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Positive solutions of the p(x) -Laplace equation with singular nonlinearity
Nonlinear Analysis 72 (2010) 4428–4437
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Positive solutions of the p(x)-Laplace equation with singular nonlinearity Jingjing Liu Department of Mathematics, Sun Yat-sen University, 510275 Guangzhou, China
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In this paper, using sub–supersolution method, we study the positive solution of the p(x)Laplacian equations −∆p(x) u = f (x, u) with Dirichlet boundary condition on a bounded domain Ω in Rn with C 1,ω boundary for some 0 < ω < 1 for a possibly singular nonlinearity f on Ω × (0, ∞). Crown Copyright © 2010 Published by Elsevier Ltd. All rights reserved.
Article history: Received 7 September 2009 Accepted 10 February 2010 MSC: 34B15 35A15 35G99 Keywords: p(x)-Laplacian Singular boundary value problem Sub–supersolution method
1. Introduction Let Ω ⊆ Rn (n ≥ 1) be a bounded domain with C 1,ω boundary for some 0 < ω < 1 and f : Ω × (0, ∞) −→ [0, ∞) be a given singular nonlinearity. In this paper we are concerned with the existence of weak solutions of the following singular boundary value problem:
( −∆p(x) u = f (x, u) u>0 u=0
in Ω , in Ω , on ∂ Ω
(1.1)
where p(x) ∈ C (Ω ), 1 < p− = infx∈Ω p(x) ≤ p+ = supx∈Ω p(x) < +∞. Let p(x) satisfy the following condition: [P ] The function p(x) is monotone when n = 1; there is a vector l ∈ Rn \ {0} such that for any x ∈ Ω , f (t ) = p(x + tl) is monotone for t ∈ Ix = {t |x + tl ∈ Ω } when n > 1. The operator −∆p(x) u = −div(|∇ u|p(x)−2 ∇ u) is called p(x)-Laplacian, which becomes p-Laplacian when p(x) ≡ p (a constant). The p(x)-Laplacian possesses more complicated nonlinearities than the p-Laplacian. For example, it is inhomogeneous, and in general, it has no first eigenvalue, in other words, the infimum of the eigenvalues of p(x)-Laplacian equals 0 (see [1]). So some techniques used in the constant exponent case cannot be carried out for the variable exponent case. Moreover, there are two cases about the infimum of the eigenvalues of p(x)-Laplacian, one is the infimum of the eigenvalue is an eigenvalue indeed, the other case is that the infimum of the eigenvalues is not an eigenvalue. In this paper, we consider the first case only. The aim of the present paper is to extend some results of [2] to the variable exponent case by using the sub–supersolution method. The regularity results and the comparison principles are the bases of the sub–supersolution method. For the
E-mail address: [email protected]. 0362-546X/$ – see front matter Crown Copyright © 2010 Published by Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.02.018
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
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regularity results in the variable exponent case, see [3–5], more precisely, for the L∞ and C 0,α regularity, see [5]; for the local C 1,α regularity of the minimizers of the corresponding integral functional, see [3]; for the global C 1,α regularity, see [4]. The present paper is organized as follows. In Section 2, we recall some basic facts about the variable exponent Lebesgue and Sobolev spaces. In Section 3 we provide basic facts and recall some lemmas that will be needed later. We also specify the conditions on the nonlinearity f that will be used throughout the paper. Section 4 contains the main result on the existence of weak solutions of the problem (1.1). Several results that can be drawn from the main theorem are also given. 2. Preliminaries 1 Let Ω be a bounded domain of Rn (n ≥ 1) with a smooth boundary ∂ Ω . Denote by S (Ω ) the set of all measurable real functions defined on Ω . Note that two measurable functions are considered as the same element of S (Ω ) when they are equal almost everywhere. ∞ L∞ + (Ω ) = {p ∈ L (Ω ) : ess inf p(x) > 1}. x∈Ω
For p ∈ L∞ + (Ω ), denote p− = p− (Ω ) = ess inf p(x),
p+ = p+ (Ω ) = ess sup p(x).
x∈Ω
x∈Ω
(Ω ) by p(x) |u(x)| dx < ∞ ,
Define the variable exponent Lebesgue space L Lp(x) (Ω ) =
Z
u ∈ S (Ω ) :
Ω
p(x)
with the norm
) Z u(x) p(x) = inf λ > 0 : λ dx ≤ 1 Ω (
kukLp(x) (Ω ) = |u|p(x)
and the variable exponent Sobolev space W 1,p(x) (Ω ) by W 1,p(x) (Ω ) = u ∈ Lp(x) (Ω ) : |∇ u| ∈ Lp(x) (Ω ) ,
with the norm
kukW 1,p(x) (Ω ) = |u|p(x) + |∇ u|p(x) . 1,p(x)
(Ω ) denotes the closure of C0∞ (Ω ) in W 1,p(x) (Ω ). |∇ u|Lp(x) (Ω ) is an equivalent norm on W01,p(x) , we denote this norm 1,p(x) by kuk. The spaces Lp(x) (Ω ), W 1,p(x) (Ω ) and W0 (Ω ) are all separable and reflexive Banach spaces. We refer to [6–8] for W0
the elementary properties of these spaces. Here, we only display some facts which will be used later. For u, v ∈ S (Ω ), we write u ≤ v if u(x) ≤ v(x) for a.e. x ∈ Ω . Define u+ (x) = max{u(x), 0} and u− (x) = max{−u(x), 0}. 0 0 Proposition 2.1 ([6]). The conjugate space of Lp(x) (Ω ) is Lp (x) (Ω ), where p(1x) + p01(x) = 1. For any u ∈ Lp(x) (Ω ) and v ∈ Lp (x) R (Ω ) , Ω |uv| dx ≤ 2 |u|p(x) |v|p0 (x) .
Proposition 2.2 ([6]). Set ρ(u) =
R
Ω
|u(x)|p(x) dx. For u, uk ∈ Lp(x) (Ω ), we have
(1) For u 6= 0, |u|p(x) = λ ⇔ ρ( λ ) = 1; (2) |u|p(x) < 1 (=1; > 1) ⇔ ρ(u) < 1 (=1; > 1); u
p−
p+
p+
p−
(3) If |u|p(x) > 1, then |u|p(x) ≤ ρ (u) ≤ |u|p(x) ; (4) If |u|p(x) < 1, then |u|p(x) ≤ ρ (u) ≤ |u|p(x) ; (5) limk→∞ |uk |p(x) = 0 ⇐⇒ limk→∞ ρ(uk ) = 0; (6) |uk |p(x) → ∞ ⇐⇒ ρ(uk ) → ∞. Similar to Proposition 2.2, we have Proposition 2.3. Set φ(u) = (1) (2) (3) (4) (5) (6)
R
Ω
|∇ u|p(x) dx. For u, uk ∈ Lp(x) (Ω ), we obtain
For u 6= 0, kuk = λ ⇔ φ( λu ) = 1; kuk < 1 (=1; > 1) ⇔ φ(u) < 1 (=1; > 1); − + If kuk > 1, then kukp ≤ φ (u) ≤ kukp ; + − If kuk < 1, then kukp ≤ φ (u) ≤ kukp ; limk→∞ kuk k = 0 ⇐⇒ limk→∞ φ(uk ) = 0; kuk k → ∞ ⇐⇒ φ(uk ) → ∞.
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3. Preliminaries 2 Throughout the paper the nonlinearity f : (0, ∞) → [0, ∞) will be assumed to be a Caratheodory function such that f (·, t ) is in C θ (Ω ) for some 0 < θ < 1. Unless specified otherwise we will also suppose that Ω ∈ Rn is a bounded domain with C 1,ω boundary for some 0 < ω < 1. The positive constants that will be used in estimations are not necessarily the same in every occurrence. In what follows, further conditions will be stipulated on f . Now, consider equation
−∆p(x) u = f (x, u) u=h
in Ω , on ∂ Ω .
(3.1)
1,p(x)
Definition 3.1. (1) Given h ∈ W 1,p(x) (Ω ), u ∈ W 1,p(x) (Ω ) is called a (weak) solution of (3.1) if u − h ∈ W0
Z
|∇ u|p−2 ∇ u · ∇ϕ = Ω
Z Ω
f (x, u)ϕ
(Ω ) and (3.2)
1,p(x)
for all ϕ ∈ W0 (Ω ). (2) u ∈ W 1,p(x) (Ω ) is called a sub-solution (respectively a supersolution) of (3.1) if u ≤ (respectively ≥) h on ∂ Ω and for 1,p(x) all ϕ ∈ W0 (Ω ) with ϕ ≥ 0,
Z Ω
|∇ u|p−2 ∇ u · ∇ϕ ≤ (respectiv ely ≥)
Z Ω
f (x, u)ϕ.
(3.3)
1,p(x)
Remark 3.2. (1) We say u ∈ W0 (Ω ) is a (weak) solution of (1.1) if u(x) > 0 in Ω and (3.2) holds. Obviously, (1.1) is a special case of (3.1). 1,p(x) (2) Following usual practice, we shall write u ≤ h on Ω to mean (u − h)+ ∈ W0 (Ω ); u ≥ h on Ω to mean (u − h)−
∈ W01,p(x) (Ω ).
