Special Issue in honor of John W. Neuberger Electronic Journal of Differential Equations, Special Issue 02 (2023), pp. 209–230. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu or https://ejde.math.unt.edu FUC̆IK SPECTRUM WITH WEIGHTS AND EXISTENCE OF SOLUTIONS FOR NONLINEAR ELLIPTIC EQUATIONS WITH NONLINEAR BOUNDARY CONDITIONS NSOKI MAVINGA, QUINN A. MORRIS, STEPHEN B. ROBINSON In Memory of John W. Neuberger Abstract. We consider the boundary value problem −∆u+ c(x)u = αm(x)u+ − βm(x)u− + f(x, u), x ∈ Ω, ∂u ∂η + σ(x)u = αρ(x)u+ − βρ(x)u− + g(x, u), x ∈ ∂Ω, where (α, β) ∈ R2, c,m ∈ L∞(Ω), σ, ρ ∈ L∞(∂Ω), and the nonlinearities f and g are bounded continuous functions. We study the asymmetric (Fuc̆ik) spectrum with weights, and prove existence theorems for nonlinear perturba- tions of this spectrum for both the resonance and non-resonance cases. For the resonance case, we provide a sufficient condition, the so-called generalized Landesman-Lazer condition, for the solvability. The proofs are based on vari- ational methods and rely strongly on the variational characterization of the spectrum. 1. Introduction We consider the partial differential equation −∆u+ c(x)u = m(x)[αu+ − βu−], x ∈ Ω, ∂u ∂η + σ(x)u = ρ(x)[αu+ − βu−], x ∈ ∂Ω, (1.1) where ∆z := ∇ ·∇z, ∂ ∂η is the outward normal derivative, (α, β) ∈ R2 are parame- ters, and c,m ∈ L∞(Ω), σ, ρ ∈ L∞(∂Ω) with c(x),m(x) ≥ 0 almost everywhere in Ω, σ(x), ρ(x) ≥ 0 almost everywhere in ∂Ω,∫ c(x) dx+ ∮ σ(x) dx > 0 and ∫ m(x) dx+ ∮ ρ(x) dx > 0, where ∫ denotes the (volume) integral on Ω and ∮ denotes the (surface) integral on ∂Ω. Throughout this paper we assume that Ω is a bounded domain in Rn (n ≥ 2) with smooth boundary ∂Ω. 2020 Mathematics Subject Classification. 35P30, 35J60, 35J66. Key words and phrases. Fuc̆ik Spectrum; resonance; nonlinear boundary condition. ©2023 This work is licensed under a CC BY 4.0 license. Published March 27, 2023. 209 210 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 We are interested in the Fuc̆ik spectrum, namely, Σ := {(α, β) ∈ R2 : (1.1) has a non-trivial solution} and our first main result provides a variational characterization of a curve in Σ. As an application of the variational characterization we consider −∆u+ c(x)u = m(x)[αu+ − βu−] + f(x, u) in Ω, ∂u ∂ν + σ(x)u = ρ(x)[αu+ − βu−] + g(x, u) on ∂Ω, (1.2) i.e. a nonlinear perturbation of (1.1). We assume nonlinearities of the form f(x, u) := m(x)f̃(u) and g(x, u) := ρ(x)g̃(u), where f̃ , g̃ : R → R are bounded continuous functions. We prove existence theorems for the non-resonance case, (α, β) 6∈ Σ, and the resonance case, (α, β) ∈ Σ. For the resonance case we assume a generalized Landesman-Lazer condition as in [4] and [8]. Our methods are built on the results in [7, 4, 8]. Section 2 provides a brief summary of the function spaces and the variational setting. In Section 3, we prove the variational characterization of a curve in Σ using a Hilbert space reduction method as in [2, 4, 8]. Section 4 contains the existence theorem for the non- resonance case. Section 5 contains the existence theorem for the resonance case. 2. Characterization of the Fuc̆ik Spectrum 2.1. Variational preliminaries. Define the (c, σ)-inner product 〈·, ·〉(c,σ) : H1(Ω)× H1(Ω)→ R by 〈u, v〉(c,σ) = ∫ ∇u · ∇v + ∫ c(x)uv + ∮ σ(x)uv, with the associated norm denoted by ‖u‖(c,σ). This norm is equivalent to the standard H1(Ω)-norm. Set 〈u, v〉(m,ρ) = ∫ m(x)uv + ∮ ρ(x)uv, ‖u‖2(m,ρ) := ∫ m(x)u2 + ∮ ρ(x)u2, for u, v ∈ H1(Ω). Let V(m,ρ) = {u ∈ H1(Ω) : ‖u‖(m,ρ) = 0}, and let H1 (m,ρ) = V ⊥(m,ρ) be the orthogonal complement with respect to the (c, σ) inner product. Then H1(Ω) = H1 (m,ρ) ⊕ V(m,ρ) (see [7]) and it further follows that H1 (m,ρ) and V(m,ρ) are (m, ρ) orthogonal. We will also make use of the norm ‖ · ‖(c,σ) on H1(Ω) and ‖ · ‖(m,ρ) on H1 (m,ρ). We also provide an alternate characterization of V(m,ρ) from [7]: taking Ω(m) := {x ∈ Ω : m(x) > 0} and ∂Ω(ρ) := {x ∈ ∂Ω : ρ(x) > 0}, we have V(m,ρ) = {u ∈ H1(Ω) : u = 0 a.e in Ω(m) and Γu = 0 a.e in ∂Ω(ρ)}, (2.1) where Γ is the trace operator on ∂Ω. Consider the functional J : H1(Ω)→ R defined by Jα,β(u) = 1 2 [‖u‖2(c,σ) − α‖u +‖2(m,ρ) − β‖u −‖2(m,ρ)]. (2.2) Then J ′α,β(u) · v = 〈u, v〉(c,σ) − α〈u, v〉(m,ρ) + (β − α)〈u−, v〉(m,ρ). (2.3) We note that critical points of Jα,β are weak solutions of (1.1). We begin with a lemma on the nature of the Fuc̆ik eigenfunctions. EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 211 Lemma 2.1. Every Fuc̆ik eigenfunction ψ is contained in H1 (m,ρ). Proof. Assume to the contrary that ψ = u + v, where u ∈ H1 (m,ρ), v ∈ V(m,ρ), and v is nonzero on a set of positive measure. Then 0 = J ′α,β(ψ) · v = 〈u+ v, v〉(c,σ) − α〈u+ v, v〉(m,ρ) + (β − α)〈(u+ v)−, v〉(m,ρ) = ‖v‖2(c,σ), because of the alternate characterization of V(m,ρ) in (2.1). Hence, v = 0 a.e. which contradicts our assumption. So all Fuc̆ik eigenfunctions are in H1 (m,ρ). � 2.2. Trivial curves. It is known (see [7]) that the problem −∆u+ c(x)u = µm(x)u, x ∈ Ω, ∂u ∂η + σ(x)u = µρ(x)u, x ∈ ∂Ω, has a simple first eigenvalue µ1 > 0 with associated eigenfunction φ1 which is of one sign in Ω. Therefore φ+ 1 = φ1 and φ−1 = 0, so that −∆φ1 + c(x)φ1 = µ1m(x)φ1 = m(x)[µ1φ + 1 − βφ − 1 ] for any β ∈ R, and similarly ∂φ1 ∂η + σ(x)φ1 = ρ(x)[µ1φ + 1 − βφ − 1 ] for any β ∈ R. Therefore C0 := {(µ1, β) : β ∈ R} ⊂ Σ. A similar argument will show that C′0 := {(α, µ1) : α ∈ R} ⊂ Σ. The curves C0 and C′0 are depicted in Figure 1. α β C1 (µ2, µ2) (µ1, µ1) C′0 C0 Figure 1. Trivial and first Fuc̆ik curves 212 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 Lemma 2.2. Σ ∩ {(α, β) ∈ R2 : α < µ1 or β < µ1} ∩ ( C0 ∪ C′0 )C = ∅ Proof. Let α < µ1 and β 6= µ1. Assume that (α, β) ∈ Σ and let ψ ∈ H1 (m,ρ) be a Fuc̆ik eigenfunction associated to (α, β). Then 0 = J ′α,β(ψ) · ψ+ = ‖ψ+‖2(c,σ) − α‖ψ +‖2(m,ρ) ≥ (µ1 − α)‖ψ+‖2(m,ρ). So, since α < µ1, it follows that ‖ψ+‖2(m,ρ) = 0, which implies that ψ+ = 0 almost everywhere. Hence, ψ = −ψ−, and hence ψ is a non-positive Steklov eigenfunction. So ψ satisfies −∆ψ + c(x)ψ = m(x)βψ; x ∈ Ω, ∂ψ ∂η + σ(x)ψ = ρ(x)βψ; x ∈ ∂Ω. But if ψ is a non-sign-changing solution, then β = µ1, a contradiction. Hence (α, β) 6∈ Σ. If β < µ1 and α 6= µ1, the argument proceeds similarly by examining the expres- sion J ′α,β(ψ) · ψ−. � 2.3. Higher curves. In what follows, we will consider the case µk < α < µk+1 and α < β. If (α, β) ∈ Σ, then (β, α) ∈ Σ, and therefore, it suffices to only consider the case α < β. The first curve C1 is depicted in Figure 1. We split the space H1 (m,ρ) = Xk ⊕ Yk where Xk = span{φ1, φ2, . . . , φk} and Yk = span{φk+1, φk+2, . . .