ENERGY IDENTITY FOR THE MAPS FROM A SURFACE WITH TENSION FIELD BOUNDED IN L p

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1 Available at: htt://ublications.ict.it IC/2009/083 United Nations Educational, Scientific and Cultural Organization and International Atomic Energy Agency THE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS ENERGY IDENTITY FOR THE MAPS FROM A SURFACE WITH TENSION FIELD BOUNDED IN L Li Jiayu Academy of Mathematics and Systems Sciences, Chinese Academy of Sciences, Beijing 00080, Peole s Reublic of China and The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy and Zhu Xiangrong 2 The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy. Abstract Let M be a closed Riemannian surface and u n a sequence of mas from M to Riemannian manifold N satisfying su( u n L 2 (M) + τ(u n ) L (M)) Λ n for some >, where τ(u n ) is the tension field of the maing u n. For the general target manifold N, if 6 5, we rove the energy identity and neckless during blowing u. If the target manifold N is a standard shere S m, we get the same results for any >. MIRAMARE TRIESTE November

2 Introduction Let (M, g) be a closed Riemannian manifold and (N, h) be a Riemannian manifold without boundary. For a maing u from M to N in W,2 (M,N), the energy density of u is defined by e(u) = 2 du 2 = Trace g u h where u h is the ull-back of the metric tensor h. The energy of the maing u is defined as where dv is the volume element of (M,g). E(u) = M e(u)dv A ma u C (M,N) is called harmonic if it is a critical oint of the energy E. By Nash embedding theorem we know that (N,h) can be isometrically into an Euclidean sace R K with some ositive integer K. Then (N,h) may be considered as a submanifold of R K with the metric induced from the Euclidean metric. Thus a ma u C (M,N) can be considered as a ma of C (M,R K ) whose image lies on N. In this sense we can get the following Euler-Lagrange equation The tension field τ(u) is defined by u = A(u)(du,du). τ(u) = M u A(u)(du,du) where A(u)(du,du) is the second fundamental form of N in R K. So u is harmonic means that τ(u) = 0. The harmonic maings are of secial interest when M is a Riemann surface. Consider a sequence of maings u n from Riemann surface M to N with bounded energies. It is clear that u n converges weakly to u in W,2 (M,N) for some u W,2 (M,N). But in general, it may not converge strongly in W,2 (M,N). When τ(u n ) = 0, i.e. u n are all harmonic, Parker in [9] roved that the lost energy is exactly the sum of some harmonic sheres which is defined as a harmonic maing from S 2 to N. This result is called energy identity. Also he roved that the images of these harmonic sheres and u(m) are connected, i.e. there is no neck during blowing u. When τ(u n ) is bounded in L 2, the energy identity is roved in [0] for the shere and [2] for the general target manifold. In [] they roved there is no neck during blowing u. For the heat flow of harmonic maings, the results can also be found in [5, 6]. When the target manifold is a shere, there are some good observations in [8] for general tension field. In this aer we rove the energy identity and neckless during blowing u for a sequence of mas u n with τ(u n ) bounded in L for some 6 5. Esecially, when the target manifold is a shere, we obtain the same results for any >. 2

3 When τ(u n ) is bounded in L for some >, the small energy regularity roved in [2] imlies that u n converges strongly in W,2 (M,N) outside a finite set of oints. For simlicity in exosition, it is no matter to assume that M is the unit disk D = D(0,) and there is only one singular oint at 0. In this aer we rove the following two theorems. Theorem Let {u n } be a sequence of maings from D to N in W,2 (D,N) with tension field τ(u n ). If (a) u n W,2 (D ) + τ(u n ) L (D ) Λ for some 6 5 ; (b) u n u strongly in W,2 (D \ {0},R K ) as n. Then there exist a subsequence of {u n } (we still denote it by {u n }) and some nonnegative integer k. For any i =,...,k, there exist oints x i n, ositive numbers r i n and a nonconstant harmonic shere w i (which we view as a ma from R 2 { } N) such that () x i n 0,ri n 0 as n ; (2) lim n ( ri n + rj rn j n ) = for any i j; rn+r i n j (3) w i is the weak limit or strong limit of u n (x i n + rnx) i in W,2 Loc (R2,N); (4) Energy identity: r i n + xi n x j n lim E(u n,d ) = E(u,D ) + n (5) Neckless: The image u(d ) k i= w i (R 2 ) is a connected set. k E(w i ); () i= Theorem 2 If the target manifold N is a standard shere S m, Theorem holds for any >. This aer is organized as follows. In section 2 we state some basic lemmas and some standard arguments in the blow-u analysis. In section 3 and section 4 we rove Theorem. In the roof, we used delicate analysis based on the ideas in [2], [7] and []. Energy identity is roved in section 3 and neckless is roved in section 4. In section 5 and section 6 we introduced new ideas in roving Theorem 2. Similarly, energy identity is roved in section 5 and neckless is roved in section 6. Throughout this aer, without illustration the letter C denotes a ositive constant which deends only on,λ and the target manifold N and may vary in different cases. Furthermore, we do not always distinguish the sequence and its subsequence. 2 Some basic lemmas and standard arguments The Lorentz sace L 2, is defined as {f : f L 2, = 0 f (t)t 2 dt < } where f (t) = inf{s > 0 : {x : f(x) > s} t} is the non-increasing rearrangement function of f. The following roerty is fundamental on R 2. 3

