Existence and non-existence of a positive solution for ( p , q )-Laplacian with singular weights

driven by nonhomogeneous operator (p, q)-Laplacian with singular weights under the Dirichlet boundary condition. We also prove that in the case where μ > 0 and with 1 < q < p < ∞ the results are completely different from those for the usual eigenvalue problem for the p-Laplacian with singular weight under the Dirichlet boundary condition, which is retrieved when μ = 0. Precisely, we show that when μ > 0 there exists an interval of eigenvalues for our eigenvalue problem.

The problem (P λ,µ ) comes, for example, from a general reaction diffusion system where D(u) = (|∇u| p−2 + µ|∇u| q−2 ).This system has a wide range of applications in physics and related sciences like chemical reaction design [2], biophysics [8] and plasma physics [18].In such applications, the function u describes a concentration, the first term on the right-hand side of (1.1) corresponds to the diffusion with a diffusion coefficient D(u); whereas the second one is the reaction and relates to source and loss processes.Typically, in chemical and biological applications, the reaction term c(x; u) has a polynomial form with respect to the concentration.
Our problem was addressed in [15] for domains with boundary C 2 and bounded weights, when only the condition (H4) holds true.These work proved that in the case where µ > 0, there exists an interval of eigenvalues.The authors proved the existence of positive solutions in resonant cases.A non-existence result is also given.Here we will assume that the boundary ∂Ω is a piecewise C 1 and singular weights m r (r = p, q) which satisfy one of the conditions (H1), (H2), (H3) or (H4).Our work represent developments of the study performed in [15] because we prove all results of this paper by considering others conditions that represent the singularity of the domain and the weights.Our main tool is the Hardy-Sobolev inequality, see Lemma 2.2 in preliminary section.Many authors have studied the nonhomogeneous operator (p, q)-Laplacian (see [12,16,21,22]).However, there are few results one the eigenvalue problems for the (p, q)-Laplacian.In [4,5], the authors established the existence of the principal eigenvalue and of a continuous family of eigenvalues for problem where g is a bounded positive weight.Eigenvalue problem for a (p, 2)-Laplacian was studied in [3].The existence of non trivial solution for the following Dirichlet equation is proved in [6] −△ p u − µ△u = λ|u| p−2 u + g(u) in Ω, u = 0 on ∂Ω, in the case where p > 2, g ∈ C 1 and λ ∈ σ(−∆ p ), where σ(−∆ p ) is the spectrum of (−∆ p ).Under the Neumann boundary condition, [13] determined the set of eigenvalues for the equation Existence and non-existence of a positive solution 149 where p > 2. In [19], M. Tanaka has completely described the generalized eigenvalue λ for which the following equation has a positive solution, where 1 < r = r * < ∞ and µ > 0.
We recall that a value λ ∈ R is an eigenvalue of problem (P λ,µ ) if and only if there exists u ∈ W 1,p 0 (Ω)\{0} such that 2) for all ϕ ∈ W 1,p 0 (Ω).u is then called an eigenfunction of λ.Letting µ → 0 + , our problem (P λ,µ ) turns into the (p−1)-homogeneous problem known as the usual weighted eigenvalue problem for the p-Laplacian with singular weight m p : Moreover, after multiplying our equation (P λ,µ ) by 1/µ and then letting µ → +∞, we obtain the (q − 1)-homogeneous equation: Nonlinear eigenvalue problem (P λ,mr ), where r = p, q and with bounded weight have been studied by several authors, for example (see [1,7,9,11,17,18]).These works proved that there exists a first eigenvalue λ 1 (r, m r ) > 0, where which is simple in the sense that two eigenfunctions corresponding to it are proportional.Moreover, the corresponding first eigenfunction φ 1 (r, m r ) can be assumed to be positive.It was also shown (see [1]) that λ 1 (r, m r ) is simple and isolated.Recently, the problem (P λ,mr ) with singular weight m r satisfying the conditions (H1), (H2), (H3) or (H4), was studied in [14].The authors use the Hardy-Sobolev inequality to characterize the first eigenvalue.In some cases it is shown that λ 1 (r, m r ) > 0 is positive simple, isolated and has a nonnegative corresponding eigenfunction φ 1 (r, m r ) ∈ L ∞ (Ω).Higher eigenvalues, in particular the second one, are also determined.
The plan of this paper is the following.In Section 2, which has a preliminary character, we collect some results concerning the first eigenvalue λ 1 (r, m r ) of problem (P λ,mr ), where r = p, q.In Section 3, we study Rayleigh quotient for our problem (P λ,µ ).In contrast to homogeneous case, we prove that if λ 1 (p, m p ) = λ 1 (q, m q ) or φ 1 (p, m p ) = kφ 1 (q, m q ) for every k > 0, then the infimum in Rayleigh quotient is not attained.We also show nonexistence results for positive solutions of the eigenvalue problem (P λ,µ ) formulated as Theorem 3.5.Our existence results for positive solutions of the eigenvalue problem (P λ,µ ) are presented in Section 4. After studying the non-resonant cases (Theorem 4.1) which prove that when µ > 0 there exists an interval of positive eigenvalues for the problem (P λ,µ ), we present the resonant cases in Theorem 4.8.

