Electronic Journal of Differential Equations, Vol. 2008(2008), No. 98, 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)
EXISTENCE OF SOLUTIONS FOR A RESONANT PROBLEM
UNDER LANDESMAN-LAZER CONDITIONS
´ C ANH NGÔ, HOANG QUOC TOAN
QUÔ
Abstract. This article shows the existence of weak solutions in W01 (Ω) to a
class of Dirichlet problems of the form
− div(a(x, ∇u)) = λ1 |u|p−2 u + f (x, u) − h
in a bounded domain Ω of RN . Here a satisfies
`
´
|a(x, ξ)| ≤ c0 h0 (x) + h1 (x)|ξ|p−1
p
for all ξ ∈ RN , a.e. x ∈ Ω, h0 ∈ L p−1 (Ω), h1 ∈ L1loc (Ω), h1 (x) ≥ 1 for a.e.
x in Ω; λ1 is the first eigenvalue for −∆p on Ω with zero Dirichlet boundary
condition and g, h satisfy some suitable conditions.
1. Introduction
Let Ω be a bounded domain in RN . In the present paper we study the existence
of weak solutions of the following Dirichlet problem
− div(a(x, ∇u)) = λ1 |u|p−2 u + f (x, u) − h
p−1
(1.1)
N
where |a(x, ξ)| ≤ c0 (h0 (x) + h1 (x)|ξ| ) for any ξ in R and a.e. x ∈ Ω, h0 (x) ≥ 0
and h1 (x) ≥ 1 for any x in Ω. λ1 is the first eigenvalue for −∆p on Ω with zero
Dirichlet boundary condition. We define X := W01,p (Ω) as the closure of C0∞ (Ω)
under the norm
1/p
Z
|∇u|p dx
.
kuk =
Ω
It is well-known that
Z
λ1 =
Z
|∇u| dx :
inf
u∈W01,p (Ω)
p
Ω
|u| dx = 1 .
p
Ω
Recall that λ1 is simple and positive. Moreover, there exists a unique positive
eigenfunction φ1 whose norm in W01,p (Ω) equals to one. Regarding the functions
f , we assume that f is a bounded Carathéodory function. We also assume that
0
p
h ∈ Lp (Ω) where p0 = p−1
.
p
In the present paper, we study the case in which h0 and h1 belong to L p−1 (Ω)
and L1loc (Ω), respectively. The problem now may be non-uniform in sense that the
functional associated to the problem may be infinity for some u in X. Hence, weak
2000 Mathematics Subject Classification. 35J20, 35J60, 58E05.
Key words and phrases. p-Laplacian; Non-uniform; Landesman-Laser type; Divergence form.
c
2008
Texas State University - San Marcos.
Submitted March 24, 2008. Published July 25, 2008.
1
2
Q. A. NGÔ, H. Q. TOAN
EJDE-2008/98
solutions of the problem must be found in some suitable subspace of X. To our
knowledge, such problems were firstly studied by [9, 16, 15]. In order to state our
main theorem, let us introduce our hypotheses on the structure of problem (1.1).
Assume that N ≥ 1 and p > 1. Ω be a bounded domain in RN having C 2
boundary ∂Ω. Consider a : RN × RN → RN , a = a(x, ξ), as the continuous
derivative with respect to ξ of the continuous function A : RN × RN → R, A =
. Assume that there are a positive real number
A(x, ξ), that is, a(x, ξ) = ∂A(x,ξ)
∂ξ
c0 and two nonnegative measurable functions h0 , h1 on Ω such that h1 ∈ L1loc (Ω),
p
h0 ∈ L p−1 (Ω), h1 (x) ≥ 1 for a.e. x in Ω.
Suppose that a and A satisfy the hypotheses below
(A1) |a(x, ξ)| ≤ c0 (h0 (x) + h1 (x)|ξ|p − 1) for all ξ ∈ RN , a.e. x ∈ Ω.
