Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 12, pp. 1–24. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS WITH SUPERLINEAR REACTION AND NO SYMMETRIES NIKOLAOS S. PAPAGEORGIOU, CALOGERO VETRO, FRANCESCA VETRO Abstract. We study a semilinear Robin problem driven by the Laplacian with a parametric superlinear reaction. Using variational tools from the critical point theory with truncation and comparison techniques, critical groups and flow invariance arguments, we show the existence of seven nontrivial smooth solutions, all with sign information and ordered. 1. Introduction Let Ω ⊆ RN be a bounded domain with a C2-boundary ∂Ω. In this article we study the parametric semilinear Robin problem −∆u(z) + ξ(z)u(z) = fλ(z, u(z)) in Ω, ∂u ∂n + β(z)u = 0 on ∂Ω, λ ∈ (0, λ̃). (1.1) In this problem the potential function ξ ∈ L∞(Ω) and ξ(z) ≥ 0 for a.a. z ∈ Ω. The reaction fλ(z, x) is parametric with λ > 0 being the parameter, it is measurable in (z, x) ∈ Ω × R and it is C1 in the x-variable. We assume that fλ(z, ·) exhibits superlinear growth as x → ±∞, without satisfying the common in such cases Ambrosetti-Rabinowitz condition (the AR-condition for short). In the boundary condition ∂u ∂n denotes the normal derivative of u and it is defined via the Green’s identity (see Papageorgiou-Rădulescu-Repovš [17, p. 35]. We know that if u ∈ C1(Ω), then ∂u ∂n = (∇u, n)RN with n(·) being the outward unit normal on ∂Ω. The boundary coefficient β ∈W 1,∞(∂Ω) and β(z) ≥ 0 for all z ∈ ∂Ω. Our aim is to prove a multiplicity theorem for problem (1.1) providing sign infor- mation for all the solutions, without imposing any symmetry conditions on fλ(z, ·). Using variational tools coming from the critical point theory, with truncation and comparison techniques, the use of critical groups and flow invariance arguments, we show that for all small values of the parameter λ > 0, problem (1.1) has at least seven nontrivial smooth solutions all with sign information (two positive, two negative and three nodal (sign changing) solutions). Moreover, the constant sign and nodal solutions are ordered. 2010 Mathematics Subject Classification. 35J20, 35J61. Key words and phrases. Constant sign and nodal solutions; superlinear reaction; critical point theory; critical groups; flow invariance. c©2021 Texas State University. Submitted February 15, 2021. Published March 10, 2021. 1 2 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 The study of superlinear elliptic problems was initiated with the seminal paper by Ambrosetti-Rabinowitz [2] and continued with the works of Wang [20], Miyagaki- Souto [12] (semilinear Dirichlet problems) and by Fang-Liu [5], Li-Yang [11], Sun [18], Aizicovici-Papageorgiou-Staicu [1] (nonlinear Dirichlet problems driven by the p-Laplacian). More recently, Mugnai-Papageorgiou [14] and Papageorgiou- Rădulescu [16], studied superlinear problems driven by nonlinear nonhomogeneous differential operators. All the aforementioned works produce at most three non- trivial solutions without providing sign information for all of them. Also, only Miyagaki-Souto [12] deal with a parametric problem with fλ(z, x) = λf(z, x) and show that for all λ > 0 the problem has a nontrivial solution. Actually their as- ymptotic hypotheses on f(z, ·) as x → ±∞ and as x → 0 (see hypotheses (f1) an (f4)) make the presence of the parameter λ > 0 irrelevant and for this reason their existence result is global in λ > 0. We also mention the works of Castro- Cossio-Vélez [4] and of Papageorgiou-Papalini [15], which study Dirichlet problems with an asymptotically linear reaction and prove the existence of seven nontrivial solutions without any symmetry conditions on f(z, ·). The methods of proof differ in the two papers. Our approach here is closer to that of [15]. 2. Mathematical background - hypotheses The main spaces in the analysis of problem (1.1) are the Sobolev space H1(Ω), the Banach space C1(Ω) and the boundary Lebesgue spaces Lr(∂Ω), 1 ≤ r < ∞. By ‖ · ‖ we denote the norm of the Sobolev space H1(Ω). We have ‖u‖ = [‖u‖22 + ‖∇u‖22]1/2 for all u ∈ H1(Ω). The space C1(Ω) is an ordered Banach space with positive (order) cone C+ = {u ∈ C1(Ω) : u(z) ≥ 0 for all z ∈ Ω}. This cone has a nonempty interior given by intC+ = {u ∈ C+ : u(z) > 0 for all z ∈ Ω}. On ∂Ω we consider the (N−1)-dimensional Hausdorff (surface) measure σ(·). From the theory of Sobolev spaces, we know that there exists a unique continuous linear map γ̂0 : H1(Ω)→ L2(∂Ω) such that γ̂0(u) = u ∣∣ ∂Ω for all u ∈ H1(Ω) ∩ C(Ω). This map is known as the “trace map” and it extends the notion of boundary values to all Sobolev functions. The trace map is actually compact into Lr(∂Ω) for all 1 ≤ r < 2(N−1) N−2 if N ≥ 3 and into Lr(∂Ω) for all 1 ≤ r <∞ if N = 1, 2. Moreover, the trace map is not surjective. In fact we have im γ̂0 = H1/2,2(∂Ω) and ker γ̂0 = H1 0 (Ω). In what follows, for the sake of notational simplicity we drop the use of the trace map γ̂0. All restrictions of Sobolev functions on ∂Ω are understood in the sense of traces. Let u, v : Ω → R be two measurable functions such that u(z) ≤ v(z) for a.a. z ∈ Ω. We introduce the following order intervals in the Sobolev space H1(Ω): [u, v] = {h ∈ H1(Ω) : u(z) ≤ h(z) ≤ v(z) for a.a. z ∈ Ω}, [u) = {h ∈ H1(Ω) : u(z) ≤ h(z) for a.a. z ∈ Ω}, intC1(Ω)[u, v] = the interior in C1(Ω) of [u, v] ∩ C1(Ω). EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 3 Also, for u ∈ H1(Ω) we define u± = max{±u, 0}. We know that u± ∈ H1(Ω), u = u+ − u−, |u| = u+ + u−. Given f : Ω×R→ R a measurable function, by Nf we denote the Nemytski operator corresponding to f(·, ·); that is, for all u : Ω→ R measurable Nf (u)(·) = f(·, u(·)). Note that z → f(z, u(z)) is measurable. Our hypotheses on the