Some elliptic problems with nonlinear boundary conditions

November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 SOME ELLIPTIC PROBLEMS WITH NONLINEAR BOUNDARY CONDITIONS C. MORALES-RODRIGO AND A. SU´ AREZ∗ Dpto. Ecuaciones Diferenciales y An´alisis Num´erico, C/ Tarfia s/n, 41080, Univ. of Sevilla, Seville, Spain E-mails: [email protected] and suar[email protected] In memory of Prof. J. Esquinas This paper concerns with some elliptic equations with non-linear boundary conditions. Sub-supersolution and bifurcation methods are used in order to obtain existence, uniqueness or multiplicity of positive solutions. 1. Introduction In this paper we study positive solutions of some nonlinear elliptic problems with mixed nonlinear boundary conditions. Throughout it, we consider the following assumptions: (1) Ω ⊂IRN,N≥1, is a bounded domain with boundary ∂Ω of class C2. Moreover, ∂Ω := Γ0∪Γ1, where Γ0and Γ1denote two disjoint open and closed sets in the relative topology of ∂Ω. (2) Lis a uniformly elliptic differential operator in Ω of the form L:= − N X i,j=1 aij ∂2 ∂xi∂xj + N X i=1 bi ∂ ∂xi +c, (1) with coefficients aij =aji ∈ C2,α(Ω), bi∈ C1,α(Ω) and c∈ Cα(Ω), α∈(0,1). ∗Supported by the Spanish Ministry of Science and Technology under grant BFM200306446. 1 November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 2 (3) We define the mixed boundary operator, B, by Bu := ½uon Γ0, Buon Γ1,(2) where the operator B:= ∂ν+bwith ν∈ C1(Γ1,IRN) an outward pointing nowhere tangent vector-field and b∈ C1,α(Γ1). In this paper we study the following problems where ais a positive or negative regular function on Γ1and 0 < q < 1< p, r. We first study an elliptic equation with a logistic term on the boundary    Lu = 0 in Ω, u= 0 on Γ0, Bu=µu +a(x)uron Γ1, (3) where µ∈IR will be regarded as bifurcation parameter. We do not know previous works in which (3) was analyzed. We characterize the existence, uniqueness and stability of positive solution in terms of the parameter µ (see Theorem 5.2). Second, we study of the sublinear-superlinear equation    −∆u=λu −upin Ω, ∂u ∂n =uron ∂Ω, (4) where nis the outward normal vector-field of Ω. (The case −urinstead ur has been studied in Ref. 8.) Equation (4) has attracted a lot of attention in the last years with λ= 0, see Refs. 6,10,18,21,22 and 26, among others, where basically the equation and its corresponding parabolic problem were analyzed in the particular case λ= 0, and in Refs. 28,29 where the local bifurcation was studied. We complete this study giving existence, nonexistence and stability results in function of λ(see Theorem 5.3). Finally, we study the concave-convex equation    Lu =λm(x)uqen Ω, ∂u ∂n =a(x)uren ∂Ω. (5) where m∈ C(Ω) is nonnegative and non-trivial. (5) was studied previously in Ref. 14 when Lu =−∆u+u, and m≡a≡1 by variational methods. When a < 0 we prove that there exists a positive solution of (5) if and only if λ > 0. If a > 0 we complete and improve the results of Ref. 14 (see Theorems 5.4 and 5.5). November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 3 In order to study these equations we employ mainly sub-supersolution and bifurcation methods. We present in Sect. 2 results related with principal eigenvalues associated to these problems. In Sec. 3 we prove a general result of bifurcation from the trivial solution when the bifurcation parameter appears in both equation and boundary. As consequence, we can use it for equations (3) and (4). For the study of (5) we need a different result of bifurcation, where the parameter is in front of a non-linear term. In Sec. 4 we present results concerning to uniqueness, stability and a-priori bounds of positive solutions for general equations with nonlinear boundary conditions. Finally, in Sec. 5 we apply the results to the cited equations. 