Multiplicity of positive weak solutions to subcritical singular elliptic Dirichlet problems

We study a superlinear subcritical problem at infinity of the form −∆u = a (x) u−α + f (λ, x, u) in Ω, u = 0 on ∂Ω, u > 0 in Ω, where Ω is a bounded domain in Rn, 0 ≤ a ∈ L∞ (Ω) , and 0 < α < 3. Under suitable assumptions on f , we prove that there exists Λ > 0 such that this problem has at least one weak solution in H1 0 (Ω) if and only if λ ∈ [0, Λ] ; and also that there exists Λ∗ such that for any λ ∈ (0, Λ∗), at least two solutions exist.

The singular biparametric bifurcation problem −∆u = g (u) + λ |∇u| p + µh (•, u) in Ω, u = 0 on ∂Ω, u > 0 in Ω was studied, by Ghergu and Rȃdulescu, in [24].Dupaigne, Ghergu and Rȃdulescu [19] treated Lane-Emden-Fowler equations with convection term and singular potential.Rȃdulescu [38] studied blow-up boundary solutions for logistic equations, and for Lane-Emden-Fowler equations, with a singular nonlinearity, and a subquadratic convection term.The existence of positive solutions to the inequality Lu ≥ K (x) u p on the punctured ball Ω = B r (0) \ {0} was investigated by Ghergu, Liskevich and Sobol [22] for a second order linear elliptic operator L. Singular initial value parabolic problems involving the p-Laplacian were treated by Bougherara and Giacomoni [3], and concentration phenomena for singularly perturbed elliptic problems on an annulus were studied by Manna and Srikanth [36].
Godoy and Kaufmann [33] stated sufficient conditions for the existence of positive solutions to problems of the form −∆u = Ku −α − λMu −γ in Ω, u = 0 on ∂Ω, where Ω is a smooth bounded domain in R n , K and M are nonnegative functions on Ω, α > 0, γ > 0, and λ > 0 is a real parameter.
Kaufmann and Medri [34] obtained existence and nonexistence results for positive solutions of one dimensional singular problems of the form − (u ) p−2 u = m (x) u −γ in Ω, u = 0 on ∂Ω, where Ω ⊂ R is a bounded open interval, p > 1, γ > 0, and m : Ω → R is a function that may change sign in Ω.
Aranda and Godoy [2], obtained multiplicity results for positive solutions in W in Ω, u = 0 on ∂Ω, for the case when Ω is a C 2 bounded and strictly convex domain in R n , 1 < p ≤ 2; and g, h are locally Lipschitz functions on (0, ∞) and [0, ∞) respectively, with g nonincreasing, and allowed to be singular at the origin; and h nondecreasing, with subcritical growth at infinity, and satisfying inf s>0 s −p+1 h (s) > 0.
Recently Saoudi, Agarwal and Mursaleen [39], obtained a multiplicity result for positive solutions of problems of the form Additional references, and a comprehensive treatment of the subject, can be found in [23], [38], see also [15].
For b ∈ L ∞ (Ω) such that b + ≡ 0, λ 1 (b) will denote the positive principal eigenvalue for −∆ in Ω, with Dirichlet boundary condition, and weight function b (see Remark 2.2 below).
The aim of this work is to prove the following three theorems concerning the existence, and the multiplicity, of weak solutions to problem (1.1).
Our approach follows that in [2], however, there are significant differences between the two works.Here we are concerned with weak solutions in [2].Also, in this paper we do not assume that Ω is convex, and we do not require that f (λ, x, s) be a local Lipschitz function.
It is a well known fact that, when a is Hölder continuous on Ω, and min Ω a > 0, the classical solution of −∆u = au −α in Ω, u = 0 on ∂Ω, u > 0 in Ω, belongs to H 1 0 (Ω) if, and only if, α < 3 (see theorem 2 in [35]).It is therefore reasonable, in order to obtain weak solutions in H 1 0 (Ω) to problem (1.1), we restrict ourselves to the case when the singular term of the nonlinearity has the form au −α , with a nonnegative and nonidentically zero function in L ∞ (Ω) , and 0 < α < 3.
