Upper and lower solutions method and a second order three-point singular boundary value problem

Computers and Mathematics with Applications 56 (2008) 1059–1070 www.elsevier.com/locate/camwa

Upper and lower solutions method and a second order three-point singular boundary value problem Qiumei Zhang a,b,∗ , Daqing Jiang b a School of Science, Changchun University, Changchun 130022, PR China b Department of Mathematics, Northeast Normal University, Changchun 130024, PR China

Received 26 July 2007; accepted 7 January 2008

Abstract The singular boundary value problem  00 u + q(t)g(t, u, u 0 ) = 0, t ∈ (0, 1), η ∈ (0, 1), γ ∈ (0, 1] u(0) = 0, u(1) = γ u(η), is studied in this paper.The singularity may appear at t = 0 and the function g may be superlinear at u = ∞ and change sign. The existence of solutions is obtained via an upper and lower solutions method. c 2008 Elsevier Ltd. All rights reserved.

Keywords: Singular boundary value problem; Upper and lower solutions; Existence; Superlinear

1. Introduction Motivated by the study of multi-point boundary value problems for linear second order ordinary differential equations, Gupta [1] studied certain three point boundary value problems for nonlinear ordinary differential equations. Since then, more general nonlinear multi-point boundary value problems have been studied by several authors using the Leray–Schauder theorem, a nonlinear alternative of Leray–Schauder or coincidence degree theory. We refer the reader to [2–7] for some existence results of nonlinear multi-point boundary value problems. Ma [8] proved the existence of positive solutions for the three point boundary value problem  00 u + b(t)g(u) = 0, t ∈ (0, 1) u(0) = 0, u(1) = αu(η), where η ∈ (0, 1), 0 < α < η1 , b ≥ 0 and g ≥ 0 is either superlinear or sublinear by the simple application of a fixed point theorem in cones. Recently, there are many papers dealing with the three-point boundary value problems when g is independent of u 0 , for example, see [9–12]. ∗ Corresponding author at: School of Science, Changchun University, Changchun 130022, PR China.

E-mail address: [email protected] (Q. Zhang). c 2008 Elsevier Ltd. All rights reserved. 0898-1221/$ - see front matter doi:10.1016/j.camwa.2008.01.033

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In this paper, we study the singular three point boundary value problem  00 u + q(t)g(t, u, u 0 ) = 0, t ∈ (0, 1), η ∈ (0, 1), γ ∈ (0, 1] u(0) = 0, u(1) = γ u(η). The singularity may appear at u = 0, t = 0, and the function g may be superlinear at u = ∞ and change sign. Some basic results on the singular two point boundary value problems were obtained in [13–15], in all these papers the arguments rely on the assumption that g(t, u) is positive. This implies that the solutions are concave. Recently, some authors have studied the case when g is allowed to change sign by applying the modified upper and lower solutions method, for example, see [16]. The present work is a direct extension of some results on the singular two point boundary value problems. In the same way as in [16], our technique relies essentially on a modified method of the upper and lower solutions method for singular three point boundary value problems which we believe is well adapted to this type of problem. 2. Upper and lower solutions Consider the three-point boundary value problem  00 u + f (t, u, u 0 ) = 0, t ∈ (0, 1), η ∈ (0, 1), γ ∈ (0, 1] u(0) = A, u(1) − γ u(η) = B.

(2.1)

Suppose the following conditions are satisfied: (A1) f : (0, 1] × R 2 → R is a continuous function satisfying a Nagumo condition on the set Dαβ := {(t, u, v) : t ∈ (0, 1], α(t) ≤ u ≤ β(t), v ∈ R}, for given α, β ∈ C([0, 1], R) are such that α(t) ≤ β(t), for all t ∈ [0, 1]. β A function f : (0, 1] × R 2 → R is said to satisfy a Nagumo condition on the set Dα , if there exist a function + + h ∈ C( (0, 1], (0, ∞)), and a function H ∈ C(R , [1, ∞)), R = [0, +∞), such that | f (t, u, v)| ≤ h(t)(1 + |tv|)H (|tv|)

on Dαβ ,

(2.2)

and ∞

Z

ds = ∞. H (s)

(2.3)

lim t 2 h(t) = 0,

(2.4)

0

Moreover, t→0+

and 1

Z

th(t)dt < ∞.

