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On the existence of positive solutions for 2n -order singular boundary value problems
Nonlinear Analysis 64 (2006) 2553 – 2561 www.elsevier.com/locate/na
On the existence of positive solutions for 2n-order singular boundary value problems夡 Zengqin Zhao School of Mathematics Sciences, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China Received 1 July 2005; accepted 9 September 2005
Abstract By applying upper and lower solutions method, the existence of C 2n−2 [0, 1] and C 2n−1 [0, 1] positive solution for a class of 2n-order singular differential equations boundary value problems is obtained. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34B15; 34B16; 34B18 Keywords: Higher order differential equation; Singular boundary value problem; Positive solution; Upper and lower solution; Decreasing function
1. Introduction and the main results In this paper, we consider the existence of positive solutions for the following 2n-order singular boundary value problem (BVP) (−1)n u(2n) (t) = f (t, u), u(2k) (0) = u(2k) (1) = 0,
t ∈ (0, 1), k = 0, 1, 2, . . . , n − 1,
(1) (2)
where n2, and f : (0, 1) × (0, +∞) −→ [0, +∞). In the case of n = 1 and n = 2, the existence and multiplicity of positive solutions for the problem BVP (1), (2) have been widely studied [1,3,4,7]. When n is an arbitrary natural number boundary value problem has attracted considerable attention [5,6] in recent years. In particular, 夡 Research supported by the National Natural Science Foundation of China (10471075) and the Natural Science Foundation of Shandong Province of China (02BS119). E-mail address: [email protected].
0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.09.003
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Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
Yao [6] show the existence of N symmetric positive solutions, when f (t, u) is nondecreasing in u, here N is a natural number. In this paper, we only require that f (t, u) is decreasing with respect to u, and f (t, u) may be singular at u = 0, t = 0 and/or 1. By using upper and lower solution method, and the properties of Green’s function, we establish a new existence criterion due to Shauder fixed point theorem. Denote I = [0, 1], J = (0, 1), R + = (0, +∞), s(1 − t), 0 s < t 1, e(t) = t (1 − t), t ∈ I, G(t, s) = (3) t (1 − s), 0 t s 1. Obviously e(t) = G(t, t), and e(s)e(t) G(t, s)e(t),
t ∈ I.
(4)
Definition 1. By a nonzero solution, also called a C (2n−2) (I ) solution of the BVP (1), (2), we mean a function w(t) ∈ C (2n−2) (I )∩C (2n) (J ) satisfying the BVP (1), (2) with w(t) not identically zero on J. w(t) is called a C (2n−1) (I ) solution, we mean that w (2n−1) (0+) and w (2n−1) (1 − 0) exist. w(t) is called a positive solution of the BVP (1), (2) if w(t) is a solution of (1), (2) and w(t) > 0 for any t ∈ J . Definition 2. A function (t) is called a lower solution, if (t) ∈ C (2n−2) (I ) ∩ C (2n) (J ) and satisfying (−1)n (2n) (t)f (t, (t)), t ∈ J, (−1)[k−1] (2k) (0)0, (−1)[k−1] (2k) (1)0, k = 0, 1, 2, . . . , n − 1. Upper solution is defined by reversing the above inequality signs. If there exist a lower solution (t) and an upper solution (t) for the BVP (1), (2) such that (t) (t), then ((t), (t)) is called a couple of upper and lower solution for the BVP (1), (2). Where (−1)[k−1] denotes −1 to the power k − 1, (2k) denotes the 2kth derivative of . In order to prove the main result, we need the following Lemma which is obtained in [2]. Lemma 1.1 (Maximal principle). Let 0 a < b, if x ∈ C (2n−2) [a, b] ∩ C (2n) (a, b) satisfies (−1)k x (2k) (a)0, and
(−1)n x (2n) (t)0, t
(−1)k x (2k) (b)0,
k = 0, 1, 2, . . . , n − 1,
∈ (a, b), then x(t) 0, t ∈ [a, b].
The main result in this paper is the following. Theorem 1.1. Assume that (H1) f (t, u) ∈ C(J × R + , [0, +∞)), and f (t, u) is decreasing with respect to u; 1 (H2) f (t, ) ≡ / 0, 0 t (1 − t)f (t, t (1 − t)) dt < ∞, ∀ > 0; (H3) There exists a function a(t), such that a(t)kt(1 − t), t ∈ I (k > 0 is a certain real number), 1 1 ··· G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s) a1 (t) = 0
0
× f (s, a(s)) ds ds1 · · · dsn−1 a(t),
(5)
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
a2 (t) =
1
2555
1
···
G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s) 0 × f (s, a1 (s)) ds ds1 · · · dsn−1 a(t), 0
(6)
where G(t, s) is defined as (3). Then the singular 2n-order BVP (1), (2) has a C (2n−2) (I ) positive solution w(t), and mt(1 − t)w(t)Mt(1 − t),
∃M m > 0,
∀t ∈ I .
