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Green’s function and positive solutions for higher-order ODE
Applied Mathematics and Computation 136 (2003) 379–393 www.elsevier.com/locate/amc
GreenÕs function and positive solutions for higher-order ODE Xiaojing Yang Department of Mathematics, Tsinghua University, Beijing 100084, People’s Republic of China
Abstract GreenÕs function for ðk; n kÞ conjugate boundary value problem ð1Þnk xðnÞ ðtÞ ¼ aðtÞf ðxðtÞÞ;
a < t < b;
ðiÞ
0 6 i 6 k 1;
ðjÞ
0 6 j 6 n k 1;
x ðaÞ ¼ 0; x ðbÞ ¼ 0;
is obtained and its integral representation is given. As applications, we obtain sufficient conditions for the existence of one, two or three positive solutions. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: GreenÕs function; Positive solutions; Fixed points
1. Introduction GreenÕs function plays an important role in solving boundary value problems of ordinary differential equations. But to obtain explicit formula of GreenÕs function is not always an easy job. In this paper, we consider the ðk; n kÞ conjugate boundary value problem: ð1Þnk xðnÞ ðtÞ ¼ aðtÞf ðxðtÞÞ;
a < t < b;
ðiÞ
0 6 i 6 k 1;
ðjÞ
0 6 j 6 n k 1;
x ðaÞ ¼ 0; x ðbÞ ¼ 0;
E-mail address: [email protected] (X. Yang). 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 2 ) 0 0 0 5 6 - 5
ð1Þ
380
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
where a; f are continuous, 1 6 k < n. First, we obtain the explicit formula of GreenÕs function for the following boundary value problem: ð1Þ
nk ðnÞ
x ðtÞ ¼ 0;
xðiÞ ðaÞ ¼ 0;
a < t < b; ð2Þ
0 6 i 6 k 1;
xðjÞ ðbÞ ¼ 0 0 6 j 6 n k 1: Then by applying the properties of GreenÕs function, we give sufficient conditions for the existence of at least one, two or three positive solutions to problem (1). By GreenÕs function Gðt; sÞ of the ðk; n kÞ conjugate boundary value problem (1) we mean that Gðt; sÞ satisfies ð1Þ
nk
GðnÞ ðt; sÞ ¼ dðt; sÞ;
ðiÞ
a < t < b; a < s < b; ð3Þ
G ðaÞ ¼ 0;
0 6 i 6 k 1;
GðjÞ ðbÞ ¼ 0;
0 6 j 6 n k 1;
where GðmÞ ðt; sÞ denotes the mth-order derivatives with respect to t and dðxÞ is the Dirac delta function. It is very useful to study the existence of positive solutions to various boundary value problems by deducing related inequalities from GreenÕs functions, see, for example [1–5].
2. Green’s function Now we can state our main result of this paper. Theorem 1. The Green function for the boundary value problem (2) is given by 8 a 6 t 6 s 6 b; < P ðt; sÞ; n1 nk t s Gðt; sÞ ¼ ð4Þ ð1Þ : P ðt; sÞ þ ; a 6 s 6 t 6 b; ðn 1Þ! b a where P ðt; sÞ ¼
nk1 X
nkj1
ððs aÞ=ðb aÞÞ ððt aÞ=ðb aÞÞ ðk 1Þ!j!ðn k j 1Þ! j¼0 k1 Z b bs s a j ds: ba ba a
kþj
1 ðb aÞ ð5Þ
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
381
Moreover, Green’s function Gðt; sÞ can be further reduced to the following integral form: 8 R ððtaÞðbsÞÞ=ðbaÞ k1 nk1 1 > s ðs þ s tÞ ds; n1 > 0 ðk1Þ!ðnk1Þ!ðbaÞ > < a 6 t 6 s 6 b; ð6Þ Gðt; sÞ ¼ R ððbtÞðsaÞÞ=ðbaÞ nk1 k1 1 > s ðs þ t sÞ ds; > n1 0 > ðk1Þ!ðnk1Þ!ðbaÞ : a 6 s 6 t 6 b: Corollary 1. Let k 1 ¼ n k 1 in Theorem 1 and Corollary 2 below, we have for n ¼ 2k and ( P ðt; sÞ; a 6 t 6 s 6 b; ts 2k1 Gðt; sÞ ¼ ð7Þ ð1Þk P ðt; sÞ þ ð2k1Þ! ba ; a 6 s 6 t 6 b; with P ðt; sÞ ¼
k1j kþj k1 X ððs aÞ=ðb aÞÞ ððt aÞ=ðb aÞÞ 1 b a ðk 1Þ!j!ðk j 1Þ! j¼0 k1 Z b bs s a j ds: ba ba a
Further, Gðt; sÞ can be given by integral form 8 R ððtaÞðbsÞÞ=ðbaÞ k1 k1 1 > s ðs þ s tÞ ds; 2 2k1 > 0 ðbaÞ ððk1Þ!Þ > < a 6 t 6 s 6 b; Gðt; sÞ ¼ R ððbtÞðsaÞÞ=ðbaÞ nk1 k1 1 > s ðs þ t sÞ ds; > 2 2k1 0 > : ððk1Þ!Þ ðbaÞ a 6 s 6 t 6 b: Corollary 2. Let a ¼ 0; b ¼ 1 in Theorem 1, we obtain ( P ðt; sÞ; 0 6 t 6 s 6 1; Gðt; sÞ ¼ n1 ð1Þnk P ðt; sÞ þ ðn1Þ! ðt sÞ ; 0 6 s 6 t 6 1;
ð8Þ
ð9Þ
with P ðt; sÞ ¼
nk1 X j¼0
snkj1 tkþj ðk 1Þ!j!ðn k j 1Þ!
Z
1
ð1 sÞk1 ðsÞj ds:
0
In this case, the integral form of Gðt; sÞ is given by ( R tð1sÞ k1 1 s ðs þ s tÞnk1 ds; ðk1Þ!ðnk1Þ! 0 Gðt; sÞ ¼ R sð1tÞ nk1 k1 1 s ðs þ t sÞ ds; ðk1Þ!ðnk1Þ! 0
0 6 t 6 s 6 1; 0 6 s 6 t 6 1:
ð10Þ
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X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Proof of Theorem 1. Let hðtÞ 2 ½a; b . Then, it is well known from [2] that linear ðk; n kÞ conjugate boundary value problem: ð1Þ
nk ðnÞ
x ðtÞ ¼ hðtÞ;
a 6 t 6 b;
ðiÞ
0 6 i 6 k 1;
ðjÞ
0 6 j 6 n k 1;
x ðaÞ ¼ 0; x ðbÞ ¼ 0;
ð11Þ
has a unique solution xðtÞ, which is given by xðtÞ ¼
Z
b
Gðt; sÞhðsÞ dx;
ð12Þ
a 6 t 6 b:
a
First we assume n k 6 k. Since x 2 C n ½0; 1 , we have the Taylor expansion xðtÞ ¼
Z n1 X ai ð1Þnk t i n1 ðt aÞ þ ðt sÞ hðsÞ ds; i! ðn 1Þ! a i¼0
a 6 t 6 b:
ð12Þ
It follows from (11), that a0 ¼ a1 ¼ ¼ ak1 ¼ 0, and kj
ak ðb aÞ ðk jÞ!
