Existence of Positive Solutions for Boundary Value Problems of Second-order Functional Differential Equations. Abstract.

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1 Existence of Positive Solutions for Boundary Value Problems of Second-order Functional Differential Equations Daqing Jiang Northeast Normal University Changchun 1324, China Peixuan Weng 1 South China Normal University Guangzhou 51631, China Abstract We use a fixed point index theorem in cones to study the existence of positive solutions for boundary value problems of second-order functional differential equations of the form y (x) + r(x)f(y(w(x))) =, < x < 1, αy(x) βy (x) = ξ(x), a x, γy(x) + δy (x) = η(x), 1 x b; where w(x) is a continuous function defined on [, 1] and r(x) is allowed to have singularities on [, 1]. The result here is the generalization of a corresponding result for ordinary differential equations. Keywords: functional differential equation, solution, superlinear and sublinear 1991 AMS classification: 34K1, 34B15 boundary value problem, positive 1 Introduction Boundary value problems (abbr. as BVP) associated with second order diffferential equations have a long history and many different methods and techniques have been used and developed in order to obtain various qualitative properties of the solutions (see [1-5, 8, 11, 15, 17]). In recent years, accompanied by the development of theory of functional differential equations (abbr. as FDE), many authors have paid attention to BVP of second order FDE such as [p(t)x (t)] = f(t, x t, x(t)), or x (t) = f(t, x(t), x(σ(t)), x (t), x (τ(t))) (for example see [6-7, 1, 12-14, 16]). As pointed out by the authors of [6], the background for these problems lies in many areas of physics, applied mathematics and variational problems of control theory. 1 Supported by NSF of China and Guangdong Province EJQTDE, 1998 No. 6, p.1

2 In this paper, we investigate the existence of positive solutions for BVP of secondorder FDE with the form: y (x) + r(x)f(y(w(x))) =, < x < 1, αy(x) βy (x) = ξ(x), a x, γy(x) + δy (x) = η(x), 1 x b. Here we assume that (P 1 ) w(x) is a continuous function defined on [, 1] satisfying c = inf{w(x); x 1} < 1, d = sup{w(x); x 1} >. (1.1) Thus E := {x [, 1]; w(x) 1} is a compact set and mes E > ; (P 2 ) ξ(x) and η(x) are continuous functions defined on [a, ] and [1, b] respectively, where a := min{, c}, b := max{1, d}; furthermore, ξ() = η(1) = ; ξ(x) as β = ; α x e β s ξ(s)ds as β > ; η(x) as δ = ; x 1 e γ δ s η(s)ds as δ >. For the case of w(x) x, BVP (1.1) is related to two point BVP of ODE for which Erbe and Wang [4] have got the following theorem: Theorem A. [4] Assume that w(x) x, ξ(x), η(x) and (A 1 ) f C([, + ), [, + )); (A 2 ) r C([, 1], [, + )) and r(x) for any subinterval of [, 1]; (A 3 ) α, β, γ, δ, ρ := γβ + αγ + αδ >. Then any one of the following is a sufficient condition for the existence of at least one positive solution of BVP (1.1): (1) lim = + and lim = (sublinear case); v v v + v (2) lim = and lim = + (superlinear case). v v v + v Motivated by [4], in this paper we shall extend the results of [4] to BVP (1.1). Firstly, we have the following hypotheses: (H 1 ) α, β, γ, δ, ρ := γβ + αγ + αδ > ; (H 2 ) r(x) is a measurable function defined on [, 1], and < < E E h(x)r(x)dx φ 1 (x)r(x)dx 1 1 h(x)r(x)dx < +, φ 1 (x)r(x)dx < +, where E is defined as in (P 1 ); h(x) : [, 1] [, 1] is defined by 1, δβ >, x, β =, δ >, h(x) = 1 x, β >, δ =, x(1 x), β = δ = and φ 1 (x) (φ 1 (x) >, x (, 1)) is the eigenfunction related to the smallest eigenvalue λ 1 (λ 1 > ) of the eigenvalue problem φ = λφ, αφ() βφ () =, γφ(1) + δφ (1) = ; EJQTDE, 1998 No. 6, p.2

