A note on the inviscid limit of the incompressible MHD equations

A NOTE ON THE INVISCID LIMIT OF THE INCOMPRESSIBLE MHD EQUATIONS

arXiv:1612.04645v2 [math.AP] 8 May 2017

JINLU LI AND ZHAOYANG YIN Abstract. In this paper, we prove that as the viscosity and resistivity go to zero, the solution of the Cauchy problem for the incompressible MHD equations converges to the solution of the ideal MHD equations in the same topology with the initial data. Our proof mainly depends on the method introduced by the paper [5] and the constructions of the incompressible MHD equations.

1. Introduction and main result In the paper, we consider the following Cauchy problem of the incompressible MHD equations:   ∂t u + u · ∇u − µ∆u + ∇P = b · ∇b, (1.1) ∂t b + u · ∇b − ν∆b = b · ∇u,  divu = divb = 0, (u, b)| = (u , b ), t=0 0 0

where the unknowns are the vector fields u : R × Rd → Rd , b : R × Rd → Rd and the scalar function P . Here, u and b are the velocity and magnetic, respectively, while P denotes the pressure, µ ≥ 0 is the viscosity coefficient and ν ≥ 0 is the magnetics diffusive coefficient. We expect the above equations converge to the following ideal MHD equations in the corresponding Besov spaces of the initial data when µ and ν tend to 0:   ∂t u + u · ∇u + ∇P = b · ∇b, (1.2) ∂t b + u · ∇b = b · ∇u,  divu = divb = 0, (u, b)| = (u , b ). t=0 0 0

The MHD system is a well-known model which governs the dynamics of the velocity and magnetic fields in electrically conducting fluids such as plasmas, liquid metals, and salt water. The vanishing viscosity limit problem is one of the challenging topics in fluid dynamics. It has been studied by many authors in H s space (see [1, 4, 9, 10, 11, 12]), where the main approaches were the energy arguments and depending strongly on the treatments of the trilinear forms which lead to the (µ, ν) independent estimates. Most of their paper just prove in the lower regularity comparing the topology for the initial data. Therefore, we want to solve the inviscid limit of the incompressible MHD equations in the same topology with the initial data. Our proof mainly relies on the method introduced by the paper [5] and the constructions of the incompressible MHD equations. Then, our argument can state as follows: 2010 Mathematics Subject Classification. 76W05. Key words and phrases. Incompressible MHD equations, inviscid limit. 1

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J. LI AND Z. YIN

Theorem 1.1. Let d ≥ 2 and s > 1 + d2 . Suppose that un0 ∈ H s (Rd ) goes to u0 ∈ H s (Rd ) in H s (Rd ) and bn0 ∈ H s (Rd ) goes to b0 ∈ H s (Rd ) in H s (Rd ) when n goes to infinity. Let µn , νn ≥ 0 and µn , νn go to 0 when n goes to ∞. If divu0 = divb0 = divun0 = divbn0 = 0, then there exists a positive T > 0 independent of n such that (un , bn ) ∈ C([0, T ]; H s (Rd )) be the solution of  n n n n n n  ∂t u + u · ∇u − µn ∆u + ∇Pn = b · ∇b , (1.3) ∂t bn + un · ∇bn − νn ∆bn = bn · ∇un ,  divun = divbn = 0, (un , bn )| = (un , bn ), t=0 0 0 and (u, b) ∈ C([0, T ]; H s(Rd )) be the solution of (1.2) with initial data (u0 , b0 ). Moreover, there holds s d = 0. lim ||un − u||L∞ T (H (R ))

n→∞

If the viscosity coefficient is equal to the magnetics diffusive coefficient, that is ν = µ, then the system (1.1) becomes   ∂t u + u · ∇u − µ∆u + ∇P = b · ∇b, (1.4) ∂t b + u · ∇b − µ∆b = b · ∇u,  divu = divb = 0, (u, b)| = (u , b ). t=0 0 0 Then, we have a generalized conclusion as follows:

Theorem 1.2. Let d ≥ 2. Assume that (s, p, r) satisfies s>

d d + 1, p ∈ [1, ∞], r ∈ (1, ∞) or s = + 1, p ∈ [1, ∞], r = 1. p p

(1.5)

s s s Suppose that un0 ∈ Bp,r (Rd ) goes to u0 ∈ Bp,r (Rd ) in Bp,r (Rd ) when n goes to infinity. Let µn ≥ 0 and µn goes to 0 when n goes to ∞. If divu0 = divb0 = divun0 = divbn0 = 0, then there exists a positive T > 0 independent of n such that (un , bn ) ∈ s C([0, T ]; Bp,r (Rd )) be the solution of  n n n n n n  ∂t u + u · ∇u − µn ∆u + ∇Pn = b · ∇b , (1.6) ∂t bn + un · ∇bn − µn ∆bn = bn · ∇un ,  divun = divbn = 0, (un , bn )| = (un , bn ), t=0 0 0 s and (u, b) ∈ C([0, T ]; Bp,r (Rd )) be the solution of (1.2) with initial data (u0 , b0 ). Moreover, there holds

lim ||un − u||L∞ s d = 0. T (Bp,r (R ))

n→∞

s Remark 1.3. We prove local well-posedness in Bp,r where (s, p, r) satisfies (1.5), improving the previous result in [7].

Notations. Given a Banach space X, we denote its norm by k · kX . Since all spaces of functions are over Rd , for simplicity, we drop Rd in our notations of function spaces if there is no ambiguity. The symbol A . B denotes that there exists a constant c¯0 > 0 independent of A and B, such that A ≤ c¯0 B. The symbol A ≃ B represents A . B and B . A.


