Electronic Journal of Differential Equations, Vol. 2007(2007), No. 61, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
BLOW-UP SOLUTIONS FOR N COUPLED SCHRÖDINGER
EQUATIONS
JIANQING CHEN, BOLING GUO
Abstract. It is proved that blow-up solutions to N coupled Schrödinger equations
N
X
iϕjt + ϕjxx + µj |ϕj |p−2 ϕj +
βkj |ϕk |pk |ϕj |pj −2 ϕj = 0
k6=j, k=1
exist only under the condition that the initial data have strictly negative energy.
1. Introduction
In this paper, we consider the existence of blow-up solutions of the N coupled
Schrödinger equations
N
X
iϕjt + ϕjxx + µj |ϕj |p−2 ϕj +
βkj |ϕk |pk |ϕj |pj −2 ϕj = 0,
k6=j, k=1
ϕj (x, t)
√
t=0
= ψj (x),
(1.1)
x ∈ R,
where i = −1, ϕj = ϕj (x, t) : R × R+ → C, j, k ∈ {1, . . . , N } and µj , βkj ∈ R.
System of this kind appears in several branches of physics, such as in the study
of interactions of waves with different polarizations [3] or in the description of
nonlinear modulations of two monochromatic waves [9].
When p = 4, pj = 2, and pk = 2, the solution ϕj of (1.1) denotes the jth
component of the beam in Kerr-like photo refractive media [1]. The constants βkj
is the interaction between the kth and the jth component of the beam. As βkj > 0,
the interaction is attractive while the interaction is repulsive if βkj < 0. Moreover,
the system (1.1) is integrable and there are various analytical and numerical results
on solitary wave solutions of the general N coupled Schrödinger equations [6, 8].
When 2 < p < 6, 2 ≤ pk + pj < 6 and N = 2, the existence and stability
of standing wave, which is a trivial global solution, of (1.1) have been studied
by Cipolatti et al [5]. Also when 2 < p < 6 and 2 ≤ pk + pj < 6, for any
2000 Mathematics Subject Classification. 35Q55, 35B35.
Key words and phrases. Blow-up solutions; coupled Schrödinger equations.
c
2007
Texas State University - San Marcos.
Submitted May 29, 2006. Published April 22, 2007.
J. Chen was supported by grant 10501006 from the Youth Foundation of NSFC,
by the China Post-Doc Science Foundation, and by the Program NCETFJ.
1
2
J. CHEN, B. GUO
EJDE-2007/61
j, k ∈ {1, . . . , N }, we know from [4] that for any
→
−
→
−
ϕ (x, 0) = (ϕ1 (x, 0), . . . , ϕN (x, 0)) = ψ = (ψ1 (x), . . . , ψN (x)) ∈ (H 1 (R))N ,
→
Equation (1.1) admits a unique global solution −
ϕ ∈ C(R , (H 1 (R))N ).
+
The main purpose here is to prove the existence of blow-up solutions of (1.1)
only under the condition of the initial data with strictly negative energy. The main
result is the following theorem.
β
=
Theorem 1.1. Let p = 6, pk + pj = 6 and µj ≥ 0, βkj > 0 with θkj := pkj
j
→
−
βjk
pk := θjk , pk , pj ≥ 2. If E( ψ ) < 0(for the definition of E, see Proposition 2.1),
→
−
then the solution of (1.1) with initial data ψ must blow up in finite time.
We emphasize that when N = 1, i.e. no coupling terms, the blow up problem
has been studied extensively, see e.g. [10, 7, 4]. But as far as we know, there is
no blow-up result to the N coupled Schrödinger equations. The main contribution
here is to overcome the additional difficulties created by the coupling terms and
then prove Theorem 1.1.
This paper is organized as follows. In Section 2, we give some preliminaries and
derive a variant of virial identity which generalizes some previous works for the
single equation. Section 3 is devoted to the proof of Theorem 1.1.
R
Notation. As above and henceforth, the integral R . . . dx is simply denoted by
R
. . . . For any t, the function x 7→ ϕj (x, t) is simply denoted by ϕj (t). f denotes
the complex conjugate of f . fx and ft denote the derivative of f with respect to
x and t, respectively. By f (m) we denote the mth order derivatives of f . k · kLq
denotes the norm in Lq (R) or (Lq (R))N which will be understood from the context.
Re denotes the real part and Im the imaginary part.