4 Lemma 3 For any δ > 0, we have Proof: q > 2 there holds f L 2 (D δ ) C f L 2, (D δ ); f L 2, (D δ ) C q δ q 2 q f L q (R 2 ),q > 2. We only rove the last inequality. By the definition of rearrangement function, when f L 2, (D δ ) = ( So we comlete the roof of this lemma. 0 Dδ 0 0 (fχ Dδ ) (t)t 2dt f (t)t 2dt C q δ 2 q f q. (f ) q (t)dt) πδ 2 q ( 0 t q 2(q ) dt) q q Lemma 4 ([5] P42 Theorem 3.3.0) For each m 2, the sace W, (R m ) is continuously embedded in L m m, (R m ). As a corollary, when m = 2 we have Lemma 5 The Riesz otential is bounded from the Hardy sace H (R 2 ) to L 2, (R 2 ), i.e. where R i (x) = x i x 2, i =,2. R i f L 2, (R 2 ) C f H (R 2 ) Remark: The roof is contained in [5] P4-P42 (The roof of Theorem 3.3.8). For convenience we illustrate the roof here. and Let Φ be the Newtonian otential of f, then Hélein in [5] P42 showed that Φ W 2, (R 2 ) dφ L 2, (R 2 ) C f H (R 2 ). As the Riesz otential of f is the artial derivative of Φ, i.e. R i f = i (N f) = i Φ, we obtain the desired result. Lemma 6 ([5],P37 Theorem 3.3.4) If f L 2, (R 2 ), then f is continuous and f f(0) C 0 (D ) C f L 2, (D ). We also need the following result on the Hardy sace H (R 2 ). Lemma 7 ([]) If f,g W,2 (R 2 ),then f g = f 2 g 2 f g belongs to the Hardy sace H (R 2 ) and furthermore f g H C f 2 g 2. (2) 4

5 Now we recall the regular theory for the maing with small energy on the unit disk and the tension field in L ( > ). Lemma 8 Let ū be the mean value of u on the disk D. There exists a ositive constant ǫ N that 2 deends only on the target manifold such that if E(u,D ) ǫ 2 N then where >. that u ū W 2, (D 2 ) C( u L 2 (D ) + τ(u) ) (3) As a direct consequence of (3) and the Sobolev embedding W 2, (R 2 ) C 0 (R 2 ), we have u Osc(D2 ) = su x,y D 2 u(x) u(y) C( u L 2 (D ) + τ(u) ). (4) The lemma has been roved in [2]. Remark. In [2] they roved this lemma for the mean value of u on the unit disk. Note D u(x)dx D 2 u(x)dx C u D D L 2 (D ). 2 So we can use the mean value of u on D in this lemma. 2 Remark 2. Suose we have a sequence of maings u n from the unit disk D to N with u n W,2 (D ) + τ(u n ) L (D ) Λ for some >. D, A oint x D is called an energy concentration oint (blow-u oint) if for any r,d(x,r) where ǫ N is given in this lemma. sue(u n,d(x,r)) > ǫ 2 N n If x D isn t an energy concentration oint, then we can find a ositive number δ such that E(u n,d(x,δ)) ǫ 2 N, n. Then it follows from Lemma 8 that we have a uniformly W 2, (D(x, δ 2 ))-bound for u n. Because W 2, is comactly embedded into W,2, there is a subsequence of u n (denoted by u n ) and u W 2, (D(x, δ 2 )) such that lim n u n = u in W,2 (D(x, δ 2 )). So u n converges to u strongly in W,2 (D ) outside a finite set of oints. Under the assumtions in the theorems, by the standard blow-u argument, i.e. rescalling u n suitable and reeated, we can obtain some nonnegative integer k. For any i =,..., k, there exist a oint x i n, a ositive number ri n and a nonconstant harmonic shere wi satisfying (), (2) and (3) of Theorem. By the standard induction argument in [2] we only need to rove the theorems in the case that there is only one bubble. 5

6 In this case we may assume that w is the strong limit of the sequence u n (x n +r n x) in W,2 Loc (R2 ). It does nothing to assume that x n = 0. Set w n (x) = u n (r n x). As the energy identity is equivalent to lim lim E(u n,d \ D δ ) = E(u,D ), δ 0 n lim lim lim E(u n,d δ \ D δ 0 n R rnr) = 0. (5) To rove the set u(d ) and w(r 2 ) is connected, it is enough to show that lim lim lim δ 0 n R su u n (x) u n (y) = 0. (6) x,y D δ \D rnr 3 Energy identity for the general target manifold In this section, we rove the energy identity for the general target manifold when 6 5. Assume that there is only one bubble w which is the strong limit of u n (r n ) in W,2 Loc (R2 ). Let ǫ N be the constant in Lemma 8. Furthermore, by the standard argument of blow-u analysis we can assume that for any n, E(u n,d rn ) = By the argument in [2], we can show su E(u n,d(x,r)) = ǫ2 N rr n,d(x,r) D 4. (7) Lemma 9 ([2]) If τ(u n ) is bounded in L for some >, then the tangential energy on the neck domain equals to zero, i.e. lim δ 0 R n x 2 θ u 2 dx = 0. D δ \D rnr (8) Proof: The roof is the same as that in [2], we sketch it. For any ǫ > 0, take δ,r such that for any n, E(u,D 4δ ) + E(w,R 2 \ D R ) + δ 4( ) < ǫ 2. It is no matter to suose that r n R = 2 jn,δ = 2 j 0. When n is big enough, for any j 0 j j n, there holds (see [2]) E(u n,d 2 j \ D 2 j) < ǫ 2. For any j, set h n (2 j ) = 2π S u n (2 j,θ)dθ and h n (t) = h n (2 j ) + (h n (2 j ) h n (2 j )) ln(2j t) ln 2,t [2 j,2 j ]. It is easy to check that d 2 h n (t) dt 2 + t dh n (t) dt = 0,t [2 j,2 j ]. 6