In Section 2 we consider, for ε ≥ 0, and 0 We show that, under the assumptions H1)-H3), this problem has a unique weak solution u ε ∈ H 1 0 (Ω) , and that its associated solution operator S ε , defined by S ε (ζ) := u ε , satisfies S ε (P) ⊂ P, where In Section 3 we obtain an a priori bound for the L ∞ norm of the bounded solutions of −∆u = a (u + ε) −α + f (λ, •, u) in Ω, u = 0 on ∂Ω, u > 0 in Ω This is achieved by adapting, to our singular setting, the well known Gidas-Spruck blow up technique.
In Section 4, we consider problem (1.1); which we rewrite as u = S 0 ( f (λ, •, u)).We use the properties of S 0 , and a classical fixed point theorem for nonlinear eigenvalue problems, to prove that, for any λ small enough, (1.1) has at least one positive weak solution in H 1 0 (Ω) ∩ L ∞ (Ω) ; moreover, the solution set for this problem (i.e., the set of the pairs (λ, u) that solve it) contains an unbounded subcontinuum (i.e., an unbounded connected subset) emanating from (0, S 0 (0)).Using this subcontinuum, and the a priori estimate obtained in Section 3, we prove that, for every λ positive small enough, there exist at least two positive weak solutions of (1.1).Finally, a number Λ with the properties stated in Theorem 1.1 is obtained by using the sub and supersolution method (as well as the properties of the operator S), applied to the approximating problems

Preliminary results
We assume, from now on, that Ω is a bounded domain in R n with C 2 boundary, and that α and a satisfy the conditions H1)-H3) in the statement of Theorem 1.1.The next two remarks collect some well known facts from the linear theory of elliptic problems.
We recall that λ ∈ R is called a principal eigenvalue for −∆ in Ω, with homogeneous Dirichlet boundary condition and weight function b, if the problem −∆u = λbu in Ω, u = 0 on ∂Ω has a solution φ (called a principal eigenfunction) such that φ > 0 in Ω. Remark 2.2.Let us mention some properties of principal eigenvalues and principal eigenfunctions (for a proof of i)-iii), see e.g., [17], also [30]), and [29]).If Ω is a C 1,1 domain in R n , b ∈ L ∞ (Ω) and b + ≡ 0 then: i) There exists a unique positive principal eigenvalue for −∆ in Ω, with homogeneous Dirichlet boundary condition and weight function b, denoted by λ 1 (b); its associated eigenspace is one dimensional and it is included in C 1 Ω .Moreover, λ 1 (b) has the following variational characterization: Furthermore, for each positive eigenfunction φ associated to λ 1 (b) , and for δ positive and small enough, there are positive constants c 1 , c 2 such that c In particular, φ γ is integrable if, and only if, γ > −1.
We recall also that λ , and the corresponding solution operator 0 Ω is bounded and strongly positive, i.e., if ρ ∈ L ∞ (Ω) and 0 ≤ ρ ≡ 0 then u belongs to the interior of the positive cone of C 1 0 Ω where iv) Let ρ be a nonnegative function in L ∞ loc (Ω) such that ρϕ ∈ L 1 (Ω) for any ϕ ∈ H 1 0 (Ω).If ρ ≡ 0 in Ω, λ > 0, and if u ∈ H 1 0 (Ω) ∩ C Ω satisfies, for some positive constant c, u ≥ cd Ω in Ω and, in weak sense, To prove this assertion we can proceed as in the proof of Proposi- tion 2.4 in [29], where a similar result was proved for Neumann problems.Indeed, let v := − ln u and let w ∈ C ∞ c (Ω).Since u ≥ cd Ω in Ω and w has compact support we have u −1 w 2 ∈ H 1 0 (Ω).Taking u −1 w 2 as a test function in (2.2), a computation gives , and since λ Ω bϕ 2 j ≤ Ω ∇ϕ j 2 , we get λ Ω bϕ 2 ≤ Ω |∇ϕ| 2 , and so Then, by the variational characterization of λ 1 (b) , we obtain λ ≤ λ 1 (b) .We will need the following comparison principle.Lemma 2.3.Let U be a bounded domain in R n and ε ≥ 0. Let u and v be two positive functions in Proof.We proceed by contradiction.Let V := {x ∈ U : u (x) > v (x)} and suppose that V is nonempty.Thus u − v ∈ H 1 (V) and u = v on ∂V.Then u − v ∈ H 1 0 (V) (see e.g., Theorem 8.17 and also Remark 19 in [5]).Let ϕ j j∈N be a sequence in . Now, using suitable mollifiers, we obtain a sequence Remark 2.4.The following forms of the comparison principle hold: if ε ≥ 0, and if u, v are two functions in which are positive a.e. in Ω and satisfy that, for any nonnegative and if, in addition, u − v ≤ 0 on ∂Ω (i.e., (u If a and u are functions defined on Ω, we will write χ {u>0} au −α to denote the function in the following sense: for any ϕ ∈ H to the above problem (in the sense stated in i)) and there exists a positive constant c, independent of ζ, such that u ≥ cd Ω a.e. in Ω.
