DISCRETE AND CONTINUOUS Website: http://AIMsciences.org DYNAMICAL SYSTEMS Supplement Volume 2005 pp. 940–946

WAVE MAP WITH POTENTIAL AND HYPERSURFACE FLOW

Jian Zhai, Jianping Fang and Lanjun Li

Department of Mathematics, Zhejiang University, Hangzhou 310027, PRC

Abstract. The simplified equation of the dynamics of weak ferromagnets magneti- zation is related to wave maps with potential. The global existence and behavior of solutions as parameter ² → 0 are obtained.

1. Introduction. We consider the equation   ∂2u 1  − ∆u − B(u)(∂ u, ∂αu) + (W (u)  ∂t2 α ²2 u − (W (u) · n(u))n(u)) = 0 in Rn × R+ (1.1)  u  u : Rn+1 → N. Here B is the second fundamental form of N and W : N → R is a smooth function in a neighborhood of N. When N = S2, (1.1) is a simplified equation of the dynamics of weak ferromagnets magnetization. The ferromagnetic theory states that below a critical temperature, a sufficiently large ferromagnetic body breaks up into small uniformly magnetized regions separated by thin transition layers which are often called Bloch walls or domain walls ([1], [8]). The equation (1.1) can be regarded as wave map equation with potential. The existence of wave maps and development of singularities are researched in [2], [3], [4], [5] etc. We interest in simple wave solutions of (1.1). In this paper, the Bloch wall motions are governed by hyperbolic equation. It is difficult to prove the convergence of solutions to (2.2) (see section 2), since the third term may change its sign. In Theorem 4.1 (see section 4), we shall prove that the limit of the solutions of the equation (2.2), in some sense, satisfies following Hamilton-Jacobi equation 2 2 (rt) − |∇r| + 1 = 0. (1.2) It means that the Bloch wall motions are governed by the Hamilton-Jacobi equation (1.2) as the parameter ² → 0.

2. Simple wave. The solutions of equation (2.1) are called the simple waves of the equation (1.1) ( Rss + B(R)(Rs,Rs) − Wu(R) + (Wu(R) · n(R))n(R) = 0 (2.1) R(−∞) = R−,R(∞) = R+.

2000 Mathematics Subject Classification. Primary: 35L70,35Q60,35F20,78A25. Key words and phrases. wave maps with potential; Hamilton-Jacobi equation; weak ferromagnets.

940 WAVE MAP WITH POTENTIAL AND HYPERSURFACE FLOW 941

We want to seek the simple wave solutions of (1.1) by the simple waves. Let u(x, t) = R(r(x, t)/²) in (1.1). We get the equation for r(x, t) (ln|R (r(x, t)/²)|2) r − ∆r + s s ((r )2 − |∇r|2 + 1) = 0. (2.2) tt 2² t 2 2 2 tr Hereafter, we take N = S , W (u) = 2(u2+u3) and u3 ≡ 0, and R± = (±1, 0, 0) . Then R(s) = (q1(s), q2(s), 0) e4s − 1 ±2e2s q (s) = , q (s) = . (2.3) 1 e4s + 1 2 1 + e4s and 2q (r/²) r − ∆r − 1 ((r )2 − |∇r|2 + 1) = 0. (2.4) tt ² t

3. Global existence. By a modified method for the wave maps, we have following existence result. For s ∈ N0, we let Hs(Rn,N) = {v ∈ Hs,2(Rn, Rm); v ∈ N} and 2 n ˙ s+1 n m 1 n s+1 n (¯v, vˆ) ∈ Lloc(R ,TN) :v ¯ ∈ H (R , R ),W (¯v) ∈ L (R ) Hc (R ,TN) = { }. vˆ ∈ Hs(Rn, Rm), spt(∇v,¯ vˆ) ⊂⊂ Rn

