# BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE ... - CAPDE BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE ... - CAPDE

2 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2(see Remark 2.4 for further discussion), we study systems of the form⎧⎪⎨∆u = g(u − v) in Ω(2)∆v = f(v − βu) in Ω⎪⎩u = v = ∞ on ∂Ωwhere Ω is a smoothly bounded domain of Euclidean space, f, g are nondecreasing,nonnegative C 1 functions such that f = g = 0 on R − and β > 0 is a parameter.Solutions are sought in the class C 2 (Ω) and the boundary condition is to beunderstood aslim u(x) = lim v(x) = +∞x→x 0 x→x 0for all x 0 ∈ ∂Ω.We study existence, first order asymptotics and uniqueness of solutions respectivelyin Sections 2, 3, 4. In Section 5, we study in more detail a list of relevant examples.Here is a summary of our main results.Theorem 1.1. Let Ω denote a smoothly bounded domain of Euclidean space, f, g :R → R nondecreasing, nonnegative C 1 functions such that f = g = 0 on R − andβ > 0. There exists a solution of the system (2) if and only if the following threeconditions hold• f satisfies the Keller-Osserman condition (1),• g satisfies the Keller-Osserman condition (1),• β < 1.The asymptotics of solutions is obtained at the price of a technical assumptionon the nonlinearities commonly found in the literature (see e.g. ). More precisely,let∫ ∞dtφ(u) = √ ,2F (t)whereF (t) =u∫ tWe assume in what follows that f satisfies(3)lim inft→∞φ(at)φ(t)0f(s) ds.> 1 ∀a ∈ (0, 1).Examples are given by f(u) = e u or f(u) = u p , p > 1. A counter-example is givenby f(u) = u (ln(1 + u)) 2p , p > 1.For 0 < β < 1 and β < θ ≤ 1, we let w θ > 0 denote the minimal solution to⎧ ( )⎨θ − β∆w θ = f w θ in Ω(4)θ⎩w θ = +∞on ∂Ω.(see e.g.  for the notion of minimal blow-up solution). Then, we have the followingresult.Theorem 1.2. Make the same assumptions as in Theorem 1.1. Assume in additionthat f satisfies (3).a) If f is smaller than g at infinity in the sense that for any ε > 0,(5)f(t)limt→+∞ g(ɛt) = 0

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 3then for any solution (u, v) of the system (2)(6)limx→∂Ωuw 1= 1,limx→∂Ωvw 1= 1,where w 1 is the minimal nonnegative solution to (4) with θ = 1.b) If f is of the order of g at infinity in the sense that for some θ 0 ∈ (β, 1),(7)if θ ∈ (β, θ 0 ), thenif θ ∈ (θ 0 , 1), theng((1 − θ)t)lim inf θt→+∞ f((θ − β)t) ≥ 1g((1 − θ)t)lim sup θt→+∞ f((θ − β)t) ≤ 1then for any solution (u, v) of the system (2)(8)limx→∂Ωuw θ0= 1 θ 0, limx→∂Ωvw θ0= 1,where w θ0 is the minimal solution to (4) with θ = θ 0 .Remark 1.3. Condition (7) looks rather unpleasant. Nevertheless, its validity canbe easily checked on examples. If e.g. f(t) = g(t) = t p , by the Intermediate Valueg((1−θTheorem there exists θ 0 ∈ (β, 1) such that lim t→+∞ θ 0)t)0 f((θ = θ 0−β)t) 0 (1−θ0)p(θ 0−β)= 1.pSince the quantity θ g((1−θ)t)f((θ−β)t) is nonincreasing in θ, (7) follows. If f(t) = g(t) = et ,then letting θ 0 = (1 + β)/2, we have lim t→+∞ θ g((1−θ)t)f((θ−β)t) = +∞ for θ < θ 0, whilethe limit is equal to 0 if θ > θ 0 . Observe that in this case, though (7) holds, thereis no value of θ for which the limit is equal to 1.Remark 1.4. Note that when condition (5) holds, the first order asymptotics ofboth u and v is independent of the nonlinearity g. The influence of g can be detectedin the next terms of the asymptotic expansion of the solution. See Example 1 inSection 5.On the contrary, when condition (7) holds, f and g already interplay in the valueof the constant θ 0 of the leading asymptotics of u and v. See Example 2 in Section5.As a consequence of Theorem 1.2, we obtain under an extra convexity assumption:Corollary 1.5. Assume that f satifies (3) and either condition (5) or (7) holds.a) If f and g are convex then the system (2) has a unique solution.b) Assume Ω is a ball, f, g nondecreasing everywhere and f(t)t, g(t)tare nondecreasingin a neighborhood of +∞. Then the system (2) has a unique solution.Notation. For functions m, n : [0, ∞) → [0, ∞) we say that m ∼ nat infinity ifm(t)limt→∞ n(t) = 1and use similar notation when m, n are defined near t 0 and lim t→t0m(t)n(t) = 1.

4 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ22. ExistenceThe proof of the existence of solutions in Theorem 1.1 follows a standard schemewhere one first solves the system with a finite boundary condition m and thenlets m → +∞. The former step can be carried out in a more general setting asdescribed next. Consider the system⎧⎪⎨∆u = g(u, v) in Ω(9)∆v = f(v, u) in Ω⎪⎩u = v = ∞ on ∂Ω,where f and g are two nonnegative C 1 functions such that f(0, 0) = g(0, 0) = 0 and∂g/∂v ≤ 0, ∂f/∂u ≤ 0 (the system is then called cooperative).Proposition 2.1. Given m > 0 the system⎧⎪⎨∆u = g(u, v)(10)∆v = f(u, v)⎪⎩u = v = min Ωin Ωon ∂Ωadmits a unique minimal nonnegative solution (u, v).In the previous statement minimality refers to the following property: take anyopen set ω ⊆ Ω and ū, ¯v ∈ C 2 (ω) satisfying⎧∆ū ≤ g(ū, ¯v) in ω⎪⎨ ∆¯v ≤ f(ū, ¯v) in ω(11)ū ≥ 0, ¯v ≥ 0 in ω⎪⎩ū ≥ u, ¯v ≥ v on ∂ω.Then,u ≤ ū, v ≤ ¯v in ω.Proof.We use the method of sub and supersolutions. Choose a > 0, b > 0 sufficientlylarge such that the functions(12)(13)Defineand for k ≥ 1(14)We claim thatandu ↦→ g(u, v) − au, v ↦→ f(u, v) − bv are decreasing for 0 ≤ u, v ≤ m.u 0 ≡ 0, v 0 ≡ 0⎧⎪⎨∆u k − au k = g(u k−1 , v k−1 ) − au k−1∆v k − bv k = f(u k−1 , v k−1 ) − bv k−1⎪⎩u k = v k = m0 ≤ u k−1 ≤ u k ≤ m in Ω0 ≤ v k−1 ≤ v k ≤ m in Ωin Ωin Ωon ∂Ω

