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Positive solutions of second-order m -point boundary value problems with changing sign singular nonlinearity
Applied Mathematics Letters 20 (2007) 629–636 www.elsevier.com/locate/aml
Positive solutions of second-order m-point boundary value problems with changing sign singular nonlinearity Xinguang Zhang a,c,∗ , Huichao Zou b , Lishan Liu a a Department of Mathematics, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China b Department of Mathematics and Informational Science, Ludong University, Yantai 264025, Shandong, People’s Republic of China c Department of Mathematics and Informational Science, Yantai University, Yantai 264005, Shandong, People’s Republic of China
Received 25 February 2006; received in revised form 6 August 2006; accepted 22 August 2006
Abstract In this work, some new existence results of positive solutions for a class of singular m-point boundary value problems with changing sign nonlinearity are obtained, which improve on many known results. c 2006 Elsevier Ltd. All rights reserved.
Keywords: m-Point boundary value problem; Semipositone; Positive solutions; Cone
1. Introduction In this work, we consider the following second-order m-point boundary value problem: 0 0 t ∈ (0, 1), ( p(t)x (t)) − q(t)x(t) + λ f (t, x(t)) = 0, m−2 m−2 X X 0 0 ax(0) − bp(0)x (0) = α x(ξ ), cx(1) + d p(1)x (1) = βi x(ξi ), i i i=1
(1.1)
i=1
where λ > 0 is a parameter, a, c ∈ [0, ∞), b, d ∈ (0, ∞), ξi ∈ (0, 1), αi , βi ∈ [0, ∞) for (i ∈ {1, 2, . . . , m − 2}) are given constants. p ∈ C 1 ([0, 1], (0, ∞)), q ∈ C([0, 1], (0, ∞)) and f ∈ C((0, 1) × [0, ∞), (−∞, +∞)) satisfies −r (t) ≤ f (t, x) ≤ z(t)h(x) with r, z ∈ C((0, 1), (0, ∞)), h ∈ C([0, ∞), (0, ∞)). The form of (1.1) for differential equations arises from many fields of applied mathematics and physics, and can describe a great deal of nonlinear problems; see [1,2]. If p ≡ 1, q ≡ 0, λ = 1, αi , βi = 0, (for i = 1, 2, . . . , m − 2), m-point boundary value problem (1.1) reduces to the two-point boundary value problem 00 x (t) + f (t, x(t)) = 0, t ∈ (0, 1), (1.2) ax(0) − bx 0 (0) = cx(1) + d x 0 (1) = 0.
∗ Corresponding author at: Department of Mathematics, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China.
E-mail addresses: [email protected] (X. Zhang), [email protected] (H. Zou), [email protected] (L. Liu). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.08.005
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In the case where f is nonnegative, i.e. the positone case, (1.2) has been intensively studied; see [3–7]. Recently, Ma [8] and Ma and Thompson [9] obtained many excellent results about the existence of positive solutions and eigenvalue intervals for the more general m-point boundary value problem (1.1), but they only considered the case of the nonlinearity taking on nonnegative values and being nonsingular. Moreover their method cannot deal with the nonlinearity taking on negative values. In this work, we consider the existence of positive solutions for the BVP (1.1) when the nonlinearity f may change sign and f (t, x) may be singular at t = 0, 1. Such problems where the nonlinearity is restricted to nonnegative values are known as semipositone problems in the literature and they arise naturally in chemical reactor theory [10]. The constant λ is usually called the Thiele modulus, and in applications one is interested in showing the existence of positive solutions for λ > 0 small. However, because the semipositone problem is different from the positone problem, one can easily find that a former prior bound estimation method is not applicable for (1.1). So we will use a different method of proof. In particular, we will use Krasnoselskii’s fixed point theorem to prove our main result. For the convenience of the reader, we now state Krasnoselskii’s fixed point theorem for cones. Lemma 1.1 ([11]). Let X be a real Banach space, Q ⊂ X be a cone. Assume Ω1 , Ω2 are two bounded open subsets of X with 0 ∈ Ω1 , Ω 1 ⊂ Ω2 , and let T : Q ∩ (Ω 2 \ Ω1 ) → Q be a completely continuous operator such that either (1) kT uk ≤ kuk, u ∈ Q ∩ ∂Ω1 and kT uk ≥ kuk, u ∈ Q ∩ ∂Ω2 , or (2) kT uk ≥ kuk, u ∈ Q ∩ ∂Ω1 and kT uk ≤ kuk, u ∈ Q ∩ ∂Ω2 . Then T has a fixed point in Q ∩ (Ω 2 \ Ω1 ). This work is organized as follows. In Section 2, we present some lemmas that are used to prove our main result. In Section 3, by using Lemma 1.1, we will complete the proof of our main result. 2. Preliminaries and some lemmas In the rest of the work, we adopt the following assumptions: (H1 ) p ∈ C 1 ([0, 1], (0, ∞)), q ∈ C([0, 1], (0, ∞)). (H2 ) a, c ∈ [0, ∞), b, d ∈ (0, ∞) with ac + ad + bc > 0, αi , βi ∈ [0, ∞) for i ∈ {1, . . . , m − 2}. (H3 ) For any (t, x) ∈ (0, 1) × (0, ∞), f (t, x) satisfies −r (t) ≤ f (t, x) ≤ z(t)h(x) with r, z ∈ C((0, 1), (0, ∞)), h ∈ C([0, ∞), (0, ∞)). The following lemma, which can be found in paper [8] or [9], plays an important role in proving our main result. Lemma 2.1. Let (H1 ) and (H2 ) hold. Let ψ and φ be the solutions of the linear problems ( p(t)ψ 0 (t))0 − q(t)ψ(t) = 0, t ∈ (0, 1), ψ(0) = b, p(0)ψ 0 (0) = a, and
( p(t)φ 0 (t))0 − q(t)φ(t) = 0, t ∈ (0, 1), φ(1) = d, p(1)φ 0 (1) = −c,
respectively. Then (i) ψ is strictly increasing on [0, 1], and ψ(t) > 0 on [0, 1]. (ii) φ is strictly decreasing on [0, 1], and φ(t) > 0 on [0, 1]. As in [9], set m−2 X αi ψ(ξi ) − i=1 Γ = m−2 X βi ψ(ξi ) ρ − i=1
ρ− αi φ(ξi ) i=1 , m−2 X − βi φ(ξi ) m−2 X
i=1
φ(t) ρ = p(t) 0 φ (t)
ψ(t) . ψ 0 (t)
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Then, by Liouville’s formula, we have φ(0) ψ(0) = constant. ρ = p(0) 0 φ (0) ψ 0 (0) Now we define 1 φ(t)ψ(s), G(t, s) = ρ φ(s)ψ(t),
0 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ 1.
