GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 6, 1999, 567-590

MULTIPLE SOLUTIONS OF GENERALIZED MULTIPOINT CONJUGATE BOUNDARY VALUE PROBLEMS

PATRICIA J. Y. WONG AND RAVI P. AGARWAL

Abstract. We consider the boundary value problem

y^{(n)}(t) = P (t, y), t ∈ (0, 1)
y^{(j)}(t_{i}) = 0, j = 0, …, n_{i} − 1, i = 1, …, r,

where r ≥ 2, n_{i} ≥ 1 for i = 1, …, r, ∑_{i=1}^{r} n_{i} = n and 0 = t_{1} < t_{2} < ⋯ < t_{r} = 1. Criteria are offered for the existence of double and triple ‘positive’ (in some sense) solutions of the boundary value problem. Further investigation on the upper and lower bounds for the norms of these solutions is carried out for special cases. We also include several examples to illustrate the importance of the results obtained.

1. Introduction

Let 0 = t_{1} < t_{2} < ⋯ < t_{r} = 1 be r (≥ 2) fixed points and let n_{i} (≥ 1), i = 1, …, r, be integers with ∑_{i=1}^{r} n_{i} = n. In this paper we shall consider the multipoint conjugate boundary value problem

\begin{matrix}y^{(n)}(t) = P (t, y), t ∈ (0, 1), \\ y^{(j)}(t_{i}) = 0, j = 0, …, n_{i} − 1, i = 1, …, r, \end{matrix}  (M)

where P is continuous at least in the interior of the domain of definition.

We shall define a positive solution y of (M) as follows: y ∈ C^{(n)}(0, 1) is a nontrivial function that fulfills (M) and for each 1 ≤ i ≤ r − 1, (−1)^{n_{i+1}+···+n_{r}} y is nonnegative on [t_{i}, t_{i+1}]. Our first task is to develop criteria for the existence of double positive solutions of (M). Next, we shall consider two special cases of (M), namely,

y^{(3)}(t)=q(t)[|y(t)|^{α} + |y(t)|^{β}], t ∈ (0, 1), y(0)=y(c)=y(1)=0 (M1)

567 and

1991 Mathematics Subject Classification. 34B15.

Key words and phrases. Generalised multipoint conjugate boundary value problem, multiple solution.

1072-947X/99/1100-0567$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Plenum Publishing Corporation

y^{(3)}(t) = q(t)e^{γ|y(t)|}, t ∈ (0, 1), y(0) = y(c) = y(1) = 0,  (M2)

where 0 < c < 1, 0 ≤ α < 1 < β, γ > 0 and q is continuous at least in the interior of the domain of definition. It is noted that the importance of (M1) is well illustrated in [1, 2] and that of (M2) is related to the analysis of diffusion of heat generated by positive temperature-dependent sources [3]. In addition to providing conditions for the existence of double positive solutions of (M1) and (M2), we also derive upper and lower bounds for the norms of these solutions. Finally, we shall examine the existence of triple positive solutions of a ‘separable’ case of (M), namely,

\begin{matrix}y^{(n)}(t) = b(t)f(y), t ∈ (0, 1), \\ y^{(j)}(t_{i}) = 0, j = 0, …, n_{i} − 1, i = 1, …, r, \end{matrix}  (M3)

where b and f are continuous at least in the interior of the domain of definition. The criterion established will also provide estimates on the norms of these positive solutions.

The present work is motivated by the fact that a multipoint boundary value problem of the type (M) models various dynamical systems with n degrees of freedom in which n states are observed at n times, see Meyer [4]. In particular, when n = r = 2 the boundary value problem (M) describes a vast spectrum of nonlinear phenomena which include gas diffusion through porous media, nonlinear diffusion generated by nonlinear sources, thermal self ignition of a chemically active mixture of gases in a vessel, catalysis theory, chemically reacting systems, adiabatic tubular reactor processes, as well as concentration in chemical or biological problems, e.g., see [5–11]. It is important to note that in most of these models, only positive solutions are meaningful. Recently, special cases of (M), namely, second order problem (n = r = 2) and two-point problem (r = 2), have been tackled by several authors [12–15]. Further, related investigations on other boundary value problems such as Sturm–Liouville type, focal type, Lidstone type as well as (n, p) type are documented in the monographs [3, 16–18, 1]. Our results, besides complement and extend the literature to multipoint problem, also improve the work of Eloe and Henderson [19].

The paper is organized as follows. In Section 2, we shall state the fixed point theorems required, and provide some properties of certain Green’s functions which are needed later. By defining an appropriate Banach space and cone, in Section 3 we shall establish existence criteria for double positive solutions of (M). The cases (M1) and (M2) are respectively treated in Sections 4 and 5. Finally, in Section 6 we shall discuss the existence of triple positive solutions of (M3).

2. Preliminaries

Let B be a Banach space equipped with the norm ∥ · ∥. Definition 2.1. Let C(⊂ B) be a nonempty closed convex set. We say

that C is a cone provided the following conditions are satisfied:

(a) If y ∈ C and ℓ ≥ 0, then ℓy ∈ C;

(b) If y ∈ C and −y ∈ C, then y = 0. Definition 2.2. Let C(⊂ B) be a cone. A map ψ is a nonnegative

continuous concave functional on C if the following conditions are satisfied: (a) ψ : C → [0, ∞) is continuous;

(b) ψ(ℓx + (1 − ℓ)y) ≥ ℓψ(x) + (1 − ℓ)ψ(y) for all x, y ∈ C and 0 ≤ ℓ ≤ 1. With C and ψ defined as above, we shall introduce the following nota-

tions. For k, ℓ, m > 0, we shall denote

C(k) = {y ∈ C | ∥y∥ < k} and C(ψ, ℓ, m) = {y ∈ C | ψ(y) ≥ ℓ, ∥y∥ ≤ m}.

The following fixed point theorems are needed later.

Theorem 2.1 ([20]). Let C(⊂ B) be a cone. Assume Ω_{1}, Ω_{2} are open subsets of B with 0 ∈ Ω_{1}, Ω̄_{1} ⊂ Ω_{2}, and let

S : C ∩ (Ω̄_{2}\Ω_{1}) → C

be a completely continuous operator such that either (a) ∥Sy∥ ≤ ∥y∥, y ∈ C ∩ ∂Ω_{1}, and ∥Sy∥ ≥ ∥y∥, y ∈ C ∩ ∂Ω_{2}, or

(b) ∥Sy∥ ≥ ∥y∥, y ∈ C ∩ ∂Ω_{1}, and ∥Sy∥ ≤ ∥y∥, y ∈ C ∩ ∂Ω_{2}. Then S has a fixed point in C ∩ (Ω̄_{2}\Ω_{1}).