(2.5)

0

Here we call a function α(t) a lower solution for problem (2.1), if α ∈ C([0, 1], R) ∩ C 2 ((0, 1), R), tα 0 (t) ∈ C([0, 1], R), and  00 α + f (t, α, α 0 ) ≥ 0, for t ∈ (0, 1), (2.6) α(0) ≤ A, α(1) − γ α(η) ≤ B. Similarly, we call a function β(t) an upper solution for problem (2.1), if β ∈ C([0, 1], R) ∩ C 2 ((0, 1), R), tβ 0 (t) ∈ C([0, 1], R), and  00 β + f (t, β, β 0 ) ≤ 0, for t ∈ (0, 1), (2.7) β(0) ≥ A, β(1) − γβ(η) ≥ B. A function u(t) is said to be a solution to (2.1), if it is both a lower and an upper solution to (2.1). Then we have the following result.

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Theorem 1. Assume that (A1) is satisfied. Let α, β be, respectively, a lower solution and an upper solution for problem (2.1) such that α(t) ≤ β(t) on [0, 1]. Then problem (2.1) has at least one solution u(t) such that α(t) ≤ u(t) ≤ β(t),

for t ∈ [0, 1].

Consider now the modified boundary value problem  00 u + f ∗ (t, u(t), tu 0 (t)) = 0, for t ∈ (0, 1), u(0) = A, u(1) − γ u(η) = B,

((2.1)∗ )

where  α(t) − u   f (t, α(t), v) + h(t), if u < α(t),   1 + u2 f 1 (t, u, v) = f (t, u, v), if α(t) ≤ u ≤ β(t),    f (t, β(t), v) + β(t) − u h(t), if u > β(t),  1 + u2 and    1  N , if z < −N , f t, u, −  1   t      1 f ∗ (t, u, z) = f 1 t, u, z , if − N ≤ z ≤ N ,  t       1   f 1 t, u, N , if z > N . t Here we choose N so large that N > max{ sup |tα 0 (t)|, sup |tβ 0 (t)|} + ν, t∈[0,1]

ν := max{|β(1) − α(0)|, |α(1) − β(0)|},

t∈[0,1]

and N

Z ν

ds > E, H (s)

where 1

Z E :=

th(t)dt + (1 + sup t 2 h(t))( max β(t) − min α(t)).

0

t∈[0,1]

t∈[0,1]

t∈[0,1]

Lemma 2.1. Assume that (2.4) and (2.5) hold, then the following boundary value problem  00 y = −h(t), 0 < t < 1, y(0) = 0, y(1) − γ y(η) = 0

(2.8)

(2.9)

has a unique solution y(t) ∈ C([0, 1], R + ) ∩ C 2 ((0, 1), R), t y 0 (t) ∈ C([0, 1], R), which can be represented as Z 1 y(t) = G(t, s)h(s)ds, 0 ≤ t ≤ 1, 0

where G(t, s) is Green’s function of the BVP −y 00 = 0, y(0) = 0, y(1) = γ y(η), which is explicitly given by: when 0≤s≤η  s[1 − t − γ (η − t)]  ; s ≤ t,  − γη G(t, s) = t[1 − s1 − γ (η − s)]   ; s > t; 1 − γη

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when η < s ≤ 1  s(1 − t) + γ η(t − s)  ;  1 − γη G(t, s) = t (1 − s)   ; s > t. 1 − γη

s ≤ t,

Proof. Uniqueness. The proof of the uniqueness of a solution is standard and hence omitted here. Existence. Let Z 1 y(t) := G(t, s)h(s)ds,



0≤t ≤1

0

i.e., Z η Z 1 Z t t[1 − s − γ (η − s)] t (1 − s) s[1 − t − γ (η − t)]   h(s)ds + h(s)ds + h(s)ds;   1 − γ η 1 − γ η  t η 1 − γη   0 0 ≤ t ≤ η, Z t Z 1 y(t) = Z η s(1 − t) + γ η(t − s) t (1 − s) s[1 − t − γ (η − t)]    h(s)ds + h(s)ds + h(s)ds;   1 − γη 1 − γη 1 − γη  η t  0 η < t ≤ 1.

(2.10)

Then we have Z η Z 1 Z t 1 − s − γ (η − s) 1−s s(γ − 1)   h(s)ds + h(s)ds + h(s)ds; 0 < t ≤ η,  1 − γ η 1 − γ η 1 − γη 0 t η 0 Z t Z 1 y (t) = Z η s(γ − 1) γη − s 1−s    h(s)ds + h(s)ds + h(s)ds; η < t ≤ 1. 1 − γη 0 η 1 − γη t 1 − γη y 00 (t) = −h(t) for all t ∈ (0, 1). R1 Rt Since 0 th(t)dt < ∞, then limt→0+ 0 sh(s)ds = 0, so we have Z η 1 − s − γ (η − s) h(s)ds. y(0) = lim t 1 − γη t→0+ t R1 (η−s) If 0 1−s−γ h(s)ds < ∞, then y(0) = 0. 1−γ η R 1 1−s−γ (η−s) If 0 h(s)ds = ∞, then by (2.4) we obtain 1−γ η Rη y(0) = lim

t→0+

t

1−s−γ (η−s) h(s)ds 1−γ η

1/t

= lim t 2 h(t) t↓0+

1 − γ η + t (γ − 1) = 0. 1 − γη

We also have y(1) − γ y(η) Z η Z 1 sγ (1 − η) γ η(1 − s) = h(s)ds + h(s)ds − γ 1 − γ η 1 − γη 0 η

η

Z 0

s(1 − η) h(s)ds + 1 − γη

1

Z η

η(1 − s) h(s)ds 1 − γη

! = 0.

