Nonlinear Analysis 68 (2008) 1316–1331www.elsevier.com/locate/na

Higher order regularity results in 3D for the Landau–Lifshitzequation with an exchange field

Ivan Cimrak∗, Roger Van Keer

Department of Mathematical Analysis, Ghent University, Galglaan 2, 9000 Ghent, Belgium

Received 1 October 2004; accepted 7 December 2006

Abstract

We derive and prove higher order regularity results for the Landau–Lifshitz (LL) equation describing the evolution of spinfields in continuum ferromagnets. These results are based on the weight κ(s) = min 1, s which helps to get nonlinear parts undercontrol.c© 2007 Elsevier Ltd. All rights reserved.

Keywords: Landau–Lifshitz equation; Micromagnetism; Ferromagnets

1. Introduction

We study the LL equation

mt = γ m × Heff − αγ m × (m × Heff), m |t=0 = m0, (1)

where α is a positive constant, called the Gilbert damping constant, and γ denotes the gyromagnetic factor. Theunknown m stands for a spin vector of magnetization and Heff denotes the effective field.

For the purposes of mathematical analysis, we can skip the gyromagnetic factor γ . It disappears after a timerescaling of the LL equation.

For the sake of simplicity, we consider the effective field Heff of a form that corresponds to the pure isotropic casewithout any external field. Then, Heff = ∆m. However it would be no serious obstacle to add also terms describingcrystalline anisotropy, magnetostatic self-energy or external magnetic field.

After these assumptions, the problem we are interested in reads as

mt = m × ∆m − αm × (m × ∆m) in R+× Ω , (2)

∇m · ν = 0 on R+× ∂Ω , (3)

m(0, .) = m0 in Ω . (4)

∗ Corresponding author. Tel.: +32 9 2644920; fax: +32 9 2644987.E-mail addresses: ivan.cimrak@ugent.be (I. Cimrak), roger.vankeer@ugent.be (R. Van Keer).URLs: http://cage.ugent.be/∼cimo (I. Cimrak), http://cage.ugent.be/∼rvk (R. Van Keer).

0362-546X/$ - see front matter c© 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2006.12.023

I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331 1317

Simple multiplication of (2) gives us directly that the modulus of m remains constant in time. Thus, we can usethroughout the text the inequality

‖m‖L∞ ≤ C. (5)

Then, (2) according to [2] becomes equivalent to

mt − α∆m = m × ∆m + α|∇m|2m. (6)

2. Regularity results

First we begin with some notation. With ∂pt f we understand the p-th time derivative of the time-dependent function

f . In the following text, we use the symbol (., .) for the standard L2(Ω) product. The norm in the space X is denotedby ‖.‖X . The norm in L p(Ω) is denoted by ‖.‖p for ∞ ≥ p ≥ 1. The scalar product in R3 (Rm , respectively) isdenoted by 〈., .〉3 (〈., .〉m , respectively). We omit the index m in the expression 〈., .〉m when it is clear how manycomponents the function inside the scalar product has.

Throughout the text, C and ε denote generic real numbers, which can change if necessary.Carbou and Fabrie in [2] have proved the following result for the solution to the LL equation:

Theorem 1. Supposing m0 ∈ W 2,2(Ω), there exists a positive T0 such that for the solution m to the problem (2)–(4)the following estimate is valid:

maxt∈(0,T0)

‖m(t)‖W 2,2(Ω) ≤ C. (7)

Remark 1. Using (14) we can get

maxt∈(0,T0)

‖∇m(t)‖4 ≤ C, (8)

since m is bounded in the W 2,2(Ω) norm.

In [21] the author has derived and proved new-type regularity results for the exact solution of the LL equationintroducing a time weight κ(s) = min1, s. This weight helps to get the highly nonlinear terms under control atthe beginning of the time interval, on which we solve the problem (2)–(4). This reduces requirements on the regularityof initial data. The estimates were proved for the first and the second time derivatives of the exact solution, as well asfor the gradient and the second space derivatives of the exact solution. The author obtained these regularity results intwo dimensions. In this paper we consider the case of three dimensions and time derivatives ∂

pt m, ∇∂

pt m and ∆∂

pt m

for arbitrary value of p, not only for p = 1, 2 as was considered for 2D in [21]. These results constitute a first step inhigher order analysis of the LL equation in three dimensions.

The following theorem is the main result of this paper:

Theorem 2. For the solution m to the problem (2)–(4), taking T0 from Theorem 1, the following estimates are validfor any positive p ∈ Z:

maxt∈(0,T0)

κ p

‖∂p+1t m‖2

+

(∫ T0

0κ2p

‖∇∂p+1t m‖

22ds

)1/2

≤ C(p, α,Ω), (9)

maxt∈(0,T0)

κ p

‖∆∂pt m‖2

≤ C(p, α,Ω), (10)

maxt∈(0,T0)

κ

2p+12 ‖∇∂

p+1t m‖2

+

(∫ T0

0κ2p+1

‖∆∂p+1t m‖

22ds

)1/2

≤ C(p, α,Ω), (11)

(∫ T0

0κ2p+1

‖∂p+2t m‖

22

)1/2

≤ C(p, α,Ω), (12)

where κ is a time weight equal to κ(s) = min1, s and C(p, α,Ω) is a constant depending only on p, α and Ω .

From now on, we do not explicitly write the dependence of the constant C(p, α,Ω) on p, α and Ω .

1318 I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331

In [21] the author analyses a number of numerical schemes concerning the LL equation in two dimensions.Estimates of a similar type to those in Theorem 2 are used to estimate higher order nonlinear terms arising from theTaylor series. This kind of analysis was necessary for proving the convergence in time of several numerical schemes.

First theoretical results concerning the regularity of the LL equation can be found in [24]. The author studies thesystem of Maxwell’s equations coupled with the LL equation considering both the exchange field and the anisotropyfield. The case without Maxwell’s equations was studied simultaneously in [1] and [13]. The authors showed theexistence of a global weak solution for the LL equation. Moreover, they give an example in which the LL equationdoes not have a unique solution. In that case, the initial condition does not belong to W 2,2(Ω) space.

The combination of the LL equation with Maxwell’s equation, considering Neumann boundary conditions, wasstudied in [12] and [14]. They proved the global existence of a weak solution in two or three space dimensions.

The single LL equation in two dimensions was discussed in [13] and in more detail in [4] and [6]. They proved thatany weak solution with finite energy is unique and smooth apart from at most finitely many points in space-time. Theauthor of [5] devoted his paper to the study and localization of these singularity points.