Now we give a comparison principle as follows. Lemma 3.3. Let g (x, t ) : Ω × R → R be measurable and nondecreasing in t. Let u, v ∈ W 1.p(x) (Ω ) satisfy
−∆p(x) u + g (x, u) ≤ −∆p(x) v + g (x, v) (x ∈ Ω ). If u ≤ v on ∂ Ω , then u ≤ v on Ω . Proof. It is an obvious result according to Proposition 2.3. in [9] by using reduction to absurdity.
Lemma 3.4 (See [1]). If n = 1, then the infimum of the eigenvalues of p(x)-Laplacian > 0 if and only if the function p(x) is monotone. Lemma 3.5 (See [1]). Let n > 1, if there is a vector l ∈ Rn \ {0} such that for any x ∈ Ω , f (t ) = p(x + tl) is monotone for t ∈ Ix = {t |x + tl ∈ Ω }, then he infimum of the eigenvalues of p(x)-Laplacian > 0. p(x)
In this paper we shall use the notation q(x) for the Höder conjugate of p(x) > 1, that is q(x) := p(x)−1 . Also we let p∗ (x) be the Sobolev conjugate exponent of p(x), namely,
Np(x) p∗ (x) = N − p(x) ∞
if p(x) < N , if p(x) ≥ N .
Before we state conditions on f that will be needed in the paper, let us identify a class G of functions g : (0, ∞) → (0, ∞) and a class B of non-negative functions b in Lq(x) (Ω ) such that the following conditions hold:
[G1 ] There is ρ ≥ 1 and a positive constant C such that g (t ) ≤ C for all t ≥ ρ . [G2 ] One of the conditions (1), (2), or (3) below holds, where h(t ) = tg (t ), t > 0: (1) h(t ) ≤ C for all 0 < t < δ and some positive constants C and δ . (2) h is nondecreasing. (3) (i) g is nonincreasing. (ii) For each ∈ (0, 1) there is a constant Cθ ≥ 1 such that g (θ t ) ≤ Cθ g (t ) for all t > 0. p∗
(iii) There is ω ∈ C01 (Ω ) with ω > 0 on Ω such that bg (ω) ∈ L p∗ −1 (Ω ).
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
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We are now ready to list the conditions on f that will be used in this paper. In stating the conditions, we use the notation
γs (x) = sup{f (x, t ) : t ≥ s} (x ∈ Ω ) for any s > 0.
[F1 ] γs (x) ∈ Lq(x) (Ω ) for each s > 0. [F2 ] There is (g , b) ∈ G × B that satisfies conditions [G1 ]–[G2 ], a measurable function a for which {x ∈ Ω : 0 < a(x) ≤ 1} has a positive measure, and 0 ≤ a ≤ b} such that (1) f (x, s) ≥ a(x) for (x, s) ∈ Ω × (0, ∞), (2) f (x, s) ≤ b(x)g (s) for (x, s) ∈ Ω × [ρ, ∞), (3) f (x, s)g (t ) ≤ Cf (x, t )g (s) for x ∈ Ω , 0 < s < t and some constant C > 0. We start with the following lemma. Lemma 3.6. Let f satisfy [F1 ] and h ∈ W 1,p(x) (Ω ) with h > 0 on Ω . Let ψsub be a sub-solution and ψsup a supersolution of (3.1) on Ω . If 0 < infΩ ψsub ≤ ψsub ≤ ψsup on Ω then there is a solution u ∈ W 1,p(x) (Ω ) of (3.1) such that ψsub ≤ u ≤ ψsup a.e. on Ω. Proof. The lemma is a consequence of [10, Theorem 4.14]. For completeness we include the short proof. Take 0 < l = infΩ ψsub and let F : Ω × R → R be defined by F (x, t ) =
f (x, l) f (x, t )
if t < l, if t ≥ l.
Then F (x, ·) is Hölder continuous on R for each x ∈ Ω and |F (x, t )| ≤ γl (x) on Ω . Consider now the Dirichlet problem
−∆p(x) u = F (x, u)
in Ω , on ∂ Ω .
u=h
Note that ψsub is a sub-solution, and ψsup is a supersolution of this problem. Since γl ∈ Lq(x) (Ω ), it follows by [10, Theorem 4.14] that this problem admits a solution u ∈ W 1,p(x) (Ω ) such that ψsub ≤ u ≤ ψsup . Finally we note that u is a solution of (3.1) as claimed because of u ≥ l. 4. Existence of weak solutions 1,p(x)
We now show the existence of a solution of (1.1) in W0
(Ω ). 1,p(x)
Theorem 4.1. Suppose that f satisfies [F1 ]–[F2 ]. Then problem (1.1) admits a solution in W0
(Ω ).
Proof. By replacing g and b by C −1 g and Cb respectively, if necessary, we can assume that g (t ) ≤ 1 for all t ≥ ρ in condition [G1 ]. For k = 1, 2, . . ., let ψk be the solution of
−∆p(x) u(x) = ak (x) u(x) = k−1
in Ω , on ∂ Ω
(4.1)
where for k = 1, 2, . . .,
ak (x) = min a(x),
k+1
k
.
Likewise, let ψ∞ be the solution of (4.1) with k = ∞. Note that since ak (x) ∈ L∞ (Ω ) the problem (4.1) does indeed have a solution. By the comparison lemma, we have 0 ≤ ψ∞ ≤ ψk ≤ ψ1 for all k = 1, 2, . . . and ψk ≥ k−1 on Ω for all k = 1, 2, . . .. By the Strong Maximum Principle [11], we note that ψ∞ > 0 on Ω . So, we can get −∆p(x) ψk ≤ a in Ω , and ψk = k−1 on ∂ Ω . For each k ∈ Z + let us now consider the Dirichlet problem
−∆p(x) u(x) = f (x, u(x)) u(x) = k−1
in Ω , on ∂ Ω .
By (1) of [F2 ] we observe that
−∆p(x) ψk − f (x, ψk ) ≤ −∆p(x) ψk − a(x) ≤ 0 and thus ψk is a sub-solution of (BVPk ) for all k.
(BVPk )
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Now, let ψ be a solution of
−∆p(x) u(x) = b(x) u(x) = ρ
in Ω , on ∂ Ω .
(4.2)
Note that (4.2) has a solution since b ∈ Lq(x) (Ω ). By comparison principle, we note that ψk ≤ ψ for all k = 1, 2, . . . and ψ ≥ ρ on Ω . Then by (2) of [F2 ] we have
−∆p(x) ψ − f (x, ψ) ≥ −∆p(x) ψ − b(x) = 0. Thus ψ is a supersolution of (BVPk ) on Ω for any k = 1, 2, . . .. Let us fix a positive integer m. By Lemma 3.6 let um be a solution of (BVPm ) such that ψm ≤ um ≤ ψ on Ω . Note that um is a supersolution of (BVPm+1 ) and therefore, again by Lemma 3.6, there is a solution um+1 of (BVPm+1 ) such that ψm+1 ≤ um+1 ≤ um on Ω . Continuing in this manner, we construct a sequence {uk } of solutions of (BVPk ) such that for all k ≥ m
ψ∞ ≤ uk+1 ≤ uk ≤ · · · ≤ um ≤ ψ in Ω . Let us also note that uk ≥ k−1 on Ω . We define u(x) = lim uk (x) k→∞
(x ∈ Ω ).
(4.3)
We use ϕk := uk − 1/k as a test function for the solution uk on Ω (by condition [F1 ]). We obtain
Z Ω
|∇ uk |p(x) =
Z Ω
|∇ uk |p(x)−2 ∇ uk · ∇(uk − /k) =
Z Ω
f (x, uk )(uk − /k) ≤
Z Ω
f (x, uk )uk .
(4.4)
We now use condition [F2 ] to show that there is a positive constant C , independent of k, such that
Z Ω
f (x, uk )uk ≤ C .
(4.5)
First we note that conditions (2) and (3) of [F2 ] imply that f (x, uk )uk =
f (x, uk )uk g ( uk )
g (uk )uk ≤ C
f (x, ψ) g (ψ)
g (uk )uk .
If (1) of [G2 ] holds, then this together with [G1 ] implies f (x, ψ) g (ψ)
g (uk )uk ≤ Mb max{ρ, uk } ≤ Mbψ
for some positive constant M. If (2) of [G2 ] holds, then recalling g (ψ) ≤ 1, it follows from (2) of [F2 ] that f (x, ψ) g (ψ)
g (uk )uk ≤ f (x, ψ)ψ ≤ bψ.
Therefore, since ψ ∈ Lp(x) (Ω ) and b ∈ Lq(x) (Ω ), the estimate (4.5) holds if either (1) or (2) of [G2 ] holds. Suppose now that (3) of [G2 ] holds. By [11] we note that ψ∞ > 0 on Ω , and ∂ψ∞ /∂ν > 0 on ∂ Ω , where ν is the unit inner normal. Thus, we have infΩ (ψ∞ /ω) > 0, where ω is as in condition 3(iii) of [G2 ]. That is θ ω ≤ ψ∞ on Ω for some 0 < θ ≤ 1. Then on using 3(i) and 3(ii) of [G2 ], we estimate f (x, ψ) g (ψ)
g (uk )uk ≤
f (x, ψ) g (ψ)
g (θ ω)ψ ≤ Cθ bg (ω)ψ.
