Existence of a cooperative elliptic system involving pucci operator

Acta Mathematica Scientia 2010,30B(1):137–147 http://actams.wipm.ac.cn

EXISTENCE OF A COOPERATIVE ELLIPTIC SYSTEM INVOLVING PUCCI OPERATOR∗



Yang Jianfu (

)

Department of Mathematics, Jiangxi Normal University, Nanchang 330022, China E-mail: jfyang [email protected]


)

Yu Xiaohui (

Institute for Advanced Study, Shenzhen University, Shenzhen 518060, China E-mail: yuxiao [email protected]

Abstract

The authors study the existence of solutions for the nonlinear elliptic system

 −M (D u) = f (u, v)  −M (D v) = g(u, v)  u ≥ 0, v ≥ 0  + λ,Λ

2

+ λ,Λ

2

u=v=0

in

Ω,

in

Ω,

in

Ω,

on

∂Ω,

where Ω is a bounded convex domain in RN , N ≥ 2. It is shown that under some assumptions on f and g, the problem has at least one positive solution (u, v). Key words fixed point index; nonlinear elliptic system; Pucci operator 2000 MR Subject Classification

1

35J50; 35B52

Introduction

The main objective of this article is to investigate the existence of solutions for elliptic systems involving the Pucci extremal operator ⎧ 2 ⎪ −M+ ⎪ λ,Λ (D u) = f (u, v) ⎪ ⎪ ⎪ ⎨ −M+ (D2 v) = g(u, v) λ,Λ ⎪ ⎪ u ≥ 0, v ≥ 0 ⎪ ⎪ ⎪ ⎩ u=v=0

in Ω, in Ω,

(1.1)

in Ω, on ∂Ω,

where Ω ⊂ RN , N ≥ 2 is a bounded convex domain. The Pucci extremal operator M+ λ,Λ is defined as follows. For any two positive constants λ ≤ Λ, let A be a symmetric matrix whose ∗ Received

July 28, 2007; revised September 17, 2007. This work is supported by National Natural Sciences Foundations of China (10571175, 10631030).

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eigenvalues belong to [λ, Λ]. Denoted by Aλ,Λ the class of all such matrices. For any symmetric matrix M , the Pucci extremal operators are defined by M+ λ,Λ (M ) = sup Aij Mij , A∈Aλ,Λ

If e1 , · · ·, en are eigenvalues of M , then,   ei + λ ei , M+ λ,Λ (M ) = Λ ei >0

ei <0

M− λ,Λ (M ) =

inf

A∈Aλ,Λ

M− λ,Λ (M ) = λ

 ei >0

Aij Mij .

ei + Λ



(1.2)

ei .

(1.3)

ei <0

We recall that a continuous pair (u, v) : Ω × Ω → R × R is a viscosity super-solution (subsolution) of (1.1) if, for all (φ, ϕ) ∈ C 2 (Ω) × C 2 (Ω) and x0 ∈ Ω, y0 ∈ Ω such that u − φ, v − ϕ have local minima (maxima) at x0 , y0 , respectively, there hold ⎧ ⎨ M+ (D2 φ(x )) + f (u(x ), v(x )) ≤ (≥)0, 0 0 0 λ,Λ ⎩ M+ (D2 ϕ(y0 )) + g(u(y0 ), v(y0 )) ≤ (≥)0. λ,Λ

If (u, v) is both a viscosity super-solution and a viscosity sub-solution of (1.1), we say (u, v) is a viscosity solution of (1.1). Recently, an equation problem involving Pucci extremal operator ⎧ + 2 ⎪ ⎪ ⎨ −Mλ,Λ (D u) = f (x, u) in Ω, ⎪ ⎪ ⎩

u≥0

in Ω,

u=0

on ∂Ω

(1.4)

