Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 53, 1-13; http://www.math.u-szeged.hu/ejqtde/

POSITIVE SOLUTIONS OF THE (n − 1, 1) CONJUGATE BOUNDARY VALUE PROBLEM

BO YANG

Abstract. We consider the (n−1, 1) conjugate boundary value problem. Some upper estimates to positive solutions for the problem are obtained. We also establish some explicit sufficient conditions for the existence and nonexistence of positive solutions of the problem.

1. Introduction

In this paper we consider the (n − 1, 1) conjugate boundary value problem

u^{(n)}(t) + g(t)f(u(t)) = 0, 0 ≤ t ≤ 1,  (1.1)
u^{(i)}(0) = u(1) = 0, i = 0, 1, ⋯, n − 2.  (1.2)

Throughout this paper, we assume that

(H1) n ≥ 3 is a fixed integer, and f : [0, ∞) → [0, ∞) and g : [0, 1] → [0, ∞) are continuous functions.

The Green’s function G : [0, 1] × [0, 1] → [0, ∞) for the problem (1.1)-(1.2) is given by

G(t, s) = \frac{1}{(n − 1)!} \{ \begin{matrix}t^{n−1}(1 − s)^{n−1} − (t − s)^{n−1}, & t ≥ s, \\ t^{n−1}(1 − s)^{n−1}, & s ≥ t. \end{matrix}

And the problem (1.1)-(1.2) is equivalent to the integral equation

u(t) = ∫_{0}^{1} G(t, s)g(s)f(u(s)) ds, 0 ≤ t ≤ 1.  (1.3)

The (n − 1, 1) conjugate problem and its various generalizations have been considered by many authors. For example, in 1997, Eloe and Henderson [1] considered the problem (1.1)-(1.2) and obtained some existence results for positive solutions to the problem. If n = 2, then the problem (1.1)-(1.2) reduces to the second order problem

u′′(t) + g(t)f(u(t)) = 0, 0 ≤ t ≤ 1,
u(0) = u(1) = 0,

which has been extensively studied by many authors, including Graef and Yang [2]. EJQTDE, 2010 No. 53, p. 1

2000 Mathematics Subject Classification. 34B18.

Key words and phrases. Positive solutions; conjugate boundary value problem; upper and lower estimates; first eigenvalue.

In 2009, Webb [4] considered a nonlocal version of the (n − 1, 1) conjugate problem. He obtained a lower estimate for the Green’s function G(t, s), based on which a lower estimate to positive solutions for the problem (1.1)-(1.2) can be proved (see Lemmas 2.8 and 2.9 below). However, to our knowledge, no satisfactory upper estimates to positive solutions for the problem (1.1)-(1.2) have been obtained in the literature. We know that upper and lower estimates for positive solutions of boundary value problems have important applications. For example, once we find some a priori upper and lower estimates for positive solutions of a certain boundary value problem, we can use them together with the Krasnosell’skii fixed point theorem to derive a set of existence and nonexistence conditions for positive solutions of the problem (See [5] for a paper taking this approach). With this motivation, we in this paper make a further study of positive solutions to the problem (1.1)-(1.2). Our main goal is to develop some new upper estimates for positive solutions of the problem (1.1)-(1.2). Here, by a positive solution, we mean a solution u(t) such that u(t) > 0 on (0, 1). Since the case n = 2 is a well-studied case, we in this paper assume that n ≥ 3.

Throughout the paper, we let X = C[0, 1] be equipped with the supremum norm

∥v∥ = max_{t∈[0,1]} |v(t)|, for all v ∈ X.

Obviously X is a Banach space. Also, we define

F_{0} = lim sup_{x→0^{+}} \frac{f(x)}{x}, f_{0} = lim inf_{x→0^{+}} \frac{f(x)}{x},

F_{∞} = lim sup_{x→+∞} \frac{f(x)}{x}, f_{∞} = lim inf_{x→+∞} \frac{f(x)}{x}.

These constants will be used later in the statements of our existence and nonexistence theorems.