(A2) There exists a constant k1 > 0 such that
ξ + ψ 1
1
A x,
≤ A(x, ξ) + A(x, ψ) − k1 h1 (x)|ξ − ψ|p
2
2
2
for all x, ξ, ψ, that is, A is p-uniformly convex
(A3) A is p-subhomogeneous, that is,
0 ≤ a(x, ξ)ξ ≤ pA(x, ξ)
for all ξ ∈ RN , a.e. x ∈ Ω.
(A4) There exists a constant k0 ≥ 1/p such that
A(x, ξ) ≥ k0 h1 (x)|ξ|p
for all ξ ∈ RN , a.e. x ∈ Ω.
(A5) A(x, 0) = 0 for all x ∈ Ω.
A special case of (1.1) is the following equation involving the p-Laplacian operator
− div(|∇u|p−2 ∇u) = λ1 |u|p−2 u + f (x, u) − h
(1.2)
We refer the reader to [9, 11, 12, 15, 16] for more examples. We suppose also that
there exists
(H1)
lim f (x, s) = f−∞ (x) ,
lim f (x, s) = f +∞ (x)
x→−∞
x→+∞
for almost every x ∈ Ω.
As is well-known, under (H1), problem (1.2) may not have solutions. In [3, 5, 4],
the existence of solutions of (1.2) is shown provided that one of the following two
conditions are satisfied
(H2)
Z
Z
Z
f +∞ (x)φ1 (x) dx <
h(x)φ1 (x) dx <
f−∞ (x)φ1 (x) dx.
Ω
Ω
Ω
(H2’)
Z
Z
f−∞ (x)φ1 (x) dx <
Ω
Z
h(x)φ1 (x) dx <
Ω
f +∞ (x)φ1 (x) dx.
Ω
It should be noticed that though conditions (H2) and (H2’) look rather similar, the
existence proof in [4] is different in each case. Indeed, under (H2) the functional
associated with the problem is coercive and achieves a minimum, whereas under
(H2’) the functional has the geometry of the saddle point theorem. In the present
EJDE-2008/98
EXISTENCE OF SOLUTIONS FOR A RESONANT PROBLEM
3
paper, we only consider the case (H2), the case (H2’) is still an open question due
to the fact that there are some difficulties in verifying geometric conditions of the
saddle point theorem.
We also point out that in that papers, the property pA(x, ξ) = a(x, ξ) · ξ, which
may not hold under our assumptions by (A4), play an important role in the arguments. In this paper, we shall extend some results in [3, 5, 4] in two directions: one
is from p-Laplacian operators to general elliptic operators in divergence form and
the other is to the case on non-uniform problem.
Since problem (1.1) may be non-uniform, then we must consider the problem in
a suitable subspace of X. In fact, we consider the following subspace of W01,p (Ω)
Z
1,p
p
E = u ∈ W0 (Ω) :
h1 (x)|∇u| dx < +∞ .
(1.3)
Ω
R
The space E can be endowed with the norm kukE = ( Ω h1 (x)|∇u|p dx)1/p . As in
[9], it is known that E is an infinite dimensional Banach space. We say that u ∈ E
is a weak solution for problem (1.1) if
Z
Z
Z
Z
a(x, ∇u)∇φ dx − λ1
|u|p−2 uφ dx −
f (x, u)φ dx +
hφ dx = 0
Ω
Ω
Ω
Ω
for all φ ∈ E. Letting
Z
F (x, t) =
J(u) =
λ1
p
t
f (x, s) ds,
Z0
Z
|u|p dx +
F (x, u) dx −
Ω
Ω
Z
Λ(u) =
A(x, ∇u) dx,
Z
hu dx,
Ω
Ω
I(u) = Λ(u) − J(u)
for all u ∈ E. The following remark plays an important role in our arguments.
Remark 1.1.
(i) kuk ≤ kukE for all u ∈ E since h1 (x) ≥ 1.
(ii) By (A1), A satisfies the growth condition
|A(x, ξ)| ≤ c0 (h0 (x)|ξ| + h1 (x)|ξ|p )
for all ξ ∈ RN , a.e. x ∈ Ω.
(iii) By (ii) above and (A4), it is easy to see that
E = u ∈ W01,p (Ω) : Λ(u) < +∞ = u ∈ W01,p (Ω) : I(u) < +∞ .
