L2-Concentration of Blow-Up Solutions for Two-Coupled Nonlinear Schrodinger¨ Equations with Harmonic Potential1

Indian J. Pure Appl. Math., 43(1): 49-70, February 2012 °c Indian National Science Academy

L2-CONCENTRATION OF BLOW-UP SOLUTIONS FOR TWO-COUPLED NONLINEAR SCHRODINGER¨ EQUATIONS WITH HARMONIC POTENTIAL1

Zhong-Xue Lu¨∗, Zuhan Liu∗,∗∗ and Changcheng Yao∗

∗School of Mathematical Science, Xuzhou Normal University, Xuzhou, 221116, Peoples’ Republic of China ∗∗School of Mathematics, Yangzhou University, Yangzhou 225002, Peoples’ Republic of China e-mail: [email protected]

(Received 28 August 2010; after ﬁnal revision 27 December 2011; accepted 28 December 2011)

In this paper, we consider the blow-up solutions of Cauchy problem for two- coupled nonlinear Schrodinger¨ equations with harmonic potential. We establish the lower bound of blow-up rate. Furthermore, the L2 concentration for radially symmetric blow-up solutions is obtained.

Key words : L2 concentration; nonlinear Schrodinger¨ equations; harmonic potential; Bose-Einstein condensates.

1This work is supported by the Natural Science Foundation of China (No. 10771181; 11071206) and NSF of Jiangsu Province (No. BK2010172). 50 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO

1. INTRODUCTIONAND MAIN RESULTS

In this paper, we consider two-coupled nonlinear Schrodinger¨ equations with harmonic potential

( 1 1 1 ω2 2 2 1 1 2 2 2 1 2 iψt = − 2 ∆ψ + 2 (x1 + x2)ψ − (v11|ψ | + v12|ψ | )ψ in R × R+, 2 1 2 ω2 2 2 2 1 2 2 2 2 2 iψt = − 2 ∆ψ + 2 (x1 + x2)ψ − (v12|ψ | + v22|ψ | )ψ in R × R+, (1.1)

1 1 2 2 ψ (x, 0) = ψ0(x), ψ (x, 0) = ψ0(x). (1.2)

1 2 2 2 1 2 Here ψ (x, t): R × R+ → C, ψ (x, t): R × R+ → C with ψ0(x), ψ0(x) 2 being the initial data. ∆ is the Laplace operator on R . vij, i, j = 1, 2 are coupling constants. The system (1.1) arises in the Hartree-Fock theory for a double condensate, i.e., a binary mixture of Bose-Einstein condensates in two different hyperﬁne states |1i and |2i ([1]). Physically, the solution ψj denotes the corre- sponding macroscopic wave function of the jth (j=1, 2) component. v11, v22 and v12 are the intraspecies and interspecies scattering lengths. The sign of the scattering length v12 determines whether the interactions of states |1i and |2i are repulsive or attractive. When v12 < 0, the interactions are repulsive ([2]); when v12 > 0, they are attractive.

On the Cauchy problem (1.1) and (1.2), researching sharp condition and limit behaviour for the blow-up solution are very important topics. Since the cubic non- linearities are physically relevant, the equation with a cubic nonlinearity occurs in various chapters of physics, including nonlinear optics, superconductivity, and plasma physics. In space dimension n = 2, the cubic nonlinearity is critical. In fact, the power 1 + 4/n, where n is the space dimension, is the one for which the virial identity gives immediately the Zakharov-Glassey condition for the blow-up (negative initial energy). For nonlinearity power p < 3 in space dimension n = 2, blow-up in ﬁnite time never occurs(see, Ginibre-Velo [3]).

For single nonlinear Schrodinger¨ equation, many authors studied sharp condition for global existence of solutions (see [4-19]). ‘Mass concentration’ of blow-up NONLINEAR SCHRODINGER¨ EQUATIONS 51 solutions in the critical power nonlinear case is quite different from the supercritical case(see [20-26]). In particular, in [24-26], authors established the relation between the mass concentration of radially symmetric blow-up solutions and the ground state equation of some elliptic equation.

For coupled nonlinear Schrodinger¨ systems, Fanelli and Montefusco [27] gave the sharp thresholds of blow-up solution for the case without a harmonic potential term. Lu¨ and Liu [28, 29] gave the sharp thresholds of blow-up solution for the case with a harmonic potential term. They prove that the L2 norm of the gradient of solution blows up in a finite time. More precisely, applying a consequence of the standard Hardy’s inequality to any solution of coupled nonlinear Schrodinger¨ system, and by the mass is conserved and the L2−norm of xψ vanish in a finite time, they obtain the L2−norm of the gradient needs necessarily to blow up in a finite time. But for ‘mass concentration’, to our knowledge, there is no related result in the literature.