m−1 1 n Theorem 3.1. Let n ≥ 1 and N = S . For (ϕ, ψ) ∈ Hc (R ,TN), there exists 1 a global weak solution u of (1.1) of class H˙ . Theorem 3.1 can be proved by a penalty method developed in [3]. We first n m consider the relaxation of the equation (1.1) for uλ : R × R → R  2  ∂ uλ  − ∆u + λ(|u |2 − 1)u  ∂t2 λ λ λ 1 (3.1) + (W (u ) − (W (u ) · u )u ) = 0 in Rn × R+  2 u λ u λ λ λ  ²  n uλ(x, 0) = ϕ(x), ∂tuλ(x, 0) = ψ(x), ∀x ∈ R . Let Z 1 2 1 2 Eλ(uλ)(t) := |∂tuλ(x, t)| + |∇uλ(x, t)| Rn 2 2 1 λ + W (u ) + (|u |2 − 1)2dx. (3.2) ²2 λ 4 λ d We have Eλ(uλ)(t) = 0 and dt Z 1 2 1 2 1 Eλ(uλ)(t) = Eλ(uλ)(0) = |ψ| + |∇ϕ| + 2 W (ϕ)dx (3.3) Rn 2 2 ² uniformly for λ and t. Then we have Claim 1. There is a subsequence (denoted also by uλ) such that ∞ 2 n ∂tuλ * ∂tu weak -* in L (R; L (R )) ∞ 2 n ∂xi uλ * ∂xi u weak -* in L (R; L (R )) (3.4) 2 n uλ −→ u strongly in Lloc(R ), uniformly in t. 942 JIAN ZHAI, JIANPING FANG AND LANJUN LI

By Fatou’s lemma, for any t, Z Z 2 2 2 2 1 (|u(x, t)| − 1) dx ≤ lim inf (|uλ(x, t)| − 1) dx ≤ lim sup Eλ(uλ)(t) = 0, Rn λ→∞ Rn λ→∞ λ so we get Claim 2. u(x, t) ∈ Sm−1. Note that the first equation of (3.1) is equivalent to ∂2u 1 ( λ − ∆u + (W (u ) − (W (u ) · u )u )) × u = 0. (3.5) ∂t2 λ ²2 u λ u λ λ λ λ That is 1 ∂α(∂ u × u ) + W (u ) × u = 0. (3.6) α λ λ ²2 u λ λ Since α α ∂ (∂αuλ × uλ) → ∂ (∂αu × u) in the sense of distributions and

Wu(uλ) × uλ → Wu(u) × u 1 in Lloc, we obtain that u(x, t) weakly solves (1.1). ∞ n We show that the initial conditions are also satisfied. For any φ ∈ C0 (R × R), Z ∞ Z α dt (∂αuλ × uλ − ∂αu × u)∂ φdx n 0 Z RZ ∞ 1 1 + dt { 2 Wu(uλ) × uλ − 2 Wu(u) × u}φdx n ² ² Z0 R

= (ψ(x) × ϕ(x) − ∂tu(x, 0) × ϕ(x))φ(x, 0)dx → 0 as λ → ∞. Rn Then n (∂tu(x, 0) − ψ(x)) × ϕ(x) = 0, ∀a.e.x ∈ R . (3.7) But

(∂tu(x, 0) − ψ(x)) · ϕ(x) = 0, (3.8) so (3.7)-(3.8) imply n ∂tu(x, 0) = ψ(x) ∀a.e.x ∈ R . (3.9) For any f ∈ L1(R) and g ∈ L2(Rn), Z ∞ Z f(t)dt (∂tuλ(x, t) − ∂tu(x, t))g(x)dx → 0, as λ → ∞, 0 Rn then Z 2 n lim (∂tuλ(x, t) − ∂tu(x, t))g(x)dx = 0, ∀g ∈ L (R ) t↓0 Rn and as t ↓ 0 2 n ∂tuλ(x, t) → ψ(x) strongly in L (R ). Thus as t ↓ 0 2 n ∂tu(x, t) * ψ(x) weakly in L (R ). (3.10) WAVE MAP WITH POTENTIAL AND HYPERSURFACE FLOW 943