(iv) C0∞ (Ω) ⊂ E since |∇u| is in Cc (Ω) for any u ∈ C0∞ (Ω) and h1 ∈ L1loc (Ω).
(v) By (A4) and Poincaré inequality, we see that
Z
Z
Z
1
λ1
A(x, ∇u) dx ≥
|∇u|p dx ≥
|u|p dx,
p Ω
p Ω
Ω
for all u ∈ W01,p (Ω).
Now we describe our main result.
Theorem 1.2. Assume conditions (A1)–(A5), (H1)–(H2) are fulfilled. Then problem (1.1) has at least a weak solution in E.
4
Q. A. NGÔ, H. Q. TOAN
EJDE-2008/98
2. Auxiliary results
Due to the presence of h1 , the functional Λ may not belong to C 1 (E, R). This
means that we cannot apply the Minimum Principle directly. In this situation, we
need some modifications.
Definition 2.1. Let F be a map from a Banach space Y to R. We say that F
is weakly continuous differentiable on Y if and only if following two conditions are
satisfied
(i) For any u ∈ Y there exists a linear map DF(u) from Y to R such that
lim
t→0
F(u + tv) − F(u)
= hDF(u), vi
t
for every v ∈ Y .
(ii) For any v ∈ Y , the map u 7→ hDF(u), vi is continuous on Y .
1
(Y ) the set of weakly continuously differentiable functionals on
Denote by Cw
1
(Y ) where we denote by C 1 (Y ) the set of all
Y . It is clear that C 1 (Y ) ⊂ Cw
1
(Y ), we put
continuously Fréchet differentiable functionals on Y . Now let F ∈ Cw
kDF(u)k = sup{|hDF(u), hi : |h ∈ Y, khk = 1}
for any u ∈ Y , where kDF(u)k may be +∞.
Definition 2.2. We say that F satisfies the Palais-Smale condition if any sequence
{un } ⊂ Y for which F(un ) is bounded and limn→∞ kDF(un )k = 0 possesses a
convergent subsequence.
The following theorem is our main ingredient.
1
(Y ) where Y is a
Theorem 2.3 (The Minimum Principle, see [13]). Let F ∈ Cw
Banach space. Assume that
(i) F is bounded from below, c = inf F,
(ii) F satisfies Palais-Smale condition.
Then there exists u0 ∈ Y such that F(u0 ) = c.
The proof of Theorem 2.3 is similar to the proof of Theorem 3.1 in [6] where
we need a modified Deformation Lemma which is proved in [16, Theorem 2.2].
For simplicity of notation, we shall denote DF(u) by F 0 (u). The following lemma
concerns the smoothness of the functional Λ.
Lemma 2.4 ([9]).
(i) If {un } is a sequence weakly converging to u in X, denoted by un * u, then
Λ(u) ≤ lim inf n→∞ Λ(un ).
(ii) For all u, z ∈ E
u+z
1
1
Λ
≤ Λ(u) + Λ(z) − k1 ku − zkpE .
2
2
2
(iii) Λ is continuous on E.
(iv) Λ is weakly continuously differentiable on E and
Z
0
hΛ (u), vi =
a(x, ∇u)∇v dx
Ω
for all u, v ∈ E.
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EXISTENCE OF SOLUTIONS FOR A RESONANT PROBLEM
5
(v) Λ(u) − Λ(v) ≥ hΛ0 (v), u − vi for all u, v ∈ E.
The following lemma concerns the smoothness of the functional J. The proof is
standard and simple, so we omit it.
Lemma 2.5.
(i) If un * u in X, then limn→∞ J(un ) = J(u).
(ii) J is continuous on E.
(iii) J is weakly continuously differentiable on E and
Z
Z
Z
hJ 0 (u), vi = λ1
|u|p−2 uv dx +
f (x, u)v dx −
hv dx
Ω
Ω
Ω
for all u, v ∈ E.
3. Proofs
We remark that the critical points of the functional I correspond to the weak
solutions of (1.1). Throughout this paper, we sometimes denote by ”const” a
positive constant.
Lemma 3.1. The functional I satisfies the Palais-Smale condition on E provided
(H2) holds.
