Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 251, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SEMICLASSICAL SOLUTIONS FOR LINEARLY COUPLED
SCHRÖDINGER EQUATIONS
SITONG CHEN, XIANHUA TANG
Abstract. We consider the system of coupled nonlinear Schrödinger equations
−ε2 ∆u + a(x)u = Hu (x, u, v) + µ(x)v,
x ∈ RN ,
−ε2 ∆v + b(x)v = Hv (x, u, v) + µ(x)u,
x ∈ RN ,
1
N
u, v ∈ H (R ),
C(RN )
where N ≥ 3, a, b, µ ∈
and Hu , Hv ∈ C(RN × R2 , R). Under conditions
that a0 = inf a = 0 or b0 = inf b = 0 and |µ(x)|2 ≤ θa(x)b(x) with θ ∈ (0, 1)
and some mild assumptions on H, we show that the system has at least one
nontrivial solution provided that 0 < ε ≤ ε0 , where the bound ε0 is formulated
in terms of N, a, b and H.
1. Introduction
In this article, we study the existence of semiclassical solutions of the system of
coupled nonlinear Schrödinger equations
−ε2 ∆u + a(x)u = Hu (x, u, v) + µ(x)v,
x ∈ RN ,
−ε2 ∆v + b(x)v = Hv (x, u, v) + µ(x)u,
x ∈ RN ,
1
(1.1)
N
u, v ∈ H (R ),
where z := (u, v) ∈ R2 , N ≥ 3, a, b, µ ∈ C(RN , R) and H, Hu , Hv ∈ C(RN × R2 , R).
Systems of this type arise in nonlinear optics [1].
In the past several years, there are many papers about the semiclassical solutions
of the nonlinear perturbed Schrödinger equation
−ε2 ∆u + V (x)u = f (x, u),
u ∈ H 1 (RN )
under various hypotheses on the potential and the nonlinearity (see [2, 6, 7, 12, 13,
14, 16, 19, 22, 25]).
However, by Kaminow [17], we know that single-mode optical fibers are not really
“single-mode”, but actually bimodal due to the presence of birefringence. And
recently, different authors focused their attention on coupled nonlinear Schrödinger
2000 Mathematics Subject Classification. 35J20, 58E50.
Key words and phrases. Nonlinear Schrödinger equation; semiclassical solution;
coupled system.
c
2014
Texas State University - San Marcos.
Submitted October 23, 2014. Published December 1, 2014.
1
2
S. CHEN, X. H. TANG
EJDE-2014/251
systems (see [3, 4, 8, 9, 10, 11, 18]) which describe physical phenomena (see, e.g.,
[1, 5, 15]).
In a recent article, [8], Chen and Zou studied the system of nonlinear Schrödinger
equations
−ε2 ∆u + a(x)u = f (u) + µv, x ∈ RN ,
−ε2 ∆v + b(x)v = g(v) + µu,
u, v > 0
in RN ,
x ∈ RN ,
(1.2)
u, v ∈ H 1 (RN ),
where N, a and b are the same as in (1.1). Under the assumptions
(i) there exists a constant a0 > 0 such that a(x), b(x) ≥ a0 and 0 ≤ µ < a0 ;
g(s)
(ii) f, g ∈ C(RN , R) and lims→0 f (s)
s = s = 0;
∗
(iii) there exists a constant p0 ∈ (1, 2 − 1) such that
lim sup
s→+∞
f (s)
< +∞,
sp0
Rs
lim sup
s→+∞
g(s)
< +∞;
sp0
Rs
f (t) dt
s2
g(t) dt
(iv) either lim sups→+∞ 0
= +∞ or lim sups→+∞ 0 s2
= +∞.
They proved that (1.2) has a positive solution for sufficiently small ε > 0 and all
µ ∈ (0, µ1 ] for some µ1 ∈ (0, a0 ).
Obviously, if a0 = 0, their arguments become invalid due to the fact that 0 ≤ µ <
a0 can not be satisfied. To the best of our knowledge, the existence of semiclassical
solutions to system (1.1), under the assumption of a0 = inf a = 0 or b0 = inf b = 0,
has not ever been studied by variational methods. In addition, as the nonlinearity is
non-autonomous and dependent on u and v, the problem will become more complex.
Motivated by [8, 20, 24, 26], we shall choose the case a0 = inf a = 0 or b0 =
inf b = 0 as the objective of the present paper.
