BIFURCATION OF POSITIVE SOLUTIONS FOR NONLINEAR NONHOMOGENEOUS ROBIN AND NEUMANN PROBLEMS WITH COMPETING NONLINEARITIES. Nikolaos S.

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1 DISCRETE AND CONTINUOUS doi: /dcds DYNAMICAL SYSTEMS Volume 35, Number 10, October 2015 pp BIFURCATION OF POSITIVE SOLUTIONS FOR NONLINEAR NONHOMOGENEOUS ROBIN AND NEUMANN PROBLEMS WITH COMPETING NONLINEARITIES Nikolaos S. Papageorgiou National Technical University, Department of Mathematics Zografou Campus, Athens 15780, Greece Vicenţiu D. Rădulescu Department of Mathematics, Faculty of Sciences King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia and Institute of Mathematics Simion Stoilow of the Romanian Academy P.O. Box 1-764, Bucharest, Romania (Communicated by Manuel del Pino) Abstract. In this paper we deal with Robin and Neumann parametric elliptic equations driven by a nonhomogeneous differential operator and with a reaction that exhibits competing nonlinearities (concave-convex nonlinearities). For the Robin problem and without employing the Ambrosetti-Rabinowitz condition, we prove a bifurcation theorem for the positive solutions for small values of the parameter > 0. For the Neumann problem with a different geometry and using the Ambrosetti-Rabinowitz condition we prove bifurcation for large values of > Introduction. Let R N be a bounded domain with C 2 -boundary. In this paper, we study the following nonlinear, nonhomogeneous parametric Robin problem: div a(du(z)) = f(z, u(z), ) in, u (z) + β(z)u(z) p 1 = 0 on, n a u > 0, 1 < p <. (P ) Hence a : R N R N is a continuous and strictly monotone map, which satisfies certain other regularity and growth conditions, listed in hypotheses H(a) below. These conditions are general enough, to incorporate in our setting various differential operators of interest, such as the p-laplacian (1 < p < ). u Also, denotes the conormal derivative defined by n a u n a = (a(du), n) R N with n(z) being the outward unit normal at z. The reaction f(z, x, ) is a parametric function with > 0 being the parameter and (z, x) f(z, x, ) is Carathéodory (that 2010 Mathematics Subject Classification. 35J66, 35J70, 35J92. Key words and phrases. Competing nonlinearities, nonhomogeneous differential operator, positive solutions, bifurcation phenomena, Robin and Neumann problems. V.D. Rădulescu is supported by CNCS grant PCE-47/

2 5004 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU is, for all x R the mapping z f(z, x, ) is measurable and for a.a. z the map x f(z, x, ) is continuous). We assume that f(z,, ) exhibits competing nonlinearities, namely near the origin, it has a concave term ( that is, a strictly (p 1)- sublinear term), while near +, the reaction is convex term (that is, x f(z, x, ) is (p 1)-superlinear). A special case of our reaction, is the following function: with f(z, x, ) = f(x, ) = x q 1 + x r 1 for all x 0, Np 1 < q < p < r < p if p < N = N p + if N p. This reaction is encountered in the literature in the context of equations driven by the Laplacian (that is, p = 2) or by the p-laplacian (1 < p < ). Our aim is to investigate the existence, nonexistence and multiplicity of positive solutions as the parameter > 0 varies. So, we prove two bifurcation type results, describing the set of positive solutions of (P ) as the parameter > 0 changes, when the reaction exhibits the competing effects of concave (that is, sublinear) and convex (that is, superlinear) nonlinearities. In the first theorem the bifurcation occurs near zero. More precisely, under general hypotheses we show that there exists > 0 such that the following properties hold: (a) for all (0, ), problem (P ) has at least two positive solutions; (b) for = problem (P ) has at least one positive solution; (c) for all > problem (P ) has no positive solution. In the second case, we assume that β 0 (Neumann boundary condition) and we consider the problem div a(du(z)) = f 0 (z, u(z)) u(z) p 1 in, u n (z) = 0 on, (S ) u > 0 in. We obtain a different geometry and we establish that the bifurcation occurs for large values of the parameter > 0. More precisely, under natural assumptions on f 0 we show that there exists > 0 such that (a) for every > problem (S ) has at least two positive solutions; (b) for = problem (S ) has at least one positive solution; (c) for every (0, ) problem (S ) has no positive solution. The first work concerning positive solutions for problems with concave and convex nonlinearities, was that of Ambrosetti, Brezis and Cerami [2]. They studied semilinear equations driven by the Dirichlet Laplacian and with a reaction of the form (1). Their work was extended to equations driven by the Dirichlet p-laplacian by Garcia Azorero, Manfredi and Peral Alonso [10] and by Guo and Zhang [14]. We also refer to the contributions of de Figueiredo, Gossez and Ubilla [7], [8] to concave-convex type problems and general nonlinearities for the Laplacian, resp. p-laplacian case. Extensions to equations involving more general reactions, were obtained by Gasinski and Papageorgiou [13], Hu and Papageorgiou [15] and Rădulescu and Repovš [22]. Other problems with competition phenomena, can be found in the works of Cîrstea, Ghergu and Rădulescu [4] (problems with singular terms) and of Kristaly

3 BIFURCATION OF POSITIVE SOLUTIONS 5005 and Moroşanu [16] (problems with oscillating reaction). Finally we mention the recent work of Papageorgiou and Rădulescu [20], who studied a Robin problem driven by the p-laplacian and with a logistic reaction and proved multiplicity theorems for all large values of the parameter > 0, producing also nodal solutions. We stress that the differential operator in (P ) is not homogeneous and this is a source of difficulties in the analysis of the problem, since many of the methods and techniques developed in the aforementioned papers do not work here. It appears that our results in the present paper are the first bifurcation-type theorems for nonhomogeneous elliptic equations. 2. Mathematical background. Let X be a Banach space and X its topological dual. By, we denote the duality brackets for the pair (X, X). Given ϕ C 1 (X), we say that ϕ satisfies the Cerami condition (the C-condition), if the following is true: Every sequence {u n } n 1 X such that {ϕ(u n )} n 1 R is bounded and (1 + u n )ϕ (u n ) 0 in X as n, admits a strongly convergent subsequence. This is a compactness-type condition on the function ϕ which compensates for the fact that the space X need not be locally compact (being in general infinite dimensional). It is more general than the more common Palais-Smale condition. Nevertheless, the C-condition suffices to prove a deformation theorem and from it derive the minimax theory of the critical values of ϕ. One of the main results in that theory, is the so-called mountain pass theorem of Ambrosetti and Rabinowitz [3]. Here we state it in a slightly more general form. Theorem 2.1. Let X be a Banach space, ϕ C 1 (X) satisfies the C-condition, u 0, u 1 X with u 1 u 0 > ρ > 0 max{ϕ(u 0 ), ϕ(u 1 )} < inf[ϕ(u) : u u 0 = ρ] = m ρ and c = inf max ϕ(γ(t)) with Γ = {γ C([0, 1], X) : γ(0) = u 0, γ(1) = u 1 }. Then γ Γ 0 t 1 c m ρ and c is a critical value of ϕ. Let η C 1 (0, ) and assume that 0 < ĉ tη (t) η(t) c 0 and c 1 t p 1 η(t) c 2 (1 + t p 1 ) for all t > 0 (1) The hypotheses on the map a( ) are the following: with c 1, c 2 > 0, 1 < p <. H(a) : a(y) = a 0 ( y )y for all y R N, with a 0 (t) > 0 for all t > 0 and (i) a 0 C 1 (0, ), t a 0 (t)t is strictly increasing on (0, ), a 0 (t)t 0 as t 0 + and a lim 0(t)t t 0 + a 0 (t) > 1; (ii) a(y) c 3 η( y ) y for some c 3 > 0, all y R N \{0}; (iii) η( y ) ξ 2 ( a(y)ξ, ξ) y R N for all y R N \{0}, all ξ R N ;

