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Existence of solution for a degenerate p(x) -Laplacian equation in RN
J. Math. Anal. Appl. 345 (2008) 731–742
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J. Math. Anal. Appl. www.elsevier.com/locate/jmaa
Existence of solution for a degenerate p (x)-Laplacian equation in R N Claudianor O. Alves 1 Universidade Federal de Campina Grande, Unidade Acadêmica de Matemática e Estatística, Cep 58109-970, Campina Grande, PB, Brazil
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 23 January 2008 Available online 30 April 2008 Submitted by V. Radulescu
Using variational methods we establish the existence of nontrivial solutions for the following class of problems
Keywords: Variational methods p (x)-Laplacian Degenerate equations
where p , q, a, K : R N → R are continuous functions satisfying some conditions and the set {x ∈ R N ; a(x) = 0} is not empty. © 2008 Elsevier Inc. All rights reserved.
− div a(x)|∇ u | p(x)−2 ∇ u + |u | p(x)−2 u = K (x)|u |q(x)−2 u in R N ,
1. Introduction In this paper, we are interested in the existence of nontrivial solution for the following class of problems
− div a(x)|∇ u | p (x)−2 ∇ u + |u | p (x)−2 u = K (x)|u |q(x)−2 u in R N ,
where p , q, a, K : R → R are continuous functions verifying the following hypotheses: N
(H1 ) 1 < p − = infRN p (x) p (x) p + = supRN p (x) < N in R N . (H2 ) There exists R ∗ > 0 such that p (x) m in R N , p (x) ≡ m for all x ∈ B cR ∗ (0). (H3 ) 1 < q− = infRN q(x) q(x) q+ = supRN q(x) in R N , p + < q− and p (x) q(x) p ∗ (x) =
Np (x) N − p (x)
in R N .
The notation h g means that infRN { g (x) − h(x)} > 0. Relating to functions a and K , we assume the following conditions:
(H4 ) a(x) 0 in R N , a ≡ 0, a ∈ L ∞ (R N ), and there exists θ > 0 such that 1 x ∈ R N ; a(x) = 0 ⊂ B θ (0), inf a(x) > 0 and ∈ L r (x) B θ (0) a
R N \ B θ (0)
for some function r (x) ∈
L ∞ (R N ) ∩ C (R N ) satisfying
Nq(x) p (x) N + p (x)q(x) − Nq(x)
1
r (x) and
1 + r (x) r (x)
p (x).
E-mail address: [email protected]. Partially supported by FAPESP 2007/03399-0, CNPq 472281/2006-2 and 620025/2006-9.
0022-247X/$ – see front matter doi:10.1016/j.jmaa.2008.04.060
© 2008 Elsevier Inc.
All rights reserved.
(P )
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C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
(H5 ) There exist η, β > 0 such that
η < K (x) < β, ∀x ∈ RN . Equations with variable exponents appear in various mathematical models, for example:
• Electrorheological fluids: see Acerbi and Mingione [1,2], Antontsev and Rodrigues [5] and Ruzicka [18]. • Nonlinear Darcy’s law in porous medium: see Antontsev and Shmarev [6,7]. • Image processing: see Chambolle and Lions [8] and Chen, Levine and Rao [11]. The study of L p (x) and W 1, p (x) spaces done by Kovacik and Rákosník [19] has generated a large amount of research of equations involving variable exponents, for example, Alves and Souto [3], Fan [15], El Hamidi [12], Chabrowski and Fu [10], Mihailescu, Pucci and Radulescu [21] and references therein. In [9], Chabrowski has considered the problem ( P ) for the particular case where p (x) = 2 and q(x), r (x) are constant functions verifying the hypotheses (H3 )–(H5 ). In that paper, the author established the existence of a solution to ( P ) by minimization together Lagrange Multiplier Theorem. We cite the papers of Murthy and Stampacchia [20], Musso and Passaseo [17], Passaseo [22] and reference therein, where other types of degenerate elliptic problems were also considered. In this paper, motivated by [9] and some arguments developed by Alves and Souto [3], we complete the study made in [9], in the sense that we are considering the same equation in R N , but with variable exponents. However, we would like to stress that the method used in [9] cannot be applied here because the operator and nonlinearity are not homogeneous. 1, p (x) This paper is organized in the following way: In Section 2, we define the space W a(x) (R N ) and establish an embedding result for this space. In Section 3, we show the existence of solutions for two different behaviors of q(x) at infinity. 1, p ( x )
2. The space W a(x)
RN ) (R
The appropriate Sobolev space to study problem ( P ) is the space W a(x) (R N ), defined as a completion of C 0∞ (R N ) with respect to norm 1, p (x)
u = ∇ u p (x),a(x) + u p (x) where
∇ u p (x),a(x) = inf α > 0,
∇ u p (x) a(x) dx 1
RN
α
and
p (x) u
u p (x) = inf α > 0, dx 1 . α
RN
1, p (x)
Using similar arguments developed in [13,14,19], it is possible to prove that W a(x) (R N ) is separable and reflexive. Since a is bounded from above, we have the following continuous embedding
1, p (x)
W 1, p (x) R N → W a(x)
RN . 1, p (x)
Another important norm on W a(x) (R N ) is
∇ u p (x) u p (x)
u ∗ = inf α > 0, a(x) + dx 1 α α RN
which is equivalent to . 1, p (x) The next lemma establishes other embeddings involving W a(x) (R N ), which are key points in all this work. Lemma 2.1. Suppose that (H1 )–(H5 ) hold. Then, the following continuous embeddings hold
R N → W 1,s(x) B R (0) ,
1, p (x)
W a(x) and
1, p (x)
W a(x)
R N → L q(x) R N .
∀ R > 0 where s(x) =
p (x)r (x) 1 + r (x)
(2.1)
(2.2)
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Proof. We begin fixing R > 1 such that
x ∈ R N ; a(x) = 0 ⊂ B R −1 (0)
and
inf
R N \ B R −1 (0)
a(x) > 0.
Since Nq(x) p (x) N + p (x)q(x) − Nq(x)
r (x)
a straightforward calculation leads to Ns(x)
q(x)
N − s(x)
.
This together with some results found in [13,14] yields
W 1,s(x) B R (0) → L q(x) B R (0) .
(2.3)
In the what follows, φ R is a C 1 function verifying
• φ R (x) = 1, ∀x ∈ B R (0); • φ R (x) = 0, ∀x ∈ R N \ B R +1 (0) and
• 0 φ R (x) 1, ∀x ∈ R N . Fixing
Θ = ∇ u p (x),a(x), B R +1 (0) + u p (x), B R +1 (0) , we get
∇ u s(x) Θ dx B R (0)
Hence
∇(u φ R ) s(x) Θ dx.
B R +1 (0)
∇ u s(x) Θ dx c 1
B R (0)
∇ u s(x) Θ dx + c 1
B R +1 (0)
s(x) u dx Θ
B R +1 (0)
for some positive constant c 1 , because φ R is a smooth function. By virtue of preceding inequality together with a ∈ L ∞ (R N ), it follows that there exists c 2 > 0 such that
s(x) ∇ u s(x) r (x) 1 dx c 1 a 1+r(x) ∇ u dx + c 2 r (x) Θ Θ a 1+r (x) RN
B R (0)
RN
s(x) u 1 dx. r (x) Θ a 1+r (x)
From Hölder’s inequality
s(x) s(x) −s(x) ∇ u s(x) s(x) u −s(x) a p(x) p(x)r(x) , dx c 1 a p(x) p(x)r(x) a p(x) ∇ u + c 2 Θ Θ p(x) Θ p(x) s(x) s(x) s(x)
B R (0)
that is,
s(x) s(x) ∇ u s(x) s(x) u . dx c 3 a p(x) ∇ u + c 3 Θ Θ p(x) Θ p(x) s(x)
B R (0)
Since
s(x)
s(x) ∇ u s(x) a p(x) Θ p(x) , s(x)
it follows that
s(x) u Θ p(x) 1, s(x)
∇ u s(x) s+ Θ dx 2c 3 (1 + 2c 3 ) .