kþ1j
n1
akþ1 ðb aÞ an1 ðb aÞ þ
þ ðk þ 1 jÞ! ðn 1 jÞ! Z b nk ð1Þ n1j ðb sÞ hðsÞ ds ¼ 0 þ ðn 1 jÞ! a þ
ð13Þ
for 0 6 j 6 n k 1. By using CramerÕs rule, we obtain akþj ¼ Djþ1 =D;
j ¼ 0; 1; . . . ; n k 1;
where D is the determinant of coefficients for the system (13) and Dm , 1 6 m 6 n k is the determinant that the mth column of D is replaced by ðh1 ; . . . ; hnk Þs , where nk1
ð1Þ hi ¼ ðn iÞ!
Z
b
ðb sÞ
ni
hðsÞ ds;
i ¼ 1; . . . ; n k:
ð14Þ
a
We are going to calculate D and Dm =D. For i ¼ 1; 2; . . . ; n k 1, let the ith row of D minus the ði þ 1Þst row multiplied by ðb aÞ, then we have by induction,
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
ðn2Þ!ðnk1ÞðbaÞk k! ðn3Þ!ðnk1ÞðbaÞk1 ðk1Þ! .. D ¼ . 2knþ2 k!ðnk1ÞðbaÞ ð2knþ2Þ! k!ðbaÞ2knþ1 ð2knþ1Þ!
ðn2Þ!ðnk2ÞðbaÞkþ1 ðkþ1Þ! ðn3Þ!ðnk2ÞðbaÞk k!
.. .
k!ðnk2ÞðbaÞ2knþ3 ð2knþ3Þ! k!ðbaÞ2knþ2 ð2knþ2Þ!
¼ ðn k 1Þ!ðn k 2Þ! 2!ðb aÞ
383 n2
ðb aÞ
.. .
ðb aÞ .. . ðb aÞk
k!ðbaÞkþ1 ðk1Þ!
ðnkÞk
¼
n3
nk1 Y
!
0 0 .. . 0 k ðb aÞ
i! ðb aÞ
ðnkÞk
:
i¼1
Similarly, we can obtain for j ¼ 0; 1; . . . ; n k 1, Z b s a nkj1 Djþ1 ðk þ jÞ! ¼ D a ðk 1Þ!j!ðn k j 1Þ! b a Pj j1 i ððb sÞ=ðb aÞÞkþi hðsÞ ds i¼0 Cj ð1Þ kþi ," ! # nk1 nk1 Y Y jþ1 i! i! ðb aÞ ; i6¼nkj1
i¼1
where Cji ¼ j!=ði!ðj iÞ!Þ is the binomial coefficient. Hence, we obtain Z b s a nm1 m! m ðb aÞ am ¼ a ðk 1Þ!ðm kÞ!ðn m 1Þ! b a k mk i i mki bs X Cmk ð1Þ bs hðsÞ ds ba ba kþi i¼0
ð15Þ
for m ¼ k; . . . ; n 1. Let j ¼ m k (j ¼ 0; 1; . . . ; n k 1) and P ðt; sÞ ¼
s a nk1 t a jþk 1 ðk 1Þ!j!ðn k j 1Þ! b a ba i¼0 ji j kþi X Cji ð1Þ bs : ba kþi i¼0 nk1 X
Then from (12)–(16) and the arbitrariness of hðsÞ, we conclude that 8 a 6 t 6 s 6 b; < P ðt; sÞ; nk n1 ð1Þ ts Gðt; sÞ ¼ : P ðt; sÞ þ ; a 6 s 6 t 6 b: ðn 1Þ! b a
ð16Þ
ð17Þ
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X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Since kþi k1 Z b j1 j X Cji ð1Þ bs 1 bs s a j ¼ ds; ba ba s ba ba kþi i¼0 we have k1 nk1 Z b X 1 bs 1 P ðt; sÞ ¼ b a ðk 1Þ! s ba j!ðn k j 1Þ! j¼0 t a s a j s a nkj1 ds ba ba ba ba k1 nk1 Z b t a k X j 1 bs ¼ Cnk1 b a ðk 1Þ!ðn k 1Þ! s ba j¼0 t a s a j s a nkj1 ds ba ba ba ba k1 Z b t a k 1 bs ¼ b a ðk 1Þ!ðn k 1Þ! s ba h s a t a s a ink1 ds ba ba ba ba k1 Z b 1 ta bs ¼ ðk 1Þ!ðn k 1Þ! s ba ba nk1 st ta bs ds : þ ba ba ba ba t a k
For fixed t, let u ¼ ððt aÞðb sÞÞ=ðb aÞ, then the above expression equals Z
1 ðk 1Þ!ðn k 1Þ!ðb aÞ
n1
ððtaÞðbsÞÞ=ðbaÞ
nk1
uk1 ðu þ s tÞ
du:
0
ð18Þ Let v ¼ u þ s t and s; t fixed. Then (18) becomes Z
1 ðk 1Þ!ðn k 1Þ!ðb aÞn1 ¼
ððbtÞðsaÞÞ=ðbaÞ
st
1 n1
Z
ðk 1Þ!ðn k 1Þ!ðb aÞ 0 Z st k1 vnk1 ðt s þ vÞ dv : 0
vnk1 ðt s þ vÞk1 dv
ððbtÞðsaÞÞ=ðbaÞ
vnk1 ðt s þ vÞ
k1
dv
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
385
Let v ¼ ðs tÞk. Then Z
1 ðk 1Þ!ðn k 1Þ!ðb aÞ ¼ ¼
ð1Þ
k1
ðs tÞ
n1
nk
ðt sÞ
vnk1 ðt s þ vÞ
k1
dv
k1
dk
0
Z
n1
ðk 1Þ!ðn k 1Þ!ðb aÞ ð1Þ
st
n1
ðk 1Þ!ðn k 1Þ!ðb aÞ
n1
knk1 ð1 kÞ
0
Z
n1
1
1
knk1 ð1 kÞk1 dk:
ð19Þ
0
By using integration by parts, it is easy to obtain Z 1 ðk 1Þ!ðn k 1Þ! k1 : knk1 ð1 kÞ dk ¼ ðn 1Þ! 0 Therefore (19) equals ð1Þnk ðt sÞn1
ðt > sÞ:
n1
ðn 1Þ!ðb aÞ
ð20Þ
Combining the results of (17)–(20), we obtain 8 R ððtaÞ=ðbsÞÞ=ðbsÞ k1 nk1 1 > s ðs þ s tÞ ds; n1 > 0 > < ðk1Þ!ðnk1Þ!ðbaÞ a 6 t 6 s 6 b; Gðt; sÞ ¼ R ððbtÞðsaÞÞ=ðbaÞ nk1 k1 1 > s ðs þ t sÞ ds; > n1 0 > ðk1Þ!ðnk1Þ!ðbaÞ : a 6 s 6 t 6 b: For the case k < n k, we can get the same results. Theorem 1 is thus proved.