This shows that y(t) is a positive solution to (2.9), and y ∈ C([0, 1], [0, ∞)) ∩ C 2 ((0, 1), R). Moreover, since Z η Z 1 Z t st (γ − 1) t[1 − s − γ (η − s)] t (1 − s)   h(s)ds + h(s)ds + h(s)ds; 0 < t ≤ η,  1 − γη 1 − γη η 1 − γη Zt t Z 1 t y 0 (t) = Z0 η st (γ − 1) t (γ η − s) t (1 − s)    h(s)ds + h(s)ds + h(s)ds; η < t ≤ 1. 1 − γ η 1 − γ η 1 − γη 0 η t

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i.e., Z η Z 1 Z t s[t (γ − 1) + (1 − γ η)] t[1 − s − γ (η − s)] t (1 − s)   h(s)ds + h(s)ds + h(s)ds;   1 − γη 1 − γη  t η 1 − γη   0 0 < t ≤ η, Z t Z 1 t y 0 (t) ≤ Z η t (γ η − s) + s(1 − γ η) t (1 − s) s[t (γ − 1) + (1 − γ η)]    h(s)ds + h(s)ds + h(s)ds;   1 − γ η 1 − γ η 1 − γη  η t  0 η < t ≤ 1. i.e., |t y 0 (t)| ≤ y(t)

for all t ∈ (0, 1],

we have t y 0 (t) ∈ C([0, 1], R), and lim t y 0 (t) = 0.

(2.11)

t→0+

Let us define an operator Φ : X → X by Z 1 B − A(1 − γ ) t+ G(t, s) f ∗ (s, u(s), tu 0 (s))ds, (Φu)(t) = A + 1 − γη 0

(2.12)

while X = {u ∈ C([0, 1], R), tu 0 (t) ∈ C([0, 1], R), limt→0+ tu 0 (t) = 0 with the norm |u|1 } is the Banach space, and |u|1 := max{|u|0 , |tu 0 |0 },

|u|0 := sup{|u(t)| : 0 ≤ t ≤ 1}.

Without loss of generality, we assume that A = 0, B = 0. In order to prove the existence of the solution to (2.1)∗ , we need the following lemma. Lemma 2.2. Φ is continuous from X to X and Φ(X ) is a compact subset of X . Proof. Similar to the proof of Lemma 2.1, from the definition of f ∗ , we have from (2.12) that Z 1 |(Φu)(t)| ≤ max (1 + v)H (v) G(t, s)h(s)ds =: C y(t), t ∈ [0, 1], 0≤v≤N

|t (Φu)0 (t)| ≤ C y(t),

(2.13)

0

t ∈ (0, 1),

(2.14)

and lim t (Φu)0 (t) = 0.

(2.15)

t→0+

So we have Φu ∈ C([0, 1], R) ∩ C 2 ((0, 1), R), t (Φu)0 (t) ∈ C([0, 1], R) and |Φu|1 ≤ C|y|0 .

(2.16)

This shows that Φ(X ) is a bounded subset of X . Noting the facts that y(0) = 0 and the continuity of y(t) on [0, 1], we have from (2.13) that for any ε > 0, one can find a δ1 > 0 (independent with u) such that 0 < δ1 < 18 and ε , t ∈ [0, 2δ1 ]. 2 On the other hand, from (2.16), one has (Φu)(t) <

|(Φu)0 (t)| ≤ L , Let δ2 =

ε 2L ,

(2.17)

t ∈ [δ1 , 1].

then for t1 , t2 ∈ [δ1 , 1], |t1 − t2 | < δ2 , we have

|(Φu)(t1 ) − (Φu)(t2 )| ≤ L|t1 − t2 | <

ε . 2

(2.18)

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Define δ = min{δ1 , δ2 }, then by using (2.17) and (2.18), we obtain that |(Φu)(t1 ) − (Φu)(t2 )| < ε,

