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Local existence of multiple positive solutions to a singular cantilever beam equation
J. Math. Anal. Appl. 363 (2010) 138–154
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Local existence of multiple positive solutions to a singular cantilever beam equation Qingliu Yao Department of Applied Mathematics, Nanjing University of Finance and Economics, Nanjing 210003, PR China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 30 April 2009 Available online 28 July 2009 Submitted by D.L. Russell
By constructing suitable cone and control functions, we prove some local existence theorems of positive solutions for a singular fourth-order two-point boundary value problem. In mechanics, the problem is called cantilever beam equation. Furthermore, we improve a famous method appeared in the studies of singular boundary value problems. The approximation theorem of completely continuous operators and the Guo–Krasnosel’skii fixed point theorem of cone expansion–compression type play important parts in this work. © 2009 Elsevier Inc. All rights reserved.
Keywords: Singular ordinary differential equation Boundary value problem Positive solution Existence Multiplicity
1. Introduction The purpose of this paper is to study the local existence and multiplicity of positive solutions for the following nonlinear fourth-order two-point boundary value problem
(P )
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0,
and its simplified form
( Pˆ )
u (4) (t ) = q(t ) f t , u (t ), u (t ) ,
u (4) (t ) = q(t ) f t , u (t ) ,
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0,
where q : (0, 1) → [0, +∞) is continuous and q(t ) > 0, 0 < t < 1. In this paper, the function u ∗ ∈ C [0, 1] is called positive solution of the problem ( P ) if u ∗ is a solution of ( P ) and u ∗ (t ) > 0, 0 < t 1. In addition, let 0 < α < β 1 and u = max0t 1 |u (t )| for u ∈ C [0, 1]. We consider singular problem ( P ). In other words, the nonlinear term f (t , u , v ) is allowed to be singular at t = 0, t = 1 and u = 0, v = 0. Because of the extensive applications in mechanics and engineering [1–3], the nonlinear fourth-order boundary value problems have received wide attentions. In mechanics, the problem ( P ) is called cantilever beam equation. The equation describes the deflection of an elastic beam fixed at the left and freed at the right. There are some papers discussed the existence of solutions of ( P ) [4–12]. Because the problem ( P ) is a (2, 4 − 2) focal boundary value problem, most results on the problem ( P ) appear in the literature on the ( p , m − p ) focal boundary value problems.
E-mail address: [email protected]. 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.07.043
©
2009 Elsevier Inc. All rights reserved.
Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
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When f (t , u , v ) = f (t , u ), some existence theorems of positive solutions have been established in the papers [4–9] for the problem ( Pˆ ). For example, the following local existence theorem is a special case of Theorem 4.2 in Agarwal [9]. Theorem 1.1. Assume that (a1) f : [0, 1] × (0, +∞) → [0, +∞) is continuous. (a2) f (t , u ) g (u ) + h(u ), (t , u ) ∈ [0, 1] × (0, +∞), where h : [0, +∞) → [0, +∞) is continuous, g : (0, +∞) → (0, +∞) is 1 continuous and 0 q(t ) g (t 2 ) dt < +∞. (a3) g (u ) is nonincreasing on (0, +∞) and there exists constant M > 0 such that
g (u 1 u 2 ) M g (u 1 ) g (u 2 ),
(u 1 , u 2 ) ∈ (0, +∞) × (0, +∞).
(a4) hg((uu)) is nondecreasing on (0, +∞). (a5) There exists θ ∈ C [α , β] such that θ(t ) > 0, α t β and
q(t ) f (t , u ) θ(t ) g (u ) + h(u ) ,
(t , u ) ∈ [α , β] × [0, +∞).
(a6) There exists a > 0 such that
1
G (t , s)q(s) g s2 ds
M max
0t 1 0
a g (a) + h(a)
.
(a7) There exists b > a such that
β G (t , s)θ(s) ds
max
0t 1
α
bg (bα 2 ) g (b) g (bα 2 ) + g (b)h(bα 2 )
.
Then problem ( Pˆ ) has two positive solutions u ∗1 , u ∗2 ∈ C 3 [0, 1] ∩ C 4 (0, 1) and 0 < u ∗1 < a < u ∗2 b. In Theorem 1.1, G (t , s) is the Green function of related homogeneous linear problem, see Section 2. The remarkable characterization of Theorem 1.1 is that the nonlinear term f (t , u ) is allowed to be singular at u = 0. Theorem 1.1 obtains not only the existence of positive solutions, but presents also an effective method in order to study various singular boundary value problems. However, the assumptions (a4) and (a5) are very stern. For example, the assumption (a5) implies that f (t , u ) > 0, α t β , 0 < u < +∞. Therefore, Theorem 1.1 is unapplicable to some singular problems ( Pˆ ). The purpose of this paper is to extend Theorem 1.1 and improve the method used in the theorem. Moreover, we will cancel the assumptions (a3)–(a5). The motivation comes from the localization method arisen in papers [8,9,12–25]. By the method, we can determine the existence of positive solutions by considering the properties of nonlinear term on some bounded sets. Consequently, the localization method is a powerful tool for many nonlinear boundary value problems. Very recently, in papers [26–28], we consider some conditions that ensure applications of the localization method to singular boundary value problems. In this paper, we will improve the localization method by applying the results in [26–28]. After that, we will study the local existence and multiplicity of positive solutions of the problem ( P ) by the improved method. In this paper, we will use the following assumptions. (H1) f : (0, 1) × (0, +∞) × (0, +∞) → [0, +∞) is continuous. (H2) f (t , u , v ) g (t , u , v ) + h(t , u , v ), (t , u , v ) ∈ (0, 1) × (0, +∞) × (0, +∞), where g : (0, 1) × (0, +∞) × (0, +∞) → [0, +∞) is continuous, h : [0, 1] × [0, +∞) × [0, +∞) → [0, +∞) is continuous. 1 r (H3) 0 q(t )γr12 (t ) dt < +∞ for any 0 < r1 < r2 , where
r2 r1 (t )
γ
= max g (t , u , v ) + h(t , u , v ):
1 4
r1 t u r2 , r1 t v r2 . 2
(H4) g (t , u 1 , v 1 ) g (t , u 2 , v 2 ), 0 < t < 1, 0 < u 1 u 2 , 0 < v 1 v 2 . (H5) There exist α μ < ν β , r¯ > 0 and η > 0 such that
f (t , u , v ) η,
(t , u , v ) ∈ [μ, ν ] × (0, r¯] × (0, r¯]. r
The assumption (H2) implies that f (t , u , v ) may be singular at u = 0, v = 0, t = 0, t = 1 and γr12 (t ) is continuous on (0, 1). The outline of this paper is as follows. In Section 2, we construct a suitable cone and transfer the problem ( P ) into an integral equation. After that, we prove that the related integral operator is completely continuous on some proper sets
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by using the approximation theorem of completely continuous operators. In Section 3, we establish the local existence of n positive solutions under the assumptions (H1)–(H3) by applying the Guo–Krasnosel’skii fixed point theorem of cone expansion–compression type, where n is an arbitrary positive integer. In Section 4, we study the problem ( P ) under the assumptions (H1)–(H5). In Section 5, we consider the simplified problem ( Pˆ ). In Section 6, we verify that Theorem 1.1 is a corollary of main results. Moreover, five examples illustrate that our extension is true and our results are applicable to more complicated cases. All results are independent of the existence of upper and lower solutions. 2. Preliminaries Let C 1 [0, 1] be the Banach space with norm |||u ||| = max{u , u } and let K be the cone of nonnegative functions in C 1 [0, 1] with following form
K = u ∈ C 1 [0, 1]: u (0) = 0, u (t )
1 2
u t 2 , u (t ) u t , 0 t 1 .
Write Ω(r ) = {u ∈ K : |||u ||| < r }, ∂Ω(r ) = {u ∈ K : |||u ||| = r }. Lemma 2.1. If u ∈ K , then u = |||u ||| and
1 |||u ||| 2
u |||u |||.
Proof. Since u (t ) 0 and u (0) = 0, we have
t u = max
1
u (s) ds =
0t 1 0
1
u (s) ds 0
max u (s) ds = u .
0s1 0
Since u (t ) u t, 0 t 1, we have
1 u =
1
u (s) ds u 0
s ds =
1 2
u .
0
Hence u = |||u ||| and
1 |||u ||| 2
u |||u |||. 2
Let G (t , s) be the Green function of the homogeneous linear problem u (4) (t ) = 0, u (0) = u (0) = u (1) = u (1) = 0, that is,
1 G (t , s) =
(3t 2 s 6 1 (3ts2 6
− t 3 ), 0 t s 1, − s3 ), 0 s t 1.
Computing the partial derivatives of G (t , s) with respect to t, we have
∂ G (t , s) = ∂t
1
s2 − 12 (s 2 1 2 s , 2
s − t, ∂ G (t , s) = ∂t2 0, 2
− t )2 , 0 t s 1, 0 s t 1,
−1, ∂ G ( t , s ) = ∂t3 0, 3
0 t s 1, 0 s < t 1.
Lemma 2.2. (1) G , ∂∂t G : [0, 1] × [0, 1] → [0, +∞) are continuous. (2) G (t , s) > 0, ∂∂t G (t , s) > 0, 0 < t , s < 1. (3) max0t 1 ∂∂t G (t , s) = ∂∂t G (s, s) = 12 s2 , 0 s 1. (4) ∂∂t G (t , s) t ∂∂t G (s, s), 0 t , s 1. (5) max0t 1 G (t , s) = G (1, s) = 16 s2 (3 − s), 0 s 1. (6) G (t , s) t 2 G (1, s), 0 t , s 1. 2 3 (7) G (0, s) ≡ 0, ∂∂t G (0, s) ≡ 0, ∂∂t 2 G (1, s) ≡ 0, ∂∂t 3 G (1, s) ≡ 0.
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2
Proof. The proofs are direct.
Define the operator T as follows
1 ( T u )(t ) =
G (t , s)q(s) f s, u (s), u (s) ds,
0 t 1.
0
We will transfer the problem ( P ) into the integral equation
1 u (t ) =
G (t , s)q(s) f s, u (s), u (s) ds,
0 t 1,
0
and prove that the integral operator T is completely continuous on the each bounded closed set of K \{0}. For this end, we need the following approximation theorem of completely continuous operators. Lemma 2.3. Let X be a Banach space, let V ⊂ X be a bounded closed subset, let T k : V → X , k = 1, 2, . . ., be a sequence of completely continuous operators. If the operator T : V → X and supu ∈ V T k u − T u → 0 (k → ∞), then T : V → X is a completely continuous operator. Lemma 2.4 gives basic properties of the operator T . Lemma 2.4. Suppose that (H1)–(H3) are satisfied and 0 < r1 < r2 . Then (1) T : Ω(r2 )\Ω(r1 ) → K is completely continuous. (2) For any u ∈ Ω(r2 )\Ω(r1 ),
t 1 ( T u )(t ) = 0
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds, ∂s
0 t 1.
0
γrr12 (t ) as in the assumption (H3). For every k = 2, 3, . . . , let ⎧ r r ⎪ min{γr12 (t ), kt γr12 ( 1k )}, 0 t 1k , ⎪ ⎨ r2 r2 1 γr1 k (t ) = γr1 (t ), t k−k 1 , k ⎪ ⎪ ⎩ r2 r 2 k −1 min{γ (t ), k(1 − t )γ ( )}, k−1 t 1.
Proof. Let
r1
r
r1
r
k
k
r
Then [γr12 ]k ∈ C [0, 1] and [γr12 ]k (0) = [γr12 ]k (1) = 0. By the assumption (H3),
1
q(t )
1 0
r
q(t )γr12 (t ) dt < +∞. So
γrr12 (t ) − γrr12 k (t ) dt → 0 (k → ∞).
0
Construct the function f k (t , u , v ) as follows, for 0 t 1,
f k (t , u , v ) = min f (t , u , v ),
γrr12 k (t ) ,
1 4
r1 t 2 u < ∞, r1 t v < +∞;
⎧ 1 ⎪ f (t , u , r1 t ), r t 2 u < ∞, 0 v r1 t , ⎪ 4 1 ⎨ k f k (t , u , v ) = f k (t , 14 r1 t 2 , v ), 0 u 14 r1 t 2 , r1 t v < +∞, ⎪ ⎪ ⎩ f (t , 1 r t 2 , r t ), 0 u 1 r t 2 , 0 v r t . 1 1 k 4 1 4 1
By the definition, f k : [0, 1] × [0, +∞) × [0, +∞) → [0, +∞) is continuous and f k (0, u , v ) ≡ 0, f k (1, u , v ) ≡ 0. By (H2) and (H3), for any u ∈ K ,
1 0
q(t ) f k t , u (t ), u (t ) dt
1
r
q(t )γr12 (t ) dt < +∞. 0
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Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
Define the operator T k as follows
1 ( T k u )(t ) =
G (t , s)q(s) f k s, u (s), u (s) ds,
0 t 1.