This paper is organized as follows. In Section 2, we obtain some new upper estimates to positive solutions to the (n − 1, 1) conjugate problem, and discuss some lower estimates from the literature. In Section 3, we establish some explicit sufficient conditions for the existence and nonexistence of positive solutions to the problem.

2. Upper and Lower Estimates for Positive Solutions

Throughout the paper we define the constants

p = \frac{(n − 1)^{n−1}}{(n − 1)^{n−1} + (n − 2)^{n−2}} , q = \frac{n − 2}{n − 1}.

EJQTDE, 2010 No. 53, p. 2 We also define the functions w_{1} : [0, 1] → [0, +∞) and w_{2} : [0, 1] → [0, ∞) by

w_{1}(t) = \{ \begin{matrix}(n − 1)^{n−1}(n − 2)^{2−n}(t^{n−2} − t^{n−1}), & if t ≥ p, \\ t^{n−1}, & if t ≤ p, \end{matrix}

and

w_{2}(t) = \{ \begin{matrix}(n − 1)^{n−1}(n − 2)^{2−n}(t^{n−2} − t^{n−1}), & if t ≤ q, \\ 1, & if t ≥ q. \end{matrix}

The functions w_{1}(t) and w_{2}(t) will be used to estimate positive solutions of the problem (1.1)- (1.2). It is easy to see that ∥w_{2}∥ = 1 and that both w_{1}(t) and w_{2}(t) are continuous functions. An equivalent definition for the function w_{1}(t) is

w_{1}(t) = min{t^{n−1}, (n − 1)^{n−1}(n − 2)^{2−n}(t^{n−2} − t^{n−1})}, 0 ≤ t ≤ 1.

The function w_{1}(t) first appeared in [4]. It can be shown that

w_{1}(t) ≤ w_{2}(t), 0 ≤ t ≤ 1.  (2.1)

The verification of (2.1) is straightforward and is therefore left to the reader. Lemma 2.1. If u ∈ C^{n}[0, 1] satisfies (1.2), and

u^{(n)}(t) ≤ 0 for 0 ≤ t ≤ 1,  (2.2)

then u(t) ≥ 0 for 0 ≤ t ≤ 1. Proof. It is well known that the Green’s function G(t, s) has the property

G(t, s) > 0 for (t, s) ∈ (0, 1) × (0, 1).

If u^{(n)}(t) ≤ 0 on [0, 1], then

u(t) = ∫_{0}^{1} G(t, s)(−u^{(n)}(s))ds ≥ 0, 0 ≤ t ≤ 1.

The proof of the lemma is complete. □

The next lemma was proved by Eloe and Henderson in [1]. Lemma 2.2. If u ∈ C^{n}[0, 1] satisfies (1.2) and (2.2), and u(t_{0}) > 0 for some t_{0} ∈ (0, 1), then

u(t) > 0 for 0 < t < 1,  (2.3)

and there exists c ∈ (0, 1) such that u′(c) = 0,

u′(t) > 0 for 0 < t < c, and u′(t) < 0 for c < t < 1.  (2.4)

In other words, c is the unique zero of u′ in (0, 1).

EJQTDE, 2010 No. 53, p. 3

The next lemma provides a new upper estimate for functions satisfying (1.2), (2.2), and (2.3).

Lemma 2.3. Suppose that u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3). Let c be the unique zero of u′ in (0, 1), then

u(t) ≤ β(t)∥u∥ for 0 ≤ t ≤ 1,  (2.5)

where

β(t) = \frac{t^{n−2}}{c^{n−1}} ((n − 1)c − (n − 2)t), 0 ≤ t ≤ 1.

Proof. We see from Lemma 2.2 that u(c) = ∥u∥. The inequality in (2.5) is trivial for t = 0, t = c, and t = 1. We need only to show that the inequality holds for 0 < t < c and c < t < 1. Let x ∈ (0, c) ∪ (c, 1) be a fixed number. Define

h(t) = u(t) − β(t)u(c) − (u(x) − β(x)u(c))\frac{t^{n−2}(t − c)^{2}}{x^{n−2}(x − c)^{2}} , 0 ≤ t ≤ 1.

