Upper and lower solutions method for second order nonlinear four point boundary value problems

UPPER AND LOWER SOLUTIONS METHOD FOR SECOND ORDER NONLINEAR FOUR

POINT BOUNDARY VALUE PROBLEMS

Rahmat Ali Khan, Juan J. Nieto, and Rosana Rodr´ıguez-L´ opez

Abstract. We develop the upper and lower solutions method for the four point problem relative to second order differential equations in order to obtain the existence of solution.

1. Introduction

We consider the following second order nonlinear differential equation with four point boundary conditions

(1)

½ x

(t) = f (t, x(t), x

(t)), t ∈ I = [a, b], x(a) = x(c), x(b) = x(d),

where a < c ≤ d < b and f : I × R

→ R is continuous.

The method of upper and lower solutions is used to prove the ex- istence of solution to (1). To obtain a sequence which converges to a solution for this equation, approximated problems are considered. We extend some results of [5].

Note that (1) is a problem at resonance since any constant function is a solution for the linear equation x

= 0 with the four-point boundary condition.

Received June 22, 2005.

2000 Mathematics Subject Classification: Primary 34A34, 34B10, 34B15.

Key words and phrases: second-order differential equations, four point problems, upper and lower solutions.

The research of R. A. Khan has been supported by Ministry of Science and

Technology, Pakistan. The research of J. J. Nieto and R. Rodr´ıguez-L´opez has

been partially supported by Ministerio de Educaci´on y Ciencia and FEDER, project

MTM2004 - 06652 - C03 - 01, and by Xunta de Galicia and FEDER, project

PGIDIT02PXIC20703PN.

In this paper, the type of upper and lower solutions of (1) admissible is more general, whereas in [5], upper and lower solutions were considered constant functions.

In Section 2, the terms of upper and lower solutions of (1) are defined and an existence result for problem (1) is given, considering the conti- nuity of f and the existence of well-ordered upper and lower solutions.

In [1], it is considered different four-point boundary condition so that the linear part is invertible (see Lemma 2.2 for the expression of the corresponding Green’s function).

Also, in [2], it is studied a four-point boundary value problem for second-order nonlinear ordinary differential equations with the nonlin- earity independent of the derivative. Positive solutions of nonlinear four- point boundary value problems have been studied recently [3, 4].

2. Main results

We consider the spaces of functions C(I) and C

(I) furnished, re- spectively, with the norms

kxk = sup{|x(t)| : t ∈ I}, kxk

= kxk + kx

k.

Definition 1. A function α ∈ C

[a, b] is said to be a lower solution to problem (1) if

α

(t) ≥ f (t, α(t), α

(t)), t ∈ [a, b], α(a) ≤ α(c), α(d) ≥ α(b).

Similarly, β ∈ C

[a, b] is an upper solution to problem (1) if β

(t) ≤ f (t, β(t), β

(t)), t ∈ [a, b],

β(a) ≥ β(c), β(d) ≤ β(b).

This formulation includes the constant case, considered in [5], since functions

α(t) = r

, β(t) = r

, t ∈ [a, b],

where r

, r

∈ R, verify the boundary conditions in the previous def-

inition. So that the existence result we present is a generalization of

Lemma 2 in [5].

Theorem 2.1. Suppose that α, β ∈ C

[a, b] are, respectively, lower and upper solutions to problem (1), with α ≤ β on I = [a, b] and that there exists K ∈ R, K > 0 such that

Z

a

|f (t, x, y)| dt ≤ K, ∀x ∈ [α(t), β(t)], y ∈ R.

Then problem (1) has at least one solution u verifying that α(t) ≤ u(t) ≤ β(t), t ∈ [a, b].