Since ψ ∈ W 1,p(x) (Ω ) ⊆ Lp (Ω ), the assumption in 3(iii) of [G2 ] shows that estimate (4.5) holds in this case as well. ∗
1,p(x)
(Ω ). We pick a subsequence, still denoted (Ω ), strongly in Lp(x) (Ω ) and pointwise a.e. on Ω . Note that {ψk } converges 1,p(x) to u ∈ W0 (Ω ), and that {uk } converges weakly to u in W 1,p(x) (Ω ), strongly in Lp(x) (Ω ) and a.e. on Ω . Now let Ω0 b Ω . On Ω0 we have Thus estimates (4.4) and (4.5) show that the sequence ψk is bounded in W0 1,p(x)
by {ψk }, such that it converges weakly in W0
|(f (x, uk ) − f (x, uj ))(uk − uj )| ≤ 4ψl (x)ψ, where l = minΩ¯ ψ∞ , and hence proceeding as in [12] one can show that
Z Ω0
(|∇ uk |p(x)−2 ∇ uk − |∇ uj |p(x)−2 ∇ uj ) · ∇(uk − uj ) → 0
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
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as k, j → ∞. Then we can get
Z Ω0
|∇ uk − ∇ uj |p(x) → 0 k, j → ∞
(4.6)
by using the inequality (see [13])
[(|ξ |p−2 ξ − |η|p−2 η)(ξ − η)] · (|ξ |p + |η|p )(2−p)/p ≥ (p − 1)|ξ − η|p , p 1 (|ξ |p−2 ξ − |η|p−2 η)(ξ − η) ≥ |ξ − η|p , p ≥ 2.
1 < p < 2;
2
p(x)
Since {uk } converges strongly to u in L (Ω0 ), the limits (4.6) shows that the sequence {uk } is Cauchy in W 1,p(x) (Ω0 ), and hence converges to u in W 1,p(x) (Ω0 ). In conclusion, given any compact set Ω0 ⊂ Ω , we can find a subsequence of {uk } that converges to u strongly in W 1,p(x) (Ω0 ). Let us take note of the following estimates. If p(x) ≥ 2, then by the Hölder inequality Propositions 2.1 and 2.3, we have
Z
|∇ uk − ∇ u|(|∇ uk | + |∇ u|)p(x)−2 dx ≤ 2k∇ uk − ∇ ukLp(x) (Ω ) k(|∇ uk | + |∇ u|)p(x)−2 kLq(x) (Ω0 ) 0
Ω0
1
−
= 2k∇ uk − ∇ ukLp(x) (Ω0 ) (k(|∇ uk | + |∇ u|)p(x)−2 kqLq(x) (Ω ) ) q− 0 Z 1− (p(x)−2)p(x) q p ( x )− 1 ≤ 2kuk − ukW 1,p(x) (Ω0 ) (|∇ uk | + |∇ u|) dx . Ω0
Here we consider k(|∇ uk | + |∇ u|)
p(x)−2
Then we will give an estimate about ( get
Z Ω0
(|∇ uk | + |∇ u|)
(p(x)−2)p(x) p(x)−1 dx
kLq(x) (Ω0 ) ≥ 1 and we can just change q− to q+ if k(|∇ uk | + |∇ u|)p(x)−2 kLq(x) (Ω0 ) ≤ 1.
R
Ω0
(|∇ uk | + |∇ u|)
Z = Ω0
1 (p(x)−2)p(x) p(x)−1 dx q−
)
. Let m(x) =
(p(x)−2)p(x) , p(x)−1
p(x)−1
n(x) = p(x)−2 then we can
(|∇ uk | + |∇ u|)m(x) dx
≤ C k(|∇ uk | + |∇ u|)m(x) kLn(x) (Ω0 ) 1
= C (k(|∇ uk | + |∇ u|)m(x) knLn(x) (Ω ) ) n0 0 Z 10 n m(x)·n(x) ≤C (|∇ uk | + |∇ u|) dx 0
Ω0
Z
p(x)
=C Ω0 + +1
≤ 2p
(|∇ uk | + |∇ u|) Z
C Ω0
n
dx
(|∇ uk |p(x) + |∇ u|p(x) )dx
p+ uk Lp(x) (Ω ) 0
≤ C k∇ k
10 10 n
−
+
−
+ k∇ uk kpLp(x) (Ω ) + k∇ ukpLp(x) (Ω ) + k∇ ukpLp(x) (Ω 0
0
10 n
0)
≤C where
n o 1 1 0 ± (k(|∇ uk | + |∇ u|)m(x) knLn(x) (Ω ) ) n0 = min (k(|∇ uk | + |∇ u|)m(x) knLn(x) (Ω ) ) n± 0
0
and in the last inequality, we have used the boundedness of uk in W
Z Ω0
|∇ uk − ∇ u|(|∇ uk | + |∇ u|)p(x)−2 dx ≤ C kuk − ukW 1,p(x) (Ω0 ) . p(x)
Similarly, let d(x) = p(x)−1 we can get
Z Ω0
1,p(x)
|∇ uk − ∇ u|p(x)−1 dx ≤ C k|∇ uk − ∇ u|p(x)−1 kLd(x) (Ω0 ) 10 0 d = C k|∇ uk − ∇ u|p(x)−1 kdLd(x) (Ω ) 0
(Ω0 ). So we can get that (4.7)
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J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
Z
(p(x)−1)d(x)
≤C Ω0
|∇ uk − ∇ u|
Z
p(x)
=C Ω0
|∇ uk − ∇ u|
10 d
dx
10 d
dx
+
−
≤ C k∇ uk − ∇ ukpLp(x) (Ω ) + k∇ uk − ∇ ukpLp(x) (Ω
10 d
0)
0
10 + − d , ≤ C k∇ uk − ∇ ukpW 1,p(x) (Ω ) + k∇ uk − ∇ ukpW 1,p(x) (Ω ) 0
0
(4.8)
where
1± 10 d p(x)−1 d± p(x)−1 d0 kLd(x) (Ω ) k|∇ uk − ∇ u| = min k|∇ uk − ∇ u| kLd(x) (Ω ) d . 0
0
We note that the inequality above is (4.8). We now recall some useful inequalities (see [14]) that hold for all ξ , η ∈ Rn :
p−2 | + |η|)p−2 |ξ | ξ − |η|p−2 η ≤ C |ξ − η|(|ξ p−1 C |ξ − η|
if p ≥ 2, if 1 < p ≤ 2
(4.9)
where C is a positive constant independent of ξ and η. The estimates (4.7) and (4.8) together with the inequalities (4.9) show that
Z lim
k→∞
Ω0
|∇ uk |p−2 ∇ uk − |∇ u|p−2 ∇ u = 0.
(4.10)
Now let ϕ ∈ C0∞ (Ω ) with supp ϕ ⊆ Ω0 b Ω . It then follows from the limit (4.10) that
Z Ω
|∇ uk |p(x)−2 ∇ uk · ∇ϕ →
Z
|∇ u|p(x)−2 ∇ u · ∇ϕ.
(4.11)
Ω
Since |f (x, uk )ϕ| ≤ C γl (x) on Ω0 and γl (x) ∈ L1 (Ω ), it follows that
Z Ω
f (x, uk )ϕ →
Z Ω
f (x, u)ϕ.
(4.12)
Therefore (4.11) and (4.12) show that the identity (3.2) holds for all ϕ ∈ C0∞ (Ω ). 1,p(x)
We proceed to show that it also holds for any ϕ ∈ W0 1,p(x) W0
(Ω ). We choose a sequence {ηk } of non-negative functions in
C0 (Ω ) such that ηk → |ω| in (Ω ). By going to a subsequence if necessary, we can assume that ηk → |ω| a.e. on Ω . Then, by the Fatou lemma and the Höder inequality and Proposition 2.3 we see that ∞
Z Z f (x, u)ω ≤ f (x, u)|ω| Ω Ω Z ≤ lim inf f (x, u)ηk k→∞ Z Ω = lim |∇ u|p(x)−2 ∇ u · ∇ηk k→∞
Ω
≤ (kuk
p+
= (kuk
p+
−
1,p(x)
W0
1,p(x)
W0
(Ω ) (Ω )
+ k uk p
1,p(x)
W0
+ k uk
(Ω )
)M lim kηk kW 1,p(x) (Ω )
(Ω )
) kωkW 1,p(x) (Ω )
p− 1,p(x)
W0
k→∞
0
M
0
1,p(x)
where M is an appropriate positive constant according to the computation. Now if ϕ ∈ W0 ω := ϕk − ϕ in the above inequality shows that
Z lim
k→∞
Ω
f (x, uk )ϕ =
Z Ω
f (x, u)ϕ.
We also have
Z lim Ω
|∇ u|p(x)−2 ∇ u · ∇ϕk =
Z
|∇ u|p(x)−2 ∇ u · ∇ϕ. Ω
(Ω ), and ϕk → ϕ , then taking
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437 1,p(x)
It follows that (3.2) holds for all ϕ ∈ W0 Ω.