3. Positive solutions Before stating the existence results of positive solutions to boundary value problem (1), we need the following lemmas. Lemma 1. Let k
nk
pðtÞ ¼
1 ðt aÞ ðb tÞ n n1 ðb aÞ
qðsÞ ¼
1 ðs aÞ ðb sÞ : n ðk 1Þ!ðn k 1Þ! ðb sÞ
; nk
Then for t; s 2 ½a; b , we have pðtÞqðsÞ 6 Gðt; sÞ 6 qðsÞ:
k
386
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Proof. Applying the binomial formula to the integrands in (6) and integrating term by term, we obtain Gðt; sÞ 8 h ini1 Pnk1 i i ðtaÞðbsÞ > 1 > Cnk1 ðs tÞ ðn i 1Þ; > i¼0 ba > ðk1Þ!ðnk1Þ!ðbaÞn1 > > > < a 6 t 6 s 6 b; ¼ h ini1 > Pk1 i i ðbtÞðsaÞ > 1 > C ðt sÞ ðn i 1Þ; > ðk1Þ!ðnk1Þ!ðbaÞn1 i¼0 k1 ba > > > : a 6 s 6 t 6 b: ð21Þ Replacing n 1 i by n 1 in (21) and applying the binomial formula, we obtain 8 k nk1 > ðtaÞðbsÞ ðsaÞðbtÞ 1 > ; > 2 2 ðk1Þ!ðnk1Þ!ðn1Þ > ðbaÞ ðbaÞ > > < a 6 t 6 s 6 b; Gðt; sÞ P k1 nk > ðtaÞðbsÞ ðsaÞðbtÞ 1 > > ; 2 2 > ðk1Þ!ðnk1Þ!ðn1Þ ðbaÞ ðbaÞ > > : a 6 s 6 t 6 b: 8 ta k bt nk bs k sa nk1 1 > ; > ðk1Þ!ðnk1Þ!ðn1Þ ba ba ba ba > > < a 6 t 6 s 6 b; P ta k bt k bs k sa nk 1 > > ; > ðk1Þ!ðnk1Þ!ðn1Þ ba ba ba ba > : a 6 s 6 t 6 b: ¼ pðtÞqðsÞ: On the other hand, replacing n 1 j by 1 in (21) and applying the binomial formula again, we obtain 8 ta k ta nk1 bs k sa nk1 1 > ; > ðk1Þ!ðnk1Þ! ba ba ba ba > > < a 6 t 6 s 6 b; Gðt; sÞ 6 ta k1 bt nk bs k1 sa nk 1 > > ; > ðk1Þ!ðnk1Þ! ba ba ba ba > : a6s6t6b ( ta qðsÞ; a 6 t 6 s 6 b; 6 sa bt qðsÞ; a 6 s 6 t 6 b; bs 6 qðsÞ:
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Lemma 1 is proved.
387
The following fixed point theorem due to Krasnoselskii [6] will be used. Lemma 2. Let E be a Banach space, and let K E be a cone in E. Assume X1 ; X2 are open subsets of E with 0 2 X1 ; X1 X2 , and let T : K \ ðX2 n X1 Þ ! K be a completely continuous operator such that either (i) kTxk 6 kxk 8x 2 K \ oX1 , and kTxk P kxk 8x 2 K \ oX2 or (ii) kTxk P kxk 8x 2 K \ oX1 , and kT k 6 kxk 8x 2 K \ oX2 . Then T has a fixed point in K \ ðX2 n X1 Þ. The following lemma is a simple consequence of [7, Theorems 1.2 and 1.3]. Lemma 3. Let E be a Banach space, K be a cone in E, let 0 < r1 < r2 < r3 be positive numbers, D ¼ fx 2 K j r1 6 kxk 6 r3 g, and let T : D ! K be a compact continuous operator such that (a) x 2 D, kxk ¼ r1 , Tx ¼ kx ) k P 1; (b) x 2 D, kxk ¼ r2 , Tx ¼ kx ) k < 1; (c) x 2 D, kxk ¼ r3 , Tx ¼ kx ) k P 1; (d) inf kxk¼r1 kTxk > 0, inf kxk¼r3 kTxk > 0. Then T has two fixed points x1 and x2 such that 0 < kx1 k < r2 < kx2 k: Now, we can state the main results of this section. Theorem 2. Assume a : ða; bÞ ! ½0; 1Þ and f : ½0; 1Þ ! ½0; 1Þ are continuous and let pðtÞ; qðsÞ be defined in Lemma 1 and a; q-satisfy the following inequalities: Z b 0< aðsÞqðsÞ ds ¼ M < 1: ð22Þ a
Moreover, we assume that (i) f0 ¼ 0 and f1 ¼ þ1 (suplinear) or (ii) f0 ¼ þ1 and f1 ¼ 0 (sublinear), where f0 ¼ limþ f ðxÞ=x; x!0
f1 ¼ lim f ðxÞ=x: x!þ1
Then the boundary value problem (1) has at least one positive solution. Proof. Let E be the space of all continuous functions defined on ½a; b , that is, E ¼ C½a; b . Let K be the cone in E defined by K ¼ fxðtÞ 2 C½a; b j xðtÞ P pðtÞkxk for all t 2 ½a; b }, where kxk ¼ maxa 6 t 6 b jxðtÞjg. Define a mapping T : K ! E as
388
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
ðTxÞðtÞ ¼
Z
b
Gðt; sÞaðsÞf ðxðsÞÞ ds;
a 6 t 6 b:
ð23Þ
a
Then any solution of (1) is a fixed point of T, what we need now is to show that T has a fixed point in K. Let x 2 K, then from Lemma 1 and the assumption on f ðxÞ, we have Z b Z b pðtÞ qðsÞaðsÞf ðxðsÞÞ ds 6 TxðtÞ 6 qðsÞaðsÞf ðxðsÞÞ ds; a 6 t 6 b; a
a
ð24Þ which means T ðKÞ K. Let KR ¼ fx 2 K j kxk 6 Rg and MR ¼ max0 6 x 6 R f ðxÞ. Then for x 2 KR , we have by (24) Z b kTxk 6 MR qðsÞbðsÞ ds; a
which implies the boundedness of T ðKR Þ. From the expression of Gðt; sÞ in (6), it is not difficult to show that there exists a constant c0 > 0 such that oGðt; sÞ C0 qðsÞ=ðb sÞ; a 6 s 6 t 6 b; ot 6 C0 qðsÞ=s a; a 6 t 6 s 6 b: Hence oGðt; sÞ ot qðsÞgðxðsÞÞ ds a Z t Z b qðsÞ qðsÞ bðsÞ ds þ bðsÞ ds ¼ QðsÞ 6 C0 MR sa a bs t
jðTxÞ0 ðtÞj 6
Z
b
and Z a
b
QðsÞds ¼ 2C0 MR
Z
b
qðsÞbðsÞ ds; a
which implies that T ðKR Þ is a compact set in K. In addition the continuity of aðtÞf ðxÞ implies the continuity of T. Now the Ascoli–Arzela theorem claims that T is a compact continuous mapping. If f is suplinear, then it follows from (22) that there exists d 2 ða; ðða þ bÞ=2ÞÞ such that Z bd 1 aðsÞqðsÞ ds ¼ M > 0: 2 aþd
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
389