All the authors above consider the LL equation with an exchange field.One can find numerical analysis and simulation of micromagnetic phenomena in [3] and [17]. These authors

propose methods which deal with non-convexity of a minimizing functional using Young measures. The same authorshave written a survey paper about numerical methods for micromagnetics, see [18]. A different approach was usedin [10] and [25]. They introduced a numerical scheme for the LL equation and compared this scheme with otherknown schemes.

In the paper [15] the authors study long-time convergence of solutions to Maxwell–LL equations. The authorsin [16] and [19] are interested in the numerical modeling of absorbing ferromagnetic materials. They propose anumerical scheme which conserves the magnitude of magnetization but they do not prove any error estimates in time.In [22] and [23] a similar problem is studied. The authors suggest a new numerical scheme conserving the magnitudeof magnetization and they also prove error estimates in time. They consider the LL equation in a simplified formconsidering a demagnetizing and anisotropy field but without an exchange field. Additional results on this schemewere mentioned in [8] and [9].

We mention also a nice numerical analysis [20] for the full Maxwell–LL system, where the LL equation isconsidered with an exchange field. They prove the existence of a new class of Lyapunov functions for the continuousproblem, and next for a variational formulation of the continuous problem. The authors also show a special result oncontinuous dependence.

3. Analysis tools

Before we prove Theorem 2 we sum up necessary inequalities which we need in our proofs.

Lemma 1 (Special Cases of Sobolev Inequalities). Let Ω be a bounded domain with ∂Ω in C1, and let u be anyfunction in W 2,2(Ω) ∩ L2(Ω). Consider the case of spatial dimension n = 3. Then the following hold:

‖u‖L4 ≤ C‖u‖3/4W 1,2‖u‖

1/4L2 ≤ C‖u‖L2 + C‖∇u‖

3/4L2 ‖u‖

1/4L2 , (13)

‖∇u‖L4 ≤ C‖∇u‖3/4W 1,2‖∇u‖

1/4L2 ≤ C‖∇u‖L2 + C‖∆u‖

3/4L2 ‖∇u‖

1/4L2 , (14)

‖u‖L∞ ≤ C‖u‖W 1,4 ≤ C‖u‖L4 + C‖∇u‖L4 . (15)

Proof. The inequalities are a special case of extended Sobolev inequalities which can be found in [11, Theorem 10.1].

By X p and Yp we denote the following sets:

X p = (i, j, k) : i, j, k ∈ Z, 0 ≤ i ≤ p, 0 ≤ j ≤ p, 0 ≤ k ≤ p, i + j + k = p,

Yp = (i, j) : i, j ∈ Z, 0 ≤ i ≤ p, 0 ≤ j ≤ p, i + j = p.

I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331 1319

Define A = 〈∇m, ∇m〉m and B = m × ∆m. For the time derivations of A and B we can write

∂pt A =

∑X p

〈∇∂ it m, ∇∂

jt m〉∂k

t m, (16)

∂pt B =

∑Yp

∂ it m × ∆∂

jt m. (17)

We can formally take the r -th derivative in time of the LL equation to get

∂r+1t m − α∆∂r

t m = α∂rt A+ ∂r

t B. (18)

4. Proof of Theorem 2

We use the notation P(p) for the statement of Theorem 2 for specific value of p. We will prove the theorem bymathematical induction. Note that P(0) was proved in [7]. Then we prove P(p) using P(0), . . . ,P(p − 1). We proveall four inequalities (9)–(12) sequentially. This means that in the proof of (9) we use P(m) for m ≤ p − 1 only,whereas in the proof of (10) we can use (9) from P(p) too. Finally, in the proof of (12) we will use (9)–(11) fromP(p).

Proof of Inequality (9). We take the (p + 1)-th derivative of the LL equation (18). Then we multiply the result byκ2p∂

p+1t m, and finally we integrate it over Ω to get

κ2p(∂p+2t m, ∂

p+1t m) − ακ2p(∆∂

p+1t m, ∂

p+1t m) = ακ2p(∂

p+1t A, ∂

p+1t m) + κ2p(∂

p+1t B, ∂

p+1t m). (19)

Because of the following identity:

κ2p(∂p+2t m, ∂

p+1t m) =

12∂t (κ

2p‖∂

p+1t m‖

22) − pκ2p−1

‖∂p+1t m‖

22, (20)

together with integration by parts in the term ακ2p(∆∂p+1t m, ∂

p+1t m), we can deduce

12∂t (κ

2p‖∂

p+1t m‖

22) + ακ2p

‖∇∂p+1t m‖

22

= pκ2p−1‖∂

p+1t m‖

22 + ακ2p(∂

p+1t A, ∂

p+1t m) + κ2p(∂

p+1t B, ∂

p+1t m). (21)

In Lemmas 2 and 3 we estimate absolute value of terms κ2p(∂p+1t A, ∂

p+1t m) and κ2p(∂

p+1t B, ∂

p+1t m).

Lemma 2. For any fixed ε the following estimate holds:∫ T0

0|κ2p(∂

p+1t A, ∂

p+1t m)| ≤ C + ε

∫ T0

0κ2p

‖∇∂p+1t m‖

22 + Cε

∫ T0

0κ2p

‖∂p+1t m‖

22.

Proof. From the definition of A we have

|κ2p(∂p+1t A, ∂

p+1t m)| ≤ κ2p

∑X p+1

|(〈∇∂ it m, ∂

jt m〉∂k

t m, ∂p+1t m)|. (22)

We divide the proof of the lemma into several steps. All possible choices of values k, i, j , such that i + j + k = p +1,can be put into four classes; see Table 1. We study these cases separately.

Class 1aWhen k = p + 1 in (22), we have the following estimate using (7):

|(〈∇m, ∇m〉∂p+1t m, ∂

p+1t m)| ≤ ‖∇m‖

24‖∂

p+1t m‖

24 ≤ C‖∂

p+1t m‖

24. (23)

1320 I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331

Table 1Classes of indexes i, j, k when 2i − 1 + 2 j − 1 + 2k = 2p

Class Value ofk i j

1a p + 1 0 01b 0 p + 1 (or 0) 0 (or p + 1)1c p 0 (or 1) 1 (or 0)1d ≤p − 1 ≤p ≤p

Now we first apply (13) and then the Young inequality with exponents 4/3 and 4 to get

|(〈∇m, ∇m〉∂p+1t m, ∂

p+1t m)| ≤ Cε‖∂

p+1t m‖

22 + Cε‖∂

p+1t m‖

1/22 ‖∇∂

p+1t m‖

3/22

≤ Cε‖∂p+1t m‖

22 + ε‖∇∂

p+1t m‖

22.