4435
(Ω ). Thus u ∈ W01,p(x) (Ω ) is a solution of (1.1) such that ψ∞ ≤ u ≤ ψ0 on
As an immediate corollary we have the following, which reduces to Corollary 3.2 of [2] when p(x) ≡ p. Corollary 4.2. Let g : (0, ∞) → (0, ∞) be nonincreasing, and assume that g ∈ L1 (0, δ) for some δ > 0. If f (x, t ) = b(x)g (t ) for some non-trivial and non-negative b ∈ Lq(x) (Ω ), then (1.1) has a solution. Proof. It is enough to realize that if g satisfies the hypothesis, then tg (t ) ≤ C for all 0 < t < δ and some positive constant C . Thus, g satisfies both the conditions [G1 ] and (1) of [G2 ]. Therefore f satisfies both the conditions [F1 ] − [F2 ]. The corollary then follows from Theorem 4.1. For the next result, we will need the following condition on g : (0, ∞) → (0, ∞), where g is C α , 0 < α < 1.
[G3 ] limt →0+ g (t ) = ∞. Consider G(t ) :=
d
Z
g (s)ds,
(4.13)
t
where 0 < d ≤ ∞ is chosen such that the integral (4.13) is finite for 0 < t < d. We define ψ : [0, d] → [0, ψ(d)] to be the increasing function
ψ(t ) =
t
Z
1 G(s)1/p
−
0
ds.
(4.14)
Let c = ψ(d) and ϕ : [0, c ] → [0, d] be the inverse of ψ . We note some properties of ϕ in the following remark. Remark 4.3. (1) ϕ satisfies
− −p− |ϕ 0 |p −2 ϕ 00 = g (ϕ) ϕ(t ) > 0 ϕ(0) = 0.
in (0, c ), in (0, c ],
(2) From (1) we see that ϕ 0 is decreasing on (0, c ) and therefore ϕ 0 (0+) ∈ (0, ∞]. Furthermore ϕ is increasing on (0, c ]. (3) If, in addition g is nonincreasing, then on invoking Lemma 2.1 of [15] we see that lim
t →0+
G(t ) g (t )
= 0.
Therefore, in this case
−ϕ 0 (t ) G(ϕ(t )) 1 = · 0 00 ϕ (t ) g (ϕ(t )) ϕ (t ) is bounded on (0, c ]. Remark 4.4. (1) Let ψ(d − δ) = c − ε where δ, ε > 0. Then we can get
ϕ 0 (c − ε) =
1
ψ 0 (d
− δ)
=
1 G(d − δ)
1/p−
= Rd d−δ
1 g (s)ds
.
It follows that ϕ 0 (c − ε) → ∞ when δ → 0. So there is a δ0 > 0 such that ϕ 0 (c − ε0 ) > 1, where c − ε0 = ψ(d − δ0 ). (2) By Lemmas 3.4 and 3.5 and p(x) satisfies the condition [P ], we can define the first eigenvalue λ1 = the infimum of the eigenvalues >0 from [1]. By Remark 4.4(2), let z be an positive eigenfunction of −∆p(x) on Ω corresponding to the first eigenvalue λ1 . Let 1,α
z1 (x) = µz (x) so that 0 < z1 (x) < c − ε0 for all x ∈ Ω where µ is a positive constant. It is known that z (x), z1 (x) ∈ C0 (Ω ) for some 0 < α < 1 and that |∇ z1 | 6= 0 on ∂ Ω by the regularity results of eigenfunction. Theorem 4.5. Let f (x, t ) = b(x)g (t ) where b ∈ L∞ (Ω ) with infΩ b > 0 and g satisfies 3(i), 3(ii) of [G2 ] and [G3 ]. Then problem (1.1) has a solution if and only if ϕ(z1 )g (ϕ(z1 )) ∈ L1 (Ω ).
4436
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
Proof. Suppose ϕ(z1 )g (ϕ(z1 )) ∈ L1 (Ω ). Since ω := ϕ(z1 ) ∈ C01 (Ω ) it follows that conditions [G1 ] and 3(iii) of [G2 ] hold. Therefore the hypotheses of Theorem 4.1 are satisfied by f , and thus problem (1.1) has a solution. For the converse, let ω = βϕ(z1 ) where β ≥ 1 to be determined shortly. Then
|∇ω|p(x)−2 ∇ω = β p(x)−1 (ϕ 0 (z1 ))p(x)−1 |∇ z |p(x)−2 ∇ z · µp(x)−1 . Let η ∈ C0∞ (Ω ) and η ≥ 0. By Remarks 4.3(2) and 4.4(1) we can know that ϕ 0 (z1 ) ≥ 1. By Remark 4.3(1), (3), and recalling that z is an eigenfunction of the p(x)-Laplacian, direct computation shows that
Z
|∇ω|p(x)−2 ∇ω · ∇η = Ω
Z Ω
(β p(x)−1 |∇ z |p(x)−2 ∇ z · µp(x)−1 · ∇((ϕ 0 (z1 ))p(x)−1 η)
− β p(x)−1 µp(x)−2 (p(x) − 1) · |∇ z |p(x) ϕ 0 (z1 )p(x)−2 ϕ 00 (z1 )η)dx Z |∇ z |p(x)−2 ∇ z · ∇((ϕ 0 (z1 ))p(x)−1 η)dx ≥ M1 Ω Z − − M2 (p(x) − 1)|∇ z |p(x) ϕ 0 (z1 )p −2 ϕ 00 (z1 )ηdx Z Ω Z p(x) − 1 |∇ z |p(x)−2 ∇ z · ∇((ϕ 0 (z1 ))p(x)−1 η)dx + M2 = M1 |∇ z |p(x) g (ϕ(z1 ))ηdx p− Ω Ω Z Z p(x) − 1 p(x)−1 0 p(x)−1 |∇ z |p(x) g (ϕ(z1 ))ηdx = λ1 M1 z (ϕ (z1 )) ηdx + M2 − Ω
Ω
p
where M1 , M2 are positive constants determined by our computation and M1 , M2 → ∞ when β → ∞. Consequently we get
Z
|∇ω|p(x)−2 ∇ω · ∇η − Ω
Z Ω
Z
|∇ω|p(x)−2 ∇ω · ∇η − bg (ϕ)ηdx Z 0 p(x)−1 p(x) − 1 p(x)−1 (ϕ (z1 )) p(x) = ηg (ϕ(z )) λ1 M1 z + M2 |∇ z | − b dx g (ϕ(z )) p− Ω " ! # Z 0 p− −1 p(x) − 1 p(x)−1 (ϕ (z1 )) p(x) ≥ ηg (ϕ(z )) M z + |∇ z | − b dx g (ϕ(z )) p− Ω Z 1 −ϕ 0 (z1 ) p(x) − 1 p(x)−1 p(x) = ηg (ϕ(z )) M z · − 00 + |∇ z | − b dx p ϕ (z1 ) p− Ω
bg (ω)η ≥
Ω
where M = min{λ1 M1 , M2 }. Note that, since |∇ z | 6= 0 on ∂ Ω , we have 1 −ϕ 0 (z1 ) p(x) − 1 + |∇ z |p(x) inf z p(x)−1 · − 00 Ω p ϕ (z1 ) p−
>0
and thus, by choosing β ≥ 1 sufficiently big, we see that
Z
|∇ω|p(x)−2 ∇ω · ∇η ≥ Ω
Z Ω
bg (ϕ(z ))η ≥
Z Ω
bg (ω)η
for all η ∈ C0∞ (Ω ) with η ≥ 0. A straightforward application of the Fatou lemma shows that the above inequality holds for 1,p(x)
all non-negative η ∈ W0 u(x) ≤ βϕ(z (x))
(Ω ). Thus, if u is a solution of (1.1) by the comparison principle, Lemma 3.3, we have
(x ∈ Ω ).
Using ϕ := u in (3.2) we note that
(inf) Ω
Z Ω
g (u)u ≤
Z Ω
bg (u)u =
Z Ω
|∇ u|p(x)−2 < ∞.
Therefore since, by condition 3(ii) of [G2 ], Cβ ϕ(z )g (ϕ(z )) ≤ βϕ(z )g (βϕ(z )) ≤ ug (u) on Ω for some positive constant Cβ , we have
Z Ω
ϕ(z )g (ϕ(z )) < ∞.
The proof is complete.
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
4437
Acknowledgements The authors would like to thank professor Xianling Fan for clear suggestions and the reviewer(s) for clear valuable comments and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
X.L. Fan, Q.H. Zhang, D. Zhao, Eigenvalues of p(x)-Laplacian Dirichlet problem, J. Math. Anal. Appl. 302 (2005) 306–317. A. Mohammed, Positive solutions of the p-Laplace equation with singular nonlinearity, J. Math. Anal. Appl. 352 (2009) 234–245. E. Acerbi, G. Mingione, Regularity results for a class of functionals with nonstandard growth, Arch. Ration. Mech. Anal. 156 (2001) 121–140. X.L. Fan, Global C 1,α regularity for variable exponent elliptic equations in divergence form, preprint. E-preprint available at Chinese Science-paper Online, 200601-192, 2006, http://www.paper.edu.cn/process/download.jsp? file=200601-192. X.L. Fan, D. Zhao, A class of De Giorgi type and Höder continuity, Nonlinear Anal. 36 (1996) 295–318. X.L. Fan, D. Zhao, On the spaces Lp(x) and W m,p(x) , J. Math. Anal. Appl. 263 (2001) 424–446. P. Harjulehto, P. Hästö, An overview of variable exponent Lebesgue and Sobolev spaces, in: D. Herron (Ed.), Future Trends in Geometric Function Theory, RNC Workshop, in: Rep. Univ. Jyväkylä Dep. Math. Stat., vol. 92, 2003, pp. 85–94. O. Ková˘cik, J. Rákosník, On spaces Lp(x) (Ω ) and W k,p(x) (Ω ), Czechoslovak Math. J. 41 (116) (1991) 592–618. X.L. Fan, On the sub-supersolution method for p(x)-Laplacian equations, J. Math. Anal. Appl. 330 (2007) 665–682. J. Diaz, Nonlinear Partial Differential Equations and Free Boundaries, Volume I, Elliptic Equations, in: Pitman Res. Notes Math., vol. 106, Pitman, Boston, 1985. X.L. Fan, Y.Z. Zhao, Q.H. Zhang, A strong maximum principle for p(x)-Laplace equations, Chinese J. Contemp. Math. 24 (3) (2003) 277–282. K. Perera, E. Silva, On singular p-Laplacian problems, Differential Integral Equations 20 (1) (2007) 105–120. X.L. Fan, Q.H. Zhang, Existence of solutions for p(x)-Laplacian Dirichlet problems, Nonlinear Anal. 52 (2003) 1843–1852. G. Dinca, P. Jebelean, J. Mawhin, Variational and topological methods for Dirichlet problems with p-Laplacian, Port. Math. (N.S.) 58 (2001) 339–378. A.C. Lazer, P.J. McKenna, Asymptotic behavior of solutions of boundary blowup problems, Differential Integral Equations 7 (1994) 1001–1019.