Before presenting the main results, we introduce the following assumptions.
(A0) a(x) ≥ a(0) = 0, b(x) ≥ 0 and there exist a0 , b0 > 0 such that the sets
Aa0 := {x ∈ RN : a(x) < a0 } and Bb0 := {x ∈ RN : b(x) < b0 } have finite
measure;
(A1) there exists a constant θ ∈ (0, 1) such that |µ(x)|2 ≤ θa(x)b(x), for all
x ∈ RN ;
(B0) a(x) ≥ 0, b(x) ≥ b(0) = 0 and there exist a0 , b0 > 0 such that the sets
Aa0 := {x ∈ RN : a(x) < a0 } and Bb0 := {x ∈ RN : b(x) < b0 } have finite
measure;
(H1) there exist constants p ∈ (2, 2∗ ) and C > 0 such that
|H(x, z)| ≤ C(|z| + |z|p ),
∀(x, z) ∈ RN × R2 ;
(H2) Hz (x, z) · z = o(|z|2 ), as |z| → 0, uniformly in x ∈ RN ;
(H3) lim|z|→∞ |H(x,z)|
= ∞ uniformly in x ∈ RN ;
|z|2
(H4) there exist c0 > 0, T0 > 0 and q ∈ (2, 2∗ ) such that
H(x, u, 0) ≥ c0 |u|q ,
∀x ∈ RN , u ∈ [−T0 , T0 ]
and
−2 4−N
u
Z
h
|x|≤h
H(λ−1/2 x, u/h, 0) dx ≥
(N 2 + 2)ωN
,
2N (1 − 2−N )2
for all h ≥ 1, λ ≥ 1, u ≥ hT0 ; here and in the sequel, ωN = meas(B1 (0)) =
2π N/2 /N Γ(N/2);
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SEMICLASSICAL SOLUTIONS
3
(H4’) there exist c0 > 0, T0 > 0 and q ∈ (2, 2∗ ) such that
H(x, 0, v) ≥ c0 |v|q ,
∀x ∈ RN , v ∈ [−T0 , T0 ]
and
v −2 h4−N
Z
H(λ−1/2 x, 0, v/h) dx ≥
|x|≤h
(N 2 + 2)ωN
,
2N (1 − 2−N )2
for all h ≥ 1, λ ≥ 1, v ≥ hT0 ;
(H5) H(x, z) := 12 Hz (x, z) · z − H(x, z) ≥ 0 for all (x, z) ∈ RN × R2 , and there
exist c1 > 0 and κ > max{1, N/2} such that
(1 − θ)m0
Hz (x, z) · z
≥
⇒ |Hz (x, z) · z|κ ≤ c1 |z|2κ H(x, z),
|z|2
3
where m0 := min{a0 , b0 };
(H6’) there exist c0 > 0 and q ∈ (2, 2∗ ) such that H(x, u, 0) ≥ c0 |u|q for all
(x, u) ∈ RN × R;
(H6”) there exist c0 > 0 and q ∈ (2, 2∗ ) such that H(x, 0, v) ≥ c0 |v|q for all
(x, v) ∈ RN × R.
Remark 1.1. It is easy to check that (H6’) and (H6”) imply (H4) and (H4’) with
T0 =
1/(q−2)
N2 + 2
,
−N
2
2c0 (1 − 2 )
respectively, but (H4), (H4’) can not yield (H6’), (H6”). We give the following
nonlinear example to illustrate it. Let
H(x, u, v) = (|u|2 + |v|2 ) ln(1 + |u| + |v|).
Clearly, H satisfies both (H4) and (H4’) with
ln(1 + T0 ) =
N2 + 2
2 1 − 2−N
2 ,
but neither (H6’) nor (H6”).
Example 1.2. Let q ∈ (2, 2∗ ). Then it is easy to see that following two functions
satisfy (H1)–(H3) and (H6’):
q/2
H(x, u, v) = a1 |u|q + a2 |v|q , H(x, u, v) = ζ(x) |u|2 + |v|2
,
where a1 , a2 > 0 and ζ ∈ C(RN ) with 0 < inf RN ζ ≤ supRN ζ < +∞.
