or positive solutions to a semipositone elastic beam equation

Nonlinear Analysis 66 (2007) 138–150 www.elsevier.com/locate/na

Existence of n solutions and/or positive solutions to a semipositone elastic beam equation Qingliu Yao Department of Applied Mathematics, Nanjing University of Finance and Economics, Nanjing 210003, China Received 25 September 2005; accepted 10 November 2005

Abstract The existence of n solutions and/or positive solutions is established for a nonlinear fourth-order twopoint boundary value problem where n is an arbitrary natural number and the nonlinear term has a nonpositive lower bound. In mechanics, the problem describes deformation of the elastic beam whose two ends are simply supported. The main results show that the problem has at least n solutions and/or positive solutions provided that the “heights” of the nonlinear term are appropriate on some bounded sets. c 2005 Elsevier Ltd. All rights reserved.  Keywords: Nonlinear differential equation; Boundary value problem; Solution and positive solution; Existence; Multiplicity

1. Introduction In this paper we consider the solutions and positive solutions of the following nonlinear fourth-order two-point boundary value problem with second derivative:  (4) u (t) = f (t, u(t), u  (t)), 0 ≤ t ≤ 1, (P) u(0) = u(1) = u  (0) = u  (1) = 0. Here, the solution u ∗ of (P) is called positive if u ∗ (t) > 0, 0 < t < 1. In mechanics, the problem (P) describes deformation of an elastic beam whose two ends are simply supported. Throughout this paper, we use the notation u = max0≤t ≤1 |u(t)|, u ∈ C[0, 1] and assume that     5 1 f : [0, 1] × − M, +∞ × −∞, M → [−M, +∞) 384 8 E-mail address: [email protected]. c 2005 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter  doi:10.1016/j.na.2005.11.016

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is continuous, where M ≥ 0 is a constant. This assumption implies that the problem (P) is semipositone. If M = 0, then f : [0, 1] × [0, +∞) × (−∞, 0] → [0, +∞). The existence and multiplicity of positive solutions for (P) with M = 0 have been widely studied by many authors (see [1–14]). Recently, Liu [10] proved the following theorems. Theorem 1.1. Let M = 0. If there exist two positive constants r1 = r2 such that (1) max{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r1 ] × [−r1 , 0]} ≤ 6r1 , (2) min{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r2 ] × [−r2 , − 14 r2 ]} ≥ 32 3 r2 , then problem (P) has at least one positive solution u ∗ satisfying min{r1 , r2 } ≤ (u ∗ )  ≤ max{r1 , r2 }. Theorem 1.2. Let M = 0. If min f0 = min f ∞ = +∞ and there exists a constant r > 0 such that max{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r ] × [−r, 0]} ≤ 6r, then problem (P) has at least two positive solutions u ∗1 , u ∗2 satisfying 0 < (u ∗1 )  < r < (u ∗2 )  < +∞. Theorem 1.3. Let M = 0. If max f 0 = max f ∞ = 0 and there exists a constant r > 0 such that    32 1 ≥ r, min f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r ] × −r, − r 4 3 then problem (P) has at least two positive solutions u ∗1 , u ∗2 satisfying 0 < (u ∗1 )  < r < (u ∗2 )  < +∞. The symbols used in Theorems 1.2 and 1.3 are as follows: f (t, u, v) , −v f (t, u, v) max f ∞ = lim , max sup −v→+∞ 0≤t ≤1 0≤u<+∞ −v f (t, u, v) min f 0 = lim min inf , −v→0+ 0≤t ≤1 0≤u<+∞ −v f (t, u, v) . min f ∞ = lim min inf −v→+∞ 0≤t ≤1 0≤u<+∞ −v max f 0 =

lim

max

sup

−v→0+ 0≤t ≤1 0≤u<+∞

If M > 0, to the best of our knowledge, there are no results concerning with the existence and multiplicity of positive solutions for (P). The purpose of this paper is to establish the existence of n solutions and/or positive solutions for the problem (P) with M ≥ 0, where n is an arbitrary natural number. We will extend and improve the method used in [5,9–18]. In Section 2 we will introduce two control functions and give a geometric interpretation. In Section 3 we will transform the problem (P) to a fixed point equation by applying the corresponding Green function. After that, we will establish a

