Generalized eigenvalue problem for an interface elliptic equation

Available online at www.sciencedirect.com ScienceDirect Journal of Differential Equations 390 (2024) 494–524 www.elsevier.com/locate/jde Generalized eigenvalue problem for an interface elliptic equation Braulio B.V. Maia a, Mónica Molina-Becerra b, Cristian Morales-Rodrigo c, Antonio Suárez c,∗ aUniversidade Federal Rural da Amazonia, Campus de Capitao-Poco, PA, Brazil bDpto. Matemática Aplicada II, Escuela Politécnica Superior, Univ. de Sevilla, Calle Virgen de Africa, 7, 41011, Sevilla, Spain cDpto. Ecuaciones Diferenciales y Análisis Numérico, IMUS, Fac. Matemáticas, Univ. de Sevilla, C/ Tarfia s/n, 41013, Sevilla, Spain Received 8 June 2023; revised 1 November 2023; accepted 10 February 2024 Available online 20 February 2024 Abstract In this paper we deal with an eigenvalue problem in an interface elliptic equation. We characterize the set of principal eigenvalues as a level set of a concave and regular function. As application, we study a problem arising in population dynamics. In these problems each species lives in a subdomain, and they interact in a common border, which acts as a geographical barrier; but unlike previous results, we consider the case of different growth rates in each subdomain. ©2024 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http://creativecommons .org /licenses /by -nc -nd /4 .0/). MSC: 35A16; 35B09; 35J15; 35J60 Keywords: Interface; Principal eigenvalue; Positive solutions *Corresponding author. E-mail addresses: [email protected] (B.B.V. Maia), [email protected] (M. Molina-Becerra), [email protected] (C. Morales-Rodrigo), [email protected] (A. Suárez). https://doi.org/10.1016/j.jde.2024.02.015 0022-0396/©2024 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http://creativecommons .org /licenses /by -nc -nd /4 .0/). B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Fig. 1. A possible configuration of the domain =1∪2∪. 1. Introduction Recently, the following semilinear interface problems have been analyzed ⎧ ⎨ ⎩ −ui=λfi(x, ui)in i,i=1,2, ∂νui=γi(u2−u1)on , ∂νu2=0on, (1) where is a bounded domain of IR Nwith =1∪2∪, with isubdomains, with internal interface  =∂1, and  =∂2\, νiis the outward normal to i, and we call ν:= ν1=−ν2(see Fig. 1where we have illustrated an example of ). In (1), uirepresents the density of a species inhabiting in i, and they interact on under the so called Kedem-Katchalsky conditions (see [7]), and it means that the flux is proportional to the jump of the function through (see [3], [4], [5], [6] and references therein). Here, fi: i×IR → IR a r e regular functions, γi>0 stands for the proportional coefficient of the jump and 1/λ, λ >0, is a real parameter representing the diffusion coefficient of the species, the same in both subdomains. It seems natural to consider two different diffusion coefficients, one for each species, that is, a problem as ⎧ ⎨ ⎩ −ui=λifi(x, ui)in i, ∂νui=γi(u2−u1)on , ∂νu2=0on, (2) with λi∈IR . Although mathematically it makes sense to consider λias a real parameter, its usual meaning is that 1/λiis the diffusion coefficient in i, λibeing a positive parameter in such a case. As a first step