2. Preliminaries
Throughout this paper, we always assume that the conditions of Theorem 1.1
hold. The following proposition is useful in what follows.
→
−
Proposition 2.1. For any ψ = (ψ1 (x), . . . , ψN (x)) ∈ (H 1 (R))N , there is T > 0
→
and a unique solution −
ϕ ∈ C([0, T ), (H 1 (R))N ) satisfying (1.1). Moreover, there
holds the following conservation laws:
Z
Z
2
|ϕj (t)| ≡ |ψj |2 ,
→
E(−
ϕ (t)) =
N Z X
j=1
X
2
|ϕjx |2 − µj |ϕj |p − 2
θkj
p
Z
(2.1)
→
−
|ϕk |pk |ϕj |pj ≡ E( ψ ). (2.2)
k<j
→
Proof. The existence of the local solution −
ϕ follows from [4]. We only sketch
the proof on the conservative laws. Firstly, multiplying (1.1) by ϕj , integrating
over R and taking imaginary part, we obtain (2.1). Secondly, it is deduced from
multiplying (1.1) by ϕjt , integrating over R and taking real part that
Z X βkj Z
1
µj
− |ϕjx |2 + |ϕj |p +
|ϕk |pk (|ϕj |pj )t = 0.
(2.3)
2
p
pj
t
k6=j
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BLOW-UP SOLUTIONS
3
Similarly, for (1.1) with k instead of j, we have
Z X βjk Z
1
µk
2
p
− |ϕkx | +
+
|ϕj |pj (|ϕk |pk )t = 0.
|ϕk |
2
p
pk
t
(2.4)
j6=k
From (2.3) and (2.4) it follows that
Z N Z X
X
µj
1
θkj
|ϕk |pk |ϕj |pj = 0.
− |ϕjx |2 + |ϕj |p +
2
p
t
t
j=1
(2.5)
k<j
Then (2.2) holds.
Next we derive a variant of virial identity.
Lemma 2.2. Let ϕj be a local smooth solution
R x of (1.1) with ϕj (x, 0) = ψj (x). For
real function φ ∈ W 3,∞ (R), define Φ(x) = 0 φ(y)dy. Then
N
X
Z
Im
φψj ψ jx −
j=1
N
X
Z
Im
φϕj (t)ϕjx (t)
j=1
Z tn X
Z
N Z
N Z
N
1X
2−pX
2
2 (3)
=
2
|ϕjx | φx −
|ϕj | φ +
µj |ϕj |p φx
2
p
0
j=1
j=1
j=1
Z
o
X
− (p − 2)
θkj |ϕk |pk |ϕj |pj φx dτ,
(2.6)
k<j
and
Z
Φ|ϕj |2 =
Z
Φ|ψj |2 − 2
Z tZ
Im φϕj ϕjx dx dτ.
(2.7)
0
Proof. Let ϕj be a smooth solution of (1.1). Firstly, multiplying (1.1) by φϕjx ,
integrating over R and taking the real part, we obtain
Z Z
X
µj
1
φ(|ϕjx |2 )x + φ(|ϕj |p )x +
θkj |ϕk |pk (|ϕj |pj )x φ = 0.
− Im φϕjt ϕjx +
2
p
k6=j
(2.8)
From
Z
− Im
Z
Im
Z
Z
d
Im φϕj ϕjx + Im φϕj ϕjxt ,
dt
Z
Z
= − Im φx ϕjt ϕj + Im φϕjt ϕjx ,
φϕjt ϕjx = −
φϕj ϕjxt
we obtain
Z
− Im
φϕjt ϕjx
1 d
Im
=−
2 dt
Z
1
φϕj ϕjx − Im
2
Z
φx ϕjt ϕj .
It is deduced from (2.8) and (2.9) that
Z
Z
Z 1 d
1
1
−
Im φϕj ϕjx − Im φx ϕjt ϕj +
φ(|ϕjx |2 )x
2 dt
2
2
X
µj
+ φ(|ϕj |p )x +
θkj |ϕk |pk (|ϕj |pj )x φ = 0.
p
k6=j
(2.9)
(2.10)
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J. CHEN, B. GUO
EJDE-2007/61
For (1.1) with k instead of j, we obtain by a similar argument that
Z Z
Z
1 d
1
1
−
Im φϕk ϕkx − Im φx ϕkt ϕk +
φ(|ϕkx |2 )x
2 dt
2
2
X
µk
φ(|ϕk |p )x +
θjk |ϕj |pj (|ϕk |pk )x φ = 0.
