Existence and approximation of solutions for nonlinear second-order four-point boundary value problems

Mathematical and Computer Modelling 50 (2009) 1348–1359

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Existence and approximation of solutions for nonlinear second-order four-point boundary value problems Libo Wang a,b,∗ , Minghe Pei a , Weigao Ge b a

Department of Mathematics, Beihua University, Ji’lin 132013, PR China

b

Department of Mathematics, Beijing Institute of Technology, Beijing 100081, PR China

article

info

Article history: Received 9 April 2008 Received in revised form 26 October 2008 Accepted 12 November 2008 Keywords: The method of upper and lower solutions Leray–Schauder degree theory One-sided Nagumo condition The generalized approximation method

abstract Applying the method of upper and lower solutions, Leray–Schauder degree theory and one-sided Nagumo condition, we obtain the existence and uniqueness results for a class of nonlinear second-order four-point boundary value problems. By the generalized approximation method, a monotone iteration sequence which converges uniformly to the unique solution of the nonlinear problem and converges quadratically to the unique solution of the linear problem is also obtained. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction The existence and the approximation of solutions to nonlinear boundary value problems have been given in a number of papers (see [1–3,5–8] and the references therein). In [2], using the generalized quasilinearization method, Khan obtained a monotone sequence of solutions for linear problems converging uniformly and quadratically to a solution of the following problem

−x00 = f (t , x, x0 ), t ∈ [0, 1] p0 x(0) − q0 x0 (0) = a, p1 x(1) + q1 x0 (1) = b, where f : [0, 1] × R2 → R is continuous and a, b ∈ R such that p0 , q0 , p1 , q1 > 0. Motivated by [2], we consider a more general nonlinear second-order four-point boundary value problem x00 = f (t , x, x0 ),

a < t < b,

φ(x(a)) − [x (a)] = g (x(c )), 0

l

(1)

ψ(x(b)) + [x (b)] = h(x(d)), 0

q

(2)

subject to one-sided Nagumo condition, weaker than the general Nagumo condition given in [2]. Through out we suppose f : [a, b] × R2 → R is continuous, φ, ψ, g , h : R → R are continuous, l, q ∈ N are odd integers and a < c ≤ d < b. Then the problem discussed in [2] is a special case of BVP (1)–(2). Using the method of upper and lower solutions and Leray–Schauder degree theory, we get the existence and uniqueness results for BVP (1)–(2) and by the use of the generalized approximation method obtain a monotone iteration sequence which converges uniformly to the unique solution of BVP (1)–(2). We also prove that, if (2) is replaced by the following special case

∗

Corresponding author at: Department of Mathematics, Beihua University, Ji’lin 132013, PR China. E-mail addresses: [email protected], [email protected] (L. Wang).

0895-7177/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2008.11.018

L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

p0 x(a) − x0 (a) = g (x(c )),

p1 x(b) + x0 (b) = h(x(d)),

1349

(3)

where p0 > 0, p1 > 0, then the monotone iteration sequence converges quadratically to the unique solution of BVP (1) (3). This work is organized as follows. In Section 2, the existence and the uniqueness results of solutions of BVP (1)–(2) are discussed. In Section 3, we study the approximation of solutions of BVP (1)–(2) and BVP (1) (3). As applications of our results, two examples are given in the last section. 2. Upper and lower solutions Definition 1. Function α(t ) ∈ C 2 [a, b] is said to be a lower solution of BVP (1)–(2) if

α 00 (t ) ≥ f (t , α(t ), α 0 (t )),

t ∈ [a, b]

(4)

and

φ(α(a)) − [α 0 (a)]l ≤ g (α(c )),

(5)

ψ(α(b)) + [α (b)] ≤ h(α(d)).

(6)

0

q

Function β(t ) ∈ C [a, b] is said to be an upper solution of BVP (1)–(2) if it satisfies the reversed inequalities. 2

Definition 2. A function f : [a, b] × R2 → R is said to satisfy one-sided Nagumo condition on [a, b] relative to α, β if there exists ω ∈ C ([0, +∞), (0, +∞)), such that f (t , x, y) ≤ ω(|y|),

(t , x, y) ∈ [a, b] × [α, β] × R

(7)

and +∞

Z 0

s

ω(s)

ds = +∞.

(8)

Lemma 1. Let α, β ∈ C [a, b] satisfy α(t ) ≤ β(t ), t ∈ [a, b], and let f : [a, b] × R2 → R be a continuous function that satisfies one-sided Nagumo condition on [a, b] relative to α, β . Then for every ρ > 0 there exists K > 0 (depending on α(t ), β(t ), ω and ρ ) such that for every solution x(t ) of x00 = f (t , x, x0 ),

a
with x0 (a) ≤ ρ,

x0 (b) ≥ −ρ

(9)

and

α(t ) ≤ x(t ) ≤ β(t ),

t ∈ [a, b],

(10)

we have

kx 0 k∞ < K . Proof. Define the non-negative number

 β(b) − α(a) β(a) − α(b) r := max , . b−a b−a 

We have two cases to consider. Case 1. ρ ≥ r. Suppose; by contradiction |x0 (t )| > ρ, ∀t ∈ (a, b). If x0 (t ) > ρ, ∀ t ∈ (a, b), then we can get the following contradiction:

β(b) − α(a) ≥ x(b) − x(a) =

b

Z

x0 (t )dt > a

Z

b

rdt ≥ β(b) − α(a).

a

If x0 (t ) < −ρ, ∀ t ∈ (a, b), a similar contradiction can be derived. So, there is t ∈ (a, b) such that |x0 (t )| ≤ ρ . From (8), we can take K1 > ρ such that K1