4 5006 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU (iv) if G 0 (t) = some ˆξ > 0; t 0 a 0 (s)sds for all t 0, then pg 0 (t) a 0 (t)t 2 ˆξ for all t 0, (v) there exists τ (1, p) such that t G 0 (t 1/τ G 0 (t) ) is convex on (0, ), lim t 0 + t τ = 0 and a 0 (t)t 2 τg 0 (t) ct p for some c > 0, all t > 0. Remark 1. These conditions on a( ) are motivated by the regularity results of Lieberman [17] and the nonlinear maximum principle of Pucci and Serrin [21]. According to the above conditions, the potential function G 0 ( ) is strictly convex and strictly increasing. We set G(y) = G 0 ( y ) for all y R N. Then the function y G(y) is convex and differentiable on R N \{0}. We have G(y) = G 0( y ) y y = a 0( y )y = a(y) for all y R N \{0}, G(0) = 0. So, G( ) is the primitive of the map a( ). Because G(0) = 0 and y G(y) is convex, from the properties of convex functions, we have G(y) (a(y), y) R N for all y R N. (2) The next lemma summarizes the main properties of the map a( ). They follow easily from hypotheses H(a) above. Lemma 2.2. If hypotheses H(a)(i), (ii), (iii) hold, then (a) y a(y) is continuous and strictly monotone, hence maximal monotone too; (b) a(y) c 4 (1 + y p 1 ) for some c 4 > 0, all y R N ; (c) (a(y), y) R N c 1 p 1 y p for all y R N. Lemma 2.2 together with (1) and (2), lead to the following growth estimates for the primitive G( ). c 1 Corollary 1. If hypotheses H(a)(i), (ii), (iii) hold, then p(p 1) y p G(y) c 5 (1 + y p ) for some c 5 > 0, all y R N. Example 1. The following maps a(y), satisfy hypotheses H(a) above: (a) a(y) = y p 2 y with 1 < p <. This map corresponds to the p-laplace operator defined by p u = div ( Du p 2 Du) for all u W 1,p (). (b) a(y) = y p 2 y + µ y q 2 y with 1 < q < p < and µ > 0. This map corresponds to the (p, q)-differential operator defined by p u + µ q u for all u W 1,p (). Such differential operators arise in many physical applications (see Papageorgiou and Rădulescu [18], [19] and the references therein). (c) a(y) = (1 + y 2 ) p 2 2 y with 1 < p <. This map corresponds to the generalized p-mean curvature differential operator defined by ] div [(1 + Du 2 ) p 2 2 Du for all u W 1,p ().

5 BIFURCATION OF POSITIVE SOLUTIONS 5007 (d) a(y) = y p 2 y + y p 2 y with 1 < p <. 1 + y p The hypotheses on the boundary weight map β( ) are the following: H(β) : β C 1,α ( ) with α (0, 1) and β(z) 0 for all z. In the analysis of problem (P ) in addition to the Sobolev space W 1,p (), we will also use the Banach space C 1 (). This is an ordered Banach space, with positive cone C + = {u C 1 () : u(z) 0 for all z }. This cone has a nonempty interior given by int C + = {u C + : u(z) > 0 for all z }. In the Sobolev space W 1,p (), we use the norm u = [ u p p + Du p p] 1/p for all u W 1,p (). To distinguish, we use to denote the norm of R N. If on we use the (N 1)-dimensional Hausdorff measure σ( ) (the surface measure on ), then we can define the Lebesgue spaces L q ( ), 1 q. We know that there exists a unique continuous, linear map γ 0 : W 1,p () L p ( ), known as the trace map, such that γ 0 (u) = u for all u C 1 (). In fact γ 0 is compact. We have im γ 0 = W 1 p,p ( ) ( 1 p + 1 ) p = 1 and ker γ 0 = W 1,p 0 (). In the sequel, for the sake of notational simplicity, we drop the use of the trace map γ 0, with the understanding that all restrictions of elements of W 1,p () on, are defined in the sense of traces. Suppose f 0 : R R is a Carathéodory function with subcritical growth in the x R variable, that is f 0 (z, x) a 0 (z)(1 + x r 1 ) for a.a. z, all x R, with a 0 L () +, 1 < r < p. We set F 0 (z, x) = C 1 -functional ϕ 0 : W 1,p () R defined by ϕ 0 (u) = G(Du)dz + 1 p x 0 f 0 (z, s)ds and consider the β(z) u p dσ F 0 (z, u)dz for all u W 1,p (). The next proposition, was proved by Papageorgiou and Rădulescu [20] for G(y) = 1 p y p for all y R N. The proof remains valid in the present more general setting, using Lemma 2.2, Corollary 1 and the regularity result of Lieberman [17] [p. 320]. Proposition 1. Assume that u 0 W 1,p () is a local C 1 ()-minimizer of ϕ 0, that is, there exist ρ 0 > 0 such that ϕ 0 (u 0 ) ϕ 0 (u 0 + h) for all h C 1 () with h C 1 () ρ 0. Then u 0 C 1,η () for some η (0, 1) and it is also a local W 1,p ()-minimizer of ϕ 0, that is, there exists ρ 1 > 0 such that ϕ 0 (u 0 ) ϕ 0 (u 0 + h) for all h W 1,p () with h ρ 1.

6 5008 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU Let A : W 1,p () W 1,p () be the nonlinear map defined by A(u), y = (a(du), Dy) R N dz for all u, y W 1,p () (3) The following, is a particular case of a more general result due to Gasinski and Papageorgiou [12]. Proposition 2. If A : W 1,p () W 1,p () is defined by (3), then A is demicontinuous and of type (S) +, that is, if u n u in W 1,p () w and then u n u in W 1,p (). lim sup A(u n ), u n u 0, n In the sequel, by N we denote the Lebesgue measure on R N. Also, if x R, then we set x ± = max{±x, 0} and for u W 1,p (), we define u ± ( ) = u( ) ±. We know that u ± W 1,p () and u = u + + u, u = u + u. Also, if h : R R is a measurable function (for example a Carathéodory function), then we define N h (u)( ) = h(, u( )) for all u W 1,p (), (the Nemytskii operator corresponding to the function h). 3. Bifurcation near zero for the Robin problem. In this section, we deal with competition phenomena that give rise to bifurcation of the problem solutions, when the parameter > 0 is near zero. This situation includes the classical equations with concave and convex nonlinearities. The hypotheses on the reaction f(z, x, ) are the following: H 1 : f : R (0, ) R is a function such that for all (z, x) [0, + ), f(z, x, ) is nondecreasing, for all > 0 f(z, 0, ) = 0 for a.a. z and (i) (z, x) f(z, x, ) is a Carathéodory function on [0, + ); (ii) f(z, x, ) a (z)(1 + x r 1 ) for a.a. z, all x 0, with a L () +, p < r < p ; x F (z, x, ) (iii) if F (z, x, ) = f(z, s, )ds, then lim = + uniformly for a.a. x + z ; 0 (iv) there exists ϑ = ϑ() ((r p) max x p { } N p, 1, p ) such that f(z, x, )x pf (z, x, ) 0 < γ 0 lim inf x + x ϑ uniformly for a.a. z ; (v) there exists 1 < µ = µ() < q = q() < τ (see hypothesis H(a)(v)) and γ = γ() > µ, δ 0 = δ 0 () (0, 1] such that c 6 x p f(z, x, )x qf (z, x, ) ξ (z)x µ + cx γ for a.a. z, all x [0, δ 0 ], with c 6 = c 6 () > 0, τ = τ() > 0, ξ L () + and ξ 0 as 0 +. Remark 2. Since we are interested to find positive solutions and the above hypotheses concern the positive semiaxis R + = [0, + ), without any loss of generality