In this paper, motivated by Fanelli and Montefusco [27] and Li and Zhang [26], on the basis of results on the existence of blow-up solution [27-29], we investigate the L2 concentration for radially symmetric blow-up solutions of (1.1) and (1.2). As we will see, we prove that in small neighborhood of origin, i.e., |x| < a(t), t → 1 2 T (T is the maximal existence time), concentrated mass is large than ku kL2 + 2 2 1 2 ku kL2 . Where (u , u ) is a ground state solution of some elliptic system.

On the existence and coupling properties of ground-states for these systems, we refer readers to Maia-Montefusco-Pellacci [30], Ambrosetti-Colorado [31], Lin- Wei [32, 33], Sirakov [34] and the references therein. Note that the deﬁnition of ground state given by Maia, Montefusco-Pellacci in [30] or given by Ambrosetti -Colorado in [31] is quite different from that introduced by Lin-Wei in [32, 33]. In this paper, we are concerned with the deﬁnition introduced by Sirakov in [34]. More precisely, a least energy solution of elliptic systems is called a ground state of this systems, if it is with two nonnegative components.

In what follows, we give our main result. We deﬁne a space H by

H := H1(R2) ∩ {ψ : |x|ψ ∈ L2(R2)} 52 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO with the inner product Z < ψ, φ >:= ∇ψ · ∇φ¯ + ψφ¯ + |x|2ψφ,¯ R2 for all ψ, φ ∈ H. The norm of H is denoted by k · kH . By the standard technique (see e.g. Ginibro and Velo [3-5] and Cazenave [35]), it is easy to prove that:

1 2 1 2 Assume that ψ0, ψ0 ∈ H, then there exists a solution ψ , ψ of the Cauchy problem (1.1) and (1.2) in C([0,T ],H) for some T ∈ (0, ∞], T = +∞ or 1 2 2 2 1 2 T < +∞ with kψ kH + kψ kH → ∞ as t → T . Furthermore ψ (x, t), ψ (x, t) satisﬁes Z 1 2 2 2 N (ψ , ψ ) := |ψ1(x, t)| + |ψ2(x, t)| ≡ C1, (1.3) R2

1 2 E(ψ , ψ ) ≡ C2, (1.4) where Z 1 ω2 E(ψ1, ψ2) := (|∇ψ1|2 + |∇ψ2|2) + |x|2(|ψ1|2 + |ψ1|2) 2 R2 2 1 − (v |ψ1|4 + v |ψ2|4) − v |ψ1|2|ψ2|2, (1.5) 2 11 22 12 with C1,C2 as constants.

The main result of this paper is the following theorem.

Theorem 1.1 (L2- concentration). — Let ψ1, ψ2 ∈ C([0,T ),H) be a solution of the Cauchy problem (1.1) and (1.2) such that (ψ1, ψ2) blows up at ﬁnite time t = T . If a(t) is a decreasing function from [0,T ) to R+ such that a(t) → 0(t → T ) and (T − t)1/4/a(t) → 0(t → T ), then

1 2 2 2 1/2 lim inf(kψ (t)k 2 + kψ (t)k 2 ) t→T L (|x|

We organize the paper as follows: In Section 2, we establish the lower bound of blow-up rate. In Section 3, we prove the L2 concentration for radially symmetric blow-up solutions.

2. BLOW-UP RATE

We consider the Cauchy problem of two-coupled nonlinear Schrodinger¨ equations without harmonic potential ( iφ1 = − 1 ∆φ1 − (v |φ1|2 + v |φ2|2)φ1 in R2 × R , t 2 11 12 + (2.1) 2 1 2 1 2 2 2 2 2 iφt = − 2 ∆φ − (v12|φ | + v22|φ | )φ in R × R+, 1 1 2 2 φ (x, 0) = φ0(x), φ (x, 0) = φ0(x). (2.2)

We ﬁrst recall some results on the Cauchy problem (2.1) and (2.2) (see [27]).

1 2 1 1 2 Assume that φ0, φ0 ∈ H , then there exists a solution (φ , φ ) of the Cauchy problem (2.1)-(2.2) in C([0,T ],H1) for some T ∈ (0, ∞], T = +∞ or T < +∞ 1 2 2 2 1 2 and limt→T − kφ kH1 + kφ kH1 = ∞. Furthermore (φ (x, t), φ (x, t)) satisﬁes Z ∗ 1 2 2 2 N (φ , φ ) := |φ1(x, t)| + |φ2(x, t)| ≡ C1, (2.3) R2

∗ 1 2 E (φ , φ ) ≡ C2, (2.4) where Z ∗ 1 2 1 1 2 2 2 1 1 4 2 4 1 2 2 2 E (φ , φ ) := (|∇φ | +|∇φ | )− (v11|φ | +v22|φ | )−v12|φ | |φ | , 2 R2 2 (2.5) with C1,C2 as constants.