Proof. Consider m ∈ N, m > 1 fixed and define f

(t, x, y)

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



f (t, β(t), β

(t)) +

, if β(t) +

≤ x, f (t, β(t), y) +

h

f (t, β(t), β

(t)) − f (t, β(t), y) +

i

×m(x − β(t)), if β(t) < x < β(t) +

, f (t, x, y), if α(t) ≤ x ≤ β(t),

f (t, α(t), y) − h

f (t, α(t), α

(t)) − f (t, α(t), y) +

i

×m(x − α(t)), if α(t) −

< x < α(t), f (t, α(t), α

(t)) +

, if x ≤ α(t) −

.

Note that f

is continuous on I × R

. We consider the problem (2)

½ x

(t) = f

(t, x(t), x

(t)), t ∈ I = [a, b], x(a) = x(c), x(b) = x(d).

If x is a solution to (2) such that α ≤ x ≤ β, then x is a solution to (1).

Take r = max

β − min

α. If min

α = max

β, then α = β is a constant solution to (1). We can assume, therefore, that r > 0, and define σ =

, then σ > 0. Choose R ≥ 0 such that

|f (t, α(t), α

(t))|, |f (t, β(t), β

(t))| ≤ R, t ∈ I.

Take M

> 0 large enough such that 0 ≤ R ≤ M

σ.

Therefore,

−M

σ ≤ −R ≤ f (t, α(t), α

(t)), f (t, β(t), β

(t)) ≤ R ≤ M

σ, t ∈ I.

Define bounded and continuous functions K

(t, x) =

 



MR1σ

, for x ≥ α(t) +

, m(x − α(t)), for α(t) ≤ x ≤ α(t) +

1σ

,

0, for x ≤ α(t),

K

(t, x) =

 



0, for x ≥ β(t),

m(x − β(t)), for β(t) −

≤ x ≤ β(t),

−

, for x ≤ β(t) −

. Taking

φ(t, x) = [K

(t, x) + K

(t, x)]M

σ, (t, x) ∈ I × R,

then φ is continuous and bounded on I × R and satisfies −R ≤ φ ≤ R on I × R.

Denote

p(t, x) = max{α(t), min{x, β(t)}}.

To prove existence of solution to (2), consider problem (3)

½ x

(t) = ¯ f

(t, x(t), x

(t), λ), t ∈ I = [a, b], x(a) = x(c), x(b) = x(d),

where

f ¯

(t, x, y, λ) = λf

(t, x, y) + (1 − λ) {φ(t, x) − p(t, x) + x} , and λ ∈ [0, 1].

Note that if x is a solution to (3) for λ = 1 and x(t) ∈ [α(t), β(t)],

∀t ∈ I, then

x

(t) = f

(t, x(t), x

(t)) = f (t, x(t), x

(t)) and x is a solution to (1).

If λ = 0, problem (3) is reduced to (4)

½ x

(t) = φ(t, x(t)) − p(t, x(t)) + x(t), t ∈ I, x(a) = x(c), x(b) = x(d),

and, for λ = 1, it is (2). Since φ(t, x)−p(t, x) is continuous and bounded and the linear problem

½ x

(t) = x(t), t ∈ I, x(a) = x(c), x(b) = x(d),

has only the trivial solution, then problem (4) has a solution, i.e., prob- lem (3) has a solution for λ = 0.

For λ ∈ [0, 1], any solution u to problem (3) satisfies that α(t) − 1

m ≤ u(t) ≤ β(t) + 1

m , t ∈ I.

Indeed, let λ ∈ [0, 1] and u a solution to (3). Consider the function v(t) = u(t) − β(t) −

and suppose that max

v(t) = v(t

) > 0. Using the boundary conditions, we obtain that

v(a) = u(a) − β(a) − 1

m ≤ u(c) − β(c) − 1

m = v(c), v(d) ≥ v(b), so that we can assume that t

∈ (a, b). Then

v(t

) > 0, v

(t

) = 0, v

(t

) ≤ 0, that is,

u(t

) > β(t

) + 1

m , u

(t

) = β

(t

).

