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Bifurcation of Lotka–Volterra competition model with nonlinear boundary conditions
Accepted Manuscript Bifurcation of Lotka–Volterra competition model with nonlinear boundary conditions Yang Zhang, Mingxin Wang PII: DOI: Reference:
S0893-9659(14)00221-3 http://dx.doi.org/10.1016/j.aml.2014.07.001 AML 4581
To appear in:
Applied Mathematics Letters
Received date: 21 May 2014 Revised date: 1 July 2014 Accepted date: 2 July 2014 Please cite this article as: Y. Zhang, M. Wang, Bifurcation of Lotka–Volterra competition model with nonlinear boundary conditions, Appl. Math. Lett. (2014), http://dx.doi.org/10.1016/j.aml.2014.07.001 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
*Manuscript Click here to view linked References
Bifurcation of Lotka-Volterra competition model with nonlinear boundary conditions∗ Yang Zhang
Mingxin Wang†
Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, P. R. China
Abstract In this paper, we discuss the bifurcation of semi-trivial solutions for the Lotka-Volterra competition model with nonlinear boundary conditions over a smooth bounded domain. Applying the Crandall-Rabinowitz local bifurcation theorem [6], we prove the existence of a smooth curve bifurcating from the appropriate semi-trivial branch. Key words: Competition model; Nonlinear boundary conditions; Positive solutions; Bifurcation. AMS subject classifications (2000): 35J55, 35J65, 35B32, 35B35.
1
Introduction
In this paper, we study the following problem with nonhomogeneous Neumann boundary conditions. −d1 ∆u = u(a1 − b1 u − c1 v), x ∈ Ω, −d2 ∆v = v(a2 − b2 u − c2 v), x ∈ Ω, (1.1) ∂u + f (u) = 0, x ∈ ∂Ω, ∂ν ∂v + g(v) = 0, x ∈ ∂Ω, ∂ν where Ω ⊆ RN is a bounded domain with smooth boundary. ai , bi , ci (i = 1, 2) are positive constants. f, g ∈ C 2 (R) are nondecreasing functions with f (0) = g(0) = 0 and f 0 (0) = g 0 (0) > 0. Recently, bifurcation of nonlinear boundary problems were discussed in some biological models. Cantrell, Cosner and Mart´ınez [3, 4] considered the following logistic model: ut = d∆u + ru(1 − u), x ∈ Ω, α(u) ∂u + (1 − α(u))u = 0, x ∈ ∂Ω, ∂ν
where α(u) is assumed to be smooth, nonnegative and nondecreasing and satisfy α([0, 1]) ⊆ [0, 1]. This problem have two trivial solutions u = 0 and u = 1. Using λ = r/d as bifurcation parameter, ∗ †
This work was supported by NSFC Grant 11371113 E-mail address: [email protected]
1
2 they show that there is a positive solution bifurcating from the trivial solution. The other related studies to bifurcation of nonlinear boundary problems, we refer to, for instance [8, 9, 10] and references cited therein. Inspired by all the above works, a nonlinear Neumann boundary value problem of competition type is considered in this paper. We analyze the structure of the continuum of positive solutions emanating from the semi-trivial state at a unique bifurcation value, depending on parameters of the problem. The results in this paper extend the previous ones obtained by J. Blat and K. J. Brown [2], for a competition problem of under Dirichlet boundary conditions, to a wide class of competition problems with nonlinear Neumann boundary conditions. Considerable amounts of study have been attracted by competition model with Dirichlet boundary condition, we mention only [5] and [11]; many other works can be found from the reference of these papers. ¯ For the given continuous Throughout this paper, we shall consider the solutions u, v ∈ C(Ω). ¯ functions q, b : Ω → R, let λ1 (d, q, b) be the principal eigenvalue of the following eigenvalue problem
where b(x) ≥ 0.
−d∆u + q(x)u = λu, ∂u + b(x)u = 0, ∂ν
x ∈ Ω,
(1.2)
x ∈ ∂Ω,
Lemma 1.1 [1] Suppose that f and g are increasing and convex functions with f (0) = g(0) = 0. Then (1.1) has two semi-trivial solutions (θ1 , 0) and (0, θ2 ), where θ1 is the unique positive solution of −d1 ∆u = u(a1 − b1 u), ∂u + f (u) = 0, ∂ν
x ∈ Ω,
(1.3)
x ∈ ∂Ω,
and θ2 is the unique positive solution of
−d2 ∆v = v(a2 − c2 u), ∂v + g(u) = 0, ∂ν
x ∈ Ω,
(1.4)
x ∈ ∂Ω.