0
Then T k : K → C [0, 1] is completely continuous by the continuity of f k (t , u , v ) and the Arzela–Ascoli theorem. And, for 0 t 1,
1
( T k u ) (t ) =
∂ G (t , s)q(s) f k s, u (s), u (s) ds. ∂t
0
Hence, ( T k (·))
: K → C [0, 1] is a completely continuous operator, too. Let u ∈ Ω(r2 )\Ω(r1 ). Then r1 |||u ||| r2 . From Lemma 2.1, we get 1 4
r1t 2
1
r1 t u t u (t ) r2 ,
u t 2 u (t ) r2 ,
2
1 r t2 4 1
By the definition of f k (t , u , v ), for 0 < t < 1, r
0 f (t , u , v ) − f k (t , u , v ) γr12 (t ) −
1 r 2 1
u r2 , r1 u r2 . Thus,
0 t 1.
u r2 , r1 t v r2 ,
γrr12 k (t ).
It follows that
0 f t , u (t ), u (t ) − f k t , u (t ), u (t ) γr12 (t ) − r
γrr12 k (t ), 0 < t < 1.
Applying the above-mentioned results, we obtain that
1 sup u ∈Ω(r2 )\Ω(r1 )
T u − T k u =
sup u ∈Ω(r2 )\Ω(r1 )
max
0t 1
0
1
sup u ∈Ω(r2 )\Ω(r1 )
max
G (t , s)q(s)
0t 1
γrr12 (s) − γrr12 k (s) ds
0
1 max G (t , s)
q (s)
0t ,s1
γrr12 (s) − γrr12 k (s) ds → 0.
0
By Lemma 2.3, T : Ω(r2 )\Ω(r1 ) → C [0, 1] is a completely continuous operator. Similarly,
1 sup
max
u ∈Ω(r2 )\Ω(r1 )
0t 1
∂ G (t , s)q(s) f s, u (s), u (s) − f k s, u (s), u (s) ds → 0. ∂t
0
It follows that
1
( T u ) (t ) =
∂ G (t , s)q(s) f s, u (s), u (s) ds, ∂t
0 t 1,
0
and ( T (·))
: Ω(r2 )\Ω(r1 ) → C [0, 1] is completely continuous. By Lemma 2.2(7), we have ( T u )(0) = 0. So
t 1 ( T u )(t ) = 0
0
G (t , s)q(s) f s, u (s), u (s) − f k s, u (s), u (s)
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds, ∂s
0 t 1.
ds
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143
On the other hand, by Lemma 2.2(3) and (4),
t 1 ( T u )(t ) 0
∂ s G (τ , τ )q(τ ) f τ , u (τ ), u (τ ) dτ ∂s
0
1
2
1 max
2
0s1 0
1
= t 2 max
0t 1 0
( T u ) (t )
1 t
s ds max
0s1
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds ∂s
0
∂ 1 G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds t 2 ∂s 2
0
t 1
2
1
0
1
= t
t
1 1 0
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds ∂s
0
∂ 1 G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds = T u t 2 , ∂s 2
0
∂ G (s, s)q(s) f s, u (s), u (s) ds t max ∂t 0t 1
0
1
∂ G (t , s)q(s) f s, u (s), u (s) ds = ( T u ) t . ∂t
0
It implies that T : Ω(r2 )\Ω(r1 ) → K . Therefore, T : Ω(r2 )\Ω(r1 ) → K is a completely continuous operator.
2
Our approach is based on the following Guo–Krasnosel’skii fixed point theorem of cone expansion–compression type. Lemma 2.5. Let X be a Banach space, and let K ⊂ X be a cone in X . Assume Ω1 , Ω2 are bounded open subset of K with 0 ∈ Ω1 ⊂ Ω 1 ⊂ Ω2 , and let F : Ω 2 \Ω1 → K be a completely continuous operator such that (1) F u u , u ∈ ∂Ω1 and F u u , u ∈ ∂Ω2 ; or (2) F u u , u ∈ ∂Ω1 and F u u , u ∈ ∂Ω2 . Then F has a fixed point in Ω 2 \Ω1 . 3. Local existence theorems under (H1)–(H3) We introduce the following control functions and constants:
Φ(t , r ) = max g (t , u , v ) + h(t , u , v ): Ψ (t , r ) = min f (t , u , v ):
1 4
ψ(r ) = min f (t , u , v ): α t β, A=
1 max
0t 1
4
2
rt u r , rt v r ,
rt 2 u r , rt v r ,
ϕ (r ) = max h(t , u , v ): 0 t 1,
1
∂ G (t , s)q(s) ds ∂t
1 4
rt 2 u r , rt v r ,
1 4
rt 2 u r , rt v r ,
−1 ,
0
B=
β max
0t 1
1 ζr (t ) = max g (t , u , v ): rt 2 u r , rt v r ,
α
∂ G (t , s)q(s) ds ∂t
−1 ,
4
1 I (r ) = max
0t 1
∂ G (t , s)q(s)ζr (s) ds. ∂t
0 6 If q(t ) ≡ 1, then A = 6, B = β 3 − . α3 1 If the assumption (H3) is satisfied, then 0 q(t )ζr (t ) dt < +∞ for any r > 0. We obtain the following local existence theorems. Among others, Theorem 3.1 is a basic existence criterion of positive solution for the problem ( P ).
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Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
Theorem 3.1. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied: (b1) max0t 1
1
∂
0 ∂t
β G (t , s)q(s)Φ(s, a) ds a and max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, b) ds b.
β 1 (b2) max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, a) ds a and max0t 1 0 ∂∂t G (t , s)q(s)Φ(s, b) ds b.
Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗ ||| b. Proof. Without loss of generality, we prove only the case (b1). If u ∈ ∂Ω(a), then |||u ||| = a. So 12 a u a, u = a by Lemma 2.1. From this,
1
1
u t 2 u (t ) a, at u t u (t ) a, 0 t 1, 4 2 g t , u (t ), u (t ) + h t , u (t ), u (t ) Φ(t , a), 0 < t < 1. at 2
Since T u ∈ K and Lemma 2.4(2), we obtain
||| T u ||| = ( T u ) = max
1
0t 1
∂ G (t , s)q(s) f s, u (s), u (s) ds ∂t
0
1 max
0t 1
∂ G (t , s)q(s) g s, u (s), u (s) + h s, u (s), u (s) ds max ∂t 0t 1
0
1
∂ G (t , s)q(s)Φ(s, a) ds ∂t
0
a = |||u |||. If u ∈ ∂Ω(b), then |||u ||| = b. It means
1 b 2
u b, u = b. So,
1
bt u (t ) b, 0 t 1, bt 2 u (t ) b, 4 f t , u (t ), u (t ) Ψ (t , b), α t β. It follows that
||| T u ||| = ( T u ) max
0t 1
β α
∂ G (t , s)q(s) f s, u (s), u (s) ds max ∂t 0t 1
β α
∂ G (t , s)q(s)Ψ (s, b) ds b = |||u |||. ∂t
By Lemmas 2.4(1) and 2.5, the operator T has one fixed point u ∗ ∈ Ω(b)\Ω(a). It implies that
u ∗ (t ) =
t 1 0
∂ G (s, τ )q(τ ) f τ , u ∗ (τ ), u ∗ (τ ) dτ ds, ∂t
0 t 1.
0
Computing the first, second, third and fourth derivatives of u ∗ (t ), we get
u∗
u
( i )
1 (t ) =
∗ (4 )
∂i G (t , s)q(s) f s, u ∗ (s), u ∗ (s) ds, i ∂t
0 t 1, i = 1, 2, 3,
0
(t ) = q(t ) f t , u ∗ (t ), u ∗ (t ) ,
0 < t < 1.
According to Lemma 2.2(7), u ∗ (0) = (u ∗ ) (0) = (u ∗ ) (1) = (u ∗ ) (1) = 0. Therefore, u ∗ (t ) is a solution of the problem ( P ) and a |||u ∗ ||| b. Since u ∗ (t ) 14 at 2 > 0, 0 < t 1, the solution u ∗ (t ) is positive. Since (u ∗ ) (t ) at > 0, 0 < t 1, we see that u ∗ (t ) is strictly increasing on [0, 1]. u ∗ ∈ C 3 [0, 1] ∩ C 4 (0, 1) is obvious. 2 In the real problems, the computations of
1 max
0t 1 0
∂ G (t , s)q(s)Φ(s, r ) ds, ∂t
β max
0t 1
α
∂ G (t , s)q(s)Ψ (s, r ) ds ∂t
are more lengthy and tedious. For convenience, we give the following results.
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145
Theorem 3.2. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied: (c1) ϕ (a) (a − I (a)) A, ψ(b) bB. (c2) ψ(a) aB, ϕ (b) (b − I (b)) A. Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗ ||| b. Proof. Without loss of generality, we prove only the case (c1). By the definition of Φ(t , a), for 0 t 1,
Φ(t , a) max g (t , u , v ):
1 4
2
at u a, at v a + max h(t , u , v ):
1 4
2
at u a, at v a
ζa (t ) + ϕ (a). Applying the inequality, we get
1 max
0t 1
∂ G (t , s)q(s)Φ(s, a) ds max ∂t 0t 1
0
1
∂ G (t , s)q(s)ϕ (a) ds + max ∂t 0t 1
0
1
∂ G (t , s)q(s)ζa (s) ds ∂t
0
a − I (a) A max
1
0t 1
∂ G (t , s)q(s) ds + I (a) = a − I (a) A A −1 + I (a) = a. ∂t
0
By the definitions of Ψ (t , b), Ψ (t , b) ψ(b),
β max
0t 1
α
∂ G (t , s)q(s)Ψ (s, b) ds max ∂t 0t 1
By Theorem 3.1(b1), the proof is completed.
α t β . So β α
∂ G (t , s)q(s)ψ(b) ds bB max ∂t 0t 1
β α
∂ G (t , s)q(s) ds = b. ∂t
2
Theorem 3.3. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist three positive numbers a < b < c such that one of the following conditions is satisfied: (d1) ϕ (a) (a − I (a)) A, ψ(b) > bB, ϕ (c ) (c − I (c )) A. (d2) ψ(a) aB, ϕ (b) < (b − I (b)) A, ψ(c ) c B. Then problem ( P ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗1 ||| < b < |||u ∗2 ||| c. Proof. We prove only the case (d2). Applying the assumptions ψ(a) aB, ϕ (b) < (b − I (b)) A and imitating the proof of Theorem 3.2(c2), we assert that ( P ) has one strictly increasing positive solution u ∗1 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗1 ||| < b. Similarly, ( P ) has another positive solution u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) which is strictly increasing and b < |||u ∗2 ||| c. 2 In the same way, we can prove following theorems. Theorem 3.4. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist four positive numbers a < b < c < d such that one of the following conditions is satisfied: (e1) ϕ (a) (a − I (a)) A, ψ(b) > bB, ϕ (c ) < (c − I (c )) A, ψ(d) dB. (e2) ψ(a) aB, ϕ (b) < (b − I (b)) A, ψ(c ) > c B, ϕ (d) (d − I (d)) A. Then problem ( P ) has at least three strictly increasing positive solutions u ∗1 , u ∗2 , u ∗3 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that
a u ∗1 < b < u ∗2 < c < u ∗3 d.
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Denote the integer part of c by [c ]. Generally, we have the following existence theorem on n positive solutions. Theorem 3.5. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist n + 1 positive numbers a1 < a2 < · · · < an+1 such that one of the following conditions is satisfied:
ϕ (a2k−1 ) < (a2k−1 − I (a2k−1 )) A, k = 1, . . . , [ n+2 2 ], and ψ(a2k ) > a2k B, k = 1, . . . , [ n+2 1 ]. 2 ], and ϕ (a2k ) < (a2k − I (a2k )) A, k = 1, . . . , [ n+2 1 ]. (f2) ψ(a2k−1 ) > a2k−1 B, k = 1, . . . , [ n+ 2 (f1)
Then problem ( P ) has at least n strictly increasing positive solutions uk∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1), k = 1, 2, . . . , n, such that ak < |||uk∗ ||| < ak+1 . 4. The existence results under (H1)–(H5) In this section, let J (r ) = max0t 1
1
∂
0 ∂t
G (t , s)q(s) g (s, 14 rs2 , rs) ds.
If the assumption (H4) is satisfied, then ζr (t ) = g (t , 14 rt 2 , rt ), J (r ) is a nonincreasing function and J (r ) = I (r ). If limmax{u , v }→0 infμt ν g (t , u , v ) > 0, then the assumption (H5) is satisfied. Theorem 4.1. Suppose that (1) The assumptions (H1)–(H3) and (H5) are satisfied. 1 (2) There exists a > 0 such that max0t 1 0 ∂∂t G (t , s)q(s)Φ(s, a) ds a. Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗ ||| a. Proof. Let
ν = Cμ
ν max
0t 1
μ
∂ G (t , s)q(s) ds ∂t
−1
,
ν −1 1 . b = min r¯ , a, η C μ 2
By (H5), we have ν, f (t , u , v ) bC μ
μ t ν , 0 < u b, 0 < v b.
