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Fixed point theorem of Leggett–Williams type and its application
J. Math. Anal. Appl. 299 (2004) 254–260 www.elsevier.com/locate/jmaa
Fixed point theorem of Leggett–Williams type and its application Mirosława Zima Institute of Mathematics, University of Rzeszów, Rejtana 16 A, 35-310 Rzeszów, Poland Received 8 July 2003 Available online 18 August 2004 Submitted by T. Krisztin
Abstract We establish a norm-type cone expansion and compression fixed point theorem for completely continuous operators. Our theorem is then applied to prove the existence of positive solution of second order three-point boundary value problem. 2004 Elsevier Inc. All rights reserved. Keywords: Positive solution; Fixed point theorem; Cone
1. Introduction One of the most frequently used tools for proving the existence of positive solutions to the integral equations and boundary value problems is Krasnoselskii’s theorem on cone expansion and compression and its norm-type version due to Guo (see [6]). For applications of Krasnoselskii’s theorem to various problems see for example [2,4,5,10,12,13] and the references therein. In [9] Leggett and Williams obtained a generalization of Krasnoselskii’s original result and applied their fixed point theorem to the nonlinear equation modelling certain infectious diseases. In the present paper we state and prove a fixed point theorem of
E-mail address: [email protected]. 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.07.002
M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
255
Leggett–Williams type and give an application of our result to the following second order three-point boundary value problem: x (t) + f (t, x(t)) = 0, (1) x(0) = 0, αx(η) = x(1), where t ∈ [0, 1], f is a continuous function, η ∈ (0, 1), α 0, and 1 − αη > 0.
2. Fixed point theorem We begin this section with reviewing some facts on cone theory in Banach spaces. A nonempty subset P of a Banach space E is called a cone if P is convex, closed and (i) µx ∈ P for all x ∈ P and µ 0, (ii) x, −x ∈ P implies x = θ . It is known that every cone P induces a partial order in E by xy
if and only if
y − x ∈ P.
We will write x y if y − x ∈ / P and x ≺ y if y − x ∈ P \ {θ }. We say that P is a normal cone in E if there exists a positive number γ such that θ xy
implies x γ y.
(2)
The smallest γ satisfying (2) is called the normal constant of P . Clearly, for any normal cone we have γ 1. For u0 ∈ P \ {θ } let P (u0 ) = {x ∈ P : λu0 x for some λ > 0}. Moreover, for R > 0 put P¯R = {x ∈ P : x R}. In 1980, Leggett and Williams proved the following generalization of Krasnoselskii’s theorem (see [9, Theorem 2]). Proposition 1. Let F : P¯R → P be a completely continuous operator with F (θ ) = θ . Suppose there exist r, 0 < r < R, and u0 ∈ P \ {θ } such that F x x
if x ∈ P (u0 ) and x = r.
Suppose further that for each ε > 0 (1 + ε)x F x
if x ∈ P and x = R.
Then F has a fixed point in P with r x R. In [1] and [6] one can find some refinements of Proposition 1. Here we recall the result from [1]. Proposition 2 [1, Theorem 1.3]. Let P be a cone in a Banach space E and r1 , r2 > 0, r1 = r2 , with R = max{r1 , r2 }, r = min{r1 , r2 }. Let F : P¯R → P be a completely continuous operator such that: (iii) x F x for x ∈ P and x = r1 ,
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M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
