Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 252, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
NONTRIVIAL SOLUTIONS FOR ASYMMETRIC
PROBLEMS ON RN
RUICHANG PEI, JIHUI ZHANG
Abstract. We consider the elliptic equation
−∆u + V (x)u = f (x, u),
x ∈ Rn ,
u ∈ H 1 (RN ), N ≥ 2,
C(RN )
where V (x) ∈
and V (x) ≥ V0 > 0 for all x ∈ RN . The nonlinear term
f exhibits an asymmetric growth at +∞ and −∞ in RN (N ≥ 2). Namely,
it is linear at −∞ and superlinear at +∞. However, it need not satisfy the
Ambrosetti-Rabinowitz condition on the positive semiaxis. Some existence
results for nontrivial solution are established by using the minimax methods
combined with the improved Moser-Trudinger inequality.
1. Introduction
In this article we consider the semilinear elliptic equation
− ∆u + V (x)u = f (x, u),
x ∈ Rn , u ∈ H 1 (RN ), N ≥ 2,
N
(1.1)
N
where V (x) ∈ C(R ) and V (x) ≥ V0 > 0 for all x ∈ R . This equation arises from
many physical and chemical problems. By using the variational methods, there
were many papers studying the existence and multiplicity of solutions for problem
(1.1). Most of them treated the superlinear case (see [1, 2, 3]) and some deal with
the asymptotically linear case (see [4, 5, 6, 7, 8, 9]).
The main difficulty in dealing with this class of problem is the lack of compactness
due to the fact that the domain is unbounded. This was overcome in [10] by
assuming that the potential V is coercive. Such condition was generalized in [2] by
assuming
(V1) for every M > 0, µ({x ∈ RN : V (x) ≤ M }) < ∞, with µ denoting the
Lebesgue measure in RN .
Actually, the above hypothesis implies that the eigenvalue problem
−∆u + V (x)u = λu, x ∈ RN ,
possesses a sequence of positive eigenvalues: 0 < λ1 < λ2 < λ3 · · · < λk < · · · → ∞
with finite multiplicity for each λk . The principal eigenvalue λ1 is simple with
positive eigenfunction ϕ1 , and eigenfunction ϕk corresponding to λk (k ≥ 2) is
sign-changing.
2000 Mathematics Subject Classification. 35J60, 35J20, 35B38.
Key words and phrases. Schrödinger equation; asymmetric problems; one side resonance;
subcritical exponential growth.
c
2014
Texas State University - San Marcos.
Submitted October 5, 2014. Published December 4, 2014.
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R. PEI, J. ZHANG
EJDE-2014/252
In the present paper, motivated by [11, 12, 13, 14, 15, 16], our main purpose is to
establish existence results of nontrivial solution for problem (1.1) with N ≥ 2 when
the nonlinear term exhibits an asymmetric behavior as t ∈ R approaches +∞ and
−∞. More precisely, we assume that for a.e. x ∈ RN , f (x, .) grows superlinearly
at +∞, while at −∞ it has a linear growth. To our knowledge, this asymmetric
problem is rarely considered by other people.
In case of N ≥ 3, we noticed that almost all of the above mentioned works
involve the nonlinearity term f (x, u) of a subcritical (polynomial) growth, say,
(SCP) There exist positive constants c1 and c2 and q0 ∈ (1, 2∗ − 1) such that
|f (x, t)| ≤ c1 + c2 |t|q0
for all t ∈ R and x ∈ RN ,
where 2∗ = 2N/(N − 2) denotes the critical Sobolev exponent.
One of the main reasons to assume this condition (SCP) is that they can use the
Sobolev compact embedding theory.
Over the years, many researchers studied problem (1.1) by trying to drop the
condition (AR) (see [2]), see for instance [5, 6, 7, 8, 9].