potential term ξ(·) and the boundary coefficient β(·) are the following: (H1) ξ ∈ L∞(Ω), β ∈W 1,∞(∂Ω), ξ(z) ≥ 0 for a.a. z ∈ Ω, β(z) ≥ 0 for all z ∈ ∂Ω and ξ 6≡ 0 or β 6≡ 0. We introduce the C1-functional γ : H1(Ω)→ R defined by γ(u) = ‖∇u‖22 + ∫ Ω ξ(z)u2 dz + ∫ ∂Ω β(z)u2dσ for all u ∈ H1(Ω). On account of hypotheses (H1) and using Mugnai-Papageorgiou [13, Lemma 4.11] and Gasiński-Papageorgiou [8, Proposition 2.4], we have c0‖u‖2 ≤ γ(u) for some c0 > 0, all u ∈ H1(Ω). (2.1) In our arguments we will employ the spectrum of the operator u → −∆u + ξ(z)u with the Robin boundary condition. So, we consider the linear eigenvalue problem −∆u(z) + ξ(z)u(z) = λ̂u(z) in Ω, ∂u ∂n + β(z)u = 0 on ∂Ω. (2.2) Using the spectral theorem for linear, compact self-adjoint operators, we show that the spectrum of (2.2) consists of a sequence {λ̂n}n∈N of distinct eigenvalues such that λ̂n → +∞. Also there is a corresponding sequence {ûn}n∈N ⊆ H1(Ω) of eigenfunctions which form an orthonormal basis of H1(Ω). By E(λ̂n) we denote the eigenspace corresponding to the eigenvalue λ̂n, n ∈ N. These items have the following properties: (a) E(λ̂n) is finite dimensional and E(λ̂n) ⊆ C1(Ω) for all n ∈ N (see Wang [19]). (b) Each eigenspace E(λ̂n) has the so-called “Unique Continuation Property” (UCP for short), namely if u ∈ E(λ̂n) and vanishes on a set of positive Lebesgue measure, then u ≡ 0. (c) H1(Ω) = ⊕n∈NE(λ̂n). (d) Let λ̂1 = inf [ γ(u) ‖u‖22 : u ∈ H1(Ω), u 6= 0 ] > 0, (2.3) λ̂n = inf [ γ(u) ‖u‖22 : u ∈ Ĥn, u 6= 0 ] = sup [ γ(u) ‖u‖22 : u ∈ Hn, u 6= 0 ] , (2.4) where Hn = ⊕nk=1E(λ̂k), Ĥn = ⊕k≥nE(λ̂k), n ∈ N. The infimum in (2.3) is realized on E(λ̂1), while both the infimum and the supremum in (2.4) are realized on E(λ̂n). The elements of E(λ̂1) do not change sign and λ̂1 > 0 is simple. By û1 we denote the positive, L2-normalized (that is, ‖û1‖2 = 1) eigenfunction corresponding to λ̂1. We have that û1 ∈ intC+. On the 4 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 other hand the eigenfunctions corresponding to an eigenvalue λ̂n > 0, n ≥ 2, are all nodal (sign changing). For details, we refer to Gasiński-Papageorgiou [7]. Using the orthogonality of the eigenspaces, (2.3) and (2.4) and the UCP, we obtain the following result. Lemma 2.1. (a) If n ∈ N, ϑ ∈ L∞(Ω), ϑ(z) ≥ λ̂n for a.a. z ∈ Ω and the inequality is strict on a set of positive Lebesgue measure, then we can find c1 > 0 such that c1‖u‖2 ≤ ∫ Ω ϑ(z)u2 dz − γ(u) for all u ∈ Hn. (b) If n ∈ N, ϑ ∈ L∞(Ω), ϑ(z) ≤ λ̂n for a.a. z ∈ Ω and the inequality is strict on a set of positive Lebesgue measure, then we can find c2 > 0 such that c2‖u‖2 ≤ γ(u)− ∫ Ω ϑ(z)u2 dz for all u ∈ Ĥn. Let X be a Banach space and ϕ ∈ C1(X). We say that ϕ(·) satisfies the “C- condition”, if it has the property Every sequence {un}n∈N ⊆ X such that {ϕ(un)}n∈N ⊆ R is bounded and (1 + ‖un‖X)ϕ′(un) → 0 in X∗ as n → ∞, admits a strongly convergent subsequence. This is a compactness-type condition on the functional ϕ(·) which compensates for the fact that the ambient space X need not be locally compact (in most cases X is infinite dimensional). Using this compactness-type condition on ϕ(·), we can prove a deformation theorem from which follow the minimax theorems of critical point theory (see [17, Sections 5.3, 5.4]). Consider the following two sets: Kϕ = {u ∈ X : ϕ′(u) = 0} (the critical set of ϕ), ϕc = {u ∈ X : ϕ(u) ≤ c} for any c ∈ R. If B ⊆ A ⊆ X, then by Hk(A,B), k ∈ N0, we denote the kth-relative singular homology group for the pair (A,B) with integer coefficients. If u ∈ Kϕ is isolated, then the critical groups of ϕ at u are defined by Ck(ϕ, u) = Hk(ϕc ∩ U,ϕc ∩ U \ {u}) for all k ∈ N0, where U is a neighborhood of u such that Kϕ ∩ ϕc ∩ U = {u}. The excision property implies that this definition is independent of the choice of the isolating neighborhood. If ϕ ∈ C1(X) satisfies the C-condition and ϕ ∣∣ Kϕ is bounded below, then for c < infu∈Kϕ ϕ(u), the critical groups of ϕ at infinity are defined by Ck(ϕ,∞) = Hk(X,ϕc) for all k ∈ N0. [17, Corollary 5.3.13, p. 392] implies that this definition is independent of the choice of c < infKϕ ϕ. Suppose that Kϕ is finite and introduce the following quantities: M(t, u) = ∑ k≥0 rank Ck(ϕ, u)tk, P (t,∞) = ∑ k≥0 rank Ck(ϕ,∞)tk for all t ∈ [0, 1], all u ∈ Kϕ. EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 5 The Morse relation says that∑ u∈Kϕ M(t, u) = P (t,∞) + (1 + t)Q(t), (2.5) with Q(t) being a formal series in t with nonnegative, integer coefficients (see [17]). Now we introduce the hypotheses on the reaction fλ(z, x). (H2) fλ : Ω × R → R (λ ∈ (0, λ̃)) is a measurable function such that for all λ ∈ (0, λ̃), a.a. z ∈ Ω fλ(z, ·) ∈ C1(R), fλ(z, 0) = 0 and (i) |(fλ)′x(z, x)| ≤ aλ(z) + c|x|r−2x for a.a. z ∈ Ω, all x ≥ 0, with aλ ∈ L∞(Ω), c > 0, 2 < r < 2∗ = { 2N N−2 if N ≥ 3 +∞ if N = 1, 2 and ‖aλ‖∞ → 0 as λ→ 0+; (ii) if Fλ(z, x) = ∫ x 0 fλ(z, s)ds, then limx→±∞ Fλ(z,x) x2 = +∞ uniformly for a.a. z ∈ Ω; (iii) if eλ(z, x) = fλ(z, x)x − 2Fλ(z, x), then there exists ηλ ∈ L1(Ω) such that eλ(z, x) ≤ eλ(z, y) + ηλ(z) for a.a. z ∈ Ω, all 0 ≤ x ≤ y or y ≤ x ≤ 0; (iv) there exists m ∈ N such that (fλ)′x(z, 0) = lim x→0 fλ(z, x) x uniformly for a.a. z ∈ Ω, all λ ∈ (0, λ̃), (fλ)′x(z, 0) ∈ [λ̂m, λ̂m+1] for a.a. z ∈ Ω, (fλ)′x(·, 0) 6≡ λ̂m, (fλ)′x(·, 0) 6≡ λ̂m+1. Remark 2.2. Hypotheses (H2)(ii),(iii) imply that for all λ ∈ (0, λ̃) lim x→0 fλ(z, x) x = +∞ uniformly for a.a. z ∈ Ω. So, the reaction of problem (1.1) is superlinear. However, fλ(z, ·) need not satisfy the AR-condition as it is often the case in the literature (see Wang [20]). Recall that the AR-condition says that there exist M > 0 and p > 2 such that 0 < pFλ(z, x) ≤ fλ(z, x)x for a.a. z ∈ Ω, all |x| ≥M, 0 < ess inf Ω Fλ(·,±M). Integrating the first inequality and