Z ρ

s

ω(s)

ds > max β(t ) − min α(t ). t ∈[a,b]

t ∈[a,b]

If |x (t )| ≤ ρ, ∀ t ∈ [a, b], then we have |x0 (t )| ≤ K1 , ∀ t ∈ [a, b]. If not, we can take t1 ∈ [a, b) such that x0 (t1 ) < −ρ or t1 ∈ (a, b] such that x0 (t1 ) > ρ . Suppose the first case holds. From (9), there exists t¯1 ∈ (t1 , b] such that x0 (t¯1 ) = −ρ, x0 (t ) < −ρ, ∀ t ∈ [t1 , t¯1 ). Applying a convenient change of variable and from (7), we have 0

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L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

− x 0 ( t1 )

Z

s

ω(s)

−x0 (t¯1 )

Z

t1

ds = t¯1

Z

t1

= t¯1

Z

t1

≤ t¯1

−x0 (t ) [−x00 (t )]dt ω(−x0 (t )) f (t , x(t ), x0 (t )) 0 x (t )dt ω(−x0 (t )) x0 (t )dt = x(t1 ) − x(t¯1 )

≤ max β(t ) − min α(t ) < t ∈[a,b]

t ∈[a,b]

K1

Z ρ

s

ω(s)

ds.

Hence x0 (t1 ) > −K1 . Since t1 can be taken arbitrarily as long as x0 (t1 ) < −ρ , we can conclude that for every t ∈ [a, b) such that x0 (t ) > −K1 . By a similar way, it can be proved that x0 (t ) < K1 for every t ∈ (a, b] such that x0 (t ) > ρ . Therefore

|x0 (t )| < K1 ,

∀ t ∈ [a, b].

Case 2. r > ρ . Take K2 > r such that K2

Z r

s

ds > max β(t ) − min α(t ).

ω(s)

t ∈[a,b]

t ∈[a,b]

From (9), there exists t ∈ [a, b] such that |x0 (t )| ≤ r. If |x0 (t )| ≤ r for every t ∈ [a, b], then it is trivial that |x0 (t )| < K2 . If |x0 (t )| ≤ r is not true for every t ∈ [a, b], we can take t2 ∈ [a, b) such that x0 (t2 ) < −r or t2 ∈ (a, b] such that x0 (t2 ) > r. Suppose the first case holds. From (9), we can take t¯2 ∈ (t2 , b] such that x0 (t¯2 ) = −r , x0 (t ) < −r , ∀t ∈ [t2 , t¯2 ). By a similar way in case 1, we have

Z

− x 0 ( t2 ) −x0 (t¯2 )

s

ω(s)

ds <

K2

Z r

s

ω(s)

ds.

So x0 (t2 ) > −K2 . Arguing in a similar way in case 1 we have

|x0 (t )| < K2 ,

∀ t ∈ [a, b].

Taking K = max{K1 , K2 }, we have kx0 k∞ < K .



Remark. The proof of Lemma 1 is similar as the proof of [4, Lemma 2]. Lemma 2. Boundary value problem x00 = xω(|x0 |), x(a) = [x (a)] , 0

l

(11) x(b) = −[x (b)] 0

q

(12)

has only the trivial solution, where ω ∈ C (R0 , (0, +∞)). +

Proof. Assume, by contradiction, let x0 (t ) be a nontrivial solution of BVP (11)–(12). Then there exist some t ∈ [a, b] such that x0 (t ) > 0 or x0 (t ) < 0. Suppose the first case holds. Define x0 (t1 ) := max x0 (t ) > 0. t ∈[a;b]

If t1 ∈ (a, b), then x00 (t1 ) = 0 and x000 (t1 ) ≤ 0. From (11) we have the following contradiction: 0 ≥ x000 (t1 ) = x0 (t1 )ω(|x00 (t1 )|) > 0. If t1 = a, then x00 (a) ≤ 0, which contradicts (12). If t1 = b, then x00 (b) ≥ 0, which contradicts (12). The similar proof is true for the second case. Thus BVP (11)-(12) has only the trivial solution. Lemma 3. Assume that

(A1 ) α(t ), β(t ) are any lower and upper solutions of BVP (1)–(2); (A2 ) f (t , x, y) is continuous on [a, b] × R2 and increasing in x on D = {(t , x, y) ∈ [a, b] × R2 : min{α(t ), β(t )} ≤ x ≤ max{α(t ), β(t )}};

(A3 ) φ(x) is continuously differentiable on R, g (x) is continuous on R, g 0 (x) exists and satisfies min{φ 0 (x) : x ∈ [min{α(a), β(a)}, max{α(a), β(a)}]}

≥ max{g 0 (x) : x ∈ [min{α(c ), β(c )}, max{α(c ), β(c )}]} > 0;



L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

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(A4 ) ψ(x) is continuously differentiable on R, h(x) is continuous on R, h0 (x) exists and satisfies min{ψ 0 (x) : x ∈ [min{α(b), β(b)}, max{α(b), β(b)}]}

≥ max{h0 (x) : x ∈ [min{α(d), β(d)}, max{α(d), β(d)}]} > 0. Then

α(t ) ≤ β(t ),

t ∈ [a, b].