Class 1bWhen i = p +1 or j = p +1, we can estimate the corresponding terms in (22) as follows, using (8) and the Young

inequality:

|(〈∇∂p+1t m, ∇m〉m, ∂

p+1t m)| ≤ ‖∇∂

p+1t m‖2‖∇m‖4‖m‖L∞‖∂

p+1t m‖4

≤ C‖∇∂p+1t m‖2‖∂

p+1t m‖4

≤ ε‖∇∂p+1t m‖

22 + Cε‖∂

p+1t m‖

24. (24)

For the term Cε‖∂p+1t m‖

24 we can use the same technique as in (23). Then we get

|(〈∇∂p+1t m, ∇m〉m, ∂

p+1t m)| ≤ ε‖∇∂

p+1t m‖

22 + Cε‖∂

p+1t m‖

22.

Class 1cIn this case we have k = p and i + j = 1. Without loss of generality we assume i = 0. Then we have

κ2p|(〈∇∂ i

t m, ∇∂j

t m〉∂kt m, ∂

p+1t m)| ≤ κ2p

‖∇m‖4‖∇mt‖4‖∂pt m‖4‖∂

p+1t m‖4

≤ κ2p‖∇mt‖

24‖∂

pt m‖

24 + ‖∂

p+1t m‖

24,

where we used the boundedness of m in W 2,2 norm and the Young inequality. The term ‖∂p+1t m‖

24 was already

successfully estimated in (23) and (24). We now apply the embeddings W 2,2 → W 1,4 and W 1,2 → L4 to get

κ2p|(〈∇∂ i

t m, ∇∂j

t m〉∂kt m, ∂

p+1t m)| ≤ Cκ(‖∆mt‖

22 + ‖∇mt‖

22)κ

2p−1(‖∇∂pt m‖

22 + ‖∂

pt m‖

22)

+ Cε‖∂p+1t m‖

22 + ε‖∇∂

p+1t m‖

22. (25)

We use P(p − 1) and P(0) to get

κ2p−1‖∇∂

pt m‖

22 ≤ C, κ‖∇mt‖

22 ≤ C

κ2p−1‖∂

pt m‖

22 ≤ κ2p−2

‖∂pt m‖

22 ≤ C,

which can be applied in (25) to arrive at

κ2p|(〈∇∂ i

t m, ∇∂j

t m〉∂kt m, ∂

p+1t m)| ≤ C + Cκ‖∆mt‖

22 + Cε‖∂

p+1t m‖

22 + ε‖∇∂

p+1t m‖

22.

After the time integration we use (11) for P(0) which completes the case that k = p.

Class 1dWe have k ≤ p − 1. For the terms on the right-hand side of (22) we use

D1 := κ2p|(〈∇∂ i

t m, ∇∂j

t m〉∂kt m, ∂

p+1t m)|

≤ κ2p‖∇∂ i

t m‖2‖∇∂j

t m‖4‖∂kt m‖L∞‖∂

p+1t m‖4

≤ κ2p‖∇∂ i

t m‖22‖∇∂

jt m‖

24‖∇∂k

t m‖24 + κ2p

‖∂p+1t m‖

24,

I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331 1321

Table 2Classes of indexes i, j when 2i − 1 + 2 j − 1 = 2p

Class Value ofi j

2a 0 p + 12b p + 1 02c ≤p ≤p

where we have used the Young inequality and the embedding W 1,4 → L∞. We use estimates which we have alreadyproved in (23) for the term κ2p

‖∂p+1t m‖

24. Note that 2i − 1 + 2 j − 1 + 2k = 2p. Now we use the embedding

W 2,2 → W 1,4 to get

D1 ≤ κ2i−1‖∇∂ i

t m‖22κ

2 j−1(‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22) + κ2k(‖∇∂k

t m‖22 + ‖∆∂k

t m‖22)

+ Cεκ2p

‖∂p+1t m‖

22 + εκ2p

‖∇∂p+1t m‖

22. (26)

Since k ≤ p − 1, i ≤ p, j ≤ p, using P(k),P(i − 1) and P( j − 1), we have

κ2i−1‖∇∂ i

t m‖22 ≤ C, κ2 j−1

‖∇∂j

t m‖22 ≤ C,

κ2k‖∇∂k

t m‖22 ≤ κ2k−1

‖∇∂kt m‖

22 ≤ C, κ2k

‖∆∂kt m‖

22 ≤ C,

which can be applied in (26) to get

D1 ≤ C + Cκ2 j−1‖∆∂

jt m‖

22 + Cεκ

2p‖∂

p+1t m‖

22 + εκ2p

‖∇∂p+1t m‖

22.

After time integration we use (11) from P( j − 1) for the term κ2 j−1‖∆∂

jt m‖

22, which completes the proof of the

lemma.

Lemma 3. For any fixed ε the following estimate holds:∫ T0

0|κ2p(∂

p+1t B, ∂

p+1t m)| ≤ C + ε

∫ T0

0κ2p

‖∇∂p+1t m‖

22 + Cε

∫ T0

0κ2p

‖∂p+1t m‖

22.

Proof. The definition of B gives us

|κ2p(∂p+1t B, ∂

p+1t m)| ≤ κ2p

∑Yp+1

|(∂ it m × ∆∂

jt m, ∂

p+1t m)|. (27)

Now we make another distribution of possibilities for the values of i, j such that i + j = p + 1. See Table 2.

Class 2aWhen i = p + 1 we have

(∂p+1t m × ∆m, ∂

p+1t m) = 0.

Class 2bWhen j = p + 1 we can estimate

|(m × ∆∂p+1t m, ∂

p+1t m)| = |(∂

p+1t m × m,∆∂

p+1t m)|

≤ |(∇∂p+1t m × m, ∇∂

p+1t m)| + |(∂

p+1t m × ∇m, ∇∂

p+1t m)|,

where we have applied integration by parts. The boundary terms vanish. In the next step we use (8) and (13)

|(m × ∆∂p+1t m, ∂

p+1t m)| ≤ |(∂

p+1t m × ∇m, ∇∂

p+1t m)|

≤ ‖∂p+1t m‖4‖∇m‖4‖∇∂

p+1t m‖2

≤ C(‖∂p+1t m‖2 + ‖∂

p+1t m‖

1/42 ‖∇∂

p+1t m‖

3/42 )‖∇∂

p+1t m‖2.

