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Abstract and Applied Analysis Positive solutions of higher order quasilinear elliptic equations
Positive solutions of higher order quasilinear elliptic equations
Montenegro, MarceloHow much do you like this book?
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7
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2002
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Abstract and Applied Analysis
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10.1155/s1085337502204030
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POSITIVE SOLUTIONS OF HIGHER ORDER QUASILINEAR ELLIPTIC EQUATIONS MARCELO MONTENEGRO Received 25 February 2002 The higher order quasilinear elliptic equation −∆(∆ p (∆u)) = f (x,u) subject to Dirichlet boundary conditions may have unique and regular positive solution. If the domain is a ball, we obtain a priori estimate to the radial solutions via blowup. Extensions to systems and general domains are also presented. The basic ingredients are the maximum principle, Moser iterative scheme, an eigenvalue problem, a priori estimates by rescalings, sub/supersolutions, and Krasnosel’skiı̆ fixed point theorem. 1. Introduction We are interested in studying the higher order quasilinear elliptic equation −∆ ∆ p (∆u) = f (x,u) in Ω, u = 0 on ∂Ω, (1.1) where Ω ⊂ RN , N ≥ 2, is a smooth bounded domain and ∆ p u = div(∇u p−2 ∇u), p > 1. Throughout the paper, it is useful to split (1.1) as a system of three equations −∆u1 = u2 , −∆ p u2 = u3 in Ω, −∆u3 = f x,u1 , u1 = u2 = u3 = 0 (1.2) on ∂Ω. There has been some interest in the study of polyharmonic operators, corresponding to p = 2 here, see [4, 6, 7, 9, 15]. These references testify the wide range of applications of higher order elliptic operators. A critical exponent problem involving ∆(∆u p−2 ∆u) was studied in [14], see also [11] for an account on these issues involving polyharmonic operators. Systems dealing with quasilinear equations in radial form were treated in [2, 3]. They used a blowup method to Copyright © 2002 Hindawi Publishing Corporation Abstract and Applied Analysis 7:8 (2002) 423–452 2000 Mathematics Subject Classification: 35J55, 35A05, 35J60 URL: http://dx.doi.org/10.1155/S1085337502204030 424 Higher order quasilinear elliptic equations obtain a priori estimates and proved the existence of a solution by degree theoretical arguments. We also take advantage of this general strategy. Here we are concerned with the existence, nonexistence, uniqueness, and regularity of positive solutions to (1.1) whenever p > 1 and p = 2. A; nother goal is to treat systems which are, roughly speaking, a perturbation of (1.2). In this introductory part, we give some examples of our main results, technical assumptions for dealing with general situations are left to other sections. Problem (1.1) has a variational formulation, so that weak solutions correspond to critical points of the functional I(u) = 1 p Ω ∇(∆u) p dx − Ω F(x,u)dx (1.3) defined in the Sobolev space 1,p 1,p Ᏹ p (Ω) = u ∈ W 2,p (Ω) ∩ W0 (Ω) : ∆u ∈ W0 (Ω), 1 < p < ∞ , (1.4) s where F(x,s) = 0 f (x,t)dt. In Theorem 2.2, we employ the socalled Moser iterative scheme to (1.2), in order to regularize the weak solutions of (1.1). The eigenvalue problem −∆ ∆ p (∆u) = Λρ(x)u p−2 u in Ω, u = 0 on ∂Ω (1.5) will help to formulate conditions under which solutions of (1.1) appear. There is ρ a first, positive isolated eigenvalue Λ1 of the weighted problem (1.5), this is the content of Proposition 3.2. The radial form of problem (1.1) is interesting because it is possible to obtain an a priori bound for solutions by means of a blowup process, the key step is a Pohozaev identity in the whole RN , see Theorem 4.1. Notice that the radial ground states of −∆ ∆ p (∆u) = uq in RN (1.6) may fail to be suﬃciently smooth at x = 0, therefore, it is not possible to apply directly, for instance, the general program of [10]. We proceed by approximation, writing an integral relation in the annulus A defined by 0 < R1 < x < R2 . A solution of (1.6) and some of its derivatives are bounded near 0 and exhibit rapid decay at ∞. This fact allows to take the limits R1 → 0 and R2 → ∞, so we obtain RN N − 3p q+1 N − u x dx = 0. q+1 p (1.7) Therefore, positive radial solutions of (1.6) defined in the whole RN cease to exist if N > 3p and p − 1 < q < pN/(N − 3p) − 1. We use this information to Marcelo Montenegro 425 obtain the a priori estimate for positive radial solutions of problem (1.1). In fact, it is possible to work with a class of systems of radial equations that includes (1.1), we pursue this approach in Proposition 5.1. We apply Theorem 5.2 due to Krasnosel’skiı̆ to obtain a positive radial solution. The following example is a consequence of Theorem 5.3 and illustrates the preceding comments, notice the relation with the spectral problem (1.5). Example 1.1. Suppose that for i = 1,2,3 each function gi : [0,R] × [0,+∞) → [0,+∞) is continuous and gi (r,t) ≤ a t βi + 1 (1.8) for r ∈ [0,R], t ≥ 0 and constants a > 0, 0 < β1 , β2 < 1, 0 < β3 < q, p − 1 < q < pN/(N − 3p) − 1, and N > 3p. We also assume that a1 t + g1 (r,t) ≤ a1 + λ t, a2 t + g2 (r,t) ≤ a2 + µ t, q q a3 t + g3 (r,t) ≤ a3 t + γt (1.9) p −1 for r ∈ [0,R] and 0 < t ≤ δ, where λ,µ,γ > 0, ai > 0, and (a1 + λ) p−1 (a2 + µ)(a3 + γ) < Λ11 . The solutions of the system −∆u1 = a1 u2 + g1 r,u2 , −∆ p u2 = a2 u3 + g2 r,u3 in BR , q −∆u3 = a3 u1 + g3 r,u1 , u1 = u2 = u3 = 0 (1.10) on ∂BR are a priori bounded, and in fact there is a C 1 positive weak solution. One of our aims is to extend results obtained for (1.1) to more general systems of the form −∆u1 = f1 x,u1 ,u2 ,u3 , −∆ p u2 = f2 x,u1 ,u2 ,u3 in Ω, −∆u3 = f3 x,u1 ,u2 ,u3 , (1.11) u1 = u2 = u3 = 0 on ∂Ω, which may not have a straightforward variational structure and Ω is not a ball. For instance, if we replace the ball BR in Example 1.1 by a smooth bounded 426 Higher order quasilinear elliptic equations domain Ω, by Lemma 6.1, we see that there is a nonnegative (maybe identically zero) solution to the corresponding problem in Ω. Essentially, the solution comes up by reducing the problem to the verification of the homotopic invariance