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 5Indeed, the property is straightforward if k = 1. Take k ≥ 2 and assume byinduction that u k−2 ≤ u k−1 , v k−2 ≤ v k−1 in Ω. Then,∆(u k − u k−1 ) − a(u k − u k−1 ) = g(u k−1 , v k−1 ) − g(u k−2 , v k−2 ) − a(u k−1 − u k−2 )≤ g(u k−1 , v k−2 ) − g(u k−2 , v k−2 ) − a(u k−1 − u k−2 )≤ 0in Ωand hence u k − u k−1 ≥ 0 in Ω. The remaining inequalities are obtained similarly.In particular, the limitsu = limk→∞ u k,v = limk→∞ v kexist, solve (10) and satisfy0 ≤ u ≤ m, 0 ≤ v ≤ m in Ω.Let us show now that the solution constructed in this way does not depend on a, bas long as these parameters satisfy (12). For this we argue as follows: suppose thatu, v are constructed using a, b and ũ, ṽ are constructed with ã, ˜b satisfying (12).Let u k , v k denote the sequences defined by (13), (14). Arguing by induction we seethat if ũ ≥ u k−1 , ṽ ≥ v k−1 then∆(ũ − u k ) − a(ũ − u k ) = g(ũ, ṽ) − g(u k−1 , v k−1 ) − a(ũ − u k−1 )≤ g(ũ, v k−1 ) − g(u k−1 , v k−1 ) − a(ũ − u k−1 )≤ 0and then ũ − u k ≥ 0 in Ω. Note that u ↦→ g(u, v) − au and v ↦→ f(u, v) − bv aremonotone in the appropriate regions because u, v and ũ, ṽ are between 0 and m.Similarly, ṽ − v k ≥ 0 in Ω and thus ũ ≥ u, ṽ ≥ v in Ω. By symmetry we obtain theconverse inequality and we deduce that ũ = u, ṽ = v.Minimality. Let ω ⊂ Ω be open and ū, ¯v ∈ C(ω) satisfy (11). Choose a, b largeenough so that g(u, v) − au is decreasing in u and f(u, v) − bv is decreasing in v forall u, v in the range 0 ≤ u, v ≤ M with M ≥ m, M ≥ max ω ū and M ≥ max ω ¯v.Consider u k , v k defined by (13), (14). Now we show that ū ≥ u k , ¯v ≥ v k in ωfor all k. By induction, if ū ≥ u k−1 , ¯v ≥ v k−1 in ω then∆(ū − u k ) − a(ū − u k ) ≤ g(ū, ¯v) − g(u k−1 , v k−1 ) − a(ū − u k−1 )≤ g(ū, v k−1 ) − g(u k−1 , v k−1 ) − a(ū − u k−1 )≤ 0in ωand hence ū − u k ≥ 0 in ω.□Proof of Theorem 1.1Necessary conditions. Suppose that (u, v) is a solution to (2) and for given γ > 0set w = min(γu, v). Let χ A denote the characteristic function of a set A. By Kato’s

6 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2inequality,∆w ≤ γ∆uχ [γuv]= γg(u − v)χ [γuv]((≤ γg((1 − γ)u)χ [γuv]γ( )((1 − γ= γg w χ [γuv]γγ( ( ) ((1 − γ≤ max γg w , f 1 − β ) ))w =: h 1 (w).γγHence w is a supersolution to the single equation ∆u = h 1 (u) in Ω with u = +∞on ∂Ω. Therefore this problem admits a solution and hence h 1 must satisfy theKeller-Osserman condition (1) (see e.g. ). Choosing γ = 1 implies that f satisfies(1) and β < 1. Then, choosing γ = β < 1 implies that g satisfies (1).Sufficient conditions. Consider the minimal solution (u m , v m ) to the truncatedproblem⎧⎪⎨∆u = g(u − v) in Ω(15)∆v = f(u − βv) in Ω⎪⎩u = v = m on ∂Ωwhere m > 0. Such a solution can easily be constructed by the method of sub andsupersolutions, see Proposition 2.1. Let γ ∈ (β, 1) and setThen,w m = max(γu m , v m ).∆w m ≥ γ∆u m χ [γum>v m] + ∆v m χ [γumv m] + f(v m − βu m )χ [γumv m] + f 1 − β ) )w m χ [γum 0. Hence (u m ), (v m ) remain bounded oncompact sets of Ω as m → ∞, and by standard elliptic estimates they converge- upto a subsequence - in Cloc 2 (Ω) to a solution of (2).□Remark 2.2. The proof of Theorem 1.1 implies that whenever solutions exist, oneof them is minimal in the class of nonnegative solutions. Moreover this solution(u, v) satisfies(16)βu ≤ v ≤ u in Ω.Indeed let us show that the minimal nonnegative solution (u m , v m ) to (15) satisfiesv m ≤ u m in Ω. For this let us recall that u m = lim k→∞ u m,k , v m = lim k→∞ v m,kwhere u m,k , v m,k are defined recursively by (14) starting with the trivial solutions,

8 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2Lemma 3.2. Suppose f ∼ g at infinity and that f satisfies (3). Let u be anysolution to (19) and v any solution to (19) with nonlinearity replaced by g. Thenulimx→∂Ω v = 1.Proof. Let G(t) = ∫ t0 g(s) ds, φ g(u) = ∫ ∞√dt, ψug = φ −12G(t)g . Let φ f and ψ f denotethe corresponding functions associated to f. By Lemma 3.1 it suffices to prove thatψ f (δ)limδ→0 ψ g (δ) = 1.Since f ∼ g at infinity we also have F ∼ G at infinity and therefore φ f ∼ φ g atinfinity. It follows from this and the fact that φ f satisfies the condition (3) that φ gsatisfies this condition too.Let m > 1. Condition (3) on φ g implies that there exists η > 1 and δ 0 > 0 suchthat(20)Since φ f ∼ φ g at infinity we haveHence taking δ 1 > 0 small,ψ g (δ) ≤ m ψ g (ηδ) ∀ 0 < δ < δ 0 .limδ→0Since ψ g is non-increasing we deduceand by (20)It follows thatδφ g (ψ f (δ)) = 1.δ ≤ ηφ g (ψ f (δ)) < δ 0 ∀ 0 < δ < δ 1 .ψ g (δ) ≥ ψ g (ηφ g (ψ f (δ))) ∀ 0 < δ < δ 1ψ g (δ) ≥ 1 m ψ f (δ) ∀ 0 < δ < δ 1 .lim supδ→0and since m > 1 was arbitrary, thatlim supδ→0ψ f (δ)ψ g (δ) ≤ mψ f (δ)ψ g (δ) ≤ 1.The corresponding inequality for the liminf is prove by reversing the roles of ψ fand ψ g .□The next lemma asserts that under the condition (3) the boundary behaviorof solutions to (19) depends continuously on multiplicative perturbations of thenonlinearity.Lemma 3.3. Assume f satisfies (3). Given γ > 0 let u γ denote any solution ofThen∆u γ = f(γu γ ) in Ω, u γ = +∞ on ∂Ω.u γlim sup lim sup ≤ 1 ≤ lim infγ→1 x→∂Ω u 1 γ→1lim infx→∂Ωu γu 1

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 9Proof. Given γ > 0 let f γ (u) = f(γu), F γ (t) = ∫ t0 f γ(s) ds = 1 γ F (γt), φ γ(u) =√dt= √ 12Fγ(t) γφ(γu) and ψ γ = (φ γ ) −1 . Note that ψ γ (δ) = 1 γ ψ(√ γδ).∫ ∞uBy Lemma 3.1 it is enough to establishψ γ (δ)lim sup lim supγ→1 δ→0 ψ(δ)≤ 1 ≤ lim infγ→1lim infδ→0ψ γ (δ)ψ(δ) .Let m > 1 and δ 0 > 0, η > 1 be such that (20) holds. Then if √ γ ≤ η it followsthat1γ ψ(δ) ≤ m γ ψ(ηδ) ≤ m γ ψ(√ γδ) = mψ γ (δ) ∀0 < δ < δ 0and thereforelim infδ→0As m > 1 is arbitrary we deduceψ γ (δ)ψ(δ) ≥ 1γm ∀0 < γ < η2 .lim infγ→1lim infδ→0ψ γ (δ)ψ(δ) ≥ 1.Similarly, let m > 1 and δ 0 > 0, η > 1 such that (20) holds. If √ γ ≥ 1 ηfor δ > 0 small and thereforeHenceψ γ (δ) = 1 γ ψ(√ γδ) ≤ 1 γ ψ(δ/η) ≤ m γ ψ(δ)lim supδ→0ψ γ (δ)ψ(δ) ≤ m γ∀γ ≥ 1 η 2 .ψ γ (δ)lim sup lim supγ→1 δ→0 ψ(δ) ≤ 1.we haveProof of Theorem 1.2, part a). Let (u, v) be any solution to (2) and w 1 be theminimal nonnegative solution to (4) with θ = 1. For simplicity we write w = w 1 .First we note that we have(21)w ≤ v ≤ u.Indeed for the minimal solution (u, v), we always have u ≥ v by (16). Consequently,∆v = f(v − βu) ≤ f((1 − β)v)so v is a supersolution of (4) and since w is the minimal nonnegative solution itfollows that w ≤ v. Letz θ = max(θu, v),whereBy Kato’s inequality we havewith h θ given by(22)β < θ < 1.∆z θ ≥ h θ (z)( ( ) ( ))1 − θ θ − βh θ (w) = min θg w , f w .θθ□