Proof. Suppose that α(t ) > β(t ) for some t ∈ [a, b]. Then there exists t0 ∈ [a, b] such that max [α(t ) − β(t )] = α(t0 ) − β(t0 ) > 0.

t ∈[a,b]

We have three cases to consider. Case 1. If t0 ∈ (a, b), then α 0 (t0 ) − β 0 (t0 ) = 0 and α 00 (t0 ) − β 00 (t0 ) ≤ 0. From (A1 ) and (A2 ), we can get the following contradiction: 0 ≥ α 00 (t0 ) − β 00 (t0 ) ≥ f (t0 , α(t0 ), α 0 (t0 )) − f (t0 , β(t0 ), β 0 (t0 ))

= f (t0 , α(t0 ), β 0 (t0 )) − f (t0 , β(t0 ), β 0 (t0 )) > 0. Case 2. If t0 = a, then α(a)−β(a) > 0 and α 0 (a)−β 0 (a) ≤ 0. By the mean value theorem, there exists ξ1 ∈ [β(a), α(a)], ξ2 ∈ R, ξ3 ∈ [min{α(c ), β(c )}, max{α(c ), β(c )}] such that

{φ(α(a)) − [α 0 (a)]l } − {φ(β(a)) − [β 0 (a)]l } = φ 0 (ξ1 )[α(a) − β(a)] − lξ2 l−1 [α 0 (a) − β 0 (a)] and g (α(c )) − g (β(c )) = g 0 (ξ3 )[α(c ) − β(c )]. From (A3 ) and

{φ(α(a)) − [α 0 (a)]l } − {φ(β(a)) − [β 0 (a)]l } ≤ g (α(c )) − g (β(c )), we have

α(a) − β(a) ≤ α(c ) − β(c ), which implies

α(c ) − β(c ) = α(a) − β(a) = max [α(t ) − β(t )] > 0. t ∈[a,b]

Replace t0 = a by t0 = c, and then we can get a contradiction as in case 1. Case 3. If t0 = b, by using the analogous technique of case 2, we can get a contradiction as in case 1. Hence we obtain

α(t ) ≤ β(t ),

t ∈ [a, b]. 

Theorem 1. Assume that

(A1 ) There exist lower and upper solutions of BVP (1)–(2), α(t ), β(t ), respectively, such that α(t ) ≤ β(t ), t ∈ [a, b]; (A2 ) (A3 ) (A4 ) (A5 ) (A6 )

f (t , x, y) is continuous on [a, b] × R2 ; f (t , x, y) satisfies one-sided Nagumo condition on [a, b] relative to α, β ; φ, ψ are continuous on R; g (x) is continuous on R and nondecreasing on [α(c ), β(c )]; h(x) is continuous on R and nondecreasing on [α(d), β(d)].

Then BVP (1)–(2) has at least one solution x(t ) ∈ C 2 [a, b] such that

α(t ) ≤ x(t ) ≤ β(t ),

t ∈ [a, b].

Proof. Define w(t , x) = max{α(t ), min{x, β(t )}}. For λ ∈ [0, 1], we consider the auxiliary equation x00 (t ) = λf (t , w(t , x(t )), x0 (t )) + [x(t ) − λw(t , x(t ))]ω(|x0 (t )|),

(13)

where ω is determined by one-sided Nagumo condition, with the boundary condition x(a) = λ[g (w(c , x(c ))) − φ(w(a, x(a))) + w(a, x(a))] + [x0 (a)]l ,

(14)

x(b) = λ[h(w(d, x(d))) − ψ(w(b, x(b))) + w(b, x(b))] − [x (b)] .

(15)

0

q

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L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

Then we can select M0 > 0 such that for every t ∈ [a, b],

−M0 < α(t ) ≤ β(t ) < M0 ,

(16)

f (t , α(t ), 0) − [M0 + α(t )]ω(0) < 0,

(17)

f (t , β(t ), 0) + [M0 − β(t )]ω(0) > 0,

(18)

g (α(c )) − φ(α(a)) + α(a) > −M0 , g (β(c )) − φ(β(a)) + β(a) < M0 ,

(19)

h(α(d)) − ψ(α(b)) + α(b) > −M0 , h(β(d)) − ψ(β(b)) + β(b) < M0 .

(20)

In the following, we shall complete the proof by four steps. Step1. Every solution x(t ) of BVP (13)-(15) satisfies

|x(t )| < M0 ,

t ∈ [a, b],

(21)

and independently of λ ∈ [0, 1]. We suppose the estimate is not true. Then there exists t ∈ [a, b] such that x(t ) ≥ M0 or x(t ) ≤ −M0 . Suppose the first case holds. Define max x(t ) := x(t0 )(≥ M0 > 0).

t ∈[a,b]

If t0 ∈ (a, b), then x0 (t0 ) = 0 and x00 (t0 ) ≤ 0. For λ ∈ (0, 1], by (13), (16) and (18), we have the following contradiction: 0 ≥ x00 (t0 )

= λf (t0 , w(t0 , x(t0 )), x0 (t0 )) + [x(t0 ) − λw(t0 , x(t0 ))]ω(|x0 (t0 )|) = λf (t0 , w(t0 , x(t0 )), 0) + [x(t0 ) − λβ(t0 )]ω(0) ≥ λ{f (t0 , β(t0 ), 0) + [M0 − β(t0 )]ω(0)} > 0 and for λ = 0, we have 0 ≥ x00 (t0 ) = x(t0 )ω(0) ≥ M0 ω(0) > 0. If t0 = a, then max x(t ) = x(a)(≥ M0 > 0),

t ∈[a,b]

and x0 (a) ≤ 0. For λ = 0, by (14) we have the following contradiction: 0 < M0 ≤ x(a) = [x0 (a)]l ≤ 0. For λ ∈ (0, 1], by (14),(A5 ), (16) and (19) we can get the following contradiction: M0 ≤ x(a) = λ[g (w(c , x(c ))) − φ(w(a, x(a))) + w(a, x(a))] + [x0 (a)]l ≤ λ[g (β(c )) − φ(β(a)) + β(a)] < M0 . If t0 = b, then max x(t ) = x(b)(≥ M0 > 0),