(7)
2. Proof of the main result Let E = denote the Banach space C(I). Define the set P and the operator T in E as follows: P = {u(t) ∈ E | there exists a positive number ku such that u(t) ku e(t), t ∈ I },
1
Tu(t) =
1
···
0
0
(8)
G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s)
× f (s, u(s)) ds ds1 · · · dsn−1 ,
∀u(s) ∈ P ,
(9)
where G(t, s), e(t) are defined as in (3). Obviously e(t) ∈ P , then P is nonempty. For all b(t) ∈ P , by the definition of P, there exists a positive number kb such that b(t) kb e(t), t ∈ I . It following from (H2) that
1
1
e(s)f (s, b(s)) ds
0
0
e(s)f (s, kb e(s)) ds < ∞.
By (9) and (4) we obtain
1
Tb(t)
1
···
0
0
e(sn−1 )e(sn−2 ) · · · e(s)f (s, b(s)) ds ds1 · · · dsn−1 < ∞.
Let B = maxt∈I b(t). By condition (H2) and the continuity of f (t, u), we know f (s, B) ds > 0. Thus,
1
1
e(s)f (s, b(s)) ds
0
1 0
e(s)
e(s)f (s, B) ds > 0.
0
On the other hand, by (4) we obtain
1
Tb(t)e(t)
0
= k(Tb) e(t), where k(Tb) = Tb ∈ P ,
1 0
1
··· 0
e(sn−1 )2 · · · e(s1 )2 e(s)f (s, b(s)) ds ds1 · · · dsn−1
∀t ∈ I , 1
e(s)f (s, b(s)) ds · (
∀b ∈ P .
0
e()2 d)n−1 . So Tb is well defined, (10)
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By simple computation we obtain 1 t sn−1 ··· Tb(t) = (1 − t) 0
0
1 0
G(sn−1 , sn−2 ) · · · G(s1 , s)
× f (s, b(s)) ds ds1 · · · dsn−1 1 1 1 G(sn−1 , sn−2 ) · · · G(s1 , s) (1 − sn−1 ) ··· +t t
0
0
× f (s, b(s)) ds ds1 · · · dsn−1 ; 1 1 d(2k) (Tb(t)) k · · · G(t, sn−k−1 ) · · · G(s1 , s) = (−1) dt (2k) 0 0 × f (s, b(s)) ds ds1 · · · dsn−k−1 , t ∈ I, k = 1, 2, . . . , n − 1;
(11)
(−1)n (Tb)(2n) (t) = f (t, b(t)),
(12)
t ∈ J.
Thus, Tb ∈ C (2n−2) (I ) ∩ C (2n) (J ),
(13)
(Tb)(2k) (0) = (Tb)(2k) (1) = 0,
k = 1, 2, . . . , n − 1.
(14)
By conditions (5), (6) and (9), since operator T is decreasing, we know a(t)a2 (t) = (Ta1 )(t), (Ta1 )(t)(Ta)(t) = a1 (t),
a(t)a1 (t) = (Ta)(t), t ∈ I , t ∈ I.
(15) (16)
By (8) we know a(t) ∈ P . From (5), (10) we obtain Ta(t), Ta1 (t) ∈ P . (12), (15) and (16) imply (−1)n (Ta1 )(2n) (t) − f (t, (Ta1 )(t))(−1)n (Ta1 )(2n) (t) − f (t, a1 (t)) = 0,
(17)
(−1)n (Ta)(2n) (t) − f (t, (Ta)(t))(−1)n (Ta)(2n) (t) − f (t, a(t)) = 0.
(18)
(14) implies that Ta1 (t), Ta(t) satisfy condition (2). Thus it follows from (13), (16), (17) and (18) that (H (t), Q(t)) = (Ta1 (t), Ta(t))
(19)
is a couple of upper and lower solution of BVP (1), (2), and H (t), Q(t) ∈ C (2n−2) (I ) ∩ C (2n) (J ).
H (t), Q(t) ∈ P ,
Define the function F and the operator A in E = C(I ) by f (t, H (t)) if u < H (t), F (t, u) = f (t, u) if H (t) u Q(t), f (t, Q(t)) if u > Q(t), 1 1 Au(t) = ··· G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s) 0
0
× F (s, u(s)) ds ds1 · · · dsn−1 ,
∀u ∈ E.
(20)
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It follows from the assumption that F : J × R → R + is continuous. Consider 2n-order differential equation (−1)n u(2n) (t) = F (t, u(t)),
t ∈ J.
(21)
Obviously, a fixed point of the operator A are a solution of the singular boundary value problem of (21), (2). By H (t) ∈ P we know that there exists a positive number kH such that H (t)kH e(t), t ∈ I . 1 From the condition (H2) we obtain that 0 e(s)f (s, H (s)) ds is a finite number. Denote 1 M= e(s)f (s, H (s)) ds. (22) 0
For all u(t) ∈ E, from (4) and (22) we obtain 1 1 1 1 Au(t) e(sn−1 ) e(sn−2 ) · · · e(s1 ) e(s)F (s, u(s)) ds ds1 · · · dsn−1 , 0 0 0 0 1 1 1 1 e(sn−2 ) · · · e(s1 ) e(s)F (s, H (s)) ds ds1 · · · dsn−1 , e(sn−1 ) 0
0
M
1
0
[n−1]
e() d
0
.
