Dynamic Systems and Applications 15 (2006) 119-132

EXISTENCE OF AT LEAST TWO SOLUTIONS OF SECOND ORDER NONLINEAR THREE POINT BOUNDARY VALUE PROBLEMS

RAHMAT ALI KHAN AND J. R. L. WEBB

Department of Mathematics, University of Glasgow, Glasgow G12 8QW, UK.

Dedicated to John Baxley on the occasion of his retirement from Wake Forest University.

ABSTRACT. We study existence of solutions of some second order nonlinear three point boundary value problems depending on a parameter of the type

−x′′ = f(t, x, x′) − rϕ(t), t ∈ (0, 1)

x(0) = 0, x(1) = δx(η), where η ∈ (0, 1) and 0 < δη < 1.

We show that for a certain range of the parameter there are no solutions while for some other range of values there are at least two solutions. We use upper and lower solutions and topological degree arguments. AMS (MOS) Subject Classification. 34B10, 34B15.

1. INTRODUCTION

In this paper, we study existence and multiplicity results for boundary value problems (BVPs) of the form

(1.1)  −x′′(t) = f(t, x, x′) − rϕ(t), t ∈ I = [0, 1],

subject to the three-point boundary conditions

x(0) = 0, x(1) = δx(η), η ∈ (0, 1), 0 < δη < 1,

where f : I × ℝ^{2} → ℝ and ϕ : I → ℝ are continuous and r is a parameter. Such BVPs have been extensively studied in recent years by many authors by applying Leray-Schauder degree theory for example, [5, 6, 7]. The study of (multiple) positive solutions has been done using fixed point theory in cones, mainly when f does not depend on x′, see, for example [9, 12].

Existence of at least two solutions for other boundary conditions have been studied in many papers, for example [2, 3, 4, 10, 11]. In [11], existence of at least two solutions for (1.1) with ϕ(t) = 1 subject to the four-point boundary conditions is

Received November 26, 2004              1056-2176 $15.00 \begin{matrix}c \\ ◯ \end{matrix}Dynamic Publishers, Inc.

120 RAHMAT ALI KHAN AND J. R. L. WEBB

studied in the presence of constant lower and upper solutions, and it is assumed that the set of all solutions is bounded above. In contrast, here we study the problem (1.1) not only with different boundary conditions (three-point boundary conditions) but also assume the existence of lower and upper solutions that are not necessarily constants. Moreover, we do not require the set of solutions to be bounded. In [4], ϕ is taken to be the normalized positive eigenfunction associated with the first eigenvalue λ = 1 of the linear problem

x′′ + λx = 0, x(0) = 0, x(π) = 0,

and existence of at least two solutions for (1.1) with the Dirichlet boundary conditions is studied. It is assumed that the set of all solutions is bounded above and the nonlinearity f(t, x, x′) is bounded for bounded x. In this paper, we allow ϕ to be any positive continuous bounded function and relax the condition on f.

By different methods, existence of at least three solutions of the two point boundary value problem

x′′ + f(t, x, x′) = 0, x(0) = 0, x(1) = 0,

in the presence of two lower and two upper solutions is studied in [8].

The existence of at least one positive solution for some three-point boundary value problems, in the case that the nonlinearity f is non-negative, is given in [5], by using a fixed point theorem in a cone under different hypotheses.

The purpose of this paper is to study how the number of solutions changes as the parameter r varies and to show the existence of no solution and the existence of at least two solutions for certain ranges of values of r. We employ upper and lower solutions, which are not necessarily constant, and use some topological degree arguments.

We give a class of realistic examples to show that our hypotheses can easily be satisfied. Some of the conditions in [5] are more restrictive, in particular (H_{3}) − (H_{5}) of [5] can never be satisfied in the type of example we give.

2. EXISTENCE RESULTS

Consider the problem

\begin{matrix}− x′′(t) = f(t, x, x′) − rϕ(t), t ∈ I = [0, 1], \\ x(0) = 0, x(1) = δx(η), η ∈ (0, 1), 0 < δη < 1. \end{matrix} ; (2.1)

We seek a solution x via the integral equation

(2.2)  x(t) = ∫_{0}^{1} G(t, s)[f(s, x(s), x′(s)) − rϕ(s)] ds,

where,

G(t, s) = \begin{cases}G_{1}(t, s), & 0 < η ≤ s \\ G_{2}(t, s), & 0 < s ≤ η, \end{cases}

G_{1}(t, s) = \frac{1}{1 − δη} \begin{cases}(1 − s)t, & 0 ≤ t ≤ η ≤ s ≤ 1 \\ (1 − δη)s + (δη − s)t, & η ≤ s ≤ t ≤ 1, \end{cases}

G_{2}(t, s) = \frac{1}{1 − δη} \begin{cases}[(1 − s) + δ(s − η)]t, & 0 ≤ t ≤ s ≤ η \\ [(1 − t) + δ(t − η)]s, & 0 ≤ s ≤ η ≤ t ≤ 1. \end{cases}

We recall the concept of upper and lower solutions for the BVP (2.1).