Proposition 2.1 — Let φ1, φ2 ∈ H1, and (φ1, φ2) be a solution of the Cauchy 0 0 R 1 2 i 2 problem (2.1)-(2.2) in C([0,T ],H ). Put Ji(t) := R2 |x| |φ | , i = 1, 2, and J(t) = J1(t) + J2(t). Then one has Z J 0(t) = −4I ∇φ¯1 · (φ1x) + ∇φ¯2 · (φ2x) (2.6) R2 54 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO

J 00(t) = 16E∗(φ1, φ2). (2.7)

1 2 1 2 For some initial data (φ0, φ0), the solution (φ , φ ) of (2.1) and (2.2) blows up in ﬁnite time. We can get the following lower estimate of the blow-up order of 1 2 2 2 1/2 (k∇ψ kL2 + k∇ψ kL2 ) .

Lemma 2.2 — Let (φ1, φ2) be a solution of the Cauchy problem (2.1) and (2.2) in C([0,T ),H1) such that (φ1, φ2) blows up at ﬁnite time t = T . Then, there exists an L > 0 such that

1 2 2 2 1/2 −1/4 (k∇φ kL2 + k∇φ kL2 ) ≥ L(T − t) , t ∈ [0,T ).

PROOF : Let Ã ! Ã ! Ã ! φ f (φ) v |φ |2φ + v |φ |2φ φ = 1 f(φ) = 1 = 11 1 1 12 2 1 , and 2 2 φ2 f2(φ) v12|φ1| φ2 + v22|φ2| φ2

then we have

|f(φ) − f(ψ)| ≤ C(|φ|2 + |ψ|2)|φ − ψ|, (2.8)

2 2 kf(φ) − f(ψ)kL4/3 ≤ C(kφkL4 + kψkL4 )kφ − ψkL4 , (2.9) and 2 k∇f(φ)kL4/3 ≤ CkφkL4 kφ − ψkL4 . (2.10)

∞ 2 Let θ ∈ C0 (C , R) be such that θ(z) = 1 for |z| ≤ 1. Set

g1(ψ) = θ(ψ)f(ψ),

g2(ψ) = (1 − θ(ψ))f(ψ), one easily veriﬁes that

|g1(φ) − g1(ψ)| ≤ C|φ − ψ|, (2.11)

2 2 |g2(φ) − g2(ψ)| ≤ C(|φ| + |ψ| )|φ − ψ|. (2.12) NONLINEAR SCHRODINGER¨ EQUATIONS 55

From Holder’s¨ inequality, we deduce that

kg1(φ) − g1(ψ)kL2 ≤ Ckφ − ψkL2 , (2.13)

2 2 kg2(φ) − g2(ψ)kL4/3 ≤ C(kφkL4 + |ψ|L4 )kφ − ψkL4 (2.14) and from Remark 1.3.1(vii) in [35] that

k∇g1(φ)kL2 ≤ Ck∇φkL2 , (2.15)

2 k∇g2(φ)kL4/3 ≤ Ck∇φkL4 k∇φkL4 . (2.16)

We will prove the theorem by a ﬁxed point argument.

Fix M, T > 0, to be chosen later. Consider the set

E = {φ ∈ L∞((0,T ),H1(R2) × H1(R2)) ∩ L4((0,T ),W 1,4(R2) × W 1,4(R2));

kφkL∞((0,T ),H1(R2)×H1(R2)) ≤ M, kφkL4((0,T ),W 1,4(R2)×W 1,4(R2)) ≤ M} equipped with the distance

d(φ, ψ) = kφ − ψkL4((0,T ),L4) + kφ − ψkL∞((0,T ),L2).

We easily claim that (E, d) is a complete metric space. We wish to ﬁnd a condition on M and T which imply that F, given by

Z t F = S(t)ϕ + i S(t − s)g(φ(s))ds. 0 is a strict contraction on E. Where S(t) is the unitary group eit∆ determined by the linear Schrodinger¨ equation and ϕ ∈ H1(R2) × H1(R2).