Since Lemma 1.1, there exist two semi-trivial solution branches Γ(θ1 ,0) and Γ(0,θ2 ) , we denote Γ(θ1 ,0) := {(a2 , θ1 , 0)}, Γ(0,θ2 ) := {(a1 , 0, θ2 )}.
2
Main results
In the following, let a2 be a bifurcation parameter and fixed other parameters, or let a1 be a bifurcation parameter and fixed other parameters. Define ¯ ∩ C 2 (Ω), X := C 1 (Ω)
Y1 := C(Ω) × C(Ω),
Y2 := C(∂Ω) × C(∂Ω).
We have the following result about the existence of a C 1 smooth curve bifurcating from the semitrivial solution branches Γ(θ1 ,0) and Γ(0,θ2 ) .
3 Theorem 2.1 Suppose that f and g are convex functions. The following hold: (a) If a2 = a ˜2 := λ1 (d2 , b2 θ1 , g 0 (0)), then (˜ a2 , θ1 , 0) is a bifurcation point from the semi-trivial branch Γ(θ1 ,0) . Moreover, the solution set of (1.1) near (˜ a2 , θ1 , 0) consists precisely of the curves Γ(θ1 ,0) and Σ1 := {(a2 (s), u(s), v(s)) : s ∈ I = (−η, η) ⊂ R} , where a2 (s) = λ0 + y1 (s), u(s) = θ1 + sφ + sξ1 (s), v(s) = sψ + sζ1 (s) are C 1 functions such that y1 (0) = ξ1 (0) = ζ1 (0) = 0, (φ, ψ) is the solution of the linearization of (1.1) at (θ1 , 0) corresponding to a ˜2 . (b) If a1 = a ˜1 := λ1 (d1 , c1 θ2 , f 0 (0)), then (˜ a1 , 0, θ2 ) is a bifurcation point from the semi-trivial branch Γ(0,θ2 ) . Moreover, the solution set of (1.1) near (˜ a1 , 0, θ2 ) consists precisely of the curves Γ(0,θ2 ) and Σ2 := {(a1 (s), u(s), v(s)) : s ∈ I = (−η, η) ⊂ R} , where a1 (s) = a ˜1 + y2 (s), u(s) = sφ˜ + sξ2 (s), v(s) = θ2 + sψ˜ + sζ2 (s) are C 1 functions such that ˜ ψ) ˜ is the solution of the linearization of (1.1) at (0, θ2 ) corresponding y2 (0) = ξ2 (0) = ζ2 (0) = 0, (φ, to a ˜1 . Proof. We prove (a) in several steps: Step 1 Prove dim Ker(G(u,v) (˜ a2 , θ1 , 0)) = 1. Define a nonlinear mapping G : R × X × X → Y1 × Y2 by ∂v ∂u + f (u), + g(v) . G(a2 , u, v) = [−d1 ∆u − (a1 − b1 u − c1 v)u, −d2 ∆v − (a2 − b2 u − c2 v)v] , ∂ν ∂ν Let G(a2 , u, v) := G1 (a2 , u, v), G2 (a2 , u, v) . For any (u, v) ∈ X × X, we have
and
G1(u,v) (a2 , θ1 , 0)[(u, v)] =
−d1 ∆u − (a1 − 2b1 θ1 )u + c1 θ1 v −d2 ∆v − (a2 − b2 θ1 )v + b2 θ1 v
∂u 0 (θ )u + f 1 ∂ν . G2(u,v) (a2 , θ1 , 0)[(u, v)] = ∂v 0 + g (0)v ∂ν
The linearized G(λ, u, v) = 0 at (θ1 , 0) can be written as
−d1 ∆u − (a1 − 2b1 θ1 )u + cθ1 v = 0, −d2 ∆v − (a2 − b2 θ1 )v = 0, ∂u + f 0 (θ1 )u = 0, ∂ν ∂v + g 0 (0)v = 0, ∂ν
x ∈ Ω,
x ∈ Ω, x ∈ ∂Ω, x ∈ ∂Ω.
(2.1)
4 Now, we consider the following problem −d2 ∆v − (a2 − b2 θ1 )v = 0, ∂v + g 0 (0)v = 0, ∂ν
x ∈ Ω,
(2.2)
x ∈ ∂Ω.