Definition 2.1. Let α ∈ C^{2}(I). We say that α is a strict lower solution of the BVP (2.1), if

\begin{matrix}−α′′(t) < f(t, α(t), α′(t)) − rϕ(t), t ∈ [0, 1] \\ α(0) ≤ 0, α(1) ≤ δα(η). \end{matrix}

A strict upper solution β ∈ C^{2}(I) of the BVP (2.1) is defined similarly by reversing the inequalities.

Definition 2.2. A continuous function ω : [0, ∞) → (0, ∞) is called a Nagumo function if

∫_{λ}^{∞} \frac{s ds}{ω(s)} = ∞,

where λ ≥ 0. Let R(α, β) = {(t, x, y) ∈ I × ℝ^{2} : α(t) ≤ x ≤ β(t)}. We say that f ∈ C[I × ℝ × ℝ] satisfies a Nagumo condition on R(α, β), if there exists a Nagumo function ω such that for every t ∈ I and x ∈ [α, β],

|f(t, x, s)| ≤ ω(|s|), s ∈ ℝ.

Now we state and prove the following theorem which establish the existence of solutions.

Theorem 2.3. Assume that

(a_{1}) r̄ ∈ ℝ and α, β ∈ C^{2}(I) are strict lower and upper solutions of (2.1) with r = r̄,

such that α < β on (0, 1]. (a_{2}) f : I × ℝ^{2} → ℝ is continuous, satisfies a Nagumo condition on R(α, β) and is

such that

\begin{matrix}f(t, α(t), x′) ≥ f(t, α(t), α′(t)) for x′ ≤ α′(t), t ∈ I \\ f(t, β(t), x′) ≤ f(t, β(t), β′(t)) for x′ ≥ β′(t), t ∈ I. \end{matrix}

(a_{3}) ϕ : I → ℝ is continuous, positive and is bounded on I.

Then there exists a solution x(t) of (2.1) (with r = r̄) such that α(t) < x(t) < β(t), t ∈ I.

122 RAHMAT ALI KHAN AND J. R. L. WEBB

Proof. Let max{δ|α(η)|, δ|β(η)|} = λ. Since f satisfies a Nagumo condition on R(α, β), it follows that for every t ∈ I and x ∈ [α(t), β(t)], there exist a Nagumo function ω : [0, ∞) → (0, ∞) and a constant N(> λ) depending on α, β and ω such that

|f(t, x, x′)| ≤ ω(|x′|) on R(α, β) and
(2.3)  ∫_{λ}^{N} \frac{s ds}{ω(s) + M} > max_{t∈I} β(t) − min_{t∈I} α(t),

where M = r̄max{ϕ(t) : t ∈ I}. Let C > max {N, ∥α′∥, ∥β′∥} and define

q(y) = max { − C, min{y, C}},

a retraction onto [−C, C]. Then sgn q(y) = sgn y and q(y) = y for |y| ≤ C. Let ε > 0 be a fixed small number and consider the modified problem

\begin{matrix}− x′′ = F (t, x, x′) − r̄ϕ(t), t ∈ I, \\ x(0) = 0, x(1) = δx(η), 0 < δη < 1, η ∈ (0, 1), \end{matrix} ; (2.4)

where,

F (t, x, x′) = \begin{cases}f(t, β(t), β′(t)), & if x ≥ β(t) + ε, \\ f(t, β(t), q(x′(t))) + [f(t, β(t), β′(t)) \\ −f(t, β(t), q(x′(t)))]\frac{x−β(t)}{ε}, & if β(t) ≤ x < β(t) + ε, \\ f(t, x, q(x′(t))), & if α(t) < x < β(t), \\ f(t, α(t), q(x′(t))) + [f(t, α(t), α′(t)) \\ −f(t, α(t), q(x′(t)))]\frac{α(t)−x}{ε}, & if α(t) − ε < x ≤ α(t), \\ f(t, α(t), α′(t)), & if x ≤ α(t) − ε. \end{cases}

The modified problem (2.4) is equivalent to the integral equation

x(t) = ∫_{0}^{1} G(t, s)[F (s, x(s), x′(s)) − r̄ϕ(s)] ds.