(2.1)
It is clear that 0 ≤ G(t, s) ≤ G(s, s),
0 ≤ s, t ≤ 1.
(2.2)
Lemma 2.2. For any t ∈ [0, 1], we have G(t, s) ≥ θ G(s, s), where b d , . θ = min φ(0) ψ(1)
(2.3)
Proof. By (2.1) and Lemma 2.1, for any t ∈ [0, 1], we have φ(t) d , 0 ≤ s ≤ t ≤ 1, , 0 ≤ s ≤ t ≤ 1, G(t, s) φ(s) φ(0) = ≥ b ψ(t) G(s, s) , 0 ≤ t ≤ s ≤ 1. , 0 ≤ t ≤ s ≤ 1, ψ(1) ψ(s) n o d b Let θ = min φ(0) , ψ(1) . Then G(t, s) ≥ θ G(s, s). R1 Lemma 2.3. Let (H1 ) and (H2 ) hold. Assume that Γ 6= 0 and 0 G(s, s)r (s)ds < +∞. Then the problem 0 0 t ∈ (0, 1), ( p(t)x (t)) − q(t)x(t) + λr (t) = 0, m−2 m−2 X X 0 αi x(ξi ), cx(1) + d p(1)x 0 (1) = βi x(ξi ), ax(0) − bp(0)x (0) = i=1
(2.4)
i=1
has a unique solution ) (Z 1 G(t, s)r (s)ds + A(r )ψ(t) + B(r )φ(t) , w(t) = λ
(2.5)
0
where ρ− αi φ(ξi ) i=1 m−2 X − βi φ(ξi )
Z 1 m−2 X αi G(ξi , s)r (s)ds 1 i=1 0 A(r ) = m−2 Z 1 Γ X βi G(ξi , s)r (s)ds 0
m−2 X
i=1
(2.6)
i=1
and m−2 X αi ψ(ξi ) − 1 i=1 B(r ) = m−2 X Γ βi ψ(ξi ) ρ − i=1
G(ξi , s)r (s)ds αi 0 i=1 , m−2 X Z 1 βi G(ξi , s)r (s)ds 0
m−2 X
Z
1
(2.7)
i=1
with w(t) ≤ ληθ , where θ is defined by (2.3), R1 (1 + Aψ(1) + Bφ(0)) 0 G(s, s)r (s)ds η= θ
(2.8)
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and m−2 X αi 1 i=1 A = m−2 Γ X βi i=1
m−2 X αi ψ(ξi ) − 1 i=1 B= m−2 X Γ βi ψ(ξi ) ρ −
ρ− αi φ(ξi ) i=1 , m−2 X − βi φ(ξi ) m−2 X
i=1
αi i=1 . m−2 X βi
m−2 X
i=1
i=1
R1
Proof. Using 0 G(s, s)r (s)ds < +∞, according to Lemma 2.2 in [8] or [9], it is easy to prove that (2.5) holds. By Lemma 2.1 and (2.2), (2.6) and (2.7), we have ) (Z Z 1 Z 1 1 G(s, s)r (s)ds G(s, s)r (s)ds + Bφ(t) G(s, s)r (s)ds + Aψ(t) w(t) ≤ λ 0
0
0
≤ λ (1 + Aψ(1) + Bφ(0))
Z
1
G(s, s)r (s)ds ≤ ληθ.
0
Now we shall denote by X = C[0, 1] the space of all continuous functions u: [0, 1] → R. This is a real Banach space when it is endowed with the usual maximal norm kxk = maxt∈[0,1] |x(t)| for x ∈ C[0, 1]. Let C + [0, 1] = {x | x ∈ X, x(t) ≥ 0}, P = {x | x ∈ C + [0, 1], x(t) ≥ θkxk for t ∈ [0, 1]}. Clearly P is a cone of C[0, 1]. Next we consider the following approximately singular nonlinear boundary value problem: ( p(t)x 0 (t))0 − q(t)x(t) + f (t, [x(t)]∗ ) + r (t) = 0, t ∈ (0, 1), m−2 m−2 X X 0 0 ax(0) − bp(0)x (0) = α x(ξ ), cx(1) + d p(1)x (1) = βi x(ξi ), i i i=1
i=1
where [x(t)]∗ = max{x(t) − w(t), 0} and w(t) is given by (2.5) which is the solution of the BVP (2.4). In this work, we will also use the following conditions. Pm−2 Pm−2 (H4 ) Γ < 0, ρ − i=1 αi φ(ξi ) > 0, ρ − i=1 βi ψ(ξi ) > 0. (H5 ) Z 1 G(s, s)[r (s) + z(s)]ds < +∞. 0
(H6 ) limx→+∞
f (t,x) x
= +∞ for t uniformly on [ 21 ξ1 , ξ1 ].
