- Email: [email protected]
Positive solutions for nonhomogeneous boundary value problems in banach spaces
Applied Mathematics and Computation 217 (2010) 3504–3510
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Positive solutions for nonhomogeneous boundary value problems in banach spaces q Weihua Jiang a,⇑, Bin Wang b a b
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018 Hebei, PR China Department of Basic Courses, Hebei Professional and Technological College of Chemical and Pharmaceutical Engineering, Shijiazhuang, 050026 Hebei, PR China
a r t i c l e
i n f o
a b s t r a c t
Keywords: Banach space Integral operator Strict set contraction Nonhomogeneous boundary value problem
By using Darbo fixed point theorem, the existence and nonexistence of fixed point are obtained for the following integral operator
T k yðtÞ ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds þ kt v :
0
Using these results, we investigate the existence and nonexistence of positive solutions for second-order nonhomogeneous local and nonlocal boundary value problems in Banach spaces. 2010 Elsevier Inc. All rights reserved.
1. Introduction The nonlinear ordinary differential equations with homogeneous and nonhomogeneous boundary conditions have been investigated extensively. We refer the readers to [1–9] and the references cited therein. Using Shauder fixed point theorem, Ma [4] and Guo et al. [5] studied the existence of solutions for the second-order differential equation
u00 ðtÞ þ aðtÞf ðt; uÞ ¼ 0; under the nonhomogeneous three-point boundary conditions
uð0Þ ¼ 0;
uð1Þ auðgÞ ¼ b
and multi-point boundary conditions
uð0Þ ¼ 0;
uð1Þ
m 2 X
ki uðni Þ ¼ b;
i¼1
respectively. Recently, the theory of ordinary differential equations in Banach spaces has become a new important branch (see, for example [10–13]). And the existence of solutions for boundary value problems of ordinary differential equations in Banach spaces has been studied by many authors (see, for example [14–20]). Guo and Lakshmikantham [14] and Liu [15] studied the existence of positive solutions for homogeneous second-order two-point boundary value problem
q This work is supported by the Natural Science Foundation of China (10875094) (10701032), the Foundation of Hebei Education Department (2008153) and the Foundation of Hebei University of Science and Technology (XL200814). ⇑ Corresponding author. E-mail addresses: [email protected] (W. Jiang), [email protected] (B. Wang).
0096-3003/$ - see front matter 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2010.09.023
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
x00 ðtÞ ¼ f ðt; xÞ;
3505
t 2 ½0; 1;
xð0Þ ¼ xð1Þ ¼ h and second-order three-point boundary value problem
y00 ðtÞ þ aðtÞf ðyÞ ¼ h;
t 2 ð0; 1Þ;
yð1Þ ¼ byðgÞ;
yð0Þ ¼ h;
in Banach space E, respectively, where h is the zero of E, 0 < g < 1. To the best of our knowledge, no papers have considered the existence of solutions for nonhomogeneous boundary value problems in Banach spaces. We will fill this gap in the literature. In this paper, we study the existence and nonexistence of fixed points in Banach space C[I, E] for the integral operator
T k yðtÞ ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds þ ktv ;
t 2 ½0; 1;
0
where E is a Banach space with cone P, v 2 P, kvk = 1, k > 0 and f : [0, 1] P ? P being continuous. Using these results, we investigate the existence and nonexistence of positive solutions for the differential equation
y00 ðtÞ þ f ðt; yÞ ¼ h;
t 2 ½0; 1;
ð1:1Þ
under the following boundary conditions
yð0Þ ¼ h;
yð1Þ ¼ kv
yð0Þ ¼ h;
yð1Þ
ð1:2Þ
and m2 X
ki yðni Þ ¼ kv ;
ð1:3Þ
i¼1
respectively, where ki > 0, i = 1, 2, . . . , m 2, 0 < n1 < n2 < < nm2 < 1. 2. Preliminaries Let P be a normal cone of a real Banach space E with norm k k. The normal constant is denoted by N, i.e. h 6 u 6 v implies kuk 6 N kvk. The norm in space C[I, E] is defined by kxkc = maxt2Ikx(t)k, where I = [0, 1]. Obviously, C[I, E] is a Banach space and Q = {x 2 C[I, E]jx(t) P h for t 2 I} is a cone of C[I, E]. The closed balls in spaces E and C[I, E] are denoted by Sl = {x 2 Ejjjxk 6 l}(l > 0) and Bl = {x 2 C[I, E] jjjxkc 6 l}(l > 0), respectively. y being a positive fixed point of Tk means that Tky = y and y 2 Q. y is a positive solution of problem (1.1) and (1.2) (or (1.1) and (1.3)) if it satisfies (1.1) and (1.2) (or (1.1) and (1.3)) and y 2 Q. For convenience, we list the following conditions: (H1) G(, s) and G(t, ) are continuous on I and there exists a constant e > 0 such that
0 6 Gðt; sÞ 6 e;
t; s 2 I:
(H2) There exist constants 0 < c 6 a < b < 1 such that
Gðt; sÞ P cGðs; sÞ;
t 2 ½a; b;
s; s 2 I:
Rb (H3) There exists t0 2 [a, b] such that m :¼ a Gðt 0 ; sÞds > 0. (H4) f 2 C[I P, P], f(t, h) = h, "t 2 I. For any l > 0, f(t, x) is uniformly continuous and bounded on I (P \ Sl) and there exists a 1 constant Ll with 0 6 Ll < 2e such that
aðf ðt; DÞÞ 6 Ll aðDÞ; 8t 2 I; D P \ Sl ; where a denotes the Kuratowski measure of non-compactness. ðt;uÞk (H5) lim supkuk!0 sup kf kuk ¼ 0. t2I
(H6) There exists u 2 P* such that u(u) > 0 for any u > h and
lim inf inf kuk!1
t2I
uðf ðt; uÞÞ ¼ 1; uðuÞ
where P* is the dual cone of P([20]). (H7) For any t 2 I, u, v 2 P and u 6 v, f(t, u) 6 f(t, v). ðt;uÞk (H8) lim supkuk!1 supt2I kf kuk ¼ 0. Lemma 2.1. Suppose (H1) and (H4) hold. Then, for any l > 0, operator Tk is a strict set contraction on Q \ Bl.
