Combining local and non-local terms in a nonlinear elliptic problem

Combining local and non-local terms in a nonlinear elliptic problem1 Francisco Julio S.A. Corrˆ ea1and Antonio Su´ arez2 1. Universidade Federal de Campina Grande Centro de Ciˆencias e Tecnologia Unidade Acadˆemica de Matem´atica e Estat´ıstica CEP:58.109-970, Campina Grande - PB - Brazil 2. Dpto. de Ecuaciones Diferenciales y An´alisis Num´erico Fac. de Matem´aticas, Univ. de Sevilla Calle Tarfia s/n, 41012-Sevilla, Spain E-mail addresses: [email protected], [email protected], Abstract In this paper we study the existence, uniqueness, multiplicity and stability of positive solution of a non-linear elliptic problem that combines local and non-local terms taking the form of an integral in space. The proofs are mainly based on fixed point theorems, bifurcation techniques, sub-supersolutions and continuation arguments. AMS Classification. ??. Keywords. Non-local elliptic equations, bifurcation techniques, a priori bounds. 1 Introduction Throughout this work we consider the following problem    −∆u=λup+ZΩ uβin Ω, u= 0 on ∂Ω, (1.1) where Ω ⊂IRNis a bounded regular domain, λ∈IR and p, β > 0. During recent years the so called non-local elliptic problems have attracted the attention of a lot of researchers due two main aspects: Firstly due to their mathematical importance. The presence of non-local terms provokes some difficulties which, sometimes, do not appear in the local ones. So, the behaviour of these problems may be, in general, distinct of their local counterpart. Secondly, these problems arise from practical motivations from Biology, Physics, Heat Transfer, Mechanics and so on, which makes their studies particularly interesting. See, for instance, the review paper [8]. In particular, in problem (1.1) there exists a combination of a local and a non-local terms in additive way. Observe that while for λ= 0 equation (1.1) is a non-local elliptic equation, when λ < 0 there is a competition between both terms. It is interesting to study 1AS have been supported by the Spanish Ministry of Science and Technology under Grant MTM200912367. August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez the behaviour of the set of positive solutions of (1.1) depending of the size of pand βand of course of the sign of λ. Problem (1.1) has been previously analyzed in [14] and [12], at least to our knowledge, only the case λ≤0, β > 1 and p≥1. In both works, the parabolic problem related to (1.1) was studied. In particular, both works showed the value p=βrepresents a critical blow-up exponent. Indeed, they proved that if β > p or β=pand λ > −|Ω|, the blow-up can occur in finite time. However, when β < p or β=pand λ≤ −|Ω|all the solutions are global and bounded. With respect to the elliptic problem (1.1), the authors proved the existence of positive solution for λsmall in the particular case λ < 0, p > β > 1. In this paper, we complete this study, and give results for all the values of pand β. Before proceeding to the statement of the main results, we need to introduce some notation. Given regular, non-negative and non-trivial functions a, b and m, we denote by λ1(−∆ + m;a, b) the principal eigenvalue of the following integro-differential eigenvalue problem −∆u+m(x)u−a(x)ZΩ b(x)u=λu in Ω, u= 0 on ∂Ω, (1.2) (see Section 2 for a detailed study of this problem). Denote also λ1:= λ1(−∆; 0,0) and σ1:= λ1(−∆; 1,1). We use the principal eigenvalues of (1.2) to characterize the stability of the solutions with respect to the parabolic counterpart problem. We say that a positive solution u0of (1.1) is stable (resp. unstable) if the principal eigenvalue of the linearization of (1.1) around u0 is positive (resp. negative), i.e., λ1(−∆−λpup−1 0;β;uβ−1 0)>0 (resp. <0.) We also say that u0is neutrally stable if it is zero. Observe that pand βcan be less than one, and so the eigenvalue problem (1.2) can have singular terms. We