has been studied extensively. In [3], a Liouville type theorem was established for non-negative viscosity super-solutions for (1.4). It is known that the Pucci extremal operators are not in divergence form, the variational method cannot be applied. To find solutions of (1.4), one turns to the topological method. It is then necessary to establish a priori bound for solutions of (1.4). This was done in [8] and [9] by blow up arguments, where Liouville type theorems are involved. Solutions of (1.4) were then found in [8], [9] by topological method. On the other hand, the same argument was used in [5] for semilinear elliptic system ⎧ ⎪ −Δu = f (u, v) in Ω, ⎪ ⎪ ⎪ ⎪ ⎨ −Δv = g(u, v) in Ω, (1.5) ⎪ ⎪ u ≥ 0, v ≥ 0 in Ω, ⎪ ⎪ ⎪ ⎩ u=v=0 on ∂Ω. The key point is to establish Liouville type theorem and use blow up argument. We apply the method to problem (1.1) to show the existence of solutions. Let 2,N + 2 ¯ μ+ 1 = sup{μ ∈ R : ∃φ ∈ Wloc (Ω) ∩ C(Ω), such that φ > 0 and Mλ,Λ (D φ) + μφ ≤ 0 in Ω}. + 2 ¯ It was proved in [1] that μ+ 1 > 0 and that there a positive function ϕ1 ∈ C (Ω) ∩ C(Ω), such + + that (μ1 , ϕ1 ) is a solution of ⎧ + 2 ⎪ ⎪ ⎨ Mλ,Λ (D u) + μu = 0 in Ω,

⎪ ⎪ ⎩

u≥0

in

u=0

on ∂Ω.

Ω,

(1.6)

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We assume, in this article, that (A1) f, g ≥ 0 are locally Lipschitz continuous, and there exists an r0 > 0, such that + f (u, v) < μ+ 1 u and g(u, v) < μ1 v if 0 ≤ |u| + |v| ≤ r0 . (A2) f, g can be written for large |u| + |v| as f (u, v) = auα11 + bv α12 + h1 (u, v),

g(u, v) = cuα21 + dv α22 + h2 (u, v),

where a, b, c, d ≥ 0, αij ≥ 0, i, j = 1, 2, and |h1 (u, v)| ≤ C1 (1 + |u|β11 + |v|β12 ),

|h2 (u, v)| ≤ C2 (1 + |u|β21 + |v|β22 )

with 0 ≤ βij < αij , i, j = 1, 2. (A3) There are positive numbers ξ1 , ξ2 > 0 and a positive constant C > 0, such that 2 ξ1 ξ2 > (μ+ 1 ) , and f (u, v) ≥ ξ1 u − C uniformly for v ∈ R+ , g(u, v) ≥ ξ2 v − C

uniformly for u ∈ R+ .

(A4) The system (1.1) is cooperative, that is, for fixed u ≥ 0, f (u, v1 ) ≤ f (u, v2 ) if v1 ≤ v2 , and for fixed v ≥ 0, g(u1 , v) ≤ f (u2 , v) if u1 ≤ u2 . We say (1.1) is weakly coupled if α11 > 1, α22 > 1, and α12 <

α22 − 1 α11 , α11 − 1

α21 <

α11 − 1 α22 , α22 − 1

α22 <

α12 + 1 α21 . α21 + 1

and it is strongly coupled if α12 α21 > 0 and α11 <

α21 + 1 α12 , α12 + 1

Our main results are as follows. Theorem 1.1 Suppose (A1)–(A4) hold. If (1.1) is weakly coupled and α11 ≤

α , α−2

α22 ≤

α , α−2

λ where α = Λ (N − 1) + 1, then, (1.1) has at least one positive solution. Theorem 1.2 Suppose (A1)–(A4) hold. If (1.1) is strongly coupled and one of the following conditions holds: (i) 0 < α12 α21 ≤ 1; 12 21 (ii) α12 α21 > 1, but α2+2α > α − 2 or α2+2α > α − 2; 12 α21 −1 12 α21 −1 2+2α12 2+2α21 α (iii) α12 α21 −1 = α − 2 and α12 α21 −1 = α − 2, that is, α12 = α12 = α−2 , then, problem (1.1) has at least one positive solution. − Remark 1.3 Similar results can be obtained for Mλ,Λ in the same way. The functions α12 α11 α11 α12 f (u, v) = au + b(v − r0 )+ and g(u, v) = c(u − r0 )+ + dv with a, b, c, d > 0 satisfy the conditions (A1)–(A4). To prove our main results, we first establish a priori bound for solutions of (1.1) by the blow up argument. It is then necessary to establish Liouville type theorem related to problem (1.1). This is done in Section 2 and Section 3. We will finish the proofs of Theorem 1.1 and Theorem 1.2 in Section 4.

