J. Math. Anal. Appl. 361 (2010) 131–138
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Solutions for the fractional Landau–Lifshitz equation Boling Guo a , Ming Zeng b,∗ a b
Center for Nonlinear Studies, Institute of Applied Physics and Computational Mathematics, PO Box 8009-28, Beijing 100088, China College of Applied Sciences, Beijing University of Technology, PingLeYuan 100, Chaoyang District, Beijing 100022, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 25 February 2009 Available online 9 September 2009 Submitted by T. Witelski
This article considers the dynamic equation of a reduced model for thin-film micromagnetics deduced by A. DeSimone, R.V. Kohn and F. Otto in [A. DeSimone, R.V. Kohn, F. Otto, A reduced theory for thin-film micromagnetics, Comm. Pure Appl. Math. 55 (2002) 1–53]. To derive the existence of weak solutions under periodical boundary condition, the authors first prove the existence of smooth solutions for the approximating equation, then prove the convergence of the viscosity solution when the viscosity term vanishes, which implies the existence of solutions for the original equation. © 2009 Elsevier Inc. All rights reserved.
Keywords: Landau–Lifshitz equation Fractional derivative Fourier series Viscosity vanishing method
1. Introduction Landau–Lifshitz equation is fundamental in describing the behavior of magnetization field inside ferromagnetic material. Usually it can be written as [2]:
δE δE ∂u + βu × = −α u × u × , ∂t δu δu
(1.1)
where the R 3 -valued function u defined in ferromagnetic sample Ω ⊂ R 3 denotes the magnetization; δE δ u is the variation of functional E:
2
E (u ) =
|∇ u | dx + Ω
α 0 and β > 0;
|∇Φ|2 dx.
φ(u ) dx +
(1.2)
R3
Ω
Here the right-hand side of (1.2) are called the exchange energy, the anisotropic energy and the energy of the stray field respectively, the last term can be also called the magnetostatic energy [3]. φ is a given function of u; Φ and u satisfy the following Maxwell equation:
∇Φ · ∇ζ dx = R3
u · ∇ζ dx,
ζ ∈ C 0∞ R 3 .
(1.3)
Ω
When the magnetostatic energy is ignored, there are numerous papers discussing (1.1), see the monographs of Guo and Ding [4,5] for a bibliography. Otherwise the problem is more difficult, recently many physicists and mathematicians focus on its simplification. One widely explored theme is to address the problem by considering its limiting behavior in various asymptotic regimes. Several regimes are well understood, see [6–11]. In 2002, with the following assumptions:
*
Corresponding author. E-mail address:
[email protected] (M. Zeng).
0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.09.009
©
2009 Elsevier Inc. All rights reserved.
132
B. Guo, M. Zeng / J. Math. Anal. Appl. 361 (2010) 131–138
(i) Ω is a cylindrical domain of thickness t with cross section Ω :
Ω = Ω × (0, t ); (ii) u does not depend on the thickness direction x3 ; (iii) t l, where l is the diameter of the cross section Ω , DeSimone et al. [1] derived the following convergent magnetostatic energy through Fourier transforms:
1
|∇Φ|2 dx = ∇ · u 2 − 1 2
H
2
(R2)
R3
=
− 1 2 Λ 2 (∇ · u ) dx,
R2 1 2
where u = (u 1 , u 2 ) and Λ = (− ) . Considering this reduced problem, they obtained some important features of the ground-state magnetization pattern in the associated thin-film limit. In this paper we will only consider the magnetostatic energy since the other two terms are dissipative. Then δδ uE =
(− )α u (α = 12 ). However, from mathematical point of view, it is of interest to consider 0 < α < 1 and u defined on N-variables. Consider the following problem:
⎧ ut = u × (− )α u in Ω × (0, T ), ⎪ ⎪ ⎪ ⎪ N ⎨
u x+ ki e i , t = u (x, t ) in R N × (0, T ), ⎪ ⎪ i =1 ⎪ ⎪ ⎩ u (x, 0) = u 0 in R N ,
(1.4)
where 0 < α 1; Ω is the unit cube in R N : Ω = (0, 1) × (0, 1) × · · · × (0, 1); ki ∈ Z and e i is the i-th basis vector of R N
N
α (Ω) which will be specified in the next section. with 1 i N; u 0 ∈ H per The main result of this paper is
Theorem 1.1. Let 0 <
α (Ω) and |u (x)| = 1 a.e. in R N . Then there exists a global weak solution u for (1.1) such α < 1, u 0 ∈ H per 0
that for any T > 0, there holds u ∈ L ∞ (0, T ; H α (Ω)) and |u (x, t )| = 1 a.e. in R N × [0, T ). ‘Weak solution’ means that for any Φ(x, t ) ∈ C ∞ ( R N × [0, T ]) such that Φ is periodical in space and Φ(x, T ) = 0, u satisfies the following integral equation:
u Φt dx dt + Ω×(0, T )
u 0 Φ(·, 0) dx =
α
α
(− ) 2 u × Φ · (− ) 2 u dx dt .