If a2 = a ˜2 = λ1 (d2 , b2 θ1 , g 0 (0)), let ψ be the eigenvalue function corresponding to λ1 (d2 , b2 θ1 , g 0 (0)). Taking it into the first equation of (2.1), we have x ∈ Ω, −d1 ∆u − (a1 − 2b1 θ1 )u + cθ1 ψ = 0, (2.3) ∂u + f 0 (θ )u = 0, x ∈ ∂Ω. 1 ∂ν We would like to prove that the solution of (2.1) is unique. If not, (2.3) have two solutions φ1 and φ2 , set φ = φ1 − φ2 6≡ 0, it is easy to see that φ satisfies x ∈ Ω, −d1 ∆φ − (a1 − 2b1 θ1 )φ = 0, (2.4) ∂φ + f 0 (θ1 )φ = 0, x ∈ ∂Ω. ∂ν
Thus 0 is an eigenvalue of (2.4) and 0 > λ1 (d1 , −a1 +2b1 θ1 , f 0 (θ1 )), where λ1 (d1 , −a1 +2b1 θ1 , f 0 (θ1 )) is the principal eigenvalue of (2.4), denote φ∗ > 0 is the eigenvalue function responding to λ1 (d1 , −a1 + 2b1 θ1 , f 0 (θ1 )). Therefore φ∗ satisfies x ∈ Ω, −d1 ∆φ∗ − (a1 − 2b1 θ1 )φ∗ = 0, (2.5) ∗ ∂φ + f 0 (θ1 )φ∗ = 0, x ∈ ∂Ω. ∂ν Multiplying the first equation of (1.3) by φ∗ , integrating over Ω, we derive Z −d1 φ∗ ∆θ1 − (a1 − b1 θ1 )θ1 φ∗ dx = 0.
(2.6)
Ω
Multiplying the first equation of (2.5) by φ∗ , integrating over Ω, we have Z Z −d1 θ1 ∆φ∗ − (a1 − 2b1 θ1 )θ1 φ∗ dx = λ1 φ∗ θ1 dx. Ω
(2.7)
Ω
In view of (2.6) and (2.7), it is not difficult to see Z Z d1 θ1 ∆φ∗ − d1 φ∗ ∆θ1 − b1 θ12 φ∗ dx = − λ1 φ∗ θ1 dx. Ω
Ω
Taking account (2.8), Green’s identity and f is convex function in R, we know Z Z Z Z ∂φ∗ ∂θ1 ∗ − λ1 φ∗ θ1 dx = d1 θ1 dS − d1 φ dS − b1 θ12 φ∗ dx ∂ν ∂ν Ω ∂Ω ∂Ω Ω Z Z Z = d1 −f 0 (θ1 )φ∗ θ1 dS + d1 f (θ1 )φ∗ dS − b1 θ12 φ∗ ∂Ω ∂Ω Ω Z Z = d1 [f (θ1 ) − f 0 (θ1 )θ1 ]φ∗ dS − b1 θ12 φ∗ dx ∂Ω
< 0.
Ω
(2.8)
5 This is a contradiction to 0 > λ1 (d1 , −a1 + 2b1 θ1 , f 0 (θ1 )). Thus (2.3) has a unique solution φ. It is obvious that a ˜2 is the principal eigenvalue of (2.2), and ψ is the simple eigenfunction corresponding to a ˜2 . Thus (φ, ψ) is the solution of (2.1), it follows that Ker G(u,v) (˜ a2 , θ1 , 0) = span(φ, ψ) and dim Ker G(u,v) (˜ a2 , θ1 , 0) = 1.
