Hilbert’s projective metric and nonlinear elliptic problems

Hilbert’s projective metric and nonlinear elliptic problems

Nonlinear Analysis 71 (2009) 2576–2584 Contents lists available at ScienceDirect Nonlinear Analysis journal homepage: www.elsevier.com/locate/na Hi...

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Nonlinear Analysis 71 (2009) 2576–2584

Contents lists available at ScienceDirect

Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

Hilbert’s projective metric and nonlinear elliptic problems Min-Jei Huang a,∗ , Chao-Ya Huang a , Tzong-Mo Tsai b a

Department of Mathematics, National Tsing Hua University, Hsinchu 30043, Taiwan

b

Department of Electrical Engineering, Mingchi University of Technology, Taishan, Taipei 24301, Taiwan

article

info

Article history: Received 10 September 2007 Accepted 15 January 2009

abstract This paper deals with the eigenvalue problems for a class of positive nonlinear operators. The projective metric technique is used to prove the existence of a unique fixed point of the operator. We then show how to apply our results to the study of nonlinear elliptic problems. © 2009 Elsevier Ltd. All rights reserved.

MSC: primary 47H07 Keywords: Projective metric Eigenvalue problem Fixed point Cone Concave operator Elliptic problem Positive solution

1. Introduction In a recent paper [1], we study the eigenvalue problems for a class of positive nonlinear operators defined on a cone in a Banach space. To be precise, let K be a closed (not necessarily solid) cone in a real Banach space X , and let d be  the Hilbert projective metric in K + = K \ {0}. For a given f ∈ K + and r > 0, let Kf = x ∈ K + : d(x, f ) < ∞ and





Ef ,r = x ∈ Kf : kxk = r . We consider the eigenvalue problem Tx = λx,

x ∈ Ef ,r , λ > 0, (1.1) and the associated fixed-point equation Tx = x for compact cone mappings T : K → K which are of the form T = T1 + Tp , where Tj is increasing j-concave for j = 1, p with 0 < p < 1. Assuming T1 f ∈ Kf and Tp f ∈ Kf for some f ∈ K + , and T K + ⊆ Kf , we prove that for each r > 0, there exists a unique pair (xr , λr ) ∈ Ef ,r × (0, ∞) such that Txr = λr xr . We also prove that T has a unique fixed point in K + under the additional assumption that tf − T1 f ∈ K for some t < 1. Much of the work of [1] is motivated by results which appear in Nussbaum [2] and Zeidler [3]. We apply a theorem of Zeidler to prove the existence of solutions to the problem (1.1), and then use the techniques of Nussbaum to obtain uniqueness results. We next show that the solutions of (1.1) have some continuity and monotonicity properties which ensure that the fixed-point equation Tx = x has a unique solution. We should mention that results concerning this type of problem have been obtained in [4,5], and that other related work appears in [6–8]. The purpose of this paper is to establish some results analogous to those of [1] for compact cone mappings which are f -increasing and strictly 1-concave. To demonstrate the applicability of our abstract results, we give, in the last two sections of the paper, some simple applications to nonlinear elliptic problems and related boundary-value problems. Although the examples we have given can be generalized in several directions, we do not attempt maximum generality here.



Corresponding author. E-mail address: [email protected] (M.-J. Huang).

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.01.093

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2. Preliminaries By a closed cone K in a real Banach space(X , k·k) we mean a closed subset of X such that (i) aK + K ⊆ K for all a ≥ 0 and (ii) K ∩ (−K ) = {0}. Notice that K induces a partial ordering on X by x ≤ y if and only if y − x ∈ K . The partial ordering is clearly reflexive and transitive. We will write x < y if x ≤ y and x 6= y. For x, y ∈ K + = K \ {0}, we let M (x, y) = inf {λ : x ≤ λy}, or M (x, y) = ∞ if the set is empty, and m(x, y) = sup {µ : µy ≤ x}. The projective distance between x and y is defined by d(x, y) = log

M (x, y) m(x, y)

.