7 BIFURCATION OF POSITIVE SOLUTIONS 5009 we may assume that f(z, x, ) = 0 for a.a. z, all x 0 and all > 0. Note that hypotheses H 1 (ii), (iii) imply that f(z, x, ) lim x + x p 1 = + uniformly for a.a. z. Therefore, f(z,, ) is (p 1)-superlinear near +. However, we do not employ the AR-condition (unilateral version). We recall (see [3]), that f(z,, ) satisfies the (unilateral) AR-condition, if there exist η = η() > p and M = M() > 0 such that (a) 0 < ηf (z, x, ) f(z, x, )x for a.a. z, all x M, (b) ess inf F (, M, ) > 0. (4) Integrating (4a) and using (4b), we obtain a weaker condition, namely that c 7 x η F (z, x, ) for a.a. z, all z M and some c 7 > 0. (5) Evidently (5) implies { the} much weaker hypothesis H 1 (iii). In (4) we may assume N that η > (r p) max p, 1. Then we have f(z, x, )x pf (z, x, ) x η f(z, x, )x ηf (z, x, ) (η p)f (z, x, ) = x η + x η (η p)c 7 for a.a. z, all x M (see (4a) and (5)). So, we see that the AR-condition implies hypothesis H 1 (iv). This weaker superlinearity condition, incorporates in our setting (p 1)-superlinear nonlinearities with slower growth near +, which fail to satisfy the AR-condition (see the examples below). Finally note that hypothesis H 1 (v) implies the presence of a concave nonlinearity near zero. Example 2. The following functions satisfy hypotheses H 1. For the sake of simplicity, we drop the z-dependence: f 1 (x, ) = x q 1 + x r 1 for all x 0, with 1 < q < p < r < p x q 1 ( x η 1 if x [0, 1] f 2 (x, ) = x p 1 ln x + 1 ) ( + 1 ) x ν 1 if 1 < x p p with q, ν (1, p) and η > p { x q 1 if x [0, ρ()] f 3 (x, ) = x r 1 + η() if ρ() < x with 1 < q < p < r < p, η() = ρ() p 1 ρ() r 1 and ρ() 0 + as 0 +. Note that f 2 (, ) does not satisfy the AR-condition. We introduce the following Carathéodory function ˆf(z, x, ) = f(z, x, ) + (x + ) p 1 for all (z, x, ) R (0, + ).

8 5010 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU x Let ˆF (z, x, ) = 0 R defined by ˆϕ (u) = G(Du)dz + 1 p u p p + 1 p ˆf(z, s, )ds and consider the C 1 -functional ˆϕ : W 1,p () β(z)(u + ) p dσ ˆF (z, u, )dz for all u W 1,p (). Proposition 3. If hypotheses H(a), H(β) and H 1 hold and > 0, then the functional ˆϕ satisfies the C-condition. Proof. Let {u n } n 1 W 1,p () be a sequence such that From (7) we have ˆϕ (u n ) M 1 for some M 1 > 0, all n 1 (6) (1 + u n ) ˆϕ (u n ) 0 in W 1,p () as n. (7) ˆϕ (u n ), h ɛ n h 1 + u n for all h W 1,p (), all n 1, A(u n), h + with ɛ n 0 + as n, u n p 2 u n hdz + β(z)(u + n ) p 1 hdσ ˆf(z, u n, )hdz ɛ n h for all n 1. (8) 1 + u n In (8), first we choose h = u n W 1,p (). Using Lemma 2.2, we have c 1 p 1 Du n p p + u n p p ɛ n for all n 1, u n 0 in W 1,p () as n. (9) From (6), (9) and hypothesis H 1 (i), we have pg(du + n )dz + β(z)(u + n ) p dσ Also, in (8) we choose h = u + n W 1,p () and obtain (a(du + n ), Du + n ) R N dz β(z)(u + n ) p dσ + Adding (10) and (11), we have [ pg(du + n ) (a(du + n ), Du + ] n ) R N dz + pf (z, u + n, )dz M 2 (10) for some M 2 > 0, all n 1. f(z, u + n, )u + n dz ɛ n for all n 1. (11) [ f(z, u + n, )u + n pf (z, u + n, ) ] dz M 3 for some M 3 > 0, all n 1, [ f(z, u + n, )u + n pf (z, u + n, ) ] dz M 3 + ˆξ for all n 1 (12) (see hypothesis H(a)(iv)). By virtue of hypotheses H 1 (ii), (iv), we can find γ 1 (0, γ 0 ) and c 8 = c 8 (γ 1, ) > 0 such that f(z, x, )x pf (z, x, ) γ 1 x ϑ c 8 for a.a. z, all x 0.

9 BIFURCATION OF POSITIVE SOLUTIONS 5011 We use this unilateral growth estimate in (12) and obtain γ 1 u + n ϑ ϑ M 4 for some M 4 > 0, all n 1, {u + n } n 1 L ϑ () is bounded. (13) First assume that N p. From hypothesis H 1 (iv) it is clear that without any loss of generality, we may assume that ϑ r < p. Then we can find t [0, 1) such that 1 r = 1 t ϑ + t p. (14) From the interpolation inequality (see, for example, Gasinski and Papageorgiou [11] [p. 905]), we have u + n r u + n 1 t ϑ u+ n t p c 9 u + n t for some c 9 > 0, all n 1 (see (13) and use the Sobolev embedding theorem), u + n r r c 10 u + n tr for all n 1, with c 10 = c p 9 > 0. (15) By virtue of hypothesis H 1 (ii) we have f(z, x, )x a (z)(x + x r ) for a.a z, all x 0. (16) In (8) we choose h = u + n W 1,p (). Then Du + n p p + β(z)(u + n ) p dσ f(z, u + n, )u + n dz ɛ n for all n 1, Du + n p p c 11 (1 + u + n r r) for some c 11 > 0, all n 1 (see (16) and H(β)), c 12 (1 + u + n tr ) for some c 12 > 0, all n 1 (see (15)), Du + n p p + u + n p ϑ c 13(1 + u + n tr ) for some c 13 > 0, all n 1 (17) Since ϑ r < p, we know that u u ϑ + Du p (see (13)) is an equivalent norm on W 1,p () (see, for example, Gasinski and Papageorgiou [11] [p. 227]). So, from (17) we obtain u + n p c 14 (1 + u + n tr ) for some c 14 > 0, all n 1. (18) The hypothesis on ϑ (see H 1 (iv)) and (14), imply that tr < p. So, from (18) we infer that {u + n } n 1 W 1,p () is bounded. (19) If N = p, then p =, while from the Sobolev embedding theorem, we know that W 1,p () is embedded (compactly) in L s () for all s [1, ). So, in the above argument, we need to replace p = by s > r large such that tr = s(r µ) s µ < p (see (14) with p replaced by s > r). Then the previous argument works and leads again to (19). From (9) and (19) it follows that {u n } n 1 W 1,p () is bounded. So, we may assume that u n w u in W 1,p () and u n u in L r () and in L p ( ). (20)

10 5012 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU In (8) we choose h = u n u W 1,p (), pass to the limit as n and use (20). Then lim A(u n), u n u = 0, n u n u in W 1,p (), ˆϕ satisfies the C condition. Proposition 4. If hypotheses H(a), H(β) and H 1 hold, then there exists + > 0 such that for every (0, + ) there exists ρ > 0 for which we have inf [ ˆϕ (u) : u = ρ ] = ˆm > 0 = ˆϕ (0). Proof. Hypotheses H 1 (ii), (v) imply that for every > 0, we can find c 15 = c 15 () > 0 such that F (z, x, ) ξ (z) µ (x+ ) µ + c 15 [(x + ) γ + (x + ) r ] for a.a. z, all x R. (21) Then for u W 1,p (), we have ˆϕ (u) = G(Du)dz + 1 p u p p + 1 p c 1 p(p 1) Du p p + 1 p u p p + 1 p β(z)(u + ) p dσ ˆF (z, u, )dz β(z)(u + ) p dσ ξ u + µ µ µ c 15 u + r r c 15 u + γ γ 1 p u+ p p (see Corollary 1 and (21)). (22) It is clear that in hypothesis H 1 (v) we can always assume γ p and that µ > 1 is small enough so that p 1 µ 1 µ r and γ 1 µ r. By Young s inequality with µ 1 ɛ > 0 (see, for example, Gasinski and Papageorgiou [11] [p. 913]), we have u p = u u p 1 ɛ µ u µ + 1 ( (p 1)µ u µ 1 1 ɛµ µ + 1 ) µ = 1 ɛ µ u µ + µ 1 ɛµ u r u γ = u u γ 1 ɛ µ u µ + 1 (γ 1)µ u µ 1 ɛµ ɛ µ u µ + µ 1 ɛµ u r for all u W 1,p () with u 1 (recall that µ < γ, p p ). Using these bounds in (22), we obtain ˆϕ (u) c 16 u p c 17 [( ξ + ɛ) u µ + (1 + c ɛ ) u r ] for some c 16, c 17, c ɛ > 0 = [ c 16 c 17 ( ( ξ + ɛ) u µ p + (1 + c ɛ ) u r p)] u p (23) for all u W 1,p () with u 1. Let k ɛ (t) = ( ξ + ɛ) t µ p + (1 + c ɛ )t r p. Evidently k ɛ C 1 (0, ) and since µ < p < r we have k ɛ (t) + as t 0 + and as t +.