Note that Z 1 2 1 2 1 lim sup |∂tu(x, t)| + |∇ϕ(x)| + 2 W (ϕ)dx n 2 2 ² t→0 R Z 1 2 1 2 1 ≤ lim sup |∂tu(x, t)| + |∇u(x, t)| + 2 W (u(x, t))dx t→0 Rn 2 2 ² ≤ lim sup lim inf Eλ(uλ)(t) λ→∞ Z t→0 1 2 1 2 1 ≤ |ψ| + |∇ϕ| + 2 W (ϕ)dx. Rn 2 2 ² So Z Z 1 2 1 2 lim sup |∂tu(x, t)| dx ≤ |ψ(x)| dx. (3.11) t→0 2 Rn 2 Rn From (3.10)-(3.11), we get as t ↓ 0 2 n ∂tu(x, t) → ψ(x), strongly in L (R ).

4. Convergence. Define Z ² ∂u(x, t) 2 ² 2 1 E²(u)(t) = | | + |∇u(x, t)| + W (u)dx. Rn 2 ∂t 2 ² Lemma 4.1. Assume that u is a classical solution of (1.1). Then d E (u)(t) = 0. (4.1) dt ²

Assume that R(r(x, t)/²) is a solution of (1.1) with the initial data R(r0(x)/²) satisfying r² E (R(r(x, t)/²)) ≡ E (R( ))(0) ≤ C . (4.2) ² ² ² 2 Then, we have Z 2 2 2 ²C2 ≥ 2q2(r/²)((rt) + |∇r| + 1)dx ZRn 2 2 2 = 2(1 − q1(r/²))((rt) + |∇r| + 1)dx (4.3) ZRn 2 ≥ 2(1 − q1(r/²))dx. Rn For s ∈ (0, 1), we define n 2 ² s ωs(², t) := {x ∈ R | 1 − q1(r (x, t)/²) ≥ ² } (0 < s < 1), (4.4) T Let χ²(x, t) be the characteristic function of the set ∪t=0(ωs(², t) × {t}) and √ ² 1 + 1 − ²s r+(²) = ln √ . (4.5) 4 1 − 1 − ²s Thus we get Lemma 4.2. For any t > 0, n n 2 s 1−s L [ωs(², t)] = |{x ∈ R | 1 − q1(r(x, t)/²) ≥ ² }| ≤ ² C2/2 (4.6) with 0 < s < 1. Moreover, ² x ∈ ωs(², t) ⇐⇒ −r+(²) ≤ r (x, t) ≤ r+(²). (4.7) 944 JIAN ZHAI, JIANPING FANG AND LANJUN LI