Step 2 To show that Codim R(G(u,v) (˜ a2 , θ1 , 0)) = 1. According to the definition of Fredholm index, we have ind G(u,v) (˜ a2 , θ1 , 0) = dim Ker G(u,v) (˜ a2 , θ1 , 0) − Codim R G(u,v) (˜ a2 , θ1 , 0) . In the following, we prove that ind G(u,v) (˜ a2 , θ1 , 0) = 0. Write a2 , θ1 , 0) − (σI, 0). a2 , θ1 , 0) + σI, G2(u,v) (˜ a2 , θ1 , 0) = G1(u,v) (˜ a2 , θ1 , 0), G2(u,v) (˜ G1(u,v) (˜ By [10, Remark2.5, 5, Case 3], G1(u,v) (˜ a2 , θ1 , 0), G2(u,v) (˜ a2 , θ1 , 0) satisfies Agmon’s condition. In view of [10, Theorem2.7], G1(u,v) (˜ a2 , θ1 , 0) + σI, G2(u,v) (˜ a2 , θ1 , 0) is an isomorphism and hence Fredholm with index zero for all σ with large |σ|. Fix such a σ, since the mapping 2 2 u ∈ W 2,p (Ω) 7→ (σu, 0) ∈ (Lp (Ω))2 × ∂W 2,p (Ω) is compact. Applying [7, Theorem V.2.1], it deduces ind G1(u,v) (˜ a2 , θ1 , 0) + σI, G2(u,v) (˜ a2 , θ1 , 0) = ind G1(u,v) (˜ a2 , θ1 , 0), G2(u,v) (˜ a2 , θ1 , 0) = 0, which implies Codim R G(u,v) (˜ a2 , θ1 , 0) = 1. Step 3 We would like to prove
Ga2 (u,v) (˜ a2 , θ1 , 0)[(φ, ψ)] 6∈ R G(u,v) (˜ a2 , θ1 , 0) . In fact, if (h1 , h2 , h3 , h4 ) ∈ R G(u,v) (˜ a2 , θ1 , 0) , there exits (u, v) ∈ X × X, such that −d1 ∆u − (a1 − 2b1 θ1 )u + c1 θ1 ψ = h1 , x ∈ Ω, −d2 ∆v − (˜ a2 − b2 θ1 )v = h2 , x ∈ Ω, (2.9) ∂u + f 0 (θ1 )u = h3 , x ∈ ∂Ω, ∂ν ∂v + g 0 (0)v = h , x ∈ ∂Ω. 4 ∂ν Recall that a ˜2 = λ1 (d2 , b2 θ1 , g 0 (0)) and ψ is the eigenfunction corresponding to a ˜2 , we have −d2 ∆ψ − (˜ a2 − b2 θ1 )ψ = 0, x ∈ Ω, (2.10) ∂ψ + g 0 (0)ψ = 0, x ∈ ∂Ω. ∂ν For (h1 , h2 , h3 , h4 ) ∈ R G(u,v) (˜ a2 , θ1 , 0) , in view of (2.9), (2.10) and Green formula, we have Z Z h2 ψdx = [−d2 ∆v − (˜ a2 − b2 θ1 )v]ψdx Ω Ω Z Z Z ∂v = −d2 ψdS + d2 DvDψdx − (˜ a2 − b2 θ1 )vψdx ∂ν Z∂Ω ZΩ Z Ω = −d2 h4 ψdS − d2 v∆ψdx − (˜ a2 − b2 θ1 )vψdx Ω Ω Z∂Ω = −d2 h4 ψdS. ∂Ω
6 Through simple calculation, we obtain a2 , θ1 , 0)[(φ, ψ)] G1a2 (u,v) (˜ and a2 , θ1 , 0)[(φ, ψ)] G2a2 (u,v) (˜ Thus
Z
Ω
0
=
−ψ
0
=
˜ 2 ψdx < 0 = −d2 h
0
Z
:=
:=
˜1 h ˜2 h
˜3 h ˜4 h
.
˜ 4 ψdS. h
∂Ω
This is a contradiction. We are able to conclude that Ga2 (u,v) (˜ a2 , θ1 , 0)[(φ, ψ)] 6∈ R G(u,v) (˜ a2 , θ1 , 0) .
Finally, applying the Crandall and Rabinowitz local bifurcation theorem, we conclude the desired result. The proof of (b) is similar to the proof of (a), we omit it. This completes the proof.