It is easy to verify the following facts:

h(0) = h′(0) = ⋯ = h^{(n−3)}(0) = 0,

h(c) = h′(c) = 0, h(x) = 0.

h^{(n)}(t) = u^{(n)}(t) − (u(x) − β(x)u(c))\frac{n!}{x^{n−2}(x − c)^{2}} , 0 ≤ t ≤ 1.

We take two cases to continue the proof.

Case I: 0 < x < c. In this case, because h(0) = h(x) = h(c) = 0, there exist t_{1} ∈ (0, x) and s_{1} ∈ (x, c) such that h′(t_{1}) = h′(s_{1}) = 0.

Because h′(0) = h′(t_{1}) = h′(s_{1}) = h′(c) = 0, there exist t_{2} ∈ (0, t_{1}), s_{2} ∈ (t_{1}, s_{1}), and r_{2} ∈ (s_{1}, c) such that h′′(t_{2}) = h′′(s_{2}) = h′′(r_{2}) = 0.

Because h′′(0) = h′′(t_{2}) = h′′(s_{2}) = h′′(r_{2}) = 0, there exist t_{3} ∈ (0, t_{2}), s_{3} ∈ (t_{2}, s_{2}), and r_{3} ∈ (s_{2}, r_{2}) such that h′′′(t_{3}) = h′′′(s_{3}) = h′′′(r_{3}) = 0.

If we continue this procedure, we can show that for each i = 2, 3, ⋯, n − 2, there exist t_{i}, s_{i}, and r_{i} such that 0 < t_{i} < s_{i} < r_{i} < c and h^{(i)}(t_{i}) = h^{(i)}(s_{i}) = h^{(i)}(r_{i}) = 0.

In particular, we have h^{(n−2)}(t_{n−2}) = h^{(n−2)}(s_{n−2}) = h^{(n−2)}(r_{n−2}) = 0. This implies that there exist t_{n−1} ∈ (t_{n−2}, s_{n−2}) and s_{n−1} ∈ (s_{n−1}, r_{n−2}) such that h^{(n−1)}(t_{n−1}) = h^{(n−1)}(s_{n−1}) = 0. Because h^{(n−1)}(t_{n−1}) = h^{(n−1)}(s_{n−1}) = 0, there exists t_{n} ∈ (t_{n−1}, s_{n−1}) such that h^{(n)}(t_{n}) = 0, which implies that

0 = u^{(n)}(t_{n}) − (u(x) − β(x)u(c))\frac{n!}{x^{n−2}(x − c)^{2}} .

The above equation implies that

u(x) − β(x)u(c) = \frac{u^{(n)}(t_{n})}{n!}x^{n−2}(x − c)^{2} ≤ 0.

EJQTDE, 2010 No. 53, p. 4 In summary, if 0 < x < c then u(x) ≤ β(x)u(c). Hence, the inequality in (2.5) holds for 0 < t < c.

Case II: c < x < 1. In this case, because h(0) = h(c) = h(x) = 0, there exist t_{1} ∈ (0, c) and s_{1} ∈ (c, x) such that h′(t_{1}) = h′(s_{1}) = 0.

Because h′(0) = h′(t_{1}) = h′(c) = h′(s_{1}) = 0, there exist t_{2} ∈ (0, t_{1}), s_{2} ∈ (t_{1}, c), and r_{2} ∈ (c, s_{1}) such that h′′(t_{2}) = h′′(s_{2}) = h′′(r_{2}) = 0. If we continue from here and follow the same lines as in Case I, we can show that u(x) ≤ β(x)u(c) for c < x < 1. The proof in Case II is now complete.

In summary, we have u(t) ≤ β(t)u(c) for t ∈ (0, c) ∪ (c, 1). The proof of the lemma is complete. □

As a by-product of Lemma 2.3, we have

Lemma 2.4. Suppose that u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3). Let c be the unique zero of u′ in (0, 1). Then q ≤ c ≤ 1.

Proof. By Lemma 2.3 we have

0 ≤ u(t) ≤ β(t)u(c), 0 ≤ t ≤ 1.