2. Some Preliminaries Results: eigenvalues problems Along this paper, we use the positive cone P:= {u∈ C1(Ω) : u≥0, u 6= 0 in Ω ∪Γ1, Bu = 0 on ∂Ω}, and we say that uis positive if u∈Pand that uis strongly positive if u∈int(P) := {u∈P:u > 0 in Ω ∪Γ1, ∂u/∂n < 0 on Γ0}, where nis the outward normal vector-field of Ω. On the other hand, the mixed operator B+m,m∈ C(Γ1), means a similar operator to (2) with b+minstead of bin B. Finally, given two functions u, v we write (u, v)>0 if u, v ≥0 and some of the inequalities non-trivial. Consider the eigenvalue problem ½Lϕ =λϕ in Ω, Bϕ = 0 on ∂Ω. H. Amann 2proved the existence of a unique simple eigenvalue, the principal eigenvalue, whose associated eigenfunction can be chosen strongly positive in Ω. We denote this eigenvalue by σ1[L, B]. σ1[L, D] and σ1[L, N] stand for the principal eigenvalues under Dirichlet and Neumann homogeneous boundary conditions, respectively. Some properties of σ1[L, B] have been studied in details by S. CanoCasanova and J. L´opez-G´omez 9(see also Ref. 4), we state some of them. Proposition 2.1. (1) σ1[L, B]>0if and only if there exists a positive supersolution of (L, B, Ω), i.e., a positive function usuch that Lu ≥0in Ωand Bu ≥0on ∂Ωwith some inequality strict. (2) The map q∈L∞(Ω) 7→ σ1[L+q, B]is increasing and continuous. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 4 (3) The map m∈ C(Γ1)7→ σ1[L, B +m]is increasing and continuous. (4) Suppose Γ16=∅and consider a sequence bn∈ C(Γ1)such that limn→+∞minΓ1bn= +∞.Then, lim n→+∞σ1[L, B +bn] = σ1[L, D]. (5) Suppose Γ16=∅, then σ1[L, B]< σ1[L, D]. Consider now the eigenvalue problem    Lϕ =λm(x)ϕin Ω, ϕ= 0 on Γ0, Bϕ=λr(x)ϕon Γ1. (6) We suppose the following condition m∈ Cα(Ω), r ∈ C1,α(Γ1),∃µ≥0 such that (c+µm, b +µr)>0. (7) The following result provides us the existence of principal eigenvalue of (6). The second paragraph gives a characterization of the principal eigenvalue of (6) when m≡0, i.e., an eigenvalue problem at the boundary, the classical Steklov problem. In our acknowledge this result is new, although it nearly follows by the results on Ref. 9(see Ref. 15 where a particular result is obtained.) Theorem 2.1. Assume (m, r)>0. Then: (1) Under condition (7), the eigenvalue problem (6) has a unique principal eigenvalue, γ1[L, B], it is simple and its associated eigenfunction can be chosen strongly positive in Ω. (2) If m≡0and r > 0, then, the principal eigenvalue exists for (6), denoted by λ1[L, B], if and only if σ1[L, D]>0. Moreover, its associated eigenfunction can be chosen strongly positive in Ω. Proof: The first paragraph follows with the same kind of arguments used in Theorem 2.2 of H. Amann 3where Γ0=∅. It is clear that λ1is a principal eigenvalue of (6) with m≡0 if and only if µ(λ1) = 0 where µ(λ) := σ1[L, B −λr(x)]. We know by Proposition 2.1 that limλ→−∞ µ(λ) = σ1[L, D], µ(λ) is a decreasing and continuous function. So, it suffices to prove that limλ→+∞µ(λ) = −∞. Suppose the contrary, then limλ→+∞µ(λ) = −l. Take k∈IR large enough such that k+c(x)>0 and k > l then, first part November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 5 of the Theorem can be applied to the eigenvalue problem    Lϕ +kϕ =eµϕ in Ω, ϕ= 0 on Γ0, Bϕ=λr(x)ϕon Γ1. Hence, there is a principal eigenvalue e λ1that verifies 0 = eµ(e λ1) = µ(e λ1)+k. This is a contradiction. ¤ The following result will be very useful along this work. Lemma 2.1. Assume (7) and (m, r)>0. Then γ1[L, B]>0⇐⇒ σ1[L, B]>0 Proof: We know by Theorem 2.1 that γ1[L, B] exists, and it is the unique zero of the application µ(σ) = σ1[L−σm, B −σr]. Since µis a decreasing function, then µ(0) >0 implies µ(σ0) = 0 for σ0=γ1[L, B]>0 and the contrary. ¤ 3. Bifurcation Results for Equations with Nonlinear Boundary Consider the nonlinear equation    Lu =λm(x)u+f(x, u) in Ω, u= 0 on Γ0, Bu=λr(x)u+g(x, u) on Γ1, (8) where f∈ Cα(Ω ×IR), g∈ C1,α(Γ1×IR), such that f(x, 0) = 0 ∀x∈Ω, g(x, 0) = 0 ∀x∈Γ1,(9) (m, r)>0 and satisfy condition (7) and λis a bifurcation parameter. Remark 3.1. Due to the condition (7) we can assume, adding µm and µr to both sides of (8), that (c, b)>0. Now, we reduce the equation (8) to a suitable equation for compact operators. Define Cα Γ0(Ω) = {v∈ Cα(Ω) : v|Γ0= 0}(analogously it can defined C2,α Γ0(Ω)) and the map K1:Cα Γ0(Ω) → C2,α Γ0(Ω) by, given f,K1(f) = uwhere uis the unique solution of the problem ½Lu =fin Ω, Bu = 0 on ∂Ω. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 6 We can extend this operator to CΓ0(Ω). Thanks to elliptic regularity results, this new operator, denoted again by K1, is compact as operator from CΓ0(Ω) to CΓ0(Ω). We define now K2:C(Γ1)→ C2,α Γ0(Ω) by, given g,K2(g) = u with uthe unique solution of the problem    Lu = 0 in Ω, u= 0 on Γ0, Bu=gon Γ1. Again, it can be proved that the operator K2:CΓ0(∂Ω) → CΓ0(Ω) is compact. Denote by γ:C(Ω) → C(Γ1) the trace operator. Following the same kind of arguments that in Ref. 3, Lemma 4.1, and denoting M,R,Fand G by the Nemitski operators associated to m(x)u,r(x)u,fand grespectively, we have Proposition 3.1. usatisfies u=K1[λM(u) + F(u)] + K2[λR(γ(u)) + G(γ(u))] if and only if uis a classical solution of (8). Since we are only interested in non-negative solutions of (8), we rewrite (8) as a problem with only non-negative solutions. Let u+= max{u, 0}. Lemma 3.1. If uis a solution of    Lu =λm(x)u++f(x, u+)in Ω, u= 0 on Γ0, Bu=λr(x)u++g(x, u+)on Γ1, (10) then u≥0. Proof: Suppose that the problem (10) possesses solution usuch that there exists a connected component Ω1⊂Ω of the set Ω0={x∈Ω : u(x)<0} such that u < 0 in Ω1. Observe that ∂Ω1∩Γ16=∅. Indeed, if Ω1⊂Ω, then Lu = 0 in Ω1,u= 0 on ∂Ω1. Since c≥0, then by the maximum principle u≡0 in Ω1. Hence, ∂Ω1∩Γ16= ∅. Due to Lu ≥0 in Ω1and c≥0 then, by the maximum principle, the minimum of umust be attained on ∂Ω1. As u < 0 in Ω1and u= 0 in ∂Ω1∩Γ0then, minimum must be attained on ∂Ω1∩Γ1, but in such points we have ∂u ∂ν =−b(x)u≥0, contradicting Hopf’s Lemma (see Lemma 3.4 in Ref. 16). ¤ Remark 3.2. Lemma 3.1 is still true if f(x, 0) ≥0 and g(x, 0) ≥0. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 7 Consider the maps Φλ,Φt λ:CΓ0(Ω) → CΓ0(Ω) defined by Φλ(u) = u−K1[λM(u+) + F(u+)] −K2[λR(γ(u+)) + G(γ(u+))], Φt λ(u) = u−tK1[λM(u+) + F(u+)] −tK2[λR(γ(u+)) + G(γ(u+))], t ≥0. Thanks to Proposition 3.1 and Lemma 3.1, uis a classical nonnegative solution of (8) if and only if Φλ(u) = 0 in CΓ0(Ω). Assume that lim s→0+ f(x, s) s= 0 unif. in Ω, lim s→0+ g(x, s) s= 0 unif. on Γ1. (11) Finally, denote by γ1:= γ1[L, B],and ξ1its strongly positive eigenfunction associated. Lemma 3.2. Let Λ⊂IR be a compact interval such that λ<γ1for all λ∈Λ. Then, there exists δ > 0such that Φt λ(u)6= 0 ∀u∈ CΓ0(Ω) with kukC(Ω) =kuk ∈ (0, δ),∀λ∈Λand ∀t∈[0,1]. Proof: Suppose the contrary, that there exist λn, tn∈IR and un∈ CΓ0(Ω) such that λn→λ,tn→t,kunk → 0 and Φtn λn(un) = 0. By Lemma 3.1, un≥0 and dividing by kunkwe obtain vn=tnK1µλnM(un) + F(un) kunk¶+tnK2µλnR(γ(un)) + G(γ(un)) kunk¶, (12) where vn=un kunk. Thanks to (11) we have