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basic existence criterion, that is Theorem 3.1, by constructing a suitable cone and using the Krasnosel’skii fixed point theorem of cone expansion–compression type. In Section 4 we will prove Theorems 4.3 and 4.4 which ensures the existence of n solutions and/or positive solutions. In Section 5 we will show that Theorem 1.1 is a corollary of Theorem 3.1, and Theorems 1.2 and 1.3 are corollaries of Theorem 4.1. Moreover, we will illustrate that our conclusions cannot be derived from Theorems 1.1–1.3, even if M = 0, by two examples. In this paper, we assume that the nonlinear term f is continuous and bounded below. This implies that f is not necessarily nonnegative, monotone, superlinear or sublinear. And we do not require the existence of lower and upper solutions. Our results show that the problem (P) has at least n solutions and/or positive solutions provided that the “heights” of f are appropriate on some bounded sets of its domain. Moreover, the existence and multiplicity are independent of the growth of f outside these bounded sets. 2. Preliminaries 5 u  }. Write Let C 2 [0, 1] be the Banach space with norm |u| = max{u, 48 2 [0, 1] = {u ∈ C 2 [0, 1] : min u(t) ≥ 0, u(0) = u(1) = u  (0) = u  (1) = 0}. C+ 0≤t ≤1

The function u ∈ C[0, 1] is said to be concave (convex) if u(λt1 + (1 − λ)t2 ) ≥ λu(t1 ) + (1 − λ)u(t2 ),

λ, t1 , t2 ∈ [0, 1].

u(λt1 + (1 − λ)t2 ) ≤ λu(t1 ) + (1 − λ)u(t2 ),

λ, t1 , t2 ∈ [0, 1].

Let q(t) = min{t, 1 − t}. Obviously, if u is a nonnegative concave function, then u(t) ≥ uq(t), 0 ≤ t ≤ 1. 1 4 1 3 1 t − 12 t + 24 t and u 0 (t) = Mq (t). Since Write p(t) = 24 √  √   1 1+ 3 1− 3 1 1 2  t− t (1 − t)(1 + t − t ), p (t) = t− t− , p(t) = 24 6 2 2 2 we have p(t) ≥ 0, 0 ≤ t ≤ 1, and  p = max0≤t ≤1 p(t) = p( 12 ) = 1 p (t) = − t (1 − t), 2

p(4) (t) = 1,

5 384 .

Since

p(0) = p(1) = p (0) = p (1) = 0,

we have p (t) ≤ 0, 0 ≤ t ≤ 1, and  p  = max0≤t ≤1(− p (t)) = 18 . From this, we see that p is concave and p is convex. Lemma 2.1.

5 384 q(t)

≤ p(t) ≤

1 24 q(t), 0

≤ t ≤ 1.

Proof. Since p is a nonnegative concave function, we have p(t) ≥  pq(t) =

5 q(t), 384

On the other hand, since p (0) = p(t) ≤ Thus, p(t) ≤

1 t, 24

0≤t ≤

1 24 q(t), 0

0 ≤ t ≤ 1.

1 24 ,

1 p (1) = − 24 , we have

1 ; 2

≤ t ≤ 1.

p(t) ≤ 

1 (1 − t), 24

1 ≤ t ≤ 1. 2

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141

Our approach is based on the following Krasnosel’skii fixed point theorem. Lemma 2.2. Let X be a Banach space, and K ⊂ X be a cone in X. Assume Ω1 , Ω2 are bounded open subsets of K with 0 ∈ Ω1 ⊂ Ω 1 ⊂ Ω2 , and let F : K → K be a completely continuous operator such that either (1) Fu ≤ u, u ∈ ∂Ω1 , and Fu ≥ u, u ∈ ∂Ω2 , or (2) Fu ≥ u, u ∈ ∂Ω1 , and Fu ≤ u, u ∈ ∂Ω2 . Then F has a fixed point in Ω 2 \Ω1 . To apply the Krasnosel’skii theorem to study the existence of solutions, we need to construct a suitable cone. It is easy to check that the following set K is a cone in C 2 [0, 1]: 2 K = {u ∈ C+ [0, 1] : u(t) ≥ q(t)u, −u  (t) ≥ q(t)u  , 0 ≤ t ≤ 1}.