Since f0 ¼ 0, there exists q > 0 such that f ðxÞ 6 ex for 0 < x 6 p, where e satisfies Z b 0
For x 2 K with kxk ¼ q we obtain from Lemma 1 that Z b kTxk 6 eq qðsÞbðsÞ < q ¼ kxk; a
which implies that kTxk < kxk for x 2 K \ oXq , where Xq ¼ fx 2 C½a; b j kxk < qg. From f1 ¼ þ1, there exists R; R > q > 0 such that f ðxÞ P NR x for x P C1 R, where C1 ¼
min
aþd 6 t 6 bd
pðtÞ
and NR > 0 satisfies Z bd C1 NR qðsÞaðsÞ ds > 1:
ð26Þ
aþd
For any x 2 K with kxk > R, we have mind 6 t 6 bd xðtÞ P C1 R and hence by Lemma 1 and (26) we get Z bd ðTxÞ P pðtÞ qðsÞaðsÞf ðxðsÞÞ ds aþd
Z
bd
qðsÞaðsÞ ds > R;
P C1 NR R
t 2 ½a þ d; b d ;
aþd
which implies that kTxk > R ¼ kxk
for all x 2 K \ oXR :
Now, by the first part of Lemma 2, T has a fixed point x 2 K \ ðXR n Xq Þ. Similarly we can prove that there exists 0 < q1 < R1 such that T has a fixed point x1 2 K \ ðXR ; nXq1 Þ if f is sublinear. Since a fixed point of T in K is a positive solution of (1), in both cases, we have shown that (1) has at least one positive solution. Remark 1. A solution xðtÞ of (1) is said to be a positive solution of (1) if x 2 C n ða; bÞ \ C n1 ½a; b and xðtÞ > 0 for t 2 ða; bÞ. Theorem 2 also implies that aðtÞ may be singular at t ¼ a or t ¼ b. Theorem 3. Suppose conditions (i), (ii) of Theorem 1 are replaced by f0 ¼ f1 ¼ þ 1 and there exists q > 0 such that for 0 6 x 6 q; 0 6 f ðxÞ < M 1 q, or
390
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
equivalently f : ½0; p ! ½0; M 1 qÞ. Then the boundary value problem (1) has at least two positive solutions. Proof. Similar to the proof of Theorem 2, we need only to prove that the mapping T has at least two fixed point is K. From the proof of Theorem 2, TK K and T is completely continuous. We examine now conditions (a)–(d) of Lemma 3. Since f0 ¼ 1 there exists e > 0 such that for 0 6 x 6 e; f ðxÞ > 2MC11 , where C1 ¼
min
aþd 6 t 6 bd
pðtÞ
and Z
bd
O< aþd
1 1 qðsÞaðsÞ ds ¼ M ¼ 2 2
Z
b
qðsÞaðsÞ ds < 1: a
Let x 2 K; kxk ¼ e and Tx ¼ kx, for k > 0. Then Z b kTxk ¼ max Gðt; sÞaðsÞf ðxðsÞÞ ds a6t6b
P P
a
Z
bd
max
aþd 6 t 6 bd
dþa Z bd
max
aþd 6 t 6 bd
Z
Gðt; sÞaðsÞf ðxðsÞÞ ds pðtÞqðsÞaðsÞf ðxðsÞÞ ds
dþa
bd
qðsÞaðsÞ ds > e:
P C1 Me dþa
This implies that k > 1. Similarly, we can prove that there exists R > q > e such that for x 2 K with kxk ¼ R we have kTxk > kxk ¼ R. Hence for Tx ¼ lx; kxk ¼ R; x 2 K, we have l > 1. Now let x 2 K; kxk ¼ q, then f ðxÞ < M 1 p and Z b Z b kTxk 6 qðsÞaðsÞf ðxðsÞÞ ds < qðsÞaðsÞ ds M 1 q ¼ p: a
a
This implies if Tx ¼ kx, then k < 1. Let r1 ¼ e < r2 ¼ q < r3 ¼ R; D ¼ fx 2 K j e 6 kxk 6 Rg. Then the above discussion shows that conditions (a)–(d) of Lemma 3 are fulfilled. Therefore Lemma 3 implies that there exist two positive solutions x1 ðtÞ, x2 ðtÞ such that 0 < kx1 k < q < kx2 k:
Let K be a cone in Banach space E.
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
391
Definition 1. A map a : K ! ½0; 1Þ is said to be a nonnegative continuous concave functional if a is continuous and aðkx þ ð1 kÞyÞ P kaðxÞ þ ð1 kÞaðyÞ
8x; y 2 K and 0 6 k 6 1:
Definition 2. Let 0 < a < b be positive numbers and a be a nonnegative continuous functional on K, define convex sets Kr and Sða; a; bÞ, respectively, by Kr ¼ fx 2 K j kxk < rg and Sða; a; bÞ ¼ fx 2 k j 6 aðxÞ; kxk 6 bg: Lemma 4 (Legget–Williams fixed point theorem [8]). Let T : K C ! K C be a completely continuous and a be a nonnegative continuous concave functional on K such that aðxÞ 6 kxk for all x 2 K C . Suppose there exists 0 < u < b < d 6 C such that (C1) fx 2 Sða; b; dÞ j aðxÞ > bg 6¼ ; and aðTxÞ > b for x 2 Sða; b; dÞ, (C2) kTxk < a for kxk 6 a, and (C3) aðTxÞ > b for x 2 Sða; b; CÞ with kTxk > d. Then T has at least three fixed points x1 , x2 and x3 satisfying kx1 k < a;
b < aðx2 Þ and
kx3 k > a
with aðx3 Þ < b:
Theorem 4. Consider the ðk; n kÞ conjugate boundary value problem ð1Þnk xðnÞ ðtÞ ¼ bðtÞf ðxðtÞÞ;
a < t < b;
ð27Þ 0 6 i 6 k 1; 0 6 j 6 n k 1; Rb where b 2 Cð½a; b ; ½0; 1ÞÞ, f 2 Cð½0; 1Þ; ½0; 1ÞÞ. If 0 < a qðsÞbðsÞ ds ¼ M < 1, and there exist positive numbers 0 < A < B < C < D such that (i) f ½0; A ½0; M 1 AÞ; (ii) f ½0; D ½0; M 1 D ; (iii) f ½B; C ð2BM 1 C11 ; M 1 D . Then problem (27) has at least three positive solutions x1 ðtÞ, x2 ðtÞ and x3 ðtÞ such that ðiÞ
ðjÞ
x ðaÞ ¼ x ðbÞ ¼ 0;
kx1 k < A;
B<
min
aþd 6 t 6 bd
x2 ðtÞ
and kx3 k > A with minaþd 6 t 6 bd x3 ðtÞ < B; where d 2 ða; ðða þ bÞ=2ÞÞ is such that Z bd 12 and C1 ¼ min pðtÞ; qðsÞbðsÞ ds ¼ aþd 6 t 6 bd M aþd and pðsÞ; qðsÞ are given by Lemma 1.