7 Consider h n (x) = h n ( x ) as a ma from R 2 to R K, then h n = 0 in R 2. Set P j = D 2 j \D 2 j we have (u n h n ) = u n h n = u n = A(u n ) + τ(u n ),x P j. (9) Taking the inner roduct of this equation with u n h n and integrating over P j, we get that (u n h n ) 2 dx = P j (u n h n )(A(u n ) + τ(u n ))dx + P j (u n h n )(u n h n ) r ds. P j Note that by the definition, h n (2 j ) is the mean value of {2 j } S and (h n ) r is indeendent of θ. So the integral of (u n h n )(h n ) r on P j vanishes. When j 0 < j < j n, by Lemma 8 we have u n h n C 0 (P j ) u n h n (2 j ) C 0 (P j ) + u n h n (2 j ) C 0 (P j ) 2 u n Osc(Pj ) C( u n L 2 (P j P j P j+ ) + 2 2( )j τ(u n ) ) C(ǫ + 2 2( )j ) C(ǫ + δ 2( ) ) Cǫ. Summing j for j 0 < j < j n, we have (u n h n ) 2 dx = (u n h n ) 2 dx D δ \D 2rnR j 0 <j<j P j n u n h n ( A(u n ) + τ(u n ) )dx j 0 <j<j P j n + (u n h n )(u n h n ) r ds j 0 <j<j P j n Cǫ( ( u n 2 + τ(u n ) )dx + u n ds) D 2δ \D 2rnR D 2δ D 2rnR Cǫ( u n 2 dx + δ 2( ) + ǫ) D 2δ \D 2rnR Cǫ. (0) Here we use the inequality D 2δ D 2rnR u n ds Cǫ, which can be derived from the Sobolev trace embedding theorem. As h n (x) is indeendent of θ, it can be shown that x 2 θ u n 2 dx (u n h n ) 2 dx Cǫ. D 2δ \D 2rnR D 2δ \D 2rnR So this lemma is roved. It is left to show that the normal energy on the neck domain also equals to zero. We need the following Pohozaev equality which was first roved by Lin-Wang [7]. 7

8 Lemma 0 (Pohozaev equality, [7], lemma 2.4, P374) Let u be a solution to then there holds u + A(u)(du,du) = τ(u), D t ( r u 2 r 2 θ u 2 )ds = 2 t As a direct corollary, integrating it over [0,δ] we have D δ ( r u 2 r 2 θ u 2 )dx = δ 0 D t τ (x u)dx. () 2 τ (x u)dxdt. (2) t D t Proof: Multilying both sides of the equation by x u and integrating it over D t, we get u 2 dx t r u 2 ds + x u 2 dx = τ (x u)dx. D t D t 2 D t D t Note that Hence, x u 2 dx = u 2 dx + t u 2 ds. 2 D t D t 2 D t D t ( r u 2 2 u 2 )ds = t As u 2 = r u 2 + r 2 θ u 2, we roved this lemma. D t τ (x u)dx. Now we use this equality to estimate the normal energy on the neck domain. We rove the following lemma. Lemma If τ(u n ) is bounded in L for some 6 5, then for δ small enough, there holds ( r u n 2 x 2 θ u 2 )dx Cδ D δ where C deends on, Λ, the target manifold N and the bubble w. Proof: Take ψ C 0 (D 2) satisfying that ψ = in D, then (ψu n ) = ψa(u n )(du n,du n ) + ψτ n + 2 ψ u n + u n ψ. Set g n = ψa(u n )(du n,du n ) + ψτ n + 2 ψ u n + u n ψ. When x <, xi y i i u n (x) = R i g n (x) = x y 2g n(y)dy. Let Φ n be the Newtonian otential of ψτ n, then Φ n = ψτ n. The corresonding Pohozaev equality is D δ ( r Φ n 2 r 2 θ Φ n 2 )dx = Here i Φ n (x) = R i (ψτ n )(x) = x i y i x y 2 (ψτ n )(y)dy. As τ n is bounded in L ( > ), there holds Φ n 2 dx Cδ 4( ) D δ Φ n δ 0 2 ψτ n (x Φ n )dxdt. (3) t D t Cδ 4( ) τ n 2 Cδ 4( ).