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Positive solutions of the p(x)-Laplace equation with singular nonlinearity Jingjing Liu Department of Mathematics, Sun Yat-sen University, 510275 Guangzhou, China
article
abstract
info
In this paper, using sub–supersolution method, we study the positive solution of the p(x)Laplacian equations −∆p(x) u = f (x, u) with Dirichlet boundary condition on a bounded domain Ω in Rn with C 1,ω boundary for some 0 < ω < 1 for a possibly singular nonlinearity f on Ω × (0, ∞). Crown Copyright © 2010 Published by Elsevier Ltd. All rights reserved.
Article history: Received 7 September 2009 Accepted 10 February 2010 MSC: 34B15 35A15 35G99 Keywords: p(x)-Laplacian Singular boundary value problem Sub–supersolution method
1. Introduction Let Ω ⊆ Rn (n ≥ 1) be a bounded domain with C 1,ω boundary for some 0 < ω < 1 and f : Ω × (0, ∞) −→ [0, ∞) be a given singular nonlinearity. In this paper we are concerned with the existence of weak solutions of the following singular boundary value problem:
( −∆p(x) u = f (x, u) u>0 u=0
in Ω , in Ω , on ∂ Ω
(1.1)
where p(x) ∈ C (Ω ), 1 < p− = infx∈Ω p(x) ≤ p+ = supx∈Ω p(x) < +∞. Let p(x) satisfy the following condition: [P ] The function p(x) is monotone when n = 1; there is a vector l ∈ Rn \ {0} such that for any x ∈ Ω , f (t ) = p(x + tl) is monotone for t ∈ Ix = {t |x + tl ∈ Ω } when n > 1. The operator −∆p(x) u = −div(|∇ u|p(x)−2 ∇ u) is called p(x)-Laplacian, which becomes p-Laplacian when p(x) ≡ p (a constant). The p(x)-Laplacian possesses more complicated nonlinearities than the p-Laplacian. For example, it is inhomogeneous, and in general, it has no first eigenvalue, in other words, the infimum of the eigenvalues of p(x)-Laplacian equals 0 (see [1]). So some techniques used in the constant exponent case cannot be carried out for the variable exponent case. Moreover, there are two cases about the infimum of the eigenvalues of p(x)-Laplacian, one is the infimum of the eigenvalue is an eigenvalue indeed, the other case is that the infimum of the eigenvalues is not an eigenvalue. In this paper, we consider the first case only. The aim of the present paper is to extend some results of [2] to the variable exponent case by using the sub–supersolution method. The regularity results and the comparison principles are the bases of the sub–supersolution method. For the
E-mail address: [email protected]. 0362-546X/$ – see front matter Crown Copyright © 2010 Published by Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.02.018
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
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regularity results in the variable exponent case, see [3–5], more precisely, for the L∞ and C 0,α regularity, see [5]; for the local C 1,α regularity of the minimizers of the corresponding integral functional, see [3]; for the global C 1,α regularity, see [4]. The present paper is organized as follows. In Section 2, we recall some basic facts about the variable exponent Lebesgue and Sobolev spaces. In Section 3 we provide basic facts and recall some lemmas that will be needed later. We also specify the conditions on the nonlinearity f that will be used throughout the paper. Section 4 contains the main result on the existence of weak solutions of the problem (1.1). Several results that can be drawn from the main theorem are also given. 2. Preliminaries 1 Let Ω be a bounded domain of Rn (n ≥ 1) with a smooth boundary ∂ Ω . Denote by S (Ω ) the set of all measurable real functions defined on Ω . Note that two measurable functions are considered as the same element of S (Ω ) when they are equal almost everywhere. ∞ L∞ + (Ω ) = {p ∈ L (Ω ) : ess inf p(x) > 1}. x∈Ω
For p ∈ L∞ + (Ω ), denote p− = p− (Ω ) = ess inf p(x),
p+ = p+ (Ω ) = ess sup p(x).
x∈Ω
x∈Ω
(Ω ) by p(x) |u(x)| dx < ∞ ,
Define the variable exponent Lebesgue space L Lp(x) (Ω ) =
Z
u ∈ S (Ω ) :
Ω
p(x)
with the norm
) Z u(x) p(x) = inf λ > 0 : λ dx ≤ 1 Ω (
kukLp(x) (Ω ) = |u|p(x)
and the variable exponent Sobolev space W 1,p(x) (Ω ) by W 1,p(x) (Ω ) = u ∈ Lp(x) (Ω ) : |∇ u| ∈ Lp(x) (Ω ) ,
with the norm
kukW 1,p(x) (Ω ) = |u|p(x) + |∇ u|p(x) . 1,p(x)
(Ω ) denotes the closure of C0∞ (Ω ) in W 1,p(x) (Ω ). |∇ u|Lp(x) (Ω ) is an equivalent norm on W01,p(x) , we denote this norm 1,p(x) by kuk. The spaces Lp(x) (Ω ), W 1,p(x) (Ω ) and W0 (Ω ) are all separable and reflexive Banach spaces. We refer to [6–8] for W0
the elementary properties of these spaces. Here, we only display some facts which will be used later. For u, v ∈ S (Ω ), we write u ≤ v if u(x) ≤ v(x) for a.e. x ∈ Ω . Define u+ (x) = max{u(x), 0} and u− (x) = max{−u(x), 0}. 0 0 Proposition 2.1 ([6]). The conjugate space of Lp(x) (Ω ) is Lp (x) (Ω ), where p(1x) + p01(x) = 1. For any u ∈ Lp(x) (Ω ) and v ∈ Lp (x) R (Ω ) , Ω |uv| dx ≤ 2 |u|p(x) |v|p0 (x) .
Proposition 2.2 ([6]). Set ρ(u) =
R
Ω
|u(x)|p(x) dx. For u, uk ∈ Lp(x) (Ω ), we have
(1) For u 6= 0, |u|p(x) = λ ⇔ ρ( λ ) = 1; (2) |u|p(x) < 1 (=1; > 1) ⇔ ρ(u) < 1 (=1; > 1); u
p−
p+
p+
p−
(3) If |u|p(x) > 1, then |u|p(x) ≤ ρ (u) ≤ |u|p(x) ; (4) If |u|p(x) < 1, then |u|p(x) ≤ ρ (u) ≤ |u|p(x) ; (5) limk→∞ |uk |p(x) = 0 ⇐⇒ limk→∞ ρ(uk ) = 0; (6) |uk |p(x) → ∞ ⇐⇒ ρ(uk ) → ∞. Similar to Proposition 2.2, we have Proposition 2.3. Set φ(u) = (1) (2) (3) (4) (5) (6)
R
Ω
|∇ u|p(x) dx. For u, uk ∈ Lp(x) (Ω ), we obtain
For u 6= 0, kuk = λ ⇔ φ( λu ) = 1; kuk < 1 (=1; > 1) ⇔ φ(u) < 1 (=1; > 1); − + If kuk > 1, then kukp ≤ φ (u) ≤ kukp ; + − If kuk < 1, then kukp ≤ φ (u) ≤ kukp ; limk→∞ kuk k = 0 ⇐⇒ limk→∞ φ(uk ) = 0; kuk k → ∞ ⇐⇒ φ(uk ) → ∞.
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J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
3. Preliminaries 2 Throughout the paper the nonlinearity f : (0, ∞) → [0, ∞) will be assumed to be a Caratheodory function such that f (·, t ) is in C θ (Ω ) for some 0 < θ < 1. Unless specified otherwise we will also suppose that Ω ∈ Rn is a bounded domain with C 1,ω boundary for some 0 < ω < 1. The positive constants that will be used in estimations are not necessarily the same in every occurrence. In what follows, further conditions will be stipulated on f . Now, consider equation
−∆p(x) u = f (x, u) u=h
in Ω , on ∂ Ω .
