Stability of half-degree point defect profiles for 2-D nematic liquid crystal

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by  Zhiyuan Geng

STABILITY OF HALF-DEGREE POINT DEFECT PROFILES FOR 2-D
NEMATIC LIQUID CRYSTAL
ZHIYUANGENG, WEI WANG, PINGWEN ZHANG,ANDZHIFEI ZHANG
6
1
0 Abstract. In this paper, we prove the stability of half-degree point defect profiles in R2 for the
2 nematic liquid crystal within Landau-deGennes model.

n
a
J

2
1 1. Introduction

Defects in liquid crystal are known as the places where the degree of symmetry of the nematic
]
P order increases so that the molecular direction cannot be well defined. The most striking feature
A
of liquid crystal is a variety of visual defect patterns. Predicting the profiles of defect as well as
.
h stability is thus of great practical importance and theoretical interest. We mention some works
t [2, 15, 22, 25] on the defects based on the topological properties of the order parameter manifolds.
a
m There exist three commonly used continuum theories describing the nematic liquid crystal:
[ Oseen-Frank model, Ericksen model and Landau-de Gennes model. In the Oseen-Frank model,

the state of nematic liquid crystals is described by a unit-vector filed which represents the mean
1
local orientation of molecules, and defects are interpreted as all singularities of this vector field
v
5 [9, 10, 6, 18]. However, the core structure of defects in nematic liquid crystals, such as the disclina-
4 tion lines observed in experiments, cannot be represented by the usual director field and requires
8
description by Landau-de Gennes model [4]. In this model, the state of nematic liquid crystals is
2
0 described by a 3 3 order tensor Q belonging to
×
.
1
= Q :Q M3 3, Q = QT, trQ = 0 .
0 ×
Q ∈
6
n o
1 For Q , one can find s,b R,n,m S2 with n m= 0 such that
: ∈ Q ∈ ∈ ·
v 1 1
i Q = s(n n I)+b(m m I),
X ⊗ − 3 ⊗ − 3
r where I is 3 3 identity matrix. The local physical properties of nematic liquid crystals depend on
a
×
the degree of symmetry of order tensor Q. Specifically, there are three different states:
(1) s = b = 0, which describes the isotropic distribution;
(2) s = 0,b = 0, which corresponds to the uniaxial distribution;
6
(3) s = 0,b = 0, which describes the biaxial distribution.
6 6
Configuration of nematic liquid crystals corresponds to local minimizers of Landau-de Gennes
energy functional, whose simplest form is given by
L
(1.1) [Q]= Q(x)2 +f (Q(x)) dx,
LG B
F 2|∇ |
ZΩn o
where L > 0 is a material-dependent elastic constant, and f is the bulk energy density, which can
B
be taken as follows
a2 b2 c2
f (Q) = tr(Q2) tr(Q3)+ tr(Q2)2,
B
− 2 − 3 4
1
2 ZHIYUANGENG,WEIWANG,PINGWENZHANG,ANDZHIFEIZHANG
where a2,b2,c2 are material-dependent and non-zero constants, which may depend on temperature.
A well-known fact is that f (Q) attains its minimum on a manifold given by
B
N
1
= Q :Q = s+(n n I), n R3, n = 1 ,
N ∈ Q ⊗ − 3 ∈ | |
n o
where s+ = b2+√b4+4a2c2. It is easy to see that is a smooth submanifold of , homemorphic to
4c2 N Q
the real projective plane RP2, and contained in the sphere Q : Q = 2s+ . Critical points
∈ Q | | 3
of Landau-de Gennes functional satisfy the Euler-Lagrangenequation q o
1
(1.2) L∆Q= a2Q b2(Q2 Q 2I)+c2Q Q 2.