that the terms inside K1and K2 are uniformly bounded in Ω and on Γ1, respectively. Since K1and K2are compact operators, then the sequence vnis a relatively compact in C(Ω). Therefore, we can suppose that vn→vin C(Ω). By (11), we have F(un) kunk→0 in C(Ω),G(γ(un)) kunk→0 on C(Γ1). Passing to the limit in (12), we conclude that v=t[λK1(M(v)) + λK2(R(γ(v)))]. Thanks to un≥0, kvnk= 1 and by the maximum principle, vis a strongly positive function in Ω. Due this fact λt =γ1but this is not possible because λt < γ1by the choice of the set Λ. ¤ We are going to use the following notation: for R > 0, let BR={u∈ CΓ0(Ω) : kuk< R}. Then, deg(Φλ, BR,0) stands for the degree of Φλon BRwith respect to 0, and i(Φλ, u0,0) denotes the index of the solution u0 of the equation Φλ(u) = 0. Corollary 3.1. If λ < γ1, then i(Φλ,0,0) = 1. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 8 Proof: If λ > 0 consider the interval Λ = [0, λ] in the contrary case consider Λ = [λ, 0]. Thanks to the Lemma 3.2, we know that ∃δ > 0 such that ∀u∈ CΓ0(Ω) with kuk ∈ (0, δ) we have Φt λ(u)6= 0, ∀t∈[0,1]. Therefore by homotopy invariance of the degree we obtain i(Φλ,0,0) = deg(Φ1 λ= Φλ, Bδ,0) = deg(Φ0 λ=I, Bδ,0) = 1. ¤ Lemma 3.3. Let λ > γ1. Then, there exists δ > 0such that ∀u∈ CΓ0(Ω) with kuk ∈ (0, δ),Φλ(u)6=τξ1,∀τ≥0. Proof: Assume that there exist sequences τn≥0, un∈ CΓ0(Ω) such that kunk → 0 and Φλ(un) = τnξ1. Thanks to Proposition 3.1 and similar arguments that we have employed in Lemma 3.1, we have that un>0 is a classical solution of the problem    Lun=λm(x)un+f(x, un) + γ1τnm(x)ξ1in Ω, un= 0 on Γ0, Bun=λr(x)un+g(x, un) + γ1τnr(x)ξ1on Γ1. Since by Remark 3.1 we can assume that (b, c)>0, positive constants are supersolutions of (L, B, Ω), and so by Proposition 2.1 it follows that σ1[L, B]>0, and so that by Lemma 2.1, γ1>0. Thanks to conditions (11), we obtain    Lun> λm(x)un−εunin Ω, un= 0 on Γ0, Bun> λr(x)un−εunon Γ1, Hence, unis strict positive supersolution of (L−λm(x)+ε, B−λr(x)+ε, Ω), and then δε(λ) = σ1[L−λm(x) + ε, B −λr(x) + ε]>0.(13) On the other hand, we know that γ1is the unique zero of the continuous and decreasing function δ(λ) = σ1[L−λm(x), B −λr(x)]. Since λ>γ1 then δ(λ)<0. Moreover, by Proposition 2.1, we infer that exists ε > 0 such that δε(λ)<0, contradicting (13). ¤ Corollary 3.2. If λ > γ1, then i(Φλ,0,0) = 0. Proof: Let ε∈(0, δ) where δis given Lemma 3.3. Since Φλis bounded on Bε, then by Lemma 3.3, there exists a > 0 such that Φλ(u)6=taξ1, ∀u∈Bε,∀t∈[0,1]. Hence, i(Φλ,0,0) = deg(Φλ, Bε,0) = deg(Φλ−aξ1, Bε,0) = 0. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 9 ¤ Let C ⊂ IR×CΓ0(Ω) be the closure of the set of positive solutions of (8). Then, Theorem 3.1. Assume that (m, r)>0, (7), (9) and (11). γ1is a bifurcation point from the trivial solution, and it is the only one for positive solutions. Moreover, there exists an unbounded continuum C0⊂ C of positive solutions emanating from (γ1,0). Proof: The result follows by Corollaries 3.1 and 3.2 and Ref. 5, Proposition 3.5. We only remark that the uniqueness of γ1follows with the same kind of arguments as in the proof of Lemma 3.2. ¤ Remark 3.3. (1) Assume that there exist constants c1, c2∈IR such that lim s→0+ f(x, s) s=c1unif. in Ω, lim s→0+ g(x, s) s=c2unif. on Γ1. Then, we can apply the above result to the problem L1u=λm(x)u+ f1(x, u) in Ω, u= 0 on Γ0and B2u=λr(x)u+g2(x, u) on Γ1, where L1=L−c1,B2=B − c2,f1(x, u) = f(x, u)−c1uand g2(x, u) = g(x, u)−c2u, and so f1and g2satisfy (11) (2) The case that m > 0, r≡0 (i.e. the bifurcation parameter only in the equation) can be included in the Theorem 3.1. Indeed, if