using the second, we show that c3|x|p ≤ Fλ(z, x) for a.a. z ∈ Ω, all |x| ≥M , some c3 > 0. Therefore the AR-condition dictates that fλ(z, ·) eventually has at least (p − 1)- polynomial growth. This excludes from consideration nonlinearities with “slower” growth near ±∞ (see the example below). Here instead of the AR-condition, we employ the less restrictive quasimonotonicity hypothesis (H2)(iii). This condition is a slight generalization of a hypothesis employed by Li-Yang [11] (see also [14]). There are easy ways to verify this condition. So, if there exists M > 0 such that for a.a. z ∈ Ω x 7→ fλ(z, x) x is nondecreasing on [M,+∞), 6 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 x 7→ fλ(z, x) x is nonincreasing on (−∞,−M ], then the quasimonotonicity condition (H2)(iii) is satisfied. Hypothesis (H2)(iv) implies nonuniform nonresonance with respect to any spectral interval. Example 2.3. The following function satisfies hypotheses (H2) but not the AR- condition. For the sake of simplicity, we drop the z-dependence: fλ(x) =  x ln |x| − λ sin ( η λx ) if x < −1, λ sin ( η λx ) if |x| ≤ 1 x ln |x|+ λ sin ( η λx ) if 1 < x, with η ∈ (λ̂m, λ̂m+1), m ∈ N. Finally note that on account of (H2)(i), we have |fλ(z, x)| ≤ âλ(z) + ĉ |x|r−1 for a.a. z ∈ Ω, all x ∈ R (2.6) with âλ ∈ L∞(Ω), ĉ > 0 and ‖âλ‖∞ → 0 as λ→ 0+. 3. Solutions of constant sign On account of hypothesis (H2)(iv), we can find δ0 > 0 such that fλ(z, x)x ≥ 0 for a.a. z ∈ Ω, all |x| ≤ δ0, all λ ∈ (0, λ̃). (3.1) Proposition 3.1. If hypotheses (H1), (H2)(i), (H2)(iv) hold, then there exists λ∗ ∈ (0, λ̃) such that for all λ ∈ (0, λ∗) problem (1.1) has a positive solution u0 ∈ intC+ and a negative solution v0 ∈ − intC+ such that ‖v0‖∞, ‖u0‖∞ ≤ δ0. Proof. We consider the auxiliary Robin problem −∆u(z) + ξ(z)u(z) = 1 in Ω, ∂u ∂n + β(z)u = 0 on ∂Ω. (3.2) We consider the operator V ∈ L(H1(Ω), H1(Ω)∗) defined by 〈V (u), h〉 = ∫ Ω (∇u,∇h)RNdz + ∫ Ω ξ(z)uh dz + ∫ ∂Ω β(z)uh dσ for all u, h ∈ H1(Ω). From (2.1) it is clear that V (·) is strongly monotone (hence maximal monotone too), and V (·) is coercive. It follows that V (·) is surjective [17, Corollary 2.8.7, p. 135]. So, there exists ũ ∈ H1(Ω) \ {0} such that V (ũ) = 1. The strong monotonicity of V (·) implies that this solution is unique. We have∫ Ω (∇ũ,∇h)RNdz + ∫ Ω ξ(z)ũh dz + ∫ ∂Ω β(z)ũhdσ = ∫ Ω h dz (3.3) for all h ∈ H1(Ω) which implies that ũ is the unique solution of (3.2). In (3.3) we use the test function h = −ũ− ∈ H1(Ω) and obtain that ‖∇ũ−‖22 + ∫ Ω ξ(z)(ũ−)2 dz + ∫ ∂Ω β(z)(ũ−)2dσ ≤ 0, ⇒ c0‖ũ−‖2 ≤ 0 (see (2.1)), ⇒ ũ ≥ 0, ũ 6= 0. EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 7 From the regularity theory of Wang [19], we have ũ ∈ C+ \ {0}. Moreover, from (3.2) and hypotheses (H1), we have ∆ũ(z) ≤ ‖ξ‖∞ũ(z) for a.a. z ∈ Ω, ⇒ ũ ∈ intC+ (by Hopf’s lemma). We claim that there exists λ∗ ∈ (0, λ̃) such that, if λ ∈ (0, λ∗), then we can find ξ̃ = ξ̃(λ) ∈ ( 0, δ0 ‖ũ‖∞ ) such that ‖âλ‖∞ + ĉ (ξ̃ ‖ũ‖∞)r−1 < ξ̃ (3.4) with âλ ∈ L∞(Ω) and ĉ > 0 as in (2.6) and δ0 > 0 as in (3.1). Arguing by contradiction, suppose that we can find λn → 0+ such that ξ ≤ ‖âλn‖∞ + ĉ(ξ‖ũ‖∞)r−1 for all n ∈ N, all ξ ∈ ( 0, δ0 ‖ũ‖∞ ) . Passing to the limit as n→∞, we obtain 1 ≤ ĉ ξr−2‖ũ‖r−1 ∞ for all ξ ∈ ( 0, δ0 ‖ũ‖∞ ) . Letting ξ → 0+, we have a contradiction. This proves (3.4). Fix λ ∈ (0, λ∗) and let u = ξ̃ ũ ∈ intC+. We consider the following truncation of the reaction f̂+ λ (z, x) = { fλ(z, x+) if x ≤ u(z), fλ(z, u(z)) if u(z) < x. (3.5) This is a Carathéodory function. We set F̂+ λ (z, x) = ∫ x 0 f̂+ λ (z, s)ds and consider the C1-functional ϕ̂+ λ : H1(Ω)→ R defined by ϕ̂+ λ (u) = 1 2 γ(u)− ∫ Ω F̂+ λ (z, u)dz for all u ∈ H1(Ω). From (2.1) and (3.5) we see that ϕ̂+ λ (·) is coercive. Also, the Sobolev embedding theorem and the compactness of the trace operator, imply that ϕ̂+ λ (·) is sequentially weakly lower semicontinuous. Then by the Weierstrass-Tonelli theorem, we know that there exists u0 ∈ H1(Ω) such that ϕ̂+ λ (u0) = inf[ϕ̂+ λ (u) : u ∈ H1(Ω)]. (3.6) On account of hypothesis (H2)(iv), we see that given ε > 0, we can find δ = δ(ε) > 0 such that F+ λ (z, x) ≥ 1 2 [(fλ)′x(z, 0)− ε]x2 for a.a. z ∈ Ω, all |x| ≤ δ. (3.7) We can always take δ ≤ minΩ u (recall u ∈ intC+). Choose t ∈ (0, 1) small so that tû1(z) ∈ (0, δ] for all z ∈ Ω (recall û1 ∈ intC+). (3.8) We have ϕ̂+ λ (tû1) = t2 2 [ γ(û1)− ∫ Ω F̂+ λ (z, tû1)dz ] ≤ t2 2 [ ∫ Ω ( λ̂1 − (fλ)′x(z, 0) ) û2 1dz + ε ] 8 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 (see (3.5), (3.7), (3.8) and recall ‖û1‖2 = 1). Note that∫ Ω ( (fλ)′x(z, 0)− λ̂1 ) û2 1dz > 0 (since û1 ∈ intC+). So, choosing ε > 0 small, we obtain ϕ̂+ λ (tû1) < 0, ⇒ ϕ̂+ λ (u0) < 0 = ϕ̂+ λ (0) (see (3.6)), ⇒ u0 6= 0. From (3.6) we have (ϕ̂+ λ )′+(u0) = 0 ⇒ 〈γ′(u0), h〉 = ∫ Ω f̂+ λ (z, u0)h dz for all h ∈ H1(Ω). (3.9) In (3.9) first we choose h = −u−0 ∈ H1(Ω). Then γ(u−0 ) = 0, ⇒ c0‖u−0 ‖2 ≤ 0 (see (2.1)), ⇒ u0 ≥ 0, u0 6= 0. Next in (3.9) we use the test function h = (u0 − u)+ ∈ H1(Ω). Then 〈γ′(u0), (u0 − u)+〉 = ∫ Ω f̂+ λ (z, u0)(u0 − u)+ dz = ∫ Ω fλ(z, u)(u0 − u)+ dz (see (3.5)) ≤ ∫ Ω [âλ(z) + ĉ ur−1](u0 − u)+ dz (see (2.6)) ≤ ∫ Ω ξ̃(u− u0)+ dz (see (3.4)) = 〈γ′(u), (u0 − u)+〉 (since u = ξ̃ ũ), ⇒ u0 ≤ u. So, we have proved that u0 ∈ [0, u], u0 6= 0. (3.10) From (3.10), (3.5) and (3.9), we infer that −∆u0 + ξ(z)u0 = fλ(z, u0) a.e. in Ω, ∂u0 ∂n + β(z)u0 = 0 on ∂Ω, implies that u0 ∈ intC+ \{0} is a solution of (1.1), λ ∈ (0, λ∗) (see Wang [19]). We have ‖u‖∞ = ξ̃ ‖ũ‖∞ ≤ δ0 ⇒ u(z) ≤ δ0 for all z ∈ Ω. So, from (3.1) we have −∆u0(z) + ξ(z)u0(z) ≥ 0 for a.a. z ∈ Ω, ⇒ ∆u0(z) ≤ ‖ξ‖∞u0(z) for a.a. z ∈ Ω, ⇒ u0 ∈ intC+ (by Hopf’s lemma). EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 9 Also we have ∆(u− u0)(z) ≤ ‖ξ‖∞(u− u0)(z) for a.a. z ∈ Ω (see (3.4)), ⇒ u− u0 ∈ intC+ (again Hopf’s lemma). So, finally we can say that u0 ∈ intC1(Ω)[0, u]. (3.11) In a similar fashion working on the negative semiaxis with v = −ξ̃ ũ ∈ − intC+, we produce a negative solution v0 such that v0 ∈ − intC+, ‖v0‖∞ ≤ δ0, v0 ∈ intC1(Ω)[v, 0]. � Now using these two solutions and the full set of hypotheses (H2), we will gen- erate two more constant sign smooth solutions. Proposition 3.2. If (H1), (H2) hold, then for all λ ∈ (0, λ∗), problem (1.1) has two more constant sign solutions û ∈ intC+ and v̂ ∈ − intC+ such that û−u0 ∈ intC+ and v0 − v̂ ∈ intC+. Proof. Let u0 ∈ intC+ be the positive solution of problem (1.1) (λ ∈ (0, λ∗)) produced in Proposition 3.1. We introduce the following truncation of fλ(z, ·): g+ λ (z, x) = { fλ(z, u0(z)) if x ≤ u0(z), fλ(z, x) if u0(z) < x. (3.12) This is a Carathéodory function. We set G+ λ (z, x) = ∫ x 0 g+ λ (z, s)ds and consider the C1-functional ψ+ λ : H1(Ω)→ R defined by ψ+ λ (u) = 1 2 γ(u)− ∫ Ω G+ λ (z, u)dz for all u ∈ H1(Ω). Using (3.8) and the regularity theory of Wang [19], we obtain easily that Kψ+ λ ⊆ [u0) ∩ intC+. (3.13) From (3.12) and (3.13) we see that we may assume that Kψ+ λ ∩ [0, u] = {u0}. (3.14) Otherwise we already have a second positive smooth solution for problem (1.1). We introduce the following truncation of g+ λ (z, ·): ĝ+ λ (z, x) = { g+ λ (z, x) if x ≤ u(z), g+ λ (z, u(z)) if u(z) < x. (3.15) This is a Carathéodory function. We set Ĝ+ λ (z, x) = ∫ x 0 ĝ+ λ (z, s)ds and consider the C1-functional ψ̂+ λ : H1(Ω)→ R defined by ψ̂+ λ (u) = 1 2 γ(u)− ∫ Ω Ĝ+ λ (z, u)dz for all u ∈ H1(Ω). From (2.1) and (3.15) it is clear that ψ̂+ λ (·) is coercive. Also it is sequentially weakly lower semicontinuous. So, we can find ũ0 ∈ H1(Ω) such that ψ̂+ λ (ũ0) = inf[ψ̂+ λ (u) : u ∈ H1(Ω)]. (3.16) 10 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 Using (3.12) and (3.15), we can easily show that Kψ̂+ λ ⊆ [u0, u] ∩ intC+. (3.17) Also we have ψ+ λ ∣∣ [0,u] = ψ̂+ λ ∣∣ [0,u] and (ψ+ λ )′ ∣∣ [0,u] = (ψ̂+ λ )′ ∣∣ [0,u] . (3.18) From (3.16) we have ũ0 ∈ Kψ̂+ λ . From (3.17), (3.18) and (3.14) it follows that ũ0 = u0 ∈ intC+. Since u0 ∈ intC1(Ω)[0, u] (see (3.11)), from (3.16) and (3.18) it follows that u0 is a local C1(Ω)-minimizer of ψ+ λ , ⇒ u0 is a local H1(Ω)-minimizer of ψ+ λ (see Brezis-Nirenberg [3]). (3.19) From (3.13) and (3.12), we see that we can assume that Kψ+ λ is finite. (3.20) Otherwise we already have a whole sequence of distinct positive smooth solutions of (1.1), all strictly bigger than u0 and so we are done. Then (3.19), (3.20) and Papageorgiou-Rădulescu-Repovš [17, Theorem 5.7.6, p. 449] imply that we can find ρ ∈ (0, 1) small such that ψ+ λ (u0) < inf[ψ+ λ (u) : ‖u− u0‖ = ρ] = m+ λ . (3.21) Hypothesis (H2)(ii) implies that if u ∈ intC+, then ψ+ λ (tu)→ −∞ as t→ +∞. (3.22) Claim: ψ+ λ satisfies the C-condition. We consider a sequence {un}n∈N ⊆ H1(Ω) such that |ψ+ λ (un)| ≤ c3 for some c3 > 0, all n ∈ N, (3.23) (1 + ‖un‖)(ψ+ λ )′(un)→ 0 in H1(Ω)∗ as n→∞. (3.24) From (3.24) we have∣∣〈γ′(un), h〉 − ∫ Ω g+ λ (z, un)h dz ∣∣ ≤ εn ‖h‖ 1 + ‖un‖ (3.25) for all h ∈ H1(Ω), with εn → 0+. In (3.25) we use the test function h = −u−n ∈ H1(Ω). Then γ(u−n ) ≤ c4‖u−n ‖ for some c4 > 0, all n ∈ N (see (3.12)), ⇒ {u−n }n∈N ⊆ H1(Ω) is bounded (see (2.1)). (3.26) Using (3.26) in (3.23), we obtain γ(u+ n )− ∫ Ω 2G+ λ (z, u+ n )dz ≤ c5 for some c5 > 0, all n ∈ N. (3.27) On the other hand, if in (3.25) we use the test function h = u+ n ∈ H1(Ω) then − γ(u+ n ) + ∫ Ω g+ λ (z, u+ n )u+ n dz ≤ εn for all n ∈ N. (3.28) EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 11 We add (3.27) and (3.28) and obtain∫ Ω [g+ λ (z, u+ n )u+ n − 2G+ λ (z, u+ n )]dz ≤ c6 for some c6 > 0, all n ∈ N, ⇒ ∫ Ω [fλ(z, u+ n )u+ n − 2Fλ(z, u+ n )]dz ≤ c7 for some c7 > 0, all n ∈ N, ⇒ ∫ Ω eλ(z, u+ n )dz ≤ c7 for all n ∈ N. (3.29) Using (3.29) we will show that {u+ n }n∈N ⊆ H1(Ω) is bounded. We argue by contradiction. So, suppose that at least for a subsequence, we have ‖u+ n ‖ → ∞ as n→∞. (3.30) Let yn = u+ n ‖u+ n ‖ , n ∈ N. Then ‖yn‖ = 1, yn ≥ 0 for all n ∈ N and so we may assume that yn w−→ y in H1(Ω) and yn → y in Lr(Ω), yn → y in L2(∂Ω), y ≥ 0. (3.31) First we assume that y 6= 0. Let Ω+ = {z ∈ Ω : y(z) > 0}. Then from (3.31) we see that |Ω+|N > 0 with | · |N denoting the Lebesgue measure on RN . We have u+ n (z)→ +∞ for a.a. z ∈ Ω+, ⇒ Fλ(z, u+ n (z)) u+ n (z)2 → +∞ for a.a. z ∈ Ω+ (see hypothesis (H2)(ii)), ⇒ Fλ(z, u+ n (z)) ‖u+ n ‖2 → +∞ for a.a. z ∈ Ω+. Then using Fatou’s lemma, we have∫ Ω+ Fλ(z, u+ n ) ‖u+ n ‖2 dz → +∞, ⇒ ∫ Ω+ G+ λ (z, u+ n ) ‖u+ n ‖2 dz → +∞ (see (3.12)). (3.32) On account of hypotheses (H2)(i),(ii), we have Fλ(z, x) ≥ −c8 for a.a. z ∈ Ω, all x ∈ R, some c8 > 0, ⇒ G+ λ (z, x) ≥ −c9 for a.a. z ∈ Ω, all x ∈ R, some c9 > 0 (see (3.12)). (3.33) So, we have∫ Ω G+ λ (z, u+ n ) ‖u+ n ‖2 dz = ∫ Ω+ G+ λ (z, u+ n ) ‖u+ n ‖2 dz + ∫ Ω\Ω+ G+ λ (z, u+ n ) ‖u+ n ‖2 dz ≥ ∫ Ω+ G+ λ (z, u+ n ) ‖u+ n ‖2 dz − c9|Ω|N (see (3.33)), ⇒ ∫ Ω G+ λ (z, u+ n ) ‖u+ n ‖2 dz → +∞ (see (3.32)). (3.34) From (3.23) and (3.26), we have∫ Ω G+ λ (z, u+ n ) ‖u+ n ‖2 dz ≤ c10 + 1 2 γ(u+ n ) for some c10 > 0, all n ∈ N. (3.35) Comparing (3.34) and (3.35) we have a contradiction. 