Nonlinear Schr¨odinger Equations
Tai-Chia Lin
∗†and Juncheng Wei
‡2006.7.25
Abstract
Conventionally, to learn wave collapse and optical turbulence, one must study finite-time blow-up solu-tions of one-component self-focusing nonlinear Schr¨odinger equasolu-tions (NLSE). Here we consider simultaneous blow-up solutions of two-component system of self-focusing NLSE. By studying the associated self-similar solutions, we prove two components of solutions blow up at the same time. These self-similar solutions may come from solitary wave solutions with multi-bumps forming abundant geometric patterns which cannot be found in one-component self-focusing NLSE. Our results may provide the first step to investigate optical turbulence in two-component system of NLSE.
Keywords: two-component system, solitary wave, wave collapse, nonlinear optics
1
Introduction
Here we study solutions of two-component system of nonlinear Schr¨odinger equations given by i∂tΦ + 4Φ + µ1|Φ|2Φ + β|ψ|2Φ = 0, i∂tψ + 4ψ + µ2|ψ|2ψ + β|Φ|2ψ = 0, x ∈ Rn, t > 0, Φ = Φ(x, t) , ψ = ψ(x, t) ∈ C , Φ(x, t) , ψ(x, t) → 0 as |x| → +∞, t > 0, (1.1)
with initial data
Φ|t=0= Φ0∈ H0s(Rn; C), ψ|t=0= ψ0∈ H0s(Rn; C), s > 2 , (1.2) where µj > 0’s are positive constants, n ≤ 3, and β ∈ R is a coupling constant. The system (1.1) has applications in many physical problems, especially in nonlinear optics. Physically, the solution (Φ, ψ) denotes the two-component beam in Kerr-like photorefractive media(cf. [1]). The positive constant µjis for self-focusing in the j-th component of the beam. The coupling constant β is the interaction between two components of the beam. As β > 0, the interaction is attractive, but the interaction is repulsive if β < 0. When the spatial dimension is one i.e. n = 1, the system (1.1) is integrable, and there are many analytical and numerical results on solitary wave solutions of the general N coupled nonlinear Schr¨odinger equations(cf. [2], [6], [7], [8]). However, when the spatial dimension is two and three i.e. n = 2, 3, there are only few results on solitary wave solutions of general N coupled nonlinear Schr¨odinger equations. One may refer to [11] for high dimensional solitary wave solutions of three coupled nonlinear Schr¨odinger equations.
From physical experiment(cf. [12]), two dimensional photorefractive screening solitons and a two dimensional self-trapped beam were observed. It is natural to believe that there are two dimensional multi-component solitons and self-trapped beams. To obtain solitary wave solutions of the system (1.1), we may set Φ(x, t) =
ei λ1tu(x) and ψ(x, t) = ei λ2tv(x). Then we may transform the system (1.1) to steady-state two coupled
nonlinear Schr¨odinger equations given by 4u − λ1u + µ1u3+ βuv2= 0 in Rn, 4v − λ2v + µ2v3+ βu2v = 0 in Rn, u, v > 0 in Rn, u, v(x) → 0, as |x| → +∞ , (1.3) where λj, µj > 0 are positive constants, n ≤ 3, and β is a coupling constant. From [10], the existence of ground state (i.e. least energy) solutions of the system (1.3) may depend on the coupling constant β. When β is positive
∗Department of Mathematics, National Taiwan University, Taipei, Taiwan 106. email : [email protected] †National Center of Theoretical Sciences, National Tsing Hua University, Hsinchu, Taiwan
‡Department of Mathematics, The Chinese University of Hong Kong, Shatin, Hong Kong. email: [email protected]
but sufficiently small, the system (1.3) has a ground state solution (u, v) which is radially symmetric. On the other hand, as β becomes negative, there is no ground state solution of the system (1.3). Here we show the existence and the configuration of bound state solutions if β is negative and |β| is small enough. Moreover, bound state solutions of the problem (1.3) have multi-bumps forming abundant geometric patterns, provided the ratio λ1/λ2is sufficiently small. One may refer to Theorem 1.2 and 1.3 for the details.
Conventionally, solutions of one-component self-focusing nonlinear Schr¨odinger equations may blow up at finite time (cf. [17]). Such a blow-up behavior may result in wave collapse and optical turbulence (cf. [4], [5] and [15]). Due to the positive sign of µj’s, the system (1.1) is of two-component self-focusing nonlinear Schr¨odinger equations having an increasing tendency for the solution to be trapped in regions of highest intensity. Consequently, it is natural to believe that the system (1.1) may have blow-up solutions which may produce wave collapse and optical turbulence. Here we prove simultaneous blow-up on two components of the system (1.1) by studying the associated self-similar solutions. These self-similar solutions have multi-bumps forming abundant geometric patterns which cannot be found in one-component nonlinear Schr¨odinger equations. Therefore the wave collapse of the system (1.1) is more complex than that of one-component self-focusing nonlinear Schr¨odinger equations. This may provide the first step to investigate optical turbulence in two-component system of nonlinear Schr¨odinger equations.
Now we state a theorem which may support the simultaneous blow-up behavior as follows:
Theorem 1.1. Let β > −√µ1µ2 if n = 3 ,and β is arbitrary if n = 2. Assume the initial condition (Φ0, ψ0) ∈
H1(Rn; C) satisfying Z
Rn
|x|2(|Φ
0|2+ |ψ0|2) dx < ∞ and one of the conditions as follows :
(i) H(Φ0, ψ0) < 0 , (ii) H(Φ0, ψ0) = 0,and Z Rn n X j=1 xj[(iΦ0· ∂xjΦ0) + (iψ0· ∂xjψ0)]dx < 0 , (iii) H(Φ0, ψ0) > 0,and Z Rn n X j=1 xj[(iΦ0·∂xjΦ0)+(iψ0·∂xjψ0)]dx ≤ − p H(Φ0, ψ0)( Z Rn |x|2(|Φ0|2+|ψ0|2)dx) 1 2,
where H is the Hamiltonian of (1.1) defined by H(Φ, ψ) = Z Rn (|∇Φ|2+ |∇ψ|2)dx −1 2 Z Rn (µ1|Φ|4+ µ2|ψ|4)dx −β Z Rn |Φ|2|ψ|2dx . (1.4)
Besides, the dot ”·” denotes complex inner product defined by (a · b) = 12(¯ab + a¯b) ∈ R for a, b ∈ C,where ¯a is
the complex conjugate of a. Then there exists a time t∗< ∞ such that lim
t↑t∗k∇ΦkL2(Rn)= limt↑t∗k∇ψkL2(Rn)= ∞ . (1.5)
Furthermore, either lim
t↑t∗kΦkL∞(Rn) = ∞ or limt↑t∗kψkL∞(Rn) = ∞ , provided kΦ0kL2(Rn) and kψ0kL2(Rn) are
strictly positive.