− − − 3| | | |
The Landau-de Gennes energy (1.1) and Euler-Lagrange equation (1.2) are widely used to study
the behavior of defects, see [1, 3, 7, 21] and references therein. However, there still exist many
challenging problems in understandingthe mechanism which generates defects and predicting their
profiles as well as stability, see [11]for many conjectures. Theradial symmetric solution in a ball or
in R3, named hedgehog solution, is regarded as a potential candidate profile for the isolated point
defect in 3-D region. The property and stability of this solution are well studied and it is shown
that the radial symmetric solution are not stable for large a2 and stable for small a2 [13]. We also
refer [26, 20, 16, 12] and references therein for related works.
In this paper, we are concerned with a class of point defects in R2, which correspond to “radial”
solutions of the Euler-Lagrange equation (1.2). Here “radial” means that the eigenvectors of Q
don’t change along the radial direction. Precisely speaking, we study the solution with the form
(1.3) Q(r,ϕ) = u(r)F +v(r)F ,
1 2
where (r,ϕ) is the polar coordinate in R2, and
coskϕ sinkϕ 0 1 0 0
−
(1.4) F =2nn I = sinkϕ coskϕ 0 , F = 3e e I = 0 1 0 .
1 2 2 3 3
−  −  ⊗ −  − 
0 0 0 0 0 2
   
The boundary condition on these solutions is taken to be
1 k k
(1.5) lim Q(r,ϕ) = s (n(ϕ) n(ϕ) I), n(ϕ) = (cos ϕ,sin ϕ,0),
+
r + ⊗ − 3 2 2
→ ∞
which has degree k about origin as an RP2-valued map. Here k Z 0 . Note that if we assume
2 ∈ \{ }
theinvariance of Q along thedefect line, then disclination linein 3-D domain can beideally treated
as a point defect in 2-D domain.
In [14], Ignat, Nguyen, Slastikov and Zarnescu proved the existence of the radial solution for
any non-zero integer k. Moreover, the solution is also a local minimizer of the reduced functional
(2.3). An important question is whether the radical solution they constructed is a local minimizer
of the energy . This problem was also partially answered in [14], where the instability result
LG
F
is proved for k > 1. However, the question of whether the k-radially symmetric solutions (1.3)
| |
subject to (1.5) for k = 1 are stable remains open.
±
The goal of this paper is to give a positive answer to this question. Precise result will be stated
in next section. We remark that this problem is somewhat analogous to the stability of radial
solutions of the Ginzburg-Landau equation(see [17, 23, 19, 8, 24] for example).
STABILITY OF HALF-DEGREE POINT DEFECT PROFILES FOR 2-D NEMATIC LIQUID CRYSTAL 3
2. The stability of radially symmetric solution with k = 1
±
We make the following rescaling
c2 2 b2
Q = Q, x = ,
b2 L c
r
and let t = ab24c2. Then Landau-deeGennes eneregy functional (1.1) is rescaled into the form(drop
the tildes):
1 t 1 1
[Q] = Q(x)2 tr(Q2) tr(Q3)+ tr(Q2)2 dx.
LG
F R2 2|∇ | − 2 − 3 4
Z n o
Therefore, without loss of generality, we may take L = b = c = 1 and a2 = t > 0.
In such case, substituting (1.3) into (1.2), (u,v) satisifes the following ODE system(see [5, 11]):
u + u′ k2u= u[ t+2v+(6v2 +2u2)],
(2.1) ( v′′′′+ vrr′ =−vr2[−t−v−+(6v2 +2u2)]+ 13u2,
together with the boundary conditions
s+ s+
(2.2) u(0) = 0, v (0) = 0, u( ) = , v(+ ) = ,
′
∞ 2 ∞ − 6
where s+ = 1+√1+24t. The system (2.1) can be also viewed as the Euler-Lagrange equation of the
4
functional
1 ∂u ∂v 2k2u2 t
E(u,v) = ∞ 2( )2+6( )2+ (2u2+6v2)
2 ∂r ∂r r2 − 2
Z0 n
(cid:2) 1 (cid:3)
(2.3) 2v(v2 u2)+ (2u2+6v2)2 rdr,
− − 4
o
which is a reduced form of (1.1) under the assumption (1.3).