, and so χ {u>0} au −α ∈ L 1 loc (Ω).Also χ {u>0} au −α ≡ 0 and −∆u ≥ χ {u>0} au −α in D (Ω).Thus, by Remark 2.1 iii), there exists a positive constant c (in principle, depending perhaps on u) such that u ≥ cd Ω χ {u>0} au −α d Ω in Ω.Then, for some positive constant c , we have u ≥ c d Ω in Ω and so Let w be a solution of (2.4), in the sense of i), corresponding to ζ = 0.By Remark 2.4 we have u ≥ w in Ω, and, as above, we have w ≥ cd Ω in Ω for some constant c > 0. Since c is independent of ζ, the last assertion of ii) holds.In particular, u is positive in Ω.Now, the uniqueness assertion follows from Remark 2.4.
and there exists a positive constant c 3 such that , then the solution u to problem (2.5), given by Lemma 2.6, satisfies u ≥ cd Ω in Ω for some positive constant c independent of ε and ζ.
Proof.By Lemma 2.6, u > 0 a.e. in Ω.Let w be as in the proof of Lemma 2.5.Thus there exists a positive constant c such that w ≥ cd Ω in Ω.As in Lemma 2.5 we have u ≥ w in Ω.Thus u ≥ cd Ω in Ω.Since c is independent of ε and ζ, the lemma follows.
Remark 2.8.Let us recall the Hardy inequality (see e.g., [5], p. 313): There exists a positive constant c such that Lemma 2.9.Let ζ ∈ L ∞ (Ω) be such that ζ ≥ 0, and let ε ∈ (0, 1] (respectively ε = 0), and let u be the solution to problem given by Lemma 2.6 (resp.by Lemma 2.5, in the sense stated there).Then: Proof.Let λ 1 be the principal eigenvalue for −∆ on Ω, with weight function 1 and let ϕ 1 be the corresponding positive principal eigenfunction normalized by For δ positive and small enough there exists a positive constant c δ such that |∇ϕ 1 | ≥ c δ in A δ , and, by diminishing c δ if necessary, we can assume that ϕ 1 ≥ c δ in Ω δ .To see i), we consider first the case when 1 < α < 3 and ε = 0. Clearly ϕ (2.7) and thus (2.8) (2.9) Ω in Ω, for a positive constant c depending only on M, α, and Ω, therefore i) holds when ε = 0.The proof of i) for the case ε ∈ (0, 1] reduces to the previous one.Indeed, Remark 2.4 gives u ≤ u 0 in Ω, where u 0 is the solution (given by Lemma 2.5) to problem (2.6) and corresponding to ε = 0.
Proof.Let Ω be a subdomain of Ω such that Ω ⊂ Ω; and let Ω be a subdomain of Ω such that Ω ⊂ Ω ⊂ Ω ⊂ Ω.By Lemmas 2.7 and 2.9 there exist positive constants c 1 , c 2 and Then, by [28,Theorem 8.24], u |Ω ∈ C β Ω for some β ∈ (0, 1).Since this holds for any domain in Ω, and so u is continuous on ∂Ω.Then u ∈ C Ω .Lemma 2.11.Assume 1 < α < 3, and let ζ ∈ L ∞ (Ω) be such that ζ ≥ 0. Let u be the solution to problem (2.5) given by Lemma 2.5 (in the sense stated there).Then there exists a positive constant c Proof.We consider first the case when a := inf Ω a > 0. Let λ 1 be the principal eigenvalue for −∆ in Ω with homogeneous Dirichlet boundary condition and weight function a, and let ϕ 1 be the corresponding positive principal eigenfunction, normalized by in the weak sense of Lemma 2.5, (i.e., with test functions in H 1 0 (Ω) ∩ L ∞ (Ω)).We have also, again in the weak sense of Lemma 2.5, −∆u ≥ au −α in Ω.Then, by Lemma 2.3, Ω in Ω for some positive constant c independent of ζ.Thus the lemma holds when inf Ω a > 0.