Remark 2.1. If λ > 0 and (H4 ) hold, then w given in (2.5) satisfies w(t) ≥ 0 for any t ∈ [0, 1]. In fact, this is a direct conclusion of Lemma 2.3 of [9]. For any fixed x ∈ C + [0, 1], set L = maxt∈[0,1] x(t) and f ∗ (t, x) = f (t, [x(t)]∗ ) + r (t). Applying (H2 ), (H3 ) and noticing that [x(t)]∗ ≤ x(t) ≤ L , we have Z 1 0
G(t, s) f (s, x)ds ≤ ∗
Z
1
G(s, s) z(s)h([x(s)]∗ ) + r (s) ds 0 Z 1 ≤ max h(τ ) + 1 G(s, s)(r (s) + z(s))ds 0≤τ ≤L
< +∞.
0
(2.9)
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In addition, by (2.2), for i = 1, . . . , m − 2, we know Z 1 Z 1 G(s, s)(r (s) + z(s))ds. G(ξi , s)(r (s) + z(s))ds ≤ 0
0
Thus,
A(r + z) =
1 Γ
≤
1 Γ
Z 1 m−2 m−2 X X αi G(ξi , s)(r (s) + z(s))ds ρ − αi φ(ξi ) i=1 0 i=1 Z 1 m−2 m−2 X X βi G(ξi , s)(r (s) + z(s))ds − βi φ(ξi ) 0 i=1 i=1 m−2 m−2 X X αi ρ − αi φ(ξi ) Z 1 i=1 i=1 G(s, s)(r (s) + z(s))ds m−2 m−2 X X 0 βi − βi φ(ξi ) i=1
i=1
< +∞,
(2.10)
and
B(r + z) =
1 Γ
≤
1 Γ
m−2 X αi ψ(ξi ) − i=1 m−2 X βi ψ(ξi ) ρ − i=1 m−2 X αi ψ(ξi ) − i=1 m−2 X βi ψ(ξi ) ρ − i=1
αi G(ξi , s)(r (s) + z(s))ds 0 i=1 m−2 X Z 1 βi G(ξi , s)(r (s) + z(s))ds 0 i=1 m−2 X αi Z 1 i=1 G(s, s)(r (s) + z(s))ds m−2 X 0 βi m−2 X
Z
1
i=1
< +∞.
(2.11)
Therefore, by (2.9)–(2.11), we can define an operator T : C + [0, 1] → C + [0, 1] by ) (Z 1
T x(t) = λ
G(t, s) f ∗ (s, x)ds + A( f ∗ )ψ(t) + B( f ∗ )φ(t) ,
0
where Z 1 m−2 X αi G(ξi , s) f ∗ (s, x(s))ds 1 0 i=1 A( f ∗ ) = m−2 Z 1 Γ X β G(ξi , s) f ∗ (s, x(s))ds i 0 i=1
ρ− αi φ(ξi ) i=1 m−2 X − βi φ(ξi ) m−2 X
i=1
and m−2 X αi ψ(ξi ) − 1 ∗ i=1 B( f ) = m−2 X Γ βi ψ(ξi ) ρ − i=1
αi G(ξi , s) f ∗ (s, x(s))ds 0 i=1 . Z m−2 1 X ∗ βi G(ξi , s) f (s, x(s))ds 0
m−2 X
i=1
Z
1
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Lemma 2.4. Suppose that (H1 )–(H5 ) hold. Then T : P → P is a completely continuous operator. Proof. For any x ∈ P, (2.2) and Lemma 2.1 imply that (Z 1
T x(t) ≤ λ
)
G(s, s) f ∗ (s, x)ds + A( f ∗ )ψ(1) + B( f ∗ )φ(0) .
(2.12)
0
On the other hand, from Lemma 2.2 and the monotonicity of ψ and φ, we have Z 1 b d G(s, s) f ∗ (s, x)ds + λ A( f ∗ )ψ(1) T x(t) ≥ λθ + B( f ∗ )φ(0) ψ(1) φ(0) 0 ) (Z 1 G(s, s) f ∗ (s, x)ds + A( f ∗ )ψ(1) + B( f ∗ )φ(0) . ≥ λθ
(2.13)
0
Then, (2.12) and (2.13) yield that T x(t) ≥ θkT xk which implies that T : P → P. According to the Ascoli–Arzela theorem, we can easily check that T : P → P is a completely continuous operator. 3. Main result Theorem. Suppose that (H1 )–(H6 ) hold. Then there exists λ∗ > 0 such that the m-point boundary value problem (1.1) has at least one positive solution for any λ ∈ (0, λ∗ ). Proof. By Lemma 2.4, we know T is a completely continuous operator; let Ω1 = {x ∈ C[0, 1]: kxk < η}, where η is given in (2.8). Choose R 1 G(s, s)r (s)ds 0 ∗ λ = min 1, , R θ max h(τ ) + 1 01 G(s, s) [z(s) + r (s)] ds 0≤τ ≤η
where θ is defined by (2.3). Then for any x ∈ P ∩ ∂Ω1 , notice that 0 ≤ [x(s)]∗ ≤ x(s) ≤ kxk = η; we have Z 1 T x(t) ≤ λ G(s, s) z(s)h([x(s)]∗ ) + r (s) ds 0
+ λ (Aψ(1) + Bφ(0))
Z
1
G(s, s) z(s)h([x(s)]∗ ) + r (s) ds 0 Z 1 ≤ λ (1 + Aψ(1) + Bφ(0)) max h(τ ) + 1 G(s, s) [z(s) + r (s)] ds 0≤τ ≤η
0
≤ η = kxk. Therefore, kT xk ≤ kxk, x ∈ P ∩ ∂Ω1 . Now choose a real number K > 0 such that Z ξ1 λK θ min G(t, s)ds. 1≤ η + 1 21 ξ1 ≤t≤ξ1 21 ξ1 By (H6 ), for any t ∈ [ 21 ξ1 , ξ1 ], there exists a constant N > 0 such that f (t, x) > K, x
x > N.