3

2. Preliminaries In this section we collect some preliminary definitions and lemmas. For more details we refer the readers to [2]. Let χ : Rd → [0, 1] be a radial, non-negative, smooth and radially decreasing function which is supported in B , {ξ : |ξ| ≤ 43 } and χ ≡ 1 for |ξ| ≤ 45 . Let ϕ(ξ) = χ( 2ξ )−χ(ξ). Then ϕ is supported in the ring C , {ξ ∈ Rd : 43 ≤ |ξ| ≤ 83 }. For ˙ q u = F −1 (ϕ(2−q ·)F u), u ∈ S ′ , q ∈ Z, we define the Littlewood-Paley operators: ∆ ˙ q u for q ≥ 0, ∆q u = 0 for q ≤ −2 and ∆−1 u = F −1 (χF u), and Sq u = ∆q u = ∆ F −1 χ(2−q ξ)F u . Here we use F (f ) or fb to denote the Fourier transform of f . s s We define the standard vector-valued Besov spaces Bp,r and B˙ p,r of the functions d d u : R → R with finite norms which are defined by s kukBp,r , (2js k∆j ukLp )j∈Z ℓr , ˙ j ukLp )j∈Z r . , (2js k∆ kukB˙ p,r s ℓ

Remark 2.1. Let s ≥ 0 and 1 ≤ p, r ≤ ∞. There holds

s ; ≃ ||(Id − ∆−1 )u||Bp,r ||(Id − ∆−1 )u||B˙ p,r s

s , . ||u||Bp,r ||u||B˙ p,r s

if

s > 0;

0 ||u||Lp . ||u||Bp,1 . ||u||B˙ p,1 0 .

Next we recall nonhomogeneous Bony’s decomposition from [2]. uv = Tu v + Tv u + R(u, v), with Tu v ,

X j

Sj−1 u∆j v,

R(u, v) ,

X X j

∆j u∆k v.

|k−j|≤1

This is now a standard tool for nonlinear estimates. Now we use Bony’s decomposition to prove some nonlinear estimates which will be used for the estimate of pressure term. Remark 2.2. Note that Sj−1u∆j v is supported in the 2j C˜ where the annulus C˜ , 1 {ξ ∈ Rd : 12 ≤ |ξ| ≤ 12 }. 3 Remark 2.3. Let f be a smooth function on Rd \{0} which satisfies f (λξ) = λm f (ξ). Then, we have s−m ≤ C||u|| ˙ s , ||f (D)u||B˙ p,r Bp,r and s−m ≤ C||Tu v||B s . ||f (D)Tu v||Bp,r p,r

The main properties of the paraproduct and remainder are described below. Lemma 2.4. Let s ∈ R and 1 ≤ p, r ≤ ∞. Then there exists a constant C, depending only on d, p, r, s, such that s . s kTu vkBp,r ≤ C||u||L∞ ||v||Bp,r

Let s > 0 and 1 ≤ p, r ≤ ∞. Then there exists a constant C, depending only on d, p, r, s, such that s , s kR(u, v)kBp,r ≤ CkukL∞ kvkBp,r or s −1 kvk s+1 . kR(u, v)kBp,r ≤ CkukB∞,r Bp,∞

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J. LI AND Z. YIN

Let s > 1 and 1 ≤ p, r ≤ ∞. Then there exists a constant C, depending only on d, p, r, s, such that s s−1 . kR(u, v)kBp,r ≤ C(||u||L∞ + ||∇u||L∞ )kvkBp,r

Lemma 2.5. Let s > 0 and 1 ≤ p, r ≤ ∞. Then there exists a constant C, depending only on d, p, r, s, such that s s s + ||v||L∞ ||u||Bp,r ||uv||Bp,r ≤ C ||u||L∞ ||v||Bp,r .

Lemma 2.6. Assume (σ, p, r) satisfies (1.5). Then there exists a constant C, deσ pending only on d, p, r, σ, such that for all u, f ∈ Bp,r with div u = 0, σ−1 ≤ C||u|| σ−1 ||f ||B σ . ||u · ∇f ||Bp,r Bp,r p,r

Proof. Due to div u = 0, then we decompose the term u · ∇f into u · ∇f =

d X

Tui ∂i f +

d X i=1

i=1

T∂i f ui +

d X

∂i R(ui , f ).

i=1

According to Lemma 2.4, we have σ−1 ||f ||B σ , σ−1 ≤ C||u||L∞ ||∇f ||B σ ≤ C||u||Bp,r ||Tui ∂i f ||Bp,r p,r p,r

σ−1 ||f ||B σ , σ−1 ≤ C||∇f ||L∞ ||u||B σ ≤ C||u||Bp,r ||T∂i f ui ||Bp,r p,r p,r

i σ σ−1 ≤ C||R(u , f )||B σ ≤ C||u||L∞ ||f ||Bp,r ||∂i R(ui , f )||Bp,r p,r σ−1 ||f ||B σ . ≤ C||u||Bp,r p,r

Combining this results, we complete the proof of this lemma.