b≥0 then (7) is verified. If b < 0 or changes sign we can perform a change u=eMψvwhere ψis the function that appears on Ref. 20, Proposition 3.4, and the original problem is transformed into a similar new problem where the new b, say e b > 0. (3) It is also possible to cover the case m≡0, r > 0 (i.e. the bifurcation parameter only at the boundary). Indeed, if σ1[L, D]≤0 then it can be proved that bifurcation from the trivial solution does not occur. Now, assume σ1[L, D]>0. By Proposition 2.1 there exists µr with µenough big such that σ1[L, B +µr]>0. Then, there exists a unique solution h > 0 in Ω of the problem    Lh = 1 in Ω, h= 1 on Γ0, (B+µr)h= 0 on Γ1. Now, we perform the change u=hv, which transforms the original problem into a new problem where the new c,ec > 0. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 16 (1) Respect bifurcation direction: (a) If p < r (resp. p>r) then bifurcation direction is supercritical (resp. subcritical). (b) If p=rbifurcation direction is supercritical (resp. subcritical) for |Ω|>|∂Ω|(resp. |Ω|<|∂Ω|). (2) If p=rand |Ω| ≤ |∂Ω|, (4) does not have positive solutions for λ≥0. (3) If p < 2r−1, (4) does not have positive solutions for λlarge enough. (4) If p≤rand λ≤0every positive solution is unstable. (5) If p < 2r−1and r < N N−2then every positive solution is bounded in L∞norm. (6) If p > 2r−1, there exists solution for all λ≥0. Proof: Due to Theorem 3.1, we have a unbounded continuum C0of positive (4) emanating from (σ1[−∆,N] = 0,0). We study the bifurcation direction. Consider λn→σ1[−∆,N] and its solutions associated un. Then, multiplying the equation by ϕ1=c > 0, the eigenfunction associated to the eigenvalue σ1[−∆,N], we obtain (σ1[−∆,N]−λn)ZΩ unϕ1dx =Z∂Ω ur nϕ1dσ −ZΩ up nϕ1dx. (24) Assume for example that p < r, multiply (24) by kunk−p C(Ω) and taking into account that un kunkC(Ω) →ϕ1in C(Ω) (see the proof of Lemma 3.2), it follows Sg(σ1[−∆,N]−λn) = Sg µ−ZΩ ϕp+1 1dx¶, hence, σ1[−∆,N]< λn. All results related to local bifurcation can be proved by the same way. Let ua positive solution of (4) with p=r. Then, if we multiply the equation (4) by 1/ur, and integrating by parts, we get −rZΩ u−r−1|∇u|2− |∂Ω|+|Ω|=λZΩ u1−r. Then, paragraph (2) follows. Assume that the problem (4) has a positive solution for every λ > 0. Consider the parabolic problem      wt−∆w=−wpin Ω ×(0, T), ∂w ∂n =wron ∂Ω×(0, T), w(x, 0) = w0in Ω. (25) November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 17 We know by Ref. 6, Theorem 2.3, that if p < 2r−1 then all positive solutions of (25) blow-up in finite time for w0with large L∞norm. Take uλa solution of (4), if we prove that uλis supersolution of (25) for large λ, then uλ(x)> w(x, t) for all t∈(0, T) which is a contradiction. In order to prove this, we only need that uλ> w0. It is clear that for λ > 0uλis supersolution of the problem    −∆v=λv −vpin Ω, ∂v ∂n = 0 on ∂Ω. (26) As solutions of (26) are, for λ > 0, λ1/(p−1) then uλ> λ1/(p−1).(27) Now, there exists λ > 0 large enough such that kw0k∞< λ1/(p−1) < uλ, this concludes paragraph (3). Let ua positive solution, we are going to prove that under condition p≤rthis solution is unstable. For that, thanks to Theorem 4.2, we have to show that σ1[−∆−λ+pup−1,N − rur−1]<0.(28) For this fact we choose as subsolution, u=uq, where qwill be fixed later. We have that ∂u ∂n −rur−1u= (q−r)uq+r−1on ∂Ω, and in Ω, (−∆−λ+pup−1)u=q(1 −q)uq−2|∇u|2+λuq(q−1) + up+q−1(p−q). Choosing qsuch that p≤q≤r, it follows (28), so that paragraph (4). By (27), uattains its maximum on ∂Ω. So, paragraph (5) follows by Theorem 4.4. For the last paragraph we only need to find a sub-supersolution of (4) for every λ≥0. We choose as subsolutiona u=εe−δφ1 where ε, δ > 0 can be chosen later and φ1is the positive eigenfunction associated to σ1=σ1[−∆, D] with kφ1k∞= 1/2. After some calculations we obtain ∇u=−εδe−δφ1∇φ1,∆u=−εδe−δφ1(−δ|∇φ1|2+ ∆φ1). aThis subsolution appears on Ref. 18 for the particular case λ= 0. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 18 Thanks to Hopf’s Lemma, it follows that max x∈∂Ω|∂φ1 ∂n |=C1, C1>0.(29) Since, φ1= 0 on ∂Ω, we only need to verify on the boundary that δC1≤εr−1.(30) In the equation we must check that −δ2|∇φ1|2−δσ1φ1+εp−1e−δφ1(p−1) ≤λ. (31) Observe that if λ > 0, we only need to choose εand δpositive and small enough for that (30) and (31) hold. So that, we are going to study the case λ= 0. From (30) we choose δ=εr−1 C1, so that (31) transforms into −ε2(r−1) C2 1 |∇φ1|2−εr−1 C1 σ1φ1+εp−1e−εr−1 C1φ1(p−1) ≤0.(32) Now, due to φ1= 0 on ∂Ω, but on the boundary ∂φ1/∂n < 0, there exist some constants C2, C3>0 such that |∇φ1| ≥ C2en Ω1:= {x∈Ω : φ1(x)≤C3}. (33) In this way, in Ω1for that the condition (32) must be fulfilled we need that p−1>2(r−1) thus p+ 1 >2r. On the other hand, in Ω \Ω1we need that p−1> r −1. The supersolution follows by Ref. 22, it was used also in Ref. 18, in both cases for the particular case λ= 0. We choose u:= MA[2 −(1 −φ1)B]C, where M > 0 will be chosen large and A, B and Cwill be fixed later. Observe that ∇u=BCMA[2 −(1 −φ1)B]C−1(1 −φ1)B−1∇φ1, ∆u=BCMA[2 −(1 −φ1)B]C−2(1 −φ1)B−2· ©(C−1)(1 −φ1)B|∇φ1|2+ [2 −(1 −φ1)B]∆φ1(1 −φ1)− [2 −(1 −φ1)B](B−1)|∇φ1|2ª. Taking into account (29), on the boundary it must be verified (observe that φ1= 0 and so that [2 −(1 −φ1)B] = 1): −BCMA(1−r)≥1 C1 .(34) November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 19 For the equation, we need that λ≤ −BC(C−1)[2 −(1 −φ1)B]−2(1 −φ1)2(B−1)|∇φ1|2 −BC[2 −(1 −φ1)B]−1∆φ1(1 −φ1)B−1 +BC(B−1)[2 −(1 −φ1)B]−1(1 −φ1)B−2|∇φ1|2 +MA(p−1)[2 −(1 −φ1)B]C(p−1). (35) Take A > 0, C=−1/C1and B=Mb, with bto be fixed later. With this choice, for condition (34) it will be needed b+A(1 −r)≥0.(=⇒b > 0).(36) Now, we study the term (35). First term in the right hand tends to −∞ or zero (the term (1 −φ1)2(B−1) can tend to zero). Second term is similar, we remind that −∆φ1=σ1[−∆, D]φ1. Third term tends to −∞ with order M2band the last one to +∞with order MA(p−1), so we have to impose A(p−1) >2b. This last inequality and (36) are possible because p+1 >2r. ¤ Remark 5.2. Except paragraphs (2) and (3), Theorem 5.3 is true for more general operators Land B. 5.3. The concave-convex equation Finally, we consider (5). Assume the following conditions c > c0>0 en Ω,with c0∈IR.(37) We distinguish two different cases: anegative and positive. Theorem 5.4. Assume that a < 0. The problem (5) has a positive solution vλif and only if λ > 0. For λ > 0, it is the unique positive solution, it is l. a. s. and lim λ→0+kvλk∞= 0.(38) Proof: Thanks to the maximum principle (5) does not posses nonnegative solutions for λ≤0. By Theorem 3.2, there exists a continuum C0emanating from (0,0) supercritically. On the other hand, v=Mϕ1is, for M large enough, supersolution of (5) where ϕ1is the positive eigenfunction associated to σ1[L, N]. This is true because σ1[L, N]>0, which it follows by (37). Since Mcan be chosen large enough that Mϕ1> vλfor λ > 0 small, where vλdenotes the solution of the problem founded by bifurcation. Then, we have a family of supersolutions such that a solution belonging to November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 20 the continuum is smaller than the supersolution. Now, adapting the proof of