}. We further define Y = Yk⊕V(m,ρ) so thatH1 = Xk⊕Y . We begin with an estimate which will be crucial for several lemmas later. Lemma 2.3. Let (αi, βi) ∈ R2 for i = 1, 2 satisfy the previous hypotheses, and let si = βi − αi. Let xi ∈ Xk and yi ∈ Y for i = 1, 2. Then( J ′α2,β2 (x2 + y2)− J ′α1,β1 (x1 + y1) ) · (x2 − x1) ≤ −δ‖x2 − x1‖2(c,σ) + s2 ( ‖x2 − x1‖(m,ρ) + ‖y2 − y1‖(m,ρ) ) ‖y2 − y1‖(m,ρ) + |α2 − α1|‖x1‖(m,ρ)‖x2 − x1‖(m,ρ) + |s2 − s1|‖x1 + x2‖(m,ρ)‖x2 − x1‖(m,ρ), where δ = α2 µk − 1. Proof. First we show that J ′αi,βi(xi + yi)(x2 − x1) = 〈xi + yi, x2 − x1〉(c,σ) − αi〈xi + yi, x2 − x1〉(m,ρ) + si〈(xi + yi) − , x2 − x1〉(m,ρ) = 〈xi, x2 − x1〉(c,σ) − αi〈xi, x2 − x1〉(m,ρ) + si〈(xi + yi) − , x2 − x1〉(m,ρ), EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 213 by the (c, σ)- and (m, ρ)-orthogonality of Xk and Y . Then utilizing the previous expression, we have( J ′α2,β2 (x2 + y2)− J ′α1,β1 (x1 + y1) ) · (x2 − x1) = ‖x2 − x1‖2(c,σ) − 〈α2x2 − α1x1, x2 − x1〉(m,ρ) + 〈s2(x2 + y2)− − s1(x1 + y1)−, x2 − x1〉(m,ρ) = ‖x2 − x1‖2(c,σ) − α2‖x2 − x1‖2(m,ρ) − (α2 − α1)〈x1, x2 − x1〉(m,ρ) + s2〈(x2 + y2)− − (x1 + y1)−, x2 − x1〉(m,ρ) + (s2 − s1)〈(x1 + y1)−, x2 − x1〉(m,ρ) (2.4) By the variational characterization of µk and the definition of Xk, we have that ‖x2 − x1‖2(c,σ) − α2‖x2 − x1‖2(m,ρ) ≤ ( 1− α2 µk ) ‖x2 − x1‖2(c,σ) = −δ‖x2 − x1‖2(c,σ). Since f(t) = t− is non-increasing, we have that v−1 −v − 2 and v1−v2 have opposite sign for all v1, v2 ∈ H1. Furthermore, |f(t2)− f(t1)| ≤ |t2 − t1|. Hence, s2〈(x2 + y2)− − (x1 + y1)−, x2 − x1〉(m,ρ) = s2〈(x2 + y2)− − (x1 + y1)−, (x2 + y2)− (x1 + y1)〉(m,ρ) + s2〈(x2 + y2)− − (x1 + y1)−, y1 − y2〉(m,ρ) ≤ s2〈|(x2 + y2)− − (x1 + y1)−|, |y1 − y2|〉(m,ρ) ≤ s2〈|(x2 + y2)− (x1 + y1)|, |y1 − y2|〉(m,ρ) = s2〈|(x2 − x1)− (y2 − y1)|, |y1 − y2|〉(m,ρ) ≤ s2 ( ‖x2 − x1‖(m,ρ) + ‖y2 − y1‖(m,ρ) ) ‖y2 − y1‖(m,ρ). Using Hölder’s inequality, we estimate the remaining two terms as∣∣(α2 − α1)〈x, x2 − x1〉(m,ρ) ∣∣ ≤ |α2 − α1|‖x1‖(m,ρ)‖x2 − x1‖(m,ρ) and∣∣(s2 − s1)〈(x1 + y1)−, x2 − x1〉(m,ρ) ∣∣ ≤ |s2 − s1|‖x1 + x2‖(m,ρ)‖x2 − x1‖(m,ρ). Combining the previous estimates into (2.4) yields the desired result. � Lemma 2.4. For a fixed y ∈ Y , Jα,β(x + y) is concave on Xk and moreover, for any x1, x2 ∈ Xk,( J ′α,β(x2 + y)− J ′α,β(x1 + y) ) · (x2 − x1) ≤ −δ‖x2 − x1‖2(c,σ). Proof. Take y1 = y2 = y, α1 = α2 = α, and β1 = β2 = β in Lemma 2.3. Then s1 = s2 = β − α, and the inequality reduces to( J ′α,β(x2 + y)− J ′α,β(x1 + y) ) · (x2 − x1) ≤ −δ‖x2 − x1‖2(c,σ), as desired. If we further set x1 = 0 and x2 = x, we observe that( J ′α,β(x+ y)− J ′α,β(y) ) · x ≤ −δ‖x‖2(c,σ), and hence Jα,β(x+ y) is concave on Xk. � 214 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 Since Jα,β is concave on Xk, for any fixed y ∈ Y , we define rα,β(y) ∈ Xk to be the unique maximizer of Jα,β restricted to Xk + y, namely Jα,β(rα,β(y) + y) = max x∈Xk Jα,β(x+ y). (2.5) We now establish several properties of the function rα,β(y) which will be helpful later. Lemma 2.5. The function rα,β(y) is homogeneous (i.e., rα,β(ty) = trα,β(y) for all t ≥ 0.) Proof. For any t > 0, we have that Jα,β(rα,β(ty)+ty) ≥ Jα,β(x+ty) for all x ∈ Xk. By the homogeneity of Jα,β , we therefore have Jα,β ( rα,β(ty) t + y ) ≥ Jα,β ( x t + y ) for all x ∈ Xk. But this implies that rα,β(ty) t = rα,β(y), and therefore rα,β is homogeneous. For t = 0, we need only to show rα,β(0) = 0. Clearly Jα,β(0) = 0. We will show that Jα,β(x) < 0 for all x ∈ Xk\{0}, and therefore, 0 = maxx∈Xk Jα,β(x) = rα,β(0). Since ‖x‖2(c,σ) ≤ µk‖x‖ 2 (m,ρ) (see [7, Corollary 2.2]), we observe that Jα,β(x) = 1 2 ( ‖x‖2(c,σ) − α‖x +‖2(m,ρ) − β‖x −‖2(m,ρ) ) ≤ 1 2 ( µk‖x‖2(m,ρ) − α‖x +‖2(m,ρ) − β‖x −‖2(m,ρ) ) ≤ 1 2 ( µk‖x‖2(m,ρ) − α‖x +‖2(m,ρ) − α‖x −‖2(m,ρ) ) ≤ 1 2 (µk − α)‖x‖2(m,ρ) < 0, for all x ∈ Xk\{0}. Hence, rα,β(0) = 0, and therefore rα,β(ty) = trα,β(y) for all t ≥ 0 and y ∈ Y . � Lemma 2.6. For each y 6= 0, rα,β(y) + y changes sign. Proof. Suppose to the contrary that u = rα,β(y) + y is nonnegative and strictly positive on some set of positive measure, say Ω1. Since u ∈ H1, u = v+ Σknφn for kn = 〈u, φn〉(m,ρ) and some v ∈ V(m,ρ). We note that k1 = 〈u, φ1〉(m,ρ) > 0 since φ1 > 0 on Ω1 and u 6∈ V(m,ρ). Since φ1 ∈ Xk ∀k ≥ 1 and rα,β(y) maximizes Jα,β on Xk, we have 0 = J ′α,β(u) · φ1 = 〈u, φ1〉(c,σ) − α〈u, φ1〉(m,ρ) + (β − α)〈u−, φ1〉(m,ρ) = 〈u, φ1〉(c,σ) − α〈u, φ1〉(m,ρ) = k1‖φ1‖2(c,σ) − αk1‖φ1‖2(m,ρ) = k1(µ1 − α)‖φ1‖2(m,ρ) < 0, which is a contradiction. An identical contradiction can be reached in the case that we assume u is nonpositive and strictly negative on some set of positive measure. Hence, rα,β(y) + y must change sign for y 6= 0. � To be precise about the result of the following lemma, let us consider the space Ỹ , which is the set of points in Y endowed with the topology generated by ‖·‖(m,ρ). EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 215 Lemma 2.7. rα,β(y) is locally Lipschitz continuous as