Substituting t = 1 into the above inequality gives

0 ≤ \frac{u(c)}{c^{n−1}} ((n − 1)c − (n − 2)).

This implies that (n − 1)c − (n − 2) ≥ 0. Hence c ≥ (n − 2)/(n − 1). The proof is complete. □

Lemma 2.4 is interesting in its own right. The following lower estimate was given in [1].

Lemma 2.5. Suppose that u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3). Let c be the unique zero of u′ in (0, 1). Then

u(t) ≥ u(c) min{(t/c)^{n−1}, (1 − t)/(1 − c)}, 0 ≤ t ≤ 1.  (2.6)

Combining Lemmas 2.4 and 2.5, we get

Lemma 2.6. If u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3), then

u(t) ≥ ∥u∥γ(t), 0 ≤ t ≤ 1,  (2.7)

where γ(t) = min{t^{n−1}, (n − 1)(1 − t)}, 0 ≤ t ≤ 1.

Proof. Let c be the unique zero of u′ in (0, 1). By Lemma 2.4, we have (n − 2)/(n − 1) ≤ c ≤ 1. For 0 ≤ t ≤ c we have

u(t) ≥ ∥u∥(t/c)^{n−1} ≥ ∥u∥t^{n−1} ≥ ∥u∥γ(t).

EJQTDE, 2010 No. 53, p. 5 For c ≤ t ≤ 1 we have

u(t) ≥ ∥u∥\frac{1 − t}{1 − c} ≥ ∥u∥\frac{1 − t}{1 − (n − 2)/(n − 1)} = ∥u∥(n − 1)(1 − t) ≥ ∥u∥γ(t).

The proof is complete. □

Both Lemmas 2.5 and 2.6 provide a lower estimate to functions satisfying (1.2), (2.2), and (2.3). The difference is that the lower estimate in Lemma 2.5 depends on the unique zero c of u′ in (0, 1), while the lower estimate in Lemma 2.6 does not. If we know where c is, then (2.6) is a better estimate that (2.7). However, if we don’t know where c is, then we have to do with (2.7).

Lemma 2.7. If u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3), then

u(t) ≤ ∥u∥w_{2}(t), 0 ≤ t ≤ 1.  (2.8)

Proof. It is obvious that u(t) ≤ ∥u∥w_{2}(t) for q ≤ t ≤ 1. We need only to show that u(t) ≤ ∥u∥w_{2}(t) for 0 ≤ t ≤ q. Let c be the unique zero of u′ in (0, 1). We have q ≤ c and u(c) = ∥u∥.

If we define

h(t) = w_{2}(t)u(c) − u(t), 0 ≤ t ≤ q,

then we have h(0) = h′(0) = h′′(0) = ⋯ = h^{(n−3)}(0) = 0, and

h^{(n)}(t) = −u^{(n)}(t) ≥ 0, 0 ≤ t ≤ q.  (2.9)

We also note that

h(q) = w_{2}(q)u(c) − u(q) = u(c) − u(q) ≥ 0,
h′(q) = w′_{2}(q)u(c) − u′(q) = −u′(q) ≤ 0.

To prove the lemma, it suffices to show that h(t) ≥ 0 for 0 ≤ t ≤ q. Assume the contrary that h(t_{0}) < 0 for some t_{0} ∈ (0, q). Because h(0) = 0 > h(t_{0}) and h(t_{0}) < 0 ≤ h(q), there exist t_{1} ∈ (0, t_{0}) and s_{1} ∈ (t_{0}, q) such that h′(t_{1}) < 0 and h′(s_{1}) > 0.

Because h′(0) = 0 > h′(t_{1}), h′(t_{1}) < 0 < h′(s_{1}), and h′(s_{1}) > 0 ≥ h′(q), there exist t_{2} ∈ (0, t_{1}), s_{2} ∈ (t_{1}, s_{1}), and r_{2} ∈ (s_{1}, q) such that h′′(t_{2}) < 0, h′′(s_{2}) > 0, and h′′(r_{2}) < 0.