3506
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
Proof. Define the operator A as follows
AyðtÞ ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds:
0
By (H1), (H4) and Lebesgue Dominated Convergence theorem, we get that A : Q ? Q is continuous, uniformly bounded on Q \ Bl and for any s 2 [0, 1]
lim t!s
Z
1
jGðt; sÞ Gðs; sÞjds ¼ 0:
ð2:1Þ
0
As [0, 1] is compact, the limit in (2.1) is uniform in s 2 [0, 1]. For any x 2 Q \ Bl, t1, t2 2 I, we have
Z
kAxðt 1 Þ Axðt 2 Þk 6 M
1
jGðt1 ; sÞ Gðt 2 ; sÞjds;
0
where M ¼ supt2I;y2Q \Bl kf ðt; yðtÞÞk. From (H4) we know that M < +1. This means that the functions A(Q \ Bl) = {Axjx 2 Q \ Bl} are equicontinuous. By the same method used in [14], we can obtain that A is a strict set contraction on Q \ Bl. Since Tk(S) = A(S) + ktv, for any S Q \ Bl, we get a(Tk(S)) = a(A(S)). This implies that Tk is a strict set contraction. The proof is completed. h Lemma 2.2. Suppose (H1), (H2) and (H4) hold. Then, for y 2 Q, we have
T k yðtÞ P cT k yðsÞ;
t 2 ½a; b;
s 2 I:
Since the proof of this lemma is easy, we omit it. Lemma 2.3 (Darbo [10]). Let D be a bounded, closed and convex subset of E. If operator A : D ? D is a strict set contraction, then A has a fixed point in D.
3. Fixed point theorems of operator Tk In this section, we always suppose that (H1), (H2) and (H4) hold. Let
K ¼ fy 2 Q jyðtÞ P cyðsÞ; t 2 ½a; b; s 2 Ig: Obviously, K Q is a cone of C[I, E]. By Lemma 2.2, we get that Tk(Q) K. So, we have y 2 K if y 2 Q is a fixed point of Tk. Lemma 3.1. Suppose (H5) holds. Then Tk has at least one fixed point in K for sufficiently small k 2 (0, 1). Proof. By (H5), we get that there exists a constant r > 0 such that
1 kuk; 2e
kf ðt; uÞk 6
kuk 6 r;
t 2 I:
ð3:1Þ
Let
K r ¼ fy 2 Kjjjykc 6 rg: For y 2 Kr and 0 < k 6 2r , from (3.1) we get
kT k ykc 6
Z
1
ekf ðs; yðsÞÞkds þ k 6 r:
0
This implies Tk(Kr) Kr. It follows from Lemmas 2.1 and 2.3 that Tk has at least one fixed point in Kr. The proof is completed. h Lemma 3.2. Suppose (H3) and (H6) hold. Then Tk has no fixed point in K for k large enough. Proof. Otherwise, there exist constants 0 < k1 < k2 < < kn < with limn?1kn = +1 such that T kn has at least one fixed point yn 2 K, i.e.