Theorem 2.2 ([2, 21]). Let C(⊂ B) be a cone and let ν > 0 be given. Assume that ψ is a nonnegative continuous concave functional on C such that ψ(y) ≤ ∥y∥ for all y ∈ C̄(ν), and let S : C̄(ν) → C̄(ν) be a completely continuous operator. Suppose that there exist numbers k, ℓ, m, where 0 < k < ℓ < m ≤ ν, such that

(a) {y ∈C(ψ, ℓ, m) | ψ(y) > ℓ}≠∅, and ψ(Sy)>ℓ for all y ∈C(ψ, ℓ, m);

(b) ∥Sy∥ < k for all y ∈ C̄(k);

(c) ψ(Sy) > ℓ for all y ∈ C(ψ, ℓ, ν) with ∥Sy∥ > m. Then S has (at least) three fixed points y_{1}, y_{2} and y_{3} in C̄(ν). Further

we have

\begin{matrix}y_{1} ∈ C(k), y_{2} ∈ \{y ∈ C(ψ, ℓ, ν)|ψ(y) > ℓ\}, \\ y_{3} ∈ C̄(ν)\(C(ψ, ℓ, ν) ∪ C̄(k)). \end{matrix}  (2.1)

To obtain a solution for (M), we require a mapping whose kernel G(t, s) is the Green’s function of the boundary value problem

y^{(n)}(t) = 0, t ∈ [0, 1] y^{(j)}(t_{i}) = 0, j = 0, …, n_{i} − 1, i = 1, …. (2.2)

We shall use the following notation. For each 1 ≤ i ≤ r − 1, we shall denote α_{i} = ∑_{j=i+1}^{r} n_{j} and I_{i} = [(3t_{i} + t_{i+1})/4, (t_{i} + 3t_{i+1})/4]. Further, for each s ∈ (0, 1), we define

∥G(·, s)∥ = sup_{t∈[0,1]} |G(t, s)|.  (2.3)

It is well known [3, 22–24] that

(−1)^{α_{i}}G(t, s) > 0, (t, s) ∈ (t_{i}, t_{i+1}) × (0, 1), i = 1, …, r − 1. (2.4)

In view of (2.3) we readily obtain

(−1)^{α_{i}}G(t, s)≤∥G(·, s)∥, (t, s)∈[t_{i}, t_{i+1}] × [0, 1], i=1, …, r−1. (2.5)

Moreover, we have the following lemma which improves the result of Eloe and Henderson [23].

Lemma 2.1 ([24]). For (t, s) ∈ I_{i} × (0, 1), i = 1, …, r − 1, we have

(−1)^{α_{i}}G(t, s) ≥ L_{i}∥G(·, s)∥,

where 0 < L_{i} ≤ 1 is a constant defined by

L_{i} = min \{ min \{g(\frac{3t_{i} + t_{i+1}}{4} ), g(\frac{t_{i} + 3t_{i+1}}{4} )\}/ max_{t∈[0,1]} g(t),

min \{h(\frac{3t_{i} + t_{i+1}}{4} ), h(\frac{t_{i} + 3t_{i+1}}{4} )\}/ max_{t∈[0,1]} h(t)\},  (2.6)

where g(t) = ∏_{j=1}^{r−1} |t − t_{j}|^{n_{j}}(1 − t)^{n_{r}−1} and h(t) = t^{n_{1}−1} ∏_{j=2}^{r} |t − t_{j}|^{n_{j}}.

Lemma 2.2. Let 0 < c < 1 be fixed. Consider the following special case of (2.2):

cy^{(3)}(t) = 0, t ∈ [0, 1], y(0) = y(c) = y(1) = 0.  (2.7)

Let G_{1} denote the Green’s function of (2.7). We have

∥G_{1}(·, s)∥≤ϕ(s)≡\begin{cases}\frac{1}{2(1−c)}(1−s)^{2} max\{c, 1−c\}, & c≤s ≤ 1 \\ \frac{1}{2c}s^{2} max\{c, 1−c\}, & 0≤s ≤ c. \end{cases}  (2.8)

Further, the constants L_{1} and L_{2} in (2.6) are respectively given as

L_{1} =min \{\frac{3c^{2}}{16} / max \{\frac{c^{2}}{4} , 1−c\}, \frac{c(4−3c)}{16}/ max \{\frac{(1−c)^{2}}{4} , c\}\},  (2.9)
L_{2} =min\{\frac{(3c+1)(1−c)}{16}/max\{\frac{c^{2}}{4} , 1−c\}, \frac{3(1−c)^{2}}{16}/max\{\frac{(1−c)^{2}}{4} , c\}\}. (2.10)

Proof. Here n = r = 3, n_{1} = n_{2} = n_{3} = 1, t_{1} = 0, t_{2} = c and t_{3} = 1. The explicit expression of the Green’s function G_{1} is given by [25]

−G_{1}(t, s)=\begin{cases}\frac{1}{2(1−c)}(1−s)^{2}t(t−c), & c≤s≤1, 0≤t≤s \\ \frac{1}{2(1−c)}(1−s)^{2}t(t−c)− \frac{1}{2}(t−s)^{2}, & c≤s≤1, s≤t≤1 \\ \frac{1}{2c}s^{2}(t−c)(1−t)+ \frac{1}{2}(s−t)^{2}, & 0≤s≤c, 0≤t≤s \\ \frac{1}{2c}s^{2}(t−c)(1−t), & 0≤s≤c, s≤t≤1 \end{cases} (2.11)

from which we find

∥G_{1}(·, s)∥ ≤ \begin{cases}\frac{1}{2(1−c)}(1−s)^{2}s max\{s−c, c\}, & c≤s≤1, 0≤t≤s \\ \frac{1}{2(1−c)}(1−s)^{2}(1−c), & c≤s≤1, s≤t≤1 \\ \frac{1}{2c}s^{2}c, & 0≤s≤c, 0≤t≤s \\ \frac{1}{2c} s^{2} max\{c−s, 1−c\}(1−s), & 0≤s≤c, s≤t≤1 \end{cases} ≤

≤ \begin{cases}\frac{1}{2(1−c)}(1−s)^{2} max\{c, 1−c\}, & c≤s≤1, 0≤t≤1 \\ \frac{1}{2c}s^{2} max\{c, 1−c\}, & 0≤s≤c, 0≤t≤1 \end{cases} = ϕ(s).

The constants L_{1} and L_{2} are obtained by direct computation.

For clarity, we shall list the conditions that are needed later. In these conditions it is assumed that the functions f : IR → [0, ∞) and a, b : (0, 1) → IR are continuous.

(A1) If |u_{1}| ≥ |u_{2}|, then f(u_{1}) ≥ f(u_{2}).

(A2) For (t, u) ∈ (0, 1) × IR, a(t) ≤ P (t, u)/f(u) ≤ b(t).

(A3) The function a is nonnegative and not identically zero on any nondegenerate subinterval of (0, 1); and there exists a number 0 < ρ ≤ 1 such that a(t) ≥ ρb(t) for t ∈ (0, 1).

(A4) ∫_{0}^{1} ∥G(·, s)∥b(s)ds < ∞.

(A5) Functions a, b : [0, 1] → IR are continuous; and (A2) and (A3) hold with (0, 1) replaced by [0, 1].

Further, we introduce the notation f_{0} = lim_{|u|→0^{+}} f(u)/|u| and f_{∞} = lim_{|u|→∞} f(u)/|u|.

3. Double Positive Solutions of (M)

For y ∈ C[0, 1] let

θ = ∫_{0}^{1} ∥G(·, s)∥b(s)f(y(s))ds  (3.1)

and

Γ = ∫_{0}^{1} ∥G(·, s)∥a(s)f(y(s))ds.  (3.2)

If (A2) and (A3) hold, then it is clear that

θ ≥ Γ ≥ ρθ ≥ 0.  (3.3)

Further, we define the constants

σ_{i} = ρL_{i}, i = 1, …, r − 1,  (3.4)

where L_{i} is given in (2.6) and ρ appears in (A3). It is noted that 0 < σ_{i} ≤ 1 for each 1 ≤ i ≤ r − 1.

Let the Banach space B = {y | y ∈ C[0, 1]} be equipped with the norm ∥y∥ = sup_{t∈[0,1]} |y(t)|. Define

C = {y ∈ B | for each 1 ≤ i ≤ r − 1, (−1)^{α_{i}}y(t) ≥ 0 for t ∈ [t_{i}, t_{i+1}]

and min_{t∈I_{i}}(−1)^{α_{i}}y(t) ≥ σ_{i}∥y∥}.

Clearly, C is a cone in B.

We define the operator S : C → B by

Sy(t) = ∫_{0}^{1} G(t, s)P (s, y(s))ds, t ∈ [0, 1].  (3.5)

To obtain a positive solution of (M), we shall seek a fixed point of the operator S in the cone C.