3 (H 3 ) f(y) is a nonnegative continuous function defined on [, + ). Under the assumptions (H 1 ) (H 3 ), we allow that r(x) on some subset of E, and r(x) has some kind of singularities on [, 1]. For example, if β = δ =, then φ 1 (x) = sin πx and r(x) = x m (1 x) n, < m < 2, < n < 2, satisfies (H 2 ). To the best of the authors knowledge, there has not been much work done about the positive solutions for the singular boundary value problems with deviating arguments, although they have importance in applications. 2 Main Theorem First, we give the following definitions of solution and positive solution of BVP (1.1). Definition. y(x) is said to be a solution of BVP (1.1) if it satisfies the following: 1. y(x) is nonnegative and continuous on [a, b]; 2. y(x) = y(a; x) as x [a, ], where y(a; x) : [a, ] [, + ) is defined by { α e β x ( 1 α y(a; x) = β x e β s ξ(s)ds + y() ), β >, 1 ξ(x), β = ; (2.1) α 3. y(x) = y(b; x) as x [1, b], where y(b; x) : [1, b] [, + ) is defined by e ( γ δ x 1 x y(b; x) = δ 1 e γ δ s η(s)ds + e γ δ y(1) ), δ >, 1 η(x), δ = ; (2.2) γ 4. while δβ >, y (x) exists and is absolutely continuous on [, 1]; while β =, δ >, y (x) exists and is locally absolutely continuous on (, 1]; while β >, δ =, y (x) exists and is locally absolutely continuous on [, 1); while β = δ =, y (x) exists and is locally absolutely continuous on (, 1); 5. y (x) = r(x)f(y(w(x))) for x (, 1) almost everywhere. Furthermore, a solution y(x) of (1.1) is called a positive solution if y(x) > for x (, 1). Suppose that y(x) is a solution of BVP (1.1), then it could be expressed as and Green s function y(x) = G(x, t) := y(a; x), a x, 1 G(x, t)r(t)f(y(w(t)))dt, x 1, y(b; x), 1 x b, { 1 ρ 1 ρ (δ + γ γx)(β + αt), t x 1, (δ + γ γt)(β + αx), x t 1, (2.3) EJQTDE, 1998 No. 6, p.3

4 where ρ is given in (H 1 ). It is obvious that < G(x, t) G(t, t) for (x, t) (, 1) (, 1). By an elementary calculation, one can find constants λ and B such that λbh(x)h(t) G(x, t) Bh(t), (x, t) [, 1] [, 1], (2.4) where h(x) is decided by (H 2 ). By using (2.3) and (2.4), we know that for every solution y(x) of BVP (1.1), one has { y [,1] B 1 h(t)r(t)f(y(w(t)))dt, (2.5) y(x) λ y [,1] h(x), x [, 1], where y [,1] := sup{ y(x) ; x 1}. Choose σ (, 1 ) such that 4 h(s)r(s)ds >, φ 1 (s)r(s)ds >, (2.6) E σ E σ where E σ := {x E; σ w(x) 1 σ}. In this paper, we always assume that σ satisfies (2.6). Then we have from (2.5) that y(x) λ y [,1] h(x) λσ o y [,1], x [σ, 1 σ], here 1, δβ >, σ o = σ, β >, δ = σ(1 σ), δ = β =. or β =, δ >, (2.7) The following theorem is our main result. Theorem 1. If (P 1 ), (P 2 ), (H 1 ) (H 3 ) are satisfied, then any one of the following is a sufficient condition for the existence of at least one positive solution of BVP(1.1): (H 4 ) lim inf > kλ v + v 1, lim sup < qλ v 1 ; v + (H 5 ) lim inf > kλ v + v 1, lim sup < qλ v 1, ξ(x), η(x) ; v + where k > is large enough such that kλσ o E σ r(x)φ 1 (x)dx and q > is small enough such that 1 φ 1 (x)dx, 1 1 σ q r(x)φ 1 (x)dx λσ o φ 1 (x)dx σ (λ and σ o are defined as in (2.4) and (2.7) respectively). Corollary. Using the following (H 6 ) or (H 7 ) instead of (H 4 ) or (H 5 ), the conclusion of Theorem 1 is true. (H 6 ) lim = +, lim = (sublinear); v v v + v (H 7 ) lim v v =, lim v + v = + (superlinear), ξ(x), η(x). EJQTDE, 1998 No. 6, p.4