Then

u(t

) > β(t

) + 1

m ≥ α(t

) + 1

m ≥ α(t

) + R mM

σ , so that

K

(t

, u(t

)) = R

M

σ , K

(t

, u(t

)) = 0, and, using the definition of φ and the hypotheses on β,

v

(t

) = u

(t

) − β

(t

)

= λf

(t

, u(t

), u

(t

))

+(1 − λ) {φ(t

, u(t

)) − p(t

, u(t

)) + u(t

)} − β

(t

)

≥ λf

(t

, u(t

), u

(t

))

+(1 − λ) {φ(t

, u(t

)) − p(t

, u(t

)) + u(t

)}

−f (t

, β(t

), β

(t

))

= λ

·

f (t

, β(t

), β

(t

)) + u(t

) − β(t

) 1 + |u(t

) − β(t

)|

¸

+(1 − λ) [R − β(t

) + u(t

)] − f (t

, β(t

), β

(t

))

= (1 − λ) £

R − f (t

, β(t

), β

(t

)) ¤

+ (u(t

) − β(t

))

×

· λ

1 + |u(t

) − β(t

)| + (1 − λ)

¸

> (1 − λ) £

R − f (t

, β(t

), β

(t

)) ¤

≥ 0,

which is a contradiction. This proves that v ≤ 0 on I, so that u(t) ≤ β(t) +

, t ∈ I.

Also, we can prove that u(t) ≥ α(t) −

, for all t ∈ I. Defining v(t) = u(t) − α(t) +

, if min

v(t) = v(t

) < 0, then

v(a) ≥ v(c), v(d) ≤ v(b),

so that we can assume that t

∈ (a, b). Then

v(t

) < 0, v

(t

) = 0, v

(t

) ≥ 0, which imply

u(t

) < α(t

) − 1

m , u

(t

) = α

(t

).

Hence

u(t

) < α(t

) − 1

m ≤ β(t

) − 1

m ≤ β(t

) − R mM

σ , and, therefore,

K

(t

, u(t

)) = 0, K

(t

, u(t

)) = − R M

σ . Using the definition of φ and the hypotheses satisfied by α,

v

(t

) = u

(t

) − α

(t

)

≤ λf

(t

, u(t

), u

(t

))

+(1 − λ) {φ(t

, u(t

)) − p(t

, u(t

)) + u(t

)}

−f (t

, α(t

), α

(t

))

= λ

·

f (t

, α(t

), α

(t

)) + u(t

) − α(t

) 1 + |u(t

) − α(t

)|

¸

+(1 − λ) [−R − α(t

) + u(t

)] − f (t

, α(t

), α

(t

))

= −(1 − λ) £

R + f (t

, α(t

), α

(t

)) ¤

+ (u(t

) − α(t

))

×

· λ

1 + |u(t

) − α(t

)| + (1 − λ)

¸

< −(1 − λ) £

R + f (t

, α(t

), α

(t

)) ¤

≤ 0,

obtaining a contradiction again. This proves that u(t) ≥ α(t) − 1 m , for all t ∈ I.

The boundary conditions on u imply that there exists ˜t ∈ (a, b) with u

(˜t) = 0, so that, integrating the equation in (3), we obtain, for t > ˜t, that

u

(t) = u

(t) − u

(˜t) = Z

˜t

u

(s) ds

= Z

˜t

£ λf

(s, u(s), u

(s)) + (1 − λ) {φ(s, u(s)) − p(s, u(s)) + u(s)} ¤

ds,

so that

|u

(t)| ≤ Z

˜t

|λf

(s, u(s), u

(s)) + (1 − λ) {φ(s, u(s)) − p(s, u(s)) + u(s)}| ds

≤ Z

a

|λf

(s, u(s), u

(s)) + (1 − λ) {φ(s, u(s)) − p(s, u(s)) + u(s)}| ds.