can now state our main results, which depend on the size of pand β. First, it is clear that if (p, β) = (1,1) then (1.1) is an eigenvalue problem and it possesses positive solution if λ=σ1. So, we assume that (p, β)6= (1,1). In the case p= 1 we can obtain: Theorem 1.1. Assume that p= 1. Then, there exists a unique positive solution of (1.1) for λ<λ1, and no positive solutions for λ≥λ1. Moreover, the solution is stable for β < 1 and unstable for β > 1. Finally, lim λ→λ1 kuk∞=(0when β > 1, +∞when β < 1.and lim λ→−∞ kuk∞=(+∞when β > 1, 0when β < 1. In Figure 1 we have represented the bifurcation diagrams corresponding to the case p= 1. Case 1 represents the solutions of (1.1) when β < 1 and and Case 2 shows the case β > 1. Observe that we have a bifurcation from zero when β > 1 and a bifurcation from infinity when β < 1 at λ=λ1. In the case p < 1, we get: Theorem 1.2. Assume that p < 1. 2 Combining local and non-local terms August 14, 2010 uu 1 1 Case Case Figure 1: Bifurcation diagrams for equation (1.1) for p= 1. a) Assume also that β= 1. (a) If σ1>0there exists a positive solution of (1.1) if, and only if, λ > 0. The solution is unique and stable. Moreover, lim λ→0kuλk∞= 0,lim λ→∞ kuλk∞=∞. (b) If σ1= 0 there exists a positive solution of (1.1) if, and only if, λ= 0. There are infinite positive solutions and they are neutrally stable. (c) If σ1<0there exists a positive solution of (1.1) if, and only if, λ < 0. The solution is unique and unstable. Moreover, lim λ→0kuλk∞= 0,lim λ→−∞ kuλk∞=∞. b) Assume also that β > 1. There exists a value λ > 0such that there exists a positive solution of (1.1) if and only if λ≤λ. There is a unique and unstable positive solution for λ≤0and at least two solutions, uλ 1< uλ 2, for λ > 0and small, uλ 1is stable and uλ 2unstable. Moreover, lim λ→−∞ kuλk∞=∞and lim λ→0kuλ 1k∞= 0. c) Assume now that β < 1. (a) If β < p there exists a positive solution of (1.1) for all λ∈IR. The solution is unique and stable. Moreover, lim λ→−∞ kuλk∞= 0 and lim λ→∞ kuλk∞=∞. (b) If β=pthere exists λ0<0such that there exists positive solution if and only if λ>λ0. In fact, λ0∈(−|Ω|,−RΩϕp 1), being ϕ1>0the eigenfunction associated to λ1such that kϕ1k∞= 1. Furthermore, the solution is unique and stable and lim λ↓λ0 kuλk∞= 0 and lim λ→∞ kuλk∞=∞. 3 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez (c) If p<βthere exists λ0<0such that there exists positive solution if and only if λ≥λ0. Moreover, for λ≥0the solution is unique and stable, and for λ negative and small there exist at least two positive solutions, uλ 1< uλ 2,uλ 1is unstable and uλ 2stable. Moreover, lim λ→0kuλ 2k∞= 0 and lim λ→∞ kuλk∞=∞. In Figure 2 we have drawn the bifurcation diagrams of (1.1) corresponding to the case p < 1. Cases 1, 2 and 3 represent the solutions when β= 1 and σ1>0, σ1= 0 and σ1<0, respectively. Case 4 shows the case β > 1, and when β < 1 we have the Cases 5, 6 and 7 when β > p,β=pand β < p, respectively. u u u u u Case 1 Case 2 Case 3 Case 4 u Case 6 Case 5 u Case 7 Figure 2: Bifurcation diagrams for equation (1.1) for p < 1. Let us compare some of our results with the well-known ones of the local equation −∆u=λup+uβ. In the case p= 1 the existence results are rather similar to the local case. However, for the case β > 1 in the non-local case we do not need impose the condition β < (N+2)/(N−2) 4 Combining local and non-local terms August 14, 2010 to obtain the existence of a priori bounds. Moreover, in this case we show that the solution is unstable (similar to the local case) but the solution is unique, unlike the local case. With respect to the case p < 1 we would like to point out that in the non-local case any non-negative and non-trivial solution is positive in all Ω. This contrasts with the local case in which for λnegative could exist non-negative and non-trivial solutions that vanishes in a part of Ω, the dead core. Observe that in the case p < 1< β the result