Lemma 2.7. Assume (σ, p, r) satisfies (1.5). Then there exists a constant C, deσ pending only on d, p, r, σ, such that for all u, v ∈ Bp,r with div u = div v = 0, σ σ ||∇(−∆)−1 div (u · ∇v)||Bp,r ≤ C (||u||L∞ + ||∇u||L∞ )||v||Bp,r σ . + (||v||L∞ + ||∇v||L∞ )||u||Bp,r

Proof. It is easy to get

σ ||∇(−∆)−1 div (u · ∇v)||Bp,r

−1 div (u · ∇v)||Lp .||∇(−∆)−1 div (Id − ∆−1 )(u · ∇v)||B˙ p,r σ + ||∇(−∆) −1 div div (u ⊗ v)||Lp , I + II. .||∇(−∆)−1 (Id − ∆−1 )(∇u · ∇v)||B˙ p,r σ + ||∇(−∆)


5

According to Remark 2.1 and Lemma 2.4, we have I . ||

d X

∇(−∆)−1 (Id − ∆−1 )[T∂j ui (∂i v j ) + T∂i vj (∂j ui )]||B˙ p,r σ

i,j=1

+ ||

d X

∇(−∆)−1 ∂i ∂j (Id − ∆−1 )R(ui , v j )||B˙ p,r σ

i,j=1

.

d X

σ−1 ||(Id − ∆−1 )[T∂j ui (∂i v j ) + T∂i vj (∂j ui )]||B˙ p,r

i,j=1

+

d X

σ+1 ||(Id − ∆−1 )R(ui , v j )||B˙ p,r

i,j=1

.

d X

σ−1 + ||[T∂j ui (∂i v j ) + T∂i vj (∂j ui )]||Bp,r

i,j=1

d X

σ+1 ||R(ui , v j )||Bp,r

i,j=1

σ + (||v||L∞ + ||∇v||L∞ )||u||B σ , .||∇u||L∞ ||v||Bp,r p,r

and 1 II.||∇(−∆)−1 div div (u ⊗ v)||B˙ p,1 0 .||u ⊗ v||B ˙ 1 .||u ⊗ v||Bp,1 p,1 σ .||u||L∞ ||v||B σ + ||v||L∞ ||u||B σ . .||u ⊗ v||Bp,r p,r p,r

This completes the proof of this lemma.

Lemma 2.8. Assume (σ, p, r) satisfies (1.5). Then there exists a constant C, deσ pending only on d, p, r, σ, such that for all u, v ∈ Bp,r with div u = div v = 0, σ−1 ≤ C||u|| σ−1 ||v||B σ , ||∇(−∆)−1 div (u · ∇v)||Bp,r Bp,r p,r

or σ−1 ≤ C||v|| σ−1 ||u||B σ . ||∇(−∆)−1 div (u · ∇v)||Bp,r Bp,r p,r

Proof. Due to the fact that div (u · ∇v) = div (v · ∇u), then we have σ−1 ||∇(−∆)−1 div (u · ∇v)||Bp,r

−1 σ−1 + ||∇(−∆) div ∆−1 (u · ∇v)||Lp .||∇(−∆)−1 div (Id − ∆−1 )(u · ∇v)||B˙ p,r −1 σ−1 + ||∇(−∆) div (u · ∇v)||Lp , III + IV. .||(Id − ∆−1 )(u · ∇v)||B˙ p,r

Due to Remarks 2.1-2.3, Lemma 2.4 and Lemma 2.6, we get σ−1 .||u · ∇v|| σ−1 .||u|| σ−1 ||v||B σ , III.||(Id − ∆−1 )(u · ∇v)||Bp,r Bp,r Bp,r p,r

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J. LI AND Z. YIN

and IV.||

d X

0 ∇(−∆)−1 ∂i (Tuj ∂j v i + T∂j vi uj )||Bp,1

i,j=1

+ ||

d X

∇(−∆)−1 ∂i ∂j R(ui , uj )||B˙ p,1 0

i,j=1

.

d X

0 + ||Tuj ∂j v i + T∂j vi uj ||Bp,1

.

||R(ui , v j )||B˙ p,1 1

i,j=1

i,j=1 d X

d X

i

j

σ−1 + ||Tuj ∂j v + T∂j vi u ||Bp,r

1 ||R(ui , v j )||Bp,1

i,j=1

i,j=1 σ−1 ||∇v|| .||u||Bp,r

d X

L∞

+ ||u||

L∞

σ + ||u||L∞ ||v|| 1 ||v||Bp,r Bp,1

σ−1 ||v||B σ . .||u||Bp,r p,r

This completes this proof of this lemma.

Combining the results of Lemmas 2.6-2.8, we have Lemma 2.9. Assume (s, p, r) satisfies (1.5). Then 1) there exists a constant C, depending only on d, p, r, s, such that for all u, f ∈ s Bp,r with div u = 0, s−1 ≤ Ckuk s−1 kf kB s . ku · ∇f kBp,r Bp,r p,r 2) there exists a constant C, depending only on d, p, r, s, such that for all u, v ∈ s Bp,r with div u = div v = 0, s s s + kvkC 0,1 kukBp,r k∇(−∆)−1 div (u · ∇v)kBp,r ≤ C kukC 0,1 kvkBp,r , s−1 ≤ C min(kuk s−1 kvkB s , kvk s−1 kukB s ), k∇(−∆)−1 div (u · ∇v)kBp,r Bp,r Bp,r p,r p,r

where ||f ||C 0,1 = ||f ||L∞ + ||∇f ||L∞ . Lemma 2.10. [6] Let σ ∈ R and 1 ≤ p, r ≤ ∞. Let v be a vector field over Rd . Assume that σ > −d min{1 − p1 , 1p }. Define Rj = [v · ∇, ∆j ]f . There a constant C = C(p, σ, d) such that   σ , if ||f ||Bp,r σ < 1 + pd , C||∇v|| dp   ∞ ∩L B  p,∞ jσ (2 ||Rj ||Lp )j≥−1 r ≤ C||∇v|| d +1 ||f ||Bσ , if σ = 1 + pd , r > 1, p,r ℓ p  B p,∞   C||∇v|| σ−1 ||f || σ , otherwise. Bp,r Bp,r

We need an estimate for the transport-diffusion equation which is uniform with respect to the viscocity. Consider the following equation: ( ∂t f + v · ∇f − ε∆f = g, (2.1) f (0) = f0 , where ε ≥ 0, v : R × Rd → Rd , f0 : Rd → RN , and g : R × Rd → RN are given.


