WAVE MAP with POTENTIAL and HYPERSURFACE FLOW Jian Zhai
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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 simpli¯ed 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 8 > @2u 1 > ¡ ¢u ¡ B(u)(@ u; @®u) + (W (u) < @t2 ® 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 simpli¯ed equation of the dynamics of weak ferromagnets magnetization. The ferromagnetic theory states that below a critical temperature, a su±ciently 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 di±cult 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, satis¯es following Hamilton-Jacobi equation 2 2 (rt) ¡ jrrj + 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(¡1) = R¡;R(1) = R+: 2000 Mathematics Subject Classi¯cation. 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) (lnjR (r(x; t)=²)j2) r ¡ ¢r + s s ((r )2 ¡ jrrj2 + 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 ¡ jrrj2 + 1) = 0: (2:4) tt ² t 3. Global existence. By a modi¯ed method for the wave maps, we have following existence result. For s 2 N0, we let Hs(Rn;N) = fv 2 Hs;2(Rn; Rm); v 2 Ng and s+1 2 n _ n m 1 n s+1 n (¹v; v^) 2 Lloc(R ;TN) :v ¹ 2 H (R ; R );W (¹v) 2 L (R ) Hc (R ;TN) = f g: v^ 2 Hs(Rn; Rm); spt(rv;¹ v^) ½½ Rn m¡1 1 n Theorem 3.1. Let n ¸ 1 and N = S . For ('; Ã) 2 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 ¯rst n m consider the relaxation of the equation (1.1) for u¸ : R £ R ! R 8 2 > @ u¸ > ¡ ¢u + ¸(ju j2 ¡ 1)u <> @t2 ¸ ¸ ¸ 1 (3:1) + (W (u ) ¡ (W (u ) ¢ u )u ) = 0 in Rn £ R+ > 2 u ¸ u ¸ ¸ ¸ > ² : n u¸(x; 0) = '(x);@tu¸(x; 0) = Ã(x); 8x 2 R : Let Z 1 2 1 2 E¸(u¸)(t) := j@tu¸(x; t)j + jru¸(x; t)j Rn 2 2 1 ¸ + W (u ) + (ju j2 ¡ 1)2dx: (3:2) ²2 ¸ 4 ¸ d We have E¸(u¸)(t) = 0 and dt Z 1 2 1 2 1 E¸(u¸)(t) = E¸(u¸)(0) = jÃj + jr'j + 2 W (')dx (3:3) Rn 2 2 ² uniformly for ¸ and t. Then we have Claim 1. There is a subsequence (denoted also by u¸) such that 1 2 n @tu¸ *@tu weak -* in L (R; L (R )) 1 2 n @xi u¸ *@xi u weak -* in L (R; L (R )) (3:4) 2 n u¸ ¡! 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 (ju(x; t)j ¡ 1) dx · lim inf (ju¸(x; t)j ¡ 1) dx · lim sup E¸(u¸)(t) = 0; Rn ¸!1 Rn ¸!1 ¸ so we get Claim 2. u(x; t) 2 Sm¡1. Note that the ¯rst equation of (3.1) is equivalent to @2u 1 ( ¸ ¡ ¢u + (W (u ) ¡ (W (u ) ¢ u )u )) £ u = 0: (3:5) @t2 ¸ ²2 u ¸ u ¸ ¸ ¸ ¸ That is 1 @®(@ u £ u ) + W (u ) £ u = 0: (3:6) ® ¸ ¸ ²2 u ¸ ¸ Since ® ® @ (@®u¸ £ u¸) ! @ (@®u £ u) in the sense of distributions and Wu(u¸) £ u¸ ! Wu(u) £ u 1 in Lloc, we obtain that u(x; t) weakly solves (1.1). 1 n We show that the initial conditions are also satis¯ed. For any Á 2 C0 (R £ R), Z 1 Z ® dt (@®u¸ £ u¸ ¡ @®u £ u)@ Ádx n 0 Z RZ 1 1 1 + dt f 2 Wu(u¸) £ u¸ ¡ 2 Wu(u) £ ugÁdx n ² ² Z0 R = (Ã(x) £ '(x) ¡ @tu(x; 0) £ '(x))Á(x; 0)dx ! 0 as ¸ ! 1: Rn Then n (@tu(x; 0) ¡ Ã(x)) £ '(x) = 0; 8a:e:x 2 R : (3:7) But (@tu(x; 0) ¡ Ã(x)) ¢ '(x) = 0; (3:8) so (3.7)-(3.8) imply n @tu(x; 0) = Ã(x) 8a:e:x 2 R : (3:9) For any f 2 L1(R) and g 2 L2(Rn), Z 1 Z f(t)dt (@tu¸(x; t) ¡ @tu(x; t))g(x)dx ! 0; as ¸ ! 1; 0 Rn then Z 2 n lim (@tu¸(x; t) ¡ @tu(x; t))g(x)dx = 0; 8g 2 L (R ) t#0 Rn and as t # 0 2 n @tu¸(x; t) ! Ã(x) strongly in L (R ): Thus as t # 0 2 n @tu(x; t) *Ã(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 j@tu(x; t)j + jr'(x)j + 2 W (')dx n 2 2 ² t!0 R Z 1 2 1 2 1 · lim sup j@tu(x; t)j + jru(x; t)j + 2 W (u(x; t))dx t!0 Rn 2 2 ² · lim sup lim inf E¸(u¸)(t) ¸!1 Z t!0 1 2 1 2 1 · jÃj + jr'j + 2 W (')dx: Rn 2 2 ² So Z Z 1 2 1 2 lim sup j@tu(x; t)j dx · jÃ(x)j dx: (3:11) t!0 2 Rn 2 Rn From (3.10)-(3.11), we get as t # 0 2 n @tu(x; t) ! Ã(x); strongly in L (R ): 4. Convergence. De¯ne Z ² @u(x; t) 2 ² 2 1 E²(u)(t) = j j + jru(x; t)j + 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) + jrrj + 1)dx ZRn 2 2 2 = 2(1 ¡ q1(r=²))((rt) + jrrj + 1)dx (4:3) ZRn 2 ¸ 2(1 ¡ q1(r=²))dx: Rn For s 2 (0; 1), we de¯ne n 2 ² s !s(²; t) := fx 2 R j 1 ¡ q1(r (x; t)=²) ¸ ² g (0 < s < 1); (4:4) T Let ²(x; t) be the characteristic function of the set [t=0(!s(²; t) £ ftg) and p ² 1 + 1 ¡ ²s r+(²) = ln p : (4:5) 4 1 ¡ 1 ¡ ²s Thus we get Lemma 4.2. For any t > 0, n n 2 s 1¡s L [!s(²; t)] = jfx 2 R j 1 ¡ q1(r(x; t)=²) ¸ ² gj · ² C2=2 (4:6) with 0 < s < 1. Moreover, ² x 2 !s(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 p q ( § ) = § 1 ¡ ²s: 1 ² Noting (2.3) we have p ² 1 + 1 ¡ ²s r+ = ln p ; 4 1 ¡ 1 ¡ ²s and r¡ = ¡r+. We shall use the notation denoting a smooth bounded domain of Rn and [t0; t1] the subset of [0;T ]. Recall that Â(x; t) is the characteristic function of the T set [ (!s(²; t) £ ftg). That is, t=0 ( T 1; 8(x; t) 2 [t=0(!s(²; t) £ ftg) Â(x; t) = T 0; 8(x; t) 62 [t=0(!s(²; t) £ ftg): Theorem 4.1.