Because h′′(0) = 0 > h′′(t_{2}), h′′(t_{2}) < 0 < h′′(s_{2}), and h′′(s_{2}) > 0 > h′′(r_{2}), there exist t_{3} ∈ (0, t_{2}), s_{3} ∈ (t_{2}, s_{2}), and r_{3} ∈ (s_{2}, r_{2}) such that h′′′(t_{3}) < 0, h′′′(s_{3}) > 0, and h′′′(r_{3}) < 0.

If we continue this procedure, then finally we can show that there exist t_{n−2}, s_{n−2}, and r_{n−2} such that 0 < t_{n−2} < s_{n−2} < r_{n−2} < q and

h^{(n−2)}(t_{n−2}) < 0, h^{(n−2)}(s_{n−2}) > 0, and h^{(n−2)}(r_{n−2}) < 0.

EJQTDE, 2010 No. 53, p. 6 Because h^{(n−2)}(t_{n−2}) < 0 < h^{(n−2)}(s_{n−2}) and h^{(n−2)}(s_{n−2}) > 0 > h^{(n−2)}(r_{n−2}), there exist t_{n−1} ∈ (t_{n−2}, s_{n−2}) and s_{n−1} ∈ (s_{n−2}, r_{n−2}) such that

h^{(n−1)}(t_{n−1}) > 0, h^{(n−1)}(s_{n−1}) < 0.

Therefore, there exists t_{n} ∈ (t_{n−1}, s_{n−1}) such that h^{(n)}(t_{n}) < 0, which contradicts (2.9). The proof is complete. □

Both Lemmas 2.3 and 2.7 provide an upper estimate to functions satisfying (1.2), (2.2), and (2.3). The difference is that the upper estimate in Lemma 2.3 depends on the unique zero c of u′ in (0, 1), while the upper estimate in Lemma 2.7 does not.

The next lemma was proved by Webb in [4].

Lemma 2.8. For t, s ∈ [0, 1] we have

w_{1}(t)Φ_{0}(s) ≤ G(t, s) ≤ Φ_{0}(s),

where

Φ_{0}(s) = \frac{(τ(s))^{n−2}s(1 − s)^{n−1}}{(n − 1)!} , τ(s) = \frac{s}{1 − (1 − s)^{(n−1)/(n−2)}} .

Using Lemma 2.8, we can easily establish the following lower estimate.

Lemma 2.9. If u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3), then

u(t) ≥ w_{1}(t)∥u∥, 0 ≤ t ≤ 1.  (2.10)

Proof. On one hand, for 0 ≤ t ≤ 1 we have

u(t) = ∫_{0}^{1} G(t, s)(−u^{(n)}(s))ds ≤ ∫_{0}^{1} Φ_{0}(s)(−u^{(n)}(s))ds.

This means that

∥u∥ ≤ ∫_{0}^{1} Φ_{0}(s)(−u^{(n)}(s))ds.

On the other hand, for 0 ≤ t ≤ 1, we have

u(t) = ∫_{0}^{1} G(t, s)(−u^{(n)}(s))ds
≥ w_{1}(t) ∫_{0}^{1} Φ_{0}(s)(−u^{(n)}(s))ds
≥ w_{1}(t)∥u∥.

The proof is complete. □ EJQTDE, 2010 No. 53, p. 7

The estimates (2.7) and (2.10) are of the same type, that is, both are independent of c. The two estimates are similar but (2.10) is a little better than (2.7). Both estimates are listed here because they are obtained by different methods, and both methods are useful in finding estimates.

Theorem 2.10. If u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3), then

w_{2}(t)∥u∥ ≥ u(t) ≥ w_{1}(t)∥u∥, 0 ≤ t ≤ 1,  (2.11)

and

u(t) ≤ u(p)w_{2}(t)/w_{1}(p), 0 ≤ t ≤ 1.  (2.12)

In particular, if u ∈ C^{n}[0, 1] is a positive solution to the boundary value problem (1.1)–(1.2), then u(t) satisfies (2.11) and (2.12).