9 By (3), it can be shown that for any δ > 0, δ ψτ n (x Φ n )dxdt Φ n 2 dx Cδ 4( ). (4) 0 t D t D δ For δ small enough, we have ( r u n 2 r 2 θ u n 2 )dx D δ δ 2 = τ n (x u n )dxdt 0 t D t δ δ 2 τ n (x Φ n )dxdt + 2 xτ n (u n Φ n )(x) dxdt 0 t D t 0 t D t Cδ 4( ) δ + 2 xτ n (u n Φ n )(x) ( D δ x t dt)dx Cδ 4( ) + 2 τ n (u n Φ n )(x) x ln dx. (5) D δ x x For any j > 0, set ϕ j (x) = ψ( 2 2 j δ ) ψ( x 2 2 j δ ). When 2 j δ x < 2 j δ, we have xi y i i (u n Φ n )(x) = x y 2(g n(y) ψτ n (y))dy ψa(un )(du n,du n ) + 2 ψ u n + u n ψ (y) x y ψa(un )(y) dy + C ( u n + u n )(y)dy x y < y <2 ϕj A(u n )(y) (ψ ϕj )A(u n )(y) dy + dy + C x y x y ϕj A(u n )(y) A(un )(y) dy dy + + C x y x ϕj A(u n )(y) x y dy + C x. (6) When δ > 0 is small enough and n is big enough, for any j > 0 we claim that ϕ j A(u n ) C(2 j δ) 4( ) (7) 2 where the constant C deends only on, Λ, the bubble w and the target manifold N. Take δ > 0 and R(w) which deends on w such that E(u,D 8δ ) ǫ2 N 8 ; E(w,R2 \ D R(w) ) ǫ2 N 8. The standard blow-u analysis (see [2]) show that for any j with 8r n R(w) 2 j δ and n big enough, there holds By (7), when 2 j δ < rn 6, there holds E(u n,d 2 4 j δ \ D 2 3 j δ) ǫ2 N 3. E(u n,d 2 4 j δ \ D 2 3 j δ) ǫ2 N 4. 9

10 So when 2 j δ < rn 6 or 2 j δ 8r n R(w), by Lemma 8, we have ϕ j A(u n ) 2 C u n 2 L 2 2 (D 2 3 j δ \D 2 2 j δ ) C u n u n,j 2 W 2, (D 2 3 j δ \D 2 2 j δ ) C[(2 j δ) 4( ) u n 2 L 2 (D 2 4 j δ \D 2 4 j δ ) + τ(u n) 2 ] C(2 j δ) 4( ) where u n,j is the mean of u n on D 2 3 j δ \ D 2 2 j δ. On the other hand, when rn 6 2 j δ 8r n R(w), we can find no more than CR(w) 2 balls with radius rn 2 to cover D 2 3 j δ \ D 2 2 j δ, i.e. D 2 3 j δ \ D 2 2 j δ m i= D(y i, r n 2 ). Denote B i = D(y i, rn 2 ) and 2B i = D(y i,r n ). By (7), for any i with i m there holds Using Lemma 8 we have ϕ j A(u n ) 2 E(u n,2b i ) ǫ2 N 4. C u n 2 L C( u n i= 2 2 (D 2 3 j δ \D 2 2 j δ ) L 2 (B i ) ) 2 C u n 2 2 i= L2 (B i ) C u n u n,i 2 W 2, (B i ) C i= i= ((r n ) 4( ) u n 2 L 2 (2B i ) + τ(u n) 2 ) Cm((2 j δ) 4( ) + ) C(2 j δ) 4( ) where u n,i is the mean of u n over B i and the constant C deends only on, Λ, the bubble w and the target manifold N. So we roved the claim (7). By (6) and (7), when > we get that τ n (u n Φ n )(x) x ln D δ x dx 2 j δ< x <2 j δ C 2 j δ< x <2 j δ τ n (u n Φ n )(x) x ln x dx τ n ( x + ϕj A(u n )(y) dy) x ln x y x dx 0

11 C( τ n ln D δ x dx + τ n ( 2 j δ< x <2 j δ C( ln + 2 j δ ln 2j L (D δ ) δ ϕj A(u n )(y) y C(δ 2 (ln δ ) + ϕj A(u n )(y) x y dy) x ln x dx) dy ) τ n 2 j δ ln 2j δ ϕ ja(u n ) 2 ). (8) 3 2 Here we use the fact that the fraction integral oerator I(f) = f is bounded from Lq (R 2 ) to L 2q 2 q (R 2 ) for < q < 2. When 6 5, i.e. ϕ j A(u n ) From (8) and (9) we have 2, by (7) there holds C(2 j δ) 5 6 ϕ j A(u n ) C(2 j δ) 5 6 4( ) 2 τ n (u n Φ n )(x) x ln D δ x dx C(δ 2 (ln δ ) + 2 j δ ln 2j δ ϕ ja(u n ) 2 ) 3 2 C(δ + C(δ + δ 2( ) ln δ )) 2 j δ ln 2j δ (2 j δ) 2 ) C(2 j δ) 2. (9) Cδ. (20) It is clear that (5) and (20) imly that ( r u n 2 r 2 θ u n 2 )dx Cδ. (2) D δ Now we use these lemmas to rove the energy identity. Note that w is harmonic, from Lemma 0 we see that D R ( r w 2 r 2 θ w 2 )dx = 0 for any R > 0. It is easy to see that lim lim ( r u n 2 r 2 θ u n 2 )dx = lim ( r w 2 r 2 θ w 2 )dx = 0. R n D rnr R D R Letting δ 0 in (2), we obtain lim lim lim ( r u n 2 r 2 θ u n 2 )dx δ 0 R n D δ \D rnr lim lim ( r u n 2 r 2 θ u n 2 )dx + lim lim ( r u n 2 r 2 θ u n 2 )dx δ 0 n D δ R n D rnr = 0. Using Lemma 9 we obtain that the normal energy also vanishes on the neck domain. So the energy identity is roved.