ν, It follows that Ψ (t , b) bC μ
β max
0t 1
α
μ t ν and
∂ G (t , s)q(s)Ψ (s, b) ds max ∂t 0t 1
ν μ
∂ ν C ν − 1 = b . G (t , s)q(s)Ψ (s, b) ds bC μ μ ∂t
By Theorem 3.1(b2), the problem ( P ) has one positive solution u ∗ ∈ K and b |||u ∗ ||| a.
2
Theorem 4.2. Suppose that (1) The assumptions (H1)–(H5) are satisfied. (2) There exists a > 0 such that ϕ (a) (a − J (a)) A. Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗ ||| a. Proof. Since I (a) = J (a), we have
1 max
0t 1 0
ϕ (a) (a − J (a)) A = (a − I (a)) A. It follows that
∂ G (t , s)q(s)Φ(s, a) ds max ∂t 0t 1
1 0
∂ G (t , s)q(s)ϕ (a) ds + max ∂t 0t 1
1
∂ 1 G (t , s)q(s) g s, as2 , as ds ∂t 4
0
a − I (a) A A −1 + J (a) = a − I (a) + I (a) = a.
By Theorem 4.1, the problem ( P ) has one positive solution u ∗ ∈ K and b |||u ∗ ||| a.
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Theorem 4.3. Suppose that (1) The assumptions (H1)–(H3) and (H5) are satisfied. 1 β (2) There exist 0 < a < b such that max0t 1 0 ∂∂t G (t , s)q(s)Φ(s, a) ds < a and max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, b) ds b. Then problem ( P ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗1 ||| < a < |||u ∗2 ||| b.
β
Proof. By the proof of Theorem 4.1, there exists 0 < c < a such that max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, c ) ds c. Imitating the proofs of Theorem 3.1, the problem ( P ) has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) and c |||u ∗1 ||| < a < |||u ∗2 ||| b. 2 Theorem 4.4. Suppose that (1) The assumptions (H1)–(H5) are satisfied. (2) There exist 0 < a < b such that ϕ (a) < (a − J (a)) A, ψ(b) bB. Then problem ( P ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗1 ||| < a < |||u ∗2 ||| b. Proof. By Theorem 4.2, ( P ) has one strictly increasing positive solutions u ∗1 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗1 ||| < a by applying the assumption (H5) and the inequality ϕ (a) < (a − J (a)) A. Since ϕ (a) < (a − J (a)) A and ψ(b) bB, we see that ( P ) has another strictly increasing positive solution u ∗2 ∈ K ∩ 3 C [0, 1] ∩ C 4 (0, 1) such that a < |||u ∗2 ||| b by imitating the proof of Theorem 3.2(c1). 2 Theorem 4.5. Suppose that (1) The assumptions (H1)–(H5) are satisfied. (2) There exist three positive numbers a < b < c such that
ϕ (a) < a − J (a) A ,
ψ(b) > bB ,
ϕ (c ) c − J (c ) A .
Then problem ( P ) has at least three strictly increasing positive solutions u ∗1 , u ∗2 , u ∗3 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that
0 < u ∗1 < a < u ∗2 < b < u ∗3 c . Obviously, we can write similar results for arbitrary positive integer n. For short, we omit them. 5. The existence results on the simplified problem ( Pˆ ) In this section, we consider the simplified problem ( Pˆ ). We will use the following assumptions. (H1) f : (0, 1) × (0, +∞) → [0, +∞) is continuous. (H2) f (t , u ) g (t , u ) + h(t , u ), (t , u ) ∈ (0, 1) × (0, +∞), where g : (0, 1) × (0, +∞) → [0, +∞) is continuous, h : [0, 1] × [0, +∞) → [0, +∞) is continuous. 1 r (H3) 0 q(t )γˆr12 (t ) dt < +∞ for any 0 < r1 < r2 , where
γˆrr12 (t ) = max g (t , u ) + h(t , u ): r1t 2 u r2 . (H4) g (t , u 1 ) g (t , u 2 ), 0 < t < 1, 0 < u 1 u 2 . (H5) There exist α μ < ν β , r¯ > 0 and η > 0 such that
f (t , u ) η,
(t , u ) ∈ [μ, ν ] × (0, r¯].
The following control functions and constants are applied in this section.
ˆ t , r ) = max g (t , u ) + h(t , u ): rt 2 u r , Φ(
Ψˆ (t , r ) = min f (t , u ): rt 2 u r ,
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ϕˆ (r ) = max h(t , u ): 0 t 1, rt 2 u r ,
ˆ r ) = min f (t , u ): α t β, rt 2 u r , ψ( −1 −1 1 β ˆ = max A G (t , s)q(s) ds , Bˆ = max G (t , s)q(s) ds , 0t 1
0t 1
0
α
ζˆr (t ) = max g (t , u ): rt 2 u r , ˆI (r ) = max
1
0t 1
G (t , s)q(s)ζˆr (s) ds,
0
ˆJ (r ) = max
1
G (t , s)q(s) g s, rs2 ds.
0t 1 0
24 ˆ = 8, Bˆ = If q(t ) ≡ 1, then A . 4(β 3 −α 3 )−(β 4 −α 4 ) If the assumption (H4) is satisfied, then ζˆr (t ) = g (t , rt 2 ), ˆJ (r ) is nonincreasing and ˆI (r ) = ˆJ (r ). Define the operator Tˆ and the cone Kˆ as follows
( Tˆ u )(t ) =
1
G (t , s)q(s) f s, u (s) ds,
0 t 1,
0
Kˆ = u ∈ C [0, 1]: u (t ) u t 2 , 0 t 1 . ˆ r ) = {u ∈ Kˆ : u < r }, ∂ Ω( ˆ r ) = {u ∈ Kˆ : u = r }. Write Ω(
ˆ r2 )\Ω( ˆ r1 ) → Kˆ is completely continuIf the assumptions (H1) –(H3) are satisfied, we can prove that the operator Tˆ : Ω( ous for any 0 < r1 < r2 by imitating the proof of Lemma 2.4 and applying Lemma 2.2(5) and (6). Further, we can prove the following local existence theorems. Theorem 5.1. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied: (h1) max0t 1
1 0
β
β
ˆ s, a) ds a, max0t 1 ˆ G (t , s)q(s)Φ( α G (t , s)q(s)Ψ (s, b) ds b.
(h2) max0t 1 α G (t , s)q(s)Ψˆ (s, a) ds a, max0t 1
1 0
ˆ s, b) ds b. G (t , s)q(s)Φ(
Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a u ∗ b. Theorem 5.2. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied:
ˆ ψ( ˆ ˆ b) b B. (i1) ϕˆ (a) (a − ˆI (a)) A, ˆ ˆ ϕˆ (b) (b − ˆI (b)) A. ˆ a) a B, (i2) ψ( Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a u ∗ b. Theorem 5.3. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist three positive numbers a < b < c such that one of the following conditions is satisfied:
ˆ ψ( ˆ ˆ ϕˆ (c ) (c − ˆI (c )) A. ˆ b) > b B, (j1) ϕˆ (a) (a − ˆI (a)) A, ˆ ψ( ˆ ϕˆ (b) < (b − ˆI (b)) A, ˆ ˆ a) a B, ˆ c ) c B. (j2) ψ( Then problem ( Pˆ ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a u ∗1 < b < u ∗2 c.
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Theorem 5.4. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist four positive numbers a < b < c < d such that one of the following conditions is satisfied:
ˆ ψ( ˆ ψ( ˆ ϕˆ (c ) < (c − ˆI (c )) A, ˆ ˆ b) > b B, ˆ d) d B. (k1) ϕˆ (a) (a − ˆI (a)) A, ˆ ψ( ˆ ˆ ϕˆ (b) < (b − ˆI (b)) A, ˆ ϕˆ (d) (d − ˆI (d)) A. ˆ a) a B, ˆ c ) > c B, (k2) ψ( Then problem ( Pˆ ) has at least three strictly increasing positive solutions u ∗1 , u ∗2 , u ∗3 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that
a u ∗1 < b < u ∗2 < c < u ∗3 d. Imitating the proofs of Theorems 4.1–4.5, we can obtain the following existence results. Theorem 5.5. Suppose that (1) The assumptions (H1) –(H3) and (H5) are satisfied. 1 ˆ s, a) ds a. (2) There exists a > 0 such that max0t 1 0 G (t , s)q(s)Φ( Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗ a. Theorem 5.6. Suppose that (1) The assumptions (H1) –(H5) are satisfied. ˆ (2) There exists a > 0 such that ϕˆ (a) (a − ˆJ (a)) A. Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗ a. Theorem 5.7. Suppose that (1) The assumptions (H1) –(H3) and (H5) are satisfied. 1 ˆ s, a) ds < a and max0t 1 β G (t , s)q(s)Ψˆ (s, b) ds b. (2) There exist 0 < a < b such that max0t 1 0 G (t , s)q(s)Φ( α Then problem ( Pˆ ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < a < u ∗2 b. Theorem 5.8. Suppose that (1) The assumptions (H1) –(H5) are satisfied. ˆ ψ( ˆ ˆ b) b B. (2) There exist 0 < a < b such that ϕˆ (a) < (a − ˆJ (a)) A, Then problem ( Pˆ ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < a < u ∗2 b. 6. Several remarks Remark 6.1 shows that our work extends Theorem 1.1. Remark 6.1. Theorem 1.1 is a corollary of Theorem 5.7. First of all, the assumptions (H1) and (H2) are satisfied by (a1)–(a2). Let 0 < r1 < r2 . By (a2)–(a4), we have r2 r1 (t )
γ
h (u ) 2 = max g (u ) + h(u ): r1 t u r2 = max g (u ) 1 + : r1 t u r2 g (u ) h(r2 ) h (u ) +1 . : r1 t 2 u r2 M g (r1 ) g t 2 max g r1t 2 1 + g (u ) g (r2 )
2
By the condition (a2),
1
r2 r1 (t ) dt
q(t )γ 0
1 h(r2 ) +1 M g (r1 ) q(t ) g t 2 dt < +∞. g (r2 ) 0
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So the assumption (H3) is satisfied. Since q ∈ C (0, 1), q(t ) > 0, 0 < t < 1, and the condition (a5), we see that the assumption (H5) is satisfied. Secondly, according to the conditions (a2)–(a4), we have
h (u ) 2 2 ˆ : at u a Φ(t , a) = max g (u ) + h(u ): at u a = max g (u ) 1 + g (u ) h(a) h (u ) + 1 = M g t 2 g (a) + h(a) . : at 2 u a M g (a) g t 2 max g at 2 1 + g (u ) g (a)
From the condition (a6), we get
1 max
0t 1
ˆ s, a) ds M g (a) + h(a) max G (t , s)q(s)Φ(
1
0
a
G (t , s)q(s) g s2 ds g (a) + h(a) ·
0t 1
g (a) + h(a)
0
= a.
α t β , we have
q(t )Ψˆ (t , b) = min q(t ) f (u ): bt 2 u b min θ(t ) g (u ) + h(u ) : bt 2 u b h (u ) h(bα 2 ) θ(t ) g (b) 1 + : bt 2 u b = θ(t ) min g (u ) 1 + g (u ) g (bα 2 ) 2 2 g (b) g (bα ) + g (b)h(bα ) = θ(t ). g (bα 2 )
Thirdly, according to (a1)–(a5), for
Applying the condition (a7), we get
β max
0t 1
G (t , s)q(s)Ψˆ (s, a) ds
g (b) g (bα 2 ) + g (b)h(bα 2 ) g (bα 2 )
α
β max
0t 1
g (b) g (bα 2 ) + g (b)h(bα 2 ) g (bα 2 )
·
G (t , s)θ(s) ds α
bg (bα 2 ) g (b) g (bα 2 ) + g (b)h(bα 2 )
= b.
By Theorem 5.7, the problem ( Pˆ ) has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ C 3 [0, 1] ∩ C 4 (0, 1) and 0 < u ∗1 < a < u ∗2 b. Remarks 6.2 and 6.3 show that our work improves the method used in papers [8,9,15]. Remark 6.2. The following example shows that the existence of twin positive solutions is possible for the problem ( Pˆ ) even if f (t , u ) = f (u ) and the assumption (a4) is not satisfied. Consider the singular problem
⎧ ⎨ u (4) (t ) = √ 1 + 36e −u (t ) + 1 min u 2 (t ), 15625000 , 3 4 u (t ) u (t ) ⎩ u (0) = u (0) = u (1) = u (1) = 0.
0 < t < 1,
In this example, q(t ) ≡ 1, f (t , u ) = f (u ) = g (u ) + h(u ), where h(u ) = 36e −u + singular at u = 0. ˆ = 8, Bˆ = Let α = 23 , β = 1. Then A
1 4
min{u 2 ,
15625000 }, u
1 g (u ) = √ 3 . So f (u ) is u
≈ 11.9264. Obviously, the assumptions (H1) –(H5) are satisfied. By Lemma 2.2(5), max0t ,s1 G (t , s) = G (1, 1) = 13 . So
1
ˆJ (8) = max
G (t , s) g 8s
0t 1
2
1944 163
ds
1
1
2 0t ,s1
0
0
We have 2502 = 62500 =
15625000 . 250
1
ds
√ 3
max G (t , s)
s2
1
= . 2
Direct computations give that
ϕˆ (8) = max 36e−u + u 2 : 0 u 8 36 + 16 = 52 < 60 8 − ˆJ (8) Aˆ , 4
ˆ 243) = min f (u ): ψ( =
1 4
2 3
1 t 1, 243t 2 u 243 = min f (u ): 108 u 243 min u 2 : 108 u 243
· 108 = 2916 > 2898.11 ≈ 243 Bˆ . 2
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By Theorem 5.8, the problem has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < 8 < u ∗2 < 243. h (u )
h (u )
√ = 0, we see that g (u ) is not nondecreasing. So the assumption (a4) is not Since limu →+∞ g (u ) = limu →+∞ 15625000 4 4 u3 satisfied. Therefore, the existence conclusion cannot be derived from Theorem 1.1.