To prove the lemma in the general case, consider the solution θ to the problem be a solution, in the sense of Lemma 2.5, of problem (2.4) corresponding to ζ = 0.By Lemma 2.3 we have u ≥ w in Ω and, by Lemma 2.9, there exists a positive constant c 2 such that w ≤ c 2 d Ω in Ω.Now, for ε ∈ (0, 1)  and β ∈ (0, 1) , we have, in the weak sense of Lemma 2.5, (2.10) where α−1 α+1 +α+1−β .Thus, from (2.10) and (2.11), we get on ∂Ω and so, by the weak maximum principle, we have, for ε small enough, (w a.e. in Ω.By taking lim ε→0 + in this inequality we get, for any β ∈ (0, 1) , a.e. in Ω. (2.12) By taking lim β→0 + in (2.12), using that lim which ends the proof of the lemma.Lemma 2.12.Let ζ be a nonnegative function in L ∞ (Ω) , and let u ∈ H 1 0 (Ω) ∩ L ∞ (Ω) be the solution (in the sense of Lemma 2.5) to (2.4).Then, for some positive constant c, u Ω in Ω if 1 ≤ α < 3.Moreover, u is the unique weak solution, in the usual Proof.To see that u is a weak solution of (2.4), i.e., that, for any we consider first the case 0 < α ≤ 1.Let ψ ∈ H 1 0 (Ω) and, for j ∈ N and x ∈ Ω, let Also, by Lemma 2.7, there exists a positive constant c such that u ≥ cd Ω in Ω, and so, for any j ∈ N, with c = a ∞ c −α ; applying Hardy's inequality we get (2.17) for some positive constant c.Since ψ j j∈N converges to ψ in H 1 0 (Ω) and a.e. in Ω, Lebesgue's dominated convergence theorem gives (au −α + ζ) ψ ∈ L 1 (Ω) and (2.14).Thus u is a weak solution (in the usual H 1 0 (Ω) sense) of (2.4), it satisfies u ≥ cd Ω in Ω and, by Lemma 2.5, u is the unique weak solution to (2.4).Thus i) and ii) holds when 0 < α ≤ 1.
Consider now the case 1 < α < 3. To see that u is a weak solution of (2.13) we proceed as in the case 0 < α ≤ 1, except that instead of (2.16) we use now that, by Lemma 2.11, there exists a positive constant c such that u ≥ cd Ω .Thus, with c a constant independent of j.Since α < 3 we have d 2 < ∞ and so, by Hardy's inequality, c d Then, as in the case 0 < α ≤ 1, Lebesgue's dominated convergence theorem gives (au −α + ζ) ψ ∈ L 1 (Ω) and (2.14).Thus u is a weak solution (in the usual H 1 0 (Ω) sense) of (2.13), it satisfies u ≥ cd 2 1+α Ω a.e. in Ω and, by the comparison principle in Remark 2.4, u is the unique weak solution (in the usual H 1 0 (Ω) sense) to problem (2.13).
Proof.For j ∈ N, let u j := S ε j ζ j .Since u j is a weak solution of −∆u j = a u j + ε j −α + ζ j in Ω, u j = 0 on ∂Ω, and using u j as a test function, we get with c independent of j, which proves the lemma when 0 < α ≤ 1.Let us consider now the case 1 < α < 3. The function z := S 0 (0) is a weak solution of −∆z = az −α in Ω, z = 0 on ∂Ω and, by Lemma 2.10, z ∈ C Ω .Also −∆u j ≥ au −α j in Ω, u j = 0 on ∂Ω, and then, by Lemma 2.3, u j ≥ z in Ω.By Lemma 2.9, Ω for any j, with c a positive constant independent of j.Thus, from (2.19), we have Ω .
Proof.i) and ii) follow directly from Lemma 2.3.To prove iii) it is enough to show that if , therefore there exist u ∈ H 1 0 (Ω) , and a subsequence u j k k∈N , such that u j k k∈N converges strongly in L 2 (Ω) to u, and ∇u j k k∈N converges weakly in L 2 (Ω, R n ) to ∇u.Taking a further subsequence if necessary, we can assume that u j k k∈N converges to u, a.e. in Ω.