Here we have assumed that neither Φ nor ψ may blow up earlier than the other. From Theorem 1.1, the system (1.1) may have a solution (Φ, ψ) such that both k∇ΦkL2(Rn)and k∇ψkL2(Rn)blow up at the same time but we don’t know whether kΦkL∞(Rn) and kψkL∞(Rn)blow up simultaneously.
To get synchronous blow-up for both kΦkL∞(Rn)and kψkL∞(Rn), we study self-similar solutions of the system (1.1) in the critical case n = 2. We may generalize the idea of Rozanova (cf. [16]) to the system (1.1) by setting Φ(x, t) = A1(x, t) eiφ1(x,t), ψ(x, t) = A2(x, t) eiφ2(x,t), (1.6) where A1(x, t) = u(ξ) exp µ − Z t 0 a(τ )dτ ¶ , A2(x, t) = v(ξ) exp µ − Z t 0 a(τ )dτ ¶ , (1.7) and φj(x, t) = a(t)|x| 2 4 + γj(t) , γ 0 j(t) = λjexp µ −2 Z t 0 a(τ )dτ ¶ , j = 1, 2 . (1.8)
Here u and v are real-valued functions, λj’s are positive constants, ξ = (ξ1, · · · , ξn) ∈ Rn is defined by ξ = x exp µ − Z t 0 a(τ )dτ ¶ , x = (x1, · · · , xn) ∈ Rn, (1.9)
and a(·) is defined by an ordinary differential equation given by
a0(t) + a2(t) = 0 , ∀ t > 0 , (1.10)
with initial data
a(0) = a0< 0 . (1.11)
By (1.6)-(1.10), we may transform the system (1.1) into ½
4u − λ1u + µ1u3+ βuv2= 0 in Rn,
4v − λ2v + µ2v3+ βu2v = 0 in Rn, (1.12)
where 4 is the Laplacian corresponding to ξ-coordinates denoted as 4 = n X j=1 ∂2 ξj. Moreover, (1.10) and (1.11) imply a(t) = a0 a0t + 1 → −∞ as t ↑ t∗= −1/a0. (1.13)
Hence by (1.6), (1.7) and (1.13), we obtain a simultaneous blow-up solution with the blow-up time t∗= −1/a0. The configuration of such a solution is governed by the system (1.12) which is equivalent to the system (1.3).
To solve the system (1.3) and (1.12), we study the following problem: 4u − λ1u + µ1u3+ βuv2= 0 in Rn, 4v − λ2v + µ2v3+ βu2v = 0 in Rn, u, v > 0 in Rn, u, v ∈ H1(Rn), (1.14) where λ1, λ2, µ1, µ2> 0 and β < 0. For self-similar solutions of the system (1.1), we are particularly interested in the case of n = 2. The energy functional of the problem (1.14) is defined by
E[u, v] = 1 2 Z Rn |∇u|2+λ1 2 Z Rn u2−µ1 4 Z Rn u4+1 2 Z Rn |∇v|2+λ2 2 Z Rn v2−µ2 4 Z Rn v4−β 2 Z Rn u2v2. (1.15)
To find a least energy (ground state) solution of the problem (1.14), we consider the following minimization problem:
C = inf
(u,v)∈N
u,v≥0,uv6=0
E[u, v] , (1.16)
where N is the associated Nehari manifold given by
N = ( (u, v) ∈ H1(Rn) × H1(Rn) ¯ ¯ ¯ Z Rn |∇u|2+ λ1 Z Rn u2= µ1 Z Rn u4+ β Z Rn u2v2, Z Rn |∇v|2+ λ2 Z Rn v2= µ2 Z Rn v4+ β Z Rn u2v2 ) . (1.17) In [10], we proved
Theorem A. There exists β0 ∈ (0,√µ1µ2) such that for β < 0, the minimum C of (1.16) is not attained.
However, for β ∈ (0, β0), the minimum C of (1.16) is attained.
A natural question is : are there another bound state solutions of the problem (1.14) for β < 0? In this paper, we shall show amazing rich structures of bound state solutions for β < 0. Without loss of generality, we assume that λ16= λ2. Note that when λ1= λ2, β0∈ (−√µ1µ2, 0), a radially symmetric bound state of the type ¡c
1
c2
¢
wλ1,1 exists. Hereafter, wλ,µ denotes the unique solution of
4w − λw + µw3= 0, w(0) = max y∈Rnw(y) , w > 0, w ∈ H1(Rn). (1.18)
It is obvious that wλ,µ(y) = q
λ µw1,1(
√
λy). Hence from now on, we may assume that
λ1< λ2. (1.19)
Now we state our main result as follows:
Theorem 1.2. Let the spatial dimension n = 2. Assume that there exists a positive integer k ≥ 2 such that √ λ1 √ λ2 < sinπ k. (1.20)
Then there exists β0< 0 such that for β0< β < 0, problem (1.14) has a bound state solution (uβ, vβ) satisfying
(1) (
uβ ¡
ye2πki¢= uβ(y), uβ(y) = uβ(y),
vβ ¡
ye2πki ¢
= vβ(y), vβ(y) = vβ(y),
(1.21)
where i =√−1, y = (y1, y2) and y = (y1, −y2).
(2) As β → 0−,
uβ(y) = wλ1,µ1(y) + O(|β|
τ), vβ(y) = k−1X j=0 wλ2,µ2(y − ξ β j) + O(|β|τ),
whereξβ0, ..., ξβk−1® forms a regular k−polygon and |ξβj| ∼ log 1 |β|· 1 2¡√λ2sin(π/k) − √ λ1 ¢ (1.22)
and τ is a positive number.
For any positive integer k ≥ 2, one may find a bound state solution (uβ, vβ) such that vβ has k bumps forming a regular k−polygon around the single bump of uβ by reducing the ratio
√ λ1/
√
λ2 and |β|. This may provide abundant geometric patterns for multi-bumps of solitary and self-similar solutions of the system (1.1). In particular, as k = 6, the geometric pattern of multi-bumps can be sketched below:
±° ²¯ v ±° ²¯ v ¢¢ ¢¢ A A A A ¢¢ ¢¢ A A A A ±° ²¯ v ±° ²¯ u v ±° ²¯ v A A A A ¢¢ ¢¢ A A A A ¢¢ ¢¢ ±° ²¯ v v ±° ²¯ v Remark.
(1) Theorem 1.2 can be regarded as results for ”bifurcation from β = 0”. If λ1= λ2, computations show that
bifurcation point for β may be a finite number.
(2) More complicated patterns, such as concentric polygons, honeycombs, etc can also be constructed by similar arguments with specific symmetry properties.
