MHD Equations with Regularity in One Direction

Definition 1. Let (u₀, b₀) ∈ L²(R³) satisfying ∇·u₀ = ∇·b₀ = 0, T > 0 be given. A measurable pair (u₀, b₀) on (0, T) is said to be a weak solution to (1) provided that

• (1)

  (u, b) ∈ L^∞(0, T; L²(R³))∩L²(0, T; H¹(R³));

• (2)

  (1)_1,2 are satisfied in the sense of distributions;

• (3)

  the energy inequality is given as

  $$ ()

•  

  for all t ∈ [0, T].

Lemma 2. Suppose that $$; then, one has

where C is a generic constant independent of f and g; 1 ≤ α, q ≤ ∞, and 1 < r ≤ ∞ satisfy

Proof . Consider the following

where in the last inequality we have used Hölder inequality with and the Gagliardo-Nirenberg inequality.

Theorem 3. Let (u₀, b₀) ∈ L²(R³) satisfying ∇·u₀ = ∇·b₀ = 0, T > 0 be given. Assume that the measurable pair (u, b) is a weak solution as in Definition 1 on (0, T). If

then (u, b) ∈ C^∞((0, T) × R³).

Proof. For any ɛ ∈ (0, T), we can find a δ ∈ (0, ɛ) such that

since (u, b) ∈ L²(0, T; H¹(R³)) as in Definition 1. Our strategy is to show that under condition (10) the weak solution is in fact strong; that is, which would imply the smoothness of the solution via standard energy estimates and Sobolev embeddings. Due to the arbitrariness of ɛ, we complete the proof.

To prove (12), we multiply (1)₁ by −Δu and (1)₂ by −Δb to get

Integration by parts formula together with the divergence-free conditions ∇·u = ∇·b = 0 yields

where we use the summation convention; that is, the repeated index (say, i here) is automatically summed over {1,2, 3}.

Substituting (14) into (13) and using a simple Cauchy-Schwarz inequality, we obtain

where 0 < ɛ ≪ 1 is to be determined later on.

Due to the Calderón-Zygmund inequality,

(R_j being the Riesz transform) we may take that the ɛ in (15) equals 1/2C₁ to get

To further bound I₁, I₂, we introduce some notations. Denote

Then, by (10), we have Invoking Lemma 2 with q as in (10), r as in (18), and α as in (19), we may estimate I₁ as where C₁ is as in (16). Applying Hölder inequality with (21) becomes when (qγ − 3)/2q(r − 1) = 0, or when (qγ − 3)/2q(r − 1) ∈ (0,1), or when (qγ − 3)/2q(r − 1) = 1.

Similarly, I₂ can be dominated as

when (qγ − 3)/2q(r − 1) = 0, or when (qγ − 3)/2q(r − 1) ∈ (0,1), or when (qγ − 3)/2q(r − 1) = 1.

Thus, if q = 3/γ, then, combing (23) and (26), we deduce from (17) that

Applying Gronwall inequality and noting that u ∈ L²(δ, T; H¹(R³)) as in Definition 1, we obtain (12) as desired.

If 3/γ < q < 1/(γ − 1), we gather (24) and (27) into (17) to get

To deduce (12) by applying Gronwall inequality to (30), we need only to show that This is indeed true. First, simple interpolation inequality together with the fact that yields Second, the integrability indices as in (31) satisfy and α ∈ (2,6), by (18) and (19).

If, however, q = 1/(γ − 1), then (qγ − 3)/2q(r − 1) = 1, and, combining (25) and (28), we deduce from (17) that

where we recall p = 2q/(qγ − 3) from (10). Applying Gronwall inequality and noting that u ∈ L^∞(δ, T; L²(R³)) as in Definition 1, we obtain (12) as desired.

The proof of Theorem 3 is completed.