References [1] I. Ahn and L. G. Li, Positive solutions of certain elliptic systems with density-dependent diffusions, Proc. Roy. Soc. Edinburgh Sect. A 125(1995), 1031-1050. [2] J. Blat and K. J. Brown, Bifurcation of steady-state solutions in predator-prey and competition systems, Proc. Roy. Soc. Edinburgh Sect. A 97(1984), 21-34. [3] R. S. Cantrell, C. Cosner, On the effects of nonlinear boundary conditions in diffusive logistic equations on bounded domains, J. Differential Equations 231(2)(2006), 768-804. [4] R. S. Cantrell, C. Cosner and S. Mart´ınez, Global bifurcation of solutions to diffusive logistic equations on bounded domains subject to nonlinear boundary conditions, Proc. Roy. Soc. Edinburgh Sect. A 139(1)(2009), 45-56. [5] G. Z. Lin and M. Pedersen Coexistence of a general elliptic system in population dynamics, Computers & Mathematics with Applications. 48(3)(2004), 617-628. [6] M. G. Crandall and P. H. Rabinowitz, Bifurcation from simple eigenvalues, J. Func. Anal. 8(1971), 321-340. [7] S. Goldberg, Unbounded Linear Operators: Theory and Applications, Dover Publications.com. 2006. [8] P. Liu and J. P. Shi, Bifurcation from trivial solutions for steady state reaction-diffusion equations with nonlinear boundary condition. Preprint, 2014. [9] G. F. Madeira and A. S. Nascimento, Bifurcation of stable equilibria and nonlinear flux boundary condition with indefinite weight, J. Differential Equations 251(11)(2011), 3328-3247. [10] J. P. Shi and X. F. Wang, On global bifurcation for quasilinear elliptic systems on bounded domains, J. Differential Equations 246(7)(2009), 2788-2813. [11] A. Schiaffino and A. Tesei, Competition systems with dirichlet boundary conditions, J. Math. Biology. 15(1982), 93-105.
S0893-9659(14)00221-3 http://dx.doi.org/10.1016/j.aml.2014.07.001 AML 4581
To appear in:
Applied Mathematics Letters
Received date: 21 May 2014 Revised date: 1 July 2014 Accepted date: 2 July 2014 Please cite this article as: Y. Zhang, M. Wang, Bifurcation of Lotka–Volterra competition model with nonlinear boundary conditions, Appl. Math. Lett. (2014), http://dx.doi.org/10.1016/j.aml.2014.07.001 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
*Manuscript Click here to view linked References
Bifurcation of Lotka-Volterra competition model with nonlinear boundary conditions∗ Yang Zhang
Mingxin Wang†
Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, P. R. China
Abstract In this paper, we discuss the bifurcation of semi-trivial solutions for the Lotka-Volterra competition model with nonlinear boundary conditions over a smooth bounded domain. Applying the Crandall-Rabinowitz local bifurcation theorem [6], we prove the existence of a smooth curve bifurcating from the appropriate semi-trivial branch. Key words: Competition model; Nonlinear boundary conditions; Positive solutions; Bifurcation. AMS subject classifications (2000): 35J55, 35J65, 35B32, 35B35.
1
Introduction
In this paper, we study the following problem with nonhomogeneous Neumann boundary conditions. −d1 ∆u = u(a1 − b1 u − c1 v), x ∈ Ω, −d2 ∆v = v(a2 − b2 u − c2 v), x ∈ Ω, (1.1) ∂u + f (u) = 0, x ∈ ∂Ω, ∂ν ∂v + g(v) = 0, x ∈ ∂Ω, ∂ν where Ω ⊆ RN is a bounded domain with smooth boundary. ai , bi , ci (i = 1, 2) are positive constants. f, g ∈ C 2 (R) are nondecreasing functions with f (0) = g(0) = 0 and f 0 (0) = g 0 (0) > 0. Recently, bifurcation of nonlinear boundary problems were discussed in some biological models. Cantrell, Cosner and Mart´ınez [3, 4] considered the following logistic model: ut = d∆u + ru(1 − u), x ∈ Ω, α(u) ∂u + (1 − α(u))u = 0, x ∈ ∂Ω, ∂ν
where α(u) is assumed to be smooth, nonnegative and nondecreasing and satisfy α([0, 1]) ⊆ [0, 1]. This problem have two trivial solutions u = 0 and u = 1. Using λ = r/d as bifurcation parameter, ∗ †
This work was supported by NSFC Grant 11371113 E-mail address: [email protected]
1
2 they show that there is a positive solution bifurcating from the trivial solution. The other related studies to bifurcation of nonlinear boundary problems, we refer to, for instance [8, 9, 10] and references cited therein. Inspired by all the above works, a nonlinear Neumann boundary value problem of competition type is considered in this paper. We analyze the structure of the continuum of positive solutions emanating from the semi-trivial state at a unique bifurcation value, depending on parameters of the problem. The results in this paper extend the previous ones obtained by J. Blat and K. J. Brown [2], for a competition problem of under Dirichlet boundary conditions, to a wide class of competition problems with nonlinear Neumann boundary conditions. Considerable amounts of study have been attracted by competition model with Dirichlet boundary condition, we mention only [5] and [11]; many other works can be found from the reference of these papers. ¯ For the given continuous Throughout this paper, we shall consider the solutions u, v ∈ C(Ω). ¯ functions q, b : Ω → R, let λ1 (d, q, b) be the principal eigenvalue of the following eigenvalue problem
where b(x) ≥ 0.