0
Hence A(E) is bounded. It is easy to see A: E → E is continuous. Since G(t, sn−1 ) is continuous on I × I , it is uniformly continuous. Thus ∀ > 0, ∃ > 0 such 1 that when |t1 − t2 | < , we have |G(t1 , sn−1 ) − G(t2 , sn−1 )| < · M −1 ( 0 e() d)[2−n] . For all x ∈ E, we have 1 [n−2] 1 |Ax(t2 ) − Ax(t1 )| |G(t1 , sn−1 ) − G(t2 , sn−1 )|M e() d dsn−1 0
1
M
0
[n−2]
1
e() d
0
|G(t1 , ) − G(t2 , )| d < .
0
This implies A(E) is relative compact. By Schauder fixed point theorem, A has at least one fixed point w, that is, w(t) satisfies w =Aw. Thus w ∈ C (2n−2) (I ) ∩ C (2n) (J ). In a similar way as (11), by (20) we obtain 1 1 k (2k) w (t) = (−1) ··· G(t, sn−k−1 ) · · · G(s1 , s)F (s, w(s)) ds ds1 · · · dsn−k−1 , 0
0
t ∈ I, k = 1, 2, . . . , n − 1. w(2k) (0) = w (2k) (1) = 0,
k = 1, 2, . . . , n − 1.
(−1)n w (2n) (t) = F (t, w(t)),
t ∈ J.
(23)
Now we prove, H (t) w(t)Q(t),
t ∈ I.
(24)
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Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
Let z(t) = Q(t) − w(t), t ∈ I . By (14) and (19), w(t) satisfying (2), we have z(2k) (0) = z(2k) (1) = 0,
k = 0, 1, 2, . . . , n − 1.
(25)
From the definition of F and (15), (16) we obtain f (t, Q(t))F (t, x(t)) f (t, H (t)), f (t, a1 (t))f (t, Q(t)),
∀x(t) ∈ C(J ),
f (t, a(t))f (t, H (t)),
t ∈ J.
Thus, f (t, a1 (t))F (t, x(t)) f (t, a(t)),
∀x(t) ∈ C(J ).
This and (12) together with (19) and (23) imply (−1)n z(2n) (t) = (−1)n Q(2n) (t) − (−1)n w (2n) (t) = f (t, a(t)) − F (t, w(t)) 0, t ∈ J . This together (25) imply that z(t) 0 on I by using the maximal principle. Hence Q(t) w(t) on I. In a similar way, we have w(t)H (t), t ∈ I . Thus (24) holds. Hence F (t, w(t))=f (t, w(t)). Then w(t) is a C (2n−2) (I ) positive solution of the singular boundary value problem (1), (2). Suppose w(t) be a positive solution of (1), (2). Thus, (−1)n w (2n) (t) 0, t ∈ J, (26) w (2k) (0) = w (2k) (1) = 0, k = 0, 1, 2, . . . , n − 1. Let y(t) = w (t). Using (26) we have (−1)[n−1] y (2n−2) (t)0, t ∈ J y (2k) (0) = y (2k) (1) = 0, k = 0, 1, 2, . . . , n − 2. It follows from Lemma 1.1 that y(t)0, so w (t)0, t ∈ I , and w (t) is decreasing on I. If w (0)0, then w (t) 0, t ∈ I . Thus, w(t) is decreasing on I. By w(0) = 0 we know w(t)0, t ∈ I . In contradiction with w(t) > 0, hence w (0) > 0. Thus, lim
t→0+
w(t) w (t) = lim = w (0) > 0. t (1 − t) t→0+ 1 − 2t
Thus, there exist positive numbers M1 , m1 and 1 (1 < 21 ) such that m1
w(t) M1 , t (1 − t)
∀t ∈ [0, 1 ].
If w (1)0, since w (t) is decreasing, we know that w (t) 0, t ∈ I . Thus w(t) is increasing on I. By w(1) = 0 we know that w(t)0, t ∈ I . This is contradiction with w(t) > 0, hence w (1) < 0. So lim
t→1−
w(t) w (t) = lim = −w (1) > 0. t (1 − t) t→1− 1 − 2t
Thus, there exist positive numbers M2 , m2 and 2 (1 > 2 > 21 ) such that m2
w(t) M2 , t (1 − t)
∀t ∈ [2 , 1].
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
2559
w(t)/t (1 − t) is positive on [1 , 2 ]. Thus, there exist positive numbers M3 , m3 such that m3
w(t) M3 , t (1 − t)
∀t ∈ [1 , 2 ].
Choose m = min{mi , i = 1, 2, 3}, M = max{Mi , i = 1, 2, 3}, then mt(1 − t)w(t) Mt(1 − t),
t ∈ I,
that is (7) holds.