1322 I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331

Now we apply the Young inequality first with both exponents equal to 2 and next with the exponents taking the value4/3 and 4 respectively. We get

|(m × ∆∂p+1t m, ∂

p+1t m)| ≤ Cε‖∂

p+1t m‖

22 + ε‖∇∂

p+1t m‖

22.

Class 2cWhen i ≤ p and j ≤ p we have

|(∂ it m × ∆∂

jt m, ∂

p+1t m)| = |(∂

p+1t m × ∂ i

t m,∆∂j

t m)|

≤ |(∇∂p+1t m × ∂ i

t m, ∇∂j

t m)| + |(∂p+1t m × ∇∂ i

t m, ∇∂j

t m)|

≤ ‖∇∂p+1t m‖2‖∂

it m‖4‖∇∂

jt m‖4 + ‖∂

p+1t m‖4‖∇∂ i

t m‖2‖∇∂j

t m‖4

≤ ‖∇∂p+1t m‖2(‖∂

it m‖

22 + ‖∇∂ i

t m‖22)

1/2(‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22)

1/2

+ (‖∂p+1t m‖

22 + ‖∇∂

p+1t m‖

22)

1/2‖∇∂ i

t m‖2(‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22)

1/2,

where we have used the embeddings W 1,2 → L4 and W 2,2 → W 1,4. Now from Young’s inequalities we get

κ2p|(∂ i

t m × ∆∂j

t m, ∂p+1t m)| ≤ εκ2p

‖∇∂p+1t m‖

22 + Cεκ

2p(‖∂ it m‖

22 + ‖∇∂ i

t m‖22)(‖∇∂

jt m‖

22 + ‖∆∂

jt m‖

22)

+ εκ2p‖∂

p+1t m‖

22 + Cεκ

2p‖∇∂ i

t m‖22(‖∇∂

jt m‖

22 + ‖∆∂

jt m‖

22)

≤ εκ2p‖∇∂

p+1t m‖

22 + εκ2p

‖∂p+1t m‖

22

+ Cεκ2i−1(‖∂ i

t m‖22 + ‖∇∂ i

t m‖22)κ

2 j−1(‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22).

Note that 2i − 1 + 2 j − 1 = 2(i + j) − 2 = 2p. Since j ≤ p and i ≤ p, we have from P( j − 1) and P(i − 1) that

κ2i−1‖∂ i

t m‖22 ≤ κ2i−2

‖∂ it m‖

22 ≤ C κ2i−1

‖∇∂ it m‖

22 ≤ C,

κ2 j−1‖∇∂

jt m‖

22 ≤ C.

Thus, we get

κ2p|(∂ i

t m × ∆∂j

t m, ∂p+1t m)| ≤ εκ2p

‖∇∂p+1t m‖

22 + εκ2p

‖∂p+1t m‖

22 + C + Cεκ

2 j−1‖∆∂

jt m‖

22.

We integrate the previous inequality in time. Then we use (11) from P( j − 1) to estimate the term κ2 j−1‖∆∂

jt m‖

22.

This completes the proof of the lemma.

Proof of Inequality (9) (Continuation). Now we can pass to the proof of inequality (9). From Lemmas 2 and 3 wecan estimate the terms |ακ2p(∂

p+1t A, ∂

p+1t m)| and |κ2p(∂

p+1t B, ∂

p+1t m)| in (21). After time integration we find

κ2p‖∂

p+1t m‖

22 + α

∫ T

0κ2p

‖∇∂p+1t m‖

22

≤ p∫ T

0κ2p−1

‖∂p+1t m‖

22 + α

∫ T

0|κ2p(∂

p+1t A, ∂

p+1t m)| +

∫ T

0|κ2p(∂

p+1t B, ∂

p+1t m)|

≤ p∫ T

0κ2p−1

‖∂p+1t m‖

22 + (α + 1)

[T .C + Cε

∫ T

0κ2p

‖∂p+1t m‖

22 + ε

∫ T

0κ2p

‖∇∂p+1t m‖

22

].

Taking ε sufficiently small and using P(p − 1) for the term∫ T

0 κ2p−1‖∂

p+1t m‖

22, we get

κ2p‖∂

p+1t m‖

22 +

∫ T

0κ2p

‖∇∂p+1t m‖

22 ≤ C(α, p) + C(α, p)

∫ T

0κ2p

‖∂p+1t m‖

22.

After applying Gronwall’s lemma we complete the proof of inequality (9) for P(p).

Proof of Inequality (10). Take the p-th derivative of the LL equation and multiply it by −κ2p∆∂pt m. We find

−κ2p(∂p+1t m,∆∂

pt m) + ακ2p

‖∆∂pt m‖

22 = −ακ2p(∂

pt A,∆∂

pt m) − κ2p(∂

pt B,∆∂

pt m).

I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331 1323

Table 3Classes of indexes i, j, k when 2i + 2 j + 2k = 2p

Class Value ofk i j

3a p 0 03b 0 p (or 0) 0 (or p)3c ≤p − 1 ≤p − 1 ≤p − 1

Using Young’s inequality we get

ακ2p‖∆∂

pt m‖

22 ≤ ακ2p

|(∂pt A,∆∂

pt m)| + κ2p

|(∂pt B,∆∂

pt m)| + Cεκ

2p‖∂

p+1t m‖

22 + εκ2p

‖∆∂pt m‖

22.

We take ε small. Since we have proved (9) for P(p), we can estimate the term κ2p‖∂

p+1t m‖

22 by C to get

ακ2p‖∆∂

pt m‖

22 ≤ ακ2p

|(∂pt A,∆∂

pt m)| + κ2p

|(∂pt B,∆∂

pt m)| + Cε . (28)

The terms κ2p|(∂

pt A,∆∂

pt m)| and κ2p

|(∂pt B,∆∂

pt m)| are estimated in Lemmas 4 and 5.

Lemma 4. For any fixed ε the following estimate holds:

κ2p|(∂

pt A,∆∂

pt m)| ≤ C + εκ2p

‖∆∂pt m‖

22.

Proof. From the definition of A we have the following:

|κ2p(∂pt A,∆∂

pt m)| ≤ κ2p

∑X p

|(〈∇∂ it m, ∇∂

jt m〉∂k

t m,∆∂pt m)|. (29)

We again split all possible combinations of values i, j, k into several classes; see Table 3.