Since F, ϕ and G are continuous and bounded, this integral equation has a fixed point by Schauder’s fixed point theorem. Hence, the BVP (2.4) has a solution. Moreover,

\begin{matrix}−α′′(t) & < f(t, α(t), α′(t)) − r̄ϕ(t) = F (t, α(t), α′(t)) − r̄ϕ(t), t ∈ I \\ α(0) ≤ 0, α(1) ≤ δα(η), \end{matrix}

\begin{matrix}−β′′(t) & > f(t, β(t), β′(t)) − r̄ϕ(t) = F (t, β(t), β′(t)) − r̄ϕ(t), t ∈ I \\ β(0) ≥ 0, β(1) ≥ δβ(η), \end{matrix}

so that α and β are strict lower and upper solutions of (2.4). We show that any solution x of (2.4) satisfies α(t) < x(t) < β(t), t ∈ (0, 1]. For this, set v(t) = α(t) − x(t), t ∈ I. Then, v ∈ C^{2}(I) and the boundary conditions imply that

(2.5)  v(0) ≤ 0, v(1) ≤ δv(η).

Let max{v(t) : t ∈ I} = v(t_{0}) ≥ 0. If t_{0} ∈ (0, 1), then v′(t_{0}) = 0 and v′′(t_{0}) ≤ 0. However,

v′′(t_{0}) = α′′(t_{0})−x′′(t_{0}) > −F (t_{0}, α(t_{0}), α′(t_{0}))+r̄ϕ(t)+F (t_{0}, α(t_{0}), α′(t_{0}))−r̄ϕ(t) = 0,

a contradiction. Thus, v(t) has no nonnegative local maximum. If t_{0} = 1, then v(1) ≥ 0 and v′(1) ≥ 0. If v(1) = 0, then the boundary condition v(1) ≤ δv(η) implies that v(η) ≥ 0 and hence v(1) cannot be the maximum of v(t). Thus, v(1) > 0. Since v(0) ≤ 0, it follows that there exists t_{1} ∈ [0, η) ⊂ I such that v(t_{1}) = 0 and v′(t_{1}) ≥ 0. We claim that

v′(t) ≥ 0 on [t_{1}, 1].

If not, then there exists a t_{2} ∈ (t_{1}, 1) such that v′(t_{2}) < 0. But then v(t) must have a positive local maximum at some t ∈ (t_{1}, t_{2}), which is a contradiction. Hence, we have v′(t) ≥ 0 on [t_{1}, 1], which implies that v(t) ≥ 0 on [t_{1}, 1]. The condition v′(t) ≥ 0 on [t_{1}, 1] yields q(x′(t)) ≤ α′(t), t ∈ [t_{1}, 1]. Therefore,

\begin{aligned}F (t, x, x′) & = f(t, α(t), q(x′(t))) + [f(t, α(t), α′(t)) − f(t, α(t), q(x′(t)))]\frac{v(t)}{ε} \\ & ≥ f(t, α(t), α′(t)). \end{aligned}

Since v is increasing on [t_{1}, 1], there exists t_{2} such that 0 ≤ v(t) ≤ ε for t ∈ [t_{1}, t_{2}] and v(t) > ε for t ∈ (t_{2}, 1]. For t ∈ [t_{1}, t_{2}], we have

v′′(t) = α′′(t) − x′′(t) > −f(t, α(t), α′(t)) + r̄ϕ(t) + F (t, x, x′) − r̄ϕ(t)
≥ −f(t, α(t), α′(t)) + f(t, α(t), α′(t)) = 0.

For t ∈ (t_{2}, 1], we have F (t, x, x′) = f(t, α(t), α′(t)), and so

v′′(t) > −f(t, α(t), α′(t)) + r̄ϕ(t) + f(t, α(t), α′(t)) − r̄ϕ(t) = 0.

Hence, v′′(t) ≥ 0 for t ∈ [t_{1}, 1], so v is convex on [t_{1}, 1]. Moreover,

v(η) < v(1)\frac{η − t_{1}}{1 − t_{1}} < \frac{v(η)}{η} \frac{(η − t_{1})}{(1 − t_{1})},

which leads to η > 1, a contradiction.

Thus, the maximum of v must occur at t_{0} = 0. Since v(0) ≤ 0 and v(t) has no nonnegative local maximum on (0, 1], it follows that v(t) < 0 on (0, 1].