12 4 Neckless for the general target manifold In this section we use the method in [] to rove the neckless during blowing u. For any ǫ > 0, take δ,r such that E(u,D 4δ ) + E(w,R 2 \ D R ) + δ 4( ) < ǫ 2. Suose r n R = 2 jn,δ = 2 j 0. When n is big enough, the standard blow-u analysis show that for any j 0 j j n, 2π E(u n,d 2 j \ D 2 j) < ǫ 2. For any j 0 < j < j n, set L j = min{j j 0,j n j}. Now we estimate the norm u n L 2 (P j ). Denote P j,t = D 2 t j \ D 2 t j and take h n,j,t similar to h n in the last section but h n,j,t (2 ±t j ) = S u n(2 ±t j,θ)dθ. By an argument similar to the one used in deriving (0), we have, for any 0 < t L j, r 2 θ u n 2 dx Cǫ( P j,t P j,t u n 2 dx + (2 t j ) 2( ) ) + Set f j (t) = P j,t u n 2 dx, a simle comutation shows that f j (t) = ln 2(2t j u n 2 ds + 2 t j {2 t j } S P j,t u n h n,j,t u n ds. (22) {2 t j } S u n 2 ds). As h n,j,t is indeendent of θ and h n,j,t is the mean value of u n at the two comonents of P j,t, by Poincaré inequality we get P j,t u n h n,j,t u n ds = ( u n h n,j,t u n ds {2 t j } S + u n h n,j,t u n ds {2 t j } S u n h n,j,t 2 ds) 2 ( u n 2 ds) 2 {2 t j } S {2 t j } S +( u n h n,j,t 2 ds) 2( u n 2 ds) 2 {2 t j } S {2 t j } S C(2 t j u n 2 ds + 2 t j u n 2 ds) {2 t j } S {2 t j } S Cf j(t). On the other hand, by a similar argument as we did in obtaining (2), we have ( r u n 2 r 2 θ u n 2 )dx C((2 t j ) + (2 t j ) ) C(2 t j ). (23) P j,t Since u 2 = r u 2 +r 2 θ u 2 = 2r 2 θ u 2 +( r u 2 r 2 θ u 2 ), by (22) and (23) we have f j (t) 2 r 2 θ u n dx + P j,t ( r u n 2 r 2 θ u n 2 )dx P j,t Cǫ(f j (t) + (2 t j ) 2( ) ) + Cf j (t) + C(2t j ) C(ǫf j (t) + 2 ( )j 2 ( )t + f j(t)). 2

13 Take ǫ small enough and denote ǫ = ln 2, then for some ositive constant C big enough there holds f j (t) C f j(t) + Ce ǫj e ǫt 0. It is no matter to assume that ǫ > C, then we have Integrating this inequality over [2,L j ], we get f j (2) C(e L j C f j (L j ) + e ǫj Lj Note that j L j, there holds (e t C f j (t)) + Ce ǫj e (ǫ C )t 0. e (ǫ C )t dt) C(e L j C f j (L j ) + e ǫj e (ǫ C )L j ). f j (2) C(e L j C fj (L j ) + e j C ). Since the energy identity has been roved in the last section, we can take δ small such that the energy on the neck domain is less than ǫ 2 which imlies that f j (L j ) < ǫ 2. So we get f j (2) C(e L j C ǫ 2 + e j C ). Using Lemma 8 on the domain P j = D 2 j \ D 2 j when j < j n, we obtain u n Osc(Pj ) C( u n L 2 (P j P j P j+ ) + 2 2( )j τ(u n ) ) C(f j (2) + e 2ǫj ). Summing j from j 0 to j n we obtain that u n Osc(Dδ \D 2rnR ) j n j=j 0 u n Osc(Pj ) j n C (f j (2) + e 2ǫ j ) j=j 0 j n C (e L j C ǫ 2 + e j C + e 2ǫ j ) j=j 0 C( e i C ǫ 2 j + C ) i=0 C(ǫ 2 + e j 0 C ) C(ǫ 2 + δ C ). j=j 0 e Here we use the assumtion that ǫ > C. So we roved that there is no neck during the blowing u. 3

14 5 Energy identity for the shere The notations are the same as before. At first we give the estimate of L 2, norm of u n on the neck domain. Lemma 2 Suose that u is a ma from the unit disk D to N in W,2 (D,N) with τ(u) L for some >. Assume that ǫ > 0 is small enough, if r, δ satisfies D 2t \D t u 2 < ǫ 2 for any r < t < 4δ. Then we have u L 2, (D δ \D 4r ) = sua {x D δ \ D 4r : u(x) > a} 2 C(ǫ + δ 2( ) ) (24) a>0 where C only deends on u 2, τ(u) and the target manifold. Proof: Assume δ = 2 N r. Then for any i N + 2 there holds E(u,D 2 i r \ D 2 i r) < ǫ. Take ψ C0 (D x 2) satisfying that ψ = in D. Set θ i (x) = ψ( 2 i+2 r ) ψ( x 2 i 2 r ) and u i = u(x)dx D 2 i+3 \D r 2 i 2 r D 2 i+3 r \D 2 i 2 r, then For x D 2 i+ r \ D 2 i r, there holds (θ i (u u i )) = θ i u + 2 θ i u + (u u i ) θ i = θ i A(u) + θ i τ(u) + 2 θ i u + (u u i ) θ i. u(x) = (θ i (u u i ))(x) = R (θ i A(u) + θ i τ(u) + 2 θ i u + (u u i ) θ i )(x) R (θ i A(u))(x) + R (θ i τ(u))(x) + R (2 θ i u + (u u i ) θ i )(x) = I (x) + I 2 (x) + I 3 (x). By Sobolev inequality it can be shown that when x D 2 i+ r \ D 2 i r with i 2, I 3 (x) R(x y)(2 θ i u + (u u i ) θ i )(y) dy C (D 2 i+3 r \D 2 i+2 r ) (D 2 i r \D 2 i 2 r ) x y (2 i r u + 2 2i r 2 u u i )(y)dy C (2 i r) ((2 i r) u + (2 i r) 2 u u i )(y)dy D 2 i+3 r \D 2 i 2 r C(2 i r) 2 D 2 i+3 r \D 2 i 2 r u (y)dy C x E(u,D 2 i+3 r \ D 2 i 2 r) Cǫ x. (25) 4