yn ðtÞ ¼ T kn yn ðtÞ ¼
Z 0
1
Gðt; sÞf ðs; yn ðsÞÞds þ kn t v :
ð3:2Þ
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
3507
Obviously, yn(t) P kntv P h. Since P is a normal cone, we have
kyn kc P kyn ð1Þk P
kn ! þ1; N
n ! 1:
By (H6), we get that there exists a constant R > 0 such that
uðf ðt; uÞÞ >
2
cm
uðuÞ; kuk > R; t 2 I:
ð3:3Þ
Take n large enough such that kyn kc > Nc R. Then, for t 2 [a, b], we have
kyn ðtÞk P
c
kyn kc > R:
N
By (3.2) and (3.3), we have
uðyn ðtÞÞ ¼
Z
1
Gðt; sÞuðf ðs; yn ðsÞÞÞds þ kn t uðv Þ P
Z
0
b
Gðt; sÞuðf ðs; yn ðsÞÞÞds P
a
2 cm
Z
b
Gðt; sÞuðyn ðsÞÞds:
a
Take t = t0. For s 2 [a, b], from yn(s) P cyn(t0) we get
2 cm
uðyn ðt0 ÞÞ P
Z a
b
Gðt0 ; sÞ cuðyn ðt0 ÞÞds ¼ 2uðyn ðt0 ÞÞ:
This is a contradiction since yn(t0) > h. The proof is completed. h Theorem 3.1. Suppose (H3), (H5)–(H7) hold. Then there exists k* > 0 such that Tk has at least one fixed point in K for k 2 (0, k*) and no fixed point in K for k 2 (k*, 1). Proof. Take
K ¼ fk > 0jT k has at least one fixed point in Kg; and k* = supK. By Lemmas 3.1 and 3.2, we get 0 < k* < +1. For any 0 < k < k*, there exists k1 2 K with k < k1 < k* such that T k1 has a fixed point y1 2 K. Set
K 1 ¼ fy 2 KjyðtÞ 6 y1 ðtÞ; t 2 Ig: Obviously, K1 is bounded, closed and convex in C[I, E]. From Lemma 2.1 we know that Tk is a strict set contraction on K1. For y 2 K1, by (H7), we have
Tky ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds þ ktv 6
0
Z
1
Gðt; sÞf ðs; y1 ðsÞÞds þ k1 t v ¼ y1 :
0
This means Tk(K1) K1. From Lemma 2.3 we get that Tk has at least one fixed point in K1. The proof is completed. h Theorem 3.2. Suppose (H8) holds. Then Tk has at least one fixed point in K for any k 2 (0, 1). Proof. It follows from (H8) that there exists R0 > 0 such that
kf ðt; uÞk <
1 kuk; 3e
u 2 P;
kuk > R0 ;
t 2 I:
ð3:3Þ
By (H4), we get that there exists M > 0 such that
kf ðt; uÞk < M;
t 2 I;
u 2 P;
kuk 6 R0 :
ð3:4Þ
Take
R ¼ maxfR0 ; 3k; 3Meg;
K R ¼ fy 2 Kjjjykc 6 Rg:
For y 2 KR, from (3.3) and (3.4) we get
kT k yðtÞk 6
Z 0
1
Gðt; sÞkf ðs; yðsÞÞkds þ k ¼
Z I1
Gðt; sÞkf ðs; yðsÞÞkds þ
Z
1 Gðt; sÞkf ðs; yðsÞÞkds þ k 6 Me þ kykc þ k 6 R; 3 InI1
where I1 = {s 2 Ijjjy(s)k 6 R0}. This implies Tk(KR) KR. By Lemma 2.3, we get that Tk has at least one fixed point in KR. h
3508
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
4. Positive solutions to the problem (1.1) and (1.2) We can easily get the following lemmas. Lemma 4.1. y(t) is a positive solution to the problem (1.1) and (1.2) if and only if y(t) 2 Q is a fixed point of the following operator
Tky ¼
Z
1
G0 ðt; sÞf ðs; yðsÞÞds þ kt v ;
0
where
G0 ðt; sÞ ¼
tð1 sÞ; 0 6 t 6 s 6 1; sð1 tÞ; 0 6 s < t 6 1:
Lemma 4.2. Take 0 < a < 12 < b < 1; t 0 ¼ 12 ; c ¼ að1 bÞ < a; e ¼ 14. Then G0(t, s) satisfies conditions (H1)–(H3). By Theorems 3.1 and 3.2, we can get the following theorems. Theorem 4.1. Suppose (H4)–(H7) hold. Then, for v 2 P, kvk = 1, there exists k* > 0 such that the problem (1.1) and (1.2) has at least one positive solution for k 2 (0, k*) and no positive solution for k 2 (k*, 1). Theorem 4.2. Suppose (H4) and (H8) hold. Then, for v 2 P, kvk = 1, the problem (1.1) and (1.2) has at least one positive solution for any k 2 (0, 1). 5. Positive solutions to the problem (1.1) and (1.3) In this section, we will always suppose that 0 <
Pm2 i¼1
ki ni < 1.