obtained is rather similar to the case of the local equation studied in [1]. However, again in our case we do not need to impose the condition β < (N+ 2)/(N−2). Also, the result obtained in the case p < β < 1 is similar to the local equation analyzed in [7]. Let us remark that to obtain the existence results in the previous results, we can not use the variational methods due to the equation (1.1) has not a variational structure. In fact, we have used basically a fixed point argument and the sub-supersolution method to obtain above results. For the case p > 1 we are not able to use the fixed point argument. So, we have introduced our equation (1.1) in a more general equation, see equation (4.18), and use bifurcation methods and classical results from [2]. For that, we need to obtain a priori bounds of positive solutions of (1.1). This is not a trivial problem. We distinguish two cases. When λ < 0 we obtain a priori bounds except in the case β=p≥1. The case λ > 0 is harder. Basically, we have used to different arguments: boot-strapping and blow-up arguments to obtain the results. For the case λ > 0 we have proved that if p < 1,∀β > 0 or p= 1, β > 1,(1.3) or, 1<p<(N+ 2)/(N−2),∀β > 0,(1.4) or p≥(N+ 2)/(N−2),and β > (N/2)(p−1),(1.5) then there exist a priori bounds of (1.1). Observe that (1.4) is the classical restriction in the local case. On the other hand, (1.3) means that when pis small, we obtain a priori bounds for all the values of β; while (1.5) gives a priori bounds when βis large, even when pis greater that critical exponent (N+ 2)/(N−2). Moreover, these results are optimal in some way, because for λnegative and β=p > and for λpositive and β < 1 = pwe prove that there exist a bifurcation from infinity for some λ, and so a priori bounds do not exist. Moreover, we show that for p=β > (N+ 2)/(N−2) there is not positive solution for λlarge. Theorem 1.3. Assume that p > 1. a) Assume that β= 1. (a) Suppose that σ1>0. If there exists a positive solution then λ > 0. If λ > 0and p < (N+ 2)/(N−2) then there exists at least a positive solution. The solution is unstable and lim λ→0kuλk∞=∞and lim λ→∞ kuλk∞= 0.(1.6) (b) If σ1= 0 there exists a positive solution if, and only if, λ= 0. There are infinite positive solutions and they are neutrally stable. 5 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez (c) If σ1<0there exists a positive solution if, and only if, λ < 0. The solution is unique and stable and and lim λ→−∞ kuλk∞= 0,lim λ→0kuλk∞=∞.(1.7) b) Assume also that β > 1. (a) If β > p there exists a unique and unstable positive solution for λ≤0. If, moreover, there exist a priori bounds, there exists positive solutions for λ > 0 and it is unstable. Moreover, lim λ→−∞ kuλk∞=∞. (b) If β=p, there exists λ0<0such that (1.1) possesses positive solution for λ∈(λ0,0] and lim λ→λ0 kuk∞= +∞.(1.8) Moreover, this solution is unique and unstable. If, moreover, there exist a priori bounds, there exists positive solutions for λ > 0 and it is unstable. (c) If β < p, there exists λ0<0such that (1.1) possesses positive solution for λ∈[λ0,0]. Moreover, if λis small and negative, there exist at least two positive solutions, uλ 1< uλ 2,uλ 1is unstable and uλ 2stable and lim λ↑0kuλ 2k∞= +∞. If, moreover, there exist a priori bounds, there exists positive solutions for λ > 0 and it is unstable. c) Assume now that β < 1. There exists a unique and stable positive solution of (1.1) for λ≤0. Assume now the existence of a a priori bounds. There exists λ > 0such that there exists a positive solution if, and only if, λ≤λ. Moreover, λ > 0and small there exist at least two positive solutions uλ 1, uλ 2,uλ 1is stable and lim λ→−∞ kuλk∞= +0,lim λ↑0kuλ 2k∞= +∞. In Figure 3 we have represented the bifurcation diagrams of (1.1) corresponding to the case p > 1. Cases 1, 2 and 3 represent the solutions when β= 1 and σ1>0, σ1= 0 and σ1<0, respectively. Cases 4, 5 and 6 show the cases β > p,β=pand β < p, respectively. Finally, Case 7 represents β < 1. An outline of the paper is: in Section 2 we study the eigenvalue problem and some preliminaries results; Section 3 is devoted to obtain a priori bounds of positive solutions of (1.1) and in the last Section we prove Theorems 1.1, 1.2 and 1.3. 