3. Further discussions
Corollary 3.1. Suppose that in Theorem 1.1 the conditions (H1) and (H3) hold, the condition (H2) is strengthened become (H4) f (t, ) ≡ / 0,
1 0
f (t, t (1 − t)) dt < ∞, ∀ > 0;
Then the positive solution w(t) of the problem (1), (2) is a C (2n−1) (I ) solution. Proof. By (7) and f (t, u) is decreasing with respect to u, we know that f (t, w(t)) f (t, mt(1 − t)). From (H4) we know that f (t, w(t)) is integrate on J, that is, the derivable function w(2n) (t) of w(2n−1) (t) is integrate on J. Thus w (2n−1) (0+), w(2n−1) (1 − 0) exist, that is, w(t) ∈ C (2n−1) (I ) ∩ C (2n) (J ). If f (t, u) is nonsingular at u = 0. Then for all u 0, f (t, u) f (t, 0), t ∈ J , and we have the following corollary. Corollary 3.2. Let (H1 ) f (t, u): J × [0, +∞) → [0, +∞) is continuous, and is decreasing with respect to u; 1 / 0, ∀ 0, 0 t (1 − t)f (t, 0) dt < ∞; (H2 ) f (t, ) ≡ Then the singular 2n-order BVP (1), (2) has a C (2n−2) (I ) positive solution w(t), and mt(1 − t)w(t)Mt(1 − t),
∃M m > 0, ∀t ∈ I .
(27)
In (8) the set P is replaced by P1 = {u(t) ∈ E | u(t)0, t ∈ I }, then (Tb) ∈ P1 holds, and (12), (13) and (14) hold. Let a(t) ≡ 0, then by conditions (H1 ) and (H2 ), we know that (15), (16) hold. The rest of proof is similar in Theorem 1.1.
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Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
Corollary 3.3. Suppose that in Corollary 3.2 the condition (H1 ) holds, the condition (H2 ) is strengthened become / 0, ∀ 0, (H4 ) f (t, ) ≡
1 0
f (t, 0) dt < ∞,
Then the positive solution w(t) of the problem (1), (2) is a C (2n−1) (I ) solution. If f (t, u) is nonsingular at t, u, then corollary.
1 0
f (t, 0) dt < ∞ holds. Then we have the following
Corollary 3.4. If f (t, u) ∈ C(I × [0, +∞), [0, +∞)) is decreasing with respect to u, and f (t, ) ≡ / 0, ∀0. Then the problem (1), (2) has a positive solution, and (27) holds. Remark 1. If f (t, u) is nonnegative, decreasing and not identically zero, we obtain the existence of positive solution without other conditions. In order to show the validity of our results, we study the following sixth-order boundary value problem (6) u (t) + a0 (t) + ni=1 ai (t)u−i = 0, t ∈ J, (28) u(0) = u(1) = u(2) (0) = u(2) (1) = u(4) (0) = u(4) (1) = 0, where n a0 (t) and ai (t) are nonnegative and continuous on J, 0 < i < 1, (i = 1, 2, . . . , n), / 0 on I. i=0 ai (t) ≡ If
1 n (29) t (1 − t) a0 (t) + ai (t)t −i (1 − t)−i dt < + ∞, 0
i=1
then the boundary value of (28) has a positive solution w(t) ∈ C (4) (I ) ∩ C (6) (J ), and mt(1 − t)w(t)Mt(1 − t), If 0
1
a0 (t) +
n
∃M m > 0.
ai (t)t
−i
(1 − t)
−i
dt < + ∞,
(30)
i=1
then the positive solution of (28) is a C (5) (I ) positive solution. Proof. Let f (t, u) = a0 (t) + ni=1 ai (t)u−i , t ∈ J . It is easy to prove that (H1) and (H2) are satisfied in Theorem 1.1 under the condition (29). (H1) and (H4) are satisfied in Corollary 3.1 under the condition (30). Denote =max1 i n {i }. Then f (t, u) f (t, ru)r − f (t, u) holds for all positive numbers r < 1. By Te, T 2 e ∈ P , we know that there exist positive numbers k1 , k2 such that Te k1 e, T 2 e k2 e. Choose a positive number r0 such that r0 (min{1, k1 , k2 })1/(1−
2)
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561 1−2
Thus k1 r0
1−2
r0 , k2 r0
2561
; Consequently,
T (r0 e) Te k1 e r0 e, −
T (r0 e) r0 Te,
2
2
T 2 (r0 e)r0 T 2 e r0 k2 e r0 e.
We take a(t) = r0 e(t), then the condition (H3) of Theorem 1.1 is satisfied. Thus, this completes the proof. References [1] J. Henderson, H.B. Thompson, Existence of multiple solutions for second order boundary value problems, J. Differential Equations 166 (2000) 443–454. [2] L. Kong, Q. Kong, Even order nonlinear eigenvalue problems on a measure chain, Nonlinear Anal. 52 (2003) 1891–1909. [3] Y.S. Liu, Multiple positive solutions of nonlinear singular boundary value problem for fourth-order equations, Appl. Math. Lett. 17 (2004) 747–757. [4] Z.L. Wei, Positive solution of fourth order singular boundary value problem, Acta Math. Sinica 42 (4) (1999) 715–722 (in Chinese). [5] X.J. Yang, Green’s function and positive solutions for higher-order ODE, Appl. Math. Comput. 136 (2003) 379–393. [6] Q.L. Yao, Monotone iterative technique and positive solutions of Lidstone boundary value problems, Appl. Math. Comput. 131 (2002) 477–485. [7] Q.L. Yao, Existence and iteration of n symmetric positive solutions for a singular two-point boundary value problem, Comput. Math. Appl. 47 (2004) 1195–1200.