(iv) there exists u0 ∈ P \ {θ } such that F x x for x ∈ P (u0 ) and x = r2 . Then F has at least one fixed point x ∗ ∈ P with r x ∗ R. Notice that the assumption (iii) can be replaced by (v) F x x for x ∈ P and x = r1 (see, for example, [6]). In this paper we are interested in replacing (iv) by the condition of the norm type, namely x F x for x ∈ P (u0 ) and x = r2 . Remark. In [3], Anderson and Avery obtained a generalization of Krasnoselskii’s fixed point theorem of the norm type. They replaced the norm type assumptions by conditions formulated in the terms of two functionals. In our considerations we will use the following theorem on fixed point index for completely continuous operators in cones. Proposition 3 [6, Lemma 2.3.2]. Let Ω be a bounded open set in a Banach space E and P ⊂ E be a cone. Let A : P ∩ Ω¯ → P and B : P ∩ ∂Ω → P be completely continuous operators. Suppose that (v) infx∈P ∩∂Ω Bx > 0, (vi) x − Ax = tBx for x ∈ P ∩ ∂Ω and t 0. Then iP (A, P ∩ Ω) = 0, where iP (A, P ∩ Ω) denotes the fixed point index of A on P ∩ Ω with respect to P . For the details of the fixed point index theory in cones see, for example, [6]. Now we can state and prove our main result. Theorem 1. Let P be a normal cone in E and γ be the normal constant of P . Assume that Ω1 and Ω2 are bounded open sets in E such that θ ∈ Ω1 and Ω¯ 1 ⊂ Ω2 . Let F : P ∩ (Ω¯ 2 \ Ω1 ) → P be a completely continuous operator and u0 ∈ P \ {θ }. If either (vii) γ x F x for x ∈ P (u0 ) ∩ ∂Ω1 and F x x for x ∈ P ∩ ∂Ω2 , or (viii) F x x for x ∈ P ∩ ∂Ω1 and γ x F x for x ∈ P (u0 ) ∩ ∂Ω2 is satisfied, then F has a fixed point in the set P ∩ (Ω¯ 2 \ Ω1 ). Proof. We can assume that F has no fixed points on P ∩ ∂Ω1 and P ∩ ∂Ω2 . Suppose that (vii) is satisfied. It is known that if F x x for x ∈ P ∩ ∂Ω2 , then iP (F, P ∩ Ω2 ) = 1
(3)
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257
(see [6]). Next, we will prove that for t 0 and x ∈ P ∩ ∂Ω1 , x − F x = t (u0 + F x).
(4)
Suppose, for the contrary, that there exist x0 ∈ P ∩ ∂Ω1 and t0 > 0 such that x0 − F x0 = t0 (u0 + F x0 ).
(5)
Observe that t0 u0 t0 (u0 + F x0 ) + F x0 . This, together with (5), gives x0 ∈ P (u0 ). Moreover, from (5) we get θ x0 − F x0 − t0 F x0 . Hence θ (1 + t0 )F x0 x0 . Since P is a normal cone, we obtain (1 + t0 )F x0 γ x0 , and, in consequence, F x0 < γ x0 , which contradicts (vii). Furthermore, for x ∈ P ∩ ∂Ω1 , θ ≺ u0 u0 + F x. Hence 0 < u0 γ u0 + F x, because the cone P is normal. Therefore inf
x∈P ∩∂Ω1
u0 + F x > 0.
(6)
Taking A = F for x ∈ P ∩ Ω1 and Bx = u0 + F x for x ∈ P ∩ ∂Ω1 in Proposition 3, we get in view of (4) and (6), iP (F, P ∩ Ω1 ) = 0.
(7)
By (3), (7), and the additivity of the fixed point index we obtain iP F, P ∩ (Ω2 \ Ω¯ 1 ) = 1, which implies that F has a fixed point in the set P ∩ (Ω2 \ Ω¯ 1 ). If (viii) holds the proof is similar. 2
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M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
3. Existence of positive solution of (1) In this section we will study the existence of positive solution of (1). Similar problems have been considered for example in [7,8,11]. Particularly, in [8] Infante and Webb proved the existence of multiple nonzero (not necessarily positive) solutions, when either 0 α < 1 − η or α < 0. It is easy to verify that the kernel of the integral equation related to (1) is given by 1−α 1 − 1−αη s t, 0 t s 1, s η, 1−α 1 − 1−αη t s, 0 s t 1, (8) k(t, s) = 1−s t, 0 t s 1, 1−αη η s. s − s−αη t, 0 s t 1, 1−αη Observe that for η ∈ (0, 1), α 0 and 1 − αη > 0 we have
Mk(s, s) k(t, s) 0
(9)
t,s∈[0,1]
and
k(t, s) m(t)k(s, s),
(10)
t,s∈[0,1]
where
α(1 − η) M = max 1, and m(t) = min{t, 1 − t} 1 − αη
for t ∈ [0, 1]. Suppose that: (1◦ ) f : [0, 1] × [0, ∞) → [0, ∞) is a continuous function and there exists r1 > 0 such that 1 −1 f (t, x) M k(s, s) ds r1 0
for t ∈ [0, 1] and x ∈ [0, r1 ]; (2◦ ) there exist t0 ∈ (0, 1], a > 0, r2 > 0, r2 = r1 , and the continuous functions g : [0, 1] → [0, ∞), h : (0, r2 ] → [0, ∞) such that f (t, x) g(t)h(x) for t ∈ [0, 1] and x ∈ (0, r2 ], h(x)/x a is nonincreasing on (0, r2 ] and h(r2 ) Ma
1 k(t0 , s)g(s)ma (s) ds r2 . 0
Theorem 2. If the assumptions (1◦ ) and (2◦ ) are satisfied, then the problem (1) has a positive solution on [0, 1].