The function d is known as the Hilbert projective metric. We shall use the following properties of M, m and d in this paper. (a) (b) (c) (d)

0 ≤ m(x, y) ≤ M (x, y) ≤ ∞. m(x, y)y ≤ x ≤ M (x, y)y if M (x, y) < ∞. M (x, y)m(y, x) = 1 if M (x, y) < ∞. d(λx, µy) = d(x, y) for all λ, µ > 0. For a proof of these statements and a more detailed discussion on the projective metric, see Bushell [6]. For f ∈ K + , we define Kf = x ∈ K + : d(x, f ) < ∞ .





It is then easy to see that aKf + Kf ⊆ Kf for all a ≥ 0. We say that K is a solid cone if K˚ , the interior of K , is not empty. Notice that if K is a solid cone, and if f ∈ K˚ , then Kf = K˚ . The set Kf is particularly useful in proving the existence of a unique positive solution for certain nonlinear elliptic equations and related boundary-value problems (see [4,1]). For r > 0, we define E f ,r = x ∈ K f : k x k = r .





It is known that Ef ,r , d forms a metric space. Furthermore, if the norm of X is monotonic on K , that is, 0 ≤ x ≤ y implies

 



kxk ≤ kyk, then Ef ,r , d is a complete metric space for all r > 0 (see [4] and [5]). In this paper we always assume the norm under consideration is monotonic on K . Definition 2.1. Let T : K → K , and let f ∈ K + . We say that (a) T is increasing if 0 ≤ x ≤ y implies Tx ≤ Ty. (b) T is f -increasing if for any x, y ∈ K with x < y, there is a positive number α (x, y) depending on x and y such that Tx + α (x, y) f ≤ Ty.

Evidently, every f -increasing operator is increasing. Also, T is f -increasing if and only if T is g-increasing for any g ∈ Kf . The main consideration of this paper is the study of a class of cone mappings which satisfy a certain concavity type condition. To be precise we make the following definition. Definition 2.2. Let T : K → K , and let p ≥ 0. We say that (a) T is p-concave if T (tx) ≥ t p Tx for all x ∈ K and 0 < t < 1; or equivalently, T (sx) ≤ sp Tx for all x ∈ K and s > 1. (b) T is strictly 1-concave if T (tx) > tTx for all x ∈ K and 0 < t < 1. We note that if T is strictly 1-concave, then T is 1-concave and T (0) > 0. 3. Eigenvalue problems In this section we consider the eigenvalue problem (1.1). The following existence result is proved in [1]. Theorem 3.1. Suppose that T : K → K is a compact, increasing p-concave operator with 0 < p ≤ 1, and that T K + ⊆ Kf for some f ∈ K + . Then for each r > 0, there exist xr ∈ Ef ,r and λr > 0 such that Txr = λr xr .



We can obtain the uniqueness result if T is assumed to be f -increasing and 1-concave. Theorem 3.2. Let f ∈ K + , and let T : K → K be an f -increasing 1-concave operator so that T K + ⊆ Kf . Then



d (Tx, Ty) < d (x, y)

for all x, y ∈ Ef ,r with x 6= y.

In particular, T has at most one eigenvector in Ef ,r .

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Proof. Let x, y ∈ Ef ,r , and suppose that x 6= y. Then we have m(x, y)y < x < M (x, y)y, where m(x, y) ≤ 1 ≤ M (x, y). Since T is f -increasing and 1-concave, there are positive numbers α 0 and β 0 depending on x and y such that



Tx ≥ T (m(x, y)y) + α 0 f ≥ m(x, y)Ty + α 0 f M (x, y) Ty ≥ T (M (x, y) y) ≥ Tx + β 0 f .

Since Ty ∈ Kf , f ≥ m(f , Ty)Ty, where m(f , Ty) > 0. It follows that Tx ≥ m(x, y)Ty + α Ty M (x, y) Ty ≥ Tx + β Ty



for some α, β > 0. This gives that



m(Tx, Ty) ≥ m(x, y) + α M (Tx, Ty) ≤ M (x, y) − β.