Proof. Let {un } be a sequence in E and β be a real number such that
|I(un )| ≤ β
for all n
(3.1)
and
I 0 (un ) → 0 in E ? .
(3.2)
We prove that {un } is bounded in E. We assume by contradiction that kun kE → ∞
as n → ∞.
Let vn = un /kun kE for every n. Thus {vn } is bounded in E. By Remark 1.1(i),
we deduce that {vn } is bounded in X. Since X is reflexive, then by passing to a
subsequence, still denoted by {vn }, we can assume that the sequence {vn } converges
weakly to some v in X. Since the embedding X ,→ Lp (Ω) is compact then {vn }
converges strongly to v in Lp (Ω).
Dividing (3.1) by kun kpE together with Remark 1.1(v), we deduce that
!
Z
Z
Z
Z
λ
u
F
(x,
u
)
1
1
n
n
|∇vn |p dx −
|vn |p dx −
h
≤ 0.
lim sup
p dx +
p dx
p Ω
p Ω
n→+∞
Ω kun kE
Ω kun kE
Since, by the hypotheses on p, f , h and {un },
!
Z
Z
F (x, un )
un
lim sup
h
= 0,
p dx +
p dx
n→+∞
Ω kun kE
Ω kun kE
while
Z
lim sup
n→+∞
we have
Z
Ω
Z
|v|p dx,
Ω
p
Z
|∇vn | dx ≤ λ1
lim sup
n→+∞
|vn |p dx =
Ω
|v|p dx.
Ω
Using the weak lower semi-continuity of norm and Poincaré inequality, we get
Z
Z
Z
p
p
λ1
|v| dx ≤
|∇v| dx ≤ lim inf
|∇vn |p dx
Ω
Ω
n→+∞
Ω
6
Q. A. NGÔ, H. Q. TOAN
Z
≤ lim sup
n→+∞
EJDE-2008/98
|∇vn |p dx ≤ λ1
Ω
Z
|v|p dx.
Ω
Thus, these inequalities are indeed equalities. Besides, {vn } converges strongly to
v in X and
Z
Z
|∇v|p dx = λ1
|v|p dx.
Ω
Ω
This implies, by the definition of φ1 , that v = ±φ1 .
On the other hand, by means of (3.1), we deduce that
Z
Z
Z
p
−βp ≤ −p
A(x, ∇un ) dx + λ1
|un | dx + p
F (x, un ) dx
Ω
Ω
ZΩ
−p
hun dx
(3.3)
Ω
≤ βp.
In view of (3.2), there exists a sequence of positive real numbers {εn }n such that
εn → 0 as n → +∞ and
Z
Z
Z
p
−εn kun kE ≤
a(x, ∇un )∇un dx − λ1
|un | dx −
f (x, un )un dx+
Ω
Ω
ZΩ
(3.4)
hun dx
Ω
≤ εn kun kE .
Letting

 F (x, s)
g(x, s) =
s
f (x, 0)
if s 6= 0,
if s = 0.
We then consider the following two cases.
Case 1: Suppose that vn → −φ1 . Letting n → +∞. Since un (x) → −∞, it follows
that
f (x, un (x)) → f−∞ (x), a.e x ∈ Ω,
g(x, un (x)) → f−∞ (x), a.e x ∈ Ω.
Therefore, the properties of f and F , the Lebesgue theorem then imply
Z
Z
lim
(f (x, un )vn − pg(x, un )vn ) dx = (p − 1)
f−∞ (x)φ1 (x) dx.
n→+∞
Ω
(3.5)
Ω
On the other hand, by summing up (3.3) and (3.4), we get
Z
Z
(pF (x, un ) − f (x, un )un ) dx + (1 − p)
hun dx
Ω
Ω
Z
≥
(a(x, ∇un )∇un − pA(x, ∇un )) dx+
ZΩ
Z
(pF (x, un ) − f (x, un )un ) dx + (1 − p)
hun dx
Ω
Ω
≥ −βp − εn kun kE ,
and after dividing by kun kE , we obtain
Z
Z
βp
− εn .