Now we may use Theorem 1.2 to observe simultaneous blow-up solutions of the system (1.1). As the spatial dimension n = 2, we may set xβj(t) = ξjβ exp
µZ t 0
a(τ ) dτ
¶
, for j = 0, 1, · · · , k − 1, where ξjβ’s are obtained in Theorem 1.2 and a(·) is defined in (1.13). Then by (1.6)-(1.9), we have kψkL∞(Rn)∼ |ψ(xβj(t), t)| =
v(ξβj)/(a0t + 1) → ∞ and kΦkL∞(Rn) ∼ |Φ(0, t)| = u(0)/(a0t + 1) → ∞ as t ↑ t∗ = −1/a0. This may provide simultaneous blow-up solutions of the system (1.1). Here we have used the fact from Theorem 1.2 that u(0) and v(ξβj)’s are strictly positive numbers, provided β < 0 and |β| is small enough.
Theorem 1.2 can also be extended to n = 3. When n = 3, the geometric patterns are very important. We only consider two geometric structures: cube and tetrahedra
¿¿ ¿¿ ¿¿ ¿¿ ¿¿ ¿¿ x !!!! !! c c c c c ¶¶ ¶¶ ¶¶ e e e e e e e x
Theorem 1.3. Let n = 3, and r λ1
λ2
<
( √
3
3 for the cube,
√
3
2 for the tetrehedra. Then for β < 0 and |β| small, problem (1.14) has a solution (uβ, vβ) such that
uβ≈ wλ1,µ1(y) + O(|β| τ), vβ≈ k X j=1 wλ2,µ2(y − ξj) + O(|β| τ), (1.23) whereξ1, ..., ξk ®
forms regular cube or tetrahedra.
Remark. It is natural to believe that solutions with multi-bumps forming geometric patterns like octahedron, dodecahedron, and icosahedron (i.e. the other three regular polyhedra) can also be constructed by similar methods.
The rest of this paper is organized as follows: In Section 2, we provide the proof of Theorem 1.1. Theorem 1.2 is proved in Sections 3-5, and Theorem 1.3 is proved in Section 6.
Acknowledgements: The research of the first author is partially supported by a research Grant (No. NSC 94-2115-M-002-019) from NSC of Taiwan. The research of the second author is partially supported by an Earmarked Grant from RGC of HK.
2
Proof of Theorem 1.1
Here we may generalize ideas for single scalar nonlinear Schr¨odinger equations (cf. [17]) to the system (1.1). To prove Theorem 1.1, we need the following lemma:
Lemma 2.1. Let V (t) = RRn|x|2(|Φ|2+ |ψ|2)dx, ∀t ≥ 0, where(Φ, ψ) is the regular solution of (1.1). Then
∀t > 0, d2 dt2V (t) = 8H + 8β(1 − n 2) Z Rn|Φ| 2|ψ|2dx − 2(n − 2) Z Rn(µ1|Φ| 4+ µ 2|ψ|4)dx, (2.1)
where H is the Hamiltonian of (1.1) defined by (1.4). Proof. By direct calculation and (1.1),it is easy to check that
−d dtV (t) = −4 Z Rn n X j=1 xj[(iΦ · ∂xjΦ) + (iψ · ∂xjψ)]dx . (2.2) Here we have used integration by parts. Besides,the dot ”·” denotes complex inner product defined by (a · b) = 1
2(¯ab + a¯b) for a, b ∈ C,where ¯a is the complex conjugate of a. Moreover,by (2.2), −d2 dt2V (t) = −4 Z Rn d X j=1 xj[(i∂tΦ · ∂xjΦ) + (iΦ · ∂xj∂tΦ)]dx −4 Z Rn n X j=1 xj[(i∂tψ · ∂xjψ) + (iψ · ∂xj∂tψ)]dx . (2.3)
Hence by (1.1), (2.3) and integration by parts, −d 2 dt2V (t) = −4 Z Rn n X j=1 xj £ Φ · ∂xj(4Φ + µ1|Φ|2Φ + β|ψ|2Φ) ¤ dx +4 Z Rn n X j=1 xj £ (4Φ · ∂xjΦ) + µ1(Φ · ∂xjΦ)|Φ|2+ β(Φ · ∂xjΦ)|ψ|2 ¤ dx +4 Z Rn n X j=1 xj £ (4ψ · ∂xjψ) + µ2(ψ · ∂xjψ)|ψ|2+ β(ψ · ∂xjψ)|Φ|2 ¤ dx −4 Z Rn n X j=1 xj[ψ · ∂xj(4ψ + µ2|ψ|2ψ + β|Φ|2ψ)]dx = 4 Z Rn n X j=1 xj £ (4Φ · ∂xjΦ) + µ1(Φ · ∂xjΦ)|Φ|2+ β(Φ · ∂xjΦ)|ψ|2 ¤ dx +4 Z Rn n X j=1 ∂xj(xjΦ) · (4Φ + µ1|Φ|2Φ + β|ψ|2Φ)dx +4 Z Rn n X j=1 xj £ (4ψ · ∂xjψ) + µ2(ψ · ∂xjψ)|ψ| 2+ β(ψ · ∂ xjψ)|Φ| 2¤dx +4 Z Rn n X j=1 ∂xj(xjψ) · (4ψ + µ2|ψ|2ψ + β|Φ|2ψ)dx = 8 Z Rn n X j=1 xj £ (4Φ · ∂xjΦ) + µ1(Φ · ∂xjΦ)|Φ|2+ β(Φ · ∂xjΦ)|ψ|2 ¤ dx +4n Z Rn Φ · (4Φ + µ1|Φ|2Φ + β|ψ|2Φ)dx +8 Z Rn n X j=1 xj £ (4ψ · ∂xjψ) + µ2(ψ · ∂xjψ)|ψ|2+ β(ψ · ∂xjψ)|Φ|2 ¤ dx +4n Z Rn ψ · (4ψ + µ2|ψ|2ψ + β|Φ|2ψ)dx i.e. −d2 dt2V (t) = 8 Z Rn n X j=1 xj[(4Φ · ∂xjΦ) + µ1(Φ · ∂xjΦ)|Φ|2 +β(Φ · ∂xjΦ)|ψ|2]dx +4n Z Rn Φ · (4Φ + µ1|Φ|2Φ + β|ψ|2Φ)dx +8 Z Rn n X j=1 xj[(4ψ · ∂xjψ) + µ2(ψ · ∂xjψ)|ψ|2 +β(ψ · ∂xjψ)|Φ|2]dx +4n Z Rn ψ · (4ψ + µ2|ψ|2ψ + β|Φ|2ψ)dx (2.4)