If (A2) and (A3) hold, then in view of (2.4) it is clear that for t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1,

(−1)^{α_{i}}Uy(t) ≤ (−1)^{α_{i}}Sy(t) ≤ (−1)^{α_{i}}V y(t),  (3.6)

where

Uy(t) = ∫_{0}^{1} G(t, s)a(s)f(y(s))ds and V y(t) = ∫_{0}^{1} G(t, s)b(s)f(y(s))ds.

Now we shall state two lemmas whose proof is available in [26].

Lemma 3.1. Let (A1)–(A4) hold. Then the operator S is compact on the cone C.

Remark 3.1. From the proof of Lemma 3.1 we observe that the conclusion of Lemma 3.1 still holds if the conditions (A1)–(A4) are replaced by (A5).

Lemma 3.2. Let (A2) and (A3) hold. Then the operator S maps C into itself.

Theorem 3.1. Let (A5) or (A1)–(A4) hold and let w > 0 be given. Suppose that f satisfies

f(u) ≤ w[ ∫_{0}^{1} ∥G(·, s)∥b(s)ds]^{−1}, |u| ≤ w.  (3.7)

(a) If f_{0} = ∞, then (M) has a positive solution y_{1} such that

0 < ∥y_{1}∥ ≤ w.  (3.8)

(b) If f_{∞} = ∞, then (M) has a positive solution y_{2} such that

∥y_{2}∥ ≥ w.  (3.9)

(c) If f_{0} = f_{∞} = ∞, then (M) has two positive solutions y_{1} and y_{2} such that

0 < ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥.  (3.10)

Proof. (a) Let

A = [σ_{1} ∫_{I_{1}}(−1)^{α_{1}}G(t_{2}/2, s)a(s)ds]^{−1}.  (3.11)

Since f_{0} = ∞, there exists 0 < r < w such that

f(u) ≥ A|u|, 0 < |u| ≤ r.  (3.12)

First, let y ∈ C be such that ∥y∥ = r. Using (3.6), (3.12) and (3.11) successively, we get

(−1)^{α_{1}}Sy(t_{2}/2) ≥ ∫_{0}^{1}(−1)^{α_{1}}G(t_{2}/2, s)a(s)f(y(s))ds ≥
≥ ∫_{0}^{1}(−1)^{α_{1}}G(t_{2}/2, s)a(s)A|y(s)|ds ≥
≥ ∫_{I_{1}}(−1)^{α_{1}}G(t_{2}/2, s)a(s)Aσ_{1}∥y∥ds = ∥y∥.

This immediately implies

∥Sy∥ ≥ ∥y∥.  (3.13)

If we set Ω_{1} = {y ∈ B | ∥y∥ < r}, then (3.13) holds for y ∈ C ∩ ∂Ω_{1}.

Next, let y ∈ C be such that ∥y∥ = w. Then, employing (3.6), (2.5) and (3.7), we find for t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1,

(−1)^{α_{i}}Sy(t) ≤ ∫_{0}^{1} ∥G(·, s)∥b(s)f(y(s))ds ≤ w = ∥y∥.

Consequently,

∥Sy∥ ≤ ∥y∥.  (3.14)

If we set Ω_{2} = {y ∈ B | ∥y∥ < w}, then (3.14) holds for y ∈ C ∩ ∂Ω_{2}.

Having obtained (3.13) and (3.14), it follows from Theorem 2.1 that S has a fixed point y_{1} ∈ C ∩ (Ω̄_{2}\Ω_{1}) such that r ≤ ∥y_{1}∥ ≤ w. Clearly, this y_{1} is a positive solution of (M) fulfilling (3.8).

(b) As seen in the proof of Case (a), condition (3.7) leads to (3.14). Hence, if we set Ω_{1} = {y ∈ B | ∥y∥ < w}, then (3.14) holds for y ∈ C ∩∂Ω_{1}.

Next, noting that f_{∞} = ∞, we may choose T > w such that

f(u) ≥ A|u|, |u| ≥ T,  (3.15)

where A is defined in (3.11). Let T_{1} = max {2w, T/min_{1≤j≤r−1} σ_{j}} and let y ∈ C be such that ∥y∥ = T_{1}. Then we have for s ∈ I_{1},

|y(s)| = (−1)^{α_{1}}y(s) ≥ σ_{1}∥y∥ ≥ σ_{1} · \frac{T}{min_{1≤j≤r−1} σ_{j}} ≥ T.

So in view of (3.15) it follows that

f(y(s)) ≥ A|y(s)|, s ∈ I_{1}.  (3.16)

Applying (3.6), (3.16), and (3.11), we again find (−1)^{α_{1}}Sy(t_{2}/2) ≥ ∥y∥. Therefore (3.13) holds. By setting Ω_{2} = {y ∈ B | ∥y∥ < T_{1}} we have (3.13) for y ∈ C ∩ ∂Ω_{2}.

Now that we have obtained (3.14) and (3.13), it follows from Theorem 2.1 that S has a fixed point y_{2} ∈ C ∩ (Ω̄_{2}\Ω_{1}) such that w ≤ ∥y_{2}∥ ≤ T_{1}. It is clear that this y_{2} is a positive solution of (M) satisfying (3.9).

(c) This follows from Cases (a) and (b).

Theorem 3.2. Let (A5) or (A1)–(A4) hold and let w > 0 be given. Suppose that f satisfies

f(u)≥w[ ∑_{i=1}^{r−1} ∫_{I_{i}}(−1)^{α_{1}}G(t_{2}/2, s)a(s)ds]^{−1}, min_{1≤j≤r−1} σ_{j}w ≤|u|≤w. (3.17)

(a) If f_{0} = 0, then (M) has a positive solution y_{1} such that (3.8) holds.

(b) If f_{∞} = 0, then (M) has a positive solution y_{2} such that (3.9) holds.

(c) If f_{0} = f_{∞} = 0, then (M) has two positive solutions y_{1} and y_{2} such that (3.10) holds.

Proof. (a) Define

Q = [ ∫_{0}^{1} ∥G(·, s)∥b(s)ds]^{−1}.  (3.18)

Since f_{0} = 0, there exists 0 < r < w such that

f(u) ≤ Q|u|, 0 < |u| ≤ r.  (3.19)

First, let y ∈ C be such that ∥y∥ = r. Then the application of (3.6), (2.5), (3.19) and (3.18) yields for t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1,

(−1)^{α_{i}}Sy(t) ≤ ∫_{0}^{1} ∥G(·, s)∥b(s)f(y(s))ds≤∫_{0}^{1} ∥G(·, s)∥b(s)Q∥y∥ds = ∥y∥.

Hence (3.14) follows immediately. Set Ω_{1} = {y ∈ B | ∥y∥ < r}; then (3.14) holds for y ∈ C ∩ ∂Ω_{1}.

Next, let y ∈ C be such that ∥y∥ = w. Noting that for s ∈ I_{i}, 1 ≤ i ≤ r−1,

min_{1≤j≤r−1} σ_{j}w ≤ σ_{i}∥y∥ ≤ |y(s)| ≤ w,

it follows from (3.6) and (3.17) that

(−1)^{α_{1}}Sy(t_{2}/2) ≥ ∫_{0}^{1}(−1)^{α_{1}}G(t_{2}/2, s)a(s)f(y(s))ds ≥

≥ ∑_{i=1}^{r−1} ∫_{I_{i}}(−1)^{α_{1}}G(t_{2}/2, s)a(s)f(y(s))ds ≥ w = ∥y∥.

Thus we get (3.13). By setting Ω_{2} = {y ∈ B | ∥y∥ < w} we see that (3.13) holds for y ∈ C ∩ ∂Ω_{2}.