(1.5)
Ω×(0, T )
Ω
The rest of this paper is divided into three parts. Section 2 introduces some notations and Young’s inequality in discrete form; Section 3 proves Theorem 1.1; Section 4 introduces a compactness theorem which is critical in Section 2. 2. Notations and Young’s inequality In this section we first present the definition of (− )α through Fourier series. The readers may consult Chapter 3 of [12] for a more detailed introduction. 2π in·x ˆ where ˆf (n) = Ω f (x)e −2π in·x dx. Note that n is a multi index: n = If f ∈ L 2 (Ω), then f ∼ n∈ Z N f (n)e (n1 , n2 , . . . , n N ) where ni ∈ Z (1 i N). For any nonnegative multi index m, m = (m1 , m2 , . . . , m N ) where mi ∈ Z and mi 0 (1 i N), formally we have N
(2π ini )mi e 2π in·x .
Dm f =
n∈ Z N
(2.1)
i =1
Motivated by (2.1), we infer the following relation (see also [13]):
(− )α f = (2π )2α
|n|2α ˆf (n)e 2π in·x .
(2.2)
n∈ Z N 2α Besides, f should belong to the space H per (Ω).
2α H per (Ω)
2 2 4α ˆ = f f ∈ L (Ω) and |n| f (n) < ∞ . n∈ Z N
(2.3)
B. Guo, M. Zeng / J. Math. Anal. Appl. 361 (2010) 131–138
133
2α The norm of H per (Ω) is defined as
α f H per 2α (Ω) = f 2 + (− ) f 2 .
2α If f and g belong to H per (Ω), then combining (2.1)–(2.3) and Parseval’s identity we conclude the following equation:
(− )α f · g dx = Ω
(− )α1 f · (− )α2 g dx,
(2.4)
Ω
where α1 and α2 are nonnegative and α1 + α2 = α . Next we introduce Young’s inequality in discrete form. Recall that the standard Young’s inequality is the following. Lemma 2.1. If F ∈ L p ( R N ) (1 p ∞) and G ∈ L 1 ( R N ), then F ∗ G ∈ L p ( R N ) and F ∗ G p F p G 1 .
For two series
n∈ Z N
F (x) = f n
f n and
when x ∈
n∈ Z N
g n , define their continuous extensions as
N [ni , ni + 1) i =1
and
G (x) = g n
when x ∈
N [ni , ni + 1). i =1
Immediately we have the following trivial proposition:
f n l p = F p .
(2.5)
We also notice that when y = n,
F ∗ G ( y) =
F (x)G (n − x) dx =
f n1 g n2 .
(2.6)
n1 + n2 = n
RN
N
In fact when y ∈ i =1 [ni , ni + 1), (2.6) still holds. Thus we conclude the following Young’s inequality in discrete form. Lemma 2.2. If { f n } ∈ l p and { g n } ∈ l1 , then { Remark. Here {
n1 +n2 =n
n1 +n2 =n
f n1 g n2 } ∈ l p and
n1 +n2 =n
f n1 g n2 p f n p g n 1 .
f n1 g n2 } should be regarded as the n-th term of a series.
Moreover, the same idea can be used to deduce the following lemma. Lemma 2.3. If { ˆf (n)} and { gˆ (n)} are Fourier series of f and g, then
f g (n) =
ˆf (n1 ) gˆ (n2 ).
n1 + n2 = n
3. Proof of the theorem 3.1. A prior estimates The viscosity vanishing method is used to solve (1.4). Consider the following equation:
⎧ u u ⎪ × (− )α u − β × u + ε u in Ω × (0, T ), ut = ⎪ ⎪ max ( 1 , | u |) max ( 1, |u |) ⎪ ⎪ ⎨ N
ki e i , t = u (x, t ) in R N × (0, T ), ⎪ ⎪u x + ⎪ ⎪ ⎪ i =1 ⎩ u (x, 0) = u 0 in Ω.