11 BIFURCATION OF POSITIVE SOLUTIONS 5013 Therefore we can find t 0 > 0 such that k ɛ (t 0 ) = min t>0 k ɛ (t), (kɛ ) (t 0 ) = (µ p)( ξ + ɛ)t µ p (r p)(1 + c ɛ )t r p 1 [ ] 1 (p µ)( ξ r µ + ɛ) t 0 = t 0 () =. (r p)(1 + c ɛ ) Then we have k ɛ (t) χ(ɛ) as 0 + with χ(ɛ) 0 + as ɛ 0 +. We choose ɛ > 0 small such that χ(ɛ) < 1 2 find + = + (ɛ) > 0 such that 0, c 16 c 17. Then for such an ɛ > 0, we can k ɛ (t 0 ) < c 16 c 17 and t 0 () 1 for all (0, + ) (see hypothesis H 1 (v)) Then by virtue of (23), we have ˆϕ (u) ˆm > 0 = ˆϕ (0) for all u W 1,p () with u = ρ = t 0 () 1. Note that as a direct consequence of hypothesis H 1 (iii), we have: Proposition 5. If hypotheses H(a), H(β) and H 1 hold, > 0 and u int C +, then ˆϕ (tu) as t. We introduce the following sets: S = { > 0 : problem (P ) admits a positive solution}, S() = the set of positive solutions of (P ). We can show that S is nonempty, as well as a useful structural property of the solution set S(). Proposition 6. If hypotheses H(a), H(β) and H 1 hold, then S and for every S S() int C +. Proof. Let + > 0 be as postulated by Proposition 4 and let (0, + ). Propositions 3, 4 and 5 permit the use of Theorem 2.1 (the mountain pass theorem) on the functional ˆϕ. So, we can find u 0 W 1,p () such that ˆϕ (u 0 ) = 0 and ˆϕ (0) = 0 < ˆm ˆϕ (u 0 ). (24) From the inequality in (24) we see that u 0 0. From the inequality in (24), we have A(u 0 ), h + u 0 p 2 u 0 hdz + β(z)(u + 0 )p 1 hdσ = ˆf(z, u 0, )hdz (25) for all h W 1,p (). In (25) we choose h = u 0 W 1,p (). Using Lemma 2.2, we have c 1 p 1 Du 0 p p + u 0 p p 0, u 0 0, u 0 0.

12 5014 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU Therefore (25) becomes A(u 0 ), h + β(z)u p 1 0 hdσ = f(z, u 0, )dz for all h W 1,p (). (26) In what follows by, 0 we denote the duality brackets for the pair (W 1,p (), W 1,p 0 ()) (recall that 1 p + 1 p = 1 and W 1,p () = W 1,p 0 () ). From the representation theorem for the elements of the dual space W 1,p () (see, for example, Gasinski and Papageorgiou [11] [p. 212]), we have div a(du 0 ) W 1,p () (see Lemma 2.2). Performing integration by parts, we have A(u 0 ), h = div a(du 0 ), h 0 for all h W 1,p 0 () W 1,p (). Using this equation in (26) and recalling that h = 0 for all h W 1,p 0 (), we obtain div a(du 0 ), h = f(z, u 0, )hdz for all h W 1,p 0 () W 1,p (), div a(du 0 (z)) = f(z, u 0 (z), ) for a.a. z. (27) Note that f(, u 0 ( ), ) L r () where 1 r + 1 r = 1 (see hypothesis H 1(ii)). Since p < r, we have W 1,r 0 () W 1,p 0 () continuously and densely. Then W 1,p () W 1,r () continuously and densely (see, for example, Gasinski and Papageorgiou [11] [p. 141]). Then from (27) we see that we can apply the nonlinear Green s identity (see, for example, Gasinski and Papageorgiou [11] [p. 210]) and have u0 A(u 0 ), h + (div a(du 0 ))hdz =, h (28) n a for all h W 1,r () W 1,p (). Here by, we denote the duality brackets for the pair (W 1 r,r ( ), W 1 r,r ( )). Returning to (26) and using (28), we obtain u0 div a(du 0 ), h +, h + β(z)u p 1 0 hdσ = f(z, u 0, )hdz n a for all h W 1,r () u0, h + β(z)u p 1 0 hdσ = 0 for all h W 1,r () (see (27)). (29) n a But we know that if γ 0 is the trace map on W 1,p (), then im ( γ 0 W 1,r ()) = W 1 r,r ( ). So, from (29), it follows that u 0 n a + β(z)u p 1 0 = 0 on. (30) From (27) and (30) it follows that u 0 S() and so (0, + ) S. From Winkert [23] we have that u 0 L (). Then we can apply the regularity result of Lieberman [17] [p. 320] and infer that u 0 C +, u 0 0. Hypotheses H 1 (ii), (v) imply that given ρ > 0, we can find ξ ρ > 0 such that f(z, x, ) + ξ ρ x p 1 0 for a.a. z, all x [0, ρ]. (31)

13 BIFURCATION OF POSITIVE SOLUTIONS 5015 Let ρ = u 0 and let ξ ρ > 0 be as postulated by (31). Then div a(du 0 (z)) + ξ ρ u 0 (z) p 1 = f(z, u 0 (z), ) + ξ ρ u 0 (z) p 1 0 for a.a. z (see (27) and (31)), div a(du 0 (z)) ξ ρ u 0 (z) p 1 a.e. in, u 0 int C + (see Pucci and Serrin [21] [pp. 111, 120]) S() int C +. The next proposition establishes a useful property of the set S. Proposition 7. If hypotheses H(a), H(β) and H 1 hold and S, then (0, ] S. Proof. Since S, we can find u S() int C +. Let η (0, ) and consider the following truncation-perturbation of the reaction in problem (P η ): 0 if x < 0 k η (z, x) = f(z, x, η) + x p 1 if 0 x u (z) (32) f(z, u (z), η) + u (z) p 1 if u (z) < x. This is a Carathéodory function. We set K η (z, x) = the C 1 -functional ˆψ η : W 1,p () R defined by ˆψ η (u) = G(Du)dz + 1 p u p p + 1 p x 0 k η (z, s)ds and consider β(z)(u + ) p dσ K η (z, u)dz for all u W 1,p (). From Corollary 1, hypothesis H(β) and (32), it is clear that ˆψ η is coercive. Also, from the Sobolev embedding theorem and the compactness of the trace map γ 0 into L p ( ), we see that ˆψ η is sequentially weakly lower semicontinuous. So, from the Weierstrass theorem we can find u η W 1,p () such that [ ˆψ η (u η ) = inf ˆψ(u) : u W ()] 1,p. (33) Let ξ (0, δ 0 (η)] and ξ min u (see hypothesis H 1 (v) and recall that u int C + ). Then ˆψ η (ξ) ξp p β L ( ) ξq c 6 q N (see (32)). Since q < p (see hypothesis H 1 (v)), by taking ξ (0, 1) even smaller if necessary, we will have From (33) we have ˆψ η(u η ) = 0, A(u η ), h + ˆψ η (ξ) < 0 ˆψ η (u η ) < 0 = ˆψ η (0) (see (33)), hence u η 0. u η p 2 u η hdz + β(z)(u + η ) p 1 hdσ = k η (z, u η )hdz(34) for all h W 1,p ().