On the existence of positive solutions for 2n-order singular boundary value problems夡 Zengqin Zhao School of Mathematics Sciences, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China Received 1 July 2005; accepted 9 September 2005
Abstract By applying upper and lower solutions method, the existence of C 2n−2 [0, 1] and C 2n−1 [0, 1] positive solution for a class of 2n-order singular differential equations boundary value problems is obtained. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34B15; 34B16; 34B18 Keywords: Higher order differential equation; Singular boundary value problem; Positive solution; Upper and lower solution; Decreasing function
1. Introduction and the main results In this paper, we consider the existence of positive solutions for the following 2n-order singular boundary value problem (BVP) (−1)n u(2n) (t) = f (t, u), u(2k) (0) = u(2k) (1) = 0,
t ∈ (0, 1), k = 0, 1, 2, . . . , n − 1,
(1) (2)
where n2, and f : (0, 1) × (0, +∞) −→ [0, +∞). In the case of n = 1 and n = 2, the existence and multiplicity of positive solutions for the problem BVP (1), (2) have been widely studied [1,3,4,7]. When n is an arbitrary natural number boundary value problem has attracted considerable attention [5,6] in recent years. In particular, 夡 Research supported by the National Natural Science Foundation of China (10471075) and the Natural Science Foundation of Shandong Province of China (02BS119). E-mail address: [email protected].
0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.09.003
2554
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
Yao [6] show the existence of N symmetric positive solutions, when f (t, u) is nondecreasing in u, here N is a natural number. In this paper, we only require that f (t, u) is decreasing with respect to u, and f (t, u) may be singular at u = 0, t = 0 and/or 1. By using upper and lower solution method, and the properties of Green’s function, we establish a new existence criterion due to Shauder fixed point theorem. Denote I = [0, 1], J = (0, 1), R + = (0, +∞), s(1 − t), 0 s < t 1, e(t) = t (1 − t), t ∈ I, G(t, s) = (3) t (1 − s), 0 t s 1. Obviously e(t) = G(t, t), and e(s)e(t) G(t, s)e(t),
t ∈ I.
(4)
Definition 1. By a nonzero solution, also called a C (2n−2) (I ) solution of the BVP (1), (2), we mean a function w(t) ∈ C (2n−2) (I )∩C (2n) (J ) satisfying the BVP (1), (2) with w(t) not identically zero on J. w(t) is called a C (2n−1) (I ) solution, we mean that w (2n−1) (0+) and w (2n−1) (1 − 0) exist. w(t) is called a positive solution of the BVP (1), (2) if w(t) is a solution of (1), (2) and w(t) > 0 for any t ∈ J . Definition 2. A function (t) is called a lower solution, if (t) ∈ C (2n−2) (I ) ∩ C (2n) (J ) and satisfying (−1)n (2n) (t)f (t, (t)), t ∈ J, (−1)[k−1] (2k) (0)0, (−1)[k−1] (2k) (1)0, k = 0, 1, 2, . . . , n − 1. Upper solution is defined by reversing the above inequality signs. If there exist a lower solution (t) and an upper solution (t) for the BVP (1), (2) such that (t) (t), then ((t), (t)) is called a couple of upper and lower solution for the BVP (1), (2). Where (−1)[k−1] denotes −1 to the power k − 1, (2k) denotes the 2kth derivative of . In order to prove the main result, we need the following Lemma which is obtained in [2]. Lemma 1.1 (Maximal principle). Let 0 a < b, if x ∈ C (2n−2) [a, b] ∩ C (2n) (a, b) satisfies (−1)k x (2k) (a)0, and
(−1)n x (2n) (t)0, t
(−1)k x (2k) (b)0,
k = 0, 1, 2, . . . , n − 1,
∈ (a, b), then x(t) 0, t ∈ [a, b].
The main result in this paper is the following. Theorem 1.1. Assume that (H1) f (t, u) ∈ C(J × R + , [0, +∞)), and f (t, u) is decreasing with respect to u; 1 (H2) f (t, ) ≡ / 0, 0 t (1 − t)f (t, t (1 − t)) dt < ∞, ∀ > 0; (H3) There exists a function a(t), such that a(t)kt(1 − t), t ∈ I (k > 0 is a certain real number), 1 1 ··· G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s) a1 (t) = 0
0
× f (s, a(s)) ds ds1 · · · dsn−1 a(t),
(5)
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
a2 (t) =
1
2555
1
···
G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s) 0 × f (s, a1 (s)) ds ds1 · · · dsn−1 a(t), 0
(6)
where G(t, s) is defined as (3). Then the singular 2n-order BVP (1), (2) has a C (2n−2) (I ) positive solution w(t), and mt(1 − t)w(t)Mt(1 − t),
∃M m > 0,
∀t ∈ I .
(7)
2. Proof of the main result Let E = denote the Banach space C(I). Define the set P and the operator T in E as follows: P = {u(t) ∈ E | there exists a positive number ku such that u(t) ku e(t), t ∈ I },
1
Tu(t) =
1
···
0
0
(8)
G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s)
× f (s, u(s)) ds ds1 · · · dsn−1 ,
∀u(s) ∈ P ,
(9)
where G(t, s), e(t) are defined as in (3). Obviously e(t) ∈ P , then P is nonempty. For all b(t) ∈ P , by the definition of P, there exists a positive number kb such that b(t) kb e(t), t ∈ I . It following from (H2) that
1
1
e(s)f (s, b(s)) ds
0
0
e(s)f (s, kb e(s)) ds < ∞.