(3.1)
(3.2) (3.3)
134
B. Guo, M. Zeng / J. Math. Anal. Appl. 361 (2010) 131–138 1 1 α (Ω) through ε are viscosity coefficients; u 0 ∈ H per (Ω) temporarily, indeed, H per (Ω) can be replaced by H per
Here β and
density argument in the end; We utilize the auxiliary function max(11,|u |) to derive a better a prior estimates for u which will vanish once |u | 1. Taking the scalar product with u in (3.1) and integrating over Ω , we have
1 d
|u |2 dx + ε
2 dt Ω
|∇ u |2 dx = 0.
(3.4)
Ω
Integrating (3.4) over [0, t ], we have
u (·, t ) C , 2
for any 0 t < T .
(3.5)
Taking the scalar product with β u in (3.1), we have
β u · ut = β
u max(1, |u |)
× (− )α u · u + ε β| u |2 ,
(3.6)
taking the scalar product with (− )α u in (3.1), we also have
(− )α u · ut = ε · (− )α u − β
u max(1, |u |)
× u · (− )α u ,
(3.6)–(3.7) and integrating over Ω leads to
−
β d
|∇ u |2 dx −
2 dt
1 d
2 dt
Ω
(− ) α2 u 2 dx = β ε
Ω
(3.7)
u · (− )α u dx.
| u |2 dx − ε Ω
(3.8)
Ω
According to (2.4), we know
α +1 2
u · (− )α u dx = −(− ) 2 u 2 .
(3.9)
Ω
Consequently the following estimate holds
t
t u 22 dt
βε
+ε
0
(− ) α+2 1 u 2 dt + β ∇ u 2 + 1 (− ) α2 u 2 = β ∇ u 0 2 + 1 (− ) α2 u 2 . 2 2 2 2 2 2
2
2
2
(3.10)
0
3.2. Approximation We are looking for approximate solutions un (x, t ) of Eq. (2.1) under the form:
un (x, t ) =
ϕn (t )e2π in·x ,
|n|n
where
ϕn are R 3 -valued vectors, such that for any |n| n, there holds ∂ un un un × (− )α un + β × un − ε un , e 2π in·x = 0, − ∂t max(1, |un |) max(1, |un |)
(3.11)
where ·,· denotes the inner product in L 2 (Ω) and the initial value condition for (3.2) is
un (x, 0) =
n
1 ϕi (0)e i (x) → u 0 in H per (Ω).
i =1
The solvability of (3.2) can be inferred from ODE theory. Now we establish estimates on un . In the sequel the notation ‘C ’ denotes a constant independent of β and e 2π in·x with un in (3.11), we find
1 d
|un |2 dx + ε
2 dt
ε . Replacing
Ω
|∇ un |2 dx = 0.
(3.12)
Ω
Integrating (3.12) over [0, t ], we have
un (t ) C , 2
for any 0 t < T .
(3.13)
B. Guo, M. Zeng / J. Math. Anal. Appl. 361 (2010) 131–138
135
We also have
(− ) α2 un (t ) + β ∇ un (t ) C , 2 2
for any 0 t < T ,
(3.14)
and
T βε
un (t )2 dt C .
(3.15)
2
0
The bound of unt 2 follows directly from (3.11). Therefore, fixing β and ε and denoting Q T by Ω × (0, T ), we can extract from {un } a subsequence (still denoted by {un }) such that
un → u β,ε weakly in L 2 ( Q T ); 1 un → u β,ε weakly ∗ in L ∞ 0, T ; H per (Ω) ; un → u β,ε β,ε
unt → ut
strongly in L 2 ( Q T ) and a.e.; weakly in L 2 ( Q T ).
Taking the limit n → ∞ in (3.11), we deduce that for any Fourier series ψ with finite degree and scalar function C ∞ [0, T ], there holds
β,ε
ut
· ψ ϕ dx dt =
QT
u β,ε max(1, |u β,ε |)
QT
u β,ε
× (− )α u β,ε · ψ ϕ − β
max(1, |u β,ε |)
ϕ∈
× u β,ε · ψ ϕ + ε u β,ε · ψ ϕ dx dt .
It follows from the density that the above equation also holds if we replace ψ ϕ by φ for any φ ∈ L 2 ( Q T ), i.e.
β,ε
ut
· φ dx dt =
QT
QT
u β,ε max(1, |u β,ε |)
× (− )α u β,ε · φ − β
u β,ε max(1, |u β,ε |)
× u β,ε · φ + ε u β,ε · φ d xdt .