Thus, d (Tx, Ty) ≤ log

M (x, y) − β m(x, y) + α

< log

M ( x, y ) m(x, y)

= d (x, y) .

Finally, suppose that Tx = λx and Ty = µy, where x, y ∈ Ef ,r and λ, µ > 0. We claim that x = y. For, if x 6= y, we would have d (x, y) > d (Tx, Ty) = d (λx, µy) = d (x, y) which is impossible. Hence x = y. This completes the proof of the theorem.



As an immediate consequence of Theorems 3.1 and 3.2, we obtain



Theorem 3.3. Let f ∈ K + , and let T : K → K be a compact, f -increasing 1-concave operator so that T K + ⊆ Kf . Then, for each r > 0, there exists a unique pair (xr , λr ) ∈ Ef ,r × (0, ∞) such that Txr = λr xr . Moreover, we have

 kf k  m(Tf , f ) ≤ λr ≤ M (Tf , f ) r kf k   m(Tf , f ) ≤ λr ≤ M (Tf , f ) r

for r ≤ kf k (3.1)

for r ≥ kf k .

Proof. The existence of a unique pair (xr , λr ) ∈ Ef ,r × (0, ∞) with Txr = λr xr follows immediately from Theorems 3.1 and 3.2. It remains to prove (3.1). Since m(xr , f )f ≤ xr ≤ M (xr , f )f ,

(3.2)

and since the norm is monotonic, we have m(xr , f ) ≤ r kf k ≤ M (xr , f ). Consider, first, the case r ≤ kf k. Then m(xr , f ) ≤ 1 ≤ M (xr , f )r −1 kf k, so by applying T to (3.2), we obtain −1

Txr ≥ m(xr , f )Tf ≥ m(xr , f )m(Tf , f )f and Txr ≤ M (xr , f )r −1 kf k T r kf k−1 f ≤ M (xr , f )r −1 kf k Tf



≤ M (xr , f )r −1 kf k M (Tf , f )f . Since Txr = λr xr , the first part of (3.1) follows. On the other hand, for r ≥ kf k, we have m(xr , f )r −1 kf k ≤ 1 ≤ M (xr , f ). Thus, by applying T to (3.2), we obtain Txr ≥ m(xr , f )r −1 kf k T r kf k−1 f ≥ m(xr , f )r −1 kf k Tf



≥ m(xr , f )r −1 kf k m(Tf , f )f and Txr ≤ M (xr , f )Tf ≤ M (xr , f )M (Tf , f ) f . This proves the second part of (3.1) because Txr = λr xr .



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Theorem 3.4. Under the hypotheses of Theorem 3.3, (a) the mapping r 7→ xr ∈ Kf is continuous in the norm topology. (b) the mapping r → 7 λr ∈ (0, ∞) is continuous and monotone decreasing. Proof. The proofs of Theorem 3.6 (b) and (c) in [1] carry through without change to prove that xr and λr are continuous. To prove the monotonicity of λr , we suppose that xr 6≥ xs . Then m (xr , xs ) < 1 so that

λr xr = Txr ≥ T (m (xr , xs ) xs ) ≥ m (xr , xs ) Txs = λs m (xr , xs ) xs . It follows that λr ≥ λs . Thus, λr < λs implies xr ≥ xs . Since the norm is monotonic, it follows that r ≥ s. This shows that λr is monotone decreasing in r .  Theorem 3.5. If the 1-concave in Theorem 3.3 is replaced by strictly 1-concave, then (a) λr is a strictly decreasing function of r ; and (b) limr →0+ λr = ∞. Proof. (a) By Theorem 3.4(b), λr is monotone decreasing in r . To prove (a), it suffices to show that λr is one-to-one. For this, suppose that λr = λs . We will show that r = s by proving that xr = xs . Assume first that m (xr , xs ) < 1. Then, since T is strictly 1-concave,

λr xr = Txr ≥ T (m (xr , xs ) xs ) > m (xr , xs ) Txs = λr m (xr , xs ) xs so that xr > m (xr , xs ) xs . Now, since T is f -increasing and xs ∈ Kf , T is xs -increasing. So, there exists a positive number α depending on xr and xs such that