Since (q − 2)N − 2q < 0, we can let h0 ≥ 1 be such that
n N 2 + 2(N + 2) oq/(q−2)
(q − 2)ωN
[(q−2)N −2q]/(q−2)
h0
2N q(qc0 )2/(q−2) (N + 2)(1 − 2−N )2
(2κ−N )/2
(1 − θ)κ m0
=
3κ c1 (γ2∗ γ0 )N
(1.3)
,
where γ0 and γ2∗ are embedding constants, see (2.1) and (2.2). If a and b satisfy
(A0), we can choose λ0 > 1 such that
sup
λ1/2 |x|≤2h0
|a(x)| ≤ h−2
0 ,
∀λ ≥ λ0 ,
(1.4)
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S. CHEN, X. H. TANG
EJDE-2014/251
if a and b satisfy (B0), we can choose λ0 > 1 such that
|b(x)| ≤ h−2
0 ,
sup
λ1/2 |x|≤2h
∀λ ≥ λ0 .
(1.5)
0
Letting ε−2 = λ, (1.1) is rewritten as
−∆u + λa(x)u = λHu (x, u, v) + λµ(x)v,
x ∈ RN ,
−∆v + λb(x)v = λHv (x, u, v) + λµ(x)u,
x ∈ RN ,
1
(1.6)
N
u, v ∈ H (R ).
Let
Z
1
2
(|∇u|2 + |∇v|2 + λa(x)|u|2 + λb(x)|v|2 ) dx
Z
Z
−λ
H(x, z) dx − λ
µ(x)uv dx, z = (u, v).
Φλ (z) =
RN
RN
(1.7)
RN
Obviously, the solutions of (1.1) are the critical points of Φε−1/2 (z); the solutions
of (1.6) are the critical points of Φλ (z).
We are now in a position to state the main results of this paper.
Theorem 1.3. Assume that a, b, µ and H satisfy (A0), (A1), (H1)–(H5). Then
−1/2
for 0 < ε ≤ λ0 , (1.1) has a solution zε = (uε , vε ) such that
(2κ−N )/2
0 < Φε−1/2 (zε ) ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
εN −2 ,
(2κ−N )/2
Z
H(x, zε ) dx ≤
RN
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
εN .
Theorem 1.4. Assume that a, b, µ and H satisfy (A1), (B0), (H1)–(H3), (H4’),
−1/2
(H5). Then for 0 < ε ≤ λ0 , (1.1) has a solution zε = (uε , vε ) such that
(2κ−N )/2
0 < Φε−1/2 (zε ) ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
εN −2 ,
(2κ−N )/2
Z
H(x, zε ) dx ≤
RN
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
εN .
Theorem 1.5. Assume that a, b, µ and H satisfy (A0), (A1), (H1)–(H5). Then
for λ ≥ λ0 , (1.6) has a solution zλ = (uλ , vλ ) such that
(2κ−N )/2
0 < Φλ (zλ ) ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
λ1−N/2 ,
(2κ−N )/2
Z
H(x, zλ ) dx ≤
RN
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
λ−N/2 .
Theorem 1.6. Assume that a, b, µ and H satisfy (A1), (B0), (H1)–(H3), (H4’),
(H5). Then for λ ≥ λ0 , (1.6) has a solution zλ = (uλ , vλ ) such that
(2κ−N )/2
0 < Φλ (zλ ) ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
(2κ−N )/2
Z
H(x, zλ ) dx ≤
RN
λ1−N/2 ,
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
λ−N/2 .
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SEMICLASSICAL SOLUTIONS
5
The rest of the article is organized as follows. In Section 2, we provide some
preliminaries and lemmas. In Section 3, we give the proofs of Theorems 1.3–1.6.
2. Preliminaries
Let
E = (u, v) ∈ H 1 (RN ) × H 1 (RN ) :
Z
[a(x)|u|2 + b(x)|v|2 ] dx < +∞ ,
RN
Z
[|∇u|2 + λa(x)|u|2 + |∇v|2 + λb(x)|v|2 ] dx
kzkλ† =
1/2
,
∀z = (u, v) ∈ E.
RN
Analogous to the proof of [23, Lemma 1], by using (A0) or (B0) and the Sobolev
inequality, one can show that there exists a constant γ0 > 0 independent of λ such
that
kzkH 1 (RN ) ≤ γ0 kzkλ† , ∀z ∈ E, λ ≥ 1.
(2.1)
This shows that (E, k · kλ† ) is a Banach space for λ ≥ 1. Furthermore, by the
Sobolev embedding theorem, we have
kzks ≤ γs kzkH 1 (RN ) ≤ γs γ0 kzkλ† ,
∀z ∈ E, λ ≥ 1, 2 ≤ s ≤ 2∗ ,
(2.2)
here and in the sequel, we denote by k · ks the usual norm in space Ls (RN ).
