Domains 1025 Choosing 1 and 2 small enough we get .

(4.4)
The Gronwall inequality yields Integrating (4.4) with respect to time and using the above inequality along with the Sobolev inequality ends the proof.

Auxiliary Estimates
We start this section with a simple observation that follows from the Gagliardo-Nirenberg inequality.
Remark 5.1. If we set p = 4, r = q = 2, j = 0 and m = 1 in Lemma 3.11, then If additionallyz| ∂Ω = 0 or Ω z dx = 0, then (see e.g. [ Below we prove a couple of results related to integrals corresponding to non-linear terms. We saw in Remark 3.6 that every integral involving a non-linear term splits into four integrals. These four integrals are estimated separately in a series of Lemmas 5.3-5.6.
Integrating by parts and using divw = 0 andw · n = 0 we conclude the proof. Lemma 5.3. Suppose that a ∈ H 1 (Ω), b, w ∈ H 2 (Ω) and div b = 0. Then Proof. In light of the Hölder and Young inequalities To estimate the L 4 -norms we use Remark 5.1 and the Young inequality To conclude the proof we use Remark 3.4.

Lemma 5.4. Suppose that a, b ∈ {v, H}. Then
Proof. Using the Hölder and the Young inequalities we get By the interpolation inequality between L p -spaces . Remark 3.8, the embedding theorem and the Cauchy inequality yield . We conclude the proof using Lemma 3.10 and Remark 3.4.

Lemma 5.5. Suppose that a, b ∈ {v, H}. Then
Proof. The proof is identical to the proof of Lemma 5.4. Proof. Sinceā 3 = 0 we get from the application of the Hölder and the Young inequalities where ∇ = [∂ x1 , ∂ x2 ]. By Remark 3.8 we get 4 3 L2(Ω) . Application of the Young inequality along with Lemma 3.10 and Remark 3.4 ends the proof.

Higher Order Estimates
for any t ∈ (t 0 , T ).

Assumption (ii).
In light of definition of Φ we easily see that this assumption is met.
Assumption (iii). This follows immediately from Lemma 6.1.