Dynamic Systems and Applications 15 (2006) 97-110

POSITIVE SOLUTIONS OF A NONLINEAR THIRD ORDER EIGENVALUE PROBLEM

JOHN R. GRAEF^{1} AND BO YANG^{2}

1Department of Mathematics, University of Tennessee at Chattanooga Chattanooga, TN 37403 USA. E-mail: John-Graef@utc.edu

2Department of Mathematics, Kennesaw State University Kennesaw, GA 30144 USA. E-mail: byang@kennesaw.edu

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

ABSTRACT. The authors consider the boundary value problem consisting of the third order nonlinear ordinary differential equation

u′′′(t) = λq(t)f(u), 0 < t < 1,

together with the boundary conditions

u(0) = u′(p) = u′′(1) = 0,

where λ > 0 is a parameter, and p ∈ (\frac{1}{2}, 1) is a constant. They give existence and nonexistence results for positive solutions of this problem. Examples are included to show that the results are sharp.

AMS (MOS) Subject Classification. 34B15.

1. INTRODUCTION

We consider the boundary value problem consisting of the third order nonlinear ordinary differential equation

(E)  u′′′(t) = λq(t)f(u), 0 < t < 1,

together with the boundary conditions

(B)  u(0) = u′(p) = u′′(1) = 0.

Throughout the paper, we will assume that

(H1) f : [0, ∞) → [0, ∞) and q : [0, 1] → [0, ∞) are continuous,

Received September 10, 2004             1056-2176 $15.00 \begin{matrix}c \\ ◯ \end{matrix}Dynamic Publishers, Inc.

(H2) ∫_{0}^{1}q(t)dt > 0,

(H3) λ > 0 is a parameter, and p ∈ (\frac{1}{2}, 1) is a constant. If f(0) = 0, then the boundary value problem (E)–(B) always has the trivial solution, but here we are only interested in positive solutions, i.e., a solution u(t) such that u(t) > 0 on (0, 1).

Due to their important role in both theory and applications, boundary value problems for ordinary differential equations have generated a great deal of interest over the years, and excellent surveys of known results can be found in the monographs by Agarwal [1] and Agarwal, O’Regan, and Wong [2]. For examples of the variety of applications that such problems have in such areas as physics, biology, and chemistry, we refer the reader to the classic works of Love [14], Prescott [18], and Timoshenko [20] on elasticity, the monographs by Mansfield [17] and Soedel [19] on deformation of structures, and the work of Dulácska [7] on the effects of soil settlement. Recent contributions to the literature include the papers of Anderson and Davis [3], Cao and Ma [4], Davis et al. [5, 6], Eloe [8], Eloe and Henderson [9], Graef and Yang [10, 11], Henderson and Wang [12], Ma [15, 16], Webb [21], and Wong and Agarwal [22].

The problem (E)–(B) considered here is often referred to as a right focal problem. In a very nice paper, Anderson and Davis [3] studied this same problem and obtained intervals on λ in which the problem (E)–(B) above has at least one positive solution. In addition to using Krasnosel’skii’s Fixed Point Theorem, they also discuss the application of the Legget–Williams Theorem and the Five Functionals Fixed Point Theorem to the determination of the existence of positive solutions to boundary value problems. Here, we use Krasnosel’skii’s Fixed Point Theorem and obtain better intervals for existence of positive solutions for the problem (E)–(B) than are obtained by Anderson and Davis. This will be illustrated by examples at the end of the paper.

The organization of the paper is as follows. In the next section, we define the Green’s function for the problem (E)–(B) and give some preliminary lemmas that provide estimates for the positive solutions of the problem. Section 3 contains our existence and nonexistence results for positive solutions of (E)–(B), and the final section of the paper contains some examples and discussion.

2. GREEN’S FUNCTIONS AND SOME LEMMAS

For convenience, we begin this section by stating Krasnosel’skii’s Fixed Point Theorem [13], which is the main tool used to prove our results.

Theorem K. Let B be a Banach space over the reals, and let P ⊂ B be a cone in B. Assume Ω_{1} and Ω_{2} are bounded open subsets of B with 0 ∈ Ω_{1} ⊂ \overline{Ω_{1}} ⊂ Ω_{2}, and let

L : P ∩ (\overline{Ω_{2}} − Ω_{1}) → P

THIRD ORDER EIGENVALUE PROBLEM 99

be a completely continuous operator such that either

(i) ∥Lu∥ ≤ ∥u∥ if u ∈ P ∩ ∂Ω_{1} and ∥Lu∥ ≥ ∥u∥ if u ∈ P ∩ ∂Ω_{2}, or (ii) ∥Lu∥ ≥ ∥u∥ if u ∈ P ∩ ∂Ω_{1} and ∥Lu∥ ≤ ∥u∥ if u ∈ P ∩ ∂Ω_{2}. Then L has a fixed point in P ∩ (\overline{Ω_{2}} − Ω_{1}).

