A Two-Dimensional Landau-Lifshitz Model in Studying Thin Film Micromagnetics

and Applied Analysis 3 linear equation, which requires special tricks and techniques. In the convergence process, a compensated compactness principle is applied. The rest of this paper is organized as follows. Section 2 is devoted to studying 1.5 . More precisely, we first study the penalized equation. In order to do this, we consider the corresponding linear equation and get its formal solution and well-posedness, then we get the existence of a unique mild solution of the penalized equation using semigroup theory. Second, we get the existence of a weak solution of 1.5 by passing to the limit in the penalized equation. The key point in the convergence process relies on a compensated compactness principle. In Section 3, we get existence of weak solution of 1.4 in Theorem 3.1 by passing to the limit in 1.5 as ε → 0. 2. Approximation Equations In this section, we always suppose that T2 R2/ 2πZ 2 is the flat torus. We prove existence of a weak solution of the following equations: ∂m ∂t εΔm ∇ −Δ −1/2divmε ε∣∇mε∣2mε − ∇ −Δ −1/2divmε ·mεmε, in T2 × 0, ∞ , 2.1 m x, 0 m0 x , on T2, 2.2 m : T2 × 0,∞ −→ R2, ∣mε∣ 1 a.e. in T2. 2.3 Denote Lm −εΔmε − ∇ −Δ −1/2divmε. Note that the corresponding energy is E m ε ∫ Ω|∇m|dx ∫ R2 |ξ · m̂χΩ|/|ξ| dξ. The variation of the self-induced energy is lim η→ 0 ∫ R2 ∣ ∣ξ · ̂ mχΩ ηv ∣ ∣2 − ∣ξ · m̂χΩ ∣ ∣2 |ξ|η dξ ∫ R2 2iξ · m̂χΩ |ξ|1/2 iξ · v̂ |ξ|1/2 2 ∫ R2 ( −Δ divmχΩ )( −Δ −1/4divvdx 2 ∫ R2 −Δ divmχΩ divv dx 2 ∫ R2 − ∇ −Δ divmχΩ ·v dx. Equation 2.1 can be written as ∂m ∂t −Lmε Lmε ·mεmε. 2.4 It is very easy to prove that 2.1 is equivalent to m × ∂m ε ∂t m × Lm 0. 2.5 4 Abstract and Applied Analysis The equivalence follows from the following. Lemma 2.1. In the classical sense, m is a solution of 2.1 – 2.3 if and only if m is a solution of 2.5 . Proof. Suppose that m is a solution of 2.1 – 2.3 . By the vector cross product formula a × b × c a · c b − a · b c, 2.6 we have ∂m ∂t −Lmε Lmε ·mεmε ( Lm ·mεmε − mε ·mεLmε m × mε × Lmε. 2.7 By the cross product of m and 2.7 , we have m × ∂m ε ∂t m × mε × mε × Lmε −mε × Lm. 2.8 This proves that m satisfies 2.5 . Suppose that m is a solution of 2.5 . Then by the cross product of m and 2.5 , we obtain m × ( m × ∂m ε ∂t ) m × mε × Lmε 0. 