Having obtained (3.14) and (3.13), it follows from Theorem 2.1 that S has a fixed point y_{1} ∈ C ∩ (Ω̄_{2}\Ω_{1}) such that r ≤ ∥y_{1}∥ ≤ w. Clearly, this y_{1} is a positive solution of (M) satisfying (3.8).

(b) It is seen in the proof of Case (a) that condition (3.17) gives rise to (3.13). So if we set Ω_{1} = {y ∈ B | ∥y∥ < w}, then (3.13) holds for y ∈ C ∩ ∂Ω_{1}.

Next, let Q be defined as in (3.18). Since f_{∞} = 0, we may choose T > w such that

f(u) ≤ Q|u|, |u| ≥ T.  (3.20)

There are two cases to consider, namely, f is bounded and f is unbounded.

Case 1. Suppose that f is bounded. Then, there exists some J > 0 such that

f(u) ≤ J, u ∈ IR.  (3.21)

We define

T_{1} = max {2w, J ∫_{0}^{1} ∥G(·, s)∥b(s)ds}.

Let y ∈ C be such that ∥y∥ = T_{1}. Using (3.6), (2.5), and (3.21), we find for t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1,

(−1)^{α_{i}}Sy(t) ≤∫_{0}^{1}∥G(·, s)∥b(s)f(y(s))ds≤∫_{0}^{1}∥G(·, s)∥b(s)Jds ≤ T_{1} = ∥y∥.

Hence (3.14) follows immediately.

Case 2. Suppose that f is unbounded. Then there exists T_{1} > max{2w, T } such that

f(u) ≤ max{f(T_{1}), f(−T_{1})}, |u| ≤ T_{1}.  (3.22)

Let y ∈ C be such that ∥y∥ = T_{1}. Then applying (3.6), (2.5), (3.22), (3.20) and (3.18) successively gives for t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1,

(−1)^{α_{i}}Sy(t) ≤ ∫_{0}^{1} ∥G(·, s)∥b(s)f(y(s))ds ≤
≤ ∫_{0}^{1} ∥G(·, s)∥b(s) max{f(T_{1}), f(−T_{1})}ds ≤
≤ ∫_{0}^{1} ∥G(·, s)∥b(s)QT_{1}ds = T_{1} = ∥y∥.

Therefore we have (3.14) immediately.

In Cases 1 and 2, if we set Ω_{2} = {y ∈ B | ∥y∥ < T_{1}}, then (3.14) holds for y ∈ C ∩ ∂Ω_{2}.

Now that we have obtained (3.13) and (3.14), it follows from Theorem 2.1 that S has a fixed point y_{2} ∈ C ∩ (Ω̄_{2}\Ω_{1}) such that w ≤ ∥y_{2}∥ ≤ T_{1}. Obviously, this y_{2} is a positive solution of (M) such that (3.9) holds.

(c) This is immediate by Cases (a) and (b).

Example 3.1. Consider the boundary value problem

y^{(3)} =t(2y^{2}+1)+t^{2}(y^{2} + 1), t ∈ (0, 1), y(0)=y(0.5)=y(1)=0.

Here, n = r = 3. Take f(y) = y^{2} + 1. Then f_{0} = f_{∞} = ∞ and

\frac{P (t, y)}{f(y)} = t \frac{2y^{2} + 1}{y^{2} + 1} + t^{2}.

Thus we may take a(t) = t + t^{2} and b(t) = 2t + t^{2}. The condition (A5) is fulfilled with ρ = 1/2. Since f(u) ≤ w^{2} + 1 for |u| ≤ w, by Lemma 2.2 (c = 0.5) we have

∫_{0}^{1} ϕ(s)b(s)ds ≥ ∫_{0}^{1} ∥G_{1}(·, s)∥b(s)ds.

For condition (3.7) to be satisfied, we impose

f(u) ≤ w^{2} + 1 ≤ w[ ∫_{0}^{1} ϕ(s)b(s)ds]^{−1} =

= 18.82w ≤ w[ ∫_{0}^{1} ∥G_{1}(·, s)∥b(s)ds]^{−1}, |u| ≤ w.

The above inequality holds if

0.05329 ≤ w ≤ 18.76.  (3.23)

Hence (3.7) is fulfilled for any w ∈ [0.05329, 18.76]. By Theorem 3.1(c), the boundary value problem has two positive solutions y_{1} and y_{2} such that 0 < ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥. In view of (3.23) we further conclude that

0 < ∥y_{1}∥ ≤ 0.05329  and  ∥y_{2}∥ ≥ 18.76.  (3.24)

Example 3.2. Consider the boundary value problem

y^{(3)} = h(t)y^{2}e^{−|y|}, t ∈ (0, 1), y(0) = y(0.5) = y(1) = 0,

where h ∈ C[0, 1] is nonnegative.

Let f(y) = y^{2}e^{−|y|}. Then we have f_{0} = f_{∞} = 0 and we may take a(t) = b(t) = h(t) so that (A5) is satisfied with ρ = 1. By Lemma 2.2 (c = 0.5) we compute that σ_{i} = ρL_{i} = L_{i} = 3/32, i = 1, 2. Our aim is to find some w > 0 such that condition (3.17) be fulfilled.

Case 1. Let h(t) = (t + 0.1)^{−10}. Suppose that w ≤ 2. Then, since f

is nondecreasing in |u| for |u| ∈ [0, 2], we have f(u) ≥ (3w/32)^{2}e^{−3w/32}, 3w/32 ≤ |u| ≤ w. Therefore (3.17) is satisfied if we set

f(u) ≥ (\frac{3w}{32} )^{2}e^{−3w/32} ≥ w[ ∑_{i=1}^{r−1} ∫_{I_{i}}(−1)^{α_{1}}G(t_{2}/2, s)a(s)ds]^{−1} =

= w[ ∑_{i=1}^{2} ∫_{I_{i}} G_{1}(0.25, s)h(s)ds]^{−1} = 0.002939w, 3w/32≤|u|≤w.

The explicit expression of the Green’s function (see (2.11)) is used in the computation of the above integrals. It can be checked that the above inequality holds if

0.3454 ≤ w ≤ 2.  (3.25)

Thus by Theorem 3.2(c) the boundary value problem has two positive solutions y_{1} and y_{2} such that 0 < ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥. Moreover, it follows from (3.25) that

0 < ∥y_{1}∥ ≤ 0.3454 and ∥y_{2}∥ ≥ 2.  (3.26)

Case 2. Let h(t) = (t^{6} + 0.1)^{−16}. Suppose that 3w/32 ≥ 2. Then, since f is nonincreasing in |u| for |u| ≥ 2, we have f(u) ≥ w^{2}e^{−w}, 3w/32 ≤ |u| ≤ w. Thus, for (3.17) to be satisfied we impose

f(u) ≥ w^{2}e^{−w} ≥ w[ ∑_{i=1}^{r−1} ∫_{I_{i}}(−1)^{α_{1}}G(t_{2}/2, s)a(s)ds]^{−1} =

= 4.001 × 10^{−14}w, 3w/32 ≤ |u| ≤ w.

The above inequality holds if

21.34 ≤ w ≤ 34.38.  (3.27)

It follows from Theorem 3.2(c) that the boundary value problem has two positive solutions y_{1} and y_{2} such that 0 < ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥. Moreover, in view of (3.27) we have

0 < ∥y_{1}∥ ≤ 21.34  and  ∥y_{2}∥ ≥ 34.38.  (3.28)

4. Double Positive Solutions of (M1)

We assume that the function q : (0, 1) → [0, ∞) is continuous and satisfies (B1) and (B2), or (B3), where

(B1) q is not identically zero on any nondegenerate subinterval of (0, 1);

(B2) ∫_{0}^{1} ∥G_{1}(·, s)∥q(s)ds < ∞, where G_{1} is given in (2.11);

(B3) q : [0, 1] → [0, ∞) is continuous and (B1) holds with (0, 1) replaced by [0, 1].