A similar estimate can be obtained for t < ˜t, so that, for all t ∈ I, we get

|u

(t)| ≤ Z

a

|f

(s, u(s), u

(s))| ds + Z

a

|φ(s, u(s)) − p(s, u(s)) + u(s)| ds.

Since α(t) − 1

m ≤ u(t) ≤ β(t) + 1

m , for t ∈ I, and −R ≤ φ(s, u(s)) ≤ R, s ∈ I, then

−R − β(s) + α(s) − 1

m ≤ φ(s, u(s)) − p(s, u(s)) + u(s)

≤ R − α(s) + β(s) + 1

m , s ∈ I, and

|φ(s, u(s)) − p(s, u(s)) + u(s)| ≤ R + β(s) − α(s) + 1

m , s ∈ I, which implies that

|u

(t)| ≤ Z

a

|f

(s, u(s), u

(s))| ds+

Z

a

·

R + β(s) − α(s) + 1 m

¸

ds, t ∈ I.

The expression α(t) − 1

m ≤ u(t) ≤ β(t) + 1

m , for t ∈ I, also provides that f

(s, u(s), u

(s)) can be equal to

f (s, β(s), u

(s)) +

·

f (s, β(s), β

(s)) − f (s, β(s), u

(s)) + u(s) − β(s) 1 + |u(s) − β(s)|

¸

× m(u(s) − β(s)), if β(s) < u(s) < β(s) + 1 m , f (s, u(s), u

(s)), if α(s) ≤ u(s) ≤ β(s), f (s, α(s), u

(s))

−

·

f (s, α(s), α

(s)) − f (s, α(s), u

(s)) + u(s) − α(s) 1 + |u(s) − α(s)|

¸

× m(u(s) − α(s)), if α(s) − 1

m < u(s) < α(s).

For β(s) < u(s) < β(s) + 1 m ,

|f

(s, u(s), u

(s))|

≤ |f (s, β(s), u

(s))|

+

¯ ¯

¯ ¯f (s, β(s), β

(s)) − f (s, β(s), u

(s)) + u(s) − β(s) 1 + |u(s) − β(s)|

¯ ¯

¯ ¯

× m (u(s) − β(s))

≤ |f (s, β(s), u

(s))|

+

¯ ¯

¯ ¯f (s, β(s), β

(s)) − f (s, β(s), u

(s)) + u(s) − β(s) 1 + |u(s) − β(s)|

¯ ¯

¯ ¯ m 1 m

≤ |f (s, β(s), u

(s))| + |f (s, β(s), β

(s))| + |f (s, β(s), u

(s))|

+ u(s) − β(s) 1 + u(s) − β(s)

< |f (s, β(s), u

(s))| + |f (s, β(s), β

(s))| + |f (s, β(s), u

(s))| + 1 m , and, for α(s) − 1

m < u(s) < α(s),

|f

(s, u(s), u

(s))|

≤ |f (s, α(s), u

(s))|

+

¯ ¯

¯ ¯f (s, α(s), α

(s)) − f (s, α(s), u

(s)) + u(s) − α(s) 1 + |u(s) − α(s)|

¯ ¯

¯ ¯

× m |u(s) − α(s)|

≤ |f (s, α(s), u

(s))|

+

¯ ¯

¯ ¯f (s, α(s), α

(s)) − f (s, α(s), u

(s)) + u(s) − α(s) 1 + |u(s) − α(s)|

¯ ¯

¯ ¯ m 1 m

≤ |f (s, α(s), u

(s))| + |f (s, α(s), α

(s))| + |f (s, α(s), u

(s))|

+

¯ ¯

¯ ¯ u(s) − α(s) 1 + |u(s) − α(s)|

¯ ¯

¯ ¯

< |f (s, α(s), u

(s))| + |f (s, α(s), α

(s))| + |f (s, α(s), u

(s))| + 1 m , so that

Z

a

|f

(s, u(s), u

(s))| ds ≤ 3 K + 1

m (b − a),

which implies

ku

k ≤ 3 K + 1

m (b − a) + Z

a

·

R + β(s) − α(s) + 1 m

¸ ds, in particular,

ku

k ≤ 3 K + (b − a) + Z

a

[R + β(s) − α(s) + 1] ds = γ, with γ independent of m.