2 2 2 2 Consider φ ∈ E. Since g1 is continuous L ×L → L ×L , it follows that g1 : 2 2 ∞ 2 2 (0,T ) → L ×L is measurable, and we deduce that g1(φ) ∈ L ((0,T ),L ×L ). 4 4/3 4/3 Similarly, we can deduce that g2(φ) ∈ L ((0,T ),L × L ). Using (2.13)- ∞ 2 2 (2.16), and Remark 1.2.2(iii) in [35], we deduce that g1(φ) ∈ L ((0,T ),L ×L ), 4 4/3 4/3 g2(φ) ∈ L ((0,T ),L × L ) and

kg1(φ)kL∞((0,T ),H1×H1) ≤ CkφkL∞((0,T ),H1×H1), 56 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO

kg2(φ)kL4((0,T ),W 1,4/3×W 1,4/3) ≤ C(kφkL∞((0,T ),L4×L4)kφkL4((0,T ),W 1,4×W 1,4),

kg1(φ) − g1(ψ)kL∞((0,T ),L2×L2) ≤ Ckφ − ψkL∞((0,T ),L2×L2),

kg2(φ) − g2(ψ)kL4((0,T ),L4/3×L4/3) 2 2 ≤ C(kφkL∞((0,T ),L4×L4) + kψkL∞((0,T ),L4×L4))(kφ − ψkL4((0,T ),L4×L4).

Using the embedding H1(R2) ,→ L4(R2) and Holder’s¨ inequality, we deduce that

kg1(φ)kL1((0,T ),H1×H1) + kg2(φ)kL4/3((0,T ),W 1,4/3×W 1,4/3) ≤ C(T + T 1/2)(1 + M 2)M, (2.17)

kg1(φ) − g1(ψ)kL1((0,T ),L2×L2) + kg2(φ) − g2(ψ)kL4/3((0,T ),L4/3×L4/3) ≤ C(T + T 1/2)(1 + M 2)d(φ, ψ). (2.18)

Then it follows from (2.17) and Strichartz’s estimates that for ϕ ∈ H1(R2) × H1(R2)

F(φ) ∈ C([0,T ],H1(R2) × H1(R2)) ∩ L4((0,T ),W 1,4(R2) × W 1,4(R2)), (2.19) and

kF(φ)kL∞((0,T ),H1(R2)×H1(R2)) + kF(φ)kL4((0,T ),W 1,4(R2)×W 1,4(R2))

1/2 2 ≤ CkϕkH1×H1 + C(T + T )(1 + M )M. (2.20)

Also we have kF(φ) − F(ψ)kL∞((0,T ),L2(R2)×L2(R2)) + kF(φ) − F(ψ)kL4((0,T ),L4(R2)×L4(R2))

≤ C(T + T 1/2)(1 + M 2)d(φ, ψ). (2.21)

Hence, for ϕ ∈ H1(R2) × H1(R2), if we set

1/2 2 CkϕkH1×H1 + C(T + T )(1 + M )M ≤ M, (2.22) NONLINEAR SCHRODINGER¨ EQUATIONS 57 and choose T small enough so that

1 C(T + T 1/2)(1 + M 2) < , 2

1 then it follows that F(φ) ∈ E and d(F(φ), F(ψ)) ≤ 2 d(φ, ψ). Namely, F is a strict contraction on E. By Banach’s fixed point theorem, F has a unique fixed point φ ∈ E. If we consider φ(t) as the initial value, where t < T ∗, it follows from (2.22) and the fixed argument that if for some M > 0,

1/2 2 Ckφ(t)kH1×H1 + C(T − t + (T − t) )(1 + M ) ≤ M, then T < T ∗. Thus for all M > 0,

∗ ∗ 1/2 2 Ckφ(t)kH1×H1 + C(T − t + (T − t) )(1 + M )M > M.

As t → T ∗, we have

∗ 1/2 2 Ckφ(t)kH1×H1 + C(T − t) )(1 + M )M > M.

Choosing for example, M = Ckφ(t)kH1×H1 , we see that

C kφ(t)k2 > . H1×H1 (T ∗ − t)1/2

Because C > 1 as t → T ∗, then 2(T ∗−t)1/2

C kφ(t)k 1 1 > . H ×H (T ∗ − t)1/4

By Young’s inequality, we complete the proof of the lemma.

Following the methods by Carles in [36], we can prove the following proposition that deals with the relation between the solution of (1.1)-(1.2) and that of (2.1)-(2.2). 58 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO

Lemma 2.3 — (1) Assume that φ1, φ2 is a solution of the Cauchy problem (2.1)-(2.2) is in C([0,T ),H). Let µ ¶ i 1 −i ω |x|2 tan ωt i x tan ωt ψ (x, t) = e 2 φ , , i = 1, 2, (2.23) cos ωt cos ωt ω

i arctan ωT then, ψ (x, t) ∈ C([0, ω ),H), i = 1, 2 is a solution of (1.1)-(1.2).