Remark 6.3. The following example shows that the existence of twin positive solutions is possible for the problem ( Pˆ ) even if f (t , u ) = f (u ) and the assumption (a5) is not satisfied. Consider the singular problem
⎧ ⎨ ⎩
u
(4 )
3(1 + cos(u (t )π ))
(t ) =
16
+ max 0,
2420000 − 8(u (t ) − 750)2
4840000
u 2 (t ) +
2[1 + cos(u (t )π )]
√ 4
u (t )
,
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0.
In this example, q(t ) ≡ 1, f (t , u ) = f (u ) = g (u ) + h(u ), where
h (u ) =
3(1 + cos(u π )) 16
2420000 − 8(u − 750)2 + max 0, u2 , 4840000
So f (u ) is singular at u = 0. ˆ = 8, Bˆ = Let α = 12 , β = 1. Then A Obviously, g (u ) 2, 0 < u
1 2
g (u ) =
2[1 + cos(u π )]
√ 4
u
.
384 41
≈ 9.3659. 4 and g (u ) √ 4 , 0 < u < +∞. Hence the assumptions (H1) –(H5) are satisfied. Moreover, u
we have
1
ˆJ (16) = max
1
G (t , s) g 16s2 ds 2 max G (t , s)
0t 1 0
ds
4
s
3
√ = .
0t ,s1
0
Since h(u ) 38 u 2 , 0 u 16 and h(u ) > 18 u 2 , 300 u 1200, we get
4 16 − ˆJ (16) Aˆ , · 162 = 96 < 8 16 − 8 8 3 1200 1 2 ˆ 1200) = min f (u ): ψ( u 1200 min u : 300 u 1200
ϕˆ (16) = max h(u ): 0 u 16 max 4
=
1 8
3
u 2 : 0 u 16 =
3
8
· 3002 = 11250 > 11239.08 ≈ 1200 Bˆ .
By Theorem 5.8, the problem has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < 16 < u ∗2 < 1200. Since f (2k − 1) = 0, k = 1, 2, . . . , 99 and k = 651, 652, . . . , the condition (a5) is not satisfied. Therefore, the existence conclusion cannot be derived from Theorem 1.1. The following Remarks 6.4–6.6 imply that our results are applicable to more complicated cases. Remark 6.4. Our results is applicable to the general problems ( P ) in which the nonlinear term contains the unknown function and its first derivative. For example, consider the general problem
⎧ ⎨ ⎩
u (4) (t ) =
u (t )u (t )
cos2 (u (t ) + u (t ))
3 + 2 cos2 (u (t ) + u (t ))
+√ 6
u (0) = u (0) = u (1) = u (1) = 0.
t (1 − t )u (t )u (t )
,
0 < t < 1,
In the example, q(t ) ≡ 1, f (t , u , v ) = g (t , u , v ) + h(t , u , v ), where
h(t , u , v ) =
uv 3 + 2 cos2 (u + v )
,
cos2 (u + v ) g (t , u , v ) = √ . 6 t (1 − t )uv
So f (t , u , v ) is singular at t = 0, t = 1 and u = v = 0. If 0 < r1 < r2 , then
√ 3
2 g (t , u , v ) , 6 2 4 r1 t (1 − t )
1 4
r1 t 2 u r2 , r1 t v r2 .
Hence the assumptions (H1)–(H3) are satisfied. In addition, let
α = 12 , β = 1, then A = 6, B =
48 . 7
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By Lemma 2.2(3), max0t ,s1 ∂∂t G (t , s) = 12 . So
∂ I (r ) max G (t , s) 0t ,s1 ∂ t
√ 1 √ 3 3 2ds 3 2 = . √ √ 6 3 2 r
r 2 s4
0
3 From this, I (4) √ 3 . Moreover, we have 2
ϕ (4) = max
2
uv
2
3 + 2 cos2 (u + v )
: 0 t 1, t u 4, 4t v 4
3 < 11.2378 ≈ 4 4 − √ 4 − I (4) A , 3 3 2 2 1 uv 2 ψ(640) min : t 1 , 160t u 640 , 640t v 640 3 + 2 cos2 (u + v ) 2
16
1 5
· 40 · 320 = 2560 > 2194.29 ≈ 640B .
By Theorem 3.2, the problem has a positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) and 4 < |||u ∗ ||| < 640. In this problem, the nonlinear term contains unknown function and its first derivative, and nonlinear term is singular at t = 0, t = 1 and u = 0, v = 0. Hence above existence conclusion cannot be derived by the method used in [15,22,24,31]. Remark 6.5. In this paper, we allow not only limmax{u , v }→0 min0
⎧ ⎨ u (4) (t ) = 1 + sin u (t )u (t ) + ⎩
1 u (t )u (t )
1
+ u (t )u (t ), 4
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0.
Here q(t ) ≡ 1, f (t , u , v ) = f (u , v ) = g (u , v ) + h(u , v ), where
h (u , v ) =
1 4
g (u , v ) = 1 + sin uv +
uv ,
1 uv
.
So the limit limmax{u , v }→0 min0
If r > 0, then
ψ(b)
7290000 432
= 16875 > 16823.08 ≈ Bb.
By Theorem 3.2, the problem has a positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) and 2 |||u ∗ ||| < 2700. Because g (u , v ) is not nonincreasing with respect to u and v, the conclusion cannot be derived by applying the method used in [28–30,32]. Remark 6.6. For ease of representation, we use the following symbols
D [r1 , r2 ] = (t , u , v ): 0 t 1,
D ◦ [r1 , r2 ] = (t , u , v ): 0 < t < 1,
1 4
r1t 2 u r2 , r1 t v r2 ,
1 4
r1t 2 u r2 , r1 t v r2 .
Theorems 3.1 and 3.2 show that the existence of positive solution depends only upon the properties of nonlinear term f (t , u , v ) on the set D [a, b] and is independent of the states of f (t , u , v ) outside D [a, b]. The following example illustrates this. Consider the discontinuous boundary value problem
( Q 1)
u (4) (t ) = f 1 t , u (t ), u (t ) ,
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0,
Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
where
153
⎧ cot t 2 −136u , (t , u , v ) ∈ ((0, 1) × (0, 3]2 )\ D ◦ [ 19 , 3], ⎪ ⎪ ⎨ 2(1+t ) √ , (t , u , v ) ∈ D ◦ [ 19 , 3], f 1 (t , u , v ) = 4 t (u + v 2 ) ⎪ ⎪ ⎩ cot( v − 3)π , (t , u , v ) ∈ (0, 1) × ((0, +∞)2 \(0, 3]2 ).
In the example, the nonlinear term f 1 (t , u , v ) is very discontinuous and is not nonnegative outside D [ 19 , 3]. Therefore, the assumptions (H1)–(H3) are not satisfied. So we cannot directly obtain the existence of positive solution by the help of the existence theorems in this paper. Now, consider the singular boundary value problem
⎧ ⎨ u (4) (t ) = 2(1 + t ) , 0 < t < 1, 4 t (u (t ) + (u )2 (t )) ⎩ u (0) = u (0) = u (1) = u (1) = 0.
( Q 2)
2(1+t ) In this example, q(t ) ≡ 1, f 2 (t , u , v ) = g (t , u , v ) = √ 4 2
t (u + v )
and h(t , u , v ) ≡ 0. So f 2 (t , u , v ) is singular at t = 0 and u =
v = 0. Obviously, the assumptions (H1)–(H3) are satisfied. Let α = 23 , β = 1. According to Lemma 2.2(3) and (4), min2/3t ,s1 ∂∂t G (t , s) tion, for any r > 0, we have
2(1 + t )
Φ(t , r ) = max 4
t (u +
v 2)
2(1 + t )
Ψ (t , r ) = min 4
t (u + v 2 )
:
:
1 4
8 81
and max0t ,s1 ∂∂t G (t , s) = 12 . In addi-
√
1+t , rt u r , rt v r √ √ 4 4 2 r + 4r t3 2 2
2
2 1+t . rt 2 u r , rt v r √ √ 4 4 2 4 t r +r 1
It follows that
1 max
0t 1 2/3
1 max
0t 1
1 ∂ 1 ∂ 1+s 1 4 10 ds 2 ds ≈ 0.1168 > , G (t , s)h(s)Ψ s, min G (t , s) √ 4 ∂t 9 81 2/3t ,s1 ∂ t 9 s 2/3
∂ ∂ 4 4 G (t , s)h(s)Φ(s, 3) ds 2 max G (t , s) ∂t 39 0t ,s1 ∂ t
0
1
1+s
√ 4
0
By Theorem 3.1, the problem ( Q 2) has one positive solution u ∗ ∈ K and Since u ∗
∈
K [ 91 , 3],
we have
1 2 t 36
u ∗ (t ) 3, 19 t
(u ∗ ) (t ) 3, 0 < t
f 1 (t , u , v ) = f 2 (t , u , v ), (t , u , v ) ∈ D [ 19 , 3], we have
f 1 t , u ∗ (t ), u ∗ (t ) = f 2 t , u ∗ (t ), u ∗ (t ) ,
s3
ds ≈ 2.7164 < 3.
|||u ∗ ||| 3. < 1. So (t , u ∗ (t ), (u ∗ ) (t )) ∈ D [ 19 , 3], 0 < t < 1. Since 1 9
0 < t < 1.
Therefore, u ∗ (t ) is positive solution of the problem
( Q 1), too. The example illustrates that the strong force imposed outside the set D [a, b] cannot change the stability of positive solution even if the nonlinear term is singular. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
R.P. Agarwal, Boundary Value Problems for Higher Order Differential Equations, World Scientific, Singapore, 1986. A.R. Aftabizadeh, Existence and uniqueness theorems for fourth-order boundary value problems, J. Math. Anal. Appl. 116 (1986) 415–426. C.P. Gupta, Existence and uniqueness theorems for the bending of an elastic beam equation, Appl. Anal. 26 (1988) 289–304. R.P. Agarwal, J. Henderson, Supperlinear and sublinear focal boundary value problems, Appl. Anal. 60 (1996) 189–200. P.J. Wong, R.P. Agarwal, On two-point right focal eigenvalue problems, Z. Anal. Anwend. 17 (1998) 691–713. R.P. Agarwal, D. O’Regan, Right focal singular boundary value problems, Z. Angew. Math. Mech. 79 (1999) 363–373. R.P. Agarwal, D. O’Regan, V. Lakshmikantham, Singular ( p , n − p ) focal and (n, p ) higher order boundary value problems, Nonlinear Anal. 42 (2000) 215–228. R.P. Agarwal, D. O’Regan, Twin solutions to singular boundary value problems, Proc. Amer. Math. Soc. 128 (2000) 2085–2094. R.P. Agarwal, Multiplicity results for singular conjugate, focal and (n, p ) problems, J. Differential Equations 170 (2001) 142–156. R. Ma, Multiple positive solutions for a semipositone fourth-order boundary value problem, Hiroshima Math. J. 33 (2003) 217–227. Q. Yao, Monotonically iterative method of nonlinear cantilever beam equations, Appl. Math. Comput. 205 (2008) 432–437. Q. Yao, Existence and multiplicity of positive solutions to a class of nonlinear cantilever beam equations, J. Systems Sci. Math. Sci. 29 (2009) 63–69 (in Chinese). R.P. Agarwal, F.H. Wong, Existence of positive solutions for non-positive higher-order BVPs, J. Comput. Appl. Math. 88 (1998) 3–14.