Let us recall some basic properties of the solution (u,v) constructed in [14]:
(H1) u > 0,v < 0,u+3v < 0, u2+3v2 < s2/3 for r (0, );
(H2) For t < 1/3: v > s+ > 1; + ∈ ∞
− 6 −6
(H3) For t > 1/3: v < s+ < 1;
− 6 −6
(H4) For t = 1/3: v = s+ = 1;
− 6 −6
(H5) u > 0, v (1+6v) 0.
′ ′
≤
Remark 2.1. In fact, (H5) was not verified in [14]. In the appendix, we will present a proof by
using (H1)-(H4) and the fact that (u,v) is a minimizer of (2.3).
The main result of this paper is stated as follows.
Theorem 2.1. Let (u,v) be a stable critical point of (2.3) for k = 1 with the properties (H1)-
±
(H4). Then the solution Q = u(r)F +v(r)F is a local minimizer of Landau-de Gennes energy
1 2
(1.1). That is, for any perturbation V H1(R2, ), it holds
∈ Q
d2 1 1 t
(2.4) (V), (Q+εV)2 Q 2 (Q+εV 2 Q 2)
I dε2 R2 2|∇ | − 2|∇ | − 2 | | −| |
Z n
1 1
tr((Q+εV)3) tr(Q3) + (Q+εV 4 Q 4) dx
− 3 − 4 | | −| | ε=0
o (cid:12)
= V 2(cid:2) t V 2 2tr(QV2)+ Q(cid:3) 2 V 2+2(tr(QV))2 dx (cid:12)0.
(cid:12)
R2 |∇ | − | | − | | | | ≥
Z n o
4 ZHIYUANGENG,WEIWANG,PINGWENZHANG,ANDZHIFEIZHANG
Moreover, the equality holds if and only if V span V ,V ,V ,V ,V , where
0 1 2 3 4
∈ { }
V = uE ,
0 2
ku
V = (u(r)E +√3v (r)E )cosϕ E sinϕ,
1 ′ 1 ′ 0 2
− r
ku
V = (u(r)E +√3v (r)E )sinϕ+ E cosϕ,
2 ′ 1 ′ 0 2
r
V = (ucoskϕ 3v)E +usinkϕE ,
3 3 4
−
V = usinkϕE (ucoskϕ+3v)E .
4 3 4
−
Here E ,E , ,E are defined in (3.2).
0 1 4
···
Remark 2.2. The above null space is generated by the invariance of the energy F under the
LG
rotation and translation, i.e.,
(2.5) (RQRT) = (Q), Q(x+x ) = (Q(x)),
LG LG LG 0 LG
F F F F
for any constant R SO(3) and constant x0 R2. (cid:0) (cid:1)
∈ ∈
Remark 2.3. This result implies that the solution (1.3) for k = 1 can be regarded as the profile
of point defects in R2 or the local profile of line defects in R3. ±
3. The second variation of Landau-de Gennes energy
To prove the stabillity, we need to compute the second variation of at critical point Q =
LG
u(r)F +v(r)F . For any V H1(R2, ), we have F
1 2
∈ Q
d2 1 1 t
(3.1) (V), (Q+εV)2 Q 2 (Q+εV 2 Q 2)
I dε2 R2 2|∇ | − 2|∇ | − 2 | | −| |
Z n
1 1
(tr((Q+εV)3) tr(Q3))+ (Q+εV 4 Q 4) dx
− 3 − 4 | | −| | ε=0
o (cid:12)
= V 2 t V 2 2tr(QV2)+c2 Q 2 V 2+2(tr(QV))2(cid:12) dx
(cid:12)
R2 |∇ | − | | − | | | |
Z n o
= V 2 t V 2 2(u tr(F V2)+v tr(F V2))
1 2
R2 |∇ | − | | − · ·
Z n
+(6v2 +2u2)V 2+2(u tr(F V)+v tr(F V))2 dx.