Preliminaries
Throughout this paper Ω will be a bounded domain of R N with piecewise C 1 boundary, 1 < q < p < ∞ and r = p or q.We will always assume for r = p, q that m r δ τ ∈ L a (Ω) with δ(x) = dist(x, ∂Ω) and m + r ≡ 0, where a, r and τ satisfy one of the conditions (H1), (H2), (H3) or (H4).
We will write u r := Ω |u| r dx 1/r for the L r (Ω)−norm and W 1,r 0 (Ω) will denote the usual Sobolev space with usual norm ∇u r .
In the sequel, we collects some results relative to the first eigenvalue λ 1 (r, m r ) defined by (1.3) and its corresponding normalized eigenfunction φ 1 (r, m r ).The following lemma concerns the Hardy-Sobolev inequality proved in [10], which characterize the first eigenvalue λ 1 (r, m r ) of problem (P λ,mr ).This inequality is our main tool in this paper.Lemma 2.2.[10] Let 0 ≤ τ ≤ 1 and s such that 1 s = 1 r − 1−τ N for r < N and for all u ∈ W 1,r 0 (Ω), where δ(x) = dist(x, ∂Ω) and C = C(N, r, τ ) > 0 is a constant.In the case s = r = p, q, no regularity on ∂Ω is needed.We give now an example of the weight m r such that m r δ τ ∈ L a (Ω) with m + r ≡ 0, where a, τ and r satisfying the condition (H2).
To use Harnack inequality as in [14] and [20], we make now the following definitions involving locally integrable weights.Let ǫ(ρ) be a smooth function defined for ρ > 0 such that Existence and non-existence of a positive solution 151 for some ρ * > 0. We denote by K x0 (ρ) an N -dimensional cube contained in Ω whose edges are of length ρ and are parallel to the coordinate axes.We define The following theorem guarantees the simplicity and isolation of λ 1 (r, m r ), where r = p, q.This result is proved by M. Montenegro and S. Lorca in [14].To ensure positiveness of φ 1 (r, m r ), the authors apply the Harnack inequality of [20].

Rayleigh quotient for the problem (P λ,µ )
This subsection concerns the Rayleigh quotient for our problem (P λ,µ ).Remark 3.1.We start by pointing out to find a solution for the problem (P λ,µ ) is equivalent to seek a solution in the case µ = 1, that is to solve the problem (P λ,1 ).Indeed, if u is a solution of (P λ,1 ), then multiplying equation (P λ,1 ) by s p−1 for s > 0 we deduce that v = su is a solution for problem (P λ,µ=s p−q ).Conversely, let u be a solution of problem (P λ,µ ).Then it follows that v = µ 1/q−p u is a solution of (P λ,1 ).
We introduce now the functionals A and B on W 1,p 0 (Ω) by for all u ∈ W 1,p 0 (Ω).
Proposition 3.2.(i) The functional A is well defined and sequently weakly lower semi-continuous.
For the proof of Proposition 3.3, we will need to use the following lemma.
Case (ii): Suppose that Ω m p |u| p dx ≤ 0 and Ω m q |u| q dx > 0. Using the definition of λ 1 (q, m q ), we also arrive at contradiction Case (iii): Suppose now that Ω m p |u| p dx > 0 and Ω m q |u| q dx > 0. It follows from the definition of λ 1 (r, m r ), where r = p, q that ∇u r r ≥ λ 1 (r, m r ) Ω m r |u| r dx.