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Liouville Type Theorems In this section, we establish a Liouville type theorem. It was proved in [3] that if 0 < p ≤ then, any viscosity solution of

α α−2 ,

⎧ ⎨ −M+ (D2 u) = up λ,Λ ⎩u ≥ 0

in

RN ,

in

RN

(2.1)

should be a trivial one. Consider the problem ⎧ + 2 p ⎪ ⎪ ⎨ Mλ,Λ (D u) + v = 0 2 q M+ λ,Λ (D v) + u = 0 ⎪ ⎪ ⎩ u ≥ 0, v ≥ 0

in

RN ,

in

RN ,

in

N

(2.2)

R ,

we have the following Liouville type theorem. Theorem 2.1 Suppose (u, v) ∈ C(RN ) × C(RN ) is a viscosity solution of (2.2), p and q satisfy one of the following conditions: (i) 0 < pq ≤ 1; (ii) pq > 1 and either 2 + 2q > α − 2 or pq − 1

2 + 2p > α − 2. pq − 1

(2.3)

2+2q α (iii) p = q = α−2 , that is, pq−1 = α − 2 and 2+2p pq−1 = α − 2. λ Then, u = 0 and v = 0, where α = Λ (N − 1) + 1. The proof of Theorem 2.1 is based on the following nonlinear Hadamard theorem in [3]. ¯r1 centered at the Lemma 2.2 Let Ω be a domain of RN containing the closed ball B origin and of radius r1 > 0. If u ∈ C(Ω) is a viscosity solution of 2 M+ λ,Λ (D u) ≤ 0,

then, the function m(r) = min u(x) {|x|≤r}

is, respectively, a concave function of log r if α = 2 and of r2−α if α = 2 with α = More precisely, for every fixed r2 < r1 and for all r2 ≤ r ≤ r1 , it results ⎧ m(r2 ) log(r1 /r) + m(r1 ) log(r/r2 ) ⎪ ⎪ , if α = 2, ⎪ ⎨ log(r1 /r2 ) m(r) ≥ ⎪ m(r2 )(r2−α − r12−α ) + m(r1 )(r22−α − r2−α ) ⎪ ⎪ , if α = 2. ⎩ r22−α − r12−α

λ Λ (N

− 1) + 1.

In the proof of Theorem 2.1, the following two lemmas are needed (see [2]). Lemma 2.3 (Comparison Principle) Let Ω be a bounded open set and let f ∈ C(Ω); if − 2 2 u1 is a super-solution and u2 is a sub-solution of M+ λ,Λ (D u) = f (x) or Mλ,Λ (D u) = f (x), ¯ respectively, in Ω and if u1 ≥ u2 on ∂Ω, then, u1 ≥ u2 in Ω.

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Lemma 2.4 (Strong Maximum Principle) Let Ω be a bounded open set and u be a + 2 2 viscosity super-solution (sub-solution) of either M− λ,Λ (D u) = 0 or Mλ,Λ (D u) = 0 in Ω. If u attains its minimum (maximum) at an interior point of Ω, then, u is a constant. Proof of Theorem 2.1 First, we note that 2 M+ λ,Λ (D u) ≤ 0,

2 M+ λ,Λ (D v) ≤ 0.

If α ≤ 2, by Theorem 3.2 of [3], both u and v are constants, then, we deduce from (2.2) that u = v = 0. The result follows. In the following, we suppose α > 2. Let m1 (r) = min u(x) and m2 (r) = min v(x). {|x|≤r}