1 2
| | · ·
o
We define
1 0 0
3 −3 1
E = 0 1 0 = F ,
0 r2  0 −03 2  √6 2
3
 
coskϕ sinkϕ 0
1 1
E = sinkϕ coskϕ 0 = F ,
1 1
√2 −  √2
0 0 0
(3.2)
 
sinkϕ coskϕ 0
1 −
E = coskϕ sinkϕ 0 ,
2
√2 
0 0 0
 
0 0 1 0 0 0
1 1
E = 0 0 0 , E = 0 0 1 .
3 4
√2  √2 
1 0 0 0 1 0
   
STABILITY OF HALF-DEGREE POINT DEFECT PROFILES FOR 2-D NEMATIC LIQUID CRYSTAL 5
A straightforward calculation shows
tr(E E ) = δi for 0 i,j 4,
i j j
≤ ≤
which implies that E is an orthonormal basis in . Thus, we can write V H1(R2, ) as
i 0 i 4
{ } ≤≤ Q ∈ Q
a linear combination of this basis in the polar coordinate
4
(3.3) V(r,ϕ) = w (r,ϕ)E (ϕ).
i i
i=0
X
Using (3.3), a direct calculation yields that
4
V 2 = w2, tr(F V) = √2w , tr(F V) = √6w ,
i 1 1 2 0
| |
i=0
X
V2 =w2E2+(w2+w2)E2 +w2E2+w2E2+(w w +w w )(E E +E E )
0 0 1 2 1 3 3 4 4 1 3 2 4 1 3 3 1
√6 √6
+(w w +w w )(E E +E E )+w w (E E +E E ) w w E w w E
1 4 2 3 1 4 4 1 3 4 3 4 4 3 0 1 1 0 2 2
− 3 − 3
√6 √6
+ w w E + w w E ,
0 3 3 0 4 4
6 6
thus we obtain
coskϕ 2√3
tr(F V2)= (w2 w2) w w +sinkϕw w ,
1 2 3 − 4 − 3 0 1 3 4
1 1
tr(F V2)= w2 w2 w2+ w2+ w2.
2 0 − 1 − 2 2 3 2 4
From the fact V 2 = (∂ V)2+ 1 (∂ V)2, we have
|∇ | r r2 ϕ
4
1
V 2 = w2 + w2 +(kw w )2+(kw +w )2+w2 +w2 .
|∇ | ir r2 0ϕ 2− 1ϕ 1 2ϕ 3ϕ 4ϕ
i=0
X (cid:0) (cid:1)
In summary, we conclude that
+ 2π 4 1
(V)= ∞ w2 + w2 +(kw w )2+(kw +w )2+w2 +w2
I ir r2 0ϕ 2− 1ϕ 1 2ϕ 3ϕ 4ϕ
Z0 Z0 nXi=0 (cid:2) (cid:3)
4
1 2
+(6v2 +2u2 t) w2 2 u( coskϕ(w2 w2) w w +sinkϕw w )
− i − 2 3 − 4 − √3 0 1 3 4
i=0
(cid:0)X (cid:1) (cid:2)
1 1
+v(w2 w2 w2+ w2 + w2) +4(uw +√3vw )2 rdrdϕ.
0 − 1 − 2 2 3 2 4 1 0
o
(cid:3)
In order to prove (V) 0, it suffices to show that for k = 1,
I ≥ | |
+ 2π 1
IA(w ,w ,w ) , ∞ w2 +w2 +w2 + [w2 +(kw w )2+(kw +w )2]
0 1 2 0r 1r 2r r2 0ϕ 2− 1ϕ 1 2ϕ
Z0 Z0 n
4
+(6v2 +2u2 t)(w2+w2+w2) uw w
− 0 1 2 − √3 0 1
(3.4) +v(w2 w2 w2)]+4(uw +√3vw )2 rdrdϕ 0,
0 − 1 − 2 1 0 ≥
o
6 ZHIYUANGENG,WEIWANG,PINGWENZHANG,ANDZHIFEIZHANG
and
+ 2π 1
IB(w ,w ) , ∞ w2 +w2 + (w2 +w2 )+(6v2 +2u2 t)(w2 +w2)
3 4 3r 4r r2 3ϕ 4ϕ − 3 4
Z0 Z0 n
(3.5) u(coskϕ(w2 w2)+2sinkϕw w ) v(w2 +w2) rdrdϕ 0.