−d∆u + q(x)u = λu, ∂u + b(x)u = 0, ∂ν
x ∈ Ω,
(1.2)
x ∈ ∂Ω,
Lemma 1.1 [1] Suppose that f and g are increasing and convex functions with f (0) = g(0) = 0. Then (1.1) has two semi-trivial solutions (θ1 , 0) and (0, θ2 ), where θ1 is the unique positive solution of −d1 ∆u = u(a1 − b1 u), ∂u + f (u) = 0, ∂ν
x ∈ Ω,
(1.3)
x ∈ ∂Ω,
and θ2 is the unique positive solution of
−d2 ∆v = v(a2 − c2 u), ∂v + g(u) = 0, ∂ν
x ∈ Ω,
(1.4)
x ∈ ∂Ω.
Since Lemma 1.1, there exist two semi-trivial solution branches Γ(θ1 ,0) and Γ(0,θ2 ) , we denote Γ(θ1 ,0) := {(a2 , θ1 , 0)}, Γ(0,θ2 ) := {(a1 , 0, θ2 )}.
2
Main results
In the following, let a2 be a bifurcation parameter and fixed other parameters, or let a1 be a bifurcation parameter and fixed other parameters. Define ¯ ∩ C 2 (Ω), X := C 1 (Ω)
Y1 := C(Ω) × C(Ω),
Y2 := C(∂Ω) × C(∂Ω).
We have the following result about the existence of a C 1 smooth curve bifurcating from the semitrivial solution branches Γ(θ1 ,0) and Γ(0,θ2 ) .
3 Theorem 2.1 Suppose that f and g are convex functions. The following hold: (a) If a2 = a ˜2 := λ1 (d2 , b2 θ1 , g 0 (0)), then (˜ a2 , θ1 , 0) is a bifurcation point from the semi-trivial branch Γ(θ1 ,0) . Moreover, the solution set of (1.1) near (˜ a2 , θ1 , 0) consists precisely of the curves Γ(θ1 ,0) and Σ1 := {(a2 (s), u(s), v(s)) : s ∈ I = (−η, η) ⊂ R} , where a2 (s) = λ0 + y1 (s), u(s) = θ1 + sφ + sξ1 (s), v(s) = sψ + sζ1 (s) are C 1 functions such that y1 (0) = ξ1 (0) = ζ1 (0) = 0, (φ, ψ) is the solution of the linearization of (1.1) at (θ1 , 0) corresponding to a ˜2 . (b) If a1 = a ˜1 := λ1 (d1 , c1 θ2 , f 0 (0)), then (˜ a1 , 0, θ2 ) is a bifurcation point from the semi-trivial branch Γ(0,θ2 ) . Moreover, the solution set of (1.1) near (˜ a1 , 0, θ2 ) consists precisely of the curves Γ(0,θ2 ) and Σ2 := {(a1 (s), u(s), v(s)) : s ∈ I = (−η, η) ⊂ R} , where a1 (s) = a ˜1 + y2 (s), u(s) = sφ˜ + sξ2 (s), v(s) = θ2 + sψ˜ + sζ2 (s) are C 1 functions such that ˜ ψ) ˜ is the solution of the linearization of (1.1) at (0, θ2 ) corresponding y2 (0) = ξ2 (0) = ζ2 (0) = 0, (φ, to a ˜1 . Proof. We prove (a) in several steps: Step 1 Prove dim Ker(G(u,v) (˜ a2 , θ1 , 0)) = 1. Define a nonlinear mapping G : R × X × X → Y1 × Y2 by ∂v ∂u + f (u), + g(v) . G(a2 , u, v) = [−d1 ∆u − (a1 − b1 u − c1 v)u, −d2 ∆v − (a2 − b2 u − c2 v)v] , ∂ν ∂ν Let G(a2 , u, v) := G1 (a2 , u, v), G2 (a2 , u, v) . For any (u, v) ∈ X × X, we have
and
G1(u,v) (a2 , θ1 , 0)[(u, v)] =
−d1 ∆u − (a1 − 2b1 θ1 )u + c1 θ1 v −d2 ∆v − (a2 − b2 θ1 )v + b2 θ1 v
∂u 0 (θ )u + f 1 ∂ν . G2(u,v) (a2 , θ1 , 0)[(u, v)] = ∂v 0 + g (0)v ∂ν
The linearized G(λ, u, v) = 0 at (θ1 , 0) can be written as
−d1 ∆u − (a1 − 2b1 θ1 )u + cθ1 v = 0, −d2 ∆v − (a2 − b2 θ1 )v = 0, ∂u + f 0 (θ1 )u = 0, ∂ν ∂v + g 0 (0)v = 0, ∂ν
x ∈ Ω,
x ∈ Ω, x ∈ ∂Ω, x ∈ ∂Ω.