{|x|≤r}

By Corollary 3.1 in [3], both m1 (r)rα−2 and m2 (r)rα−2 are increasing in r, that is, for any R ≥ r > 0, we have m1 (r) ≤

m1 (R)Rα−2 rα−2

and m2 (r) ≤

m2 (R)Rα−2 . rα−2

(2.4)

We define that, for R > r ≥ r0 fixed,  [(|x| − r)+ ]3  , ζ1 (x) = m1 (r) 1 − (R − r)3

 [(|x| − r)+ ]3  ζ2 (x) = m2 (r) 1 − . (R − r)3

Then, the minima of u − ζ1 and u − ζ2 in RN are non-positive, they are attained, respectively, r r at xrR with r ≤ xrR < R and yR with r ≤ yR < R. By the definition of viscosity solution, 2 r p r M+ λ,Λ (D ζ1 (xR )) + v (xR ) ≤ 0,

2 r q r M+ λ,Λ (D ζ2 (yR )) + u (yR ) ≤ 0.

(2.5)

+ 2

] By Lemma 3.1 of [3], D2 ζ1 (x) has two eigenvalues, one is −m1 (r) 3[(|x|−r) (R−r)3 |x| , which is of +

multiplicity N − 1; the other is −m1 (r) 6(|x|−r) (R−r)3 |x| which is simple. So, we have 2 r M+ λ,Λ (D ζ1 (xR )) = −

which implies v p (xrR ) ≤

3λm1 (r) (|xrR | − r)+ ](|xrR | − r)+ , [2 + (N − 1) (R − r)3 |xrR |

(2.6)

3λm1 (r) (|xrR | − r)+ ](|xrR | − r)+ . [2 + (N − 1) (R − r)3 |xrR |

(2.7)

r 3λm2 (r) (|yR | − r)+ r ](|yR [2 + (N − 1) | − r)+ . r| (R − r)3 |yR

(2.8)

In the same way, we have r )≤ uq (yR

r | = r. In fact, suppose on the contrary, then either u or v attains We note that |xrR | = r and |yR its minimum 0 in an interior point, so the strong maximum principle (Lemma 2.4) implies that r u = v = 0. Therefore, we might assume r < xrR < R and r < yR < R. By (2.7) and (2.8),

v p (xrR ) ≤

3λm1 (r)(N + 1) , (R − r)2

r uq (yR )≤

3λm2 (r)(N + 1) . (R − r)2

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r Since v(xrR ) ≥ m2 (R) and u(yR ) ≥ m1 (R), we have 1

m2 (R) ≤ C

1

m1 (r) p

m1 (R) ≤ C

,

2

(R − r) p

m2 (r) q 2

(R − r) q

,

(2.9)

where C is a positive constant, which may vary from line to line. From (2.4) and (2.9), we deduce 1

m2 (R) ≤ C

m1 (R) p R 2

(R − r) p r

α−2 p

1

m1 (R) ≤ C

α−2 , p

they implies 1

m1 (R) ≤ C

m1 (R) pq R 2

(R − r) pq r

and

1

m2 (R) ≤ C Choosing r =

R 2

m2 (R) pq R 2

(R − r) pq r

α−2 pq α−2 pq

α−2 pq α−2 pq

·

·

R

m2 (R) q R 2

(R − r) q r α−2 q 2

(R − r) q r R

α−2 q

α−2 p 2

(R − r) p r

α−2 p

α−2 q α−2 q

,

(2.10)

,

.

and R ≥ 2r0 , we obtain 1

m1 (R) ≤ C

m1 (R) pq R

2+2p pq

1

m2 (R) ≤ C

,

m2 (R) pq R

2+2q pq

.

If 0 < pq ≤ 1, then, we obtain R

2+2p pq

1

1

≤ Cm1 (R) pq −1 ≤ Cm1 (0) pq −1 ,

a contradiction. This proves case (i). Next, we may assume pq > 1. In this case, m1 (R) ≤

C R

2+2p pq−1

,

m2 (R) ≤

C R

2+2q pq−1

,

that is, m1 (R)Rα−2 ≤

C R

2+2p pq−1 −α+2

,

m2 (R)Rα−2 ≤

C R

2+2q pq−1 −α+2

.