In this paper, our first main results will be to study problem (1.1) in the improved
subcritical polynomial growth
(SCPI) For ε > 0, there exists positive constant C(ε) such that
∗
|f (x, t)| ≤ C(ε) + ε|t|2
−1
for all t ∈ R and x ∈ RN ,
which is weaker than (SCP).
Note that in this case, we do not have the Sobolev compact embedding anymore.
Our work is to study asymmetric problem (1.1) without the (AR)-condition in the
positive semiaxis. In fact, this condition was studied by Liu and Wang in [17] in
the case of Laplacian by the Nehari manifold approach. However, we will use the
Mountain Pass Theorem and a suitable version of the Mountain Pass Theorem to
get the nontrivial solution to problem (1.1) in the general case N ≥ 3. Our proof
of compactness condition is completely different from those in [14, 15, 16].
Let us now state our main results: Suppose that f (x, t) ∈ C(RN × R) and
satisfies:
(H1) limt→0 f (x,t)
= f0 uniformly for a.e. x ∈ RN , where f0 ∈ [0, +∞);
t
(H2) limt→−∞ f (x,t)
= l uniformly for a.e. x ∈ RN , where l ∈ [0, +∞];
t
f (x,t)
(H3) limt→+∞ t = +∞ uniformly for a.e. x ∈ RN ;
(H4) f (x,t)
is non-increasing with respect to t ≤ 0, for a.e. x ∈ RN .
t
Let H := H 1 (RN ) be the Sobolev space with the norm
Z
1/2
kukH :=
(|∇u|2 + u2 ) dx
.
RN
In our problem, the work space E is defined by
Z
E := u ∈ H :
(|∇u|2 + V (x)u2 ) dx < ∞ .
RN
Thus, E is a Hilbert space with the inner product
Z
(u, v)E :=
(∇u∇v + V (x)uv) dx
RN
and is defined by k · k the associated norm.
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NONTRIVIAL SOLUTIONS
3
We denote by | · |p the usual Lp -norm. The condition (V1 ) and the Sobolev Embedding Theorem imply that the immersion E ,→ Ls (RN , R) (N ≥ 3) is continuous
for 2 ≤ s ≤ 2∗ . Actually it is proved in [2] that this embedding is compact for
2 ≤ s < 2∗ .
Recall that a function u ∈ E is called a weak solution of (1.1) if
Z
Z
(∇u∇v + V (x)uv) dx =
f (x, u)v dx, ∀v ∈ E.
RN
RN
Seeking a weak solution of problem (1.1) is equivalent to finding a critical point
u∗ of C 1 functional
Z
1
F (x, u) dx, ∀u ∈ E,
(1.2)
I(u) := kuk2 −
2
RN
Ru
where F (x, u) = 0 f (x, s)ds. Then
Z
Z
0 ∗
∗
∗
hI (u ), vi =
(∇u ∇v + V (x)u v) dx −
f (x, u∗ )v dx = 0, ∀v ∈ E.
RN
RN
Definition 1.1. Let (E, k · kE ) be a real Banach space with its dual space (E ∗ , k ·
kE ∗ ) and I ∈ C 1 (E, R). For c ∈ R, we say that I satisfies the (P S)c condition if
for any sequence {xn } ⊂ E with
I(xn ) → c,
in E ∗ ,
DI(xn ) → 0
there is a subsequence {xnk } such that {xnk } converges strongly in E. Also, we
say that I satisfies the (C)c condition if for any sequence {xn } ⊂ E with
I(xn ) → c,
kDI(xn )kE ∗ (1 + kxn kE ) → 0,
there is subsequence {xnk } such that {xnk } converges strongly in E.
We have the following version of the Mountain Pass Theorem (see [18, 19]).