Then, for λ = 1, problem (3) has a solution u

such that α(t) − 1

m ≤ u

(t) ≤ β(t) + 1

m , t ∈ I.

Repeating this procedure for all m ∈ N, m > 1, we obtain a sequence {u

}

, where u

is a solution to problem

 



x

(t) = f

(t, x(t), x

(t)), t ∈ I, x(a) = x(c), x(b) = x(d),

and

ku

k ≤ max{kαk, kβk} + 1, ku

k ≤ γ, so that the sequence {u

} is bounded in C

(I) and

ku

(t)k = kf

(t, u

(t), u

(t))k

is bounded, so that {u

} is equicontinuous in C

(I). Thus, Arzel´a- Ascoli Theorem implies the existence of a subsequence of {u

} conver- gent in C

(I) to a function u. Since

α(t) − 1

m ≤ u

(t) ≤ β(t) + 1

m , for all t ∈ I, then

α(t) ≤ u(t) ≤ β(t), t ∈ I.

Moreover, ku

k ≤ γ, or more specifically, ku

k ≤ 3 K +

Z

a

[R + β(s) − α(s)] ds.

Function u is a solution to problem (1).

Remark 1. Theorem 2.1 extends Lemma 2 in [5], where constant

upper and lower solutions for (1) are considered.

Theorem 2.2. Suppose that α, β are, respectively, lower and upper solutions of problem (1) such that α ≤ β on I and that there exist real numbers R

, R

, R

, R

such that R

6= R

, R

6= R

,

R

, R

≤ − max{kα

k, kβ

k}, R

, R

≥ max{kα

k, kβ

k}

and

f (t, x, R

) ≥ 0, f (t, x, R

) ≤ 0, ∀t ∈ [a, b], ∀x ∈ [α(t), β(t)], f (t, x, R

) ≥ 0, f (t, x, R

) ≤ 0, for t ∈ [d, b], x ∈ [α(t), β(t)].

Then problem (1) has at least one solution satisfying, for every t ∈ [a, b], that

α(t) ≤ u(t) ≤ β(t),

min{R

, R

} ≤ u

(t) ≤ max{R

, R

}.

Proof. Suppose that R

< R

and R

> R

. Following the proof of Theorem 1 in [5], we can choose n

∈ N such that, for all n ∈ N, n ≥ n

, we have that

R

+ 2

n < R

, R

− 2 n > R

. For n ≥ n

, let

h

(t, x, y) =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



f (t, x, R

), R

< y,

f (t, x, y), R

+

≤ y ≤ R

,

f (t, x, R

+

) + w

, R

+

< y < R

+

,

f (t, x, R

), R

< y ≤ R

+

,

f (t, x, y), R

≤ y ≤ R

,

f (t, x, R

), R

−

≤ y < R

,

f (t, x, R

−

) − w

, R

−

< y < R

−

,

f (t, x, y), R

≤ y ≤ R

−

,

f (t, x, R

), y < R

,

where w

=

· f

µ

t, x, R

+ 2 n

¶

− f (t, x, R

)

¸ n

µ

y − R

− 2 n

¶ ,

w

=

· f

µ

t, x, R

− 2 n

¶

− f (t, x, R

)

¸ n

µ

y − R

+ 2 n

¶ . We check that problem

(5)

½ x

(t) = h

(t, x(t), x

(t)), t ∈ I = [a, b], x(a) = x(c), x(b) = x(d)

is under the hypotheses of Theorem 2.1. Indeed, Z

a

|h

(t, x, y)| dt ≤ K, for x ∈ [α(t), β(t)], y ∈ R, where

K = Z

a

sup{|h

(t, x, y)| : x ∈ [α(t), β(t)], y ∈ R} dt.

