MULTIPLE CRITICAL POINTS FOR NONDIFFERENTIABLE FUNCTIONALS INVOLVING HARDY POTENTIALS BENEDETTA PELLACCI1 Abstract. In this paper we study general functionals of the calculus of variations with the presence of a Hardy potential. We will improve several results obtained in the semilinear framework. We will first prove a general weak lower semicontinuity result, which will imply the existence of a minimum point whenever the functional is coercive. Then we will demonstrate existence and multiplicity results of critical points, even if our functional is not differentiable. We will apply a nonsmooth critical point theory developed in [?] and [?].
1. Introduction Let Ω be a bounded open set in RN (N ≥ 2), 0 ∈ Ω, 1 < p < N and let us define the functional J : W01,p (Ω) → R by Z Z Z λ |v|p J(v) = j(x, v, ∇v) − − G(x, v), p Ω |x|p Ω Ω where j(x, s, ξ) : Ω × R × RN → R is a measurable function with respect to x for all (s, ξ) ∈ R × RN , and of class C 1 with respect to (s, ξ) for a.e. x ∈ Ω. We also assume that for almost every x in Ω and for every s in R (1.1)
{ξ 7→ j(x, s, ξ)} is convex.
We will denote with js (x, s, ξ) and jξ (x, s, ξ) the derivatives of j(x, s, ξ) with respect to the variables s and ξ respectively. Regarding the function js (x, s, ξ), we assume that there exist positive constants R and β0 such that the following conditions are satisfied almost everywhere in Ω and for every ξ ∈ RN : js (x, s, ξ) ≤ β0 |ξ|p (1.2) for every s in R, (1.3)
js (x, s, ξ)s ≥ 0
for every s in R with |s| ≥ R.
Moreover, we suppose that there exists α0 > 0 such that the following growth conditions are satisfied for almost every x ∈ Ω and for every (s, ξ) ∈ R × RN jξ (x, s, ξ) ≤ β0 |ξ|p−1 . (1.4) (1.5)
jξ (x, s, ξ) · ξ ≥ α0 |ξ|p .
Notice that, conditions (??) and (??) imply that the function j(x, s, ξ) satisfies the following hypothesis for almost every x ∈ Ω and for every (s, ξ) ∈ R × RN (see Remark ?? for more details) (1.6) 1
α0 p β0 |ξ| ≤ j(x, s, ξ) ≤ |ξ|p . p p
Research supported by the MIUR project “Metodi Variazionali ed Equazioni Differenziali Nonlineari”. 1
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BENEDETTA PELLACCI
The function G(x, s) is the primitive with respect to s (such that G(x, 0) = 0) of a Carath´eodory (i.e. measurable with respect to x and continuous with respect to s) function g(x, s). Let H be the best constant such that the Hardy inequality holds (see [?] or [?]) Z Z |v|p 1 (1.7) ≤ |∇v|p ∀ v ∈ W01,p (Ω). p H Ω Ω |x| We will also assume that λ (1.8) < α0 . H We will prove different existence results of critical points for the functional J in dependence of different classes of functions g(x, s). Let p∗ = N p/(N − p) and p∗ 0 its conjugate exponent. As a first model example of function g, let us consider (1.9)
g(x, s) = g0 (x, s) = |s|ϑ + f (x)
∗0
0 < ϑ < p − 1, f ∈ Lp (Ω).
Hypothesis (??), Hardy inequality and the definition of g0 (x, s) imply that the functional J is well defined on W01,p (Ω). In addition, due to the dependence on s of the function j(x, s, ξ), the functional J is not differentiable on W01,p (Ω). Indeed, (??) and (??) imply that we can compute hJ 0 (u), vi for every u ∈ W01,p (Ω) and for every v ∈ W01,p (Ω) ∩ L∞ (Ω). Thus, we can say that a critical point of J is a function u ∈ W01,p (Ω), which is a distributional solution of the following problem p−2 −div(j (x, u, ∇u)) + j (x, u, ∇u) = λ |u| u + g0 (x, u) in Ω, s ξ (P0 ) |x|p u=0 on ∂Ω. We will prove the following result, as a corollary of a more general theorem (see Theorem ??). Theorem 1.1. Suppose that conditions (??), (??), (??), (??) are satisfied. Then there exists u ∈ W01,p (Ω) minimum point of the functional J. Moreover, if (??) holds, u is a solution of the problem (P0 ). Notice that conditions (??) and (??) imply that the functional J is coercive, then, in order to get the existence of a minimum we only have to prove that J is weakly lower semicontinuous on W01,p (Ω). When j(x, s, ξ) = |ξ|p /p, both the elliptic and the parabolic problem have been widely studied (see [?], [?], [?], [?] and the references therein). In [?] the elliptic problem is faced by means of variational techniques. In that work the authors study first the case in which g(x, s) = f (x) and they prove the existence of a minimum point of the associated functional without proving that the functional is weakly lower semicontinuous. Recently, in [?] it has been proved that the functional in [?] is weakly lower semicontinuous; so that, when j(x, s, ξ) = |ξ|p /p, the existence of a minimum can be proved by means of the direct methods in the calculus of variations. The proof in [?] makes use of ”concentration compactness” arguments. Here we will prove that the general functional I : W01,p (Ω) → R defined by Z Z λ |v|p (1.10) I(v) = j(x, v, ∇v) − with α0 ≥ 0, and λ ≤ Hα0 , p Ω |x|p Ω is weakly lower semicontinuous. Indeed, we will demonstrate a more general result (see Theorem ??) by means of an elementary proof. Thanks to this result we will prove the existence of a minimum point whenever the nonlinearity has a “sublinear” growth
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
3
(Theorem ??). Then we will consider “superlinear” nonlinearity. As a first simple model example of function g that we will study, let us consider Np (1.11) g(x, s) = g1 (x, s) = arctg(|s|p ) + |s|σ−2 s p<σ< . N −p We will prove the following theorem, as a corollary of a more general result (see Theorem ??). Theorem 1.2. Suppose that conditions (??), (??), (??), (??), (??) and (??) are satisfied. Moreover, let us assume that (1.12) the function ξ 7→ j(x, s, ξ) is strictly convex. We also suppose that there exists δ1 > 0 such that σj(x, s, ξ) − js (x, s, ξ)s − jξ (x, s, ξ) · ξ ≥ δ1 |ξ|p if |s| > R, (1.13) δ1 p λ and < , H σ−p where R is given by (??) and σ is given by (??). Then there exists a nontrivial distributional solution of the following problem p−2 −div(j (x, u, ∇u)) + j (x, u, ∇u) = λ |u| u + g1 (x, u) in Ω, s ξ (P1 ) |x|p u=0 on ∂Ω. We will also prove a multiplicity result of critical points if J is even. In this case the simplest model example of function g(x, s) we will study is Np (1.14) g(x, s) = g2 (x, s) = |s|σ−2 s, with p < σ < N −p and we will prove the following result as acorollary of a general Theorem (see Theorem ??). Theorem 1.3. Suppose that conditions (??), (??), (??), (??), (??), (??), (??) and (??) are satisfied. In addition, suppose that j(x, −s, −ξ) = j(x, s, ξ). Then, there exists a sequence {un } of nontrivial distributional solutions of the problem p−2 −div(j (x, u, ∇u)) + j (x, u, ∇u) = λ |u| u + g2 (x, u) in Ω, s ξ (P2 ) |x|p u=0 on ∂Ω, such that J(un ) → ∞. When j(x, s, ξ) = |ξ|p /p and g(x, s) = g2 (x, s) (defined in (??)), the existence of a positive solution in [?] is proved by means of critical point theory. Indeed conditions (??) (which is evidently satisfied when j(x, s, ξ) = |ξ|p /p) and (??) (which is assumed) imply that the geometrical assumptions of the Ambrosetti-Rabinowitz Theorem ([?]) are satisfied. Then, in [?] it is proved that a Palais-Smale sequence weakly converges to a function u ∈ W01,p (Ω) and the convergence is strong in a larger space W01,q (Ω) for every q < p. Then, this weak limit is proved to be a nontrivial solution by means of homogeneity arguments. Recently, in [?] it has been proved, when p = 2 and g = g2 that