To prove iv), consider a bounded sequence ζ j , ε j j∈N ⊂ P ∞ × [0, ∞).Taking a subse- quence if necessary, we can assume that ε j j∈N converges to some ε ∈ [0, ∞).Let {Ω r } r∈N be a sequence of subdomains of Ω such that Ω r ⊂ Ω r+1 for all r, and Ω = ∪ ∞ r=1 Ω r .Let u j = S ε j ζ j .Let B > 0 be such that ζ j ∞ ≤ B for all j.Since 0 ≤ ζ j ≤ B we have 0 ≤ u j = S ε j ζ j ≤ S 0 (B).By Lemmas 2.6 and 2.7, there exist positive constants c 1 , c 2 and τ such that, for all j, c 1 d Ω ≤ u j ≤ c 2 d τ Ω in Ω.Thus, for each r there exists a positive constant B r > 0 such that, for all j, u j|Ω r+1 L ∞ (Ω r+1 ) ≤ B r and a u j Then, by [28,Theorem 8.24], for each r there exist constants B r > 0 and γ r ∈ (0, 1) such that, for all j, u j|Ω r C γ (Ωr) ≤ B r .Then, for each r, the Ascoli-Arzelà theorem gives a subse- quence, still denoted by u j j∈N which converges uniformly in Ω r .Now, a Cantor diagonal process gives a subsequence u j k k∈N which converges uniformly on each Ω r to a function u independent of r (therefore u j k k∈N converges uniformly to u on each compact subset of Ω).Let us show that u j k k∈N is a Cauchy sequence in C Ω : Since 0 ≤ ζ j k ≤ B we have 0 ≤ S ε j k ζ j k ≤ S 0 (B).Also S 0 (B) ∈ C Ω , and S 0 (B) = 0 on ∂Ω pointwise.Since S 0 (B) ∈ C Ω we have that for any µ > 0 there exists η > 0 such that 0 Cauchy sequence in C Ω , and so u j k l∈N converges in C Ω .

A priori estimates
We assume for the whole section that H1)-H5) of Theorem 1.1 are satisfied.
for all x ∈ R n , ε > 0 and r > 0. We have also (see e.g., [5], Theorem 4.22 for any y ∈ B r (x) , r > 0 and ε > 0, Lebesgue's dominated convergence theorem gives that v (x) The following lemma is an adaptation, suitable for our purpose here, of the blow up method developed in [27], to obtain a priori estimates for the L ∞ norm of solutions to subcritical superlinear elliptic problems.For the convenience of the reader, and as our statement is somewhat different to that in Theorem 1.1 of [27], we provide a detailed proof of it.
For r > 0, and x ∈ R n , we will write B r (x) (respectively B r (x)) to denote the open (resp.closed) ball in R n of radius r and centered at x. Lemma 3.2.Let Θ be an equibounded family of nonnegative measurable functions in L ∞ (Ω) , and let G be a family of nonnegative functions in C Ω × [0, ∞) .Assume that there exist p ∈ 1, n+2 n−2 , and h ∈ C Ω , such that min Ω h > 0 and lim s→∞ g(x,s) s p = h (x) uniformly on g ∈ G and x ∈ Ω.Then there exists a constant C such that u ∞ < C whenever u ∈ H 1 0 (Ω) ∩ L ∞ (Ω) is a weak solution, for some θ ∈ Θ and g ∈ G, to the problem Proof.To prove the lemma we proceed by contradiction.Suppose that for any k ∈ N there exist θ k ∈ Θ, g k ∈ G, and a weak solution , and so u k ∈ W 2,r (Ω) ∩ W 1,r 0 (Ω) for any r ∈ (1, ∞).Thus u k ∈ C Ω , and u k is a strong solution of (3.1).Let P k ∈ Ω be such that u k ∞ = u k (P k ).Taking a subsequence if necessary, we can assume that lim k→∞ P k = P for some P ∈ Ω.