a function of R2 × Ỹ into Xk. Proof. Take xi = rαi,βi(yi). By the definition of rαi,βi(yi), we have that( J ′α2,β2 (rα2,β2 (y2) + y2)− J ′α1,β1 (rα1,β1 (y1) ) · (rα2,β2 (y2)− rα1,β1 (y1)) = 0, and hence by Lemma 2.3, we have that δ‖rα2,β2(y2)− rα1,β1(y1)‖2(c,σ) ≤ s2 ( ‖rα2,β2 (y2)− rα1,β1 (y1)‖(m,ρ) + ‖y2 − y1‖(m,ρ) ) ‖y2 − y1‖(m,ρ) + |α2 − α1|‖rα1,β1 (y1)‖(m,ρ)‖rα2,β2 (y2)− rα1,β1 (y1)‖(m,ρ) + |s2 − s1|‖rα1,β1 (y1) + rα2,β2 (y2)‖(m,ρ)‖rα2,β2 (y2)− rα1,β1 (y1)‖(m,ρ), Applying a Poincare-type inequality (see Corollary 2.2 in [7]), we obtain δ‖rα2,β2 (y2)− rα1,β1 (y1)‖2(c,σ) ≤ s2 ( 1 µ1 ‖rα2,β2 (y2)− rα1,β1 (y1)‖(c,σ) + ‖y2 − y1‖(m,ρ) ) ‖y2 − y1‖(m,ρ) + |α2 − α1|‖rα1,β1(y1)‖(m,ρ) 1 µ1 ‖rα2,β2(y2)− rα1,β1(y1)‖(c,σ) + |s2 − s1|‖rα1,β1 (y1) + rα2,β2 (y2)‖(m,ρ) 1 µ1 ‖rα2,β2 (y2)− rα1,β1 (y1)‖(c,σ), (2.6) Now, for a given y1, let c1 = ‖rα1,β1 (y1)‖(m,ρ), c2 = ‖rα1,β1 (y1) + y1‖(m,ρ), and z = ‖rα2,β2 (y2)− rα1,β1 (y1)‖(c,σ). It follows from (2.6) that δz2 ≤ ( ‖y2 − y1‖(m,ρ) + c1|α2 − α1|+ c2|s2 − s1| ) 1 µ1 z + ‖y2 − y1‖2(m,ρ) Taking γ := ( ‖y2 − y1‖(m,ρ) + c1|α2 − α1|+ c2|s2 − s1| ) , we observe that ‖y2 − y1‖(m,ρ) ≤ γ, and therefore, δz2 ≤ γ µ1 z + γ2. Therefore, z ≤ C(δ)γ, and the lemma is proven. � Note that in the case α1 = α2 = α and β1 = β2 = β, γ is independent of c1 and c2. Therefore, since C(δ) is also independent of y1 and y2, we have the following corollary. Corollary 2.8. For a given α and β, rα,β : Ỹ → Xk is globally Lipschitz continu- ous. Lemma 2.9. There exists a C > 0 such that ‖rα,β(y)‖(c,σ) ≤ C‖y‖(m,ρ). Proof. Suppose y2 = y and y1 = 0 are fixed and further suppose that α1 = α2 = α and β1 = β2 = β. Then x2 = rα2,β2 (y2) = rα,β(y) and x1 = rα1,β1 (y1) = rα,β(0) = 0. Then (2.6) reduces to δ‖rα,β(y)‖2c,σ ≤ ( 1 µ1 ‖rα,β(y)‖(c,σ) + ‖y‖(m,ρ) ) ‖y‖(m,ρ). We may solve this inequality to observe that δ‖rα,β(y)‖c,σ ≤ C(δ)‖y‖(m,ρ) where C(δ) = 1 2µ1δ + √ 1 δ + 1 4µ2 1δ 2 > 0. Note that C is a decreasing function of δ, and 216 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 therefore, if α− µk = ε > 0, then we can choose δ = ε µk < α µk − 1 = δ such that δ‖rα,β(y)‖c,σ ≤ δ‖rα,β(y)‖c,σ ≤ C(δ)‖y‖(m,ρ) ≤ C(δ)‖y‖(m,ρ) . � The function rα,β(y) also satisfies a compactness condition, namely: Lemma 2.10. Let {(αn, βn)} be a bounded sequence in R2 satisfying µk < αn < µk+1 and αn < βn and let {yn} be a bounded sequence in Y . Then there exist subsequences, again called, {(αn, βn)} and {yn} such that (αn, βn)→ (α, β) in R2, yn ⇀ y in Y , yn → y in Ỹ , and rαn,βn(yn)→ rα,β(y) in Xk. Proof. There exists a subsequence of {(αn, βn)} converging to (α, β) in R2 by the Bolzano-Weierstrauss Theorem, call it again {(αn, βn)}. Then there exists a subse- quence of {yn} converging weakly to y in Y by the fact that H1(Ω) is reflexive. We again call that subsequence {yn}. Finally, by the Rellich-Kondrachov Theorem and the compactness of the trace operator given m ∈ L∞(Ω) and ρ ∈ L∞(∂Ω), there ex- ists a subsequence of {yn} converging strongly to y in Ỹ , called again {yn}. Hence, by the continuity of rα,β established in Lemma 2.7, we have rαn,βn(yn) → rα,β(y) in X. � Finally, we observe the following property of rα,β . Lemma 2.11. If u ∈ H1(Ω) is a critical point of Jα,β, then u = rα,β(y) + y for some y ∈ Y . Proof. Since u is a critical point of Jα,β , J ′α,β(u) · v = 0 for all v ∈ H1(Ω). Since H1(Ω) = Xk ⊕ Y , we may write u = x + y where x ∈ Xk and y ∈ Y . We observe that 0 = J ′α,β(u) · x = J ′α,β(x) · x, showing that x is a critical point of Jα,β on the set y+X. But Jα,β is strictly concave on y+X and its unique maximizer is defined as rα,β(y). So x = rα,β(y) and hence u = rα,β(y) + y. � 3. Reducing the functional Motivated by Lemma 2.11, we now define the restricted functional J̃α,β : Y → R by J̃α,β(y) := Jα,β(rα,β(y) + y). We begin by establishing some properties of this new functional. Lemma 3.1. The functional J̃α,β ∈ C1(Y,R) and J̃ ′α,β(y) = J ′α,β(rα,β(y) + y) for all y ∈ Y . Proof. We will establish this claim by showing that J̃α,β(y2)− J̃α,β(y1) = J ′α,β(rα,β(y1) + y1) · (y2 − y1) + o ( ‖y2 − y1‖(m,ρ) ) . EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 217 In addition to showing that J̃α,β ∈ C1(Y,R), this will also establish that J̃ ′α,β(y) = J ′α,β(rα,β(y) + y). First, note that J̃α,β(y2)− J̃α,β(y1) = Jα,β(rα,β(y2) + y2)− Jα,β(rα,β(y1) + y1) ≤ Jα,β(rα,β(y2) + y2)− Jα,β(rα,β(y2) + y1) = J ′α,β(rα,β(y2) + y1) · (y2 − y1) + o ( ‖y2 − y1‖(m,ρ) ) = J ′α,β(rα,β(y1) + y1) · (y2 − y1) + ( J ′α,β(rα,β(y2) + y1) − J ′α,β(rα,β(y1) + y1) ) · (y2 − y1) + o ( ‖y2 − y1‖(m,ρ) ) (3.1) by the maximizing property of rα,β , the Lipschitz continuity of rα,β , and the dif- ferentiability of Jα,β . By the continuity of rα,β and J ′α,β , we note that( J ′α,β(rα,β(y2) + y1)− J ′α,β(rα,β(y1) + y1) ) · (y2 − y1) = o ( ‖y2 − y1‖(m,ρ) ) , and hence (3.1) reduces to J̃α,β(y2)− J̃α,β(y1) ≤ J ′α,β(rα,β(y1) + y1) · (y2 − y1) + o ( ‖y2 − y1‖(m,ρ) ) . A similar argument will show that J̃α,β(y2)− J̃α,β(y1) ≥ J ′α,β(rα,β(y1) + y1) · (y2 − y1) + o ( ‖y2 − y1‖(m,ρ) ) , and hence the claim is proven. � Remark 3.2. If we knew rα,β to be differentiable, this result would be a simple consequence of the chain rule. However, in general, this is not the case. Given that we have now established that J̃α,β ∈ C1(Y,R), we may improve upon Lemma 2.11. Lemma 3.3. The element y ∈ Y is a critical point of J̃α,β if and only if rα,β(y)+y is a critical point of Jα,β. Proof. First, assume that rα,β(y)+y is a critical point of Jα,β . Then J ′α,β(rα,β(y)+ y) · v = 0 for all v ∈ H1(Ω) (and in particular, for all v ∈ Y ). By Lemma 3.1, this implies that J̃ ′α,β(y) · v = 0 for all v ∈ Y , and y is a critical point of J̃α,β . Now, assume that y is a critical point of J̃α,β . As before, we then have that J ′α,β(rα,β(y)+y) ·v = 0 for all v ∈ Y . However, since rα,β(y) maximizes Jα,β(x+y) for all x ∈ Xk, we also have that J ′α,β(rα,β(y) + y) · x = 0 for all x ∈ Xk. Hence, since H1(Ω) = Xk ⊕ Y , we have J ′α,β(rα,β(y) + y) · w = 0 for all w ∈ H1(Ω). � Now, we observe a homogeneity property of J̃α,β . Lemma 3.4. The functional J̃α,β(ty) = t2J̃α,β(y) for all t ≥ 0 and y ∈ Y . The result follows immediately from the homogeneity of Jα,β and the homo- geneity of rα,β from Lemma 2.5. An important consequence of this lemma easily follows. Lemma 3.5. If y ∈ Y is a critical point of J̃α,β, then J̃α,β(y) = 0. Proof. Differentiating the identity J̃α,β(ty) = t2J̃α,β(y) with respect to t, we find that J̃ ′α,β(ty) · y = 2tJ̃α,β(y). Setting t = 1, the result immediately follows. � 218 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 As with Jα,β , it will occasionally be helpful to think of J̃α,β as a function on R2 × Y , which we denote J̃(α, β, y). Lemma 3.6. For each fixed y 6= 0, the functional J̃(α, β, y) := J̃α,β(y) is strictly decreasing in α and β. Proof. Assume that α1 ≤ α2 and β1 ≤ β2, with at least one of these inequalities strict. Then J̃(α2, β2, y) = J(α2, β2, r(α2, β2, y) + y) = 1 2 [ ‖r(α2, β2, y) + y‖2(c,σ) − α2‖ ( r(α2, β2, y) + y )+‖2(m,ρ) − β2‖ ( r(α2, β2, y) + y )−‖2(m,ρ)]. (3.2) Since r(α2, β2, y) + y is sign-changing for y 6= 0 by Lemma 2.6, it follows that ‖ (r(α2, β2, y) + y) + ‖(m,ρ)‖ (r(α2, β2, y) + y) − ‖(m,ρ) > 0, and hence, since at least one of the inequalities α1 ≤ α2 and β1 ≤ β2 is strict, we have from (3.2) that J̃(α2, β2, y) < 1 2 [ ‖r(α2, β2, y) + y‖2(c,σ) − α1‖ (r(α2, β2, y) + y) + ‖2(m,ρ) − β1‖ (r(α2, β2, y) + y) − ‖2(m,ρ) ] = J(α1, β1, r(α2, β2, y) + y). (3.3) But recalling the maximizing property of rα,β (see (2.5)), we must have that J(α1, β1, r(α2, β2, y) + y) ≤ J(α1, β1, r(α1, β1, y) + y) = J̃(α1, β1, y). (3.4) Combining (3.3) and (3.4) gives the desired result, that J̃(α2, β2, y) < J̃(α1, β1, y) for each y 6= 0. � Lemma 3.7. Given any K > 0, there exists C > 0 such that∣∣J̃(α2, β2, x)− J̃(α1, β1, x) ∣∣ ≤ C (|α2 − α1|+ |β2 − β1|) on R(K) := {(α1, α2, β1, β2, y) ∈ R4 ×H1(Ω) : max{|α1|, |α2|, |β1|, |β2|, ‖y‖(c,σ)} ≤ K}. Proof. First, we establish that the functional J is uniformly Lipschitz in α, β, and x. Note that |J(α2, β2, x)− J(α1, β1, x)| = 1 2 ∣∣∣(α2 − α1)‖x+‖2(m,ρ) + (β2 − β1)‖x−‖2(m,ρ) ∣∣∣ ≤ 1 2µ1 ‖x‖2(c,σ) (|α2 − α1|+ |β2 − β1|) ≤ 1 2µ1 K (|α2 − α1|+ |β2 − β1|) . Hence J is uniformly Lipschitz in α, β on R(K). Since J ∈ C1(H1(Ω);R), it is also uniformly Lipschitz in x on R(K) Recall from Lemma 2.7 that rα,β is locally Lipschitz in α, β. Therefore, we have that rα,β is uniformly Lipschitz in α, β on R(K). Therefore, J̃(α, β, x) = J(α, β, r(α, β, x) + x) is a composition of uniformly Lipschitz functions, and hence the claim follows. � EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 219 3.1. Minimizing in Y . By Lemma 3.3, we know that searching for critical points of Jα,β on H is equivalent to searching for critical points of J̃α,β on Y . Further, since J̃α,β is homogeneous, it is sufficient to search for critical points on the (m, ρ)-unit sphere in Y , namely SY := {y ∈ Y : ‖y‖(m,ρ) = 1}. Since we assume m, ρ ∈ L∞(Ω), SY is weakly closed in H1(Ω); that is, for any sequence {yn} ⊂ SY with yn ⇀ y in H1(Ω), we have yn → y in Y and y ∈ SY . First we note several properties of J̃α,β when restricted to SY . Lemma 3.8. J̃α,β attains a global minimum on SY . Proof. First, note that 2J̃α,β(y) = 2Jα,β(rα,β(y) + y) = ‖rα,β(y) + y‖2(c,σ) − α‖(rα,β(y) + y)+‖2(m,ρ) − β‖(rα,β(y) + y)−‖2(m,ρ) > −α‖rα,β(y) + y‖2(m,ρ) − β‖rα,β(y) + y‖2(m,ρ) > −2β‖rα,β(y) + y‖2(m,ρ). Since rα,β(y) ∈ X and y ∈ Y , 〈rα,β(y), y〉(m,ρ) = 0 and hence 2J̃α,β(y) > −2β‖rα,β(y) + y‖2(m,ρ) = −2β(‖r(y)‖2(m,ρ) + ‖y‖2(m,ρ)) > −2β( 1 µ1 ‖r(y)‖2(c,σ) + ‖y‖2(m,ρ)) > −2β( C2 µ1 ‖y‖2(m,ρ) + ‖y‖2(m,ρ)) ≥ −k(‖y‖2(m,ρ)) where k = 2β(C 2 µ1 + 1) by Corollary 2.2 (a) in [7] and Lemma 2.9. Now, take M = infSY J̃α,β(y) > −∞ (since ‖y‖(m,ρ) = 1 on SY ) and choose {yn} ⊂ SY to be a minimizing sequence, that is, J̃α,β(yn) → M . So J̃α,β(yn) is bounded. We wish to show that ‖yn‖(c,σ) is also bounded. Note first that since J̃α,β(yn) is bounded and 2J̃α,β(yn) = 2Jα,β(rα,β(yn) + yn) = ‖rα,β(yn) + yn‖2(c,σ) − α‖(rα,β(yn) + yn)+‖2(m,ρ) − β‖(rα,β(yn) + yn)−‖2(m,ρ) = ‖rα,β(yn)‖2(c,σ) + ‖yn‖2(c,σ) − α‖(rα,β(yn) + yn)+‖2(m,ρ) − β‖(rα,β(yn) + yn)−‖2(m,ρ). (3.5) We wish to show that all terms other than ‖yn‖2(c,σ) in (3.5) are bounded, and hence ‖yn‖(c,σ) must also be bounded. We recall that ‖rα,β(yn)‖2(c,σ) < C2‖yn‖2(m,ρ) = C2 by Lemma 2.9 and by the fact that {yn} ⊂ SY . We also note that ‖(rα,β(yn) + yn)+‖2(m,ρ) ≤ ‖rα,β(yn) + yn‖2(m,ρ) ≤ ‖rα,β(yn)‖2(m,ρ) + ‖yn‖2(m,ρ) 220 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 ≤ 1 µ1 ‖rα,β(yn)‖2(c,σ) + ‖yn‖2(m,ρ) ≤ C2 µ1 ‖yn‖2(m,ρ) + ‖yn‖2(m,ρ) = C2 µ1 + 1, by [7, Corollary 2.2(a)], Lemma 2.9, and the fact that {yn} ⊂ SY . An identical argument will show that ‖(rα,β(yn) + yn)−‖2(m,ρ) ≤ C2 µ1 + 1. Hence, we have shown via equation (3.5) that ‖yn‖(c,σ) is also bounded. Hence, by Lemma 2.10, we may choose a subsequence, call it again {yn}, with yn (c,σ)−−−⇀ y0, yn (m,ρ)−−−→ y0 with ‖y0‖(m,ρ) = 1, and rα,β(yn) (c,σ)−−−→ rα,β(y0). So taking the limit inferior of both sides of (3.5) as n→∞, we see that 2M = lim inf n→∞ 2J̃α,β(yn) = ‖rα,β(y0)‖2(c,σ) + lim inf n→∞ ‖yn‖2(c,σ) − α‖(rα,β(y0) + y0)+‖2(m,ρ) − β‖(rα,β(y0) + y0)−‖2(m,ρ) ≥ ‖rα,β(y0)‖2(c,σ) + ‖y0‖2(c,σ) − α‖(rα,β(y0) + y0)+‖2(m,ρ) − β‖(rα,β(y0) + y0)−‖2(m,ρ) = 2J̃α,β(y0), by the weak lower semicontinuity of the (c, σ) norm. But then M ≥ J̃α,β(y0) with y0 ∈ SY and hence we must have J̃α,β(y0) = M as desired. � Lemma 3.9. y0 is a nontrivial critical point of J̃α,β if and only if y0 ‖y0‖(m,ρ) is a critical point of J̃α,β restricted to SY and J̃α,β(y0) = 0. Proof. If y0 is a nontrivial critical point of J̃α,β , then by Lemma 3.5, J̃α,β(y0) = 0. Furthermore, since y0 is a critical point of J̃α,β , we may differentiate both sides of the equation in Lemma 3.4 with respect to y and set t = 1/‖y0‖(m,ρ) to see that 0 = 1 ‖y0‖(m,ρ) J̃ ′α,β(y0) · y = J̃ ′α,β ( y0 ‖y0‖(m,ρ) ) · y holds for all y ∈ Y . So in particular, it holds for y ∈ SY and the forward direction is established. Now, let y0/‖y0‖(m,ρ) be a critical point of J̃α,β restricted to SY and let J̃α,β(y0) = 0. Then as in the previous case, we have 0 = J̃ ′α,β ( y0 ‖y0‖(m,ρ) ) · y = 1 ‖y0‖(m,ρ) J̃ ′α,β(y0) · y for all y ∈ SY . But note that, for any ŷ ∈ Y , we may write ŷ = ty for some y ∈ SY . So, J̃ ′α,β(y0) · ŷ = J̃ ′α,β(y0) · (ty) = tJ̃ ′α,β(y0) · y = 0. So y0 is a critical point of J̃α,β as desired. � Lemma 3.10. A function u ∈ H1(Ω) is a nontrivial critical point of Jα,β if and only if u = rα,β(y0) + y0 where y0/‖y0‖(m,ρ) is a critical point of J̃α,β restricted to SY and J̃α,β(y0) = 0. EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 221 The above lemma follows from combining Lemma 3.3 and Lemma 3.9. We now define M(α, β) = miny∈SY Jα,β(y). Lemma 3.11. The function M(α, β) is Lipschitz continuous and is strictly de- creasing as a function of both α and β. Moreover, M(α, α) > 0. Proof. Let (α1, β1) and (α2, β2) be points in the plane, and let y1 and y2 be the corresponding minimizers on SY (i.e. M(αk, βk) = J̃αk,βk(yk) for k = 1, 2). Let uij = rαi,βi(yj) + yj for i, j = 1, 2. Then M(αi, βi) = J̃αi,βi(yi) ≤ J̃αi,βi(yj) = Jαi,βi(rαi,βi(yj) + yj) = Jαj ,βj (rαi,βi(yj) + yj) + 1 2 (αj − αi)‖u+ ij‖ 2 (m,ρ) + 1 2 (βj − βi)‖u−ij‖ 2 (m,ρ) ≤ Jαj ,βj (rαj ,βj (yj) + yj) + 1 2 (αj − αi)‖u+ ij‖ 2 (m,ρ) + 1 2 (βj − βi)‖u−ij‖ 2 (m,ρ) = M(αj , βj) + 1 2 (αj − αi)‖u+ ij‖ 2 (m,ρ) + 1 2 (βj − βi)‖u−ij‖ 2 (m,ρ) (3.6) by the minimizing property of yi and the maximizing property of rαj ,βj . This inequality holds in the case i = 1 and j = 2, as well as the case i = 2 and j = 1. Hence |M(α2, β2)−M(α1, β1)| ≤ c(|α2 − α1|+ |β2 − β1|) where c = 1 2 max{‖u12‖2(m,ρ), ‖u21‖2(m,ρ)}. Note that if α2 ≥ α1 and β2 ≥ β1 with at least one of the inequalities strict, then M(α2, β2) < M(α1, β1) by taking i = 2 and j = 1 in (3.6). This follows from the fact that uij must be sign-changing by Lemma 2.6. In the case α = β, for every w ∈ H1(Ω), we may write w = u + v where u ∈ H1 (m,ρ) and v ∈ V(m,ρ) have 2Jα,β(w) = 2Jα,α(w) = ‖u+ v‖2(c,σ) − α‖u+ v‖2(m,ρ) = ‖u‖2(c,σ) + ‖v‖2(c,σ) − α‖u‖ 2 (m,ρ) + ‖v‖2(m,ρ) = ‖u‖2(c,σ) − α‖u‖ 2 (m,ρ) + ‖v‖2(c,σ) = ∞∑ i=1 (µi − α)|ci|2 + ‖v‖2(c,σ) by [7, Theorem 2.1(iii)] where ci = 1 µi 〈u, φi〉(c,σ) = 〈u, φi〉(m,ρ). Since µk < α < µk+1, we note that the coefficients (µi − α) are negative for i ≤ k, positive for i ≥ k + 1, and are increasing in i. Writing u = x + y where x ∈ Xk and y ∈ Yk, we note that, if we maximize in the Xk direction, the maximum occurs when ci = 0 for all i ≤ k since (µi − α) are negative for i ≤ k. In other words, 222 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 rα,α(y) ≡ 0. Therefore, 2J̃α,α(y) = 2Jα,α(y) = ∞∑ i=k+1 (µi − α)|ci|2 for all y ∈ Yk. (3.7) We now wish to show that M(α, α) = infy∈SY J̃α,α(y) > 0. Taking f(ck+1, ck+2, . . .) = ∞∑ i=k+1 (µi − α)|ci|2, g(ck+1, ck+2, . . .) = ∞∑ i=k+1 |ci|2, we apply the method of Lagrange multipliers to find the critical points of f subject to the constraint g(ck+1, ck+2, . . .) = 1. Setting ∇f = λ∇g, we obtain 2(µi−α)ci = 2λci for i ≥ k+1. Hence, critical points occur when cj = ±1 for some j ≥ k+1 and ci = 0 for all i 6= j (corresponding to the Lagrange multiplier λ = µj − α). Since the coefficients (µi − α) are positive for i ≥ k + 1 and increasing, the minimizing choice occurs when ck+1 = ±1 and ci = 0 for all i > k+ 1. Hence, the minimizer is y = ±φk+1 and M(α, α) = J̃α,α(φk+1) + ‖v‖2(c,σ) = 1 2 (µk+1−α) + ‖v‖2(c,σ) > 0. � Not only M(α, α) > 0, we can also make an additional estimate which will later help in establishing bounds for the Fuc̆ik spectrum. Lemma 3.12. M(α, µk+1) > 0. Proof. Let y ∈ SY and let y = z + v where z ∈ Yk and v ∈ V(m,ρ). Then, by the maximizing property J̃α,µk+1 = Jα,µk+1 (rα,µk+1 (y) + y) ≥ Jα,µk+1 (y) = 1 2 ( ‖y‖2(c,σ) − α‖y +‖2(m,ρ) − µk+1‖y−‖2(m,ρ) ) ≥ 1 2 ( ‖y‖2(c,σ) − µk+1‖y‖2(m,ρ) ) = 1 2 ( ‖z‖2(c,σ) − µk+1‖z‖2(m,ρ) + ‖v‖2(c,σ) ) = 1 2 ∞∑ i=k+1 (µi − µk+1)|ci|2 + ‖v‖2(c,σ) ≥ 0. We note that, of the last two inequalities above, at least one must be strict. If the last inequality is in fact an equality, then ck+1 = 1 and ci = 0 for all i > k+ 1. But this would imply that y = ±φk+1, in which case y+ is nontrivial, and the previous inequality was strict. So M(α, µk+1) > 0. � 3.2. Variational characterization of the Fuc̆ik spectrum. All of the previous lemmas lead to the following theorem. Theorem 3.13. Let µk < α < µk+1. Then one of the following is true: (1) M(α, β) > 0 for all β ≥ α, which implies that (α, β) 6∈ Σ. (2) There is a unique β(α) > µk+1 such that M(α, β(α)) = 0, which implies that (α, β(α)) ∈ Σ but (α, β) 6∈ Σ for all α ≤ β < β(α). Lemma 3.14. The curve (α, β(α)) is Lipschitz continuous, strictly decreasing, and contains the point (µk+1, µk+1). EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 223 Proof. Consider two points (α1, β1), (α2, β2) ∈ Σ with α2 > α1. Let yi ∈ SY be a minimizer of J̃αi,βi(y). Then J̃αi,βi(yi) = 0 and J̃αi,βi(y) ≥ 0 for all y ∈ SY , and therefore by the homogeneity of J̃αi,βi , J̃αi,βi(y) ≥ 0 for all y ∈ Y . Let ui = rαi,βi(yi) + yi. Since (α1, β1) ∈ Σ, we have 0 = M(α1, β1) = 2J̃α1,β1(y1) = 2Jα1,β1 (u1) = ‖u1‖2(c,σ) − α1‖u+ 1 ‖2(m,ρ) − β1‖u−1 ‖2(m,ρ) > ‖u1‖2(c,σ) − α2‖u+ 1 ‖2(m,ρ) − β1‖u−1 ‖2(m,ρ) = 2Jα2,β1 ( rα1,β1(y1) ) ≤M(α2, β1) where