towards the study of (1), it is necessary to analyze the eigenvalue problem ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −ui=λmi(x)uiin i, ui>0ini, ∂νui=γi(u2−u1)on , ∂νu2=0on, (3) 495 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 where mi∈L∞(i), mi≡ 0in i. (3) has been analyzed in [9]in the self-adjoint case γ1=γ2. For that, the authors used variational arguments to prove the existence of principal eigenvalue as well as its main properties. The general case γ1= γ2was studied in [8]using a different argument. In [8], to study (3), the authors first analyze the problem ⎧ ⎨ ⎩ −ui+ci(x)ui=λuiin i, ∂νui=γi(u2−u1)on , ∂νu2=0on, (4) where ci∈L∞(i). They prove the existence of a unique principal eigenvalue of (4), denoted by 1(c1, c2). Hence, the study of (3)is equivalent to find the zeros of the map λ∈IR → f(λ):= 1(−λm1,−λm2). The main goal of this paper is to study the following generalized eigenvalue problem: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −ui=λimi(x)uiin i, ui>0ini, ∂νui=γi(u2−u1)on , ∂νu2=0on. (5) Motivated by [8], to study (5)we analyze the zeros of the map (λ1,λ 2)∈IR 2→ F(λ 1,λ 2):= 1(−λ1m1,−λ2m2), that is, we analyze the set C:= {(λ1,λ 2)∈IR 2:F(λ 1,λ 2)=0}. We show that Fis a regular function, concave and F(0, 0) =0. Hence, for instance, fixed λ1, there exist at most two values of λ2such that F(λ 1, λ2) =0. Moreover, due to the concavity of F, it is well known that the set {(λ1, λ2) ∈IR 2:F(λ 1, λ2) ≥0}is convex. In any case, the study of the set Cdepends strongly on the signs of mi. It is obvious that the case where the functions mihave a definite sign, for example they are positive, is a simpler case than the case where one or both of them change sign. We summarize the main results. Our first result deals with the case both minon-negative and non-trivial functions (see Fig. 2). Theorem 1.1. Assume that mi0in i, i=1, 2and define M0 i:= i\ {x∈i:mi(x) > 0} and assume that ∂M0 iis regular and M0 i⊆i.(6) Then, there exist positive values + i, i=1, 2such that: 496 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Fig. 2. Case m1and m2both non-negative and non-trivial verifying (6): we have represented in the plane λ1−λ2the curve F(λ 1, λ2) =0, as well as, the regions where Fis negative and positive. In this case, C={(λ1, λ2) :λ2=H(λ1)}. 1. Assume that λ1≥+ 1. Then, F(λ 1, λ2) <0for all λ2∈IR . 2. Assume that λ1< + 1. There exists a unique λ2:= H(λ1)such that F(λ 1, λ2) =0and F(λ 1,λ 2)<0for λ2>H(λ1), F(λ1,λ 2)>0for λ2<H(λ1). Moreover, the map λ1→ H(λ1)is continuous, decreasing, H(0) =0and lim λ1→−∞ H(λ1)=+ 2,lim λ1→+ 1 H(λ1)=−∞. The values + 1and + 2will be defined in Section 2. In the next result we analyze the case m1non-negative and non-trivial and m2changing sign (see Fig. 3). Theorem 1.2. Assume that m10in 1and verifies (6) and m2changes sign in 2. There exists λmax 1≥0such that: 1. If λ1>λ max 1, then F(λ 1, λ2) <0for all λ2∈IR . 2. If λ1=λmax 1, then there exists a unique λ2such that F(λ max 1, λ2) =0and F(λ max 1, λ2) <0 for all λ2∈IR \{λ2}. 3. For all λ1<λ max 1, there exist λ− 2=H−(λ1) <λ + 2=H+(λ1)such that F(λ 1,λ − 2)=F(λ 1,λ + 2)=0, and F(λ 1,λ 2)<0for λ2>H+(λ1)or λ2<H−(λ1), >0for λ2∈(H−(λ1), H+(λ1)). 