Lemma 5.1. Suppose that g(t, s) : [0, 1] [0, 1] ? [0, 1), g(, s) and g(t, ) are continuous on [0, 1]. If for any y 2 C[0, 1], y(t) P 0, the following inequality
min
t2½a;b
Z
1
gðt; sÞyðsÞds P c max
s2½0;1
0
Z
1
gðs; sÞyðsÞds
0
holds. Then we have
gðt; sÞ P cgðs; sÞ;
8t 2 ½a; b;
s; s 2 ½0; 1;
where 0 6 a < b 6 1, 0 < c < 1. Proof. Otherwise, there exist t0 2 [a, b], s0, s0 2 [0, 1] such that
gðt 0 ; s0 Þ cgðs0 ; s0 Þ < 0: From the continuity of g(t0, s) cg(s0, s) we get that there exists d > 0 such that
gðt 0 ; sÞ cgðs0 ; sÞ < 0;
s 2 ½s0 d; s0 þ d
\ ½0; 1:
Take
y0 ðsÞ ¼ maxfgðt 0 ; sÞ þ cgðs0 ; sÞ; 0g: s2½0;1
Obviously, y0 2 C[0, 1], y0(t) P 0. But
Z
1
gðt0 ; sÞy0 ðsÞds < c
Z
0
1
gðs0 ; sÞy0 ðsÞds:
0
A contradiction. The proof is completed. h Lemma 5.2 [6]. For h(t) 2 C[I, E], the boundary value problem
y00 þ hðtÞ ¼ h; yð0Þ ¼ h;
yð1Þ
m2 X i¼1
t 2 ½0; 1; ki yðni Þ ¼ h
ð5:1Þ
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
3509
has a unique solution
yðtÞ ¼
Z
1
G1 ðt; sÞhðsÞds;
0
where
8 Pm2 Pi1 sð1tÞ kj ðnj tÞsþ k n ðtsÞ > j¼i j¼1 j j > > P ; m2 > > k n 1 > i¼1 i i > > < 0 6 t 6 1; n 6 s 6 minfn ; tg; i ¼ 1; 2; . . . ; m 1; i1 i G1 ðt; sÞ ¼ Pm2 > t½ð1sÞ kj ðnj sÞ > j¼i > > Pm2 ; > > kn 1 > i¼1 i i > : 0 6 t 6 1; maxfni1 ; tg 6 s 6 ni ; i ¼ 1; 2; . . . ; m 1: Lemma 5.3. G1(t, s) satisfies (H1)–(H3). Proof. Take a = n1, b = 1, a < t0 < b, e = maxs,t2IG1(t, s). Obviously, 0 < e < 1. It is easy to see that G1(t, s) satisfies (H1) and (H3). By Lemma 3 in [5], Lemmas 5.1 and 5.2, we get
G1 ðt; sÞ P cG1 ðs; sÞ;
t 2 ½n1 ; 1;
s; s 2 I;
where
(
c ¼ min
26s6m2
n1 ;
Pm2
ki ð1 ni Þ Xm2 ; kn; Pm2 i¼1 i i 1 i¼1 ki ni i¼1
Ps1
) P þ m2 i¼s ki ð1 ni Þ : Pm2 1 i¼s ki ni
i¼1 ki ni
So, G1(t, s) satisfies (H2). The proof is completed. h We can easily get that the boundary value problem
y00 ðtÞ ¼ h;
t 2 ½0; 1;
yð0Þ ¼ h;
yð1Þ
m2 X
ki yðni Þ ¼ kv
i¼1
has a unique solution
yðtÞ ¼
1
kt Pm2 i¼1
ki ni
v:
This, together with Lemma 5.2, implies that y(t) is a positive solution to the problem (1.1) and (1.3) if and only if it is a positive fixed point of the following operator
Tky ¼
Z 0
1
G1 ðt; sÞf ðs; yðsÞÞds þ
1
kt Pm2 i¼1
ki ni
v:
By Theorems 3.1 and 3.2, we can get the following theorems. Theorem 5.1. Suppose (H4)–(H7) hold. Then, for v 2 P, kvk = 1, there exists k* > 0 such that the problem (1.1) and (1.3) has at least one positive solution for k 2 (0, k*) and no positive solution for k 2 (k*, 1).
Theorem 5.2. Suppose (H4) and (H8) hold. Then, for v 2 P, kvk = 1, the problem (1.1) and (1.3) has at least one positive solution for any k 2 (0, 1). References [1] J.R.L. Webb, K.Q. Lan, Eigenvalue criteria for existence of multiple positive solutions of nonlinear boundary value problems of local and nonlocal type. Topological Methods in Nonlinear Analysis, J. Juliusz Schauder Cent. 27 (2006) 91–115. [2] J.R.L. Webb, Positive solutions of some three point boundary value problems via fixed point index theory, Nonlinear Anal. 4 (2001) 4319–4332. [3] W. Jiang, The existence of positive solutions for second-order multi-point BVPs with the first derivative, J. Comput. Appl. Math. 225 (2009) 387–392. [4] R. Ma, Positive solutions for second order three-point boundary value problems, Appl. Math. Lett. 14 (2001) 1–5. [5] Y. Guo, W. Shan, W. Ge, Positive solutions for second-order m-point boundary value problems, J. Comput. Appl. Math. 151 (2003) 415–424. [6] X. Liu, J. Qiu, Y. Guo, Three positive solutions for second-order m-point boundary value problems, Appl. Math. Comput. 156 (2004) 733–742. [7] B. Ahmad, A. Alsaedi, B.S. Alghamdi, Analytic approximation of solutions of the forced Duffng equation with integral boundary conditions, Nonlinear Anal. (RWA) 9 (2008) 1727–1740. [8] X. Zhang, W. Ge, Positive solutions for a class of boundary-value problems with integral boundary conditions, Comput. Math. Appl. 58 (2009) 203–215.