λr xr = Txr ≥ T (m (xr , xs ) xs ) + α xs > [λr m (xr , xs ) + α ] xs . It follows that λr m (xr , xs ) ≥ λr m (xr , xs ) + α , which is impossible. Hence m (xr , xs ) ≥ 1, and this gives xr ≥ xs . By interchanging the roles of r and s, we also have xs ≥ xr . Therefore, xr = xs as required. (b) Suppose on the contrary that limr →0+ λr = λ < ∞. Then, since xr → 0 as r → 0+ and T is continuous, we have T (0) = lim Txr = lim λr xr = λ · 0 = 0. r →0+

r →0+

But this is impossible, since T is strictly 1-concave. Hence limr →0+ λr = ∞.



After these preliminaries, we can state and prove our main theorem.



Theorem 3.6. Let f ∈ K + , and let T : K → K be a compact, f -increasing strictly 1-concave operator so that T K + ⊆ Kf and M (Tf , f ) < 1. Then T has a unique fixed point in K + . Proof. From Theorem 3.5, (3.1) and the hypothesis M (Tf , f ) < 1, we see that lim λr = ∞ and

r →0+

lim λr < 1.

r →∞

Since λr is continuous and strictly decreasing in r , there is a unique r with λr = 1. Thus, the corresponding xr is a fixed point  of T . The uniqueness of the fixed point follows directly from the hypothesis T K + ⊆ Kf and Theorem 3.3.  4. Applications to nonlinear boundary-value problems We now illustrate how the results of Section 3 can be used in the study of certain nonlinear boundary-value problems. We consider the question of existence and uniqueness of positive solutions to the boundary-value problem x00 (t ) + a(t )ϕ (x(t )) = 0,

0 < t < 1;

(4.1)

α x(0) − β x (0) = γ x(1) + δ x (1) = 0,

(4.2)

0

0

where a(t ) is positive and continuous for 0 ≤ t ≤ 1, α, β, γ , δ ≥ 0 and ρ ≡ αγ + αδ + βγ > 0. Eq. (4.1) with ϕ (x) = xp , p > 0, is known as the generalized Emden-Fowler equation. It has received considerable attention because of its importance in gas dynamics, fluid mechanics, nuclear physics and chemically reacting systems. The cases 0 < p < 1 and p > 1 are called sublinear and superlinear, respectively. For a general survey on the generalized Emden-Fowler equation, see Wong [9]. Our purpose here is to give an existence and uniqueness result for positive solutions of (4.1)–(4.2) under hypotheses different from those used in [10,1,9]. To apply Theorem 3.6 to the boundary-value problem (4.1)–(4.2), we proceed as follows. Let X = C [0, 1] be the Banach space of continuous real-valued functions defined on [0, 1] with the norm kxk = sup {|x(t )| : 0 ≤ t ≤ 1}, and let K be the cone of nonnegative functions in X . It is evident that the norm of X is monotonic on

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K . By an elementary computation, we see that solving (4.1)– (4.2) is equivalent to finding a continuous function x(t ) such that x(t ) =

1

Z

G(t , s)a(s)ϕ (x(s)) ds,

(4.3)

0

where G(t , s) is the Green’s function G(t , s) =

 (β + α t ) (δ + γ − γ s) /ρ (β + α s) (δ + γ − γ t ) /ρ

for 0 ≤ t ≤ s ≤ 1 for 0 ≤ s ≤ t ≤ 1.