We may rewrite the first integral of (2.4) as 8 Z Rn n X j=1 xj(4Φ · ∂xjΦ)dx + 8 Z Rn n X j=1 xj(µ1 4 ∂xj|Φ| 4+β 2|ψ| 2∂ xj|Φ|2)dx (2.5)
Similarly, the third integral of (2.4) can be written as 8 Z Rn n X j=1 xj(4ψ · ∂xjψ)dx + 8 Z Rn n X j=1 xj(µ2 4 ∂xj|ψ| 4+β 2|Φ| 2∂ xj|ψ|2)dx (2.6)
Combining the second integral of (2.5) and (2.6),we obtain 8 Z Rn n X j=1 xj( µ1 4 ∂xj|Φ| 4+µ2 4 ∂xj|ψ| 4)dx + 4β Z Rn n X j=1 xj∂xj(|Φ|2|ψ|2)dx (2.7) Using integration by parts, (2.7) becomes
−2µ1n Z Rn |Φ|4dx − 2µ2n Z Rn |ψ|4dx − 4βn Z Rn |Φ|2|ψ|2dx . (2.8)
For the first integral of (2.5) and (2.6), we use integration by parts as follow: Z Rn n X j=1 xj(4Φ · ∂xjΦ)dx = Z Rn n X j=1 xj( n X k=1 ∂2 xkΦ · ∂xjΦ)dx = − Z Rn n X j=1 n X k=1 ∂xk(xj∂xjΦ) · ∂xkΦdx i.e. Z Rn n X j=1 xj(4Φ · ∂xjΦ)dx = − Z Rn n X j,k=1 ∂xk(xj∂xjΦ) · ∂xkΦdx . (2.9) Similarly, Z Rn n X j=1 xj(4ψ · ∂xjψ)dx = − Z Rn n X j,k=1 ∂xk(xj∂xjψ) · ∂xkψdx . (2.10) For the second integral of right side of (2.9) and (2.10),
− Z Rn n X j,k=1 ∂xk(xj∂xjΦ) · ∂xkΦdx = − Z Rn n X j,k=1 (δjk∂xjΦ + xj∂xj∂xkΦ) · ∂xkΦdx = − Z Rn |∇Φ|2dx − Z Rn 1 2 n X j,k=1 xj∂xj|∂xkΦ|2dx = − Z Rn |∇Φ|2dx − Z Rn 1 2 n X j=1 xj∂xj|∇Φ|2dx = (n 2 − 1) Z Rn |∇Φ|2dx i.e. − Z Rn n X j,k=1 ∂xk(xj∂xjΦ) · ∂xkΦdx = (n 2 − 1) Z Rn |∇Φ|2dx . (2.11) Similarly − Z Rn n X j,k=1 ∂xk(xj∂xjψ) · ∂xkψdx = (n 2 − 1) Z Rn |∇ψ|2dx . (2.12)
Here we have used integration by parts.
Now we put (2.11) and (2.12) into (2.9) and (2.10). Then Z Rn n X j=1 xj(4Φ · ∂xjΦ)dx = (n 2 − 1) Z Rn |∇Φ|2dx , (2.13) and Z Rn n X j=1 xj(4ψ · ∂xjψ)dx = (n 2 − 1) Z Rn |∇ψ|2dx . (2.14)
Moreover,we may put (2.7), (2.8), (2.13) and (2.14) into (2.4) and obtain −d2 dt2V (t) = (4n − 8) Z Rn (|∇Φ|2+ |∇ψ|2)dx − 2µ 1n Z Rn |Φ|4dx −2µ2n Z Rn |ψ|4dx − 4βn Z Rn |Φ|2|ψ|2dx +4n Z Rn Φ · (4Φ + µ1|Φ|2Φ + β|ψ|2Φ)dx +4n Z Rn ψ · (4ψ + µ2|ψ|2ψ + β|Φ|2ψ)dx (2.15)
Using integration by parts,we have 4n Z Rn Φ · (4Φ + µ1|Φ|2Φ + β|ψ|2Φ)dx = −4n Z Rn ¡ |∇Φ|2− µ 1|Φ|4− β|Φ|2|ψ|2 ¢ dx . (2.16) and 4n Z Rn ψ · (4ψ + µ2|ψ|2ψ + β|Φ|2ψ)dx = −4n Z Rn ¡ |∇ψ|2dx − µ 2|ψ|4− β|Φ|2|ψ|2 ¢ dx . (2.17) Combining (2.15)-(2.17), one may get
−d2 dt2V (t) = −8 Z Rn (|∇Φ|2+ |∇ψ|2)dx + 2n Z Rn (µ1|Φ|4+ µ2|ψ|4)dx + 4βn Z Rn |Φ|2|ψ|2dx = −8H − 8β(1 −n 2) Z Rn |Φ|2|ψ|2dx + 2(n − 2) Z Rn (µ1|Φ|4+ µ2|ψ|4)dx , (2.18)
where H is the Hamiltonian of (1.1) defined by
H = Z Rn (|∇Φ|2+ |∇ψ|2)dx −1 2 Z Rn (µ1|Φ|4+ µ2|ψ|4)dx − β Z Rn |Φ|2|ψ|2dx . (2.19) Therefore by (2.18) and (2.19), we may complete the proof of Lemma 2.1.
Now we want to prove Theorem 1.1 as follows:
Firstly, we claim that the Hamiltonian H is independent of time t i.e.
H = H(Φ, ψ) = H(Φ0, ψ0), ∀t > 0 . (2.20)
One may multiply the equation of Φ in (1.1) by ∂tΦ and integrate the resulting equation over R¯ n, where ¯Φ is the complex conjugate of Φ. Then using integration by parts, we obtain
i Z Rn |∂tΦ|2dx − Z Rn n X j=1 ∂xjΦ∂xj∂tΦdx +¯ Z Rn (µ1|Φ|2Φ + β|ψ|2Φ)∂tΦdx = 0¯ (2.21) Take complex conjugate on (2.21) and we have
−i Z Rn |∂tΦ|2dx − Z Rn d X j=1 ∂xjΦ∂¯ xj∂tΦdx + Z Rn (µ1|Φ|2Φ + β|ψ|¯ 2Φ)∂¯ tΦdx = 0 (2.22) Adding (2.21) and (2.22) together may give
d dt Z Rn (|∇Φ|2−µ1 2 |Φ| 4)dx − β Z Rn |ψ|2∂t|Φ|2dx = 0 (2.23)
As for (2.23), we may use the equation of ψ in (1.1) to derive
d dt Z Rn (|∇ψ|2−µ2 2 |ψ| 4)dx − β Z Rn |Φ|2∂t|ψ|2dx = 0 (2.24)
Hence by adding (2.23) and (2.24), we obtain
d
where H(Φ, ψ) is defined in (1.4). This may imply (2.20).