5 It is obvious that our corollary is an extension of Theorem A, and Theorem 1 is an improvement of Theorem A even for the case w(x) = x. We remark here that only if δβ > and r(x) is continuous on [, 1], every positive solution of BVP (1.1) belongs to C 1 [a, b] C 2 [, 1]. 3 Proof of Theorem In this section, we shall show the conclusion of Theorem 1 only for the situation β >, δ =. The arguments for the other three cases are similar. First we give a lemma which will be used later (see [5] or [11]). Lemma 1. Assume that X is a Banach space, and K X is a cone in X. Let K p = {u K; u p}. Furthermore, assume that Φ : K K is a compact map, and Φu u for u K p = {u K; u = p}. Then one has the following conclusions: 1. if u Φu for u K p, then i(φ, K p, K) = ; 2. if u Φu for u K p, then i(φ, K p, K) = 1. If u o (x) is a solution of BVP (1.1) for f, then it can be expressed as 1 e α β x α β x e β s ξ(s)ds, a x, u o (x) =, x 1, (3.1) 1 η(x), 1 x b. γ If y(x) is a solution of BVP(1.1), let u(x) = y(x) u o (x), noting that u(x) y(x) as x 1, then by using (3.1), we have e α β x u(), a x, u(x) = 1 G(x, t)r(t)f(u(w(t)) + u o(w(t)))dt, x 1, (3.2), 1 x b. Let K be a cone in the Banach space X = C[a, b] which is defined as K = {u C[a, b]; u(x) λσ o u, x [σ, 1 σ]}, where u := sup{ u(x) ; a x b} (noting that u [,1] is defined as in (2.5)). Define an operator Φ : K K by e α β x 1 G(, t)r(t)f(u(w(t)) + u o(w(t)))dt, a x, (Φu)(x) = 1 G(x, t)r(t)f(u(w(t)) + u o(w(t)))dt, x 1, (3.3), 1 x b. Then we have the following four lemmas. Lemma 2. Φ(K) K. Proof. It is obvious that (Φu)(x) (Φu)() as a x, then one has Φu = Φu [,1]. We have from (2.4) and (3.3) that (noting that h(x) = 1 x) Φu [,1] B 1 (1 t)r(t)f(u(w(t)) + u o (w(t)))dt, EJQTDE, 1998 No. 6, p.5