14 5016 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU In (34), first we choose h = u η W 1,p (). Using Lemma 2.2 and (32), we have c 1 p 1 Du η p p + u η p p 0, u η 0, u η 0. Next, in (34), we choose h = (u η u ) + W 1,p (). Then A(uη ), (u η u ) + + u p 1 η (u η u ) + dz + β(z)u p 1 η (u η u ) + dσ = [f(z, u, η) + u p 1 ](u η u ) + dz (see (32)) [f(z, u, ) + u p 1 ](u η u ) + dz (since f(z, u (z), ) is nondecreasing) = A(u ), (u η, u ) + + u p 1 (u η u ) + dz + A(uη ) A(u ), (u η u ) + + (u p 1 {u η > u } N = 0, hence u η u. So, we have proved that η β(z)u p 1 (u η u ) + dσ u p 1 )(u η u ) + dz 0 (see hypothesis H(β)) u η [0, u ] = {u W 1,p () : 0 u(z) u (z) for a.a. z }, u η 0. Then because of (32), equation (34) becomes A(u η ), h + β(z)uη p 1 hdσ = f(z, u η, η)hdz for all h W 1,p (). From this, as in the proof of Proposition 6, using the nonlinear Green s identity, we infer that u η S(η) int C +, hence η S, (0, ] S. Let = sup S. We show that is finite by strengthening the conditions on the reaction f(z, x, ). So, the new stronger hypotheses on f are the following: H 2 : f : R (0, ) R is a function such that for a.a. z and all > 0 f(z, 0, ) = 0 and (i) for all (x, ) R (0, ), z f(z, x, ) is measurable, while for a.a. z, (x, ) f(z, x, ) is continuous; (ii) f(z, x, ) a (z)(1 + x r 1 ) for a.a. z, all x 0, all > 0, with a L (), a bounded on bounded sets in (0, ) and p < r < p ; (iii) if F (z, x, ) = x 0 f(z, s, )ds, then lim x + F (z, x, ) = + uniformly for a.a. z ; ( { } ) N (iv) there exists ϑ = ϑ() (r p) max p, 1, p such that f(z, x, )x pf (z, x, ) 0 < γ 0 lim inf x + x ϑ uniformly for a.a. z ; x p

15 BIFURCATION OF POSITIVE SOLUTIONS 5017 (v) there exists 1 < µ = µ() < q = q() < τ (see hypothesis H(a)(v)) and γ = γ() > µ, δ 0 = δ 0 () (0, 1) such that c 6 x q f(z, x, )x qf (z, x, ) ξ (z)x µ + τx γ for a.a. z, all 0 x δ 0 with c 6 = c 6 () > 0, c 6 () + as +, c = c() > 0, ξ L () + with ξ 0 as 0 + ; (vi) for every ρ > 0, there exists ξ ρ = ξ ρ () > 0 such that for a.a. z, x f(z, x, ) + ξ ρ x p 1 is nondecreasing on [0, ρ]; (vii) for every interval K = [x 0, ˆx] with x 0 > 0 and every > > 0, there exists d K (x 0, ) nondecreasing in with d K (x 0, ) + as + and ˆd K (x 0,, ) such that f(z, x, ) d K (x 0, ) for a.a. z, all x K f(z, x, ) f(z, x, ) ˆd K (x 0,, ) for a.a. z, all x K. Remark 3. Suppose that f(z, x, ) = g(x) + h(z, x) with g( ) continuous, nondecreasing, positive on (0, ) and h 0, h(z, ) C 1 (R) for a.a. z and h x(z, x) ξ x η 2 for a.a. z, all x > 0 and some ξ > 0, η p. Then hypotheses H 2 (vi), (vii) are satisfied. Also, the examples presented after hypotheses H 1, satisfy also the new conditions. Proposition 8. If hypotheses H(a), H(β) and H 2 hold, then <. Proof. We claim that there exists > 0 such that f(z, x, ) x p 1 for a.a. z, all x 0. (35) Indeed by virtue of hypothesis H 2 (v), we have f(z, x, ) c 6 ()x q 1 for a.a. z all x [0, δ 0 ()]. The hypothesis on c 6 ( ) implies that we can find 0 > 0 and 0 < δ 1 δ 0 ( 0 ) such that f(z, x, 0 ) c 6 ( 0 )x q 1 x p 1 for a.a. z, all x [0, δ 1 ] (36) Hypotheses H 1 (iii), (iv) imply that we can find M 5 > 0 such that f(z, x, 0 ) x p 1 for all a.a. z, all x M 5. (37) Finally, from hypothesis H 2 (vii), for K = [δ 1, M 5 ] we have f(z, x, ) d K (z, ) for a.a. z, all x [δ 1, M 5 ], all > 0. Since d K (x, ) + as +, we can find 0 such that f(z, x, ) d K (x, ) M p 1 5 x p 1 for a.a. z, all x [δ 1, M 5 ]. (38) Recalling that f(z, x, ) and c 6 ( ) are nondecreasing in > 0, from (36), (37) and (38) we conclude that (35) is true. Now, let > and assume that S. Then we can find u S() int C 1 (see Proposition 6). Let m = min u > 0. For δ > 0 we set m δ = m + δ int C +. Also, let ρ = u and let ξ ρ > 0 be as postulated by hypothesis H 2 (vi). We

16 5018 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU have div a(dm δ ) + ξ ρ (m δ ) p 1 ξ ρ m p 1 + χ(δ) with χ(δ) 0 + as δ 0 + (1 + ξ ρ )m p 1 + χ(δ) f(z, m, ) + ξ ρ m p 1 + χ(δ) (see (35)) = f(z, m, ) + ξ ρ m p 1 + [f(z, m, ) f(z, m, )] + χ(δ) f(z, m, ) + ξ ρ m p 1 ˆd K (m,, ) + χ(δ) with K = {m } (see hypothesis H 2 (vii)) f(z, m, ) + ξ ρ m p 1 for all δ > 0 small f(z, u, ) + ξ ρ u (z) p 1 (since m u (z) for all z, see hypothesis H 2 (vii)) = div a(du (z)) + ξ ρ u (z) p 1 for a.a. z (since u S()), m δ u (z) for all z, all δ > 0 small, a contradiction. This means that / S and so <. Proposition 9. If hypotheses H(a), H(β) and H 2 hold and η (0, ), then problem (P η ) admits at least two distinct positive solutions u 0, û int C +, u 0 û. Proof. Let η, (0, ) with η < and let u S() int C +. From the proof of Proposition 7, we know that by using a suitable truncation-perturbation of the reaction of problem (P η ) (see (32)), we can find u 0 [0, u ] S(η), which is a minimizer of the corresponding truncated energy functional ˆψη (see the proof of Proposition 7). For δ > 0, let u δ 0 = u 0 + δ int C + and for ρ = u, let ξ ρ > 0 be as postulated by hypothesis H 2 (vi). We have div a(du δ 0(z)) + ξ ρ u δ 0(z) p 1 div a(du 0 (z)) + ξ ρ u 0 (z) p 1 + χ(δ) with χ(δ) 0 + as δ 0 + = f(z, u 0 (z), η) + ξ ρ u 0 (z) p 1 + χ(δ) (since u 0 S(η)) = f(z, u 0 (z), ) + ξ ρ u 0 (z) p 1 + [f(z, u 0 (z), η) f(z, u 0 (z), )] + χ(δ) f(z, u (z), ) + ξ ρ u (z) p 1 ˆd K (m 0,, η) + χ(δ) (since u 0 u, see hypothesis H 2 (vi) and with K = u 0 (), m 0 = inf K) f(z, u (z), ) + ξ ρ u (z) p 1 for δ > 0 small, = div a(du (z)) + ξ ρ u (z) p 1 a.e. in (since u S()), u δ 0 u for δ > 0 small, u u 0 int C +. So, we have proved that u 0 int C1 () [0, u ]. (39)