of degree in cones. For that matter, we obtain a priori estimates by performing a certain scaling that resembles the blowup method used to prove Proposition 5.1. The third equation of (1.10) behaves like q > p − 1 for large values of u1 . A diﬀerent behavior at infinity is also treated in the present paper, namely for q ≤ p − 1, see Example 1.2 below. Some additional conditions taking into account the monotonicity of the functions fi permit us to truncate the problem between a positive subsolution and a supersolution, and actually obtain a positive solution, see Theorem 6.2. The next example fits in the general hypotheses of Theorem 6.2 and is diﬀerent, in nature, from the previous one. Example 1.2. The system −∆u1 = uα2 , β −∆ p u2 = u3 in Ω, (1.12) γ −∆u3 = u1 , u1 = u2 = u3 = 0 on ∂Ω admits a positive solution, provided that 0 < α, β ≤ 1, 0 < γ ≤ p − 1, and αβγ < p − 1. A more general situation occurs when the nonlinearities depend on u1 , u2 , and u3 . The following example is also a consequence of Theorem 6.2. Example 1.3. The system has a positive solution α α α α α α −∆u1 = a11 u1 11 + a12 u2 12 + a13 u3 13 , −∆ p u2 = a21 u1 21 + a22 u2 22 + a23 u3 23 α −∆u3 = a31 u1 31 + a32 uα2 32 + a33 uα3 33 , in Ω, (1.13) u1 = u2 = u3 = 0 on ∂Ω provided that ai j ≥ 0, a12 ,a23 ,a31 > 0, 0 < α11 ,α33 < 1, 0 < α13 < 1/(p − 1), 0 < α21 ,α22 ,α32 < p − 1, 0 < α12 ,α23 ≤ 1, 0 < α31 ≤ p − 1, and α12 α23 α31 < p − 1. The next example is an application of Theorem 6.3, the righthand side nonlinearities have a diﬀerent behavior from the previous ones. But even in this situation, it is possible to combine the ideas of Lemma 6.1 in order to get a priori estimate in a suitable homotopy path, similarly to Theorem 5.3. We finalize by applying Theorem 5.2. Marcelo Montenegro 427 Example 1.4. Let gi : Ω × [0,+∞) → [0,+∞), i = 1,2,3, be bounded continuous functions such that limsup g1 (x,t) < λ < liminf g1 (x,t), t →+∞ t →0+ limsup g2 (x,t) < µ < liminf g2 (x,t), t →+∞ t →0+ (1.14) limsup g3 (x,t) < γ < liminf g3 (x,t), t →+∞ t →0+ ρ uniformly for x ∈ Ω. If λ p−1 µγ = Λ1 , then the system −∆u1 = g1 x,u2 u2 , −∆ p u2 = g2 x,u3 u3 in Ω, p −1 −∆u3 = g3 x,u1 ρ(x)u1 , u1 = u2 = u3 = 0 (1.15) on ∂Ω possesses a positive weak solution. It follows from Theorem 7.1 that the systems (1.12) and (1.13) have a unique positive weak solution. 2. Regularity of weak solutions The space Ᏹ p (Ω) is normed by u Ᏹ p (Ω) = ( Ω ∇(∆u) p dx)1/ p . In what follows, we obtain embeddings which follow from the continuity of the mappings 1,p ∆ : Ᏹ p (Ω) → W0 (Ω) and ∆−1 : Lν (Ω) → W 2,ν (Ω) for 1 < ν < +∞ and from the 1,p classical Sobolev embeddings W0 (Ω) Lν (Ω) and W 2,ν (Ω) Lτ (Ω). Lemma 2.1. (a) The embedding Ᏹ p (Ω) W 2,ν (Ω) is continuous for ν ∈ [1, pN/ (N − p)] if p < N, or for ν ∈ [1,+∞) if p ≥ N and is compact for ν ∈ if 3p < N, or for τ ∈ [1,+∞) if 3p ≥ N and is compact for τ ∈ [1, p∗ ) if 3p < N, or for ∗ τ ∈ [1,+∞) if 3p ≥ N, where p∗ = pN/(N − 3p). But Ᏹ p (Ω) L p (Ω) is not compact. The already defined functional I in (1.3) is of class C 1 if one assumes that f (x,t) ≤ c t q + 1 , (2.1) for some constant c > 0 and for 0 < q ≤ p∗ − 1 if 3p < N and 0 < q < +∞ if 3p ≥ N. The derivative of I is given by I (u)ϕ = Ω ∇(∆u) p−2 ∇(∆u) · ∇(∆ϕ)dx − Ω f (x,u)ϕdx. (2.2) We employ a variant of Moser iterative scheme to conclude that weak solutions of (1.1) are regular. If 3p ≥ N, a weak solution of (1.1) belongs to C 3 (Ω) by a simple application of Lemma 2.1 and L p estimates. 428 Higher order quasilinear elliptic equations Theorem 2.2. Let u ∈ Ᏹ p (Ω) be a weak solution of (1.1). If q < p∗ − 1 and 3p < N, then u ∈ C 3 (Ω). Proof. It is convenient to rewrite (1.1) in the system form (1.2). In this way, we 1,p 1,p∗ /(p∗ −1) (Ω) denote u = u1 and we claim that there are u2 ∈W0 (Ω) and u3 ∈ W0 such that (u1 ,u2 ,u3 ) is a weak solution of the system (1.2). Indeed, u1 ∈ Ᏹ p (Ω) is a critical point of I, then Ω ∇ ∆u1 p−2 ∇ ∆u1 · ∇(∆ψ)dx = Ω f x,u1 ψ dx, (2.3) 1,p for every ψ ∈ Ᏹ p (Ω). Set u2 = −∆u1 ∈ W0 (Ω). Then − Ω ∇u2 p−2 ∇u2 · ∇(∆ψ)dx = for every ψ ∈ Ᏹ p (Ω). Since f (x,u1 ) ∈ L p −∆u3 = f x,u1 admits a unique solution u3 ∈ W 2,p ∗ 1,p∗ the problem ∗ ∗ /(p∗ −1) (Ω) ∩ W 1,p /(p −1) (Ω). 0 ∇u3 · ∇ψ dx = (2.4) u3 = 0 on ∂Ω Ω f x,u1 ψ dx, Ω /(p∗ −1) (Ω), in Ω, for every ψ ∈ W0 ∗ Ω f x,u1 ψ dx, (2.5) Hence, (2.6) (Ω), implying − Ω u3 ∆ψ dx = 1,p∗ ∗ for every ψ ∈ W 2,p (Ω) ∩ W0 Ω Ω f x,u1 ψ dx, (2.7) (Ω). From (2.4) and (2.7), we conclude that ∇u2 p−2 ∇u2 · ∇ϕdx = Ω u3 ϕdx, (2.8) for every ϕ ∈ C0∞ (Ω). Thus, (u1 ,u2 ,u3 ) is a weak solution of system (1.2). Now, we prove its regularity. Define the sequence j if u2 (x) ≥ j, u2 j (x) = u2 (x) if − j < u2 (x) < j, − j if u2 (x) ≤ − j. (2.9) Marcelo Montenegro 429 1,p For any given β ≥ 0, we have u2 j β u2 j ∈ W0 (Ω) and − Ω β u2 j u2 j ∆ p u2 dx ≤ c Ω q β+1 (−∆)−1 (−∆)−1 u2 + 1 u2 dx. (2.10) Suppose that u2 ∈ L pk (Ω) for some pk ≥ pN/(N − p). If 2pk ≥ N or 2pk (q + 1) ≥ Nq, it is easy to verify that u1 ∈ Lα (Ω) for every α ∈ [1,+∞), so we are done. Else, we claim that u2 ∈ L pk+1 (Ω), where pk+1 = N βk + p , N−p βk = pk − (q + 1) N − 2pk . N (2.11) Indeed, since pk ≥ pN/(N − p) and q < p∗ − 1, it follows that βk ≥ 0. There holds − Ω β k u2 j u2 j ∆ p u2 dx ≥ cu2 j βkp+p , L k+1 (2.12) with c > 0 independent of j, see [8]. Using L p estimates, we obtain (−∆)−1 (−∆)−1 u2 q L pk N/((N −2pk )q−2pk ) q ≤ c u 2 L p k + 1 . (2.13) Noting that (βk + 1)/ pk + ((N − 2pk )q − 2pk )/ pk N = 1 and applying Young inequality in (2.10), we get Ω q+β +1 q β +1 (−∆)−1 (−∆)−1 u2 u2 k dx ≤ c u2 L pk k + 1 . (2.14) Therefore, βk +p k +1 u2 j p ≤ c u2 q+β + 1 pk k+1 L L (2.15) with c > 0 not depending on j. Thus, βk +p k +1 u2 p ≤ liminf u2 j βkp+p ≤ c u2 q+β + 1 , pk k+1 k+1 L L L j →+∞ (2.16) proving the claim. Let p0 = pN/(N − p), we are going to show that 2pk ≥ N or 2pk (q + 1) ≥ Nq for some k ∈ N. Observe that pk ≥ p0 for every k ∈ N arguing 430 Higher order quasilinear elliptic equations by induction, since pk ≥ p0 implies βk ≥ 0. Note also that, pk is an increasing sequence, by induction and because pk+2 − pk+1 = N + 2(q + 1) pk+1 − pk . N−p (2.17) Suppose on the contrary that 2pk < N and 2pk (q + 1) < Nq for every k ∈ N. Then pk converges to L ≥ p0 . Using (2.11) and taking the limit L = lim pk+1 = k→+∞ pN N lim βk + N − p k→+∞ N−p (2.18) pN N − 2L = L− (q + 1) + , N−p N N−p N we see that L = N(q + 1 − p)/(p + 2(q + 1)) ≥ pN/(N − p), implying that q + 1 ≥ p∗ , a contradiction. 