(2) Assume that ψi, i = 1, 2 is a solution of (1.1)-(1.2) is in C([0, τ),H) with π τ ∈ (0, 2ω ). Let

2 2 1 i |x| ω t ) i 2 1+(ωt)2 i φ (x, t) = 1 e ψ (1 + (ωt)2) 2 Ã ! x arctan ωt 1 , , i = 1, 2, (2.24) (1 + (ωt)2) 2 ω

i tan ωτ then φ (x, t) ∈ C([0, ω ),H), i = 1, 2 is a solution of (2.1)-(2.2).

On the basis of Lemma 2.3, we get following lemma directly.

Lemma 2.4 — (1) Assume that ψ1, ψ2 is a solution of the Cauchy problem π (1.1)-(1.2) is in C([0,T ),H), where T ∈ (0, 2ω )(T is maximal existence time). Let Ã ! 2 2 1 i |x| ω t ) x arctan ωt i 2 1+(ωt)2 i φ (x, t) = 1 e ψ 1 , , i = 1, 2, (1 + (ωt)2) 2 (1 + (ωt)2) 2 ω then

i tan ωT (1) φ (x, t) ∈ C([0, ω ),H), i = 1, 2 is a solution of (2.1)-(2.2), where tan ωT ω is the maximal existence time.

µ ¶ i 1 −i ω |x|2 tan ωt i x tan ωt (2)ψ (x, t) = e 2 φ , , t ∈ [0,T ), i = 1, 2. cos ωt cos ωt ω

With the lemmas above, we can prove the following theorem. NONLINEAR SCHRODINGER¨ EQUATIONS 59

Theorem 2.5 — Let ψ1, ψ2 be a solution of the Cauchy problem (1.1) and (1.2) is in C([0,T ),H) such that (ψ1, ψ2) blows up at ﬁnite time t = T . Then, there exists an M > 0 such that

1 2 2 2 1/2 −1/4 (k∇ψ kL2 + k∇ψ kL2 ) ≥ M(T − t) , t ∈ [0,T ). (2.25)

PROOF : By Lemma 2.4, let φi, i = 1, 2 be deﬁned by (2.24), then, φi ∈ tan ωt tan ωt C([0, ω ),H), i = 1, 2 is a blow-up solution for (2.1)-(2.2), where ω ) is the maximal existence time, and µ ¶ i 1 −i ω |x|2 tan ωt i x tan ωt ψ (x, t) = e 2 φ , , t ∈ [0,T ), i = 1, 2. cos ωt cos ωt ω

π Then, for t ∈ [0,T ),T ∈ (0, 2ω ), we have 1 tan ωt tan ωt k∇ψi(t)k2 = k − iωx sin ωtφi(·, ) + ∇φi(·, )k2 L2 cos2(ωt) ω ω L2 1 tan ωt tan ωt ≥ k∇φi(·, )k2 − ω2kxφi(·, )k2 cos2(ωt) ω L2 ω L2 tan ωt tan ωt ≥ k∇φ(·, )k2 − ω2kxφi(·, )k2 . ω L2 ω L2

Hence, tan ωt tan ωt k∇ψ1(t)k2 + k∇ψ2(t)k2 ≥ k∇φ1(·, )k2 + k∇φ2(·, )k2 L2 L2 ω L2 ω L2 tan ωt tan ωt − ω2kxφ1(·, )k2 − ω2kxφ2(·, )k2 . (2.26) ω L2 ω L2 We claim that there is a constant C > 0, such that tan ωt tan ωt ω2kxφ1(·, )k2 + ω2kxφ2(·, )k2 ≤ C, t ∈ [0,T ). (2.27) ω L2 ω L2

In fact, consider Z tan ωT J(t) = |x|2(|φ1(x, t)|2 + |φ2(x, t)|2), t ∈ [0, ). R2 ω 60 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO

By Proposition 2.2, we have that J 00(t) = 16E∗(φ1, φ2); E∗(φ1, φ2) is a constant. Consider the analytical identity,

Z t? J(t?) = J(0) + j0(0)t? + J 00(t? − s)ds = J(0) + j0(0)t? + 8E∗(φ1, φ2)(t?)2. 0

? tan ωt π Let t = ω , t ∈ [0,T ), 0 < T < 2ω , we have

tan ωt tan ωt kxφ1(·, )k2 + kxφ2(·, )k2 = J(0) + j0(0)t? + 8E(φ1, φ2)(t?)2. ω L2 ω L2

This equation implies (2.27).