Proposition 1.2. Let E be a real Banach space and suppose that I ∈ C 1 (E, R)
satisfies the condition
max{I(0), I(u1 )} ≤ α < β ≤ inf I(u),
kuk=ρ
for some α < β, ρ > 0 and u1 ∈ E with ku1 k > ρ. Let c ≥ β be characterized by
c = inf max I(γ(t)),
γ∈Γ 0≤t≤1
where Γ = {γ ∈ C([0, 1], E), γ(0) = 0, γ(1) = u1 } is the set of continuous paths
joining 0 and u1 . Then, there exists a sequence {un } ⊂ E such that
I(un ) → c ≥ β and (1 + kun k)kI 0 (un )kE ∗ → 0
as n → ∞.
Theorem 1.3. Let N ≥ 3 and assume that f has the improved subcritical polynomial growth on RN (condition (SCPI)) and satisfies (H1)–(H3). If f0 < λ1 < l <
∞, then problem (1.1) has at least one nontrivial solution.
Remark 1.4. In view of conditions (SCPI), (H2) and (H3), problem (1.1) with
the improved subcritical polynomial growth is called asymmetric. Hence, Theorem
1.3 is completely different from results contained in [4, 5, 6, 7, 8, 9].
Theorem 1.5. Let N ≥ 3 and assume that f has the improved subcritical polynomial growth on RN (condition (SCPI)) and satisfies (H1)–(H3). If f0 < λ1 = l and
limt→−∞ [f (x, t)t − 2F (x, t)] = −∞ uniformly for a.e. x ∈ RN , then problem (1.1)
has at least one nontrivial solution.
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R. PEI, J. ZHANG
EJDE-2014/252
Remark 1.6. When l = λ1 , problem (1.1) is called resonant at negative infinity.
This case is completely new. Here, we also give an example for f (x, t). It satisfies
our conditions (H1)–(H3) and (SCPI).
Example Define
(
f (x, t) =
g(t)t,
t ≤ 0,
g(t)t + h(t), t > 0,
where g(t) ∈ C(R), g(0) = 0; g(t) ≥ 0, t ∈ R; h(t) ∈ C[0, +∞); limt→+0 h(t)
t = 0;
h(t)
=
+∞.
Moreover,
there
exists
t
>
0
such
that
limt→+∞ th(t)
=
0;
lim
0
t→+∞ t
2∗ −1
g(t) ≡ λ1 for all |t| ≥ t0 .
Theorem 1.7. Let N ≥ 3 and assume that f has the improved subcritical polynomial growth on RN (condition (SCPI)) and satisfies (H1)–(H4). If f0 < λ1 and
l = +∞, then problem (1.1) has at least one nontrivial solution.
Remark 1.8. When l = +∞, problem (1.1) is generalized superlinearity at negative
infinity.
In the case N = 2, we have 2∗ = +∞. In this case, every polynomial growth is
admitted. Hence, one is led to look for a function g(s) : R → R+ with maximal
growth such that
Z
g(u) dx < ∞.
sup
u∈H,kukH ≤1
RN
It was shown by Trudinger [20], Moser [21] and Ruf [22] that the maximal growth
is of exponential type. So, we must redefine the subcritical (exponential) growth in
this case as follows:
(SCE): f has subcritical (exponential) growth on RN , i.e., For ε > 0, there exists
positive constant C ∗ (ε) such that
|f (x, t)| ≤ C ∗ (ε) + ε exp(α|t|2 )
for all t ∈ R, x ∈ RN and α > 0.
When N = 2 and f has the subcritical (exponential) growth (SCE), our work is
still to study asymmetric problem (1.1) without the (AR)-condition in the positive
semiaxis. To our knowledge, this problem is rarely studied by other people. Hence,
our results are completely new and our methods are skillful since we skillfully combined Mountain Pass Theorem with Moser-Trudinger inequality. Our results are as
follows:
Theorem 1.9. Let N = 2 and assume that f has the subcritical exponential growth
on RN (condition (SCE)) and satisfies (H1)–(H3). If f0 < λ1 < l < ∞, then
problem (1.1) has at least one nontrivial solution.