Class 3aWhen k = p in (29) we have the following estimate:

|(〈∇m, ∇m〉∂pt m,∆∂

pt m)| ≤ ‖∇m‖

24‖∂

pt m‖L∞‖∆∂

pt m‖2

≤ Cε‖∇∂pt m‖

24 + ε‖∆∂

pt m‖

22, (30)

where we have used the boundedness of m in the W 2,2 norm, the embedding W 1,4 → L∞ and Young’s inequality.Using first (14) and next Young’s inequality with coefficients 4 and 3/4 we find

κ2p|(〈∇m, ∇m〉∂

p+1t m,∆∂

p+1t m)| ≤ Cεκ

2p‖∇∂

pt m‖

22 + Cεκ

2p‖∇∂

pt m‖

1/22 ‖∆∂

pt m‖

3/22 + εκ2p

‖∆∂pt m‖

22

≤ Cεκ2p

‖∇∂pt m‖

22 + εκ2p

‖∆∂pt m‖

22. (31)

From P(p − 1) we have

κ2p‖∇∂

pt m‖

22 ≤ κ2p−1

‖∇∂pt m‖

22 ≤ C.

Then we arrive at

κ2p|(〈∇m, ∇m〉∂

p+1t m,∆∂

p+1t m)| ≤ Cε + εκ2p

‖∆∂pt m‖

22. (32)

Class 3bWhen i = p or j = p we can estimate the corresponding terms in (29) as follows:

|(〈∇∂pt m, ∇m〉m,∆∂

pt m)| ≤ ‖∇∂

pt m‖4‖∇m‖4‖m‖L∞‖∆∂

pt m‖2

≤ C‖∇∂pt m‖4‖∆∂

pt m‖2

≤ Cε‖∇∂pt m‖

24 + ε‖∆∂

pt m‖

22.

We end up with the same terms as on the right-hand side of (30).

Class 3cWhen i ≤ p − 1, j ≤ p − 1 and k ≤ p − 1, on the right-hand side of (29) the following argument can be used:

1324 I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331

Table 4Classes of indexes i, j when 2i + 2 j = 2p

Class Value ofi j

4a 0 p4b p 04c ≤p − 1 ≤p − 1

E1 := κ2p|(〈∇∂ i

t m, ∇∂ it m〉∂k

t m,∆∂pt m)|

≤ κ2p‖∇∂ i

t m‖4‖∇∂j

t m‖4‖∂kt m‖L∞‖∆∂

pt m‖2

≤ Cεκ2p

‖∇∂ it m‖

24‖∇∂

jt m‖

24‖∇∂k

t m‖24 + εκ2p

‖∆∂pt m‖

22,

where we invoked the embedding W 1,4 → L∞ and Young’s inequality.Note that 2i + 2 j + 2k = 2p. Now we use the embedding W 1,2 → L4 to get

E1 ≤ Cεκ2i (‖∇∂ i

t m‖22 + ‖∆∂ i

t m‖22)κ

2 j (‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22)κ

2k(‖∇∂kt m‖

22 + ‖∆∂k

t m‖22)

+ εκ2p‖∆∂

pt m‖

22. (33)

Since i ≤ p − 1, j ≤ p − 1 and k ≤ p − 1 using P(i),P( j),P(k) we have

κ2i‖∆∂ i

t m‖22 ≤ C, κ2i

‖∇∂ it m‖

22 ≤ κ2i−1

‖∇∂ it m‖

22 ≤ C,

κ2 j‖∆∂

jt m‖

22 ≤ C, κ2 j

‖∇∂j

t m‖22 ≤ κ2 j−1

‖∇∂j

t m‖22 ≤ C,

κ2k‖∆∂k

t m‖22 ≤ C, κ2k

‖∇∂kt m‖

22 ≤ κ2k−1

‖∇∂kt m‖

22 ≤ C,

which can be applied in (33) to get

E1 ≤ C + εκ2p‖∆∂

p+1t m‖

22.

This completes the proof of the lemma.

Lemma 5. For any fixed ε the following estimate holds:

κ2p|(∂

pt B,∆∂

pt m)| ≤ C + εκ2p

‖∆∂pt m‖

22.

Proof. From the definition of B we have

|κ2p(∂pt B,∆∂

pt m)| = κ2p

∑Yp

|(∂ it m × ∆∂

jt m,∆∂

pt m)|. (34)

We use Table 4 in order to proceed in several steps.

Class 4aWhen j = p in (34) we have

(m × ∆∂pt m,∆∂

pt m) = 0.

Class 4bWhen i = p we can estimate

|(∂pt m × ∆m,∆∂

pt m)| ≤ ‖∂

pt m‖L∞‖∆m‖2‖∆∂

pt m‖2

≤ Cε‖∇∂pt m‖

24 + ε‖∆∂

pt m‖

22.

where we invoked the boundedness of m in W 2,2, the embedding W 1,4 → L∞, and Young’s inequality. Using (14)and then Young’s inequality with exponents 4 and 4/3, we arrive at

κ2p|(∂

pt m × ∆m,∆∂

pt m)| ≤ Cεκ

2p‖∇∂

pt m‖

22 + Cεκ

2p‖∇∂

pt m‖

1/22 ‖∆∂

pt m‖

3/22 + εκ2p

‖∆∂pt m‖

22

≤ Cεκ2p

‖∇∂pt m‖

22 + εκ2p

‖∆∂pt m‖

22.

I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331 1325

Since P(p − 1) we have

κ2p‖∇∂

pt m‖

22 ≤ κ2p−1

‖∇∂pt m‖

22 ≤ C.

Thus we find

κ2p|(∂

pt m × ∆m,∆∂

pt m)| ≤ Cε + εκ2p

‖∆∂pt m‖

22.

Class 4cWhen i ≤ p − 1 and j ≤ p − 1, one has

|(∂ it m × ∆∂

jt m,∆∂

pt m)| ≤ ‖∂ i

t m‖L∞‖∆∂j

t m‖2‖∆∂pt m‖2

≤ Cε‖∇∂ it m‖

24‖∆∂

jt m‖

22 + ε‖∆∂

pt m‖

22,

where we have used the embedding W 1,4 → L∞ and Young’s inequality. Now we apply the embedding W 2,2 →

W 1,4 to get

|(∂ it m × ∆∂

jt m,∆∂

pt m)| ≤ Cε(‖∇∂ i

t m‖22 + ‖∆∂ i

t m‖22)‖∆∂

jt m‖

22 + ε‖∆∂

pt m‖

22

≤ Cε‖∇∂ it m‖

22‖∆∂

jt m‖

22 + Cε‖∆∂ i

t m‖22‖∆∂

jt m‖

22 + ε‖∆∂

pt m‖

22

Since i ≤ p − 1 and j ≤ p − 1, we can deduce from P(i) and P( j) that

κ2i‖∇∂ i

t m‖22 ≤ κ2i−1

‖∇∂ it m‖

22 ≤ C, κ2i

‖∆∂ it m‖

22 ≤ C,

κ2 j‖∆∂

jt m‖

22 ≤ C.