392
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Proof. Let K ¼ fx 2 C½a; b ; xðtÞ P 0 and minaþd 6 t 6 bd xðtÞ P C1 kxk, where C1 ¼ mind 6 t 6 bd pðtÞ 2 ð0; 1Þ; pðtÞ is given by Lemma 1}. Then from Lemma 1, we have Z b TxðtÞ Gðt; sÞbðsÞf ðxðsÞÞ ds a
and min
aþd 6 t 6 bd
Z b min pðtÞ qðsÞbðsÞf ðxðsÞÞ ds aþd 6 t 6 bd a Z b qðsÞbðsÞf ðxðsÞÞ ds ¼ C1 a Z b Gðt; sÞbðsÞf ðxðsÞÞ ds P C1 ¼ C1 kxk:
TxðtÞ P
a
This implies that TK K. Let KD ¼ fx 2 K; kxk < Dg. We show that T K D K D . In fact, by Lemma 1, Gðt; sÞ 6 qðsÞ and by (ii), f ð½0; DÞ ½0; M 1 D , we have therefore Z b kTxk ¼ max TxðtÞ 6 qðsÞbðsÞf ðxðsÞÞ ds a6t6b a Z b qðsÞbðsÞ ds DM 1 ¼ D: 6 a
If x 2 KD and kxk 6 A, then from (i) f ðxÞ 6 M 1 A and Z b kTxk < qðsÞbðsÞ ds M 1 A ¼ A; a
which implies (C2) of Lemma 4. Since xðtÞ ððB þ CÞ=2Þ 2 Sða; B; CÞ, where aðxÞ ¼ minaþd 6 t 6 bd xðtÞ, then a is a nonnegative continuous concave functional on K and we have fx 2 Sða; B; CÞ j aðxÞ > Bg 6¼ ;. For x 2 Sða; B; CÞ, we have by (iii), aðTxÞ ¼
min
TxðtÞ Z b ¼ min Gðt; sÞbðsÞf ðxðsÞÞ ds aþd 6 t 6 bd a Z bd Z bd P C1 qðsÞf ðxðsÞÞ ds > C1 qðsÞbðsÞ ds 2C11 M 1 B ¼ B: aþd 6 t 6 bd
aþd
This proves (C1) of Lemma 4.
aþd
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
393
Now let x 2 Sða; B; DÞ, then minaþd 6 t 6 bd xðtÞ P B and Z bd aðTxÞ ¼ min TxðtÞ P C1 qðsÞbðsÞf ðxðsÞÞ ds aþd 6 t 6 bd
Z
aþd
bd
> C1
qðsÞbðsÞ ds2M 1 C11 B ¼ B:
aþd
This implies that (C3) of Lemma 4 is fulfilled. Combining the above results, Theorem 4 is proved.
References [1] P.W. Eloe, J. Henderson, Positive solutions for ðn 1; 1Þ conjugate boundary value problems, Nonlinear Anal. 28 (1997) 1669–1680. [2] P.W. Eloe, J. Henderson, Singular nonlinear ðk; n kÞ conjugate boundary value problems, J. Differential Equations 133 (1979) 136–151. [3] R.P. Agarwal, D. OÕRegen, Positive solutions for ðp; n pÞ conjugate boundary value problems, J. Differential Equations 150 (1998) 462–473. [4] R.P. Agarwal, M. Bohner, P.J.Y. Womg, Positive solutions and eigenvalues of conjugate boundary value problems, Proc. Edinburgh Math. Soc. 42 (1999) 349–374. [5] P.J.Y. Wong, R.P. Agarwal, Eigenvalues of Lidstone boundary value problems, Appl. Math. Comput. 104 (1999) 15–31. [6] M.A. Krasnoselskii, Positive Solutions of Operator Equations, Noordhoff, Groningen, 1964. [7] J.A. Gatica, H.L. Smith, Fixed point theorems in cone with applications, J. Math. Anal. Appl. 61 (1997) 58–71. [8] R. Leggett, L. Williams, Multiple positive fixed points of nonlinear operators on ordered Banach spaces, Indiana Univ. Math. J. 28 (1979) 673–688.
GreenÕs function and positive solutions for higher-order ODE Xiaojing Yang Department of Mathematics, Tsinghua University, Beijing 100084, People’s Republic of China
Abstract GreenÕs function for ðk; n kÞ conjugate boundary value problem ð1Þnk xðnÞ ðtÞ ¼ aðtÞf ðxðtÞÞ;
a < t < b;
ðiÞ
0 6 i 6 k 1;
ðjÞ
0 6 j 6 n k 1;
x ðaÞ ¼ 0; x ðbÞ ¼ 0;
is obtained and its integral representation is given. As applications, we obtain sufficient conditions for the existence of one, two or three positive solutions. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: GreenÕs function; Positive solutions; Fixed points
1. Introduction GreenÕs function plays an important role in solving boundary value problems of ordinary differential equations. But to obtain explicit formula of GreenÕs function is not always an easy job. In this paper, we consider the ðk; n kÞ conjugate boundary value problem: ð1Þnk xðnÞ ðtÞ ¼ aðtÞf ðxðtÞÞ;
a < t < b;
ðiÞ
0 6 i 6 k 1;
ðjÞ
0 6 j 6 n k 1;
x ðaÞ ¼ 0; x ðbÞ ¼ 0;
E-mail address: [email protected] (X. Yang). 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 2 ) 0 0 0 5 6 - 5
ð1Þ
380
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
where a; f are continuous, 1 6 k < n. First, we obtain the explicit formula of GreenÕs function for the following boundary value problem: ð1Þ
nk ðnÞ
x ðtÞ ¼ 0;
xðiÞ ðaÞ ¼ 0;
a < t < b; ð2Þ
0 6 i 6 k 1;
xðjÞ ðbÞ ¼ 0 0 6 j 6 n k 1: Then by applying the properties of GreenÕs function, we give sufficient conditions for the existence of at least one, two or three positive solutions to problem (1). By GreenÕs function Gðt; sÞ of the ðk; n kÞ conjugate boundary value problem (1) we mean that Gðt; sÞ satisfies ð1Þ
nk
GðnÞ ðt; sÞ ¼ dðt; sÞ;
ðiÞ
a < t < b; a < s < b; ð3Þ
G ðaÞ ¼ 0;
0 6 i 6 k 1;
GðjÞ ðbÞ ¼ 0;
0 6 j 6 n k 1;
where GðmÞ ðt; sÞ denotes the mth-order derivatives with respect to t and dðxÞ is the Dirac delta function. It is very useful to study the existence of positive solutions to various boundary value problems by deducing related inequalities from GreenÕs functions, see, for example [1–5].