Similarly, we can show that x(t) < β(t), t ∈ (0, 1].

124 RAHMAT ALI KHAN AND J. R. L. WEBB

Now, it remains to show that every solution x of (2.4) with α(t) < x < β(t) on I satisfies |x′(t)| < C, t ∈ I. Since f(t, x, x′) satisfies a Nagumo condition on R(α, β) and x ∈ [α, β], there exists a Nagumo function ω such that

|F (t, x, x′) − r̄ϕ(t)| = |f(t, x, q(x′(t))) − r̄ϕ(t)| ≤ ω(|q(x′(t))|) + M = ω̃(|x′|) on I,

where ω̃(s) = ω(q(s)) + M for s ≥ 0. Now

∫_{λ}^{∞} \frac{s ds}{ω̃(s)} = ∫_{λ}^{C} \frac{s ds}{ω(s) + M} + ∫_{C}^{∞} \frac{s ds}{ω(C) + M} = ∞,

which implies that F − r̄ϕ(t) satisfies a Nagumo condition on R(α, β). Further, in view of (2.3), we have

∫_{λ}^{C} \frac{s ds}{ω̃(s)} ≥ ∫_{λ}^{N} \frac{s ds}{ω(s) + M} > max_{t∈I} β(t) − min_{t∈I} α(t),

and hence, by a standard argument as in the proof of theorem (1.4.1) of [1], we conclude that |x′(t)| < C on I, which implies that x(t) is a solution of (2.1).

3. MULTIPLICITY RESULTS

In this section, we establish multiplicity results for the boundary value problem (2.1). Firstly, we introduce some notation and make some observations. Let k > max{∥α∥, ∥β∥}, and assume that

(3.1)  f(t, k, 0) − r̄ϕ(t) < 0 for every t ∈ I.

Let

σ = max{|f(t, k, 0)| : t ∈ I},
c = min{ϕ(t) : t ∈ I}, d = max{ϕ(t) : t ∈ I};

then σ, c, d > 0. Define r_{0} = −σ/c; then r_{0} < r̄ and

f(t, k, 0) − r_{0}ϕ(t) ≥ 0 for every t ∈ I.

For r ≤ r_{0}, we have

(3.2)  f(t, k, 0) − rϕ(t) ≥ 0 for every t ∈ I.

Let

(3.3)  r_{1} = sup{r ∈ [r_{0}, r̄) : f(t, k, 0) − rϕ(t) ≥ 0 on I};

then r_{0} ≤ r_{1} < r̄, and for every r ∈ (r_{1}, r̄],

(3.4)  f(t, k, 0) − rϕ(t) < 0 for every t ∈ I.

Let D_{α} = {x ∈ C^{1}(I) : α(t) < x < k, t ∈ I}. We prove the following theorem.

THREE POINT BVPS 125

Theorem 3.1. Assume that (a_{1}), (a_{2}) and (a_{3}) of Theorem 2.3 hold and that the assumption in equation (3.1) holds. Then, there exists r_{2} ≤ r_{1} such that (2.1) has no solution in D_{α} for r < r_{2} and (2.1) has at least two solutions in D_{α} for r ∈ (r_{2}, r̄].

Proof. In view of (a_{1}), we have

−α′′(t) < f(t, α(t), α′(t)) − r̄ϕ(t) ≤ f(t, α(t), α′(t)) − rϕ(t) for r ≤ r̄

and

−β′′(t) > f(t, β(t), β′(t)) − r̄ϕ(t) ≥ f(t, β(t), β′(t)) − rϕ(t) for r ≥ r̄.

Thus, α is a strict lower solution of (2.1) for every r ≤ r̄ and β is a strict upper solution of (2.1) for every r ≥ r̄. Let

R^{∗} = sup{r ∈ ℝ : −α′′(t) < f(t, α(t), α′(t)) − rϕ(t)} ≥ r̄,
r^{∗} = inf{r ∈ ℝ : −β′′(t) > f(t, β(t), β′(t)) − rϕ(t)} ≤ r̄.