6 Combining local and non-local terms August 14, 2010 u u u u u Case 1 Case 2 Case 3 Case 4 u Case 6 Case 5 u Case 7 Figure 3: Bifurcation diagrams for equation (1.1) for p > 1. 2 The eigenvalue problem and preliminaries results In this section we study a non-local and singular eigenvalue problem, which appears when one linearizes around a positive solution of (1.1). Specifically, we study the following problem    −∆u+m(x)u−a(x)ZΩ b(x)u=σu in Ω, u= 0 on ∂Ω, (2.1) where m∈C1(Ω), a∈C(Ω) and b∈C1(Ω) and verify: for some α∈(−1,1) and γ < 1 (Hm)|∂im|d(x, ∂Ω)2−αare bounded for all x∈Ω and i= 1, ..., N; (Hb) there exists K > 0 such that b(x)≤Kd(x, ∂Ω)−γ, where d(x, ∂Ω) := dist(x, ∂Ω) The next result was proved in [4]: 7 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez Theorem 2.1. Assume that mverifies (Hm),a∈C1(Ω) ∩C(Ω), is a non-negative and non-trivial function, b∈C1(Ω) is a non-negative and non-trivial function and it verifies (Hb). Then, there exists a principal eigenvalue of (2.1), denoted by λ1(−∆ + m;a;b), which has an associated positive eigenfunction ϕ1∈C2(Ω) ∩C1,δ 0(Ω) for some δ∈(0,1), and ∂ϕ1 ∂n <0on ∂Ω,(2.2) where ndenotes the outward unit normal vector. Moreover, λ1(−∆ + m;a;b)is simple, and it is the unique eigenvalue having an associated eigenfunction without change of sign. In the following result we give a criteria to ascertain the sign of λ1(−∆ + m;a;b), see also [4]: Proposition 2.2. a) Assume that there exists a positive function u∈C2(Ω)∩C1,δ 0(Ω), δ∈(0,1), such that −∆u+m(x)u−a(x)ZΩ b(x)u > 0in Ω. Then, λ1(−∆ + m;a;b)>0. b) Assume that there exists a positive function u∈C2(Ω) ∩C1,δ 0(Ω),δ∈(0,1), such that −∆u+m(x)u−a(x)ZΩ b(x)u < 0in Ω. Then, λ1(−∆ + m;a;b)<0. Along the paper, we are going to denote by λ1:= λ1(−∆; 0; 0) and σ1:= λ1(−∆; 1; 1). The next result characterizes the sign of λ1(−∆ + m;a;b) on terms of the solution of the problem (−∆ζ+m(x)ζ=b(x) in Ω, ζ= 0 on ∂Ω. (2.3) Thanks to Proposition 2.5 in [11] if λ1(−∆ + m; 0; 0) >0 there exists a unique positive solution ζ∈C2(Ω) ∩C1,δ 0(Ω), δ∈(0,1), of (2.3). Lemma 2.3. Assume that λ1(−∆ + m; 0; 0) >0. Then, sgn (λ1(−∆ + m;a;b)) = sgn 1−ZΩ a(x)ζ. Proof. Denote by ϕ1a positive eigenfunction associated to λ1(−∆ + m;a;b). Multiplying (2.3) by ϕ1, and integrating we obtain that 1−ZΩ a(x)ζZΩ b(x)ϕ1=λ1(−∆ + m;a;b)ZΩ ϕ1ζ. This concludes the result. 8 Combining local and non-local terms August 14, 2010 The next result shows the monotony of the principal eigenvalue with respect to the domain. Lemma 2.4. Consider a sub-domain Ω0⊂Ω, and that the functions a, b and mverify the conditions of Theorem 2.1. Denote by λ0and λ1the principal eigenvalues λ1(−∆+m;a;b) in Ω0and Ω, respectively. Then, λ1< λ0. Proof. Consider ϕ∗ 0the adjoint positive eigenfunction associated to λ0, that is −∆ϕ∗ 0+m(x)ϕ∗ 0−b(x)ZΩ0 a(x)ϕ∗ 0=λ0ϕ∗ 0in Ω0,ϕ∗ 0= 0 on ∂Ω0. Then, prolonging ϕ∗ 0by zero at Ω, and multiplying by ϕ1, a positive eigenfunction associated to λ1, we get ZΩ0 a(x)ϕ∗ 0ZΩ0 b(x)ϕ1−ZΩ b(x)ϕ1+Z∂Ω0 ∂ϕ∗ 0 ∂n ϕ1= (λ1−λ0)ZΩ0 ϕ∗ 0ϕ1, whence, using (2.2), we deduce that λ1< λ0. With respect to the monotony on the potentials, we have: Lemma 2.5. Assume that m1≤m2,a1≥a2and b1≥b2. Then, λ1(−∆ + m1;a1;b1)≤λ1(−∆ + m2;a2;b2). Proof. Let ϕ > 0 an eigenfunction associated to λ1(−∆ + m1;a1;b1). Then −∆ϕ+m2ϕ−λ1(−∆+m1;a1;b1)ϕ−a2ZΩ b2ϕ= (m2−m1)ϕ+a1ZΩ b1ϕ−a2ZΩ b2ϕ≥0, and so, by Proposition 2.2, λ1(−∆ + m2−λ1(−∆ + m1;a1;b1); a2;b2)≥0, that is, λ1(−∆ + m2;a2;b2)≥λ1(−∆ + m1;a1;b1). Finally, the following result will be very useful during the work: Lemma 2.6. Assume a > 0in Ω. It holds that lim λ→+∞λ1(−∆ + m;λa;b) = −∞. Proof. Consider a