(2.19)

for t1 , t2 ∈ [0, 1], |t1 − t2 | < δ. This shows that {(Φu)(t) : u ∈ X } is equicontinuous on [0, 1]. Similarly, it follows from (2.14), and the fact |[t (Φu)0 (t)]0 | ≤ L 1 ,

t ∈ [δ1 , 1) ⊂ (0, 1),

we obtain that |t1 (Φu)0 (t1 ) − t2 (Φu)0 (t2 )| < ε,

(2.20) {(Φu)0 (t)

for t1 , t2 ∈ [0, 1], |t1 − t2 | < δ. This shows that : u ∈ X } is equicontinuous on [0, 1]. We can obtain the continuity of Φ in a similar way above. In fact, if u n , u ∈ X and |u n − u|1 → 0 as n → ∞, then we have Z 1 G(t, s)h(s)ds =: 2C y(t), t ∈ [0, 1], (2.21) |(Φu n )(t) − (Φu)(t)| ≤ 2 max (1 + v)H (v) 0≤v≤N

|t (Φu n ) (t) − t (Φu) (t)| ≤ 2C y(t), 0

0

0

t ∈ (0, 1),

(2.22)

and lim t (Φu n )0 (t) = 0.

(2.23)

t→0+

Noting the facts that y(0) = 0 and the continuity of y(t) on [0, 1], then for any ε > 0, one can find a δ1 > 0 (independent with u n ) such that 0 < δ1 < 81 and |(Φu n )(t) − (Φu)(t)| + |t (Φu n )0 (t) − t (Φu)0 (t)| < ε, On the other hand, from the continuity of

f ∗,

t ∈ [0, δ1 ].

(2.24)

one has

|(Φu n )(t) − (Φu)(t)| + |(Φu n )0 (t) − (Φu)0 (t)| → 0,

t ∈ [δ1 , 1],

as n → ∞. This together with (2.24) implies that |Φu n − Φu|1 → 0 as n → ∞. Therefore, Φ : X → X is completely continuous. The proof of Lemma 2.2 is complete.

(2.25) 

Lemma 2.3. Let u(t) be a solution to (2.1)∗ . Then the following two statements hold: (i) α(t) ≤ u(t) ≤ β(t) for all t ∈ [0, 1], and (ii) supt∈[0,1] |tu 0 (t)| ≤ N , i.e., u(t) is a solution to (2.1). Proof. We first prove that u(t) ≤ β(t) on [0, 1]. Let x(t) := u(t) − β(t). Assume that u(t) > β(t) for some t ∈ [0, 1]. Since u(0) = 0 ≤ β(0), then, x(0) ≤ 0, x(1) = u(1) − β(1) ≤ γ u(η) − γβ(η) = γ x(η). Let σ ∈ [0, 1] such that x(σ ) = maxt∈[0,1] x(t), then x(σ ) > 0. Since x(0) ≤ 0, x(1) ≤ γ x(η), then σ ∈ (0, 1). So there exists an interval (a, σ ] ⊂ (0, 1) such that x(t) > 0 in (a, σ ], and x(a) = 0,

x(σ ) = max x(t) > 0, t∈[0,1]

x 0 (σ ) = 0,

x 00 (σ ) ≤ 0.

It follows from (2.26), we have (notice that |σ u 0 (σ )| = |σβ 0 (σ )| < N ) u 00 (σ ) = − f ∗ (σ, u(σ ), tu 0 (σ )) = − f 1 (σ, u(σ ), u 0 (σ ))   β(σ ) − u(σ ) = − f (σ, β(σ ), β 0 (σ )) + h(σ ) 1 + u 2 (σ ) > β 00 (σ ) and hence x 00 (σ ) > 0, which is a contradiction.

(2.26)

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In very much the same way, we can prove that u(t) ≥ α(t) on [0, 1]. Part (i) is thus proved. Since ν := max{|β(1) − α(0)|, |α(1) − β(0)|}, there exists a τ ∈ (0, 1) such that |u 0 (τ )| ≤ ν, thus |τ u 0 (τ )| ≤ ν. Assume that part (ii) is not satisfied. Without loss of generality, we assume that there exist 0 ≤ t1 < τ and t0 ∈ (t1 , τ ] such that t1 u 0 (t1 ) = N ,

t0 u 0 (t0 ) = ν,

ν ≤ tu 0 (t) ≤ N , t ∈ [t1 , t0 ].