(3.1)
635
X. Zhang et al. / Applied Mathematics Letters 20 (2007) 629–636 N (η+1) θ
n
Choose R = max λ(η + 1), η + 1,
o
. Let Ω2 = {x ∈ C[0, 1]: kxk < R}. Then for any x ∈ P ∩ ∂Ω2 , we have λη λη x(t) − w(t) ≥ x(t) − ληθ ≥ x(t) − x(t) ≥ 1 − x(t) R R 1 λη x(t) ≥ ≥ 1− x(t) ≥ 0, t ∈ [0, 1]. λ(η + 1) η+1
And so, min (x(t) − w(t)) ≥
t∈[ 21 ξ1 ,ξ1 ]
min
t∈[ 12 ξ1 ,ξ1 ]
1 θR x(t) ≥ ≥ N. η+1 η+1
(3.2)
Therefore from (3.1) and (3.2), for any x ∈ P ∩ ∂Ω2 , we have ) (Z 1 ∗ ∗ ∗ min T x(t) = min λ G(t, s) f (s, x)ds + A( f )ψ(t) + B( f )φ(t) 1 2 ξ1 ≤t≤ξ1
1 2 ξ1 ≤t≤ξ1
≥λ
1
min
1 2 ξ1 ≤t≤ξ1
≥ λK ≥
0
Z
G(t, s) f ∗ (s, x)ds ≥ λ
0 ξ1
Z min
1 2 ξ1 ≤t≤ξ1
λK θ η+1
1 2 ξ1
Z min
1 2 ξ1 ≤t≤ξ1
Z min
1 2 ξ1 ≤t≤ξ1
G(t, s)[x(s) − w(s)]ds ≥
ξ1 1 2 ξ1
G(t, s)[ f (s, [x(s)]∗ ) + r (s)]ds
λK θ η+1
Z min
1 2 ξ1 ≤t≤ξ1
ξ1 1 2 ξ1
G(t, s)ds R
ξ1 1 2 ξ1
G(t, s)dskxk ≥ kxk.
Thus kT xk ≥ kxk, x ∈ P ∩ ∂Ω2 . By Lemma 1.1, T has a fixed point x such that η ≤ kxk ≤ R. Since kxk ≥ η, we have x(t) − w(t) ≥ ηθ − ληθ ≥ (1 − λ)ηθ ≥ 0.
(3.3)
By direct computation and combining (3.3), we have 0 0 t ∈ (0, 1), ( p(t)x (t)) − q(t)x(t) + λ [ f (t, x(t) − w(t)) + r (t)] = 0, m−2 m−2 X X 0 αi x(ξi ), cx(1) + d p(1)x 0 (1) = βi x(ξi ). ax(0) − bp(0)x (0) = i=1
( p(t)x 0 (t))0
i=1
Let u(t) = x(t) − w(t). Then = and x 0 (t) = u 0 (t) − w 0 (t). Notice that w is a solution of (2.4); then 0 0 ( p(t)u (t)) − q(t)u(t) + λ f (t, u(t)) = 0, t ∈ (0, 1), m−2 m−2 X X 0 0 au(0) − bp(0)u (0) = α u(ξ ), cu(1) + d p(1)u (1) = βi u(ξi ). i i i=1
( p(t)u 0 (t))0
+ ( p(t)w 0 (t))0
i=1
Thus u(t) is a positive solution of the singular semipositone m-point boundary value problem (1.1). The proof is completed. Acknowledgements We are grateful for the referee’s valuable suggestions. The first and third authors were supported financially by the National Natural Science Foundation of China (10471075) and the Natural Science Foundation of Shandong Province of China (Y2003A01, 02BS119). References [1] M. Moshinsky, Sobre los problemas de condiciones a la frontiera en una dimension de caracteristicas discontinuas, Bol. Mat. Mexicana 7 (1950) 1–25.