+
p
(2.11)
j6=k
Secondly, multiplying the complex conjugate of (1.1) by ϕj φx , integrating by parts
and taking the real part, we get that
Z Z
X
1
−φx |ϕjx |2 + |ϕj |2 φ(3) +µj φx |ϕj |p +
− Im φx ϕjt ϕj =
βkj |ϕk |pk |ϕj |pj φx .
2
k6=j
(2.12)
Similarly,
Z
Z X
1
− Im φx ϕkt ϕk =
−φx |ϕkx |2 + |ϕk |2 φ(3) +µk φx |ϕk |p +
βjk |ϕj |pj |ϕk |pk φx .
2
j6=k
(2.13)
We now obtain from (2.10)–(2.13) that
Z
Z
Z
Z
p−2
d
1
2 (3)
2
Im φϕj ϕjx − 2 φx |ϕjx | +
|ϕj | φ +
µj φx |ϕj |p
−
dt
2
p
Z
X
X 2βkj Z
+
βkj |ϕk |pk |ϕj |pj φx +
|ϕk |pk (|ϕj |pj )x φ = 0
pj
k6=j
(2.14)
k6=j
and
Z
Z
Z
Z
1
p−2
d
2
2 (3)
Im φϕk ϕkx − 2 φx |ϕkx | +
|ϕk | φ +
µk φx |ϕk |p
−
dt
2
p
Z
(2.15)
X
X 2βjk Z
+
βjk |ϕj |pj |ϕk |pk φx +
|ϕj |pj (|ϕk |pk )x φ = 0.
pk
j6=k
j6=k
It follows that
N
d X
Im
−
dt j=1
+
N
p−2X
p
Z
φϕj ϕjx − 2
N Z
X
j=1
Z
µj
φx |ϕj |p + (p − 2)
j=1
N
1X
φx |ϕjx | +
2 j=1
2
X
Z
θkj
Z
|ϕj |2 φ(3)
(2.16)
|ϕk |pk |ϕj |pj φx = 0.
k<j
Hence (2.6) holds. Finally, multiplying the complex conjugate of (1.1) by Φϕj ,
integrating by parts and taking the imaginary part, we obtain
Z
Z
−Re Φϕj ϕjt + Im Φϕj ϕjxx = 0,
which implies
d
dt
Z
2
Φ|ϕj | = −2 Im
So (2.7) easily follows. The proof is complete.
Z
φϕj ϕjx .
(2.17)
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5
3. Proof of Theorem 1.1
In this section, we will borrow an idea from [7, 10] to prove Theorem 1.1. Firstly
we introduce two lemmas from [10].
Lemma 3.1 ([10, Lemma 2.1]). Let u ∈ H 1 (R) and ρ be a real valued function in
W 1,∞ (R). Then for any r > 0, we have
1/2
1/2
kρukL∞ (|x|>r) ≤ kukL2 (|x|>r) 2kρ2 ux kL2 (|x|>r) + ku(ρ2 )x kL2 (|x|>r)
.
(3.1)
Lemma 3.2 ([10, Lemma 2.3]). Let v(x) be in L2 . We define R(x) such that
R(x) = |x| for |x| < 1 and R(x) = 1 for |x| > 1. Put vε (x) = ε−1/2 v(x/ε) for
ε > 0. Then for any δ > 0, there exists an ε0 > 0 such that kRvε kL2 ≤ δ for
0 < ε < ε0 .
We are now in a position to prove the theorem. Observe that p = 6, pj + pk = 6
for j, k ∈ {1, . . . , N } and the solution ϕj (x, t) of (1.1) has the following scaling
invariance. More precisely, if we put
ϕεj (x, t) = ε−1/2 ϕj (x/ε, t/ε2 ),
ϕεk (x, t) = ε−1/2 ϕk (x/ε, t/ε2 )
(3.2)
for ε > 0, then ϕεj and ϕεk also satisfy (1.1) and (1.1) with k instead of j and with
initial data ϕεj (x, 0) = ψεj = ε−1/2 ψj (x/ε) and ϕεk (x, 0) = ψεk = ε−1/2 ψk (x/ε),
respectively. The proof is divided into two steps. In the first step, we show that if
→
−
→
−
→
−
−E( ψ ) is large and k ψ kL2 (|x|>1) is small (but k ψ kL2 (|x|<1) may be large), then
→
k−
ϕ (t)kL2 (|x|>1) is small for all t > 0.