12 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 Now assume that y = 0. With k > 0, let vn = (2k)1/2yn, n ∈ N. We have vn w−→ 0 in H1(Ω) and vn → 0 in L2(Ω) and in L2(∂Ω) (see (3.31)). (3.36) Let tn ∈ [0, 1] be such that ψ+ λ (tnu + n ) = max[ψ+ λ (tu+ n ) : 0 ≤ t ≤ 1]. (3.37) On account of (3.30), we can find n0 ∈ N such that 0 < (2k)1/2 1 ‖u+ n ‖ ≤ 1 for all n ∈ N, n ≥ n0. (3.38) We have ψ+ λ (tnu + n ) ≥ ψ+ λ (vn) = k − ∫ Ω G+ λ (z, vn)dz for all n ≥ n0 (see (3.38)). (3.39) From (3.36) we see that ∫ Ω G+ λ (z, vn)dz → 0 as n→∞. So, from (3.39) we see that we can find n1 ∈ N, n1 ≥ n0 such that ψ+ λ (tnu + n ) ≥ k 2 for all n ≥ n1. Since k > 0 is arbitrary, it follows that ψ+ λ (tnu + n )→ +∞ as n→∞. (3.40) We know that ψ+ λ (0) = 0 and ψ+ λ (u+ n ) ≤ c11 for some c11 > 0, all n ∈ N (3.41) (see (3.23), (3.26)). From (3.40) and (3.41) it follows that we can find n2 ∈ N such that tn ∈ (0, 1) for all n ≥ n2. (3.42) From (3.37) and (3.42), we infer that d dt ψ+ λ (tu+ n ) ∣∣∣ t=tn = 0 for n ≥ n2, ⇒ 〈(ψ+ λ )′(tnu + n ), tnu + n 〉 = 0 for n ≥ n2 (by the chain rule and see (3.42)), ⇒ γ(tnu + n ) = ∫ Ω g+ λ (z, tnu + n )(tnu + n )dz for all n ≥ n2. (3.43) From (3.12) we see that∫ Ω g+ λ (z, tnu + n )(tnu + n )dz ≤ ∫ Ω fλ(z, tnu + n )(tnu + n )dz + c12 (3.44) for some c12 > 0, all n ∈ N. Also from (3.42) and hypothesis (H2)(iii), we have∫ Ω eλ(z, tnu + n ) dz ≤ ∫ Ω eλ(z, u+ n ) dz + ‖ηλ‖1 ≤ c13 for some c13 > 0, all n ≥ n2 (see (3.29)), ⇒ ∫ Ω fλ(z, tnu + n )(tnu + n )dz ≤ c13 + ∫ Ω 2Fλ(z, tnu + n )dz for all n ≥ n2. (3.45) EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 13 We return to (3.44) and using (3.45) and (3.12), we obtain∫ Ω g+ λ (z, tnu + n )(tnu + n )dz ≤ c14 + ∫ Ω 2G+ λ (z, tnu + n ) dz (3.46) for some c14 > 0, all n ≥ n2. If in (3.43) we use (3.46), then 2ψ+ λ (tnu + n ) ≤ c14 for all n ≥ n2. (3.47) Comparing (3.40) and (3.47), we have a contradiction. Therefore {u+ n }n∈N ⊆ H1(Ω) is bounded, hence {un}n∈N ⊆ H1(Ω) is bounded (see (3.26)). So, we can assume that un w−→ u in H1(Ω) and un → u in Lr(Ω) and in L2(∂Ω). (3.48) In (3.25), we choose h = un − u ∈ H1(Ω), pass to the limit as n → ∞ and use (3.48). Then lim n→∞ ∫ Ω (∇un,∇un −∇u)RNdz = 0, ⇒ ‖∇un‖2 → ‖∇u‖2, ⇒ un → u in H1(Ω) (see (3.48) and use the Kadec-Klee property, see [7], p. 911). This proves the Claim. On account of (3.21), (3.22) and the Claim, we can apply the mountain pass theorem. So, we can find û ∈ H1(Ω) such that û ∈ Kψ+ λ ⊆ [u0) ∩ intC+ (see (3.13)), m+ λ ≤ ψ + λ (û) (see (3.21)). So, û 6= u0 (see (3.21)), u0 ≤ û and û ∈ intC+ is the second positive solution of (1.1) (λ ∈ (0, λ∗)), distinct from u0. If ρ = ‖û‖∞, then on account of hypothesis (H2)(i), we can find ξ̂ρ > 0 such that f ′x(z, x)x2 ≥ −ξ̂ρx2 for a.a. z ∈ Ω, all |x| ≤ ρ, ⇒ x→ f(z, x) + ξ̂ρx is nondecreasing on [−ρ, ρ]. We have −∆û+ [ξ(z) + ξ̂ρ]û = fλ(z, û) + ξ̂ρû ≥ fλ(z, u0) + ξ̂ρu0 (recall u0 ≤ û) = −∆u0 + [ξ(z) + ξ̂ρ]u0, ⇒ û− u0 ∈ intC+ (by Hopf’s lemma). Similarly using this time v0 ∈ − intC+ and reasoning as above (working this time on the negative semiaxis), we produce a second negative solution v̂ of problem (1.1) (λ ∈ (0, λ∗)), such that v̂ ∈ − intC+, v0 − v̂ ∈ intC+. � We will show that there exist extremal constant sign solutions, that is, a smallest positive solution uλ∗ ∈ intC+ and a biggest negative solution vλ∗ ∈ − intC+. We will use these extremal (barrier) solutions in Section 4 in order to produce nodal solutions. 14 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 We introduce the following two sets: S+(λ) = set of positive solutions of (1.1), S−(λ) = set of negative solutions of (1.1). We already know from Proposition 3.1, that assuming (H2)(i), (H2)(iv), ∅ 6= S+(λ) ⊆ intC+, ∅ 6= S−(λ) ⊆ − intC+, for all λ > 0 small. From hypotheses (H2)(i), (H2)(iv), we see that given ε > 0, we can find c15 = c15(ε) > 0 such that fλ(z, x)x ≥ [(fλ)′x(z, 0)− ε]x2 − c15|x|r for a.a. z ∈ Ω, all x ∈ R. (3.49) Motivated by the unilateral growth restriction on fλ(z, ·), we consider the aux- iliary Robin problem −∆u+ ξ(z)u = [(fλ)′x(z, 0)− ε]u− c15|u|r−2u in Ω, ∂u ∂n + β(z)u = 0 on ∂Ω. (3.50) Proposition 3.3. If (H1) hold and λ > 0, then for all ε > 0 small problem (3.50) has a unique positive solution wλ+ ∈ intC+ and since the equation is odd wλ− = −wλ+ ∈ − intC+ is the unique negative solution of (3.50). Proof. First we show the existence of a positive solution. To this end, we introduce the C1-functional τ+ λ : H1(Ω)→ R defined by τ+ λ (u) = 1 2 γ(u) + c15 r ‖u+‖rr − 1 2 ∫ Ω [(fλ)′x(z, 0)− ε](u+)2dz for all u ∈ H1(Ω). Since r > 2, this functional is coercive. Also, it is sequentially weakly lower semi- continuous. So, we can find wλ+ ∈ H1(Ω) such that τ+ λ (wλ+) = inf[τ+ λ (u) : u ∈ H1(Ω)]. (3.51) Let t ∈ (0, 1). We have τ+ λ (tû1) = t2 2 γ(û1) + c15t r r ‖û1‖rr − t2 2 ∫ Ω [(fλ)′x(z, 0)− ε]û2 1dz = t2 2 [ ∫ Ω [λ̂1 − (fλ)′x(z, 0)]û2 1dz + ε ] + c15t r r ‖û1‖rr. (3.52) Hypothesis (H2)(iv) and the fact that û1 ∈ intC+ imply that ϑλ = ∫ Ω [(fλ)′x(z, 0)− λ̂1]û2 1dz > 0. So, from (3.52) and choosing ε > 0 small, we have τ+ λ (tû1) ≤ c16t r − c17t 2 for some c16, c17 > 0. Since r > 2, choosing t ∈ (0, 1) small we have τ+ λ (tû1) < 0, ⇒ τ+ λ (wλ+) < 0 = τ+ λ (0) (see (3.51)), ⇒ wλ+ 6= 0. From (3.51), we have (τ+ λ )′(wλ+) = 0, EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 15 ⇒ 〈γ′(wλ+), h〉 = ∫ Ω ( [(fλ)′x(z, 0)− ε](wλ+)+ − c15(wλ+)r−1 ) h dz (3.53) for all h ∈ H1(Ω). In (3.53) we use the test function h = −(wλ+)− ∈ H1(Ω). Then γ((wλ+)−) = 0, ⇒ wλ+ ≥ 0, wλ+ 6= 0 (see (2.1)). (3.54) Therefore wλ+ is a positive solution of (3.50) (λ > 0 and ε > 0 small) and from the regularity theory of Wang [19], we have wλ+ ∈ C+ \ {0} (see (3.54)). From the equation we have ∆wλ+ ≤ [‖ξ‖∞ + c15‖wλ+‖r−2 ∞ ]wλ+ in Ω, ⇒ wλ+ ∈ intC+. (3.55) Next we show the uniqueness of the positive solution of (3.50). Suppose that ŵλ+ ∈ H1(Ω) is another positive solution. Again we have ŵλ+ ∈ intC+ (see (3.55)). Let t > 0 be the biggest positive real such that tŵλ+ ≤ wλ+ (see [17], p. 274). (3.56) Suppose that 0 < t < 1. Let ρλ = ‖wλ+‖∞. On account of hypothesis (H2)(i), we see that we can find ξ̂ρλ > 0 such that for a.a. z ∈ Ω the function x→ [(fλ)′x(z, 0)− ε]x− c15x r−1 + ξ̂ρλx is nondecreasing on [0, ρλ]. We have −∆(tŵλ+) + [ξ(z) + ξ̂ρλ ](tŵλ+) = t [ ((fλ)′x(z, 0)− ε)ŵλ+ − c15(ŵλ+)r−1 + ξ̂ρλŵ λ + ] ≤ ((fλ)′x(z, 0)− ε)(tŵλ+)− c15(tŵλ+)r−1 + ξ̂ρλ(tŵλ+) (since 0 < t < 1 and r > 2) ≤ ((fλ)′x(z, 0)− ε)wλ+ − c15(wλ+)r−1 + ξ̂ρλŵ λ + (see (3.56)) = −∆wλ+ + [ξ(z) + ξ̂ρλ ]wλ+, ⇒ ∆(wλ+ − tŵλ+) ≤ [‖ξ‖∞ + ξ̂ρλ ](wλ+ − tŵλ+), ⇒ wλ+ − tŵλ+ ∈ intC+ (by Hopf’s lemma). But this contradicts the maximality of t > 0 in (3.56). Hence 1 ≤ t and so we have ŵλ+ ≤ wλ+. Interchanging the roles of ŵλ+ and wλ+ in the above argument, we obtain wλ+ ≤ ŵλ+ and so we conclude that wλ+ = ŵλ+. This proves the uniqueness of the positive solution of problem (3.50) (λ > 0, ε > 0 small). Problem (3.50) is odd. Therefore wλ− = −wλ+ ∈ − intC+ is the unique negative solution of (3.50). � These solutions provide bounds for the sets S+(λ) and S−(λ). Proposition 3.4. If (H1), (H2)(i), (H2)(iv) hold and λ ∈ (0, λ∗), then wλ+ ≤ u for all u ∈ S+(λ) and v ≤ wλ− for all v ∈ S−(λ). 