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BENEDETTA PELLACCI
the usual Palais-Smale condition holds. This result has been demonstrated in [?] for any 1 < p < N and for g = g2 . Thus, in [?], it has also been proved a multiplicity result. Also in that paper the Palais-Smale condition is proved by using the homogeneity of the nonlinearities involved. Indeed, this property permits to compare in a suitable way the functional and its differential. Here, we can still check that the geometrical assumptions of the Ambrosetti-Rabinowitz Theorem hold as in the previous case, (see the proof of Theorems ?? and ??), but our functional is not of class C 1 , so that we cannot apply the classical Mountain-Pass Theorem. Then, we will apply an abstract critical point theory developed in [?] and [?] that is set up for continuous functions defined on a complete metric space. In addition, we do not have any homogeneity property. This is a consequence not only of the general dependence on s of the function g(x, s) (see hypotheses (??) and (??) in Section ??) but, mainly because of the general form of j(x, s, ξ). Indeed, we get an additional term (i.e. js (x, u, ∇u)) in the derivative, which is not present in the functional. Neverthless, we will still prove that the Palais-Smale condition holds. So that, if λ is below the critical value λ = α0 H, the problem is actually compact (see Theorems ?? and ??). We will give two different proofs of the Palais-Smale condition. One is direct, the other makes use of a compactness result (Theorem ??) which is in the spirit of the classical ones, and that can also be useful in other situations. For general integrals of the calculus of variations several existence results of critical point have been proved by means of different techniques. In [?] it has been demonstrated a version of the Ambrosetti-Rabinowitz Theorem for functionals which are not derivable along every directions and this abstract result has been applied in [?], [?], [?], [?] in order to get different existence theorems of a critical point for integrands j(x, s, ξ) bounded or not with respect to s. In [?], [?], [?], [?], [?], [?], [?] it has been used the abstract critical point theory developed in [?], [?]. In all these papers it has not been considered the presence of a Hardy potential. In particular in [?] it is studied the case in which (??) is not satisfied, in [?] some new simpler proof of the results in [?] and [?] are given. The key point in the proof of the Palais-Smale condition in [?], [?] is to prove that a cluster point of a Palais-Smale sequence is in L∞ (Ω) before proving that it is a critical point. If j(x, s, ξ) has a general form and it is not supposed to be bounded from above with respect to s, and p = 2, existence and multiplicity results of unbounded critical points have been recently proved in [?]. In this work the growth conditions imposed on the nonlinearity g(x, s) imply that the solutions are not bounded. For example, a model nonlinearity treated in the nonsymmetric case is g(x, s) = d(x)arctgs2 + |s|σ−2 s, 2 < σ < 2N/(N − 2), N with d ∈ L 2 (Ω). The presence of the function d(x) implies that a solution is not expected to be bounded, even if we expect it to be in every Lq (Ω) for q < ∞. In our case, the situation is even worse, because of the presence of the Hardy potential. Indeed, the N function 1/|x|p 6∈ L p (Ω), so that, even if the nonlinearity is g(x, s) = g2 (x, s) = |s|σ−2 s the solutions will not be bounded (see [?] for a proof).
The paper is organized as follows. In the following Section we will set the general problem and we will state the main results we will prove (see Theorems ??, ?? and ??). In Section ?? we recall the abstract results of nonsmooth critical point theory we will use. In Section ?? we will prove some important facts concerning the functional I defined in (??). Namely, we will first prove a key lemma (see Lemma ??) and then, we will prove our weakly lower semicontinuity result (Theorem ??). Besides, we will also prove a compactness theorem in the spirit of the classical ones (see Theorem ??). In
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
5
Section ?? we will show that every bounded Palais-Smale sequence is relatively compact in W01,p (Ω) (see Theorem ??). In Section ?? we will give the complete proof of our results. 2. Setting of the problem and main results Let Ω be a bounded open set in RN (N ≥ 2). Throughout the paper we will denote 0 by k · kq , k · k, k · k−1,p0 the standard norms in the spaces Lq (Ω), W01,p (Ω) and W −1,p (Ω) respectively. We will prove the existence of distributional solutions of the following problem p−2 −div(j (x, u, ∇u)) + j (x, u, ∇u) = λ |u| u + g(x, u) in Ω, s ξ (P ) |x|p u=0 on ∂Ω, where jξ (x, s, ξ) satisfies conditions (??), (??) and js (x, s, ξ) satisfies (??) and (??). In order to prove existence of solutions of (P ) we will use variational methods, so that we will study the functional J : W01,p (Ω) → R defined by Z Z Z Z λ |v|p J(v) = I(v) − G(x, v) = j(x, v, ∇v) − − G(x, v), p Ω |x|p Ω Ω Ω Notice that, condition (??) and Hardy inequality imply that I is well defined on W01,p (Ω). Moreover, conditions (??), (??) imply that there exists hI 0 (u), vi for every u ∈ W01,p (Ω) and for every v ∈ W01,p (Ω) ∩ L∞ (Ω). The function G(x, s) is the primitive with respect to s (G(x, 0) = 0) of a Carath´eodory function g(x, s). Let p∗ = N p/(N − p) and p∗ 0 its conjugate exponent. In the coercive case we will assume that g(x, s) satisfies the following growth condition for every s ∈ R and almost everywhere in Ω Assume that for every ε > 0 there exists fε (x) such that (2.1)
|g(x, s)| ≤ ε|s|p−1 + fε (x),
∗0
fε (x) ∈ Lp (Ω), ∀s ∈ R, a.e. in Ω,
we will prove the following result. Theorem 2.1. Assume (??), (??), (??), (??). Then, there exists u ∈ W01,p (Ω) a minimum point of the functional J. Moreover, if (??) holds u is a distributional solution of Problem (P ). Remark 2.2. In Theorem ?? we have not supposed condition (??), because from (??) and (??) we can deduce that jξ (x, s, ξ) satisfies an analogous growth condition. More precisely, if we assume that (??) holds, then we get ∀v ∈ RN : |v| ≤ 1 =⇒ j(x, s, ξ + |ξ|v) ≥ j(x, s, ξ) + jξ (x, s, ξ) · v|ξ|. This, and (??) yield jξ (x, s, ξ) · v|ξ| ≤
β0 β0 |ξ + v|ξ||p ≤ 2p |ξ|p . p p
From the arbitrariness of v we obtain (2.2)
|jξ (x, s, ξ)| ≤ 2p
β0 p−1 |ξ| . p
In the superlinear nonsymmetric case we will assume that g(x, s) satisfies the following growth condition for every s ∈ R and almost everywhere in Ω. (2.3)
|g(x, s)| ≤ a(x)|s|p−1 + b|s|r−1 ,
N
with 1 < r < p∗ , a(x) ∈ L p (Ω), b > 0
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BENEDETTA PELLACCI
Regarding the functions g(x, s) we will also assume that there exist t, a0 (x), a(x) ∈ b0 (x), b(x) ∈ Lm (Ω), with m = p∗ /(p∗ − t) and k(x) ∈ L∞ (Ω) with k(x) > 0 almost everywhere, such that
L1 (Ω), (2.4)
σG(x, s) ≤ g(x, s)s + a0 (x) + b0 (x)|s|t
(2.5)
σ
σ > p, 1 < t < p ,
t
G(x, s) ≥ k(x)|s| − a(x) − b(x)|s| ,
where σ is the same in (??). We will prove the following Theorem. Theorem 2.3. Assume conditions (??), (??), (??), (??), (??), (??), (??), (??), (??), (??). Moreover, let us suppose that (2.6)
lim
s→0
g(x, s) =0 |s|p−1
a.e. in Ω.