Hence we get
Against the assumption in our reasoning by contradiction.✷ Proof: [Proof of Proposition 3.3.]By contradiction, we suppose that there exists u ∈ W 1,p 0 (Ω) such that B(u) > 0 and A(u) B(u) = λ * .Using Lemma 3.4, we give We argue by considering the three cases in the proof of Lemma 3.4.Case (i): By (3.4), (3.7) and Ω m q |u| q dx ≤ 0, we have Existence and non-existence of a positive solution

155
We deduce that Ω m p |u| p dx and ∇u q = 0.

Non-existence results
This subsection studies a non-existence results for the problem (P λ,1 ) , so for the problem (P λ,µ ).This work is inspired from [15].The following theorem is the main result of this section.Theorem 3.5.One assumes the same conditions as for Theorem 2.5.
Proof: [Proof of Theorem 3.5.]Assume by contradiction that there exists a nontrivial solution u of problem (P λ,1 ).Then, for every s > 0, we have that v = su is a non-trivial solution of problem (P λ,s p−q ) (see Remark 3.1).Choose s p−q = p/q and then act with su as test function on the problem (P λ,s p−q ).We arrive at 0 < pA(su) = pλB(su). (3.8) From the estimate (3.8) and according to Lemma 3.4, we obtain This contradiction yields the first assertion of the theorem.
The second part of the Theorem 3.5 follows by Proposition 3.3.✷
Remark 4.2.The proof of Theorem 4.1 reduces to provide a non-trivial critical point of the functional I λ,mp,mq defined for all u ∈ W 1,p 0 (Ω) by where u + = max{u, 0} and A, B are the functionals defined by (3.1) and (3.2).This non-trivial critical point u of I λ,mp,mq is a non-negative solution of the problem (P λ,1 ).We can check that u ∈ L ∞ (Ω) (see Remark 1.7 in [14]).Then the Harnack inequality of [20] can be applied to ensure positiveness of u.The argument will be separately developed in two cases: (a) λ 1 (q, m q ) < λ < λ 1 (p, m p ). (b)λ 1 (p, m p ) < λ < λ 1 (q, m q ).In case (a), we apply the minimum principle and in case (b), we use the mountain pass theorem.
Existence and non-existence of a positive solution 157 Proof of case (a).By Proposition 3.2, A is sequently weakly lower semicontinuous and B is weakly continuous.It follows that I λ,mp,mq is sequently weakly lower semi-continuous.It is remains to show that I λ,mp,mq is coercive and bounded from below.We distinguish two cases: (i) For u ∈ W 1,p 0 (Ω) such that Ω m p (u + ) p dx ≤ 0. A calculation similar to that in the proof of Proposition 3.2 for r = q gives where C, C ′ , C ′′ > 0 are the constants given respectively by the Hardy-Sobolev inequality (see Lemma 2.2), the compact embedding W 1,q 0 (Ω) ⊂ L q (Ω) and the continuous embedding which is possible due to the assumption in case (a).By the definition of λ 1 (p, m p ) we have Then taking into account (4.2), we derive Since q < p, it follows from (4.1) and (4.3) that the functional I λ,mp,mq is coercive and bounded from below.Consequently, by minimum principle, there exists a global minimizer u 0 of I λ,mp,mq .Finally, u 0 = 0, indeed it suffices to prove that I λ,mp,mq (u 0 ) = min W 1,p 0 (Ω) I λ,mp,mq < 0. For sufficiently small k > 0, we have I λ,mp,mq (kφ 1 (q, m q )) = k q k p−q p ∇φ 1 (q, m q ) p p − λk p−q p Ω m p φ p 1 (q, m p )dx + λ 1 (q, m q ) − λ q .