Remark 1.10. In view of the conditions (H2), (H3) and (SCE), problem (1.1) is
called asymmetric subcritical exponential problem. Hence, Theorem 1.9 is completely different from results contained in [1, 2, 3, 4, 5, 6, 7, 8, 9].
Theorem 1.11. Let N = 2 and assume that f has the subcritical exponential
growth on RN (condition (SCE)) and satisfies (H1)–(H3). If f0 < λ1 = l and
limt→−∞ [f (x, t)t − 2F (x, t)] = −∞ uniformly for a. e. x ∈ RN , then problem (1.1)
has at least one nontrivial solution.
Remark 1.12. When l = λ1 , problem (1.1) is called resonant at negative infinity.
This case is completely new.
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NONTRIVIAL SOLUTIONS
5
Theorem 1.13. Let N = 2 and assume that f has the subcritical exponential
growth on RN (condition (SCE)) and satisfies (H1)–(H4). If f0 < λ1 and l = +∞,
then problem (1.1) has at least one nontrivial solution.
2. Preliminaries
Lemma 2.1. Let N ≥ 3 and ϕ1 > 0 be a λ1 -eigenfunction with kϕ1 k = 1 and
assume that (H1)–(H3) and (SCPI) hold. If f0 < λ1 < l ≤ +∞, then
(i) There exist ρ, α > 0 such that I(u) ≥ α for all u ∈ E with kuk = ρ,
(ii) I(tϕ1 ) → −∞ as t → +∞.
Proof. By (SCPI) and (H1)–(H3), for any ε > 0, there exist A1 = A1 (ε), B1 =
B1 (ε) such that for all (x, s) ∈ RN × R,
∗
1
F (x, s) ≤ (f0 + ε)|s|2 + A1 |s|2 .
(2.1)
2
Choose ε > 0 such that (f0 + ε) < λ1 . By (2.1), the continuous imbedding and the
∗
∗
Sobolev inequality: |u|22∗ ≤ Kkuk2 , we obtain
∗
f0 + ε 2
1
kuk2 −
|u|2 − A1 |u|22∗
2
2
∗
f0 + ε
1
≥ (1 −
)kuk2 − A1 Kkuk2 .
2
λ1
So, part (i) is proved if we choose kuk = ρ > 0 small enough.
On the other hand, by the definition of I and (H2) with l > λ1 , we have
I(u) ≥
1
I(tϕ1 )
≤ (λ1 − l)|ϕ1 |22 < 0.
t2
2
By a slight modification to the proof above, we can prove (ii) if l = +∞.
lim
t→−∞
Lemma 2.2 ([20, 21, 22]). Let u ∈ H. Then
Z
sup
(exp α|u|2 − 1) dx ≤ C
u∈H,kukH ≤1
for α ≤ 4π 2 .
RN
The inequality is sharp: for any α > 4π 2 the corresponding supremum is +∞.
Lemma 2.3. Let N = 2 and ϕ1 > 0 be a λ1 -eigenfunction with kϕ1 k = 1 and
assume (H1)–(H3) and (SCE) hold. If f0 < λ1 < l ≤ +∞, then
(i) There exist ρ, α > 0 such that I(u) ≥ α for all u ∈ H with kuk = ρ,
(ii) I(tϕ1 ) → −∞ as t → +∞.
Proof. By (SCE) and (H1)–(H3), for any ε > 0, there exist A1 = A1 (ε), B1 = B1 (ε),
κ > 0 and q > 2 such that for all (x, s) ∈ RN × R,
1
F (x, s) ≤ (f0 + ε)|s|2 + A1 (exp(κ|s|2 ) − 1)|s|q .