15 As Riesz otential is bounded from L (R 2 ) to L 2, (R 2 ), we get that for any a > 0, So there holds {I (x) + I 2 (x) > a} Ca 2 I + I 2 2 L 2, (R 2 ) Ca 2 θ i ( A(u) + τ(u) ) 2 Ca 2 ( θ i ( u 2 + τ(u) )(x)dx) 2 Ca 2 (( θ i u(x) 2 dx) 2 + ( θ i τ(u)(x) dx) 2 ) Ca 2 (ǫ 2 θ i u 2 (x)dx + (2 i r) 4( ) ). = {x D δ \ D 4r : u(x) > 2a} N i=2 N i=2 N i=2 {x D 2 i+ r \ D 2 i r : u(x) > 2a} N {x D 2 i+ r \ D 2 i r : I (x) + I 2 (x) > a} + {I (x) + I 2 (x) > a} + {x D : Cǫ x > a} N Ca 2 i=2 Ca 2 (ǫ 2 (ǫ 2 N ( i=2 Ca 2 (ǫ 2 + δ 4( ) ). i=2 θ i u 2 (x)dx + (2 i r) 4( ) ) + Cǫ2 a 2 θ i ) u 2 (x)dx + +δ 4( ) + ǫ 2 ) {x D 2 i+ r \ D 2 i r : I 3 (x) > a} which imlies (24). holds Now we estimate the norm u n L 2, (D ). Take ψ C 0 (D 2) and ψ = in D, then there (ψu n ) = ψ u n + 2 u n ψ + u n ψ. Using the fact u n = we can rewrite the equation as u i n = u i n u j n 2 + τ i (u n ) = (u j n u i n u i n u j n) u j n + τ i (u n ). Set (F n ) i j = ψ(uj n u i n u i n u j n) and χ(x) = ψ( x 2 ), then we have (ψu n ) = F n u n + ψτ(u n ) + 2 u n ψ + u n ψ = F n (χu n ) + ψτ(u n ) + 2 u n ψ + u n ψ. Some direct comutations show that div(f n ) i j = ψ(uj n ui n ui n uj n ) + ψ(uj n ui n ui n uj n ) = ψ(u j nτ i (u n ) u i nτ j (u n )) + ψ(u j n u i n u i n u j n) C( τ(u n ) + u n ). (26) 5

16 Let G n be the Newtonian otential of (div F n ). As div F n C 0 (D 2), we have In R 2 there holds G n (x) = N (div F n ) = N div F n, div(f n G n ) = divf n div( N divf n ) = divf n div (N divf n ) = 0. So there exists H n such that F n G n = H n = ( 2 H n, H n ) The oerator N is the Riesz otential which is bounded from L (R 2 ) to L 2 2 (R 2 ), so by (26) when < < 2, it can be shown that G n 2 2 = N div F n 2 2 C div F n C( τ(u n ) L (D 2 ) + u n L 2 (D 2 )). (27) Set φ(x) = ψ( x 4 ). It is no matter to assume that D 4 H n = 0, then we have (φh n ) 2 C H n L 2 (D 4 ) C( F n L 2 (D 4 ) + G n L 2 (D 4 )) C( u n 2 + G n 2 ) 2 C( u n 2 + τ(u n ) ) C. (28) Set f n = ψτ(u n ) + 2 u n ψ + u n ψ + G n (χu n ), now we can obtain that (ψu n ) = F n (χu n ) G n (χu n ) + f n = H n (χu n ) + f n = (φh n ) (χu n ) + f n. (29) By (28) and Lemma 7 we get that (φh n ) (χu n ) H C (φh n ) 2 (χu n ) 2 C. (30) On the other hand, there holds f n = ψτ(u n ) + 2 u n ψ + u n ψ + G n (χu n ) C( τ(u n ) L (D 2 ) + u n L (D 2 ) + + G n (χu n ) ) C( + G n 2 (χu n ) 2 ) 2 C. (3) 6