If u ∈ K , then u(t) ≥ 0 and u  (t) ≤ 0, 0 ≤ t ≤ 1. Let G(t, s) be the Green function of the second-order boundary value problem −u  (t) = 0, 0 ≤ t ≤ 1; u(0) = u(1) = 0; that is,  t (1 − s), 0 ≤ t ≤ s ≤ 1, G(t, s) = s(1 − t), 0 ≤ s ≤ t ≤ 1. After simple calculations, we get

1 1 G(t, s)ds = , max 0≤t ≤1 0 8

max

3/4

0≤t ≤1 1/4

G(t, s)ds =

3 . 32

Lemma 2.3. If u ∈ K , then 2 |u| ≤ u ≤ |u|, 5

8|u| ≤ u   ≤

48 |u|. 5

Proof. Since u(0) = u(1) = 0, we have

1 u(t) = G(t, s)[−u  (s)]ds, 0 ≤ t ≤ 1. 0

It follows that



1 G(t, s)[−u  (s)]ds ≥ max G(t, s)q(s)u  ds 0≤t ≤1 0 0≤t ≤1 0 

1  1  1  , s q(s)ds = u , ≥ u  G 2 24 0

1

1 1   G(t, s)|u (s)|ds ≤ u  max G(t, s)ds = u  . u ≤ max 0≤t ≤1 0 0≤t ≤1 0 8 u = max

1

1 Thus, u ≥ 24 u   and u   ≥ 8u. 5 u  }, we have |u| = u or |u| = Since |u| = max{u, 48 5  48 u , then

|u| ≥ u ≥

2 1  u  = |u|. 24 5

5  48 u .

If |u| =

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If |u| = u, then 48 |u| ≥ u   ≥ 8u = 8|u|. 5 Therefore, 25 |u| ≤ u ≤ |u| and 8|u| ≤ u   ≤

48 5 |u|.



In this paper, the following constants and control functions are applied:  −1

1 1 384 , A = max G(t, s)G(s, τ )dτ ds = 0≤t ≤1 0 5 0  −1

3/4 48 512 B= max , G(t, s)ds = 5 0≤t ≤1 1/4 5 ⎧ ⎫ 5 ⎪ ⎪ ⎪ ⎪ 0 ≤ t ≤ 1, − M ≤ u ≤ r, ⎨ ⎬ 384 , ϕ(r ) = max f (t, u, v) + M : ⎪ ⎪ 1 48 ⎪ ⎪ ⎩ ⎭ − r ≤v≤ M 5 8 ⎧ ⎫ 1 3 1 5 ⎪ ⎪ ⎪ ≤t ≤ , r− M ≤ u ≤ r,⎪ ⎨ ⎬ 4 4 10 384 . ψ(r ) = min f (t, u, v) + M : ⎪ ⎪ 1 48 ⎪ ⎪ ⎩ ⎭ − r ≤ v ≤ M − 2r 5 8 Obviously, 0 < A < B. In geometry, ϕ(r ) expresses the “maximum height” of f + M on the bounded set [0, 1] × 5 1 [− 384 M, r ] × [− 48 5 r, 8 M], ψ(r ) expresses the “minimum height” of f + M on the bounded set 1 5 1 r − 384 M, r ] × [− 48 [ 14 , 34 ] × [ 10 5 r, 8 M − 2r ]. The computations of ϕ(r ) and ψ(r ) are easy. If M = 0, then    48 ϕ(r ) = max f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r ] × − r, 0 , 5        48 1 3 1 , r, r × − r, −2r . ψ(r ) = min f (t, u, v) : (t, u, v) ∈ × 4 4 10 5 3. Existence Theorem 3.1 shows that the problem (P) has at least one solution or positive solution provided that the “heights” of f on some bounded sets are appropriate. Theorem 3.1. Assume that there exist two positive numbers r1 , r2 such that ϕ(r1 ) ≤ r1 A, ψ(r2 ) ≥ r2 B. Then problem (P) has at least one solution u ∗ satisfying u ∗ + u 0 ∈ K and min{r1 , r2 } ≤ |u ∗ + u 0 | ≤ max{r1 , r2 }. Besides that, if min{r1 , r2 } >

5 48 M

(particularly, M = 0), then u ∗ is a positive solution.