2. Green’s function Now we can state our main result of this paper. Theorem 1. The Green function for the boundary value problem (2) is given by 8 a 6 t 6 s 6 b; < P ðt; sÞ; n1 nk t s Gðt; sÞ ¼ ð4Þ ð1Þ : P ðt; sÞ þ ; a 6 s 6 t 6 b; ðn 1Þ! b a where P ðt; sÞ ¼
nk1 X
nkj1
ððs aÞ=ðb aÞÞ ððt aÞ=ðb aÞÞ ðk 1Þ!j!ðn k j 1Þ! j¼0 k1 Z b bs s a j ds: ba ba a
kþj
1 ðb aÞ ð5Þ
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
381
Moreover, Green’s function Gðt; sÞ can be further reduced to the following integral form: 8 R ððtaÞðbsÞÞ=ðbaÞ k1 nk1 1 > s ðs þ s tÞ ds; n1 > 0 ðk1Þ!ðnk1Þ!ðbaÞ > < a 6 t 6 s 6 b; ð6Þ Gðt; sÞ ¼ R ððbtÞðsaÞÞ=ðbaÞ nk1 k1 1 > s ðs þ t sÞ ds; > n1 0 > ðk1Þ!ðnk1Þ!ðbaÞ : a 6 s 6 t 6 b: Corollary 1. Let k 1 ¼ n k 1 in Theorem 1 and Corollary 2 below, we have for n ¼ 2k and ( P ðt; sÞ; a 6 t 6 s 6 b; ts 2k1 Gðt; sÞ ¼ ð7Þ ð1Þk P ðt; sÞ þ ð2k1Þ! ba ; a 6 s 6 t 6 b; with P ðt; sÞ ¼
k1j kþj k1 X ððs aÞ=ðb aÞÞ ððt aÞ=ðb aÞÞ 1 b a ðk 1Þ!j!ðk j 1Þ! j¼0 k1 Z b bs s a j ds: ba ba a
Further, Gðt; sÞ can be given by integral form 8 R ððtaÞðbsÞÞ=ðbaÞ k1 k1 1 > s ðs þ s tÞ ds; 2 2k1 > 0 ðbaÞ ððk1Þ!Þ > < a 6 t 6 s 6 b; Gðt; sÞ ¼ R ððbtÞðsaÞÞ=ðbaÞ nk1 k1 1 > s ðs þ t sÞ ds; > 2 2k1 0 > : ððk1Þ!Þ ðbaÞ a 6 s 6 t 6 b: Corollary 2. Let a ¼ 0; b ¼ 1 in Theorem 1, we obtain ( P ðt; sÞ; 0 6 t 6 s 6 1; Gðt; sÞ ¼ n1 ð1Þnk P ðt; sÞ þ ðn1Þ! ðt sÞ ; 0 6 s 6 t 6 1;
ð8Þ
ð9Þ
with P ðt; sÞ ¼
nk1 X j¼0
snkj1 tkþj ðk 1Þ!j!ðn k j 1Þ!
Z
1
ð1 sÞk1 ðsÞj ds:
0
In this case, the integral form of Gðt; sÞ is given by ( R tð1sÞ k1 1 s ðs þ s tÞnk1 ds; ðk1Þ!ðnk1Þ! 0 Gðt; sÞ ¼ R sð1tÞ nk1 k1 1 s ðs þ t sÞ ds; ðk1Þ!ðnk1Þ! 0
0 6 t 6 s 6 1; 0 6 s 6 t 6 1:
ð10Þ
382
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Proof of Theorem 1. Let hðtÞ 2 ½a; b . Then, it is well known from [2] that linear ðk; n kÞ conjugate boundary value problem: ð1Þ
nk ðnÞ
x ðtÞ ¼ hðtÞ;
a 6 t 6 b;
ðiÞ
0 6 i 6 k 1;
ðjÞ
0 6 j 6 n k 1;
x ðaÞ ¼ 0; x ðbÞ ¼ 0;
ð11Þ
has a unique solution xðtÞ, which is given by xðtÞ ¼
Z
b
Gðt; sÞhðsÞ dx;
ð12Þ
a 6 t 6 b:
a
First we assume n k 6 k. Since x 2 C n ½0; 1 , we have the Taylor expansion xðtÞ ¼
Z n1 X ai ð1Þnk t i n1 ðt aÞ þ ðt sÞ hðsÞ ds; i! ðn 1Þ! a i¼0
a 6 t 6 b:
ð12Þ
It follows from (11), that a0 ¼ a1 ¼ ¼ ak1 ¼ 0, and kj
ak ðb aÞ ðk jÞ!
kþ1j
n1
akþ1 ðb aÞ an1 ðb aÞ þ
þ ðk þ 1 jÞ! ðn 1 jÞ! Z b nk ð1Þ n1j ðb sÞ hðsÞ ds ¼ 0 þ ðn 1 jÞ! a þ
ð13Þ
for 0 6 j 6 n k 1. By using CramerÕs rule, we obtain akþj ¼ Djþ1 =D;
j ¼ 0; 1; . . . ; n k 1;
where D is the determinant of coefficients for the system (13) and Dm , 1 6 m 6 n k is the determinant that the mth column of D is replaced by ðh1 ; . . . ; hnk Þs , where nk1
ð1Þ hi ¼ ðn iÞ!
Z
b
ðb sÞ
ni
hðsÞ ds;
i ¼ 1; . . . ; n k:
ð14Þ
a
We are going to calculate D and Dm =D. For i ¼ 1; 2; . . . ; n k 1, let the ith row of D minus the ði þ 1Þst row multiplied by ðb aÞ, then we have by induction,
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
ðn2Þ!ðnk1ÞðbaÞk k! ðn3Þ!ðnk1ÞðbaÞk1 ðk1Þ! .. D ¼ . 2knþ2 k!ðnk1ÞðbaÞ ð2knþ2Þ! k!ðbaÞ2knþ1 ð2knþ1Þ!
ðn2Þ!ðnk2ÞðbaÞkþ1 ðkþ1Þ! ðn3Þ!ðnk2ÞðbaÞk k!