497 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Fig. 3. Case m10 and verifying (6), m2changing sign and 2m2<0. Fig. 4. Case m10 and verifying (6), m2changing sign and 2m2>0(left)and2m2=0 (right). Moreover, the map λ1→ H+(λ1)(resp. H−(λ1)) is continuous, decreasing (resp. increasing) and lim λ1→−∞ H±(λ1)=± 2and lim λ1→λmax 1 H±(λ1)=λ2. 4. Finally, (a) If 2m2<0, then λmax 1>0and λ2>0. (b) If 2m2>0, then λmax 1>0and λ2<0. (c) If 2m2=0, then λmax 1=λ2=0. Remark 1.3. 1. In Fig. 4we have represented the cases m10in 1verifying (6), m2changes sign in 2, 2m2>0 and 2m2=0. 498 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Fig. 5. Cases michanging sign. 1m1<0 and: 2m2<0 (left), 2m2>0 (center) and 2m2=0 (right). 2. Of course, by symmetry, a similar result holds for m1changing sign in 1and m2non- negative, non-trivial and verifying (6). Finally, we deal with the case of both michanging sing. Theorem 1.4. Assume that michanges sign in i. Then, there exists a closed curve C⊂IR 2, such that F(λ 1, λ2) =0if and only if (λ1, λ2) ∈C. Moreover, F(λ 1,λ 2)>0if and only if (λ1,λ 2)∈int(C), and F(λ 1,λ 2)<0if and only if (λ1,λ 2)∈Ext(C). Remark 1.5. The form and structure of Cdepends strongly on the sign of the integrals of mi. In all the cases, (0, 0) ∈C. In the following result, we complete the above Theorem, see Fig. 5. Theorem 1.6. Assume that michanges sign for i=1, 2. There exist λmin 1≤0 ≤λmax 1such that 1. If λ1<λ min 1or λ1>λ max 1, then F(λ 1, λ2) <0for all λ2∈IR . 2. If λ1=λmax 1(resp. λ1=λmin 1) then there exists a unique λ2(resp. λ2) such that F(λ max 1, λ2) =0(resp. F(λ min 1, λ2) =0) and F(λ max 1, λ2) <0(resp. F(λ min 1, λ2) <0) for all λ2∈IR \{λ2}(resp. λ2∈IR \{λ2}). 3. If λ1∈(λmin 1, λmax 1)there exist unique λ− 2=H−(λ1) <λ + 2=H+(λ1)such that F(λ 1, λ± 2) = 0. Moreover, F(λ 1,λ 2)<0for λ2>H+(λ1)or λ2<H−(λ1), >0for λ2∈(H−(λ1), H+(λ1)). Finally, lim λ1→λmin 1 H±(λ1)=λ2,and lim λ1→λmax 1 H±(λ1)=λ2. 499 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 We apply these results to the nonlinear problem ⎧ ⎨ ⎩ −ui=λimi(x)ui−upi iin i, ∂νui=γi(u2−u1)on , ∂νu2=0on, (7) with λi∈IR , mi∈L∞(i)and pi>1. Again, we consider the general case λi∈IR and mi∈L∞(i)but of course from a biological point of view only some cases are interesting. For example, when m1≡1in 1, λ1represents the growth rate of the species u1, which could be positive or negative. When m1changes sign, only the case λ1>0 should be considered. We prove (Theorem 5.1) that (7) possesses a positive solution if and only if F(λ 1,λ 2)<0. Moreover, in such a case, the solution is the unique positive solution. Hence, we can give the following consequences: 1. Assume that m1and m2are non-negative and non-trivial functions. (a) For λ1large (λ1> + 1), there exists a positive solution for all λ2∈IR . (b) For λ1< + 1, there exists a value λ2=H(λ1)such that (7) possesses a positive solution for λ2>H(λ1). In both cases, for λ1>0we have that there exists a positive solution for negative growth rate (λ2) of u2. In the case without interface, this is not possible, that is, even if the population has negative growth in one part of the domain, the interface effect makes it possible for the species to persist throughout the domain. 