By Lemma 2.2 , there is a constant C > 0, such that

tan ωt tan ωt C k∇φ1(·, )k2 + k∇φ2(·, )k2 ≥ , t ∈ [0,T ) ω L2 ω L2 (tan ωT − tan ωt)2 (2.28).

From (2.26)-(2.28),

tan ωt tan ωt C k∇φ1(·, )k2 +k∇φ2(·, )k2 +C ≥ , t ∈ [0,T ). ω L2 ω L2 (tan ωT − tan ωt)2

π Because cos ωt ≥ cos ωT , for t ∈ [0,T ), T ∈ (0, 2ω ), we have

1 tan ωt 2 2 tan ωt 2 C k∇φ (·, )k 2 + k∇φ (·, )k 2 ≥ − C, t ∈ [0,T ). ω L ω L sin2(ω(T − t))

2 cos (ω(T −t)) ∼ 1 t → T Because sin2(ω(T −t)) (ω(T −t))2 , as , we have

tan ωt tan ωt C k∇φ1(·, )k2 + k∇φ2(·, )k2 ≥ , t ∈ [0,T ), as t → T. ω L2 ω L2 (T − t)2

This implies (2.25). NONLINEAR SCHRODINGER¨ EQUATIONS 61

3. CONCENTRATION BEHAVIOUR

In this section, we shall prove Theorem 1.1.

First, we recall the existence theorem of a ground state solution of the following system by Sirakov in [34],

( 1 1 1 2 2 2 1 2 ∆u − u + (v11|u | + v12|u | )u = 0, in R , 2 2 1 2 2 2 2 2 (3.1) ∆u − u + (v12|u | + v22|u | )u = 0, in R

Lemma 3.1 ([34]) — Let 0 < v12 < min{v11, v22} > 0, then systems (3.1) has 1 2 a ground state solution u0, u0.

Moreover, by [35], we have following sharp vector-valued Gagliardo-Nirenberg inequality.

2 i Lemma 3.2 — Let vij > 0, i, j = 1, 2, and v11v22 > v12, then, for any ψ ∈ H1(R2), i = 1, 2,

Z ¡ 1 4 2 4 1 2 2 2¢ v11|ψ | + v22|ψ | + 2v12|ψ | |ψ | R2 R R 2 (|∇ψ1|2 + |∇ψ2|2) (|ψ1|2 + |ψ2|2) R2 R R2 ≤ 1 2 2 2 . (3.2) R2 (|u0| + |u0| )

1 2 Here (u0, u0) is a ground state solution of the system (3.1).

We need an auxiliary function ρ(x): ρ(x) = ρ(|x|) being a radially symmetric 1 2 nonnegative function in C0 (R ), such that ( 1, r = |x| < 1/2, ρ(x) = (3.3) 0, r = |x| > 1, and 0 ≥ ρ0(r) ≥ −8. 62 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO

Lemma 3.3 ([25]) — Let v(x) be radially symmetric function in H1. Then, for any R > 0

2 −1 kv(x)kL∞(|x|>R) ≤ C0R k∇vkL2(|x|>R)kvkL2(|x|>R), (3.4)

where C0 does not depend on R and v(x).

Lemma 3.4 — Let all the assumptions in Theorem 1.1 be satisﬁed, and we put 1 2 2 2 1/2 β(t) = (k∇ψ kL2 + k∇ψ kL2 ) .

Then there exist two positive constants M1 and M2 such that

(kψ1(t)k2 + kψ2(t)k2 )1/2 lim sup L2 L2 ≤ M . (3.5) 1 2 2 2 1/2 2 t→T (kψ (t)k 2 + kψ (t)k 2 ) L (|x|

PROOF : Let M1 be a large positive constant to be determined later.

From (1.5), we have

Z Z 2 1 1 4 2 4 1 2 2 2 1 2 β (t) ≤ (v11|ψ | + v22|ψ | ) + v12 |ψ | |ψ | + 2E(ψ , ψ ) 2 2 2 R Z R v + v ≤ 11 22 (|ψ1|4 + |ψ2|4) + 2E(ψ1, ψ2) 2 2 Z R β(t) β(t) ≤ C (|ρ( x)ψ1|4 + |(1 − ρ( x))ψ1|4) 2 M M Z R 1 1 β(t) β(t) +C (|ρ( x)ψ2|4 + |(1 − ρ( x))ψ2|4) + 2E(ψ1, ψ2) R2 M1 M1 β(t) 1 4 β(t) 1 4 = Ckρ( x)ψ kL4 + k(1 − ρ( x))ψ kL4 M1 M1 β(t) 2 4 β(t) 2 4 +Ckρ( x)ψ kL4 + Ck(1 − ρ( x))ψ kL4 M1 M1 +2E(ψ1, ψ2). (3.6) NONLINEAR SCHRODINGER¨ EQUATIONS 63