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[2] S. Timoshenko, Theory of Elastic Stability, McGraw-Hill, New York, 1961. [3] J. Henderson, H.Y. Wang, Positive solutions for nonlinear eigenvalue problems, J. Math. Anal. Appl. 208 (1997) 252–259. [4] L.H. Erbe, H.Y. Wang, On the existence of positive solutions of ordinary differential equations, Proc. Amer. Math. Soc. 120 (3) (1994) 743–748. [5] L.H. Erbe, S.C. Hu, H.Y. Wang, Multiple positive solutions of some boundary value problems, J. Math. Anal. Appl. 203 (1994) 640–648. [6] H. Dang, K. Schmitt, Existence of positive solutions of semilinear elliptic equations in annular domains, Differential Integral Equations 7 (3) (1994) 747–758. [7] K. Lan, J.R.L. Webb, Positive solutions for semilinear differential equations with singularities, J. Differential Equations 148 (1998) 407–421. [8] R.Y. Ma, Multiple positive solutions for nonlinear m-point boundary value problems, Appl. Math. Comput. 148 (2004) 249–262. [9] R.Y. Ma, B. Thompson, Positive solutions for nonlinear m-point boundary eigenvalue problems, J. Math. Anal. Appl. 297 (2004) 24–37. [10] R. Aris, Introduction to the Analysis of Chemical Reactors, Prentice-Hall, Englewood Cliffs, NJ, 1965. [11] D.J. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Inc., New York, 1988.
Positive solutions of second-order m-point boundary value problems with changing sign singular nonlinearity Xinguang Zhang a,c,∗ , Huichao Zou b , Lishan Liu a a Department of Mathematics, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China b Department of Mathematics and Informational Science, Ludong University, Yantai 264025, Shandong, People’s Republic of China c Department of Mathematics and Informational Science, Yantai University, Yantai 264005, Shandong, People’s Republic of China
Received 25 February 2006; received in revised form 6 August 2006; accepted 22 August 2006
Abstract In this work, some new existence results of positive solutions for a class of singular m-point boundary value problems with changing sign nonlinearity are obtained, which improve on many known results. c 2006 Elsevier Ltd. All rights reserved.
Keywords: m-Point boundary value problem; Semipositone; Positive solutions; Cone
1. Introduction In this work, we consider the following second-order m-point boundary value problem: 0 0 t ∈ (0, 1), ( p(t)x (t)) − q(t)x(t) + λ f (t, x(t)) = 0, m−2 m−2 X X 0 0 ax(0) − bp(0)x (0) = α x(ξ ), cx(1) + d p(1)x (1) = βi x(ξi ), i i i=1
(1.1)
i=1
where λ > 0 is a parameter, a, c ∈ [0, ∞), b, d ∈ (0, ∞), ξi ∈ (0, 1), αi , βi ∈ [0, ∞) for (i ∈ {1, 2, . . . , m − 2}) are given constants. p ∈ C 1 ([0, 1], (0, ∞)), q ∈ C([0, 1], (0, ∞)) and f ∈ C((0, 1) × [0, ∞), (−∞, +∞)) satisfies −r (t) ≤ f (t, x) ≤ z(t)h(x) with r, z ∈ C((0, 1), (0, ∞)), h ∈ C([0, ∞), (0, ∞)). The form of (1.1) for differential equations arises from many fields of applied mathematics and physics, and can describe a great deal of nonlinear problems; see [1,2]. If p ≡ 1, q ≡ 0, λ = 1, αi , βi = 0, (for i = 1, 2, . . . , m − 2), m-point boundary value problem (1.1) reduces to the two-point boundary value problem 00 x (t) + f (t, x(t)) = 0, t ∈ (0, 1), (1.2) ax(0) − bx 0 (0) = cx(1) + d x 0 (1) = 0.
∗ Corresponding author at: Department of Mathematics, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China.
E-mail addresses: [email protected] (X. Zhang), [email protected] (H. Zou), [email protected] (L. Liu). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.08.005
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In the case where f is nonnegative, i.e. the positone case, (1.2) has been intensively studied; see [3–7]. Recently, Ma [8] and Ma and Thompson [9] obtained many excellent results about the existence of positive solutions and eigenvalue intervals for the more general m-point boundary value problem (1.1), but they only considered the case of the nonlinearity taking on nonnegative values and being nonsingular. Moreover their method cannot deal with the nonlinearity taking on negative values. In this work, we consider the existence of positive solutions for the BVP (1.1) when the nonlinearity f may change sign and f (t, x) may be singular at t = 0, 1. Such problems where the