→
−
In the second step, for any initial data ψ with negative energy, we use the scaling
→
−
→
−
transform (3.2) to choose ε > 0 so small that −E( ψ ε ) ( ψ ε = (ψε1 , . . . , ψεN )) is
→
−
sufficiently large and k ψ ε kL2 (|x|>1) is small enough. Then the proof of the second
step is reduced to the first step and we complete the proof.
Let φ : [0, ∞) → R+ be a function with bounded third order derivatives and be
such that


s,
0 ≤ |s| < 1,

√


3


1 < s < 1 + 33 ,
s − (s − 1) ,
√
φ(s) = s − (s + 1)3 ,
−(1 √
+ 33 ) < s < −1,


0

smooth, φ < 0, 1 + 33 ≤ |s| < 2,




0,
2 ≤ |s|.
Putting Φ(x) =
Rx
0
→
−
φ(y)dy and E0 = E( ψ ), we have the following proposition.
Proposition 3.3. Let ϕj (t) be a solution of (1.1) in C([0, T ), H 1 (R)) with ϕj (0) =
ψj . Put a0 = 3/(16M ). If ϕj (t) satisfies
N
X
j=1
kϕj (t)k4L2 (|x|>1) ≤ 2a0 ,
0 ≤ t < T,
(3.3)
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J. CHEN, B. GUO
EJDE-2007/61
then we have
−
N
X
Z
Im
φϕj (t)ϕjx (t) +
j=1
N
X
Z
φψj ψ jx
Im
j=1
≤ 2E0 + 4M (1 + M )2
N
X
kψj k6L2
j=1
where M = kφxx kL∞ + kφ(3) kL∞ +
PN
k,j=1
βkj +
N
MX
+
kψj k2L2 t,
2 j=1
PN
j=1
(3.4)
µj .
Proof. From the energy conserved identity
−
N Z
X
→
|ϕjx |2 = E(−
ϕ (t)) −
|x|<1
j=1
N Z
X
j=1
1
+
3
N
X
Z
|ϕjx |2
|x|>1
6
|ϕj | + 2
µj
j=1
X
Z
θkj
|ϕk |pk |ϕj |pj ,
k<j
we obtain by (2.6) that
−
N
X
Z
Im
φϕj (t)ϕjx (t) +
j=1
N
X
Z
Im
N
Z tn
N Z
X
=
2E0 −
0
−
j=1
N Z
1X
2
φψj ψ jx
j=1
j=1
2X
µj
2 1 − φx |ϕjx | +
3 j=1
|x|>1
2 (3)
|ϕj | φ
+4
X
Z
θkj
2
Z
1 − φx |ϕj |6
o
1 − φx |ϕk |pk |ϕj |pj dτ.
k<j
By Lemma 3.1 with ρ(x) = (1 − φx )1/4 and Hölder inequality, we obtain
Z
(1 − φx )|ϕj |6 ≤ kϕj k2L2 (|x|>1) kρϕj k4L∞ (|x|>1)
|x|>1
2
≤ kϕj k4L2 (|x|>1) 2kρ2 ϕjx kL2 (|x|>1) + kϕj (ρ2 )x kL2 (|x|>1)
(3.5)
≤ 8kϕj k4L2 (|x|>1) kρ2 ϕjx k2L2 (|x|>1) + 2kϕj k6L2 (|x|>1) k(ρ2 )x k2L∞ (|x|>1) .
√
2
3 for
On the other hand,
we
have
from
the
definition
of
φ
and
ρ
that
|(ρ
)
|
≤
x
√
√
1 < |x| < 1 + 1/ 3. For |x| > 1 + 1/ 3, we also have |(ρ2 )x | ≤ 21 kφxx kL∞ . It
√
follows that |(ρ2 )x | ≤ 3(1 + 12 kφxx kL∞ ). So
Z
(1 − φx )|ϕj |6
|x|>1
(3.6)
1
4
2
2
2
6
∞
≤ 8kϕj kL2 (|x|>1) kρ ϕjx kL2 (|x|>1) + 6(1 + kφxx kL ) kϕj kL2 (|x|>1) .