We note that G(t , s) is positive on (0, 1) × (0, 1). Concerning ϕ we make the following assumptions. (A1) ϕ(x) is positive, continuous and strictly increasing for x ≥ 0. (A2) ϕ(λx) > λϕ(x) for 0 < λ < 1 and x ≥ 0. Simple examples of such ϕ are given by ϕ(x) = log (x + q) with q > 1 or ϕ(x) = (1 + xn )1/n with n > 0. Notice that (A2) implies in particular that ϕ(x)/x is a strictly decreasing function of x, and so limx→∞ [ϕ(x)/x] exists. As an application of Theorem 3.6, we have Theorem 4.1. Let the assumptions (A1) and (A2) hold. If lim

ϕ(x) x

x→∞

8ρ 2

<

 , (γ + 2δ) 4βρ + α 2 (γ + 2δ) kak 

(4.4)

then the boundary-value problem (4.1)–(4.2) has a unique positive solution. Proof. Let T : K → K be defined by 1

Z

(Tx) (t ) =

G(t , s)a(s)ϕ (x(s)) ds 0

so that (4.3) is now x = Tx. From the assumptions (A1) and (A2), we see that T is compact and strictly 1-concave. Let f (t ) = 8k

1

Z

G(t , s)ds = 0

4k 

ρ

 β (γ + 2δ) + α (γ + 2δ) t − ρ t 2 ,

where k > 0 is a constant. Then f (t ) is positive for 0 < t < 1 and satisfies the boundary conditions (4.2). We claim that T is f -increasing. To prove this, let x, y ∈ K with x > y. Then

(Tx − Ty) (t ) =

1

Z

G(t , s)a(s) [ϕ (x(s)) − ϕ (y(s))] ds 0

is positive for 0 < t < 1, by assumption (A1). We shall prove that Tx − Ty ∈ Kf . We consider four cases as follows. (a) f (0) > 0. In this case, β (γ + 2δ) > 0 so that

(Tx − Ty) (0) =

β ρ

1

Z

(δ + γ − γ s) a(s) [ϕ (x(s)) − ϕ (y(s))] ds > 0. 0

(b) f (1) > 0. In this case, δ (α + 2β) > 0 so that

(Tx − Ty) (1) =

δ ρ

1

Z

(β + α s) a(s) [ϕ (x(s)) − ϕ (y(s))] ds > 0. 0

(c) f (0) = 0. In this case, β (γ + 2δ) = 0. Since ρ > 0, we must have β = 0 and so ρ = α (γ + δ). It follows that (Tx − Ty) (0) = 0. We compute Z Z t α 1 a(s) [ϕ (x(s)) − ϕ (y(s))] ds (Tx − Ty)0 (t ) = (δ + γ − γ s) a(s) [ϕ (x(s)) − ϕ (y(s))] ds − ρ 0 0 and f 0 (t ) =

4k

ρ

[α (γ + 2δ) − 2ρ t] .

Hence, by l’Hôpital’s rule, we have lim

t →0+

(Tx − Ty) (t ) (Tx − Ty)0 (t ) = lim f (t ) f 0 (t ) t →0+ Z 1 1 = (δ + γ − γ s) a(s) [ϕ (x(s)) − ϕ (y(s))] ds 4k (γ + 2δ) 0 > 0.

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(d) f (1) = 0. In this case, δ (α + 2β) = 0. Since ρ > 0, we must have δ = 0 and so ρ = γ (α + β). It follows that (Tx − Ty) (1) = 0. Hence, lim

t →1−

(Tx − Ty) (t ) (Tx − Ty)0 (t ) = lim f (t ) f 0 (t ) t →1− Z 1 1 = (β + α s) a(s) [ϕ (x(s)) − ϕ (y(s))] ds 4k (α + 2β) 0 > 0.

All this proves that Tx − Ty ∈ Kf , so T is f -increasing. From this we also deduce that T K + ⊆ Kf because T (0) ∈ Kf and Kf + Kf ⊆ Kf . By Theorem 3.6 and our previous remarks, the final stage of the proof is to show that M (Tf , f ) < 1. This will be true if k is large enough. Note that



 kf k = f

α (γ + 2δ) 2ρ



= kη,

where

η=

1

ρ2

  (γ + 2δ) 4βρ + α 2 (γ + 2δ) .

(4.5)

We compute

(Tf ) (t ) =

1

Z

G(t , s)a(s)ϕ (f (s)) ds 0

≤ kak ϕ (kf k)

1

Z

G(t , s)ds 0

=

k ak 8k

ϕ (kη) f (t ).