16 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 Proof. Let u ∈ S+(λ) ⊆ intC+ and for ε > 0 small as in Proposition 3.3, we consider the Carathéodory function ϑ+ λ (z, x) defined by ϑ+ λ (z, x) = { [(fλ)′x(z, 0)− ε](x+)− c15(x+)r−1 if x ≤ u(z), [(fλ)′x(z, 0)− ε]u(z)− c15u(z)r−1 if u(z) < x. (3.57) We set Θ+ λ (z, x) = ∫ x 0 ϑ+ λ (z, s)ds and introduce the C1-functional b+λ : H1(Ω)→ R defined by b+λ (u) = 1 2 γ(u)− ∫ Ω Θ+ λ (z, u)dz for all u ∈ H1(Ω). From (2.1) and (3.57) we see that b+λ (·) is coercive. Also, it is sequentially weakly lower semicontinuous. So, we can find ŵλ+ ∈ H1(Ω) such that b+λ (ŵλ+) = inf[b+λ (u) : u ∈ H1(Ω)]. (3.58) As in the proof of Proposition 3.3, we have b+λ (ŵλ+) < 0 = b+λ (0) ⇒ ŵλ+ 6= 0. From (3.58), we have (b+λ )′(ŵλ+) = 0 ⇒ 〈γ′(ŵλ+), h〉 = ∫ Ω ϑ+ λ (z, ŵλ+)h dz for all h ∈ H1(Ω). (3.59) Here first we choose h = −(ŵλ+)− ∈ H1(Ω) and obtain c0‖(ŵλ+)−‖2 ≤ 0 (see (2.1)), ⇒ ŵλ+ ≥ 0, ŵλ+ 6= 0. Then in (3.59) we use the test function h = (ŵλ+ − u)+ ∈ H1(Ω). We have 〈γ′(ŵλ+), (ŵλ+ − u)+〉 = ∫ Ω ( [(fλ)′x(z, 0)− ε]u− c15u r−1 ) (ŵλ+ − u)+dz (see (3.57)) ≤ ∫ Ω fλ(z, u)(ŵλ+ − u)+dz (see (3.49)) = 〈γ′(u), (ŵλ+ − u)+〉 (since u ∈ S+(λ)), ⇒ ŵλ+ ≤ u. So, we can say that ŵλ+ ∈ [0, u], ŵλ+ 6= 0. (3.60) Then from (3.60), (3.57) and (3.59) it follows that ŵλ+ is a positive solution ŵλ+ of (3.50). Proposition 3.3 implies that ŵλ+ = wλ+. Therefore wλ+ ≤ u for all u ∈ S+(λ). In a similar fashion we show that v ≤ wλ− for all v ∈ S−(λ). � These bounds lead to the existence of extremal constant sign solutions. Proposition 3.5. If (H1), (H2)(i), (H2)(iv) hold and λ ∈ (0, λ∗), then problem (1.1) has a smallest positive solution uλ∗ ∈ intC+ (that is, uλ∗ ≤ u for all u ∈ S+(λ)) and a biggest negative solution vλ∗ ∈ − intC+ (that is, v ≤ vλ∗ for all v ∈ S−(λ)). EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 17 Proof. From Filippakis-Papageorgiou [6], we know that S+(λ) is downward directed (that is, if u1, u2 ∈ S+(λ), then we can find u ∈ S+(λ) such that u ≤ u1, u ≤ u2). Then using Hu-Papageorgiou [10, Lemma 3.10, p. 178], we can find a decreasing sequence {un}n∈N ⊆ H1(Ω) such that inf n∈N un = inf S+(λ). We have 〈γ′(un), h〉 = ∫ Ω fλ(z, un)h dz for all h ∈ H1(Ω), (3.61) ŵλ+ ≤ un ≤ u1 for all n ∈ N (see Proposition 3.4). (3.62) If in (3.61) we choose the test function h = un ∈ H1(Ω) and use (3.62) and (2.6), we infer that {un}n∈N ⊆ H1(Ω) is bounded. We assume that un w−→ uλ∗ in H1(Ω) and un → uλ∗ in L2(Ω) and in L2(∂Ω). (3.63) In (3.61) we choose h = un − uλ∗ ∈ H1(Ω), pass to the limit as n → ∞ and use (3.63). Then lim n→∞ ∫ Ω (∇un,∇un −∇uλ∗)RNdz = 0, ⇒ ‖∇un‖2 → ‖∇uλ∗‖2, ⇒ un → uλ∗ in H1(Ω) (by the Kadec-Klee property of Hilbert spaces). So, in the limit as n→∞, we have 〈γ′(uλ∗), h〉 = ∫ Ω fλ(z, uλ∗)h dz for all h ∈ H1(Ω), ŵλ+ ≤ uλ∗ (see (3.62)). Therefore uλ∗ ∈ S+(λ) and uλ∗ = inf S+(λ). Similarly we produce vλ∗ ∈ S−(λ) ⊆ intC+ such that vλ∗ = supS−(λ). Note that S−(λ) is upward directed (that is, if v1, v2 ∈ S−(λ), then we can find v ∈ S−(λ) such that v1 ≤ v, v2 ≤ v). � 4. Nodal solutions In this section we show the existence of nodal (sign changing) solutions for prob- lem (1.1) (λ ∈ (0, λ∗)). We produce three nodal solutions. Our strategy to obtain these nodal solutions, is the following. Using truncations, we focus on the order interval [vλ∗ , u λ ∗ ]. On account of the extremality of uλ∗ and vλ∗ , any solution of (1.1) in this order interval distinct from uλ∗ , v λ ∗ and 0, will be nodal. The first nodal solution is obtained using the critical point theory (the mountain pass theorem). The second is established using the theory of critical groups. Finally for the third nodal solution, we employ a flow invariance argument using the gradient flow. For this strategy to work, we need to slightly strengthen hypothesis (H2)(iv). So, the new conditions on fλ(z, x) are the following: (H2’) Same as hypotheses (H2) only now in (H2)(iv), m ∈ N and m ≥ 2. 18 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 We start implementing this strategy. So, we introduce the following truncation of fλ(z, ·): ĝλ(z, x) =  fλ(z, vλ∗ (z)) if x < vλ∗ (z), fλ(z, x) if vλ∗ (z) ≤ x ≤ uλ∗(z), fλ(z, uλ∗(z)) if uλ∗(z) < x. (4.1) This is a Carathéodory function. Also, we consider the positive and negative trun- cations of ĝλ(z, ·), namely the Carathéodory functions ĝ±λ (z, x) defined by ĝ±λ (z, x) = ĝλ(z,±x±). (4.2) We set Ĝλ(z, x) = ∫ x 0 ĝλ(z, s)ds, Ĝ±λ (z, x) = ∫ x 0 ĝ±λ (z, s)ds and consider the C1- functionals jλ, j ± λ : H1(Ω)→ R defined by jλ(u) = 1 2 γ(u)− ∫ Ω Ĝλ(z, u)dz, j±λ (u) = 1 2 γ(u)− ∫ Ω Ĝ±λ (z, u)dz for all u ∈ H1(Ω). Using (4.1), (4.2), the regularity theory of Wang [19] and the extremality of uλ∗ and vλ∗ , we have the following result. Proposition 4.1. If (H1), (H2)(i), (H2)(iv) hold and λ ∈ (0, λ∗), then Kjλ ⊆ [vλ∗ , u λ ∗ ] ∩ C1(Ω), Kj+λ = {0, uλ∗}, Kj−λ = {0, vλ∗ }. It is clear from this proposition