(pg(x, un ) − f (x, un )vn ) dx + (1 − p)
hvn dx ≥ −
ku
n kE
Ω
Ω
(3.6)
EJDE-2008/98
EXISTENCE OF SOLUTIONS FOR A RESONANT PROBLEM
7
0
Since h ∈ Lp and kvn − (−φ1 )kX → 0, we obtain
Z
Z
lim
hvn dx = −
hφ1 dx.
n→+∞
Ω
(3.7)
Ω
In (3.6), taking lim inf to both sides together with (3.5) and (3.7), we deduce
Z
Z
(1 − p)
f−∞ (x)φ1 (x) dx − (1 − p)
hφ1 dx ≥ 0,
Ω
Ω
which gives
Z
(p − 1)
Z
hφ1 (x) dx ≥ (p − 1)
Ω
f−∞ (x)φ1 (x) dx
Ω
which yields, since p > 1,
Z
Z
hφ1 (x) dx ≥
f−∞ (x)φ1 (x) dx
Ω
Ω
which contradicts (H2).
Case 2: Suppose that vn → φ1 . Letting n → +∞. Since un (x) → ∞,
f (x, un (x)) → f +∞ (x), a.e x ∈ Ω,
g(x, un (x)) → f +∞ (x), a.e x ∈ Ω.
Therefore, by Lebesgue theorem,
Z
Z
lim
(f (x, un )vn − pg(x, un )vn ) dx = (1 − p)
f +∞ (x)φ1 (x) dx.
n→+∞
Ω
(3.8)
Ω
On the other hand, by summing up (3.3) and (3.4), we get
Z
Z
(pF (x, un ) − f (x, un )un ) dx + (p − 1)
hun dx
Ω
Ω
Z
≤
(pA(x, ∇un ) − a(x, ∇un )∇un ) dx+
ZΩ
Z
pF (x, un ) − f (x, un )un dx + (p − 1)
hun dx
Ω
Ω
≤ βp + εn kun kE ,
and after dividing by kun kE , we obtain
Z
Z
βp
− εn .
−
(pg(x, un ) − f (x, un )vn ) dx + (p − 1)
hvn dx ≤ −
ku
n kE
Ω
Ω
(3.9)
0
Since h ∈ Lp and kvn − φ1 kX → 0,
Z
Z
lim
hvn dx =
hφ1 dx.
n→+∞
Ω
Ω
In (3.9), using (3.8), (3.10) and by taking lim sup to both sides, we deduce
Z
Z
(p − 1)
hφ1 (x) dx ≤ (p − 1)
f +∞ (x)φ1 (x) dx
Ω
Ω
which yields, since p > 1,
Z
Z
hφ1 (x) dx ≤
Ω
which contradicts (H2).
Ω
f +∞ (x)φ1 (x) dx
(3.10)
8
Q. A. NGÔ, H. Q. TOAN
EJDE-2008/98
From the two cases above, {un } is bounded in E. By Remark 1.1(i), we deduce
that {un } is bounded in X. Since X is reflexive, then by passing to a subsequence,
still denote by {un }, we can assume that the sequence {un } converges weakly to
some u in X. We shall prove that the sequence {un } converges strongly to u in E.
We observe by Remark 1.1(iii) that u ∈ E. Hence {kun − ukE } is bounded.
Since {kI 0 (un − u)kE ? } converges to 0, then hI 0 (un − u), un − ui converges to 0.
By the hypotheses on f and h, we deduce that
Z
lim
|un |p−2 un (un − u) dx = 0,
n→+∞ Ω
Z
lim
f (x, un )(un − u) dx = 0,
n→+∞ Ω
Z
h(un − u) dx = 0.
lim
n→+∞
Ω
On the other hand,
hJ 0 (un ), un − ui
Z
Z
Z
= λ1
|un |p−2 un (un − u) dx +
f (x, un )(un − u) dx +
h(un − u) dx.