In view of the definition of the norm k · kλ† , we can re-write Φλ in the form
Z
Z
1
H(x, z) dx − λ
µ(x)uv dx, ∀z ∈ E.
(2.3)
Φλ (z) = kzk2λ† − λ
2
RN
RN
It is easy to see that Φλ ∈ C 1 (E, R) and
Z
hΦ0λ (z), z̃i =
[∇u · ∇ũ + ∇v · ∇ṽ + λa(x)uũ + λb(x)vṽ] dx
RN
Z
−λ
[Hu (x, z)ũ + Hv (x, z)ṽ] dx
N
ZR
−λ
µ(x)(uṽ + vũ)] dx, ∀z = (u, v), z̃ = (ũ, ṽ) ∈ E.
(2.4)
RN
As in [20], we let
ϑ(x) :=

1

,


 h0N −1
h0
 1−2−N


0,
|x| ≤ h0 ,
[|x|−N − (2h0 )−N ],
h0 < |x| ≤ 2h0 ,
(2.5)
|x| > 2h0 .
Then ϑ ∈ H 1 (RN ), moreover,
Z
N 2 ωN
2
k∇ϑk2 =
|∇ϑ(x)|2 dx ≤
hN −4 ,
(N + 2)(1 − 2−N )2 0
RN
Z
2ωN
2
kϑk2 =
|ϑ(x)|2 dx ≤
hN −2 .
−N )2 N 0
(1
−
2
N
R
(2.6)
(2.7)
Let eλ (x) = ϑ(λ1/2 x). Then we can prove the following lemma which is used for
our proofs.
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S. CHEN, X. H. TANG
EJDE-2014/251
Lemma 2.1. Let H(x, z) ≥ 0, for all (x, z) ∈ RN × R2 . Suppose that (A0), (A1),
(H1)–(H4) are satisfied. Then
(2κ−N )/2
sup{Φλ (seλ , 0) : s ≥ 0} ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
λ1−N/2 ,
∀λ ≥ λ0 .
(2.8)
Proof. From (H4), (1.3), (1.4) ,(1.7), (2.5), (2.6) and (2.7), we obtain
Φλ (seλ , 0)
Z
Z
s2
(|∇eλ |2 + λa(x)|eλ |2 ) dx − λ
H(x, seλ , 0) dx
=
2 RN
RN
Z
h s2 Z
i
(|∇ϑ|2 + a(λ−1/2 x)|ϑ|2 ) dx −
= λ1−N/2
H(λ−1/2 x, sϑ, 0) dx
2 RN
RN
h s2 k∇ϑk22 + kϑk22 sup |a(λ−1/2 x)|
≤ λ1−N/2
2
|x|≤2h0
Z
i
−
H(λ−1/2 x, s/h0 , 0) dx
(2.9)
|x|≤h0
≤λ
1−N/2
h s2
(k∇ϑk22
2
∀s ≥ 0, λ ≥ λ0 ,
+
2
h−2
0 kϑk2 )
s2
2
(k∇ϑk22 + h−2
0 kϑk2 ) −
2
Z
Z
i
H(λ−1/2 x, s/h0 , 0) dx ,
−
|x|≤h0
H(λ−1/2 x, s/h0 , 0) dx
|x|≤h0
2
s2
(N + 2)ωN N −4
2
≤ [k∇ϑk22 + h−2
h
] ≤ 0, ∀s ≥ h0 T0 , λ ≥ λ0
0 kϑk2 −
2
N (1 − 2−N )2 0
and
Z
s2
−2
2
2
(k∇ϑk2 + h0 kϑk2 ) −
H(λ−1/2 x, s/h0 , 0) dx
2
|x|≤h0
≤
≤
s2
c0 ωN q N −q
2
(k∇ϑk22 + h−2
s h0
0 kϑk2 ) −
2
N
2 q/(q−2)
(q − 2)(k∇ϑk22 + h−2
0 kϑk2 )
−q 2/(q−2)
2q( qc0NωN hN
)
0
n
(q − 2)ωN
N 2 + 2(N + 2) oq/(q−2) [(q−2)N −2q]/(q−2)
≤
h0
2N q(qc0 )2/(q−2) (N + 2)(1 − 2−N )2
(2.10)
(2.11)
(2κ−N )/2
=
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
,
∀0 ≤ s ≤ h0 T0 ,
λ ≥ λ0 .