t ∈[a,b]

and x0 (b) ≥ 0. For λ = 0, by (15) we have the following contradiction: 0 < M0 ≤ x(b) = −[x0 (b)]q ≤ 0. For λ ∈ (0, 1], by (15), (A6 ), (16) and (20) we can get the following contradiction: M0 ≤ x(b) = λ[h(w(d, x(d))) − ψ(w(b, x(b))) + w(b, x(b))] − [x0 (b)]q ≤ λ[h(β(d)) − ψ(β(b)) + β(b)] < M0 . Thus x0 (t ) < M0 for t ∈ [a, b]. In a similar way, we can prove that x0 (t ) < M0 , t ∈ [a, b]. Step 2. There exists M1 > 0 such that every solution x(t ) of BVP (13)–(15) satisfies

|x0 (t )| < M1 ,

t ∈ [a, b],

independently of λ ∈ [0, 1]. Consider the function Fλ : [a, b] × R2 → R defined by Fλ (t , x, y) = λf (t , w(t , x), y) + [x − λw(t , x)]ω(|y|). In the following, we show that Fλ satisfies one-sided Nagumo condition on [a, b] relative to −M0 , M0 , independently of λ ∈ [0, 1]. In fact, since f satisfies one-sided Nagumo condition on [a, b] relative to −M0 , M0 , Fλ (t , x, y) = λf (t , w(t , x), y) + [x − λw(t , x)]ω(|y|) ≤ [2M0 + 1]ω(|y|) := ω∗ (|y|).

L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

1353

Furthermore, we obtain +∞

Z

s

ω ∗ ( s)

0

Z

+∞

ds =

s

(2M0 + 1)ω(s)

0

ds = +∞,

thus Fλ satisfies one-sided Nagumo condition on [a, b] relative to −M0 , M0 , independently of λ ∈ [0, 1]. Let ρ := max{[K1 + 1

1

G + 2M1 ] l , [K2 + H + 2M1 ] q }, where G=

max

x∈[−M0 ,M0 ]

K1 =

max

|g (x)|,

x∈[−M0 ,M0 ]

|φ(x)|,

H =

max

x∈[−M0 ,M0 ]

K2 =

max

|h(x)|,

x∈[−M0 ,M0 ]

|ψ(x)|.

From (14) and (15), every solution x(t ) of BVP (13)–(15) satisfies 1

1

x0 (a) = {x(a) − λ[g (w(c , x(c ))) − φ(w(a, x(a))) + w(a, x(a))]} l ≤ [K1 + G + 2M0 ] l ≤ ρ, 1

1

x0 (b) = {λ[h(w(d, x(d))) − ψ(w(b, x(b))) + w(b, x(b))] − x(b)} q ≥ −[K2 + H + 2M0 ] q ≥ −ρ. Define

α0 (t ) = −M0 ,

β0 (t ) = M0 ,

t ∈ [a, b].

In view of Step 1 and applying Lemma 1, then there exists M1 > 0 (independently of λ) such that |x0 (t )| < M1 , t ∈ [a, b]. Step 3. For λ = 1, BVP (13)-(15) has at least one solution x1 (t ). Define the operators L : C 2 [a, b] ⊂ C 1 [a, b] → C [a, b] × R2 by Lx = (x00 , x(a), x(b)). and Nλ : C 1 [a, b] → C [a, b] × R2 by Nλ x = (λf (t , w(t , x(t )), x0 (t )) + [x(t ) − λw(t , x0 (t ))]ω(|x0 (t )|), Bλ , Cλ ), where Bλ = λ[g (w(c , x(c ))) − φ(w(a, x(a))) + w(a, x(a))] + [x0 (a)]l , Cλ = λ[h(w(d, x(d))) − ψ(w(b, x(b))) + w(b, x(b))] − [x0 (b)]q . As L−1 is compact, we can define the completely continuous operator Tλ : C 1 [a, b] → C 1 [a, b] by Tλ (x) = L−1 Nλ (x). Consider the set

Ω = {x ∈ C 1 [a, b] : kxk∞ < M0 , kx0 k∞ < M1 }. By Step 1 and 2, the degree deg(I − Tλ , Ω , 0) is well defined for every λ ∈ [0, 1] and by homotopy invariance, we get deg(I − T0 , Ω , θ ) = deg(I − T1 , Ω , θ ). As the equation x = T0 (x) has only the trivial solution from Lemma 2, by degree theory, deg(I − T1 , Ω , θ ) = deg(I − T0 , Ω , θ ) = ±1. Hence, the equation x = T1 (x) has at least one solution. That is, the problem x00 (t ) = f (t , w(t , x(t )), x0 (t )) + [x(t ) − w(t , x(t ))]ω(|x0 (t )|)

(22)

with the boundary condition x(a) = [g (w(c , x(c ))) − φ(w(a, x(a))) + w(a, x(a))] + [x0 (a)]l ,

(23)

x(b) = [h(w(d, x(d))) − ψ(w(b, x(b))) + w(b, x(b))] − [x (b)]

(24)

0

has at least one solution x1 (t ) ∈ Ω .

q

1354

L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

Step 4. x1 (t ) is a solution of BVP (1)–(2). In fact, the solution x1 (t ) of the above problem will also be a solution of BVP (1)–(2) since it satisfies

α(t ) ≤ x(t ) ≤ β(t ),

t ∈ [a, b].