we note that the last inequality is strict since α2 > α1 and u−1 is nontrivial by Lemma 2.6. Since M(α, β) is strictly decreasing in β by Lemma 3.11 and M(α2, β2) = 0, we must have β2 < β1, which shows that β(α) is strictly decreasing as desired. Now, we consider M(α2, β1) ≤ Jα2,β1(u2) = Jα2,β1(u2)− Jα2,β2(u2) = 1 2 (β2 − β1)‖u−2 ‖2(m,ρ), since Jα2,β2 (u2) = 0. Hence M(α2, β1) ≤ 1 2 (β2 − β1)‖u−2 ‖2(m,ρ) < 0 since β1 > β2. Thus, we may rearrange the inequality to observe that |β2 − β1| = β1 − β2 ≤ 2 1 ‖u−2 ‖2(m,ρ) (−M(α2, β1)) = 2 1 ‖u−2 ‖2(m,ρ) |M(α2, β1)| = 2 1 ‖u−2 ‖2(m,ρ) |M(α2, β1)−M(α1, β1)| ≤ 2c 1 ‖u−2 ‖2(m,ρ) |α2 − α1|, by the fact that M(α1, β1) = 0 and the Lipschitz estimate for M(α, β) from Lemma 3.11. Hence, β(α) is Lipschitz continuous as desired. � 4. Nonresonance problem We are interested in the existence of weak solutions of (1.2) where (α, β) ∈ R2 is such that µk < α < µk+1 and α ≤ β < β(α). Since we consider a fixed k in this section, we set X = Xk for notational convenience. By the characterization of the Fuc̆ik Spectrum, we know that (α, β) 6∈ Σ. All properties of f, g,m, ρ, c, and σ are as outlined in Section 1. Consider the functional associated with (1.2) defined by Iα,β := I : H1(Ω)→ R; with I(u) = Jα,β(u)− [ ∫ F (x, u) + ∮ G(x, u) ] , (4.1) 224 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 where ∫ denotes the (volume) integral on Ω, ∮ denotes the (surface) integral on ∂Ω, and F (x, u) = ∫ u 0 m(x)f̃(ξ)dξ, G(x, u) = ∫ u 0 ρ(x)g̃(ξ)dξ, and Jα,β(u) is defined in (2.2). Then I ′(u) · v = J ′α,β(u)v − [ ∫ f(x, u)v + ∮ g(x, u)v ] for all v ∈ H1(Ω). So, a critical point of I is a weak solution of (1.2). Theorem 4.1. Assume that µk < α < µk+1, α ≤ β < β(α), the nonlinearities f and g are bounded continuous functions, and m ∈ L∞(Ω) and ρ ∈ L∞(∂Ω), then problem (1.2) has at least one weak solution. We will use a variational argument to prove Theorem 4.1. To do so, we first prove some lemmas which will be needed in the sequel. The first lemma shows the last two terms in I have at most linear growth. Lemma 4.2. There is a positive constant κ such that∣∣ ∫ F (x, u) + ∮ G(x, u) ∣∣ ≤ κ‖u‖(m,ρ) for all u ∈ H1(Ω), (4.2) where κ is independent of u. Proof. Since f̃ and g̃ are bounded then there exist constant C1 and C2 such that |f̃(u)| ≤ C1 and |g̃(u)| ≤ C2. Therefore, |F (x, u)| ≤ C1m(x)|u| and |G(x, u)| ≤ C2ρ(x)|u|. Using these estimates, Hölder inequality, and the fact that m is bounded, we obtain that∣∣ ∫ F (x, u) ∣∣ ≤ ∫ C1m(x)|u| ≤ C̃1 (∫ (m(x)|u|)2 )1/2 ≤ κ1 ∫ m(x)u2 ≤ κ1‖u‖(m,ρ), where κ1 = C̃1‖m‖∞. Similarly, ∣∣ ∮ G(x, u) ∣∣ ≤ κ2‖u‖(m,ρ). Thus,∣∣∣ ∫ F (x, u) + ∮ G(x, u) ∣∣∣ ≤ κ‖u‖(m,ρ) for all u ∈ H1(Ω). � The next lemma shows the geometry of I. Lemma 4.3. The functional I is such that (1) I(u)→ −∞ as ‖u‖(c,σ) →∞, for u ∈ X; that is, I is anti-coercive on X. (2) I is bounded below when restricted to Y, where Y := {rα,β(y) + y : y ∈ Y }. Proof. We shall first prove that I is anti-coercive when restricted to X. Using the fact that ‖x‖2(c,σ) ≤ µk‖x‖ 2 (m,ρ) for all x ∈ X, and α ≤ β < β(α), it follows that Jα,β(x) = 1 2 [‖x‖2(c,σ) − α‖x +‖2(m,ρ) − β‖x −‖2(m,ρ)] ≤ 1 2 [‖x‖2(c,σ) − α‖x‖ 2 (m,ρ)] ≤ 1 2 [‖x‖2(c,σ) − α µk ‖x‖2(c,σ)] = 1 2 (1− α µk )‖x‖2(c,σ) Then using Lemma 4.2, we obtain I(x) ≤ −η‖x‖2(c,σ) + κ‖x‖(m,ρ) + C, EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 225 where η = 1 2 ( αλk − 1) > 0. Since µ1‖x‖(m,ρ) ≤ ‖x‖(c,σ) (see [7, Corrolary 2.18]), then I(x) ≤ −η‖x‖2(c,σ) + κ µ1 ‖x‖(c,σ) + C. So, I(u) → −∞ as ‖u‖(c,σ) →∞ for u ∈ X. Thus I is anti-coercive on X. Now, we shall prove that I is bounded below when restricted to Y. By the assumption β < β(α) and Theorem 3.13 , it follows that miny∈SY J̃α,β(y) = M(α, β) and M(α, β) > 0. Then for y 6= 0 and y ∈ Y , we have that Jα,β(rα,β(y) + y) = J̃α,β(y) = ‖y‖2(m,ρ)J̃α,β ( y ‖y‖(m,ρ) ) ≥ ε‖y‖2(m,ρ) where ε = M(α, β). Since rα,β is Lipschitz continuous, as in Lemma 2.7, we have that ‖rα,β(y)‖(c,σ) ≤ C‖y‖(m,ρ) for some C > 0, and we see that I(u) ≥ ε‖y‖2(m,ρ) − κ‖u‖(m,ρ) = ε‖y‖2(m,ρ) − κ‖rα,β(y) + y‖(m,ρ) ≥ ε‖y‖2(m,ρ) − κ ( ‖rα,β(y)‖(m,ρ) + ‖y‖(m,ρ) ) ≥ ε‖y‖2(m,ρ) − κ (C + 1) ‖y‖(m,ρ) (4.3) Thus, I is bounded below when restricted to Y. � As a consequence of the results above, there exists some R > 0 sufficiently large such that sup {x∈X:‖x‖(c,σ)=R} I(x) < inf u∈Y I(u). The next lemma shows the linking property of I. Let BR = {x ∈ X : ‖x‖(c,σ) ≤ R} and ∂BR = {x ∈ X : ‖x‖(c,σ) = R}. Lemma 4.4. Let γ : BR ⊂ X → H1(Ω) be a continuous function such that γ|∂BR(x) = x. Then γ(BR) ∩ Y 6= ∅. Proof. Let x ∈ BR and let write γ(x) = γX(x) + γY (x), where γX(x) ∈ X and γY (x) ∈ Y . One can see that for all x ∈ ∂BR, γX(x) = x and γY (x) = 0. To show that γ(BR) ∩ Y 6= ∅, it suffices to show that there is an x ∈ BR such that γX(x) = rα,β(γY (x)). Let H : BR → X defined by H(x) = γX(x) − rα,β(γY (x)). We shall show that there is x ∈ BR such that H(x) = 0. Notice that H is continuous and for all x ∈ ∂BR, H(x) = x 6= 0. Therefore, the Brouwer degree deg(H,BR, 0) is well defined. Now, consider the homotopy h(x, t) = tH(x) + (1− t)x. Note that for x ∈ ∂BR we have h(x, t) = tx + (1 − t)x = x 6= 0. Hence, deg(H,BR, 0) = deg(Id,BR, 0) = 1, where Id represents the identity map. Thus H(x) = 0 has a solution in BR. � To prove Theorem 4.1 using the saddle point theorem of Rabinowitz, it suf- fices first to show I satisfies the Palais-Smale condition (PS) which builds some compactness into the functional I. Lemma 4.5. I satisfies the Palais-Smale condition (PS). Proof. Let {un} be a sequence in H1(Ω) such that {I(un)} is bounded and I ′(un)→ 0 as n→∞. We will show that {un} has a convergent subsequence. In view of the assumptions on the nonlinearities f and g, it suffices to first show that the