− 3 − 4 3 4 − 3 4 ≥
o
The following lemma shows that C (R2 0 ) is dense in H1(R2). Thus, we may assume V
c∞
C (R2 0 ), hence w C (R2 0 ). \{ } ∈
c∞ i c∞
\{ } ∈ \{ }
Lemma 3.1. C (R2 0 ) is dense in H1(R2).
c∞
\{ }
Proof. Since C (R2) is dense in H1(R2), it suffices to show that any C (R2) function can be
c∞ c∞
approximated by C (R2 0 ). For this, we introduce a smooth cut-off function χ(r) defined by
c∞
\{ }
0 r 2
χ(r) = ≤ −
1 r 1.
(cid:26) ≥ −
Foranyu∈ Cc∞(R2),letuN(x)= u(x)χ(lnN|x|)forN ≥ 1. Obviously,uN ∈ Cc∞(R2\{0}). Moreover,
ln x ln x
ku−uNkH1 ≤ u(1−χ( N| |)) L2 + ∇u(1−χ( N| |)) L2
(cid:13) 1 u ln x(cid:13) (cid:13) (cid:13)
(cid:13)+ N x χ′( N| (cid:13)|)) L2(cid:13). (cid:13)
| |
(cid:13) (cid:13)
(cid:13) (cid:13)
It is easy to see that the first two terms on the right hand side tend to zero as N + . While,
→ ∞
the third term is bounded by
C u(x)2 1 C
2
N(cid:16)Ze−2N≤|x|≤e−N | |x|2| dx(cid:17) ≤ √NkukL∞,
which tends to zero as N + . (cid:3)
→ ∞
4. Some important integral identities
In this section, let us derive some important integral identities, which will play crucial roles in
our proof. In the sequel, we assume that η C ((0,+ ),R).
c∞
∈ ∞
Using (2.1), we deduce that
(η) := ∞ (vη)2 +(6v2 +2u2 t v)(vη)2 rdr
r
A − −
Z0 n o
v 1
= ∞ (vη)2 +(v + ′ u2)vη2 rdr
r ′′ r − 3
Z0 n o
1
=(vv η2r) + ∞(v2η2 vu2η2)rdr
′ |∞0 r − 3
Z0
1
(4.1) = ∞ (vη )2 vu2η2 rdr,
r
− 3
Z0 n o
STABILITY OF HALF-DEGREE POINT DEFECT PROFILES FOR 2-D NEMATIC LIQUID CRYSTAL 7
and
k2
(η) := ∞ (uη)2 +(6v2 +2u2 t+2v+ )(uη)2 rdr
B r − r2
Z0 n o
u
= ∞[(η u+uη)2 +uη2(u + ′)]rdr
′ ′ ′′
r
Z0
=(uuη2r) + ∞u2η2rdr
′ |∞0 r
Z0
(4.2) = ∞(uη )2rdr.
r
Z0
Taking derivative to (2.1) gives
u 2u 2u
u + ′′ ′ + = u[ t+2v+6v2 +6u2]+2uv (1+6v),
′′′ ′ ′
r − r2 r3 −
v v 2uu
v + ′′ ′ = v [ t 2v+18v2 +2u2]+ ′(1+6v).