(25)

Suppose, by contradiction, that there exists t ∈ [a, b] such that x1 (t ) > β(t ) and define max [x1 (t ) − β(t )] := x1 (t1 ) − β(t1 ) > 0.

t ∈[a,b]

If t1 ∈ (a, b), then x01 (t1 ) = β 0 (t1 ) and x001 (t1 ) ≤ β 00 (t1 ). By (22), Definition 1, condition (A1 ) and (A2 ), we have the contradiction: 0 ≥ x001 (t1 ) − β 00 (t1 )

≥ f (t1 , w(t1 , x1 (t1 )), x01 (t1 )) + [x1 (t1 ) − w(t1 , x1 (t1 ))]ω(|x01 (t1 )|) − f (t1 , β(t1 ), β 0 (t1 )) ≥ f (t1 , β(t1 ), β 0 (t1 )) + [x1 (t1 ) − β(t1 )]ω(|x01 (t1 )|) − f (t1 , β(t1 ), β 0 (t1 )) = [x1 (t1 ) − β(t1 )]ω(|x01 (t1 )|) > 0. If t1 = a, we have max [x1 (t ) − β(t )] := x1 (a) − β(a) > 0

t ∈[a,b]

and x01 (a) − β 0 (a) ≤ 0. By (23), condition (A5 ) and Definition 1, we have the contradiction:

β(a) < x1 (a) = [g (w(c , x1 (c ))) − φ(w(a, x1 (a))) + w(a, x1 (a))] + [x01 (a)]l ≤ g (β(c )) − φ(β(a)) + β(a) + [β 0 (a)]l ≤ β(a). If t1 = b, we have max [x1 (t ) − β(t )] := x1 (b) − β(b) > 0

t ∈[a,b]

and x01 (b) − β 0 (b) ≥ 0. By (24), condition (A6 ) and Definition 1, we have the contradiction:

β(b) < x1 (b) = [h(w(d, x1 (d))) − ψ(w(b, x1 (b))) + w(b, x1 (b))] − [x01 (b)]q ≤ [h(β(d)) − ψ(β(b)) + β(b)] − [β 0 (b)]q ≤ β(b). Thus x1 (t ) ≤ β(t ),

t ∈ [a, b].

Using an analogous technique, it can be obtained that x1 (t ) ≥ α(t ) for t ∈ [a, b]. Therefore, x1 (t ) is in fact a solution of BVP (1)–(2).  Theorem 2. Assume that

(A1 ) There exist lower and upper solutions of BVP (1)–(2), α(t ), β(t ), respectively, such that α(t ) ≤ β(t ), (A2 ) (A3 ) (A4 ) (A5 )

t ∈ [a, b];

f (t , x, x ) is continuous on [a, b] × R2 and fx0 (t , x, x0 ) > 0; f (t , x, x0 ) satisfies one-sided Nagumo condition on [a, b] relative to α, β ; φ(x), ψ(x) are continuous on R, φ 0 (x), ψ 0 (x) exist and satisfy φ 0 (x) > 0 on [α(a), β(a)] and ψ 0 (x) > 0 on [α(b), β(b)]; g 0 (x), h0 (x) exists and satisfies 0

min

x∈[α(a),β(a)]

φ 0 (x) ≥

max

x∈[α(c ),β(c )]

g 0 ( x) > 0 ,

min

x∈[α(b),β(b)]

ψ 0 ( x) ≥

Then BVP (1)–(2) has a unique solution x(t ) ∈ C 2 [a, b] such that

α(t ) ≤ x(t ) ≤ β(t ), t ∈ [a, b].

max

x∈[α(d),β(d)]

h0 (x) > 0.

L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

1355

Proof. From Theorem 1, it is easy to prove the existence of solutions for BVP (1)–(2). Let x1 (t ), x2 (t ) are any two solutions of BVP (1)–(2). Suppose that there exists t ∈ [a, b] such that x1 (t ) > x2 (t ) and define max [x1 (t ) − x2 (t )] := x1 (t1 ) − x2 (t1 ) > 0.

t ∈[a,b]

We have three cases to consider. Case 1. If t1 ∈ (a, b), we have x01 (t1 ) = x02 (t1 ) and x001 (t1 ) ≤ x002 (t1 ). From (1) and condition (A2 ), we can get the following contradiction: 0 ≥ x001 (t1 ) − x002 (t1 ) = f (t1 , x1 (t1 ), x01 (t1 )) − f (t1 , x2 (t1 ), x02 (t1 ))

= f (t1 , x1 (t1 ), x02 (t1 )) − f (t1 , x2 (t1 ), x02 (t1 )) > 0. Case 2. If t1 = a, then x01 (a) ≤ x02 (a), from (2), condition (A4 ), (A5 ), we have

φ 0 (ξ1 )(x1 (a) − x2 (a)) ≤ g 0 (ξ2 )(x1 (c ) − x2 (c )), where ξ1 ∈ [x2 (a), x1 (a)], ξ2 ∈ [x2 (c ), x1 (c )]. So we have x1 (a) − x2 (a) ≤ x1 (c ) − x2 (c ). and x1 (c ) − x2 (c ) = x1 (a) − x2 (a) = max [x1 (t ) − x2 (t )], t ∈[a,b]

then x1 (c ) = x2 (c ) and x1 (c ) ≤ x2 (c ). From (1) and condition (A2 ), we can get the following contradiction: 0

0

00

00

0 ≥ x001 (c ) − x002 (c ) = f (c , x1 (c ), x01 (c )) − f (c , x2 (c ), x02 (c ))

= f (c , x1 (c ), x02 (c )) − f (c , x2 (c ), x02 (c )) > 0. Case 3. If t1 = b, by using the analogous technique of Case 2, we can get contradiction. Thus we have x1 (t ) ≤ x2 (t ),

t ∈ [a, b].