Proof. Since u 0 (t) = Mq(t), 0 ≤ t ≤ 1, we have 0 ≤ u 0 (t) ≤

5 M, 384

1 − M ≤ u 0 (t) ≤ 0, 8

0 ≤ t ≤ 1.

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Consider the fourth-order two-point boundary value problem  (4) u (t) = f (t, u(t) − u 0 (t), u  (t) − u 0 (t)) + M, 0 ≤ t ≤ 1, u(0) = u(1) = u  (0) = u  (1) = 0.

(P1 )

The problem (P1 ) is equivalent to the integral equation

1 1 G(t, s)G(s, τ )[ f (τ, u(τ ) − u 0 (τ ), u  (τ ) − u 0 (τ )) + M]dτ ds. u(t) = 0

0

2 [0, 1] and 0 ≤ t ≤ 1, Define the operator T as follows: for u ∈ C+

1 1 (T u)(t) = G(t, s)G(s, τ )[ f (τ, u(τ ) − u 0 (τ ), u  (τ ) − u 0 (τ )) + M]dτ ds. 0

0

Computing the second derivative of (T u)(t), we get

1  G(t, s)[ f (s, u(s) − u 0 (s), u  (s) − u 0 (s)) + M]ds. (T u) (t) = − 0

Noticing that −

5 M ≤ u(t) − u 0 (t) < +∞, 384

−∞ < u  (t) − u 0 (t) ≤

1 M, 8

0 ≤ t ≤ 1,

2 [0, 1] → C 2 [0, 1] is defined and continuous. we assert that T : C+ + It is easy see that G(0, s) = G(1, s) = 0, 0 ≤ s ≤ 1, and

max G(t, s) = G(s, s),

G(t, s) ≥ q(t)G(s, s),

0≤t ≤1

0 ≤ t, s ≤ 1.

Let u ∈ K . Then (T u)(0) = (T u)(1) = (T u) (0) = (T u) (1) = 0, and, for 0 ≤ t ≤ 1,



1 1

G(s, s)G(s, τ )[ f (τ, u(τ ) − u 0 (τ ), u  (τ ) − u 0 (τ )) + M]dτ ds 0 0

1 1 ≥ q(t) max G(t, s)G(s, τ )[ f (τ, u(τ ) − u 0 (τ ),

(T u)(t) ≥ q(t)

0≤t ≤1 0

0

u  (τ ) − u 0 (τ )) + M]dτ ds = q(t)T u;

1  −(T u) (t) ≥ q(t) G(s, s)[ f (s, u(s) − u 0 (s), u  (s) − u 0 (s)) + M]ds 0

1 G(t, s)[ f (s, u(s) − u 0 (s), u  (s) − u 0 (s)) + M]ds ≥ q(t) max 0≤t ≤1 0

= q(t)(T u) . Thus, T : K → K . Applying the Arzela–Ascoli theorem, we can prove that T, (T (·)) : K → C[0, 1] are completely continuous. Therefore, T : K → K is completely continuous.

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Since A < B, it is easy to check that r1 = r2 . Without loss of generality, we assume r1 < r2 . Let Ω1 = {u ∈ K : |u| < r1 },

Ω2 = {u ∈ K : |u| < r2 }.

If u ∈ ∂Ω1 , then |u| = r1 . So, u ≤ r1 and u   ≤ 0 ≤ u(t) ≤ r1 ,

−

48 r ≤ u  (t) ≤ 0, 5

48 5 r1 .

This implies that

0 ≤ t ≤ 1.