Then we find

κ2p|(∂ i

t m × ∆∂j

t m,∆∂p+1t m)| ≤ εκ2p

‖∆∂p+1t m‖

22 + Cεκ

2i‖∇∂ i

t m‖22κ

2 j‖∆∂

jt m‖

22

+ Cεκ2i

‖∆∂ it m‖

22κ

2 j‖∆∂

jt m‖

22

≤ εκ2p‖∆∂

p+1t m‖

22 + Cε,

which completes the proof of the lemma.

Proof of Inequality (10) (Continuation). To finish the proof of inequality (10) we simply apply Lemmas 4 and 5 inthe relation (28) and we take ε sufficiently small.

Proof of Inequality (11). Take the (p + 1)-th derivative of the LL equation. Then, we multiply the result by−κ2p+1∆∂

p+1t m getting

−κ2p+1(∂p+2t m,∆∂

p+1t m) + ακ2p+1

‖∆∂p+1t m‖

22 = −ακ2p+1(∂

p+1t A,∆∂

p+1t m)

− κ2p+1(∂p+1t B,∆∂

p+1t m).

In the first term on the left-hand side we apply integration by parts. On account of the analogy with (20) we get

12∂t

(κ2p+1

‖∇∂p+1t m‖

22

)+ ακ2p+1

‖∆∂p+1t m‖

22 = −κ2p+1α(∂

p+1t A,∆∂

p+1t m) − κ2p+1(∂

p+1t B,∆∂

p+1t m)

+2p + 1

2κ2p

‖∇∂p+1t m‖

22. (35)

We again estimate the terms containing ∂p+1t A and ∂

p+1t B in separate lemmas.

Lemma 6. For any fixed ε the following estimate holds:∫ T0

0|κ2p+1(∂

p+1t A,∆∂

p+1t m)| ≤ C + ε

∫ T0

0κ2p+1

‖∆∂p+1t m‖

22 + Cε

∫ T0

0κ2p+1

‖∇∂p+1t m‖

22.

1326 I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331

Table 5Classes of indexes i, j, k when 2i + 2 j − 1 + 2k = 2p + 1

Class Value ofk i j

5a p + 1 0 05b 0 p + 1 (or 0) 0 (or p + 1)5c ≤p ≤p ≤p

Proof. Starting from the definition of A we have

|κ2p+1(∂p+1t A,∆∂

p+1t m)| ≤ κ2p+1

∑X p+1

|(〈∇∂ it m, ∇∂

jt m〉∂k

t m,∆∂p+1t m)|. (36)

For the distribution of values of i, j, k see Table 5.Class 5a

When k = p + 1 in (36) we have the estimate

|(〈∇m, ∇m〉∂p+1t m,∆∂

p+1t m)| ≤ ‖∇m‖

24‖∂

p+1t m‖L∞‖∆∂

p+1t m‖2

≤ Cε‖∇∂p+1t m‖

24 + ε‖∆∂

p+1t m‖

22, (37)

where we have used the embedding W 1,4 → L∞. We continue, using first (14) and next Young’s inequality withcoefficients 4 and 3/4 to get

|(〈∇m, ∇m〉∂p+1t m,∆∂

p+1t m)| ≤ Cε‖∇∂

p+1t m‖

22 + Cε‖∇∂

p+1t m‖

1/22 ‖∆∂

p+1t m‖

3/22 + ε‖∆∂

p+1t m‖

22

≤ Cε‖∇∂p+1t m‖

22 + ε‖∆∂

p+1t m‖

22.

Class 5bWhen i = p + 1 or j = p + 1, using (8) we can estimate the corresponding terms in (36) as follows:

|(〈∇∂p+1t m, ∇m〉m,∆∂

p+1t m)| ≤ ‖∇∂

p+1t m‖4‖∇m‖4‖m‖L∞‖∆∂

p+1t m‖2

≤ C‖∇∂p+1t m‖4‖∆∂

p+1t m‖2

≤ Cε‖∇∂p+1t m‖

24 + ε‖∆∂

p+1t m‖

22.

We end up with the same terms as on the right-hand side of (37). Thus we obtain the same estimates.Class 5c

When i ≤ p, j ≤ p and k ≤ p, we can use the following argument on the right-hand side of (22)

F1 := κ2p+1|(〈∇∂ i

t m, ∇∂ it m〉∂k

t m,∆∂p+1t m)|

≤ κ2p+1‖∇∂ i

t m‖4‖∇∂j

t m‖4‖∂kt m‖L∞‖∆∂

p+1t m‖2

≤ Cεκ2p+1

‖∇∂ it m‖

24‖∇∂

jt m‖

24‖∇∂k

t m‖24 + εκ2p+1

‖∆∂p+1t m‖

22,

where we have used the embedding W 1,4 → L∞ and Young’s inequality. Note that 2i + 2 j − 1 + 2k = 2p + 1. Nowfrom the embedding W 1,2 → L4 we get

F1 ≤ Cεκ2i (‖∇∂ i

t m‖22 + ‖∆∂ i

t m‖22)κ

2 j−1(‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22)κ

2k(‖∇∂kt m‖

22 + ‖∆∂k

t m‖22)

+ εκ2p+1‖∆∂

p+1t m‖

22. (38)

Since i ≤ p, j ≤ p and k ≤ p, using P(i − 1),P( j − 1),P(k − 1), we have

κ2i‖∇∂ i

t m‖22 ≤ κ2i−1

‖∇∂ it m‖

22 ≤ C,

κ2k‖∇∂k

t m‖22 ≤ κ2k−1

‖∇∂kt m‖

22 ≤ C, κ2 j−1

‖∇∂j

t m‖22 ≤ C.

Inequality (10) was already proved for P(p) too. Thus we get

κ2i‖∆∂ i

t m‖22 ≤ C, κ2k

‖∆∂kt m‖

22 ≤ C.

I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331 1327

Table 6Classes of indexes i, j when 2i + 2 j − 1 = 2p + 1

Class Value ofi j

6a 0 p + 16b p + 1 06c ≤p ≤p

We can proceed in (38) to get

F1 ≤ C + Cκ2 j−1‖∆∂

jt m‖

22 + εκ2p

‖∆∂p+1t m‖

22.

We integrate the previous inequality in time. Then using (11) from P( j − 1) we estimate the term κ2 j−1‖∆∂

jt m‖

22.

This completes the proof of the lemma.