Then, for every r ∈ [r^{∗}, R^{∗}], α and β are strict lower and upper solutions of (2.1) and are such that α < β on (0, 1]. Moreover, f(t, x, x′) − rϕ(t) satisfies a Nagumo condition on R(α, β) for every r ∈ [r^{∗}, R^{∗}]. Hence, by Theorem 2.3, there exists a solution x ∈ D_{α} of (2.1) such that

α(t) < x(t) < β(t), t ∈ (0, 1],

for every r ∈ [r^{∗}, R^{∗}] and |x′(t)| < C on I, where C is a constant. Define the modification of f by

F (t, x, x′) = \begin{cases}f(t, k, x′), & if x ≥ k, \\ f(t, x, x′) & if α(t) < x < k, \\ f(t, α(t), x′), & if x ≤ α(t). \end{cases}

Consider the modified problem

\begin{matrix}− x′′(t) = F (t, x, x′) − rϕ(t), t ∈ I, \\ x(0) = 0, x(1) = δx(η), 0 < δη < 1, η < 1. \end{matrix} ; (3.5)

We note that any solution x of (3.5) with α < x < k on I, is a solution of (2.1).

We will show that for every r ∈ (r_{1}, r̄], any solution x of (3.5) does satisfy α < x < k on I and hence is a solution of (2.1). It will then suffice to show that (3.5) has at least two solutions in D_{α}. Let x be a solution of (3.5) for some r ∈ (r_{1}, r̄]. Firstly, we show that α < x on I. Assume that

max{v(t) = α(t) − x(t) : t ∈ I} = v(t_{0}) ≥ 0.

As in the proof of Theorem 2.3, it follows that t_{0} ∈ (0, 1). Hence, v′(t_{0}) = 0 and v′′(t_{0}) ≤ 0. However, we have

\begin{aligned}−v′′(t_{0}) & = −α′′(t_{0}) + x′′(t_{0}) \\ & < f(t_{0}, α(t_{0}), α′(t_{0})) − r̄ϕ(t) − f(t_{0}, α(t_{0}), α′(t_{0})) + rϕ(t) ≤ 0, \end{aligned}

a contradiction. Secondly, we show that x < k on I. Assume that

max{x(t) : t ∈ I} = x(t_{0}) ≥ k;

then x′(t_{0}) = 0, x′′(t_{0}) ≤ 0. However, by (3.4), we have

−x′′(t_{0}) = f(t_{0}, k, 0) − rϕ(t) < 0,

a contradiction. Hence, α < x < k on I. On the other hand, we note that for every r ∈ (r_{1}, r̄],

\begin{aligned}F (t, α(t), α′(t)) − rϕ(t) & = f(t, α(t), α′(t)) − rϕ(t) \\ & ≥ f(t, α(t), α′(t)) − r̄ϕ(t) > −α′′(t), t ∈ I, \end{aligned}

and in view of (3.4),

F (t, k, 0) − rϕ(t) = f(t, k, 0) − rϕ(t) < 0, t ∈ I,

which imply that α, k are strict lower and upper solutions of (3.5). Moreover, for every (t, x, y) ∈ R(α, k), we have

|F (t, x, x′) − rϕ(t)| = |f(t, x, x′) − rϕ(t)| ≤ ω̂(|x′|),

where ω̂(s) = ω(s) + M_{r} for s ≥ 0, M_{r} = max{|rϕ(t)| : t ∈ I, r ∈ [r_{1}, r̄]}. Thus, F −rϕ satisfies a Nagumo condition on R(α, k). Hence, by Theorem 2.3, the boundary value problem (3.5) has a solution in D_{α} for every r ∈ (r_{1}, r̄]. Let r_{2} = inf {r < r̄ : (3.5) has a solution in D_{α}}; then (3.5) has no solution in D_{α} for r < r_{2}. We show that r_{2} is finite. Let x_{2} ∈ D_{α} be a solution of (3.5) for r = r_{2}. Then, there exists a constant C depending on k and ω̂ such that |x′_{2}(t)| < C on I. Let

m = min {F (t, x, x′) : α(t) < x < k, |x′(t)| < C, t ∈ I}.

Then,

−x′′_{2}(t) ≥ m − r_{2}ϕ(t), t ∈ I,

which implies that

(3.6)  x′′_{2}(t) ≤ −m + r_{2}ϕ(t) ≤ −m + r_{2}c, t ∈ I.

Since x_{2} ∈ C^{1}(I), by the mean value theorem, there exists ξ ∈ (0, 1) such that

(3.7)  x′_{2}(ξ) = x_{2}(1) − x_{2}(0) = δx_{2}(η).

If η ∈ [ξ, 1), then integrating (3.6) from ξ to t and using (3.7), we obtain

(3.8) x′_{2}(t) ≤ x′_{2}(ξ) − m(t − ξ) + r_{2}c(t − ξ) < δx_{2}(η) − m + r_{2}c(t − η), for t ≥ ξ.