Proof of case (b).
We organize the proof of this case in several lemmas.In the sequel, we design by o(1) a quantity tending to 0 as n → ∞.Lemma 4.3.Suppose that m r δ τ ∈ L a (Ω) and m + r ≡ 0 (r = p, q), where a, r and τ satisfy one of the conditions (H1), (H2), (H3) or (H4).In addition, we assume that then the functional I λ,mp,mq satisfies the Palais-Smale condition on W 1,p 0 (Ω).
Proof: Let (u n ) ⊂ W 1,p 0 (Ω) be a sequence such that Let us first show that the sequence u n is bounded.It is sufficient only to prove the boundedness of u n p because where α, β and C are respectively the constants in inequalities u q ≤ α u p , ∇u q ≤ β ∇u p (since Ω bounded and q < p) and u δ τ b(r) ≤ C ∇u r (in each case (H1), (H2), (H3) or (H4), see the proof of Proposition 3.2) and C r = λC m r δ τ a (r = p, q).Suppose by contradiction that u n p → ∞ and let v n := un un p .The sequence v n is bounded in W 1,p 0 (Ω).Indeed, dividing (4.4) by u n p p , we have Since p > 1, the inequality (4.5) implies the boundedness of v n in W 1,p 0 (Ω).For a subsequence, v n → v weakly in W 1,p 0 (Ω).By the compact embedding W 1,p 0 (Ω) ⊂ L r (Ω) (r = p, q), we have v n → v strongly in L r (Ω) (r = p, q).First we, observe that v − ≡ 0 in Ω.In fact, acting with −u − n as test function, we have as test function, we have 1), (4.7) because q < p, u n p → +∞, v n is bounded in W 1,p 0 (Ω) and converges to v strongly in L r (Ω) (r=p,q).Thus by (4.7) and (S + ) property of (−∆ p ) on W 1,p 0 (Ω), we deduce that v n → v strongly in W 1,p 0 (Ω).For any ϕ ∈ W 1,p 0 (Ω), by taking ϕ/ u n p−1 p as test function, we obtain (4.8) Passing to the limit in (4.8), we see that v is a non-negative and non-trivial solution of problem (P λ,mp ) (note v ≥ 0 and ∇v p = 1).According to the Harnack inequality (see Remark 1.7 in [14]), we have v > 0 in Ω.This implies that λ = λ 1 (p, m p ) because any positive eigenvalue other than λ 1 (p, m p ) has no positive eigenfunctions (see Theorem 2.5).Therefore, we obtain a contradiction since we assumed λ = λ 1 (p, m p ). Hence u n is bounded in W 1,p 0 (Ω).For a subsequence, u n → u weakly in W 1,p 0 (Ω) and u n → u strongly in L r (Ω) (r = p, q).We claim now that u n → u strongly in W 1,p 0 (Ω).It suffices to prove that Abdellah Zerouali and Belhadj Karim Using Hölder inequality and for r = p, q, we have (4.10)Moreover, (4.9) and (4.10) imply that ∇v n r → ∇v r (for r = p, q).Thus u n → u strongly in W 1,p 0 (Ω).✷ Lemma 4.4.Suppose m r δ τ ∈ L a (Ω) and m + r ≡ 0 (r = p, q), where a, r and τ satisfy one of the conditions (H1), (H2), (H3) or (H4) If λ < λ 1 (q, m q ), then there exist α > 0 and β > 0 such that I λ,mp,mq (u) ≥ α whenever u q = β. (4.11) To prove the Lemma 4.4, we need the following lemma.
Then there exists α(d) > 0 such that ∇u p ≤ α(d) u q , for all u ∈ X(d).
Proof: Suppose, by contradiction, that Because of ∇u n p = 0, we set v n := un ∇un p .Thus, by (4.12), v n → 0 strongly in L q (Ω).Since ∇v n p = 1, the sequence v n is bounded in W 1,p 0 (Ω).For a subsequence, v n → v weakly in W 1,p 0 (Ω).By the compact embedding W 1,p 0 (Ω) ⊂ L r (Ω) (r = p, q), we have v n → v strongly in L r (Ω) (r = p, q).Hence, v n q → v q and v n p → v q .By uniqueness of limit, we deduce that v r = 0.It follows that v = 0.As u n ∈ X(d), we obtain where C > 0 is a constant that may differ in each case (H1), (H2), (H3) or (H4) (see the proof of Proposition 3.2).
Proof: For sufficiently large R > 0, taking into account that q < p and λ 1 (p, m p ) < λ, we have