Thus, for 0 ≤ t ≤ 1, 5 48 1 − r1 ≤ u  (t) − u 0 (t) ≤ M, M ≤ u(t) − u 0 (t) ≤ r1 , 384 5 8 f (t, u(t) − u 0 (t), u  (t) − u 0 (t)) + M ≤ ϕ(r1 ) ≤ r1 A.

−

It follows that



1 1

G(t, s)G(s, τ )[ f (τ, u(τ ) − u 0 (τ ), u  (τ ) − u 0 (τ )) + M]dτ ds

0 1 1 ≤ r1 A max G(t, s)G(s, τ )dτ ds = r1 ,

T u = max

0≤t ≤1 0

0≤t ≤1 0

1

0

(T u)  = max

G(t, s)[ f (s, u(s) − u 0 (s), u  (s) − u 0 (s)) + M]ds

1 48 384 1 · = r1 . G(t, s)ds = r1 · ≤ r1 A max 0≤t ≤1 0 5 8 5 0≤t ≤1 0

Therefore, |T u| ≤ r1 = |u|. If u ∈ ∂Ω2 , then |u| = r2 . By Lemma 2.3, 25 r2 ≤ u ≤ r2 and 8r2 ≤ u   ≤ implies 2 r2 q(t) ≤ q(t)u ≤ u(t) ≤ r2 , 5 8r2 q(t) ≤ q(t)u   ≤ −u  (t) ≤

0 ≤ t ≤ 1, 48 r2 , 5

0 ≤ t ≤ 1.

Thus, 2 1 r2 = r2 min q(t) ≤ min u(t) ≤ max u(t) ≤ r2 , 1/4≤t ≤3/4 10 5 1/4≤t ≤3/4 1/4≤t ≤3/4 48 − r2 ≤ min u  (t) ≤ max u  (t) ≤ −8r2 min q(t) = −2r2 . 1/4≤t ≤3/4 1/4≤t ≤3/4 5 1/4≤t ≤3/4 It follows that, for

1 4

≤ t ≤ 34 ,

1 5 r2 − M ≤ u(t) − u 0 (t) ≤ r2 , 10 384 1 48 − r2 ≤ u  (t) − u 0 (t) ≤ M − 2r2 , 5 8 f (t, u(t) − u 0 (t), u  (t) − u 0 (t)) + M ≥ ψ(r2 ) ≥ r2 B. From this, we get

48 5 r2 .

This

Q. Yao / Nonlinear Analysis 66 (2007) 138–150

(T u)  ≥ max



3/4

0≤t ≤1 1/4

≥ r2 B max



145

G(t, s)[ f (s, u(s) − u 0 (s), u  (s) − u 0 (s)) + M]ds 3/4

0≤t ≤1 1/4

G(t, s)ds =

48 r2 . 5

5 It follows that |T u| ≥ 48 (T u)  ≥ r2 = |u|. By Lemma 2.2, we assert that the operator T has at least one fixed point u¯ ∈ K such that r1 ≤ |u| ¯ ≤ r2 . This implies that the problem (P1 ) has at least one solution u¯ ∈ K such that r1 ≤ |u| ¯ ≤ r2 . Let u ∗ (t) = u(t) ¯ − u 0 (t), 0 ≤ t ≤ 1. We need to check that u ∗ is a solution of the problem (P). In fact, since T u¯ = u, ¯ we have

¯ = (T u)(t) ¯ u ∗ (t) + u 0 (t) = u(t)

1 1 G(t, s)G(s, τ )[ f (τ, u(τ ¯ ) − u 0 (τ ), u¯  (τ ) − u 0 (τ )) + M]dτ ds =

0

=

0 1 1

0

It follows that ∗

G(t, s)G(s, τ ) f (τ, u ∗ (τ ), (u ∗ ) (τ ))dτ ds + u 0 (t).

0



u (t) = 0

1 1

G(t, s)G(s, τ ) f (τ, u ∗ (τ ), (u ∗ ) (τ ))dτ ds.

0 ≤ t ≤ 1.