2
It is deduced from
Z
Z
pk
pk
pj
1 − φx |ϕk | |ϕj | ≤
6
pj
1 − φx |ϕk | +
6
6
Z
1 − φx |ϕj |6
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BLOW-UP SOLUTIONS
7
that
2E0 −
N Z
X
j=1
−
1
2
N Z
X
N
2X
2 1 − φx |ϕjx |2 +
µj
3 j=1
|x|>1
|ϕj |2 φ(3) + 4
j=1
X
Z
Z
1 − φx |ϕj |6
1 − φx |ϕk |pk |ϕj |pj
θkj
k<j
Z
N
N Z
1X
2X
6
|ϕj |2 φ(3)
µj
≤ 2E0 −
1 − φx |ϕj | −
2 1 − φx |ϕjx | +
3 j=1
2 j=1
j=1 |x|>1
Z
Z
2X
2X
βjk
βkj
+
1 − φx |ϕk |pk +
1 − φx |ϕj |pj .
3
3
N Z
X
2
j<k
k<j
(3.7)
Using (3.6) and the choice of M , we obtain that
−
N
X
Z
Im
φϕj (t)ϕjx (t) +
Z t
2E0 + 4M (1 + M )2
0
≤
Z t
Z
Im
φψj ψ jx
j=1
j=1
=
N
X
N
X
kϕj k6L2 (|x|>1) +
j=1
2
2E0 + 4M (1 + M )
0
N
X
kϕj k6L2
j=1
= 2E0 + 4M (1 + M )2
N
X
kψj k6L2 +
j=1
N
MX
kϕj k2L2 (|x|>1) dτ
2 j=1
N
MX
+
kϕj k2L2 dτ
2 j=1
(3.8)
N
MX
kψj k2L2 t.
2 j=1
The proof is complete.
Proof of Theorem 1.1. We assume the solution ϕj (t) of (1.1) exists for all t ≥ 0
and then derive a contradiction. The proof is divided into two steps.
→
−
→
Step 1. In this step, we assume the initial data −
ϕ (0) = ψ satisfies
η = −2E0 − 4M (1 + M )2
N
X
kψj k6L2 −
j=1
4
N Z
X
Φ|ψj |2
j=1
N
2 4 X
η
N
MX
kψj k2L2 > 0,
2 j=1
2
kψjx k2L2 + 1 ≤ a0 ,
(3.9)
(3.10)
j=1
where M and a0 are defined as in Proposition 3.3.
We first prove that if the initial data ϕj (0) = ψj satisfies (3.9) and (3.10), then
ϕj (t) satisfies (3.3) for all t ≥ 0. We prove this by contradiction. Since η > 0 and
1 ≤ 2Φ(x) for |x| > 1, we have from (3.10) that
N
X
j=1
kψj k4L2 (|x|>1) ≤ a0 .
(3.11)
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J. CHEN, B. GUO
EJDE-2007/61
Define T0 as
N
X
T0 = sup{t > 0;
kϕj (s)k4L2 (|x|>1) ≤ 2a0 , 0 ≤ s < t}.
j=1
By (3.11) we know that T0 > 0. If T0 = +∞, then we are done. Assuming now
that T0 < +∞, the continuity in L2 of ϕj (t) implies
N
X
kϕj (T0 )k4L2 (|x|>1) = 2a0 .
(3.12)
j=1
As ϕj (t) satisfies all the assumptions in Proposition 3.3 on [0, T0 ), we get from
(2.7), (3.9) and Proposition 3.3 that for 0 < t < T0 ,
Z t
N Z
N Z
N Z
X
X
X
Im
Φ|ϕj (t)|2 ≤
Φ|ψj |2 − 2
φϕj ϕjx dx dτ
j=1
0
j=1
≤
N Z
X
2
Φ|ψj | − 2t Im
j=1
j=1
N Z
X
(3.13)
2
φψj ψ jx − ηt .
j=1
This inequality yields
N Z
N Z
2
X
X
1
Φ|ϕj (t)|2 ≤ −η t + Im
φψj ψ jx
η
j=1
j=1
+
N Z
N Z
2 X
X
1
φψj ψ jx +
Im
Φ|ψj |2 .
η
j=1
j=1
Noticing that
Im
N Z
X
φψj ψ jx
2
≤2
N Z
X
2
Z
|φψj |
|ψjx |2
(3.14)
j=1
j=1
and the fact of φ2 ≤ 2Φ, we deduce that
N Z
N
N Z
4 X
X
X
2
2
Φ|ϕj (t)| ≤
kψjx kL2 + 1
Φ|ψj |2 ,
η
j=1
j=1
j=1
0 ≤ t < T0 .