Proof: Proof of case (i): Since m + p ≡ 0, the Lebesgue mesure of {x ∈ Ω; m p (x) > 0} is positive.Thus there exists an n 0 ∈ N such that (m p − 1/n 0 ) ≡ 0. For n ≥ n 0 , we define, as in [15], the functional I n on W 1,p 0 (Ω) by Using the strict monotonicity of the first eigenvalue of problem (P λ,mp ) (see [14]), we obtain Thus we are able to apply Theorem 3.5 obtaining a positive solution u n of the problem We may assume that u n is a global minimizer of I n and I n (u n ) < 0 (see the case (a) in the proof of Theorem 3.5).In addition, observing that I n ≤ I n0 provided n ≥ n 0 , we infer that for all n ≥ n 0 , We claim that if u n is bounded in W 1,p 0 (Ω), then u n is a bounded Palais-Smale sequence of I λ,mp,mq .Indeed, there exists c ∈ R such that I λ,mp,mq (u n ) → c because I n (u n ) is a convergent sequence and ∇u n p is bounded.On the other hand, since I ′ n (u n ) = 0, we have As ∇u n p is bounded, we obtain I ′ λ (u n (W 1,p 0 (Ω)) * → 0. This completes the proof of our claim.We prove now the boundedness of u n in W 1,p 0 (Ω) by way of contradiction.Suppose that ∇u n p → ∞ and let v n := u n / ∇u n p .For a subsequence, v n → v weakly in W 1,p 0 (Ω) and v n → v strongly in L p (Ω).Following the same steps of the proof of Lemma 4.3, we show that v is a positive solution of problem (P λ,mp ).This entails v is a positive eigenfunction corresponding to λ 1 (p, m p ). Thus the simplicity of λ 1 (p, m p ) implies that v = φ 1 (p, m p ).The facts that I n (u n ) < 0 for all n ≥ n 0 and u n is a critical point of I n result in Passing to the limit, we obtain Proof of case (ii): As in the proof of case (i), we can choose n 0 ∈ N such that (m q − 1/n 0 ) = 0.For n ≥ n 0 , we define the functional J n on W 1,p 0 (Ω) by Using the strict monotonicity of the first eigenvalue of problem (P λ,mq ), we obtain λ 1 (q, m q − 1/n) > λ 1 (q, m q ) = λ, for any n ≥ n 0 .Thus we may apply Theorem 3.5 obtaining a positive solution u n of the problem −△ p u − △ q u = λ[m p (x)|u| p−2 u + (m q (x) − 1/n)|u| q−2 u] in Ω, u = 0 on ∂Ω.
Following the pattern of the proof of case (i), this time proceeding as in case (b) in the proof of Theorem 3.5, we deduce that J n (u n ) > 0 for all n ≥ n 0 .For the boundedness of u n in W 1,p 0 (Ω), proceeding as in the proof of case (i) and the contradiction follows from the condition λ = λ 1 (q, m q ) > λ 1 (p, m p ).The bounded sequence u n is a Palais-Smale sequence for the functional I λ,mp,mq as can be seen from the estimate , where c is a positive constant independent of n.It follows that u n has a subsequence converging to some critical point u 0 of I λ,mp,mq .In order to complete the proof, due to the Harnack inequality, it suffices to justify that u 0 = 0. We assume, by contradiction, that u n → 0 strongly in W 1,p 0 (Ω).Set v n := un ∇un p .Then, for a subsequence, v n → v weakly in W 1,p 0 (Ω), weakly in W 1,q 0 (Ω) and strongly in L p (Ω).It is easy to see that v ≥ 0 in Ω.Using vn−v ∇un q−1 p as test function, we obtain .
Consequently, recalling that v n → v strongly in W 1,q 0 (Ω), we have ∇v q = lim ∇v n q = lim ∇u n q ∇u n p ≥ 1 α(d)[λ 1 (q, 1)] −1/q > 0, thus proving our claim.For any ϕ ∈ W 1,p (Ω), by using ϕ ∇un q−1 p as test function we show, as in proof of case (i), that v is a nonnegative nontrivial solution of (P λ,mq ).The simplicity of λ 1 (q, m q ) guarantees that v = φ 1 (q, m q ) is a positive eigenfunction corresponding to λ 1 (q, m q ).Using that J n (u n ) > 0 for all n ≥ n 0 in conjunction with (4.22), we have By passing to the limit inferior we obtain ∇φ 1 (q, m q ) p p − λ Ω m p [φ 1 (q, m q )] p dx ≤ 0.

. 6 )
Because p > 1, the inequality (4.6) guarantees the boundedness of ∇(u − n ) p and so ∇v − n p = ∇(u − n ) p / u n p → 0. Thus v − ≡ 0 holds, hence v ≥ 0 in Ω. Existence and non-existence of a positive solution 159 Now, by taking (v n − v)/ u n p−1 p

Lemma 4 . 5 .
Suppose τ , a and p as in Lemma 4.4 and let b be such that b = ap a−p if a < ∞ and b

. 13 )
Existence and non-existence of a positive solution 161 By Lemma 2.2 and under one of the properties (H1), (H2), (H3) or (H4), the sequence un δ τ b ∇un p is bounded (see the proof of Proposition 3.2).Passing to the limit in (4.13), we get 1 d ≤ 0. This contradiction completes the proof of Lemma 4.5.✷ Proof: [Proof of Lemma 4.4.]According to Lemma 4.5, choose