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R.P. Agarwal, F.H. Wong, Existence of solutions to (k, n − k − 2) boundary value problems, Appl. Math. Comput. 104 (1999) 33–50. R.P. Agarwal, D. O’Regan, Twin solutions to singular Dirichlet problems, J. Math. Anal. Appl. 240 (1999) 433–445. Q. Yao, Z. Bai, Existence of positive solutions of BVP for u (4) (t ) − λh(t ) f (u (t )) = 0, Chinese Ann. Math. Ser. A 20 (1999) 575–578 (in Chinese). P.J.Y. Wong, R.P. Agarwal, Multiple solutions for a system of (ni , p i ) boundary value problems, Z. Anal. Anwend. 19 (2000) 511–528. D.R. Anderson, J.M. Davis, Multiple solutions and eigenvalues for third-order right focal boundary value problems, J. Math. Anal. Appl. 267 (2002) 135–157. Q. Yao, The existence and multiplicity of positive solutions for a third-order three-point boundary value problem, Acta Math. Appl. Sin. Engl. Ser. 19 (2003) 117–122. P.J. Torres, Existence of one-signed periodic solutions of some second-order differential equations via a Krasnoselskii fixed point theorem, J. Differential Equations 190 (2003) 643–662. Q. Yao, Positive solutions for eigenvalue problems of fourth-order elastic beam equations, Appl. Math. Lett. 17 (2004) 237–243. K.Q. Lan, Multiple positive solutions of conjugate boundary value problems with singularities, Appl. Math. Comput. 147 (2004) 461–474. Q. Yao, Existence of n solutions and/or positive solutions to a semipositone elastic beam equation, Nonlinear Anal. 66 (2007) 138–150. Z. Liu, J.S. Ume, S.M. Kang, Positive solutions of a singular nonlinear third order two-point boundary value problem, J. Math. Anal. Appl. 326 (2007) 589–601. J.R.L. Webb, G. Infante, D. Franco, Positive solutions of nonlinear fourth-order boundary value problems with local and nonlocal boundary conditions, Proc. Roy. Soc. Edinburgh Sect. A 138 (2008) 427–446. Q. Yao, Positive solutions of nonlinear elastic beam equation rigidly fastened on the left and simply supported on the right, Nonlinear Anal. 69 (2008) 1570–1580. Q. Yao, Existence and multiplicity of positive solutions to a singular elastic beam equation rigidly fixed at both ends, Nonlinear Anal. 69 (2008) 2683–2694. Q. Yao, Positive solution of singular third-order three-point boundary value problems, J. Math. Anal. Appl. 354 (2009) 207–212. D. O’Regan, Solvability of some fourth (and higher) order singular boundary value problems, J. Math. Anal. Appl. 161 (1991) 78–116. S. Stanek, Positive solutions of singular positone Dirichlet boundary value problems, Math. Comput. Modelling 33 (2001) 341–351. Q. Yao, Positive solution of a class of singular sublinear two-point boundary value problems, Acta Math. Appl. Sin. 24 (2001) 522–526 (in Chinese). Z. Wei, A class of fourth order singular boundary value problems, Appl. Math. Comput. 153 (2004) 865–884.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Local existence of multiple positive solutions to a singular cantilever beam equation Qingliu Yao Department of Applied Mathematics, Nanjing University of Finance and Economics, Nanjing 210003, PR China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 30 April 2009 Available online 28 July 2009 Submitted by D.L. Russell
By constructing suitable cone and control functions, we prove some local existence theorems of positive solutions for a singular fourth-order two-point boundary value problem. In mechanics, the problem is called cantilever beam equation. Furthermore, we improve a famous method appeared in the studies of singular boundary value problems. The approximation theorem of completely continuous operators and the Guo–Krasnosel’skii fixed point theorem of cone expansion–compression type play important parts in this work. © 2009 Elsevier Inc. All rights reserved.
Keywords: Singular ordinary differential equation Boundary value problem Positive solution Existence Multiplicity
1. Introduction The purpose of this paper is to study the local existence and multiplicity of positive solutions for the following nonlinear fourth-order two-point boundary value problem
(P )
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0,
and its simplified form
( Pˆ )
u (4) (t ) = q(t ) f t , u (t ), u (t ) ,
u (4) (t ) = q(t ) f t , u (t ) ,
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0,
where q : (0, 1) → [0, +∞) is continuous and q(t ) > 0, 0 < t < 1. In this paper, the function u ∗ ∈ C [0, 1] is called positive solution of the problem ( P ) if u ∗ is a solution of ( P ) and u ∗ (t ) > 0, 0 < t 1. In addition, let 0 < α < β 1 and u = max0t 1 |u (t )| for u ∈ C [0, 1]. We consider singular problem ( P ). In other words, the nonlinear term f (t , u , v ) is allowed to be singular at t = 0, t = 1 and u = 0, v = 0. Because of the extensive applications in mechanics and engineering [1–3], the nonlinear fourth-order boundary value problems have received wide attentions. In mechanics, the problem ( P ) is called cantilever beam equation. The equation describes the deflection of an elastic beam fixed at the left and freed at the right. There are some papers discussed the existence of solutions of ( P ) [4–12]. Because the problem ( P ) is a (2, 4 − 2) focal boundary value problem, most results on the problem ( P ) appear in the literature on the ( p , m − p ) focal boundary value problems.
E-mail address: [email protected]. 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.07.043
©
2009 Elsevier Inc. All rights reserved.
Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
139
When f (t , u , v ) = f (t , u ), some existence theorems of positive solutions have been established in the papers [4–9] for the problem ( Pˆ ). For example, the following local existence theorem is a special case of Theorem 4.2 in Agarwal [9]. Theorem 1.1. Assume that (a1) f : [0, 1] × (0, +∞) → [0, +∞) is continuous. (a2) f (t , u ) g (u ) + h(u ), (t , u ) ∈ [0, 1] × (0, +∞), where h : [0, +∞) → [0, +∞) is continuous, g : (0, +∞) → (0, +∞) is 1 continuous and 0 q(t ) g (t 2 ) dt < +∞. (a3) g (u ) is nonincreasing on (0, +∞) and there exists constant M > 0 such that
g (u 1 u 2 ) M g (u 1 ) g (u 2 ),
(u 1 , u 2 ) ∈ (0, +∞) × (0, +∞).
(a4) hg((uu)) is nondecreasing on (0, +∞). (a5) There exists θ ∈ C [α , β] such that θ(t ) > 0, α t β and
q(t ) f (t , u ) θ(t ) g (u ) + h(u ) ,
(t , u ) ∈ [α , β] × [0, +∞).
(a6) There exists a > 0 such that
1
G (t , s)q(s) g s2 ds
M max
0t 1 0
a g (a) + h(a)
.
(a7) There exists b > a such that
β G (t , s)θ(s) ds
max
0t 1
α
bg (bα 2 ) g (b) g (bα 2 ) + g (b)h(bα 2 )
.
Then problem ( Pˆ ) has two positive solutions u ∗1 , u ∗2 ∈ C 3 [0, 1] ∩ C 4 (0, 1) and 0 < u ∗1 < a < u ∗2 b. In Theorem 1.1, G (t , s) is the Green function of related homogeneous linear problem, see Section 2. The remarkable characterization of Theorem 1.1 is that the nonlinear term f (t , u ) is allowed to be singular at u = 0. Theorem 1.1 obtains not only the existence of positive solutions, but presents also an effective method in order to study various singular boundary value problems. However, the assumptions (a4) and (a5) are very stern. For example, the assumption (a5) implies that f (t , u ) > 0, α t β , 0 < u < +∞. Therefore, Theorem 1.1 is unapplicable to some singular problems ( Pˆ ). The purpose of this paper is to extend Theorem 1.1 and improve the method used in the theorem. Moreover, we will cancel the assumptions (a3)–(a5). The motivation comes from the localization method arisen in papers [8,9,12–25]. By the method, we can determine the existence of positive solutions by considering the properties of nonlinear term on some bounded sets. Consequently, the localization method is a powerful tool for many nonlinear boundary value problems. Very recently, in papers [26–28], we consider some conditions that ensure applications of the localization method to singular boundary value problems. In this paper, we will improve the localization method by applying the results in [26–28]. After that, we will study the local existence and multiplicity of positive solutions of the problem ( P ) by the improved method. In this paper, we will use the following assumptions. (H1) f : (0, 1) × (0, +∞) × (0, +∞) → [0, +∞) is continuous. (H2) f (t , u , v ) g (t , u , v ) + h(t , u , v ), (t , u , v ) ∈ (0, 1) × (0, +∞) × (0, +∞), where g : (0, 1) × (0, +∞) × (0, +∞) → [0, +∞) is continuous, h : [0, 1] × [0, +∞) × [0, +∞) → [0, +∞) is continuous. 1 r (H3) 0 q(t )γr12 (t ) dt < +∞ for any 0 < r1 < r2 , where
r2 r1 (t )
γ
= max g (t , u , v ) + h(t , u , v ):
1 4
r1 t u r2 , r1 t v r2 . 2
(H4) g (t , u 1 , v 1 ) g (t , u 2 , v 2 ), 0 < t < 1, 0 < u 1 u 2 , 0 < v 1 v 2 . (H5) There exist α μ < ν β , r¯ > 0 and η > 0 such that
f (t , u , v ) η,
(t , u , v ) ∈ [μ, ν ] × (0, r¯] × (0, r¯]. r
The assumption (H2) implies that f (t , u , v ) may be singular at u = 0, v = 0, t = 0, t = 1 and γr12 (t ) is continuous on (0, 1). The outline of this paper is as follows. In Section 2, we construct a suitable cone and transfer the problem ( P ) into an integral equation. After that, we prove that the related integral operator is completely continuous on some proper sets
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Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
by using the approximation theorem of completely continuous operators. In Section 3, we establish the local existence of n positive solutions under the assumptions (H1)–(H3) by applying the Guo–Krasnosel’skii fixed point theorem of cone expansion–compression type, where n is an arbitrary positive integer. In Section 4, we study the problem ( P ) under the assumptions (H1)–(H5). In Section 5, we consider the simplified problem ( Pˆ ). In Section 6, we verify that Theorem 1.1 is a corollary of main results. Moreover, five examples illustrate that our extension is true and our results are applicable to more complicated cases. All results are independent of the existence of upper and lower solutions. 2. Preliminaries Let C 1 [0, 1] be the Banach space with norm |||u ||| = max{u , u } and let K be the cone of nonnegative functions in C 1 [0, 1] with following form
K = u ∈ C 1 [0, 1]: u (0) = 0, u (t )
1 2
u t 2 , u (t ) u t , 0 t 1 .
Write Ω(r ) = {u ∈ K : |||u ||| < r }, ∂Ω(r ) = {u ∈ K : |||u ||| = r }. Lemma 2.1. If u ∈ K , then u = |||u ||| and
1 |||u ||| 2
u |||u |||.
Proof. Since u (t ) 0 and u (0) = 0, we have
t u = max
1
u (s) ds =
0t 1 0
1
u (s) ds 0
max u (s) ds = u .
0s1 0
Since u (t ) u t, 0 t 1, we have
1 u =
1
u (s) ds u 0
s ds =
1 2
u .
0
Hence u = |||u ||| and
1 |||u ||| 2
u |||u |||. 2
Let G (t , s) be the Green function of the homogeneous linear problem u (4) (t ) = 0, u (0) = u (0) = u (1) = u (1) = 0, that is,
1 G (t , s) =
(3t 2 s 6 1 (3ts2 6
− t 3 ), 0 t s 1, − s3 ), 0 s t 1.
Computing the partial derivatives of G (t , s) with respect to t, we have
∂ G (t , s) = ∂t
1
s2 − 12 (s 2 1 2 s , 2
s − t, ∂ G (t , s) = ∂t2 0, 2
− t )2 , 0 t s 1, 0 s t 1,
−1, ∂ G ( t , s ) = ∂t3 0, 3
0 t s 1, 0 s < t 1.
Lemma 2.2. (1) G , ∂∂t G : [0, 1] × [0, 1] → [0, +∞) are continuous. (2) G (t , s) > 0, ∂∂t G (t , s) > 0, 0 < t , s < 1. (3) max0t 1 ∂∂t G (t , s) = ∂∂t G (s, s) = 12 s2 , 0 s 1. (4) ∂∂t G (t , s) t ∂∂t G (s, s), 0 t , s 1. (5) max0t 1 G (t , s) = G (1, s) = 16 s2 (3 − s), 0 s 1. (6) G (t , s) t 2 G (1, s), 0 t , s 1. 2 3 (7) G (0, s) ≡ 0, ∂∂t G (0, s) ≡ 0, ∂∂t 2 G (1, s) ≡ 0, ∂∂t 3 G (1, s) ≡ 0.
Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
141
2
Proof. The proofs are direct.
Define the operator T as follows
1 ( T u )(t ) =
G (t , s)q(s) f s, u (s), u (s) ds,
0 t 1.
0
We will transfer the problem ( P ) into the integral equation
1 u (t ) =
G (t , s)q(s) f s, u (s), u (s) ds,
0 t 1,
0
and prove that the integral operator T is completely continuous on the each bounded closed set of K \{0}. For this end, we need the following approximation theorem of completely continuous operators. Lemma 2.3. Let X be a Banach space, let V ⊂ X be a bounded closed subset, let T k : V → X , k = 1, 2, . . ., be a sequence of completely continuous operators. If the operator T : V → X and supu ∈ V T k u − T u → 0 (k → ∞), then T : V → X is a completely continuous operator. Lemma 2.4 gives basic properties of the operator T . Lemma 2.4. Suppose that (H1)–(H3) are satisfied and 0 < r1 < r2 . Then (1) T : Ω(r2 )\Ω(r1 ) → K is completely continuous. (2) For any u ∈ Ω(r2 )\Ω(r1 ),
t 1 ( T u )(t ) = 0
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds, ∂s
0 t 1.