The Green’s function G : [0, 1]×[0, 1] → [0, ∞) for the problem (E)–(B) is defined by (see [3])

G(t, s) = \begin{cases}\frac{1}{2}t(2s − t), & t ≤ s ≤ p, \\ \frac{1}{2}s^{2}, & s ≤ t and s ≤ p, \\ \frac{1}{2}t(2p − t), & t ≤ s and s ≥ p, \\ \frac{1}{2}t(2p − t) + \frac{1}{2}(t − s)^{2}, & t ≥ s ≥ p. \end{cases}

Clearly, we see that

(2.1)  G(t, s) > 0 for t, s ∈ (0, 1).

Notice that solving the problem (E)–(B) is equivalent to finding a solution of the integral equation

(2.2)  u(t) = λ ∫_{0}^{1} G(t, s)q(s)f(u(s))ds, 0 ≤ t ≤ 1.

We define the function a : [0, 1] → [0, ∞) by

a(t) = \frac{t(2p − t)}{p^{2}}

and we define the constants

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}.

The following two lemmas will be used to prove our main results.

Lemma 2.1. If u(t) is a positive solution of the problem (E)–(B), then

u(p) = ∥u∥,

where ∥u∥ = max_{t∈[0,1]} u(t).

Proof. Let u(t) be a positive solution of the problem (E)–(B). Since u′′(1) = 0 and u′′′(t) ≥ 0 for t ∈ [0, 1], we have

u′′(t) ≤ 0 for t ∈ [0, 1].

100 JOHN R. GRAEF AND BO YANG

Because u′(p) = 0, we see that

u′(t) ≥ 0 on [0, p] and u′(t) ≤ 0 on [p, 1].

Therefore, u(t) attains its maximum at p, which completes the proof of the lemma.

Lemma 2.2. If u(t) is a positive solution of the problem (E)–(B), then

(2.3)  u(t) ≥ ∥u∥a(t), t ∈ [0, 1].

Proof. Let u(t) be a positive solution of the problem (E)–(B). Without loss of generality, we may assume that ∥u∥ = u(p) = 1. Define

y(t) = u(t) − \frac{2t}{p} + \frac{t^{2}}{p^{2}} ;

then

y′(t) = u′(t) − \frac{2}{p} + \frac{2t}{p^{2}},

y′′(t) = u′′(t) + \frac{2}{p^{2}} ,

and

y′′′(t) = u′′′(t) ≥ 0.

Clearly, we have

y(0) = 0, y(p) = 0, y′(p) = 0.

By the Mean Value Theorem, there is a number r_{1} ∈ (0, p) such that y′(r_{1}) = 0. Now y′(t) is concave upward, so

y′(t) ≥ 0 on [0, r_{1}] and y′(t) ≤ 0 on [r_{1}, p].

Therefore, we have

y(t) ≥ 0 on [0, p].

Again by the Mean Value Theorem, there exists r_{2} ∈ (r_{1}, p) such that y′′(r_{2}) = 0. Since y′′(t) is increasing, we have y′′(t) ≥ 0 on [p, 1]. Therefore,

y′(t) ≥ 0 on [p, 1] and y(t) ≥ 0 on [p, 1].

This completes the proof of the lemma.

Remark 2.3. Our estimation (2.3) for the positive solutions of the boundary value problem (E)–(B) is better than the estimate (2.6) obtained in [3]. As a consequence, we are able to obtain better results here than those in [3]. We will return to this point later in the paper.

Our final lemma gives additional information about the size of solutions of the problem (E)–(B).

Lemma 2.4. If u(t) is a positive solution of the problem (E)–(B), then

u(t) ≥ ∥u∥ min{t, 1 − t}, t ∈ [0, 1].

Proof. The proof of this lemma follows easily from Lemma 2.2 and the fact that

a(t) ≥ min{t, 1 − t}, t ∈ [0, 1].