6 thus we have (Φu)(x) λb 1 (1 x)(1 t)r(t)f(u(w(t)) + u o(w(t)))dt, λ(1 x) Φu [,1], = λ(1 x) Φu, x [, 1]. Thus (Φu)(x) λσ o Φu, σ x 1 σ, i.e. Φ(K) K. Lemma 3. Φ : K K is completely continuous. Proof. We can obtain the continuity of Φ from the continuity of f. In fact, if u n, u K and u n u as n, then we have from (3.3) that for a x b, (Φu n )(x) (Φu)(x) max f(u n(w(x)) + u o (w(x))) f(u(w(x)) + u o (w(x))) B 1 x 1 (1 t)r(t)dt, which implies that Φu n Φu as n. Suppose that A K is a bounded set, and there exists a constant M 1 > such that u M 1 for u A. Let u o = M 2, then u + u o M 1 + M 2 = M for u A. We have from (3.3) that 1 Φu B max (1 t)r(t)dt. (3.4) v M Thus Φ(A) is bounded in K. Now it is easy to see that Φu C 1 [a, 1) C[a, b], and (Φu) (x) = 1 x β+α (β + αt)r(t)f(u(w(t)) + u o(w(t)))dt + α 1 β+α x (1 t)r(t)f(u(w(t)) + u o(w(t)))dt, x < 1, (Φu) (x) = α e α β x 1 β G(, t)r(t)f(u(w(t)) + u o(w(t)))dt, a x. For u A, x 1, we have (3.5) 1 (Φu)(x) F (x) := max G(x, t)r(t)dt, x 1. (3.6) v M Noting the facts that F (1) = and the continuity of F (x) on [, 1], we have from (3.6) that for any ɛ >, one can find a δ 1 > (independent with u) such that, < δ 1 < 1 4 and (Φu)(x) < ɛ 2, 1 2δ 1 < x < 1. (3.7) On the other hand, for x [, 1 δ 1 ] one has (Φu) (x) max v M { 1 δ1 1+δ 1 δ 1 max 1 v M r(t)dt + 1 (1 t)r(t)dt} (1 t)r(t)dt = L 1. For x [a, ], one has from (3.5) and (3.4) that (Φu) (x) α β (Φu)(x) α 1 β B max r(t)(1 t)dt = L 2. v M EJQTDE, 1998 No. 6, p.6

7 Let δ 2 = ɛ, then for x max{l 1,L 2 } 1, x 2 [a, 1 δ 1 ], x 1 x 2 < δ 2, we have (Φu)(x 1 ) (Φu)(x 2 ) max{l 1, L 2 } x 1 x 2 < ɛ. (3.8) Define δ o = min{δ 1, δ 2 }, then by using (3.7) (3.8) and the fact that (Φu)(x) for x [1, b], we obtain that (Φu)(x 1 ) (Φu)(x 2 ) < ɛ, for x 1, x 2 [a, b], x 1 x 2 < δ o, which implies that Φ(A) is equicontinuous. In view of the Arzela-Ascoli lemma, we know that Φ(A) is a compact set; thus Φ : K K is completely continuous. Lemma 4. (H 6 ) implies that there exists r o, R o : < r o < R o such that i(φ, K r, K) =, < r r o ; i(φ, K R, K) = 1, R R o. Proof. By using the first equality of (H 6 ) we can choose r o > such that Mv, v r o, where M satisfies λ 2 T Bσ o M > 2 and T = (1 s)r(s)ds. E σ If u K r ( < r r o ), one has u(x) λσ o u = λσ o r, x [σ, 1 σ]. (3.9) Then we obtain ( noting that u o (x) as x [, 1]) This leads to (Φu)( 1 2 ) = 1 G( 1 2, s)r(s)f(u(w(s)) + u o(w(s)))ds 1 2 λb E σ (1 s)r(s)f(u(w(s)))ds 1 2 λ2 Bσ o rt M > r = u. Φu > u, u K r. Thus we have from Lemma 1 i(φ, K r, K) =, for < r r o. On the other hand, the second equality of (H 6 ) leads to: for ɛ >, there is a R > r o + u o such that where ɛ satisfies Choose ɛv, v > R, (3.1) 1 ɛb(1 + u o ) (1 s)r(s)ds < 1 2. (3.11) 1 R o > 1 + 2B max{; v R + u o } (1 s)r(s)ds. (3.12) EJQTDE, 1998 No. 6, p.7