′′′ ′
r − r2 − − 3
Therefore, we have
(η) := ∞ (v η)2 +(v η)2(18v2 +2u2 t 2v) rdr
′ r ′
C − −
Z0 n o
v v 2uu
= ∞ (v η+v η )2+v η2(v + ′′ ′ ′(1+6v)) rdr
′′ ′ ′ ′ ′′′
r − r2 − 3
Z0 n o
(v η)2 2uuv (1+6v)
= ∞ (v′η′)2 ′ ′ ′ η2 rdr+(rv′′v′η2) ∞
− r2 − 3 0
Z0 n (v η)2 2uuv (1+6v) o (cid:12)
(4.3) = ∞ (v η )2 ′ ′ ′ η2 rdr, (cid:12)
′ ′
− r2 − 3
Z0 n o
and
k2
(η) := ∞ (uη)2 +(uη)2(6v2 +6u2 t+2v+ ) rdr
D ′ r ′ − r2
Z0 n o
u u 2u
= ∞ (u η+uη )2+uη2 u + ′′ ′ + 2uv (1+6v) rdr
′′ ′ ′ ′ ′′′ ′
r − r2 r3 −
Z0 n (cid:16) (cid:17)o
(uη)2 2uuη2
= ∞ (u′η′)2 ′ + ′ 2uu′v′(1+6v)η2 rdr+(ru′′u′η2) ∞
− r2 r3 − 0
Z0 n (uη)2 2uuη2 o (cid:12)
(4.4) = ∞ (uη )2 ′ + ′ 2uuv (1+6v)η2 rdr. (cid:12)
′ ′ ′ ′
− r2 r3 −
Z0 n o
In addition, we have
uη k2 (uη)2
(η) := ∞ ( )2+(6v2+2u2 t+2v+ ) rdr
E r r − r2 r2
Z0 n o
uη u uη2 u
= ∞[( ′ +( )η)2+ (u + ′)]rdr
′ ′′
r r r2 r
Z0
u η2 u
= ∞ (rηr)2+ r4(2ruu′−u2) rdr+(r)′uη2 ∞0
Z0 (cid:16) u η2 (cid:17) (cid:12)
(4.5) = ∞ ( η )2+ (2ruu u2) rdr. (cid:12)
r r r4 ′−
Z0 (cid:16) (cid:17)
8 ZHIYUANGENG,WEIWANG,PINGWENZHANG,ANDZHIFEIZHANG
5. Proof of Theorem 2.1
This section is devoted to the proof of Theorem 2.1.
5.1. Non-negativity of IB(w ,w ).
3 4
Proposition 5.1. For any w ,w C (R2 0 ) and k = 1, we have
3 4 c∞
∈ \{ } | |
+ 2π 1
IB(w ,w ):= ∞ w2 +w2 + (w2 +w2 )+(6v2 +2u2 t)(w2+w2)
3 4 3r 4r r2 3ϕ 4ϕ − 3 4
Z0 Z0 n
u(coskϕ(w2 w2)+2sinkϕw w ) v(w2 +w2) rdrdϕ 0.
− 3 − 4 3 4 − 3 4 ≥
o
Proof. Let z = w +iw , i = √ 1. Then we have
3 4
−
∂ z 2 = (∂ w )2+(∂ w )2, ∂ z 2 = (∂ w )2+(∂ w )2, z 2 = w2+w2,
| r | r 3 r 4 | ϕ | ϕ 3 ϕ 4 | | 3 4
coskϕ(w2 w2)+2sinkϕw w = Re(coskϕ isinkϕ)(w +iw )2 = Re(e ikϕz2).
3 − 4 3 4 − 3 4 −
Thus, we can rewrite IB(w ,w ) as
3 4
2π 1
(5.1) IB(z) = ∞ ∂ z 2 + ∂ z 2+(6v2 +2u2 t v)z 2 uRe(e ikϕz2)rdrdϕ.
| r | r2| ϕ | − − | | − −
Z0 Z0
Assume that
+
∞
z(r,ϕ) = z (r)eimϕ,
m
m=
X−∞
we have
+ +
∞ ∞
Re(e ikϕz2) =Re e ikϕ e(m+l)ϕz z = Re ei(m+l k)ϕz z .