nonlinearity is restricted to nonnegative values are known as semipositone problems in the literature and they arise naturally in chemical reactor theory [10]. The constant λ is usually called the Thiele modulus, and in applications one is interested in showing the existence of positive solutions for λ > 0 small. However, because the semipositone problem is different from the positone problem, one can easily find that a former prior bound estimation method is not applicable for (1.1). So we will use a different method of proof. In particular, we will use Krasnoselskii’s fixed point theorem to prove our main result. For the convenience of the reader, we now state Krasnoselskii’s fixed point theorem for cones. Lemma 1.1 ([11]). Let X be a real Banach space, Q ⊂ X be a cone. Assume Ω1 , Ω2 are two bounded open subsets of X with 0 ∈ Ω1 , Ω 1 ⊂ Ω2 , and let T : Q ∩ (Ω 2 \ Ω1 ) → Q be a completely continuous operator such that either (1) kT uk ≤ kuk, u ∈ Q ∩ ∂Ω1 and kT uk ≥ kuk, u ∈ Q ∩ ∂Ω2 , or (2) kT uk ≥ kuk, u ∈ Q ∩ ∂Ω1 and kT uk ≤ kuk, u ∈ Q ∩ ∂Ω2 . Then T has a fixed point in Q ∩ (Ω 2 \ Ω1 ). This work is organized as follows. In Section 2, we present some lemmas that are used to prove our main result. In Section 3, by using Lemma 1.1, we will complete the proof of our main result. 2. Preliminaries and some lemmas In the rest of the work, we adopt the following assumptions: (H1 ) p ∈ C 1 ([0, 1], (0, ∞)), q ∈ C([0, 1], (0, ∞)). (H2 ) a, c ∈ [0, ∞), b, d ∈ (0, ∞) with ac + ad + bc > 0, αi , βi ∈ [0, ∞) for i ∈ {1, . . . , m − 2}. (H3 ) For any (t, x) ∈ (0, 1) × (0, ∞), f (t, x) satisfies −r (t) ≤ f (t, x) ≤ z(t)h(x) with r, z ∈ C((0, 1), (0, ∞)), h ∈ C([0, ∞), (0, ∞)). The following lemma, which can be found in paper [8] or [9], plays an important role in proving our main result. Lemma 2.1. Let (H1 ) and (H2 ) hold. Let ψ and φ be the solutions of the linear problems ( p(t)ψ 0 (t))0 − q(t)ψ(t) = 0, t ∈ (0, 1), ψ(0) = b, p(0)ψ 0 (0) = a, and
( p(t)φ 0 (t))0 − q(t)φ(t) = 0, t ∈ (0, 1), φ(1) = d, p(1)φ 0 (1) = −c,
respectively. Then (i) ψ is strictly increasing on [0, 1], and ψ(t) > 0 on [0, 1]. (ii) φ is strictly decreasing on [0, 1], and φ(t) > 0 on [0, 1]. As in [9], set m−2 X αi ψ(ξi ) − i=1 Γ = m−2 X βi ψ(ξi ) ρ − i=1
ρ− αi φ(ξi ) i=1 , m−2 X − βi φ(ξi ) m−2 X
i=1
φ(t) ρ = p(t) 0 φ (t)
ψ(t) . ψ 0 (t)
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Then, by Liouville’s formula, we have φ(0) ψ(0) = constant. ρ = p(0) 0 φ (0) ψ 0 (0) Now we define 1 φ(t)ψ(s), G(t, s) = ρ φ(s)ψ(t),
0 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ 1.
(2.1)
It is clear that 0 ≤ G(t, s) ≤ G(s, s),
0 ≤ s, t ≤ 1.
(2.2)
Lemma 2.2. For any t ∈ [0, 1], we have G(t, s) ≥ θ G(s, s), where b d , . θ = min φ(0) ψ(1)
(2.3)
Proof. By (2.1) and Lemma 2.1, for any t ∈ [0, 1], we have φ(t) d , 0 ≤ s ≤ t ≤ 1, , 0 ≤ s ≤ t ≤ 1, G(t, s) φ(s) φ(0) = ≥ b ψ(t) G(s, s) , 0 ≤ t ≤ s ≤ 1. , 0 ≤ t ≤ s ≤ 1, ψ(1) ψ(s) n o d b Let θ = min φ(0) , ψ(1) . Then G(t, s) ≥ θ G(s, s). R1 Lemma 2.3. Let (H1 ) and (H2 ) hold. Assume that Γ 6= 0 and 0 G(s, s)r (s)ds < +∞. Then the problem 0 0 t ∈ (0, 1), ( p(t)x (t)) − q(t)x(t) + λr (t) = 0, m−2 m−2 X X 0 αi x(ξi ), cx(1) + d p(1)x 0 (1) = βi x(ξi ), ax(0) − bp(0)x (0) = i=1
(2.4)
i=1
has a unique solution ) (Z 1 G(t, s)r (s)ds + A(r )ψ(t) + B(r )φ(t) , w(t) = λ
(2.5)
0
where ρ− αi φ(ξi ) i=1 m−2 X − βi φ(ξi )
Z 1 m−2 X αi G(ξi , s)r (s)ds 1 i=1 0 A(r ) = m−2 Z 1 Γ X βi G(ξi , s)r (s)ds 0
m−2 X
i=1
(2.6)
i=1
and m−2 X αi ψ(ξi ) − 1 i=1 B(r ) = m−2 X Γ βi ψ(ξi ) ρ − i=1
G(ξi , s)r (s)ds αi 0 i=1 , m−2 X Z 1 βi G(ξi , s)r (s)ds 0
m−2 X
Z
1
(2.7)
i=1
with w(t) ≤ ληθ , where θ is defined by (2.3), R1 (1 + Aψ(1) + Bφ(0)) 0 G(s, s)r (s)ds η= θ
(2.8)
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and m−2 X αi 1 i=1 A = m−2 Γ X βi i=1
m−2 X αi ψ(ξi ) − 1 i=1 B= m−2 X Γ βi ψ(ξi ) ρ −
ρ− αi φ(ξi ) i=1 , m−2 X − βi φ(ξi ) m−2 X
i=1
αi i=1 . m−2 X βi
m−2 X
i=1
i=1
R1
Proof. Using 0 G(s, s)r (s)ds < +∞, according to Lemma 2.2 in [8] or [9], it is easy to prove that (2.5) holds. By Lemma 2.1 and (2.2), (2.6) and (2.7), we have ) (Z Z 1 Z 1 1 G(s, s)r (s)ds G(s, s)r (s)ds + Bφ(t) G(s, s)r (s)ds + Aψ(t) w(t) ≤ λ 0
0
0
≤ λ (1 + Aψ(1) + Bφ(0))
Z
1
G(s, s)r (s)ds ≤ ληθ.