Moreover, α, β are, respectively, lower and upper solutions to problem (5), since

α

(t) ≥ f (t, α(t), α

(t)) = h

(t, α(t), α

(t)), t ∈ I, β

(t) ≤ f (t, β(t), β

(t)) = h

(t, β(t), β

(t)), t ∈ I,

where we use that R

≤ α

(t), β

(t) ≤ R

. Using Theorem 2.1, we get that problem (5) has a solution u

satisfying

α(t) ≤ u

(t) ≤ β(t), t ∈ I.

Now, we give estimates for u

. Since u

(a) = u

(c), u

(d) = u

(b), then there exist a

∈ (a, c) and b

∈ (d, b) with u

(a

) = 0, u

(b

) = 0.

Suppose that max{u

(t) : t ∈ [a, b

]} = u

(z

) > R

+

. Then z

6=

b

and there exists (t

, t

) ⊂ (a, b

), with u

(t

) = R

, u

(t

) = R

+

and R

≤ u

(t) ≤ R

+

, for all t ∈ (t

, t

). Integrating u

in the interval (t

, t

) and using the hypotheses on f , we obtain

0 > − 1

n = R

− µ

R

+ 1 n

¶

= u

(t

) − u

(t

) = Z

2

tⁿ₁

u

(s) ds

= Z

2

tⁿ₁

h

(s, u

(s), u

(s)) ds = Z

2

tⁿ₁

f (s, u

(s), R

) ds ≥ 0,

which is a contradiction. If min{u

(t) : t ∈ [a, b

]} = u

(z

) < R

−

,

then z

6= b

and there exists (t

, t

) ⊂ (a, b

), with u

(t

) = R

,

u

(t

) = R

−

and R

−

≤ u

(t) ≤ R

, for all t ∈ (t

, t

). Integrating again, we get

0 < 1

n = R

− µ

R

− 1 n

¶

= u

(t

) − u

(t

) = Z

2

tⁿ₁

u

(s) ds

= Z

2

tⁿ₁

h

(s, u

(s), u

(s)) ds = Z

2

tⁿ₁

f (s, u

(s), R

) ds ≤ 0, another contradiction. We have proved that

R

− 1

n ≤ u

(t) ≤ R

+ 1

n , for all t ∈ [a, b

].

Now, suppose that max{u

(t) : t ∈ [b

, b]} = u

(z

) > R

+

. Then z

∈ (b

, b] and there exists (t

, t

) ⊂ (b

, b), with u

(t

) = R

, u

(t

) = R

+

, and R

≤ u

(t) ≤ R

+

, for all t ∈ (t

, t

), so that

0 < 1

n = R

+ 1

n − R

= u

(t

) − u

(t

) = Z

2

tⁿ₁

u

(s) ds

= Z

2

tⁿ₁

h

(s, u

(s), u

(s)) ds = Z

2

tⁿ₁

f (s, u

(s), R

) ds ≤ 0, which is absurd and, similarly, if min{u

(t) : t ∈ [b

, b]} = u

(z

) <

R

−

, then z

∈ (b

, b] and there exists (t

, t

) ⊂ (b

, b), with u

(t

) = R

−

, u

(t

) = R

and R

−

≤ u

(t) ≤ R

, for all t ∈ (t

, t

). Thus

0 > − 1

n = R

− 1

n − R

= u

(t

) − u

(t

) = Z

2

tⁿ₁

u

(s) ds

= Z

2

tⁿ₁

h

(s, u

(s), u

(s)) ds = Z

2

tⁿ₁

f (s, u

(s), R

) ds ≥ 0, which is again a contradiction. Therefore,

R

− 1

n ≤ u

(t) ≤ R

+ 1

n , t ∈ [b

, b], R

− 1

n < R

− 1

n ≤ u

(t) ≤ R

+ 1

n < R

+ 1

n , t ∈ [a, b

], that is,

min{R

, R

}− 1

n = R

− 1

n ≤ u

(t) ≤ R

+ 1

n = max{R

, R

}+ 1 n , t ∈ I.