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Proof. Consider the Banach space C[0, 1] and the set m(t) x, t ∈ [0, 1] . P = x ∈ C[0, 1]: x(t) M It is easy to show that P is a normal cone in C[0, 1] (with γ = 1). Obviously, for x, y ∈ C[0, 1], x y if and only if x(t) y(t) for every t ∈ [0, 1]. Fix u0 (t) ≡ 1 on [0, 1]. Then m(t) x, t ∈ [0, 1] . P (u0 ) = x ∈ C[0, 1]: x(t) > M For x ∈ P and t ∈ [0, 1] define the integral operator
1 (F x)(t) =
k(t, s)f s, x(s) ds,
0
where the kernel k is given by (8). Clearly, every fixed point of F is a solution of the problem (1). For t ∈ [0, 1] and x ∈ P we have in view of (9) and (10)
1
1 k(s, s)f s, x(s) ds F x M
0
and
1 (F x)(t) m(t)
k(s, s)f s, x(s) ds.
0
Thus for every t ∈ [0, 1], m(t) F x, M which means that F : P → P . Let Ω1 = {x ∈ C[0, 1]: x < r1 } and Ω2 = {x ∈ C[0, 1]: x < r2 }. We can assume that r1 < r2 . Clearly F is a completely continuous operator on P ∩ Ω¯ 2 . By (1◦ ) and (9), for x ∈ P ∩ ∂Ω1 and t ∈ [0, 1], we get 1 −1
1 (F x)(t)
(F x)(t) M
k(s, s) M 0
r1 ds = r1 .
k(τ, τ ) dτ 0
Hence F x x for x ∈ P ∩ ∂Ω1 . Let x ∈ P (u0 ) ∩ ∂Ω2 . Then x = r2 and x(t) ∈ (0, r2 ] for every t ∈ [0, 1]. By (2◦ ), we obtain
1 (F x)(t0 )
k(t0 , s)g(s)h x(s) ds =
0
h(r2 ) a r2
1 k(t0 , s)g(s)
h(x(s)) a x (s) ds x a (s)
0
1
h(r2 ) k(t0 , s)g(s)x (s) ds > a r2
0
1
a
k(t0 , s)g(s) 0
m(s)x M
a ds
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M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
h(r2 ) = Ma
1 k(t0 , s)g(s)ma (s) ds r2 . 0
This gives F x x for x ∈ P (u0 ) ∩ ∂Ω2 . By Theorem 1, F has a fixed point x ∗ ∈ P such that r1 x ∗ r2 . This ends the proof of Theorem 2. 2 Example. Consider the problem (1) with 2 2 f (t, x) = 16(t + 1)x (x − 3) , (t, x) ∈ [0, 1] × [0, 3], (t + 1)(x − 3), (t, x) ∈ [0, 1] × (3, ∞) and α = η = 1/2. It is easy to verify that the assumptions of Theorem 2 are satisfied for r1 = 1/60, r2 = 1, t0 = 1, a = 2, g(t) = 16(t + 1), and h(x) = x 2 (x − 3)2 . Therefore the problem (1) has a positive solution x ∗ defined on [0, 1] such that 1/60 x ∗ 1.
Acknowledgment The author expresses her thanks to the referee for helpful comments and for drawing the author’s attention to the paper [3].