Thus, M (Tf , f ) ≤

k ak 8k

ϕ (kη) .

(4.6)

From the hypothesis (4.4), we see that ϕ (kη) < 8k/ kak for k sufficiently large, so M (Tf , f ) < 1 by (4.6). This completes the proof of the theorem.  We remark that the conditions of Theorem 4.1 are satisfied if ϕ(x) = log (x + q) with q > 1 or if ϕ(x) = (1 + xn )1/n and kak < 8/η, where n > 0 and η is given by (4.5). We should mention that in [11], Bandle and Kwong study the more general equation x00 (t ) + F (t , x(t )) = 0,

0 < t0 < t < t1

(4.7)

subject to one of the following sets of boundary conditions: x(t0 ) = x(t1 ) = 0.

(4.8)

x (t0 ) = x(t1 ) = 0.

(4.9)

x(t0 ) = x (t1 ) = 0.

(4.10)

0

0

By using shooting techniques, they prove that if the reaction term F (t , x) satisfies (F1) F (t , x) is continuous in t and continuously differentiable in x; F (t ,x) (F2) lim x = ∞ uniformly on every [t0 , t1 ]; x→∞ F (t ,x) x x→0

(F3) lim

≤ 0 uniformly on every [t0 , t1 ];

then all three boundary-value problems (4.7)–((4.8)/ (4.9)/ (4.10) have positive solutions for any given t0 and t1 . The conditions (F1)–(F3) for existence are quite different from our (A1), (A2) and (4.4) for uniqueness. In our case, F (t , x) = a(t )ϕ(x), where a(t ) is positive so (A1) and (A2) imply that lim

x→∞

F ( t , x) x

< ∞ and

lim

x→0

F (t , x) x

= ∞ for all t .

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5. Applications to nonlinear elliptic problems Let Ω be the open unit ball in Rn , n ≥ 3, with center the origin. We consider the positive solutions of the Dirichlet problem



∆u + a (|x|) ϕ (u) = 0 in Ω ; u = 0 on ∂ Ω .

(5.1)

When a (|x|) = 1 and ϕ (u) = up , it is well known that (5.1) has no positive solution if p ≥ (n + 2) / (n − 2), and that the positive solution of (5.1) is unique if p > 1 (Gidas-Ni-Nirenberg [12], Ni-Nussbaum [13]). In the case when 0 < p < 1, Bushell [4] proves that (5.1) has a unique positive radial solution. Here we assume that a(r ) is positive and continuous for 0 ≤ r ≤ 1, and that ϕ (u) satisfies the assumptions (A1) and (A2) of Section 4. Our result is as follows. Theorem 5.1. The problem (5.1) has a unique positive radial solution if lim

u→∞

ϕ(u) u

<

2n

k ak

.

(5.2)

Proof. Assuming solutions to be functions of r, the radial distance from the origin, (5.1) reduces to n−1 0 u (r ) + a(r )ϕ (u(r )) = 0, r u(1) = 0.

(

u00 (r ) +

0 < r < 1;

(5.3)

The Green’s function for the problem (5.3) is

  1    s − sn−1 for 0 ≤ r ≤ s ≤ 1 n − 2 G(r , s) =   1   s (s/r )n−2 − sn−1 for 0 ≤ s ≤ r ≤ 1, n−2 which is positive on [0, 1) × (0, 1). Solving (5.3) is equivalent to finding a continuous function u such that u = Tu, where Z 1 Tu ( r ) = G(r , s)a(s)ϕ (u(s)) ds. (5.4) ( ) 0

Now let K denote the cone of nonnegative functions in C [0, 1], and define T : K → K by (5.4). By our assumptions on a and ϕ, T is compact and strictly 1-concave. In order to apply Theorem 3.6 to the fixed-point equation Tu = u, we shall take f (r ) = 2nk

1

Z

G(r , s)ds = k 1 − r 2 ,



0

where k > 0 is a constant. We next prove that T is f -increasing. For this, let x, y ∈ K with x > y. Then