B R (0)
s(x)
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From this
s(x) ∇u (1 + 2c )Θ dx 1, 3
B R (0)
and, therefore,
∇ u s(x) (1 + 2c 3 )Θ (1 + 2c 3 ) u .
(2.4)
Since there exists c 4 > 0 verifying
u s(x), B R (0) c 4 u p (x), B R +1 (0) ,
(2.5)
combining (2.4) and (2.5), we get
∇ u W 1,s(x) ( B R (0)) c 5 u for some positive constant c 5 . At this point (2.1) is proved. Now, we shall prove (2.2). To this end, we use the following inequality
u q(x), B R −1 (0) c 6 u φ R −1 q(x), B R (0) which together with the continuous embedding W 1,s(x) ( B R (0)) → L q(x) ( B R (0)) imply
u q(x), B R −1 (0) c 6 u φ R −1 W 1,s(x) ( B R (0)) and thus
u q(x), B R −1 (0) c 7 u W 1,s(x) ( B R (0)) .
(2.6)
Consequently, (2.6) and (2.1) combined imply that there exists c 8 > 0 such that
u q(x), B R −1 (0) c 8 u .
(2.7)
γ > 1 large enough, we have p (x) ∇u dx 1, γ u
On the other hand, for
B cR −1 (0)
from where follows that
∇ u p (x), B cR −1 (0) γ u .
(2.8)
Here we have used the assumption infRN \ B R −1 (0) a(x) > 0. Estimate (2.8) and the definition of the norm imply that there exists c 9 > 0 verifying
u W 1, p(x) ( B c
R −1 (0))
c 9 u .
Now, recalling that there exists c 10 > 0 such that
u q(x), B cR −1 (0) c 10 u W 1, p(x) ( B c
R −1 (0))
,
the last two inequalities yield
u q(x), B cR −1 (0) c 11 u .
(2.9)
From (2.7) and (2.9), we finish the proof of the embedding (2.2).
2
3. The variational framework 1, p (x)
1, p (x)
In this paper, a function u ∈ W a(x) (R N ) is a solution of ( P ) if for each φ ∈ W a(x) (R N ) the equality below holds
a(x)|∇ u | p (x)−2 ∇ u ∇φ + |u | p (x)−2 u φ − K (x)|u | p (x)−2 u φ dx = 0.
(3.1)
RN 1, p (x)
To find a solution to ( P ), we shall study the critical points of the functional I : W a(x) (R N ) → R given by
I (u ) = RN
p (x) 1 a(x)∇ u + |u | p (x) dx − p (x)
RN
K (x) q(x)
|u |q(x) dx.
C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
735
Since q− > p + , by using standard arguments, see for example [3] and [15], it is possible to prove that I verifies the Mountain 1, p (x)
Pass Geometry and that there exists a (PS)d sequence {un } ⊂ W a(x) (R N ), that is, I (u n ) → d
I (u n ) → 0
and
where
(3.2)
d = inf max I g (t ) > 0
(3.3)
g ∈Γ 0t 1
and
1, p (x) Γ = g ∈ C [0, 1], W a(x) R N : g (0) = 0 and I g (1) 0 .
Lemma 3.1. Let {un } be a (PS)c sequence of I . Then, {un } has a subsequence, still denoted by {un }, such that:
• • • • •
{un } is a bounded sequence with the weak limit denoted by u; 1, p (x) 1,m un → u in W a(x),loc (R N ) and W a(x),loc (R N ); N un (x) → u (x) a.e. in R ; p (x) q(x) un (x) → u (x) in L loc (R N ) and L loc (R N ); 1,m c un → u in W loc ( B R ), for all R > θ .
Proof. From (3.2), there exist C > 0 and n0 ∈ N such that I (u n ) −
1 I (un )un C + un , q−
∀n n0 ,
or equivalently
a(x)|∇ un | p (x) + |un | p (x) dx k1 + k2 un ,
∀n n0
RN
for some positive constants k1 and k2 , because p + < q− . As and ∗ are equivalent norms, if un 1, we get
u n p − k 1 + k 3 u n 1, p (x)
from where follows that {un } is bounded. The existence of a weak limit to {un } follows from the reflexivity of W a(x) (R N ). For each R > 0, the embedding 1, p (x)
W a(x)
1,m
B R (0) → W a(x) B R (0)
(3.4)
is continuous. In fact, the inequality
p (x) ∇ u m 1 + ∇u u u leads to
∇ u m
u
a(x)
B R (0)
a(x) dx +
B R (0)
B R (0)
∇ u p (x)
u
a(x)
a(x) dx + 1
B R (0)
and thus
m1
m
a(x)|∇ u | dx
C u ,
1, p (x)
∀u ∈ W a(x)
B R (0) .
(3.5)
B R (0)
Arguing as above,
m
m1
a(x)|u | dx
C u ,
1, p (x)
∀u ∈ W a(x)
B R (0) .
(3.6)
B R (0)
Then, (3.5) and (3.6) combined give (3.4). Now, denoting
P n (x) = a(x) |∇ un | p (x)−2 ∇ un − |∇ u | p (x)−2 ∇ u , ∇ un − ∇ u + |un | p (x)−2 − |u | p (x)−2 (un − u )(x) 0,
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for each R > 0 we have the following inequality:
P n dx I (un )(φ R un ) − I (un )(φ R u ) − |x| R
RN
φ R |un | p (x) dx +
− RN
a(x)|∇ un | p (x)−2 un ∇ un ∇φ R dx +
φ R |un | p (x)−2 uun dx +
RN
|un |q(x) φ R −
RN
a(x)|∇ un | p (x)−2 u ∇ un ∇φ R dx
RN
|un |q(x)−2 un u φ R + on (1)
RN
which implies
lim
n→∞
P n dx = 0.
|x| R
Since and ∗ are equivalents, the last limit leads to un → u
1, p (x)
in W a(x)
B R (0) , ∀ R > 0.
(3.7)
The same type of arguments can be used to prove that un → u
1,m
in W loc B cR (0) , ∀ R > θ.
(3.8)
The other limits follow from (2.1) and (3.7).
2
4. Existence of solutions In this section, we shall show the existence of a solution for ( P ). We divide this section into two subsections considering different situations involving the behavior of q(x) at infinity. From now on, the functions a(x) and K (x) verify the following conditions:
(H6 ) a(x) a∞ on R N , where a∞ = lim|x|→∞ a(x) with the strict inequality on a set of positive measure in R N . (H7 ) K (x) K ∞ on R N , where K ∞ = lim|x|→∞ K (x). Hereafter, let us denote by I ∞ , the Euler–Lagrange functional associated to the problem
−a∞ m u + um−1 = K ∞ |u |s−2 u in R N , u 0, u ≡ 0 and u ∈ W 1,m R N ,
( P ∞)
and by c ∞ the mountain pass level related to I ∞ . Since a∞ and K ∞ are positive constants, the existence of ground state solution to ( P ∞ ) follows by using the same arguments explored for the case a∞ = K ∞ = 1. Moreover, it is possible to prove the existence of a positive radially symmetric ground state w of ( P ∞ ) verifying
w (x), ∇ w (x) → 0 as |x| → ∞.
(4.1)
More details about the above informations can be found in [4] and [16]. Lemma 4.1. Let w be a ground state solution of ( P ∞ ). Then, there exists η∗ > 0 such that
w ∞ < 1,
∀ K ∞ η∗ .