2. Assume that m1is non-negative and non-trivial and m2changes sign. Then, if λ1is large, then there exists positive solution for all λ2∈IR . However, for λ1<λ max 1, then there exists positive solution for λ2<H−(λ1)or λ2>H+(λ1). 3. Assume that m1and m2change sign. There exist λmin 1<λ max 1such that for λ1>λ max 1or λ1<λ min 1, (7) possesses a positive solution for all λ2∈IR . However, for λ1∈(λmin 1, λmax 1), there exist H−(λ1) <H+(λ1)such that (7) possesses a positive solution only for λ2< H−(λ1)or λ2>H+(λ1). An outline of the paper is: in Section 2we include some preliminary results related to scalar eigenvalue problems. Section 3is devoted to show some general properties of F(λ 1, λ2). The main results concerning to the eigenvalue problem (5)are proved in Section 4. Finally, in Section 5, we analyze (7). 2. Preliminary results 2.1. Scalar eigenvalue problem In this section we recall some results concerning to scalar eigenvalue problems, see [2]for example. Here Gis a C2,α, α∈(0, 1), domain of IR N, ∂G =1∪2, where 1and 2are two disjoint open and closed subsets of ∂G and νis the outward unit normal vector field. For c∈L∞(G), 500 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 h ∈C(1), g∈C(2), we denote by σG 1(− +c; N+h, N+g) the principal eigenvalue of the problem ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −φ +c(x)φ =λφ in G, ∂φ ∂ν +hφ =0on1, ∂φ ∂ν +gφ =0on2, and by σG 1(− +c; N+h, D) that of the problem ⎧ ⎪ ⎨ ⎪ ⎩ −φ +c(x)φ =λφ in G, ∂φ ∂ν +hφ =0on1, φ=0on2. We will quote some important properties of σG 1(− +c; N+h, N+g) and σG 1(− +c; N+ h, D). We denote the boundary operator B(φ) =∂νφ+hφ =0on1, ∂νφ+gφ =0on2,or B(φ) =∂νφ+hφ =0on1, φ=0on2. Proposition 2.1. 1. The map c∈L∞(G) → σG 1(− +c; B)is continuous and increasing. 2. It holds that σG 1(−+c;N+h, N +g) < σG 1(−+c;N+h, D). 3. Assume that there exists a positive supersolution, that is, a positive function u∈W2,p(G), p>N, such that −u +c(x)u≥0in G,B(u) ≥0on ∂G, and some strict inequalities, then σG 1(−+c;B)>0. Now, we define μ(λ) := σG 1(−−λc;B), λ ∈IR . The main properties of μ(λ) are stated in the next result. Proposition 2.2. 1. Assume that c≡ 0in G. Then, λ ∈IR → μ(λ) is regular and concave. 501 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 2. Assume that c0in G, define C0:= G\ {x∈G:c(x) > 0},(8) and assume that C0⊆G. (9) The map λ → μ(λ) is decreasing and lim λ→+∞ μ(λ) =−∞,lim λ→−∞ μ(λ) =σC0 1(−;D). 3. Assume that cchanges sign, then lim λ→±∞ μ(λ) =−∞. Moreover, there exists λ0∈IR such that μ(λ0) =0, μ(λ) >0for λ <λ 0and μ(λ) <0for λ >λ 0. We can describe exactly the sign of the map μ(λ). Corollary 2.3. 1. Assume that c0and the set C0satisfies (9). Then, there exists a unique zero of the map μ(λ), we denote it by λ+ 1(G, c; B), and as consequence, μ(λ)⎧ ⎨ ⎩ >0if λ<λ + 1(G, c;B), =0if λ=λ+ 1(G, c;B), <0if λ>λ + 1(G, c;B). 2. Assume that cchanges sign. (a) If μ(λ0) <0, then μ(λ) <0for all λ ∈IR . (b) If μ(λ0) =0, then λ0is the unique zero of the map μ(λ). (c) If μ(λ0) >0, then there exist two zeros of the map μ(λ), we call them λ− 1(G, c; B) < λ0<λ + 1(G, c; B). As a consequence, μ(λ)⎧ ⎨ ⎩ >0if λ∈(λ− 1(G, c;B), λ+ 1(G, c;B)), =0if λ=λ− 1(G, c;B)or λ=λ+ 1(G, c;B), <0if λ<λ − 1(G, c;B)or λ>λ + 1(G, c;B). 