2+2q If 2+2p pq−1 − α + 2 > 0 or pq−1 − α + 2 > 0, we obtain a contradiction to the fact that α−2 α−2 m1 (R)R and m2 (R)R are increasing. This proves case (ii). 2+2q α Finally, we consider the case 2+2p pq−1 − α + 2 = 0 and pq−1 − α + 2 = 0, that is, p = q = α−2 . In this case, both m1 (R)Rα−2 and m2 (R)Rα−2 are bounded above. Fixing R1 ≥ 0, γ1 > 0, and γ2 ∈ R, for |x| ≥ R1 , we define

Γ(x) = γ1

log(1 + |x|) + γ2 . |x|α−2

A direct calculation shows that Γ(x) is decreasing and convex in |x| for R1 large enough. Moreover, 2 M+ λ,Λ (D Γ(x)) = −γ1 Λ



−γ Λ(α − 2) α−3 1 1 ≥ + . α−1 2 α−2 (1 + |x|)|x| (1 + |x|) |x| |x|α

(2.11)

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For R2 > R1 , we may choose γ1 > 0, γ2 ∈ R, such that Γ(R1 ) ≤ m1 (R1 ),

Γ(R2 ) = m1 (R2 ).

In fact, this can be done by choosing 0 < γ1 ≤

m1 (R1 ) log(1+R1 ) R1α−2

− m1 (R2 ) −

log(1+R2 ) R2α−2

,

γ2 = m1 (R2 ) − γ1

log(1 + R2 ) . R2α−2

Apparently, Γ ≤ u on {|x| = R1 } ∪ {|x| = R2 } and (2.11) holds in {R1 < |x| < R2 }. On the other hand, if |x| ≥ R1 , u(x) ≥ m1 (|x|) ≥

m1 (R1 )R1α−2 , |x|α−2

v(x) ≥ m2 (|x|) ≥

m2 (R1 )R1α−2 , |x|α−2

so, we have 2 M+ λ,Λ (D u) ≤ −

Since p = q =

α α−2 ,

 m (R )Rα−2 p 2 1 1 |x|α−2

2 and M+ λ,Λ (D v) ≤ −

 m (R )Rα−2 q 1 1 1 . |x|α−2

we have 2 M+ λ,Λ (D u) ≤ −

α  m (R )Rα−2 α−2 C 2 1 1 = − α, α−2 |x| |x|

for |x| > R1 . Choosing γ1 small, we obtain 2 M+ λ,Λ (D u) ≤ −

C 2 ≤ M+ λ,Λ (D Γ(x)), |x|α

for all R1 ≤ |x| ≤ R2 . The comparison principle, see Lemma 2.3, yields that Γ ≤ u in {R1 ≤ |x| ≤ R2 }. Let R2 → +∞, then, γ2 → 0. We obtain u(x) ≥ γ1

log(1 + |x|) , |x|α−2

that is, m1 (R)Rα−2 ≥ γ1 log(1 + R), for all R ≥ R1 . This contradicts to the fact that m1 (R)Rα−2 is bounded.

3

A Priori Bounds

In this section, we establish the L∞ bounds for solutions of (1.1). First, we show a maximum principle for cooperative systems defined on domains with small volume. Using the result and the assumption that Ω is convex, we will prove that the maximum points of solutions would not approach to the boundary of domain. Finally, we see that after blowing up, the limit function satisfies a system in RN , and by Liouville type theorem, we get a contradiction and the result follows. Proposition 3.1 Let Ω be a bounded domain in RN . Suppose that u, v are continuous ¯ in Ω and satisfy ⎧ 2 ⎪ M− ⎪ ⎨ λ,Λ (D u) + c1 (x)u + c2 (x)v ≤ 0 in Ω, 2 M− in Ω, λ,Λ (D v) + c3 (x)u + c4 (x)v ≤ 0 ⎪ ⎪ ⎩ u, v ≥ 0 on ∂Ω,

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where ci ≤ A for some A > 0, i = 1, 2, 3, 4, and c2 ≥ 0, c3 ≥ 0. If the measure of Ω is small enough, then u ≥ 0, v ≥ 0 in Ω. Proof We argue by contradiction. Suppose the conclusion does not hold, we may assume that u < 0 in some open set ω and u = 0 on ∂ω. Then, we have + − 2 M− λ,Λ (D u) + (c1 − c1 )u + c2 v ≤ 0

in ω.