Proof. For each K ∞ > 0, we consider the function 1
s+1−m U∞ = K∞ w.
By a direct computation, we can prove that U ∞ is a solution of
−a∞ m U + U m−1 = U s in R N , and that there exists C 1 > 0 verifying
U ∞ H 1 (RN ) C 1 ,
∀ K ∞ 0.
Using Moser iteration technique, the last inequality implies that there exists C 2 > 0 satisfying
U ∞ ∞ C 2 ,
∀ K ∞ 0.
Hence, there exists η∗ such that
w ∞ < 1,
∀ K ∞ η∗
and this completes the proof of lemma.
2
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4.1. Constant at infinity In this subsection, we shall study the case where q(x) is equal to a constant at infinity. The next proposition establishes an important relation involving the minimax levels d and c ∞ . Proposition 4.1. Suppose (H1 )–(H5 ) and that q(x) satisfies a.e. in R N
q(x) s
for all |x| R ∗ .
and q(x) ≡ s
( Q 1)
If I admits a (PS)d sequence converging weakly to 0, then d c ∞ . Proof. For each R > max{θ, R ∗ , R ∗ }, let us consider ψ R (x) = 1 − φ R (x), where φ R was introduced in the proof of Lemma 2.1, and tn, R > 0 verifying
I ∞ tn, R (ψ R un ) = max I ∞ t (ψ R un ) . t 0
From the definition of c ∞
c ∞ I ∞ tn, R (ψ R un ) = I (un ) + I ∞ tn, R (ψ R un ) − I (un ) . Assuming for a moment the claim below Claim 4.1. lim R →+∞ [lim supn→+∞ ( I ∞ (tn, R (ψ R un )) − I (un ))] = 0. Letting n → +∞ and after R → +∞, we obtain the inequality c ∞ lim I (un ) = d, n→+∞
2
which completes the proof of the proposition. Proof of Claim 4.1. First of all, we observe that
I ∞ tn, R (ψ R un ) − I (un )
tnm, R
=
m
m
a∞ ∇(ψ R un ) dx −
RN
+
K (x) q(x)
a(x) p (x)
RN
|un |q(x) dx −
RN
K∞
tns , R s
m t n, R |∇ un | p (x) dx + |ψ R un |m dx − m
RN
|ψ R un |s dx .
1 p (x)
|un | p (x) dx
RN
RN
In the next, we shall show that each expression in the above parentheses goes to zero, when n → +∞ and after R → +∞. We shall assume for a moment the following claim. Claim 4.2. There exists R 0 > 0 such that lim sup tn, R = 1, n→+∞
∀R R0.
Observe that
K (x) q(x)
RN
|un |q(x) dx −
K∞ s
tns , R |ψ R un |s dx
RN
K∞ ( K (x) − K ∞ ) |un |q(x) dx + 1 − tns , R q(x) s
= RN
K∞
− tns , R R |x| R +1
s
q(x)
|ψ R un |
|un |s dx +
|x|> R +1
R →+∞
lim sup n→+∞
K (x) q(x)
RN
|un |q(x) dx
dx.
Combining (H6 ), (H7 ), Lemma 2.1 and Claim 4.2, it follows that lim
|x| R +1
K∞ q(x)
|un |q(x) dx − RN
K∞ s
tns , R |ψ R un |s dx
= 0.
(4.2)
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Arguing as above, it is possible to prove that
lim
R →+∞
and
n→+∞
lim
m
RN
tnm, R m
n→+∞
a(x) p (x)
|∇ un | p (x) dx = 0
(4.3)
RN
lim sup
R →+∞
m ∇(ψ R un ) dx −
a∞
tnm, R
lim sup
|ψ R un |m dx −
RN
1 p (x)
|un | p (x) dx = 0.
(4.4)
RN
Thus, from (4.2)–(4.4) the proof of Claim 4.1 is completed.
2
Proof of Claim 4.2. Recalling that q(x) satisfies Np (x)
p (x) q(x) p ∗ (x) =
N − p (x)
,
we have
|un |q(x) dx → 0
|x| R
for all R > R ∗ . Thus
|un |q(x) dx = |x| R
|un |s dx 0,
|x| R
because, otherwise, we have the limit
|un |q(x) dx → 0 RN
which yields d = 0, and this is a contradiction. Now, taking tn, R > 0 verifying
m
a∞ ∇(ψ R un ) dx +
RN
m |ψ R un |m dx = K ∞ tns− ,R
RN
|ψ R un |s dx
(4.5)
RN
and using the fact that {ψ R un } is bounded in W 1,m (R N ), by a straightforward computation, it is possible to check that for each R > 0 fixed, the sequence {tn, R } is bounded. In what follows, we denote by t R the following real number lim sup tn, R = t R . n→+∞
On the other hand, combining (3.2) and (4.5) we easily derive
a(x)|∇ un | p (x) dx − RN
m
a∞ ∇(ψ R un ) dx +
RN
q(x)
=
K (x)|un |
m dx − tns− ,R
RN
RN
|un | p (x) dx −
|ψ R un |m dx
RN
s
K ∞ |ψ R un | dx + on (1).
RN
This together with Lemma 2.1 yield
0 = K ∞ 1 − t sR−m where
αR
α R = lim supn→+∞
|x| R |un | dx. Since d > 0, we derive
|un |s dx = 0,
lim
n→+∞
|x| R
and
lim
n→+∞
RN
(4.6)
|un |s dx = 0.
s
∀ R > 0,
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Hence, lim
R →+∞
α R = α > 0.
Thus, the last limit together with (4.6) imply that there exists R 0 > 0 such that t R = 1,
∀R R0,
which completes the proof of Claim 4.2.
2
Lemma 4.2. Assume that (H1 )–(H7 ) and ( Q 1 ) hold. Then, d < c ∞ . Proof. Let w be a positive radially symmetric ground state solution for ( P ∞ ), and consider the sequences xn = (0, 0, . . . , n), w n (x) = w (x − xn ) and tn > 0 satisfying I (tn w n ) = max I (t w n ). t 0
From the definition of d,
d I (tn w n ) = I ∞ (tn w n ) + I (tn w n ) − I ∞ (tn w n ) , that is,
d c∞ +
a∞
|x| R ∗
|tn ∇ w n | p (x) |tn ∇ w n |m − + p (x) m
|x| R ∗
|tn w n | p (x) |tn w n |m − + p (x) m
|x| R ∗
K∞
|tn w n |s s
−
|tn w n |q(x) . q(x)
Fixing an index n large enough such that |∇(tn w n )| < 1 and |tn w n | < 1 in B R ∗ (0), and by using the fact that f (s) = s−1 τ s is decreasing in (0, +∞), if 0 < τ < 1, for that n, we can conclude that
a∞ |x| R ∗
and
|tn ∇ w n | p (x) |tn ∇ w n |m − , p (x) m
K∞ |x| R ∗
|tn w n |s s
|tn w n |q(x) q(x)
−
are less than zero. Hence d < c ∞ .
|x| R ∗
|tn w n | p (x) |tn w n |m − p (x) m
2
Theorem 4.1. Assume that (H1 )–(H7 ) and ( Q 1 ) hold. Then, problem ( P ) has a solution. 1, p (x)
Proof. For each φ ∈ W a(x) (R N ), the (PS)d sequence {un } satisfies
a(x)|∇ un | p (x)−2 ∇ un ∇φ + |un | p (x)−2 un φ dx =
RN
K (x)|un |q(x)−2 un φ dx + on (1).
RN
Taking the limit as n → +∞ and by using Lemma 3.1, it follows that
a(x)|∇ u | p (x)−2 ∇ u ∇φ + |un | p (x)−2 u φ dx =
RN
K (x)|u |q(x)−2 u φ dx.