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Proposition 2.11. Let d ≥ 2, 1 ≤ p ≤ ∞, 1 ≤ r ≤ ∞. Assume that divv = 0 and s > −1. There exists a constant C, depending only on d, p, r, s, such that for any smooth solution f of (2.1) we have Z t s s dτ s ≤ ||f0 ||Bp,r + ||f ||L∞ ||g(τ )||Bp,r t (Bp,r ) 0 Z t s )dτ. s +C (2.2) + ||∇f ||L∞ ||v||Bp,r (||∇v||L∞ ||f ||Bp,r 0

Proof. First, applying ∆j to (2.1) yields ∂t ∆j f + v · ∇∆j f − ε∆∆j f = ∆j g + Rj ,

∆j f |t=0 = ∆j f0 .

(2.3)

If p ∈ [1, ∞), we multiply (2.3) by |∆j f |p−1sgn(∆j f ) and then integrate by parts to get Z Z 1d 1 p p−2 divv|∆j f |p dx ∆∆j f |∆j f | ∆j f dx ≤ ||∆j f ||Lp − ε p dt p d d R ZR + Rj |∆j f |p−2∆j f dx. Rd

Using the fact (see Proposition 2.1 in [3]) that for all j ≥ −1, Z −ε ∆∆j f |∆j f |p−2 ∆j f dx ≥ 0, Rd

we have

Z

t

||∆j f (t)||Lp ≤ ||∆j f0 ||Lp + ||∆j g(τ )||Lp dτ 0 Z t 1 + ||Rj (τ )||Lp + ||divv(τ )||L∞ ||∆j f (τ )||Lp dτ. p 0

(2.4)

If p = ∞, the above inequality also holds by the maximum principle. According to the fact (see Lemma 2.100, [2]), for s > −1, divv = 0, we have js (2 ||Rj ||Lp )j≥−1 r ≤ C(||∇v||L∞ ||f ||Bs + ||∇f ||L∞ ||∇v|| s−1 ). (2.5) Bp,r p,r ℓ

Multiplying both sides of (2.4) by 2js and taking ℓr norm, it follows from Minkovshi’s inequality that Z t s s s dτ ||f (t)||Bp,r ≤ ||f0 ||Bp,r + ||g(τ )||Bp,r 0 Z t s )dτ. s +C + ||∇f ||L∞ ||v||Bp,r (||∇v||L∞ ||f ||Bp,r 0

This completes the proof of this lemma.

3. Some usefull estimates In this section, we will establish some usefull estimates for smooth solutions of (1.1) and (1.4), which is the key component in the proof of Theorems 1.1 and 1.2. This estimates can be stated as follows.

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J. LI AND Z. YIN

Lemma 3.1. Let d ≥ 2 and s > 1 + d2 . Suppose that (u1, b1 ) ∈ C([0, T ]; H s) and (u2 , b2 ) ∈ C([0, T ]; H s+1) are two solutions of (1.1) with initial data (u10 , b10 ) and (u20 , b20 ) respectively. Denote δu = u1 − u2 and δb = b1 − b2 . Then, we have ||δu(t)||2H s−1 + ||δb(t)||2H s−1 ≤ (||δu0||2H s−1 + ||δb0 ||2H s−1 )eA(t) , and

||δu(t)||2H s + ||δb(t)||2H s ≤ ||δu0||2H s + ||δb0 ||2H s Z t 2 2 2 2 2 2 +C (||u ||H s+1 + ||b ||H s+1 )(||δu||H s−1 + ||δb||H s−1 )dτ eA(t) , 0

with

A(t) = C

Z

t

(1 + ||u1||H s + ||u2||H s + ||b1 ||H s + ||b2 ||H s )dτ.

0

Proof. It is easy to show that  1 2 1 2  ∂t δu + u · ∇δu + δu · ∇u − µ∆δu + ∇P = b · ∇δb + δb · ∇b , ∂t δb + u1 · ∇δb + δu · ∇b2 − ν∆δb = b1 · ∇δu + δb · ∇u2 ,  divδu = divδb = 0, (δu, δb)| = (δu , δb ). t=0 0 0

(3.1)

1d (||∆j δu||2L2 + ||∆j δb||2L2 ) ≤ K1 + K2 + K3 + K4 + K5 , 2 dt

(3.2)

Now, we apply ∆j to (3.1), and take the inner product with (∆j δu, ∆j δb) and integrate by parts to have

where

Z

1

Z

K1 = − [∆j , u · ∇]δu · ∆j δu dx, K2 = − [∆j , u1 · ∇]δb · ∆j δb dx, d d R Z R Z K3 = [∆j , b1 · ∇]δb · ∆j δu dx, K4 = [∆j , b1 · ∇]δu · ∆j δb dx, d d R ZR K5 = ∆j (δb · ∇b2 − δu · ∇u2 )∆j δu dx, d ZR K6 = ∆j (δb · ∇u2 − δu · ∇b2 )∆j δb dx. Rd

On the one hand, according to Lemma 2.10, it is easy to estimate |K1 | . ||[∆j , u1 · ∇]δu||L2 ||∆j δu||L2 . 2−2j(s−1) c2j ||∇u1||H s−1 ||δu||2H s−1 , |K2 | . ||[∆j , u1 · ∇]δb||L2 ||∆j δb||L2 . 2−2j(s−1) c2j ||∇u1||H s−1 ||δb||2H s−1 , |K3 | . ||[∆j , b1 · ∇]δb||L2 ||∆j δu||L2 . 2−2j(s−1) c2j ||∇b1 ||H s−1 ||δu||H s−1 ||δb||H s−1 , |K4 | . ||[∆j , b1 · ∇]δu||L2 ||∆j δb||L2 . 2−2j(s−1) c2j ||∇b1 ||H s−1 ||δu||H s−1 ||δb||H s−1 , |K5 | . 2−2j(s−1) c2j (||b2 ||H s ||δu||H s−1 ||δb||H s−1 + ||u2||H s ||δu||2H s−1 ), |K6 | . 2−2j(s−1) c2j (||b2 ||H s ||δu||H s−1 ||δb||H s−1 + ||u2||H s ||δb||2H s−1 ).