Arguing in a similar way, it can be obtained that x1 (t ) ≥ x2 (t ) for t ∈ [a, b]. Hence we obtain x1 (t ) ≡ x2 (t ), t ∈ [a, b]. This completes the proof of the theorem.  3. Approximation of solutions Theorem 3. Assume that

(A1 ) There exist lower and upper solutions of BVP (1)–(2), α(t ), β(t ) ∈ C 2 [a, b], respectively, such that α(t ) ≤ β(t ), t ∈ [a, b]; (A2 ) (A3 ) (A4 ) (A5 ) (A6 )

f (t , x, y) is continuous on [a, b] × R2 and satisfies one-sided Nagumo condition on [a, b] relative to α, β ; fx (t , x, y) > 0 and fxx (t , x, y) is continuous on [a, b] × [mint ∈[a,b] α(t ), maxt ∈[a,b] β(t )] × R; φ(x), ψ(x) are continuous on R and φ 0 (x) > 0, ψ 0 (x) > 0, φ 00 (x) ≤ 0, ψ 00 (x) ≤ 0; g (x), h(x) are continuous on R and g 0 (x) > 0, g 00 (x) ≤ 0 on [α(c ), β(c )], h0 (x) > 0, h00 (x) ≤ 0 on [α(d), β(d)]; minx∈[α(a),β(a)] φ 0 (x) ≥ maxx∈[α(c ),β(c )] g 0 (x), minx∈[α(b),β(b)] ψ 0 (x) ≥ maxx∈[α(d),β(d)] h0 (x).

Then, there exists a monotone nondecreasing sequence {αn } of solutions converging uniformly to the unique solution of BVP (1)–(2). Proof. By the proof of Theorem 1, we know every solution x of BVP (1)–(2), satisfies

|x0 (t )| < M1 ,

t ∈ [a, b].

Let C > max{kα k, kβ 0 k, M1 }, define p(x0 ) = max{−C , min{x, C }}. Consider boundary value problem 0

 00 x = f (t , x, p(x0 )), φ(x(a)) − [x0 (a)]l = g (x(c )), ψ(x(b)) + [x0 (b)]q = h(x(d)).

(26)

Note that any solution x ∈ C 1 [a, b] of BVP (26) such that |x0 (t )| < C , is a solution of BVP (1)–(2). Moreover, we note that for t ∈ [a, b] and α(t ) ≤ x ≤ β(t ), we have f (t , x, p(x0 )) ≤ ω(|p(x0 )|) := ω(| ¯ x0 |),

x0 ∈ R

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L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

where ω( ¯ s) = ω(p(s)) for s ≥ 0. Since +∞

Z 0

s

ω( ¯ s)

C

Z ds = 0

s

ω( ¯ s)

+∞

Z ds + C

s

ω(C )

ds = +∞,

f (t , x, p(x0 )) satisfies one-sided Nagumo condition on [a, b] relative to α, β . It follows that a solution x of BVP (26) with α(t ) ≤ x ≤ β(t ), t ∈ [a, b], does satisfy |x0 (t )| ≤ C on [a, b], and hence, is a solution of BVP (1)–(2). Define F (t , x, x0 ; y) = f (t , y, p(x0 )) + fx (t , y, p(x0 ))(x − y) +

Mf 2

(x − y)2 ,

(27)

Φ (x, x0 ; y) = φ(y) + φ 0 (y)(x − y) − [x0 ]l ,

(28)

Ψ (x, x ; y) = ψ(y) + ψ (y)(x − y) + [x ] ,

(29)

G(x; y) = g (y) + g (y)(x − y),

(30)

H (x; y) = h(y) + h0 (y)(x − y),

(31)

0

0

0 q

0

where Mf = max{|fxx (t , x, x0 )| : (t , x, x0 ) ∈ [a, b] × [ min α(t ), max β(t )] × [−C , C ]}, t ∈[a,b]

t ∈[a,b]

Moreover, F is continuous and bounded on

[a, b] × [α(t ), β(t )] × R × [ min α(t ), max β(t )]. t ∈[a,b]

t ∈[a,b]

From Lemma 1 and the proof of Theorem 1, there exists a constant C1 > 0 such that any solution x of

 x00 = F (t , x, x0 ; y), t ∈ [a, b], Φ (x(a), x0 (a); y(a)) = G(x(c ); y(c )), Ψ (x(b), x0 (b); y(b)) = H (x(d); y(d)), for y fixed, α(t ) ≤ y ≤ β(t ), t ∈ [a, b] with the property α(t ) ≤ x ≤ β(t ), t ∈ [a, b] must satisfy |x0 (t )| ≤ C1 on [a, b]. Let α0 = α and consider the four-point problem

 x00 = F (t , x, x0 ; α0 ), t ∈ [a, b], Φ (x(a), x0 (a); α0 (a)) = G(x(c ); α0 (c )), Ψ (x(b), x0 (b); α (b)) = H (x(d); α (d)). 0 0

(32)

From (A1 ) and (27), we have F (t , α0 , α00 ; α0 ) = f (t , α0 , α00 ) ≤ α000 ,

Φ (α0 (a), α00 (a); α0 (a)) = φ(α0 (a)) − [α00 (a)]l ≤ g (α0 (c )) = G(α0 (c ); α0 (c )), Ψ (α0 (b), α00 (b); α0 (b)) = ψ(α0 (b)) + [α00 (b)]q ≤ h(α0 (d)) = H (α0 (d); α0 (d)). From (A3 ) and (27), we have F (t , β, β 0 ; α0 ) − f (t , β, β 0 ) = f (t , α0 , β 0 ) + fx (t , α0 , β 0 )(β − α0 ) +