The conclusion of Lemma 2.1 follows from (2.9), (2.10) and (2.11).
We can prove the following lemma in the same way as Lemma 2.1.
Lemma 2.2. Let H(x, z) ≥ 0 for all (x, z) ∈ RN × R2 . Suppose that (A1), (B0),
(H1)–(H3), (H4’) are satisfied. Then
(2κ−N )/2
sup{Φλ (0, seλ ) : s ≥ 0} ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
λ1−N/2 ,
∀λ ≥ λ0 .
(2.12)
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SEMICLASSICAL SOLUTIONS
7
Applying the mountain-pass lemma without the (PS) condition, by standard
arguments, we can prove the following two lemmas.
Lemma 2.3. Let H(x, z) ≥ 0 for all (x, z) ∈ RN × R2 . Suppose that (A0), (A1),
(H1)–(H4) are satisfied. Then there exist a constant dλ ∈ (0, sups≥0 Φλ (seλ , 0)] and
a sequence {zn } ⊂ E satisfying
Φλ (zn ) → dλ ,
kΦ0λ (zn )kE ∗ (1 + kzn kλ† ) → 0.
(2.13)
Lemma 2.4. Let H(x, z) ≥ 0 for all (x, z) ∈ RN × R2 . Suppose (A1), (B0), (H1)–
(H3), (H4’) are satisfied. Then there exist a constant dλ ∈ (0, sups≥0 Φλ (0, seλ )]
and a sequence {zn } ⊂ E satisfying
Φλ (zn ) → dλ ,
kΦ0λ (zn )kE ∗ (1 + kzn kλ† ) → 0.
(2.14)
Lemma 2.5. Suppose that (A0), (A1), (H1)–(H5) are satisfied. Then any sequence
{zn } ⊂ E satisfying (2.13) is bounded in E.
Proof. We argue by contradiction for proving boundedness of {zn }. Suppose that
kzn kλ† → ∞. Let z̃n = zn /kzn kλ† := (ũn , ṽn ). Then kz̃n kλ† = 1. In view of (A1),
we obtain
Z
Z p
2λ
µ(x)ũn ṽn dx ≤ 2θλ
a(x)b(x)|ũn ṽn | dx
N
RN
Z R
(2.15)
2
2
≤ θλ
[a(x)ũn + b(x)ṽn ] dx ≤ θ.
RN
If
Z
δ := lim sup sup
n→∞ y∈RN
|z̃n |2 dx = 0,
B(y,1)
then by Lions’ concentration compactness principle [21] or [27, Lemma 1.21], z̃n →
(0, 0) in Ls (RN ) for 2 < s < 2∗ . Set
zn · Hz (x, zn )
(1 − θ)m0 ≤
Ωn := x ∈ RN :
,
2
|zn |
3
D := Aa0 ∪ Bb0 .
Hence, from (A0) and the Hölder inequality it follows that
Z
|Hz (x, zn ) · zn |
λ
dx
kzn k2λ†
Ωn
Z
|Hz (x, zn ) · zn |
=λ
|z̃n |2 dx
|zn |2
Ωn
Z
(1 − θ)λm0
≤
|z̃n |2 dx
3
Ωn
Z
Z
(1 − θ)λm0
(1 − θ)λm0
2
≤
|z̃n | dx +
|z̃n |2 dx
3
3
RN \D
D
≤
1−θ
(1 − θ)λm0 [meas(D)]1/(N +1)
kz̃n k2λ† +
3
3
Z
N/(N +1)
×
|z̃n |2(N +1)/N dx
D
1−θ
=
+ o(1).
3
(2.16)
8
S. CHEN, X. H. TANG
From (2.3), (2.4) and (2.13), there holds
Z
dλ + o(1) = λ
EJDE-2014/251
H(x, zn ) dx.
(2.17)
RN
Let κ0 = κ/(κ − 1), then 2 < 2κ0 < 2∗ . By (H5), (2.17) and the Hölder inequality,
one obtains
Z
|Hz (x, zn ) · zn |
dx
λ
kzn k2λ†
N
R \Ωn
Z
|Hz (x, zn ) · zn |
=λ
|z̃n |2 dx
|zn |2
N
R \Ωn
hZ
|H (x, z ) · z | κ i1/κ Z
1/κ0
z
n
n
(2.18)
2κ0
dx
≤λ
|z̃
|
dx
n
|zn |2
RN \Ωn
RN \Ωn
Z
1/κ Z
1/κ0
0
≤ λ c1
H(x, zn ) dx
|z̃n |2κ dx
RN \Ωn
≤λ
1−1/κ
RN
1/κ
[c1 dλ + o(1)]
kz̃n k22κ0
= o(1).