.. .
k!ðnk2ÞðbaÞ2knþ3 ð2knþ3Þ! k!ðbaÞ2knþ2 ð2knþ2Þ!
¼ ðn k 1Þ!ðn k 2Þ! 2!ðb aÞ
383 n2
ðb aÞ
.. .
ðb aÞ .. . ðb aÞk
k!ðbaÞkþ1 ðk1Þ!
ðnkÞk
¼
n3
nk1 Y
!
0 0 .. . 0 k ðb aÞ
i! ðb aÞ
ðnkÞk
:
i¼1
Similarly, we can obtain for j ¼ 0; 1; . . . ; n k 1, Z b s a nkj1 Djþ1 ðk þ jÞ! ¼ D a ðk 1Þ!j!ðn k j 1Þ! b a Pj j1 i ððb sÞ=ðb aÞÞkþi hðsÞ ds i¼0 Cj ð1Þ kþi ," ! # nk1 nk1 Y Y jþ1 i! i! ðb aÞ ; i6¼nkj1
i¼1
where Cji ¼ j!=ði!ðj iÞ!Þ is the binomial coefficient. Hence, we obtain Z b s a nm1 m! m ðb aÞ am ¼ a ðk 1Þ!ðm kÞ!ðn m 1Þ! b a k mk i i mki bs X Cmk ð1Þ bs hðsÞ ds ba ba kþi i¼0
ð15Þ
for m ¼ k; . . . ; n 1. Let j ¼ m k (j ¼ 0; 1; . . . ; n k 1) and P ðt; sÞ ¼
s a nk1 t a jþk 1 ðk 1Þ!j!ðn k j 1Þ! b a ba i¼0 ji j kþi X Cji ð1Þ bs : ba kþi i¼0 nk1 X
Then from (12)–(16) and the arbitrariness of hðsÞ, we conclude that 8 a 6 t 6 s 6 b; < P ðt; sÞ; nk n1 ð1Þ ts Gðt; sÞ ¼ : P ðt; sÞ þ ; a 6 s 6 t 6 b: ðn 1Þ! b a
ð16Þ
ð17Þ
384
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Since kþi k1 Z b j1 j X Cji ð1Þ bs 1 bs s a j ¼ ds; ba ba s ba ba kþi i¼0 we have k1 nk1 Z b X 1 bs 1 P ðt; sÞ ¼ b a ðk 1Þ! s ba j!ðn k j 1Þ! j¼0 t a s a j s a nkj1 ds ba ba ba ba k1 nk1 Z b t a k X j 1 bs ¼ Cnk1 b a ðk 1Þ!ðn k 1Þ! s ba j¼0 t a s a j s a nkj1 ds ba ba ba ba k1 Z b t a k 1 bs ¼ b a ðk 1Þ!ðn k 1Þ! s ba h s a t a s a ink1 ds ba ba ba ba k1 Z b 1 ta bs ¼ ðk 1Þ!ðn k 1Þ! s ba ba nk1 st ta bs ds : þ ba ba ba ba t a k
For fixed t, let u ¼ ððt aÞðb sÞÞ=ðb aÞ, then the above expression equals Z
1 ðk 1Þ!ðn k 1Þ!ðb aÞ
n1
ððtaÞðbsÞÞ=ðbaÞ
nk1
uk1 ðu þ s tÞ
du:
0
ð18Þ Let v ¼ u þ s t and s; t fixed. Then (18) becomes Z
1 ðk 1Þ!ðn k 1Þ!ðb aÞn1 ¼
ððbtÞðsaÞÞ=ðbaÞ
st
1 n1
Z
ðk 1Þ!ðn k 1Þ!ðb aÞ 0 Z st k1 vnk1 ðt s þ vÞ dv : 0
vnk1 ðt s þ vÞk1 dv
ððbtÞðsaÞÞ=ðbaÞ
vnk1 ðt s þ vÞ
k1
dv
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
385
Let v ¼ ðs tÞk. Then Z
1 ðk 1Þ!ðn k 1Þ!ðb aÞ ¼ ¼
ð1Þ
k1
ðs tÞ
n1
nk
ðt sÞ
vnk1 ðt s þ vÞ
k1
dv
k1
dk
0
Z
n1
ðk 1Þ!ðn k 1Þ!ðb aÞ ð1Þ
st
n1
ðk 1Þ!ðn k 1Þ!ðb aÞ
n1
knk1 ð1 kÞ
0
Z
n1
1
1
knk1 ð1 kÞk1 dk:
ð19Þ
0
By using integration by parts, it is easy to obtain Z 1 ðk 1Þ!ðn k 1Þ! k1 : knk1 ð1 kÞ dk ¼ ðn 1Þ! 0 Therefore (19) equals ð1Þnk ðt sÞn1
ðt > sÞ:
n1
ðn 1Þ!ðb aÞ
ð20Þ
Combining the results of (17)–(20), we obtain 8 R ððtaÞ=ðbsÞÞ=ðbsÞ k1 nk1 1 > s ðs þ s tÞ ds; n1 > 0 > < ðk1Þ!ðnk1Þ!ðbaÞ a 6 t 6 s 6 b; Gðt; sÞ ¼ R ððbtÞðsaÞÞ=ðbaÞ nk1 k1 1 > s ðs þ t sÞ ds; > n1 0 > ðk1Þ!ðnk1Þ!ðbaÞ : a 6 s 6 t 6 b: For the case k < n k, we can get the same results. Theorem 1 is thus proved.
3. Positive solutions Before stating the existence results of positive solutions to boundary value problem (1), we need the following lemmas. Lemma 1. Let k
nk
pðtÞ ¼
1 ðt aÞ ðb tÞ n n1 ðb aÞ
qðsÞ ¼
1 ðs aÞ ðb sÞ : n ðk 1Þ!ðn k 1Þ! ðb sÞ
; nk
Then for t; s 2 ½a; b , we have pðtÞqðsÞ 6 Gðt; sÞ 6 qðsÞ:
k
386
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Proof. Applying the binomial formula to the integrands in (6) and integrating term by term, we obtain Gðt; sÞ 8 h ini1 Pnk1 i i ðtaÞðbsÞ > 1 > Cnk1 ðs tÞ ðn i 1Þ; > i¼0 ba > ðk1Þ!ðnk1Þ!ðbaÞn1 > > > < a 6 t 6 s 6 b; ¼ h ini1 > Pk1 i i ðbtÞðsaÞ > 1 > C ðt sÞ ðn i 1Þ; > ðk1Þ!ðnk1Þ!ðbaÞn1 i¼0 k1 ba > > > : a 6 s 6 t 6 b: ð21Þ Replacing n 1 i by n 1 in (21) and applying the binomial formula, we obtain 8 k nk1 > ðtaÞðbsÞ ðsaÞðbtÞ 1 > ; > 2 2 ðk1Þ!ðnk1Þ!ðn1Þ > ðbaÞ ðbaÞ > > < a 6 t 6 s 6 b; Gðt; sÞ P k1 nk > ðtaÞðbsÞ ðsaÞðbtÞ 1 > > ; 2 2 > ðk1Þ!ðnk1Þ!ðn1Þ ðbaÞ ðbaÞ > > : a 6 s 6 t 6 b: 8 ta k bt nk bs k sa nk1 1 > ; > ðk1Þ!ðnk1Þ!ðn1Þ ba ba ba ba > > < a 6 t 6 s 6 b; P ta k bt k bs k sa nk 1 > > ; > ðk1Þ!ðnk1Þ!ðn1Þ ba ba ba ba > : a 6 s 6 t 6 b: ¼ pðtÞqðsÞ: On the other hand, replacing n 1 j by 1 in (21) and applying the binomial formula again, we obtain 8 ta k ta nk1 bs k sa nk1 1 > ; > ðk1Þ!ðnk1Þ! ba ba ba ba > > < a 6 t 6 s 6 b; Gðt; sÞ 6 ta k1 bt nk bs k1 sa nk 1 > > ; > ðk1Þ!ðnk1Þ! ba ba ba ba > : a6s6t6b ( ta qðsÞ; a 6 t 6 s 6 b; 6 sa bt qðsÞ; a 6 s 6 t 6 b; bs 6 qðsÞ:
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Lemma 1 is proved.