sequence {un} is bounded with respect to ‖ · ‖(m,ρ), that is, there exists a constant K such 226 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 that ‖u‖(m,ρ) < K. Suppose by contradiction that ‖un‖(m,ρ) →∞ as n→∞. Let vn = un/‖un‖(m,ρ). Then I(un) ‖un‖2(m,ρ) = Jα,β(vn)− 1 ‖un‖2(m,ρ) [ ∫ F (x, un) + ∮ G(x, un) ] . Taking the limit, we have that I(un)/‖un‖2(m,ρ) → 0 since {I(un)} is bounded, and 1 ‖un‖2(m,ρ) [ ∫ F (x, un) + ∮ G(x, un)] → 0 because of the estimate (4.2). Hence, I(un)/‖un‖2(m,ρ) and [ ∫ F (x, un) + ∮ G(x, un)]/‖un‖2(m,ρ) are bounded. Also note that ‖v±n ‖(m,ρ) ≤ 1. From the definition of Jα,β it follows that ‖vn‖(c,σ) is bounded. Using the fact that H1(Ω) is reflexive, the Sobolev compact embedding, and the continuity of the trace operator, we obtain that there exists a subsequence vn that converges weakly to v0 in H1(Ω) and that converges strongly to v0 in L2(Ω) (also in L2(∂Ω)). Since m and ρ are bounded functions and using the continuity of the norm ‖ · ‖(m,ρ), we obtain that ‖vn‖(m,ρ) → ‖v0‖(m,ρ). Thus, ‖v0‖(m,ρ) = 1 since ‖vn‖(m,ρ) = 1. Now, for any w ∈ H1(Ω), I ′(un) ‖un‖(m,ρ) · w = 〈vn, w〉(c,σ) − α〈v+ n , w〉(m,ρ) + β〈v−n , w〉(m,ρ) − 1 ‖un‖(m,ρ) [ ∫ m(x)f̃(un)w + ∮ ρ(x)g̃(un)w] Using the boundedness of the nonlinearities f and g, and of the weights m and ρ, we have that 1 ‖un‖(m,ρ) [ ∫ m(x)f̃(un)w + ∮ ρ(x)g̃(un)w]→ 0. Since vn → v0 strongly in L2(Ω) and L2(∂Ω), 〈v+ n , w〉(m,ρ) → 〈v+ 0 , w〉(m,ρ) and 〈v−n , w〉(m,ρ) → 〈v−0 , w〉(m,ρ). By the weak convergence of vn in H1(Ω), we see that 〈vn, w〉(c,σ) → 〈v0, w〉(c,σ). We also note that I′(un) ‖un‖(m,ρ) · w → 0 as n→∞. Hence, 0 = 〈v0, w〉(c,σ) − α〈v+ 0 , 〉(m,ρ) − β〈v − 0 , w〉(m,ρ) for all w ∈ H1(Ω). Thus v0 is a nontrivial weak solution of (1.1). This leads to a contradiction since (α, β) 6∈ Σ. Thus, {un} are bounded with respect to ‖ · ‖(m,ρ). Let us analyze carefully the functional I. I(un) = 1 2 [‖un‖2(c,σ) − α‖u + n ‖2(m,ρ) − β‖u − n ‖2(m,ρ)]− [ ∫ F (x, un) + ∮ G(x, un) ] Since I(un) is bounded and using the fact that {un} is bounded with respect to ‖· ‖(m,ρ) and the estimate (4.2), we have that ‖u+ n ‖(m,ρ), ‖u−n ‖(m,ρ), and [ ∫ F (x, un)+∮ G(x, un)] are all bounded. Thus, ‖un‖(c,σ) must be bounded. Therefore there exists a subsequence un that converges weakly to u in H1(Ω) and converges strongly to u in L2(Ω) (also in L2(∂Ω)). Since m and ρ are bounded functions and using the continuity of the norm ‖ · ‖(m,ρ), we obtain that ‖un‖(m,ρ) → ‖u‖(m,ρ). Now, consider I ′(un).(un − u) = 〈un, (un − u)〉(c,σ) − α〈u+ n , (un − u)〉(m,ρ) + β〈u−n , (un − u)〉(m,ρ) − [ ∫ m(x)f̃(un)(un − u) + ∮ ρ(x)g̃(un)(un − u) ] EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 227 By the assumption I ′(un) → 0 in ( H1(Ω) )∗ it follows that I ′(un).(un − u) → 0. Since ‖u+ n ‖(m,ρ), ‖u−n ‖(m,ρ), f̃ , and g̃ are all bounded, and ‖un‖(m,ρ) → ‖u‖(m,ρ), we have that − α〈u+ n , (un − u)〉(m,ρ) + β〈u−n , (un − u)〉(m,ρ) − [ ∫ m(x)f(un)(un − u) + ∮ ρ(x)g(un)(un − u) ] → 0. Therefore 〈un, (un − u)〉(c,σ) → 0. Thus ‖un‖2(c,σ) − 〈un, u〉(c,σ) → 0. Since un converges weakly to u in H1(Ω), we have that 〈un, u〉(c,σ) → ‖u‖2(c,σ). Hence, ‖un‖(c,σ) → ‖u‖(c,σ). Thus, un (c,σ)−→ u in H1(Ω). � Proof of Theorem 4.1. The functional I satisfies the Palais-Smale condition due to Lemma 4.5, and by Lemma 4.4, I satisfies the linking property. Set c = inf γ∈Γ sup u∈BR∩X I(γ(u)), where R is a sufficiently large constant, and Γ = {γ ∈ C ( BR ∩ X;H1(Ω) ) : γ|∂BR∩X(x) = x}. Then by the Saddle Point Theorem [9], it follows that c is a critical value of I. Thus problem (1.2) has a weak solution. � 5. Resonance Problem In this section, we again assume (α, β) ∈ R2 with µk < α < µk+1. However we now assume that β = β(α) so that (α, β) ∈ Σ by the characterization in Theorem 3.13. Again for notational convenience we take X = Xk. Most arguments from the previous section still apply, with the exception of Lemmas 4.3 part 2 and 4.5; namely that I is bounded from below and that I satisfies (PS). This is not surprising, as the case that (α, β) ∈ Σ corresponds to the case µ = µk+1 in the Fredholm alternative. We expect in such cases that solutions only exist when a generalized orthogonality condition is met. In establishing existence of solutions in the non-resonance cases, we will need a generalized Landesman-Lazer condition, namely Definition 5.1. If for any sequence {un} ⊂ H1(Ω) such that ‖un‖(m,ρ) →∞ and un ‖un‖(m,ρ) (m,ρ)−−−→ ψ, where ψ is a Fuc̆ik eigenfunction associated with (α, β), we have lim n→∞ ∫ Ω F (x, un) + ∫ ∂Ω G(x, un) = −∞. (5.1) Theorem 5.2. Assume that µk < α < µk+1, β = β(α), the nonlinearities f and g are bounded continuous functions, and m ∈ L∞(Ω) and ρ ∈ L∞(∂Ω), then problem (1.2) has at least one weak solution provided that condition (5.1) holds. Lemma 5.3. If (5.1) is satisfied, then I is bounded below on Y. Proof. Suppose to the contrary that there exists a sequence {un} ⊂ Y with I(un)→ −∞. Since {un} ⊂ Y, we may write un = rα,β(yn) + yn. Taking inequality (4.3) with ε = M(α, β) = 0, we observe that since I(un) → −∞, we must have ‖un‖(m,ρ) →∞. But since ‖un‖2(m,ρ) = ‖rα,β(yn) + yn‖2(m,ρ) = ‖rα,β(yn)‖2(m,ρ) + ‖yn‖2(m,ρ) 228 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 ≤ 1 µ1 ‖rα,β(yn)‖2(c,σ) + ‖yn‖2(m,ρ) ≤ C2 µ1 ‖yn‖2(m,ρ) + ‖yn‖2(m,ρ) = (C2 µ1 + 1 ) ‖yn‖2(m,ρ), we observe that ‖yn‖(m,ρ) →∞. Thus, no subsequence of {un} lies in a set of the form {u ∈ Y : u = rα,β(y) + y, J̃α,β(y) ≥ c‖y‖} for some c > 0, since if such a subsequence existed, this would imply I(u) → ∞ by (4.3). Therefore, J̃α,β(yn) → 0 and by the homogeneity of J̃α,β , J̃α,β ( yn ‖yn‖(m,ρ) ) → 0. Since M(α, β) = 0, {yn/‖yn‖(m,ρ)} ⊂ SY is a minimizing sequence of J̃α,β . As in the proof of Lemma 3.8, this implies that ‖yn/‖yn‖(m,ρ)‖(c,σ) is