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W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
[9] M. Feng, B. Du, W. Ge, Impulsive boundary value problems with integral boundary conditions and one-dimensional p-Laplacian, Nonlinear Anal. (TMA) 70 (2009) 3119–3126. [10] D. Guo, V. Lakshmikantham, X. Liu, Nonlinear Integral Equation in Abstract spaces, Kluwer Academic Publishers, Dordrecht, 1996. [11] K. Demling, Ordinary Differential Equations in Banach Spaces, Springer, Berlin, 1977. [12] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, San Diego, 1988. [13] V. Lakshmikantham, S. Leela, Nonlinear Differential Equations in Abstract Spaces, Pergamon Press, Oxford, 1981. [14] D. Guo, V. Lakshmikantham, Multiple solutions of two-point boundary value problem of ordinary differential equations in Banach space, J. Math. Anal. Appl. 129 (1988) 211–222. [15] B. Liu, Positive solutions of second-order three-point boundary value problems with change of sign in Banach spaces, Nonlinear Anal. (TMA) 64 (2006) 1336–1355. [16] W. Jiang, Positive solutions for a high-order multi-point boundary-value problem in Banach spaces, Electron. J. Differ. Equa. 70 (2008) 1–11. [17] D. Guo, Existence of positive solutions for nth-order nonlinear impulsive singular integro-differential equations in Banach spaces, Nonlinear Anal. 68 (2008) 2727–2740. [18] Z. Wang, L. Liu, Y. Wu, The unique solution of boundary value problems for nonlinear second-order integral-differential equations of mixed type in Banach spaces, Comput. Math. Appl. 54 (2007) 1293–1301. [19] M. Feng, H. Pang, A class of three-point boundary-value problems for second-order impulsive integro-differential equations in Banach spaces, Nonlinear Anal. (TMA) 70 (2009) 64–82. [20] M. Feng, D. Ji, Weigao Ge, Positive solutions for a class of boundary-value problem with integral boundary conditions in Banach spaces, J. Comput. Appl. Math. 222 (2008) 351–363.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Positive solutions for nonhomogeneous boundary value problems in banach spaces q Weihua Jiang a,⇑, Bin Wang b a b
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018 Hebei, PR China Department of Basic Courses, Hebei Professional and Technological College of Chemical and Pharmaceutical Engineering, Shijiazhuang, 050026 Hebei, PR China
a r t i c l e
i n f o
a b s t r a c t
Keywords: Banach space Integral operator Strict set contraction Nonhomogeneous boundary value problem
By using Darbo fixed point theorem, the existence and nonexistence of fixed point are obtained for the following integral operator
T k yðtÞ ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds þ kt v :
0
Using these results, we investigate the existence and nonexistence of positive solutions for second-order nonhomogeneous local and nonlocal boundary value problems in Banach spaces. 2010 Elsevier Inc. All rights reserved.
1. Introduction The nonlinear ordinary differential equations with homogeneous and nonhomogeneous boundary conditions have been investigated extensively. We refer the readers to [1–9] and the references cited therein. Using Shauder fixed point theorem, Ma [4] and Guo et al. [5] studied the existence of solutions for the second-order differential equation
u00 ðtÞ þ aðtÞf ðt; uÞ ¼ 0; under the nonhomogeneous three-point boundary conditions
uð0Þ ¼ 0;
uð1Þ auðgÞ ¼ b
and multi-point boundary conditions
uð0Þ ¼ 0;
uð1Þ
m 2 X
ki uðni Þ ¼ b;
i¼1
respectively. Recently, the theory of ordinary differential equations in Banach spaces has become a new important branch (see, for example [10–13]). And the existence of solutions for boundary value problems of ordinary differential equations in Banach spaces has been studied by many authors (see, for example [14–20]). Guo and Lakshmikantham [14] and Liu [15] studied the existence of positive solutions for homogeneous second-order two-point boundary value problem
q This work is supported by the Natural Science Foundation of China (10875094) (10701032), the Foundation of Hebei Education Department (2008153) and the Foundation of Hebei University of Science and Technology (XL200814). ⇑ Corresponding author. E-mail addresses: [email protected] (W. Jiang), [email protected] (B. Wang).