So, + − + − 2 − M− λ,Λ (D u) ≤ c1 u − c2 v ≤ c1 u + c2 v

in ω.

Note that u = 0 on ∂ω, applying the Alexandroff, Bakelman Pucci estimate, see Theorem 3.6 in [2], we obtain − − 1/N sup u− ≤ Cdiam ω c+ (sup u− + sup v − ), 1 u + c2 v LN (ω) ≤ Cdiamω|ω| ω

Ω

Ω

namely, sup u− ≤ Cdiam ω |ω|1/N (sup u− + sup v − ). Ω

Ω

Ω

Reasoning in the same way, we have sup v − ≤ Cdiamω|ω|1/N (sup u− + sup v − ). Ω

Ω

Ω

The result then follows provided that |Ω| is small enough. Lemma 3.2 Suppose conditions in Theorem 1.1 hold, then, there exists ε > 0 depending only on Ω, such that, for all x ∈ Ωε := {x ∈ Ω : d(x, ∂Ω) ≤ ε}, there exists a y ∈ Ω \ Ωε such that u(y) ≥ u(x), v(y) ≥ v(x). Proof Given β ∈ R and γ ∈ S 1 , we define Σβ = {x ∈ Ω : x · γ > β},

Tβ = {x ∈ Ω : x · γ = β}.

Let xβ = 2(β − x · γ)γ + x be the reflection point of x ∈ Σβ with respect to the plane Tβ . Let wβ = u(xβ ) − u(x), zβ = v(xβ ) − v(x), then, we have 2 M− λ,Λ (D wβ ) + c1 (x)wβ + c2 (x)zβ ≤ 0,

2 M− λ,Λ (D zβ ) + c3 (x)wβ + c4 (x)zβ ≤ 0 in Σβ ,

and wβ ≥ 0, zβ ≥ 0 on ∂Σβ . Since f, g are locally Lipschitz continuous, we have that ci , i = 1, 2, 3, 4 are well-defined and bounded. Moreover, the fact that the system is cooperative implies that c2 ≥ 0, c3 ≥ 0. Applying Proposition 3.1, we get that wβ ≥ 0, zβ ≥ 0 in Σβ if the measure of Σβ is small enough. Let us now fix ε > 0 small depending only on the geometry of Ω so that for any x ∈ Ωε , there exists γ ∈ S 1 , such that x ∈ Σβ , Σβ having small measure, and xβ ∈ Ω \ Ωε , such that u(xβ ) ≥ u(x) and v(xβ ) ≥ v(x). The assertion follows. Proposition 3.3 Under the conditions in Theorem 1.1, there exists a constant C > 0, such that for solutions of (1.1), there holds

u L∞(Ω) ≤ C,

v L∞ (Ω) ≤ C.

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Proof We use blow up arguments. Suppose, by contradiction, that there exists a sequence of solutions {(un , vn )} of (1.1), such that max{ un L∞ (Ω) , vn L∞ (Ω) } → ∞ as n → ∞. We may assume, without loss of generality, that un L∞ (Ω) → ∞ and that 1/β

1/β

un L∞ 1(Ω) ≥ vn L∞ 2(Ω) for some β1 , β2 > 0 which are to be determined later. Let xn ∈ Ω be the point un assuming its maximum. Lemma 3.2 implies that xn ∈ Ω \ Ωε . Let λn > 0, such that λβn1 un L∞ (Ω) = 1, then λn → 0 as n → ∞. Let Ωn = {y ∈ RN : λn y + xn ∈ Ω} and define u1n , vn1 : Ωn → R by u1n (y) = λβn1 un (λn y + xn ),

vn1 (y) = λβn2 vn (λn y + xn ).