2.2. Interface eigenvalue problem First, we fix some notations that will be used throughout the paper. For convenience, we write u =(u1, u2)with uidefined in iand similarly c =(c1, c2). In order to simplify the notation we write the boundary conditions as 502 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 which yields that F0=σM0 2 1(−; D) =σ0. A similar reasoning can be carried out when σ1 1(− −a∗m1; N+γ1) <σM0 2 1(−; D). This finishes the proof.  4. Proofs of the main results The main idea of the proof can be summarized as follows. Instead of looking for solutions of F(λ 1, λ2) =0in the general form (λ1, λ2), we look for solutions in the particular form λ2=μλ1, for all μ ∈IR . Hence, the following map plays an essential role in our study. Given μ ∈IR , we define fμ(λ1):= 1(−λ1m1,−λ1μm2)=F(λ 1,λ 1μ). In the following result we state that, for λ1= 0, it is equivalent to solve F(λ 1, λ2) =0to fμ(λ1) =0. Specifically, we have: Proposition 4.1. Assume that F(λ 0 1, λ0 2) =0and λ0 1= 0, then fμ0(λ0 1) =0for μ0=λ0 2/λ0 1. Conversely, if fμ0(λ0 1) =0then F(λ 0 1, λ0 2) =0for λ0 2=μ0λ0 1. In what follows, we explore the particular case λ1=0. Proposition 4.2. Assume that λ1=0and denote g(λ2):= F(0,λ 2)=1(0,−λ2m2). The map λ2→ g(λ2)is regular, concave, g(0) =0and sign(g(0)) =sign⎛ ⎜ ⎝− 2 m2⎞ ⎟ ⎠.(22) Moreover, 1. If m20in 2and M0 2verifies (6), then λ2→ g(λ2)is decreasing and lim λ2→+∞ g(λ2)=−∞ and lim λ2→−∞ g(λ2)=min{σ1 1(−;N+γ1), σ M0 2 1(−;D)}. In this case, g(λ2) >0for λ2<0and g(λ2) <0if λ2>0. 2. If m2changes sign in 2, then lim λ2→±∞ g(λ2)=−∞. Moreover, (a) If 2m2=0, then g(0) =0and λ2=0is the unique root of g(λ2) =0. As a consequence, g(λ2) <0for λ2= 0. 509 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 (b) If 2m2<0, then g(0) >0and there exists λ+ 2>0such that g(λ+ 2) =0. In this case, g(λ2)>0if λ2∈(0,λ + 2), <0if λ2<0or λ>λ + 2. (c) If 2m2>0, then g(0) <0and there exists λ− 2<0such that g(λ− 2) =0. Hence, g(λ2)>0if λ2∈(λ− 2,0), <0if λ2<λ − 2or λ2>0. Proof. To begin with, the regularity of gfollows by the regularity of the function F. On the other hand, by Proposition 3.1 follows that g(λ) is concave. It is obvious that g(0) =F(0, 0) = 1(0, 0) =0. On the other hand, taking m =(0, m2)in Proposition 3.17 in [8], we conclude (22). Finally, observe that by (13)we have g(λ2)<σ2 1(−−λ2m2;N+γ2,N), (23) whence we deduce that lim λ2→+∞ g(λ2) =−∞from Proposition 2.2 2. and 3. 1. Assume that m20in 2. In this case, gis decreasing. Moreover, by Proposition 3.5, taking an=0, we conclude that lim λ2→−∞ g(λ2)=min{σ1 1(−;N+γ1), σ M0 2 1(−;D)}. 2. Assume that m2changes sign. Then, using (23) and Proposition 2.2 3., we deduce that lim λ2→−∞ g(λ2)=−∞. Now, from the sign of g(0)in (22), we conclude the result.  In the next result, we study in detail the map λ1→ fμ(λ1). Proposition 4.3. Fix μ ∈IR . Then, λ1→ fμ(λ1)is regular, concave, fμ(0) =0and sign(f μ(0)) =−sign⎛ ⎜ ⎝γ2 1 m1+μγ1 2 m2⎞ ⎟ ⎠.