Proof. Suppose u ∈ C^{n}[0, 1] satisfies (1.2), (2.2), and (2.3). By Lemmas 2.7 and 2.9, the inequalities in (2.11) hold. Note that u(p) ≥ w_{1}(p)∥u∥. For 0 ≤ t ≤ 1 we have

u(t) ≤ w_{2}(t)∥u∥ = w_{2}(t)w_{1}(p)∥u∥/w_{1}(p) ≤ w_{2}(t)u(p)/w_{1}(p).

Thus we proved (2.12).

If u is a positive solution to the problem (1.1)–(1.2), then u(t) satisfies (1.2), (2.2), and (2.3). By the first half of the theorem, u also satisfies (2.11) and (2.12). The proof of the theorem is now complete. □

We have shown that, for functions u satisfying (1.2), (2.2), and (2.3), there are several upper and lower estimates of different types — the lower estimates (2.6), (2.7), (2.10), the upper estimates (2.5), (2.8), (2.12), and the “natural” upper estimate

u(t) ≤ ∥u∥, 0 ≤ t ≤ 1.

These upper and lower estimates can be used in different situations. In the next section, we will show how to use the upper estimate (2.10) and the lower estimate (2.8) to establish some explicit existence and nonexistence conditions for positive solution of the problem (1.1)-(1.2).

3. Nonexistence and Existence Results

We begin by fixing some notations. First, we define

P = {v ∈ X | v(t) ≥ 0 on [0, 1]}.

EJQTDE, 2010 No. 53, p. 8 Clearly, P is a positive cone of the Banach space X. Define the operator T : P → X and its associated linear operator L : X → X by

T u(t) = ∫_{0}^{1} G(t, s)g(s)f(u(s))ds, 0 ≤ t ≤ 1, for all u ∈ P,
Lu(t) = ∫_{0}^{1} G(t, s)g(s)u(s)ds, 0 ≤ t ≤ 1, for all u ∈ X.

It is well known that T : P → X and L : X → X are completely continuous operators. It is easy to see that T (P) ⊂ P and L(P) ⊂ P. Now the integral equation (1.3) is equivalent to the equality

T u = u, u ∈ P.

In order to solve the problem (1.1)–(1.2), we only need to find a fixed point u of T in P such that u ≠ 0. We also define the constants

A = ∫_{0}^{1} G(p, s)g(s)w_{1}(s) ds and B = ∫_{0}^{1} G(p, s)g(s)w_{2}(s) ds.

Now we give some explicit sufficient conditions for the nonexistence of positive solutions.

Theorem 3.1. If (B/w_{1}(p))f(x) < x for all x ∈ (0, +∞), then the problem (1.1)–(1.2) has no positive solutions.

Proof. Assume the contrary that u(t) is a positive solution of the problem (1.1)–(1.2). Then u ∈ P, u(t) > 0 for 0 < t < 1, and

u(p) = ∫_{0}^{1} G(p, s)g(s)f(u(s)) ds
< w_{1}(p)B^{−1} ∫_{0}^{1} G(p, s)g(s)u(s) ds
≤ w_{1}(p)B^{−1}(u(p)/w_{1}(p)) ∫_{0}^{1} G(p, s)g(s)w_{2}(s)ds
= u(p),

which is a contradiction. □

Theorem 3.2. If Af(x) > x for all x ∈ (0, +∞), then the problem (1.1)–(1.2) has no positive solutions.

The proof of Theorem 3.2 is similar to that of Theorem 3.1 and is therefore omitted. The next theorem is from [4].

EJQTDE, 2010 No. 53, p. 9 Theorem 3.3. Let r(L) be the radius of the spectrum of L. Then r(L) > 0 and L has a positive eigenfunction ϕ ∈ P \ {0} corresponding to the principal eigenvalue r(L) of L. Let μ_{1} = 1/r(L). If either

0 ≤ F_{0} < μ_{1} and μ_{1} < f_{∞} ≤ ∞

or

0 ≤ F_{∞} < μ_{1} and μ_{1} < f_{0} ≤ ∞,

then the problem (1.1)-(1.2) has at least one positive solution.