10 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2Let w θ be the minimal solution to (4) and ˜w θ be the maximal solution to∆ ˜w θ = h θ ( ˜w θ ) in Ω, ˜w θ = +∞ on ∂Ω.Then z θ ≤ ˜w θ in Ω. Note that under condition (5), we have h θ (w) = ffor large w. It follows from Lemma 3.2 that˜w θlim = 1x→∂Ω w θand thereforez θw θ≤ 1.lim supx→∂Ωfor any β < θ < 1. It follows from the previous inequality thatz θwlim sup ≤ lim sup lim supx→∂Ω x→∂Ω w θ x→∂ΩLetting now θ → 1 and using Lemma 3.3 we deduce thatzlim sup lim supθ→1 x→∂ΩThis together with (21) yields the conclusion.w θwz θw ≤ 1.≤ lim supx→∂Ωw θw .Proof of Theorem 1.2, part b). We use Kato’s inequality withz = max(θu, v),(θ−βθwhereβ < θ < θ 0 .We have∆z θ ≥ h θ (z)( )with h θ given by (22). By assumption (7), given ε > 0, h θ (t) ≥ (1 − ε)f θ−βθ t fort large. In particular, there exists a neighborhood V of ∂Ω, V ⊂ Ω such that( ) θ − β∆z θ ≥ (1 − ε)f z θ .θLet w ε,θ denote the maximal solution of⎧( )⎨θ − β∆w ε,θ = (1 − ε)f w ε,θ(23)θ⎩w = +∞Then, z ≤ w ε,θ in V . By Lemma 3.3(24) lim sup lim supε→0,θ→θ 0 x→∂Ωw ε,θw ≤ 1,where w is the minimal solution of (4) with θ = θ 0 . Thus,z θ(25) lim sup lim supθ→θ 0 x→∂Ω w ≤ 1,in Von ∂VStarting with θ > θ 0 , considering z θ = min(θu, v), one proves similarly thatw ε,θ(26) lim inf lim infε→0,θ→θ 0 x→∂Ω w ≥ 1,where now w ε,θ is the minimal solution of (23), whencez θ(27) lim inf lim infθ→θ 0 x→∂Ω w ≥ 1.)w□

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 11Collecting (25) and (27), the theorem is proved.□4. UniquenessIn this section, we prove Corollary 1.5, which states the uniqueness of the solutionsof (2) provided that f, g are nondecreasing, nonnegative C 1 functions suchthat f = g = 0 on R − that satisfy (3), that either condition (5) or (7) holds, andthata) either f, g are convex functions;b) or Ω is a ball and f(t)t, g(t)tnon decreasing in a neighborhood of +∞, andf, g nondecreasing everywhere.We begin with the proof of the uniqueness result assuming that f, g are convexfunctions.Proof of Corollary 1.5, part a).Let ε > 0. Consider (u, v) the minimal BBUS solution and (u 1 , v 1 ) another solutionto (2). Actually, by (6) or (8) we have that (1 + ε)u > u 1 ≥ u, (1 + ε)v > v 1 ≥ vin a neighborhood of ∂Ω.Therefore, since by convexity f(t)tis a supersolution of (2):⎧⎨⎩, g(t)tare increasing functions ((1+ε)u, (1+ε)v)∆(1 + ε)u ≤ g((1 + ε)u − (1 + ε)v) in Ω∆(1 + ε)v ≤ f((1 + ε)v − β(1 + ε)u) in Ω(1 + ε)u = (1 + ε)v = ∞ on ∂ΩTherefore, w := u 1 − (1 + ε)u, z := v 1 − (1 + ε)v satisfy w ≤ 0, z ≤ 0 in aneighborhood of ∂Ω, and∆w − g ′ (ξ 1 )w + g ′ (ξ 1 )z ≥ 0 in Ω,∆z − f ′ (ξ 2 )z + βf ′ (ξ 2 )w ≥ 0 in Ω,for some ξ 1 ≥ 0, ξ 2 ≥ 0. Since g ′ (ξ 1 ) ≥ 0, f ′ (ξ 2 ) ≥ 0 and (−1 + β)f ′ (ξ 2 ) ≤ 0we can apply the maximum principle for cooperative systems (see for example theappendix of this paper or , Theorem 3.15 and its following Remark) to concludethat w ≤ 0 and z ≤ 0 in Ω. Letting ε → 0, we obtain the desired inequality.□We now prove the uniqueness result relaxing the hypotheses on f, g if Ω is a ball.This result compares with the uniqueness result in . We begin with a lemmaconcerning the boundary behavior of the minimal solution (u, v) of our problem,and that is interesting by itself, since it is true for general domains and gives someinsight of the boundary behavior of solutions.Lemma 4.1. Let u, v denote the minimal boundary blow-up solution of (2). Then,(28) lim (u(x) − v(x)) = +∞,x→∂Ω(29) lim (v(x) − βu(x)) = +∞.x→∂Ω

12 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2Proof. We establish (28), the other limit being similar. Fix A > 0. Consider theproblem(30)(31)(32)∆u A,m = g(u A,m − v A,m ),∆v A,m = f(v A,m − βu A,m ),u A,m = m + A, v A,m = mon ∂Ω.For a given A > 0, if m is large enough then (m + A, m) is a supersolution tothis problem and by the classical iterative method described in Proposition 2.1, wecan construct a solution to this approximated problem. As we did in the proof ofTheorem 1.1, (u A,m , v A,m ) converges to the minimal BBUS solution when m →+∞.In addition, we have that(33)(34)∆(u A,m − v A,m ) ≤ g(u A,m − v A,m ),u A,m − v A,m = Aon ∂Ω.Therefore u A,m − v A,m is a supersolution to a single equation problem. Set w A forthe solution to(35)(36)Therefore, everywhere in Ω∆w A = g(w A ) in Ω,w A = Aon ∂Ω.(37) w A (x) ≤ u A,m (x) − v A,m (x).We now let m → +∞, then A → +∞ that leads to(38) w(x) ≤ u(x) − v(x),where w is the minimal BBUS solution to (35).□We now complete the proof of the uniqueness result.Proof of Corollary 1.5, part b).To fix ideas, assume that Ω is the unit ball. Consider (u, v) the minimal solutionto (2) and (u 1 , v 1 ) the maximal solution (obtained as the limit (u 1 , v 1 ) =lim δ→0 (u δ , v δ ) where (u δ , v δ ) is the minimal solution in the ball with radius 1 − δ).Then it suffices to show that u ≡ u 1 and v ≡ v 1 . It is worth to observe that u, v,u 1 , v 1 are radial functions.Consider r ∈ (0, 1). Let Ω r be the ball of radius r. Then for each x ∈ Ω r thereexist ξ, ξ ′ ∈ R such that(39) ∆(u 1 − u) = g(u 1 − v 1 ) − g(u − v) = g ′ (ξ)(u 1 − u) − g ′ (ξ)(v 1 − v),(40) ∆(v 1 − v) = f(v 1 − βu 1 ) − f(v − βu) = f ′ (ξ ′ )(v 1 − v) − βf ′ (ξ ′ )(u 1 − u).By the maximum principle for cooperative systems, we havesup(u 1 − u) ≤ max(u 1 (r) − u(r), v 1 (r) − v(r))Ω rsup(v 1 − v) ≤ max(u 1 (r) − u(r), v 1 (r) − v(r)).Ω r