ball B⊂Ω such that b≥b0>0 in Band such that λB 1(−∆+m; 0; 0) > 0. By Lemma 2.4 λΩ 1(−∆ + m;λa;b)< λB 1(−∆ + m;λa;b). We are going to prove that λB 1(−∆+m;λa;b)→ −∞ as λ→+∞. Indeed, since λB 1(−∆+ m; 0; 0) >0 there exists a unique positive solution, denoted by e, of the equation −∆e+m(x)e=b(x) in B,e= 0 on ∂B. 9 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez Proposition 3.6. Assume that 0<β<p<(N+ 2)/(N−2),p > 1, and λ∈Λ, with Λ⊂IR+compact such that 0/∈Λ. Then, there exists a priori bound of positive solutions of (1.1). Proof. We use again a Gidas-Spruck argument. With the same notation that Proposition 3.4 we get −∆wn=λwp n+M−p nZΩ uβ nin Ωn. But, M−p nZΩ uβ n≤M−p+β n|Ω| → 0, and so passing to the limit we again obtain −∆w=λwpin IRNor IRN +. Finally, we analyze the case p≤1. Observe that when p= 1 > β we will show that there exists bifurcation from infinity at λ=λ1>0 (see Theorem 1.1). So, we study the case p < 1 and β≤1. Proposition 3.7. Assume that p < 1and β≤1, and λ∈Λ, with Λ⊂IR+compact such that 0/∈Λ. Then, there exists a priori bound of positive solutions of (1.1). Proof. Assume that there exists a sequence λn→λ0>0 and positive solutions unof (1.1) such that kunk∞→ ∞. Denote by wn:= un kunk∞ . It is clear that wnverifies −∆wn=λnwp nkunkp−1 ∞+kunkβ−1 ∞ZΩ wβ nin Ω, wn= 0 on ∂Ω. Then, wn→win C2(Ω) being wa solution of −∆w= 0 if β < 1−∆w−ZΩ w= 0 if β= 1. In the firs case, it is clear that w≡0. In the second one, since there exists positive solution for λn>0 we get that σ1>0, and then w≡0. In both cases, we arrive at contradiction because kwk∞= 1. In the following result, we show that (1.1) does not possess classical positive solutions for β=p > (N+ 2)/(N−2) and λlarge. Proposition 3.8. Assume that Ωis bounded and starshaped with respect to some point x0∈Ω,β=p > (N+ 2)/(N−2) and λ > C(N), for some positive constant depending on N. Then, (1.1) does not possess positive solution. 16 Combining local and non-local terms August 14, 2010 Proof. We are going to use a Pohozaev’s argument, see for instance Chapter 1.5 in [12]. Multiplying (1.1) by x· ∇uwe get N−2 2ZΩ |∇u|2+1 2Z∂Ω ∂u ∂n 2 x·n=Nλ p+ 1 ZΩ up+1 +NZΩ uβZΩ u, and then 0<1 2Z∂Ω ∂u ∂n 2 x·n=λN p+ 1 −N−2 2ZΩ up+1 +N+ 2 2ZΩ uβZΩ u. By H¨older inequality, we get (using β=p) 0<ZΩ uβZΩ u≤C(Ω) ZΩ up+1. Hence 0<1 2Z∂Ω ∂u ∂n 2 x·n≤λN p+ 1 −N−2 2+C(Ω)N+ 2 2ZΩ up+1, an absurdum for λlarge. With a completely analogous argument, we can prove: Corollary 3.9. Assume that Ωis bounded and starshaped with respect to some point x0∈Ω,β=p < (N+ 2)/(N−2) and λ < −C(N), for some positive constant depending on N. Then, (1.1) does not possess positive solution. 4 Proof of the main results In this section we prove the main results of the paper stated in Section 1. Firstly, observe that if (p, β) = (1,1) then (1.1) is an eigenvalue problem, and so there exist positive solutions if, and only if, λ=σ1. Recall that sgn(σ1) = sgn(1 −RΩe) where eis defined in (3.1). So, from now on we assume that (p, β)6= (1,1). Also, for λ= 0 and β6= 1 there exists a unique positive solution u=eZΩ uβ=⇒u=eZΩ eβ1/(1−β) . (the case β= 1 is an eigenvalue problem). Moreover, by Proposition 2.9, uis stable for β < 1 and unstable for β > 1. So, we assume λ6= 0. 4.1 Some useful results A first attempt to study (1.1) is consider R=ZΩ uβ, 17 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez and then we have to study the equation (−∆u=λup+Rin Ω, u= 0 on ∂Ω,(4.1) and after that, to find a point fixed of R=ZΩ uβ R⇐⇒ 1 = ZΩ wRβ≡h(R) (4.2) being uRa positive solution of (4.1) and wR=uR/R1/β and so positive solution of (−∆w=λR(p−1)/βwp+R(β−1)/β in Ω, w= 0 on ∂Ω.(4.3) In the following result we study in detail the map R7→ h(R). Proposition 4.1. Assume R > 0. a) Assume p= 1. Then (4.3) possesses a positive solution, denoted by wR, if, and only if, λ<λ1. The solution is unique. Moreover, wR=eλRβ−1 β,(4.4) being eλthe unique positive solution of ((−∆−λ)eλ= 1 in Ω, eλ= 0 on ∂Ω.