We observe that u1n (0) = 1, 0 ≤ u1n (y) ≤ 1, and 0 ≤ vn1 (y) ≤ 1 for every y ∈ Ωn . ¯ ε , we might assume that xn → x0 ∈ Ω \ Ωε . Then, (u1n , vn1 ) Due to the compactness of Ω satisfies 2 1 β1 +2−β1 α11 −M+ au1n (y)α11 + λβn1 +2−β2 α12 bvn1 (y)α12 λ,Λ (D un (y)) = λn −β2 1 1 1 +λβn1 +2 h1 (λ−β n un (y), λn vn (y)) in

Ωn ,

(3.1)

and 2 1 β2 +2−β1 α21 cu1n (y)α21 + λβn2 +2−β2 α22 dvn1 (y)α22 −M+ λ,Λ (D vn (y)) = λn −β2 1 1 1 +λβn2 +2 h2 (λ−β n un (y), λn vn (y)

in Ωn .

(3.2)

On the other hand, by assumption (A3), we have −β2 1 β1 +2 1 1 + λβn1 +2−β1 β11 + λβn1 +2−β2 β12 ), λβn1 +2 h1 (λ−β n un (y), λn vn (y)) ≤ C(λn

and −β2 1 β2 +2 1 1 λβn2 +2 h2 (λ−β + λβn2 +2−β1 β21 + λβn2 +2−β2 β22 ). n un (y), λn vn (y)) ≤ C(λn

Since the system is weakly coupled, we may choose β1 =

2 α11 −1 , β2

=

2 α22 −1 ,

then,

β1 + 2 − β1 α11 = 0, β1 + 2 − β2 α12 > 0, and β2 + 2 − β1 α21 > 0, β2 + 2 − β2 α22 = 0. By the assumption that βij < αij , we obtain β1 + 2 − β1 β11 > 0, β1 + 2 − β2 β12 > 0, and β2 + 2 − β1 β21 > 0, β2 + 2 − β2 β22 > 0. The elliptic regularity theory then implies that, up to a uniformly on compact sets of RN , and (u1 , v 1 ) satisfies ⎧ 2 1 1 α11 ⎪ −M+ in ⎪ λ,Λ (D u ) = a(u ) ⎪ ⎪ ⎪ ⎨ −M+ (D2 v 1 ) = d(v 1 )α22 in λ,Λ ⎪ ⎪ 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 in ⎪ ⎪ ⎪ ⎩ u(0) = 1.

subsequence, u1n → u1 , vn1 → v 1 RN , RN , RN ,

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This contradicts to Theorem 4.1 in [3]. The assertion follows. Proposition 3.4 Suppose the conditions in Theorem 1.2 hold, then, there exists a constant C > 0, such that for solutions (u, v) of (1.1) there holds u L∞ (Ω) ≤ C, v L∞ (Ω) ≤ C. Proof The proof is similar to that of Proposition 3.3. We sketch it. Choose in (3.1) and (3.2) that 2(α12 + 1) 2(α21 + 1) , β2 = . β1 = α12 α21 − 1 α12 α21 − 1 Since the system is strongly coupled, we obtain β1 + 2 − β1 α11 > 0, β1 + 2 − β2 α12 = 0, and β2 + 2 − β1 α21 = 0, β2 + 2 − β2 α22 > 0. By the assumption that βij < αij , we obtain β1 + 2 − β1 β11 > 0, β1 + 2 − β2 β12 > 0, and β2 + 2 − β1 β21 > 0, β2 + 2 − β2 β22 > 0. We might assume that

u1n

→ u1 , vn1 → v 1 uniformly on compact sets of RN and (u1 , v 1 ) satisfies ⎧ 2 1 1 α12 ⎪ −M+ in RN , ⎪ λ,Λ (D u ) = b(v ) ⎪ ⎪ ⎪ ⎨ −M+ (D2 v 1 ) = c(u1 )α21 in RN , λ,Λ ⎪ ⎪ 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 in RN , ⎪ ⎪ ⎪ ⎩ u(0) = 1.