Here comes the natural question — how can we find the value of μ_{1}? In general, there is no explicit formula for finding μ_{1}, and some kind of approximation has to be made. One approach is to use a numerical method to find an approximation for μ_{1}. Another approach is to develop some theoretic upper and lower bounds for μ_{1}. Both approaches are interesting. The following upper and lower bounds for μ_{1} were given in [4].

Theorem 3.4. Let r(L) be the radius of the spectrum of L. Let μ_{1} = 1/r(L). Then

m < μ_{1} < M,

where

m := ( sup_{0≤t≤1} ∫_{0}^{1} G(t, s)g(s)ds)^{−1} ,

M := inf_{0≤a<b≤1} ( inf_{a≤t≤b} ∫_{a}^{b} G(t, s)g(s)ds)^{−1} .

With the newly found upper estimate (2.8) from Section 2, we can now improve the lower bound m for μ_{1}. First, we introduce some notations. For each n ≥ 1, we let θ_{n} = T^{n}w_{2} and σ_{n} = T^{n}w_{1}. In other words, we define

θ_{1}(t) = ∫_{0}^{1} G(t, s)g(s)w_{2}(s)ds, 0 ≤ t ≤ 1,
θ_{2}(t) = ∫_{0}^{1} G(t, s)g(s)θ_{1}(s)ds, 0 ≤ t ≤ 1,
· · · · · · ,
σ_{1}(t) = ∫_{0}^{1} G(t, s)g(s)w_{1}(s)ds, 0 ≤ t ≤ 1,
σ_{2}(t) = ∫_{0}^{1} G(t, s)g(s)σ_{1}(s)ds, 0 ≤ t ≤ 1,
· · · · · · .

EJQTDE, 2010 No. 53, p. 10 Next, for each n ≥ 1, we define the constants

m_{n} = ( sup_{(n−2)/(n−1)≤t≤1} θ_{n}(t))^{−1/n} ,

M_{n} = ( sup_{(n−2)/(n−1)≤t≤1} σ_{n}(t))^{−1/n} .

Theorem 3.5. Let r(L) be the radius of the spectrum of L and let μ_{1} = 1/r(L). For each n ≥ 1, we have m_{n} ≤ μ_{1} ≤ M_{n}.

Proof. Let ϕ ∈ P \ {0} be a positive eigenfunction corresponding to the principal eigenvalue r(L) of L. By Lemma 2.2, we have ϕ(t) > 0 for 0 < t < 1. Let n ≥ 1. We have

ϕ = μ_{1}T ϕ = μ_{1}^{2}T^{2}ϕ = ⋯ = μ_{1}^{n}T^{n}ϕ.

For each 0 ≤ t ≤ 1 we have

ϕ(t) = μ_{1}^{n}(T^{n}ϕ)(t)
≤ ∥ϕ∥μ_{1}^{n}(T^{n}w_{2})(t)
= ∥ϕ∥μ_{1}^{n}θ_{n}(t)
≤ ∥ϕ∥μ_{1}^{n} sup_{0≤t≤1} θ_{n}(t).

This implies that ∥ϕ∥ ≤ ∥ϕ∥μ_{1}^{n}∥θ_{n}∥. Thus we have

μ_{1} ≥ ∥θ_{n}∥^{−1/n}.

By Lemma 2.4, the maximum of θ_{n}(t) must occur at a point in the interval [(n − 2)/(n − 1), 1]. Therefore we have

∥θ_{n}∥ = sup_{(n−2)/(n−1)≤t≤1} θ_{n}(t).

Thus we have proved that m_{n} ≤ μ_{1}. In a similar fashion, we can show that μ_{1} ≤ M_{n}. The proof is complete. □

The next example was first considered in [4]. We now reconsider it to illustrate some of our results.