Theorem 4.1. Let w > 0 be given. Suppose that

∫_{0}^{1} ϕ(s)q(s)ds ≤ \frac{w}{w^{α} + w^{β}} ,  (4.1)

where ϕ is defined in (2.8). Then the boundary value problem (M1) has two positive solutions y_{1} and y_{2} such that 0 < ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥.

Proof. Let f(u) = |u|^{α} + |u|^{β}. Then f_{0} = f_{∞} = ∞. Further, we may take a(t) = b(t) = q(t). Clearly, f(u) ≤ w^{α} + w^{β} for |u| ≤ w. So to ensure that (3.7) is satisfied we apply Lemma 2.2 and impose

w^{α} + w^{β} ≤ w[ ∫_{0}^{1} ϕ(s)b(s)ds]^{−1} ≤ w[ ∫_{0}^{1} ∥G_{1}(·, s)∥b(s)ds]^{−1}

which is exactly condition (4.1). The conclusion now follows from Theorem 3.1(c).

Example 4.1. Consider the boundary value problem (M1) with c = 0.2. Let w = 1. Then condition (4.1) reduces to

∫_{0}^{1} ϕ(s)q(s)ds ≤ 1/2.  (4.2)

By Theorem 4.1 the boundary value problem has double positive solutions y_{1} and y_{2} such that 0 < ∥y_{1}∥ ≤ 1 ≤ ∥y_{2}∥ if q(t) fulfills (4.2). Examples of such q(t) include q(t) = 1, t + 1, sin^{2}(t + 1).

The next result offers upper and lower bounds for the norms of two positive solutions of (M1).

Theorem 4.2. We define

q_{i} = inf_{t∈I_{i}} q(t), i = 1, 2,  (4.3)

Q_{1}(u) = L_{1}^{u}q_{1} \frac{c^{3}(92 − 39c)}{3072}, Q_{2}(u) = L_{2}^{u}q_{2} \frac{13(1 − c)^{3}(3c + 1)}{3072},

w_{1} = [max\{Q_{1}(α), Q_{2}(α)\}]^{\frac{1}{1−α}} and w_{2} = [max\{Q_{1}(β), Q_{2}(β)\}]^{\frac{1}{1−β}} .

Let w > 0 be given. Suppose that (4.1) holds. Then the boundary value problem (M1) has double positive solutions y_{1} and y_{2} such that

(a) if w < min{w_{1}, w_{2}}, then 0 < ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥ ≤ min{w_{1}, w_{2}};

(b) if min{w_{1}, w_{2}} < w < max{w_{1}, w_{2}}, then min{w_{1}, w_{2}} ≤ ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥ ≤ max{w_{1}, w_{2}};

(c) if w > max{w_{1}, w_{2}}, then max{w_{1}, w_{2}} ≤ ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥.

Proof. Since (4.1) is satisfied, it follows from Theorem 4.1 that (M1) has double positive solutions y_{3} and y_{4} such that

0 < ∥y_{3}∥ ≤ w ≤ ∥y_{4}∥.  (4.4)

Let C_{1} be a cone in B defined by

C_{1} ={y ∈B|(−1)^{α_{i}}y(t)≥0, t∈[t_{i}, t_{i+1}], min_{t∈I_{i}}(−1)^{α_{i}}y(t)≥L_{i}∥y∥, i=1, 2}, (4.5)

where L_{i}, i = 1, 2, are given in (2.9) and (2.10). Define the operator T : C_{1} → B by

T y(t) = ∫_{0}^{1} G_{1}(t, s)q(s)[|y(s)|^{α} + |y(s)|^{β}]ds, t ∈ [0, 1].

To obtain a positive solution of (M1) we shall seek a fixed point of T in the cone C_{1}.

First, we shall show that T (C_{1}) ⊆ C_{1}. For this, let y ∈ C_{1}. Clearly, for i = 1, 2, (−1)^{α_{i}}T y(t) is nonnegative on [t_{i}, t_{i+1}]. Further, for t ∈ [t_{i}, t_{i+1}], i = 1, 2, we have

(−1)^{α_{i}}T y(t) ≤ ∫_{0}^{1} ∥G_{1}(·, s)∥q(s)[|y(s)|^{α} + |y(s)|^{β}]ds

which implies

∥T y∥ ≤ ∫_{0}^{1} ∥G_{1}(·, s)∥q(s)[|y(s)|^{α} + |y(s)|^{β}]ds.  (4.6)

Now, applying Lemma 2.2 and (4.6), we find for t ∈ I_{i}, i = 1, 2,

(−1)^{α_{i}}T y(t) ≥ ∫_{0}^{1} L_{i}∥G_{1}(·, s)∥q(s)[|y(s)|^{α} + |y(s)|^{β}]ds ≥ L_{i}∥T y∥.

Thus min_{t∈I_{i}}(−1)^{α_{i}}T y(t) ≥ L_{i}∥T y∥, i = 1, 2, and so T y ∈ C_{1}. Also, the standard arguments yield that T is completely continuous.

Let y ∈ C_{1} be such that ∥y∥ = w. Then, using Lemma 2.2 and (4.1), we find for t ∈ [t_{i}, t_{i+1}], i = 1, 2,

(−1)^{α_{i}}T y(t) ≤ ∫_{0}^{1} ∥G_{1}(·, s)∥q(s)[|y(s)|^{α} + |y(s)|^{β}]ds ≤
≤ ∫_{0}^{1} ϕ(s)q(s)(w^{α} + w^{β})ds ≤ w = ∥y∥.

Therefore

∥T y∥ ≤ ∥y∥.  (4.7)

By setting Ω = {y ∈ B | ∥y∥ < w}, we have (4.7) for y ∈ C_{1} ∩ ∂Ω. Now, let y ∈ C_{1}. It follows that

∥T y∥ = sup_{\begin{matrix}t∈[t_{i},t_{i+1}], \\ i=1,2 \end{matrix}} ∫_{0}^{1} |G_{1}(t, s)|q(s)[|y(s)|^{α} + |y(s)|^{β}]ds ≥

≥ sup_{\begin{matrix}t∈[t_{i},t_{i+1}], \\ i=1,2 \end{matrix}} ∫_{I_{i}} |G_{1}(t, s)|q(s)[(L_{i}∥y∥)^{α} + (L_{i}∥y∥)^{β}]ds ≥

≥ max { ∫_{I_{1}} |G_{1}(c/4, s)|q_{1}[(L_{1}∥y∥)^{α} + (L_{1}∥y∥)^{β}]ds,

∫_{I_{2}} |G_{1}((3c + 1)/4, s)|q_{2}[(L_{2}∥y∥)^{α} + (L_{2}∥y∥)^{β}]ds}.