(2.2)
2
Choose ε > 0 such that (f0 + ε) < λ1 . By (2.2), the Holder inequality and the
Moser-Trudinger embedding, we obtain
Z
1
f0 + ε 2
2
I(u) ≥ kuk −
|u|2 − A1
(exp(κ|u|2 ) − 1)|u|q dx
2
2
RN
Z
1/r
1
f0 + ε
|u| 2
2
≥ (1 −
)kuk − A1
(exp(κrkuk2H (
) − 1)) dx
2
λ1
kukH
RN
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R. PEI, J. ZHANG
×
Z
EJDE-2014/252
1/r0
0
|u|r q dx
Ω
f0 + ε
1
)kuk2 − Ckukq ,
≥ (1 −
2
λ1
where r > 1 sufficiently close to 1, kukH ≤ σ and κrσ 2 < 4π 2 . So, part (i) is proved
if we choose kuk = ρ > 0 small enough.
On the other hand, by the definition of I and (H2) with l > λ1 , we have
1
λ1 − l
I(tϕ1 )
≤ (λ1 − l)|ϕ1 |22 =
<0
2
t→−∞
t
2
2λ1
By a slight modification to the proof above, we can prove (ii) if l = +∞.
lim
Lemma 2.4. For the functional I defined by (1.2), if un (x) ≤ 0 a.e. x ∈ RN ,
n ∈ N and
hI 0 (un ), un i → 0 quadas n → ∞,
then there exists subsequence, still denoted by {un }, such that
1 + t2
+ I(un ) for all t ≥ 0 and n ∈ N.
2n
Proof. This lemma is essentially due to [7]. For the sake of completeness, we prove
it here. By hI 0 (un ), un i → 0 as n → ∞, for a suitable subsequence, we may assume
that
Z
1
1
0
2
− < hI (un ), un i = kun k −
f (x, un (x))un dx <
for all n.
(2.3)
n
n
RN
We claim that for any t ≥ 0 and n ∈ N,
Z
1
t2
+
{ f (x, un (x))un − F (x, un (x))} dx.
(2.4)
I(tun ) ≤
2n
RN 2
I(tun ) ≤
Indeed, for any t ≥ 0, at fixed x ∈ RN and n ∈ N if we set
1
h(t) = t2 f (x, un )un (x) − F (x, tun (x)),
2
then
h0 (t) = tf (x, un )un (x) − f (x, tun )un (x)
= tun (x){f (x, un ) − f (x, tun (x))/t}
(
≥ 0 for 0 < t ≤ 1
≤ 0 for t ≥ 1
by (H4); hence h(t) ≤ h(1) for all t ≥ 0. Therefore,
Z
1
I(tun ) = t2 kun k2 −
F (x, tun (x)) dx
2
N
Z R
Z
1 2 1
f (x, un (x))un (x) dx} −
F (x, tun (x)) dx
< t { +
2
n
RN
RN
Z
t2
1
≤
+
{ t2 f (x, un (x))un (x) − F (x, tun (x))} dx
2n
2
N
R
Z
2
t
1
≤
+
{ f (x, un (x))un (x) − F (x, un (x))} dx
2n
2
N
R
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NONTRIVIAL SOLUTIONS
and our claim (2.4) is proved.
On the other hand,
Z
1
I(un ) = kun k2 −
F (x, un (x)) dx
2
N
Z R
Z
1
1
≥ {− +
f (x, un (x))un (x) dx} −
F (x, un (x)) dx;
2
n
RN
RN
that is,
Z
1
1
+ I(un ).
{ f (x, un (x))un (x) − F (x, un (x))} dx ≤
2
2n
N
R
Combining (2.4) and (2.5), we find that
I(tun ) ≤
1 + t2
+ I(un ) for all t ≥ 0 and n ∈ N.
2n
7
(2.5)
(2.6)
3. Proofs of the main results
We prove only Theorems 1.3, 1.5, 1.7 and 1.9. Others followed from these results.
Proof of Theorem 1.3. By Lemma 2.1, the geometry conditions of Mountain Pass
Theorem hold. So, we only need to verify condition (PS). Let {un } ⊂ E be a (PS)
sequence such that for every n ∈ N,
Z
1
kun k2 −
(3.1)
F (x, un ) dx ≤ c,
2
RN
Z
Z
Z
∇un ∇v dx +
V (x)un v dx −
f (x, un )v dx ≤ εn kvk, v ∈ E, (3.2)
RN
RN
RN
where c > 0 is a positive constant and {εn } ⊂ R+ is a sequence which converges to
zero.