Lemma 7. For any fixed ε the following estimate holds:∫ T0

0|κ2p+1(∂

p+1t B,∆∂

p+1t m)| ≤ C + ε

∫ T0

0κ2p+1

‖∆∂p+1t m‖

22 + Cε

∫ T0

0κ2p+1

‖∇∂p+1t m‖

22.

Proof. From the definition of B we have the following:

|κ2p+1(∂p+1t B,∆∂

p+1t m)| ≤ κ2p+1

∑Yp+1

|(∂ it m × ∆∂

jt m,∆∂

p+1t m)|. (39)

The Classes 6a–6c are described in Table 6.

Class 7aWhen j = p + 1 in (39) we have

(m × ∆∂p+1t m,∆∂

p+1t m) = 0.

Class 7bWhen i = p + 1 we can estimate

|(∂p+1t m × ∆m,∆∂

p+1t m)| ≤ ‖∂

p+1t m‖L∞‖∆m‖2‖∆∂

p+1t m‖2

≤ Cε‖∇∂p+1t m‖

24 + ε‖∆∂

p+1t m‖

22,

where we have invoked the boundedness of m in the W 2,2 norm, the embedding W 1,4 → L∞, and Young’s inequality.Using (14) and Young’s inequality with exponents 4 and 4/3, we arrive at

|(∂p+1t m × ∆m,∆∂

p+1t m)| ≤ Cε‖∇∂

p+1t m‖

22 + Cε‖∇∂

p+1t m‖

1/22 ‖∆∂

p+1t m‖

3/22 + ε‖∆∂

p+1t m‖

22

≤ Cε‖∇∂p+1t m‖

22 + ε‖∆∂

p+1t m‖

22.

Class 7cWhen i ≤ p and j ≤ p we get

|(∂ it m × ∆∂

jt m,∆∂

p+1t m)| ≤ ‖∂ i

t m‖L∞‖∆∂j

t m‖2‖∆∂p+1t m‖2

≤ Cε‖∇∂ it m‖

24‖∆∂

jt m‖

22 + ε‖∆∂

p+1t m‖

22,

where we have invoked the embedding W 1,4 → L∞ and Young’s inequality. Now we apply the embeddingW 2,2 → W 1,4

|(∂ it m × ∆∂

jt m,∆∂

p+1t m)| ≤ Cε(‖∇∂ i

t m‖22 + ‖∆∂ i

t m‖22)‖∆∂

jt m‖

22 + ε‖∆∂

p+1t m‖

22

≤ Cε‖∇∂ it m‖

22‖∆∂

jt m‖

22 + Cε‖∆∂ i

t m‖22‖∆∂

jt m‖

22 + ε‖∆∂

p+1t m‖

22.

1328 I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331

Since i ≤ p and j ≤ p we can deduce from P(i − 1) and from (10) for P(i) and P( j) that

κ2i‖∇∂ i

t m‖22 ≤ κ2i−1

‖∇∂ it m‖

22 ≤ C, κ2i

‖∆∂ it m‖

22 ≤ C,

κ2 j‖∆∂

jt m‖

22 ≤ C.

Again notice that 2i + 2 j − 1 = 2p + 1. Then we get

κ2p+1|(∂ i

t m × ∆∂j

t m,∆∂p+1t m)| ≤ εκ2p+1

‖∆∂p+1t m‖

22 + Cεκ

2i‖∇∂ i

t m‖22κ

2 j−1‖∆∂

jt m‖

22

+ Cεκ2i

‖∆∂ it m‖

22κ

2 j−1‖∆∂

jt m‖

22

≤ εκ2p+1‖∆∂

p+1t m‖

22 + Cεκ

2 j−1‖∆∂

jt m‖

22.

We integrate the previous inequality in time and using (11) from P( j − 1) we estimate the term κ2 j−1‖∆∂

jt m‖

22. This

completes the proof of the lemma.

Proof of Inequality (11) (Continuation). We apply the results from Lemmas 6 and 7 in (35) to estimate the terms|ακ2p+1(∂

p+1t A,∆∂

p+1t m)| and |κ2p+1(∂

p+1t B,∆∂

p+1t m)|. After time integration we find

κ2p+1‖∇∂

p+1t m‖

22 + α

∫ T

0κ2p+1

‖∆∂p+1t m‖

22

≤2p + 1

2

∫ T

0κ2p

‖∇∂p+1t m‖

22 + α

∫ T

0|κ2p+1(∂

p+1t A,∆∂

p+1t m)| +

∫ T

0|κ2p+1(∂

p+1t B,∆∂

p+1t m)|

≤2p + 1

2

∫ T

0κ2p

‖∇∂p+1t m‖

22 + (α + 1)

[T .C + Cε

∫ T

0κ2p+1

‖∇∂p+1t m‖

22 + ε

∫ T

0κ2p+1

‖∆∂p+1t m‖

22

].

Taking ε sufficiently small and using P(p) for the term κ2p‖∇∂

p+1t m‖

22, we get

κ2p+1‖∇∂

p+1t m‖

22 +

∫ T

0κ2p+1

‖∆∂p+1t m‖

22 ≤ C(α, p) + C(α, p)

∫ T

0κ2p+1

‖∇∂p+1t m‖

22.

After applying Gronwall’s lemma we complete the proof of inequality (11) for P(p).

Proof of Inequality (12). Take the (p+1)-th derivative of the LL equation. Then multiply the result by κ2p+1∂p+2t m.

Integration by parts gives

κ2p+1‖∂

p+2t m‖

22 + ακ2p+1(∇∂

p+1t m, ∇∂

p+2t m) = ακ2p+1(∂

p+1t A, ∂

p+2t m) + κ2p+1(∂

p+1t B, ∂

p+2t m).

On account of the analogy with (20) we get

κ2p+1‖∂

p+2t m‖

22 +

α

2∂t (κ

2p+1‖∇∂

p+1t m‖

2) = κ2p+1(α(∂

p+1t A, ∂

p+2t m) + (∂

p+1t B, ∂

p+2t m)

)+ α

2p + 12

κ2p‖∇∂

p+1t m‖

22. (40)

We will estimate the terms containing ∂p+1t A and ∂

p+1t B separately in Lemmas 8 and 9.

Lemma 8. For any fixed ε the following estimate holds:∫ T0

0|κ2p+1(∂

p+1t A, ∂

p+2t m)| ≤ C + ε

∫ T0

0κ2p+1

‖∂p+2t m‖

22.