(4.5) b) Assume p < 1. Then (4.3) possesses a unique positive solution, denoted by wR, for all λ∈IR. Moreover, the map R∈(0,∞)7→ h(R)is continuous and derivable. For λ > 0 lim R→0h(R) = ∞. (a) When β= 1. i. If λ > 0,R7→ h(R)is decreasing and lim R→∞ h(R) = ZΩ e. ii. If λ < 0,R7→ h(R)is increasing and lim R→0h(R) = 0,lim R→∞ h(R) = ZΩ e. (b) When β > 1, i. If λ > 0 lim R→∞ h(R) = ∞. 18 Combining local and non-local terms August 14, 2010 ii. If λ < 0,R7→ h(R)is increasing and lim R→0h(R)=0,lim R→∞ h(R) = ∞. (c) When β < 1and λ > 0,R7→ h(R)is decreasing, and for all λ lim R→∞ h(R)=0. Moreover, i. If β < p and λ < 0,R7→ h(R)is decreasing, and for all λ lim R→0h(R) = ∞. ii. If β=p, the map R7→ h(R)is decreasing and lim R→0h(R) = (∞if λ > 0, ρ0(λ)if λ < 0, where ρ0(λ)∈(−1/λ)ZΩ ϕβ 1,(−1/λ)|Ω|,(4.6) ϕ1is the positive eigenfunction associated to λ1such that kϕ1k∞= 1, and ρ0(λ)is a non-decreasing function in λfor λ < 0. iii. If β > p, lim R→0h(R) = (∞if λ > 0, 0if λ < 0. c) Assume p > 1and λ < 0, then there exists a unique positive solution, denoted by wR, of (4.3). Moreover, the map R∈(0,∞)7→ h(R)is continuous and derivable. (a) When β= 1. The map R7→ h(R)is decreasing and lim R→0h(R) = ZΩ e, lim R→∞ h(R)=0. (b) When β > 1. i. If β > p,R7→ h(R)is increasing and lim R→0h(R)=0,lim R→∞ h(R)=+∞. ii. If β=p, the map R7→ h(R)is increasing lim R→0h(R)=0,lim R→∞ h(R) = ρ0(λ), with ρ0(λ)as in (4.6). iii. If β < p lim R→0h(R) = 0,lim R→∞ h(R) = 0. 19 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez (c) When β < 1. The map R7→ h(R)is decreasing and lim R→0h(R)=+∞lim R→∞ h(R)=0. Proof. a) Assume that p= 1, then wRverifies (−∆−λ)wR=R(β−1)/β and the result of paragraph a) is obtained easily. For the other cases, it is clear that (w,w) = (0, Ke) is a pair of sub-supersolution of (4.3) for Kverifying K≥λKpepR(p−1)/β +R(β−1)/β.(4.7) It is enough to take Klarge in any case. On the other hand, for λ≥0 and p < 1 the uniqueness follows by [3] and for λ < 0 thanks to λR(p−1)/βwpis a decreasing map in R. The continuity and derivability of the map R7→ wRis standard. If λ≥0 we have that −∆w≥R(β−1)/β and so wR≥R(β−1)/βe. (4.8) Analogously, if λ≤0 we have that wR≤R(β−1)/βe. (4.9) Moreover, if λ≥0 and p < 1 we have that −∆w≥λwpR(p−1)/β and then wR≥R−1/βλ1/(1−p)w1,(4.10) where w1is the unique positive solution of (2.6). Also, by the maximum principle for λ < 0 we get kwRkp ∞≤R(β−p)/β −λ.(4.11) Finally, εϕ1is subsolution of (4.3), kϕ1k∞= 1, if ελ1≤λεpϕp 1R(p−1)/β +R(β−1)/β.(4.12) b) Assume that p < 1. Then, it is clear by (4.10) that for λ > 0 lim R→0h(R)=+∞, and for β > 1 by (4.8) lim R→∞ h(R)=+∞. Take now λ≤0, then it is clear by (4.9) that lim R→0h(R) = 0 if β > 1, lim R→∞ h(R) = 0 if β < 1, and by (4.11) lim R→0h(R) = 0 if β > p, lim R→∞ h(R) = 0 if β < p. 20 Combining local and non-local terms August 14, 2010 Now, consider β > 1 and λ < 0. Observe that the map R7→ λR(p−1)/β +R(β−1)/β is increasing, and so R7→ wRalso. In this case, for (4.12) is enough ελ1−λεpR(p−1)/β =R(β−1)/β.(4.13) From this equality we deduce that ε(R)→ ∞ as R→ ∞, and then h(R)→ ∞. For β < 1 we have to distinguish several cases. If β < p and λ < 0 then again by (4.13), we get that ε(R)→ ∞ as R→0. For the case β=pwe have that εp(R)→ −1 λas R→0. Moreover, by (4.11) we deduce that h(R)≤−1 λ|Ω|. For λ > 0, for (4.7) is enough K−λkekp ∞KpR(p−1)/β =R(β−1)/β.(4.14) If p, β < 1 it is clear that K(R)→0 as R→ ∞. Finally, for β= 1 and λ > 0 observe that e≤wR≤K(R)e and K(R)→1 as R→ ∞ by (4.14). For λ < 0, wR≤eand ε(R)eis subsolution if ε−λεpCRp−1= 1.(4.15) It is clear that ε(R)→1 as R→ ∞. Observe also that in the particular case β=p,wRverifies −∆w=R(β−1)/β(λwp+ 1) and then, by the maximum principle λwp(x)+1≥0 for all x∈Ω. (4.16) Indeed, if λ≥0 then (4.16) is clear . If λ < 0 observe that λwp(x)+1≥λkwkp ∞+ 1 ≥0. Then, if R1< R2and β < 1, we get that wR2is sub-solution of (4.3) for R=R1, and then wR2< wR1. This proves that R7→ h(R) is decreasing. Hence, there exists the following limit lim R↓0h(R) := ρ0(λ). Moreover, if λ1< λ2<0, wλ1,R is subsolution of the equation (4.3) with λ=λ2, and so wλ1,R < wλ2,R. Taking limit we have that ρ0(λ1)≤ρ0(λ2). 