The Gagliardo-Nirenberg inequality implies that

β(t) 1 4 β(t) 1 2 β(t) 1 2 kρ( x)ψ kL4 ≤ Ckρ( x)ψ kL2 k∇{ρ( x))ψ }kL2 M1 M1 M1 1 2 β(t) 1 β(t) 1 2 ≤ Ckψ kL2 k∇ρ( x)ψ + ρ( x)∇ψ kL2 M1 M1 1 4 Ckψ kL2 2 1 2 1 2 ≤ β (t) + Ckψ k 2 k∇ψ k . (3.7) 2 L L2(|x|< M1 ) M1 β(t)

Similarly, we have

β(t) 2 4 β(t) 2 2 β(t) 2 2 kρ( x)ψ kL4 ≤ Ckρ( x)ψ kL2 k∇{ρ( x))ψ }kL2 M1 M1 M1 2 2 β(t) 2 β(t) 2 2 ≤ Ckψ kL2 k∇ρ( x)ψ + ρ( x)∇ψ kL2 M1 M1 2 4 Ckψ kL2 2 2 2 2 2 ≤ β (t) + Ckψ k 2 k∇ψ k . (3.8) 2 L L2(|x|< M1 ) M1 β(t)

By Lemma 3.3,

β(t) 1 4 1 4 k{1 − ρ( x)}ψ k 4 ≤ kψ k L L4(| β(t) x|> 1 ) M1 M1 2 1 2 1 2 ≤ Ckψ k ∞ M1 kψ k 2 M1 L (|x|> 2β(t) ) L (|x|> 2β(t) ) Cβ(t) ≤ k∇ψ1k2 kψ1k3 L2(|x|> M1 ) L2(|x|> M1 ) M1 2β(t) 2β(t) Ckψ1k3 ≤ L2 β2(t). (3.9) M1

Similarly, we get

2 3 β(t) 2 4 Ckψ kL2 2 k{1 − ρ( x)}ψ kL4 ≤ β (t). (3.10) M1 M1 64 ZHONG-XUE LU,¨ ZUHAN LIU AND CHANGCHENG YAO

From (3.6)-(3.10), we have

2 1 2 1 2 β (t) ≤ C1kψ kL2 k∇ψ k 2 M1 L (|x|< β(t) ) 2 2 2 2 +C1kψ kL2 k∇ψ k 2 M1 L (|x|< β(t) ) ½ 1 4 2 4 1 3 2 3 ¾ C2kψ kL2 + kψ kL2 C3kψ kL2 + kψ kL2 2 1 2 + 2 + β (t) + 2E(ψ , ψ ) M1 M1 1 2 2 2 1 2 2 2 ≤ C1(kψ kL2 + kψ kL2 )(k∇ψ k 2 M1 + k∇ψ k 2 M1 ) L (|x|< β(t) ) L (|x|< β(t) ) ½ 1 4 2 4 1 3 2 3 ¾ C2kψ kL2 + kψ kL2 C3kψ kL2 + kψ kL2 2 + 2 + β (t) M1 M1 +2E(ψ1, ψ2) (3.11)

If we choose M1 so large that

1 4 2 4 1 3 2 3 C2kψ kL2 + kψ kL2 C3kψ kL2 + kψ kL2 1 2 + ≤ M1 M1 2 then by (3.11), we have

2 1 2 2 2 1 2 β (t) ≤ 2C1(kψ kL2 + kψ kL2 )(k∇ψ k 2 M1 L (|x|< β(t) ) 2 2 1 2 +k∇ψ k 2 M1 ) + 2E(ψ , ψ ). (3.12) L (|x|< β(t) )

1 2 Because k∇ψ kL2 + k∇ψ kL2 → ∞ as t → T , (3.12) and (1.4) implies that

1 2 2 2 k∇ψ k 2 M1 + k∇ψ k 2 M1 → ∞(t → T ). L (|x|< β(t) ) L (|x|< β(t) )

This fact and (3.12) show (3.5).

We next show the following proposition concerning the relation between the 1 2 2 2 1/2 2 blow-up order of (k∇ψ kL2 + k∇ψ kL2 ) and the rate of L -concentration.