(3.15)
Since 1 ≤ 2Φ(x) for |x| > 1, (3.10) and (3.15) imply
N
N Z
X
2 X
2
2
kϕj (t)kL2 (|x|>1) ≤ 2
Φ|ϕj (t)|2
j=1
j=1
N Z
N
X
2 4 X
2
2
≤4
Φ|ψj |
kψjx k2L2 + 1
η j=1
j=1
≤ a0 ,
0 ≤ t < T0 .
This and the continuity in L2 of ϕj (t) yield
N
X
j=1
kϕj (T0 )k4L2 (|x|>1) ≤ a0 ,
(3.16)
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BLOW-UP SOLUTIONS
9
→
−
→
which contradicts to (3.12). So if the initial data −
ϕ (0) = ψ satisfies (3.9) and
(3.10), then ϕj (t) satisfies (3.3) for all t ≥ 0.
Therefore, since all the assumptions in Proposition 3.3 hold with T = ∞, ϕj (t)
PN R
satisfies (3.3) with T0 = ∞, which implies that j=1 Φ|ϕj (t)|2 goes to negative
→
−
→
in finite time. This is a contradiction. Hence if the initial data −
ϕ (0) = ψ satisfies
→
−
(3.9) and (3.10), then ϕ (t) must blow up in finite time.
Step 2. In this step, we prove the theorem for all the initial data with negative energy. The main idea is to use the scaling invariance of the (1.1). In the first place, for
ε > 0, let ϕεj (x, t) = ε−1/2 ϕj (x/ε, t/ε2 ). Put ϕεj (x, 0) = ψεj (x) = ε−1/2 ψj (x/ε).
Then ϕεj is also a solution of (1.1) with initial data ψεj in C([0, +∞), H 1 (R)).
Moreover, ϕεj (t) satisfies
kϕεj (t)kL2 = kψεj kL2 = kψj kL2 , t ≥ 0;
→
−
−
E(→
ϕ (t)) = ε−2 E( ψ ), t ≥ 0.
ε
(3.17)
(3.18)
In the second place, we show that there exists an ε > 0 such that
N
N
X
→
−
MX
kψεj k2L2 > 0;
ηε = −2E( ψ ε ) − 4M (1 + M )2
kψεj k6L2 −
2
j=1
j=1
N Z
N
X
2 4 X
2
2
4
Φ|ψεj |
kψεjx k2L2 + 1 ≤ a0 .
ηε j=1
j=1
(3.19)
(3.20)
Using (3.18), (3.19) follows by choosing ε > 0 such that
N
N
−1
X
MX
kψj k2L2
ε2 < −2E0 4M (1 + M )2
.
kψj k6L2 +
2 j=1
j=1
(3.21)
Now we have from (3.17) and (3.18) that for some ε1 > 0 and 0 < ε < ε1 ,
N
4X
kψεjx k2L2 ≤ C0 (ε1 ),
η j=1
C0 (ε1 ) denotes positive constant C0 depending on ε1 . On the other hand, Lemma
3.2 implies that there exists an ε2 > 0 with ε2 < ε1 and
N Z
X
j=1
2
Φ|ψεj | ≤ 2
N
X
j=1
kRψεj k2L2 ≤
1
1
(C0 (ε1 ) + 1)−1 a02
4
(3.22)
for 0 < ε < ε2 , where R is defined as in Lemma 3.2.
→
−
→
Thus if 0 < ε < ε2 and satisfying (3.21), then −
ϕ ε (0) = ψ satisfies (3.19)
and (3.20). Therefore the proof of the theorem in the general case is reduced to
Step 1 when we consider ϕεj (x, t) instead of ϕj (x, t). The proof of Theorem 1.1 is
complete.
Acknowledgement. The authors want to thank the anonymous referee for the
helpful comments.
10
J. CHEN, B. GUO
EJDE-2007/61
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Jianqing Chen
Department of Mathematics, Fujian Normal University, Fuzhou 350007, China
Institute of Applied Physics and Computational Mathematics, PO Box 8009, Beijing
100088, China
E-mail address: jqchen@fjnu.edu.cn
Boling Guo
Institute of Applied Physics and Computational Mathematics, PO Box 8009, Beijing
100088, China
E-mail address: gbl@mail.iapcm.ac.cn