3. EXISTENCE AND NONEXISTENCE OF POSITIVE SOLUTIONS

We begin by introducing some notation. Define

A = ∫_{0}^{1} G(p, s)q(s)a(s) ds

and

B = ∫_{0}^{1} G(p, s)q(s) ds,

let B = C[0, 1] with the norm

∥x∥ = max_{t∈[0,1]} |x(t)|, x ∈ B,

and let

P = {x ∈ B | x(t) ≥ 0, a(t)x(p) ≤ x(t) ≤ x(p), and x is concave downward on [0, 1]}.

We see that B is a Banach space and P is a positive cone in B. Define the operator T : P → B by

T u(t) = λ ∫_{0}^{1} G(t, s)q(s)f(u(s))ds, 0 ≤ t ≤ 1.

It is well known that T : P → B is a completely continuous operator, and by the same argument as is used in the proof of Lemma 2.2, we can prove that T (P) ⊂ P.

Now, solving the integral equation (2.2) is equivalent to finding a solution of

T u = u, u ∈ P.

That is, in order to solve the problem (E)–(B) we only need to find a fixed point of the operator T in P.

We now give our first existence result.

Theorem 3.1. If

(Af_{∞})^{−1} < λ < (BF_{0})^{−1},

then the boundary value problem (E)–(B) has at least one positive solution.

102 JOHN R. GRAEF AND BO YANG

Proof. Choose ϵ > 0 so that (F_{0} + ϵ)λB ≤ 1. Then, there exists H_{1} > 0 such that

f(x) ≤ (F_{0} + ϵ)x for 0 < x ≤ H_{1}.

So if u ∈ P with ∥u∥ = H_{1}, we have

(T u)(p) = λ ∫_{0}^{1} G(p, s)q(s)f(u(s)) ds ≤ λ(F_{0} + ϵ) ∫_{0}^{1} G(p, s)q(s)u(s) ds
≤ λ(F_{0} + ϵ)∥u∥ ∫_{0}^{1} G(p, s)q(s)ds ≤ λ(F_{0} + ϵ)∥u∥B ≤ ∥u∥,

which means ∥T u∥ ≤ ∥u∥. So, if we let

Ω_{1} = {u ∈ B| ∥u∥ < H_{1}},

then

∥T u∥ ≤ ∥u∥, for u ∈ P ∩ ∂Ω_{1}.

Now choose c ∈ (0, 1/4) and ϵ > 0 such that

λ > ((f_{∞} − ϵ) ∫_{c}^{1−c} G(p, s)q(s)a(s) ds)^{−1} .

There exists H_{3} > 0 such that

f(x) ≥ (f_{∞} − ϵ)x for x ≥ H_{3}.

Let H_{2} = max{H_{3}/c, 2H_{1}} and let u ∈ P with ∥u∥ = H_{2}. Then by Lemma 2.4, for c ≤ t ≤ 1 − c, we have

u(t) ≥ min{t, 1 − t}∥u∥ ≥ cH_{2} ≥ H_{3}.

Thus, if u ∈ P with ∥u∥ = H_{2}, then

(T u)(p) ≥ λ ∫_{c}^{1−c} G(p, s)q(s)f(u(s))ds ≥ λ(f_{∞} − ϵ) ∫_{c}^{1−c} G(p, s)q(s)u(s)ds
≥ λ(f_{∞} − ϵ)∥u∥ ∫_{c}^{1−c} G(p, s)q(s)a(s) ds ≥ ∥u∥,

which means ∥T u∥ ≥ ∥u∥. Letting

Ω_{2} = {u ∈ B| ∥u∥ < H_{2}},

we have

∥T u∥ ≥ ∥u∥, for u ∈ P ∩ ∂Ω_{2}.

Hence, condition (i) of Theorem K is satisfied, so there exists a fixed point of T in P. This completes the proof of the theorem.

We also have the following existence result.

Theorem 3.2. If

(Af_{0})^{−1} < λ < (BF_{∞})^{−1},

then the boundary value problem (E)–(B) has at least one positive solution.

THIRD ORDER EIGENVALUE PROBLEM 103

Proof. Choosing ϵ > 0 such that (f_{0} − ϵ)λA ≥ 1, we see that there exists H_{1} > 0 such that

f(x) ≥ (f_{0} − ϵ)x for 0 < x ≤ H_{1}.