This proves that the sequence {u

}

is bounded and equicontinuous

in C

(I), so that there exists a subsequence convergent in C

(I) to a

function u satisfying

α(t) ≤ u(t) ≤ β(t), R

≤ u

(t) ≤ R

. In consequence, u is a solution to equation

x

(t) = h(t, x(t), x

(t)), t ∈ I, x(a) = x(c), x(b) = x(d), where

h(t, x, y) =

 



f (t, x, R

), y > R

, f (t, x, y), R

≤ y ≤ R

, f (t, x, R

), y < R

,

but R

≤ u

(t) ≤ R

, t ∈ I, then u is a solution to (1) with α(t) ≤ u(t) ≤ β(t),

min{R

, R

} = R

≤ u

(t) ≤ R

= max{R

, R

}.

If R

< R

, R

< R

is not true, then the proof can be completed following a similar argument.

The following result studies the case where R

= R

, R

= R

. Theorem 2.3. Suppose that α, β are, respectively, lower and up- per solutions of problem (1) such that α ≤ β on I and that there exist real numbers R

, R

, such that R

≤ − max{kα

k, kβ

k}, R

≥ max{kα

k, kβ

k} and

f (t, x, R

) ≥ 0, f (t, x, R

) ≤ 0, ∀t ∈ [a, b], ∀x ∈ [α(t), β(t)], f (t, x, R

) ≥ 0, f (t, x, R

) ≤ 0, for t ∈ [d, b], x ∈ [α(t), β(t)].

Then problem (1) has at least one solution satisfying α(t) ≤ u(t) ≤ β(t), R

≤ u

(t) ≤ R

, t ∈ I.

Proof. Let h(t, x, y) =

 



f (t, x, R

), R

< y, f (t, x, y), R

≤ y ≤ R

, f (t, x, R

), y < R

,

 

 = f (t, x, q(y)), where

q(y) = max{R

, min{y, R

}}.

In this case, for problem (6)

½ x

(t) = h(t, x(t), x

(t)) = f (t, x(t), q(x

(t))), t ∈ I = [a, b],

x(a) = x(c), x(b) = x(d),

the hypotheses of Theorem 2.1 hold. Indeed, Z

a

|h(t, x, y)| dt ≤ K, for x ∈ [α(t), β(t)], y ∈ R, where

K = Z

a

sup{|f (t, x, y)| : x ∈ [α(t), β(t)], y ∈ [R

, R

]} dt.

The conditions on R

and R

allow to prove that

α

(t) ≥ f (t, α(t), α

(t)) = h(t, α(t), α

(t)), t ∈ I, β

(t) ≤ f (t, β(t), β

(t)) = h(t, β(t), β

(t)), t ∈ I,

so that functions α, β are, respectively, lower and upper solutions to problem (6).

Applying Theorem 2.1, we get that problem (6) has a solution u satisfying

α(t) ≤ u(t) ≤ β(t), t ∈ I.

We prove that R

≤ u

(t) ≤ R

, t ∈ I. The boundary conditions imply that u(a) = u(c), u(d) = u(b), then there exist a

∈ (a, c) and b

∈ (d, b) with u

(a

) = 0, u

(b

) = 0.