Case a): P ∈ Ω: For k large enough we have P k − P < d and σ k < 1.Thus σ k y + P k − P ≤ σ k y + P k − P < 2d.Then, taking a further subsequence if necessary, we can assume that . Also, lim k→∞ σ k = 0, and so, for R > 0, there exists k (R) Our assumptions on Θ and G imply that there exists a positive constant c such that f k (x, s) ≤ c (s p + 1) for any (x, s) ∈ Ω × [0, ∞) and k ∈ N.Then, for y ∈ B 2R (0) and k ≥ k (R) , we have, for some positive constant c independent of y and k, 0), the standard inner elliptic estimates (as stated, e.g., in [8, Proposition 4.1.2]),im- ply that v k W 2,r (B R (0)) ≤ c r for any r > n and k ≥ k (R), with c r a positive constant in- dependent of k.Therefore there exists a subsequence, still denoted by {v k } k∈N , that con- verges in C 1,γ B R (0) for some γ ∈ (0, 1).Let {R l } l∈N be an increasing sequence such that lim l→∞ R l = ∞.A Cantor diagonal process gives a further subsequence, still denoted by Note that, for each l, and for k large enough, −∆v From our assumptions on the family G we have f k (x, s) = θ k (x) + s p h (x) + s p ψ k (x, s) , with lim s→∞ ψ k (x, s) = 0 uniformly on x ∈ Ω and k ∈ N.Then, for R > 0, y ∈ B R (0) and k ≥ k (R), and so, taking into account that lim s→∞ ψ k (x, s) = 0 uniformly on x ∈ Ω and k ∈ N, we get, for y ∈ B R (0), Then, by elliptic regularity theory (see e.g., [8,Proposition 4.1.2]),v ∈ W 2,r loc (R n ) for any r ∈ (1, ∞), and v satisfies, in strong sense, −∆v = h (P) v p in R n .Let η := (h (P)) 1 1−p , and let w := ηv.Thus η > 0, w ∈ W 2,r loc (R n ) for any r ∈ (1, ∞) , and w is a bounded positive strong solution to the problem −∆w = w p in R n .Moreover, for each open ball B ⊂ R n we have w p ∈ C γ (U) for some γ ∈ (0, 1).Then, by [28, Theorem 9.19], w ∈ C 2 (R n ).But Theorem 1.2 in [27] says that such a solution w does not exist.Contradiction.

Case b): P ∈ ∂Ω:
Since Ω is a C 2 domain, there exists an open ball B = B r (P) with radius r > 0, centered at P; and a one to one mapping x n > 0} (see e.g., [28, p. 94]).After compositions with a suitable translation, and with a linear endomorphism, we can assume Φ (P) = 0, and that Φ (P) (the Jacobian matrix of Φ at P) is an orthogonal matrix.Diminishing B and D if necessary, we can also assume that where each and so, in particular, A (P) = I.
For k large enough P k ∈ B ∩ Ω.For such k, let Note that for k sufficiently large, δ k = Φ (P k ) , e n where e n = (0, . . . , 0, 1).Then, taking a subsequence if necessary, we can assume that so, taking a further subsequence, we can assume that where α i,m.k (y) := a i,m (σ k y + Φ (P k )) , and β m.k (y Since v k q is well defined on B σ −1 kq δ kq (0) and v k q (0) = 1, the same arguments of the case a) apply to obtain a positive and bounded Now, A (P) is a symmetric and positive matrix, and then there exists an invertible matrix B such that BA (P) B t = I.Let T : R n → R n be defined by Ty = yB t .Thus w := (h (P)) [27].
Observe also that inf k∈N σ −1 k δ k > 0. For, if not, taking a subsequence, we can assume lim k→∞ σ −1 k δ k = 0 and, taking a further subsequence if necessary, we can also assume that where, as before, 0 denotes the origin of R n−1 .
From (3.4) we have, for y ∈ where a i,m,k (y constant independent of k.Also, the ellipticity constants of A k and a modulus of continuity of its coefficients can be chosen independent of k. for some constant c independent of k and, as in case a), there exists a positive constant c independent of k such that ϕ k L ∞ (Q) ≤ c .Let r > n.By elliptic regularity up to the boundary (see e.g., [28,Theorem 9.13]), there exists a positive constant c r such that v k W 2,r (Q) ≤ c r for any k.Then there exists a positive constant k δ k is bounded from above and from below by positive constants, and so, taking a subsequence if necessary, we can assume lim k→∞ σ −1 be given by w k (y) := v k (y − y k ) , with y k := 0 , σ −1 k δ k , where 0 denotes the origin in R n−1 .Thus, w k satisfies, for where α l,m,k (y By repeating compactness arguments used in the case a), and taking into account that, for y ∈ R n + , lim k→∞ α l,m,k (y) = a l,m (P) , lim l→∞ β m,l (y) = 0, and A (P) = I we obtain a subsequence, still denoted {w k } k∈N , that converges in R n + to a function w ∈ C 2 (R n + ) such that w > 0 in R n + , w (0 , τ) = 1, and − ∆w (y) = h (P) w p (y denotes the open ball in R n−1 of radius R and centered at the origin.Let r > n.As above, by elliptic regularity up to the boundary [28,Theorem 9.13], we have w k W 2,r (U R ) ≤ c r for some positive constant c r independent of k.Thus, taking a further subsequence, still denoted {w k } k∈N , we have that {w k } k∈N con- verges uniformly on U. Now, by considering an increasing sequence of radius R j j∈N such that lim j→∞ R j = ∞, a Cantor diagonal process gives a further subsequence, still denoted by {w k } k∈N , which converges uniformly on K, for each compact subset K ⊂ R n + .Then w belongs to C 2 (R n + ) ∩ C R n + and, since w k = 0 on ∂R n + for each k, we also have also w = 0 on ∂R n + .Therefore w := h (P) + and w > 0 in R n + , which contradicts Theorem 1.3 in [27].