From (2.11) we have

\begin{matrix}|G_{1}(c/4, s)| = \frac{3}{8}(1 − \frac{c}{4})s^{2} − \frac{1}{2}(s − \frac{c}{4})^{2}, s ∈ I_{1}, \\ |G_{1}((3c + 1)/4, s)| = \frac{3c + 1}{32}(1 − s)^{2}, s ∈ I_{2}, \end{matrix}  (4.8)

which upon substituting into the above inequality yields

∥T y∥ ≥ max {Q_{1}(α)∥y∥^{α} + Q_{1}(β)∥y∥^{β}, Q_{2}(α)∥y∥^{α} + Q_{2}(β)∥y∥^{β}}. (4.9)

Let y ∈ C_{1} be such that ∥y∥ = w_{1}. It follows from (4.9) that

∥T y∥ ≥ max {Q_{1}(α)∥y∥^{α}, Q_{2}(α)∥y∥^{α}} =

= max {Q_{1}(α), Q_{2}(α)}∥y∥^{α−1}∥y∥ = ∥y∥.  (4.10)

If we set Ω_{1} = {y ∈ B | ∥y∥ < w_{1}}, then (4.10) holds for y ∈ C_{1} ∩ ∂Ω_{1}. Now that we have obtained (4.7) and (4.10), it follows from Theorem 2.1 that T has a fixed point y_{5} such that

min{w_{1}, w} ≤ ∥y_{5}∥ ≤ max{w_{1}, w}.  (4.11)

On the other hand, if we let y ∈ C_{1} be such that ∥y∥ = w_{2}, then from (4.9) we get

∥T y∥ ≥ max {Q_{1}(β)∥y∥^{β}, Q_{2}(β)∥y∥^{β}} = ∥y∥.  (4.12)

Take Ω_{2} = {y ∈ B | ∥y∥ < w_{2}}, then (4.12) holds for y ∈ C_{1} ∩ ∂Ω_{2}. Having obtained (4.7) and (4.12), by Theorem 2.1 we conclude that T has a fixed point y_{6} such that

min{w_{2}, w} ≤ ∥y_{6}∥ ≤ max{w_{2}, w}.  (4.13)

Now, a combination of (4.4), (4.11) and (4.13) yields our result. More

precisely, in Case (a) we may take y_{1} = y_{3} and y_{2} = \{ \begin{matrix}y_{5}, & w_{1} ≤ w_{2} \\ y_{6}, & w_{1} ≥ w_{2} \end{matrix} . In

Case (b) it is clear that y_{1} = \{ \begin{matrix}y_{5}, & w_{1} ≤ w_{2} \\ y_{6}, & w_{1} ≥ w_{2} \end{matrix} and y_{2} = \{ \begin{matrix}y_{6}, & w_{1} ≤ w_{2} \\ y_{5}, & w_{1} ≥ w_{2} \end{matrix} .

Finally, in Case (c) we shall take y_{1} = \{ \begin{matrix}y_{6}, & w_{1} ≤ w_{2} \\ y_{5}, & w_{1} ≥ w_{2} \end{matrix} and y_{2} = y_{4}.

Example 4.2. Consider the boundary value problem

y^{(3)} = (t + 1)(|y|^{0.1} + |y|^{3}), t ∈ (0, 1), y(0) = y(0.3) = y(1) = 0.

Here α = 0.1, β = 3, c = 0.3 and q(t) = t + 1. Condition (4.1) is the same as

\frac{w}{w^{0.1} + w^{3}} ≥ ∫_{0}^{1} ϕ(s)q(s)ds = 0.09718

which is satisfied for any w ∈ [0.07505, 3.151].

Further, with q_{1} = 1.075, q_{2} = 1.475, L_{1} = 0.02411 and L_{2} = 0.1188, we compute

w_{1} = [max{Q_{1}(α), Q_{2}(α)}]^{1/(1−α)} = [Q_{2}(α)]^{1/(1−α)} = 2.836 × 10^{−4}

and

w_{2} = [max{Q_{1}(β), Q_{2}(β)}]^{1/(1−β)} = [Q_{2}(β)]^{1/(1−β)} = 382.9.

Since w ∈ (w_{1}, w_{2}), by Theorem 4.2(b) the boundary value problem has two positive solutions y_{1} and y_{2} such that

2.836 × 10^{−4} ≤ ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥ ≤ 382.9.  (4.14)

Taking into account the range of w, (4.14) leads to

2.836 × 10^{−4} ≤ ∥y_{1}∥ ≤ 0.07505 and 3.151 ≤ ∥y_{2}∥ ≤ 382.9. (4.15)

5. Double Positive Solutions of (M2)

As in Section 4, it is assumed that the function q : (0, 1) → [0, ∞) is continuous and satisfies (B1) and (B2), or (B3).

Theorem 5.1. Let w > 0 be given. Suppose that

∫_{0}^{1} ϕ(s)q(s)ds ≤ we^{−γw}.  (5.1)

Then the boundary value problem (M2) has two positive solutions y_{1} and y_{2} such that

0 < ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥.  (5.2)

Proof. Let f(u) = e^{γ|u|} and a(t) = b(t) = q(t). Noting that f(u) ≤ e^{γw} for |u| ≤ w, an argument as in the proof of Theorem 4.1 yields the conclusion.

Example 5.1. Consider the boundary value problem

y^{(3)} = q(t)e^{|y|}, t ∈ (0, 1), y(0) = y(0.7) = y(1) = 0.

Let w = 1/3 be given. Then condition (5.1) reduces to

∫_{0}^{1} ϕ(s)q(s)ds ≤ \frac{1}{3} e^{−1/3}.  (5.3)

By Theorem 5.1, for those q(t) which fulfill (5.3), the boundary value problem has two positive solutions y_{1} and y_{2} such that

0 < ∥y_{1}∥ ≤ \frac{1}{3} ≤ ∥y_{2}∥.

Some examples of such q(t) are q(t) = (2t+12)^{−1}, (t^{2}+3)/2, [ cos^{2}(t+2)]/2.

Once again we shall establish upper and lower bounds for the norms of two positive solutions of (M2).

Theorem 5.2. Let k, ℓ (k ≠ ℓ) be given integers in the set {0, 2, 3, … }. We define

R_{1}(u) = q_{1} \frac{(γL_{1})^{u}}{u!} \frac{c^{3}(92 − 39c)}{3072}, R_{2}(u) = q_{2} \frac{(γL_{2})^{u}}{u!} \frac{13(1 − c)^{3}(3c + 1)}{3072},

w_{1} = [max\{R_{1}(k), R_{2}(k)\}]^{\frac{1}{1−k}} and w_{2} = [max\{R_{1}(ℓ), R_{2}(ℓ)\}]^{\frac{1}{1−ℓ}} ,

where q_{i}, i = 1, 2, are given in (4.3). Let w > 0 be given. Suppose that (5.1) holds. Then, the boundary value problem (M2) has double positive solutions y_{1} and y_{2} such that conclusions (a)–(c) of Theorem 4.2 hold.

Proof. Since (5.1) is fulfilled, by Theorem 5.1 the boundary value problem (M2) has double positive solutions y_{3} and y_{4} such that (4.4) holds.

To proceed, let C_{1} be a cone in B defined by (4.5) and let the operator T̄ : C_{1} → B be defined by

T̄y(t) = ∫_{0}^{1} G_{1}(t, s)q(s)e^{γ|y(s)|}ds, t ∈ [0, 1].

To obtain a positive solution of (M2), we shall seek a fixed point of T̄ in the cone C_{1}. Using an argument as in the proof of Theorem 4.2, it can be verified that T̄(C_{1}) ⊆ C_{1} and T̄ is completely continuous.

Let y ∈ C_{1} be such that ∥y∥ = w. Applying Lemma 2.2 and (5.1), we get for t ∈ [t_{i}, t_{i+1}], i = 1, 2,

(−1)^{α_{i}} T̄y(t) ≤ ∫_{0}^{1} ϕ(s)q(s)e^{γ}|y(s)|ds ≤ ∫_{0}^{1} ϕ(s)q(s)e^{γw}ds ≤ w = ∥y∥.

Hence

∥T̄y∥ ≤ ∥y∥.  (5.4)

If we set Ω = {y ∈ B | ∥y∥ < w}, then (5.4) holds for y ∈ C_{1} ∩ ∂Ω. Next, let y ∈ C_{1}. We find that

∥T̄y∥ ≥ sup_{\begin{matrix}t∈[t_{i},t_{i+1}] \\ i=1,2 \end{matrix}} nt_{I_{i}}|G_{1}(t, s)|q_{i}e^{γL_{i}∥y∥}ds.