17 BIFURCATION OF POSITIVE SOLUTIONS 5019 Recall that u 0 is a minimizer of the functional ˆψ (see the proof of Proposition 7). Note that ˆψ [0,u ] = ˆϕ [0,u ] (see (32)) u 0 is a local C 1 () minimizer of ˆϕ (see (39)), u 0 is a local W 1,p () minimizer of ˆϕ (see Proposition 1). Next, we consider the following truncation-perturbation of the reaction in problem (P η ): { f(z, u0 (z), η) + u γ η (z, x) = 0 (z) p 1 if x u 0 (z) f(z, x, η) + x p 1 (40) if u 0 (z) < x. This is a Carathéodory function. Let Γ η (z, x) = C 1 -functional σ η : W 1,p () R defined by σ η (u) = G(Du)dz + 1 p u p p + 1 p Note that x σ η = ˆϕ η + ˆξ η with ˆξ η R (see (40)), 0 γ η (z, s)ds and consider the β(z)(u + ) p dσ = Γ η (z, u)dz for all u W 1,p (). σ η satisfies the C condition (see Proposition 3). (41) Moreover, Proposition 5 implies that if u int C +, then σ η (tu) as t +. (42) Claim 1. We may assume that u 0 is a local minimizer of σ η. Recall that u 0 u. Then using u, we truncate γ η (z, ) as follows: { γη (z, x) if x u ˆγ η (z, x) = (z) γ η (z, u (z)) if u (z) < x This is a Carathéodory function. We set ˆΓ η (z, x) = the C 1 -functional ˆσ η : W 1,p () R defined by ˆσ η (u) = G(Du)dz + 1 p u p p + 1 p x 0 (43) ˆγ η (z, s)ds and consider β(z)(u + ) p dσ ˆΓ η (z, u)dz for all u W 1,p (). From (43), Corollary 1 and hypothesis H(β), we see that the functional ˆσ η is coercive. Also, it is sequentially weakly lower semicontinuous. So, by the Weierstrass theorem, we can find û 0 W 1,p () such that ˆσ η (û 0 ) = inf[ˆσ η (u) : u W 1,p ()], ˆσ η(û 0 ) = 0, A(û 0 ), h + û 0 p 2 û 0 hdz + β(z)(û + 0 )p 1 hdσ = ˆγ η (z, û 0 )hdz (44) for all h W 1,p ().

18 5020 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU In (44), first we choose h = (u 0 û 0 ) + W 1,p (). Then A(û0 ), (u 0 û 0 ) + + û 0 p 2 û 0 (u 0 û 0 ) + dz + + β(z)(û + 0 )p 1 (u 0 û 0 ) + dσ [ ] = f(z, u 0, η) + u p 1 0 (u 0 û 0 ) + dz (recall that u 0 u = A(u 0 ), (u 0 û 0 ) + + u p 1 0 (u 0 û 0 ) + dz + and see (43) and (40)) β(z)u p 1 0 (u 0 û 0 ) + dσ (since u 0 S(η)), A(u 0 ) A(û 0 ), (u 0 û 0 ) + + (u p 1 0 û 0 p 2 û 0 ) (u 0 û 0 ) + dz 0 (see hypothesis H(β)), {u 0 > û 0 } N = 0, hence u 0 û 0. Next in (44) we choose (û 0 u ) + W 1,p (). We obtain A(û0 ), (û 0 u ) + + û p 1 0 (û 0 u ) + dz + β(z)û p 1 0 (û 0 u ) + dσ = [f(z, u, η) + u p 1 ](û 0 u ) + dz (see (43) and (40)) [ ] f(z, u, ) + u p 1 (û 0 u ) + dz (see hypothesis H 2 (vii)) = A(u ), (û 0 u ) + + u p 1 (û 0 u ) + dz + β(z)u p 1 (û 0 u ) + dσ (since u S()), A(û 0 ) A(u ), (û 0 u ) + + (û p 1 0 u p 1 )(û 0 u ) + dz 0 (see hypothesis H(β)), {û 0 > u } N = 0, hence û 0 u. So, we have proved that û 0 [u 0, u ] = {u W 1,p () : u 0 (z) u(z) u (z) a.e. in }. If û 0 u 0, then by virtue of (43) and (40), we see that û 0 S(η) int C +, u 0 û 0, u 0 û 0 and so we are done, since this is the desired second positive solution of problem (P η ). Hence, we may assume that û 0 = u 0 int C +. Recall that u u 0 int C + (see (39)) and ˆσ η [0,u ] = σ η [0,u ] (see (43)). Therefore u 0 is a local C 1 () minimizer of σ η, u 0 is a local W 1,p () minimizer of σ η (see Proposition 1). This proves the Claim. Reasoning as above, we can show that K ση [u 0, ) = {u W 1,p () : u 0 (z) u(z) a.e. in }. (45)

19 BIFURCATION OF POSITIVE SOLUTIONS 5021 Then from (40) we see that the elements of K ση are positive solutions of problem (P η ). Therefore, we may assume that K ση is finite of otherwise we already have an infinity of positive solutions for problem (P η ). The finiteness of K ση and the Claim imply that we can find ρ (0, 1) small such that σ η (u 0 ) < inf[σ η (u) : u u 0 = ρ] = m η ρ (46) (see Aizicovici, Papageorgiou and Staicu [1] (proof of Proposition 29)). Then (41), (42) and (46) imply that we can use Theorem 2.1 (the mountain pass theorem). So, we can find û W 1,p () such that û K ση and σ η (u 0 ) < m η ρ σ η (û). (47) From (47) it follows that û u 0 and û S(η) int C +, u 0 û (see (45)). Next we examine what happens in the critical case =. To this end, note that hypotheses H 2 (ii), (v) imply that we can find c 18 = c 18 () > 0 such that f(z, x, ) c 6 x q 1 c 18 x r 1 for a.a. z, all z 0. (48) This unilateral growth estimate on the reaction f(z,, ) leads to the following auxiliary Robin problem: div a(du(z)) = c 6 u(z) q 1 c 18 u(z) r 1 in, u (z) + β(z)u(z) p 1 (49) = 0 on, u > 0. n 0 For this problem we have the following existence and uniqueness result. Proposition 10. If hypotheses H(a) and H(β) hold, the problem (49) admits a unique positive solution ū int C +. Proof. First we show the existence of a positive solution for problem (49). To this end let ξ + : W 1,p () R be the C 1 -functional defined by ξ + (u) = G(Du)dz + 1 p u p p + 1 β(z)(u + ) p dσ + c 18 p r u+ r r c 6 q u+ q q for all u W 1,p (). Using Corollary 1 and hypothesis H(β), we have ξ + (u) c 1 p(p 1) Du p p + 1 p u p p + c 18 r u+ r r c 19 ( u + q r + u + p r) for some c 19 > 0 (recall q < p < r). Because q < p < r, it follows that ξ + is coercive. Also, it is sequentially weakly lower semicontinuous. So, we can find ū W 1,p () such that ξ + (ū) = inf[ξ + (u) : u W 1,p ()]. (50) Exploiting the fact q < p < r, by choosing ξ (0, 1) small, we have ξ + (ξ ) < 0, ξ + (ū) < 0 = ξ + (0) (see (50)), hence ū 0.

20 5022 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU From (50) we have ξ +(ū) = 0, A(ū), h (ū ) p 1 hdz + β(z)(ū + ) p 1 hdσ = c 6 (ū + ) q 1 hdz c 18 (ū + ) r 1 hdz for all u W 1,p (). (51) Let, h = ū W 1,p () in (51). Then, we see that ū 0, ū 0. So, (51) becomes A(ū), h + β(z)ū p 1 hdσ = c 6 ū q 1 hdz c 18 ū r 1 hdz for all h W 1,p (), ū int C + is a solution of (49) (see the proof of Proposition 6). So, we have established the existence of positive solutions for problem (49). Next we show the uniqueness of this positive solution. To this end, let e : L τ () R = R {+ } be the integral functional defined by G(Du 1/τ )dz + 1 β(z)u p/τ dσ if u 0, u 1/τ W 1,p () e(u) = p + otherwise Let u 1, u 2 dom e = {u W 1,p () : e(u) < } (the effective domain of the functional e) and let t [0, 1]. We define y = ((1 t)u 1 + tu 2 ) 1/τ, v 1 = u 1/τ 1, v 2 = u 1/τ 2. Using Lemma 1 of Diaz and Saa [5], we have Dy(z) [(1 t) Dv 1 (z) τ + t Dv 2 (z) τ ] 1/τ for a.a. z, G 0 ( Dy(z) ) G 0 (((1 t) Dv 1 (z) τ + t Dv 2 (z) τ )) (since G 0 is increasing) (1 t)g 0 ( Dv 1 (z) ) + tg 0 ( Dv 2 (z) ) for a.a. z (see hypothesis H(a)(v)), G(Dy(z)) (1 t)g(du 1 (z) 1/τ ) + tg(du 2 (z) 1/τ ) for a.a. z, u G(Du 1/τ )dz is convex. Since p > τ and β 0 (see hypothesis H(β)), we see that u 1 β(z)u p/τ dσ p is a convex functional. Therefore, e is convex and also via Fatou s lemma, we have that e is lower semicontinuous. We already have ū int C + a positive solution of problem (49). Let ȳ W 1,p () be another positive solution. As above, we can show that ȳ int C +. Then for all h C 1 () and for t small, we have ū τ + th, ȳ τ + th dom e.