that we can assume that Kjλ ⊆ C1(Ω) is finite. (4.3) Otherwise we already have a sequence of distinct nodal smooth solutions and so we are done. Proposition 4.2. If (H1), (H2)(i), (H2)(iv) hold and λ ∈ (0, λ∗), then uλ∗ ∈ intC+ and vλ∗ ∈ − intC+ are local minimizers of jλ(·). Proof. From (4.1), (4.2) and (2.1), it is clear that j+ λ (·) is coercive. Also, it is sequentially weakly lower semicontinuous. So, we can find ũλ∗ ∈ H1(Ω) such that j+ λ (ũλ∗) = inf[j+ λ (u) : u ∈ H1(Ω)]. (4.4) As in the proof of Proposition 3.1, using (3.49), we show that j+ λ (ũλ∗) < 0 = j+ λ (0), ⇒ ũλ∗ 6= 0. We have that ũλ∗ ∈ Kj+λ \{0} (see (4.4)). Hence from Proposition 4.1 it follows that ũλ∗ = uλ∗ ∈ intC+. (4.5) It is clear from (4.1) and (4.2) that jλ ∣∣ C+ = j+ λ ∣∣ C+ . Then from (4.4) and (4.5) we infer that uλ∗ is a local C1(Ω)-minimizer of jλ, ⇒ uλ∗ is a local H1(Ω)-minimizer of jλ (see Brezis-Nirenberg [3]). Similarly for vλ∗ ∈ − intC+, using this time the functional j−λ (·). � EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 19 We may assume that jλ(vλ∗ ) ≤ jλ(uλ∗). (4.6) The analysis remains the same if the opposite inequality holds. Proposition 4.3. If (H1), (H2) hold and λ ∈ (0, λ∗), then Ck(jλ, 0) = δk,dmZ for all k ∈ N0 with dm = dimHm = dim⊕mk=1E(λ̂k). Proof. Let ϕλ : H1(Ω) → R be the energy (Euler) functional for problem (1.1) defined by ϕλ(u) = 1 2 γ(u)− ∫ Ω Fλ(z, u)dz for all u ∈ H1(Ω). Evidently ϕλ ∈ C2(H1(Ω)). We consider the homotopy hλ(t, u) defined by hλ(t, u) = (1− t)ϕλ(u) + tjλ(u) for all (t, u) ∈ [0, 1]×H1(Ω). Suppose we could find {tn}n∈N ⊆ [0, 1], {un}n∈N ⊆ H1(Ω) such that tn → t, un → 0 in H1(Ω) and (hλ)′u(tn, un) = 0 for all n ∈ N. (4.7) From the equality in (4.7), we have that 〈γ′(un), h〉 = (1− tn) ∫ Ω fλ(z, un)h dz + tn ∫ Ω ĝλ(z, un)h dz, for all h ∈ H1(Ω), all n ∈ N, imply −∆un + ξ(z)un = (1− tn)fλ(z, un) + tnĝλ(z, un) in Ω, ∂un ∂n + β(z)un = 0 on ∂Ω. (4.8) From (4.8) and the regularity theory of Wang [19], we know that there exist α ∈ (0, 1) and c16 > 0 such that un ∈ C1,α(Ω), ‖un‖C1,α(Ω) ≤ c16 for all n ∈ N. (4.9) Then from (4.9) and the compact embedding of C1,α(Ω) into C1(Ω), we have un → 0 in C1(Ω) (see (4.7)), ⇒ un ∈ [vλ∗ , u λ ∗ ] for all n ≥ n0, ⇒ {un}n≥n0 ⊆ Kjλ (see Proposition 4.1), which contradicts (4.3). Therefore (4.7) can not happen and the homotopy invari- ance property of critical groups (see [17, p. 505]) implies that Ck(ϕλ, 0) = Ck(jλ, 0) for all k ∈ N0. (4.10) Recall that ϕλ ∈ C2(H1(Ω)) and 〈ϕ′′λ(u)h, v〉 = 〈γ(h), v〉 − ∫ Ω (fλ)′x(z, u)hv dz for all u, h, v ∈ H1(Ω). Then on account of Proposition 4.3, the Morse index of u = 0 is dm = dimHm = dim⊕mk=1E(λ̂k). By [17, Proposition 6.2.6, p. 479], we have Ck(ϕλ, 0) = δk,dmZ for all k ∈ N0, ⇒ Ck(jλ, 0) = δk,dmZ for all k ∈ N0 (see (4.10)). � 20 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 Now we are ready to produce the first two nodal solutions. As we already mentioned the first will be obtained using critical point theory, while for the second we will use the theory of critical groups. At this point we need to use the stronger hypotheses (H2’). Proposition 4.4. If (H1), (H2’) hold and λ ∈ (0, λ∗), then problem (1.1) has at least two nodal solutions y0, ŷ ∈ C1(Ω) such that y0, ŷ ∈ intC1(Ω)[v λ ∗ , u λ ∗ ]. Proof. Using (4.3), (4.6), Proposition 4.2 and [17, Theorem 5.7.6, p. 449], we know that we can find ρ ∈ (0, 1) small such that jλ(vλ∗ ) ≤ jλ(uλ∗) < inf[jλ(u) : ‖u− uλ∗‖ = ρ] = mλ, ‖vλ∗ − uλ∗‖ > ρ. (4.11) Also, since jλ(·) is coercive (see (4.1)), we know that jλ(·) satisfies the C-condition (4.12) (see [17, Proposition 5.1.15, p. 369]). Then (4.11), (4.12) permit the use of the mountain pass theorem. So, we can find y0 ∈ H1(Ω) such that y0 ∈ Kjλ ⊆ [vλ∗ , u λ ∗ ] ∩ C1(Ω) (see Proposition 4.1), mλ ≤ jλ(y0). (4.13) From (4.13) and (4.11) we see that y0 ∈ [vλ∗ , u λ ∗ ] ∩ C1(Ω) is a solution of (1.1), y0 6∈ {uλ∗ , vλ∗ }. (4.14) The solution y0 is a critical point of jλ of mountain pass type. Hence [17, Theorem 6.5.8, p. 527] implies that C1(jλ, y0) 6= 0. (4.15) From Proposition 4.3, we know that Ck(jλ, 0) = δk,dmZ for all k ∈ N0, with dm ≥ 2. (4.16) Comparing (4.15) and (4.16), we see that y0 6= 0 and so y0 ∈ C1(Ω) is a nodal solution of problem (1.1). Let ρ = max{‖vλ∗ ‖∞, ‖uλ∗‖∞}. Using (H2’) (i), (H2’)(iv), we can find ξ̃ρ > 0 such that for a.a. z ∈ Ω x→ fλ(z, x) + ξ̃ρx is nondecreasing on [−ρ, ρ]. (4.17) We have −∆y0 + [ξ(z) + ξ̃ρ]y0 = fλ(z, y0) + ξ̃ρy0 ≤ fλ(z, uλ∗) + ξ̃ρu λ ∗ (see (4.14), (4.17)) = −∆uλ∗ + [ξ(z) + ξ̃ρ]u λ ∗ , ⇒ ∆(uλ∗ − y0) + [‖ξ‖∞ + ξ̃ρ](u λ ∗ − y0) in Ω, ⇒ uλ∗ − y0 ∈ intC+. Similarly we show that y0 − vλ∗ ∈ intC+. Therefore we conclude that y0 ∈ intC1(Ω)[v λ ∗ , u λ ∗ ]. (4.18) As in the proof of Proposition 4.3, using the homotopy hλ(t, u) and (4.18), we show that Ck(ϕλ, y0) = Ck(jλ, y0) for all k ∈ N0. (4.19) EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 21 Since ϕλ ∈ C2(H1(Ω)), from (4.19), (4.15) and [17, Proposition 6.5.9, p. 529], we have Ck(ϕλ, y0) = δk,1Z for all k ∈ N0, ⇒ Ck(jλ, y0) = δk,1Z for all k ∈ N0 (see (4.19)). (4.20) From Proposition 4.2 we know that uλ∗ , v λ ∗ are local minimizers of jλ(·). Hence Ck(jλ, u λ ∗) = Ck(jλ, v λ ∗ ) = δk,0Z for all k ∈ N0. (4.21) Also, since jλ(·) is coercive (see (4.1)), we have Ck(jλ,∞) = δk,0Z for all k ∈ N0 (4.22) (see [17, Proposition 6.2.24, p. 491]). Suppose that Kjλ = {0, y0, u λ ∗ , v λ ∗ }. From (4.16), (4.20), (4.21), (4.22) and the Morse relation with t = −1 (see (2.5)), we have (−1)dm + (−1)1 + 2(−1)0 = (−1)0, ⇒ (−1)dm = 0, a contradiction. Therefore, there exists ŷ ∈ Kjλ ⊆ [vλ∗ , u λ ∗ ] ∩ C1(Ω) (see Proposition 4.1), with ŷ 6∈ {0, y0, u λ ∗ , v λ ∗ }. Hence ŷ ∈ C1(Ω) is a second nodal solution of (1.1) distinct from y0. Moreover, as we did for y0 earlier in this proof, we