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 13This ensures that the functionis nondecreasing in (0, 1).Assume that(41)r ↦→ M(r) := max(u 1 (r) − u(r), v 1 (r) − v(r))u(0) < u 1 (0) or v(0) < v 1 (0)for if u(0) = u 1 (0) and v 1 (0) = v(0) by uniqueness for the system of ODEs wewould have u ≡ u 1 and v ≡ v 1 in (0, 1) and the proof is over.By Lemma 4.1 there is R 0 such that min (u − v, v − βu) (t) ≥ a for all t ≥ R 0 ,where a is such that both f(t)t, g(t)tnon decreasing for t ≥ a.We argue in a slightly different way in the following cases:(42)(43)(44)there exists r ∈ (R 0 , 1) such that u 1 (r) − u(r) > v 1 (r) − v(r)there exists r ∈ (R 0 , 1) such that u 1 (r) − u(r) < v 1 (r) − v(r)u 1 (r) − u(r) = v 1 (r) − v(r) for all r ∈ (R 0 , 1)To begin with we assume that (42) holds. In this case choose R 1 ∈ (R 0 , 1) suchthat(45)Defineand take ε > 0 small enough such thatand by (41)u 1 (R 1 ) − u(R 1 ) > v 1 (R 1 ) − v(R 1 ).w := u 1 − (1 + ε)u, z := v 1 − (1 + ε)vw(R 1 ) > z(R 1 )w(R 1 ) > 0 or z(R 1 ) > 0.Thus in particular w(R 1 ) > 0. We choose r ε > R 1 close to 1, such that w(r ε ) < 0and z(r ε ) < 0. This is possible by (6) or (8).In the annulus {r : R 1 < r < r ε } we then have{∆ ((1 + ε)u − u1 ) ≤ g((1 + ε)u − (1 + ε)v) − g(u 1 − v 1 )∆ ((1 + ε)v − v 1 ) ≤ f((1 + ε)v − β(1 + ε)u) − f(v 1 − βu 1 ).Therefore, (w, z) satisfy in the annulus∆w − g ′ (ξ 1 )w + g ′ (ξ 1 )z ≥ 0 ,∆z − f ′ (ξ 2 )z + βf ′ (ξ 2 )w ≥ 0 .By the maximum principle in the annulus R 1 ≤ r ≤ r ε we haveand hencemax(w, z) ≤ max(w(R 1 ), z(R 1 )) = w(R 1 ).u 1 (r) − (1 + ε)u(r) ≤ u 1 (R 1 ) − (1 + ε)u(R 1 ).Note that as ε → 0, r ε can be taken to approach 1. We then let ε → 0 to obtainu 1 (r) − u(r) ≤ u 1 (R 1 ) − u(R 1 ) for R 1 ≤ r < 1. By continuity M(r) = u 1 (r) − u(r)in some interval around R 1 . Since r ↦→ M(r) is nondecreasing we deduce thatM(r) = λ = const in some interval of the form [R 1 , R 1 + σ] with σ > 0. Fromequation (39) we also have g(u 1 −v 1 ) = g(u−v) in that interval. But g(t) is strictlyincreasing for t ≥ a and u(r) − v(r) ≥ a, u 1 (r) − v 1 (r) ≥ a for r ≥ R 0 if we choose

14 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2R 0 close enough to 1. Hence v 1 − v = λ = const in [R 1 , R 1 + σ]. This contradicts(45).A similar argument rules out the case (43) and therefore we are in the situation(44). The previous argument with R 1 replaced by R 0 then yieldswhich impliesmax(w, z) ≤ max(w(R 0 ), z(R 0 )) R 0 ≤ r ≤ r ε ,u 1 (r) − (1 + ε)u(r) ≤ max(u 1 (R 0 ) − (1 + ε)u(R 0 ), v 1 (R 0 ) − (1 + ε)v(R 0 ))for R 0 ≤ r ≤ r ε . Letting ε → 0 we haveM(r) = u 1 (r) − u(r) ≤ M(R 0 ) R 0 ≤ r < 1,and since M is nondecreasing we conclude that M is constant in [R 0 , 1). Thusu 1 − u = λ and v 1 − v = µ are constant in [R 0 , 1). Going back to the system wethen have in the annulus R 0 < r < 10 = ∆(u 1 − u) = g(u 1 − v 1 ) − g(u − v),0 = ∆(v 1 − v) = f(v 1 − βu 1 ) − f(v − βu).Since f, g are strictly increasing functions in the appropriate range we then haveu 1 − v 1 = u − v and v 1 − βu 1 = v − βu in R 0 < r < 1. Hence λ = µ = 0 andtherefore u 1 − u = 0 and v 1 − v = 0 in [0, 1). This completes the proof. □5. ExamplesExample 1. The first example falls in case a) of Theorem 1.2.⎧⎪⎨∆u = e u−v − 1 in Ω(46)∆v = (v − βu) p in Ω⎪⎩u = v = ∞ on ∂Ω,where p > 1, 0 < β < 1.By Theorem 1.1, 1.2 and Corollary 1.5, Problem (46) has a unique solution (u, v),which we know the leading order asymptotics of. We investigate here how the firstequation affects the next terms in the asymptotic expansions of u, v. We do thisfor simplicity in the case p > 3.Proposition 5.1. Assume 0 < β < 1 and p > 3. Let d denote the function distanceto the boundary. Then the unique solution (u, v) of (46) has the behavioru = cd −α + e 1 log d + f 1 + O(d ε )v = cd −α + e 2 log d + f 2 + O(d ε )where ε > 0 is suitably small and the constants are uniquely determined by theequations(47)α = 2p − 1(48)(49)(1 − β) p c p−1 = α(α + 1)e 1 − e 2 = −α − 2 e 2 − βe 1 = 0(50)e f1−f2 = cα(α + 1) cα(α + 1)p f 2 − βf 1(1 − β)c = −e 2.