(3.16)
Lemma 3.1. If u β,ε satisfies (3.16), then
β,ε u 1 a.e. in Ω × (0, T ). β,ε
Proof. Following the idea introduced in [14], we choose φ = u β,ε − min(1, |u β,ε |) |uu β,ε | in (3.16). Then we have
β,ε 2 u 1 −
1 d 2 dt
1
1
u β,ε · ∂t u β,ε
|u β,ε |
2 |u β,ε |1
dx
|u β,ε |
|u β,ε |1
=
dx − ε |u β,ε |1
|u β,ε · ∇ u β,ε |2 dx − ε |u β,ε |3
β,ε 2 ∇ u 1 −
|u β,ε |1
1
|u β,ε |
dx.
(3.17)
u β,ε We can also choose ℵ as a test function, where ℵ is a characteristic function of the set {|u β,ε | 1}, then we have |u β,ε |
1
u β,ε · ∂t u β,ε
|u β,ε |
2 |u β,ε |1
=
ε
2 |u β,ε |1
|u β,ε · ∇ u β,ε |2 ε dx − 2 |u β,ε |3
β,ε Since ∂ u∂ n · u β,ε =
1 d
2 dt |u β,ε |1
dx
1 |∂ u β,ε |2 2 ∂n
β,ε 2 u 1 −
β,ε 2 ∇ u
|u β,ε |1
1
dx + |u β,ε |
∂ u β,ε β,ε · u dS . ∂n
|u β,ε |=1
0 on the boundary set {u β,ε = 1}, it follows from (3.17) and (3.18) that 1
|u β,ε |
dx 0,
which implies that |u β,ε | 1 a.e. in Ω × (0, T ).
2
(3.18)
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B. Guo, M. Zeng / J. Math. Anal. Appl. 361 (2010) 131–138
If β is fixed, and ε tends to 0, (3.10) allows us to extract from {u β,ε } a subsequence (still denoted by {u β,ε }) such that {u β,ε } → u β weakly ∗ in L ∞ (0, T ; H 1 (Ω)). We will show that
β u = 1,
a.e. in Ω × (0, T ).
(3.19)
In fact, for any t > 0, it is easy to see that
β,ε 2 u (t ) dx −
Ω
β,ε 2 u (0) dx + ε
t
β,ε 2 ∇ u dx dt = 0.
(3.20)
0 Ω
Ω
When ε → 0, u β,ε (t ) → u β strongly in L 2 (Ω), ∇ u β,ε 2 is uniformly bounded and |u β,ε (0)| is always equal to 1, therefore ε → 0 in (3.20) leads to
β 2 u (t ) − 1 dx = 0.
(3.21)
Ω
However, |u β | 1 a.e., then (3.19) follows from (3.21). To consider the limiting case of (3.16), first we fix β , then for any φ ∈ C ∞ ( Q¯ T ) such that φ is periodical in space and φ(·, T ) = 0, we have
β,ε
ut
· φ dx dt =
QT
u β,ε × (− )α u β,ε · φ − β u β,ε × u β,ε · φ + ε u β,ε · φ dx dt .
(3.22)
QT
The left-hand side of (3.22) can be rewritten as
−
u
β,ε
u β,ε (·, 0) · φ(·, 0) dx.
· φt dx dt −
QT
(3.23)
Ω
ε → 0, (3.23) converges to − u β · φt dx dt − u 0 · φ(·, 0) dx.
When
QT
(3.24)
Ω
The limiting case of the first term of the right-hand side of (3.22) will be discussed in the next section. For the second term on the right-hand side of (3.22), it can be rewritten as
β u β,ε × ∇ u β,ε · ∇φ dx dt ,
(3.25)
QT
ε → 0, (3.25) converges to
from (3.10) it follows that when
β u β × ∇ u β · ∇φ dx dt .
(3.26)
QT
Similarly we know that the last term on the right-hand side of (3.22) vanishes as ε → 0. Next we consider the limiting case β → 0. It follows from (3.10) that there exists a subsequence of {u β } (still denoted by {u β }) such that
uβ → u
α (Ω) ; weakly ∗ in L ∞ 0, T ; H per
uβ → u
strongly in L 2 ( Q T ) and a.e.
Besides, β ∇ u β 22 C . Therefore, when β → 0, (3.24) converges to
− QT
u · φt dx dt +
u (·, T ) · φ(·, T ) dx − Ω
u 0 · φ(·, 0) dx.
(3.27)
Ω
To estimate (3.26), we have
! ! β u β × ∇ u β · ∇φ dx dt β β∇ u β |∇φ| dx dt . QT
QT
Using Cauchy’s inequality, the right-hand side of (3.28) tends to 0 as β → 0.