Proof. From the definition of A we have the following:

|κ2p+1(∂p+1t A, ∂

p+2t m)| ≤ κ2p+1

∑X p+1

|(〈∇∂ it m, ∇∂

jt m〉∂k

t m, ∂p+2t m)|. (41)

I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331 1329

For all terms in (41) we estimate

|(〈∇∂ it m, ∇∂

jt m〉∂k

t m, ∂p+2t m)| ≤ ‖∇∂ i

t m‖4‖∇∂j

t m‖4‖∂kt m‖L∞‖∂

p+2t m‖2

≤ Cε‖∇∂ it m‖

24‖∇∂

jt m‖

24‖∇∂k

t m‖24 + ε‖∂

p+2t m‖

22,

where we have used the embedding W 1,4 → L∞ and Young’s inequality. Further, we apply the embeddingW 2,2 → W 1,4 to find

|(〈∇∂ it m, ∇∂

jt m〉∂k

t m, ∂p+2t m)| ≤ Cε(‖∇∂ i

t m‖22 + ‖∆∂ i

t m‖22)(‖∇∂

jt m‖

22 + ‖∆∂

jt m‖

22)

× (‖∇∂kt m‖

22 + ‖∆∂k

t m‖22) + ε‖∂

p+2t m‖

22.

Without loss of generality we can assume that i ≤ j ≤ k. Then i ≤ p and j ≤ p. From (9), (10) and (11), which isalready proved for P(p) too, we have

κ2i‖∆∂ i

t m‖22 ≤ C, κ2i

‖∇∂ it m‖

22 ≤ κ2i−1

‖∇∂ it m‖

22 ≤ C,

κ2 j‖∆∂

jt m‖

22 ≤ C, κ2 j

‖∇∂j

t m‖22 ≤ κ2 j−1

‖∇∂j

t m‖22 ≤ C,

κ2k−1‖∇∂k

t m‖22 ≤ C.

Then, because 2i + 2 j + 2k − 1 = 2p + 1, we have

κ2p+1|(〈∇∂ i

t m, ∇∂j

t m〉∂kt m, ∂

p+2t m)| ≤ Cεκ

2i (‖∇∂ it m‖

22 + ‖∆∂ i

t m‖22)κ

2 j (‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22)

× κ2k−1(‖∇∂kt m‖

22 + ‖∆∂k

t m‖22) + εκ2p+1

‖∂p+2t m‖

22,

≤ Cε + Cεκ2k−1

‖∆∂kt m‖

22 + ε‖∂

p+2t m‖

22.

We integrate the previous relation in time and use (11) to estimate the term κ2k−1‖∆∂k

t m‖22, which is already proved

for P(p) too. This completes the proof of the lemma.

Lemma 9. For any fixed ε the following estimate holds:∫ T0

0|κ2p+1(∂

p+1t B, ∂

p+2t m)| ≤ C + ε

∫ T0

0κ2p+1

‖∂p+2t m‖

22.

Proof. From the definition of B we get

|κ2p+1(∂p+1t B, ∂

p+2t m)| ≤ κ2p+1

∑Yp+1

|(∂ it m × ∆∂

jt m, ∂

p+2t m)|. (42)

For all terms in (42) we have

|(∂ it m × ∆∂

jt m, ∂

p+2t m)| ≤ ‖∂ i

t m‖L∞‖∆∂j

t m‖2‖∂p+2t m‖2

≤ Cε‖∇∂ it m‖

24‖∆∂

jt m‖

22 + ε‖∂

p+2t m‖

22,

where we have used the embedding W 1,4 → L∞ and Young’s inequality. Now we apply the embedding W 2,2 →

W 1,4:

|(∂ it m × ∆∂

jt m, ∂

p+2t m)| ≤ Cε(‖∇∂ i

t m‖22 + ‖∆∂ i

t m‖22)‖∆∂

jt m‖

22 + ε‖∂

p+2t m‖

22

≤ Cε(‖∇∂ it m‖

22 + ‖∆∂ i

t m‖22)(‖∇∂

jt m‖

22 + ‖∆∂

jt m‖

22) + ε‖∂

p+2t m‖

22.

Without loss of generality we may assume that i ≤ j . Then i ≤ p and from (10) for P(p) and from P(p −1) we have

κ2i‖∇∂ i

t m‖22 ≤ κ2i−1

‖∇∂ it m‖

22 ≤ C,

κ2i‖∇∂ i

t m‖22 ≤ κ2i−1

‖∇∂ it m‖

22 ≤ C, κ2i

‖∆∂ it m‖

22 ≤ C.

Then we get

κ2p+1|(∂ i

t m × ∆∂j

t m, ∂p+2t m)| ≤ εκ2p+1

‖∂p+2t m‖

22

1330 I. Cimrak, R. Van Keer / Nonlinear Analysis 68 (2008) 1316–1331

+ Cεκ2i (‖∇∂ i

t m‖22 + ‖∆∂ i

t m‖22)κ

2 j−1(‖∇∂j

t m‖22 + ‖∆∂

jt m‖

22)

≤ Cε + εκ2p+1‖∂

p+2t m‖

22 + Cεκ

2 j−1‖∆∂

jt m‖

22.

After time integration we use (11) from P( j − 1) to estimate the terms κ2 j−1‖∆∂

jt m‖

22. This completes the proof of

the lemma.

Proof of Inequality (12) (Continuation). We apply the results from Lemmas 8 and 9 into (40) and estimate the terms|ακ2p+1(∂

p+1t A, ∂

p+2t m)| and |κ2p+1(∂

p+1t B, ∂

p+2t m)|. Then after time integration we obtain∫ T0

0κ2p+1

‖∂p+2t m‖

22 + ακ2p+1

‖∇∂p+1t m‖

22

≤ α2p + 1

2

∫ T

0κ2p

‖∇∂p+1t m‖

22 + α

∫ T

0|κ2p+1(∂

p+1t A, ∂

p+2t m)| +

∫ T

0|κ2p+1(∂

p+1t B, ∂

p+2t m)|

≤ α2p + 1

2

∫ T

0κ2p

‖∇∂p+1t m‖

22 + (α + 1)

[T .C + ε

∫ T

0κ2p+1

‖∂p+2t m‖

22

].

Taking ε sufficiently small and using P(p) for the term κ2p‖∇∂

p+1t m‖

22, we get∫ T0

0κ2p+1

‖∂p+2t m‖

22 ≤ C(α, p).

This completes the proof of inequality (12) for P(p). Consequently, the proof of Theorem 2 is now completed.

Acknowledgement

The work of Ivan Cimrak was supported by the IUAP/V - P5/34 project of Ghent University.

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