By (9) and (4) we obtain
1
Tb(t)
1
···
0
0
e(sn−1 )e(sn−2 ) · · · e(s)f (s, b(s)) ds ds1 · · · dsn−1 < ∞.
Let B = maxt∈I b(t). By condition (H2) and the continuity of f (t, u), we know f (s, B) ds > 0. Thus,
1
1
e(s)f (s, b(s)) ds
0
1 0
e(s)
e(s)f (s, B) ds > 0.
0
On the other hand, by (4) we obtain
1
Tb(t)e(t)
0
= k(Tb) e(t), where k(Tb) = Tb ∈ P ,
1 0
1
··· 0
e(sn−1 )2 · · · e(s1 )2 e(s)f (s, b(s)) ds ds1 · · · dsn−1
∀t ∈ I , 1
e(s)f (s, b(s)) ds · (
∀b ∈ P .
0
e()2 d)n−1 . So Tb is well defined, (10)
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By simple computation we obtain 1 t sn−1 ··· Tb(t) = (1 − t) 0
0
1 0
G(sn−1 , sn−2 ) · · · G(s1 , s)
× f (s, b(s)) ds ds1 · · · dsn−1 1 1 1 G(sn−1 , sn−2 ) · · · G(s1 , s) (1 − sn−1 ) ··· +t t
0
0
× f (s, b(s)) ds ds1 · · · dsn−1 ; 1 1 d(2k) (Tb(t)) k · · · G(t, sn−k−1 ) · · · G(s1 , s) = (−1) dt (2k) 0 0 × f (s, b(s)) ds ds1 · · · dsn−k−1 , t ∈ I, k = 1, 2, . . . , n − 1;
(11)
(−1)n (Tb)(2n) (t) = f (t, b(t)),
(12)
t ∈ J.
Thus, Tb ∈ C (2n−2) (I ) ∩ C (2n) (J ),
(13)
(Tb)(2k) (0) = (Tb)(2k) (1) = 0,
k = 1, 2, . . . , n − 1.
(14)
By conditions (5), (6) and (9), since operator T is decreasing, we know a(t)a2 (t) = (Ta1 )(t), (Ta1 )(t)(Ta)(t) = a1 (t),
a(t)a1 (t) = (Ta)(t), t ∈ I , t ∈ I.
(15) (16)
By (8) we know a(t) ∈ P . From (5), (10) we obtain Ta(t), Ta1 (t) ∈ P . (12), (15) and (16) imply (−1)n (Ta1 )(2n) (t) − f (t, (Ta1 )(t))(−1)n (Ta1 )(2n) (t) − f (t, a1 (t)) = 0,
(17)
(−1)n (Ta)(2n) (t) − f (t, (Ta)(t))(−1)n (Ta)(2n) (t) − f (t, a(t)) = 0.
(18)
(14) implies that Ta1 (t), Ta(t) satisfy condition (2). Thus it follows from (13), (16), (17) and (18) that (H (t), Q(t)) = (Ta1 (t), Ta(t))
(19)
is a couple of upper and lower solution of BVP (1), (2), and H (t), Q(t) ∈ C (2n−2) (I ) ∩ C (2n) (J ).
H (t), Q(t) ∈ P ,
Define the function F and the operator A in E = C(I ) by f (t, H (t)) if u < H (t), F (t, u) = f (t, u) if H (t) u Q(t), f (t, Q(t)) if u > Q(t), 1 1 Au(t) = ··· G(t, sn−1 )G(sn−1 , sn−2 ) · · · G(s1 , s) 0
0
× F (s, u(s)) ds ds1 · · · dsn−1 ,
∀u ∈ E.
(20)
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
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It follows from the assumption that F : J × R → R + is continuous. Consider 2n-order differential equation (−1)n u(2n) (t) = F (t, u(t)),
t ∈ J.
(21)
Obviously, a fixed point of the operator A are a solution of the singular boundary value problem of (21), (2). By H (t) ∈ P we know that there exists a positive number kH such that H (t)kH e(t), t ∈ I . 1 From the condition (H2) we obtain that 0 e(s)f (s, H (s)) ds is a finite number. Denote 1 M= e(s)f (s, H (s)) ds. (22) 0
For all u(t) ∈ E, from (4) and (22) we obtain 1 1 1 1 Au(t) e(sn−1 ) e(sn−2 ) · · · e(s1 ) e(s)F (s, u(s)) ds ds1 · · · dsn−1 , 0 0 0 0 1 1 1 1 e(sn−2 ) · · · e(s1 ) e(s)F (s, H (s)) ds ds1 · · · dsn−1 , e(sn−1 ) 0
0
M
1
0
[n−1]
e() d
0
.