0096-3003/$ - see front matter 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2010.09.023
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
x00 ðtÞ ¼ f ðt; xÞ;
3505
t 2 ½0; 1;
xð0Þ ¼ xð1Þ ¼ h and second-order three-point boundary value problem
y00 ðtÞ þ aðtÞf ðyÞ ¼ h;
t 2 ð0; 1Þ;
yð1Þ ¼ byðgÞ;
yð0Þ ¼ h;
in Banach space E, respectively, where h is the zero of E, 0 < g < 1. To the best of our knowledge, no papers have considered the existence of solutions for nonhomogeneous boundary value problems in Banach spaces. We will fill this gap in the literature. In this paper, we study the existence and nonexistence of fixed points in Banach space C[I, E] for the integral operator
T k yðtÞ ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds þ ktv ;
t 2 ½0; 1;
0
where E is a Banach space with cone P, v 2 P, kvk = 1, k > 0 and f : [0, 1] P ? P being continuous. Using these results, we investigate the existence and nonexistence of positive solutions for the differential equation
y00 ðtÞ þ f ðt; yÞ ¼ h;
t 2 ½0; 1;
ð1:1Þ
under the following boundary conditions
yð0Þ ¼ h;
yð1Þ ¼ kv
yð0Þ ¼ h;
yð1Þ
ð1:2Þ
and m2 X
ki yðni Þ ¼ kv ;
ð1:3Þ
i¼1
respectively, where ki > 0, i = 1, 2, . . . , m 2, 0 < n1 < n2 < < nm2 < 1. 2. Preliminaries Let P be a normal cone of a real Banach space E with norm k k. The normal constant is denoted by N, i.e. h 6 u 6 v implies kuk 6 N kvk. The norm in space C[I, E] is defined by kxkc = maxt2Ikx(t)k, where I = [0, 1]. Obviously, C[I, E] is a Banach space and Q = {x 2 C[I, E]jx(t) P h for t 2 I} is a cone of C[I, E]. The closed balls in spaces E and C[I, E] are denoted by Sl = {x 2 Ejjjxk 6 l}(l > 0) and Bl = {x 2 C[I, E] jjjxkc 6 l}(l > 0), respectively. y being a positive fixed point of Tk means that Tky = y and y 2 Q. y is a positive solution of problem (1.1) and (1.2) (or (1.1) and (1.3)) if it satisfies (1.1) and (1.2) (or (1.1) and (1.3)) and y 2 Q. For convenience, we list the following conditions: (H1) G(, s) and G(t, ) are continuous on I and there exists a constant e > 0 such that
0 6 Gðt; sÞ 6 e;
t; s 2 I:
(H2) There exist constants 0 < c 6 a < b < 1 such that
Gðt; sÞ P cGðs; sÞ;
t 2 ½a; b;
s; s 2 I:
Rb (H3) There exists t0 2 [a, b] such that m :¼ a Gðt 0 ; sÞds > 0. (H4) f 2 C[I P, P], f(t, h) = h, "t 2 I. For any l > 0, f(t, x) is uniformly continuous and bounded on I (P \ Sl) and there exists a 1 constant Ll with 0 6 Ll < 2e such that
aðf ðt; DÞÞ 6 Ll aðDÞ; 8t 2 I; D P \ Sl ; where a denotes the Kuratowski measure of non-compactness. ðt;uÞk (H5) lim supkuk!0 sup kf kuk ¼ 0. t2I
(H6) There exists u 2 P* such that u(u) > 0 for any u > h and
lim inf inf kuk!1
t2I
uðf ðt; uÞÞ ¼ 1; uðuÞ
where P* is the dual cone of P([20]). (H7) For any t 2 I, u, v 2 P and u 6 v, f(t, u) 6 f(t, v). ðt;uÞk (H8) lim supkuk!1 supt2I kf kuk ¼ 0. Lemma 2.1. Suppose (H1) and (H4) hold. Then, for any l > 0, operator Tk is a strict set contraction on Q \ Bl.
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Proof. Define the operator A as follows
AyðtÞ ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds:
0
By (H1), (H4) and Lebesgue Dominated Convergence theorem, we get that A : Q ? Q is continuous, uniformly bounded on Q \ Bl and for any s 2 [0, 1]
lim t!s
Z
1
jGðt; sÞ Gðs; sÞjds ¼ 0:
ð2:1Þ
0
As [0, 1] is compact, the limit in (2.1) is uniform in s 2 [0, 1]. For any x 2 Q \ Bl, t1, t2 2 I, we have
Z
kAxðt 1 Þ Axðt 2 Þk 6 M
1
jGðt1 ; sÞ Gðt 2 ; sÞjds;
0
where M ¼ supt2I;y2Q \Bl kf ðt; yðtÞÞk. From (H4) we know that M < +1. This means that the functions A(Q \ Bl) = {Axjx 2 Q \ Bl} are equicontinuous. By the same method used in [14], we can obtain that A is a strict set contraction on Q \ Bl. Since Tk(S) = A(S) + ktv, for any S Q \ Bl, we get a(Tk(S)) = a(A(S)). This implies that Tk is a strict set contraction. The proof is completed. h Lemma 2.2. Suppose (H1), (H2) and (H4) hold. Then, for y 2 Q, we have
T k yðtÞ P cT k yðsÞ;
t 2 ½a; b;
s 2 I:
Since the proof of this lemma is easy, we omit it. Lemma 2.3 (Darbo [10]). Let D be a bounded, closed and convex subset of E. If operator A : D ? D is a strict set contraction, then A has a fixed point in D.