Secondly, we use Lemma 2.1 to prove Theorem 1.1. Suppose n = 2. Then (2.1) implies
d2
dt2V (t) ≤ 8H for t > 0 . (2.25)
On the other hand, if n=3, then (2.1) becomes
d2 dt2V (t) ≤ 8H − 4β Z Rn |Φ|2|ψ|2dx − 2 Z Rn (µ1|Φ|4+ µ2|ψ|4)dx . (2.26) Hence by (2.26), (2.25) still holds if n = 3 and β > −√µ1µ2. By (2.20) and (2.25),
d2 dt2V (t) ≤ 8H(Φ0, ψ0) for t > 0 . (2.27) Consequently, by (2.2) and (2.27), V (t) ≤ 4H(Φ0, ψ0)t2+ V0(0)t + V (0) for t > 0 , (2.28) where V0(0) = 4 Z Rn n X j=1 xj[(iΦ0· ∂xjΦ0) + (iψ0· ∂xjψ0)]dx, (2.29) and V (0) = Z Rn |x|2(|Φ0|2+ |ψ0|2)dx . (2.30)
Under any of the hypotheses (i)–(iii) in Theorem 1.1, there exists a time t0 such that the right-hand side of (2.28) vanishes,and thus also t1≤ t0 such that
lim t↑t1
V (t) = 0 . (2.31)
Furthermore, from the equality, Z Rn |f |2dx = 1 n Z Rn (∇ · x)|f |2dx = −1 n Z Rn x · ∇(|f |2)dx, ∀f ∈ H1(Rn) , one may get the following inequality
kf k2 L2(Rn)≤ 2 nk∇f kL2(Rn)kxf kL2(Rn), ∀f ∈ H 1(Rn) . (2.32) Consequently, by (2.32), kΦk2 L2(Rn)≤ 2 nk∇ΦkL2(Rn)kxΦkL2(Rn)≤ 2 nk∇ΦkL2(Rn) p V (t) , (2.33) and kψk2L2(Rn)≤ 2 nk∇ψkL2(Rn)kxψkL2(Rn)≤ 2 nk∇ψkL2(Rn) p V (t) , (2.34)
On the other hand,by (1.1), it is easy to check that
d dtkΦk 2 L2(Rn)= d dtkψk 2 L2(Rn)= 0, ∀t > 0 . Thus kΦk2L2(Rn)= kΦ0k2L2(Rn), kψk2L2(Rn)= kψ0k2L2(Rn). (2.35) Hence by (2.31), (2.33), (2.34) and (2.35), there exists a time t∗≤ t1such that (1.5) holds.
Finally, we want to prove either lim t↑t∗ R
Rn|Φ|4dx = ∞ or lim t↑t∗ R
Rn|ψ|4dx = ∞. Suppose neither one of them holds i.e. both lim
t↑t∗ R
Rn|Φ|4dx < ∞ and lim t↑t∗ R
Rn|ψ|4dx < ∞. Then by H¨older inequality, we obtain lim t↑t∗ ¯ ¯ ¯ ¯12 Z Rn (µ1|Φ|4+ µ2|ψ|4)dx + β Z Rn |Φ|2|ψ|2dx ¯ ¯ ¯ ¯ < ∞ . (2.36)
Thus by (1.4), (1.5) and (2.36), we have
lim
t↑t∗H(Φ, ψ) = ∞ . (2.37)
However, (2.37) may contradict with (2.20) so either lim t↑t∗ R Rn|Φ|4dx = ∞ or lim t↑t∗ R Rn|ψ|4dx = ∞. Therefore we may complete the proof of Theorem 1.1.
3
Symmetry and Approximate solutions
In this section, we introduce function spaces with specific symmetry properties for the proof of Theorem 1.2. These spaces are defined by
(
X = {u ∈ H2(C)|u(y) = u(yei2π
k ); u(¯y) = u(y)},
Y = {u ∈ L2(C)|u(y) = u(yei2π
k ); u(¯y) = u(y)}, (3.1)
where y = (y1, y2) and ¯y = (y1, −y2). In X(or Y ), we use the standard H2-norm (or L2-norm). Note that equation (1.14) is invariant under the maps
(u, v) → µ u³yei2π k ´ , v³yei2π k ´¶ , and
(u, v) →¡u(y1, −y2), v(y1, −y2) ¢ , where i =√−1 ∈ C. Let ξ0 = (l, 0), ξj = ξ0ei 2jπ k , j = 1, 2, ..., k − 1, where c1log 1
|β| ≤ l ≤ c2log|β|1 and c1, c2 are positive constants to be determined. To approximate the solution of (1.14) with specific symmetry properties, we may define a vector-valued function by
µ ul vl ¶ = wλ1,µ1(y) k−1P j=0 wλ2,µ2(y − ξj) , (3.2)
called the approximate solution of (1.14). Note that ul ∈ X, vl ∈ X. For notation convenience, we set
wj(y) = wλj,µj(y) and w2,ξj(y) = wλ2,µ2(y − ξj) for j = 0, 1, ..., k − 1. Let
S µ u v ¶ = µ 4u − λ1u + µ1u3+ βuv2 4v − λ2v + µ1v3+ βu2v ¶ .
We first state the following lemma on the properties of wj(y):
Lemma 3.1. (1) As |y| → +∞, we have wj(|y|) = Aj(1 + O( 1 |y|))|y| −n−1 2 e− √ λj|y|, w0 j(|y|) = −Aj(1 + O( 1 |y|))|y| −n−1 2 e− √ λj|y|, (3.3)
where Aj> 0 is a positive generic constant, j=1,2.
(2) wj is nondegenerate, i.e. Kernel(∆ − λj+ 3µjw2j) ∩ H2(Rn) = span ½ ∂wj ∂y1, ..., ∂wj ∂yn ¾ . (3.4) (3) If 0 <√λ1α1< √
λ2α2, then we have for |x1− x2| >> 1, Z Rn wα1 λ1,µ1(y − x1)w α2 λ2,µ2(y − x2) ≈ w α1 1 (|x1− x2|) Z Rn wα2 2 (y)e− √ λ1α1y1dy, (3.5) wα1 λ1,µ1(y − x1)w α2 λ2,µ2(y − x2) ≤ w α1 1 (|x1− x2|)e( √ λ1α1− √ λ2α2)|y−x2|. (3.6)
Proof. (1) is well-known. (2) follows from the uniqueness of wj (cf.[14]). By (3.3), we have for |x1− x2| >> 1, z = y − x2, wα1 λ1,µ1(y − x1) = w α1 λ1,µ1(z + x2− x1) = (A1+ o(1)) α1(|z + x 2− x1|)− n−1 2 e− √ λ1α1|z+x2−x1| = wα1 λ1,µ1(|x2− x1|)e −√λ1α1<z,|x1−x2|x1−x2 >+o(|z|).
Using (3.3) and Lebesgue’s Dominated Convergence Theorem, we obtain Z Rn wα1 λ1,µ1(y − x1)w α2 λ2,µ2(y − x2) ≈ w α1 1 (|x1− x2|) Z Rn wα2 2 (y)e− √ λ1α1<y,|x1−x2|x1−x2>dy ≈ wα1 1 (|x1− x2|) Z Rn wα2 2 (y)e− √ λ1α1y1dy.