9

Integrating (3.2) over [0, t], multiplying the inequality above by 22j(s−1) and summing over j ≥ −1, we have ||δu||2H s−1 + ||δb||2H s−1 ≤ ||δu0||2H s−1 + ||δb0 ||2H s−1 Z t + A′ (τ )(||δu||2H s−1 + ||δb||2H s−1 )dτ.

(3.3)

0

On the anther hand, according to Lemma 2.10, it is easy to estimate |K1 | . ||[∆j , u1 · ∇]δu||L2 ||∆j δu||L2 . 2−2js c2j ||∇u1 ||H s−1 ||δu||2H s , |K2 | . ||[∆j , u1 · ∇]δb||L2 ||∆j δb||L2 . 2−2js c2j ||∇u1 ||H s−1 ||δb||2H s , |K3 | . ||[∆j , b1 · ∇]δb||L2 ||∆j δu||L2 . 2−2js c2j ||∇b1 ||H s−1 ||δu||H s ||δb||H s , |K4 | . ||[∆j , b1 · ∇]δu||L2 ||∆j δb||L2 . 2−2js c2j ||∇b1 ||H s−1 ||δu||H s ||δb||H s , |K5 | . 2−2js c2j (||b2 ||H s ||δu||H s ||δb||H s + ||u2||H s ||δu||2H s + ||b2 ||H s+1 ||δu||H s ||δb||H s−1 + ||u2||H s+1 ||δu||H s ||δu||H s−1 ), |K6 | . 2−2js c2j (||b2 ||H s ||δu||H s ||δb||H s + ||u2||H s ||δb||2H s + ||b2 ||H s+1 ||δu||H s−1 ||δb||H s + ||u2||H s+1 ||δb||H s ||δb||H s−1 ). Integrating (3.2) over [0, t], multiplying the inequality above by 22js and summing over j ≥ −1, we have ||δu||2H s + ||δb||2H s ≤ ||δu0 ||2H s + ||δb0 ||2H s Z t A′ (τ )(||δu||2H s + ||δb||2H s )dτ + 0 Z t +C (||u2||2H s+1 + ||b2 ||2H s+1 )(||δu||2H s−1 + ||δb||2H s−1 )dτ. (3.4) 0

This complete the proof of this lemma.

Lemma 3.2. Let d ≥ 2 and s > 1 + d2 . Suppose that (u, b) ∈ C([0, T ]; H s+2) is the solution of (1.1) with initial data (u0 , b0 ) and (v, c) ∈ C([0, T ]; H s+2) is the solution of (1.2) with initial data (v0 , c0 ). Denote ω = u − v and a = b − c. Then, we have ||ω(t)||2H s−1 + ||a(t)||2H s−1 ≤ C ||ω0 ||2H s−1 + ||a0 ||2H s−1 Z t Z t 2 2 2 ||b||2H s+1 dτ eB(t) , +µ ||u||H s+1 dτ + ν 0

and

||ω(t)||2H s

+

||a(t)||2H s

0

Z t Z t 2 2 2 2 2 ||b||2H s+2 dτ ||u||H s+2 dτ + ν ≤ C ||ω0 ||H s + ||a0 ||H s + µ 0 0 Z t + (||v||2H s+1 + ||c||2H s+1 )(||w||2H s−1 + ||a||2H s−1 )dτ eB(t) , 0

with

B(t) = C

Z

0

t

(1 + ||u||H s + ||v||H s + ||b||H s + ||c||H s )dτ.

10

J. LI AND Z. YIN

Proof. It is easy to show that   ∂t ω + u · ∇ω + ω · ∇v + ∇P = b · ∇a + a · ∇c + µ∆u, ∂t a + u · ∇a + ω · ∇c = b · ∇ω + a · ∇v + ν∆b,  divω = diva = 0, (ω, a)| = (ω , a ). t=0 0 0

The similar argument as in (3.3), we have ||ω||2H s−1

+

||a||2H s−1

≤

||ω0 ||2H s−1 +

Z

+

||a0 ||2H s−1

+ Cµ

t

+

||a||2H s

≤

t

||u||2H s+1 dτ

+ Cν

2

0

Z

+

||a0 ||2H s

t

+ Cµ

2

Z

t

||u||2H s+2 dτ

+ Cν

2

0

Z

||b||2H s+1 dτ 0

t

||b||2H s+2 dτ 0

B′ (τ )(||ω||2H s + ||a||2H s )dτ 0 Z t +C (||v||2H s+1 + ||c||2H s+1 )(||ω||2H s−1 + ||a||2H s−1 )dτ.

+

t

0

||ω0 ||2H s Z

Z

B′ (τ )(||ω||2H s−1 + ||a||2H s−1 )dτ.