show that ŷ ∈ intC1(Ω)[v λ ∗ , u λ ∗ ]. � Next using a flow invariance argument inspired from the works of Papageorgiou- Papalini [15] and He-Guo-Huang-Lei [9], we produce a third nodal solution. Proposition 4.5. If (H1), (H2’) hold and λ ∈ (0, λ∗), then problem (1.1) has a nodal solution ỹ ∈ C1(Ω) such that ỹ 6∈ intC1(Ω)[v λ ∗ , u λ ∗ ]. Proof. On H1(Ω) we consider the inner product (u, h)0 = ∫ Ω (∇u,∇h)RNdz + ∫ Ω ξ(z)uh dz + ∫ ∂Ω β(z)uhdσ for all u, h ∈ H1(Ω). Let ‖ · ‖0 denote the corresponding norm (that is, ‖u‖0 = (u, u) 1/2 0 ). It is clear from hypotheses (H1) and (2.1), that ‖ · ‖0 is equivalent to ‖ · ‖ (the usual norm on H1(Ω)). Consider the operator K : H1(Ω)→ H1(Ω) defined by K(u) = A(u) + ξ(·)u+ β(·)γ̂0(u), where A ∈ L(H1(Ω), H1(Ω)∗) is defined by 〈A(u), h〉 = ∫ Ω (∇u,∇h)RNdz for all u, h ∈ H1(Ω) and γ̂0(·) is the trace operator (see Section 2). We have that K ∈ L(H1(Ω), H1(Ω)∗) and 〈K(u), h〉 ≥ c17‖u‖20 for some c17 > 0, all u ∈ H1(Ω). Hence by Banach’s theorem, we have L = K−1 ∈ L(H1(Ω), H1(Ω)∗). We set E = L ◦ Nf . Then E : H1(Ω) → H1(Ω) is continuous and the regularity theory (see Wang [19]), implies that E(H1(Ω)) ⊆ C1(Ω). 22 N. S. PAPAGEORGIOU, C. VETRO, F. VETRO EJDE-2021/12 Moreover, the Sobolev embedding theorem and the strong maximum principle imply that E(·) is compact and is strictly increasing, (4.23) that is, v ≤ u, v 6= u ⇒ E(u)− E(v) ∈ intC+. We know that ϕλ ∈ C2(H1(Ω)). The gradient ∇ϕλ is defined by 〈ϕ′λ(u), h〉 = (∇ϕλ(u), h)0 for all u, h ∈ H1(Ω). Then we have ∇ϕλ = id−E. (4.24) We consider the negative gradient flow σλ(·, u) defined by the abstract Cauchy problem d dt σλ(t, u) = −∇ϕλ(σλ(t, u)), t ≥ 0, σλ(0, u) = u. (4.25) Using (4.24) we can rewrite (4.25) as follows d dt σλ(t, u) + σλ(t, u) = E(σλ(t, u)), t ≥ 0, σλ(0, u) = u. So, the flow is global (see [7, p. 618]) and it is given by σλ(t, u) = e−tu+ ∫ t 0 e−(t−s)E(σλ(s, u))ds. Now we introduce the set Bλ1 = {u ∈ C1(Ω) : ∃ t0 > 0 such that for all t ≥ t0 σλ(t, u) ∈ intC1(Ω)[v λ ∗ , u λ ∗ ]}. We know that 0 ∈ Bλ1 (see (4.23)) and the continuous dependence of the flow on the initial condition, implies that Bλ1 ⊆ C1(Ω) is open. Moreover, the semigroup property of the flow implies that Bλ1 is flow invariant. Claim 1: ∂Bλ1 is flow invariant. We argue indirectly. So, suppose that Claim 1 is not true. This means that we can find û ∈ ∂Bλ1 and t̂ > 0 such that σλ(t̂, û) 6∈ ∂Bλ1 . There are two possibilities. In the first σλ(t̂, û) ∈ Bλ1 (recall Bλ1 is open). But then the semigroup property of the flow implies that û ∈ Bλ1 , a contradiction since û ∈ ∂Bλ1 . In the second possibility we have σλ(t̂, û) 6∈ Bλ1 = Bλ1 ∪ ∂Bλ1 . Since û ∈ ∂Bλ1 , we can find {un}n∈N ⊆ Bλ1 such that un → û. Then σλ(t̂, un)→ σλ(t̂, û) ∈ Bλ1 , ⇒ σλ(t̂, un) 6∈ Bλ1 for all n ≥ n0. (4.26) On the other hand, the flow invariance of Bλ1 implies that σλ(t̂, un) ∈ Bλ1 for all n ∈ N and this contradicts (4.26). This proves Claim 1. Next we introduce a second set in C1(Ω), namely the set Bλ2 = {u ∈ C1(Ω) : ∃ t̃ > 0 such that σλ(t, u) ∈ intC+ for all t ≥ t̃}. This set too is open and from (4.23) we have C+ \ {0} ⊆ Bλ2 and 0 ∈ ∂Bλ2 . Moreover, as we did for ∂Bλ1 in Claim 1, we can show that ∂Bλ2 is flow invariant. Note that ∂Bλ1 ∩ ∂Bλ2 6= ∅ and so we can define ĉ0 = inf[ϕλ(u) : u ∈ ∂Bλ1 ∩ ∂Bλ2 ]. (4.27) EJDE-2021/12 MULTIPLE SOLUTIONS FOR SEMILINEAR ROBIN PROBLEMS 23 Claim 2: ĉ0 is a critical value of ϕλ. Since σλ is the negative (descent) flow for ϕλ, we have −∞ < inf{ϕλ(u) : u ∈ [vλ∗ , u λ ∗} ≤ ĉ0. Recall that ϕλ satisfies the C-condition. So, ϕλ(Kϕλ) is closed. Then if ĉ0 is not a critical value of ϕλ, we can find ε > 0 small such that (ĉ0 − ε, ĉ0 + ε) ∩ ϕλ(Kϕλ) = ∅. Let û ∈ ∂Bλ1 ∩ ∂Bλ2 such that ϕλ(û) ≤ ĉ0 + ε 2 (see (4.27)). From the deformation theorem (see [7], p. 636), by taking ε > 0 even smaller if necessary, we have ϕλ(σλ(1, û)) ≤ ĉ0 − ε 2 . (4.28) The flow invariance of ∂Bλ1 ∩ ∂Bλ2 (see Claim 1), implies that σλ(1, û) ∈ ∂Bλ1 ∩ ∂Bλ2 . (4.29) But then (4.28), (4.29) and (4.27), lead to a contradiction. This proves Claim 2. Using Claim 2, we can find ỹ ∈ ∂Bλ1∩∂Bλ2 such that ỹ ∈ Kϕλ . Then ỹ 6∈ {0, y0, ŷ} (see Proposition 4.4). Also we have ỹ 6∈ (intC+ ∪ (− intC+)), which implies that it is nodal. Moreover, we have ỹ 6∈ intC1(Ω)[v λ ∗ , u λ ∗ ]. � Therefore we can state the following multiplicity theorem for problem (1.1). Theorem 4.6. If (H1), (H2’) hold, then for all λ > 0 small, problem (1.1) has at least seven nontrivial smooth solutions: • u0, û ∈ intC+ with û− u0 ∈ intC+, • v0, v̂ ∈ − intC+ with v0 − v̂ ∈ intC+, • y0, ŷ, ỹ ∈ C1(Ω) nodal with y0, ŷ ∈ intC1(Ω)[v0, u0], ỹ 6∈ intC1(Ω)[v0, u0]. 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Nonlinear Stud., 16 (2016), 737-764. [17] N. S. Papageorgiou, V. D. Rădulescu, D. D. Repovš; Nonlinear Analysis - Theory and Meth- ods, Springer Nature, Switzerland, 2019. [18] M. Sun; Multiple solutions of a superlinear p-Laplacian equation without AR-condition, Appl. Anal., 89 (2010), 325-336. [19] X.-J. Wang; Neumann problems of semilinear elliptic equations involving critical Sobolev exponents, J. Differential Equations, 93 (1991), 283-310. [20] Z.-Q. Wang; On a superlinear elliptic equation, Ann. Inst. H. Poincaré Anal. Non Linéaire, 8 (1991), 43-57. Nikolaos S. Papageorgiou National Technical University, Department of Mathematics, Zografou campus, 15780, Athens, Greece Email address: npapg@math.ntua.gr Calogero Vetro University of Palermo, Department of Mathematics and Computer Science, Via Archi- rafi 34, 90123, Palermo, Italy Email address: calogero.vetro@unipa.it Francesca Vetro 90123, Palermo, Italy Email address: francescavetro80@gmail.com 1. Introduction 2. Mathematical background - hypotheses 3. Solutions of constant sign 4. Nodal solutions References