Then there exists a distributional solution of Problem (P ). In the symmetric case we will suppose that g(x, s) satisfies the following growth condition. ∗0 Assume that for every ε > 0 there exists aε ∈ Lp (Ω) such that (2.7)
∗ −1
|g(x, s)| ≤ aε (x) + ε|s|p
We will prove the following result. Theorem 2.4. Assume conditions (??), (??), (??), (??), (??), (??), (??) (??), (??), (??). Moreover, let us suppose that (2.8)
j(x, −s, −ξ) = j(x, s, ξ)
g(x, −s) = −g(x, s),
almost everywhere in Ω.
Then, there exists a sequence {un } ∈ W01,p (Ω) of distributional solutions of Problem (P ) such that J(un ) → +∞. Remark 2.5. In order to prove Theorems ?? and ?? we have to assume that λ satisfies (??) and (??) that is λ δ1 p < min α0 , . H σ−p Indeed, since in Theorems ?? and ?? the nonlinearity g(x, s) has a superlinear behavior we need (??) in order to prove that a Palais-Smale sequence is bounded in W01,p (Ω). While, in order to show Theorem ??, it is enough to suppose (??). This because in Theorem ?? g(x, s) has a “sublinear” behavior so that, (??) is sufficient to get the boundedness of a minimizing sequence. Notice also that, in the semilinear case, i.e. if j(x, s, ξ) = |ξ|2 /2 condition (??) becomes λ < H and pδ1 = 2δ1 = σ − 2, so that (??) is satisfied whenever λ satisfies (??). Remark 2.6. Notice that if g(x, s) satisfies condition (??) or (??), then it satisfies (??). Indeed, suppose that (??) holds, then Young inequality implies that for every ε > 0, it results ∗ |g(x, s)| ≤ ε|s|p −1 + c0ε + fε (x). Then (??) follows by taking aε (x) = c0ε + fε (x). If (??) holds, we deduce analogously that for every ε > 0 it results p∗ −1
|g(x, s)| ≤ ε|s|
+ cε +
“ ∗ ”0 p −1 0 cε [a(x)] p−1 .
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
7
Since a(x) ∈ Lr (Ω), with r = N/p, it follows that “
aε (x) = cε + c0ε [a(x)]
p∗ −1 p−1
”0
∗0
∈ Lp (Ω)
(For more details regarding hypotheses (??) and (??) see Remark ?? in Section ??). 3. Abstract critical point theory In this section we will recall some basic facts of the abstract critical point theory we will use.We refer to [?], [?] and [?], where this abstract theory has been developed. Let X be a complete metric space, we denote with B(u, δ) the open ball of center u ∈ X and of radius δ. Let us recall the definition of the weak slope for a continuous function introduced in [?, ?, ?, ?]. Definition 3.1. Let X be a complete metric space, f : X → R be a continuous function, and u ∈ X. We denote by |df |(u) the supremum of the real numbers σ in [0, +∞) such that there exist δ > 0 and a continuous map H : B(u, δ) × [0, δ] → X, such that, for every v in B(u, δ), and for every t in [0, δ] it results d(H(v, t), v) ≤ t, f (H(v, t)) ≤ f (v) − σt. The extended real number |df |(u) is called the weak slope of f at u. Whenever, f is of class C 1 |df |(u) coincides with the norm of the differential of f . The previous notion allows us to give the following definitions. Definition 3.2. Let X be a complete metric space and f : X → R a continuous function. A point u ∈ X is a critical point of f if |df |(u) = 0. We say that c ∈ R is a critical value of f if there exists a critical point u ∈ X of f with f (u) = c. Definition 3.3. Let X be a complete metric space, f : X → R a continuous function and let c ∈ R. We say that f satisfies the Palais–Smale condition at level c ((P S)c in short), if every sequence {un } in X such that |df |(un ) → 0, f (un ) → c, admits a subsequence {unk } converging in X. In the nonsymmetric case we will apply the following abstract result proved in [?]. Theorem 3.4. Let X be a Banach space, f : X → R a continuous function. First, suppose that there exist w ∈ X, η > f (0) and r > 0 such that (3.1)
f (u) > η,
∀ u ∈ X, kuk = r,
(3.2)
f (w) < η,
kwk > r.
We set Γ = {γ : [0, 1] → X, continuous γ(0) = 0, γ(1) = w} . Finally, suppose that f satisfies the Palais-Smale condition at the level c = inf sup f (γ(t)) < +∞, Γ [0,1]
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BENEDETTA PELLACCI
then, there exists a nontrivial critical point u of f such that f (u) = c. For even functions f the following existence result holds (see [?]). Theorem 3.5. Let X be a Banach space and f : X → R a continuous even function. Let us assume that there exists a strictly increasing sequence {Wh } of finite dimensional subspaces of X with the following properties: (a) there exist ρ > 0, η > f (0) and a subspace V ⊂ X of finite codimension such that ∀u ∈ V : kuk = ρ =⇒ f (u) ≥ η ;
(3.3)
(b) there exists a sequence {Rh } in ]ρ, +∞[ such that ∀u ∈ Wh : kuk ≥ Rh =⇒ f (u) ≤ f (0) ;
(3.4)
(c) f satisfies (P S)c for any c ≥ η; Then there exists a sequence {uh } of critical points of f with lim f (uh ) = +∞ .