Proof.To prove the lemma we proceed by contradiction.Suppose that there exist sequences and lim j→∞ u j ∞ = ∞.Let λ * > 0 be as given by Lemma 3.3.Thus λ j ≤ λ * for all j.
Then λ j j∈N is bounded and so, taking a subsequence if necessary, we can assume that , and, since u j = S ε j f λ j , •, u j , we have u j ∈ C Ω .Let ψ 1 and ψ 2 be nonnegative functions in 2 , ∞ , and ψ 1 + ψ 2 ≡ 1 on R. Let w j ∈ H 1 0 (Ω) ∩ L ∞ (Ω) be the solution, provided by Lemma 2.6 when ε j > 0, and by Lemma 2.12 when ε j = 0 (applied with ε = ε j and with a replaced by a ψ 1 • u j ) to the problem From Lemma 2.10, applied with ε = ε j , and with a replaced by a ψ 1 • u j , we have that w j ∈ C Ω .Notice that, by Lemma 2.14 i), w j ≤ w in Ω, where w ∈ H 1 0 (Ω) is the weak solution, given by Lemma 2.12 (applied with a replaced by a ψ 1 • u j ), to the problem (3.9)By Lemma 2.10, w ∈ C Ω , then w j j∈N is bounded in L ∞ (Ω).Also, in weak sense, −∆u j = a u j + ε j −α + f λ j , •, u j ≥ a ψ 1 • u j u j + ε j −α in Ω, u j = 0 on ∂Ω, and so, in weak sense, u j − w j = 0 on ∂Ω.
Using u j − w j − as test function in (3.10) we get − Ω ∇ u j − w j − ≥ 0, and then u j ≥ w j in Ω.
We claim that a ψ 1 • u j u j + ε j −α ϕ ∈ L 1 (Ω) for any ϕ ∈ H 1 0 (Ω) , and that there exists a nonnegative weak solution in Ω, To prove this, first observe that, for all j ∈ N, Indeed, (3.12) clearly holds when ε j > 0. If ε j = 0 and 0 < α ≤ 1 then, by Lemma 2.7 there exists a positive constant c, independent of j, such that u j for some positive constant c independent of j, and, since 1 < α < 3, we have 1 − 2α 1+α > −1, and so d . Therefore (3.12) holds for all j.We next prove that, for ϕ ∈ H 1 0 (Ω) and for all j, a ψ 1 . Indeed, from (3.12) and the Hardy inequality, we have, for some positive constant c, Then, by the Riesz theorem, there exists a weak solution z j ∈ H 1 0 (Ω) to (3.11), and, by the weak maximum principle, z j ≥ 0 a.e. in Ω.Since u j ≥ w j in Ω, from (3.8) and (3.11) we have −∆ z j − w j ≤ 0 in Ω z j − w j = 0 on ∂Ω, (3.13) and so z j ≤ w j in Ω.Also, z j ≥ 0, and w j ≤ w in Ω.Thus sup j∈N z j ∞ ≤ w ∞ < ∞.Now, u j ≥ cd Ω in Ω for some positive constant c independent of j, and then, for any domain Ω such that Ω ⊂ Ω, a ψ 1 • u j u j + ε j −α ∈ L ∞ (Ω ).Using that z j ∈ L ∞ (Ω ), and the inner elliptic estimates, we conclude that z j ∈ C (Ω).Since 0 ≤ z j ≤ w j and w j ∈ C Ω , then z j is continuous on ∂Ω and so z j ∈ C Ω .Now, with θ j ∈ L ∞ (Ω) defined by Let u j := u j − z j .Since u j ≥ w j ≥ z j in Ω, we have u j ≥ 0 in Ω.For j ∈ N let g j : Ω × [0, ∞) → R be defined by g j (x, s) := f λ j , x, s + z j (x) ψ 2 s + z j (x) .Thus u j is a weak solution in Note that θ j + g j •, u j is nonnegative and belongs to L 1 (Ω).We claim that, for j large enough, To prove our claim we proceed by contradiction.Taking a subsequence if