main theorem of Ref. 13, we have that there exists at least a positive solution for all λ > 0. Finally, for λ > 0 0 = σ1[L−λm(x)vq−1 λ,N −a(x)vr−1 λ]< σ1[L−λqm(x)vq−1 λ,N −ra(x)vr−1 λ], so the stability follows. Uniqueness follows by Theorem 4.3. ¤ Theorem 5.5. Assume that a > 0. (1) From (0,0) emanates supercritically an unbounded continuum C0of positive solutions. Moreover, it is the unique bifurcation point from the trivial solution. (2) There exists λ∗>0such that for λ > λ∗problem (5) does not have positive solutions. (3) There exists δ > 0such that there exists at most a positive solution uλof (5) such that kuλk∞≤δ. (4) Moreover, if Lis self-adjoint and r < N N−2then: (a) Pλ(C0) = (−∞,Λ], for some Λ<+∞. (b) There exists at least two positive solutions in (0,Λ). (c) There exists a unique positive solution in (0,Λ) l. a. s. Proof: Since the proof follows the same lines that Theorem 6.9 in Ref. 11, we only sketch it. The existence of the continuum C0follows by Theorem 3.2. We prove now that the bifurcation direction is supercritical. Assume that there exist λn≤0 and uλn∈ C(Ω), uλn≥0 such that (λn, uλn)→(0,0) in IR × C(Ω). Then, for n≥n0we get Luλn≤0 in Ω,∂uλn ∂n =a(x)ur λn< εaMuλnon ∂Ω, where aM= max∂Ωa. On the other hand, since σ1[L, N]>0 then, for ε > 0 small, σ1[L, N − εaM]>0, and applying the strong maximum principle we obtain that uλn≡0, a contradiction. Now, we are going to prove paragraph (2). Suppose that there exists positive solution uλof (5) for all λ, in particular for λ > 1. Let v1be the unique positive solution of    Lu =m(x)uqin Ω, ∂u ∂n = 0 on ∂Ω. (39) November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 21 Since uλ/λ is supersolution of (39) for λ > 1, then uλ> λv1for λ > 1. On the other hand, since uλis a positive solution of (5), we get 0 = σ1[L−λm(x)uq−1 λ,N − a(x)ur−1 λ]< σ1[L, N − λr−1ar−1 0vr−1 1], where a0= min∂Ωa(x). This is an absurdum. Indeed, since r > 1, we have lim λ→+∞σ1[L, N − λr−1ar−1 0vr−1 1] = −∞. Now, define Λ := sup{λ∈IR : (5) has a positive solution}. We have proved that 0 <Λ<+∞. Moreover, it is not difficult to prove the existence of a minimal solution uλfor all λ∈(0,Λ). The following result shows properties of the principal eigenvalue, denoted by γ1(λ), of the linearized around the minimal solution uλ, i.e.    Lξ −λqm(x)uq−1 λξ=γ1(λ)ξin Ω, ∂ξ ∂n −ra(x)ur−1 λξ=γ1(λ)ξon ∂Ω, (40) or equivalently, the unique zero of the map β(σ) = σ1[L−λqm(x)uq−1 λ−σ, N − ra(x)ur−1 λ−σ]. Lemma 5.1. (1) If uλis the minimal solution of (5), then γ1(λ)≥0. (2) If γ1(λ0)>0,for some λ0>0, then the set of positive solutions of (5) can be parametrized in a neighborhood of (λ0, u0)by a regular and increasing function on λ. (3) If γ1(λ0) = 0,for some λ0>0, then the set of positive solutions of (5) can be parametrized by a new parameter s∈(−ε, ε), such that (λ(s), u(s)) is a positive solution of (5) and λ(s) = λ0+s2λ2+o(s3), u(s) = uλ0+sΦ0+s2Ψ0+o(s3),(41) where Φ0is the positive eigenfunction associated to γ1(λ0)and RΩΦ0Ψ0= 0. Moreover, Sg(λ0(s)) = Sg(γ1(u(s))).(42) Finally, if Lis self-adjoint, λ2<0,(43) where λ2is defined in (41). November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 22 Remark 5.3. Except the first paragraph, the result is true for any positive solution unot necessarily being the minimal. Proof: (1) Assume that γ1<0 and denote by φ1the positive eigenfunction associated to γ1. It is not difficult to show (see Ref. 11) that uλ−αφ1is supersolution of (5), for α > 0 small. Since uλ> vλ, where vλis the unique positive solution obtained in Case 1 (a(x)<0), and vλis subsolution of (5), it follows the existence