RN
Thus u is a solution. Moreover, combining Lemma 4.2 and Proposition 4.1, we deduce that u cannot be zero.
2
4.2. Asymptotically constant at infinity In this subsection, we shall study the case where q(x) is not equal to s, but q(x) goes to s as |x| → ∞. Lemma 4.3. Suppose that q(x) satisfies q(x) s,
∀x ∈ R N
and
lim q(x) = s.
|x|→∞
If I admits a (PS)d sequence converging weakly to 0, then c c ∞ .
( Q 2)
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Proof. As in the proof of Proposition 4.1, we need to prove that Claims 4.1 and 4.2 remain valid by assuming that ( Q 2 ) holds. Reviewing the proof of Claim 4.2, we need to show that
m K (x)|un |q(x) dx − tns− ,R
lim sup n→+∞
RN
K ∞ |ψ R un |s dx = K ∞ 1 − t sR−m
αR
(4.7)
RN
for all R > 0. For each R > 0, it is possible to show that
m K (x)|un |q(x) dx − tns− ,R
RN
m K ∞ |ψ R un |s dx = K ∞ 1 − tns− ,R
|un |s dx + K ∞
|x| R
RN
|un |q(x) − |un |s dx + on (1).
RN
Repeating the same arguments used in [3, Lemma 4.2], the second term of the last equality is on (1). In this way,
m K (x)|un |q(x) dx − tns− ,R
RN
m K ∞ |ψ R un |s dx = K ∞ 1 − tns− ,R
|un |s dx + on (1)
|x| R
RN
and (4.7) follows. The limit (4.2) follows by using the same type of arguments. Therefore, Claims 4.1 and 4.2 remain valid if ( Q 2 ) holds. 2 Lemma 4.4. Assume that (H1 )–(H7 ) and ( Q 2 ) hold. Then, there is η∗ > 0 such that d < c ∞ for K ∞ η∗ . Proof. First of all, from Lemma 4.1 there exists
| w |∞ 1,
η∗ > 0 such that the ground state w of ( P ∞ ) satisfies
∀ K ∞ η∗ .
(4.8)
Taking w n (x) = w (x − xn ) and tn > 0 verifying I (tn w n ) = max I (t w n ), t 0
it follows that
d I (tn w n ) = I ∞ (tn w n ) + I (tn w n ) − I ∞ (tn w n ) which is equivalent to
d c∞ +
a∞
|x| R ∗
+
tns
K∞
s
p (x)
tn
p (x)
|∇ w n | p (x) −
tnm m
| w n | dx −
tn
q(x)
q(x)
|wn |
|x| R ∗
q(x)
s
|∇ w n |m dx +
p (x)
tn
p (x)
tnm
| w n | p (x) dx −
m
| w n |m dx
dx.
RN
Using for a moment the claim Claim 4.3. The sequence {tn } is convergent with limit equal to 1. (4.1) together with Lemma 4.1 leads to
∇(tn w n )
∞
< 1 in B R ∗ (0)
and
|tn w n |∞ < 1 in R N for n large enough. Since f (s) = s−1 τ s is decreasing in (0, +∞) if 0 < τ < 1, the terms
a∞ |x| R ∗
and
K∞
tns s
p (x)
tn
p (x)
|∇ w n |
p (x)
−
tnm m
|∇ w n |
| w n | dx −
tn
q(x)
q(x)
|wn |
RN
are negative, and then d < c ∞ .
2
dx, |x| R ∗
q(x)
s
m
dx
p (x)
tn
p (x)
|wn |
p (x)
dx −
tnm m
m
|wn |
dx
C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
Proof of Claim 4.3. For each n ∈ N, we have
p (x)
RN
so
a(x)|∇ w n | p (x) + | w n | p (x) dx =
tn
q(x)
K (x)tn
741
| w n |q(x) dx,
RN
p (x)
a∞ |∇ w n |m + | w n |m dx
tn RN
q(x)
K ∞ tn
| w n |s dx.
(4.9)
RN
The inequality (4.9) implies that {tn } is bounded, because if for some subsequence, still denoted by {tn }, we have lim tn = +∞,
n→+∞
it easy to see that
q − p+
a∞ |∇ w |m + | w |m dx tn−
RN
K ∞ | w |s dx
(4.10)
RN
for n large enough. This implies that {tn } is bounded, obtaining this way a contradiction. Now, we shall show that lim infn→+∞ tn = lim supn→+∞ tn = 1. Arguing by contradiction, if lim infn→+∞ tn > 1 holds, we have tn > 1 for n large enough, and thus (4.10) is true. On the other hand, using the equality
a∞ |∇ w |m + | w |m dx =
RN
K ∞ | w |s dx
(4.11)
RN
it follows from (4.10) and (4.11) that q − p+
1 tn−
for n large enough, and so lim infn→+∞ tn 1, which is impossible. Now, arguing again by contradiction, if lim supn→+∞ tn < 1 holds, we have tn < 1 for n large enough, and thus
q − p+
a(x + xn )|∇ w | p (x+xn ) + | w | p (x+xn ) dx tn−
RN
K ∞ | w |s dx + on (1).
(4.12)
RN
From (4.11) and (4.12), we can conclude that lim supn→+∞ tn 1, obtaining again a contradiction. From the above analysis we derive that lim tn = 1
n→+∞
and this completes the proof of Claim 4.3.
2
Theorem 4.2. Assume that (H1 )–(H7 ) and ( Q 2 ) hold and q(x) s a.e. in R N . Then, for K ∞ η∗ problem ( P ) has a solution. Proof. The proof follows from similar arguments to the ones used in the proof of Theorem 4.1.
2
Acknowledgments The author would like to thank ICMC/USP – São Carlos, and specially to the Professor Sérgio Monari for his attention and friendship. This work was done while the author was visiting that institution. The author would like to express sincere thanks to the Professor Xianling Fan and to the reviewers for their valuable comments about this subject.
References [1] E. Acerbi, G. Mingione, Regularity results for electrorheological fluids: Stationary case, C. R. Math. Acad. Sci. Paris 334 (2002) 817–822. [2] E. Acerbi, G. Mingione, Regularity results for stationary electrorheological fluids, Arch. Ration. Mech. Anal. 164 (2002) 213–259. [3] C.O. Alves, M.A.S. Souto, Existence of solutions for a class of problems in R N involving p (x)-Laplacian, Progr. Nonlinear Differential Equations Appl. 66 (2005) 17–32. [4] C.O. Alves, J.M. Bezerra do Ó, O.H. Miyagaki, On perturbations of a class of a periodic m-Laplacian equation with critical growth, Nonlinear Anal. 45 (2001) 849–863. [5] S.N. Antontsev, J.F. Rodrigues, On stationary thermo-rheological viscous flows, Ann. Univ. Ferrara Sez. VII Sci. Mat. 52 (2006) 19–36. [6] S.N. Antontsev, S.I. Shmarev, Elliptic equations and systems with nonstandard growth conditions: Existence, uniqueness and localization properties of solutions, Nonlinear Anal. 65 (2006) 722–755. [7] S.N. Antontsev, S.I. Shmarev, On localization of solutions of elliptic equations with nonhomogeneous anisotropic degeneracy, Siberian Math. J. 46 (2005) 765–782.