The similar argument as in (3.4), we have ||ω||2H s

2

(3.5)

0

This along with Gronwall’s inequality completes the proof of the lemma.

s Lemma 3.3. Let d ≥ 2 and (s, p, r) satisfies (1.5). Suppose that (u1 , b1 ) ∈ C([0, T ]; Bp,r ) 2 2 s+1 1 1 and (u , b ) ∈ C([0, T ]; Bp,r ) are two solutions of (1.4) with initial data (u0, b0 ) and (u20 , b20 ) respectively. Denote δu = u1 − u2 and δb = b1 − b2 . Then, we have ¯

A(t) s−1 + ||δb(t)|| s−1 ≤ C(||δu0 || s−1 + ||δb0 || s−1 )e , ||δu(t)||Bp,r Bp,r Bp,r Bp,r

and s s ||δu(t)||Bp,r + ||δb(t)||Bp,r

s s ≤ C ||δu0 ||Bp,r + ||δb0 ||Bp,r Z t ¯ 2 A(t) s+1 + ||b || s+1 )(||δu|| s−1 + ||δb|| s−1 )dτ e + (||u2||Bp,r , Bp,r Bp,r Bp,r 0

with ¯ A(t) =C

Z

t 0

s s s s )dτ. (||u1 ||Bp,r + ||u2||Bp,r + ||b1 ||Bp,r + ||b2 ||Bp,r

Proof. Denote u¯ = u + bi and ¯bi = ui − bi for i = 1, 2. Then, setting δ¯ u = u¯1 − u¯2 and δ¯b = ¯b1 − ¯b2 , we get   u + ¯b1 · ∇δ¯ u + δ¯b · ∇¯ u2 − µ∆δ¯ u + ∇P1 = 0, ∂t δ¯ 1 2 ¯ ¯ ¯ ¯ (3.6) ∂t δ b + u¯ · ∇δ b + δ¯ u · ∇b − µ∆δ b + ∇P2 = 0,  divδ¯ u = divδ¯ u = 0, (δ¯ u, δ¯b)|t=0 = (δ¯ u0 , δ¯b0 ). i

i

It follows from Lemma 2.9 that

¯ ′ (||δ¯ s−1 ), s−1 + ||δ¯ s−1 + ||∇P2 || s−1 ≤ A b||Bp,r u||Bp,r ||∇P1 ||Bp,r Bp,r

and ¯ ′ (||δ¯ s s s s ). ||∇P1 ||Bp,r + ||∇P2 ||Bp,r ≤A u||Bp,r + ||δ¯b||Bp,r


According to Proposition 2.11, we have s−1 ||δ¯ u||Bp,r

Z

s−1 + s−1 + ||δ¯ s−1 ≤ ||δ¯ b0 ||Bp,r u0||Bp,r + ||δ¯b||Bp,r

which along with Gronwall’s inequality yield u0|| ||δ¯ u|| s−1 + ||δ¯b|| s−1 ≤ (||δ¯ Bp,r

Bp,r

s−1 Bp,r

0

t

11

¯ ′ (||δ¯ s−1 )dτ, s−1 + ||δ¯ b||Bp,r A u||Bp,r

¯ A(t) s−1 )e . + ||δ¯b0 ||Bp,r

Similarly, we also have s ||δ¯ u||Bp,r

Z t ¯ ¯ ¯ ′ (||δ¯ s s s s s )dτ + ||δ b||Bp,r ≤||δ¯ u0||Bp,r + ||δ b0 ||Bp,r + A u||Bp,r + ||δ¯b||Bp,r 0 Z t 2 s+1 + ||b || s+1 )(||δ¯ s−1 + ||δ¯ s−1 )dτ, +C (||u2||Bp,r u||Bp,r b||Bp,r Bp,r 0

which along with Gronwall’s inequality yields s s s s ||δ¯ u||Bp,r + ||δ¯b||Bp,r ≤ ||δ¯ u0||Bp,r + ||δ¯b0 ||Bp,r Z t ¯ A(t) 2 2 ¯ s+1 + ||b || s+1 )(||δ¯ s−1 + ||δ b|| s−1 )dτ e +C . (||u ||Bp,r u||Bp,r Bp,r Bp,r 0

It is easy to see for all σ ∈ R,

σ + ||δ¯ σ σ + ||δb||B σ . ||δ¯ u||Bp,r b||Bp,r ≃ ||δu||Bp,r p,r

(3.7)

Therefore, combining the above inequalities, we obtain the corresponding results. s+2 Lemma 3.4. Let d ≥ 2 and (s, p, r) satisfies (1.5). Suppose that (u, b) ∈ C([0, T ]; Bp,r ) s+2 is the solution of (1.4) with initial data (u0 , b0 ) and (v, c) ∈ C([0, T ]; Bp,r ) is the solution of (1.2) with initial data (v0 , c0 ). Denote ω = u − v and a = b − c. Then, we have s−1 + ||a(t)|| s−1 ≤ C ||ω0 || s−1 + ||a0 || s−1 ||ω(t)||Bp,r Bp,r Bp,r Bp,r Z t Z t ¯ B(t) s+1 dτ + µ s+1 dτ e +µ ||u||Bp,r ||b||Bp,r , 0

and

s s ||ω(t)||Bp,r + ||a(t)||Bp,r

0

Z t Z t s s s+2 dτ + µ s+2 dτ ≤ C ||ω0 ||Bp,r + ||a0 ||Bp,r +µ ||u||Bp,r ||b||Bp,r 0 0 Z t ¯ B(t) s+1 + ||c|| s+1 )(||w|| s−1 + ||a|| s−1 )dτ e , + (||v||Bp,r Bp,r Bp,r Bp,r 0

with

¯ =C B(t)