Then we have (notice that H ≥ 1) Z t1 Z N d(tu 0 (t)) ds = E < 0 H (s) t0 H (tu (t)) ν Z t1 00 0 tu (t) + u (t) dt = H (tu 0 (t)) t0 Z t1 Z t1 u 0 (t) dt + th(t)(1 + tu 0 (t))dt ≤ 0 (t)) H (tu t0 t0 Z 1 ≤ th(t)dt + (1 + sup t 2 h(t))|u(t1 ) − u(t0 )| 0

t∈[0,1] 1

Z ≤

th(t)dt + (1 + sup t 2 h(t))( max β(t) − min α(t)) = E,

0

t∈[0,1]

t∈[0,1]

t∈[0,1]

which is a contradiction. Part (ii) is thus proved. By virtue of Lemma 2.2, the existence of a fixed point u in X for the operator Φ follows now from the Schauder fixed point theorem. This shows that the fixed point u is a solution to (2.1)∗ , and hence, it is a solution to (2.1) as well. The proof of Theorem 1 is complete.  Remark 2.1. In Theorem 1, if we assume that | f (t, u, v)| ≤ h(t)(1 + |v|)H (|v|)

on Dαβ

in (2.2), and Z 1 h(t)dt < ∞ 0

in (2.5), then the solution u we find belongs to C 1 ([0, 1], R). 3. Main results Let g : [0, 1] × R0+ × R → R(R0+ = (0, ∞)) be a continuous function and q ∈ C((0, 1], R0+ ). Consider the three-point boundary value problem  00 u + q(t)g(t, u, u 0 ) = 0, t ∈ (0, 1), η ∈ (0, 1), γ ∈ (0, 1] u(0) = 0, u(1) = γ u(η). Theorem 2. Assume that (H1 ) for each fixed M > 0, there exists a constant L M > 0 and ε > 0 such that  z g t, x, > L M , for all (t, x, z) ∈ (0, 1] × (0, ε] × [−M, M], t (H2 ) 1

|g(t, x, z)| ≤ (F(x) + Q(x))(1 + |t z| γ ),

on [0, 1] × (0, ∞) × R

(3.1)

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with γ > 1, F > 0 continuous and nonincreasing on (0, ∞), Q ≥ 0 continuous on [0, ∞), and Q F nondecreasing on (0, ∞). In addition, Z 1 tq(t)dt < ∞, (3.2) lim t 2 q(t) = 0, t→0+

0

1

!1

Z 0

where

1 γ

1

Z

1

[q(τ )τ γ ]γ¯ dτ

γ¯

ds < ∞,

(3.3)

s

+

1 γ¯

= 1, and 

 sup  

c∈(0,∞)

 Z

1 1+

Q(c) F(c)



1 γ

(a0 + b0 c )

0

c

du   > 1, F(u)

(3.4)

where 1

Z a0 =

tq(t)dt, 0 1

Z

1

Z

b0 = 0

1 γ

γ¯

[q(τ )τ ] dτ

!1 γ¯

ds.

(3.5)

s

Then problem (3.1) (or (1.1)) has at least one solution u ∈ C([0, 1], R + ) ∩ C 2 ((0, 1), R), tu 0 (t) ∈ C([0, 1], R). Remark 3.1. q(t) = t −m , 0 ≤ m < 2 satisfies (3.2) and (3.3). We consider now the sequence of boundary value problems ( 00 u + q(t)g(t, u, u 0 ) = 0, 0 < t < 1, 1 1−γ u(0) = , u(1) − γ u(η) = , n ∈ {1, 2, . . .} n n

((3.1)n )

It follows from Lemma 2.1, we obtain Lemma 3.1. There exists an unique solution W ∈ C([0, 1], [0, ∞)) ∩ C 2 ((0, 1), R), t W 0 (t) ∈ C([0, 1], R), with W (t) > 0 on (0, 1] to the problem  00 W + q(t) = 0, 0 < t < 1, (3.6) W (0) = 0, W (1) = γ W (η). By (3.4), one can choose K > 0 and δ > 0 (δ < K ) such that Z K du 1 > 1.   1 Q(K ) F(u) 1 + F(K ) (a0 + b0 K γ ) δ

(3.7)

For each fixed M > 0, let n 0 ∈ {1, 2, . . .} be chosen so that n10 < min{ε − m|W |1 , δ}, where W is the solution of (3.6), and 0 < m < min{L M , ε/|W |1 , M/|W |1 , 1} is chosen and fixed. Let N + = {n 0 , n 0 + 1, . . .}. We have the following claim Claim. Let αn (t) = mW (t) + n1 , t ∈ [0, 1], then αn (t) is a (strict) lower solution for problem ((3.1)n ). Proof. For the choice of m and n, we have mW (t) + n1 ≤ m|W |1 + n10 < ε, mt W 0 (t) ≤ m|W |1 ≤ M, then from (H1 )   1 1 0 g t, mW (t) + , (mt W (t)) > L M > m for all t ∈ (0, 1]. n t

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Then we obtain αn00 (t) + q(t)g(t, αn (t), αn0 (t))     1 1 00 0 + q(t)g t, mW (t) + , mW (t) = mW (t) + n n   1 00 0 = mW (t) + q(t)g t, mW (t) + , mW (t) n     1 0 = q(t) g t, mW (t) + , mW (t) − m > 0, 0 < t < 1. n We obtain αn (0) = mW (0) +

1 n

= n1 , and

αn (1) − γ αn (η) = mW (1) +

1 −γ n



mW (η) +

= m(W (1) − γ W (η)) + Thus the proof of the claim is complete.