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A. Chambolle, P.L. Lions, Image recovery via total variation minimization and related problems, Numer. Math. 76 (1997) 167–188. J. Chabrowski, Degenerate elliptic equation involving a subcritical Sobolev exponent, Port. Math. (N.S.) 53 (2) (1996) 167–177. J. Chabrowski, Y.Q. Fu, Existence of solutions for p (x)-Laplacian problems on a bounded domain, J. Math. Anal. Appl. 306 (2005) 604–618. Y. Chen, S. Levine, M. Rao, Variable exponent, linear growth functionals in image restoration, SIAM J. Appl. Math. 66 (4) (2006) 1383–1406. A. El Hamidi, Existence results to elliptic systems with nonstandard growth conditions, J. Math. Anal. Appl. 300 (2004) 30–42. X.L. Fan, J.S. Shen, D. Zhao, Sobolev embedding theorems for spaces W k, p (x) (Ω), J. Math. Anal. Appl. 262 (2001) 749–760. X.L. Fan, D. Zhao, On the spaces L p (x) (Ω) and W m, p (x) (Ω), J. Math. Anal. Appl. 263 (2001) 424–446. X.L. Fan, p (x)-Laplacian equations in R N with periodic data and nonperiodic perturbations, J. Math. Anal. Appl. 341 (2008) 103–119. E.S. Medeiros, Existência e concentração de soluções para uma classe de problemas elípticos quasilineares, PhD thesis, UNICAMP, 2001. M. Musso, D. Passaseo, Nonlinear elliptic problems approximating degenerate equations, Nonlinear Anal. 30 (8) (1997) 5071–5076. M. Ruzicka, Electrorheological Fluids: Modeling and Mathematical Theory, Lecture Notes in Math., vol. 1748, Springer-Verlag, Berlin, 2000. O. Kovacik, J. Rákosník, On spaces L p (x) and W 1, p (x) , Czechoslovak Math. J. 41 (1991) 592–618. K.V. Murthy, G. Stampacchia, Boundary value problems for some degenerate elliptic operators, Ann. Mat. Pura Appl. (4) 80 (1968) 1–122. M. Mihailescu, P. Pucci, V. Radulescu, Eigenvalue problems for anisotropic quasilinear elliptic equations with variable exponent, J. Math. Anal. Appl. 340 (2008) 687–698. [22] D. Passaseo, Some concentration phenomena in degenerate semilinear elliptic problems, Nonlinear Anal. 24 (7) (1995) 1011–1025.
Contents lists available at ScienceDirect
J. Math. Anal. Appl. www.elsevier.com/locate/jmaa
Existence of solution for a degenerate p (x)-Laplacian equation in R N Claudianor O. Alves 1 Universidade Federal de Campina Grande, Unidade Acadêmica de Matemática e Estatística, Cep 58109-970, Campina Grande, PB, Brazil
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 23 January 2008 Available online 30 April 2008 Submitted by V. Radulescu
Using variational methods we establish the existence of nontrivial solutions for the following class of problems
Keywords: Variational methods p (x)-Laplacian Degenerate equations
where p , q, a, K : R N → R are continuous functions satisfying some conditions and the set {x ∈ R N ; a(x) = 0} is not empty. © 2008 Elsevier Inc. All rights reserved.
− div a(x)|∇ u | p(x)−2 ∇ u + |u | p(x)−2 u = K (x)|u |q(x)−2 u in R N ,
1. Introduction In this paper, we are interested in the existence of nontrivial solution for the following class of problems
− div a(x)|∇ u | p (x)−2 ∇ u + |u | p (x)−2 u = K (x)|u |q(x)−2 u in R N ,
where p , q, a, K : R → R are continuous functions verifying the following hypotheses: N
(H1 ) 1 < p − = infRN p (x) p (x) p + = supRN p (x) < N in R N . (H2 ) There exists R ∗ > 0 such that p (x) m in R N , p (x) ≡ m for all x ∈ B cR ∗ (0). (H3 ) 1 < q− = infRN q(x) q(x) q+ = supRN q(x) in R N , p + < q− and p (x) q(x) p ∗ (x) =
Np (x) N − p (x)
in R N .
The notation h g means that infRN { g (x) − h(x)} > 0. Relating to functions a and K , we assume the following conditions:
(H4 ) a(x) 0 in R N , a ≡ 0, a ∈ L ∞ (R N ), and there exists θ > 0 such that 1 x ∈ R N ; a(x) = 0 ⊂ B θ (0), inf a(x) > 0 and ∈ L r (x) B θ (0) a
R N \ B θ (0)
for some function r (x) ∈
L ∞ (R N ) ∩ C (R N ) satisfying
Nq(x) p (x) N + p (x)q(x) − Nq(x)
1
r (x) and
1 + r (x) r (x)
p (x).
E-mail address: [email protected]. Partially supported by FAPESP 2007/03399-0, CNPq 472281/2006-2 and 620025/2006-9.
0022-247X/$ – see front matter doi:10.1016/j.jmaa.2008.04.060
© 2008 Elsevier Inc.
All rights reserved.
(P )
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C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
(H5 ) There exist η, β > 0 such that
η < K (x) < β, ∀x ∈ RN . Equations with variable exponents appear in various mathematical models, for example:
• Electrorheological fluids: see Acerbi and Mingione [1,2], Antontsev and Rodrigues [5] and Ruzicka [18]. • Nonlinear Darcy’s law in porous medium: see Antontsev and Shmarev [6,7]. • Image processing: see Chambolle and Lions [8] and Chen, Levine and Rao [11]. The study of L p (x) and W 1, p (x) spaces done by Kovacik and Rákosník [19] has generated a large amount of research of equations involving variable exponents, for example, Alves and Souto [3], Fan [15], El Hamidi [12], Chabrowski and Fu [10], Mihailescu, Pucci and Radulescu [21] and references therein. In [9], Chabrowski has considered the problem ( P ) for the particular case where p (x) = 2 and q(x), r (x) are constant functions verifying the hypotheses (H3 )–(H5 ). In that paper, the author established the existence of a solution to ( P ) by minimization together Lagrange Multiplier Theorem. We cite the papers of Murthy and Stampacchia [20], Musso and Passaseo [17], Passaseo [22] and reference therein, where other types of degenerate elliptic problems were also considered. In this paper, motivated by [9] and some arguments developed by Alves and Souto [3], we complete the study made in [9], in the sense that we are considering the same equation in R N , but with variable exponents. However, we would like to stress that the method used in [9] cannot be applied here because the operator and nonlinearity are not homogeneous. 1, p (x) This paper is organized in the following way: In Section 2, we define the space W a(x) (R N ) and establish an embedding result for this space. In Section 3, we show the existence of solutions for two different behaviors of q(x) at infinity. 1, p ( x )
2. The space W a(x)
RN ) (R
The appropriate Sobolev space to study problem ( P ) is the space W a(x) (R N ), defined as a completion of C 0∞ (R N ) with respect to norm 1, p (x)
u = ∇ u p (x),a(x) + u p (x) where
∇ u p (x),a(x) = inf α > 0,
∇ u p (x) a(x) dx 1
RN
α
and
p (x) u
u p (x) = inf α > 0, dx 1 . α
RN
1, p (x)
Using similar arguments developed in [13,14,19], it is possible to prove that W a(x) (R N ) is separable and reflexive. Since a is bounded from above, we have the following continuous embedding
1, p (x)
W 1, p (x) R N → W a(x)
RN . 1, p (x)
Another important norm on W a(x) (R N ) is
∇ u p (x) u p (x)
u ∗ = inf α > 0, a(x) + dx 1 α α RN
which is equivalent to . 1, p (x) The next lemma establishes other embeddings involving W a(x) (R N ), which are key points in all this work. Lemma 2.1. Suppose that (H1 )–(H5 ) hold. Then, the following continuous embeddings hold
R N → W 1,s(x) B R (0) ,
1, p (x)
W a(x) and
1, p (x)
W a(x)
R N → L q(x) R N .
∀ R > 0 where s(x) =
p (x)r (x) 1 + r (x)
(2.1)
(2.2)
C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
733
Proof. We begin fixing R > 1 such that
x ∈ R N ; a(x) = 0 ⊂ B R −1 (0)
and
inf
R N \ B R −1 (0)
a(x) > 0.