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 15Proof. The argument relies on constructing a sub and a supersolution having asuitable boundary behavior. The reader is invited to check that the sub and thesupersolution that we construct depend continuously on Ω : this means that if acomparison principle is available (for systems with standard boundary conditions),then the solution (u, v) of (46) can be compared e.g. on an increasing sequenceof domains Ω n ⊂⊂ Ω with the supersolution u n , v n blowing-up on ∂Ω n . Lettingn → ∞ and checking that u n (x), v n (x) converge pointwise to the supersolutionu, v blowing-up on ∂Ω, we obtain the desired inequality : u ≤ u, v ≤ v. Asimilar approximation by outer domains enables us to compare (u, v) with a givensubsolution.We finally note that the standard comparison principle for systems can be used,since for g(u, v) = e u−v , f(u, v) = (v − βu) p we have ∂g∂g≥ 0, ≤ 0 and∂f∂u + ∂f∂v∂f≥ 0,∂uLet δ > 0 be small and define U δ = {x ∈ Ω : d(x) < δ}.We use as a supersolution∂u + ∂g∂v≤ 0. So it remains to construct the sub and supersolution of (46).ū = cd −α + e 1 log d + f 1 + g 1 d ε and ¯v = cd −α + e 2 log d + f 2 + g 2 d εwhere 0 < ε < α and g 1 , g 2 > 0 are to be fixed later on.In U δ we have∆ū = cα(α + 1)d −α−2 − e 1 d −2 − g 1 ε(1 − ε)d ε−2 − cαd −α−1 ∆d + e 1 d −1 ∆d+ g 1 εd ε−1 ∆d∆¯v = cα(α + 1)d −α−2 − e 2 d −2 − g 2 ε(1 − ε)d ε−2 − cαd −α−1 ∆d + e 2 d −1 ∆d+ g 2 εd ε−1 ∆dand( )eū−¯v = cα(α + 1)d −α−2 = cα(α + 1)d −α−2 + cα(α + 1)d −α−2 e (g1−g2)dε − 1 .We take g 1 , g 2 of the form∂vg 1 = ta 1 , g 2 = ta 2where t ≥ 1 and a 1 , a 2 > 0 are fixed such thatUsing convexityand henceβa 1 < a 2 < a 1 .eū−¯v ≥ cα(α + 1)d −α−2 + cα(α + 1)(a 1 − a 2 )td ε−α−2∆ū − ( eū−¯v − 1 ) ≤ −cα(α + 1)(a 1 − a 2 )td ε−α−2 + Cd −α−1 + Ctd ε−2where C depends only on p, β, Ω, a 1 and a 2 .Again, using convexityin U δ(¯v − βū) p ≥ cα(α + 1)d −α−2 + cα(α + 1)p f 2 − βf 1(1 − β)c d−2 + cα(α + 1)p a 2 − βa 1(1 − β)c tdε−2 .Since p > 3 we have α ∈ (0, 1). Hence, using (50) we find∆¯v − (¯v − βū) p ≤ −cα(α + 1)p a 2 − βa 1(1 − β)c tdε−2 + Ctd ε−1 + Cd −α−1 in U δ .

16 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2Then there is δ > 0 such that ū, ¯v is a supersolution of the system in the setU δ = {x ∈ Ω : d(x) < δ} for any t ≥ 1. Having fixed δ we now select t large suchthatū ≥ u and ¯v ≥ v on d(x) = δ.It follows by comparison that u ≤ ū and v ≤ ¯v in U δ .The construction of a subsolution u, v is similar. We takeu = cd −α + e 1 log d + f 1 − a 1 td ε and v = cd −α + e 2 log d + f 2 − a 2 td εwhere 0 < ε < α, a 1 > 0, a 2 > 0 are chosen such thatβa 1 < a 2 < a 1and t > 0 is to be fixed later on. Let us introduceσ = a 2 − βa 1 > 0.Later on we will need σ to be small.Let δ > 0 be small and U δ = {x ∈ Ω : d(x) < δ}. Recall that the unique solutionu, v to (46) satisfies u ≥ 0, v ≥ 0. We take C large so that if(51)thenWe have in U δt = Cδ −α−εu ≤ 0 and v ≤ 0 at d(x) = δ.∆v = cα(α + 1)d −α−2 − e 2 d −2 + a 2 tε(1 − ε)d ε−2 − cαd −α−1 ∆d + e 2 d −1 ∆d− a 2 tεd ε−1 ∆dGiven l > 0 there is A > 0 such that(1 + h) p ≤ 1 + ph + Ah 2 ∀ − 1 ≤ h ≤ l.Using this inequality withh = f 2 − βf 1 σt(1 − β)c dα +(1 − β)c dα+εwe find(v − βu) p ≤ cα(α + 1)d −α−2 + cα(α + 1)p f 2 − βf 1σt(1 − β)c d−2 + cα(α + 1)p(1 − β)c dε−2(+ A cα(α + 1)p f ) 22 − βf 1σt(1 − β)c d−2 + cα(α + 1)p(1 − β)c dε−2 ,provided that h ∈ [−1, l]. This condition is indeed satisfied if we take l > 0 largebut fixed and then δ > 0 small, since t is given by (51). It follows that in U δ∆v − (v − βu) p ≥ −Cd −α−1 − Cεtd ε−1 + Ctσd ε−2 − ACd α−2 − ACσ 2 t 2 d 2ε+α−2 .By taking σ > 0 sufficiently small and then δ > 0 small we finally obtain∆v − (v − βu) p ≥ 0 in U δ .Now let us verify that ∆u − e u−v ≥ 0 in U δ . It will then follow that (u, v) is asubsolution of (46). First we have∆ū = cα(α + 1)d −α−2 − e 1 d −2 + a 1 tε(1 − ε)d ε−2 − cαd −α−1 ∆d + e 1 d −1 ∆d− a 1 tεd ε−1 ∆d

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 17In addition( )e u−v = cα(α + 1)d −α−2 + cα(α + 1)d −α−2 e −(a1−a2)tdε − 1 .If γ > 0 is sufficiently small thenThereforein U δ provided that(52)e x ≤ 1 + γx for − 12γ ≤ x ≤ 0.e u−v ≤ cα(α + 1)d −α−2 − γcα(α + 1)d ε−α−2 (a 1 − a 2 )t(a 1 − a 2 )td ε ≤ 12γin U δ .To this end we choose γ = κδ α with κ > 0 small. Recalling that t is given by (51)we see that (52) is satisfied in U δ . It then follows thatand so∆u − ( e u−v − 1 ) ≥ −Cd −2 − Ctd ε−1 + Ctγd −α+ε−2∆u − ( e u−v − 1 ) ≥ 0in U δin U δif we fix δ > 0 sufficiently small. By comparison we deduce that u ≥ u and v ≥ vin U δ .□Example 2. Our second example falls in case b) of Theorem 1.2. Let α > 0, β > 0and consider⎧⎪⎨∆u = e u−αv in Ω(53)∆v = e⎪⎩v−βu in Ωu = v = +∞ on ∂ΩWe shall see that the constants involved in the leading asymptotics of the solution(u, v) depend on both nonlinearities f and g. We also compute the next term in theasymptotics of (u, v) and we observe that it is independent of the geometry of ∂Ω.Existence of solutions for the system (53) does not follow directly from Theorem1.1, since the nonlinearities g = exp and f = exp do not vanish at 0. To obtainthe existence we thus need to construct a suitable subsolution.Proposition 5.2. The system (53) has a solution if and only if αβ < 1. Moreoverif αβ < 1 then (53) has a unique solution (u, v) and it satisfies(54)whereandu = −c 1 log d + e 1 + o(1), v = −c 2 log d + e 2 + o(1) as x → ∂Ω(55)e 1 = log c 1 − α log c 2d(x) = dist(x, ∂Ω)c 1 = 2 1 + α1 − αβ , c 2 = 2 1 + β1 − αβ ,(56)1 − αβ, e 2 = log c 2 − β log c 11 − αβ