21 BIFURCATION OF POSITIVE SOLUTIONS 5023 Then e( ) is Gâteaux differentiable at ū τ and ȳ τ in the direction h. Moreover, via the chain rule and the nonlinear Green s identity, we obtain e (ū p div a(dū) )(h) = ū τ 1 hdz e (ȳ p div a(dȳ) )(h) = ȳ τ 1 hdz for all h W 1,p () (recall that C 1 () is dense in W 1,p ()). The convexity of e implies the monotonicity of e. Then (ūp ȳ p ) (ūp ȳ p ) 0 div a(dū) dz ( div a(dȳ)) dz = = ū τ 1 c 6 ū q 1 c 18 ū r 1 (ū p ȳ p )dz ū τ 1 c 6 (ū q τ ȳ q τ )(ū p ȳ p )dz 0 (since q < τ < p < r) ū = ȳ. ȳ p 1 c 6 ȳ q 1 c 18 ȳ r 1 (ū p ȳ p )dz ȳ τ 1 c 18 (ū r τ ȳ r τ )(ū p ȳ p )dz This proves the uniqueness of the positive solution ū int C + of problem (49). Proposition 11. If hypotheses H(a), H(β) and H 2 hold and S, then ū u for all u S(). Proof. Let u S() and consider the following Carathéodory function: 0 if x < 0 w(z, x) = c 6 x q 1 c 18 x r 1 + x p 1 if 0 x u(z) c 6 u(z) q 1 c 18 u(z) r 1 + u(z) p 1 if u(z) < x. (52) Let W (z, x) = x 0 w(z, s)ds and consider the C 1 -functional ˆγ : W 1,p () R defined by ˆγ(u) = G(Du)dz + 1 p u p p + 1 p β(z)(u + ) p dσ W (z, u)dz for all u W 1,p (). From hypothesis H(β) and (52) it is clear that ˆγ is coercive. Also, it is sequentially weakly lower semicontinuous. Therefore, we can find ū W 1,p () such that ˆγ(ū ) = inf[ˆγ(u) : u W 1,p ()]. (53) Since q < p < r, for ξ (0, min u) (recall that u int C + ) small, we have ˆγ(ξ) < 0, ˆγ(ū ) < 0 = ˆγ(0) (see (53)), hence ū 0.

22 5024 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU From (53) we have ˆγ (ū ) = 0, A(ū ), h + ū p 2 ū hdz + β(z)(ū + ) p 1 hdσ = w(z, ū )hdz (54) for all h W 1,p (). In (54) we choose first h = ū W 1,p () and then h = (ū u) + W 1,p () and as in the proof of Proposition 7, we show that So, (54) becomes A(ū ), h + ū [0, u], ū 0 β(z)ū p 1 hdσ = c 6 ū q 1 hdz c 18 for all h W 1,p (), ū is a positive solution of the auxiliary problem (49), ū = ū int C + (see Proposition 10) ū u for all u S(). ū r 1 hdz Now we can show that the critical value is admissible, that is S. Proposition 12. If hypotheses H(a), H(β) and H 2 hold, then S and so S = (0, ]. Proof. Let { n } n 1 S such that n ( ). Then we can find u n S( n ) int C + and from the proof of Proposition 7 we know that we can assume that ˆϕ n (u n ) < 0 for all n 1, pg(du n )dz + β(z)u p ndσ Also, we have A(u n ), u n pf (z, u n, n )dz < 0 (55) for all n 1. β(z)u p ndσ + f(z, u n, n )u n dz = 0 for all n 1. (56) Adding (55) and (56), we obtain [pg(du n ) (a(du n ), Du n ) R N ] dz + [f(z, u n, n )u n pf (z, u n, n )] dz ξ 0 for all n 1, some ξ 0 > 0, [f(z, u n, n )u n pf (z, u n, n )] dz < 0 for all n 1 (57) (see hypothesis H(a)(iv)). From (57), as in the proof of Proposition 3, using hypothesis H 2 (iv), we show that {u n } n 1 W 1,p () is bounded. So, we may assume that u n w u in W 1,p () and u n u in L r () and in L p ( ) as n. (58)

23 BIFURCATION OF POSITIVE SOLUTIONS 5025 Since u n S() for all n 1, we have A(u n ), h + β(z)u p 1 n hdσ f(z, u n, n )hdz = 0 for all h W 1,p (). (59) Choosing h = u n u W 1,p () in (59), passing to the limit as n and using (58), we obtain lim n), u n u = 0, n u n u in W 1,p (). (60) So, if in (59) we pass to the limit as n and use (60), then A(u ), h + β(z)u p 1 hdσ = f(z, u, )hdz for all h W 1,p (), u 0 is a solution of problem (P ). From Proposition 11 we have ū u n for all n 1. u S( ) int C +. Therefore S and so S = (0, ]. Hence ū u and so Summarizing the situation for problem (P ), we can state the following bifurcation-type result. Theorem 3.1. If hypotheses H(a), H(β) and H 2 hold, then there exists > 0 such that (a) for all (0, ), problem (P ) has at least two positive solutions u 0, û int C +, u 0 û, u 0 û; (b) for = problem (P ) has at least one positive solution u int C + ; (c) for all > problem (P ) has no positive solution. 4. Bifurcation near infinity for the Neumann problem. In this section we deal with the Neumann problem (that is, β 0) and with a parametric reaction of the form f(z, x, ) = f 0 (z, x) x p 1 for all (z, x) [0, ). Here f 0 is a Carathéodory function which as before exhibits competing nonlinearities, namely it is (p 1)-superlinear near + and admits a concave term near zero. This time the superlinearity of f(z, ) is expressed via the AR-condition. The presence of the term x p 1 changes the geometry of the problem and hypotheses H 1 and H 2 do not hold anymore. In fact, we will show that in this case the bifurcation occurs at large values of the parameter > 0 (bifurcation near infinity). The problem under consideration, is the following: { div a(du(z)) = f0 (z, u(z)) u(z) p 1 } in, u (S ) (z) = 0 on, u > 0 n For the differential operator, we keep hypotheses H(a) as in Section 3. On the nonparametric nonlinearity f 0 (z, x), we impose the following conditions: H 3 : f 0 : R R is a Carathéodory function such that f 0 (z, 0) = 0 for a.a. z and (i) f 0 (z, x) a(z)(1 + x r 1 ) for a.a. z, all x 0, with a L () +, p < r < p ;

24 5026 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU x (ii) if F 0 (z, x) = 0 f 0 (z, s)ds, then there exist c 19 > 0 and η > p such that c 19 x η ηf 0 (z, x) f 0 (z, x)x for a.a. z, all x 0; (iii) there exists q (1, τ) (see hypothesis H(a)(v)) such that f 0 (z, x) f 0 (z, x) 0 < c 20 lim inf x 0 + x q 1 lim sup x 0 x + q 1 c 21 < uniformly for a.a. z ; (iv) for every ρ > 0, there exists ξ ρ > 0 such that for a.a. z, x f 0 (z, x) + ξ ρ x p 1 is nondecreasing on [0, ρ]. Remark 4. As in Section 3, without any loss of generality, we may assume that f 0 (z, x) = 0 for all (z, x) (, 0]. Hypotheses H 3 (ii), (iii) reveal the competing nonlinearities (concave-convex nonlinearities). Observe that in this case the superlinearity of f 0 (z, ) is expressed using a global version of the unilateral ARcondition. Example 3. The model for the nonlinearity f 0 (z, ), is the function with 1 < q < τ < p < r < p. f 0 (z, x) = f 0 (x) = x q 1 + x r 1 for all x 0 As before, we introduce the following two sets S 0 = { > 0 : problem (S ) admits a positive solution} S 0 () = the set of positive solutions of problem (S ). Proposition 13. Assume that hypotheses H(a) and H 3 hold. Then S 0 and for all > 0, S 0 () int C + and for S 0, we have [, + ) S 0. Proof. We consider the following auxiliary Neumann problem div a(du(z)) + u(z) p 2 u u(z) = 1 in, = 0 on. (61) n ( 1 Let K p : L p () L p () p + 1 ) p = 1 be the nonlinear map defined by K p (u)( ) = u( ) p 2 u( ) This is bounded (maps bounded sets to bounded sets) and continuous. Moreover, by the Sobolev embedding theorem ˆK p = K p W 1,p () is completely continuous (that w is, if u n u in W 1,p (), then ˆK p (u n ) K p (u) in L p ()). So, the map u V (u) = A(u) + ˆK p (u) is demicontinuous and of type (S) +, hence pseudomonotone (see [11]). Moreover, we have V (u), u c 1 p 1 Du p p + u p p for all u W 1,p () (see Lemma 2.2), V is coercive. But a pseudomonotone coercive operator is surjective (see, for example, Gasinski and Papageorgiou [11] [p. 336]). So, we can find ū W 1,p () such that V (ū) + ˆK p (ū) = 1. (62)