Since Nq(x) p (x) N + p (x)q(x) − Nq(x)
r (x)
a straightforward calculation leads to Ns(x)
q(x)
N − s(x)
.
This together with some results found in [13,14] yields
W 1,s(x) B R (0) → L q(x) B R (0) .
(2.3)
In the what follows, φ R is a C 1 function verifying
• φ R (x) = 1, ∀x ∈ B R (0); • φ R (x) = 0, ∀x ∈ R N \ B R +1 (0) and
• 0 φ R (x) 1, ∀x ∈ R N . Fixing
Θ = ∇ u p (x),a(x), B R +1 (0) + u p (x), B R +1 (0) , we get
∇ u s(x) Θ dx B R (0)
Hence
∇(u φ R ) s(x) Θ dx.
B R +1 (0)
∇ u s(x) Θ dx c 1
B R (0)
∇ u s(x) Θ dx + c 1
B R +1 (0)
s(x) u dx Θ
B R +1 (0)
for some positive constant c 1 , because φ R is a smooth function. By virtue of preceding inequality together with a ∈ L ∞ (R N ), it follows that there exists c 2 > 0 such that
s(x) ∇ u s(x) r (x) 1 dx c 1 a 1+r(x) ∇ u dx + c 2 r (x) Θ Θ a 1+r (x) RN
B R (0)
RN
s(x) u 1 dx. r (x) Θ a 1+r (x)
From Hölder’s inequality
s(x) s(x) −s(x) ∇ u s(x) s(x) u −s(x) a p(x) p(x)r(x) , dx c 1 a p(x) p(x)r(x) a p(x) ∇ u + c 2 Θ Θ p(x) Θ p(x) s(x) s(x) s(x)
B R (0)
that is,
s(x) s(x) ∇ u s(x) s(x) u . dx c 3 a p(x) ∇ u + c 3 Θ Θ p(x) Θ p(x) s(x)
B R (0)
Since
s(x)
s(x) ∇ u s(x) a p(x) Θ p(x) , s(x)
it follows that
s(x) u Θ p(x) 1, s(x)
∇ u s(x) s+ Θ dx 2c 3 (1 + 2c 3 ) .
B R (0)
s(x)
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C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
From this
s(x) ∇u (1 + 2c )Θ dx 1, 3
B R (0)
and, therefore,
∇ u s(x) (1 + 2c 3 )Θ (1 + 2c 3 ) u .
(2.4)
Since there exists c 4 > 0 verifying
u s(x), B R (0) c 4 u p (x), B R +1 (0) ,
(2.5)
combining (2.4) and (2.5), we get
∇ u W 1,s(x) ( B R (0)) c 5 u for some positive constant c 5 . At this point (2.1) is proved. Now, we shall prove (2.2). To this end, we use the following inequality
u q(x), B R −1 (0) c 6 u φ R −1 q(x), B R (0) which together with the continuous embedding W 1,s(x) ( B R (0)) → L q(x) ( B R (0)) imply
u q(x), B R −1 (0) c 6 u φ R −1 W 1,s(x) ( B R (0)) and thus
u q(x), B R −1 (0) c 7 u W 1,s(x) ( B R (0)) .
(2.6)
Consequently, (2.6) and (2.1) combined imply that there exists c 8 > 0 such that
u q(x), B R −1 (0) c 8 u .
(2.7)
γ > 1 large enough, we have p (x) ∇u dx 1, γ u
On the other hand, for
B cR −1 (0)
from where follows that
∇ u p (x), B cR −1 (0) γ u .
(2.8)
Here we have used the assumption infRN \ B R −1 (0) a(x) > 0. Estimate (2.8) and the definition of the norm imply that there exists c 9 > 0 verifying
u W 1, p(x) ( B c
R −1 (0))
c 9 u .
Now, recalling that there exists c 10 > 0 such that
u q(x), B cR −1 (0) c 10 u W 1, p(x) ( B c
R −1 (0))
,
the last two inequalities yield
u q(x), B cR −1 (0) c 11 u .
(2.9)
From (2.7) and (2.9), we finish the proof of the embedding (2.2).
2
3. The variational framework 1, p (x)
1, p (x)
In this paper, a function u ∈ W a(x) (R N ) is a solution of ( P ) if for each φ ∈ W a(x) (R N ) the equality below holds
a(x)|∇ u | p (x)−2 ∇ u ∇φ + |u | p (x)−2 u φ − K (x)|u | p (x)−2 u φ dx = 0.
(3.1)
RN 1, p (x)
To find a solution to ( P ), we shall study the critical points of the functional I : W a(x) (R N ) → R given by
I (u ) = RN
p (x) 1 a(x)∇ u + |u | p (x) dx − p (x)
RN
K (x) q(x)
|u |q(x) dx.
C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
735
Since q− > p + , by using standard arguments, see for example [3] and [15], it is possible to prove that I verifies the Mountain 1, p (x)
Pass Geometry and that there exists a (PS)d sequence {un } ⊂ W a(x) (R N ), that is, I (u n ) → d
I (u n ) → 0
and
where
(3.2)
d = inf max I g (t ) > 0
(3.3)
g ∈Γ 0t 1
and
1, p (x) Γ = g ∈ C [0, 1], W a(x) R N : g (0) = 0 and I g (1) 0 .
Lemma 3.1. Let {un } be a (PS)c sequence of I . Then, {un } has a subsequence, still denoted by {un }, such that:
• • • • •
{un } is a bounded sequence with the weak limit denoted by u; 1, p (x) 1,m un → u in W a(x),loc (R N ) and W a(x),loc (R N ); N un (x) → u (x) a.e. in R ; p (x) q(x) un (x) → u (x) in L loc (R N ) and L loc (R N ); 1,m c un → u in W loc ( B R ), for all R > θ .
Proof. From (3.2), there exist C > 0 and n0 ∈ N such that I (u n ) −
1 I (un )un C + un , q−
∀n n0 ,
or equivalently
a(x)|∇ un | p (x) + |un | p (x) dx k1 + k2 un ,
∀n n0
RN
for some positive constants k1 and k2 , because p + < q− . As and ∗ are equivalent norms, if un 1, we get
u n p − k 1 + k 3 u n 1, p (x)
from where follows that {un } is bounded. The existence of a weak limit to {un } follows from the reflexivity of W a(x) (R N ). For each R > 0, the embedding 1, p (x)
W a(x)
1,m
B R (0) → W a(x) B R (0)
(3.4)
is continuous. In fact, the inequality
p (x) ∇ u m 1 + ∇u u u leads to
∇ u m
u
a(x)
B R (0)
a(x) dx +
B R (0)
B R (0)
∇ u p (x)
u
a(x)
a(x) dx + 1
B R (0)
and thus
m1
m
a(x)|∇ u | dx
C u ,
1, p (x)
∀u ∈ W a(x)
B R (0) .
(3.5)
B R (0)
Arguing as above,
m
m1
a(x)|u | dx
C u ,
1, p (x)
∀u ∈ W a(x)
B R (0) .
(3.6)
B R (0)
Then, (3.5) and (3.6) combined give (3.4). Now, denoting
P n (x) = a(x) |∇ un | p (x)−2 ∇ un − |∇ u | p (x)−2 ∇ u , ∇ un − ∇ u + |un | p (x)−2 − |u | p (x)−2 (un − u )(x) 0,
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C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
for each R > 0 we have the following inequality:
P n dx I (un )(φ R un ) − I (un )(φ R u ) − |x| R
RN
φ R |un | p (x) dx +
− RN
a(x)|∇ un | p (x)−2 un ∇ un ∇φ R dx +
φ R |un | p (x)−2 uun dx +
RN
|un |q(x) φ R −
RN
a(x)|∇ un | p (x)−2 u ∇ un ∇φ R dx
RN
|un |q(x)−2 un u φ R + on (1)
RN
which implies
lim
n→∞
P n dx = 0.
|x| R
Since and ∗ are equivalents, the last limit leads to un → u
1, p (x)
in W a(x)
B R (0) , ∀ R > 0.