18 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2Proof. Regarding the existence part we let the reader check that Theorem 1.1 stillholds when f, g do not vanish at (0, 0), provided there exists a bounded subsolutionof the problem. We construct such a subsolution for (53) as follows. Take K > 0large so that|x| 2 − K ≤ 0 ∀x ∈ Ωand choose γ > 0 such thatLetwith A > 1 such that Aγ > 1. Thenand1α > γ > β.u = A(|x| 2 − K), v = Aγ(|x| 2 − K)∆u = 2NA ≥ 1, ∆v = 2NAγ ≥ 1,exp(u − αv) = exp ( A(|x| 2 − K)(1 − αγ) ) ≤ 1since |x| 2 − K ≤ 0 and 1 − αγ > 0. Similarlyexp(v − βu) = exp ( A(|x| 2 − K)(γ − β) ) ≤ 1.The fact that αβ < 1 is a necessary and sufficient condition for existence followsfrom the above discussion, Theorem 1.1 and the change of unknown ũ = u, ṽ = αv.For the rest of the proof we assume that αβ < 1. Regarding the asymptoticbehavior of solutions, first we establish that any solution (u, v) to (53) satisfies(54). For this purpose we first construct appropriate sub and supersolutions. Forδ > 0 defineU δ = { x ∈ Ω : d(x) < δ }.Let δ 0 > 0 be a small fixed number. Then there is ˜d smooth in Ω such that ˜d > 0in Ω and˜d ≡ d in U δ0 .Let 0 < σ < 1 and w be the solution to{−∆w = ˜dσ−2(57)w = 0in Ωon ∂Ω.Using comparison with appropriated powers of d one can obtain(58)for some constants a 1 , a 2 > 0.a 1 ˜dσ ≤ w ≤ a 2 ˜dσin ΩStep 1. For appropriate choices of γ ∈ (α, 1/β) and K > 0(59)u = −c 1 log ˜d + e 1 − γKw,v = −c 2 log ˜d + e 2 − Kw,form a subsolution of (68), where the constants c 1 , c 2 , e 1 , e 2 are given by (55), (56).Indeed,∆u = c 1 ˜d−2 |∇ ˜d| 2 − c 1 ˜d−1 ∆ ˜d + γK ˜d σ−2 ,whileTherefore(60)e u−αv = c 1 ˜d−2 e −(γ−α)Kw .∆u − e u−αv = c 1 ˜d−2 ( |∇ ˜d| 2 − e −(γ−α)Kw −)˜d∆ ˜d + γK ˜d σ−2 .

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 19Using the inequality(61)e −t ≤ 1 − t 3∀0 ≤ t ≤ 1and (58) we have−e −(γ−α)Kw ≥ −1 + 1 3 (γ − α)a 1K ˜d σwhenever˜d σ 1≤(γ − α)a 2 K .This holds in U δ if δ > 0 is small andδ σ 1K ≤ .(γ − α)a 2We note that from the start γ can be chosen close to α so that(62)c 1 + ‖ ˜d∆ ˜d‖ L ∞min(1, γ)≤1(γ − α)a 2.We now decrease δ > 0 further to achieve |∇ ˜d| = 1 in U δ and(63)δ 3‖∆ ˜d‖ L ∞(γ − α)a 1≤ c 1 + ‖ ˜d∆ ˜d‖ L ∞.With δ > 0 now being fixed we choose K such that(64)c 1 + ‖ ˜d∆ ˜d‖ L ∞min(1, γ)≤ Kδ σ ≤1(γ − α)a 2.Then in U δ we have by (60), (61)( )1∆u − e u−αv ≥ c ˜dσ−2 13 (γ − α)a 1K − ˜d 1−σ ∆ ˜d + γK ˜d σ−2 ≥ 0by (63) and (64). In Ω \ U δ∆u − e u−αv ≥ −c 1 ˜d−2 − c 1 ˜d−1 ∆ ˜d + γK ˜d σ−2 ≥ 0.thanks to (64). Similar calculations imply that ∆v − e v−βu ≥ 0 in Ω.Step 2. Let (u, v) denote the subsolution (59). Then for any solution (u, v) of (53)we have(65)u ≥ u, v ≥ v in Ω.To prove this statement, for ε > 0 small consider the domain Ω ε = Ω ∪ { x ∈R N : dist(x, ∂Ω) < ε }. Using Step 1, we can construct a subsolution (u ε , v ε )to (53) in the domain Ω ε . Note that u ε , v ε depend continuously on ε for ε > 0small. Substituting u by λu (for a given λ > 0) in the system (53), we obtain theequivalent form⎧⎪⎨∆u = g λ (u, v) in Ω(66)∆v = f λ (u, v) in Ω⎪⎩u = v = ∞ on ∂Ωwhere g λ (u, v) = 1 λ eλu−αv , f λ (u, v) = e v−λβu . Note that ∂g λ∂u= 1 ≥ 0, ∂g λ∂v=− α λ≤ 0 and ∂g λ∂u+ ∂g λ∂v= 1 − α λ . Similarly ∂f λ∂v= 1 ≥ 0, ∂f λ∂u= −βλ ≤ 0 and∂f λ∂u+ ∂f λ∂v= 1 − βλ. Since αβ < 1 it is possible to choose λ > 0 such that

20 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2∂g λ∂u+ ∂g λ∂v≥ 0 and ∂f λ∂u + ∂f λ∂v≥ 0. With these conditions the maximum principleholds for the system (66) and since u ε − u → −∞, v ε − v → −∞ as x → ∂Ω wededuceu ≥ u ε and v ≥ v ε in Ω.Letting ε → 0, we obtain (65).Step 3. Following the same argument as in the two previous steps one can showthat for appropriate choices of γ > 0 and K > 0ū = −c 1 log ˜d + e 1 + γKw,¯v = −c 2 log ˜d + e 2 + Kw,where w is the solution of (57), is a suspersolution of (68), and for any solution u, vof (68) we have(67)u ≤ ū and v ≤ ¯v in Ω.Step 4. Let u i , v i , i = 1, 2 be two solutions to (53). Then u 1 = u 2 and v 1 = v 2 .Indeed u i /λ, v i are also solutions to (66). Moreover by (65) and (67) and the factthat w(x) → 0 as x → ∂Ω we have that (u 1 −u 2 )/λ → 0 and v 1 −v 2 → 0 as x → ∂Ω.By the maximum principle for (66) we conclude that u 1 = u 2 and v 1 = v 2 . □Example 3. Our next system is not of the form (2). It demonstrates how our techniquecan still be used in more general settings. It also provides an example whereconditions (17) and (18) in Remark 2.3 are not equivalent. Consider p, q, r, s > 0and the system⎧∆u = upv⎪⎨q in Ω(68)∆v = vrin Ω⎪⎩u su > 0, v > 0u = v = +∞in Ωon ∂ΩProposition 5.3. Problem (68) has a solution if and only if(69)p > 1, r > 1 and (p − 1)(r − 1) > qs.Moreover, under condition (69) the system has a unique solution and it satisfies(70)where(71)u = c 1 d −γ (1 + o(1)) and v =c 2 d −λ (1 + o(1)) as x → ∂Ωand c 1 , c 2 are given by(72)(73)r − 1 + qγ = 2(p − 1)(r − 1) − sq , λ = 2 p − 1 + s(p − 1)(r − 1) − sqc (p−1)(r−1)−qs1 = (γ(γ + 1)) r−1 (λ(λ + 1)) qc (p−1)(r−1)−qs2 = (γ(γ + 1)) s (λ(λ + 1)) p−1 .Remark 5.4. Let m = p+s−1q+r−1. Observe the following distinction between the balancedsystem (m = 1) and the case m > 1