For u ∈ P with ∥u∥ = H_{1}, we then have

(T u)(p) = λ ∫_{0}^{1} G(p, s)q(s)f(u(s)) ds ≥ λ(f_{0} − ϵ) ∫_{0}^{1} G(p, s)q(s)u(s) ds
≥ λ(f_{0} − ϵ)∥u∥ ∫_{0}^{1} G(p, s)q(s)a(s) ds ≥ λA(f_{0} − ϵ)∥u∥ ≥ ∥u∥,

which implies ∥T u∥ ≥ ∥u∥. Now setting

Ω_{1} = {u ∈ B| ∥u∥ < H_{1}},

we have

∥T u∥ ≥ ∥u∥, for u ∈ P ∩ ∂Ω_{1}.

Next, we choose ϵ ∈ (0, 1) such that ((F_{∞} + ϵ)B + ϵ)λ ≤ 1 and let H_{3} > 0 be such that

f(x) ≤ (F_{∞} + ϵ)x for x ≥ H_{3}.

Set

M = max_{0≤x≤H_{3}} f(x)

and choose r ∈ (0, 1/4) such that

∫_{0}^{r} G(p, s)q(s) ds ≤ \frac{ϵ}{4M} and ∫_{1−r}^{1} G(p, s)q(s) ds ≤ \frac{ϵ}{4M} .

Let H_{2} = max{2H_{1}, H_{3}/r, 1} and let u ∈ P with ∥u∥ = H_{2}. Then u(t) ≥ min{t, 1 − t}H_{2}, so there exist β_{1} ∈ [0, r) and β_{2} ∈ (1 − r, 1] such that

0 ≤ u(t) ≤ H_{3} for t ∈ (0, β_{1}),
u(t) ≥ H_{3} for t ∈ (β_{1}, β_{2}),

and

0 ≤ u(t) ≤ H_{3} for t ∈ (β_{2}, 1).

(Note that it may happen that β_{2} = 1 and therefore (β_{2}, 1) = ∅.) Then, for u ∈ P with ∥u∥ = H_{2}, we have

(T u)(p) = λ ∫_{0}^{1} G(p, s)q(s)f(u(s))ds ≤ λ (ϵ + ∫_{β_{1}}^{β_{2}} G(p, s)q(s)f(u(s)) ds)
≤ λ (ϵ + (F_{∞} + ϵ) ∫_{β_{1}}^{β_{2}} G(p, s)q(s)u(s) ds)
≤ λ (ϵ + (F_{∞} + ϵ)H_{2} ∫_{0}^{1} G(p, s)q(s)ds)
≤ λ(ϵ + B(F_{∞} + ϵ)H_{2}) ≤ H_{2} = ∥u∥,

104 JOHN R. GRAEF AND BO YANG

which means ∥T u∥ ≤ ∥u∥. If we let

Ω_{2} = {u ∈ B| ∥u∥ < H_{2}},

then we obtain

∥T u∥ ≤ ∥u∥, for u ∈ P ∩ ∂Ω_{2}.

By Theorem K(ii), the problem (E)–(B) has at least one positive solution.

We conclude this section with some nonexistence results, that is, results which give intervals on λ for which there are no positive solutions. Our first such theorem is the following.

Theorem 3.3. If conditions (H1), (H2), and (H3) hold, and (H4) λBf(x) < x for all x ∈ (0, +∞), then the boundary value problem (E)–(B) has no positive solutions.

Proof. Assume to the contrary that x(t) is a positive solution of the problem (E)–(B). We then have

x(p) = λ ∫_{0}^{1} G(p, s)q(s)f(x(s)) ds < B^{−1} ∫_{0}^{1} G(p, s)q(s)x(s) ds
≤ B^{−1}x(p) ∫_{0}^{1} G(p, s)q(s)ds ≤ x(p),

which is a contradiction.

We also have a companion result to Theorem 3.3. The proof is quite similar to the above proof and will be omitted.

Theorem 3.4. If conditions (H1), (H2), and (H3) hold and (H5) λAf(x) > x for all x ∈ (0, +∞), then the boundary value problem (E)–(B) has no positive solutions.

The results in this section will be illustrated by examples in the next section of the paper.

4. EXAMPLES AND DISCUSSION

In the literature, the case where h_{0} = F_{0} = f_{0} = 0 and h_{∞} = F_{∞} = f_{∞} = ∞ is referred to as the superlinear case; similarly, the case F_{0} = f_{0} = ∞ and F_{∞} = f_{∞} = 0 is referred to as the sublinear case. Many authors have obtained results for just these cases. We allow the “intermediate cases” 0 < h_{0}, h_{∞} < ∞ here as well as for combinations with the above values. In addition, we do not need that lim_{x→0^{+}} f(x)/x or lim_{x→+∞} f(x)/x exist. We also do not require that q(t) not vanish identically on any subinterval of [0, 1] as is often the case.