(3.28)
B. Guo, M. Zeng / J. Math. Anal. Appl. 361 (2010) 131–138
137
4. A compactness theorem
This section focuses on the limit of Ω u β,ε × (− )α u β,ε · φ dx. Note that
u β,ε × (− )α u β,ε · φ dx = − Ω
u β,ε × φ · (− )α u β,ε dx Ω
=− Ω
=−
α α (− ) 2 u β,ε × φ · (− ) 2 u β,ε dx
α α α (− ) 2 u β,ε × φ − (− ) 2 u β,ε × φ · (− ) 2 u β,ε dx.
(4.1)
Ω α
α
¯ 2 2 Fix some t, then φ per (Ω). Define the linear operator L by Lu = (− ) (u φ) − (− ) u φ , where u and φ are regarded as scalar functions for convenience. We have the following theorem about L. ∈ C∞
α (Ω) to L 2 (Ω). Theorem 4.1. L is compact from H per
Applying Theorem 4.1 and (3.10), we have
lim lim
β→0 ε →0
=
α α α (− ) 2 u β,ε × φ − (− ) 2 u β,ε × φ · (− ) 2 u β,ε dx
Ω
α α α (− ) 2 (u × φ) − (− ) 2 u × φ · (− ) 2 u dx
Ω
=
α
α
(− ) 2 (u × φ) · (− ) 2 u dx. Ω
Thus we conclude Theorem 1.1. To prove Theorem 4.1, we need the following lemma. α (Ω) to H α (Ω), i.e. Lemma 4.2. L is a bounded linear operator from H per per α (Ω) C u H α (Ω) , Lu H per per
(4.2)
where C depends on φ . α (Ω) is compactly embedded into L 2 (Ω). Theorem 4.1 is an immediate consequence of Lemma 4.2 since H per
Proof of Lemma 4.2. First it is necessary to recall the following trivial inequality:
(a + b)α aα + bα (a, b 0), which leads to
|n1 + n2 |α |n1 |α + |n2 |α . α
α (Ω) . According to (2.2), we also have It is obvious that (− ) 2 u φ L 2 (Ω) C u H per
(− ) α2 (u φ)22 L
= (2π )α (Ω)
2 |n|2α u φ(n) .
(4.3)
n∈ Z N
It follows from Lemma 2.3 that
u φ(n) =
ˆ n2 ), uˆ (n1 )φ(
n1 + n2 = n
so
|n|α u φ(n) |n|α
uˆ (n1 )φ( ˆ n2 ) n1 + n2 = n
n1 + n2 = n
ˆ n2 ) + |n1 |α uˆ (n1 )φ(
uˆ (n1 )|n2 |α φ( ˆ n2 ). n1 + n2 = n
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B. Guo, M. Zeng / J. Math. Anal. Appl. 361 (2010) 131–138
Consequently we utilize Lemma 2.2 to conclude the following estimate:
(− ) α2 (u φ)22
L (Ω)
2 2 D α u 2 φ 21 + u 22 D α φ 1 .
(4.4)
Next we shall estimate (− ) 2 ( Lu ) L 2 (Ω) , from the definition it is sufficient to prove {|n|α Lu (n)} ∈ l2 . α
ˆ n)e 2π in·x , then u φ ∼ Since u ∼ n uˆ (n)e 2π in·x and φ ∼ n φ( So from the definition of Lu, we have
Lu (n) =
ˆ n2 ) − |n|α uˆ (n1 )φ(
n1 + n2 = n
n
ˆ
ˆ (n1 )φ(n2 )e n1 +n2 =n u
ˆ n2 ). |n1 |α uˆ (n1 )φ(
2π in·x
.
(4.5)
n1 + n2 = n
Then
Lu (n)
ˆ n2 ). |n2 |α uˆ (n1 )φ(
n1 + n2 = n
Furthermore
|n|α Lu (n)
ˆ n2 ) + |n1 |α uˆ (n1 )|n2 |α φ(
n1 + n2 = n
uˆ (n1 )|n2 |2α φ( ˆ n2 ), n1 + n2 = n
utilizing Lemma 2.2 once again, we have
(− ) α2 Lu 22
L (Ω)
2 2 2 D α u 2 D α φ 1 + u 22 D 2α φ 1 .
Since φ is smooth, the proof is finished.
(4.6)
2
Remark. A similar result also holds when Ω is replaced by R N . References [1] [2] [3] [4] [5] [6] [7] [8] [9]
[10] [11] [12] [13] [14]
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