− − m l − m l
l,m= l,m=
(cid:0) X−∞ (cid:1) (cid:0) X−∞ (cid:1)
Substituting it into (5.1), we get
IB(z) = 2π +∞ +∞ ∂ z 2+ m2 z 2+(6v2 +2u2 t v)z 2
| r m| r2 | m| − − | m|
Z0 nmX=−∞h i
u Re(z z ) rdr.
m l
−
mX+l=k o
When k = 1, we can write
∞
IB(z) = 2π M ,
m
m=1
X
where
1
M = ∞ ∂ z 2+ ∂ z 2+ (m2 z 2+(1 m)2 z 2)
m Z0 n| r m| | r 1−m| r2 | m| − | 1−m|
+(6v2 +2u2 t v)(z 2+ z 2) 2uRe(z z ) rdr.
m 1 m m 1 m
− − | | | − | − −
o
Noticing that m2 1,(1 m)2 0 for m 1, and using the following simple relations
≥ − ≥ ≥
z z Re(z z ), ∂ z 2 (∂ z )2, ∂ z 2 (∂ z )2,
m 1 m m 1 m r m r m r 1 m r 1 m
| − |≥ − | | ≥ | | | − | ≥ | − |
STABILITY OF HALF-DEGREE POINT DEFECT PROFILES FOR 2-D NEMATIC LIQUID CRYSTAL 9
we conclude that
1
M ∞ (∂ z )2+(∂ z )2+ z 2
m ≥ Z0 (cid:16) r| m| r| 1−m| r2| m|
+(6v2 +2u2 t v)(z 2+ z 2) 2uz z rdr.
m 1 m m 1 m
− − | | | − | − | || − |
(cid:17)
Thus, we only need to show that for any q ,q C ((0, ),R),
0 1 c∞
∈ ∞
q2
I˜(q ,q ) , ∞ (∂ q )2+(∂ q )2+ 1 +(6v2 +2u2 t v)(q2+q2) 2uq q rdr > 0.
0 1 r 0 r 1 r2 − − 0 1 − 0 1
Z0 n o
Let η = q /u and ζ = q /v. Then η,ζ C ((0, )). From (4.1) and (4.2), it is straightforward to
1 0 c∞
∈ ∞
obtain
I˜(q ,q ) = (ζ)+ (η) ∞ 3v(uη)2 +2vu2ζη rdr
0 1
A B −
Z0 n o
1
(5.2) = ∞ (uη )2+(vζ )2 vu2(3η+ζ)2 rdr,
′ ′
− 3
Z0 n o
which is non-negative since v < 0.
The case of k = 1 can be considered similarly. (cid:3)
−
5.2. Non-negativity of IA(w ,w ,w ). First of all, we expand w (r,ϕ) as
0 1 2 i
∞
w (r,ϕ) = µ(i)(r)cosnϕ+ν(i)sinnϕ .
i n n
n=0
X(cid:0) (cid:1)
Since ω C (R2 0 ), we may assume µ(i),ν(i) C ((0, ),R) for all n and i. Furthermore,
i c∞ n n c∞
∈ \{ } ∈ ∞
w C (R2) requires µ(0) = ν(0) = 0 for n 1.
0 c∞ n n
∈ ≥
Direct calculation shows that
2π(∂wi)2dϕ = π 2(∂µ(0i))2+ ∞ ((∂µ(ni))2+(∂νo(i))2) ,
∂r ∂r ∂r ∂r
Z0 n=1
(cid:2) X (cid:3)
2π(∂wi)2dϕ = π ∞ n2((µ(i))2+(ν(i))2),
∂ϕ n n
Z0 n=1
X
2π ∂wiw dϕ = π ∞ n(µ(j)ν(i) ν(j)µ(i)),
∂ϕ j n n − n n
Z0 n=1
X
2π
w2dϕ = π 2(µ(i))2+ ∞ ((µ(i))2+(ν(i))2) ,
i 0 n n
Z0 n=1
(cid:2) X (cid:3)
2π
w w dϕ = π 2µ(i)µ(j)+ ∞ (µ(i)µ(j)+ν(i)ν(j)) .