1 n



1−γ 1−γ = . n n



In a similar way as in [17], we have the following lemma. Lemma 3.2. Any solution u n (t) of ((3.1)n ) is an upper solution for ((3.1)n+1 ) (n ∈ N + ). Lemma 3.3. Problem ((3.1)n ) has at least one solution. Proof. We consider the singular boundary value problem  1 u 00 + q(t)(F(u) + Q(u))(1 + |tu 0 | γ ) = 0, for t ∈ (0, 1), 1−γ u(0) = 1 , u(1) − γ u(η) = . n n

(3.8)

In order to prove the existence of solutions for (3.8), we study the modified boundary value problem  1 u 00 + λq(t)(F ∗ (u) + Q(u))(1 + |tu 0 | γ ) = 0, for t ∈ (0, 1), 1−γ u(0) = 1 , u(1) − γ u(η) = , n n for each λ ∈ (0, 1), where    1 1  F , if u < , ∗ n n F (u) =   F(u), if u ≥ 1 . n Let u be a solution of ((3.8)λ ), then u is concave and u(t) ≥ n1 , t ∈ [0, 1]. Also since u(1) − u(η) − n1 , there exists t0 ∈ (0, 1) with u 0 (t) ≥ 0 on (0, t0 ) and u 0 ≤ 0 on (t0 , 1). Integrate the Eq. ((3.8)λ ) from s(0 < s ≤ t0 ) to t0 to obtain Q(u(t0 )) u (s) ≤ F (u(s)) 1 + ∗ F (u(t0 )) 0

∗



t0

 Z s

q(τ )dτ +

t0

Z

1 γ

γ¯

[q(τ )τ ] dτ

s

 γ1¯ Z

t0

u (τ )dτ 0

Q(u(t0 )) u (s) ≤ F(u(s)) 1 + F(u(t0 )) 

t0

 Z s

1 1 q(τ )dτ + (u(t0 ) − ) γ n

s

t0

Z s

1 γ

γ¯

[q(τ )τ ] dτ

 γ1¯ !

= γ (u(η) − n1 ) ≤

 γ1 !

noting that u(t) ≥ n1 , t ∈ [0, 1], thus we obtain 0

1 n

,

((3.8)λ )

,

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Q. Zhang, D. Jiang / Computers and Mathematics with Applications 56 (2008) 1059–1070

and then integrate from 0 to t0 to obtain u(t0 )

Z 1 n

  Z t0  γ1¯ ! Z Z t0 1 1 γ1 t0 du Q(u(t0 )) γ ¯ sq(s)ds + (u(t0 ) − ) [q(τ )τ γ ] dτ ≤ 1+ ds . F(u) F(u(t0 )) n 0 s 0

(3.9)

Now (3.9) imply 1 !   Z u(t0 ) Q(u(t0 )) du 1 γ 1+ ≤ a0 + u(t0 ) − b0 . 1 F(u) n F(u(t0 )) n i.e., u(t0 )

Z δ

  1 du Q(u(t0 )) . < (a0 + u(t0 ) γ b0 ) 1 + F(u) F(u(t0 ))

(3.10)

This together with (3.7) implies |u|0 = u(t0 ) 6= K .

(3.11)

Now notice if any solution u of ((3.8)λ ) that satisfies u(t) ≤ K , t ∈ [0, 1], then similar to the proof of Lemma 2.1, we have from ((3.8)λ ) that Z 1   1 Q(K ) 1 0 1+ sq(s)ds(1 + |tu 0 |0γ ). |tu |0 ≤ F n F(K ) 0 Then there exists K 1 > 0 such that |tu 0 |0 ≤ K 1 .

(3.12)

Solving ((3.8)λ ) is equivalent to the fixed point problem u = λT u, where   Z 1 1 1 Q(u(s)) (1 + |su 0 (s)| γ )ds. (T u)(t) := + G(t, s)q(s)F ∗ (u(s)) 1 + ∗ n F (u(s)) 0 It is easy to see (see Lemma 2.2) that T : X → X is completely continuous. Let U := {u ∈ X : |u|0 < K + 1, |tu 0 |0 < K 1 + 1}. The nonlinear alternative of Leray–Schauder [2, Theorem 1] guarantees that T has a fixed point β in U¯ , i.e., Eq. ((3.8)1 ) has a solution β with |β|0 ≤ K + 1, |tβ 0 |0 < K 1 + 1. Since β ≥ n1 , β is a solution to problem (3.8) as well. Observe that 1

|g(t, x, z)| ≤ (F(x) + Q(x))(1 + |t z|0γ ),

on [0, 1] × (0, ∞) × R.