h→∞
In order to prove that our functional J satisfies the (P S)c condition, it will be important the following notion (see [?]). Definition 3.6. A sequence {un } ⊂ W01,p (Ω) is a Concrete-Palais-Smale sequence at 0 level c ((CP S)c for short) if there exists yn ∈ W −1,p (Ω), such that J(un ) → c,
(3.5) (3.6)
Z Z Z |un |p−2 hJ (un ), vi = jξ (x, un , ∇un )∇v + js (x, un , ∇un )v − λ un v p Ω Ω |x| Ω Z − g(x, un )v = hyn , vi, ∀ v ∈ C0∞ (Ω), yn → 0 0
Ω
Moreover, we say that J satisfies the (CP S)c if every (CP S)c sequence is strongly compact in W01,p (Ω). Notice that if g(x, s) satisfies condition (??), the functional F : W01,p (Ω) → R defined by |v|p F (v) = − p Ω |x| Z
Z G(x, v) Ω
is of class C 1 in W01,p (Ω). Then in [?] the following result is proved. Proposition 3.7. Assume (??), (??), (??), (??), (??). Let u ∈ W01,p (Ω) be a critical point of J. Then, u is a distributional solution of Problem (P ). Moreover, if J satisfies the (CP S)c condition, then J satisfies (P S)c condition. 4. Study of the Functional I In this section we will prove some important facts concerning the functional I defined in (??). Let us first make the following remark. Remark 4.1. It readily seen that (??) and (??) imply that (??) holds. Indeed, we have Z 1 (4.1) j(x, s, ξ) = jξ (x, s, tξ) · ξdt 0
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
9
So that if (??) holds we get 1 |j(x, s, ξ)| ≤ β0 |ξ|p . p Moreover, from (??) and (??) we deduce the left hand side inequality in (??). Remark 4.2. Let us point out that the assumption of strict convexity on the function {ξ → j(x, s, ξ)} implies that for almost every x in Ω and for every s in R, we have: [jξ (x, s, ξ) − jξ (x, s, ξ ∗ )] · (ξ − ξ ∗ ) > 0,
(4.2)
for every ξ, ξ ∗ ∈ RN , with ξ 6= ξ ∗ . Let us define the following functions. For every k > 0 we define the functions Tk : R → R, Gk : R → R by (4.3)
Tk (s) := max{−k , min{s, k}},
Gk (s) = s − Tk (s).
From the definition it follows that Tk (u), Gk (u) ∈ W01,p (Ω) for every u ∈ W01,p (Ω). It will be fundamental the following lemma. Lemma 4.3. Let us consider a sequence {un } ⊂ W01,p (Ω), which weakly converges to a function u. Then, for any k ∈ R+ , it results (4.4) (4.5)
|u|p − |Gk (u)|p |un |p − |Gk (un )|p → strongly in L1 (Ω). |x|p |x|p |u|p−2 uGk (u) − |Gk (u)|p |un |p−2 un Gk (un ) − |Gk (un )|p → strongly in L1 (Ω). |x|p |x|p
Proof. Let us first notice that, it is enough to show that (??) and (??) hold for a subsequence. Indeed, if it is so, then we can prove that the conclusions are true for un by arguing by contradiction. Now, notice that (??) and Lagrange mean value Theorem imply that there exists a positive constant cp such that the following inequality holds for every k > 0 (4.6)
||un |p − |Gk (un )|p | ≤ cp k|un |p−1
a.e. in Ω.
Since un * u in W01,p (Ω), up to a subsequence, we have that un → u almost everywhere in Ω. Moreover, note that |un |p−1 /|x|p strongly converges in L1 (Ω). Then we can use Fatou Lemma applied to the nonnengative sequence |un |p−1 + |u|p−1 ||un |p − |Gk (un )|p − (|u|p − |Gk (u)|p )| cp k − |x|p |x|p To deduce that ||un |p − |Gk (un )|p − (|u|p − |Gk (u)|p )| ≤ 0, |x|p Ω
Z lim sup n→∞
which yields (??). Now, let us prove (??). We notice that |Gk (s)|2 ≤ sGk (s) ≤ |s|2 , and we get 0 ≤ |s|p−2 sGk (s) − |Gk (s)|p ≤ |s|p − |Gk (s)|p . This and (??) imply (??)
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BENEDETTA PELLACCI
Now, let us show that I is weakly lower semicontinuous. We will actually prove this for a slightly more general functional. We consider a function f (x, s, ξ) : Ω × R × RN → R, which is measurable with respect to x for every (s, ξ) and continuous with respect to (s, ξ) for almost every x. Moreover, the following conditions are satisfied (4.7)
f (x, s, 0) = 0,
(4.8)
{ξ 7→ f (x, s, ξ)}
(4.9)
f (x, s, ξ) ≥ α1 |ξ|p
is convex, α1 ≥ 0,
µ ≤ Hα1 .
(4.10)
We define the functional F : Lp (Ω) × W01,p (Ω) → R ∪ {+∞} by Z Z Z p f (x, w, ∇v) − µ |v| f (x, w, ∇v) dx < +∞ , if p F (w, v) = Ω |x| Ω Ω +∞ otherwise, If we take w = v, the Poincar´e inequality implies that W01,p (Ω) ⊂ Lp (Ω) so that F is defined on W01,p (Ω). We will prove the following result, which has been proved for f (x, s, ξ) = 1/p|ξ|p in [?] by means of ”concentration compactness” arguments. Here we will give a general elementary proof. Theorem 4.4. Let us assume that conditions (??), (??), (??), (??) are satisfied. Let us take two sequences {wn } ⊂ Lp (Ω) and {vn } ⊂ W01,p (Ω), such that wn → w strongly in Lp (Ω), and vn * v weakly in W01,p (Ω). Then Z Z |vn |p |v|p lim inf f (x, wn , ∇vn ) − µ p ≥ f (x, w, ∇v) − µ p . n→∞ Ω |x| |x| Ω Proof. First, notice that if Z f (x, wn , ∇vn ) = +∞
lim inf n→∞
Ω
we have nothing to prove. Otherwise, conditions (??), (??) and De Giorgi semicontinuity Theorem yield Z Z 0≤ f (x, w, ∇v) ≤ lim inf f (x, wn , ∇vn ) < ∞, Ω 1 L (Ω).