i j 0 0 n n n n
Z0 n=1
(cid:2) X (cid:3)
Thus, we can decompose IA(w ,w ,w ) as
0 1 2
∞
(5.3) IA(w ,w ,w )= IA +IA + IA,
0 1 2 0,01 0,2 n
n=1
X
10 ZHIYUANGENG,WEIWANG,PINGWENZHANG,ANDZHIFEIZHANG
where
(0) (1)
∂µ ∂µ
IA = ∞ ( 0 )2+(µ(0))2(18v2 +2u2 t 2v)+( 0 )2
0,01 ∂r 0 − − ∂r
Z0 n
k2 4u
+(µ(1))2(6v2 +6u2 t+2v+ )+ (1+6v)µ(0)µ(1)rdr,
0 − r2 √3 0 0
∂µ(2) k2
IA = ∞( 0 )2+(µ(2))2(6v2 +2u2 t+2v+ ) rdr,
0,2 ∂r 0 − r2
Z0 o
and
2 ∂µ(i) ∂ν(i) 4kn 2 n2
IA = ∞ ( n )2+( n )2 + (µ(1)ν(2) µ(2)ν(1))+ ((µ(i))2+(ν(i))2)
n ∂r ∂r r2 n n − n n r2 n n
Z0 nXi=0(cid:16) (cid:17) Xi=0
k2
+((µ(0))2+(ν(0))2)(18v2 +2u2 t 2v)+((µ(1))2+(ν(1))2)(6v2 +6u2 t+2v+ )
n n − − n n − r2
k2 4u
+((µ(2))2+(ν(2))2)(6v2 +2u2 t+2v+ )+ (1+6v)(µ(0)µ(1) +ν(0)ν(1)) rdr.
n n − r2 √3 n n n n
o
Thenon-negativity ofIA follows fromthefact that(u,v) is alocal minimizer ofreducedenergy
0,01
(2.3). Indeed, we have for any η,ξ H1((0, ),rdr),
∈ ∞
d2
(η,ξ) , E(u+εη,v +εξ) E(u,v)
J dε2 − ε=0
n o(cid:12)
= ∞ (∂ η)2 +(∂ ξ)2+η2(18v2 +(cid:12)2u2 t 2v)
r r (cid:12)
− −
Z0 n
k2 4u
+ξ2(6v2 +6u2 t+2v+ )+ (1+6v)ηξ rdr 0,
− r2 √3 ≥
o
which implies that
(5.4) IA 0.
0,01 ≥
It follows from (4.2) that
2
(5.5) IA = (µ(2)/u) = ∞ u∂ (µ(2)/u) rdr 0.
0,2 B 0 r 0 ≥
Z0 (cid:16) (cid:17)
It remains to prove IA 0 for all n 1, which is a consequence of the following proposition.
n
≥ ≥
Proposition 5.2. For any µ ,ν ,µ ,ν ,µ ,ν C ((0, )) and k =1, we have
0 0 1 1 2 2 c∞
∈ ∞ | |
IA(µ ,ν ,µ ,ν ,µ ,ν )
n 0 0 1 1 2 2
2 ∂µ ∂ν 4kn 2 n2
, ∞ ( i)2+( i)2 + (µ ν µ ν )+ (µ2+ν2)
∂r ∂r r2 1 2− 2 1 r2 i i
Z0 nXi=0(cid:16) (cid:17) Xi=0
k2
+(µ2+ν2)(18v2 +2u2 t 2v)+(µ2+ν2)(6v2 +6u2 t+2v+ )
0 0 − − 1 1 − r2
k2 4u
+(µ2+ν2)(6v2 +2u2 t+2v+ )+ (1+6v)(µ µ +ν ν ) rdr 0.
2 2 − r2 √3 0 1 0 1 ≥
o
Proof. From the fact that
2(µ ν µ ν ) (µ2+ν2+µ2+ν2),
1 2− 2 1 ≥ − 1 1 2 2