21 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez Finally, assume that β≤p. Observe that wRverifies −∆w=R(β−1)/β(λR(p−β)/βwp+ 1). Observe that by the maximum principle and since λ < 0 we get λR(p−β)/βwp(x)+1≥λR(p−β)/βkwkp ∞(x)+1≥0. Take R1< R2, then R(β−1)/β 1(λR(p−β)/β 1wp R1+ 1) ≥R(β−1)/β 2(λR(β−p)/β 2wp R1+ 1), and then wR1is a supersolution of the equation (4.3) with R=R2. We conclude that wR1≥wR2. c) Assume that p > 1 and λ < 0. From (4.9) we have that h(R)→0 as R→0 if β > 1 and h(R)→0 as R→ ∞ if β < 1. In this case if β < 1 it is clear that ε(R)→ ∞ as R→0 from (4.13). Now, assume β > 1. If p<βthen ε(R)→ ∞ as R→ ∞, if p=β,εp(R)→ −1/λ and for p>βwe have that K(R)→0. Finally, for β= 1, and using again (4.15), we have that h(R)→RΩeif R→0. In the following result we prove a stability result of a positive solution u0of (1.1), obtained such that u0=uR0for some R0>0, in function on the map hdefined in (4.2). Proposition 4.2. Let u0be a positive solution of (1.1) obtained such that u0=uR0for some R0>0. Then, if h0(R0)<0(resp. h0(R0)>0) then u0is stable (resp. u0is unstable). Proof. Let u0=uR0a positive solution of (1.1). Assume that h0(R0)<0, we want to show that λ1(−∆−λpup−1 0;β;uβ−1 0)>0,(4.17) (analogous argument in the case h0(R0)>0). First, observe that the map R7→ uRis increasing (uRdefined in (4.1)), and so its derivative u0 R>0 in Ω, being u0 Rthe unique solution of −∆u0 R=λpup−1 Ru0 R+ 1 in Ω, u0 R= 0 on ∂Ω. On the other hand, observe that since h0(R0)<0 and using that h(R) = (1/R)ZΩ uβ R, we get βZΩ uβ−1 R0u0 R0<ZΩ uβ R0 R0 =h(R0)=1. To prove (4.17) we use Proposition 2.2 with u=u0 R0>0. Indeed, observe that −∆u0 R0−λpup−1 0u0 R0−βZΩ uβ−1 0u0 R0= 1 −βZΩ uβ−1 0u0 R0>0, and then the stability follows. 22 Combining local and non-local terms August 14, 2010 For the case p > 1 and λ > 0 we work with the original equation. In fact, assume β≥1 and consider the following auxiliar problem:    −∆u=µu +λup+ZΩ uβin Ω, u= 0 on ∂Ω. (4.18) Lemma 4.3. Assume p > 1,β≥1and λ > 0. Then, for µ=λ1when β > 1and µ=σ1when β= 1 bifurcates from the trivial solution a non-bounded continuum Cof positive solutions of (4.18). Moreover, assuming the existence of a priori bound of (4.18) for µ∈Λ,Λa compact subset of IR, there exists a positive solution if, and only if, µ<λ1 if β > 1and µ<σ1for β= 1. Proof. First, observe that if uis a positive solution of (4.18) we have µ=λ1(−∆−λup−1; 1; uβ−1)< λ1(−∆; 1; uβ−1), that is, µ<λ1if β > 1 and µ<σ1in the case β= 1. That µ=λ1for β > 1 and µ=σ1for β= 1 is a bifurcation point from the trivial solution is consequence of the Crandall-Rabinowitz Theorem [5], see also [6]. The existence of an unbounded continuum Cfollows by the classical Rabinowitz Theorem [13]. 4.2 Proof of Theorem 1.1 Assume that p= 1. It is clear that (1.1) does not possess positive solution for λ≥λ1. By Proposition 4.1 a) there exists a unique positive solution for λ<λ1. The stability results follow by Proposition 2.9 a) and b). We study now the behaviour with respect to λ. Observe that if uis a positive solution of (1.1) we have (−∆−λ)u=ZΩ uβ, and so, u=eλZΩ eβ λ1/(1−β) , and so taking into account that for ϕ1>0 with kϕ1k∞= 1, ϕ1eigenfunction associated to λ1,1 λ1−λϕ1≤eλin Ω, we get that for β < 1, u≥1 (λ1−λ)1/(1−β)ϕ1ZΩ ϕβ 11/(1−β) and so kuk∞→ ∞ as λ→λ1. Assume that β > 1 and consider a sequence λn< λ1,λn→λ1and unthe positive solution of (4.18) for λ=λn. We know by Proposition 3.3 that kunk∞is bounded, and so passing to the limit we get that un→u0in C2(Ω) as λn→λ1, with u0positive solution for λ=λ1. Then, u0≡0. Finally, the behaviour as λ→ −∞ follows by Lemma 2.8. 