n→∞
Ω
so that f (x, w, ∇v) ∈ This fact and Hardy inequality imply that for every fixed ε > 0 we can find k0 > 0 such that for every k ≥ k0 it results Z Z |Gk (v)|p (4.11) f (x, w, ∇Gk (v)) − µ < ε, |x|p Ω Ω this inequality, (??) and condition (??) yield Z Z Z |v|p |v|p − |Gk (v)|p (4.12) f (x, w, ∇v) − µ p ≤ f (x, w, ∇Tk (v)) − µ + ε. |x| |x|p Ω Ω Ω Moreover, notice that ∇Tk (vn ) * ∇Tk (v) in Lp (Ω). This and assumptions (??) and (??) imply that we can use De Giorgi semicontinuity Theorem to deduce Z Z lim inf f (x, wn , ∇Tk (vn )) ≥ f (x, w, ∇Tk (v)). n→∞
Ω
Ω
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
11
This and Lemma ?? imply that Z Z |v|p − |Gk (v)|p f (x, w, ∇Tk (v)) − µ (4.13) |x|p Ω Ω Z Z |vn |p − |Gk (vn )|p ≤ lim inf f (x, wn , ∇Tk (vn )) − µ lim n→∞ Ω n→∞ |x|p ΩZ Z |vn |p − |Gk (vn )|p . = lim inf f (x, wn , ∇Tk (vn )) − µ n→∞ |x|p Ω Ω Moreover, notice that hypothesis (??) and Hardy inequality implies that Z Z |Gk (vn )|p f (x, wn , ∇Gk (vn )) − µ (4.14) ≥0 |x|p Ω Ω We use (??) and (??) in (??) and we get Z Z Z |v|p |Gk (vn )|p f (x, w, ∇v) − µ p ≤ ε + lim inf f (x, wn , ∇Gk (vn )) − µ n→∞ |x| |x|p Ω Ω Ω Z Z |vn |p − |Gk (vn )|p + lim inf f (x, wn , ∇Tk (vn )) − µ n→∞ |x|p Ω Ω Z Z |vn |p ≤ ε + lim inf (f (x, wn , ∇Tk (vn )) + f (x, wn , ∇Gk (vn ))) − µ p n→∞ Ω Ω |x| Z p |vn | = ε + lim inf f (x, wn , ∇vn ) − µ p n→∞ Ω |x| that is what we had to prove. Corollary 4.5. Assume conditions (??), (??) and (??). Then the functional I defined in (??) is weakly lower semicontinuous. Proof. We set f (x, s, ξ) = j(x, s, ξ), µ = λ/p, so that f (x, s, ξ) and µ satisfy (??), (??), (??) with α1 = α0 /p and (??). Then, the conclusion follows from Theorem ??. Now we prove an important theorem in order to get some compactness results. We will use the following lemma. Lemma 4.6. Assume conditions (??), (??), (??). Let us take {wn } ∈ Lp (Ω) such that wn → w in Lp (Ω) and {vn } ∈ W01,p (Ω) with vn * v weakly in W01,p (Ω) and ∇vn → ∇v almost everywhere in Ω. Then it results Z Z |vn |p |v|p f (x, w, ∇v) − µ p . (4.15) lim inf f (x, wn , ∇vn ) − µ p ≥ n→∞ |x| |x| Ω Ω Remark 4.7. Note that in this Lemma we do not assume condition (??), but we suppose the almost everywhere convergence of ∇vn to ∇v, which implies that vn → v strongly in W01,q (Ω) for every q < p. Proof. Since wn → w and ∇vn → ∇v almost everywhere, condition (??) implies that we can use Fatou Lemma to deduce that Z Z p (4.16) lim inf [f (x, wn , ∇vn ) − α1 |∇vn | ] ≥ [f (x, w, ∇v) − α1 |∇v|p ] n→∞
Ω
Ω
12
BENEDETTA PELLACCI
Moreover, since vn * v in W01,p (Ω) and wn → w in Lp (Ω) we can use Theorem ?? to obtain Z Z |vn |p |v|p p p α1 |∇vn | − µ p ≥ α1 |∇v| − µ p . (4.17) lim inf n→∞ Ω |x| |x| Ω Then, (??) and (??) yield Z Z |vn |p lim inf f (x, wn , ∇vn ) − µ p ≥ lim inf [f (x, wn , ∇vn ) − α1 |∇vn |p ] n→∞ n→∞ |x| Ω Ω Z |vn |p p α1 |∇vn | − µ p + lim inf n→∞ Ω |x| Z p |v| f (x, w, ∇v) − µ p . ≥ |x| Ω The following theorem is in the spirit of the classical results of compactness type. Theorem 4.8. Assume conditions (??) and (??) with α1 > 0. Moreover, suppose that µ satisfies (4.18)
µ < Hα1 .
Let us consider a sequence {un } ⊂ W01,p (Ω), such that (4.19)
un * u weakly in W01,p (Ω),
∇un → ∇u almost everywhere in Ω.
Moreover, u is such that (4.20)
f (x, u, ∇u) ∈ L1 (Ω).
In addition, suppose that Z Z |un |p |u|p (4.21) lim sup f (x, un , ∇un ) − µ p ≤ f (x, u, ∇u) − µ p |x| |x| n→∞ Ω Ω Then, un strongly converges to u in W01,p (Ω). Remark 4.9. Notice that, in Theorem ?? we do not suppose any growth condition from above on f (x, s, ξ), so that Theorem ?? can be useful when studying functionals not bounded from above. Remark 4.10. If we take wn = un and vn = un in Lemma ??, we get Z Z |u|p |un |p f (x, u, ∇u) − µ p . lim inf f (x, un , ∇un ) − µ p ≥ n→∞ |x| |x| Ω Ω This, together with (??) imply that Z Z |u|p |un |p (4.22) lim f (x, un , ∇un ) − µ p = f (x, u, ∇u) − µ p . n→∞ Ω |x| |x| Ω Remark 4.11. Notice that, the weak lower semicontinuity result also holds for α1 = 0 (µ ≤ 0), or for µ = Hα1 . While, when we prove the compactness result (Theorem ??) we have to assume that α1 > 0 and that (??) holds.
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
13
Proof. We will divide the proof into four steps. Step 1. We will show that the following inequalities hold Z Z (4.23) lim inf f (x, un , ∇Tk (un )) ≥ f (x, u, ∇Tk (u)), n→∞
Ω
Ω
(4.24) Z Z |Gk (un )|p |Gk (u)|p f (x, un , ∇Gk (un )) − µ lim inf f (x, u, ∇Gk (u)) − µ ≥ . n→∞ Ω |x|p |x|p Ω Indeed, hypotheses (??) and (??) imply that we can use Fatou Lemma to deduce (??). Moreover, we apply Lemma ?? with wn = un and vn = Gk (un ) and we get (??). Step 2. In this step we will prove that (4.25)
lim lim sup kGk (un )k = 0.
k→∞ n→∞
First, notice that (??) and (??) imply that (4.26)
f (x, u, ∇u) = f (x, u, ∇Tk (u)) + f (x, u, ∇Gk (u)),
for every u ∈ W01,p (Ω).