bounded. Therefore, there exists y ∈ SY such that yn/‖yn‖(m,ρ) (c,σ)−−−⇀ y and yn/‖yn‖(m,ρ) (m,ρ)−−−→ y. Using the the homogeneity of rα,β , we have that un = ‖yn‖(m,ρ) ( rα,β ( yn ‖yn‖(m,ρ) ) + yn ‖yn‖(m,ρ) ) , and ‖un‖(m,ρ) = ‖yn‖(m,ρ)‖rα,β ( yn ‖yn‖(m,ρ) ) + yn ‖yn‖(m,ρ) ‖(m,ρ). Therefore, un ‖un‖(m,ρ) = rα,β ( yn ‖yn‖(m,ρ) ) + yn ‖yn‖(m,ρ) ‖rα,β ( yn ‖yn‖(m,ρ) ) + yn ‖yn‖(m,ρ) ‖(m,ρ) . Therefore, un ‖un‖(m,ρ) (c,σ)−−−⇀ rα,β(y) + y ‖rα,β(y) + y‖(m,ρ) , un ‖un‖(m,ρ) (m,ρ)−−−→ rα,β(y) + y ‖rα,β(y) + y‖(m,ρ) . Setting rα,β(y) + y ‖rα,β(y) + y‖(m,ρ) = φ . we notice that ‖φ‖(m,ρ) = 1 and J̃α,β(φ) = 0 = M(α, β(α)). Therefore φ is a nontrivial eigenfunction associated to (α, β(α)). Since un ‖un‖(m,ρ) (m,ρ)−−−→ φ and (5.1) is satisfied, we have that limn→∞ (∫ F (x, un) + ∮ G(x, un) ) = −∞. It follows that I(un)→∞, a contradiction. The lemma is proved. � Lemma 5.4. If (5.1) is satisfied, then I satisfies (PS). Proof. The first part of the proof is identical to the proof in Lemma 4.5. Sup- pose {un} ⊂ H1(Ω) is a sequence such that I(un) is bounded, I ′(un) → 0, and ‖un‖(m,ρ) → ∞. As before, we take vn = un ‖un‖(m,ρ) and by an identical argument we show that vn (c,σ)−−−⇀ v and vn (m,ρ)−−−→ v with ‖v‖(m,ρ) = 1 and v a Fuc̆ik eigen- function associated with (α, β). In the previous case this was a contradiction, but since (α, β) ∈ Σ in this case, we have not yet reached a contradiction, and further argument is needed. EJDE-2023/SI/02 FUC̆IK SPECTRUM WITH WEIGHTS 229 Write un = xn + yn = x̃n + rα,β(yn) + yn. Then I ′(un) · x̃n = J ′α,β(un) · x̃n − ∫ f(x, un)x̃n + ∮ g(x, un)x̃n = J ′α,β(x̃n + rα,β(yn) + yn) · x̃n − ∫ f(x, un)x̃n + ∮ g(x, un)x̃n = ( J ′α,β(x̃n + rα,β(yn) + yn)− J ′α,β(rα,β(yn) + yn) ) · x̃n − ∫ f(x, un)x̃n + ∮ g(x, un)x̃n ≤ −δ‖x̃n‖2(c,σ) − ∫ f(x, un)x̃n + ∮ g(x, un)x̃n ≤ −δµ1‖x̃n‖2(m,ρ) − ∫ f(x, un)x̃n + ∮ g(x, un)x̃n by the fact that J ′α,β(rα,β(y) + y) · x = 0 for all x ∈ X and Lemma 2.3. Dividing the inequality through by ‖x̃n‖(m,ρ) gives I ′(un) · x̃n ‖x̃n‖(m,ρ) ≤ −δµ1‖x̃n‖(m,ρ) − ∫ f(x, un)x̃n + ∮ g(x, un) x̃n ‖x̃n‖(m,ρ) , but since I ′(un) → 0 and f, g are bounded, we obtain that ‖x̃n‖(m,ρ) is also bounded. It now follows that J ′α,β(un) · x̃n = I ′(un) · x̃n + ∫ f(x, un)x̃n + ∮ g(x, un)x̃n must also be bounded. Now, let h(t) = Jα,β(rα,β(yn) + yn + tx̃n). Then h′(t) = J ′α,β(rα,β(yn) + yn + tx̃n) · x̃, and we observe that h′(0) = 0 (by the definition of rα,β) and h′(t) is decreasing by the strict concavity of Jα,β on yn+X. By the Mean Value Theorem, h(1) − h(0) = h′(c) for some c ∈ (0, 1), and hence h(1) − h(0) ≥ h′(1) since h′ is decreasing. So, Jα,β(rα,β(yn) + yn + x̃n)− Jα,β(rα,β(yn) + yn) ≥ J ′α,β(rα,β(yn) + yn + x̃) · x̃n. Since Jα,β(rα,β(yn) + yn) ≥ ‖yn‖m,ρM(α, β) = 0, we then have that Jα,β(un) ≥ J ′α,β(un) · x̃n. Therefore, I(un) = Jα,β(un)− [ ∫ F (x, un) + ∮ G(x, un) ] ≥ J ′α,β(un) · x̃n − [ ∫ F (x, un) + ∮ G(x, un) ] . However, J ′α,β(un) · x̃n is bounded and [ ∫ F (x, un) + ∮ G(x, un)] → −∞ by (5.1), which contradicts the boundedness of I(un). Hence ‖un‖(m,ρ) is bounded, and the proof proceeds as in Lemma 4.5. � By a straightforward application of the Saddle Point Theorem, we now conclude that there exists a solution to the resonance problem. Acknowledgements. The authors dedicate this article to the memory of Dr. John W. Neuberger for his contributions to nonlinear analysis and to the analysis of partial differential equations. 230 N. MAVINGA, Q. A. MORRIS, S. B. ROBINSON EJDE/SI/02 References [1] G. Auchmuty; Steklov Eigenproblems and the Representation of Solutions of Elliptic Bound- ary Value Problems, Numer. Funct. Anal. Optim. 25, Nos. 3&4 (2004), 321-348. [2] A. Castro, C. Chang; A variational characterization of the Fuč́ık Spectrum and applications, Revista Colombiana de Matematicas, 44:1, 23-40,2010 [3] M. Cuesta, J.-P. Gossez; A variational approach to nonresonance with respect to the Fučik spectrum, Nonlinear Anal. 19 (1992), 487-500. [4] P. Drabék, S. B. Robinson; On the solvability of resonance problems with respect to the Fučik spectrum, J. Math. Anal. App. 418 (2014), 884-905. [5] D. G. de Figueiredo, J.-P. Gossez; On the first curve of the Fučik spectrum, Differential Integral Equations 7 (1994), 1285-1302. [6] L. C. Evans; Partial Differential Equations, American Mathematical Society, Providence, RI, 1998. [7] N. Mavinga; Generalized eigenproblem and nonlinear elliptic equations with nonlinear bound- ary conditions, in Proceedings of the Royal Society Edinburgh, 142A (2012), 137-153. [8] Q. Morris, S. B. Robinson; A Landesman-Lazer condition for the boundary-value problem −u′′ = au+−bu−+g(u) with periodic boundary conditions, Electronic Journal of Differential Equations, Conference 20 (2013), 103-117. [9] P. H. Rabinowitz; Minimax methods in critical point theory with application to differential equations, CBMS Regional Conf. Series in Math., no. 65, Amer. Math. Soc., Providence, RI, 1986. [10] M. W. Steklov; Sur les problèmes fondamentaux de la physique mathématique, Ann. Sci. Ecole Norm. Sup. 19 (1902), 455-490. Nsoki Mavinga Department of Mathematics & Statistics, Swarthmore College, Swarthmore, PA 19081- 1390, USA Email address: nmaving1@swarthmore.edu Quinn A. Morris Department of Mathematical Sciences, Appalachian State University, Boone, NC 28608, USA Email address: morrisqa@appstate.edu Stephen B. Robinson Department of Mathematics, Wake Forest University, Winston-Salem, NC 27109, USA Email address: sbr@wfu.edu 1. Introduction 2. Characterization of the Fuc̆ik Spectrum 2.1. Variational preliminaries 2.2. Trivial curves 2.3. Higher curves 3. Reducing the functional 3.1. Minimizing in Y 3.2. Variational characterization of the Fuc̆ik spectrum 4. Nonresonance problem 5. Resonance Problem Acknowledgements References