Step 1. To prove that {un } has a convergence subsequence, we first show that
it is a bounded sequence. To do this, we argue by contradiction assuming that for
a subsequence, which we follow denoting by {un }, we have
kun k → +∞
as n → ∞.
Without loss of generality, we can assume kun k > 1 for all n ∈ N and define
zn = kuunn k . Obviously, kzn k = 1 for all n ∈ N and then, it is possible to extract a
subsequence (denoted also by {zn }) such that
zn * z0
zn → z0
2
in E,
2
(3.3)
N
in L (R ),
(3.4)
N
zn (x) → z0 (x)
a.e. x ∈ R ,
(3.5)
|zn (x)| ≤ q(x)
a.e. x ∈ RN ,
(3.6)
N
where z0 ∈ E and q ∈ L (R ). Dividing both sides of (3.2) by kun k, we obtain
Z
Z
Z
f (x, un )
εn
|
∇zn ∇v dx +
V (x)zn v dx −
v dx| ≤
kvk for all v ∈ E.
ku
k
ku
N
N
N
n
nk
R
R
R
Passing to the limit we deduce from (3.3) that
Z
Z
Z
f (x, un )
v dx =
∇z0 ∇v dx +
V (x)z0 v dx
lim
n→∞ RN
kun k
RN
RN
(3.7)
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R. PEI, J. ZHANG
EJDE-2014/252
for all v ∈ E.
Now we claim that z0 (x) ≤ 0 for a.e. x ∈ RN . To verify this, let us observe that
by choosing v = z0+ = max{z0 , 0} in (3.7) we have
Z
Z
Z
f (x, un )
lim
z0 dx =
|∇z0 |2 dx +
V (x)|z0 |2 dx < +∞,
(3.8)
n→∞ Θ kun k
Θ
Θ
where Θ = {x ∈ RN |z0 (x) > 0}. On the other hand, from conditions (SCPI), (H1),
(H2), (H3), (3.5) and (3.6), we have
f (x, un (x))
z0 (x) ≥ (−K1 )q(x)z0 (x),
kun k
a.e. x ∈ Θ
for some positive constant K1 > 0, and
lim
n→∞
f (x, un (x))
f (x, un (x))
z0 (x) = lim
zn (x)z0 (x) = +∞,
n→∞
kun k
un
a.e. x ∈ Θ.
Therefore, if |Θ| > 0, by the Fatou’s Lemma, we obtain
Z
f (x, un (x))
z0 (x) dx = +∞,
lim
n→∞ Θ
kun k
which contradicts (3.8). Thus |Θ| = 0 and the claim is proved.
n )|
Clearly, z0 (x) 6≡ 0. By (H2), there exists c > 0 such that |f (x,u
≤ c for a.e.
|un |
N
x ∈ R . By using Lebesgue dominated convergence theorem in (3.7), we have
Z
Z
Z
∇z0 ∇v dx +
V (x)z0 v dx −
lz0 v dx = 0
(3.9)
RN
RN
RN
for all v ∈ E. This contradicts our assumption, i.e., l > λ1 .
Step 2. Now, we prove that {un } has a convergence subsequence. In fact, we
can suppose that
un * u in E,
un → u in Lq (RN ), ∀1 ≤ q < 2∗ ,
a.e. x ∈ RN .
un (x) → u(x)
Since f has the subcritical growth on RN , for every > 0, we can find a constant
C() > 0 such that
∗
f (x, s) ≤ C() + |s|2
then we obtain
Z
|
f (x, un )(un − u) dx|
RN
Z
Z
≤ C()
|un − u| dx + RN
|un − u| dx + N
∗
RN
|un − u| dx + C.