0
γrr12 (t ) as in the assumption (H3). For every k = 2, 3, . . . , let ⎧ r r ⎪ min{γr12 (t ), kt γr12 ( 1k )}, 0 t 1k , ⎪ ⎨ r2 r2 1 γr1 k (t ) = γr1 (t ), t k−k 1 , k ⎪ ⎪ ⎩ r2 r 2 k −1 min{γ (t ), k(1 − t )γ ( )}, k−1 t 1.
Proof. Let
r1
r
r1
r
k
k
r
Then [γr12 ]k ∈ C [0, 1] and [γr12 ]k (0) = [γr12 ]k (1) = 0. By the assumption (H3),
1
q(t )
1 0
r
q(t )γr12 (t ) dt < +∞. So
γrr12 (t ) − γrr12 k (t ) dt → 0 (k → ∞).
0
Construct the function f k (t , u , v ) as follows, for 0 t 1,
f k (t , u , v ) = min f (t , u , v ),
γrr12 k (t ) ,
1 4
r1 t 2 u < ∞, r1 t v < +∞;
⎧ 1 ⎪ f (t , u , r1 t ), r t 2 u < ∞, 0 v r1 t , ⎪ 4 1 ⎨ k f k (t , u , v ) = f k (t , 14 r1 t 2 , v ), 0 u 14 r1 t 2 , r1 t v < +∞, ⎪ ⎪ ⎩ f (t , 1 r t 2 , r t ), 0 u 1 r t 2 , 0 v r t . 1 1 k 4 1 4 1
By the definition, f k : [0, 1] × [0, +∞) × [0, +∞) → [0, +∞) is continuous and f k (0, u , v ) ≡ 0, f k (1, u , v ) ≡ 0. By (H2) and (H3), for any u ∈ K ,
1 0
q(t ) f k t , u (t ), u (t ) dt
1
r
q(t )γr12 (t ) dt < +∞. 0
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Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
Define the operator T k as follows
1 ( T k u )(t ) =
G (t , s)q(s) f k s, u (s), u (s) ds,
0 t 1.
0
Then T k : K → C [0, 1] is completely continuous by the continuity of f k (t , u , v ) and the Arzela–Ascoli theorem. And, for 0 t 1,
1
( T k u ) (t ) =
∂ G (t , s)q(s) f k s, u (s), u (s) ds. ∂t
0
Hence, ( T k (·))
: K → C [0, 1] is a completely continuous operator, too. Let u ∈ Ω(r2 )\Ω(r1 ). Then r1 |||u ||| r2 . From Lemma 2.1, we get 1 4
r1t 2
1
r1 t u t u (t ) r2 ,
u t 2 u (t ) r2 ,
2
1 r t2 4 1
By the definition of f k (t , u , v ), for 0 < t < 1, r
0 f (t , u , v ) − f k (t , u , v ) γr12 (t ) −
1 r 2 1
u r2 , r1 u r2 . Thus,
0 t 1.
u r2 , r1 t v r2 ,
γrr12 k (t ).
It follows that
0 f t , u (t ), u (t ) − f k t , u (t ), u (t ) γr12 (t ) − r
γrr12 k (t ), 0 < t < 1.
Applying the above-mentioned results, we obtain that
1 sup u ∈Ω(r2 )\Ω(r1 )
T u − T k u =
sup u ∈Ω(r2 )\Ω(r1 )
max
0t 1
0
1
sup u ∈Ω(r2 )\Ω(r1 )
max
G (t , s)q(s)
0t 1
γrr12 (s) − γrr12 k (s) ds
0
1 max G (t , s)
q (s)
0t ,s1
γrr12 (s) − γrr12 k (s) ds → 0.
0
By Lemma 2.3, T : Ω(r2 )\Ω(r1 ) → C [0, 1] is a completely continuous operator. Similarly,
1 sup
max
u ∈Ω(r2 )\Ω(r1 )
0t 1
∂ G (t , s)q(s) f s, u (s), u (s) − f k s, u (s), u (s) ds → 0. ∂t
0
It follows that
1
( T u ) (t ) =
∂ G (t , s)q(s) f s, u (s), u (s) ds, ∂t
0 t 1,
0
and ( T (·))
: Ω(r2 )\Ω(r1 ) → C [0, 1] is completely continuous. By Lemma 2.2(7), we have ( T u )(0) = 0. So
t 1 ( T u )(t ) = 0
0
G (t , s)q(s) f s, u (s), u (s) − f k s, u (s), u (s)
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds, ∂s
0 t 1.
ds
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143
On the other hand, by Lemma 2.2(3) and (4),
t 1 ( T u )(t ) 0
∂ s G (τ , τ )q(τ ) f τ , u (τ ), u (τ ) dτ ∂s
0
1
2
1 max
2
0s1 0
1
= t 2 max
0t 1 0
( T u ) (t )
1 t
s ds max
0s1
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds ∂s
0
∂ 1 G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds t 2 ∂s 2
0
t 1
2
1
0
1
= t
t
1 1 0
∂ G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds ∂s
0
∂ 1 G (s, τ )q(τ ) f τ , u (τ ), u (τ ) dτ ds = T u t 2 , ∂s 2
0
∂ G (s, s)q(s) f s, u (s), u (s) ds t max ∂t 0t 1
0
1
∂ G (t , s)q(s) f s, u (s), u (s) ds = ( T u ) t . ∂t
0
It implies that T : Ω(r2 )\Ω(r1 ) → K . Therefore, T : Ω(r2 )\Ω(r1 ) → K is a completely continuous operator.
2
Our approach is based on the following Guo–Krasnosel’skii fixed point theorem of cone expansion–compression type. Lemma 2.5. Let X be a Banach space, and let K ⊂ X be a cone in X . Assume Ω1 , Ω2 are bounded open subset of K with 0 ∈ Ω1 ⊂ Ω 1 ⊂ Ω2 , and let F : Ω 2 \Ω1 → K be a completely continuous operator such that (1) F u u , u ∈ ∂Ω1 and F u u , u ∈ ∂Ω2 ; or (2) F u u , u ∈ ∂Ω1 and F u u , u ∈ ∂Ω2 . Then F has a fixed point in Ω 2 \Ω1 . 3. Local existence theorems under (H1)–(H3) We introduce the following control functions and constants:
Φ(t , r ) = max g (t , u , v ) + h(t , u , v ): Ψ (t , r ) = min f (t , u , v ):
1 4
ψ(r ) = min f (t , u , v ): α t β, A=
1 max
0t 1
4
2
rt u r , rt v r ,
rt 2 u r , rt v r ,
ϕ (r ) = max h(t , u , v ): 0 t 1,
1
∂ G (t , s)q(s) ds ∂t
1 4
rt 2 u r , rt v r ,
1 4
rt 2 u r , rt v r ,
−1 ,
0
B=
β max
0t 1
1 ζr (t ) = max g (t , u , v ): rt 2 u r , rt v r ,
α
∂ G (t , s)q(s) ds ∂t
−1 ,
4
1 I (r ) = max
0t 1
∂ G (t , s)q(s)ζr (s) ds. ∂t
0 6 If q(t ) ≡ 1, then A = 6, B = β 3 − . α3 1 If the assumption (H3) is satisfied, then 0 q(t )ζr (t ) dt < +∞ for any r > 0. We obtain the following local existence theorems. Among others, Theorem 3.1 is a basic existence criterion of positive solution for the problem ( P ).
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Theorem 3.1. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied: (b1) max0t 1
1
∂
0 ∂t
β G (t , s)q(s)Φ(s, a) ds a and max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, b) ds b.
β 1 (b2) max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, a) ds a and max0t 1 0 ∂∂t G (t , s)q(s)Φ(s, b) ds b.
Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗ ||| b. Proof. Without loss of generality, we prove only the case (b1). If u ∈ ∂Ω(a), then |||u ||| = a. So 12 a u a, u = a by Lemma 2.1. From this,
1
1
u t 2 u (t ) a, at u t u (t ) a, 0 t 1, 4 2 g t , u (t ), u (t ) + h t , u (t ), u (t ) Φ(t , a), 0 < t < 1. at 2
Since T u ∈ K and Lemma 2.4(2), we obtain
||| T u ||| = ( T u ) = max
1
0t 1
∂ G (t , s)q(s) f s, u (s), u (s) ds ∂t
0
1 max
0t 1
∂ G (t , s)q(s) g s, u (s), u (s) + h s, u (s), u (s) ds max ∂t 0t 1
0
1
∂ G (t , s)q(s)Φ(s, a) ds ∂t
0
a = |||u |||. If u ∈ ∂Ω(b), then |||u ||| = b. It means
1 b 2
u b, u = b. So,
1
bt u (t ) b, 0 t 1, bt 2 u (t ) b, 4 f t , u (t ), u (t ) Ψ (t , b), α t β. It follows that
||| T u ||| = ( T u ) max
0t 1
β α
∂ G (t , s)q(s) f s, u (s), u (s) ds max ∂t 0t 1
β α
∂ G (t , s)q(s)Ψ (s, b) ds b = |||u |||. ∂t
By Lemmas 2.4(1) and 2.5, the operator T has one fixed point u ∗ ∈ Ω(b)\Ω(a). It implies that
u ∗ (t ) =
t 1 0
∂ G (s, τ )q(τ ) f τ , u ∗ (τ ), u ∗ (τ ) dτ ds, ∂t
0 t 1.
0
Computing the first, second, third and fourth derivatives of u ∗ (t ), we get
u∗
u
( i )
1 (t ) =
∗ (4 )
∂i G (t , s)q(s) f s, u ∗ (s), u ∗ (s) ds, i ∂t
0 t 1, i = 1, 2, 3,
0
(t ) = q(t ) f t , u ∗ (t ), u ∗ (t ) ,
0 < t < 1.
According to Lemma 2.2(7), u ∗ (0) = (u ∗ ) (0) = (u ∗ ) (1) = (u ∗ ) (1) = 0. Therefore, u ∗ (t ) is a solution of the problem ( P ) and a |||u ∗ ||| b. Since u ∗ (t ) 14 at 2 > 0, 0 < t 1, the solution u ∗ (t ) is positive. Since (u ∗ ) (t ) at > 0, 0 < t 1, we see that u ∗ (t ) is strictly increasing on [0, 1]. u ∗ ∈ C 3 [0, 1] ∩ C 4 (0, 1) is obvious. 2 In the real problems, the computations of
1 max
0t 1 0
∂ G (t , s)q(s)Φ(s, r ) ds, ∂t
β max
0t 1
α
∂ G (t , s)q(s)Ψ (s, r ) ds ∂t
are more lengthy and tedious. For convenience, we give the following results.
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Theorem 3.2. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied: (c1) ϕ (a) (a − I (a)) A, ψ(b) bB. (c2) ψ(a) aB, ϕ (b) (b − I (b)) A. Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗ ||| b. Proof. Without loss of generality, we prove only the case (c1). By the definition of Φ(t , a), for 0 t 1,
Φ(t , a) max g (t , u , v ):
1 4
2
at u a, at v a + max h(t , u , v ):
1 4
2
at u a, at v a
ζa (t ) + ϕ (a). Applying the inequality, we get
1 max
0t 1
∂ G (t , s)q(s)Φ(s, a) ds max ∂t 0t 1
0
1
∂ G (t , s)q(s)ϕ (a) ds + max ∂t 0t 1
0
1
∂ G (t , s)q(s)ζa (s) ds ∂t
0
a − I (a) A max
1
0t 1
∂ G (t , s)q(s) ds + I (a) = a − I (a) A A −1 + I (a) = a. ∂t
0
By the definitions of Ψ (t , b), Ψ (t , b) ψ(b),
β max
0t 1
α
∂ G (t , s)q(s)Ψ (s, b) ds max ∂t 0t 1
By Theorem 3.1(b1), the proof is completed.
α t β . So β α
∂ G (t , s)q(s)ψ(b) ds bB max ∂t 0t 1
β α
∂ G (t , s)q(s) ds = b. ∂t
2
Theorem 3.3. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist three positive numbers a < b < c such that one of the following conditions is satisfied: (d1) ϕ (a) (a − I (a)) A, ψ(b) > bB, ϕ (c ) (c − I (c )) A. (d2) ψ(a) aB, ϕ (b) < (b − I (b)) A, ψ(c ) c B. Then problem ( P ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗1 ||| < b < |||u ∗2 ||| c. Proof. We prove only the case (d2). Applying the assumptions ψ(a) aB, ϕ (b) < (b − I (b)) A and imitating the proof of Theorem 3.2(c2), we assert that ( P ) has one strictly increasing positive solution u ∗1 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a |||u ∗1 ||| < b. Similarly, ( P ) has another positive solution u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) which is strictly increasing and b < |||u ∗2 ||| c. 2 In the same way, we can prove following theorems. Theorem 3.4. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist four positive numbers a < b < c < d such that one of the following conditions is satisfied: (e1) ϕ (a) (a − I (a)) A, ψ(b) > bB, ϕ (c ) < (c − I (c )) A, ψ(d) dB. (e2) ψ(a) aB, ϕ (b) < (b − I (b)) A, ψ(c ) > c B, ϕ (d) (d − I (d)) A. Then problem ( P ) has at least three strictly increasing positive solutions u ∗1 , u ∗2 , u ∗3 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that
a u ∗1 < b < u ∗2 < c < u ∗3 d.