(3.7)
The same type of arguments can be used to prove that un → u
1,m
in W loc B cR (0) , ∀ R > θ.
(3.8)
The other limits follow from (2.1) and (3.7).
2
4. Existence of solutions In this section, we shall show the existence of a solution for ( P ). We divide this section into two subsections considering different situations involving the behavior of q(x) at infinity. From now on, the functions a(x) and K (x) verify the following conditions:
(H6 ) a(x) a∞ on R N , where a∞ = lim|x|→∞ a(x) with the strict inequality on a set of positive measure in R N . (H7 ) K (x) K ∞ on R N , where K ∞ = lim|x|→∞ K (x). Hereafter, let us denote by I ∞ , the Euler–Lagrange functional associated to the problem
−a∞ m u + um−1 = K ∞ |u |s−2 u in R N , u 0, u ≡ 0 and u ∈ W 1,m R N ,
( P ∞)
and by c ∞ the mountain pass level related to I ∞ . Since a∞ and K ∞ are positive constants, the existence of ground state solution to ( P ∞ ) follows by using the same arguments explored for the case a∞ = K ∞ = 1. Moreover, it is possible to prove the existence of a positive radially symmetric ground state w of ( P ∞ ) verifying
w (x), ∇ w (x) → 0 as |x| → ∞.
(4.1)
More details about the above informations can be found in [4] and [16]. Lemma 4.1. Let w be a ground state solution of ( P ∞ ). Then, there exists η∗ > 0 such that
w ∞ < 1,
∀ K ∞ η∗ .
Proof. For each K ∞ > 0, we consider the function 1
s+1−m U∞ = K∞ w.
By a direct computation, we can prove that U ∞ is a solution of
−a∞ m U + U m−1 = U s in R N , and that there exists C 1 > 0 verifying
U ∞ H 1 (RN ) C 1 ,
∀ K ∞ 0.
Using Moser iteration technique, the last inequality implies that there exists C 2 > 0 satisfying
U ∞ ∞ C 2 ,
∀ K ∞ 0.
Hence, there exists η∗ such that
w ∞ < 1,
∀ K ∞ η∗
and this completes the proof of lemma.
2
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737
4.1. Constant at infinity In this subsection, we shall study the case where q(x) is equal to a constant at infinity. The next proposition establishes an important relation involving the minimax levels d and c ∞ . Proposition 4.1. Suppose (H1 )–(H5 ) and that q(x) satisfies a.e. in R N
q(x) s
for all |x| R ∗ .
and q(x) ≡ s
( Q 1)
If I admits a (PS)d sequence converging weakly to 0, then d c ∞ . Proof. For each R > max{θ, R ∗ , R ∗ }, let us consider ψ R (x) = 1 − φ R (x), where φ R was introduced in the proof of Lemma 2.1, and tn, R > 0 verifying
I ∞ tn, R (ψ R un ) = max I ∞ t (ψ R un ) . t 0
From the definition of c ∞
c ∞ I ∞ tn, R (ψ R un ) = I (un ) + I ∞ tn, R (ψ R un ) − I (un ) . Assuming for a moment the claim below Claim 4.1. lim R →+∞ [lim supn→+∞ ( I ∞ (tn, R (ψ R un )) − I (un ))] = 0. Letting n → +∞ and after R → +∞, we obtain the inequality c ∞ lim I (un ) = d, n→+∞
2
which completes the proof of the proposition. Proof of Claim 4.1. First of all, we observe that
I ∞ tn, R (ψ R un ) − I (un )
tnm, R
=
m
m
a∞ ∇(ψ R un ) dx −
RN
+
K (x) q(x)
a(x) p (x)
RN
|un |q(x) dx −
RN
K∞
tns , R s
m t n, R |∇ un | p (x) dx + |ψ R un |m dx − m
RN
|ψ R un |s dx .
1 p (x)
|un | p (x) dx
RN
RN
In the next, we shall show that each expression in the above parentheses goes to zero, when n → +∞ and after R → +∞. We shall assume for a moment the following claim. Claim 4.2. There exists R 0 > 0 such that lim sup tn, R = 1, n→+∞
∀R R0.
Observe that
K (x) q(x)
RN
|un |q(x) dx −
K∞ s
tns , R |ψ R un |s dx
RN
K∞ ( K (x) − K ∞ ) |un |q(x) dx + 1 − tns , R q(x) s
= RN
K∞
− tns , R R |x| R +1
s
q(x)
|ψ R un |
|un |s dx +
|x|> R +1
R →+∞
lim sup n→+∞
K (x) q(x)
RN
|un |q(x) dx
dx.
Combining (H6 ), (H7 ), Lemma 2.1 and Claim 4.2, it follows that lim
|x| R +1
K∞ q(x)
|un |q(x) dx − RN
K∞ s
tns , R |ψ R un |s dx
= 0.
(4.2)
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C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
Arguing as above, it is possible to prove that
lim
R →+∞
and
n→+∞
lim
m
RN
tnm, R m
n→+∞
a(x) p (x)
|∇ un | p (x) dx = 0
(4.3)
RN
lim sup
R →+∞
m ∇(ψ R un ) dx −
a∞
tnm, R
lim sup
|ψ R un |m dx −
RN
1 p (x)
|un | p (x) dx = 0.
(4.4)
RN
Thus, from (4.2)–(4.4) the proof of Claim 4.1 is completed.
2
Proof of Claim 4.2. Recalling that q(x) satisfies Np (x)
p (x) q(x) p ∗ (x) =
N − p (x)
,
we have
|un |q(x) dx → 0
|x| R
for all R > R ∗ . Thus
|un |q(x) dx = |x| R
|un |s dx 0,
|x| R
because, otherwise, we have the limit
|un |q(x) dx → 0 RN
which yields d = 0, and this is a contradiction. Now, taking tn, R > 0 verifying
m
a∞ ∇(ψ R un ) dx +
RN
m |ψ R un |m dx = K ∞ tns− ,R
RN
|ψ R un |s dx
(4.5)
RN
and using the fact that {ψ R un } is bounded in W 1,m (R N ), by a straightforward computation, it is possible to check that for each R > 0 fixed, the sequence {tn, R } is bounded. In what follows, we denote by t R the following real number lim sup tn, R = t R . n→+∞
On the other hand, combining (3.2) and (4.5) we easily derive
a(x)|∇ un | p (x) dx − RN
m
a∞ ∇(ψ R un ) dx +
RN
q(x)
=
K (x)|un |
m dx − tns− ,R
RN
RN
|un | p (x) dx −
|ψ R un |m dx
RN
s
K ∞ |ψ R un | dx + on (1).
RN
This together with Lemma 2.1 yield
0 = K ∞ 1 − t sR−m where
αR
α R = lim supn→+∞
|x| R |un | dx. Since d > 0, we derive
|un |s dx = 0,
lim
n→+∞
|x| R
and
lim
n→+∞
RN
(4.6)
|un |s dx = 0.
s
∀ R > 0,
C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
739
Hence, lim
R →+∞
α R = α > 0.
Thus, the last limit together with (4.6) imply that there exists R 0 > 0 such that t R = 1,
∀R R0,
which completes the proof of Claim 4.2.