Suppose that max{u

(t) : t ∈ [a, b

]} = u

(z

) > R

. Then z

6= b

and there exists (t

, t

) ⊂ (a, b

), with u

(t

) = R

, u

(t

) > R

and R

≤ u

(t) ≤ u

(t

), for all t ∈ (t

, t

). Integrating u

in the interval (t

, t

) and using the hypotheses, we get

0 > R

− u

(t

) = u

(t

) − u

(t

) = Z

t1

u

(s) ds

= Z

t1

h(s, u(s), u

(s)) ds = Z

t1

f (s, u(s), R

) ds ≥ 0, getting a contradiction. If min{u

(t) : t ∈ [a, b

]} = u

(z

) < R

, then z

6= b

and there exists (t

, t

) ⊂ (a, b

), with u

(t

) = R

, u

(t

) < R

and u

(t

) ≤ u

(t) ≤ R

, for all t ∈ (t

, t

). Integrating again, we get

0 < R

− u

(t

) = u

(t

) − u

(t

) = Z

t1

u

(s) ds

= Z

t1

h(s, u(s), u

(s)) ds = Z

t1

f (s, u(s), R

) ds ≤ 0, again a contradiction. This provides that

R

≤ u

(t) ≤ R

, for all t ∈ [a, b

].

Now, suppose that max{u

(t) : t ∈ [b

, b]} = u

(z

) > R

. Then z

∈ (b

, b] and there exists (t

, t

) ⊂ (b

, b), with u

(t

) = R

, u

(t

) > R

, and R

≤ u

(t) ≤ u

(t

), for all t ∈ (t

, t

), so that

0 < u

(t

) − R

= u

(t

) − u

(t

) = Z

t1

u

(s) ds

= Z

t1

h(s, u(s), u

(s)) ds = Z

t1

f (s, u(s), R

) ds ≤ 0, which is absurd and, similarly, if min{u

(t) : t ∈ [b

, b]} = u

(z

) < R

, then z

∈ (b

, b] and there exists (t

, t

) ⊂ (b

, b), with u

(t

) < R

, u

(t

) = R

, and u

(t

) ≤ u

(t) ≤ R

, for all t ∈ (t

, t

), thus

0 > u

(t

) − R

= u

(t

) − u

(t

) = Z

t1

u

(s) ds

= Z

t1

h(s, u(s), u

(s)) ds = Z

t1

f (s, u(s), R

) ds ≥ 0, which is a contradiction.

In consequence,

R

≤ u

(t) ≤ R

, t ∈ I,

and q(u

(t)) = u

(t), t ∈ I, hence u is a solution to (1) with the properties α(t) ≤ u(t) ≤ β(t), t ∈ I,

R

≤ u

(t) ≤ R

, t ∈ I.

The proof is complete.

References

[1] Z. Bai, W. Ge, and Y. Wang, Multiplicity results for some second-order four-point boundary-value problems, Nonlinear Anal. 60 (2005), no. 3, 491–500.

[2] Z. Bai, W. Li, and W. Ge, Existence and multiplicity of solutions for four-point boundary value problems at resonance, Nonlinear Anal. 60 (2005), no. 6, 1151–

1162.

[3] B. Liu, Positive solutions of a nonlinear four-point boundary value problem, Appl.

Math. Comput. 155 (2004), no. 1, 179–203.

[4] , Positive solutions of a nonlinear four-point boundary value problem in Banach spaces, J. Math. Anal. Appl. 305 (2005), no. 1, 253–276.

[5] I. Rach unkov´a, On four point boundary value problem without growth conditions,

Czechoslovak Math. J. 49 (124) (1999), no. 2, 241–248.

Rahmat Ali Khan

Centre for Advanced Mathematics and Physics

National University of Sciences and Technology (NUST) Campus of College of Electrical and Mechanical Engineering Peshawar Road, Rawalpindi, Pakistan

E-mail: rahmat [email protected]

Juan J. Nieto and Rosana Rodr´ıguez-L´opez Departamento de An´alisis Matem´atico Facultad de Matem´aticas

Universidad de Santiago de Compostela 15782- Santiago de Compostela, Spain E-mail: [email protected]

[email protected]