0
Hence A(E) is bounded. It is easy to see A: E → E is continuous. Since G(t, sn−1 ) is continuous on I × I , it is uniformly continuous. Thus ∀ > 0, ∃ > 0 such 1 that when |t1 − t2 | < , we have |G(t1 , sn−1 ) − G(t2 , sn−1 )| < · M −1 ( 0 e() d)[2−n] . For all x ∈ E, we have 1 [n−2] 1 |Ax(t2 ) − Ax(t1 )| |G(t1 , sn−1 ) − G(t2 , sn−1 )|M e() d dsn−1 0
1
M
0
[n−2]
1
e() d
0
|G(t1 , ) − G(t2 , )| d < .
0
This implies A(E) is relative compact. By Schauder fixed point theorem, A has at least one fixed point w, that is, w(t) satisfies w =Aw. Thus w ∈ C (2n−2) (I ) ∩ C (2n) (J ). In a similar way as (11), by (20) we obtain 1 1 k (2k) w (t) = (−1) ··· G(t, sn−k−1 ) · · · G(s1 , s)F (s, w(s)) ds ds1 · · · dsn−k−1 , 0
0
t ∈ I, k = 1, 2, . . . , n − 1. w(2k) (0) = w (2k) (1) = 0,
k = 1, 2, . . . , n − 1.
(−1)n w (2n) (t) = F (t, w(t)),
t ∈ J.
(23)
Now we prove, H (t) w(t)Q(t),
t ∈ I.
(24)
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Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
Let z(t) = Q(t) − w(t), t ∈ I . By (14) and (19), w(t) satisfying (2), we have z(2k) (0) = z(2k) (1) = 0,
k = 0, 1, 2, . . . , n − 1.
(25)
From the definition of F and (15), (16) we obtain f (t, Q(t))F (t, x(t)) f (t, H (t)), f (t, a1 (t))f (t, Q(t)),
∀x(t) ∈ C(J ),
f (t, a(t))f (t, H (t)),
t ∈ J.
Thus, f (t, a1 (t))F (t, x(t)) f (t, a(t)),
∀x(t) ∈ C(J ).
This and (12) together with (19) and (23) imply (−1)n z(2n) (t) = (−1)n Q(2n) (t) − (−1)n w (2n) (t) = f (t, a(t)) − F (t, w(t)) 0, t ∈ J . This together (25) imply that z(t) 0 on I by using the maximal principle. Hence Q(t) w(t) on I. In a similar way, we have w(t)H (t), t ∈ I . Thus (24) holds. Hence F (t, w(t))=f (t, w(t)). Then w(t) is a C (2n−2) (I ) positive solution of the singular boundary value problem (1), (2). Suppose w(t) be a positive solution of (1), (2). Thus, (−1)n w (2n) (t) 0, t ∈ J, (26) w (2k) (0) = w (2k) (1) = 0, k = 0, 1, 2, . . . , n − 1. Let y(t) = w (t). Using (26) we have (−1)[n−1] y (2n−2) (t)0, t ∈ J y (2k) (0) = y (2k) (1) = 0, k = 0, 1, 2, . . . , n − 2. It follows from Lemma 1.1 that y(t)0, so w (t)0, t ∈ I , and w (t) is decreasing on I. If w (0)0, then w (t) 0, t ∈ I . Thus, w(t) is decreasing on I. By w(0) = 0 we know w(t)0, t ∈ I . In contradiction with w(t) > 0, hence w (0) > 0. Thus, lim
t→0+
w(t) w (t) = lim = w (0) > 0. t (1 − t) t→0+ 1 − 2t
Thus, there exist positive numbers M1 , m1 and 1 (1 < 21 ) such that m1
w(t) M1 , t (1 − t)
∀t ∈ [0, 1 ].
If w (1)0, since w (t) is decreasing, we know that w (t) 0, t ∈ I . Thus w(t) is increasing on I. By w(1) = 0 we know that w(t)0, t ∈ I . This is contradiction with w(t) > 0, hence w (1) < 0. So lim
t→1−
w(t) w (t) = lim = −w (1) > 0. t (1 − t) t→1− 1 − 2t
Thus, there exist positive numbers M2 , m2 and 2 (1 > 2 > 21 ) such that m2
w(t) M2 , t (1 − t)
∀t ∈ [2 , 1].
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
2559
w(t)/t (1 − t) is positive on [1 , 2 ]. Thus, there exist positive numbers M3 , m3 such that m3
w(t) M3 , t (1 − t)
∀t ∈ [1 , 2 ].
Choose m = min{mi , i = 1, 2, 3}, M = max{Mi , i = 1, 2, 3}, then mt(1 − t)w(t) Mt(1 − t),
t ∈ I,
that is (7) holds.