(Tx − Ty) (r ) =

1

Z

G(r , s)a(s) [ϕ (x(s)) − ϕ (y(s))] ds 0

is positive for 0 ≤ r < 1, by assumption (A1), and

(Tx − Ty) (1) = f (1) = 0. A calculation gives

(Tx − Ty) (r ) = −r 0

1 −n

r

Z

sn−1 a(s) [ϕ (x(s)) − ϕ (y(s))] ds 0

so that lim

r →1−

(Tx − Ty) (r ) (Tx − Ty)0 (r ) = lim f (r ) f 0 (r ) r →1− Z 1 1 = sn−1 a(s) [ϕ (x(s)) − ϕ (y(s))] ds 2k

0

> 0. 

This shows that Tx − Ty ∈ Kf , so T is f -increasing. The fact that T K + ⊆ Kf is proved in a similar way.

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It remains to verify that M (Tf , f ) < 1. We compute

(Tf ) (r ) =

1

Z

G(r , s)a(s)ϕ (f (s)) ds 0

≤ kak ϕ (kf k)

1

Z

G(r , s)ds 0

=

k ak 2nk

ϕ (k) f (r ).

So, M (Tf , f ) ≤

k ak 2nk

ϕ (k) .

From this and the condition (5.2), we conclude that M (Tf , f ) < 1 if k is large enough. The proof of the theorem is now complete.  We remark that the Green’s function for the problem (5.3) in the case n = 2 dimensions is given by G(r , s) =

for 0 < r ≤ s ≤ 1 for 0 < s ≤ r ≤ 1.

 −s log s −s log r

The proof of Theorem 5.1 carries through without change to prove that Theorem 5.1 remains true in the case n = 2. There is an extensive literature concerning positive radial solutions of the problem (5.1) in annular domains, see, e.g. [14, 11,15,16,13,17] and references therein. In the remainder of this section, we want to show how our results can be applied to this type of problem. Let Ω = {x ∈ Rn : 0 < R1 < |x| < R2 } be an annulus in Rn , n ≥ 2. Suppose that a(r ) is positive and continuous for R1 ≤ r ≤ R2 , and that ϕ (u) satisfies the assumptions (A1) and (A2). We consider the positive radial solutions of the equation

∆u + a (|x|) ϕ (u) = 0 in Ω

(5.5)

subject to the boundary conditions

 ∂u  α u + β = 0 on |x| = R1 ∂r (5.6)   γ u + δ ∂ u = 0 on |x| = R2 , ∂r where α, β, γ , δ ≥ 0 with αγ + αδ + βγ > 0, r = |x| and ∂/∂ r denotes differentiation in the outward radial direction. Assuming solutions to be functions of r, (5.5)–(5.6) reduces to n−1 0 u (r ) + a (r ) ϕ (u (r )) = 0, R1 < r < R2 ; r α u (R1 ) − β u0 (R1 ) = γ u (R2 ) + δ u0 (R2 ) = 0.

(

u00 (r ) +

(5.7)

Now we apply to (5.7) the transformation r

Z

θ 1−n dθ ,

s=

v (s) = u (r (s)) .

R1

The transformed equation is



v 00 (s) + r (s)2n−2 a (r (s)) ϕ (v (s)) = 0, 0 < s < R; αv (0) − β R11−n v 0 (0) = γ v (R) + δ R12−n v 0 (R) = 0,

(5.8)

where

Z

R2

R= R1

θ 1−n dθ =

log (R2 /R1 )  R21−n − R22−n / (n − 2)



for n = 2 for n ≥ 3.

In order to apply Theorem 4.1 to our problem, we need to replace the s-interval [0, R] by the unit interval. For this, we let t = s/R and y(t ) = v(s). Then (5.8) is transformed into



y00 (t ) + r (Rt )2n−2 R2 a (r (Rt )) ϕ (y (t )) = 0, 0 < t < 1; α y (0) − β R11−n R−1 y0 (0) = γ y (1) + δ R12−n R−1 y0 (1) = 0.