3. Eigenvalue problem We investigate the eigenvalue problem (1.5). Assume that ρ is a nonnegative and nontrivial function belonging to L∞ (Ω). Define the functionals A,B : Ᏹ p (Ω) → R by A(u) = 1 u p p Ᏹ p (Ω) , B(u) = 1 p Ω ρ(x) u+ p dx, (3.1) where u+ = max{u,0}. It is easy to verify that A and B are C 1 . Define ρ Λ1 = inf A(u). (3.2) B(u)=1 ρ Clearly, Λ1 is a positive number attained by some u ∈ Ᏹ p (Ω). Also, there exists η > 0 such that A (u)ϕ = ηB (u)ϕ for every ϕ ∈ Ᏹ p (Ω). Taking ϕ = u, we obtain ρ A(u) = ηB(u). Thus, η = Λ1 and u is a critical point of the functional J(u) = 1 p Ω ρ ∇(∆u) p dx − Λ1 p Ω ρ(x)u+ p dx. (3.3) The next comparison lemma is borrowed from [12]. Lemma 3.1. Let u,v ∈ C 1 (Ω) be functions satisfying −∆ p u ≤ −∆ p v in Ω and u ≤ v on ∂Ω in the weak sense, then u ≤ v in Ω. Furthermore, assume that ∇v ≡ 0 on ∂Ω and let η > 0 be small enough, such that the set Γ = {x ∈ Ω : ∇v(x) > η, dist(x,∂Ω) < η} is nonempty and open. Then either u ≡ v in Γ or u < v in Γ and for each x ∈ ∂Γ with u(x) = v(x), we have ∂u(x)/∂ν > ∂v(x)/∂ν. There is a first eigenvalue associated to problem (1.5), which is isolated from above and from below. Marcelo Montenegro 431 ρ Proposition 3.2. (i) If Λ = Λ1 , then (1.5) admits a positive weak solution; ρ (ii) if Λ < Λ1 , then (1.5) does not admit a positive weak subsolution; ρ (iii) if Λ > Λ1 , then (1.5) does not admit a positive weak supersolution; ρ (iv) Λ1 is isolated. Proof. It is useful to rewrite the eigenvalue problem in the following way: −∆u1 = λu2 , −∆ p u2 = µu3 in Ω, p −2 −∆u3 = γρ(x)u1 u1 , (3.4) u1 = u2 = u3 = 0 on ∂Ω. We reformulate items (i), (ii), and (iii) in terms of a surface in three parameters λ,µ,γ > 0: ρ (i) if λ p−1 µγ = Λ1 , then the system (3.4) admits a positive weak solution; ρ (ii) if λ p−1 µγ < Λ1 , then the system (3.4) does not admit a nonnegative weak subsolution with a positive component in Ω; ρ (iii) if λ p−1 µγ > Λ1 , then the system (3.4) does not admit a positive weak supersolution. Since u1 is a nontrivial critical point of J and ρ is a nonnegative function, (i) follows from the beginning of the proof of Theorem 2.2 and the strong maximum principle of [13]. We prove (ii) . Suppose on the contrary that problem (3.4) admits a nonnegative weak subsolution (v1 ,v2 ,v3 ) with a positive component and ρ ρ p −1 λ p−1 µγ < Λ1 . Choose λ0 = λ, µ0 = µ, and γ0 > γ such that λ0 µ0 γ0 = Λ1 . According to part (i) , we can take a positive eigenfunction (u1 ,u2 ,u3 ) corresponding to (λ0 ,µ0 ,γ0 ). Let Γ2 be the set associated to function u2 given in Lemma 3.1, that is, Γ2 = {x ∈ Ω : ∇u2 (x) > η}. Define the set S = {s > 0 : u1 > sv1 , u2 > sv2 , and u3 > s p−1 v3 in Γ2 }. By the strong maximum principle, S = ∅ and since one of the components of (v1 ,v2 ,v3 ) is positive, S is bounded. Let s∗ = supS. Since p −1 p −1 p −1 p −1 −∆ u3 − s∗ v3 ≥ γ0 ρ(x)u1 − s∗ γρ(x)v1 p −1 ≥ γ0 − γ ρ(x)u1 in Γ2 , (3.5) by the strong maximum principle, u3 > (s∗ + ε) p−1 v3 in Γ2 for ε > 0 small enough. Thus, −∆ p u2 + s∗ p −1 ∆ p v2 ≥ µ0 u3 − s∗ p−1 µ0 v3 ≥ µ0 s∗ 1− ∗ s +ε p −1 u3 in Γ2 (3.6) 432 Higher order quasilinear elliptic equations implies, by Lemma 3.1, that u2 > (s∗ + ε)v2 in Γ2 for ε > 0 small enough. Finally, from −∆ u1 − s∗ v1 ≥ λ0 u2 − s∗ λ0 v2 ≥ λ0 1 − s∗ u2 s∗ + ε in Γ2 , (3.7) it follows that u1 > (s∗ + ε)v1 in Γ2 for ε > 0 is small enough, contradicting the definition of s∗ . Suppose, on the contrary, that problem (3.4) possesses a positive supersolution (v1 ,v2 ,v3 ). Part (iii) follows similarly. Let λ0 = λ, µ0 = µ, and γ0 < ρ p −1 γ be such that λ0 µ0 γ0 = Λ1 . Denote (u1 ,u2 ,u3 ) a positive eigenfunction related to (λ0 ,µ0 ,γ0 ) and Γ2 the set associated to v2 as in Lemma 3.1. Define the set S = {s > 0 : v1 > su1 , v2 > su2 , and v3 > s p−1 u3 in Γ2 }. Item (iii) follows by the same steps of (ii) . We sketch the proof of item (iv), the details follow from the ideas in [1]. If v is another eigenfunction corresponding to an eigenvalue Λ, we ρ ρ ρ have Λ ≥ Λ1 , by (3.2) and (3.3). Hence Λ1 is isolated to the left. Let Λn > Λ1 be a sequence of eigenvalues corresponding to the eigenfunctions vn . Item (iii) implies that each vn must change sign. The sequence vn converges uniformly in a set of positive measure to the first eigenfunction of (1.5), a contradiction. 4. Nonexistence of radial solutions in RN In this section, we prove a result of Liouville type for (1.6). It is a fundamental step for obtaining a priori estimates in Section 5. Theorem 4.1. Let f (t) = t q−1 t with p − 1 < q < p∗ − 1 and N > 3p. Then (1.7) has no positive solution in C 3 (RN ). Proof. We rewrite (1.6) as a system of radial equations and proceed by approximation. Suppose that u is a positive solution, (1.6) transforms into − r N −1 u1 (r) = r N −1 u2 (r), p −2 − r N −1 u2 (r) u2 (r) = r N −1 u3 (r) for r > 0, − r N −1 u3 (r) = r N −1 f u1 , (4.1) where u = u1 . The existence of positive u2 and u3 is treated in Theorem 2.2. A solution (u1 ,u2 ,u3 ) ∈ (C 1 [0,+∞))3 of the system (4.1) satisfies the integral relations −r N −1 u1 (r) = p −2 −r N −1 u2 (r) u2 (r) = −r N −1 u3 (r) = r 0 r 0 r 0 sN −1 u2 (s)ds, sN −1 u3 (s)ds for r > 0, sN −1 f u1 (s) ds, (4.2) Marcelo Montenegro 433 and the following Pohozaev type identity for every constant a and 0 < R1 < R2 : R2 R1 NF u1 (r) − au1 (r) f u1 (r) + a + = 2 3p − N u (r) p r N −1 dr 2 p (−1)i φ Ri ,a,u1 Ri ,u2 Ri ,u3 Ri ,u1 Ri ,u2 Ri ,u3 Ri ) RNi −1 . i=1 (4.3) We need to detail the expression of φ in order to verify that φ(·)RNi −1 goes to 0 as R1 → 0 and R2 → +∞. Consider the functional Ᏼ = Ᏼ(x,u1 ,s) depending on x, u1 , and the third derivatives of u1 formally represented by s, 1 ∇ ∆u1 x p − F u1 x p Ᏼ= for x ∈ RN − {0}, (4.4) where F is the primitive of f . By relations (4.2) and a bootstrap argument, we conclude that (u1 ,u2 ,u3 ) ∈ (C 1 [0,+∞) ∩ C ∞ (0,+∞))3 , so Ᏼ is well defined. Noting that si jl = 0 if j = l, we obtain − A NᏴ − au1 Ᏼu1 − (a + 3)Di j j u1 Ᏼsi j j dx = xi Ᏼ − ∂A N xl Dl u1 + au1 D j j Ᏼsi j j + D j xl Dl u1 + au1 D j Ᏼsi j j j,l=1 − D j j xl Dl u1 + au1 Ᏼsi j j νi ds, (4.5) where ν is the unit outward normal vector to the boundary ∂A. Since, u1 Ᏼu1 = N −u1 f (u1 ) and i, j =1 Di j j u1 Ᏼsi j j = ∇(∆u1 ) p , the lefthand side of (4.5) reduces to A 3p − N u (r) p dx 2 p R2 3p − N u (r) p r N −1 dr, = ωN NF u1 (r) − au1 (r) f u1 (r) + a+ 2 p R1 NF u1 (r) − au1 (r) f u1 (r) + a + (4.6) where ωN is the area of the unit (N − 1)sphere. We obtain (4.3) after passing to radial coordinates and replacing u2 and u3 in (4.5). We also use the fact that (u1 ,u2 ,u3 ) is a solution of (4.1), translated in the integral relations (4.2). Write each term of the righthand side integral of (4.5) 434 Higher order quasilinear elliptic equations 2 xi 1 u (r) p − F u1 (r) xi , = 2 r p r 2 x φ2 = − xl Dl u1 + au1 = −u1 (r) l − au1 (r), r p −2 x x xi φ3 = D j j Ᏼsi j j = D j j − u2 (r) u2 (r) i i r r r x x p −2 p −2 i = − D j j u2 (r) u2 (r) + 2D j u2 (r) u2 (r) D j i r r p −2 x x + u2 (r) u2 (r)D j j i i , r r φ1 = xi Ᏼ (4.7) where r p −2 xj −N N −1 D j u2 (r) u2 (r) = − u3 (r) + (N − 1)r s u3 (s)ds , r 0 r p −2 xj xj D j j u2 (r) u2 (r) = − u3 (r) − N(N − 1)r −N −1 sN −1 u3 (s)ds r r 0 xj xj + (N − 1)r −1 u3 (r) r r r xj + u3 (r) + (N − 1)r −N sN −1 u3 (s)ds D j , r 0 φ4 = D j xl Dl u1 + au1 = D j u1 (r)r + aD j u1 (r) xj xj = −u2 (r)x j − (N − 2)u1 (r) + au1 (r) , r r p −2 xi x i xi φ5 = D j Ᏼsi j j = D j − u2 (r) u2 (r) r r r x p −2 p −2 x x i = − D j u2 (r) u2 (r) + u2 (r) u2 (r)D j i i , r r r 2 x φ6 = −D j j xl Dl u1 + au1 = −D j j u1 (r) l − aD j j u1 (r), r (4.8) where x2j x2j + (N − 2)u (r) 2 r3 r2 2 x u (r) j − (N − 2) 1 − u2 (r) , r r x2j u (r) x2j D j j u1 (r) = −Nu1 (r) 3 + 1 − u2 (r) 2 , r r r 2 p −2 x xi φ7 = Ᏼsi j j = −u2 (r) u2 (r) 2i . r r D j j u1 (r)r = N(N − 2)u1 (r) Hence φ = φ1 + φ2 φ3 + φ4 φ5 + φ6 φ7 . (4.9) Marcelo Montenegro 435 If (u1 ,u2 ,u3 ) is a positive solution of (4.1), the first member of (4.3) is positive for every 0 < R1 < R2 . Actually, choosing a = (N − 3p)/ p, then NF(t) − at f (t) = t q+1 N − a > 0 for t > 0. q+1 (4.10) Now, we intend to prove that the righthand side of (4.3) converges to zero as R1 → 0 and as R2 → +∞. We analyze the term φ(·)RN1 −1 near zero. Since ui and ui are bounded near 0 for i = 1,2,3, we get φ1 ≤ cr, φ2 ≤ c(1 + r), φ3 ≤ c 1 + r + r −1 + r −2 , φ4 ≤ c(1 + r), φ5 ≤ c 1 + r + r −1 , φ6 ≤ c 1 + r + r −1 , φ7 ≤ c, (4.11) where the constant c does not depend on r for r > 0 small enough. Thus, we obtain φ R1 ,a,u1 R1 ,u2 R1 ,u3 R1 ,u R1 ,u R1 ,u R1 RN −1 1 1 2 3 N −1 1 −2 ≤ c 1 + R1 + R− , 1 + R1 R1 (4.12) for every R1 near 0. Since N ≥ 4, we conclude that φ(·)RN1 −1 → 0 as R1 → 0. It remains to check the behavior of φ(·)RN2 −1 for large R2 . Here we use a similar strategy to [3]. Integrating by parts the first equation of system (4.1), we have −r N −1 u1 (r) = r 0 sN −1 u2 (s)ds = sN u2 (s)ss==r0 − N r 0 sN u (s)ds. N 2 (4.13) Repeating the same computation to the second and third equations of (4.1), we find −u1 (r) ≥ r u2 (r), N −u2 (r) ≥ r u3 (r) N 1/ p−1 , −u3 (r) ≥ r q u (r), N 1 (4.14) for every r ≥ 0. Therefore, u1 (r) , −u1 (r) ≤ (N − 2) r N − p u2 (r) , −u2 (r) ≤ p−1 r −u3 (r) ≤ (N − 2) u3 (r) , r (4.15) 436 Higher order quasilinear elliptic equations for r > 0. Putting the above relations together, we obtain, step by step, the following estimates for r > 0: u1 (r) ≤ cr −3p/(q+1− p) , −u1 (r) ≤ cr −3p/(q+1− p)−1 , u2 (r) ≤ cr −3p/(q+1− p)−2 , −u2 (r) ≤ cr −3p/(q+1− p)−3 , u3 (r) ≤ cr −3p/(q+1− p)−3(p−1)−1 , −u3 (r) ≤ cr (−3p/(q+1− p)−3)(p−1)−2 (4.16) , where the above constant c does not depend on r. From (4.16), we see that φ1 ≤ c r (−3p/(q+1− p)−3)p + r −3p(q+1)/(q+1− p) r, φ2 ≤ cr −3p/(q+1− p) , φ3 ≤ cr (−3p/(q+1− p)−3)(p−1)−2 , φ4 ≤ cr −3p/(q+1− p)−1 , φ5 ≤ cr (−3p/(q+1− p)−3)(p−1)−1 , φ6 ≤ cr −3p/(q+1− p)−2 , φ7 ≤ cr (−3p/(q+1− p)−3)(p−1) , (4.17) where the above constant c does not depend on r for suﬃciently large values of r. Therefore, it follows that φ R2 ,a,u1 R2 ,u2 R2 ,u3 R2 ,u R2 ,u R2 ,u R2 RN −1 ≤ cRk , (4.18) 2 2 1 2 3 where k = −3p2 /(q + 1 − p) + N − 3p < 0, since p − 1 < q < p∗ − 1. Hence, φ(·)RN2 −1 → 0 as R2 → +∞. Consequently, relation (4.3) becomes +∞ 0 N − 3p q+1 N − u1 (r)r N −1 dr = 0, q+1 p (4.19) implying u1 ≡ 0 in [0,+∞), a contradiction. 5. Existence of radial solutions We are going to prove the existence of nontrivial radial solutions for (1.1) in balls. Since our approach can be used to handle more general situations, in fact, we deduce the results for a system like (1.11) that includes (1.1), namely −∆u1 = f1 r,u1 ,u2 ,u3 , −∆ p u2 = f2 r,u1 ,u2 ,u3 in BR , −∆u3 = f3 r,u1 ,u2 ,u3 , u1 = u2 = u3 = 0 on ∂BR . (5.1) Marcelo Montenegro 437 Suppose that, for i = 1,2,3, each function fi : [0,R] × [0,+∞)3 → [0,+∞) is continuous and fi r,t1 ,t2 ,t3 = gi r,t1 ,t2 ,t3 + 3 hil r,t1 ,t2 ,t3 , (5.2) l=1 where gi and hil are nonnegative continuous functions verifying β β β gi r,t1 ,t2 ,t3 ≤ a t1 i1 + t2 i2 + t3 i3 + 1 , lim tl →+∞ hil r,t1 ,t2 ,t3 = ail tlαil (5.3) (5.4) for some constant a > 0, uniformly for r ∈ [0,R], tk ∈ [0,+∞) for every k = 1,2,3 with k = l and sup hil ·,t1 ,t2 ,t3 : tk ≥ 0, k = 1,2,3, k = l, 0 < tl < M ∈ L∞ (0,R), (5.5) for every M > 0, where a12 ,a23 ,a31 > 0, αil ,βil ≥ 0, β1l γl < γ2 , α1l γl ≤ γ2 , γ1 = 3p , α31 + 1 − p a11 ,a13 ,a21 ,a22 ,a32 ,a33 = 0, α12 = α23 = 1, β2l γl < γ3 , α2l γl ≤ γ3 , 2α31 + p + 2 γ2 = , α31 + 1 − p p − 1 < α31 < p∗ − 1, β3l γl < α31 γ1 , α3l γl ≤ α31 γ1 , 3pα31 + 2p − 2α31 − 2 γ3 = . α31 + 1 − p (5.6) We proceed to prove that solutions of (5.1) are a priori bounded. Clearly, (1.1) fits in the above setting if one assumes  f (r,t) ≤ c(t q + 1) for p − 1 < q < p∗ − 1. In the previous notation q = α31 , gi ≡ 0, h12 = t2 , h23 = t3 , h31 = f (r,t3 ), and others hil are zero. Proposition 5.1. There is a constant c > 0 such that ui C[0,R] ≤ c for every i = 1,2,3, where (u1 ,u2 ,u3 ) ∈ (C 1 [0,R])3 is a radial solution of the system (5.1), provided 3p < N and (5.3), (5.4), and (5.5) hold. Proof. We write (5.1) as a system of ordinary diﬀerential equations. A triplet (u1 ,u2 ,u3 ) ∈ (C 1 [0,R])3 is a radial solution of system (5.1) if and only if it is a radial weak solution in (C 1 (BR ))3 of the following problem: − r N −1 u1 (r) = r N −1 f1 r,u1 ,u2 ,u3 , p −2 − r N −1 u2 (r) u2 (r) = r N −1 f2 r,u1 ,u2 ,u3 , − r N −1 u3 (r) = r N −1 f3 r,u1 ,u2 ,u3 , u1 (R) = u2 (R) = u3 (R) = 0, (5.7) 438 Higher order quasilinear elliptic equations for r ∈ (0,R). And it satisfies u1 (r) = u2 (r) = u3 (r) = R sN −1 r R r R r s 1 1 0 t N −1 f1 t,u1 (t),u2 (t),u3 (t) dt ds, s t N −1 f2 t,u1 (t),u2 (t),u3 (t) dt 1/(p−1) ds, sN −1 0 s 1 t N −1 f3 t,u1 (t),u2 (t),u3 (t) dt ds, N − 1 s 0 (5.8) for r ∈ [0,R]. Indeed, if (u1 ,u2 ,u3 ) ∈ (C 1 (BR ))3 is a radial weak solution of (5.1), take ϕ ∈ C0∞ (0,R), then R 1 ∇u1 · ∇ϕdx, ωN BR 0 R 1 r N −1 f1 r,u1 ,u2 ,u3 ϕ(r)dr = f1 r,u1 ,u2 ,u3 ϕdx. ωN BR 0 r N −1 u1 (r)ϕ (r)dr = (5.9) Thus, R 0 r N −1 u1 (r)ϕ (r)dr = R 0 r N −1 f1 r,u1 ,u2 ,u3 ϕ(r)dr, (5.10) for every ϕ ∈ C0∞ (0,R), implying −(r N −1 u1 (r)) = r N −1 f1 (r,u1 ,u2 ,u3 ) for r ∈ (0,R). Conversely, take (u1 ,u2 ,u3 ) ∈ (C 1 [0,R])3 a radial solution of (5.1). Integrating the first equation of (5.7), from 0 to r, we get −u1 (r) = r 1 r N −1 0 sN −1 f1 s,u1 ,u2 ,u3 ds for r > 0. (5.11) Multiplying the above identity by x · ∇ϕ(x)/r with ϕ ∈ C0∞ (BR ) and integrating by parts on BR , we obtain BR ∇u1 · ∇ϕdx = = = BR 1 rN BR i Di BR r 0 sN −1 f1 s,u1 ,u2 ,u3 ds x · ∇ϕdx xi rN r 0 sN −1 f1 s,u1 ,u2 ,u3 ds ϕdx (5.12) f1 r,u1 ,u2 ,u3 ϕdx. The equivalence for other equations of system (5.1) is analogous. From now on, we are going to work with system (5.7). If the a priori estimate does not hold, there exists a sequence (u1k ,u2k ,u3k ) ∈ (C 1 [0,R])3 of nonnegative radial solutions of system (5.7) satisfying t jk = sup u jk (r) = u jk (0) −→ +∞ r ∈[0,R] as k −→ +∞, (5.13) Marcelo Montenegro 439 1/γ 1/γ for some j ∈ {1,2,3}. Consider the sequence λk defined by λk = t1k 1 + t2k 2 + 1/γ t3k 3 . Since γi > 0 for i = 1,2,3, we have λk −→ +∞. (5.14) γ 1/γ 1/γ Define the rescaling r = λk r and uik (r) = (1/λki )uik (r). Since u1k 1 (0) + u2k 2 (0) + 1/γ u3k 3 (0) = 1, without loss of generality, we may assume that uik (0) → ui for i = 1,2,3. In particular, 1/γ1 u1 1/γ2 + u2 1/γ3 + u3 = 1. (5.15) In addition, it is easy to see that (ũ1k , ũ2k , ũ3k ) is nonnegative and satisfy −γ2 − rN −1 u 1k (r = λk rN −1 f1 N −1 − r r γ1 γ2 γ3 ,λ u1k r ,λk u2k r ,λk u3k r , λk k p−2 r γ1 γ2 γ3 −γ3 N −1 u2k r u2k r =λk r f2 ,λk u1k r ,λk u2k r ,λk u3k r , −α31 γ1 N −1 − rN −1 u 3k r = λk r f3 λk r γ1 γ2 γ3 ,λ u1k r ,λk u2k r ,λk u3k r . λk k (5.16) From (5.4), (5.5), and (5.14), it follows that −αil γl lim λk k→+∞ hil r γ1 γ2 γ3 ,λ u1k r ,λk u2k r ,λk u3k r − ail uαlkil r = 0. λk k (5.17) Also, from (5.3), we see that gi γβ r γ1 γ2 γ3 γ β γ β ,λk u1k r ,λk u2k r ,λk u3k r ≤ a λk1 i1 + λk2 i2 + λk3 i3 + 1 . λk (5.18) Therefore, r γ1 γ2 γ3 ,λ u1k r ,λk u2k r ,λk u3k r −→ 0, λk k r γ1 γ2 γ3 −γ λk 3 g2 ,λ u1k r ,λk u2k r ,λk u3k r −→ 0, λk k r γ1 γ2 γ3 −α γ λk 31 1 g3 ,λ u1k r ,λk u2k r ,λk u3k r −→ 0, λk k −γ λk 2 g1 (5.19) uniformly for r ∈ [0,Rλk ]. Fix a constant R > 0. For large enough k ∈ N, we have Furthermore, we get R < Rλk . Hence it is possible to restrict uik to [0, R]. u ik C[0,R] ≤1 (5.20) for i = 1,2,3. We intend to apply ArzelaAscoli theorem, so we are going to show In fact, from (5.16), we that each sequence (uik ) is equicontinuous in C[0, R]. 440 Higher order quasilinear elliptic equations conclude that 1 d 2 n−1 2 u r − u r 2 dr 1k r 1k r γ1 γ2 γ3 −γ2 ,λk u1k r ,λk u2k r ,λk u3k r u1k r , = λk f 1 λk p n − 1 p p−1 d u u 2k r − 2k r − p dr r r γ1 γ2 γ3 −γ3 = λk f 2 ,λk u1k r ,λk u2k r ,λk u3k r u2k r , λk 1 d 2 n−1 2 − u r − u r 2 dr 3k r 3k r γ1 γ2 γ3 −α31 γ1 f3 ,λk u1k r ,λk u2k r ,λk u3k r u3k r . = λk λk − (5.21) By (5.17) and (5.19), there exists a constant c > 0 such that r γ1 γ2 γ3 ,λ u1k r ,λk u2k r ,λk u3k r ≤ c, λk k r γ1 γ2 γ3 −γ ,λ u1k r ,λk u2k r ,λk u3k r ≤ c, λk 3 f 2 λk k r γ1 γ2 γ3 −α γ λk 31 1 f3 ,λ u1k r ,λk u2k r ,λk u3k r ≤ c. λk k −γ λk 2 f 1 (5.22) Thus, from (5.21), we obtain 1 d 2 u r + cu1k r ≤ 0, 2 dr 1k p p−1 d u 2k r + cu 2k r ≤ 0, p dr 1 d 2 u r + cu3k r ≤ 0. 2 dr 3k (5.23) Integrating the above inequalities from 0 to r, we get r 1 2 u1k r + c u1k (t)dt ≤ 0, 2 0 r p −1 p u2k r + c u2k (t)dt ≤ 0, p 0 1 2 u r + c 2 3k r 0 (5.24) u3k (t)dt ≤ 0, implying that  u1k (r) ≤ (2c)1/2 ,  u2k (r) ≤ ((p/(p − 1))c)1/ p , and  u3k (r) ≤ 1/2 (2c) for every r ∈ [0, R]. Hence, (uik ) is equicontinuous in C[0, R] for i = 1,2,3. By ArzelaAscoli theorem, up to a subsequence, we have uik → ui in C[0, R]. Marcelo Montenegro 441 From (5.16), it follows that In particular, ui ≥ 0 in [0, R]. uik (0) − uik r = r 0 fik (s)ds, (5.25) where −γ λ k 2 s N −1 t γ1 γ γ t f1 ,λ u1k (t),λk2 u2k (t),λk3 u3k (t) dt, sN −1 0 λk k −γ3 1/(p−1) s λ t γ1 γ γ f2k (s) = Nk−1 t N −1 f2 ,λk u1k (t),λk2 u2k (t),λk3 u3k (t) dt , (5.26) s λk 0 f1k (s) = −α γ f3k (s) = λk 31 1 sN −1 s 0 t N −1 f 3 t γ1 γ γ ,λ u1k (t),λk2 u2k (t),λk3 u3k (t) dt. λk k Using (5.17) and (5.19), we conclude that r 0 r 0 r 0 f1k (s)ds −→ f2k (s)ds −→ f3k (s)ds −→ r s 1 N −1 0 s r 1 0 r 0 0 t N −1 a12 u2 (t)dt ds, s t N −1 a23 u3 (t)dt 1/(p−1) sN −1 0 s 1 t N −1 a31 uα1 31 (t)dt ds. sN −1 0 ds, (5.27) 3 Letting k → +∞ in (5.25), we get a nonnegative solution (u1 , u2 , u3 ) ∈ (C 1 [0, R]) of system − rN −1 u 1 r = a12 rN −1 u 2 r , p−2 2 (r) u − rN −1 u 2 r = a23 rN −1 u 3 r for r ∈ 0, R , α − rN −1 u 3 r = a31 rN −1 u 1 31 r . (5.28) A diagonal subsequence argument provides a nonnegative solution (u1 , u2 , u3 ) ∈ (C 1 [0,+∞))3 of (5.28) in (0,+∞). By Theorem 4.1, we conclude that ui ≡ 0 in [0,+∞) for i = 1,2,3, contradicting (5.15). Theorem 5.2 (Krasnosel’skiı̆). Let C be a cone in a Banach space X and T : C → C a continuous compact mapping with T(0) = 0. Assume that there exist t0 > 0 and 0 < r < R, such that (i) u = tTu, for all u ∈ C such that u X = r, for all t ∈ [0,1], (ii) there exists a continuous compact mapping H : C × [0,+∞) → C satisfying, (a) H(u,0) = Tu, for all u ∈ C with u X ≤ R, (b) H(u,t) = u, for all u ∈ C with u X ≤ R, for all t ≥ t0 , (c) H(u,t) = u, for all u ∈ C with u X = R, for all t ≥ 0. Then, T has a fixed point u ∈ C such