Proposition 3.5 — Let all the assumptions in Theorem 1.1 be satisﬁed, and put

1 2 2 2 1/2 β(t) = (k∇ψ kL2 + k∇ψ kL2 ) . NONLINEAR SCHRODINGER¨ EQUATIONS 65

If a(t) is a decreasing function from [0,T ) to R+ such that a(t) → 0(t → T ) 1 and β(t)a(t) → 0(t → T ), then

1 2 2 2 1/2 lim inf(kψ (t)k 2 + kψ (t)k 2 ) t→T L (|x|

PROOF : Let ρ(t) be deﬁned as in (3.3). We put

1 2 2 2 1/2 ρa(t) = ρ(x/a(t)), βa(t) = (k∇(ρaψ )kL2 + k∇(ρaψ )kL2 ) .

(i) By (1.4) and Lemma 3.3, we have

2 1 4 2 4 β (t) − (v11kψ (t)kL4(|x|a(t)/2) + v22kψ (t)kL4(|x|>a(t)/2)) 2 Z 1 2 2 2 1 2 +v12 |ψ | |ψ | + 2E(ψ , ψ ) |x|>a(t)/2 1 ≤ (v kψ1(t)k4 + v kψ2(t)k4 ) + 2E(ψ1, ψ2) 2 11 L4(|x|>a(t)/2) 22 L4(|x|>a(t)/2) v11 1 2 1 2 ≤ kψ (t)k ∞ kψ (t)k 2 L (|x|>a(t)/2) L2(|x|>a(t)/2) v22 2 2 2 2 1 2 + kψ (t)k ∞ )kψ (t)k ) + 2E(ψ , ψ ) 2 L (|x|>a(t)/2) L2(|x|>a(t)/2) v + v ≤ 11 22 (kψ1(t)k3 + kψ2(t)k3 )β(t) a(t) L2(|x|>a(t)/2) L2(|x|>a(t)/2) +2E(ψ1, ψ2). (3.14)

A simple calculation gives us

v11 + v22 1 4 v11 + v22 1 4 − kψ (t)k 4 ≤ − kψ (t)k 4 ), (3.15) 2 L 2 L (|x|

v11 + v22 2 4 v11 + v22 2 4 − kψ (t)k 4 ≤ − kψ (t)k 4 ), (3.16) 2 L 2 L (|x|

2 1 1 2 βa(t) ≤ (kρa∇ψ (t)kL2 + k∇ρaψ (t)kL2 ) 1 1 2 +(kρa∇ψ (t)kL2 + k∇ρaψ (t)kL2 ) 1 2 1 Ckψ (t)kL2 2 2 Ckψ (t)kL2 2 ≤ (k∇ψ (t)k 2 + ) + (k∇ψ (t)k 2 + ) L a(t) L a(t) 1 2 C(kψ k 2 + kψ k 2 )β(t) ≤ β2(t) + L L a(t) C(kψ1(t)k2 + kψ2(t)k2 ) + L2 L2 . (3.17) a2(t)

On the other hand, by Lemma 3.1 and 3.2 (the variational characterization of the ground state solution u1, u2 of (3.1) yields,) we have

1 4 2 4 1 2 2 v11kρaψ kL4 + v22kρaψ kL4 ) + 2v12kρaψ ρaψ kL2 1 2 2 2 2 2(kρaψ kL2 + kρaψ kL2 )βa(t) ≤ 1 2 2 2 . (3.18) ku kL2 + ku kL2

By (3.14)-(3.18), we obtain

1 2 2 2 kρaψ kL2 + kρaψ kL2 1 − 1 2 2 2 ku kL2 + ku kL2 1 4 2 4 1 2 2 v11kρaψ kL4 + v22kρaψ kL4 ) + 2v12kρaψ ρaψ kL2 ≤ 1 − 2 2βa(t) 1 2 1 1 3 2 3 E(ψ , ψ ) ≤ 2 (kψ kL2 + kψ kL2 )β(t) + 2 a(t)βa βa 1 2 C(kψ kL2 + kψ kL2 )β(t) + 2 a(t)βa 1 2 2 2 C(kψ kL2 + kψ kL2 )β(t) + 2 . (3.19) (a(t)βa)

Letting t → T in (3.19), by (1.4), Lemma 3.4 and the assumption of a(t), we obtain µ 1 2 2 2 ¶ kρaψ kL2 + kρaψ kL2 lim sup 1 − 1 2 2 2 ≤ 0, t→T ku kL2 + ku kL2 NONLINEAR SCHRODINGER¨ EQUATIONS 67 which proves Proposition 3.5.

Now, we are in a position to prove Theorem 1.1.

PROOFOF THEOREM 1.1 : By Theorem 2.5, we have

β(t) ≥ L(T − t)−1/4, t ∈ [0,T ) for some L > 0. Therefore, by Proposition 3.5 we obtain Theorem 1.1.

ACKNOWLEDGEMENT

The authors would like to thank the reviewers for their helps and patience in im- proving this paper.

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