0

In other words, u ∗ is a solution of the problem (P). Therefore, the problem (P) has at least one solution u ∗ satisfying u ∗ + u 0 ∈ K and r1 ≤ |u ∗ + u 0 | ≤ r2 . Thus, we have u ∗ + u 0  ≥ 25 r1 by Lemma 2.3. 5 1 M, then 25 r1 > 24 M. By Lemma 2.1, for 0 < t < 1, If r1 > 48 u ∗ (t) = [u ∗ (t) + u 0 (t)] − u 0 (t) = [u ∗ (t) + u 0 (t)] − M p(t)   2 1 1 ∗ Mq(t) ≥ r1 − M q(t) > 0. ≥ q(t)u + u 0  − 24 5 24 It follows that u ∗ is a positive solution.



4. Multiplicity Theorem 4.1. Assume that there exist three positive numbers r1 < r2 < r3 such that one of the following conditions is satisfied: (1) ϕ(r1 ) ≤ r1 A, ψ(r2 ) > r2 B, ϕ(r3 ) ≤ r3 A. (2) ψ(r1 ) ≥ r1 B, ϕ(r2 ) < r2 A, ψ(r3 ) ≥ r3 B. Then problem (P) has at least two solutions u ∗1 , u ∗2 satisfying u ∗1 + u 0 , u ∗2 + u 0 ∈ K and r1 ≤ |u ∗1 + u 0 | < r2 < |u ∗2 + u 0 | ≤ r3 . Besides that, (a) If r2 > (b) If r1 >

5 ∗ 48 M, then u 2 5 48 M or M =

is a positive solution. 0, then u ∗1 , u ∗2 are positive solutions.

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Proof. We prove only case (1). Since ψ : [0 + ∞) → [0, +∞) is continuous, there exist two positive numbers r1 < r˜2 < r2 < r¯2 < r3 such that ψ(˜r2 ) ≥ r˜2 B, ψ(¯r2 ) ≥ r¯2 B. Consider the pair of positive numbers {r1 , r˜2 } and {¯r2 , r3 }. By Theorem 3.1, the problem (P) has two solutions u ∗1 , u ∗2 satisfying u ∗1 + u 0 , u ∗2 + u 0 ∈ K and r1 ≤ |u ∗1 + u 0 | ≤ r˜2 < r2 < r¯2 ≤ |u ∗2 + u 0 | ≤ r3 . Similarly, we can prove following results. Theorem 4.2. Assume that there exist four positive numbers r1 < r2 < r3 < r4 such that one of the following conditions is satisfied: (1) ϕ(r1 ) ≤ r1 A, ψ(r2 ) > r2 B, ϕ(r3 ) < r3 A, ψ(r4 ) ≥ r4 B. (2) ψ(r1 ) ≥ r1 B, ϕ(r2 ) < r2 A, ψ(r3 ) > r3 B, ϕ(r4 ) ≤ r4 A. Then problem (P) has at least three solutions u ∗1 , u ∗2 , u ∗3 satisfying u ∗1 + u 0 , u ∗2 + u 0 , u ∗3 + u 0 ∈ K and r1 ≤ |u ∗1 + u 0 | < r2 < |u ∗2 + u 0 | < r3 < |u ∗3 + u 0 | ≤ r4 . Besides that, (a) If r3 > (b) If r2 > (c) If r1 >

5 ∗ 48 M, then u 3 5 ∗ 48 M, then u 2 5 48 M or M =

is a positive solution. and u ∗3 are positive solutions. 0, then u ∗1 , u ∗2 , u ∗3 are positive solutions.

In general, we have the following two existence results concerning n solutions and/or positive solutions. Theorem 4.3. Assume that there exist 2k positive numbers r1 < r2 < · · · < r2k such that one of the following conditions is satisfied: (1) ϕ(r1 ) ≤ r1 A, ψ(r2 ) > r2 B, . . . , ϕ(r2i−1 ) < r2i−1 A, ψ(r2i ) > r2i B, . . . , ϕ(r2k−1 ) < r2k−1 A, ψ(r2k ) ≥ r2k B. (2) ψ(r1 ) ≥ r1 B, ϕ(r2 ) < r2 A, . . . , ψ(r2i−1 ) > r2i−1 B, ϕ(r2i ) < r2i A, . . . , ψ(r2k−1 ) > r2k−1 B, ϕ(r2k ) ≤ r2k A. Then problem (P) has at least 2k −1 solutions u ∗i , i = 1, 2, . . . , 2k −1, satisfying u ∗i +u 0 ∈ K and r1 ≤ |u ∗1 + u 0 | < r2 ,

ri < |u ∗i + u 0 | < ri+1 ,

r2k−1 < |u ∗2k−1 + u 0 | ≤ r2k , 2 ≤ i ≤ 2k − 2.