RN
∀(x, s) ∈ RN × R,
|un − ukun |2
Z
ZR
≤ C()
,
−1
dx
RN
Z
≤ C()
−1
(|un |2
∗
−1
2∗2−1
Z
∗
2∗
) 2∗ −1 dx
RN
∗
|un − u|2
1/2∗
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NONTRIVIAL SOLUTIONS
9
R
Similarly, since un * u in E, RN |un − u| dx → 0. Since > 0 is arbitrary, we can
conclude that
Z
(f (x, un ) − f (x, u))(un − u) dx → 0 as n → ∞.
(3.10)
RN
By (3.2), we have
hI 0 (un ) − I 0 (u), (un − u)i → 0
From (3.10) and (3.11), we obtain
Z
Z
(∇un − ∇u)(∇un − ∇u) +
RN
as n → ∞.
V (x)|un − u|2 dx → 0
(3.11)
as n → ∞.
(3.12)
RN
We have un → u in E which means that I satisfies (PS).
Proof of Theorem 1.5. Since l = λ1 , obviously, Lemma 2.1 (i) holds. We only need
to show that Lemma 2.1 (ii) holds. Let u = −tϕ1 , then
Z
Z
1 2
2
2
(|∇ϕ1 | + V (x)|ϕ1 | ) dx −
F (x, −tϕ1 ) dx
I(−tϕ1 ) = t
2
N
RN
ZR
Z
1
1
(|∇ϕ1 |2 + V (x)|ϕ1 |2 ) dx −
f (x, −tϕ1 )(−tϕ1 ) dx
= t2
2
2
N
R
RN
Z
f (x, −tϕ1 )tϕ1
−
{F (x, −tϕ1 ) +
} dx.
2
N
R
Since f (x, s) = λ1 s + ◦(s) as s → −∞, we have
I(−tϕ1 ) → −∞ as t → +∞
and the claim is proved. By Proposition 1.2, there exists a sequence {un } ⊂ E such
that
Z
1
I(un ) = kun k2 −
F (x, un ) dx = c + ◦(1),
(3.13)
2
RN
(1 + kun k)kI 0 (un )kE ∗ → 0 as n → ∞.
(3.14)
Clearly, (3.14) implies
hI 0 (un ), un i = kun k2 −
Z
f (x, un (x))un dx = ◦(1).
(3.15)
RN
To complete our proof, we need to verify that {un } is bounded in E. Similar to the
proof of Theorem 1.3, we have z0 (x) ≤ 0, x ∈ Ω, z0 (x) 6≡ 0 and
Z
Z
(∇z0 ∇v + V (x)z0 v) dx −
lz0 v dx = 0
RN
RN
for all v ∈ E. By the maximum principle, z0 < 0 is an eigenfunction of λ1 then
|un (x)| → ∞ for a.e. x ∈ RN . By our assumptions, we have
lim (f (x, un (x))un (x) − 2F (x, un (x))) = −∞
n→∞
uniformly in x ∈ RN , which implies that
Z
(f (x, un (x))un (x) − 2F (x, un (x))) dx → −∞ as n → ∞.
RN
On the other hand, (3.15) implies that
2I(un ) − hI 0 (un ), un i → 2c
as n → ∞.
(3.16)
10
R. PEI, J. ZHANG
EJDE-2014/252
Thus
Z
(f (x, un )un − 2F (x, un )) dx → 2c as n → ∞,
RN
which contradicts (3.16). Hence {un } is bounded. According to the Step 2 proof
of Theorem 1.3, we have un → u in E which means that I satisfies (C)c .
Proof. Proof of Theorem 1.7] By Lemma 2.1 and Proposition 1.2 (3.13)-(3.15) hold.
We still can prove that {un } is bounded in E. Assume kun k → +∞ as n → ∞.
Similar to the proof of Theorem 1.3, we have z0 (x) ≤ 0 and when z0 (x) < 0,
un = zn kun k → −∞ as n → ∞. Let
√
√
2 c
2 cun
sn =
, wn = sn un =
.