2.9 Since |mε| 1, we have m · ∂m/∂t 0. Hence 2.9 implies ∂m ∂t −Lmε Lmε ·mεmε. 2.10 We define a local weak solution of 2.1 as follows. Definition 2.2. A vector-valued function m x, t is said to be a local weak solution of 2.1 , if m is defined a.e. in T2 × 0, T such that 1 m ∈ L∞ 0, T ;H1 T2 and ∂m/∂t ∈ L2 T2 × 0, T ; 2 |mε x, t | 1 a.e. in T2 × 0, T ; 3 2.1 holds in the sense of distribution; 4 m x, 0 m0 x in the trace sense. Abstract and Applied Analysis 5 We state our main result in this section as follows. Theorem 2.3. For everym0 x ∈ H1 T2 and |m0 x | 1, a.e. in T2, there exists a weak solution of 2.1 – 2.3 . To prove Theorem 2.3, we have to consider a penalized equation.and Applied Analysis 5 We state our main result in this section as follows. Theorem 2.3. For everym0 x ∈ H1 T2 and |m0 x | 1, a.e. in T2, there exists a weak solution of 2.1 – 2.3 . To prove Theorem 2.3, we have to consider a penalized equation. 2.1. The Penalized Equation In the spirit of 13 , we first construct weak solutions to a penalized problem, where the constraint |mε| 1 is relaxed: ∂m ∂t Lm − k2 ( 1 − ∣mk∣2 ) m 0, in T2 × 0, ∞ , 2.11 m x, 0 m0 x , on T2, 2.12 ∣ ∣m0 x ∣ ∣ 1, on T2. 2.13 Here m : T2 × 0,∞ → R2. In order to prove the existence of a mild solution of semilinear system 2.11 – 2.13 , we consider the corresponding linear equation. 2.1.1. The Corresponding Linear Equation First, we consider the corresponding linear equation of 2.11 – 2.13 in the whole space: ∂m ∂t εΔm ∇ −Δ −1/2divm k2m, in R2 × 0, ∞ , m x, 0 m0 x , on R2, 2.14 where m0 x m01 x , m02 x . While dealing with linear equation 2.14 , we just write m instead of m unless there may be some confusion. By Fourier transform in the x-variable, 2.14 are turned into m̂t ε ∣ ∣ξ ∣ ∣m̂ ( ξ ⊗ ξ |ξ| ) m̂ − k2m̂ 0, in R2 × 0, ∞ , m̂ ξ, 0 m̂0 ξ , on R2. 2.15 For each fixed ξ, the problem has a unique solution m̂ ξ, t e−B ξ t · e−A ξ m̂0 ξ , 2.16 where A ξ 1 |ξ| ( ξ2 1 ξ1ξ2 ξ1ξ2 ξ 2 2 ) , B ξ −k2 ε|ξ|2 0 0 −k2 ε|ξ|2 ) . 2.17 6 Abstract and Applied Analysis So the problem has the solution m x, t 1 4π2 ( e−B ξ t ∗e−A ξ t∗m0 x . 2.18 Now the only problem left is to find the inverse Fourier transforms of e−A ξ t and e−B ξ . First, we need to find an orthogonal matrix O ξ such that O ξ A ξ Oτ ξ is the Jordan normal form of A ξ . In fact, O ξ 1 |ξ| ( ξ2 −ξ1 ξ1 ξ2 ) . 2.19 Now we begin to calculate the inverse Fourier transform of e−A ξ t ( e−A ξ t )∨ 1 2π ∫ R2 e · ξe−A ξ dξ 1 2π ∫ R2 e · ξOτ ξ O ξ e−A ξ tOτ ξ O ξ dξ 1 2π ∫ R2 e · ξOτ ξ ( ∞ ∑ n 0 −1 tn n! O ξ A ξ Oτ ξ n ) O ξ dξ 1 2π ∫ R2 e · ξOτ ξ ( ∞ ∑ n 0 −1 tn n! ( 0 0 0 |ξ| )n) O ξ dξ 1 2π ∫ R2 e · ξOτ ξ ( I ∞ ∑ n 1 −1 tn n! ( 0 0 0 |ξ|n )) O ξ dξ 1 2π ∫ R2 e · ξOτ ξ ( I ( 0 0 0 e−|ξ|t − 1 )) O ξ dξ 1 2π ∫ R2 e · ξ ( I 1 |ξ|2 ( ξ2 1 ξ1ξ2 ξ1ξ2 ξ 2 2 ) ( e−|ξ|t − 1 ) dξ δ x I 1 2π ∫ R2 e · ξ 1 |ξ|2 (( ξ2 1 ξ1ξ2 ξ1ξ2 ξ 2 2 ) ( e−|ξ|t − 1 ) dξ. 2.20