3. Fixed point theorems of operator Tk In this section, we always suppose that (H1), (H2) and (H4) hold. Let
K ¼ fy 2 Q jyðtÞ P cyðsÞ; t 2 ½a; b; s 2 Ig: Obviously, K Q is a cone of C[I, E]. By Lemma 2.2, we get that Tk(Q) K. So, we have y 2 K if y 2 Q is a fixed point of Tk. Lemma 3.1. Suppose (H5) holds. Then Tk has at least one fixed point in K for sufficiently small k 2 (0, 1). Proof. By (H5), we get that there exists a constant r > 0 such that
1 kuk; 2e
kf ðt; uÞk 6
kuk 6 r;
t 2 I:
ð3:1Þ
Let
K r ¼ fy 2 Kjjjykc 6 rg: For y 2 Kr and 0 < k 6 2r , from (3.1) we get
kT k ykc 6
Z
1
ekf ðs; yðsÞÞkds þ k 6 r:
0
This implies Tk(Kr) Kr. It follows from Lemmas 2.1 and 2.3 that Tk has at least one fixed point in Kr. The proof is completed. h Lemma 3.2. Suppose (H3) and (H6) hold. Then Tk has no fixed point in K for k large enough. Proof. Otherwise, there exist constants 0 < k1 < k2 < < kn < with limn?1kn = +1 such that T kn has at least one fixed point yn 2 K, i.e.
yn ðtÞ ¼ T kn yn ðtÞ ¼
Z 0
1
Gðt; sÞf ðs; yn ðsÞÞds þ kn t v :
ð3:2Þ
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
3507
Obviously, yn(t) P kntv P h. Since P is a normal cone, we have
kyn kc P kyn ð1Þk P
kn ! þ1; N
n ! 1:
By (H6), we get that there exists a constant R > 0 such that
uðf ðt; uÞÞ >
2
cm
uðuÞ; kuk > R; t 2 I:
ð3:3Þ
Take n large enough such that kyn kc > Nc R. Then, for t 2 [a, b], we have
kyn ðtÞk P
c
kyn kc > R:
N
By (3.2) and (3.3), we have
uðyn ðtÞÞ ¼
Z
1
Gðt; sÞuðf ðs; yn ðsÞÞÞds þ kn t uðv Þ P
Z
0
b
Gðt; sÞuðf ðs; yn ðsÞÞÞds P
a
2 cm
Z
b
Gðt; sÞuðyn ðsÞÞds:
a
Take t = t0. For s 2 [a, b], from yn(s) P cyn(t0) we get
2 cm
uðyn ðt0 ÞÞ P
Z a
b
Gðt0 ; sÞ cuðyn ðt0 ÞÞds ¼ 2uðyn ðt0 ÞÞ:
This is a contradiction since yn(t0) > h. The proof is completed. h Theorem 3.1. Suppose (H3), (H5)–(H7) hold. Then there exists k* > 0 such that Tk has at least one fixed point in K for k 2 (0, k*) and no fixed point in K for k 2 (k*, 1). Proof. Take
K ¼ fk > 0jT k has at least one fixed point in Kg; and k* = supK. By Lemmas 3.1 and 3.2, we get 0 < k* < +1. For any 0 < k < k*, there exists k1 2 K with k < k1 < k* such that T k1 has a fixed point y1 2 K. Set
K 1 ¼ fy 2 KjyðtÞ 6 y1 ðtÞ; t 2 Ig: Obviously, K1 is bounded, closed and convex in C[I, E]. From Lemma 2.1 we know that Tk is a strict set contraction on K1. For y 2 K1, by (H7), we have
Tky ¼
Z
1
Gðt; sÞf ðs; yðsÞÞds þ ktv 6
0
Z
1
Gðt; sÞf ðs; y1 ðsÞÞds þ k1 t v ¼ y1 :
0
This means Tk(K1) K1. From Lemma 2.3 we get that Tk has at least one fixed point in K1. The proof is completed. h Theorem 3.2. Suppose (H8) holds. Then Tk has at least one fixed point in K for any k 2 (0, 1). Proof. It follows from (H8) that there exists R0 > 0 such that
kf ðt; uÞk <
1 kuk; 3e
u 2 P;
kuk > R0 ;
t 2 I:
ð3:3Þ
By (H4), we get that there exists M > 0 such that
kf ðt; uÞk < M;
t 2 I;
u 2 P;
kuk 6 R0 :
ð3:4Þ
Take
R ¼ maxfR0 ; 3k; 3Meg;
K R ¼ fy 2 Kjjjykc 6 Rg:
For y 2 KR, from (3.3) and (3.4) we get
kT k yðtÞk 6
Z 0
1
Gðt; sÞkf ðs; yðsÞÞkds þ k ¼
Z I1
Gðt; sÞkf ðs; yðsÞÞkds þ
Z
1 Gðt; sÞkf ðs; yðsÞÞkds þ k 6 Me þ kykc þ k 6 R; 3 InI1
where I1 = {s 2 Ijjjy(s)k 6 R0}. This implies Tk(KR) KR. By Lemma 2.3, we get that Tk has at least one fixed point in KR. h
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W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
4. Positive solutions to the problem (1.1) and (1.2) We can easily get the following lemmas. Lemma 4.1. y(t) is a positive solution to the problem (1.1) and (1.2) if and only if y(t) 2 Q is a fixed point of the following operator
Tky ¼
Z
1
G0 ðt; sÞf ðs; yðsÞÞds þ kt v ;
0
where
G0 ðt; sÞ ¼
tð1 sÞ; 0 6 t 6 s 6 1; sð1 tÞ; 0 6 s < t 6 1:
Lemma 4.2. Take 0 < a < 12 < b < 1; t 0 ¼ 12 ; c ¼ að1 bÞ < a; e ¼ 14. Then G0(t, s) satisfies conditions (H1)–(H3). By Theorems 3.1 and 3.2, we can get the following theorems. Theorem 4.1. Suppose (H4)–(H7) hold. Then, for v 2 P, kvk = 1, there exists k* > 0 such that the problem (1.1) and (1.2) has at least one positive solution for k 2 (0, k*) and no positive solution for k 2 (k*, 1). Theorem 4.2. Suppose (H4) and (H8) hold. Then, for v 2 P, kvk = 1, the problem (1.1) and (1.2) has at least one positive solution for any k 2 (0, 1). 5. Positive solutions to the problem (1.1) and (1.3) In this section, we will always suppose that 0 <
Pm2 i¼1
ki ni < 1.