(24) 1. If m10in 1, then lim λ1→+∞ fμ(λ1)=−∞ 510 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 2. If m1or m2changes sign, then lim λ1→±∞ fμ(λ1)=−∞. Proof. It is clear that fμ(0) =0. The regularity follows by the regularity of F, the concavity of fμ(λ1)follows by Proposition 2.5 4., and (24) follows taking m =(m1, μm2)in Proposition 3.17 in [8]. On the other hand, by (13)we get fμ(λ1)<min{σ1 1(−−λ1m1;N+γ1), σ 2 1(−−λ1μm2;N+γ2,N)}, and then lim λ1→+∞ fμ(λ1) =−∞, and if m1or m2changes sign, lim λ1→−∞ fμ(λ1) =−∞. For 2m2= 0, we define μ∗:= − γ2 1 m1 γ1 2 m2 ,(25) in such a way that f μ∗(0) =0. 4.1. Case mi0in i, i=1, 2 Observe that in this case μ∗<0. Proposition 4.4. Assume that mi0in iand M0 iverify (6)for i=1, 2. 1. If μ ≥0, the unique zero of fμ(λ1)is λ1=0. 2. If μ <0, μ = μ∗, there exists an unique λ1=h1(μ) = 0such that fμ(λ1) =0. Moreover, h1(μ) ⎧ ⎨ ⎩ <0if μ>μ ∗, =0if μ=μ∗, >0if μ<μ ∗. 3. The map μ ∈(−∞, 0) → h1(μ) is continuous and decreasing. Moreover, lim μ↑0h1(μ) =−∞,lim μ→−∞ h1(μ) =+ 1. Proof. 1. Assume that μ ≥0. Then, since λ1→ fμ(λ1)is decreasing and fμ(0) =0, the result follows. 511 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 2. Assume that μ <0. Recall that λ1→ fμ(λ1)is concave and fμ(0) =0. If μ >μ ∗then f μ(0) <0, and hence there exists a unique h1(μ) <0 such that fμ(h1(μ)) =0. Similarly, when μ <μ ∗there exists a unique h1(μ) >0 such that fμ(h1(μ)) =0. 3. We will show that μ → h1(μ) is decreasing. Take now μ1<μ 2<0. Observe that −μ1λ1> −μ2λ1if λ1>0 and −μ1λ1<−μ2λ1if λ1<0. Hence, we distinguish several cases: (a) Assume that μ∗≤μ1<μ 2<0. In this case, h1(μ2)and h1(μ1)are negative, and then we compare the functions fμ2and fμ1for negative values. Indeed, observe that fμ2(λ1) >f μ1(λ1)for λ1<0, and then h1(μ2) <h 1(μ1). (b) Assume that μ1<μ ∗<μ 2<0: in this case h1(μ2) <0 <h 1(μ1). (c) Assume that μ1<μ 2≤μ∗<0, then fμ2(λ1) <f μ1(λ1)for λ1>0, and then h1(μ2) < h1(μ1). This shows that μ → h1(μ) is decreasing. We prove now the continuity. Take μn∈(−∞, 0) →μ0<0 and consider λn:= h1(μn). Since 0 =fμn(λn) =F(λ n, μnλn), by Corollary 3.4 we conclude that λn< + 1and μnλn< + 2.(26) Hence, there exists λ1∈(−∞, +∞)such that λn→λ1. We have to show that λ1=h1(μ0). Indeed, observe that 0=fμn(λn)=1(−λnm1,−λnμnm2)→1(−λ1m1,−λ1μ0m2)=fμ0(λ1), that is, fμ0(λ1) =0. We separate now two cases: (a) μ0= μ∗: In this case, we assert, that λ1= 0. Indeed, assume that λ1=0, that is, h1(μn) →0. If, for instance, μ0>μ ∗, then there exists ρ1(μn) ∈(h1(μn), 0)such that f μn(ρ1(μn)) =0. Since h1(μn) →0, then ρ1(μn) →0, and as consequence, f μ0(0) =0, a contradiction. This shows that λ1= 0. Then, since h1(μ0)is the unique nonzero root of fμ0(λ1) =0, we have that λ1=h1(μ0). (b) μ0=μ∗: in this case fμ∗(λ1) =0 implies that λ1=0 =h1(μ∗). This concludes that λ1=h1(μ0), and hence the continuity. We claim that h1(μn)→−∞ as μn→0. (27) Assume that |h1(μn)| ≤C. Then, we can assume that, at least for a subsequence, h1(μn) → h∗ 1<0 and hence 0=1(−h1(μn)m1,−μnh1(μn)m2)→1(−h∗ 1m1,0)=0, a contradiction because 1(−h∗ 1m1, 0) >0. This proves (27). By (26), if μ →−∞we can assume that h1(μ) →h∗≤+ 1and h∗>0. Then, μh1(μ) → −∞. Since 512 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 0=fμ(h1(μ)) =F(h 1(μ), h1(μ)μ) and by Proposition 3.5 0=F(h 1(μ), h1(μ)μ) →min{σ1 1(−−λ∗m1;N+γ1), σ M0 2 1(−;D)}, it follows that 0=min{σ1 1(−−λ∗m1,N +γ1), σ M0 2 1(−, D)}. Since σM0 2 1(−, D) >0, we conclude that σ1 1(− −h∗m1, N+γ1) =0, that means that h∗=+ 1, that is lim μ→−∞ h1(μ) =+ 1. This concludes the proof.  