THIRD ORDER EIGENVALUE PROBLEM 105

If f is superlinear in the sense described above, then Theorem 3.1 applies, while if f is sublinear, then Theorem 3.2 should be used. In each of these special cases we get the existence of a positive solution of the problem (E)–(B) for every value of λ with 0 < λ < ∞.

Our first example illustrates the results obtained in Section 3 and compares our results to those in [3].

Example 4.1. Consider the boundary value problem

(e_{1})  u′′′(t) = λq(t)f(u(t)), 0 < t < 1,
(b_{1})  u(0) = u′(3/4) = u′′(1) = 0,

where

q(t) = 1 + t and f(u) = u\frac{1 + 2u}{1 + u}.

Clearly, F_{0} = f_{0} = 1 and F_{∞} = f_{∞} = 2, and it is easy to see that

G(3/4, s) = \begin{cases}\frac{1}{2}s^{2}, & s ≤ \frac{3}{4}, \\ \frac{9}{32}, & s ≥ \frac{3}{4}, \end{cases}

and a calculation using Maple yields

A = \frac{6973}{30720} and B = \frac{495}{2048}.

From Theorem 3.1 we see that if

2.2028 ≈ \frac{30720}{13946} < λ < \frac{2048}{495} ≈ 4.1373,

then the problem (e_{1})–(b_{1}) has at least one positive solution. From Theorems 3.3 and 3.4, we see that if either

λ < \frac{2048}{990} ≈ 2.0686 or λ > \frac{30720}{6973} ≈ 4.4056,

then (e_{1})–(b_{1}) has no positive solutions. This shows that our results are quite sharp.

The following results were proved by Anderson and Davis [3; Theorems 2.2 and 2.3].

Theorem AD. If h ∈ (0, 1 − p) and either

(f_{∞}K(h))^{−1} < λ < (f_{0}B)^{−1}

or

(f_{0}K(h))^{−1} < λ < (f_{∞}B)^{−1},

where

K(h) = \frac{1 − p − h}{1 − p} ∫_{p−h}^{p+h} G(p, s)q(s)ds,

then the boundary value problem (E)–(B) has at least one positive solution.

An easy computation using Maple shows that for the problem (e_{1})–(b_{1}),

K(h) = \frac{63}{64}h − \frac{147}{32}h^{2} + \frac{19}{6}h^{3} − \frac{55}{24}h^{4} + \frac{1}{2}h^{5}, 0 ≤ h ≤ \frac{1}{4},

and

max_{0≤h≤ \frac{1}{4}} K(h) ≈ 0.056985015 · · · .

So from the first part of Theorem AD, we have that the problem (e_{1})–(b_{1}) has at least one positive solution provided

8.77424 ≈ \frac{1}{2 max_{0≤h≤ \frac{1}{4}} K(h)} < λ < \frac{2048}{495} ≈ 4.13737 ⋯,

which is impossible. It is easy to see that the second part of Theorem AD does not apply to the problem (e_{1})–(b_{1}) as well.

Example 4.2. Consider the boundary value problem

(e_{2})  u′′′(t) = λq(t)f(u(t)), 0 < t < 1,
(b_{2})  u(0) = u′(3/4) = u′′(1) = 0,

where

q(t) = 1 + t and f(u) = \frac{u(1 + 3u)(3 + sin u)}{1 + u}.

We see that F_{0} = f_{0} = 3, F_{∞} = 12, f_{∞} = 6,

A = \frac{6973}{30720} and B = \frac{495}{2048}.

It is not difficult to see that

3u ≤ f(u) ≤ 12u for u ≥ 0.

Theorem 3.1 ensures that the problem (e_{2})–(b_{2}) has at least one positive solution provided

0.73427 ≈ \frac{5120}{6973} < λ < \frac{2048}{1485} ≈ 1.37912.

By Theorems 3.3 and 3.4, we see that if

λ < \frac{512}{1485} ≈ 0.34478 or λ > \frac{10240}{6973} ≈ 1.46853,

then the problem (e_{2})–(b_{2}) has no positive solutions. The results in [3] do not apply since F_{∞} ≠ f_{∞}.

In our next and final example, we show that even though our results do not apply, our technique of proof can still give information about the existence of positive solutions. The reason that our results in Theorems 3.1 and 3.2 above, as well as the results of other authors, do not apply to the following example is that f(x)/x is unbounded as both x → 0 and x → ∞.