(3.1)
1,p(x)
Definition 3.1. (1) Given h ∈ W 1,p(x) (Ω ), u ∈ W 1,p(x) (Ω ) is called a (weak) solution of (3.1) if u − h ∈ W0
Z
|∇ u|p−2 ∇ u · ∇ϕ = Ω
Z Ω
f (x, u)ϕ
(Ω ) and (3.2)
1,p(x)
for all ϕ ∈ W0 (Ω ). (2) u ∈ W 1,p(x) (Ω ) is called a sub-solution (respectively a supersolution) of (3.1) if u ≤ (respectively ≥) h on ∂ Ω and for 1,p(x) all ϕ ∈ W0 (Ω ) with ϕ ≥ 0,
Z Ω
|∇ u|p−2 ∇ u · ∇ϕ ≤ (respectiv ely ≥)
Z Ω
f (x, u)ϕ.
(3.3)
1,p(x)
Remark 3.2. (1) We say u ∈ W0 (Ω ) is a (weak) solution of (1.1) if u(x) > 0 in Ω and (3.2) holds. Obviously, (1.1) is a special case of (3.1). 1,p(x) (2) Following usual practice, we shall write u ≤ h on Ω to mean (u − h)+ ∈ W0 (Ω ); u ≥ h on Ω to mean (u − h)−
∈ W01,p(x) (Ω ).
Now we give a comparison principle as follows. Lemma 3.3. Let g (x, t ) : Ω × R → R be measurable and nondecreasing in t. Let u, v ∈ W 1.p(x) (Ω ) satisfy
−∆p(x) u + g (x, u) ≤ −∆p(x) v + g (x, v) (x ∈ Ω ). If u ≤ v on ∂ Ω , then u ≤ v on Ω . Proof. It is an obvious result according to Proposition 2.3. in [9] by using reduction to absurdity.
Lemma 3.4 (See [1]). If n = 1, then the infimum of the eigenvalues of p(x)-Laplacian > 0 if and only if the function p(x) is monotone. Lemma 3.5 (See [1]). Let n > 1, if there is a vector l ∈ Rn \ {0} such that for any x ∈ Ω , f (t ) = p(x + tl) is monotone for t ∈ Ix = {t |x + tl ∈ Ω }, then he infimum of the eigenvalues of p(x)-Laplacian > 0. p(x)
In this paper we shall use the notation q(x) for the Höder conjugate of p(x) > 1, that is q(x) := p(x)−1 . Also we let p∗ (x) be the Sobolev conjugate exponent of p(x), namely,
Np(x) p∗ (x) = N − p(x) ∞
if p(x) < N , if p(x) ≥ N .
Before we state conditions on f that will be needed in the paper, let us identify a class G of functions g : (0, ∞) → (0, ∞) and a class B of non-negative functions b in Lq(x) (Ω ) such that the following conditions hold:
[G1 ] There is ρ ≥ 1 and a positive constant C such that g (t ) ≤ C for all t ≥ ρ . [G2 ] One of the conditions (1), (2), or (3) below holds, where h(t ) = tg (t ), t > 0: (1) h(t ) ≤ C for all 0 < t < δ and some positive constants C and δ . (2) h is nondecreasing. (3) (i) g is nonincreasing. (ii) For each ∈ (0, 1) there is a constant Cθ ≥ 1 such that g (θ t ) ≤ Cθ g (t ) for all t > 0. p∗
(iii) There is ω ∈ C01 (Ω ) with ω > 0 on Ω such that bg (ω) ∈ L p∗ −1 (Ω ).
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
4431
We are now ready to list the conditions on f that will be used in this paper. In stating the conditions, we use the notation
γs (x) = sup{f (x, t ) : t ≥ s} (x ∈ Ω ) for any s > 0.
[F1 ] γs (x) ∈ Lq(x) (Ω ) for each s > 0. [F2 ] There is (g , b) ∈ G × B that satisfies conditions [G1 ]–[G2 ], a measurable function a for which {x ∈ Ω : 0 < a(x) ≤ 1} has a positive measure, and 0 ≤ a ≤ b} such that (1) f (x, s) ≥ a(x) for (x, s) ∈ Ω × (0, ∞), (2) f (x, s) ≤ b(x)g (s) for (x, s) ∈ Ω × [ρ, ∞), (3) f (x, s)g (t ) ≤ Cf (x, t )g (s) for x ∈ Ω , 0 < s < t and some constant C > 0. We start with the following lemma. Lemma 3.6. Let f satisfy [F1 ] and h ∈ W 1,p(x) (Ω ) with h > 0 on Ω . Let ψsub be a sub-solution and ψsup a supersolution of (3.1) on Ω . If 0 < infΩ ψsub ≤ ψsub ≤ ψsup on Ω then there is a solution u ∈ W 1,p(x) (Ω ) of (3.1) such that ψsub ≤ u ≤ ψsup a.e. on Ω. Proof. The lemma is a consequence of [10, Theorem 4.14]. For completeness we include the short proof. Take 0 < l = infΩ ψsub and let F : Ω × R → R be defined by F (x, t ) =
f (x, l) f (x, t )
if t < l, if t ≥ l.
Then F (x, ·) is Hölder continuous on R for each x ∈ Ω and |F (x, t )| ≤ γl (x) on Ω . Consider now the Dirichlet problem
−∆p(x) u = F (x, u)
in Ω , on ∂ Ω .
u=h
Note that ψsub is a sub-solution, and ψsup is a supersolution of this problem. Since γl ∈ Lq(x) (Ω ), it follows by [10, Theorem 4.14] that this problem admits a solution u ∈ W 1,p(x) (Ω ) such that ψsub ≤ u ≤ ψsup . Finally we note that u is a solution of (3.1) as claimed because of u ≥ l. 4. Existence of weak solutions 1,p(x)
We now show the existence of a solution of (1.1) in W0
(Ω ). 1,p(x)
Theorem 4.1. Suppose that f satisfies [F1 ]–[F2 ]. Then problem (1.1) admits a solution in W0
(Ω ).
Proof. By replacing g and b by C −1 g and Cb respectively, if necessary, we can assume that g (t ) ≤ 1 for all t ≥ ρ in condition [G1 ]. For k = 1, 2, . . ., let ψk be the solution of
−∆p(x) u(x) = ak (x) u(x) = k−1
in Ω , on ∂ Ω
(4.1)
where for k = 1, 2, . . .,
ak (x) = min a(x),
k+1
k
.
Likewise, let ψ∞ be the solution of (4.1) with k = ∞. Note that since ak (x) ∈ L∞ (Ω ) the problem (4.1) does indeed have a solution. By the comparison lemma, we have 0 ≤ ψ∞ ≤ ψk ≤ ψ1 for all k = 1, 2, . . . and ψk ≥ k−1 on Ω for all k = 1, 2, . . .. By the Strong Maximum Principle [11], we note that ψ∞ > 0 on Ω . So, we can get −∆p(x) ψk ≤ a in Ω , and ψk = k−1 on ∂ Ω . For each k ∈ Z + let us now consider the Dirichlet problem
−∆p(x) u(x) = f (x, u(x)) u(x) = k−1
in Ω , on ∂ Ω .
By (1) of [F2 ] we observe that
−∆p(x) ψk − f (x, ψk ) ≤ −∆p(x) ψk − a(x) ≤ 0 and thus ψk is a sub-solution of (BVPk ) for all k.
(BVPk )
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J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
Now, let ψ be a solution of
−∆p(x) u(x) = b(x) u(x) = ρ
in Ω , on ∂ Ω .
(4.2)
Note that (4.2) has a solution since b ∈ Lq(x) (Ω ). By comparison principle, we note that ψk ≤ ψ for all k = 1, 2, . . . and ψ ≥ ρ on Ω . Then by (2) of [F2 ] we have
−∆p(x) ψ − f (x, ψ) ≥ −∆p(x) ψ − b(x) = 0. Thus ψ is a supersolution of (BVPk ) on Ω for any k = 1, 2, . . .. Let us fix a positive integer m. By Lemma 3.6 let um be a solution of (BVPm ) such that ψm ≤ um ≤ ψ on Ω . Note that um is a supersolution of (BVPm+1 ) and therefore, again by Lemma 3.6, there is a solution um+1 of (BVPm+1 ) such that ψm+1 ≤ um+1 ≤ um on Ω . Continuing in this manner, we construct a sequence {uk } of solutions of (BVPk ) such that for all k ≥ m
ψ∞ ≤ uk+1 ≤ uk ≤ · · · ≤ um ≤ ψ in Ω . Let us also note that uk ≥ k−1 on Ω . We define u(x) = lim uk (x) k→∞
(x ∈ Ω ).
(4.3)
We use ϕk := uk − 1/k as a test function for the solution uk on Ω (by condition [F1 ]). We obtain
Z Ω
|∇ uk |p(x) =
Z Ω
|∇ uk |p(x)−2 ∇ uk · ∇(uk − /k) =
Z Ω
f (x, uk )(uk − /k) ≤
Z Ω
f (x, uk )uk .
(4.4)
We now use condition [F2 ] to show that there is a positive constant C , independent of k, such that
Z Ω
f (x, uk )uk ≤ C .
(4.5)
First we note that conditions (2) and (3) of [F2 ] imply that f (x, uk )uk =
f (x, uk )uk g ( uk )
g (uk )uk ≤ C
f (x, ψ) g (ψ)
g (uk )uk .