Ω
Ω
Ω
0
Thus limn→∞ hJ (un ), un − ui = 0. This and the fact that
hΛ0 (un ), un − ui = hI 0 (un ), un − ui + hJ 0 (un ), un − ui
give
lim hΛ0 (un ), un − ui = 0.
n→∞
By using (v) in Lemma 2.4, we get
Λ(u) − lim sup Λ(un ) = lim inf (Λ(u) − Λ(un )) ≥ lim hΛ0 (un ), u − un i = 0.
n→∞
n→∞
n→∞
This and (i) in Lemma 2.4 give
lim Λ(un ) = Λ(u).
n→∞
Now if we assume by contradiction that kun − ukE does not converge to 0 then
there exists ε > 0 and a subsequence {unm } of {un } such that
kunm − ukE ≥ ε.
By using relation (ii) in Lemma 2.4, we get
1
u nm + u
1
Λ(u) + Λ unm − Λ
≥ k1 kunm − ukpE ≥ k1 εp .
2
2
2
Letting m → ∞ we find that
u nm + u
lim sup Λ
≤ Λ(u) − k1 εp .
2
m→∞
u nm + u
We also have
converges weakly to u in E. Using (i) in Lemma 2.4 again,
2
we get
u nm + u
.
Λ(u) ≤ lim inf Λ
m→∞
2
That is a contradiction. Therefore {un } converges strongly to u in E.
Lemma 3.2. The functional I is coercive on E provided (H2) holds.
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EXISTENCE OF SOLUTIONS FOR A RESONANT PROBLEM
9
Proof. We firstly note that, in the proof of the Palais-Smale condition, we have
proved that if I(un ) is a sequence bounded from above with kun kE → ∞, then (up
to a subsequence), vn = un /kun kE → ±φ1 in X. Using this fact, we will prove that
I is coercive provided (H3 ) holds.
Indeed, if I is not coercive, it is possible to choose a sequence {un } ⊂ E such
that kun kE → ∞, I(un ) ≤ const and vn = un /kun kE → ±φ1 in X. By Remark
1.1 (v),
Z
Z
−
F (x, un ) dx +
hun dx ≤ I(un ).
(3.11)
Ω
Ω
Case 1: Assume that vn → φ1 . Dividing (3.11) by kun kE we get
Z
−
f
+∞
Z
Z
φ1 dx +
−
hφ1 dx = lim
Ω
n→+∞
Ω
Ω
F (x, un )
dx +
kun kE
!
un
h
dx
Ω kun kE
Z
I(un )
≤ lim sup
n→+∞ kun kE
const
≤ lim sup
= 0,
n→+∞ kun kE
which gives
Z
f +∞ φ1 dx +
−
Z
hφ1 dx ≤ 0,
Ω
Ω
which contradicts (H2).
Case 2: Assume that vn → −φ1 . Dividing (3.11) by kun kE we get
Z
Z
f−∞ φ1 dx −
Ω
Z
hφ1 dx = lim
n→+∞
Ω
−
Ω
F (x, un )
dx +
kun kE
!
un
h
dx
Ω kun kE
Z
I(un )
≤ lim sup
n→+∞ kun kE
const
≤ lim sup
= 0,
ku
n→+∞
n kE
which gives
Z
Z
f−∞ φ1 dx −
Ω
which contradicts (H2).
hφ1 dx ≤ 0,
Ω
Proof of Theorem 1.2. The coerciveness and the Palais-Smale condition are enough
to prove that I attains its proper infimum in Banach space E (see Theorem 2.3),
so that (1.1) has at least a weak solution in E. The proof is complete.
Acknowledgments. The authors wish to express their gratitude to the anonymous referee for a number of valuable comments and suggestions which help us
to improve the presentation of the present paper from line to line. This work is
dedicated to the first author’s father on the occasion of his fiftieth birthday.
10
Q. A. NGÔ, H. Q. TOAN
EJDE-2008/98
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Quôc Anh Ngô
Department of Mathematics, College of Science, Viêt Nam National University, Hà
Nôi, Viêt Nam
E-mail address: bookworm vn@yahoo.com
E-mail address: nqanh@vnu.edu.vn
Hoang Quoc Toan
Department of Mathematics, College of Science, Viêt Nam National University, H‘a
Nôi, Viêt Nam
E-mail address: hq toan@yahoo.com