Integrating (3.8) from η to 1, we have

x_{2}(1) − x_{2}(η) < δx_{2}(η)(1 − η) − m(1 − η) + r_{2} \frac{c(1 − η)^{2}}{2} ,

which in view of the boundary condition x_{2}(1) = δx_{2}(η), implies that

−(1 − δη)x_{2}(η) < −m(1 − η) + r_{2} \frac{c(1 − η)^{2}}{2} ,

which gives

(3.9)  r_{2} > −\frac{2}{c}[\frac{x_{2}(η)(1 − δη)}{(1 − η)^{2}} − \frac{m}{1 − η} ].

On the other hand, if η ∈ (0, ξ], then integrating (3.6) from t to ξ, we obtain

x′_{2}(ξ) − x′_{2}(t) ≤ −m(ξ − t) + r_{2}c(ξ − t) < −m + r_{2}c(η − t),

which in view of (3.7) implies that

(3.10)  −x′_{2}(t) < −δx_{2}(η) − m + r_{2}c(η − t).

Integrating (3.10) from 0 to η, we obtain

−x_{2}(η) < −δηx_{2}(η) − mη + r_{2}c\frac{η^{2}}{2} ,

which leads to

(3.11)  r_{2} > −\frac{2}{c}[\frac{x_{2}(η)(1 − δη)}{η^{2}} − \frac{m}{η}].

Let ρ = max{η, 1 − η}, then ρ > 0 and from (3.9) and (3.11), it follows that

r_{2} > −\frac{2}{c}[\frac{x_{2}(η)(1 − δη)}{ρ^{2}} − \frac{m}{ρ}] = r_{3} (say).

Hence, r_{2} is finite. Now, let r̃ ∈ [r_{2}, r̄) be such that (3.5) has a solution x̃ in D_{α}. Then for every r ∈ (r̃, r̄], we have

−α′′(t) < F (t, α(t), α′(t)) − r̄ϕ(t) ≤ F (t, α(t), α′(t)) − rϕ(t), t ∈ I,
−x̃′′(t) = F (t, x̃(t), x̃′(t)) − r̃ϕ(t) > F (t, x̃(t), x̃′(t)) − rϕ(t), t ∈ I.

These imply that α and x̃ are strict lower and upper solutions of (3.5). Moreover, for (t, x, y) ∈ R(α, x̃), we have

|F (t, x, x′) − rϕ(t)| ≤ ω̂(|x′|).

Hence, by Theorem 2.3, there exists a solution x ∈ D_{α} of (3.5) such that

α(t) < x(t) < x̃(t), t ∈ (0, 1].

Also, there exists a constant C_{1} > 0 which depends on α, x̃, ω̂ such that

|x′(t)| < C_{1} on I.

In case, r ≥ r^{∗}, the solution x of (3.5) satisfies α(t) < x(t) < β(t), t ∈ (0, 1]. Define the modification of F with respect to α, x̃ by

h(t, x, x′) = \begin{cases}F (t, x̃(t), x̃′(t)), & if x ≥ x̃(t) + ε, \\ F (t, x̃(t), x′) + [F (t, x̃(t), x̃′(t)) \\ − F (t, x̃(t), x′)]\frac{x − x̃(t)}{ε}, & if x̃(t) ≤ x < x̃(t) + ε, \\ F (t, x, x′), & if α(t) < x < x̃(t), \\ F (t, α(t), x′) + [F (t, α(t), α′(t)) \\ − F (t, α(t), x′)]\frac{α(t) − x}{ε}, & if α(t) − ε < x ≤ α(t), \\ F (t, α(t), α′(t)), & if x ≤ α(t) − ε, \end{cases}

and define

ψ(t, x) = \frac{(x̃(t) − p(x, α, x̃))F (t, α(t), α′(t)) + (p(x, α, x̃) − α(t))F (t, x̃(t), x̃′(t))}{x̃(t) − α(t)}

for t ∈ I, where

p(x, α, x̃) = max {α(t), min{x, x̃(t)}}, x ∈ ℝ.

Consider the system of equations

\begin{matrix}− x′′(t) = λh(t, x, x′) + (1 − λ)ψ(t, x) − rϕ(t), t ∈ I, r ∈ (r̃, r̄] \\ x(0) = 0, x(1) = δx(η), λ ∈ [0, 1], 0 < δη < 1, 0 < η < 1. \end{matrix} ; (3.12)

For λ = 0, the system reduces to

\begin{matrix}−x′′(t) = ψ(t, x) − rϕ(t), t ∈ I, r ∈ (r̃, r̄] \\ x(0) = 0, x(1) = δx(η). \end{matrix} ; (3.13)

This is equivalent to the integral equation

x(t) = ∫_{0}^{1} G(t, s)[ψ(s, x) − rϕ(s)] ds.