This contradicts to Theorem 2.1. The proof is completed.

4

Proofs of Theorems 1.1 and 1.2

We prove Theorems 1.1 and 1.2 in this section. The main tool is the following lemma. Lemma 4.1 ([4], [7]) Let E be a Banach space and K ⊂ E be a closed convex cone in E, denote Kr := {u ∈ K : u < r} and ∂Kr := {u ∈ K : u = r} with r > 0. Let T : K¯r → K be a compact mapping and 0 < ρ ≤ r. If the following conditions are satisfied: (i) If T x = tx for all x ∈ ∂Kρ and all t ≥ 1, then, i(T, Kρ , K) = 1. (ii) If there exists a compact mapping H : K¯ρ × [0, ∞) → K, such that (a) H(x, 0) = T x for all x ∈ ∂Kρ , (b) H(t, x) = x for all x ∈ ∂Kρ and all t ≥ 0. ¯ ρ for all t ≥ t0 . (c) There exists a t0 > 0, such that H(t, x) = x has no solution x ∈ K Then, i(T, Kρ , K) = 0. ¯ := {u ∈ C(Ω) ¯ : u = 0 on ∂Ω}, E = C0 (Ω) ¯ × C0 (Ω), ¯ and K := {(u, v) ∈ E : Let C0 (Ω) u ≥ 0, v ≥ 0}. Define T : K → K by T (u, v) = (L(f (u, v), L(g(u, v))), where L is the inverse of 2 −M+ λ,Λ (D ·). By Proposition 2.1, Theorem 2.2 in [8] and regularity estimates, L is well-defined and compact. So, T is well-defined and compact. Solutions of (1.1) corresponds to fixed points of T .

No.1

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Proof of Theorem 1.1 We need only to verify the conditions in Lemma 4.1. First, we show that condition (i) in Lemma 4.1 is satisfied. Assume by contradiction that for all r > 0 there is a t ≥ 1, (u, v) ∈ ∂Kr , such that 2 −M+ λ,Λ (D u) =

1 f (u, v), t

2 −M+ λ,Λ (D v) =

1 g(u, v). t

+ By (A1), we have f (u, v) < μ+ 1 u, g(u, v) < μ1 v for (u, v) ∈ ∂Kr0 . Hence, + 2 −M+ λ,Λ (D u) < μ1 u,

+ 2 −M+ λ,Λ (D v) < μ1 v.

This contradicts to the maximum principle (see Theorem 2.6 in [9]). Next, we show (ii) in Lemma 4.1 is satisfied. Let H((u, v), t) := T (u + t, v + t). Obviously, 2 H((u, v), 0) = T (u, v). By (A3), we can choose small ε > 0, such that (ξ1 − ε)(ξ2 − ε) ≥ (μ+ 1) . Moreover, there exists t0 > 0, such that for all t ≥ t0 , f (u + t, v + t) ≥ (ξ1 − ε)(u + t),

g(u + t, v + t) ≥ (ξ2 − ε)(u + t).

If (u, v) is a solution of (u, v) = T (u + t, v + t), then 2 −M+ λ,Λ (D u) = f (u + t, v + t) ≥ (ξ1 − ε)(u + t)

and 2 −M+ λ,Λ (D v) = g(u + t, v + t) ≥ (ξ2 − ε)(v + t). + + 2 Since (ξ1 − ε)(ξ2 − ε) ≥ (μ+ 1 ) implies that either ξ1 − ε ≥ μ1 or ξ2 − ε ≥ μ1 . By the definition + of μ1 , we have either u + t = 0 or v + t = 0, which yields t = 0. This proves (c) and (b) in (ii) for t ≥ t0 . For the case t < t0 , we can prove as Proposition 3.1 that (u, v) is uniformly bounded. This finished the proof of (b) for t < t0 . The proof of Theorem 1.1 is then completed by Lemma 4.1. The proof of Theorem 1.2 is similar to that of Theorem 1.1, we omit it.

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