Besides that, 5 (a) If ri > 48 M, i = 2, 3, . . . , 2k − 1, then u ∗i , u ∗i+1 , . . . , u ∗2k−1 are positive solutions. 5 M or M = 0, then u ∗1 , u ∗2 , . . . , u ∗2k−1 are positive solutions. (b) If r1 > 48

Theorem 4.4. Assume that there exist 2k + 1 positive numbers r1 < r2 < · · · < r2k+1 such that one of the following conditions is satisfied: (1) ϕ(r1 ) ≤ r1 A, ψ(r2 ) > r2 B, . . . , ϕ(r2i−1 ) < r2i−1 A, ψ(r2i ) > r2i B, . . . , ψ(r2k ) > r2k B, ϕ(r2k+1 ) ≤ r2k+1 A.

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147

(2) ψ(r1 ) ≥ r1 B, ϕ(r2 ) < r2 A, . . . , ψ(r2i−1 ) > r2i−1 B, ϕ(r2i ) < r2i A, . . . , ϕ(r2k ) < r2k A, ψ(r2k+1 ) ≥ r2k+1 B. Then problem (P) has at least 2k solutions u ∗i , i = 1, 2, . . . , 2k, satisfying u ∗i + u 0 ∈ K and r1 ≤ |u ∗1 + u 0 | < r2 ,

ri < |u ∗i + u 0 | < ri+1 ,

r2k < |u ∗2k + u 0 | ≤ r2k+1 , 2 ≤ i ≤ 2k − 1.

Besides that, (a) If ri > (b) If r1 >

5 ∗ ∗ ∗ 48 M, i = 2, 3, . . . , 2k, then u i , u i+1 , . . . , u 2k are positive solutions. 5 ∗ ∗ ∗ 48 M or M = 0, then u 1 , u 2 , . . . , u 2k are positive solutions.

5. Remarks and examples In the section, we will illustrate that our work extends and improves the results contained in [10] even if M = 0. Proposition 5.1. Let M = 0. 5 r) ≤ (1) If max{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r ] × [−r, 0]} ≤ 6r , then ϕ( 48

(2) If min{ f (t, u, v) : (t, u, v) ∈

[0, 1] × [0, r ] × [−r, − 14 r ]}

≥

32 3 r,

then

5 48 r A. 5 5 ψ( 48 r ) ≥ 48 r B.

Proof. By the definitions of ϕ and ψ, we have the following inequalities:       5 5 r = max f (t, u, v) : (t, u, v) ∈ [0, 1] × 0, r × [−r, 0] ϕ 48 48 ≤ max{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r ] × [−r, 0]},          5 1 3 1 5 5 ψ × r = min f (t, u, v) : (t, u, v) ∈ , r, r × −r, − r 48 4 4 96 48  24  1 ≥ min f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, r ] × −r, − r . 4 In case (1), we have   5 48 5 288 5 384 5 5 ϕ r ≤ 6r = 6 · · r= · r≤ · r = r A. 48 5 48 5 48 5 48 48 In case (2), we have   32 32 48 5 512 5 5 5 r ≥ r= · · r= · r = r B. ψ 48 3 3 5 48 5 48 48 Proposition 5.2. Let M = 0. (1) (2) (3) (4)

If If If If

max f0 = 0, then there exists r0 > 0 such that ϕ(r ) ≤ r A for any 0 < r ≤ r0 . max f∞ = 0, then there exists r0 > 0 such that ϕ(r ) ≤ r A for any r0 ≤ r < +∞. min f 0 = +∞, then there exists r0 > 0 such that ψ(r ) ≥ r B for any 0 < r ≤ r0 . min f ∞ = +∞, then there exists r0 > 0 such that ψ(r ) ≥ r B for any r0 ≤ r < +∞.