Since ψ, ϕ and G are continuous and bounded, this integral equation has a fixed point by Schauder’s fixed point theorem. Moreover,

\begin{matrix}ψ(t, α(t)) − rϕ(t) = & F (t, α(t), α′(t)) − rϕ(t) \\ ≥ F (t, α(t), α′(t)) − r̄ϕ(t) > −α′′(t), \end{matrix}

and

\begin{matrix}ψ(t, x̃(t)) − rϕ(t) = & F (t, x̃(t), x̃′(t)) − rϕ(t) \\ < F (t, x̃(t), x̃′(t)) − r̃ϕ(t) = x̃′′(t). \end{matrix}

Thus, α, x̃ are strict lower and upper solutions of (3.13). Hence, any solution of (3.13) satisfies α < x < x̃ on (0, 1]. Since ψ(t, x) − rϕ(t) is continuous and bounded, it satisfies a Nagumo condition on R(α, x̃). Hence, there exists C_{1} > 0 such that any solution x of (3.13) with α < x < x̃ satisfies |x′| < C_{1} on I.

For λ = 1, (3.12) is equivalent to

\begin{matrix}− x′′(t) = h(t, x, x′) − rϕ(t), t ∈ I, r ∈ (r̃, r̄], \\ x(0) = 0, x(1) = δx(η). \end{matrix} ; (3.14)

Again, α, x̃ are strict lower and upper solutions of (3.14) and for every t ∈ I and x ∈ [α, x̃], we have

|h(t, x, x′) − rϕ(t)| = |F (t, x, x′) − rϕ(t)| ≤ ω̂(|x′|).

This implies that h satisfies a Nagumo condition on R(α, x̃). Hence, by Theorem 2.3, the boundary value problem (3.14) has a solution x such that

α < x < x̃ on (0, 1] and |x′(t)| < C_{2} on I.

Now, for any λ ∈ [0, 1], we note that

λh(t, α(t), α′(t)) + (1 − λ)ψ(t, α(t)) − rϕ(t)

= λF (t, α(t), α′(t)) + (1 − λ)F (t, α(t), α′(t)) − rϕ(t)

= F (t, α(t), α′(t)) − rϕ(t) ≥ F (t, α(t), α′(t)) − r̄ϕ(t) > −α′′(t),

and

λh(t, x̃(t), x̃′(t)) + (1 − λ)ψ(t, x̃(t)) − rϕ(t)

= λF (t, x̃(t), x̃′(t)) + (1 − λ)F (t, x̃(t), x̃′(t)) − rϕ(t) < F (t, x̃(t), x̃′(t)) − r̃ϕ(t) = −x̃′′(t).

These imply that α, x̃ are strict lower and upper solutions of (3.12). Further, for λ ∈ [0, 1], x ∈ [α(t), x̃(t)], x′ ∈ ℝ and t ∈ I, we have

\begin{aligned}|λh(t, x, x′) + (1 − λ)ψ(t, x) − rϕ(t)| & ≤ λ|F (t, x, x′)| + (1 − λ)ψ|(t, x)| + M_{1} \\ & ≤ ω(|x′|) + M_{1} + 2M = ω̄(|x′|), \end{aligned}

where ω̄(s) = ω(s)+M_{2} if s ≥ 0, M_{2} = M_{1}+2M. It is easy to see that ω̂ is a Nagumo function. Thus, λh(t, x, x′) + (1 − λ)ψ(t, x) − rϕ(t) satisfies a Nagumo condition on R(α, x̃). Hence, by Theorem 2.3, any solution of the boundary value problem (3.12) satisfies

α < x < x̃ on (0, 1], and |x′(t)| < C_{3} on I,

where C_{3} > 0 is a constant depending on α, x̃ and ω̄.