Then, we have Z Z |un |p |Gk (un )|p f (x, un , ∇un ) − µ p f (x, un , ∇Gk (un )) − µ ≤ lim sup lim sup |x|p |x| n→∞ n→∞ Ω Ω Z Z p p |un | − |Gk (un )| + lim sup − f (x, un , ∇Tk (un )) +µ . |x|p n→∞ Ω Ω We use hypothesis (??), inequality (??) and Lemma ?? in the previous inequality and we obtain Z Z |u|p |Gk (un )|p (4.27) lim sup ≤ f (x, u, ∇u) − µ p f (x, un , ∇Gk (un )) − µ |x|p |x| n→∞ Ω Ω Z Z p p |u| − |Gk (u)| |Gk (u)|p − f (x, u, ∇Tk (u)) + µ = f (x, u, ∇Gk (u)) − µ . |x|p |x|p Ω Ω Now, hypothesis (??) and Hardy inequality imply that for every ε > 0 there exists k0 such that for every k ≥ k0 it results Z |Gk (u)|p < ε. (4.28) 0≤ f (x, u, ∇Gk (u)) − µ |x|p Ω Therefore, (??) and (??) imply that for every ε > 0 and for every fixed k ≥ k0 there exists n0 such that for every n ≥ n0 it holds Z µ |Gk (un )|p p α1 − kGk (un )k ≤ f (x, un , ∇Gk (un )) − µ < 2ε, H |x|p Ω which, together with (??), implies (??). Step 3. We will prove that for every k ≥ 0 (4.29)
lim kTk (un ) − Tk (u)k = 0.
n→∞
14
BENEDETTA PELLACCI
From (??) we get Z Z |un |p lim sup f (x, un , ∇Tk (un )) ≤ lim sup f (x, un , ∇un ) − µ p |x| n→∞ n→∞ Ω Ω Z |Gk (un )|p + lim sup − f (x, un , ∇Gk (un )) − µ |x|p n→∞ Ω Z |un |p − |Gk (un )|p + lim sup µ . |x|p n→∞ Ω Thus, (??), (??) and Lemma ?? imply Z Z (4.30) lim sup f (x, un , ∇Tk (un )) ≤ f (x, u, ∇Tk (u)). n→∞
Ω
Ω
Finally, condition (??) implies that we can apply Fatou Lemma to the sequence α1 f (x, un , ∇Tk (un )) + f (x, u, ∇Tk (u)) − p−1 |∇Tk (un ) − ∇Tk (u)|p . 2 Thanks to (??) and (??), we obtain (??). Step 4. Finally, let us prove the strong convergence of un . From (??) we get for every k>0 kun − uk ≤ kTk (un ) − Tk (u)k + kGk (un )k + kGk (u)k. For every ε > 0 we can fix k1 ≥ k0 (where k0 is fixed in (??)) such that kGk1 uk ≤ ε. From (??) and (??) we obtain that there exists n0 > 0 such that for every n ≥ n0 it results kun − uk ≤ kTk1 (un ) − Tk1 (u)k + kGk1 (un )k + ε ≤ c1 ε, which yields the conclusion. Remark 4.12. Note that in the previous proof we have also proved that Z Z |Gk (u)|p |Gk (un )|p = f (x, u, ∇Gk (u)) − µ . lim f (x, un , ∇Gk (un )) − µ n→∞ Ω |x|p |x|p Ω Indeed, this is a consequence of (??) and (??). 5. compactness result In this section we will prove a compactness result for the functional I. Namely, we will prove the following theorem. Theorem 5.1. Assume conditions (??), (??), (??), (??), (??) and (??). Let us consider 0 wn ∈ W −1,p (Ω), with wn → w and let un ∈ W01,p (Ω) be a bounded sequence such that (5.1)
hI 0 (un ), vi = hwn , vi
∀v ∈ C0∞ (Ω).
Then, un is compact in W01,p (Ω). We will give two different proof this result. In the first we will obtain the conclusion in a direct way, while in the second we will use Theorem ??. First Proof. In this proof we will not apply Theorem ??, but we will prove (??) and (??) directly. Step1. We will prove that (5.2)
lim lim sup kGk (un )k ≤ ε.
k→∞ n→∞
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
15
First, notice that assumption (??) and Fatou Lemma imply that we can take un as test function in (??). Then, we can also choose Gk (un ) (with k > R) as test function. Conditions (??) and (??) imply Z λ |un |p−2 un Gk (un ) − |Gk (un )|p p (5.3) α0 − kGk (un )k ≤ λ + hwn , Gk (un )i H |x|p Ω Notice that kGk (u)k → 0 as k → ∞ for every u ∈ W01,p (Ω). This and Hardy inequality imply that for every ε > 0 there exists k0 such that, for every k ≥ k0 it results |hw, Gk (u)i| ≤ ε, Z (5.4) |u|p−2 uGk (u) − |Gk (u)|p ≤ ε. 0 ≤ |x|p Ω 0
We take into account that un * u weakly in W01,p (Ω) and wn → w strongly in W −1,p (Ω). Moreover, we use conclusion (??) in Lemma ?? to deduce that for every ε > 0 and for every fixed k ≥ k0 there exists n0 such that |hwn , Gk (un )i| ≤ ε ∀ n ≥ n0 , Z |un |p−2 un Gk (un ) − |Gk (un )|p ≤ ε ∀ n ≥ n0 . |x|p Ω This, (??) and (??) imply that there exists c0 > 0 such that kGk (un )k ≤ c0 ε
(5.5)
∀ k ≥ k0 , ∀ n ≥ n0 .
That is equivalent to (??). Step 2. Let us first notice that conditions (??) and (??) imply that we can take as test function in (??) every v ∈ W01,p (Ω) ∩ L∞ (Ω), so that we can take Tk (un ) − Tk (u); we observe that Z |un |p−2 un (5.6) lim (Tk (un ) − Tk (u)) = 0. n→∞ Ω |x|p This implies that we can follow the same proof of Lemma 3.2 in [?] in order to get the strong convergence of Tk (un ) to Tk (u) in W01,p (Ω). Finally, from (??) and (??) we deduce that for every ε > 0 we can fix k0 > 0 such that (??) holds and Tk0 (un ) → Tk0 (u) in W01,p (Ω). Then, there exists n0 with the property kun − uk ≤ kTk0 un − Tk0 uk + kGk0 un k + kGk0 uk ≤ 3ε ,
∀ n ≥ n0 ,
as desidered. Second Proof. Step 1. Let u ∈ W01,p (Ω) such that, up to a subsequence, un * u in W01,p (Ω), un → u in Lq (Ω) for every q < N p/(N − p), and un → u almost everywhere. We will first prove that u satisfies Z Z Z |u|p−2 (5.7) jξ (x, u, ∇u)∇v + js (x, u, ∇u)v − λ v = hw, vi ∀v ∈ C0∞ (Ω). p |x| Ω Ω Ω In order to do this, let us first notice that conditions (??) and (??) imply that we can take as test function in (??) every v ∈ W01,p (Ω) ∩ L∞ (Ω). Then, by using the result of [?] we also deduce that ∇un → ∇u almost everywhere.