3. Further discussions
Corollary 3.1. Suppose that in Theorem 1.1 the conditions (H1) and (H3) hold, the condition (H2) is strengthened become (H4) f (t, ) ≡ / 0,
1 0
f (t, t (1 − t)) dt < ∞, ∀ > 0;
Then the positive solution w(t) of the problem (1), (2) is a C (2n−1) (I ) solution. Proof. By (7) and f (t, u) is decreasing with respect to u, we know that f (t, w(t)) f (t, mt(1 − t)). From (H4) we know that f (t, w(t)) is integrate on J, that is, the derivable function w(2n) (t) of w(2n−1) (t) is integrate on J. Thus w (2n−1) (0+), w(2n−1) (1 − 0) exist, that is, w(t) ∈ C (2n−1) (I ) ∩ C (2n) (J ). If f (t, u) is nonsingular at u = 0. Then for all u 0, f (t, u) f (t, 0), t ∈ J , and we have the following corollary. Corollary 3.2. Let (H1 ) f (t, u): J × [0, +∞) → [0, +∞) is continuous, and is decreasing with respect to u; 1 / 0, ∀ 0, 0 t (1 − t)f (t, 0) dt < ∞; (H2 ) f (t, ) ≡ Then the singular 2n-order BVP (1), (2) has a C (2n−2) (I ) positive solution w(t), and mt(1 − t)w(t)Mt(1 − t),
∃M m > 0, ∀t ∈ I .
(27)
In (8) the set P is replaced by P1 = {u(t) ∈ E | u(t)0, t ∈ I }, then (Tb) ∈ P1 holds, and (12), (13) and (14) hold. Let a(t) ≡ 0, then by conditions (H1 ) and (H2 ), we know that (15), (16) hold. The rest of proof is similar in Theorem 1.1.
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Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561
Corollary 3.3. Suppose that in Corollary 3.2 the condition (H1 ) holds, the condition (H2 ) is strengthened become / 0, ∀ 0, (H4 ) f (t, ) ≡
1 0
f (t, 0) dt < ∞,
Then the positive solution w(t) of the problem (1), (2) is a C (2n−1) (I ) solution. If f (t, u) is nonsingular at t, u, then corollary.
1 0
f (t, 0) dt < ∞ holds. Then we have the following
Corollary 3.4. If f (t, u) ∈ C(I × [0, +∞), [0, +∞)) is decreasing with respect to u, and f (t, ) ≡ / 0, ∀0. Then the problem (1), (2) has a positive solution, and (27) holds. Remark 1. If f (t, u) is nonnegative, decreasing and not identically zero, we obtain the existence of positive solution without other conditions. In order to show the validity of our results, we study the following sixth-order boundary value problem (6) u (t) + a0 (t) + ni=1 ai (t)u−i = 0, t ∈ J, (28) u(0) = u(1) = u(2) (0) = u(2) (1) = u(4) (0) = u(4) (1) = 0, where n a0 (t) and ai (t) are nonnegative and continuous on J, 0 < i < 1, (i = 1, 2, . . . , n), / 0 on I. i=0 ai (t) ≡ If
1 n (29) t (1 − t) a0 (t) + ai (t)t −i (1 − t)−i dt < + ∞, 0
i=1
then the boundary value of (28) has a positive solution w(t) ∈ C (4) (I ) ∩ C (6) (J ), and mt(1 − t)w(t)Mt(1 − t), If 0
1
a0 (t) +
n
∃M m > 0.
ai (t)t
−i
(1 − t)
−i
dt < + ∞,
(30)
i=1
then the positive solution of (28) is a C (5) (I ) positive solution. Proof. Let f (t, u) = a0 (t) + ni=1 ai (t)u−i , t ∈ J . It is easy to prove that (H1) and (H2) are satisfied in Theorem 1.1 under the condition (29). (H1) and (H4) are satisfied in Corollary 3.1 under the condition (30). Denote =max1 i n {i }. Then f (t, u) f (t, ru)r − f (t, u) holds for all positive numbers r < 1. By Te, T 2 e ∈ P , we know that there exist positive numbers k1 , k2 such that Te k1 e, T 2 e k2 e. Choose a positive number r0 such that r0 (min{1, k1 , k2 })1/(1−
2)
Z. Zhao / Nonlinear Analysis 64 (2006) 2553 – 2561 1−2
Thus k1 r0
1−2
r0 , k2 r0
2561
; Consequently,
T (r0 e) Te k1 e r0 e, −
T (r0 e) r0 Te,
2
2
T 2 (r0 e)r0 T 2 e r0 k2 e r0 e.
We take a(t) = r0 e(t), then the condition (H3) of Theorem 1.1 is satisfied. Thus, this completes the proof. References [1] J. Henderson, H.B. Thompson, Existence of multiple solutions for second order boundary value problems, J. Differential Equations 166 (2000) 443–454. [2] L. Kong, Q. Kong, Even order nonlinear eigenvalue problems on a measure chain, Nonlinear Anal. 52 (2003) 1891–1909. [3] Y.S. Liu, Multiple positive solutions of nonlinear singular boundary value problem for fourth-order equations, Appl. Math. Lett. 17 (2004) 747–757. [4] Z.L. Wei, Positive solution of fourth order singular boundary value problem, Acta Math. Sinica 42 (4) (1999) 715–722 (in Chinese). [5] X.J. Yang, Green’s function and positive solutions for higher-order ODE, Appl. Math. Comput. 136 (2003) 379–393. [6] Q.L. Yao, Monotone iterative technique and positive solutions of Lidstone boundary value problems, Appl. Math. Comput. 131 (2002) 477–485. [7] Q.L. Yao, Existence and iteration of n symmetric positive solutions for a singular two-point boundary value problem, Comput. Math. Appl. 47 (2004) 1195–1200.