References [1] R.P. Agarwal, D. O’Regan, Periodic solutions to nonlinear integral equations on the infinite interval modelling infectious disease, Nonlinear Anal. 40 (2000) 21–35. [2] R.P. Agarwal, D. O’Regan, P.J.Y. Wong, Positive Solutions of Differential, Difference and Integral Equations, Kluwer Academic, Dordrecht, 1999. [3] D.R. Anderson, R.I. Avery, Fixed point theorem of cone expansion and compression of functional type, J. Differ. Equations Appl. 8 (2002) 1073–1083. [4] D.R. Anderson, J.M. Davis, Multiple solutions and eigenvalues for third-order right focal boundary value problems, J. Math. Anal. Appl. 267 (2002) 135–157. [5] L.H. Erbe, H. Wang, On the existence of positive solutions of ordinary differential equations, Proc. Amer. Math. Soc. 120 (1994) 743–748. [6] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, San Diego, 1988. [7] C.P. Gupta, S.I. Trofimchuk, Existence of a solution of a three-point boundary value problem and the spectral radius of a related linear operator, Nonlinear Anal. 34 (1998) 489–507. [8] G. Infante, J.R.L. Webb, Nonzero solutions of Hammerstein integral equations with discontinuous kernels, J. Math. Anal. Appl. 272 (2002) 30–42. [9] R.W. Leggett, L.R. Williams, A fixed point theorem with application to an infectious disease model, J. Math. Anal. Appl. 76 (1980) 91–97. [10] W.C. Lian, F.H. Wong, C.C. Yeh, On the existence of positive solutions of nonlinear second order differential equations, Proc. Amer. Math. Soc. 124 (1996) 1117–1126. [11] R. Ma, Existence theorems for a second order three-point boundary value problem, J. Math. Anal. Appl. 212 (1997) 430–442. [12] M. Meehan, D. O’Regan, Positive Lp solutions of Hammerstein integral equations, Arch. Math. 76 (2001) 366–376. [13] M. Zima, On positive solutions of boundary value problems on the half-line, J. Math. Anal. Appl. 259 (2001) 127–136.
Fixed point theorem of Leggett–Williams type and its application Mirosława Zima Institute of Mathematics, University of Rzeszów, Rejtana 16 A, 35-310 Rzeszów, Poland Received 8 July 2003 Available online 18 August 2004 Submitted by T. Krisztin
Abstract We establish a norm-type cone expansion and compression fixed point theorem for completely continuous operators. Our theorem is then applied to prove the existence of positive solution of second order three-point boundary value problem. 2004 Elsevier Inc. All rights reserved. Keywords: Positive solution; Fixed point theorem; Cone
1. Introduction One of the most frequently used tools for proving the existence of positive solutions to the integral equations and boundary value problems is Krasnoselskii’s theorem on cone expansion and compression and its norm-type version due to Guo (see [6]). For applications of Krasnoselskii’s theorem to various problems see for example [2,4,5,10,12,13] and the references therein. In [9] Leggett and Williams obtained a generalization of Krasnoselskii’s original result and applied their fixed point theorem to the nonlinear equation modelling certain infectious diseases. In the present paper we state and prove a fixed point theorem of
E-mail address: [email protected]. 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.07.002
M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
255
Leggett–Williams type and give an application of our result to the following second order three-point boundary value problem: x (t) + f (t, x(t)) = 0, (1) x(0) = 0, αx(η) = x(1), where t ∈ [0, 1], f is a continuous function, η ∈ (0, 1), α 0, and 1 − αη > 0.
2. Fixed point theorem We begin this section with reviewing some facts on cone theory in Banach spaces. A nonempty subset P of a Banach space E is called a cone if P is convex, closed and (i) µx ∈ P for all x ∈ P and µ 0, (ii) x, −x ∈ P implies x = θ . It is known that every cone P induces a partial order in E by xy
if and only if
y − x ∈ P.
We will write x y if y − x ∈ / P and x ≺ y if y − x ∈ P \ {θ }. We say that P is a normal cone in E if there exists a positive number γ such that θ xy
implies x γ y.
(2)
The smallest γ satisfying (2) is called the normal constant of P . Clearly, for any normal cone we have γ 1. For u0 ∈ P \ {θ } let P (u0 ) = {x ∈ P : λu0 x for some λ > 0}. Moreover, for R > 0 put P¯R = {x ∈ P : x R}. In 1980, Leggett and Williams proved the following generalization of Krasnoselskii’s theorem (see [9, Theorem 2]). Proposition 1. Let F : P¯R → P be a completely continuous operator with F (θ ) = θ . Suppose there exist r, 0 < r < R, and u0 ∈ P \ {θ } such that F x x
if x ∈ P (u0 ) and x = r.
Suppose further that for each ε > 0 (1 + ε)x F x
if x ∈ P and x = R.