Mf

2 ≥ [fx (t , α0 , β 0 ) − fx (t , ξ , β 0 )](β − α0 ) ≥ 0,

(β − α0 )2 − f (t , β, β 0 ) ξ ∈ [α0 , β],

so

β 00 ≤ f (t , β, β 0 ) ≤ F (t , β, β 0 ; α0 ). From (A5 ), (28) and (30), we have

Φ (β(a), β 0 (a); α0 (a)) ≥ φ(β(a)) − [β 0 (a)]l ≥ g (β(c )) ≥ G(α0 (c ); α0 (c )), From (A5 ), (29) and (31), we have

Ψ (β(b), β 0 (b); α0 (b)) ≥ ψ(β(b)) + [β 0 (b)]q ≥ h(β(d)) ≥ H (α0 (d); α0 (d)), which imply that α0 and β are, respectively, lower and upper solutions of (32). Since α0 ≤ β on [a, b], hence by Theorem 1, there exists a solution α1 of (32) such that

α0 (t ) ≤ α1 (t ) ≤ β(t ), |α10 (t )| < C1 ,

t ∈ [a, b].

L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

1357

Similarly, we can show that α1 and β are lower and upper solutions of

 x00 = F (t , x, x0 ; α1 ), t ∈ [a, b], Φ (x(a), x0 (a); α1 (a)) = G(x(c ); α1 (c )), Ψ (x(b), x0 (b); α (b)) = H (x(d); α (d)). 1 1

(33)

Hence by Theorem 1, there exists a solution α2 of (33) such that

α1 (t ) ≤ α2 (t ) ≤ β(t ), |α20 (t )| < C1 ,

t ∈ [a, b].

Continuing this process, we obtain a monotone sequence {αn } satisfying

α0 (t ) ≤ α1 (t ) ≤ α2 (t ) ≤ · · · ≤ αn (t ) ≤ β(t ),

|wn0 (t )| < C1 ,

t ∈ [a, b]

(34)

where the element αn of the sequence {αn } satisfies BVP

 αn00 = F (t , αn , αn0 ; αn−1 ), t ∈ [a, b], Φ (αn (a), α 0 (a); αn−1 (a)) = G(αn (c ), αn−1 (c )), Ψ (α (b), αn0 (b); α (b)) = H (α (d), α (d)). n n −1 n n −1 n Since F (t , αn , αn0 ; αn−1 ) is bounded on [a, b], we can find a constant A > 0 such that

|F (t , αn , αn0 ; αn−1 )| < A,

t ∈ [a, b].

From

α (t ) = α (a) + 0

0

t

Z

α 00 (s)ds, a

we obtain

|αn (t ) − αn (s)| ≤ 0

0

t

Z

|F (u, αn , αn0 ; αn−1 )|du ≤ A|t − s|,

(35)

s

for any t , s ∈ [a, b], (s ≤ t ). Inequalities (34) and (35) imply that the sequences {αni }(i = 0, 1) are uniformly bounded and equi-continuous on [a, b] and hence the Arzela–Ascoli theorem guarantees the existence of subsequences and a function (j) x ∈ C 1 [a, b] with αn (j = 0, 1) converging uniformly to x(j) on [a, b] as n → ∞. Passing to the limit, we obtain F (t , αn , αn0 ; αn−1 ) → f (t , x, p(x0 )). Thus x(t ) is a solution of (26) and hence is a solution of BVP (1)–(2). The uniqueness of x(t ) follows from Theorem 2 immediately.  Corollary 1. Assume that all the conditions of Theorem 3 hold. Then, by using the analogue technique of Theorem 3, we can also prove that there exists a monotone decreasing sequence {βn } of solutions converging uniformly to the unique solution of BVP (1)–(2). Now we consider the linear problem BVP (1) (3). Theorem 4. Assume that

(A1 ) There exist lower and upper solutions of BVP (1) (3), α(t ), β(t ) ∈ C 2 [a, b], respectively, such that α(t ) ≤ β(t ), t ∈ [a, b]; (A2 ) f (t , x, y) is continuous on [a, b] × R2 and satisfies one-sided Nagumo condition on [a, b] relative to α, β ; (A3 ) fx (t , x, y) > 0 and fxx (t , x, y) is continuous on [a, b] × [mint ∈[a,b] α(t ), maxt ∈[a,b] β(t )] × R, f (t , x, x0 ) is nonincreasing in x0 for x0 ≤ −C , nondecreasing in x0 for x0 ≥ C , where (t , x) ∈ [a, b] × [mint ∈[a,b] α(t ), maxt ∈[a,b] β(t )]; (A4 ) g (x), h(x) are continuous on R and g 0 (x) > 0, g 00 (x) = 0 on [α(c ), β(c )], h0 (x) > 0, h00 (x) = 0 on [α(d), β(d)]; (A5 ) p0 ≥ maxx∈[α(c ),β(c )] g 0 (x), p1 ≥ maxx∈[α(d),β(d)] h0 (x); (A6 ) there exists k1 ∈ C [R+ , R+ ] and k2 ∈ C [[a, b], R+ ] such that |f (t , x, z1 ) − f (t , x, z2 )| ≤ k2 (t )k1 (|z1 − z2 |) on

[ a, b ] ×



 min α(t ), max β(t ) × [−C , C ],

t ∈[a,b]

t ∈[a,b]

where k1 satisfies k1 (0) = 0. Then there exists a monotone nondecreasing sequence {αn (t )} of solutions converging quadratically to the unique solution of BVP (1) (3).