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Denote the integer part of c by [c ]. Generally, we have the following existence theorem on n positive solutions. Theorem 3.5. Suppose that (1) The assumptions (H1)–(H3) are satisfied. (2) There exist n + 1 positive numbers a1 < a2 < · · · < an+1 such that one of the following conditions is satisfied:
ϕ (a2k−1 ) < (a2k−1 − I (a2k−1 )) A, k = 1, . . . , [ n+2 2 ], and ψ(a2k ) > a2k B, k = 1, . . . , [ n+2 1 ]. 2 ], and ϕ (a2k ) < (a2k − I (a2k )) A, k = 1, . . . , [ n+2 1 ]. (f2) ψ(a2k−1 ) > a2k−1 B, k = 1, . . . , [ n+ 2 (f1)
Then problem ( P ) has at least n strictly increasing positive solutions uk∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1), k = 1, 2, . . . , n, such that ak < |||uk∗ ||| < ak+1 . 4. The existence results under (H1)–(H5) In this section, let J (r ) = max0t 1
1
∂
0 ∂t
G (t , s)q(s) g (s, 14 rs2 , rs) ds.
If the assumption (H4) is satisfied, then ζr (t ) = g (t , 14 rt 2 , rt ), J (r ) is a nonincreasing function and J (r ) = I (r ). If limmax{u , v }→0 infμt ν g (t , u , v ) > 0, then the assumption (H5) is satisfied. Theorem 4.1. Suppose that (1) The assumptions (H1)–(H3) and (H5) are satisfied. 1 (2) There exists a > 0 such that max0t 1 0 ∂∂t G (t , s)q(s)Φ(s, a) ds a. Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗ ||| a. Proof. Let
ν = Cμ
ν max
0t 1
μ
∂ G (t , s)q(s) ds ∂t
−1
,
ν −1 1 . b = min r¯ , a, η C μ 2
By (H5), we have ν, f (t , u , v ) bC μ
μ t ν , 0 < u b, 0 < v b.
ν, It follows that Ψ (t , b) bC μ
β max
0t 1
α
μ t ν and
∂ G (t , s)q(s)Ψ (s, b) ds max ∂t 0t 1
ν μ
∂ ν C ν − 1 = b . G (t , s)q(s)Ψ (s, b) ds bC μ μ ∂t
By Theorem 3.1(b2), the problem ( P ) has one positive solution u ∗ ∈ K and b |||u ∗ ||| a.
2
Theorem 4.2. Suppose that (1) The assumptions (H1)–(H5) are satisfied. (2) There exists a > 0 such that ϕ (a) (a − J (a)) A. Then problem ( P ) has at least one strictly increasing positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗ ||| a. Proof. Since I (a) = J (a), we have
1 max
0t 1 0
ϕ (a) (a − J (a)) A = (a − I (a)) A. It follows that
∂ G (t , s)q(s)Φ(s, a) ds max ∂t 0t 1
1 0
∂ G (t , s)q(s)ϕ (a) ds + max ∂t 0t 1
1
∂ 1 G (t , s)q(s) g s, as2 , as ds ∂t 4
0
a − I (a) A A −1 + J (a) = a − I (a) + I (a) = a.
By Theorem 4.1, the problem ( P ) has one positive solution u ∗ ∈ K and b |||u ∗ ||| a.
2
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Theorem 4.3. Suppose that (1) The assumptions (H1)–(H3) and (H5) are satisfied. 1 β (2) There exist 0 < a < b such that max0t 1 0 ∂∂t G (t , s)q(s)Φ(s, a) ds < a and max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, b) ds b. Then problem ( P ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗1 ||| < a < |||u ∗2 ||| b.
β
Proof. By the proof of Theorem 4.1, there exists 0 < c < a such that max0t 1 α ∂∂t G (t , s)q(s)Ψ (s, c ) ds c. Imitating the proofs of Theorem 3.1, the problem ( P ) has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) and c |||u ∗1 ||| < a < |||u ∗2 ||| b. 2 Theorem 4.4. Suppose that (1) The assumptions (H1)–(H5) are satisfied. (2) There exist 0 < a < b such that ϕ (a) < (a − J (a)) A, ψ(b) bB. Then problem ( P ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗1 ||| < a < |||u ∗2 ||| b. Proof. By Theorem 4.2, ( P ) has one strictly increasing positive solutions u ∗1 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < |||u ∗1 ||| < a by applying the assumption (H5) and the inequality ϕ (a) < (a − J (a)) A. Since ϕ (a) < (a − J (a)) A and ψ(b) bB, we see that ( P ) has another strictly increasing positive solution u ∗2 ∈ K ∩ 3 C [0, 1] ∩ C 4 (0, 1) such that a < |||u ∗2 ||| b by imitating the proof of Theorem 3.2(c1). 2 Theorem 4.5. Suppose that (1) The assumptions (H1)–(H5) are satisfied. (2) There exist three positive numbers a < b < c such that
ϕ (a) < a − J (a) A ,
ψ(b) > bB ,
ϕ (c ) c − J (c ) A .
Then problem ( P ) has at least three strictly increasing positive solutions u ∗1 , u ∗2 , u ∗3 ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that
0 < u ∗1 < a < u ∗2 < b < u ∗3 c . Obviously, we can write similar results for arbitrary positive integer n. For short, we omit them. 5. The existence results on the simplified problem ( Pˆ ) In this section, we consider the simplified problem ( Pˆ ). We will use the following assumptions. (H1) f : (0, 1) × (0, +∞) → [0, +∞) is continuous. (H2) f (t , u ) g (t , u ) + h(t , u ), (t , u ) ∈ (0, 1) × (0, +∞), where g : (0, 1) × (0, +∞) → [0, +∞) is continuous, h : [0, 1] × [0, +∞) → [0, +∞) is continuous. 1 r (H3) 0 q(t )γˆr12 (t ) dt < +∞ for any 0 < r1 < r2 , where
γˆrr12 (t ) = max g (t , u ) + h(t , u ): r1t 2 u r2 . (H4) g (t , u 1 ) g (t , u 2 ), 0 < t < 1, 0 < u 1 u 2 . (H5) There exist α μ < ν β , r¯ > 0 and η > 0 such that
f (t , u ) η,
(t , u ) ∈ [μ, ν ] × (0, r¯].
The following control functions and constants are applied in this section.
ˆ t , r ) = max g (t , u ) + h(t , u ): rt 2 u r , Φ(
Ψˆ (t , r ) = min f (t , u ): rt 2 u r ,
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ϕˆ (r ) = max h(t , u ): 0 t 1, rt 2 u r ,
ˆ r ) = min f (t , u ): α t β, rt 2 u r , ψ( −1 −1 1 β ˆ = max A G (t , s)q(s) ds , Bˆ = max G (t , s)q(s) ds , 0t 1
0t 1
0
α
ζˆr (t ) = max g (t , u ): rt 2 u r , ˆI (r ) = max
1
0t 1
G (t , s)q(s)ζˆr (s) ds,
0
ˆJ (r ) = max
1
G (t , s)q(s) g s, rs2 ds.
0t 1 0
24 ˆ = 8, Bˆ = If q(t ) ≡ 1, then A . 4(β 3 −α 3 )−(β 4 −α 4 ) If the assumption (H4) is satisfied, then ζˆr (t ) = g (t , rt 2 ), ˆJ (r ) is nonincreasing and ˆI (r ) = ˆJ (r ). Define the operator Tˆ and the cone Kˆ as follows
( Tˆ u )(t ) =
1
G (t , s)q(s) f s, u (s) ds,
0 t 1,
0
Kˆ = u ∈ C [0, 1]: u (t ) u t 2 , 0 t 1 . ˆ r ) = {u ∈ Kˆ : u < r }, ∂ Ω( ˆ r ) = {u ∈ Kˆ : u = r }. Write Ω(
ˆ r2 )\Ω( ˆ r1 ) → Kˆ is completely continuIf the assumptions (H1) –(H3) are satisfied, we can prove that the operator Tˆ : Ω( ous for any 0 < r1 < r2 by imitating the proof of Lemma 2.4 and applying Lemma 2.2(5) and (6). Further, we can prove the following local existence theorems. Theorem 5.1. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied: (h1) max0t 1
1 0
β
β
ˆ s, a) ds a, max0t 1 ˆ G (t , s)q(s)Φ( α G (t , s)q(s)Ψ (s, b) ds b.
(h2) max0t 1 α G (t , s)q(s)Ψˆ (s, a) ds a, max0t 1
1 0
ˆ s, b) ds b. G (t , s)q(s)Φ(
Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a u ∗ b. Theorem 5.2. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist two positive numbers a < b such that one of the following conditions is satisfied:
ˆ ψ( ˆ ˆ b) b B. (i1) ϕˆ (a) (a − ˆI (a)) A, ˆ ˆ ϕˆ (b) (b − ˆI (b)) A. ˆ a) a B, (i2) ψ( Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a u ∗ b. Theorem 5.3. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist three positive numbers a < b < c such that one of the following conditions is satisfied:
ˆ ψ( ˆ ˆ ϕˆ (c ) (c − ˆI (c )) A. ˆ b) > b B, (j1) ϕˆ (a) (a − ˆI (a)) A, ˆ ψ( ˆ ϕˆ (b) < (b − ˆI (b)) A, ˆ ˆ a) a B, ˆ c ) c B. (j2) ψ( Then problem ( Pˆ ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that a u ∗1 < b < u ∗2 c.
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Theorem 5.4. Suppose that (1) The assumptions (H1) –(H3) are satisfied. (2) There exist four positive numbers a < b < c < d such that one of the following conditions is satisfied:
ˆ ψ( ˆ ψ( ˆ ϕˆ (c ) < (c − ˆI (c )) A, ˆ ˆ b) > b B, ˆ d) d B. (k1) ϕˆ (a) (a − ˆI (a)) A, ˆ ψ( ˆ ˆ ϕˆ (b) < (b − ˆI (b)) A, ˆ ϕˆ (d) (d − ˆI (d)) A. ˆ a) a B, ˆ c ) > c B, (k2) ψ( Then problem ( Pˆ ) has at least three strictly increasing positive solutions u ∗1 , u ∗2 , u ∗3 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that
a u ∗1 < b < u ∗2 < c < u ∗3 d. Imitating the proofs of Theorems 4.1–4.5, we can obtain the following existence results. Theorem 5.5. Suppose that (1) The assumptions (H1) –(H3) and (H5) are satisfied. 1 ˆ s, a) ds a. (2) There exists a > 0 such that max0t 1 0 G (t , s)q(s)Φ( Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗ a. Theorem 5.6. Suppose that (1) The assumptions (H1) –(H5) are satisfied. ˆ (2) There exists a > 0 such that ϕˆ (a) (a − ˆJ (a)) A. Then problem ( Pˆ ) has at least one strictly increasing positive solution u ∗ ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗ a. Theorem 5.7. Suppose that (1) The assumptions (H1) –(H3) and (H5) are satisfied. 1 ˆ s, a) ds < a and max0t 1 β G (t , s)q(s)Ψˆ (s, b) ds b. (2) There exist 0 < a < b such that max0t 1 0 G (t , s)q(s)Φ( α Then problem ( Pˆ ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < a < u ∗2 b. Theorem 5.8. Suppose that (1) The assumptions (H1) –(H5) are satisfied. ˆ ψ( ˆ ˆ b) b B. (2) There exist 0 < a < b such that ϕˆ (a) < (a − ˆJ (a)) A, Then problem ( Pˆ ) has at least two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < a < u ∗2 b. 6. Several remarks Remark 6.1 shows that our work extends Theorem 1.1. Remark 6.1. Theorem 1.1 is a corollary of Theorem 5.7. First of all, the assumptions (H1) and (H2) are satisfied by (a1)–(a2). Let 0 < r1 < r2 . By (a2)–(a4), we have r2 r1 (t )
γ
h (u ) 2 = max g (u ) + h(u ): r1 t u r2 = max g (u ) 1 + : r1 t u r2 g (u ) h(r2 ) h (u ) +1 . : r1 t 2 u r2 M g (r1 ) g t 2 max g r1t 2 1 + g (u ) g (r2 )
2
By the condition (a2),
1
r2 r1 (t ) dt
q(t )γ 0
1 h(r2 ) +1 M g (r1 ) q(t ) g t 2 dt < +∞. g (r2 ) 0
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So the assumption (H3) is satisfied. Since q ∈ C (0, 1), q(t ) > 0, 0 < t < 1, and the condition (a5), we see that the assumption (H5) is satisfied. Secondly, according to the conditions (a2)–(a4), we have
h (u ) 2 2 ˆ : at u a Φ(t , a) = max g (u ) + h(u ): at u a = max g (u ) 1 + g (u ) h(a) h (u ) + 1 = M g t 2 g (a) + h(a) . : at 2 u a M g (a) g t 2 max g at 2 1 + g (u ) g (a)
From the condition (a6), we get
1 max
0t 1
ˆ s, a) ds M g (a) + h(a) max G (t , s)q(s)Φ(
1
0
a
G (t , s)q(s) g s2 ds g (a) + h(a) ·
0t 1
g (a) + h(a)
0
= a.