Then F has a fixed point in P with r x R. In [1] and [6] one can find some refinements of Proposition 1. Here we recall the result from [1]. Proposition 2 [1, Theorem 1.3]. Let P be a cone in a Banach space E and r1 , r2 > 0, r1 = r2 , with R = max{r1 , r2 }, r = min{r1 , r2 }. Let F : P¯R → P be a completely continuous operator such that: (iii) x F x for x ∈ P and x = r1 ,
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M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
(iv) there exists u0 ∈ P \ {θ } such that F x x for x ∈ P (u0 ) and x = r2 . Then F has at least one fixed point x ∗ ∈ P with r x ∗ R. Notice that the assumption (iii) can be replaced by (v) F x x for x ∈ P and x = r1 (see, for example, [6]). In this paper we are interested in replacing (iv) by the condition of the norm type, namely x F x for x ∈ P (u0 ) and x = r2 . Remark. In [3], Anderson and Avery obtained a generalization of Krasnoselskii’s fixed point theorem of the norm type. They replaced the norm type assumptions by conditions formulated in the terms of two functionals. In our considerations we will use the following theorem on fixed point index for completely continuous operators in cones. Proposition 3 [6, Lemma 2.3.2]. Let Ω be a bounded open set in a Banach space E and P ⊂ E be a cone. Let A : P ∩ Ω¯ → P and B : P ∩ ∂Ω → P be completely continuous operators. Suppose that (v) infx∈P ∩∂Ω Bx > 0, (vi) x − Ax = tBx for x ∈ P ∩ ∂Ω and t 0. Then iP (A, P ∩ Ω) = 0, where iP (A, P ∩ Ω) denotes the fixed point index of A on P ∩ Ω with respect to P . For the details of the fixed point index theory in cones see, for example, [6]. Now we can state and prove our main result. Theorem 1. Let P be a normal cone in E and γ be the normal constant of P . Assume that Ω1 and Ω2 are bounded open sets in E such that θ ∈ Ω1 and Ω¯ 1 ⊂ Ω2 . Let F : P ∩ (Ω¯ 2 \ Ω1 ) → P be a completely continuous operator and u0 ∈ P \ {θ }. If either (vii) γ x F x for x ∈ P (u0 ) ∩ ∂Ω1 and F x x for x ∈ P ∩ ∂Ω2 , or (viii) F x x for x ∈ P ∩ ∂Ω1 and γ x F x for x ∈ P (u0 ) ∩ ∂Ω2 is satisfied, then F has a fixed point in the set P ∩ (Ω¯ 2 \ Ω1 ). Proof. We can assume that F has no fixed points on P ∩ ∂Ω1 and P ∩ ∂Ω2 . Suppose that (vii) is satisfied. It is known that if F x x for x ∈ P ∩ ∂Ω2 , then iP (F, P ∩ Ω2 ) = 1
(3)
M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
257
(see [6]). Next, we will prove that for t 0 and x ∈ P ∩ ∂Ω1 , x − F x = t (u0 + F x).
(4)
Suppose, for the contrary, that there exist x0 ∈ P ∩ ∂Ω1 and t0 > 0 such that x0 − F x0 = t0 (u0 + F x0 ).
(5)
Observe that t0 u0 t0 (u0 + F x0 ) + F x0 . This, together with (5), gives x0 ∈ P (u0 ). Moreover, from (5) we get θ x0 − F x0 − t0 F x0 . Hence θ (1 + t0 )F x0 x0 . Since P is a normal cone, we obtain (1 + t0 )F x0 γ x0 , and, in consequence, F x0 < γ x0 , which contradicts (vii). Furthermore, for x ∈ P ∩ ∂Ω1 , θ ≺ u0 u0 + F x. Hence 0 < u0 γ u0 + F x, because the cone P is normal. Therefore inf
x∈P ∩∂Ω1
u0 + F x > 0.
(6)
Taking A = F for x ∈ P ∩ Ω1 and Bx = u0 + F x for x ∈ P ∩ ∂Ω1 in Proposition 3, we get in view of (4) and (6), iP (F, P ∩ Ω1 ) = 0.