Z

t 0

s s s s )dτ. (||u||Bp,r + ||v||Bp,r + ||b||Bp,r + ||c||Bp,r

Proof. Denote u¯ = u + b, ¯b = u − b, v¯ = v + c and c¯ = v − c. Then, setting ω ¯ = u¯ − v¯ ¯ and a¯ = b − c¯, we have   ¯ + ¯b · ∇¯ ω+a ¯ · ∇¯ v + ∇P¯1 = µ∆¯ u, ∂t ω (3.8) ∂t a¯ + u¯ · ∇¯a + ω ¯ · ∇¯ c + ∇P¯2 = µ∆¯b,  div¯ ω = div¯a = 0, (¯ ω, a ¯)| = (¯ ω , ¯a ). t=0

0

0

12

J. LI AND Z. YIN

The similar arguments as in Lemma 3.3, we have Z t Z t s−1 + ||¯ s−1 ≤ ||¯ s−1 + ||¯ s−1 + Cµ s+1 dτ + Cµ s+1 dτ a0 ||Bp,r ||¯ ω ||Bp,r a||Bp,r ω0 ||Bp,r ||u||Bp,r ||b||Bp,r 0 0 Z t ¯ ′ (τ )(||¯ s−1 + ||¯ s−1 )dτ, + B ω ||Bp,r a||Bp,r 0

and ||¯a0 ||2Bp,r s

Z

Z

t

t

s s s s+2 + Cµ s+2 dτ ||¯ ω ||Bp,r + ||¯a||Bp,r ≤ ||¯ ω0||Bp,r + + Cµ ||u||Bp,r ||b||Bp,r 0 0 Z t s−1 )dτ s−1 + ||¯ s+1 + ||c|| s+1 )(||¯ a||Bp,r +C ω||Bp,r (||v||Bp,r Bp,r 0 Z t ¯ ′ (τ )(||¯ s s )dτ. + B ω ||Bp,r + ||¯a||Bp,r

0

This along with Gronwall’s inequality and (3.7) completes this proof of this lemma. 4. Proof of Theorem 1.1 and Theorem 1.2 In order to simplify the notation, we denote (u, b) by (u0 , b0 ) satisfies (1.3) or (1.6) with µ0 = ν0 = 0. First, we state the proof of Theorem 1.1. Proof of Theorem 1.1. First, according to classical results, there exist a positive Tn > 0 such that (1.3) have a solution (un , bn ) ∈ C([0, Tn ); H s ). Indeed, by Lemma 3.1 and (2.5), we have ||un ||2H s + ||bn ||2H s ≤ (||un0 ||2H s + ||bn0 ||2H s )eC

Rt

0 (||u

n ||

n C 0,1 +||b ||C 0,1 )dτ

.

Denote R = sup(||un0 ||H s +||bn0 ||H s ). Therefore, by continuity arguments, there exists n≥0

a positive T = T (s, d, R) such that n 2 n 2 n 2 ||un ||2L∞ s + ||b ||L∞ (H s ) ≤ C(||u0 ||H s + ||b0 ||H s ) ≤ C. T T (H )

Moreover, for all γ > s, we have ||un ||2H γ + ||bn ||2H γ ≤ (||un0 ||2H γ + ||bn0 ||2H γ )eC

Rt

0 (||u

n ||

n C 0,1 +||b ||C 0,1 )dτ

≤ C(||un0 ||2H γ + ||bn0 ||2H γ ). Let (unj , bnj ) ∈ C([0, T ]; H s) be the solution of  n n n n n n  ∂t uj + uj · ∇uj + µn ∆uj + ∇P1,j = bj · ∇bj , ∂t bnj + unj · ∇bnj + νn ∆bnj = bnj · ∇unj ,  divun = divbn = 0, (un , bn )| = S (un , bn ). j 0 0 j t=0 j j j

(4.1)

Then, according to Lemma 3.1, we have

||unj − un ||2H s−1 + ||bnj − bn ||2H s−1 ≤ C(||(Id − Sj )un0 ||2H s−1 + ||(Id − Sj )bn0 ||2H s−1 ),


13

which along with the fact that ||unj ||H s+1 + ||bnj ||H s+1 ≤ C2j leads to ||unj − un ||2H s + ||bnj − bn ||2H s ≤ C(||(Id − Sj )un0 ||2H s + ||(Id − Sj )bn0 ||2H s Z t + (||unj ||2H s+1 + ||bnj ||2H s+1 )(||unj − un ||2H s−1 + ||bnj − bn ||2H s−1 )dτ ) 0

≤ C(||(Id − Sj )un0 ||2H s + ||(Id − Sj )bn0 ||2H s

+ 22j ||(Id − Sj )un0 ||2H s−1 + 22j ||(Id − Sj )bn0 ||2H s−1 ) ≤ C(||(Id − Sj )un0 ||2H s + ||(Id − Sj )bn0 ||2H s ). Using the similar argument and Lemma 3.2, we can show that and ||unj − uj ||2H s + ||bnj − bj ||2H s ≤ C24j (||un0 − u0 ||2H s + ||bn0 − b0 ||2H s + µ2n + νn2 ). Therefore, combing the above inequalities, we obtain n 2 ||un − u||2L∞ s + ||b − b||L∞ (H s ) T T (H ) n 2 ≤ ||unj − uj ||2L∞ s + ||bj − bj ||L∞ (H s ) T T (H ) n n 2 + ||unj − un ||2L∞ s + ||bj − b ||L∞ (H s ) T T (H ) 2 + ||uj − u||2L∞ s + ||bj − b||L∞ (H s ) T T (H )

≤ C(||(Id − Sj )un0 ||2H s + ||(Id − Sj )bn0 ||2H s + ||(Id − Sj )u0||2H s + ||(Id − Sj )b0 ||2H s + 24j ||un0 − u0 ||2H s + 24j ||bn0 − b0 ||2H s + 24j µ2n + 24j νn2 ) ≤ C(||(Id − Sj )u0 ||2H s + ||(Id − Sj )b0 ||2H s + 24j ||un0 − u0 ||2H s + 24j ||bn0 − b0 ||2H s + 24j µ2n + 24j νn2 ). This completes the proof of Theorem 1.1. Now, we will prove Theorem 1.2. Proof of Theorem 1.2. First, according to classical results, there exist a positive s Tn > 0 such that (1.6) have a solution (un , bn ) ∈ C([0, Tn ); Bp,r ). Indeed, by Lemma 3.3 and (2.5), we have C s s s s )e ||un||Bp,r + ||bn ||Bp,r ≤ (||un0 ||Bp,r + ||bn0 ||Bp,r

Rt

0 (||u

n ||

n C 0,1 +||b ||C 0,1 )dτ

.