that r < u X < R. 442 Higher order quasilinear elliptic equations The following assumptions are satisfied by Example 1.1, f1 r,t1 ,t2 ,t3 ≤ λt2 , f2 r,t1 ,t2 ,t3 ≤ µt3 , (5.29) p −1 f3 r,t1 ,t2 ,t3 ≤ γρ(r)t1 , for every r ∈ (0,R) and 0 < t1 ,t2 ,t3 ≤ δ, where ρ ∈ L∞ (0,R), ρ ≡ 0, ρ ≥ 0, λ,µ,γ > ρ ρ 0, and λ p−1 µγ < Λ1 , where Λ1 is the first eigenvalue of (1.5). The above general assumptions are related to the existence of nontrivial solutions of (5.1). Note that (5.29) implies that (1.11) possesses the trivial solution. Equation (1.1) is included in the theorem below, in this particular situation, hyρ pothesis (5.29) is reduced to f (r,t) ≤ γρ(r)t p−1 for γ < Λ1 with λ = µ = 1. Theorem 5.3. System (1.11) possesses a nontrivial nonnegative weak solution if (5.3), (5.4), (5.5), and (5.29) are fulfilled. Proof. Consider the space X = {u = (u1 ,u2 ,u3 ) ∈ (C[0,R])3 : ui (R) = 0 for i = 1,2,3} endowed with the norm u X = u1 C[0,R] + u2 C[0,R] + u3 C[0,R] . Denote by C the cone of nonnegative functions of X. Define the mapping H : [0,+∞) × C → C by H(t,u) = v, where v = (v1 ,v2 ,v3 ) with v1 (r) = v2 (r) = v3 (r) = R r R r R r 1 ξ N −1 1 ξ N −1 1 ξ N −1 ξ 0 ξ 0 ξ 0 N −1 s ! ! ! f1 s,u1 (s),u2 (s),u3 (s) + t ds d ξ, 1/(p−1) sN −1 f2 s,u1 (s),u2 (s),u3 (s) + t ds N −1 s d ξ, (5.30) f3 s,u1 (s),u2 (s),u3 (s) + t ds d ξ, for every r ∈ [0,R]. It is easy to see that the mapping H is well defined, continuous and compact. Let T : C → C be given by T(u) = H(0,u). Then T(0) = 0. Now we seek r0 > 0 such that u = tT(u) for every t ∈ [0,1] and u ∈ C with u X = r0 . Take δ ≥ r0 . If u = tT(u) for some t ∈ [0,1] and u ∈ C with u X = r0 . Then, from Proposition 5.1 and (5.29), we conclude that (u1 ,u2 ,u3 ) ∈ (C 1 (BR ))3 and −∆u1 = t f1 r,u1 ,u2 ,u3 ≤ λu2 , −∆ p u2 = t p−1 f2 r,u1 ,u2 ,u3 ≤ µu3 in BR , p −1 −∆u3 = t f3 r,u1 ,u2 ,u3 ≤ γρ(r)u1 , u1 = u2 = u3 = 0 (5.31) on ∂BR , in the weak sense. Since u has a positive component in Ω, by Proposition 3.2(ii), we obtain a contradiction. We claim that there exist R > r0 and t0 > 0 such that H(t,u) = u for every t ≥ t0 and u ∈ C with u X ≤ R. Also H(t,u) = u in C for Marcelo Montenegro 443 the same r0 and for every t ≥ 0 when u X = R. Indeed, let t ≥ 0 and u ∈ C verify H(t,u) = u. From (5.4) and (5.5), we get 2 − c1 , f1 r,t1 ,t2 ,t3 ≥ λt f2 r,t1 ,t2 ,t3 ≥ µt3 − c1 , (5.32) p −1 f3 r,t1 ,t2 ,t3 ≥ γt1 − c1 , µ, γ > 0 and λ p−1 µγ > Λ11 . Again, by for every r ∈ [0,R] and t1 ,t2 ,t3 ≥ 0, where λ, Proposition 5.1 and (5.32), we have 2 + t − c1 , −∆u1 ≥ λu −∆ p u2 ≥ µu3 + t − c1 in BR , p −1 −∆u3 ≥ γu1 + t − c1 , u1 = u2 = u3 = 0 (5.33) on ∂BR , in the weak sense. Applying Proposition 3.2(iii), we obtain t ≤ c1 . By Proposition 5.1, we conclude that u X ≤ c. It is enough to take R > c and t0 > c1 . The con clusion follows from Krasnosel’skiı̆ theorem (Theorem 5.2). 6. Further generalizations to nonradial systems Some classes of general systems (1.11) possess positive solutions. The next lemma is a fundamental preliminary result in this direction and can be viewed as an extension of Theorem 5.3 to more general domains. We deduce an a priori estimate and the existence of a nonnegative solution by using the homotopic invariance of degree in cones. Assume that 1/(p−1) f1 x,t1 ,t2 ,t3 ≤ ε0 t1 + λt2 + ε0 t3 + c, p −1 p −1 f2 x,t1 ,t2 ,t3 ≤ ε0 t1 + ε0 t2 + µt3 + c, (6.1) p −1 p −1 f3 x,t1 ,t2 ,t3 ≤ γρ(x)t1 + ε0 t2 + ε0 t3 + c, for every x ∈ Ω and t1 ,t2 ,t3 ≥ 0, where ε0 > 0 and c > 0 are constants, ρ ∈ L∞ (Ω), ρ ρ ≥ 0, ρ ≡ 0, λ,µ,γ > 0, and λ p−1 µγ < Λ1 . If we take ε0 = c = 0 in (6.1), we recover condition (5.29) in the nonradial setting. In this way, (1.1) is included in the following preliminary result if we ρ assume f (x,t) ≤ γρ(x)t p−1 for γ < Λ1 . Lemma 6.1 can be viewed as a generalization of Theorem 5.3, unfortunately the solution maybe identically zero. Theorem 6.2 is sharper in the sense that it presents a positive solution under suitable additional conditions. 444 Higher order quasilinear elliptic equations Lemma 6.1. There is a constant ε0 > 0 such that for each c > 0 and every fi fulfilling the growth conditions (6.1), system (1.11) admits a nonnegative weak solution in (C 1 (Ω))3 . Proof. Let X = {u = (u1 ,u2 ,u3 ) ∈ (C(Ω))3 : u1 = u2 = u3 = 0 on ∂Ω} be the space endowed with the norm u X = u1 C(Ω) + u2 C(Ω) + u3 C(Ω) . Denote by C the cone of X given by C = {u ∈ X : u ≥ 0 in Ω}. Consider the mapping T f = T( f1 , f2 , f3 ) : [0,1] × C → C defined by T f (t,u) = v, where v = (v1 ,v2 ,v3 ) satisfies −∆v1 = t f1 x,u1 ,u2 ,u3 , −∆ p v2 = t f2 x,u1 ,u2 ,u3 in Ω, −∆v3 = t f3 x,u1 ,u2 ,u3 , (6.2) v1 = v2 = v3 = 0 on ∂Ω. By the maximum principle and C 1 estimates, T f is a welldefined, continuous, and compact mapping. We claim that there exists a constant ε0 > 0 such that for each 0 < c ≤ 1 and each triplet of nonnegative functions f = ( f1 , f2 , f3 ) satisfying (6.1), there is a constant M0 > 0 not depending on c and f such that u X < M0 for every u ∈ C with T f (t,u) = u for some t ∈ [0,1]. Otherwise, there exist sequences tk ∈ [0,1], uk ∈ C, ck ∈ (0,1], εk ∈ (0,+∞), and ( f1k , f2k , f3k )k verifying εk → 0, Tk (tk ,uk ) = uk , uk X → ∞, and (6.1) with ck , ( f1k , f2k , f3k )k , and εk in place of c, ( f1 , f2 , f3 ), and ε, respectively, where Tk = T( f1k , f2k , f3k ) and uk = (u1k ,u2k ,u3k ). Define ũ1k = ũ2k = u1k u1k C(Ω) + u2k C(Ω) + u2k 1/(p−1) ; u3k C(Ω) 1/(p−1) ; u1k C(Ω) + u2k C(Ω) + u3k C(Ω) ũ3k = u1k C(Ω) + u2k u3k C(Ω) + (6.3) 1/(p−1) p−1 . u3k C(Ω) 1/(p−1) Then ũ1k C(Ω) + ũ2k C(Ω) + ũ3k C(Ω) = 1. Using (6.1) and applying C 1 estimates in (6.2), up to a subsequence, we conclude that ũk converges to a function ũ in (C 1 (Ω))3 . Furthermore, it follows that ũ ≥ 0 in Ω, ũ1 C(Ω) + ũ2 C(Ω) + 1/(p−1) ũ3 C(Ω) = 1, and −∆ũ1 = f˜1 (x) ≤ λũ2 , −∆ p ũ2 = f˜2 (x) ≤ µũ3 in Ω, p −1 −∆ũ3 = f˜3 (x) ≤ γρ(x)ũ1 , ũ1 = ũ2 = ũ3 = 0 on ∂Ω, (6.4) Marcelo Montenegro 445 where each f˜i is a nonnegative function belonging to Lβ (Ω), β > 1. Since ũi > 0 in Ω for some i = 1,2,3, by Proposition 3.2(ii), we get a contradiction. Thus, the claimed constants ε0 > 0 and M0 > 0 do exist. Choose an arbitrary number c > 0 and a triplet of nonnegative functions f = ( f1 , f2 , f3 ) satisfying (6.1) with ε0 provided above. We aﬃrm that there is a constant M = M(c) > 0 not depending on f such that u X < M for every u ∈ C with T f (t,u) = u for some t ∈ [0,1]. In fact, for each ε > 0 define the functions f1ε (x,t1 ,t2 ,t3 ) = ε f1 (x,t1 /ε,t2 /ε,t3 /ε p−1 ), f2ε (x,t1 ,t2 ,t3 ) = ε p−1 f2 (x,t1 /ε,t2 /ε,t3 /ε p−1 ), f3ε (x,t1 ,t2 ,t3 ) = ε p−1 f3 (x,t1 /ε,t2 /ε, t3 /ε p−1 ) and put fε = ( f1ε , f2ε , f3ε ). Clearly, fε fulfills (6.1) with cε = max{ε p−1 ,ε}c instead of c. Since the functions u1ε = εu1 , u2ε = εu2 , and u3ε = ε p−1 u3 verify T fε (t,uε ) = uε , taking ε small enough such that 0 < cε ≤ 1, from the first part, it follows that uε X < M0 . Therefore, we conclude that u X < M. Hence, by the homotopic invariance property of the degree in cones, we obtain deg(I − T f (1, ·),BM ∩ C,0) = deg(I − T f (0, ·),BM ∩ C,0) = 0, implying that T f (1,u) = u for some u ∈ (C 1 (Ω))3 ∩ C. The above lemma is not useful to seek nontrivial weak solutions when (u1 ,u2 ,u3 ) ≡ 0 in Ω solves problem (1.11). Further assumptions will lead us to find a positive solution of (1.11). We assume that f1 x,t1 ,t2 ,t3 ≥ λ0 t2 , f2 x,t1 ,t2 ,t3 ≥ µ0 t3 , (6.5) p −1 f3 x,t1 ,t2 ,t3 ≥ γ0 ρ0 (x)t1 , for every x ∈ Ω and 0 < t1 ,t2 ,t3 ≤ δ0 , where δ0 > 0 is a constant, ρ0 ∈ L∞ (Ω), ρ p −1 ρ0 ≥ 0, ρ0 ≡ 0, λ0 ,µ0 ,γ0 > 0, and λ0 µ0 γ0 = Λ10 . ρ Notice that γ0 can be equal to Λ1 . Thus in the framework of (1.1), condition ρ (6.5) reduces to f (x,t) ≥ Λ1 ρ(x)t p−1 for small t. We also suppose some monotonicity on the functions fi , namely f1 x,t1 ,t2 ,t3 ≤ f1 x,s1 ,s2 ,s3 (6.6) for x ∈ Ω, 0 ≤ t1 = s1 ≤ δ, 0 ≤ t2 ≤ s2 , and 0 ≤ t3 ≤ s3 , f2 x,t1 ,t2 ,t3 ≤ f2 x,s1 ,s2 ,s3 (6.7) for x ∈ Ω, 0 ≤ t1 ≤ s1 , 0 ≤ t2 = s2 ≤ δ, and 0 ≤ t3 ≤ s3 , f3 x,t1 ,t2 ,t3 ≤ f3 x,s1 ,s2 ,s3 (6.8) for x ∈ Ω, 0 ≤ t1 ≤ s1 , 0 ≤ t2 ≤ s2 , and 0 ≤ t3 = s3 ≤ δ, where δ > 0. We establish the following result by performing a truncation between a subsolution and supersolution. 446 Higher order quasilinear elliptic equations Theorem 6.2. System (1.11) admits a positive weak solution in (C 1 (Ω))3 if we assume (6.1) with ε0 > 0 given in Lemma 6.1 and conditions (6.5), (6.6), (6.7), and (6.8). Proof. We first show that problem (1.11) possesses a positive subsolution. In fact, denote by (v1 ,v2 ,v3 ) a positive eigenfunction corresponding to (λ0 ,µ0 ,γ0 ). Since (tv1 ,tv2 ,t p−1 v3 ) is also an eigenfunction, we can assume vi ≤ min{δ,δ0 } in Ω for i = 1,2,3. From (6.5), we conclude that (v1 ,v2 ,v3 ) is a positive subsolution of system (1.11). We prove now that the problem has a positive solution. Define for i = 1,2,3 the Carathéodory functions fi x,t1 ,t2 ,t3 fi x,t1 ,t2 ,v3 (x) fi x,t1 ,v2 (x),t3 fi x,v1 (x),t2 ,t3 Fi x,t1 ,t2 ,t3 = f x,t ,v (x),v3 (x) i 1 2 f x,v (x),t ,v (x) i 1 2 3 fi x,v1 (x),v2 (x),t3 , if t1 ≥ v1 (x), t2 ≥ v2 (x), t3 ≥ v3 (x), if t1 ≥ v1 (x), t2 ≥ v2 (x), t3 < v3 (x), if t1 ≥ v1 (x), t2 < v2 (x), t3 ≥ v3 (x), if t1 < v1 (x), t2 ≥ v2 (x), t3 ≥ v3 (x), if t1 ≥ v1 (x), t2 < v2 (x), t3 < v3 (x), if t1 < v1 (x), t2 ≥ v2 (x), t3 < v3 (x), if t1 < v1 (x), t2 < v2 (x), t3 ≥ v3 (x), fi x,v1 (x),v2 (x),v3 (x) if t1 < v1 (x), t2 < v2 (x), t3 < v3 (x). (6.9) Clearly, each Fi verifies condition (6.1) for some suﬃciently large c > 0. Lemma 6.1 implies that the system −∆u1 = F1 x,u1 ,u2 ,u3 , −∆ p u2 = F2 x,u1 ,u2 ,u3 in Ω, −∆u3 = F3 x,u1 ,u2 ,u3 , u1 = u2 = u3 = 0 (6.10) on ∂Ω admits a nonnegative solution (u1 ,u2 ,u3 ) ∈ (C 1 (Ω))3 . We claim that u1 ≥ v1 in Ω. Otherwise, Ω− = {x ∈ Ω : u1 (x) < v1 (x)} is a nonempty open subset of Ω. Given x ∈ Ω− , consider the diﬀerence d = f1 (x,v1 (x),v2 (x),v3 (x)) − F1 (x,u1 (x), u2 (x),u3 (x)). There are four cases to be analyzed: (i) if u2 (x) ≥ v2 (x) and u3 (x) ≥ v3 (x). Since F1 (x,u1 ,u2 ,u3 ) = f1 (x,v1 ,u2 , u3 ), from (6.6), we get d ≤ 0; (ii) if u2 (x) < v2 (x) and u3 (x) ≥ v3 (x), again since F1 (x,u1 ,u2 ,u3 ) = f1 (x,v1 , v2 ,u3 ), from (6.6), we obtain d ≤ 0; (iii) if u2 (x) ≥ v2 (x) and u3 (x) < v3 (x), a similar reasoning to the second case furnishes the conclusion; (iv) if u2 (x) < v2 (x) and u3 (x) < v3 (x), by definition of d, it follows that d = 0. Therefore, f1 (x,v1 (x),v2 (x),v3 (x)) − F1 (x,u1 (x),u2 (x),u3 (x)) ≤ 0 for every x ∈ Ω− , implying ∆(u1 − v1 ) ≤ 0 in Ω− . Since u1 = v1 on ∂Ω− , using the maximum principle, we conclude that u1 ≥ v1 in Ω− , a contradiction. So, u1 ≥ v1 Marcelo Montenegro 447 in Ω. By similar ideas and Lemma 3.1, we show that u2 ≥ v2 and u3 ≥ v3 in Ω. Consequently, by definition of Fi , i = 1,2,3, the triplet (u1 ,u2 ,u3 ) is a positive solution of system (1.11). Under somewhat diﬀerent conditions, it is possible to obtain a positive solution again. The following two requirements are modifications of (6.1). We rewrite assumption (6.1) below 1/(p−1) f1 x,t1 ,t2 ,t3 ≤ a t1 + t2 + t3 p −1 f2 x,t1 ,t2 ,t3 ≤ a t1 p −1 f3 x,t1 ,t2 ,t3 ≤ a t1 p −1 + t2 p −1 + t2 +1 , + t3 + 1 , + t3 + 1 (6.11) for some constant a > 0 and every x ∈ Ω and t1 ,t2 ,t3 ≥ 0. Taking ε0 = c = 0 in (6.1), we obtain f1 x,t1 ,t2 ,t3 ≤ λ0 t2 , f2 x,t1 ,t2 ,t3 ≤ µ0 t3 , (6.12) p −1 f3 x,t1 ,t2 ,t3 ≤ γ0 ρ0 (x)t1 , for every x ∈ Ω and 0 < t1 ,t2 ,t3 ≤ δ0 , where δ0 > 0 is a constant, ρ0 ∈ L∞ (Ω), ρ p −1 ρ0 ≡ 0, ρ0 ≥ 0, λ0 ,µ0 ,γ0 > 0, λ0 µ0 γ0 < Λ10 . The following condition is a kind of nonresonance: f1 x,t1 ,t2 ,t3 ≥ λt2 − c, f2 x,t1 ,t2 ,t3 ≥ µt3 − c, (6.13) p −1 f3 x,t1 ,t2 ,t3 ≥ γρ(x)t1 − c, for every x ∈ Ω and t1 ,t2 ,t3 ≥ 0, where c > 0 is a constant, ρ ∈ L∞ (Ω), ρ ≥ 0, ρ ρ ≡ 0, λ,µ,γ > 0, and λ p−1 µγ > Λ1 . It is easy to see that the following result applies to (1.1); observe the diﬀerence between (6.5) and conditions (6.12) and (6.13). Theorem 6.3. System (1.11) admits a nontrivial nonnegative weak solution in (C 1 (Ω))3 , provided that (6.11), (6.12), and (6.13) are verified. Proof. Let X and C be as in the proof of Lemma 6.1. Define the mapping H : [0,+∞) × C → C by H(t,u) = v, where v = (v1 ,v2 ,v3 ) verifies −∆v1 = f1 x,u1 ,u2 ,u3 + t, −∆ p v2 = f2 x,u1 ,u2 ,u3 + t in Ω, −∆v3 = f3 x,u1 ,u2 ,u3 + t, v1 = v2 = v3 = 0 on ∂Ω. (6.14) 448 Higher order quasilinear elliptic equations By the maximum principle and C 1 estimates, it follows that H is well defined, continuous, compact, and