If (1) of [G2 ] holds, then this together with [G1 ] implies f (x, ψ) g (ψ)
g (uk )uk ≤ Mb max{ρ, uk } ≤ Mbψ
for some positive constant M. If (2) of [G2 ] holds, then recalling g (ψ) ≤ 1, it follows from (2) of [F2 ] that f (x, ψ) g (ψ)
g (uk )uk ≤ f (x, ψ)ψ ≤ bψ.
Therefore, since ψ ∈ Lp(x) (Ω ) and b ∈ Lq(x) (Ω ), the estimate (4.5) holds if either (1) or (2) of [G2 ] holds. Suppose now that (3) of [G2 ] holds. By [11] we note that ψ∞ > 0 on Ω , and ∂ψ∞ /∂ν > 0 on ∂ Ω , where ν is the unit inner normal. Thus, we have infΩ (ψ∞ /ω) > 0, where ω is as in condition 3(iii) of [G2 ]. That is θ ω ≤ ψ∞ on Ω for some 0 < θ ≤ 1. Then on using 3(i) and 3(ii) of [G2 ], we estimate f (x, ψ) g (ψ)
g (uk )uk ≤
f (x, ψ) g (ψ)
g (θ ω)ψ ≤ Cθ bg (ω)ψ.
Since ψ ∈ W 1,p(x) (Ω ) ⊆ Lp (Ω ), the assumption in 3(iii) of [G2 ] shows that estimate (4.5) holds in this case as well. ∗
1,p(x)
(Ω ). We pick a subsequence, still denoted (Ω ), strongly in Lp(x) (Ω ) and pointwise a.e. on Ω . Note that {ψk } converges 1,p(x) to u ∈ W0 (Ω ), and that {uk } converges weakly to u in W 1,p(x) (Ω ), strongly in Lp(x) (Ω ) and a.e. on Ω . Now let Ω0 b Ω . On Ω0 we have Thus estimates (4.4) and (4.5) show that the sequence ψk is bounded in W0 1,p(x)
by {ψk }, such that it converges weakly in W0
|(f (x, uk ) − f (x, uj ))(uk − uj )| ≤ 4ψl (x)ψ, where l = minΩ¯ ψ∞ , and hence proceeding as in [12] one can show that
Z Ω0
(|∇ uk |p(x)−2 ∇ uk − |∇ uj |p(x)−2 ∇ uj ) · ∇(uk − uj ) → 0
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
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as k, j → ∞. Then we can get
Z Ω0
|∇ uk − ∇ uj |p(x) → 0 k, j → ∞
(4.6)
by using the inequality (see [13])
[(|ξ |p−2 ξ − |η|p−2 η)(ξ − η)] · (|ξ |p + |η|p )(2−p)/p ≥ (p − 1)|ξ − η|p , p 1 (|ξ |p−2 ξ − |η|p−2 η)(ξ − η) ≥ |ξ − η|p , p ≥ 2.
1 < p < 2;
2
p(x)
Since {uk } converges strongly to u in L (Ω0 ), the limits (4.6) shows that the sequence {uk } is Cauchy in W 1,p(x) (Ω0 ), and hence converges to u in W 1,p(x) (Ω0 ). In conclusion, given any compact set Ω0 ⊂ Ω , we can find a subsequence of {uk } that converges to u strongly in W 1,p(x) (Ω0 ). Let us take note of the following estimates. If p(x) ≥ 2, then by the Hölder inequality Propositions 2.1 and 2.3, we have
Z
|∇ uk − ∇ u|(|∇ uk | + |∇ u|)p(x)−2 dx ≤ 2k∇ uk − ∇ ukLp(x) (Ω ) k(|∇ uk | + |∇ u|)p(x)−2 kLq(x) (Ω0 ) 0
Ω0
1
−
= 2k∇ uk − ∇ ukLp(x) (Ω0 ) (k(|∇ uk | + |∇ u|)p(x)−2 kqLq(x) (Ω ) ) q− 0 Z 1− (p(x)−2)p(x) q p ( x )− 1 ≤ 2kuk − ukW 1,p(x) (Ω0 ) (|∇ uk | + |∇ u|) dx . Ω0
Here we consider k(|∇ uk | + |∇ u|)
p(x)−2
Then we will give an estimate about ( get
Z Ω0
(|∇ uk | + |∇ u|)
(p(x)−2)p(x) p(x)−1 dx
kLq(x) (Ω0 ) ≥ 1 and we can just change q− to q+ if k(|∇ uk | + |∇ u|)p(x)−2 kLq(x) (Ω0 ) ≤ 1.
R
Ω0
(|∇ uk | + |∇ u|)
Z = Ω0
1 (p(x)−2)p(x) p(x)−1 dx q−
)
. Let m(x) =
(p(x)−2)p(x) , p(x)−1
p(x)−1
n(x) = p(x)−2 then we can
(|∇ uk | + |∇ u|)m(x) dx
≤ C k(|∇ uk | + |∇ u|)m(x) kLn(x) (Ω0 ) 1
= C (k(|∇ uk | + |∇ u|)m(x) knLn(x) (Ω ) ) n0 0 Z 10 n m(x)·n(x) ≤C (|∇ uk | + |∇ u|) dx 0
Ω0
Z
p(x)
=C Ω0 + +1
≤ 2p
(|∇ uk | + |∇ u|) Z
C Ω0
n
dx
(|∇ uk |p(x) + |∇ u|p(x) )dx
p+ uk Lp(x) (Ω ) 0
≤ C k∇ k
10 10 n
−
+
−
+ k∇ uk kpLp(x) (Ω ) + k∇ ukpLp(x) (Ω ) + k∇ ukpLp(x) (Ω 0
0
10 n
0)
≤C where
n o 1 1 0 ± (k(|∇ uk | + |∇ u|)m(x) knLn(x) (Ω ) ) n0 = min (k(|∇ uk | + |∇ u|)m(x) knLn(x) (Ω ) ) n± 0
0
and in the last inequality, we have used the boundedness of uk in W
Z Ω0
|∇ uk − ∇ u|(|∇ uk | + |∇ u|)p(x)−2 dx ≤ C kuk − ukW 1,p(x) (Ω0 ) . p(x)
Similarly, let d(x) = p(x)−1 we can get
Z Ω0
1,p(x)
|∇ uk − ∇ u|p(x)−1 dx ≤ C k|∇ uk − ∇ u|p(x)−1 kLd(x) (Ω0 ) 10 0 d = C k|∇ uk − ∇ u|p(x)−1 kdLd(x) (Ω ) 0
(Ω0 ). So we can get that (4.7)
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J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
Z
(p(x)−1)d(x)
≤C Ω0
|∇ uk − ∇ u|
Z
p(x)
=C Ω0
|∇ uk − ∇ u|
10 d
dx
10 d
dx
+
−
≤ C k∇ uk − ∇ ukpLp(x) (Ω ) + k∇ uk − ∇ ukpLp(x) (Ω
10 d
0)
0
10 + − d , ≤ C k∇ uk − ∇ ukpW 1,p(x) (Ω ) + k∇ uk − ∇ ukpW 1,p(x) (Ω ) 0
0
(4.8)
where
1± 10 d p(x)−1 d± p(x)−1 d0 kLd(x) (Ω ) k|∇ uk − ∇ u| = min k|∇ uk − ∇ u| kLd(x) (Ω ) d . 0
0
We note that the inequality above is (4.8). We now recall some useful inequalities (see [14]) that hold for all ξ , η ∈ Rn :
p−2 | + |η|)p−2 |ξ | ξ − |η|p−2 η ≤ C |ξ − η|(|ξ p−1 C |ξ − η|
if p ≥ 2, if 1 < p ≤ 2
(4.9)
where C is a positive constant independent of ξ and η. The estimates (4.7) and (4.8) together with the inequalities (4.9) show that
Z lim
k→∞
Ω0
|∇ uk |p−2 ∇ uk − |∇ u|p−2 ∇ u = 0.
(4.10)
Now let ϕ ∈ C0∞ (Ω ) with supp ϕ ⊆ Ω0 b Ω . It then follows from the limit (4.10) that
Z Ω
|∇ uk |p(x)−2 ∇ uk · ∇ϕ →
Z
|∇ u|p(x)−2 ∇ u · ∇ϕ.
(4.11)
Ω
Since |f (x, uk )ϕ| ≤ C γl (x) on Ω0 and γl (x) ∈ L1 (Ω ), it follows that
Z Ω
f (x, uk )ϕ →
Z Ω
f (x, u)ϕ.
(4.12)
Therefore (4.11) and (4.12) show that the identity (3.2) holds for all ϕ ∈ C0∞ (Ω ). 1,p(x)
We proceed to show that it also holds for any ϕ ∈ W0 1,p(x) W0
(Ω ). We choose a sequence {ηk } of non-negative functions in
C0 (Ω ) such that ηk → |ω| in (Ω ). By going to a subsequence if necessary, we can assume that ηk → |ω| a.e. on Ω . Then, by the Fatou lemma and the Höder inequality and Proposition 2.3 we see that ∞
Z Z f (x, u)ω ≤ f (x, u)|ω| Ω Ω Z ≤ lim inf f (x, u)ηk k→∞ Z Ω = lim |∇ u|p(x)−2 ∇ u · ∇ηk k→∞
Ω
≤ (kuk
p+
= (kuk
p+
−
1,p(x)
W0
1,p(x)
W0
(Ω ) (Ω )
+ k uk p
1,p(x)
W0
+ k uk
(Ω )
)M lim kηk kW 1,p(x) (Ω )
(Ω )
) kωkW 1,p(x) (Ω )
p− 1,p(x)
W0
k→∞
0
M
0
1,p(x)
where M is an appropriate positive constant according to the computation. Now if ϕ ∈ W0 ω := ϕk − ϕ in the above inequality shows that
Z lim
k→∞
Ω
f (x, uk )ϕ =
Z Ω
f (x, u)ϕ.