(3.17)
kun k
kun k
Since {wn } is bounded in E, it is possible to extract a subsequence (denoted also
by {wn }) such that
wn * w0
wn → w0
in E,
2
(3.18)
N
in L (R ),
(3.19)
N
wn (x) → w0 (x)
a.e. x ∈ R ,
(3.20)
|wn (x)| ≤ h(x)
a.e. x ∈ RN ,
(3.21)
where w0 ∈ E and h ∈ L2 (RN ).
If kun k → +∞ as n → ∞, then w0 (x) ≡ 0. In fact, letting Θ− = {x ∈ Ω :
w0 (x) < 0} and noticing l = +∞, from (H3) it follows that
f (x, un )
≥M
un
uniformly for all x ∈ Θ− ,
where M is a constant, large enough . Therefore, by (3.15) and (3.17), we have
4c = lim kwn k2
n→∞
Z
f (x, un )
= lim
|wn |2 dx
n→∞ RN
un
Z
f (x, un )
|wn |2 dx
≥ lim
n→∞ Θ−
un
Z
≥M
|w0 |2 dx.
Θ−
N
So w0 ≡ 0 for a.e. x ∈ R . But, if w0 ≡ 0, then
I(wn ) =
R
RN
F (x, wn ) dx → 0. Hence
1
kwn k2 + ◦(1) = 2c + ◦(1).
2
(3.22)
On the other hand, since kun k → ∞ as n → ∞, we have sn → 0 as n → ∞. From
Lemma 2.4 and (3.13), we obtain
I(wn ) = I(sn un ) ≤
1 + (sn )2
+ I(un ) ≤ c,
2n
as n → ∞.
Obviously, it contradicts (3.22). So {un } is bounded in E. According to the Step 2
proof of Theorem 1.3, we have un → u in E which means that I satisfies (C)c . EJDE-2014/252
NONTRIVIAL SOLUTIONS
11
Proof of Theorem 1.9. By Lemma 2.3, the geometry conditions of Mountain Pass
Theorem hold. So, we only need to verify condition (PS). Similar to the Step 1 proof
of Theorem 1.3, we easily know that (PS) sequence {un } is bounded in E. Next,
we prove that {un } has a convergence subsequence. Without loss of generality,
suppose that
kun k ≤ β,
un * u in E,
un → u in Lq (RN ), ∀q ≥ 1,
un (x) → u(x)
a.e. x ∈ RN .
Now, since f has the subcritical exponential growth (SCE) on RN , we can find a
constant Cβ0 > 0 such that
αN
|f (x, t)| ≤ Cβ0 (exp( 2 |t|2 ) − 1), ∀(x, t) ∈ RN × R,
2β0
where β0 = γβand γ is defined
kukH ≤ γkuk,
u ∈ E.
Thus, by the Moser-Trudinger inequality (see Lemma 2.2),
Z
|
f (x, un )(un − u) dx|
RN
Z
1/2
αN
(exp( 2 |un |2 ) − 1) dx
≤C
|un − u|2
β0
RN
Z
1/2
αN
un 2
(exp( 2 kun k2H |
≤C
| ) − 1) dx
|un − u|2
β0
kun kH
RN
≤ C|un − u|2 → 0.
Similar to the proof of Theorem 1.3, we have un → u in E which means that I
satisfies (PS).
Acknowledgements. This work was supported by the National Natural Science
Foundation of China (Grant No. 11101319) and Planned Projects for Postdoctoral
Research Funds of Jiangsu Province (Grant No. 1301038C).
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Ruichang Pei
School of Mathematics and Statistics, Tianshui Normal University, Tianshui 741001,
China
E-mail address: prc211@163.com
Jihui Zhang
School of Mathematics and Computer Sciences, Nanjing, Normal University, Nanjing
210097, China
E-mail address: zhangjihui@njnu.edu.cn