¯ = sup(||un ||Bs + ||bn ||Bs ). Therefore, by continuity arguments, there Denote R 0 0 p,r p,r n≥0

¯ such that exists a positive T = T (s, p, r, d, R) s s ). s s ≤ C(||un0 ||Bp,r + ||bn0 ||Bp,r + ||bn ||L∞ ||un ||L∞ T (Bp,r ) T (Bp,r )

Moreover, for all γ > s, we have C γ γ γ γ )e ||un ||Bp,r + ||bn ||Bp,r ≤ (||un0 ||Bp,r + ||bn0 ||Bp,r

Rt

0 (||u

n ||

n C 0,1 +||b ||C 0,1 )dτ

γ γ ). ≤ C(||un0 ||Bp,r + ||bn0 ||Bp,r

Now, we set (unj , bnj ) satisfies the following system:  n n n n n n  ∂t uj + uj · ∇uj + µn ∆uj + ∇Pj = bj · ∇bj , ∂t bnj + unj · ∇bnj + µn ∆bnj = bnj · ∇unj ,  divun = divbn = 0, (un , bn )| = (S un , S bn ). j 0 j 0 j j j j t=0

(4.2)

14

J. LI AND Z. YIN

Similar proof as in Theorem 1.1, by Lemmas 3.3-3.4, we have s s + ||bnj − bn ||L∞ ||unj − un ||L∞ T (Bp,r ) T (Bp,r ) s s ), ≤ C(||(Id − Sj )un0 ||Bp,r + ||(Id − Sj )bn0 ||Bp,r

and s s + ||bnj − bj ||L∞ ||unj − uj ||L∞ T (Bp,r ) T (Bp,r ) s s ≤ C22j (||un0 − u0 ||Bp,r + ||bn0 − b0 ||Bp,r + µn ).

Then, we can show that s s + ||bn − b||L∞ ||un − u||L∞ T (Bp,r ) T (Bp,r ) s s ≤ C(||(Id − Sj )u0 ||Bp,r + ||(Id − Sj )b0 ||Bp,r s s + 22j ||un0 − u0 ||Bp,r + 22j ||bn0 − b0 ||Bp,r + 22j µn ).

This completes the proof of Theorem 1.2. Acknowledgements. This work was partially supported by NNSFC (No. 11271382), RFDP (No. 20120171110014), MSTDF (No. 098/2013/A3), Guangdong Special Support Program (No. 8-2015) and the key project of NSF of Guangdong Province (No. 1614050000014).

References [1] H. Abidi and R. Danchin, Optimal bounds for the inviscid limit of Navier-Stokes equations, Asymptot. Anal., 38 (2004), 35-46. [2] H. Bahouri, J. Chemin and R. Danchin, Fourier Analysis and Nonlinear Partial Differential Equations, Grundlehren der mathematischen Wissenschaften, 343, Springer, Heidelberg (2011). [3] R. Danchin, Estimates in Besov spaces for transport and transport-diffusion equations with almost Lipschitz coefficients, Rev. Mat. Iberoamericana, 21 (2005), 863-888. [4] J. Daz and M. Lerena, On the inviscid and non-resistive limit for the equations of incompressible magnetohydrodynamics, Math. Models Methods Appl. Sci., 12 (2002), 1401-1419. [5] Z. Guo, J. Li and Z. Yin, On the inviscid limit of the incompressible Navier-Stokes equations, http://arxiv.org/abs/1612.01068 [math.AP], 2016. [6] J. Li and Z. Yin, Well-posedness and analytic solutions of the two-component Euler-Poincar´e system, Monatsh Math, http://dx.doi.org/10.1007/s00605-016-0927-8, 2016. [7] C. Miao and B. Yuan, Well-posedness of the ideal MHD system in critical Besov spaces, Methods Appl. Anal., 13 (2006), 89-106. [8] H. C. Pak and Y. J. Park, Existence of solution for the Euler equations in a critical Besov 1 space B∞,1 (Rn ), Comm. Partial Differential Equations, 29 (2004), 1149-1166. [9] H. Swann, The Convergence with Vanishing Viscosity of Nonstationary Navier-Stokes Flow to Ideal Flow in R3 , Trans. Amer. Math. Soc., 157 (1971), 373-397. [10] J. Wu, Viscous and inviscid magnetohydrodynamics equations, J. Anal. Math., 73 (1997), 251-265. [11] Y. Xiao, Z. Xin and J. Wu, Vanishing viscosity limit for the 3D magnetohydrodynamic system with a slip boundary condition, J. Funct. Anal., 257 (2009), 3375-3394. [12] J. Zhang, The inviscid and non-resistive limit in the Cauchy problem for 3-D nonhomogeneous incompressible magneto-hydrodynamics, Acta Math. Sci. Ser. B Engl. Ed., 31 (2011), 882-896.


15

Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, China E-mail address: [email protected] Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, China & Faculty of Information Technology, Macau University of Science and Technology, Macau, China E-mail address: [email protected]