Now we may apply Lemma 3.1 to prove Lemma 3.2. ° ° ° °S µ u v ¶ °° ° ° Y ×Y ≤ c à |β|wλ1,µ1(l) + w2 ³ 2l sinπ k ´! . (3.7) Proof. 4ul− λ1ul+ µ1u3l + βulvl2= βw1 µk−1X j=0 w2,ξj ¶2 = βw1 µ X j w2 2,ξj + X j6=m w2,ξjw2,ξm ¶ .
By Lemma 3.1, and the fact that √λ1<
√ λ2, |a| + |b| ≥ |a ± b|, we have ° °w1w22,ξj ° ° L2(R)≤ cw1(l) ° °w1w22,ξjw 2 2,ξm ° ° L∞(R)≤ ce −|y|√λ1· e−(|y−ξj|+|y−ξm|) √ λ2 ≤ ce−|y|√λ1· e−(|y−ξj|+|y−ξm|) √ λ1· e( √ λ1− √ λ2)|ξj−ξm| ≤ ce−√λ1(|l|+|y−ξm|)· e( √ λ1− √ λ2)|ξj−ξm|. Hence ° °4ul− λ1ul+ µ1u3l + βulvl2 ° ° L2(Rn)≤ c|β|w1(l) + cw1(|l|) · e −2(√λ2− √ λ1)|l|| sin π/k| ≤ c(|β| + e−σl)|w 1(l)|, (3.8) for 0 < σ < 2(√λ2− √ λ1)| sinπk|. Similarly, 4vl− λ2vl+ µ2v3l + βu2lvl= µ2 "Ãk−1 X j=0 w2,ξj !3 − k−1 X j=0 w32,ξj # + βw21 ³ X j w2,ξj ´ = O³ X j6=m w2 2,ξjw2,ξm ´ + O ³ |β|X j w2,ξjw 2 1 ´ . So ° °4vl− λ2vl+ µ2vl3+ βu2lvl ° ° L2(Rn)≤ c X j6=m w2(|ξj− ξm|) + c|β|w1(l) ≤ cX j6=m w2(2l| sinπ k|) + c|β|w1(l), (3.9)
since |ξj− ξm| = 2| sinπ(m−j)k |l ≥ 2l sinπk.
Therefore by (3.8) and (3.9), we obtain (3.7) and complete the proof of Lemma 3.2. Now we want to estimate E[ul, vl] as follows:
Lemma 3.3. For l >> 1, E[ul, vl] =λ 4−n 2 1 µ−11 I[w] + kλ 4−n 2 2 µ−12 I[w] − µ2 2 X j6=m Z Rn w2 2,ξjw2,ξm −β 2 X j Z Rn w2 1w22,ξj+ O ¡ e−2√λ2l sinπk + |β|e−2 √ λ1l¢· e−σl, (3.10) for some σ > 0.
Proof. We may use Lemma 3.1 to compute the energy E[ul, vl]. By (1.15), we have E[ul, vl] =λ 4−n 2 1 µ−11 I[w] + 1 2 Z Rn ¯ ¯ ¯ ¯∇ X j wλ2,µ2,ξj ¯ ¯ ¯ ¯ 2 +λ2 2 Z Rn µ X j wλ2,µ2,ξj ¶2 −µ2 4 Z Rn µ X j wλ2,µ2,ξj ¶4 −β 2 Z Rn w2 λ1,µ1 µ X j wλ2,µ2,ξj ¶2 =λ4−n2 1 µ−11 I[w] + kλ 4−n 2 2 µ−12 I[w] + 1 2 X j6=m µ2 Z Rn wλ2,µ2,ξjw 3 λ2,µ2,ξm −µ2 4 Z Rn ·µ X j wλ2,µ2,ξj ¶4 −X j w4λ2,µ2,ξj ¸ −β 2 Z Rn w2 λ1,µ1 µ X j w2 λ2,µ2,ξj + X j6=m wλ2,µ2,ξjwλ2,µ2,ξm ¶ =λ4−n2 1 µ−11 I[w] + kλ 4−n 2 2 µ−12 I[w] − 1 2 X j6=m µ2 Z Rn wλ2,µ2,ξjw 3 λ2,µ2,ξm −µ2 4 Z Rn ·µ X j wλ2,µ2,ξj ¶4 −X j w4 λ2,µ2,ξj − 4 X j6=m wλ2,µ2,ξjw 3 λ2,µ2,ξm ¸ −β 2 X j Z Rn w2 λ1,µ1w 2 λ2,µ2,ξj − β 2 Z Rn w2 λ1,µ1 X j6=m wλ2,µ2,ξjwλ2,µ2,ξm. (3.11) By Lemma 3.1, Z Rn w2,ξjw 3 2,ξj ≈ w2(|ξi− ξj|), Z Rn ¯ ¯ ¯ ¯ µ X j w2,ξj ¶4 −X j w2,ξ4 j − 4 X j6=m w2,jw32,m ¯ ¯ ¯ ¯ ≤ C X j6=m Z Rn w2,ξ2 jw 2 2,ξm ≤ C X j6=m ¡ w2(|ξj− ξm|) ¢3/2 ≤ Ce−3l√λ2| sinπk|, (3.12) Z Rn w2 1w2,ξjw2,ξm ≤ Z Rn w2 1e− √ λ2(|y−ξj|+|y−ξm|) ≤ Z Rn w2 1e− √ λ1(|y−ξj|+|y−ξm|)· e( √ λ1− √ λ2)|ξj−ξm| ≤ e−2√λ1l· e2l( √ λ1− √ λ2) sinπk, (3.13)
where C is a universal positive constant. Consequently, (3.10) follows from (3.11), (3.12) and (3.13). Therefore we may complete the proof of Lemma 3.3.
4
Localized Energy Method
In this section, we use the so-called “Localized Energy Method” to reduce the problem to a finite-dimensional one. Similar method has been used in the proof of Theorem 4 of [11]. For background and references on this method, we refer to [3], [9], [11] and [13].