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Proof. We prove only (1) and (4). If max f 0 = 0, then there exists r0 > 0 such that, for any − 48 5 r0 ≤ v ≤ 0, sup{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, +∞)} ≤

5 (−v)A. 48

It follows that, for any 0 < r ≤ r0 ,    48 ϕ(r ) ≤ max f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, +∞) × − r, 0 5 5 48 ≤ · · r A = r A. 48 5 If min f ∞ = +∞, then there exists r0 > 0 such that, for any −∞ < v ≤ −r0 , inf{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, +∞)} ≥

1 (−v)B. 2

It follows that, for any r0 ≤ r < +∞, ψ(r ) ≥ min{ f (t, u, v) : (t, u, v) ∈ [0, 1] × [0, +∞) × (−∞, −2r ]} 1 ≥ · 2r · B = r B. 2 Remark 5.3. From Proposition 5.1, we see that Theorem 1.1 is a simple corollary of Theorem 3.1. Remark 5.4. From Propositions 5.1 and 5.2, we see that Theorems 1.2 and 1.3 are corollaries of Theorem 4.1. Example 5.5. Consider the boundary value problem  (4) u (t) = f (u(t)), 0 ≤ t ≤ 1, u(0) = u(1) = u  (0) = u  (1) = 0, where f : [0, +∞) → [0, +∞) is defined by  1976u √ − 1904, 1 ≤ u < +∞, f (u) = 72 u, 0 ≤ u ≤ 1. Thus, M = 0 and       2025 2025 2025 2025 2025 81 ≤u≤ = B, ψ = min f (u) : = f = 40960 409600 40960 409600 16 40960 384 ϕ(1) = max{ f (u) : 0 ≤ u ≤ 1} = f (1) = 72 < = 1 · A, 5 ψ(20) = min{ f (u) : 2 ≤ u ≤ 20} = f (2) = 2048 = 20B. 2025 By Theorem 4.1, the problem has two positive solutions u ∗1 , u ∗2 ∈ K and 40960 ≤ |u ∗1 | < ∗ 1 < |u 2 | ≤ 20. 48 Since infr>0 { f (u) : 0 ≤ u ≤ r }/r = 72 > 288 5 = 5 · 6, the conclusion cannot be derived from Theorem 1.1.

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Example 5.6. Consider the boundary value problem  (4) u (t) = f (u  (t)), 0 ≤ t ≤ 1, u(0) = u(1) = u  (0) = u  (1) = 0, where f : (−∞, 0] → [0, +∞) is defined by ⎧     ⎪ ⎪ 11 v 2 sin v + 1 π  , −∞ < v ≤ −192, ⎪ ⎪  ⎪ 288  2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −192 ≤ v ≤ −20, ⎨1408, f (v) = 672 ⎪ ⎪ − v − 1280, −20 ≤ v ≤ −10, ⎪ ⎪ 5 ⎪ ⎪   ⎪ ⎪ 64   5π  ⎪ ⎪ ⎪ , −10 ≤ v ≤ 0. sin |v| √ ⎩  v  10 Thus, M = 0. After simple computations, we get   48 384 ϕ(1) = max f (v) : − ≤ v ≤ 0 ≤ f (−10) = 64 < = 1 A, 5 5 ψ(10) = min{ f (v) : −96 ≤ v ≤ −20} = f (−20) = 1408 > 1024 = 10B, ϕ(20) = max{ f (v) : −192 ≤ v ≤ 0} = f (−192) = 1408 < 1536 = 20 A. By Theorem 4.1, the problem has two positive solutions u ∗1 , u ∗2 ∈ K and 1 ≤ |u ∗1 | < 10 < |u ∗2 | ≤ 20. In the problem, the limits lim sup −v→0+

f (v) f (v) = lim sup = +∞, −v −v→+∞ −v

lim inf

−v→0+

f (v) f (v) = lim inf = 0. −v→+∞ −v −v

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