We also have
Z lim Ω
|∇ u|p(x)−2 ∇ u · ∇ϕk =
Z
|∇ u|p(x)−2 ∇ u · ∇ϕ. Ω
(Ω ), and ϕk → ϕ , then taking
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437 1,p(x)
It follows that (3.2) holds for all ϕ ∈ W0 Ω.
4435
(Ω ). Thus u ∈ W01,p(x) (Ω ) is a solution of (1.1) such that ψ∞ ≤ u ≤ ψ0 on
As an immediate corollary we have the following, which reduces to Corollary 3.2 of [2] when p(x) ≡ p. Corollary 4.2. Let g : (0, ∞) → (0, ∞) be nonincreasing, and assume that g ∈ L1 (0, δ) for some δ > 0. If f (x, t ) = b(x)g (t ) for some non-trivial and non-negative b ∈ Lq(x) (Ω ), then (1.1) has a solution. Proof. It is enough to realize that if g satisfies the hypothesis, then tg (t ) ≤ C for all 0 < t < δ and some positive constant C . Thus, g satisfies both the conditions [G1 ] and (1) of [G2 ]. Therefore f satisfies both the conditions [F1 ] − [F2 ]. The corollary then follows from Theorem 4.1. For the next result, we will need the following condition on g : (0, ∞) → (0, ∞), where g is C α , 0 < α < 1.
[G3 ] limt →0+ g (t ) = ∞. Consider G(t ) :=
d
Z
g (s)ds,
(4.13)
t
where 0 < d ≤ ∞ is chosen such that the integral (4.13) is finite for 0 < t < d. We define ψ : [0, d] → [0, ψ(d)] to be the increasing function
ψ(t ) =
t
Z
1 G(s)1/p
−
0
ds.
(4.14)
Let c = ψ(d) and ϕ : [0, c ] → [0, d] be the inverse of ψ . We note some properties of ϕ in the following remark. Remark 4.3. (1) ϕ satisfies
− −p− |ϕ 0 |p −2 ϕ 00 = g (ϕ) ϕ(t ) > 0 ϕ(0) = 0.
in (0, c ), in (0, c ],
(2) From (1) we see that ϕ 0 is decreasing on (0, c ) and therefore ϕ 0 (0+) ∈ (0, ∞]. Furthermore ϕ is increasing on (0, c ]. (3) If, in addition g is nonincreasing, then on invoking Lemma 2.1 of [15] we see that lim
t →0+
G(t ) g (t )
= 0.
Therefore, in this case
−ϕ 0 (t ) G(ϕ(t )) 1 = · 0 00 ϕ (t ) g (ϕ(t )) ϕ (t ) is bounded on (0, c ]. Remark 4.4. (1) Let ψ(d − δ) = c − ε where δ, ε > 0. Then we can get
ϕ 0 (c − ε) =
1
ψ 0 (d
− δ)
=
1 G(d − δ)
1/p−
= Rd d−δ
1 g (s)ds
.
It follows that ϕ 0 (c − ε) → ∞ when δ → 0. So there is a δ0 > 0 such that ϕ 0 (c − ε0 ) > 1, where c − ε0 = ψ(d − δ0 ). (2) By Lemmas 3.4 and 3.5 and p(x) satisfies the condition [P ], we can define the first eigenvalue λ1 = the infimum of the eigenvalues >0 from [1]. By Remark 4.4(2), let z be an positive eigenfunction of −∆p(x) on Ω corresponding to the first eigenvalue λ1 . Let 1,α
z1 (x) = µz (x) so that 0 < z1 (x) < c − ε0 for all x ∈ Ω where µ is a positive constant. It is known that z (x), z1 (x) ∈ C0 (Ω ) for some 0 < α < 1 and that |∇ z1 | 6= 0 on ∂ Ω by the regularity results of eigenfunction. Theorem 4.5. Let f (x, t ) = b(x)g (t ) where b ∈ L∞ (Ω ) with infΩ b > 0 and g satisfies 3(i), 3(ii) of [G2 ] and [G3 ]. Then problem (1.1) has a solution if and only if ϕ(z1 )g (ϕ(z1 )) ∈ L1 (Ω ).
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J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
Proof. Suppose ϕ(z1 )g (ϕ(z1 )) ∈ L1 (Ω ). Since ω := ϕ(z1 ) ∈ C01 (Ω ) it follows that conditions [G1 ] and 3(iii) of [G2 ] hold. Therefore the hypotheses of Theorem 4.1 are satisfied by f , and thus problem (1.1) has a solution. For the converse, let ω = βϕ(z1 ) where β ≥ 1 to be determined shortly. Then
|∇ω|p(x)−2 ∇ω = β p(x)−1 (ϕ 0 (z1 ))p(x)−1 |∇ z |p(x)−2 ∇ z · µp(x)−1 . Let η ∈ C0∞ (Ω ) and η ≥ 0. By Remarks 4.3(2) and 4.4(1) we can know that ϕ 0 (z1 ) ≥ 1. By Remark 4.3(1), (3), and recalling that z is an eigenfunction of the p(x)-Laplacian, direct computation shows that
Z
|∇ω|p(x)−2 ∇ω · ∇η = Ω
Z Ω
(β p(x)−1 |∇ z |p(x)−2 ∇ z · µp(x)−1 · ∇((ϕ 0 (z1 ))p(x)−1 η)
− β p(x)−1 µp(x)−2 (p(x) − 1) · |∇ z |p(x) ϕ 0 (z1 )p(x)−2 ϕ 00 (z1 )η)dx Z |∇ z |p(x)−2 ∇ z · ∇((ϕ 0 (z1 ))p(x)−1 η)dx ≥ M1 Ω Z − − M2 (p(x) − 1)|∇ z |p(x) ϕ 0 (z1 )p −2 ϕ 00 (z1 )ηdx Z Ω Z p(x) − 1 |∇ z |p(x)−2 ∇ z · ∇((ϕ 0 (z1 ))p(x)−1 η)dx + M2 = M1 |∇ z |p(x) g (ϕ(z1 ))ηdx p− Ω Ω Z Z p(x) − 1 p(x)−1 0 p(x)−1 |∇ z |p(x) g (ϕ(z1 ))ηdx = λ1 M1 z (ϕ (z1 )) ηdx + M2 − Ω
Ω
p
where M1 , M2 are positive constants determined by our computation and M1 , M2 → ∞ when β → ∞. Consequently we get
Z
|∇ω|p(x)−2 ∇ω · ∇η − Ω
Z Ω
Z
|∇ω|p(x)−2 ∇ω · ∇η − bg (ϕ)ηdx Z 0 p(x)−1 p(x) − 1 p(x)−1 (ϕ (z1 )) p(x) = ηg (ϕ(z )) λ1 M1 z + M2 |∇ z | − b dx g (ϕ(z )) p− Ω " ! # Z 0 p− −1 p(x) − 1 p(x)−1 (ϕ (z1 )) p(x) ≥ ηg (ϕ(z )) M z + |∇ z | − b dx g (ϕ(z )) p− Ω Z 1 −ϕ 0 (z1 ) p(x) − 1 p(x)−1 p(x) = ηg (ϕ(z )) M z · − 00 + |∇ z | − b dx p ϕ (z1 ) p− Ω
bg (ω)η ≥
Ω
where M = min{λ1 M1 , M2 }. Note that, since |∇ z | 6= 0 on ∂ Ω , we have 1 −ϕ 0 (z1 ) p(x) − 1 + |∇ z |p(x) inf z p(x)−1 · − 00 Ω p ϕ (z1 ) p−
>0
and thus, by choosing β ≥ 1 sufficiently big, we see that
Z
|∇ω|p(x)−2 ∇ω · ∇η ≥ Ω
Z Ω
bg (ϕ(z ))η ≥
Z Ω
bg (ω)η
for all η ∈ C0∞ (Ω ) with η ≥ 0. A straightforward application of the Fatou lemma shows that the above inequality holds for 1,p(x)
all non-negative η ∈ W0 u(x) ≤ βϕ(z (x))
(Ω ). Thus, if u is a solution of (1.1) by the comparison principle, Lemma 3.3, we have
(x ∈ Ω ).
Using ϕ := u in (3.2) we note that
(inf) Ω
Z Ω
g (u)u ≤
Z Ω
bg (u)u =
Z Ω
|∇ u|p(x)−2 < ∞.
Therefore since, by condition 3(ii) of [G2 ], Cβ ϕ(z )g (ϕ(z )) ≤ βϕ(z )g (βϕ(z )) ≤ ug (u) on Ω for some positive constant Cβ , we have
Z Ω
ϕ(z )g (ϕ(z )) < ∞.
The proof is complete.
J. Liu / Nonlinear Analysis 72 (2010) 4428–4437
4437
Acknowledgements The authors would like to thank professor Xianling Fan for clear suggestions and the reviewer(s) for clear valuable comments and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
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