necessary, we can assume that θ j + g j •, u j = 0 in Ω for all j.Then, for all j, (3.15) gives u j = 0 in Ω, and so u j = z j .Also, •, u j = 0 in Ω.Let P j ∈ Ω be such that u j P j = u j ∞ .Then f λ j , P j , u j P j = 0 for any j.Taking a further subsequence we can assume that lim j→∞ P j = P for some P ∈ Ω.Also lim →∞ λ j = λ ≥ λ 0 , and lim →∞ u j P j = ∞.Then, from the uniform convergence in H5), we get lim j→∞ u j P j −p f λ j , P j , u j P j = h (λ, P) > 0, which contradicts that f λ j , •, u j = 0 for all j.Thus (3.16) holds.From (3.16), (3.15), and the Hopf maximum principle in Remark 2.1 ii), we conclude that u j > 0 in Ω.

Proofs of the main results
We assume for the whole section that H1)-H5) of Theorem 1.1 hold.Let and for ε ≥ 0, let T ε : [0, ∞) × P → C Ω be defined by therefore the continuity of T ε follows from Lemma 2.14.The compactness of T ε is also given by Lemma 2.14, by observing that if Lemma 4.3.For any ε ≥ 0 the following statements hold: and, by Lemmas 2.6 and 2.12, the unique solution to this problem is S ε (0).Thus ζ = 0, and so ii) holds.Finally, iii) follows from the fact that T ε (0, u) = 0 for all nonnegative u ∈ C Ω .Let us recall the following result from [1].
Lemma 4.8.Let λ 0 > 0, let λ j j∈N be a sequence in [λ 0 , ∞) , and let ε j j∈N be a sequence in Then: i) w j j∈N is bounded in H 1 0 (Ω); ii) if w j k k∈N is a subsequence of w j j∈N that converges weakly in H 1 0 (Ω) to some w ∈ H 1 0 (Ω) ∩ L ∞ (Ω) , and if, in addition, lim k→∞ λ j k = λ and lim k→∞ ε j k = ε for some ε ∈ [0, 1] and λ ∈ [λ 0 , ∞) , then w is a weak solution to (3.6) and there exists a positive constant c such that w ≥ cd Ω in Ω.
Proof.Let c λ 0 be as given by Lemma 3.4.Then w j ∞ ≤ c λ 0 , which implies f λ j , •, w j ∞ ≤ sup [0,Λ # ]×Ω×[0,c λ 0 ] f , with Λ # given by Lemma 4.6 i).Since w j = S ε j f λ j , •, w j , Lemma 2.13 gives that w j j∈N is bounded in H 1 0 (Ω).Now suppose that w j k k∈N is a subsequence such that, for some w ∈ H 1 0 (Ω) , w j k k∈N converges to w strongly in L 2 (Ω) and ∇w j k k∈N converges to ∇w weakly in L 2 (Ω, R n ).Suppose also that lim k→∞ λ j k = λ and lim k→∞ ε j k = ε.Taking a further subsequence if necessary, we can assume that w j k k∈N converges to w a.e. in Ω.Now, w j k = S ε j k f λ j k , •, w j k and then, by Lemma 2.7, w j k ≥ cd Ω in Ω for some positive constant independent of k.Thus w ≥ cd Ω in Ω.Note that f λ j k , •, w j k k∈N is a bounded sequence in L ∞ (Ω) that converges pointwise to f (λ, •, w).Then, by Lemma 2.14 iv), S ε j k f λ j k , •, w j k k∈N converges to S ε ( f (λ, •, w)) in C Ω , i.e., w j k k∈N converges to S ε ( f (λ, •, w)) in C Ω .Thus w = S ε ( f (λ, •, w)) , i.e., w solves (3.6).Finally, Lemma 2.7 says that, for some positive constant c, w ≥ cd Ω in Ω.

Lemma 2 . 10 .
Let ζ be a nonnegative function belonging to L ∞ (Ω) and let