25 BIFURCATION OF POSITIVE SOLUTIONS 5027 In fact, using Lemma 2.2 and the strict monotonicity of the map x x p 2 x, x R, we see that V is strictly monotone and so ū W 1,p () is unique. Acting on (62) with ū W 1,p () and using Lemma 2.2, we have c 1 p 1 Dū p p + ū p p 0, ū 0, ū 0 So, ū 0 is the unique solution of the auxiliary problem (61) and as before the nonlinear regularity theory (see [17]) and the nonlinear maximum principle (see [21]), imply ū int C +. Let 0 < m = min ū and let 0 = 1 + N f 0 (ū) m p 1. We have A(ū), h + 0 ū p 1 hdz = A(ū), h + ū p 1 ū p 1 hdz + N f0 (ū) hdz m p 1 = hdz + f 0 (z, ū)hdz for all h W 1,p () with h 0 (63) (see (61) and recall m = min ū > 0). We introduce the following truncation of f 0 (z, ): 0 if x < 0 ˆf 0 (z, x) = f 0 (z, x) if 0 x x(z) f 0 (z, ū(z)) if ū(z) < x. This is a Carathéodory function. We set ˆF 0 (z, x) = x 0 (64) ˆf 0 (z, s)ds and consider the C 1 -conditional ˆϕ 0 : W 1,p () R defined by ˆϕ 0 (u) = G(Du)dz + 0 p u p p ˆF 0 (z, u)dz for all u W 1,p (). From (64) it is clear that ˆϕ 0 is coercive. Also, it is sequentially weakly lower semicontinuous. So, we can find u 0 W 1,p () such that ˆϕ 0 (u 0 ) = inf[ ˆϕ 0 (u) : u W 1,p ()]. (65) Using hypothesis H 3 (iii) and since 1 < q < p, we have that for ξ (0, min{1, m}) small From (65) we have ˆϕ 0 (ξ) < 0 = ˆϕ 0 (0), ˆϕ 0 (u 0 ) < 0 = ˆϕ 0 (0) (see (65)), hence u 0 0. ˆϕ 0(u 0 ) = 0, A(u 0 ) + 0 u 0 p 2 u 0 = N ˆf0 (u 0 ) in W 1,p (). (66) On (66) we act with u 0 W 1,p (). Then c 1 p 1 Du 0 p p + 0 u 0 p p 0 (see Lemma 2.2 and (64)), u 0 0, u 0 0.

26 5028 NIKOLAOS S. PAPAGEORGIOU AND VICENŢIU D. RĂDULESCU Next on (66) we act with (u 0 ū) + W 1,p (). Then A(u0 ), (u 0 ū) u p 1 0 (u 0 ū) + dz = f 0 (z, ū)(u 0 ū) + dz (see (64)) A(ū), (u 0 ū) ū p 1 (u 0 ū) + dz (see (63)) A(u0 ) A(ū), (u 0 ū) (u p 1 0 ū p 1 )(u 0 ū) + dz 0, {u 0 > ū} N = 0, hence u 0 ū. So, we have proved that u 0 [0, ū] = {u W 1,p () : 0 u(z) ū(z) a.e. in }. Then by virtue of (64), equation (66) becomes A(u 0 ) + 0 u p 1 0 = N f0 (u 0 ), u 0 is a positive solution of (P 0 ) and so 0 S 0 Moreover, as before the nonlinear regularity theory and the nonlinear maximum principle imply that u 0 int C +. Therefore, for every S 0 S 0 () int C +. Next, let S 0 and µ >. Let u S 0 () int C +. Then A(u ), h + µ u p 1 hdz A(u ), h + for all h W 1,p () with h 0. u p 1 hdz (67) Then truncating f 0 (z, ) at {0, u (z)} and reasoning as above, via the direct method and using this time (67), we obtain u µ [0, u ] S 0 (µ), hence µ S 0. Therefore we infer that [, + ) S 0. Remark 5. Note that in the above proof we have proved that, if S 0, u S 0 () int C + and µ >, then µ S 0 and we can find u µ S 0 (µ) int C + such that u µ u. In fact in the next proposition, we improve this conclusion. Proposition 14. If hypotheses H(a) and H 3 hold, S 0, u S() int C + and µ >, then we can find u µ S 0 (µ) int C + such that u u µ int C +. Proof. From Proposition 13, we already know that we can find u µ S 0 (µ) int C + such that u µ u. ( Let m µ = min u µ > 0 and let δ 0, m ) µ. We set u δ = u δ int C +. Also, 2 for ρ = u, we let ξ ρ > 0 be as postulated by hypothesis H 3 (iv). Then div a(du δ ) + (µ + ξ ρ )(u δ ) p 1 div a(du ) + (µ + ξ ρ )u p 1 χ(δ) with χ(δ) 0 + as δ 0 + = f 0 (z, u ) + (µ )u p 1 + ξ ρ u p 1 χ(δ) (since u S 0 ()) f 0 (z, u µ ) + ξ ρ uµ p 1 + (µ )m µ χ(δ) (since m µ u µ u and use hypothesis H 3 (iv)) f 0 (z, u µ ) + ξ ρ u p 1 µ for δ > 0 small

27 BIFURCATION OF POSITIVE SOLUTIONS 5029 = div a(du µ ) + (µ + ξ ρ )u p 1 µ (since u µ S 0 (µ)), u µ u δ, for all δ > 0 small, u u µ int C +. Let = inf S 0. Proposition 15. If hypotheses H(a) and H 3 hold, then > 0. Proof. Consider a sequence { n } n S 0 such that n. We can find a corresponding sequence {u n } n 1 such that u n S 0 ( n ) int C + for all n 1. We claim that {u n } n 1 can be chosen to be increasing. To see this, note that since 2 < 1, the function u 2 S 0 ( 2 ) int C + satisfies A(u 2 ), h + 1 u p 1 2 hdz A(u 2 ), h + 2 for all h W 1,p () with h 0 u p 1 2 hdz (68) Considering problem (P 1 ) and truncating f 0 (z, ) at {0, u 2 (z)}, via the direct method and using (68), as in the proof of Proposition 13, we obtain u 1 [0, u 2 ] S 0 ( 1 ). Then we have A(u 1 ), h + 2 u p 1 1 hdz = Also, we have f 0 (z, u 1 )hdz + ( 2 1 ) u p 1 1 hdz (since u 1 S 0 ( 1 )) f 0 (z, u 1 )hdz for all h W 1,p () with h 0 (recall 1 > 2 ) (69) A(u 3 ), h + 2 u p 1 3 hdz A(u 3 ), h + 3 for all h W 1,p () with h 0. u p 1 3 hdz (70) Truncating f 0 (z, ) at {u 1 (z), u 3 (z)} and using the direct method and (69), (70), we produce u 2 [u 1, u 3 ] S 0 ( 2 ). Continuing this way, we see that we can choose {u n } n 1 to be increasing. We have A(u n ), h + n u p 1 n hdz = f 0 (z, u n )hdz for all h W 1,p (), all n 1. Choose h 1 W 1,p (). Then for all n 1 we have n u p 1 n dz = f 0 (z, u n )dz, n u n p 1 p 1 c 22 u n η 1 p 1 for some c 22 > 0, (see hypothesis H 3 (ii) and recall p < η), n c 22 u 1 η p p 1 (recall u n u 1, for all n 1), c 22 u 1 η p p 1 > 0.