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 21(74)• balanced system: introduce n = r − s = p − q, then Problem (68) reads also⎧ ( u) q∆u = unin Ωv⎪⎨ ( v) s∆v = vnin Ωuu > 0, v > 0 in Ω⎪⎩u = v = +∞ on ∂ΩConsider z the BBUS for the single equation ∆z = z n . Then (z, z) is theBBUS for the balanced system.• unbalanced system: perform the change of variable w = u m . Then∆w ≥ ( w p−1v )q 1−q+ ww m = (v )q w n ,with n = r − s m = 1 − q + p−1m∆v = ( v w )s v n ,. Observe that n = 1 +(p−1)(r−1)−sqp+s−1> 1.Let again z denote the BBUS for the single equation ∆z = z n , then usingan ordering lemma (see Lemma 5.5 below), we haveu m ≤ v ≤ z.Before proving Proposition 5.3 we need to establish some preliminary results.The first one is the following comparison lemma.Lemma 5.5. Let u 1 , v 1 ∈ C 2 (Ω), u 1 , v 1 > 0 in Ω, be a subsolution to (68). Similarly,let (u 2 , v 2 ) be a supersolution to (68) and assumeu 1v 1(75)lim sup ≤ 1 and lim sup ≤ 1.x→∂Ω u 2 x→∂Ω v 2If p, q, r, s > 0 satisfy (69) then(76)u 1 ≤ u 2 and v 1 ≤ v 2 in Ω.Proof. Consider ũ i = log u i , ṽ i = log v i . Then ũ 1 , ṽ 1 satisfy{∆ũ1 + |∇ũ 1 | 2 ≥ e (p−1)ũ1−qṽ1 in Ω∆ṽ 1 + |∇ṽ 1 | 2 ≥ e (r−1)ṽ1−sũ1in Ωand ũ 2 , ṽ 2 satisfy the corresponding reversed inequalities. It is convenient to introduceone more change of variables: λU i = ũ i and V i = ṽ i where λ is such that(77)Then(78)r − 1s> λ > qp − 1 .{∆U1 + λ|∇U 1 | 2 ≥ g λ (U 1 , V 1 ) in Ω∆V 1 + |∇V 1 | 2 ≥ f λ (U 1 , V 1 )in Ωwhere g λ (u, v) = 1 λ e(p−1)λu−qv , f λ (u, v) = e (r−1)v−sλu .The inequalities (76) are equivalent to U 1 ≤ U 2 and V 1 ≤ V 2 in Ω. Supposethat one of these inequalities fail. We deal first with the case sup Ω (U 1 − U 2 ) ≥sup Ω (V 1 − V 2 ). Then sup Ω (U 1 − U 2 ) > 0 and, since lim sup x→∂Ω (U 1 − U 2 ) ≤ 0 by(75), there is x 0 ∈ Ω where U 1 −U 2 attains its maximum. Then ∇U 1 (x 0 ) = ∇U 2 (x 0 )

22 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2and ∆U 1 (x 0 ) − ∆U 2 (x 0 ) ≤ 0. Using the first inequality in (78) -and its analoguefor U 2 - we obtain for some η 1 , η 2 ≥ 00 ≥ ∆(U 1 − U 2 )(x 0 ) + λ ( |∇U 1 (x 0 )| 2 − |∇U 2 (x 0 )| 2)Since ∂g λ(η 1,η 2)∂vdeduce(79)≥g λ (U 1 (x 0 ), V 1 (x 0 )) − g λ (U 2 (x 0 ), V 2 (x 0 ))= ∂g λ(η 1 , η 2 )∂u(U 1 (x 0 ) − U 2 (x 0 )) + ∂g λ(η 1 , η 2 )(V 1 (x 0 ) − V 2 (x 0 )).∂v≤ 0 and V 1 (x 0 ) − V 2 (x 0 ) ≤ sup Ω (V 1 − V 2 ) ≤ U 1 (x 0 ) − U 2 (x 0 ) we( ∂gλ (η 1 , η 2 )0 ≥∂u+ ∂g )λ(η 1 , η 2 )(U 1 (x 0 ) − U 2 (x 0 )).∂vBut∂g λ (η 1 , η 2 )+ ∂g λ(η 1 , η 2 )= (λ(p − 1) − q)e (p−1)η1−qη2 > 0∂u∂vby (77). This gives a contradiction with (79).The remaining case, that is when sup Ω (U 1 − U 2 ) ≤ sup Ω (V 1 − V 2 ), is analogousso we skip it.□Proposition 5.3 will be obtained through a blow-up argument, using an idea from. We start out by studying the associated limiting problem. We write x ∈ R Nas x = (x 1 , x ′ ) with x 1 ∈ R and x ′ ∈ R N−1 . Let R N + = {(x 1 , x ′ ) : x 1 > 0}.Proposition 5.6. Assume condition (69) and let γ, λ, c 1 , c 2 be defined by (71)–(73). Suppose u, v ∈ C 2 (R N + ), u, v > 0 solve (68) in R N + and satisfy(80)1C x−γ 1 ≤ u ≤ Cx −γ1 andfor some C > 0. Then u ≡ c 1 x −γ1 and v ≡ c 2 x −λ1 .1C x−λ 1 ≤ v ≤ Cx −λ1Proof. We start proving that u ≤ c 1 x −γ1 in R N + . Let σ > 0 satisfyq(81)r − 1 < σ < p − 1 .sFor t > 0 setu t = tc 1 x −γ1 , v t = t σ c 2 x −λ1and note that for t > 1 the pair (u t , v t ) is a supersolution of (68) in R N + . LetNote that by (80) t 0 is well defined and(82)t 0 = inf{t > 1 : u ≤ u t and v ≤ v t in R N + }.u ≤ u t0 and v ≤ v t0 in R N + .We wish to show that t 0 ≤ 1. Assume by contradiction that t 0 > 1. Let t n be asequence such that t n → t 0 and for each n, either u ≤ u tn fails or v ≤ v tn does. Atleast one of these inequalities has to fail for infinitely many n’s, and we work outthe details in the former case. Passing to a subsequence if necessary there are x nsuch that(83)u(x n ) > u tn (x n ).We write x n = (x 1,n , x ′ n) and define r n = x 1,n and the functionsu n (y) = r γ nu(r n y + (0, x ′ n)),v n (y) = r λ nv(r n y + (0, x ′ n)).

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 23Then u n , v n also satisfy the bounds (80). By standard elliptic estimates, up to anew subsequence u n → u ∗ and v n → v ∗ uniformly on compact sets of R N + and(u ∗ , v ∗ ) is a solution to (68) in R N + and satisfies (80). From (82) we find(84)and (83) implies thatu ∗ ≤ u t0 and v ∗ ≤ v t0 in R N +u ∗ (e 1 ) = u t0 (e 1 ).This yields ∆u ∗ (e 1 ) ≤ ∆u t0 (e 1 ). On the other hand, since t 0 > 1 we have ∆u t0 v t0 (e 1 ), contradicting (84).Proof of Proposition 5.3.Step 1. Here we show that (69) is necessary for existence. Consider α, β ∈ (0, 1).Suppose that (u, v) is a solution of (68) and defineThen,Hence the functionw = min(u α , v β ).∆w ≤ αw α−1+pα w − q β χ[u α v ]. βw ↦→ max(w α−1+pα − q β−1+rβ , wβ − s α )must satisfy the Keller-Osserman condition (1). Therefore for any α, β ∈ (0, 1)This implies (69).p − 1q> α βorαβ > sr − 1 .Step 2. Condition (69) is sufficient for the existence of a solution.First we show that there exists a subsolution u, v > 0 of (68). ConsiderThen ∆u ≥ u p /v q if(85)u = a(|x| 2 + 1), v = b(|x| 2 + 1).2N ≥ a p−1 b −q A where A = supΩ(|x| 2 + 1 ) p−qand similarly, in order that ∆v ≥ v r /u s it is sufficient that2N ≥ a −s b r−1 (B, B = sup |x| 2 + 1 ) r−s.ΩSetting a s = b r−1 B/(2N) and inserting in (85) shows that it is enough that(2N) p−1+s ≥ b (p−1)(r−1)−qs B p−1 A s ,which can be achieved for small b > 0 thanks to the condition (p−1)(r−1)−qs > 0.Consider now the minimal solution (u m , v m ) to the truncated problem (10) withg(u, v) = u p /v q , f(u, v) = v r /u s where m > 0, which existence is guaranteed byProposition 2.1. Let α, β > 1 be such that(86)p − 1q> α βandαβ > sr − 1□