We have β 00 (t) + q(t)g(t, β(t), β 0 (t)) ≤ 0,

for t ∈ (0, 1),

so that β is an upper solution for problem ((3.1)n ). We have that αn := mW (t) + n1 and βn := β are respectively a lower and an upper solution for problem ((3.1)n ) 1

with αn (t) ≤ βn (t), for all t ∈ [0, 1]. Since H (z) = 1 + z γ is sublinear, then by Theorem 1 we know that there is a solution u n (t) of ((3.1)n ) such that u n ∈ C([0, 1], R + ) ∩ C 2 ((0, 1), R), tu 0n (t) ∈ C([0, 1], R), and 1 = αn (t) ≤ u n (t) ≤ βn (t), ∀t ∈ [0, 1]. n Lemma 3.3 is thus proved. Proceeding by induction using Lemma 3.2, we obtain (via Theorem 1) a sequence {u n (t)}(n ∈ N + ) of solutions to ((3.1)n ) such that mW (t) +

mW (t) +

1 ≤ u n (t) ≤ u n−1 (t), n

∀t ∈ [0, 1],

u n (0) =

1 , n

1 u n (1) − γ u n (η) = (1 − γ ) , n

n ≥ n0.

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Consider now the pointwise limit w(t) := lim u n (t), n→+∞

∀t ∈ [0, 1].

(3.13)

Now let e = [a, b] ⊂ (0, 1], then there an index n ∗ = n ∗ (e) ∈ N + , for all n ≥ n ∗ ,we have u n (t) ≤ u n ∗ (t). Set r = mint∈e (mW (t) + n1 ), then u n (t) ≥ r for t ∈ e. Moreover, by (H2 )     1 F(u n ∗ (t)) sup{|g(t, x, z)| : t ∈ e, r ≤ x ≤ u n ∗ (t)} ≤ max F(r ) 1 + (1 + (t|z|) γ ) . (3.14) t∈e Q(u n ∗ (t)) n (a) , one can obtain that Let tn ∈ (a, b) such that u 0 (tn ) = u n (b)−u b−a Z tn u n (b) − u n (a) q(t)g(s, u n (s), u 0n (s))ds, t ∈ e. + u 0n (t) = b−a t Since r ≤ u n (t) ≤ u n ∗ (t), t ∈ e, then we have   Z b 1 F(u n ∗ (t)) u n ∗ (b) + u n ∗ (a) 0 q(t)F(r ) 1 + |u n (t)| ≤ + (1 + |u 0n (t)| γ )dt, ∗ b−a Q(u n (t)) a

t ∈ e.

(3.15)

Set vn = maxt∈e |u 0n (t)|, then by (3.15) we have   Z b vn u n ∗ (b) + u n ∗ (a) F(u n ∗ (t)) ≤ + q(t)F(r ) 1 + dt =: C(a, b), 1 b−a Q(u n ∗ (t)) a 1 + vnγ which implies vn = maxt∈e |u 0n (t)| is bounded. That means u 0n (t) is bounded on e. Then, by the Ascoli-Arzela theorem it is standard to conclude that w(t) is a solution of (3.1) on the interval e = [a, b]. Since e is arbitrary, we find that w ∈ C((0, 1), R0+ ) ∩ C 2 ((0, 1), R),

and

w 00 (t) + q(t)g(t, w(t), w0 (t)) = 0,

for t ∈ (0, 1).

Also we have 1 1 = 0, w(1) − γ w(η) = lim (1 − γ ) = 0. n→+∞ n n→+∞ n The same argument as in [16] works, we can prove the continuity of w(t) at t = 0 and t = 1. The proof of Theorem 2 is complete.  w(0) = lim

We now state a consequence of Theorem 2 when g(t, u, v) =: g(t, u), and the following Corollary is just the main result of [15]. Corollary 1. Assume that g(t, u, v) =: g(t, u) and (H1 ) |g(t, x)| ≤ F(x) + Q(x) on [0, 1] × (0, ∞) with F > 0 continuous and nonincreasing on (0, ∞), Q ≥ 0 continuous on [0, ∞), and Q F nondecreasing on (0, ∞); (H2 ) there exist constants L > 0 and ε > 0 such that g(t, x) > L

for all (t, x) ∈ [0, 1] × (0, ε],

and

F(x) > L ,

x ∈ (0, ε];

(H3 ) 1

Z

lim t 2 q(t) = 0,

t→0+

tq(t)dt < ∞,

0

and 1

sup

1+

c∈(0,∞)

Q(c) F(c)

Z 0

c

du F(u)

! > b0 ,

where 1

Z b0 =

rq(r )dr.