In this reduction process, the symmetry assumption plays an important role so we focus on two spatial dimension case i.e. n = 2. Let
L µ φ ψ ¶ = S0 µ ul vl ¶ µ φ ψ ¶ = µ 4φ − λ1φ + 3µ1u2lφ + βvl2φ + 2βulvlψ 4ψ − λ2ψ + 3µ2vl2ψ + βu2lψ + 2βulvlφ ¶ ,
for¡ψφ¢∈ X × X. Let Kl= span ½µ 0 ∂vl ∂l ¶¾ . (4.1) Recall that vl= k−1P j=0 w2 ³ y − lei2πj k ´
and i =√−1. Now we consider the following linear problem:
Given f, g ∈ Y , find φ, ψ ∈ X and c ∈ R such that L à φ ψ ! = à f g ! + c à 0 ∂vl ∂l ! , Z R2 φ∂vl ∂l = 0 . (4.2)
For the problem (4.2), we have the following crucial a-priori estimates
Lemma 4.1. There exist β0> 0 such that for |β| < β0, if ¡φ
ψ ¢
satisfies (4.2), then
kφkH2+ kψkH2 ≤ C(kf kL2+ kgkL2) . (4.3)
Proof. We prove this by contradiction. Suppose there exist |βn| → 0, fn, gn, φn and ψn satisfying (4.2), such that
kfnkL2+ kgnkL2 → 0 as n → ∞ , and kφnkH2+ kψnkH2= 1 for n ∈ N . (4.4)
To avoid clumsy notation, we omit the index n. Firstly, we derive the estimate for c. Multiplying the equation of ψ by ∂vl ∂l, we obtain Z R2 ¡ 4ψ − λ2ψ + 3µ2v2lψ + βu2lψ + 2βulvlφ¢ ∂vl ∂l = Z R2 g∂vl ∂l + c Z R2 ³ ∂vl ∂l ´2 . (4.5) Since 4¡∂vl ∂l ¢
− λ2∂vl∂l + 3µ2v2l · ∂vl∂l = o(1) in L2, it is easy to check that
c = O(kgkL2) + O(kφkH2+ kψkH2)|β| = o(1). (4.6)
Now we claim that the operator
L01φ = 4φ − λ1φ + 3µ1u2lφ (4.7)
is invertible from X to Y . In fact, if L01φ = 0 and φ ∈ ∩X, then φ = 2 X j=1 cj∂wλ1,µ1 ∂yj . Since φ(¯y) = φ(y), we
obtain that c2= 0 and φ = c1∂w∂yλ1,µ11 = c1wλ01,µ1(|y|)
y1
|y|. On the other hand, since φ(yei2π/k) = φ(y), we also obtain c1= 0. Hence φ ≡ 0. Thus L01 is invertible. Moreover, since β → 0, the operator
L11φ = 4φ − λ1φ + 3µ1u2lφ + βvl2φ
can be regarded as a small perturbation of L01so the operator L11is also invertible. Hence, we may write
φ = L−111(−2βulvlψ + f ) , and obtain
kφkH2(R2)≤ ck − 2βulvlψ + f kL2(R2)→ 0. (4.8)
As for the operator L01on φ, we may define another operator on ψ by
L02ψ = 4ψ − λ2ψ + 3µ2v2lψ . Then we have L02ψ = −βu2lψ − βulvlφ + g + c∂vl ∂l =: g2. Hence kg2kL2(R2)= o(1). By (4.8), kψkH2(R2)= 1 − kφkH2(R2)≥1 2. (4.9)
Now we set
ψ0(y) ≡ ψ(ξ0+ y) → eψ0(y) in Cloc1 (R2) as l → +∞ . Then as l → +∞, ψ0 satisfies 4ψ0− λ2ψ0+ 3µ2wλ22,µ2,ξ0(ξ0+ y)ψ0 +3µ2 X j6=m ¡ wλ2,µ2,ξjwλ2,µ2,ξm ¢ (ξ0+ y)ψ0+ 3µ2 X j6=0 w2λ2,µ2,ξj(ξ0+ y)ψ0 = g2(ξ0+ y) , i.e. 4 eψ0− λ2ψe0+ 3µ2w2λ2,µ2ψf0= 0. (4.10)
Here we have used the fact that kg2kL2(R2) = o(1), wλ
2,µ2,ξ0(ξ0+ y) = wλ2,µ2(y) and w 2 λ2,µ2,ξj(ξ0+ y) = w2 λ2,µ2(ξ0− ξj+ y) → 0 as l → +∞ for j 6= 0. Hence e ψ0= c1∂wλ2,µ2 ∂y1 + c2 ∂wλ2,µ2 ∂y2 . (4.11) Moreover, 0 = Z R2 ψ(y + ξ0) ∂vl ∂l (y + ξ0) = Z R2 ψ0(y) ∂vl ∂l (y + ξ0) → k Z R2 e ψ0(y) ∂wλ2,µ2 ∂y1 (y) . Here we have used the k-symmetry property of eψ0. Thus it is obvious that c1= 0 and
e ψ0= c2∂wλ2,µ2 ∂y2 = c2w 0 λ2,µ2(|y|) y2 |y|. (4.12)
Notice that ψ0(¯y) = ψ(y) i.e. eψ0(¯y) = eψ0(y) so we obtain c2= 0. Hence eψ0≡ 0 and we have Z R2 vl4(y)ψ2(y) = Z R2 v4l(y + ξ0)ψ2(y + ξ0) = Z R2 v4l(y + ξ0)ψ20(y) → 0 .
Thus by the equation of ψ, we obtain kψkH2 ≤ k 4ψ − λ2ψkL2 → 0 which may contradict with (4.9). Therefore
we may complete the proof of Lemma 4.1.
Remark. The proof of Lemma 4.1 also explains how we use specific symmetry properties to show the kernel of L is one dimensional, and hence the problem becomes one dimensional, too.
Once Lemma 4.1 is proved, we have the following lemmas. We refer to Lemma 8, Proposition 1 and Lemma 10 of [11] for similar proofs.
Lemma 4.2. Given¡fg¢∈ Y × Y , there exists unique ¡φψ¢∈ X × X and c satisfying (4.2).
Lemma 4.3. The following nonlinear problem has a unique solution:
S à ul+ φl vl+ ψl ! = c à 0 ∂vl ∂l ! , Z R2 φl∂vl ∂l = 0, à φl ψl ! ∈ (H2(R2) ∩ X)2. (4.13) Moreover, we have ° ° ° ° µ φl ψl ¶°° ° ° H2(R2) ≤ c ° ° ° °S µ ul vl ¶°° ° ° L2(R2) (4.14)
and the map l → φl is C1.
Finally, we define M (l) =: E[ul+ φl, vl+ ψl]. Then we have
Lemma 4.4. If M (l) has a critical point l = l0, then (ul0 + φl0, vl0+ ψl0) satisfies (1.14) and the properties
5
Reduced problem
We first compute M (l):
Lemma 5.1. For 0 < |β| << 1, we have M (l) = A0− α(l) −β 2γ(l) + O ¡ e−2√λ2sin πl/k+ |β|e−2 √ λ1l¢e−σl. where A0= λ 4−n 2 1 µ−11 I[w] + kλ 4−n 2 2 µ−12 I[w] , α(l) = kµ2 Z R2 w3 2,ξ0w2,ξ1, γ(l) = k Z R2 w2 1w2,ξ2 0.
Proof. It is easy to check that
M (l) =E[ul+ φl, vl+ ψl] =E[ul, vl] + Z R2 (4ul− λ1ul+ µ1u3l + βulvl2)φl + Z R2 (4vl− λ2vl+ µ2v3l + βu2lvl)ψl+ Z R2 ¡ |φl|2+ |ψl|2 ¢ =E[ul, vl] + O ¡ e−2 √ λ2sinπlk + |β|e−2 √ λ1l¢e−σl.