16
BENEDETTA PELLACCI +
Now, consider ϕ ∈ C0∞ (Ω) such that ϕ ≥ 0 and vn = ϕe−M un , with M = β0 /α0 . When we take vn as test function in (??), we get Z Z |un |p−2 −M u+ n jξ (x, un , ∇un )∇ϕe −λ un vn p Ω |x| Ω Z + js (x, un , ∇un ) − M jξ (x, un , ∇un )∇u+ n vn = hwn , vn i. Ω +
Since |vn | ≤ ϕ and vn → ϕe−M u almost everywhere we deduce that Z Z |un |p−2 |u|p−2 + u v = uϕe−M u . lim n n p n→∞ Ω |x|p |x| Ω Then, we can follow the same proof of Theorem 3.2 in [?] in order to get (??). Step 2. First, let us fix k > 0 and define the following function M |s| if |s| ≤ R, 1 1 (5.8) ζk (s) = M R + lg R − lg |s| if R < |s| < RekM R , k k 0 if |s| > RekM R , where M = β0 /α0 . From the definition of ζk (s) it results 1 (5.9) 0 ≤ ζk (s) ≤ M R, |ζk0 (s)s| ≤ if R < |s| < RekM R . k By arguing as Theorem 3.2 in [?] we deduce that it is possible to take v = un eζk (un ) as test function in (??) and v = ueζk (u) in (??). For every fixed k > 0 we define the function (5.10) fk (x, s, ξ) = eζk (s) jξ (x, s, ξ)ξ (1 + sζk0 (s)) + js (x, s, ξ)s When we take v = un eζk (un ) as test function in (??) we get Z Z |un |p ζk (un ) e = hwn , un eζk (un ) i. (5.11) fk (x, un , ∇un ) − λ p Ω Ω |x| Since un eζk (un ) * ueζk (u) , from (??) and (??) we obtain Z Z Z Z |un |p ζk (un ) |u|p ζk (u) (5.12) lim fk (x, un , ∇un ) − λ e = f (x, u, ∇u) − λ e . k p p n→∞ Ω Ω |x| Ω Ω |x| Now, we observe that (5.13)
lim
Z h
n→∞ Ω
ζk (un )
e
i |u |p Z h i |u|p n ζk (u) −1 = e − 1 . |x|p |x|p Ω
Indeed, i |u |p Z h i |u |p n n ζk (un ) eζk (un ) − 1 = e − 1 , p p |x| |x| kM R Ω {x : |un |≤Re } and we have eζk (un ) − 1 |un |p ≤ Ck for almost every x such that |un | ≤ RekM R . This, and the almost everywhere convergence of un to u give (??). Now, (??) and (??) yield Z Z |un |p |u|p lim fk (x, un , ∇un ) − λ p = fk (x, u, ∇u) − λ p . n→∞ Ω |x| |x| Ω Z h
Let us now observe that fk (x, s, ξ) is such that fk (x, s, 0) = 0,
p
fk (x, s, ξ) ≥ α1 |ξ| ,
1 α1 = α0 1 − . k
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
17
Indeed, if |s| ≤ R or if |s| > RekM R , the value of M and (??) imply that fk (x, s, ξ) ≥ α0 |ξ|p . While, if R < |s| < RekM R , hypothesis (??) and (??) yield the value of α1 . Then, conditions (??) and (??) are satisfied for every fixed k > 1. Moreover, thanks to conditions (??) we can choose k such that also (??) is verified. Then, Theorem ?? yields the conclusion. Remark 5.2. The use of the cut-off test function ζ in the second proof on Theorem ?? is important because condition (??) is assumed only for large s. If (??) held for every s, then we could get the conclusion by taking v = un as test function. 6. Proof of the main theorems. In this section we will first prove Theorem ??. Then, we will show that J satisfies the Concrete Palais-Smale condition. Finally, we will give the proof of Theorems ??, ?? and of Theorems ??, ??, ?? in Remark ?? Proof of Theorem ??. Corollary ?? and (??) imply that J is weakly lower semicontinuous. Moreover, from (??) we deduce that J is coercive. Indeed, there exist positive constants c0 , c1 and c2 such that the following inequality holds λ ε 1 α0 − kvkp − kvkpp − c0 kfε kp∗0 kvk + c1 J(v) ≥ p H p λ 1 α0 − − c2 ε kvkp − c0 kfε kp∗0 kvk + c1 . ≥ p H Condition (??) implies that we can fix ε > 0 such that α0 − λ/H − c2 ε > 0 and we get that J is coercive. Therefore, there exists u ∈ W01,p (Ω) minimum point of J. Finally, we assume that (??) holds, we deduce (??) from (??) and (??), so that we can compute hJ 0 (u), vi = 0 for every v ∈ W01,p (Ω) ∩ L∞ (Ω). That is, u is a distributional solution of Problem (P ). Now we will show that J satisfies the Concrete-Palais-Smale condition. In order to do this, it will be enough to prove that a Concrete-Palais-Smale sequence is bounded in W01,p (Ω). Indeed, the following result holds. Theorem 6.1. Let us assume conditions (??), (??), (??), (??), (??), (??), (??). Then every bounded sequence that satisfies (??) is compact. Proof. We set wn = yn +g(x, un ). Moreover, we can argue as in [?] to show that condition 0 (??) implies that g(x, un ) → g(x, u) in W −1,p (Ω). More precisely, we have that, up to a subsequence, g(x, un ) → g(x, u) almost everywhere. In addition, condition (??) implies that for every v ∈ W01,p (Ω) the following inequality holds ∗ ∗ ε |un |p −1 + |u|p −1 |v| − |g(x, un ) − g(x, u)||v| ≥ −2aε (x)|v| ∈ L1 (Ω). Then, we can apply Fatou Lemma to deduce Z Z Z ∗ p∗ −1 p∗ −1 lim inf ε |un | + |u| |v| − |g(x, un ) − g(x, u)||v| ≥ 2ε |u|p −1 |v| ≥ 0. n→∞
Ω
Ω
Ω
18
BENEDETTA PELLACCI
Therefore, we have Z p∗ −1 p∗ −1 kvk ≤ c1 ε, lim sup |g(x, un ) − g(x, u)||v| ≤ c0 ε sup kun k + kuk n→∞
n
Ω
0
0
and this implies that g(x, un ) → g(x, u) in W −1,p (Ω). Moreover, yn → 0 in W −1,p (Ω), so that we can apply Theorem ?? to get the conclusion. Now, let us show that every Concrete-Palais-Smale sequence is bounded in W01,p (Ω). Theorem 6.2. Assume conditions (??), (??), (??), (??), (??), (??), (??), (??). Then, any Concrete-Palais-Smale sequence is bounded in W01,p (Ω). Proof. We fix ε > 0 and we consider the function ϑε : R → R defined by 0 0 ≤ s ≤ R, (1 + ε)(s − R) R ≤ s ≤ R , R = R 1+ε , ε ε ε (6.1) ϑε (s) = s Rε ≤ s, −ϑε (−s) s ≤ 0, where R is given in (??). For every u ∈ W01,p (Ω), it results |∇ϑε (u)| ≤ (1 + ε)|∇u|.