0
Now we shall denote by X = C[0, 1] the space of all continuous functions u: [0, 1] → R. This is a real Banach space when it is endowed with the usual maximal norm kxk = maxt∈[0,1] |x(t)| for x ∈ C[0, 1]. Let C + [0, 1] = {x | x ∈ X, x(t) ≥ 0}, P = {x | x ∈ C + [0, 1], x(t) ≥ θkxk for t ∈ [0, 1]}. Clearly P is a cone of C[0, 1]. Next we consider the following approximately singular nonlinear boundary value problem: ( p(t)x 0 (t))0 − q(t)x(t) + f (t, [x(t)]∗ ) + r (t) = 0, t ∈ (0, 1), m−2 m−2 X X 0 0 ax(0) − bp(0)x (0) = α x(ξ ), cx(1) + d p(1)x (1) = βi x(ξi ), i i i=1
i=1
where [x(t)]∗ = max{x(t) − w(t), 0} and w(t) is given by (2.5) which is the solution of the BVP (2.4). In this work, we will also use the following conditions. Pm−2 Pm−2 (H4 ) Γ < 0, ρ − i=1 αi φ(ξi ) > 0, ρ − i=1 βi ψ(ξi ) > 0. (H5 ) Z 1 G(s, s)[r (s) + z(s)]ds < +∞. 0
(H6 ) limx→+∞
f (t,x) x
= +∞ for t uniformly on [ 21 ξ1 , ξ1 ].
Remark 2.1. If λ > 0 and (H4 ) hold, then w given in (2.5) satisfies w(t) ≥ 0 for any t ∈ [0, 1]. In fact, this is a direct conclusion of Lemma 2.3 of [9]. For any fixed x ∈ C + [0, 1], set L = maxt∈[0,1] x(t) and f ∗ (t, x) = f (t, [x(t)]∗ ) + r (t). Applying (H2 ), (H3 ) and noticing that [x(t)]∗ ≤ x(t) ≤ L , we have Z 1 0
G(t, s) f (s, x)ds ≤ ∗
Z
1
G(s, s) z(s)h([x(s)]∗ ) + r (s) ds 0 Z 1 ≤ max h(τ ) + 1 G(s, s)(r (s) + z(s))ds 0≤τ ≤L
< +∞.
0
(2.9)
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In addition, by (2.2), for i = 1, . . . , m − 2, we know Z 1 Z 1 G(s, s)(r (s) + z(s))ds. G(ξi , s)(r (s) + z(s))ds ≤ 0
0
Thus,
A(r + z) =
1 Γ
≤
1 Γ
Z 1 m−2 m−2 X X αi G(ξi , s)(r (s) + z(s))ds ρ − αi φ(ξi ) i=1 0 i=1 Z 1 m−2 m−2 X X βi G(ξi , s)(r (s) + z(s))ds − βi φ(ξi ) 0 i=1 i=1 m−2 m−2 X X αi ρ − αi φ(ξi ) Z 1 i=1 i=1 G(s, s)(r (s) + z(s))ds m−2 m−2 X X 0 βi − βi φ(ξi ) i=1
i=1
< +∞,
(2.10)
and
B(r + z) =
1 Γ
≤
1 Γ
m−2 X αi ψ(ξi ) − i=1 m−2 X βi ψ(ξi ) ρ − i=1 m−2 X αi ψ(ξi ) − i=1 m−2 X βi ψ(ξi ) ρ − i=1
αi G(ξi , s)(r (s) + z(s))ds 0 i=1 m−2 X Z 1 βi G(ξi , s)(r (s) + z(s))ds 0 i=1 m−2 X αi Z 1 i=1 G(s, s)(r (s) + z(s))ds m−2 X 0 βi m−2 X
Z
1
i=1
< +∞.
(2.11)
Therefore, by (2.9)–(2.11), we can define an operator T : C + [0, 1] → C + [0, 1] by ) (Z 1
T x(t) = λ
G(t, s) f ∗ (s, x)ds + A( f ∗ )ψ(t) + B( f ∗ )φ(t) ,
0
where Z 1 m−2 X αi G(ξi , s) f ∗ (s, x(s))ds 1 0 i=1 A( f ∗ ) = m−2 Z 1 Γ X β G(ξi , s) f ∗ (s, x(s))ds i 0 i=1
ρ− αi φ(ξi ) i=1 m−2 X − βi φ(ξi ) m−2 X
i=1
and m−2 X αi ψ(ξi ) − 1 ∗ i=1 B( f ) = m−2 X Γ βi ψ(ξi ) ρ − i=1
αi G(ξi , s) f ∗ (s, x(s))ds 0 i=1 . Z m−2 1 X ∗ βi G(ξi , s) f (s, x(s))ds 0
m−2 X
i=1
Z
1
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Lemma 2.4. Suppose that (H1 )–(H5 ) hold. Then T : P → P is a completely continuous operator. Proof. For any x ∈ P, (2.2) and Lemma 2.1 imply that (Z 1
T x(t) ≤ λ
)
G(s, s) f ∗ (s, x)ds + A( f ∗ )ψ(1) + B( f ∗ )φ(0) .