H(0,0) = 0. Let T : C → C be given by T(u) = H(0,u). At first, we get r > 0 such that u = tT(u), for every t ∈ [0,1] and u ∈ C with u X = r. In fact, take 0 < r ≤ δ0 and suppose u = tT(u) for some t ∈ [0,1] and u ∈ C with u X = r. Then, from (6.12), we get −∆u1 = t f1 x,u1 ,u2 ,u3 ≤ λ0 u2 , −∆ p u2 = t p−1 f2 x,u1 ,u2 ,u3 ≤ µ0 u3 in Ω, p −1 −∆u3 = t f3 x,u1 ,u2 ,u3 ≤ γ0 ρ0 (x)u1 . (6.15) In particular, one of the components of u is positive in Ω. By virtue of Proposition 3.2(ii), we get a contradiction. We prove that there exist R > r and t0 > 0 such that H(t,u) = u for every t ≥ t0 and u ∈ C with u X ≤ R and also for every t ≥ 0 when u X = R. Indeed, let t ≥ 0 and u ∈ C with H(t,u) = u. From (6.13), we have −∆u1 ≥ λu2 + t − c, −∆ p u2 ≥ µu3 + t − c in Ω, (6.16) p −1 −∆u3 ≥ γρ(x)u1 + t − c. By Proposition 3.2(iii), we see that t ≤ c. Hence, we can take R > 0 such that u X ≤ R for every u ∈ C verifying H(t,u) = u for some t ∈ [0,c]. Otherwise, there are sequences tk ∈ [0,c] and uk ∈ C satisfying H(tk ,uk ) = uk and uk X → ∞. Denoting ũik the normalized functions as in the proof of Lemma 6.1, we have 1/(p−1) ũ1k C(Ω) + ũ2k C(Ω) + ũ3k C(Ω) = 1. Using (6.11), (6.13), and C 1 estimates, up to a subsequence, we conclude that ũk converges to a function ũ in (C 1 (Ω))3 . 1/(p−1) In particular, ũ ≥ 0 in Ω, ũ1 C(Ω) + ũ2 C(Ω) + ũ3 C(Ω) = 1, and −∆ũ1 ≥ λũ2 , −∆ p ũ2 ≥ µũ3 in Ω, p −1 −∆ũ3 ≥ γρ(x)ũ1 ũ1 = ũ2 = ũ3 = 0 , (6.17) on ∂Ω. Thus, ũ > 0 in Ω, contradicting (iii) of Proposition 3.2. Hence, the conclusion follows from Krasnosel’skiı̆ theorem (Theorem 5.2). 7. Uniqueness of solutions In this section, we give conditions under which problem (1.11) admits a unique positive weak solution. Essentially, we assume a certain homogeneity and Marcelo Montenegro 449 monotonicity on functions fi f1 x,st1 ,st2 ,s p−1 t3 ≥ s f1 x,t1 ,t2 ,t3 , f2 x,st1 ,st2 ,s p−1 t3 ≥ s p−1 f2 x,t1 ,t2 ,t3 , f3 x,st1 ,st2 ,s p −1 t3 ≥ s p −1 f3 x,t1 ,t2 ,t3 , (7.1) (7.2) (7.3) for every x ∈ Ω, s ∈ [0,1], and t1 ,t2 ,t3 > 0, fi x,t1 ,t2 ,t3 ≤ fi x,s1 ,s2 ,s3 , (7.4) for every x ∈ Ω, 0 < t1 ≤ s1 , 0 < t2 ≤ s2 , 0 < t3 ≤ s3 , and i = 1,2,3, f1 x1 ,t1 ,t2 ,t3 < f1 x1 ,s1 ,s2 ,s3 , (7.5) for some x1 ∈ Ω and every 0 < t1 ≤ s1 , 0 < t2 < s2 , 0 < t3 ≤ s3 , f2 x,t1 ,t2 ,t3 < f2 x,s1 ,s2 ,s3 , (7.6) for every x in a neighborhood of ∂Ω, 0 < t1 ≤ s1 , 0 < t2 ≤ s2 , 0 < t3 < s3 , and f3 x3 ,t1 ,t2 ,t3 < f3 x3 ,s1 ,s2 ,s3 , (7.7) for some x3 ∈ Ω and every 0 < t1 < s1 , 0 < t2 ≤ s2 , 0 < t3 ≤ s3 . We adopt a variant of a comparison strategy due to Krasnosel’skiı̆ [5], see also [12]. Theorem 7.1. System (1.11) admits, at most, one positive weak solution in (C 1 (Ω))3 , provided (7.1), (7.2), (7.3), (7.4), (7.5), (7.6), and (7.7) are true and if just only one of inequalities (7.1) or (7.3) holds in the strict sense for some x0 ∈ Ω and every s ∈ (0,1). Proof. Let (u1 ,u2 ,u3 ) and (v1 ,v2 ,v3 ) be two positive weak solutions of (1.11) belonging to (C 1 (Ω))3 . Define the set S = {s > 0 : v1 > su1 , v2 > su2 and v3 > s p−1 u3 in Ω}. Take s∗ = supS. Changing (u1 ,u2 ,u3 ) and (v1 ,v2 ,v3 ), if necessary, we may assume that s∗ ∈ (0,1]. We show first that v2 (y) = s∗ u2 (y) for some y ∈ Ω. Indeed, suppose on the contrary that v2 > s∗ u2 in Ω. Since p −1 −∆ p v2 + ∆ p s∗ u2 = f2 x,v1 ,v2 ,v3 − s∗ f2 x,u1 ,u2 ,u3 p −1 p −1 f2 x,u1 ,u2 ,u3 ≥ f2 x,s∗ u1 ,s∗ u2 ,s∗ u3 − s∗ ≥ 0, (7.8) 450 Higher order quasilinear elliptic equations then, by Lemma 3.1, v2 > (s∗ + ε)u2 in Ω for ε > 0 small enough. Thus, −∆ v1 − s∗ u1 = f1 x,v1 ,v2 ,v3 − s∗ f1 x,u1 ,u2 ,u3 p −1 ≥ f1 x,s∗ u1 ,s∗ u2 ,s∗ u3 − s∗ f1 x,u1 ,u2 ,u3 ≥ 0, −∆ v1 − s u1 (x1 ) = f1 x1 ,v1 ,v2 ,v3 − s∗ f1 x1 ,u1 ,u2 ,u3 > f1 x1 ,s∗ u1 ,s∗ u2 ,s∗ p−1 u3 − s∗ f1 x1 ,u1 ,u2 ,u3 ∗ (7.9) ≥ 0 imply that v1 > (s∗ + ε)u1 in Ω for ε > 0 small enough. The inequalities p −1 p −1 −∆ v3 − s∗ u3 = f3 x,v1 ,v2 ,v3 − s∗ f3 x,u1 ,u2 ,u3 p −1 p −1 f3 x,u1 ,u2 ,u3 ≥ f3 x,s∗ u1 ,s∗ u2 ,s∗ u3 − s∗ ∗ p −1 −∆ v3 − s u3 x 3 ≥ 0, p −1 f3 x3 ,u1 ,u2 ,u3 = f3 x3 ,v1 ,v2 ,v3 − s∗ > f3 x3 ,s∗ u1 ,s∗ u2 ,s∗ p−1 u3 − s∗ p−1 f3 x3 ,u1 ,u2 ,u3 ≥ 0 (7.10) furnish v3 > (s∗ + ε) p−1 u3 in Ω for ε > 0 small enough, contradicting the definition of s∗ . Hence, there is y ∈ Ω satisfying v2 (y) = s∗ u2 (y). Choose the set Γ2 associated to the function v2 as in Lemma 3.1. We show next that there exists z ∈ Γ2 such that v2 (z) = s∗ u2 (z). Take a subdomain Ω0 of Ω with smooth boundary ∂Ω0 which verifies Ω0 ⊂ Ω, ∂Ω0 ⊂ Γ2 , and y ∈ Ω0 . We claim that there is z ∈ ∂Ω0 with v2 (z) = s∗ u2 (z). Indeed, if v2 > s∗ u2 on ∂Ω0 , by continuity, we get v2 ≥ s∗ u2 + η on ∂Ω0 for some η > 0. Since −∆ p v2 ≥ −∆ p s∗ u2 = −∆ p (s∗ u2 + η) in Ω0 , then by Lemma 3.1, v2 ≥ s∗ u2 + η in Ω0 . But y ∈ Ω0 , so we arrive at a contradiction. Therefore, the claimed point z ∈ Γ2 exists. Noting that −∆ p v2 ≥ −∆ p s∗ u2 in Γ2 , v2 ≥ s∗ u2 in Γ2 and v2 (z) = s∗ u2 (z). It follows that v2 ≡ s∗ u2 in Γ2 , again by Lemma 3.1. We aﬃrm that v1 ≡ s∗ u1 and v3 ≡ s∗ p−1 u3 in Ω. In fact, suppose v1 ≡ s∗ u1 in Ω. Using (7.1), (7.4), and (7.5) and the strong maximum principle as above, we conclude that v1 > s∗ u1 in Ω. Applying (7.1), (7.4), (7.7), and the strong maximum principle once more, it follows easily that v3 > s∗ p−1 u3 in Ω. Finally, from condition (7.6), there is a point x2 ∈ Γ2 such that p −1 −∆ p v2 x2 + ∆ p s∗ u2 x2 = f2 x2 ,v1 ,v2 ,v3 − s∗ f2 x2 ,u1 ,u2 ,u3 > f2 x2 ,s∗ u1 ,s∗ u2 ,s∗ p−1 u3 − s∗ p−1 f2 x2 ,u1 ,u2 ,u3 ≥ 0, (7.11) Marcelo Montenegro 451 contradicting v2 ≡ s∗ u2 in Γ. Similarly, we see that v3 ≡ s∗ p−1 u3 in Ω. We prove that s∗ = 1. Indeed, assume that s∗ ∈ (0,1). If (7.3) holds strictly for some x0 ∈ Ω, that is, f3 x0 ,st1 ,st2 ,s p−1 t3 > s p−1 f3 x0 ,t1 ,t2 ,t3 , (7.12) for every s ∈ (0,1) and t1 ,t2 ,t3 > 0, then p−1 p −1 −∆ v3 − s∗ u3 x0 = f3 x0 ,v1 ,v2 ,v3 − s∗ f3 x0 ,u1 ,u2 ,u3 p −1 p −1 f3 x0 ,u1 ,u2 ,u3 = f3 x0 ,s∗ u1 ,s∗ u2 ,s∗ u3 − s∗ > 0, (7.13) a contradiction. If (7.1) holds strictly in some point of Ω, we proceed analogously. Therefore, we have s∗ = 1. Define the set S̃ = {s > 0 : u1 > sv1 , u2 > sv2 and u3 > s p−1 v3 in Ω} and let s̃ = sup S̃. Since s∗ s̃ ≤ 1, then s̃ ≤ 1. 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