(6.2)
Moreover, ϑε (un ) has the same sign of un . Then, Fatou Lemma implies that we can take v = ϑε (un ) as test function. We get Z Z (6.3) [jξ (x, un , ∇un )∇ϑε (un ) + js (x, un , ∇un )ϑε (un )] = g(x, un )un Ω {x : |un |>R} Z + g(x, un )(ϑε (un ) − un ) + hyn , ϑε (un )i Ω Z Z |un |p−2 un |un |p +λ (ϑ (u ) − u ) + λ . ε n n p |x|p Ω Ω |x| Notice that, Z Z g(x, un )(ϑε (un ) − un ) ≤ cε , g(x, un )(ϑε (un ) − un ) = Ω {x :Z|un |≤Rε } Z (6.4) |un |p−2 un |un |p−2 un (ϑ (u ) − u ) = (ϑε (un ) − un ) ≤ cε . λ ε n n |x|p |x|p Ω {x :|un |≤Rε } Moreover, ϑε (un ) has the same sign of un , so that, (??) and (??) imply that 0 ≤ js (x, un , ∇un )ϑε (un ) ≤ js (x, un , ∇un )un . Then, (??) and (??) yield Z Z [jξ (x, un , ∇un )∇ϑε (un ) + js (x, un , ∇un )un ] ≥ g(x, un )un {x : |un |>R}
Ω
|un |p +hyn , ϑε (un )i + λ − 2cε . p Ω |x| Z
From (??), (??) and (??) we obtain Z (6.5) [jξ (x, un , ∇un )∇un + js (x, un , ∇un )un ] ≥ −2cε {x : |un |>R}
Z Z |un |p p + g(x, un )un − εβ0 kun k + λ − (1 + ε)kyn k−1,p0 kun k. p Ω Ω |x|
MULTIPLE CRITICAL POINTS WITH HARDY POTENTIAL
19
Now, let us compute Z σJ(un ) −
[jξ (x, un , ∇un )∇un + js (x, un , ∇un )un ] . {x : |un |>R}
From the previous inequality and from (??) we deduce Z [σj(x, un , ∇un ) − jξ (x, un , ∇un )∇un − js (x, un , ∇un )un ] {x : |un |>R} Z Z + σj(x, un , ∇un ) − [σG(x, un ) − g(x, un )un ] {x : |un |≤R}
−λ
σ −1 p
Ω
Z
|p
|un ≤ εβ0 kun kp + (1 + ε)kyn k−1,p0 kun k + 2cε + c0 . p |x| Ω
Conditions (??), (??), (??), Hardy inequality yield Z Z 1 λ λ σ p σα0 − (σ − p) |∇un | + δ1 − −1 |∇un |p ≤ c0 p H H p {x : |un |≤R} {x : |un |>R} +εβ0 kun kp + (1 + ε)kyn k−1,p0 kun k + 2cε + ka0 k1 + kb0 km kun ktp∗ . The conditions (??) and (??) imply that σ λ λ σ δ2 = min α0 − , δ1 − −1 > 0. p H H p Then, we fix c0ε = 2cε + c0 + ka0 k1 . Sobolev inequality implies that (δ2 − εβ0 ) kun kp ≤ c0ε + (1 + ε)kyn k−1,p0 kun k + c1 kb0 km kun kt . Finally, we fix ε = δ2 /2β0 and we note that 1 < t < p to get the conclusion. Remark 6.3. The use of the test function ϑε has been important in the above result, because (??) and (??) are assumed for large s. If (??) and (??) were assumed for every s, then we could prove the boundedness of a Palais-Smale sequence as in the semilinear case. This test function has first been adopted in [?] (see Lemma 3.3). Finally, we prove Theorem (??) and (??). Proof of Theorem ??. First notice that J satisfies the Palais-Smale condition, indeed it satisfies the Concrete-Palais-Smale condition. Namely, if un is a (CP S)c sequence Theorem ?? implies that un is bounded and from Theorem ?? we deduce that un is compact. Therefore, in order to get the conclusion it is enough to show that the geometrical conditions of Theorem ?? are satisfied. First, notice that conditions (??) and (??) yield Z Z λ β0 1 p p α0 − kvk − G(x, v) ≤ J(v) ≤ kvk − G(x, v) (6.6) p H p Ω Ω Moreover, conditions(??) and (??) imply that for every sequence un ∈ W01,p (Ω) with un → 0 we have Z 1 (6.7) lim G(x, un ) = 0. n→∞ kun kp Ω Then, (??) is satisfied. Now, let ϕ1 be the first eigenfunction of the Laplace operator with homogenous boundary conditions and we suppose it normalized, so that kϕ1 k = 1.
20
BENEDETTA PELLACCI
In order to prove (??), it is enough to show that there exists T > 0 such that J(T ϕ1 ) < 0 and T > r . Note that, (??) and conditions (??) and (??) yield Z β0 1 c0 σ σ J(tϕ1 ) ≤ t − k(x)ϕ1 + σ kak1 − σ−p+1 kbkm ptσ−p t t Ω Since σ > p we deduce (??). Finally, Proposition ?? implies the conclusion. Remark 6.4. We have supposed condition (??) instead of (??) in the nonsymmetric case because we have to satisfy condition (??) of Theorem ??. More precisely, (??), joint with (??), is fundamental in order to show (??). Now, let us prove Theorem ??. Proof of Theorem ??. As before we have that J satisfies the Palais-Smale condition. Moreover, it is even. Then, we only have to show that hypotheses (??) and (??) are satisfied. Conditions (??) and (??) actually imply that (??) and (??) can be obtained as in Theorem 2.1 in [?]. Remark 6.5. Proofs of Theorems ??, ??, ??. Theorem ?? easily follows from Theorem ??, since g0 (x, s) satisfies (??). In order to prove Theorems ?? we only have to show that (??), (??), (??) hold, because (??) is evidently satisfied. First, note that π p−1 |s| , |g1 (x, s)| ≤ |s|σ + 1 + 2 so that (??) is satisfied with a(x) = 1 + π/2 and σ = r. Moreover, (??) holds with 0 1 π(σ − 1) + p t 1 a0 = 0 , b0 = . t 2 t Condition (??) follows by taking k(x) ≡ 1 and a = b = 0. In oder to prove Theorem ?? it is enough to show that (??) holds. This condition follows from (??) and Young inequality. Indeed, it results, ∗ −1
|g2 (x, s)| = |s|σ ≤ ε|s|p
+ Cε ,
so that (??) holds with aε (x) = Cε . Acknowledgements. The author wish to thank M. Degiovanni for his encouragment and for several stimulating and helpful discussions. References [1] B. Abdellaoui, I. Peral, Existence and nonexistence results for quasilinear elliptic equations involving p-Laplacian with a critical potential. Ann. Mat. Pura e Appl. 182 (2003) no. 3, 247-270. [2] A. Ambrosetti, P.H. Rabinowitz. Dual variational methods in critical point theory and applications. J. Funct. Anal. 14 (1973), 349-381. [3] D. Arcoya, L. Boccardo, Critical points for multiple integrals of the Calculus of Variations. Arch. Ration. Mech. Anal. 134 (1996), 249–274. [4] D. Arcoya, L. Boccardo, Some remarks on critical point theory for nondifferentiable functionals, NoDEA Nonlinear Differential Equations Appl. 6 (1999), 79–100. [5] L. Boccardo, The Bensoussan & Co. technique for the study of some critical points problems. Optimal Control and PDE, J.L. Menaldi et al. ed. Book in honour of Professor Alain Bensoussan’s 60th birthday. IOS Press (2000).
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[email protected] ` di Roma,, P.le Aldo Moro 2, I-00185 Roma, (B) Dipartimento di Matematica, Universita Italy.