24 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2which is possible thanks to (69). Now letw m = max(u α m, v β m).Then by Kato’s inequality and since α, β > 1where∆w m ≥ αw α−1+pαmw − q βmh(w) = minβ−1+rχ [uα v β ] ≥ h(w m ))− q β−1+rβ , βwβ − s α.But thanks to (86) we have α−1+pα− q β−1+rβ> 1 andβ− s α> 1, which implies thath satisfies the Keller-Osserman condition (1). Hence the problem ∆w = h(w) in Ωhas a solution w with w = ∞ on ∂Ω. It follows that w m ≤ w, which shows that u mand v m remain bounded on compact sets of Ω, and in a standard way one obtainsa solution to (68).From now on we assume that (69) holds.Step 3. For any solution (u, v) of (68) we have1C d−γ ≤ u ≤ Cd −γ 1(87)andC d−λ ≤ v ≤ Cd −λ in Ωfor some constant C > 0.We let t > 1 and σ > 0 such that (81) holds. Let 0 < ε < min(γ, λ) and defineū = t ( c 1 d −γ − k 1 d ε−γ) ,¯v = t σ ( c 2 d −λ − k 2 d ε−λ)where γ, λ, c 1 , c 2 are defined by (71)–(73) and k 1 , k 2 > 0 will be specified later on.Let δ > 0 be small such that in U δ = {x ∈ Ω : d(x) < δ}, d is smooth and |∇d| = 1.Then, in U δ ,[∆ū = tc 1 γ(1 + γ)d −2−γ1 − 11 + γ d∆d − k 1By choosing k 1 such thatc 1 γk 1 ≥ 2‖∆d‖ L ∞(γ − ε)(γ − ε + 1)and δ < 1 such thatwe haveAdditionally,Thereforeδ‖∆d‖ L ∞ ≤ 1 2∆ū ≤ tc 1 γ(1 + γ)d −2−γ in U δ .ū p¯v q = tp−σq c p 1 c−q 2 d−2−γ (1 − k 1d ε /c 1 ) p(1 − k 2 d ε /c 2 ) q .(γ − ε)(γ − ε + 1)d εc 1 γ(1 + γ)]+k 1γ − εc 1 γ(1 + γ) d1+ε ∆d[∆ū − ūp¯v q = tc 1γ(1 + γ)d −2−γ 1 − t p−1−σq (1 − k 1d ε /c 1 ) p ](1 − k 2 d ε /c 2 ) q

BOUNDARY BLOW-UP SOLUTIONS OF COOPERATIVE SYSTEMS 25Since p − 1 − σq > 0 thanks to (81) we may fix t 0 > 1 and find a uniform δ smallsuch that for all t > t 0 :∆ū − ūp¯v q ≤ 0 in U δSimilarly∆¯v − ¯vrū s ≤ 0 in U δfor all t > t 0 . By decreasing δ we also can achieveinf ū(x) > 0 and inf ¯v(x) > 0.d(x)=δ d(x)=δNow we take t 0 large enough such that ū ≥ u and ¯v ≥ v on {x ∈ Ω : d(x) = δ}. ByLemma 5.5, we deduce that u ≤ ū and v ≤ ¯v in U δ which implies the upper boundsin (87). The lower bounds are obtained similarly.Step 4. Any solution (u, v) of (68) satisfies the boundary behavior (70).Let x n ∈ Ω be such that x n → x 0 ∈ ∂Ω. Without loss of generality we mayassume that ν(x 0 ) = −e 1 . Let r n = d(x n ) and ˆx n ∈ ∂Ω be the point on ∂Ω closestto x n .Defineu n (x) = r γ nu(r n x + ˆx n ), v n (x) = r λ nv(r n x + ˆx n ).Then (u n , v n ) solves (68) in Ω n = (Ω − ˆx n )/r n . As n → ∞, Ω n approaches thehalf space R + N . Morover, letting d n(x) = dist(x, ∂Ω n ) we have d n (x) = d(r n x)/r n .Using this and (87)1C d−γ n≤ u n ≤ Cd −γnand1C d−λ n≤ v n ≤ Cd −λn in Ω n .Using standard elliptic estimates and the above inequalities, up to a subsequence,u n → u ∗ , v n → v ∗ uniformly on compact sets of R N + where (u ∗ , v ∗ ) is a solution to(68) in R N + satisfying1C x−γ 1 ≤ u ∗ ≤ Cx −γ1 and1C x−λ 1 ≤ v ∗ ≤ Cx −λ1 in R N + .By Proposition 5.6 we have u ∗ ≡ c 1 x −γ1 and v ∗ ≡ c 2 x −λ1 . Henced(x n ) γ u(x n ) = u n ((x n − ˆx n )/r n ) → u ∗ (e 1 ).It follows that lim n→∞ d(x n ) γ u(x n ) = c 1 . Similarly lim n→∞ d(x n ) λ v(x n ) = c 2 .Step 5. From Step 4 and Lemma 5.5 we deduce that (68) has a unique solution,which satisfies (70).□(88)Appendix A. Maximum principle for cooperative systemsLet Ω ⊆ R N be a bounded open set and consider the linear systemWe assume that a ij ∈ L ∞ (Ω) satisfy∆u ≥ a 11 u + a 12 v∆v ≥ a 21 u + a 22 vin Ωin Ωa 11 + a 12 ≥ 0, a 12 ≤ 0 in Ωa 21 + a 22 ≥ 0, a 21 ≤ 0 in Ω

26 JUAN DÁVILA1 , LOUIS DUPAIGNE 2 , OLIVIER GOUBET 1 AND SALOMÉ MARTÍNEZ2Theorem A.1. Suppose u, v ∈ C 2 (Ω) ∩ C(Ω) satisfy (88) andwhere M ≥ 0. Thenu ≤ M, v ≤ M on ∂Ωu ≤ M, v ≤ M in ΩProof. First we assume M = 0. Let ε > 0 andũ = u + εe Ax1 ,where A > 0 is a large constant. Thenṽ = v + εe Ax1∆ũ ≥ a 11 u + a 12 v + εA 2 e x1= a 11 ũ + a 12 ṽ + εe x1 (A 2 − a 11 − a 12 ) > a 11 ũ + a 12 ṽin Ω if A is taken large, and similarly∆ṽ > a 21 ũ + a 22 ṽ.Moreover ũ ≤ εK and ṽ ≤ εK on ∂Ω whereWe claim thatK = max∂Ω eAx1 .ũ ≤ εK and ṽ ≤ εK in Ω.Suppose that the conclusion fails and that max Ωu > εK and max Ωu ≥ max Ωv.Let x 0 ∈ Ω be a point where ũ attains its maximum. Then0 ≥ ∆ũ(x 0 ) > a 11 ũ(x 0 ) + a 12 ṽ(x 0 ) ≥ (a 11 + a 12 )ũ(x 0 ) ≥ 0which is impossible. Thus we havemaxΩLetting ε → 0 we obtain(u + εe Ax1 ) ≤ εK, max(v + εe Ax1 ) ≤ εK.max u ≤ 0,ΩΩmax v ≤ 0.ΩNow we assume M ≥ 0. Consider ũ = u − M, ṽ = v − M. Then∆ũ ≥ a 11 u + a 12 v = a 11 ũ + a 12 ṽ + M(a 11 + a 12 ) ≥ a 11 ũ + a 12 ṽby the assumptions M ≥ 0, a 11 + a 12 ≥ 0. Similarly∆ṽ ≥ a 21 ũ + a 22 ṽ.Applying the previous case we deduce ũ ≤ 0 and ṽ ≤ M which yieldsu ≤ M and v ≤ M in Ω.Acknowledgments. This work was started in Santiago in December 2006 andfinished in Amiens in January 2008. All four authors were partially supported byEcos-Conycit grant C05E04. L.D. and O.G. were partially supported by DIM andFondap Chile. They thank the mathematics department for its hospitality. J.Dwas partially supported by Fondecyt 1050725 and Fondap Chile. S.M was partiallysupported by Fondecyt 1050754, Fondap Chile and Nucleus Millenium P04-069-FInformation and Randomness. J.D. and S.M. thank LAMFA for its hospitality.□

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