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L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

Proof. From Theorem 3, we know there exists a monotone sequence {αn (t )} of solutions converging uniformly to the unique solution of BVP (1) (3). Now we show that the convergence of the sequence {αn (t )} is quadratic. Let en (t ) = x(t ) − αn (t ), t ∈ [a, b], where x(t ) is the unique solution of BVP (1) (3). Then en (t ) ≥ 0, t ∈ [a, b]. In view of the definition of p(y) and the assumptions, we have f (t , x, p(αn0 (t ))) = f (t , x, αn0 (t )),

if |αn0 (t )| ≤ C ,

f (t , x, p(αn (t ))) ≤ f (t , x, αn (t )),

if |αn0 (t )| > C ,

0

0

and e00n (t ) = x00 (t ) − αn00 (t ) = f (t , x, x0 ) − F (t , αn (t ), αn0 (t ); αn−1 (t ))

= f (t , x, x0 ) − f (t , αn−1 , p(αn0 )) − fx (t , αn−1 , p(αn0 ))(αn − αn−1 ) −

Mf 2

(αn − αn−1 )2

= f (t , x, x0 ) − f (t , x, p(αn0 )) + fx (t , αn−1 , p(αn0 ))(x − αn ) + f (t , x, p(αn0 )) − f (t , αn−1 , p(αn0 )) − fx (t , αn−1 , p(αn0 ))(x − αn−1 ) −

Mf

− f (t , αn−1 , p(αn0 )) − fx (t , αn−1 , p(αn0 ))(x − αn−1 ) −

Mf

(αn − αn−1 )2

2 ≥ f (t , x, x0 ) − f (t , x, αn0 ) + fx (t , αn−1 , p(αn0 ))(x − αn ) + f (t , x, p(αn0 )) 2

(αn − αn−1 )2 .

Therefore e00n (t ) ≥ fx (t , αn−1 (t ), p(αn0 (t )))en (t ) − k2 (t )k1 (|e0n (t )|) −

1 2

fxx (t , ξ , p(αn0 ))e2n−1 (t ) −

Mf 2

(αn (t ) − αn−1 (t ))2

≥ µen (t ) − k2 (t )k1 (|e0n (t )|) − Mf ken−1 k2∞ , with p0 en (a) − e0n (a) = r0 en (c ),

p1 en (b) + e0n (b) = r1 en (d),

where αn−1 < ξ < x, µ = min{fx (t , x, y) : (t , x, y) ∈ [a, b] × [mint ∈[a,b] α(t ), maxt ∈[a,b] β(t )] × [−C , C ]},ken−1 k∞ = maxt ∈[a,b] |en−1 (t )|, r0 ≡ g 0 (x) on [α(c ), β(c )] and r1 ≡ h0 (x) on [α(d), β(d)]. ken−1 k2∞ . It is easy to see that Let W (t ) = M λ W 00 (t ) = λW (t ) − k2 (t )k1 (|W 0 (t )|) − M ken−1 k2∞ , p0 W (a) − W 0 (a) ≥ r0 W (c ), p1 W (b) + W 0 (b) ≥ r1 W (d). From Lemma 3, we can get en (t ) ≤ W (t ) =

M

λ

ken−1 k2∞ ,

t ∈ [a, b].

Thus

ken k∞ ≤ ken−1 k2∞ .  Corollary 2. In Theorem 4, we replace f (t , x, x0 ) is nonincreasing in x0 for x0 ≤ −C , nondecreasing in x0 for x0 ≥ C by f (t , x, x0 ) is nondecreasing in x0 for x0 ≤ −C , nonincreasing in x0 for x0 ≥ C and assume all other conditions hold. Then there exists a monotone nonincreasing sequence {βn (t )} of solutions converging quadratically to the unique solution of BVP (1) (3). 4. Example Next we discuss two examples to illustrate the efficiency of the obtained results. Example 1. Consider the problem of x00 = (2 + t )x + x3 − [x0 ]4 ,

0 < t < 1,

1

1

4

2

− e−x(0) + x(0) + 1 − [x0 (0)]l =  cos

1 2



x(1) + x(1) −

where l, q are odd numbers.

1

  x

1

4

(36)

,

1 −x e + [x0 (1)]q = − e 4 4

(37)   1 2

,

(38)

L. Wang et al. / Mathematical and Computer Modelling 50 (2009) 1348–1359

1359

Let f (t , x, y) = (2 + t )x + x3 − y4 , 1 1 φ(x) = − e−x + x + 1, g (x) = x, 4 2 1 1 1 ψ(x) = cos x + x − e, h(x) = − e−x , 2 4 4 and choose α(t ) = −1, β(t ) = 0. Then it is easy to check that all the assumptions of Theorem 3 and Corollary 1 are satisfied. Therefore, from Theorem 3 and Corollary 1, we can obtain monotone increasing sequence {αn (t )} and monotone decreasing sequence {βn (t )} converging uniformly to the unique solution of BVP (36)–(38). Example 2. Consider the problem of x00 = (2 + t )x + x3 − [x0 ]4 , 2x(0) − x0 (0) = x 3x(1) + x (1) = x 0

  1 4

  1 2

0 < t < 1,

(39)

,

(40)

,

(41)

choose α(t ) = −1, β(t ) = 0. Then it is easy to check that all the assumptions of Corollary 2 are satisfied. Therefore, from Corollary 2, we can obtain the monotone decreasing sequence {βn (t )} converging quadratically to the unique solution of BVP (39)–(41). But we cannot get this result from theorem 3.1 in [2].

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