Lemma 5.1. Suppose that g(t, s) : [0, 1] [0, 1] ? [0, 1), g(, s) and g(t, ) are continuous on [0, 1]. If for any y 2 C[0, 1], y(t) P 0, the following inequality
min
t2½a;b
Z
1
gðt; sÞyðsÞds P c max
s2½0;1
0
Z
1
gðs; sÞyðsÞds
0
holds. Then we have
gðt; sÞ P cgðs; sÞ;
8t 2 ½a; b;
s; s 2 ½0; 1;
where 0 6 a < b 6 1, 0 < c < 1. Proof. Otherwise, there exist t0 2 [a, b], s0, s0 2 [0, 1] such that
gðt 0 ; s0 Þ cgðs0 ; s0 Þ < 0: From the continuity of g(t0, s) cg(s0, s) we get that there exists d > 0 such that
gðt 0 ; sÞ cgðs0 ; sÞ < 0;
s 2 ½s0 d; s0 þ d
\ ½0; 1:
Take
y0 ðsÞ ¼ maxfgðt 0 ; sÞ þ cgðs0 ; sÞ; 0g: s2½0;1
Obviously, y0 2 C[0, 1], y0(t) P 0. But
Z
1
gðt0 ; sÞy0 ðsÞds < c
Z
0
1
gðs0 ; sÞy0 ðsÞds:
0
A contradiction. The proof is completed. h Lemma 5.2 [6]. For h(t) 2 C[I, E], the boundary value problem
y00 þ hðtÞ ¼ h; yð0Þ ¼ h;
yð1Þ
m2 X i¼1
t 2 ½0; 1; ki yðni Þ ¼ h
ð5:1Þ
W. Jiang, B. Wang / Applied Mathematics and Computation 217 (2010) 3504–3510
3509
has a unique solution
yðtÞ ¼
Z
1
G1 ðt; sÞhðsÞds;
0
where
8 Pm2 Pi1 sð1tÞ kj ðnj tÞsþ k n ðtsÞ > j¼i j¼1 j j > > P ; m2 > > k n 1 > i¼1 i i > > < 0 6 t 6 1; n 6 s 6 minfn ; tg; i ¼ 1; 2; . . . ; m 1; i1 i G1 ðt; sÞ ¼ Pm2 > t½ð1sÞ kj ðnj sÞ > j¼i > > Pm2 ; > > kn 1 > i¼1 i i > : 0 6 t 6 1; maxfni1 ; tg 6 s 6 ni ; i ¼ 1; 2; . . . ; m 1: Lemma 5.3. G1(t, s) satisfies (H1)–(H3). Proof. Take a = n1, b = 1, a < t0 < b, e = maxs,t2IG1(t, s). Obviously, 0 < e < 1. It is easy to see that G1(t, s) satisfies (H1) and (H3). By Lemma 3 in [5], Lemmas 5.1 and 5.2, we get
G1 ðt; sÞ P cG1 ðs; sÞ;
t 2 ½n1 ; 1;
s; s 2 I;
where
(
c ¼ min
26s6m2
n1 ;
Pm2
ki ð1 ni Þ Xm2 ; kn; Pm2 i¼1 i i 1 i¼1 ki ni i¼1
Ps1
) P þ m2 i¼s ki ð1 ni Þ : Pm2 1 i¼s ki ni
i¼1 ki ni
So, G1(t, s) satisfies (H2). The proof is completed. h We can easily get that the boundary value problem
y00 ðtÞ ¼ h;
t 2 ½0; 1;
yð0Þ ¼ h;
yð1Þ
m2 X
ki yðni Þ ¼ kv
i¼1
has a unique solution
yðtÞ ¼
1
kt Pm2 i¼1
ki ni
v:
This, together with Lemma 5.2, implies that y(t) is a positive solution to the problem (1.1) and (1.3) if and only if it is a positive fixed point of the following operator
Tky ¼
Z 0
1
G1 ðt; sÞf ðs; yðsÞÞds þ
1
kt Pm2 i¼1
ki ni
v:
By Theorems 3.1 and 3.2, we can get the following theorems. Theorem 5.1. Suppose (H4)–(H7) hold. Then, for v 2 P, kvk = 1, there exists k* > 0 such that the problem (1.1) and (1.3) has at least one positive solution for k 2 (0, k*) and no positive solution for k 2 (k*, 1).
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