8 Thus, if u K R and R R o, then we have from (3.1)-(3.12) that That is (Φu)(x) B 1 (1 s)r(s)f(u(w(s)) + u o(w(s)))ds = B u(w(s))>r (1 s)r(s)f(u(w(s)) + u o(w(s)))ds +B u(w(s)) R (1 s)r(s)f(u(w(s)) + u o(w(s)))ds ɛb( u + u o ) 1 (1 s)r(s)ds +B max{; v R + u o } 1 < 1 u B max{; v 2 2 R + u o } 1 < 1 u + 1 R, x Φu = Φu [,1] < u, u K R,. (1 s)r(s)ds (1 s)r(s)ds Thus i(φ, K R, K) = 1 for R R o. Lemma 5. (H 7 ) implies that there exists r o, R o : < r o < R o such that i(φ, K r, K) = 1, < r r o ; i(φ, K R, K) =, R R o. Proof. Since ξ(x), η(x), we have u o (x). By the first equality of (H 7 ), one can choose a r o > such that where ɛ > satisfies ɛv, v r o, (3.13) < ɛb 1 (1 s)r(s)ds < 1 2. (3.14) If u K r, < r r o, then we have from (3.3), (3.13) and (3.14) that That is (Φu)(x) B 1 (1 s)r(s)f(u(w(s)))ds ɛb 1 (1 s)r(s)u(w(s))ds ɛb u 1 (1 s)r(s)ds < u, x 1, Φu = Φu [,1] < u, u K r. So we have the conclusion that i(φ, K r, K) = 1, < r r o. On the other hand, the second equality of (H 7 ) implies that M >, there is an R o > r o such that Mv, v > λσ o R o ; (3.15) here we choose M > such that λ 2 T Bσ o M > 2. For u K R, R R o, we have from the definition of K R that u(x) λσ o u = λσ o R, x [σ, 1 σ]. (3.16) Thus we have from (3.3),(3.15)-(3.16) that (Φu)( 1 2 ) = 1 G( 1 2, s)r(s)f(u(w(s)))ds 1 2 λb E σ (1 s)r(s)f(u(w(s)))ds 1 2 λ2 Bσ o RM E σ (1 s)r(s)ds = 1 2 λ2 Bσ o RT M > R = u, EJQTDE, 1998 No. 6, p.8

9 which leads to Φu > u u K R. Thus i(φ, K R, K) = for R R o. Now by using the above lemmas, we can show the conclusions of Theorem 1. Proof of Theorem 1. For < m < 1 < n, define f 1 (u) = u m, f 2 (u) = u n, u, so that f 1 (u) satisfies (H 6 ) and f 2 (u) satisfies (H 7 ). Define Φ i : K K (i = 1, 2) as follows: e α β x 1 G(, t)r(t)f i(u(w(t)) + u o (w(t)))dt, a x, (Φ i u)(x) = 1 G(x, t)r(t)f i(u(w(t)) + u o (w(t)))dt, x 1, (3.17), 1 x b. Then Φ i u (i = 1, 2) are completely continuous operators. One has from Lemma 4-5 that and i(φ 1, K r, K) =, < r r o ; i(φ 1, K R, K) = 1, R R o, (3.18) i(φ 2, K r, K) = 1, < r r o ; i(φ 2, K R, K) =, R R o, (3.19) Define H i (s, u) = (1 s)φu + sφ i u (i = 1, 2) so that for any s [, 1], H i is a completely continuous operator. Furthermore, for any ω > and i = 1, 2, we have H i (s 1, u) H i (s 2, u) s 1 s 2 [ Φ i u + Φu ] as s 1, s 2 [, 1], u K ω. Note that Φ i u + Φu is uniformly bounded in K ω. Thus H i (s, u) is continuous on u K ω uniformly for s [, 1]. According to Lemma in [18], we conclude that H i (s, u) is a completely continuous operater on [, 1] K ω. Suppose that (H 4 ) holds. By using the first inequality of (H 4 ) and the definition of f 1, one can find ɛ > and r 1 : < r 1 r o such that { f(u) (kλ1 + ɛ)u, u r 1, (3.2) f 1 (u) (kλ 1 + ɛ)u, u r 1. In what follows, we shall show that H 1 (s, u) u for u K r1 and s [, 1]. If this is not true, then there exist s 1 : s 1 1 and u 1 K r1 such that H 1 (s 1, u 1 ) = u 1. Note that u 1 (x) satisfies u 1 (x) = (1 s 1)r(x)f(u 1 (w(x)) + u o (w(x))) +s 1 r(x)f 1 (u 1 (w(x)) + u o (w(x))), < x < 1; and { αu1 (x) βu 1(x) =, a x, u 1 (x) =, 1 x b. (3.21) (3.22) Multiplying both sides of (3.21) by φ 1 (x) and then integrating it from to 1, after two times of integrating by parts, we get from (3.22) that λ 1 1 u 1(x)φ 1 (x)dx = 1 φ 1(x)[(1 s 1 )r(x)f(u 1 (w(x)) + u o (w(x))) +s 1 r(x)f 1 (u 1 (w(x)) + u o (w(x)))]dx. (3.23) EJQTDE, 1998 No. 6, p.9