Example 4.3. Consider the boundary value problem

(e_{3})  u′′′(t) = λq(t)f(u(t)), 0 < t < 1,
(b_{3})  u(0) = u′(3/4) = u′′(1) = 0,

where

q(t) = 1 + t, and f(u) = 1 + u^{2}.

For this problem, f_{0} = F_{0} = +∞, f_{∞} = F_{∞} = +∞, A = 6973/30720, and B = 495/2048, so Theorems 3.1, 3.2, and 3.3 do not apply here. Now f(u) = (1 + u^{2}) = [(1 − u)^{2} + 2u] ≥ 2u, so by Theorem 3.4, we see that if

λ > \frac{15360}{6973} ≈ 2.20279,

then the problem (e_{3})–(b_{3}) has no positive solutions.

For x ≥ 0, define

α(x) = ∫_{0}^{1} G(3/4, s)q(s)f(a(s)x)ds = ∫_{0}^{1} G(3/4, s)q(s)(1 + (a(s)x)^{2})ds

= \frac{495}{2048} + \frac{166585}{774144}x^{2}

and

β(x) = ∫_{0}^{1} G(3/4, s)q(s)f(x)ds = ∫_{0}^{1} G(3/4, s)q(s)(1 + x^{2})ds

= \frac{495}{2048} (1 + x^{2}) .

It is easy to show that

α(x) ≥ k_{α}x and β(x) ≥ k_{β}x for x ≥ 0,

where

k_{α} = \frac{5}{1024}\sqrt{\frac{366487}{42}} and k_{β} = \frac{495}{1024}.

Moreover, it can be seen that if

λ < k_{β}^{−1},

then

λβ(t_{0}) < t_{0},

where

t_{0} = 9\sqrt{\frac{642}{33317}}.

108 JOHN R. GRAEF AND BO YANG

Now if u ∈ P (the positive cone defined at the start of Section 3), then we have

∥T u∥ = T u(3/4) = λ ∫_{0}^{1} G(3/4, s)q(s)(1 + (u(x))^{2}) ds
≥ λ ∫_{0}^{1} G(3/4, s)q(s)(1 + (a(s)u(3/4))^{2}) ds
= λα(u(3/4)),

and

∥T u∥ = T u(3/4) = λ ∫_{0}^{1} G(3/4, s)q(s)(1 + (u(x))^{2}) ds
≤ λ ∫_{0}^{1} G(3/4, s)q(s)(1 + (u(3/4))^{2}) ds
= λβ(u(3/4)).

If u ∈ P is a positive solution of the problem (e_{3})–(b_{3}), then we have u = T u, and so

u(3/4) = T u(3/4) ≥ λα(u(3/4)) ≥ λk_{α}u(3/4),

which implies that

1 ≥ λk_{α}.

Therefore, if

λ > k_{α}^{−1} ≈ 2.1924,

then the problem (e_{3})–(b_{3}) has no positive solutions.

Claim. If

λ < k_{β}^{−1} ≈ 2.068686,

then the problem (e_{3})–(b_{3}) has at least two positive solutions.

Proof of the Claim. If λ < k_{β}^{−1}, then λβ(t_{0}) < t_{0}. By continuity, we can choose H_{2} ∈ (0, t_{0}) and H_{3} > t_{0} such that

λβ(H_{2}) < H_{2} and λβ(H_{3}) < H_{3}.

We can also choose H_{1} ∈ (0, H_{2}) and H_{4} > H_{3} such that

λα(H_{1}) > H_{1} and λα(H_{4}) > H_{4}.

Therefore, we have u ∈ P with ∥u∥ = H_{1} implies ∥T u∥ ≥ ∥u∥

u ∈ P with ∥u∥ = H_{2} implies ∥T u∥ ≤ ∥u∥,
u ∈ P with ∥u∥ = H_{3} implies ∥T u∥ ≤ ∥u∥,

and

u ∈ P with ∥u∥ = H_{4} implies ∥T u∥ ≥ ∥u∥.

Now by Krasnosel’skii’s Fixed Point Theorem, we see that the problem (e_{3})–(b_{3}) has at least two positive solutions u_{1} and u_{2}, such that H_{1} ≤ ∥u_{1}∥ ≤ H_{2} and H_{3} ≤ ∥u_{2}∥ ≤ H_{4}.

ACKNOWLEDGMENTS

The research by J. R. Graef was supported in part by the University of Tennessee at Chattanooga Center of Excellence for Computer Applications.

REFERENCES

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