Combining (2.17) with (2.18) and using (2.4), (2.13) and (2.15), we have
kzn k2λ† − hΦ0λ (zn ), zn i
kzn k2λ†
Z
Z
|Hz (x, zn ) · zn |
=λ
+
2λ
µ(x)ũn ṽn dx
kzn k2λ†
RN
RN
Z
Z
|Hz (x, zn ) · zn |
|Hz (x, zn ) · zn |
≤λ
dx + λ
dx + θ
2
kz
k
kzn k2λ†
N
n λ†
Ωn
R \Ωn
1 + o(1) ≤
(2.19)
1 + 2θ
+ o(1).
3
This contradiction shows that δ > 0.
Going
to a subsequence if necessary, we assume the existence of kn ∈ ZN such
R
that B √ (kn ) |z̃n |2 dx > 2δ . Let wn (x) = z̃n (x + kn ); then
1+ N
Z
δ
|wn |2 dx > .
(2.20)
2
√
B1+ N (0)
≤
Now we define ẑn (x) = zn (x + kn ), then ẑn /kzn kλ† = wn and kwn k2H 1 (RN ) =
kz̃n k2H 1 (RN ) . Passing to a subsequence, we have wn * w in H 1 (RN ), wn → w in
Lsloc (RN ), 2 ≤ s < 2∗ and wn → w a.e. on RN . Obviously, (2.20) implies that
w 6= (0, 0). For a.e. x ∈ {z ∈ RN : w(z) 6= (0, 0)}, we have limn→∞ |ẑn (x)| = ∞.
Hence, it follows from (H3), (2.3), (2.13), (2.15) and Fatou’s lemma that
dλ + o(1)
Φλ (zn )
= lim
2
n→∞ kzn k2
n→∞ kzn k
λ†
λ†
Z
Z
h1
i
H(x + kn , ẑn )
2
= lim
kz̃n k2λ† − λ
|w
|
dx
−
λ
µ(x)ũn ṽn dx
n
2
n→∞ 2
|ẑn |
RN
RN
Z
1+θ
H(x + kn , ẑn )
−λ
|wn |2 dx = −∞.
≤
lim inf
2
n→∞
2
|ẑ
|
N
n
R
0 = lim
This contradiction shows that {kzn kλ† } is bounded.
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SEMICLASSICAL SOLUTIONS
9
We can prove the following lemma in the same way as Lemma 2.5.
Lemma 2.6. Suppose that (A1), (B0), (H1)–(H3), (H4’), (H5) are satisfied. Then
any sequence {zn } ⊂ E satisfying (2.14) is bounded in E.
3. Proofs of main results
In this section, we give the proofs of Theorems 1.3–1.6.
Proof of Theorem 1.5. Applying Lemmas 2.1, 2.3 and 2.5, we deduce that there
exists a bounded sequence {zn } ⊂ E satisfying (2.13) with
(2κ−N )/2
dλ ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
λ1−N/2 ,
∀λ ≥ λ0 .
(3.1)
Going to a subsequence, if necessary, we can assume that zn * zλ in (E, k · kλ† )
and Φ0λ (zn ) → 0. Next, we prove that zλ 6= (0, 0).
Arguing by contradiction, suppose that zλ = (0, 0), i.e. zn * (0, 0) in E, and so
zn → (0, 0) in Lsloc (RN ), 2 ≤ s < 2∗ and zn → (0, 0) a.e. on RN . Since D is a set
of finite measure, there holds
kzn k22
Z
Z
2
|zn | dx +
=
RN \D
|zn |2 dx ≤
D
1
kzn k2λ† + o(1).
λm0
(3.2)
For s ∈ (2, 2∗ ), it follows from (2.2), (3.2) and the Hölder inequality that
2(2∗ −s)/(2∗ −2)
kzn kss ≤ kzn k2
≤ (γ2∗ γ0 )2
∗
2∗ (s−2)/(2∗ −2)
kzn k2∗
(s−2)/(2∗ −2)
(λm0 )−(2
∗
−s)/(2∗ −2)
kzn ksλ† + o(1).