17 Now from (29), (30) and (3) it follows that u n L 2, (D 2 ) (ψu n ) L 2, (φh n ) (χu n ) H + f n C. (32) For any ǫ > 0, take δ small such that E(u,D 4δ ) + δ 4( ) < ǫ 2 where u is the weak limit of u n in W,2 Loc (R2 ). The standard blow-u arguments show that (see [2]) when n,r are big enough there holds for any r n R < t < 4δ. By Lemma 2 we have D 2t \D t u n 2 < Cǫ 2 u n L 2, (D δ \D 4rnR ) C(ǫ + δ 2( ) ) Cǫ. Using (32) and the duality between the L 2, and L 2, we obtain that u n L 2 (D δ \D 4rn ) C( u n L 2, u n L 2, (D δ \D 4rnR )) 2 C ǫ which yields the energy identity. 6 Neckless for the shere As the energy identity is true for this case, for any ǫ > 0, there exists δ such that E(u n,d 4δ \ D rnr) < ǫ 2 when n,r are big enough. For simlicity suose that D 4rnR \D 2rnR u n (x) = 0. Take ψ C 0 (D 2) satisfying that ψ = in D. Set ψ r (x) = ψ( x r );ϕ n = ψ δ ψ 2rnR. Using the similar arguments in the last section, it can be shown that (ϕ n u i n) = ϕ n u i n + 2 u i n ϕ n + u i n ϕ n = ϕ n ( u i n u j n 2 + τ i (u n )) + 2 u i n ϕ n + u i n ϕ n m = ϕ n (u j n u i n u i n u j n) u j n + 2 u i n ϕ n + u i n ϕ n + ϕ n τ i (u n )) = m (u j n ui n ui n uj n ) (ϕ nu j n ) (u j n ui n ui n uj n )uj n ϕ n +2 u i n ϕ n + u i n ϕ n + ϕ n τ i (u n ) 7

18 = = (u j n u i n u i n u j n) (ϕ n u j n) (u j n) 2 u i n ϕ n +u i n ϕ n (u j n )2 + 2 u i n ϕ n + u i n ϕ n + ϕ n τ i (u n ) (u j n u i n u i n u j n) (ϕ n u j n) u i n ϕ n +2 u i n ϕ n + u i n ϕ n + ϕ n τ i (u n ) m = ψ 2δ (u j n ui n ui n uj n ) (ϕ nu j n ) + div(ui n ϕ n) + ϕ n τ i (u n ). (33) In the last equality we use the fact ψ 2δ = in the suort of ϕ n. Set (F n ) i j = ψ 2δ(u j n ui n ui n uj n ), some simle comutations show that div(f n ) i j = ψ 2δ(u j n ui n ui n uj n ) + ψ 2δ(u j n ui n ui n uj n ) It is easy to check that divf n C( τ(u n ) + δ 2( ) ( = ψ 2δ (u j nτ i (u n ) u i nτ j (u n )) + ψ 2δ (u j n u i n u i n u j n) C( τ(u n ) χ D4δ + u n χ δ D4δ \D 2δ ). (34) Let G n be the Newtonian otential of (divf n ), then D 4δ \D 2δ u n 2 dx) 2 ) C( + δ 2( ) ǫ). div(f n G n ) = divf n divn (div F n ) = divf n div (N div F n ) = 0. So there exists H n such that F n G n = H n = ( 2 H n, H n ). By the same argument as we derived (27), we have, for < < 2, G n 2 2 = N div F n 2 2 As a direct consequence there holds Set a n = G n L 2 (D 4δ ) Cδ 2( ) G n 2 2 D 2rnR \D rnr H n(x)dx D 2rnR \D rnr,b n = C div F n C( + δ 2( ) ǫ). (35) Cδ 2( ) D 4δ \D 2δ H n(x)dx D 4δ \D 2δ and L n = ψ 2δ ( ψ rnr)(h n a n ) ψ 2δ (b n a n ). As ψ 2δ ( ψ rnr) =, ψ 2δ = 0 in the suort of ϕ n, there holds L n (ϕ n u n ) = L n (ϕ n u n ) ( + δ 2( ) ǫ) C(δ 2( ) + ǫ). (36) = (ψ 2δ ( ψ rnr)(h n a n ) ψ 2δ (b n a n )) (ϕ n u n ) = ((H n a n ) (ψ 2δ ( ψ rnr)) + (ψ 2δ ( ψ rnr)) (H n a n ) (b n a n ) ψ 2δ ) (ϕ n u n ) = H n (ϕ n u n ) = H n (ϕ n u n ). 8

19 So from (33) we can get that (ϕ n u n ) = F n (ϕ n u n ) + div(u n ϕ n ) + ϕ n τ(u n ) = (F n G n ) (ϕ n u n ) + G n (ϕ n u n ) + div(u n ϕ n ) + ϕ n τ(u n ) = H n (ϕ n u n ) + G n (ϕ n u n ) + div(u n ϕ n ) + ϕ n τ(u n ) = L n (ϕ n u n ) + G n (ϕ n u n ) + div(u n ϕ n ) + ϕ n τ(u n ). (37) By the definition of L n,a n,b n, (36) and Poincaré inequality, there holds that L n 2 = [ψ 2δ ( ψ rnr)(h n a n ) ψ 2δ (b n a n )] 2 = ψ 2δ ( ψ rnr) H n + (H n a n )( ψ 2δ ψ rnr) (b n a n ) ψ 2δ 2 = ψ 2δ ( ψ rnr) H n + (H n b n ) ψ 2δ (H n a n ) ψ rnr 2 C( H n L 2 (D 4δ \D rnr ) + H n L 2 (D 4δ \D 2δ ) + H n L 2 (D 2rnR \D rnr )) C H n L 2 (D 4δ \D rnr ) C( F n L 2 (D 4δ \D rnr ) + G n L 2 (D 4δ )) C( u n L 2 (D 4δ \D rnr ) + ǫ + δ 2( ) ) C(ǫ + δ 2( ) ). (38) Take t small enough, we estimate the L 2, -norm of u n in D tδ \ D 8rnR. From (37) we know i (ϕ n u n ) = R i (ϕ n u n ) Here R i is the Riesz otential. = R i ( L n (ϕ n u n )) + R i (G n (ϕ n u n ) + ϕ n τ(u n )) + R i div(u n ϕ n ) = R i ( L n (ϕ n u n )) + R i (G n (ϕ n u n ) + ϕ n τ(u n )) +R i div(u n ψ δ ) R i div(u n ψ 2rnR). (39) For the first term, by (38), Lemma 5 and Lemma 7, we have Here we use the fact (ϕ n u n ) 2 C. R i ( L n (ϕ n u n )) L 2, C L n (ϕ n u n ) H C L n 2 (ϕ n u n ) 2 C(ǫ + δ 2( ) ). (40) For the second term, as Riesz otential is bounded from L (R 2 ) to L 2 2 (R 2 ), by Lemma 3 and (35) we can get that 9