Using the relation

e^{u} ≥ \frac{u^{k}}{k!} + \frac{u^{ℓ}}{ℓ!} , u > 0,

in the above inequality, we find

∥T̄y∥ ≥ sup_{\begin{matrix}t∈[t_{i},t_{i+1}] \\ i=1,2 \end{matrix}} ∫_{I_{i}} |G_{1}(t, s)|q_{i}[\frac{(γL_{i})^{k}}{k!} ∥y∥^{k} + \frac{(γL_{i})^{ℓ}}{ℓ!} ∥y∥^{ℓ}] ≥

≥ max \{ ∫_{I_{1}} |G_{1}(\frac{c}{4}, s)|q_{1}[\frac{(γL_{1})^{k}}{k!} ∥y∥^{k} + \frac{(γL_{1})^{ℓ}}{ℓ!} ∥y∥^{ℓ}],

∫_{I_{2}} |G_{1}(\frac{3c + 1}{4}, s)|q_{2}[\frac{(γL_{2})^{k}}{k!} ∥y∥^{k} + \frac{(γL_{2})^{ℓ}}{ℓ!} ∥y∥^{ℓ}]\}.

On substituting (4.8), it follows that

|T̄y∥ ≥ max {R_{1}(k)∥y∥^{k} + R_{1}(ℓ)∥y∥^{ℓ}, R_{2}(k)∥y∥^{k} + R_{2}(ℓ)∥y∥^{ℓ}}. (5.5)

Employing a technique as in the proof of Theorem 4.2, from (5.5) we obtain

|T̄y∥ ≥ ∥y∥  (5.6)

for y ∈ C_{1} ∩ ∂Ω_{1} as well as for y ∈ C_{1} ∩ ∂Ω_{2}, where

Ω_{1} = {y ∈ B | ∥y∥ < w_{1}}  and  Ω_{2} = {y ∈ B | ∥y∥ < w_{2}}.

Now that we have obtained (5.4) and (5.6), by Theorem 2.1 T̄ has a fixed point y_{5} satisfying

min {w_{1}, w} ≤ ∥y_{5}∥ ≤ max {w_{1}, w},  (5.7)

and also a fixed point y_{6} such that

min {w_{2}, w} ≤ ∥y_{6}∥ ≤ max {w_{2}, w}.  (5.8)

As in the proof of Theorem 4.2, the combination of (4.4), (5.7) and (5.8) readily gives rise to conclusions (a)–(c).

Example 5.2. Consider the boundary value problem

y^{(3)} = \frac{6}{exp(0.5t(1−t)|t−0.3|)}e^{|y|/2}, t∈(0, 1), y(0)=y(0.3)=y(1)=0.

With c = 0.3 and q(t) = 6[exp(0.5t(1 − t)|t − 0.3|)]^{−1}, we get L_{1} = 0.02411, L_{2} = 0.1188, q_{1} = q((13 − \sqrt{79})/30) = 5.942 and q_{2} = q((13+

\sqrt{79})/30) = 5.751. It can be checked that condition (5.1) is satisfied provided

0.5166 ≤ w ≤ 5.093.  (5.9)

Let k = 0 and ℓ = 9. We find w_{1} = [R_{2}(k)]^{1/(1−k)} = 0.01586 and w_{2} = [R_{1}(ℓ)]^{1/(1−ℓ)} = 1415. Since w ∈ (w_{1}, w_{2}), it follows from Theorem 5.2(b) that the boundary value problem has two positive solutions y_{1} and y_{2} such that 0.01586 ≤ ∥y_{1}∥ ≤ w ≤ ∥y_{2}∥ ≤ 1415. In view of (5.9) we further conclude that

0.01586 ≤ ∥y_{1}∥ ≤ 0.5166  and  5.093 ≤ ∥y_{2}∥ ≤ 1415.  (5.10)

Indeed, the boundary value problem has a positive solution given by y(t) = t(t − 0.3)(t − 1) and we note that ∥y∥ = y((13 + \sqrt{79})/30) = 0.08475 is within the range given in (5.10).

6. Triple Positive Solutions of (M3)

It is assumed that the functions f : IR → [0, ∞) and b : (0, 1) → IR are continuous and satisfy (A1), (A3)′ and (A4), or (A5)′, where

(A3)′ b is nonnegative and not identically zero on any nondegenerate subinterval of (0, 1);

(A5)′ b : [0, 1] → IR is continuous and (A3)′ holds with (0, 1) replaced by [0, 1].

With the same Banach space B (= C[0, 1]), let

C_{2} = {y ∈ B | (−1)^{α_{i}}y(t) ≥ 0, t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1}.

We note that C_{2} is a cone in B.

Let the operator V : C_{2} → B be defined by

V y(t) = ∫_{0}^{1} G(t, s)b(s)f(y(s))ds, t ∈ [0, 1].  (6.1)

To obtain a positive solution of (M3), we shall seek a fixed point of the operator V in the cone C_{2}.

By using an argument as in Section 3, we see that the operator V is compact on the cone C_{2}. Next, it is clear from (6.1) and (2.4) that if y ∈ C_{2}, then (−1)^{α_{i}}V y(t) ≥ 0 for t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1. Hence V y ∈ C_{2} and we have shown that V maps C_{2} into itself. Also, the standard arguments yield that V is completely continuous.

It is clear that

∥V y∥ = ∫_{0}^{1} ∥G(·, s)∥b(s)f(y(s))ds.  (6.2)

We define the constants

z =∫_{0}^{1}∥G(·, s)∥b(s)ds and μ= min_{1≤i≤r−1} min_{t∈I_{i}} ∫_{0}^{1}(−1)^{α_{i}}G(t, s)b(s)ds. (6.3)

Lemma 6.1. Suppose that there exists ν > 0 such that f(u) < \frac{ν}{z} for |u| ∈ [0, ν]. Then

V (C̄_{2}(ν)) ⊆ C_{2}(ν) ⊂ C̄_{2}(ν).

Proof. Let y ∈ C̄_{2}(ν). Then we have for t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1,

(−1)^{α_{i}}V y(t) ≤ ∫_{0}^{1} ∥G(·, s)∥b(s)f(y(s))ds < ∫_{0}^{1} ∥G(·, s)∥b(s) \frac{ν}{z} ds = ν.

Consequently, ∥V y∥ < ν and so V y ∈ C_{2}(ν). This immediately implies that V (C̄_{2}(ν)) ⊆ C_{2}(ν) ⊂ C̄_{2}(ν).

Theorem 6.1. Suppose that there exist numbers k, ℓ, m, where 0 < k < ℓ < m, m ≥ 1 and ℓ ≤ [ min_{1≤i≤r−1}(t_{i+1} − t_{i})/4]^{n}, such that the following conditions are satisfied:

(H1) one of the following holds:

(i) lim sup_{|u|→∞} f(u)/|u| < 1/z; (ii) there exists a number η, where η > m, such that if |u| ∈ [0, η], then

f(u) < η/z; (H2) if |u| ∈ [ℓ, m], then f(u) > ℓ/μ; (H3) if |u| ∈ [0, k], then f(u) < k/z. Then the boundary value problem (M3) has (at least) three positive solu-

tions y_{1}, y_{2} and y_{3} such that

∥y_{1}∥ < k; (−1)^{α_{i}}y_{2}(t) > ℓ, t ∈ I_{i}, 1 ≤ i ≤ r − 1;
∥y_{3}∥ > k  and  min_{1≤i≤r−1} min_{t∈I_{i}}(−1)^{α_{i}}y_{3}(t) < ℓ. (6.4)

Proof. We shall show that the conditions of Theorem 2.2 are fulfilled. First we shall prove that condition (H1) leads to the existence of a number ν, where ν > m, such that

V (C̄_{2}(ν)) ⊂ C̄_{2}(ν).  (6.5)

For this, it is clear that if (ii) holds, then by Lemma 6.1 we immediately have (6.5) where ν = η. Suppose now that (i) is satisfied. Then there exist T > 0 and ϵ < \frac{1}{z} such that

f(u)/|u| < ϵ, |u| > T.  (6.6)

Define M = max_{|u|∈[0,T]} f(u). In view of (6.6) it is obvious that

f(u) ≤ M + ϵ|u|, |u| ≥ 0.  (6.7)

Now let ν be such that

ν > max {m, M(1/z − ϵ)^{−1}}.  (6.8)

For y ∈ C̄_{2}(ν) and t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1, we use (2.5), (6.7), (6.3) and (6.8) to get

(−1)^{α_{i}}V y(t) ≤ ∫_{0}^{1} ∥G(·, s)∥b(s)f(y(s))ds ≤
≤ ∫_{0}^{1} ∥G(·, s)∥b(s)(M + ϵ|y(s)|)ds ≤
≤ ∫_{0}^{1} ∥G(·, s)∥b(s)(M + ϵν)ds =
= z(M + ϵν) < z[ν(1/z − ϵ) + ϵν] = ν.