Then by Lemma 3.3, we may complete the proof of Lemma 5.1. Let c1< 1
2(√λ2sinπk −
√ λ1)
< c2. Then we have
Lemma 5.2. The following problem
max n M (l) ¯ ¯ ¯c1log 1 |β| ≤ l ≤ c2log 1 |β| o
has a critical point lβ, i.e. M0(lβ) = 0.
Proof. By lemma 3.1 and (3.3), it is not difficult to see that
α(l) ≈ l−1/2e−2 √ λ2l sinπk, α0(l) ≈ −2 √ λ2α(l) sin π k; γ(l) ≈ l−1/2e−2√λ1l, γ0(l) ≈ −2√λ 1γ(l). Now we let lβ be such that α0(lβ) + β2γ0(lβ) = 0. Then we have
lβ≈ 1 2(√λ2sinπk − √ λ1) log 1 |β|, and M (lβ) =A0− α(lβ) − β 2γ(lβ) =A0− α(lβ) + γ(lβ) γ0(lβ)α 0(l β) ≈A0− α(lβ) + √ λ√2sinπk λ1 · α(lβ) ≈A0+ √ λ2sinπk − √ λ1 √ λ1 · l−1/2β e−2√λ2lβsinπk ≈A0+ √ λ2sinπk − √ λ1 √ λ1 ³ log 1 |β| ´−1/2 · |β| √ λ2 sinπk √ λ2 sinπk− √ λ1. Hence max M (l) ≥ M (lβ) ≥ A0+ c0 ³ log 1 |β| ´−1/2 |β|τ1, (5.1)
where τ1= √ λ2sinπk √ λ2sinπk − √ λ1 . Now that M ³ c1log 1 |β| ´ ≈A0− µ |β|2c1 √ λ2sinπk +β 2 · |β| 2c1 √ λ1 ¶ µ c1log 1 |β| ¶−1/2 ≈A0− ³ |β|2c1 √ λ2sinπk − |β|1+2c1 √ λ1 ´ µ c1log 1 |β| ¶−1/2 ≤A0− |β|2c1 √ λ2sinπk µ c1log 1 |β| ¶−1/2 if 2c1 √ λ2sin π k < 1 + 2c1 √ λ1. M ³ c2log 1 |β| ´ ≈A0− ³ |β|2c2 √ λ2sinπk − |β|1+2c2 √ λ1 ´ µ c2log 1 |β| ¶−1/2 ≤A0+ |β|1+2c2 √ λ1 µ c2log 1 |β| ¶−1/2 if 2c2 √ λ2sin π k > 1 + 2c2 √ λ1. In any case, we have
maxn(M³c1log 1 |β| ´ , M³c2log 1 |β| ´o ≤ A0+ |β|1+2c2 √ λ1 µ c2log 1 |β| ¶−1/2 if 2c2 √ λ2sin π k > 1 + 2c2 √ λ1. (5.2) Suppose M (lβ) ≤ max n (M ³ c1log|β|1 ´ , M ³ c2log|β|1 ´o
. Then comparing (5.1) and (5.2), we obtain
c0|β|τ1 ≤ c−1/22 |β|1+2c2 √ λ1. As long as we choose 1 + 2c2 √ λ1> τ1= √ λ2sinπk √ λ2sinπk − √ λ1 i.e. c2>2(√λ 1 2sinπk− √ λ1), we obtain a contradiction
right away. Therefore we may complete the proof of Lemma 5.2.
Remark. We have showed that the critical point lβ satisfying lβ∼ 1 2(√λ2sinπk − √ λ1) log 1 |β|.
6
Proof of Theorem 1.3
The proof of Theorem 1.3 depends on the choice of symmetry class. We first consider the cubic case
¿¿ ¿¿ ¿¿ ¿¿ ¿¿ ¿¿ x
Assume the cube has a center at (0, 0, 0) and eight vertices at (a, b, c)’s for a, b, c ∈ {±1}. Let L1 be the plane
y − z = 0 , L2be the plane x − z = 0, and L3 be the plane x − y = 0. Let Tj be the transformation of reflection through the plane Lj, i.e. Tj(x, y, z) = reflection of (x, y, z) to the plane Lj.
Now we set
X = {u ∈ H2(R3)|u(x, y, z) = u(−x, y, z) = u(x, −y, z) = u(x, y, −z) = u¡T
j(x, y, z) ¢
, j = 1, 2, 3} ,
and
Y = {u ∈ L2(R3)|u(x, y, z) = u(−x, y, z) = u(x, −y, z) = u(x, y, −z) = u¡Tj(x, y, z) ¢
We wish to put one spike at the center for u and eight spikes at the vertices of a cube for v. We need to determine the locations of eight spikes. Due to specific symmetry properties v(x, y, z) = v(±x, ±y, ±z), the problem of determining eight spikes can be reduced first to determine one spike in the quadrant {x > 0, y > 0, z > 0}. One may regard the problem as a three dimensional problem. Then we use the reflection symmetry
v(x, y, z) = v(Tj(x, y, z)) , j = 1, 2, 3 to further reduce the problem to an one dimensional problem. Thus, as for the proof of Theorem 1.2, we may set ξ0 = (l, l, l) as one vertex of the cube and ξj, j = 1, · · · , 7 as the other seven vertices of the cube, where l > 0 is an one-dimensional parameter. Then the problem can be reduced to an one dimensional problem like the critical point problem of Lemma 4.4 and 5.2. Therefore we may apply similar arguments to complete the proof of Theorem 1.3 for this case.
In the tetrahedra case, we may assume that the four vertices are P1 = (0, 0, 1), P2 = (1, 0, 0), P3 = (−1/2,√3/2, 0) and P4= (−1/2, − √ 3/2, 0). !!!! !! c c c c c ¶¶ ¶¶ ¶¶ e e e e e e e Q3 r Q1 r Q4 r Q2 r P2 P1 P3 P4
Let Q1 be the center of the triangle P2P3P4. We set l1 as the axis joining P1 and Q1. Let T1 be the rotation around l1 by angle 2π3 . Similarly, we can define lj, Tj, j = 2, 3, 4. Let L1 be the plane containing P1, P2 and (0, 0, 0), L2 be the plane containing P1, P3 and (0, 0, 0) and L3 be the plane containing P1, P4 and (0, 0, 0). Let T4+j be the reflection through Lj for j = 1, 2, 3. Then it is easy to see that the tetrahedra is invariant to
Tj’s. Moreover, the Laplace operator is also invariant under Tj’s. Now we set
X = {u ∈ H2(R3)|u¡T j(x, y, z) ¢ = u(x, y, z) , j = 1, · · · , 7} , and Y = {u ∈ L2(R3)|u¡T j(x, y, z) ¢ = u(x, y, z) , j = 1, · · · , 7}.
Then similar as before, the problem can be reduced to an one-dimensional problem. Therefore we may apply similar arguments to complete the proof of Theorem 1.3 for this case.
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