20 R i (G n (ϕ n u n ) + ϕ n τ(u n )) L 2, (D δ ) Cδ 2( ) R i (G n (ϕ n u n ) + ϕ n τ(u n )) Cδ 2( ) G n (ϕ n u n ) + ϕ n τ(u n )) 2 L 2 Cδ 2( ) ( G n 2 (ϕ n u n ) 2 + τ(u n )) ) 2 Cδ 2( ) ( + δ 2( ) ǫ + ) C(ǫ + δ 2( ) ). (4) For the third term, when x < δ 2, direct comutations show that R i div(u n ψ δ )(x) x i y i = R 2 x y 2div(u n ψ δ )(y)dy div(u n ψ δ )(y) C dy D 2δ \D δ x y C δ which imlies that R i div(u n ψ δ ) L 2, (D tδ ) Ct. (42) Using the assumtion that D 4rnR \D 2rnR u n (x) = 0, by Sobolev inequality, we have div(u n ψ 2rnR)(y) dy C( ((r n R) u n + (r n R) 2 u n )dx) D 4rnR \D 2rnR C( u n 2 dx) 2 Cǫ. D 4rnR \D 2rnR As R 2 div(u n ψ 2rnR)dy = 0, when x > 8r n R, the last term can be estimated as R i div(u n ψ 2rnR)(x) x i y i = R 2 x y 2div(u n ψ 2rnR)(y)dy = ( x i y i x y 2 x i x 2)div(u n ψ 2rnR)(y)dy x i y i D 4rnR \D 2rnR x y 2 x i x 2 div(u n ψ 2rnR)(y) dy Cr nr x 2 div(u n ψ 2rnR)(y) dy D 4rnR \D 2rnR Cr nrǫ x 2. So it can be shown that R i div(u n ψ 2rnR) L 2, (R 2 \D 8rnR ) Cr n Rǫ Cr n Rǫ x >8r nr 8r nr x 2( x 2 (8r n R) 2 ) 2dx t 3 2 (t 8rn R) 2 dt Cǫ. (43) 20

21 From (40), (4), (42) and (43) we can obtain that u n L 2, (D tδ \D 8rnR ) = (ϕ n u n ) L 2, (D tδ \D 8rnR ) C(ǫ + δ 2( ) + t). (44) Take t,δ small enough, we can show that there is no neck during blowing u. References [] R. Coifman; P. L. Lions; Y. Meyer; S. Semmes, Comacité ar comensation et esaces de Hardy. (French) [Comensated comactness and Hardy saces.] C. R. Acad. Sci. Paris Sér. I Math. 309 (989), no. 8, [2] W. Ding; G. Tian, Energy identity for a class of aroximate harmonic mas from surfaces. Comm. Anal. Geom. 3 (995), no. 3-4, [3] J. Eells; L. Lemaire, A reort on harmonic mas. Bull. London Math. Soc. 0 (978), no., 68. [4] J. Eells; L. Lemaire, Another reort on harmonic mas. Bull. London Math. Soc. 20 (988), no. 5, [5] F. Hélein, Harmonic mas, conservation laws and moving frames (English summary). Cambridge Tracts in Mathematics, 50. Cambridge University Press, Cambridge, [6] J. Jost, Two-dimensional geometric variational roblems. Proceedings of the ICM, Vol., 2 (986), Amer. Math. Soc., Providence, RI, 987. [7] F. Lin; C. Wang, Energy identity of harmonic ma flows from surfaces at finite singular time. Calc. Var. PDE. 6 (998), no. 4, [8] F. Lin; C. Wang, Harmonic and quasi-harmonic sheres. II. Comm. Anal. Geom. 0 (2002), no. 2, [9] T. Parker, Bubble tree convergence for harmonic mas. J. Differential Geom. 44 (996), no. 3, [0] J. Qing, On singularities of the heat flow for harmonic mas from surfaces into sheres. Comm. Anal. Geom. 3 (995), no. -2, [] J. Qing; G. Tian, Bubbling of the heat flows for harmonic mas from surfaces. Comm. Pure Al. Math. 50 (997), no. 4, [2] T. Riviére, Conservation laws for conformally invariant variational roblems. Invent. Math. 68 (2007), no., 22. 2

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