23 August 14, 2010 F. J. S. A. Corrˆea and A. Su´arez 4.3 Proof of Theorem 1.2 a) Assume p < 1 = β. Assume that σ1>0, then it is clear that λ > 0. Observe that since σ1>0, applying Lemma 2.3 with a≡b≡1 and m≡0 we get that RΩe < 1. Now, the existence and uniqueness follow by Proposition 4.1 b). The stability follows by Proposition 2.9. Finally, observe that (see (2.7)) uλ=λ1/(1−p)u1, being u1the unique solution of (1.1) for λ= 1. From here, we can deduce the behaviour as λ→0 and λ→ ∞. The other cases can be treated similarly. b) Assume p < 1< β. The existence and uniqueness in the case λ < 0 follow by Proposition 4.1. Also, the stability follows by Proposition 2.9. Now consider λ > 0. Denote eR:= R(β−1)/βe. Take R0>0 small such that ZΩ eβ R0=Rβ−1 0ZΩ eβ<1. Fix such R0>0. Now, it is clear wR0→eR0in L∞(Ω) as λ→0, hence h(R0)<1 for λ≤λ0, with λ0small. So, since lim R→0h(R) = lim R→∞ h(R) = +∞, there exist at least two positive values R1 0< R0< R2 0such that h(Ri 0) = 1, i= 1,2, and so two positive solutions uλ i=uRi 0of (1.1) for λ≤λ0with u1< u2, and h0(R1 0)<0< h0(R2 0). Thank to Proposition 4.2 we have that uλ 1is stable and uλ 2unstable. Now, we show that there does not exist positive solutions of (1.1) for λlarge. Observe that u≥λ1/(1−p)w1,(4.19) where w1is defined in (2.6), and then −∆u≥λ1/(1−p)wp 1+λ(β−1)/(1−p)ZΩ wβ−1 1u and so λ1(−∆; λ(β−1)/(1−p);wβ−1 1)>0. This is an absurdum because by Lemma 2.6 λ1(−∆; λ(β−1)/(1−p);wβ−1 1)→ −∞ as λ→ ∞. Then, we can define Λ := {λ∈IR : there exists at least a positive solution of (1.1)}. 24 Combining local and non-local terms August 14, 2010 We have proved that 0 < λ := sup Λ <∞. Thanks to the bounds by Proposition 3.3, there exists positive solution for λ=λ. Now, it is clear that if λ∈(0, λ) then (εw1, uλ) is a pair of sub-supersolution of (1.1) with εsmall and uλa positive solution of (1.1) for λ=λ. Observe that this method works for non-local equation, see for instance [9]. On the other hand, consider uλ 1for λ∈(0, λ0). Since for λ= 0 the solution is unstable, we can assure that lim λ→0kuλ 1k∞= 1. Finally, the behaviour of the solution as λ→ −∞ follows by Lemma 2.8. c) Assume p, β < 1. In this case, it is clear by Propositions 4.1 and 4.2 the existence, uniqueness and stability for λ > 0. (a) Suppose that β < p. In this case we have again by Proposition 4.1 the existence and uniqueness for all λ∈IR and the stability follows by Proposition 2.9. (b) Suppose β=p. Observe that since the map h(R) is decreasing, in case of existence of positive solution, it is unique. Moreover, by (4.6) and since ρ0(λ) is non-decreasing, there exists a unique value λ0<0 such that ρ0(λ)≤1 for λ≤λ0and ρ0(λ)>1 for λ∈(λ0,0). Hence, there exists a positive solution of (1.1) if, and only if, λ>λ0. Then, lim R→0h(R)≤1 for λ<λ0, lim R→0h(R)>1 for λ>λ0, and lim λ→λ0 kuλk∞= 0. Again, thanks to Proposition 4.2 we know that the solution is stable. (c) Suppose β > p. With a similar argument to the used in the paragraph b) we can show the existence of two positive solutions for λnegative and small. Indeed, in this case lim R→0h(R) = lim R→∞ h(R)=0, and there exist at least two positive values R1 0< R0< R2 0such that h(Ri 0) = 1, i= 1,2, and so two positive solutions uλ i=uRi 0of (1.1) for λ≤λ0with uλ 1< uλ 2, and h0(R1 0)>0> h0(R2 0), and then uλ 1is unstable and uλ 2stable. We prove now the non-existence of positive solutions for λvery negative. Indeed, observe that by (2.5) we have −λ≤Ckukβ−p ∞≤C(4.20) this last inequality by Lemma 3.1. Then, if there exists a positive solution of (1.1) we get that λ≥ −C. Again, we can define Λ := {λ∈IR : there exists at least a positive solution of (1.1)}. We know that λ:= inf Λ >−∞ and λ < 0, and using as sub-supersolution the pair (uλ, Ke) for Klarge, we prove the existence of positive solution for all λ∈(λ, 0). Finally, by (4.20) it can not occur that for a sequence (λn, un) we have λn→λ0<0 and kunk∞→ 0. Moreover, since the solution for λ= 0 we can conclude that lim λ→0kuλ 1k∞= 0. 25