Once we have studied the map μ → h1(μ), we need to analyze the map μ∈(−∞,0)→ h2(μ) := μh1(μ). Proposition 4.5. Assume that mi0in iand M0 iverify (6)for i=1, 2. The map μ ∈ (−∞, 0) → h2(μ) := μh1(μ) is continuous, increasing, h2(μ) ⎧ ⎨ ⎩ >0if μ>μ ∗, =0if μ=μ∗, <0if μ<μ ∗, lim μ→−∞ h2(μ) =−∞,(28) and lim μ→0h2(μ) =+ 2.(29) Proof. To start with, the continuity and the sign of the map h2(μ) follow directly from Proposition 4.4. Moreover, it is clear that lim μ→−∞ h2(μ) =lim μ→−∞ μh1(μ) =−∞. In order to prove (29)we can argue exactly as the proof of Proposition 4.4. Finally, using that Fis increasing, we prove that the map μ → h2(μ) is increasing. Take μ1<μ 2and assume that h2(μ1) ≥h2(μ2). Since h1(μ1) >h 1(μ2), then 0=F(h 1(μ1), h2(μ1)) < F (h1(μ2), h2(μ1)) ≤F(h 1(μ2), h2(μ2)) =0, a contradiction.  513 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Fig. 6. Case mi0in iand verifying (6)for i=1, 2. We have represented the functions μ → h1(μ) (left) and μ → h2(μ) (right). In Fig. 6we have represented the functions μ → h1(μ), h2(μ). Now, we proceed to the proof of Theorem 1.1. Proof of Theorem 1.1:(see Fig. 2) 1. Observe that by Corollary 3.4, we obtain F(λ 1,λ 2)<0ifλ1≥+ 1or λ2≥+ 2. 2. Take λ1< + 1. Then, by Proposition 4.4 there exists a unique μ =μ(λ1) <0 such that λ1=h1(μ). Take λ2=h2(μ) =μh1(μ), then F(λ 1,λ 2)=F(h 1(μ), h2(μ)) =F(h 1(μ), μh1(μ)) =fμ(h1(μ)) =0. We define the function H(λ1):= h2(h−1 1(λ1)), λ1< + 1. It is clear that His well-defined (observe that h−1 1exists due to that h1is a decreasing function), is continuous and F(λ 1,H(λ1)) =0. Moreover, since fixed λ1, the map λ2→ F(λ 1, λ2)is concave, it follows that F(λ 1, λ2) >0 for λ2<H(λ1)and F(λ 1, λ2) <0for λ2>H(λ1). Furthermore, by Propositions 4.4 and 4.5, lim λ1↑+ 1 H(λ1)=lim λ1↑+ 1 h2(h−1 1(λ1)) =lim μ→−∞ h2(μ) =−∞, and, 514 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 lim λ1↑−∞ H(λ1)=lim λ1↑−∞ h2(h−1 1(λ1)) =lim μ→0h2(μ) =+ 2. Finally, we prove that λ1→ H(λ1)is decreasing. Take λ1< λ1< + 1and consider H(λ1) and H(λ1). We are going to show that H(λ1) >H(λ1). Assume that H(λ1) ≤H(λ1), then 0=F(λ 1,H(λ1)) > F (λ1,H(λ1)) ≥F(λ1,H(λ1)) =0, a contradiction. This concludes the proof.  4.2. Case m10in 1and m2changes sign in 2 In this case, the results depend on the sign of 2m2. We detail the case  2 m2<0, similarly the other cases can be studied (see Remark 4.8). Observe that in this case μ∗=− γ2 1 m1 γ1 2 m2 >0. Proposition 4.6. Assume that m10in 1and M0 1verifies (6), m2changes sign in 2and 2m2<0. Then, for each μ = 0there exists a unique h1(μ) ∈IR such that fμ(h1(μ)) =0. Moreover, h1(μ) ⎧ ⎨ ⎩ >0if μ>μ ∗, =0if μ=μ∗, <0if μ<μ ∗. Furthermore, the map μ ∈IR \{0} → h1(μ) ∈IR is continuous, and lim μ→±∞ h1(μ) =0,lim μ→0h1(μ) =−∞. As consequence, there exists μmax >μ ∗such that max μ=0h1(μ) =h1(μmax):= λmax 1. Finally, the map μ ∈IR \{0} → h1(μ) ∈IR is increasing in (0, μmax)and decreasing in (−∞, 0) and (μmax, ∞). 