2
Lemma 4.2. Assume that (H1 )–(H7 ) and ( Q 1 ) hold. Then, d < c ∞ . Proof. Let w be a positive radially symmetric ground state solution for ( P ∞ ), and consider the sequences xn = (0, 0, . . . , n), w n (x) = w (x − xn ) and tn > 0 satisfying I (tn w n ) = max I (t w n ). t 0
From the definition of d,
d I (tn w n ) = I ∞ (tn w n ) + I (tn w n ) − I ∞ (tn w n ) , that is,
d c∞ +
a∞
|x| R ∗
|tn ∇ w n | p (x) |tn ∇ w n |m − + p (x) m
|x| R ∗
|tn w n | p (x) |tn w n |m − + p (x) m
|x| R ∗
K∞
|tn w n |s s
−
|tn w n |q(x) . q(x)
Fixing an index n large enough such that |∇(tn w n )| < 1 and |tn w n | < 1 in B R ∗ (0), and by using the fact that f (s) = s−1 τ s is decreasing in (0, +∞), if 0 < τ < 1, for that n, we can conclude that
a∞ |x| R ∗
and
|tn ∇ w n | p (x) |tn ∇ w n |m − , p (x) m
K∞ |x| R ∗
|tn w n |s s
|tn w n |q(x) q(x)
−
are less than zero. Hence d < c ∞ .
|x| R ∗
|tn w n | p (x) |tn w n |m − p (x) m
2
Theorem 4.1. Assume that (H1 )–(H7 ) and ( Q 1 ) hold. Then, problem ( P ) has a solution. 1, p (x)
Proof. For each φ ∈ W a(x) (R N ), the (PS)d sequence {un } satisfies
a(x)|∇ un | p (x)−2 ∇ un ∇φ + |un | p (x)−2 un φ dx =
RN
K (x)|un |q(x)−2 un φ dx + on (1).
RN
Taking the limit as n → +∞ and by using Lemma 3.1, it follows that
a(x)|∇ u | p (x)−2 ∇ u ∇φ + |un | p (x)−2 u φ dx =
RN
K (x)|u |q(x)−2 u φ dx.
RN
Thus u is a solution. Moreover, combining Lemma 4.2 and Proposition 4.1, we deduce that u cannot be zero.
2
4.2. Asymptotically constant at infinity In this subsection, we shall study the case where q(x) is not equal to s, but q(x) goes to s as |x| → ∞. Lemma 4.3. Suppose that q(x) satisfies q(x) s,
∀x ∈ R N
and
lim q(x) = s.
|x|→∞
If I admits a (PS)d sequence converging weakly to 0, then c c ∞ .
( Q 2)
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C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
Proof. As in the proof of Proposition 4.1, we need to prove that Claims 4.1 and 4.2 remain valid by assuming that ( Q 2 ) holds. Reviewing the proof of Claim 4.2, we need to show that
m K (x)|un |q(x) dx − tns− ,R
lim sup n→+∞
RN
K ∞ |ψ R un |s dx = K ∞ 1 − t sR−m
αR
(4.7)
RN
for all R > 0. For each R > 0, it is possible to show that
m K (x)|un |q(x) dx − tns− ,R
RN
m K ∞ |ψ R un |s dx = K ∞ 1 − tns− ,R
|un |s dx + K ∞
|x| R
RN
|un |q(x) − |un |s dx + on (1).
RN
Repeating the same arguments used in [3, Lemma 4.2], the second term of the last equality is on (1). In this way,
m K (x)|un |q(x) dx − tns− ,R
RN
m K ∞ |ψ R un |s dx = K ∞ 1 − tns− ,R
|un |s dx + on (1)
|x| R
RN
and (4.7) follows. The limit (4.2) follows by using the same type of arguments. Therefore, Claims 4.1 and 4.2 remain valid if ( Q 2 ) holds. 2 Lemma 4.4. Assume that (H1 )–(H7 ) and ( Q 2 ) hold. Then, there is η∗ > 0 such that d < c ∞ for K ∞ η∗ . Proof. First of all, from Lemma 4.1 there exists
| w |∞ 1,
η∗ > 0 such that the ground state w of ( P ∞ ) satisfies
∀ K ∞ η∗ .
(4.8)
Taking w n (x) = w (x − xn ) and tn > 0 verifying I (tn w n ) = max I (t w n ), t 0
it follows that
d I (tn w n ) = I ∞ (tn w n ) + I (tn w n ) − I ∞ (tn w n ) which is equivalent to
d c∞ +
a∞
|x| R ∗
+
tns
K∞
s
p (x)
tn
p (x)
|∇ w n | p (x) −
tnm m
| w n | dx −
tn
q(x)
q(x)
|wn |
|x| R ∗
q(x)
s
|∇ w n |m dx +
p (x)
tn
p (x)
tnm
| w n | p (x) dx −
m
| w n |m dx
dx.
RN
Using for a moment the claim Claim 4.3. The sequence {tn } is convergent with limit equal to 1. (4.1) together with Lemma 4.1 leads to
∇(tn w n )
∞
< 1 in B R ∗ (0)
and
|tn w n |∞ < 1 in R N for n large enough. Since f (s) = s−1 τ s is decreasing in (0, +∞) if 0 < τ < 1, the terms
a∞ |x| R ∗
and
K∞
tns s
p (x)
tn
p (x)
|∇ w n |
p (x)
−
tnm m
|∇ w n |
| w n | dx −
tn
q(x)
q(x)
|wn |
RN
are negative, and then d < c ∞ .
2
dx, |x| R ∗
q(x)
s
m
dx
p (x)
tn
p (x)
|wn |
p (x)
dx −
tnm m
m
|wn |
dx
C.O. Alves / J. Math. Anal. Appl. 345 (2008) 731–742
Proof of Claim 4.3. For each n ∈ N, we have
p (x)
RN
so
a(x)|∇ w n | p (x) + | w n | p (x) dx =
tn
q(x)
K (x)tn
741
| w n |q(x) dx,
RN
p (x)
a∞ |∇ w n |m + | w n |m dx
tn RN
q(x)
K ∞ tn
| w n |s dx.
(4.9)
RN
The inequality (4.9) implies that {tn } is bounded, because if for some subsequence, still denoted by {tn }, we have lim tn = +∞,
n→+∞
it easy to see that
q − p+
a∞ |∇ w |m + | w |m dx tn−
RN
K ∞ | w |s dx
(4.10)
RN
for n large enough. This implies that {tn } is bounded, obtaining this way a contradiction. Now, we shall show that lim infn→+∞ tn = lim supn→+∞ tn = 1. Arguing by contradiction, if lim infn→+∞ tn > 1 holds, we have tn > 1 for n large enough, and thus (4.10) is true. On the other hand, using the equality
a∞ |∇ w |m + | w |m dx =
RN
K ∞ | w |s dx
(4.11)
RN
it follows from (4.10) and (4.11) that q − p+
1 tn−
for n large enough, and so lim infn→+∞ tn 1, which is impossible. Now, arguing again by contradiction, if lim supn→+∞ tn < 1 holds, we have tn < 1 for n large enough, and thus
q − p+
a(x + xn )|∇ w | p (x+xn ) + | w | p (x+xn ) dx tn−
RN
K ∞ | w |s dx + on (1).
(4.12)
RN
From (4.11) and (4.12), we can conclude that lim supn→+∞ tn 1, obtaining again a contradiction. From the above analysis we derive that lim tn = 1
n→+∞
and this completes the proof of Claim 4.3.
2
Theorem 4.2. Assume that (H1 )–(H7 ) and ( Q 2 ) hold and q(x) s a.e. in R N . Then, for K ∞ η∗ problem ( P ) has a solution. Proof. The proof follows from similar arguments to the ones used in the proof of Theorem 4.1.
2
Acknowledgments The author would like to thank ICMC/USP – São Carlos, and specially to the Professor Sérgio Monari for his attention and friendship. This work was done while the author was visiting that institution. The author would like to express sincere thanks to the Professor Xianling Fan and to the reviewers for their valuable comments about this subject.
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