Hence ∥V y∥ < ν and so (6.5) follows immediately.

Let ψ : C_{2} → [0, ∞) be defined by

ψ(y) = min_{1≤i≤r−1} min_{t∈I_{i}}(−1)^{α_{i}}y(t).

Clearly, ψ is a nonnegative continuous concave functional on C_{2} such that ψ(y) ≤ ∥y∥ for all y ∈ C_{2}.

We shall now show that condition (a) of Theorem 2.2 is satisfied. For

this, note that y_{1}(t) ≡ ∏_{i=1}^{r}(t − t_{i})^{n_{i}} has the properties

(−1)^{α_{i}}y_{1}(t) ≥ 0, t ∈ [t_{i}, t_{i+1}], 1 ≤ i ≤ r − 1, ∥y_{1}∥ < 1 ≤ m

and

ψ(y_{1}) = min_{1≤i≤r−1} min_{t∈I_{i}}(−1)^{α_{i}}y_{1}(t) =

= min_{1≤i≤r−1} \{|y_{1}(\frac{3t_{i} + t_{i+1}}{4} )|, |y_{1}(\frac{t_{i} + 3t_{i+1}}{4} )|\} >

> [ min_{1≤i≤r−1}(t_{i+1} − t_{i})/4]^{n} ≥ ℓ.

Hence

y_{1} ∈ {y ∈ C_{2}(ψ, ℓ, m) | ψ(y) > ℓ} ≠ ∅.  (6.9)

Next, let y ∈ C_{2}(ψ, ℓ, m). Then ℓ ≤ ∥y∥ ≤ m and so |y(s)| ∈ [ℓ, m] for all s ∈ [0, 1]. Using this together with (H2) and (6.3), we get

ψ(V y) = min_{1≤i≤r−1} min_{t∈I_{i}}(−1)^{α_{i}}V y(t) =

= min_{1≤i≤r−1} min_{t∈I_{i}} ∫_{0}^{1}(−1)^{α_{i}}G(t, s)b(s)f(y(s))ds >

> min_{1≤i≤r−1} min_{t∈I_{i}} ∫_{0}^{1}(−1)^{α_{i}}G(t, s)b(s) (ℓ/μ) ds = ℓ.

Therefore ψ(V y) > ℓ for all y ∈ C_{2}(ψ, ℓ, m).

Moreover, it follows from Lemma 6.1 and condition (H3) that V (C̄_{2}(k)) ⊆ C_{2}(k). Hence condition (b) of Theorem 2.2 is satisfied.

It remains to verify that condition (c) of Theorem 2.2 holds. Let m > ℓ(min_{1≤i≤r−1} L_{i})^{−1} (≥ ℓ) and let y ∈ C_{2}(ψ, ℓ, ν) with ∥V y∥ > m. Applying Lemma 2.1 and (6.2), we find

ψ(V y) = min_{1≤i≤r−1} min_{t∈I_{i}} ∫_{0}^{1}(−1)^{α_{i}}G(t, s)b(s)f(y(s))ds ≥

≥ min_{1≤i≤r−1} ∫_{0}^{1} L_{i}∥G(·, s)∥b(s)f(y(s))ds =

= min_{1≤i≤r−1} L_{i}∥V y∥ > m min_{1≤i≤r−1} L_{i} > ℓ.

This shows that ψ(V y) > ℓ for all y ∈ C_{2}(ψ, ℓ, ν) with ∥V y∥ > m.

Consequently, it follows from Theorem 2.2 that the boundary value problem (M3) has (at least) three positive solutions y_{1}, y_{2}, y_{3} ∈ C̄_{2}(ν). Further, we have (2.1) which reduces to (6.4).

Example 6.1. Consider the boundary value problem

y^{(3)} = (t^{2} + 1)f(y), t ∈ (0, 1), y(0) = y(0.5) = y(1) = 0,

where

f(y) = \begin{cases}| sin y|, & |y| ≤ 0.001 \\ sin 0.001 + (|y| − 0.001)^{1/8}, & 0.001 ≤ |y| ≤ 1 \\ sin 0.001 + 0.999^{1/8}, & |y| ≥ 1 \end{cases} .

Here n = r = 3 and b(t) = t^{2} + 1. It is clear that (A5)′ is satisfied. Using (2.11), we find

z = ∫_{0}^{1} ∥G_{1}(·, s)∥b(s)ds =

= max { max_{t∈[0,0.5]} ∫_{0}^{1}|G_{1}(t, s)|b(s)ds, max_{t∈[0.5,1]} ∫_{0}^{1}|G_{1}(t, s)|b(s)ds}=0.0109

and

μ=min\{ min_{t∈[ \frac{1}{8} , \frac{3}{8} ]} ∫_{0}^{1}|G_{1}(t, s)|b(s)ds, min_{t∈[ \frac{5}{8} , \frac{7}{8} ]} ∫_{0}^{1}|G_{1}(t, s)|b(s)ds\}=0.006081.

Obviously, condition (H1) holds as

lim_{|u|→∞} \frac{f(u)}{|u|} = lim_{|u|→∞} \frac{sin 0.001 + 0.999^{1/8}}{|u|} = 0.

Next, take k = 0.001. Then (H3) is fulfilled, since for |u| ∈ [0, k],

f(u) ≤ sin 0.001 < k/z.

Finally, let m = 1 and take ℓ (≤ [ min_{1≤i≤r−1}(t_{i+1} − t_{i})/4]^{n} = 0.001953) such that condition (H2) is satisfied. Clearly, (H2) is fulfilled provided that for |u| ∈ [ℓ, m],

f(u) ≥ sin 0.001 + (ℓ − 0.001)^{1/8} > ℓ/μ.

The above inequality is satisfied if

0.001001 ≤ ℓ ≤ 0.001953.  (6.10)

By Theorem 6.1 the boundary value problem has (at least) three positive solutions y_{1}, y_{2} and y_{3}. Further, in view of (6.10), it follows from (6.4) that

\begin{matrix}∥y_{1}∥ < 0.001; |y_{2}(t)| > 0.001953, t ∈ [\frac{1}{8}, \frac{3}{8}] ∪ [\frac{5}{8}, \frac{7}{8}]; \\ ∥y_{3}∥ > 0.001 and min ; t∈[\frac{1}{8} , \frac{3}{8}]∪[\frac{5}{8} , \frac{7}{8}] |y_{3}(t)| < 0.001001. \end{matrix} (6.11)

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(Received 14.12.1998)

Authors’ addresses:

P. J. Y. Wong Division of Mathematics Nanyang Technological University, Singapore

R. P. Agarwal Department of Mathematics National University of Singapore Singapore