515 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Proof. The proof of this result is rather similar to the proof of Proposition 4.4, hence we sketch the proof. Since fμn(λ1(μn)) =F(h 1(μn), μnh1(μn)) =0, by Corollary 3.4 we get − 2<h 1(μn)μn< + 2, whence we conclude that h1(μn) →0as μn→±∞. Before proving the monotony, we claim that for any c∈IR there exist at most two values of μ such that h1(μ) =c. We argue by contradiction. Assume that for μ1<μ 2<μ 3we get h1(μi) =cfor i=1, 2, 3. Taking λi 2=cμiwe obtain 0=F(c,λ i 2), λ1 2<λ 2 2<λ 3 2, a contradiction because, fixed c, the map λ2→ F(c, λ2)is concave. Now, for instance, we show that h1(μ) is decreasing in (−∞, 0). Take μ1<μ 2<0 and assume that h1(μ1) ≤h1(μ2). Since h1(μ) →0as μ →−∞and h1(μ) →−∞as μ →0. Then, there exists c<0 such that h1(μ) =cpossesses at least three solutions. This is a contradiction and proves that h1(μ) is decreasing in (−∞, 0). With a similar argument, it can be proved that h1(μ) is decreasing in (μmax, ∞)and increasing in (0, μmax). Again, we can deduce properties of the map h2(μ) =μh1(μ). Proposition 4.7. Assume that m10in 1and M0 1verifies (6), m2changes sign in 2and 2m2<0. Then h2(μ) =μh1(μ) is continuous in μ = 0, increasing and verifies h2(μ) ⎧ ⎨ ⎩ >0if μ>μ ∗or μ<0, =0if μ=μ∗, <0if μ∈(0,μ ∗), and lim μ→0±h2(μ) =∓ 2,lim μ→±∞ h2(μ) =λ∗ 2, for some λ∗ 2∈(0, + 2). Proof. Assume that μn→0+, then since h2(μn)is bounded, at least for a subsequence, h2(μn) →λ2<0. Observe that since h1(μn) →−∞, by Proposition 3.5 0=F(h 1(μn), h2(μn)) →σ2 1(−−λ2m2;N+γ2,N)=0, and then λ2=− 2. Analogously for μn→0−. 516 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Fig. 7. Representation of the maps h1(μ) (left) and h2(μ) (right) in the case m1non-negative, non-trivial and verifies (6), m2changing sign and 2m2<0. On the other hand, assume that μn→+∞and h2(μn) →λ2<0. In this case, h1(μn) →0, and then 0=F(−h1(μn)m1,−h2(μn)m2)→F(0,−λ2m2), whence λ2=λ∗ 2. Observe that 0=F(0,−λ∗ 2m2)<σ2 1(−−λ∗ 2m2;N+γ2,N) and so λ∗ 2< + 2. Finally, with an argument similar to the one used in Proposition 4.6 we can conclude that the equation h2(μ) =cpossesses at most a unique solution. Hence, the monotony of h2(μ) follows. This completes the proof.  In Fig. 7, one may see a representation of the maps μ → h1(μ), h2(μ). Remark 4.8. 1. In the case  2 m2>0 we can obtain a similar result switching μby −μ(see Fig. 8). 2. When  2 m2=0 517 B.B.V. Maia, M. Molina-Becerra, C. Morales-Rodrigo et al. Journal of Differential Equations 390 (2024) 494–524 Fig. 8. Representations of the maps h1(μ) (left) and h2(μ) (right) in the case m1non-negative, non-trivial and verifies (6), m2changing sign and 2m2>0. Fig. 9. We have represented the maps h1(μ) (left) and h2(μ) (right) in the case m1non-negative, non-trivial and verifies (6), m2changing sign and 2m2=0. observe that f μ(0) <0for all μ(see (24)), and then h1(μ) <0for all μ. The global behavior of h1(μ) at μ =0 and μ →±∞is similar to Proposition 4.6 (see Fig. 9). Proof of Theorem 1.2:(See Fig. 3.) Assume that 2m2<0(see Fig. 7). We introduce the following notation: h1(μ) := ⎧ ⎨ ⎩ h1 1(μ) if μ<0, h2 1(μ) if μ∈(0,μ max], h3 1(μ) if μ>μ max. 1. If λ1>λ max 1, then there does not exist μ ∈IR such that λ1=h1(μ). Hence, F(λ 1, λ2) = 0for all λ2∈IR , in fact, F(λ 1, λ2) <0for all λ2∈IR . Indeed, if for some λ2we have 518