(5.9)

To simplify the notation, we now set

˜ . β˜ = β R11−n R−1 , δ˜ = δ R12−n R−1 , ρ˜ = αγ + α δ˜ + βγ 2n − 2 2n − 2 Notice that r (Rt ) R2 a (r (Rt )) ≤ R R2 kak for 0 ≤ t ≤ 1. Applying Theorem 4.1, we have proved the following 2

result.

2584

M.-J. Huang et al. / Nonlinear Analysis 71 (2009) 2576–2584

Theorem 5.2. The problem (5.5)–(5.6) has a unique positive radial solution if lim

u→∞

ϕ(u) u

< 

8ρ˜ 2

γ + 2δ˜

h



4β˜ ρ˜ + α 2 γ + 2δ˜

i

−2 2 R2n R k ak 2

.

(5.10)

Finally, we note that for the Dirichlet boundary conditions: α = γ = 1 and β = δ = 0, the RHS of (5.10) = 8/R22n−2 R2 kak; while, for the Dirichlet/Neumann (Neumann/Dirichlet) boundary conditions: α = δ = 1 and β = γ = 0 −2 2 (β = γ = 1 and α = δ = 0), the RHS of (5.10) = 2/R2n R k ak . 2 References [1] M.-J. Huang, C.-Y. Huang, T.-M. Tsai, Eigenvalue problems and fixed point theorems for a class of positive nonlinear operators, Math. Z. 257 (2007) 581–595. [2] R.D. Nussbaum, Hilbert’s projective metric and iterated nonlinear maps, Mem. Amer. Math. Soc. 75 (1988) 391. [3] E. Zeidler, Nonlinear Functional Analysis and its Applications I: Fixed-Point Theorems, Springer-Verlag, New York, 1985. [4] P.J. Bushell, The Cayley–Hilbert metric and positive operators, Linear Algebr. Appl. 84 (1986) 271–280. [5] M.-J. Huang, C.-Y. Huang, T.-M. Tsai, Applications of Hilbert’s projective metric to a class of positive nonlinear operators, Linear Algebr. Appl. 413 (2006) 202–211. [6] P.J. Bushell, Hilbert’s metric and positive contraction mappings in a Banach space, Arch. Ration. Mech. Anal. 52 (1973) 330–338. [7] A.J.B. Potter, Hilbert’s projective metric applied to a class of positive operators, ordinary and partial differential equations, Lect. Notes Math. 564 (1976) 377–382. [8] A.J.B. Potter, Applications of Hilbert’s projective metric to certain classes of non-homogeneous operators, Quart. J. Math. Oxford 28 (1977) 93–99. [9] J. Wong, On the generalized Emden–Fowler equation, SIAM Rev. 17 (1975) 339–360. [10] L.H. Erbe, H. Wang, On the existence of positive solutions of ordinary differential equations, Proc. Amer. Math. Soc. 120 (1994) 743–748. [11] C. Bandle, M.K. Kwong, Semilinear elliptic problems in annular domains, Z. Angew. Math. Phys. 40 (1989) 245–257. [12] B. Gidas, W.-M. Ni, L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phys. 68 (1979) 209–243. [13] W.-M. Ni, R.D. Nussbaum, Uniqueness and nonuniqueness for positive radial solutions of ∆u + f (u, r ) = 0, Comm. Pure Appl. Math. 38 (1985) 67–108. [14] C. Bandle, C.V. Coffman, M. Marcus, Nonlinear elliptic problems in annular domains, J. Differential Equations 69 (1987) 322–345. [15] C.V. Coffman, M. Marcus, Existence and uniqueness results for semilinear Dirichlet problems in annuli, Arch. Ration. Mech. Anal. 108 (1989) 293–307. [16] X. Garaizar, Existence of positive radial solutions for semilinear elliptic equations in the annulus, J. Differential Equations 70 (1987) 69–92. [17] H. Wang, On the existence of positive solutions for semilinear elliptic equations in the annulus, J. Differential Equations 109 (1994) 1–7.