(2.12)
0
On the other hand, from Lemma 2.2 and the monotonicity of ψ and φ, we have Z 1 b d G(s, s) f ∗ (s, x)ds + λ A( f ∗ )ψ(1) T x(t) ≥ λθ + B( f ∗ )φ(0) ψ(1) φ(0) 0 ) (Z 1 G(s, s) f ∗ (s, x)ds + A( f ∗ )ψ(1) + B( f ∗ )φ(0) . ≥ λθ
(2.13)
0
Then, (2.12) and (2.13) yield that T x(t) ≥ θkT xk which implies that T : P → P. According to the Ascoli–Arzela theorem, we can easily check that T : P → P is a completely continuous operator. 3. Main result Theorem. Suppose that (H1 )–(H6 ) hold. Then there exists λ∗ > 0 such that the m-point boundary value problem (1.1) has at least one positive solution for any λ ∈ (0, λ∗ ). Proof. By Lemma 2.4, we know T is a completely continuous operator; let Ω1 = {x ∈ C[0, 1]: kxk < η}, where η is given in (2.8). Choose R 1 G(s, s)r (s)ds 0 ∗ λ = min 1, , R θ max h(τ ) + 1 01 G(s, s) [z(s) + r (s)] ds 0≤τ ≤η
where θ is defined by (2.3). Then for any x ∈ P ∩ ∂Ω1 , notice that 0 ≤ [x(s)]∗ ≤ x(s) ≤ kxk = η; we have Z 1 T x(t) ≤ λ G(s, s) z(s)h([x(s)]∗ ) + r (s) ds 0
+ λ (Aψ(1) + Bφ(0))
Z
1
G(s, s) z(s)h([x(s)]∗ ) + r (s) ds 0 Z 1 ≤ λ (1 + Aψ(1) + Bφ(0)) max h(τ ) + 1 G(s, s) [z(s) + r (s)] ds 0≤τ ≤η
0
≤ η = kxk. Therefore, kT xk ≤ kxk, x ∈ P ∩ ∂Ω1 . Now choose a real number K > 0 such that Z ξ1 λK θ min G(t, s)ds. 1≤ η + 1 21 ξ1 ≤t≤ξ1 21 ξ1 By (H6 ), for any t ∈ [ 21 ξ1 , ξ1 ], there exists a constant N > 0 such that f (t, x) > K, x
x > N.
(3.1)
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X. Zhang et al. / Applied Mathematics Letters 20 (2007) 629–636 N (η+1) θ
n
Choose R = max λ(η + 1), η + 1,
o
. Let Ω2 = {x ∈ C[0, 1]: kxk < R}. Then for any x ∈ P ∩ ∂Ω2 , we have λη λη x(t) − w(t) ≥ x(t) − ληθ ≥ x(t) − x(t) ≥ 1 − x(t) R R 1 λη x(t) ≥ ≥ 1− x(t) ≥ 0, t ∈ [0, 1]. λ(η + 1) η+1
And so, min (x(t) − w(t)) ≥
t∈[ 21 ξ1 ,ξ1 ]
min
t∈[ 12 ξ1 ,ξ1 ]
1 θR x(t) ≥ ≥ N. η+1 η+1
(3.2)
Therefore from (3.1) and (3.2), for any x ∈ P ∩ ∂Ω2 , we have ) (Z 1 ∗ ∗ ∗ min T x(t) = min λ G(t, s) f (s, x)ds + A( f )ψ(t) + B( f )φ(t) 1 2 ξ1 ≤t≤ξ1
1 2 ξ1 ≤t≤ξ1
≥λ
1
min
1 2 ξ1 ≤t≤ξ1
≥ λK ≥
0
Z
G(t, s) f ∗ (s, x)ds ≥ λ
0 ξ1
Z min
1 2 ξ1 ≤t≤ξ1
λK θ η+1
1 2 ξ1
Z min
1 2 ξ1 ≤t≤ξ1
Z min
1 2 ξ1 ≤t≤ξ1
G(t, s)[x(s) − w(s)]ds ≥
ξ1 1 2 ξ1
G(t, s)[ f (s, [x(s)]∗ ) + r (s)]ds
λK θ η+1
Z min
1 2 ξ1 ≤t≤ξ1
ξ1 1 2 ξ1
G(t, s)ds R
ξ1 1 2 ξ1
G(t, s)dskxk ≥ kxk.
Thus kT xk ≥ kxk, x ∈ P ∩ ∂Ω2 . By Lemma 1.1, T has a fixed point x such that η ≤ kxk ≤ R. Since kxk ≥ η, we have x(t) − w(t) ≥ ηθ − ληθ ≥ (1 − λ)ηθ ≥ 0.
(3.3)
By direct computation and combining (3.3), we have 0 0 t ∈ (0, 1), ( p(t)x (t)) − q(t)x(t) + λ [ f (t, x(t) − w(t)) + r (t)] = 0, m−2 m−2 X X 0 αi x(ξi ), cx(1) + d p(1)x 0 (1) = βi x(ξi ). ax(0) − bp(0)x (0) = i=1
( p(t)x 0 (t))0
i=1
Let u(t) = x(t) − w(t). Then = and x 0 (t) = u 0 (t) − w 0 (t). Notice that w is a solution of (2.4); then 0 0 ( p(t)u (t)) − q(t)u(t) + λ f (t, u(t)) = 0, t ∈ (0, 1), m−2 m−2 X X 0 0 au(0) − bp(0)u (0) = α u(ξ ), cu(1) + d p(1)u (1) = βi u(ξi ). i i i=1
( p(t)u 0 (t))0
+ ( p(t)w 0 (t))0
i=1
Thus u(t) is a positive solution of the singular semipositone m-point boundary value problem (1.1). The proof is completed. Acknowledgements We are grateful for the referee’s valuable suggestions. The first and third authors were supported financially by the National Natural Science Foundation of China (10471075) and the Natural Science Foundation of Shandong Province of China (Y2003A01, 02BS119). References [1] M. Moshinsky, Sobre los problemas de condiciones a la frontiera en una dimension de caracteristicas discontinuas, Bol. Mat. Mexicana 7 (1950) 1–25.
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