(2.1)
4 Now, we consider the following problem −d2 ∆v − (a2 − b2 θ1 )v = 0, ∂v + g 0 (0)v = 0, ∂ν
x ∈ Ω,
(2.2)
x ∈ ∂Ω.
If a2 = a ˜2 = λ1 (d2 , b2 θ1 , g 0 (0)), let ψ be the eigenvalue function corresponding to λ1 (d2 , b2 θ1 , g 0 (0)). Taking it into the first equation of (2.1), we have x ∈ Ω, −d1 ∆u − (a1 − 2b1 θ1 )u + cθ1 ψ = 0, (2.3) ∂u + f 0 (θ )u = 0, x ∈ ∂Ω. 1 ∂ν We would like to prove that the solution of (2.1) is unique. If not, (2.3) have two solutions φ1 and φ2 , set φ = φ1 − φ2 6≡ 0, it is easy to see that φ satisfies x ∈ Ω, −d1 ∆φ − (a1 − 2b1 θ1 )φ = 0, (2.4) ∂φ + f 0 (θ1 )φ = 0, x ∈ ∂Ω. ∂ν
Thus 0 is an eigenvalue of (2.4) and 0 > λ1 (d1 , −a1 +2b1 θ1 , f 0 (θ1 )), where λ1 (d1 , −a1 +2b1 θ1 , f 0 (θ1 )) is the principal eigenvalue of (2.4), denote φ∗ > 0 is the eigenvalue function responding to λ1 (d1 , −a1 + 2b1 θ1 , f 0 (θ1 )). Therefore φ∗ satisfies x ∈ Ω, −d1 ∆φ∗ − (a1 − 2b1 θ1 )φ∗ = 0, (2.5) ∗ ∂φ + f 0 (θ1 )φ∗ = 0, x ∈ ∂Ω. ∂ν Multiplying the first equation of (1.3) by φ∗ , integrating over Ω, we derive Z −d1 φ∗ ∆θ1 − (a1 − b1 θ1 )θ1 φ∗ dx = 0.
(2.6)
Ω
Multiplying the first equation of (2.5) by φ∗ , integrating over Ω, we have Z Z −d1 θ1 ∆φ∗ − (a1 − 2b1 θ1 )θ1 φ∗ dx = λ1 φ∗ θ1 dx. Ω
(2.7)
Ω
In view of (2.6) and (2.7), it is not difficult to see Z Z d1 θ1 ∆φ∗ − d1 φ∗ ∆θ1 − b1 θ12 φ∗ dx = − λ1 φ∗ θ1 dx. Ω
Ω
Taking account (2.8), Green’s identity and f is convex function in R, we know Z Z Z Z ∂φ∗ ∂θ1 ∗ − λ1 φ∗ θ1 dx = d1 θ1 dS − d1 φ dS − b1 θ12 φ∗ dx ∂ν ∂ν Ω ∂Ω ∂Ω Ω Z Z Z = d1 −f 0 (θ1 )φ∗ θ1 dS + d1 f (θ1 )φ∗ dS − b1 θ12 φ∗ ∂Ω ∂Ω Ω Z Z = d1 [f (θ1 ) − f 0 (θ1 )θ1 ]φ∗ dS − b1 θ12 φ∗ dx ∂Ω
< 0.
Ω
(2.8)
5 This is a contradiction to 0 > λ1 (d1 , −a1 + 2b1 θ1 , f 0 (θ1 )). Thus (2.3) has a unique solution φ. It is obvious that a ˜2 is the principal eigenvalue of (2.2), and ψ is the simple eigenfunction corresponding to a ˜2 . Thus (φ, ψ) is the solution of (2.1), it follows that Ker G(u,v) (˜ a2 , θ1 , 0) = span(φ, ψ) and dim Ker G(u,v) (˜ a2 , θ1 , 0) = 1.