10 Noting that u o (w(x)) as x E σ, we obtain from (3.2) and (H 4 ) that λ 1 1 u 1(x)φ 1 (x)dx We also have E σ φ 1 (x)r(x)[(1 s 1 )f(u 1 (w(x)))) + s 1 f 1 (u 1 (w(x)))]dx E σ φ 1 (x)r(x)[(1 s 1 )(kλ 1 + ɛ)u 1 (w(x)) +s 1 (kλ 1 + ɛ)u 1 (w(x))]dx (λ 1 + ɛ k )kλσ o u 1 E σ φ 1 (x)r(x)dx (λ 1 + ɛ k ) u 1 1 φ 1(x)dx. (3.24) 1 1 λ 1 u 1 (x)φ 1 (x)dx λ 1 u 1 φ 1 (x)dx, (3.25) which together with (3.24) leads to λ 1 λ 1 + ɛ k. This is impossible. Thus H 1 (s, u) u for u K r1 and s [, 1]. In view of the homotopic invariant property of topological degree (see [9] or [18]) and (3.18) we know that i(φ, K r1, K) = i(h 1 (, ), K r1, K) = i(h 1 (1, ), K r1, K) = i(φ 1, K r1, K) =. (3.26) On the other hand, according to the second inequality of (H 4 ), there exist ɛ > and R > R o such that f(u) (qλ 1 ɛ)u, u R. If C = max u R f(u) (qλ 1 ɛ)u + 1, then we deduce that f(u) (qλ 1 ɛ)u + C, u. (3.27) Define H(s, u) = sφu, s [, 1]. We shall show that there exists a R 1 R such that H(s, u) u, s [, 1], u K, u R 1. (3.28) If s 1 [, 1], u 1 K such that H(s 1, u 1 ) = u 1, then it is similar to the argument of (3.24)-(3.25) that λ 1 1 u 1(x)φ 1 (x)dx = s 1 1 r(x)φ 1(x)f(u 1 (w(x)) + u o (w(x)))dx q(λ 1 ɛ ) u q 1 + u o 1 r(x)φ 1(x)dx + C 1 r(x)φ 1(x)dx q(λ 1 ɛ) u q 1 1 r(x)φ 1(x)dx + C 1 1 r(x)φ 1(x)dx, and λ 1 1 u 1(x)φ 1 (x)dx λ 1 λσ o u 1 1 σ σ φ 1 (x)dx λ 1 q u 1 1 r(x)φ 1(x)dx, where C 1 = q(λ 1 ɛ) u q o + C. Combining (3.29) with (3.3), we have (3.29) (3.3) u 1 C 1 ɛ = R. EJQTDE, 1998 No. 6, p.1