of a solution u<uλof (5), an absurdum because uλis the minimal solution. (2)-(3) Except (43), these two paragraphs follow by el Propositions 20.6, 20.7 and 20.8 of Ref. 1. Using (41) and the definition of Φ0, we get λ2=ZΩ λ0 q(1 −q) 2m(x)uq−2 λ0Φ3 0+Z∂Ω r(1 −r) 2a(x)ur−2 λ0Φ3 0 ZΩ m(x)uq λ0Φ0 . To determine the sign of λ2, we use the Picone’s identity, see for instance Lemma 4.1 in 19. Taking Ψ(t) = t2,v= Φ0and u=uλ0we get ZΩ λ0(1 −q)m(x)uq−2 λ0Φ3 0+Z∂Ω (1 −r)a(x)ur−2 λ0Φ3 0<0,(44) whence it follows that λ2<0. ¤ As an easy consequence we obtain: Corollary 5.1. Assume Lself-adjoint and let (λ0, u0)be a positive solution of (5) with λ=λ0>0, such that γ1(λ0) = 0 Then, there exists ε > 0 such that for each λ∈(λ0−ε, λ0), (5) has two positive solutions, one of them l. a. s and the other one linearly unstable. Moreover, there exists a neighborhood of (λ0, u0)such that (5) does not have a positive solution for λ > λ0. Now, we will prove paragraph (3) of Theorem 5.5. Assume that there exists a second solution w=uλ+v where v > 0 and kwk∞< δ. Consider v1the solution of (39). Then, there exists β > 0 such that 0 = σ1[L−m(x)vq−1 1,N]< β < σ1[L−qm(x)vq−1 1,N].(45) We claim that σ1[L−qm(x)vq−1 1,N − aMrδr−1]<0,(46) November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 23 which is an absurdum with (45). In order to prove (46), it suffices to prove that vis a positive subsolution of (L−qm(x)vq−1 1,N −aMpδp−1,Ω). Indeed, since wis solution of (5) and by the concavity of the map tq, it follows that Lv ≤λm(x)quq−1 λv. But, since λ1/(1−q)v1is subsolution of (5), then uλ> λ1/(1−q)v1, and hence (L−qm(x)vq−1 1)v < 0. On the other hand, a(x)[(uλ+v)r−ur λ]≤aMrδr−1v, whence ∂v ∂n −aMrδr−1v≤0. ¤ The following result shows that all the positive solution of (5) are unstable for λ≤0. Lemma 5.2. If uis a positive solution of (5) for λ≤0, then uis unstable. Proof: It suffices to prove that σ1[L−λqm(x)uq−1,N − a(x)rur−1]<0. First, observe that the first eigenvalue is well defined because minΩu > 0. It remains to find a positive subsolution of (L−λqm(x)uq−1,N − a(x)rur−1,Ω). It is hot hard to show that u=up λwith 1 < p ≤ris the desired subsolution. ¤ We are going to finish the proof of the Theorem. Since r < N N−2, by Theorem 4.4, we have that Pλ(C0) = (−∞,Λ]. We consider the set Γ := {(λ, uλ) : λ > 0, γ1(λ)>0}. We claim that Λ = sup Γ. By Lemma 5.1, the uniqueness of solution with small norm and Corollary 5.1, it follows that sup Γ = e λ > 0. It is clear that e λ≤Λ. Assume that e λ < Λ, then there exists λ0>e λsuch that uλ0 is supersolution of (5) for all λ≤λ0. Since we always can build small subsolutions, then there exists a solution uλfor λ < λ0. Since uλis built by the sub-supersolution, then γ1(uλ)≥0,and so, by Lemma 5.1 and Corollary 5.1 we have that there exists λ > e λsuch that γ1(uλ)>0. November 26, 2007 10:33 WSPC/Trim Size: 9in x 6in for Proceedings CM-ASuarez2 24 Now, we can continue this solution to the left, we call Γ0:= {(λ, uλ) : λ < λ} to this new set. It can occur four possibilities. First, there exists λ2∈(0,e λ) such that uλ2=uλ2, which is not possible because around a l.a.s. solution there is not another solution. Second, there exits λ3such that uλ3= 0. Recall that the unique bifurcation point is λ= 0, so this is not possible. Third, there exists uλfor all λ≤λ0, a contradiction with Lemma 5.2. Finally, there exists λ4such that γ1(uλ4) = 0, which is impossible by Lemma 5.1 and Corollary 5.1. This proves that e λ= Λ. With a similar reasoning it can be proved the uniqueness of l. a. s. positive solution. 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