Step 2 To show that Codim R(G(u,v) (˜ a2 , θ1 , 0)) = 1. According to the definition of Fredholm index, we have ind G(u,v) (˜ a2 , θ1 , 0) = dim Ker G(u,v) (˜ a2 , θ1 , 0) − Codim R G(u,v) (˜ a2 , θ1 , 0) . In the following, we prove that ind G(u,v) (˜ a2 , θ1 , 0) = 0. Write a2 , θ1 , 0) − (σI, 0). a2 , θ1 , 0) + σI, G2(u,v) (˜ a2 , θ1 , 0) = G1(u,v) (˜ a2 , θ1 , 0), G2(u,v) (˜ G1(u,v) (˜ By [10, Remark2.5, 5, Case 3], G1(u,v) (˜ a2 , θ1 , 0), G2(u,v) (˜ a2 , θ1 , 0) satisfies Agmon’s condition. In view of [10, Theorem2.7], G1(u,v) (˜ a2 , θ1 , 0) + σI, G2(u,v) (˜ a2 , θ1 , 0) is an isomorphism and hence Fredholm with index zero for all σ with large |σ|. Fix such a σ, since the mapping 2 2 u ∈ W 2,p (Ω) 7→ (σu, 0) ∈ (Lp (Ω))2 × ∂W 2,p (Ω) is compact. Applying [7, Theorem V.2.1], it deduces ind G1(u,v) (˜ a2 , θ1 , 0) + σI, G2(u,v) (˜ a2 , θ1 , 0) = ind G1(u,v) (˜ a2 , θ1 , 0), G2(u,v) (˜ a2 , θ1 , 0) = 0, which implies Codim R G(u,v) (˜ a2 , θ1 , 0) = 1. Step 3 We would like to prove
Ga2 (u,v) (˜ a2 , θ1 , 0)[(φ, ψ)] 6∈ R G(u,v) (˜ a2 , θ1 , 0) . In fact, if (h1 , h2 , h3 , h4 ) ∈ R G(u,v) (˜ a2 , θ1 , 0) , there exits (u, v) ∈ X × X, such that −d1 ∆u − (a1 − 2b1 θ1 )u + c1 θ1 ψ = h1 , x ∈ Ω, −d2 ∆v − (˜ a2 − b2 θ1 )v = h2 , x ∈ Ω, (2.9) ∂u + f 0 (θ1 )u = h3 , x ∈ ∂Ω, ∂ν ∂v + g 0 (0)v = h , x ∈ ∂Ω. 4 ∂ν Recall that a ˜2 = λ1 (d2 , b2 θ1 , g 0 (0)) and ψ is the eigenfunction corresponding to a ˜2 , we have −d2 ∆ψ − (˜ a2 − b2 θ1 )ψ = 0, x ∈ Ω, (2.10) ∂ψ + g 0 (0)ψ = 0, x ∈ ∂Ω. ∂ν For (h1 , h2 , h3 , h4 ) ∈ R G(u,v) (˜ a2 , θ1 , 0) , in view of (2.9), (2.10) and Green formula, we have Z Z h2 ψdx = [−d2 ∆v − (˜ a2 − b2 θ1 )v]ψdx Ω Ω Z Z Z ∂v = −d2 ψdS + d2 DvDψdx − (˜ a2 − b2 θ1 )vψdx ∂ν Z∂Ω ZΩ Z Ω = −d2 h4 ψdS − d2 v∆ψdx − (˜ a2 − b2 θ1 )vψdx Ω Ω Z∂Ω = −d2 h4 ψdS. ∂Ω
6 Through simple calculation, we obtain a2 , θ1 , 0)[(φ, ψ)] G1a2 (u,v) (˜ and a2 , θ1 , 0)[(φ, ψ)] G2a2 (u,v) (˜ Thus
Z
Ω
0
=
−ψ
0
=
˜ 2 ψdx < 0 = −d2 h
0
Z
:=
:=
˜1 h ˜2 h
˜3 h ˜4 h
.
˜ 4 ψdS. h
∂Ω
This is a contradiction. We are able to conclude that Ga2 (u,v) (˜ a2 , θ1 , 0)[(φ, ψ)] 6∈ R G(u,v) (˜ a2 , θ1 , 0) .
Finally, applying the Crandall and Rabinowitz local bifurcation theorem, we conclude the desired result. The proof of (b) is similar to the proof of (a), we omit it. This completes the proof.
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