P roof. (4.6) is obtained by (4.3). To prove (4.7), let r± satisfy r √ q ( ± ) = ± 1 − ²s. 1 ² Noting (2.3) we have √ ² 1 + 1 − ²s r+ = ln √ , 4 1 − 1 − ²s and r− = −r+. We shall use the notation Ω denoting a smooth bounded domain of Rn and [t0, t1] the subset of [0,T ]. Recall that χ(x, t) is the characteristic function of the T set ∪ (ωs(², t) × {t}). That is, t=0 ( T 1, ∀(x, t) ∈ ∪t=0(ωs(², t) × {t}) χ(x, t) = T 0, ∀(x, t) 6∈ ∪t=0(ωs(², t) × {t}). Theorem 4.1. Assume that r² is a solution of (2.4) which satisfies ² 2 kr kL (Ω×[t0,t1]) ≤ C in some region Ω × [t0, t1] (4.8) where C is independent of ². 2 Then, for any ϕ ∈ C (Ω × [t0, t1], R), Z 0 ² 2 ² 2 lim ϕ(x, t)w²(x, t)[(∂tr (x, t)) − |∇xr (x, t)| + 1]dxdt = 0, (4.9) ²→0 Rn×[0,T ] where for all ² ∈ (0, 1/2], ² r (x,t) ² ² + 2q1( ² )r (x, t) w²(x, t) = [ (1 − χ²(x, t)) + χ²(x, t)] r+(²) (4.10) 1 ≥ , ∀(x, t) ∈ Rn × [0,T ]. 2 2 P roof. Step 1. For any ϕ ∈ C (Ω × [t0, t1], R), from (2.4) we have Z 0 r² ϕ(x, t)(² + 2q ( )r²)[(∂ r²)2 − |∇ r²|2 + 1]dxdt 1 ² t x Z (r²)2 = ² ϕ(x, t)[(∂ − ∆ ) + 1]dxdt (4.11) tt x 2 Z (r²)2 = ² {(∂ ϕ − ∆ ϕ) + ϕ}dxdt. tt x 2 2 Using (4.8), we have for any ϕ ∈ C (Ω × [t0, t1]), as ² → 0, Z 0 r² ϕ(x, t)(² + 2q ( )r²)[(∂ r²)2 − |∇ r²|2 + 1]dxdt = O(²). (4.12) 1 ² t x Step 2. Let Z ² r ² ² 2 ² 2 I1(²) = ϕ(x, t)(²+2q1( )r )[(∂tr ) −|∇xr | +1]dxdt n T ² {(R ×[t0,t1])\∪t=0(ωs(²,t)×{t})} and Z ² r ² ² 2 ² 2 I2(²) = ϕ(x, t)(² + 2q1( )r )[(∂tr ) − |∇xr | + 1]dxdt. T ² {∪t=0(ωs(²,t)×{t})} From Step 1 (4.12),

I1(²) + I2(²) = O(²), as ² → 0. (4.13) WAVE MAP WITH POTENTIAL AND HYPERSURFACE FLOW 945

From (4.3) and Lemma 4.2, we have Z T Z ² ² 2 ² 2 |I2(²)| ≤ 2 dt |ϕ(x, t)|[² + |r |]|(∂tr ) − |∇xr | + 1|dx 0 ωs(²,t) Z T Z ² 2 ² 2 ≤ 2kϕkC0 dt [² + r+(²)][(∂tr ) + |∇xr | + 1]dx 0 ωs(²,t) (4.14) Z T Z ² 2 ² 2 ≤ 4kϕkC0 (² + r+(²)) dt [(∂tr ) + |∇xr | + 1]dx 0 ωs(²,t) 1−s ≤ 4TC2kϕkC0 r+(²)² . Note that from (4.5) O(²) → 0, as ² → 0. (4.15) r+(²) Then (4.13) and (4.14) imply I (²) lim 1 = 0. (4.16) ²→0 r+(²) On the other hand, using (4.3) again, we have Z ² 2 ² 2 | ϕ(x, t)[(∂tr ) − |∇xr | + 1]dxdt| T ∪t=0(ωs(²,t)×{t}) Z T Z ² 2 ² 2 (4.17) ≤ kϕkC0 dt [(∂tr ) + |∇xr | + 1]dx| 0 ωs(²,t) 1−s ≤ TC2kϕkC0 ² → 0, as ² → 0. From (4.16)-(4.17), we get (4.9). Note that r rq ( ) ≥ 0, ∀r ∈ R. (4.18) 1 ²

We have for all x 6∈ ωs(², t), r²(x, t) √ ² + q ( )r²(x, t) ≥ r (²) 1 − ²s 1 ² + √ (4.19) 1 − ²s 1 ≥ ² ln( )s. 4 ² (4.19) implies (4.10). √ Note that (4.8) is satisfied, for example, by r = ∓ 2x ± t + C.

Acknowledgements. This work is supported by NSFC NO.10171091.

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Received September, 2004; revised May, 2005.

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