0

Then problem (3.1) has at least one solution u ∈ C([0, 1], [0, ∞)) ∩ C 2 ((0, 1), R) with u(t) > 0 on (0, 1].

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Q. Zhang, D. Jiang / Computers and Mathematics with Applications 56 (2008) 1059–1070

4. An example Example. Consider the singular boundary value problem ( 1 u 00 + σ t −m (u −a + u b + sin(8πt))(1 + |tu 0 | γ ) = 0, t ∈ (0, 1) u(0) = 0, u(1) = γ u(η), η ∈ (0, 1), γ ∈ (0, 1], σ > 0 a constant

(4.1)

with γ > 1, 0 ≤ m, n < 2, a > 0, b ≥ 0. Set 1

g(t, u, v) = σ t −m (u −a + u b + sin(8πt))(1 + |tv| γ ), F(u) = σ u −a , Q(u) = σ (u b + 1), q(t) = t −m , Z 1 1 t (1−m) dt = a0 = , 2 − m 0 !1 Z 1 Z 1 γ¯ γ¯ ( γ1 −m) dτ ds. b0 = τ 0

s

Applying Theorem 2, we can find that BVP (4.1) has at least one positive solution if σ <

sup x∈(0,∞)

x a+1 1

(a + 1)(1 + x a + x a+b )(a0 + b0 x γ )

.

(4.2)

Obviously, (H1 )–(H2 ) in Theorem 2 are satisfied. Remark 4.1. If 0 ≤ b + side of (4.2) is infinity.

1 γ

< 1, then BVP (4.1) has at least one positive solution for all σ > 0 since the right hand

Acknowledgment The work was supported by the NSFC of China (No: 10571021) and Key Laboratory for Applied Statistics of MOE (KLAS). References [1] C.P. Gupta, Solvability of a three-point nonlinear boundary value problem for a second order ordinary differential equation, J. Math. Anal. Appl. 168 (1992) 540–551. [2] C.P. Gupta, A sharper condition for the solvability of a three-point second-order boundary value problem, J. Math. Anal. Appl. 205 (1997) 586–597. [3] R. Ma, Existence theorems for a second order three-point boundary value problem, J. Math. Anal. Appl. 212 (1997) 430–442. [4] R. Ma, Existence theorems for a second order m-point boundary value problem, J. Math. Anal. Appl. 211 (1997) 545–555. [5] W. Feng, J.R.L. Webb, Solvability of a three-point Nonlinear boundary value problems at resonance, Nonlinear Anal. TMA 30 (6) (1997) 3227–3238. [6] W. Feng, J.R.L. Webb, Solvability of a m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. [7] W. Feng, On a m-point nonlinear boundary value problem, Nonlinear Anal. TMA 30 (6) (1997) 5369–5374. [8] R. Ma, Positive solutions of a nonlinear three-point boundary value problem, Electronic J. Differential Equations 34 (1) (1999) 1–8. [9] J.R.L. Webb, Positive solutions of some three point boundary value problems via fixed point index theory, Nonlinear Anal. 47 (2001) 4319–4332. [10] Y. Guo, W. Ge, Positive solutions for three-point boundary value problems with dependence on the first derivative, J. Math. Anal. Appl. 290 (2004) 291–301. [11] R. Ma, H. Wang, Positive solutions of a nonlinear three-point boundary value problems, J. Math. Anal. Appl. 279 (2003) 216–227. [12] Xian Xu, Multiplicity results for positive solutions of some semi-positone three-point boundary value problems, J. Math. Anal. Appl. 291 (2004) 673–689. [13] D. O’Regan, Positive solutions to singular and nonsingular second-order boundary value problems, J. Math. Anal. Appl. 142 (1989) 40–52. [14] R.P. Agarwal, D. O’Regan, Nonlinear superlinear singular and nonsingular second order boundary value problems, J. Differential Equations 143 (1998) 60–95. [15] D.Q. Jiang, Upper and lower solutions for a superlinear singular boundary value problem, Comput. Math. Appl. 41 (2001) 563–569. [16] D.Q. Jiang, Upper and lower solutions method and a superlinear singular boundary value problem, Comput. Math. Appl. 44 (2002) 323–337. [17] P. Habets, F. Zanolin, Upper and lower solutions for a generalized Emden–Fowler equation, J. Math. Anal. Appl. 181 (1994) 684–700.