(7)
By (3), (7), and the additivity of the fixed point index we obtain iP F, P ∩ (Ω2 \ Ω¯ 1 ) = 1, which implies that F has a fixed point in the set P ∩ (Ω2 \ Ω¯ 1 ). If (viii) holds the proof is similar. 2
258
M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
3. Existence of positive solution of (1) In this section we will study the existence of positive solution of (1). Similar problems have been considered for example in [7,8,11]. Particularly, in [8] Infante and Webb proved the existence of multiple nonzero (not necessarily positive) solutions, when either 0 α < 1 − η or α < 0. It is easy to verify that the kernel of the integral equation related to (1) is given by 1−α 1 − 1−αη s t, 0 t s 1, s η, 1−α 1 − 1−αη t s, 0 s t 1, (8) k(t, s) = 1−s t, 0 t s 1, 1−αη η s. s − s−αη t, 0 s t 1, 1−αη Observe that for η ∈ (0, 1), α 0 and 1 − αη > 0 we have
Mk(s, s) k(t, s) 0
(9)
t,s∈[0,1]
and
k(t, s) m(t)k(s, s),
(10)
t,s∈[0,1]
where
α(1 − η) M = max 1, and m(t) = min{t, 1 − t} 1 − αη
for t ∈ [0, 1]. Suppose that: (1◦ ) f : [0, 1] × [0, ∞) → [0, ∞) is a continuous function and there exists r1 > 0 such that 1 −1 f (t, x) M k(s, s) ds r1 0
for t ∈ [0, 1] and x ∈ [0, r1 ]; (2◦ ) there exist t0 ∈ (0, 1], a > 0, r2 > 0, r2 = r1 , and the continuous functions g : [0, 1] → [0, ∞), h : (0, r2 ] → [0, ∞) such that f (t, x) g(t)h(x) for t ∈ [0, 1] and x ∈ (0, r2 ], h(x)/x a is nonincreasing on (0, r2 ] and h(r2 ) Ma
1 k(t0 , s)g(s)ma (s) ds r2 . 0
Theorem 2. If the assumptions (1◦ ) and (2◦ ) are satisfied, then the problem (1) has a positive solution on [0, 1].
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259
Proof. Consider the Banach space C[0, 1] and the set m(t) x, t ∈ [0, 1] . P = x ∈ C[0, 1]: x(t) M It is easy to show that P is a normal cone in C[0, 1] (with γ = 1). Obviously, for x, y ∈ C[0, 1], x y if and only if x(t) y(t) for every t ∈ [0, 1]. Fix u0 (t) ≡ 1 on [0, 1]. Then m(t) x, t ∈ [0, 1] . P (u0 ) = x ∈ C[0, 1]: x(t) > M For x ∈ P and t ∈ [0, 1] define the integral operator
1 (F x)(t) =
k(t, s)f s, x(s) ds,
0
where the kernel k is given by (8). Clearly, every fixed point of F is a solution of the problem (1). For t ∈ [0, 1] and x ∈ P we have in view of (9) and (10)
1
1 k(s, s)f s, x(s) ds F x M
0
and
1 (F x)(t) m(t)
k(s, s)f s, x(s) ds.
0
Thus for every t ∈ [0, 1], m(t) F x, M which means that F : P → P . Let Ω1 = {x ∈ C[0, 1]: x < r1 } and Ω2 = {x ∈ C[0, 1]: x < r2 }. We can assume that r1 < r2 . Clearly F is a completely continuous operator on P ∩ Ω¯ 2 . By (1◦ ) and (9), for x ∈ P ∩ ∂Ω1 and t ∈ [0, 1], we get 1 −1
1 (F x)(t)
(F x)(t) M
k(s, s) M 0
r1 ds = r1 .
k(τ, τ ) dτ 0
Hence F x x for x ∈ P ∩ ∂Ω1 . Let x ∈ P (u0 ) ∩ ∂Ω2 . Then x = r2 and x(t) ∈ (0, r2 ] for every t ∈ [0, 1]. By (2◦ ), we obtain
1 (F x)(t0 )
k(t0 , s)g(s)h x(s) ds =
0
h(r2 ) a r2
1 k(t0 , s)g(s)
h(x(s)) a x (s) ds x a (s)
0
1
h(r2 ) k(t0 , s)g(s)x (s) ds > a r2
0
1
a
k(t0 , s)g(s) 0
m(s)x M
a ds
260
M. Zima / J. Math. Anal. Appl. 299 (2004) 254–260
h(r2 ) = Ma
1 k(t0 , s)g(s)ma (s) ds r2 . 0
This gives F x x for x ∈ P (u0 ) ∩ ∂Ω2 . By Theorem 1, F has a fixed point x ∗ ∈ P such that r1 x ∗ r2 . This ends the proof of Theorem 2. 2 Example. Consider the problem (1) with 2 2 f (t, x) = 16(t + 1)x (x − 3) , (t, x) ∈ [0, 1] × [0, 3], (t + 1)(x − 3), (t, x) ∈ [0, 1] × (3, ∞) and α = η = 1/2. It is easy to verify that the assumptions of Theorem 2 are satisfied for r1 = 1/60, r2 = 1, t0 = 1, a = 2, g(t) = 16(t + 1), and h(x) = x 2 (x − 3)2 . Therefore the problem (1) has a positive solution x ∗ defined on [0, 1] such that 1/60 x ∗ 1.
Acknowledgment The author expresses her thanks to the referee for helpful comments and for drawing the author’s attention to the paper [3].
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