11 Define R 1 = max{r, R}, then (3.28) is true. By the homotopic invariant property of topological degree, one has i(φ, K R1, K) = i(h(1, ), K R1, K) = i(h(, ), K R1, K) = i(θ, K R1, K) = 1, (3.31) where θ is zero operator. In view of (3.26),(3.31), we obtain i(φ, K R1 \K r1, K) = 1. Thus Φ has a fixed point in K R1 \K r1. Now assume that (H 5 ) is true. The first inequality and the definition of f 2 lead to: ɛ > and R > R o such that { f(u) (kλ1 + ɛ)u, u > R, Let then we have f 2 (u) (kλ 1 + ɛ)u, u > R. C = max u R f(u) (kλ 1 + ɛ)u + max u R f 2(u) (kλ 1 + ɛ)u + 1, f(u) (kλ 1 + ɛ)u C, f 2 (u) (kλ 1 + ɛ)u C, u. (3.32) We want to show that R 1 R such H 2 (s, u) u, s [, 1], u K, u R 1. (3.33) In fact, if there are s 1 [, 1], u 1 K such that H 2 (s 1, u 1 ) = u 1, then using (3.32), it is analogous to the argument of (3.24)-(3.25) that λ 1 1 u 1(x)φ 1 (x)dx E σ φ 1 (x)r(x){(1 s 1 )[(kλ 1 + ɛ)u 1 (w(x)) C] +s 1 [(kλ 1 + ɛ)u 1 (w(x)) C]}dx (λ 1 + ɛ k )kλσ o u 1 E σ φ 1 (x)r(x)dx E σ Cφ 1 (x)r(x)dx; λ 1 1 u 1(x)φ 1 (x)dx λ 1 u 1 1 φ 1(x)dx λ 1 kλσ o u 1 E σ r(x)φ 1 (x)dx. (3.34) (3.35) (3.34)-(3.35) lead to u 1 C λσ oɛ = R. Let R 1 = max{r, R}. We obtain (3.33) and then we have i(φ, K R1, K) = i(φ 2, K R1, K) =. (3.36) On the other hand, noting that ξ(x), µ(x), one has u o (x) for x [a, b]. Define H(s, u) as above. By the second inequality of (H 5 ), there exist ɛ > and r 1 : < r 1 r o such that We could also show that f(u) (qλ 1 ɛ)u, u r 1. (3.37) H(s, u) u, s [, 1], u K r1. EJQTDE, 1998 No. 6, p.11

12 But we omit the details. Thus we obtain In view of (3.36),(3.38), we obtain i(φ, K r1, K) = i(θ, K r1, K) = 1. (3.38) i(φ, K R1 \K r1, K) = 1. Thus Φ has a fixed point in K R1 \K r1. Suppose that u is the fixed point of Φ in K R1 \K r1. Let y(x) = u(x) + u o (x). Since y(x) = u(x) for x [, 1] and < r 1 u = u [,1] = y [,1] R 1, we have from (2.5) that y(x) is the positive solution of BVP(1.1). Thus we complete the proof. Example. Let us introduce an example to illustrate the usage of our theorem. Consider the BVP: where y (x) + r(x)y 1 2 (x 1 ) =, < x < 1, 3 y(x) = sin πx, 1 x, 3 y(1) = ; r(x) = π 2 (sin πx) sin π(x 1) 1 2, x [, 1) ( 1, 1], 3 3 3, x = 1. 3 (3.39) Then w(x) = x 1 3, a = 1 3, b = 1, = v 1 2, α = γ = 1, β = δ =, E = [ 1 3, 1]. Since we have lim v + v v = v 1 2 v = 1 v 2 = +. Thus (P 1 ), (P 2 ), (H 1 ) (H 3 ), (H 6 ) are =, lim v v + satisfied and (3.39) has at least one positive solution y(x). In fact, y(x) = is a positive solution of (3.39). References { sin πx, 1 3 x, sin πx, < x 1, [1] C. Bandle and M. K. Kwong, Semilinear Elliptic Problems in Annular Domains, J Appl Math Phys, 4 (1989), [2] C. Bandle, C. V. Coffman and M. Marcus, Nonlinear Elliptic Problems in Annular Domains. J Differential Equations, 69 (1987), [3] C. V. Coffman and M. Marcus, Existence and Uniqueness Results for Semilinear Dirichlet Problems in Annuli. Arch Rational Mech Anal, 18 (1987), EJQTDE, 1998 No. 6, p.12

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