(3.3)
According to (3.2), one can obtain that
Z
Hz (x, zn ) · zn
|zn |2 dx
|zn |2
Ωn
(1 − θ)λm0
1−θ
≤
kzn k22 ≤
kzn k2λ† + o(1).
3
3
Z
Hz (x, zn ) · zn dx = λ
λ
Ωn
(3.4)
By (2.3), (2.4) and (2.13), we have
1
Φλ (zn ) − hΦ0λ (zn ), zn i = λ
2
Z
H(x, zn ) dx = dλ + o(1).
RN
(3.5)
10
S. CHEN, X. H. TANG
EJDE-2014/251
Using (H5), (3.1), (3.3) with s = 2κ/(κ − 1) and (3.5), we obtain
Z
λ
Hz (x, zn ) · zn dx
RN \Ωn
|H (x, z ) · z | κ 1/κ
z
n
n
dx
kzn k2s
|zn |2
RN \Ωn
Z
1/κ
∗
∗
≤ (γ2∗ γ0 )2·2 (s−2)/s(2 −2) λ c1
H(x, zn ) dx
≤λ
Z
RN \Ωn
−2(2∗ −s)/s(2∗ −2)
× (λm0 )
1/κ
kzn k2λ†
+ o(1)
∗
1/κ
≤ c1 (γ2∗ γ0 )N/κ λ1−1/κ dλ (λm0 )−2(2
−s)/s(2∗ −2)
kzn k2λ† + o(1)
(3.6)
1/κ
=
c1 (γ2∗ γ0 )N/κ (N −2)/2 1/κ
λ
dλ
kzn k2λ† + o(1)
(2κ−N )/2κ
m0
≤
c1 (γ2∗ γ0 )N/κ (1 − θ)κ m0
(2κ−N )/2κ
3κ c1 (γ2∗ γ0 )N
m
(2κ−N )/2
1/κ
1/κ
kzn k2λ† + o(1)
0
1−θ
=
kzn k2λ† + o(1),
3
which, together with (2.4), (2.13) and (3.4), yields
o(1) = hΦ0λ (zn ), zn i
Z
2
= kzn kλ† − λ
Z
Hz (x, zn ) · zn dx − 2λ
µ(x)un vn dx
RN
Z
Z
(3.7)
≥ (1 − θ)kzn k2λ† − λ
Hz (x, zn ) · zn dx − λ
Hz (x, zn ) · zn dx
RN
RN \Ωn
Ωn
1−θ
kzn k2λ† + o(1).
3
Consequently, it follows from (2.3) and (2.13) that
≥
0 < dλ = lim Φλ (zn ) ≤
n→∞
1+θ
lim kzn k2λ† = 0,
2 n→∞
since H(x, z) ≥ 0, ∀(x, z) ∈ RN × R2 . This contradiction shows zλ 6= (0, 0). By a
standard argument, we easily certify that Φ0λ (zλ ) = 0 and Φλ (zλ ) ≤ dλ . Then zλ
is a nontrivial solution of (1.7), moreover
Z
1
dλ ≥ Φλ (zλ ) = Φλ (zλ ) − hΦ0λ (zλ ), zλ i = λ
H(x, zλ ) dx.
(3.8)
2
RN
Proof of Theorem 1.6. Applying Lemmas 2.2, 2.4 and 2.6, we deduce that there
exists a bounded sequence {zn } ⊂ E satisfying (2.14) with
(2κ−N )/2
dλ ≤
(1 − θ)κ m0
3κ c1 (γ2∗ γ0 )N
λ1−N/2 ,
∀λ ≥ λ0 .
The rest of the proof is the same as Theorem 1.5, so we omit it.
Theorem 1.3 and 1.4 are direct consequences of Theorem 1.5 and 1.6, respectively.
We omit their proofs.
EJDE-2014/251
SEMICLASSICAL SOLUTIONS
11
Acknowledgements. This work is partially supported by the National Natural
Science Foundation of China (No: 11171351).
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Sitong Chen
School of Mathematics and Statistics, Central South University, Changsha, Hunan
410083, China
E-mail address: sitongchen2041@hotmail.com
Xianhua Tang
School of Mathematics and Statistics, Central South University, Changsha, Hunan
410083, China
E-mail address: tangxh@mail.csu.edu.cn