α t β , we have
q(t )Ψˆ (t , b) = min q(t ) f (u ): bt 2 u b min θ(t ) g (u ) + h(u ) : bt 2 u b h (u ) h(bα 2 ) θ(t ) g (b) 1 + : bt 2 u b = θ(t ) min g (u ) 1 + g (u ) g (bα 2 ) 2 2 g (b) g (bα ) + g (b)h(bα ) = θ(t ). g (bα 2 )
Thirdly, according to (a1)–(a5), for
Applying the condition (a7), we get
β max
0t 1
G (t , s)q(s)Ψˆ (s, a) ds
g (b) g (bα 2 ) + g (b)h(bα 2 ) g (bα 2 )
α
β max
0t 1
g (b) g (bα 2 ) + g (b)h(bα 2 ) g (bα 2 )
·
G (t , s)θ(s) ds α
bg (bα 2 ) g (b) g (bα 2 ) + g (b)h(bα 2 )
= b.
By Theorem 5.7, the problem ( Pˆ ) has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ C 3 [0, 1] ∩ C 4 (0, 1) and 0 < u ∗1 < a < u ∗2 b. Remarks 6.2 and 6.3 show that our work improves the method used in papers [8,9,15]. Remark 6.2. The following example shows that the existence of twin positive solutions is possible for the problem ( Pˆ ) even if f (t , u ) = f (u ) and the assumption (a4) is not satisfied. Consider the singular problem
⎧ ⎨ u (4) (t ) = √ 1 + 36e −u (t ) + 1 min u 2 (t ), 15625000 , 3 4 u (t ) u (t ) ⎩ u (0) = u (0) = u (1) = u (1) = 0.
0 < t < 1,
In this example, q(t ) ≡ 1, f (t , u ) = f (u ) = g (u ) + h(u ), where h(u ) = 36e −u + singular at u = 0. ˆ = 8, Bˆ = Let α = 23 , β = 1. Then A
1 4
min{u 2 ,
15625000 }, u
1 g (u ) = √ 3 . So f (u ) is u
≈ 11.9264. Obviously, the assumptions (H1) –(H5) are satisfied. By Lemma 2.2(5), max0t ,s1 G (t , s) = G (1, 1) = 13 . So
1
ˆJ (8) = max
G (t , s) g 8s
0t 1
2
1944 163
ds
1
1
2 0t ,s1
0
0
We have 2502 = 62500 =
15625000 . 250
1
ds
√ 3
max G (t , s)
s2
1
= . 2
Direct computations give that
ϕˆ (8) = max 36e−u + u 2 : 0 u 8 36 + 16 = 52 < 60 8 − ˆJ (8) Aˆ , 4
ˆ 243) = min f (u ): ψ( =
1 4
2 3
1 t 1, 243t 2 u 243 = min f (u ): 108 u 243 min u 2 : 108 u 243
· 108 = 2916 > 2898.11 ≈ 243 Bˆ . 2
4
Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
151
By Theorem 5.8, the problem has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < 8 < u ∗2 < 243. h (u )
h (u )
√ = 0, we see that g (u ) is not nondecreasing. So the assumption (a4) is not Since limu →+∞ g (u ) = limu →+∞ 15625000 4 4 u3 satisfied. Therefore, the existence conclusion cannot be derived from Theorem 1.1.
Remark 6.3. The following example shows that the existence of twin positive solutions is possible for the problem ( Pˆ ) even if f (t , u ) = f (u ) and the assumption (a5) is not satisfied. Consider the singular problem
⎧ ⎨ ⎩
u
(4 )
3(1 + cos(u (t )π ))
(t ) =
16
+ max 0,
2420000 − 8(u (t ) − 750)2
4840000
u 2 (t ) +
2[1 + cos(u (t )π )]
√ 4
u (t )
,
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0.
In this example, q(t ) ≡ 1, f (t , u ) = f (u ) = g (u ) + h(u ), where
h (u ) =
3(1 + cos(u π )) 16
2420000 − 8(u − 750)2 + max 0, u2 , 4840000
So f (u ) is singular at u = 0. ˆ = 8, Bˆ = Let α = 12 , β = 1. Then A Obviously, g (u ) 2, 0 < u
1 2
g (u ) =
2[1 + cos(u π )]
√ 4
u
.
384 41
≈ 9.3659. 4 and g (u ) √ 4 , 0 < u < +∞. Hence the assumptions (H1) –(H5) are satisfied. Moreover, u
we have
1
ˆJ (16) = max
1
G (t , s) g 16s2 ds 2 max G (t , s)
0t 1 0
ds
4
s
3
√ = .
0t ,s1
0
Since h(u ) 38 u 2 , 0 u 16 and h(u ) > 18 u 2 , 300 u 1200, we get
4 16 − ˆJ (16) Aˆ , · 162 = 96 < 8 16 − 8 8 3 1200 1 2 ˆ 1200) = min f (u ): ψ( u 1200 min u : 300 u 1200
ϕˆ (16) = max h(u ): 0 u 16 max 4
=
1 8
3
u 2 : 0 u 16 =
3
8
· 3002 = 11250 > 11239.08 ≈ 1200 Bˆ .
By Theorem 5.8, the problem has two strictly increasing positive solutions u ∗1 , u ∗2 ∈ Kˆ ∩ C 3 [0, 1] ∩ C 4 (0, 1) such that 0 < u ∗1 < 16 < u ∗2 < 1200. Since f (2k − 1) = 0, k = 1, 2, . . . , 99 and k = 651, 652, . . . , the condition (a5) is not satisfied. Therefore, the existence conclusion cannot be derived from Theorem 1.1. The following Remarks 6.4–6.6 imply that our results are applicable to more complicated cases. Remark 6.4. Our results is applicable to the general problems ( P ) in which the nonlinear term contains the unknown function and its first derivative. For example, consider the general problem
⎧ ⎨ ⎩
u (4) (t ) =
u (t )u (t )
cos2 (u (t ) + u (t ))
3 + 2 cos2 (u (t ) + u (t ))
+√ 6
u (0) = u (0) = u (1) = u (1) = 0.
t (1 − t )u (t )u (t )
,
0 < t < 1,
In the example, q(t ) ≡ 1, f (t , u , v ) = g (t , u , v ) + h(t , u , v ), where
h(t , u , v ) =
uv 3 + 2 cos2 (u + v )
,
cos2 (u + v ) g (t , u , v ) = √ . 6 t (1 − t )uv
So f (t , u , v ) is singular at t = 0, t = 1 and u = v = 0. If 0 < r1 < r2 , then
√ 3
2 g (t , u , v ) , 6 2 4 r1 t (1 − t )
1 4
r1 t 2 u r2 , r1 t v r2 .
Hence the assumptions (H1)–(H3) are satisfied. In addition, let
α = 12 , β = 1, then A = 6, B =
48 . 7
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By Lemma 2.2(3), max0t ,s1 ∂∂t G (t , s) = 12 . So
∂ I (r ) max G (t , s) 0t ,s1 ∂ t
√ 1 √ 3 3 2ds 3 2 = . √ √ 6 3 2 r
r 2 s4
0
3 From this, I (4) √ 3 . Moreover, we have 2
ϕ (4) = max
2
uv
2
3 + 2 cos2 (u + v )
: 0 t 1, t u 4, 4t v 4
3 < 11.2378 ≈ 4 4 − √ 4 − I (4) A , 3 3 2 2 1 uv 2 ψ(640) min : t 1 , 160t u 640 , 640t v 640 3 + 2 cos2 (u + v ) 2
16
1 5
· 40 · 320 = 2560 > 2194.29 ≈ 640B .
By Theorem 3.2, the problem has a positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) and 4 < |||u ∗ ||| < 640. In this problem, the nonlinear term contains unknown function and its first derivative, and nonlinear term is singular at t = 0, t = 1 and u = 0, v = 0. Hence above existence conclusion cannot be derived by the method used in [15,22,24,31]. Remark 6.5. In this paper, we allow not only limmax{u , v }→0 min0
⎧ ⎨ u (4) (t ) = 1 + sin u (t )u (t ) + ⎩
1 u (t )u (t )
1
+ u (t )u (t ), 4
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0.
Here q(t ) ≡ 1, f (t , u , v ) = f (u , v ) = g (u , v ) + h(u , v ), where
h (u , v ) =
1 4
g (u , v ) = 1 + sin uv +
uv ,
1 uv
.
So the limit limmax{u , v }→0 min0
If r > 0, then
ψ(b)
7290000 432
= 16875 > 16823.08 ≈ Bb.
By Theorem 3.2, the problem has a positive solution u ∗ ∈ K ∩ C 3 [0, 1] ∩ C 4 (0, 1) and 2 |||u ∗ ||| < 2700. Because g (u , v ) is not nonincreasing with respect to u and v, the conclusion cannot be derived by applying the method used in [28–30,32]. Remark 6.6. For ease of representation, we use the following symbols
D [r1 , r2 ] = (t , u , v ): 0 t 1,
D ◦ [r1 , r2 ] = (t , u , v ): 0 < t < 1,
1 4
r1t 2 u r2 , r1 t v r2 ,
1 4
r1t 2 u r2 , r1 t v r2 .
Theorems 3.1 and 3.2 show that the existence of positive solution depends only upon the properties of nonlinear term f (t , u , v ) on the set D [a, b] and is independent of the states of f (t , u , v ) outside D [a, b]. The following example illustrates this. Consider the discontinuous boundary value problem
( Q 1)
u (4) (t ) = f 1 t , u (t ), u (t ) ,
0 < t < 1,
u (0) = u (0) = u (1) = u (1) = 0,
Q. Yao / J. Math. Anal. Appl. 363 (2010) 138–154
where
153
⎧ cot t 2 −136u , (t , u , v ) ∈ ((0, 1) × (0, 3]2 )\ D ◦ [ 19 , 3], ⎪ ⎪ ⎨ 2(1+t ) √ , (t , u , v ) ∈ D ◦ [ 19 , 3], f 1 (t , u , v ) = 4 t (u + v 2 ) ⎪ ⎪ ⎩ cot( v − 3)π , (t , u , v ) ∈ (0, 1) × ((0, +∞)2 \(0, 3]2 ).
In the example, the nonlinear term f 1 (t , u , v ) is very discontinuous and is not nonnegative outside D [ 19 , 3]. Therefore, the assumptions (H1)–(H3) are not satisfied. So we cannot directly obtain the existence of positive solution by the help of the existence theorems in this paper. Now, consider the singular boundary value problem
⎧ ⎨ u (4) (t ) = 2(1 + t ) , 0 < t < 1, 4 t (u (t ) + (u )2 (t )) ⎩ u (0) = u (0) = u (1) = u (1) = 0.
( Q 2)
2(1+t ) In this example, q(t ) ≡ 1, f 2 (t , u , v ) = g (t , u , v ) = √ 4 2
t (u + v )
and h(t , u , v ) ≡ 0. So f 2 (t , u , v ) is singular at t = 0 and u =
v = 0. Obviously, the assumptions (H1)–(H3) are satisfied. Let α = 23 , β = 1. According to Lemma 2.2(3) and (4), min2/3t ,s1 ∂∂t G (t , s) tion, for any r > 0, we have
2(1 + t )
Φ(t , r ) = max 4
t (u +
v 2)
2(1 + t )
Ψ (t , r ) = min 4
t (u + v 2 )
:
:
1 4
8 81
and max0t ,s1 ∂∂t G (t , s) = 12 . In addi-
√
1+t , rt u r , rt v r √ √ 4 4 2 r + 4r t3 2 2
2
2 1+t . rt 2 u r , rt v r √ √ 4 4 2 4 t r +r 1
It follows that
1 max
0t 1 2/3
1 max
0t 1
1 ∂ 1 ∂ 1+s 1 4 10 ds 2 ds ≈ 0.1168 > , G (t , s)h(s)Ψ s, min G (t , s) √ 4 ∂t 9 81 2/3t ,s1 ∂ t 9 s 2/3
∂ ∂ 4 4 G (t , s)h(s)Φ(s, 3) ds 2 max G (t , s) ∂t 39 0t ,s1 ∂ t
0
1
1+s
√ 4
0
By Theorem 3.1, the problem ( Q 2) has one positive solution u ∗ ∈ K and Since u ∗
∈
K [ 91 , 3],
we have
1 2 t 36
u ∗ (t ) 3, 19 t
(u ∗ ) (t ) 3, 0 < t
f 1 (t , u , v ) = f 2 (t , u , v ), (t , u , v ) ∈ D [ 19 , 3], we have
f 1 t , u ∗ (t ), u ∗ (t ) = f 2 t , u ∗ (t ), u ∗ (t ) ,
s3
ds ≈ 2.7164 < 3.
|||u ∗ ||| 3. < 1. So (t , u ∗ (t ), (u ∗ ) (t )) ∈ D [ 19 , 3], 0 < t < 1. Since 1 9
0 < t < 1.
Therefore, u ∗ (t ) is positive solution of the problem
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