To use the Leray-Schauder degree theory, we define the operators

F_{r} : C_{b}^{1}(I) → C_{b}^{1}(I), Ψ : C_{b}^{1}(I) → C_{b}^{1}(I), H_{r} : C_{b}^{1}(I) → C_{b}^{1}(I),

by

F_{r}x(t) = ∫_{0}^{1} G(t, s)[F (s, x(s), x′(s)) − rϕ(s)] ds
H_{r}x(t) = ∫_{0}^{1} G(t, s)[h(s, x(s), x′(s)) − rϕ(s)] ds
Ψx(t) = ∫_{0}^{1} G(t, s)[ψ(s, x(s)) − rϕ(s)] ds,

where C_{b}^{1} = {x ∈ C^{1}(I) : x(0) = 0, x(1) = δx(η)}. Then, (3.5) and (3.12) can be written in the form

(3.15)  (I − F_{r})x = 0

and

(3.16)  (I − λH_{r} − (1 − λ)Ψ)x = 0,

respectively. Let C > max{C_{1}, C_{2}, C_{3}} and consider two open bounded sets in C^{1}(I)

Ω = {x ∈ C_{b}^{1} : |x| < k, |x′| < C on I},
Ω_{1} = {x ∈ C_{b}^{1} : |x′| < C on I, α < x < x̃ on (0, 1]}.

For any λ ∈ [0, 1], each solution x of (3.12), belongs to Ω and so x ∉ ∂Ω. For λ = 0, (3.16) is equivalent to (I − Ψ)(x)(t) = 0, and the homotopy invariance of degree implies that

d(I − Ψ, Ω) = d(I, Ω) = 1.

Again by the homotopy property of degree, we have (λ = 1), d(I − H_{r}, Ω) = 1. Since each solution of (3.14) belongs to Ω_{1}, by the excision property of degree, we have

d(I − H_{r}, Ω_{1}) = 1.

Since H_{r} = F_{r} on Ω_{1}, it follows that

d(I − F_{r}, Ω_{1}) = 1.

Now, since (3.15) has no solution for r ≤ r_{3}, it follows by the existence property that, for r ≤ r_{3},

d(I − F_{r}, Ω) = 0.

Also, for r ≤ r̄, any solution x of (3.15) belongs to Ω and so x ∉ ∂Ω. Letting r vary from r_{3} to r̄, by the homotopy property of degree, we have

d(I − F_{r}, Ω) = 0, for r ≤ r̄.

Using the additivity property of degree, we have for r ∈ (r̃, r̄]

d(I − F_{r}, Ω \ Ω_{1}) = d(I − F_{r}, Ω) − d(I − F_{r}, Ω_{1}) = 0 − 1 ≠ 0.

Therefore, the problem (3.15) has at least two solutions in D_{α} for r > r_{2}.

Example 3.2. (A class of examples) Consider the following boundary value problem

\begin{matrix}−x′′(t) & = f(t, x, x′) − r = −|x′|^{p−1}x′ + ψ(x) − r, t ∈ [0, 1], \\ x(0) = 0, x(1) = δx(η), 0 < δη < 1, \end{matrix} ; (3.17)

where 0 ≤ p ≤ 2. We take ψ : ℝ → ℝ to be continuous and to satisfy ψ(0) = 0, ψ(x) > 0 for x < 0 and −cx^{p} ≤ ψ(x) ≤ 0 for x ≥ 0, where 0 < c < 1. Take r̄ = a^{p} (a > 0), α = −at and β = at; then α, β satisfy the boundary conditions

α(0) = 0, α(1) − δα(η) < 0
β(0) = 0, β(1) − δβ(η) > 0.

Moreover, for every t ∈ (0, 1], we have

f(t, −at, −a) − r̄ = ψ(−at) > 0,

and

f(t, at, a) − r̄ = −2a^{p} + ψ(at) < 0.

Thus, α is a strict lower solution and β is a strict upper solution of (3.17). Further, we note that for t ∈ I, x ∈ [−a, a] and x′ ∈ ℝ, we have

|f(t, x, x′) − r̄| = | − |x′|^{p−1}x′ + ψ(x) − a^{p}| ≤ |x′|^{p} + a^{p}(1 + c) = ω(|x′|),

where ω(s) = s^{p} + a^{p}(1 + c). Since

∫_{0}^{∞} \frac{s ds}{ω(s)} = ∞,

it follows that ω is a Nagumo function and hence f − r̄ satisfies a Nagumo condition on R(−a, a). Now, let k ≥ a; then f(t, k, 0) − r̄ = ψ(k) − a^{p} < 0, and for every r ≤ −k^{p}, we have

f(t, k, 0) − r̄ ≥ ψ(k) + k^{p} ≥ k^{p}(1 − c) > 0 for every t ∈ I.

Hence, by Theorem 3.1, there exists r_{2} < −k^{p} such that the problem (3.17) has no solution in D_{α} for r < r_{2}, and has at least two solutions in D_{α} for r ∈ (−k^{p}, a^{p}], where D_{α} = {x ∈ C^{2} : −at < x < at on (0, 1]}.

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