Example 3.6. Consider the (3, 1) conjugate boundary value problem

u′′′′(t) + μg(t)u(t) = 0, 0 < t < 1,  (3.1)
u^{(i)}(0) = u(1) = 0, 0 ≤ i ≤ 2,  (3.2)

EJQTDE, 2010 No. 53, p. 11 where g(t) ≡ 1, 0 ≤ t ≤ 1. Here μ > 0 is a parameter. This problem (3.1)-(3.2) is equivalent to the equality

u = μLu,

where L : X → X is defined as

Lu(t) = ∫_{0}^{1} G(t, s)g(s)u(s)ds, 0 ≤ t ≤ 1, ∀u ∈ X.

Let r(L) be the radius of the spectrum of L and let μ_{1} = 1/r(L). In other words, we let μ_{1} be the smallest eigenvalue of the boundary value problem (3.1)-(3.2). According to Webb [4], we have

227.557 ≈ m ≤ μ_{1} ≤ M ≈ 2859.530.

With the software Maple, we can easily calculate the numbers m_{i} and M_{i}, i = 1, 2, 3, 4. Our calculations indicate that

m_{1} ≈ 437.107, M_{1} ≈ 2783.13,
m_{2} ≈ 598.3, M_{2} ≈ 1658.2,
m_{3} ≈ 693.5, M_{3} ≈ 1379.2,
m_{4} ≈ 751.6, M_{4} ≈ 1260.9.

By Theorem 3.5, we have m_{i} ≤ μ_{1} ≤ M_{i}, i = 1, 2, 3, 4. This example shows that our lower bounds m_{i} (1 ≤ i ≤ 4) for μ_{1} improve m significantly, and our upper bounds M_{i} (1 ≤ i ≤ 4) improve M significantly.

If we combine Theorems 3.3 and 3.5, we get the following existence result.

Theorem 3.7. If there exist positive integers k and l such that either

0 ≤ F_{0} < m_{k} and M_{l} < f_{∞} ≤ ∞

or

0 ≤ F_{∞} < m_{k} and M_{l} < f_{0} ≤ ∞,

then the problem (1.1)-(1.2) has at least one positive solution.

Example 3.8. Consider the (3, 1) conjugate boundary value problem

u′′′′(t) + g(t)f(u(t)) = 0, 0 < t < 1,  (3.3)
u^{(i)}(0) = u(1) = 0, 0 ≤ i ≤ 2,  (3.4)

where g(t) ≡ 1, 0 ≤ t ≤ 1,

f(x) = \frac{λx(1 + 9x)}{1 + x}, x ≥ 0.

EJQTDE, 2010 No. 53, p. 12 Here λ > 0 is a parameter. It is easy to see that F_{0} = f_{0} = λ and f_{∞} = F_{∞} = 9λ. We see from Example 3.6 that

m_{4} ≈ 751.6, M_{4} ≈ 1260.9.

By Theorem 3.7, we see that if

140.1 ≈ M_{4}/9 < λ < m_{4} ≈ 751.6,

then the boundary value problem (3.3)-(3.4) has at least one positive solution.

Acknowledgment

The research of B. Yang was supported by the Kennesaw State University Tenured Faculty Professional Development Full Paid Leave Program in Spring 2010. The author is grateful to the referee for the valuable comments.

References

[1] P. W. Eloe and J. Henderson, Positive solutions for (n − 1, 1) conjugate boundary value problems, Nonlinear Anal. 28 (1997), 1669–1680.

[2] J. R. Graef and B. Yang, Boundary value problems for second order nonlinear differential equations, Commun. Appl. Anal. 6 (2002) No. 2, 273–288.

[3] M. A. Krasnoselskii, Positive Solutions of Operator Equations, Noordhoff, Groningen, 1964.

[4] J. R. L. Webb, Nonlocal conjugate type boundary value problems of higher order, Nonlinear Anal. 71 (2009), 1933-1940.

[5] B. Yang, Positive solutions for the (n, p) boundary value problem, Electron. J. Qual. Theory Differ. Equ., Spec. Ed. I (2009), no. 31, 1–13.

(Received June 24, 2010)

Bo Yang Department of Mathematics and Statistics, Kennesaw State University, Kennesaw, GA 30144, USA

  E-mail address: byang@kennesaw.edu

EJQTDE, 2010 No. 53, p. 13