387
The following fixed point theorem due to Krasnoselskii [6] will be used. Lemma 2. Let E be a Banach space, and let K E be a cone in E. Assume X1 ; X2 are open subsets of E with 0 2 X1 ; X1 X2 , and let T : K \ ðX2 n X1 Þ ! K be a completely continuous operator such that either (i) kTxk 6 kxk 8x 2 K \ oX1 , and kTxk P kxk 8x 2 K \ oX2 or (ii) kTxk P kxk 8x 2 K \ oX1 , and kT k 6 kxk 8x 2 K \ oX2 . Then T has a fixed point in K \ ðX2 n X1 Þ. The following lemma is a simple consequence of [7, Theorems 1.2 and 1.3]. Lemma 3. Let E be a Banach space, K be a cone in E, let 0 < r1 < r2 < r3 be positive numbers, D ¼ fx 2 K j r1 6 kxk 6 r3 g, and let T : D ! K be a compact continuous operator such that (a) x 2 D, kxk ¼ r1 , Tx ¼ kx ) k P 1; (b) x 2 D, kxk ¼ r2 , Tx ¼ kx ) k < 1; (c) x 2 D, kxk ¼ r3 , Tx ¼ kx ) k P 1; (d) inf kxk¼r1 kTxk > 0, inf kxk¼r3 kTxk > 0. Then T has two fixed points x1 and x2 such that 0 < kx1 k < r2 < kx2 k: Now, we can state the main results of this section. Theorem 2. Assume a : ða; bÞ ! ½0; 1Þ and f : ½0; 1Þ ! ½0; 1Þ are continuous and let pðtÞ; qðsÞ be defined in Lemma 1 and a; q-satisfy the following inequalities: Z b 0< aðsÞqðsÞ ds ¼ M < 1: ð22Þ a
Moreover, we assume that (i) f0 ¼ 0 and f1 ¼ þ1 (suplinear) or (ii) f0 ¼ þ1 and f1 ¼ 0 (sublinear), where f0 ¼ limþ f ðxÞ=x; x!0
f1 ¼ lim f ðxÞ=x: x!þ1
Then the boundary value problem (1) has at least one positive solution. Proof. Let E be the space of all continuous functions defined on ½a; b , that is, E ¼ C½a; b . Let K be the cone in E defined by K ¼ fxðtÞ 2 C½a; b j xðtÞ P pðtÞkxk for all t 2 ½a; b }, where kxk ¼ maxa 6 t 6 b jxðtÞjg. Define a mapping T : K ! E as
388
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
ðTxÞðtÞ ¼
Z
b
Gðt; sÞaðsÞf ðxðsÞÞ ds;
a 6 t 6 b:
ð23Þ
a
Then any solution of (1) is a fixed point of T, what we need now is to show that T has a fixed point in K. Let x 2 K, then from Lemma 1 and the assumption on f ðxÞ, we have Z b Z b pðtÞ qðsÞaðsÞf ðxðsÞÞ ds 6 TxðtÞ 6 qðsÞaðsÞf ðxðsÞÞ ds; a 6 t 6 b; a
a
ð24Þ which means T ðKÞ K. Let KR ¼ fx 2 K j kxk 6 Rg and MR ¼ max0 6 x 6 R f ðxÞ. Then for x 2 KR , we have by (24) Z b kTxk 6 MR qðsÞbðsÞ ds; a
which implies the boundedness of T ðKR Þ. From the expression of Gðt; sÞ in (6), it is not difficult to show that there exists a constant c0 > 0 such that oGðt; sÞ C0 qðsÞ=ðb sÞ; a 6 s 6 t 6 b; ot 6 C0 qðsÞ=s a; a 6 t 6 s 6 b: Hence oGðt; sÞ ot qðsÞgðxðsÞÞ ds a Z t Z b qðsÞ qðsÞ bðsÞ ds þ bðsÞ ds ¼ QðsÞ 6 C0 MR sa a bs t
jðTxÞ0 ðtÞj 6
Z
b
and Z a
b
QðsÞds ¼ 2C0 MR
Z
b
qðsÞbðsÞ ds; a
which implies that T ðKR Þ is a compact set in K. In addition the continuity of aðtÞf ðxÞ implies the continuity of T. Now the Ascoli–Arzela theorem claims that T is a compact continuous mapping. If f is suplinear, then it follows from (22) that there exists d 2 ða; ðða þ bÞ=2ÞÞ such that Z bd 1 aðsÞqðsÞ ds ¼ M > 0: 2 aþd
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
389
Since f0 ¼ 0, there exists q > 0 such that f ðxÞ 6 ex for 0 < x 6 p, where e satisfies Z b 0
For x 2 K with kxk ¼ q we obtain from Lemma 1 that Z b kTxk 6 eq qðsÞbðsÞ < q ¼ kxk; a
which implies that kTxk < kxk for x 2 K \ oXq , where Xq ¼ fx 2 C½a; b j kxk < qg. From f1 ¼ þ1, there exists R; R > q > 0 such that f ðxÞ P NR x for x P C1 R, where C1 ¼
min
aþd 6 t 6 bd
pðtÞ
and NR > 0 satisfies Z bd C1 NR qðsÞaðsÞ ds > 1:
ð26Þ
aþd
For any x 2 K with kxk > R, we have mind 6 t 6 bd xðtÞ P C1 R and hence by Lemma 1 and (26) we get Z bd ðTxÞ P pðtÞ qðsÞaðsÞf ðxðsÞÞ ds aþd
Z
bd
qðsÞaðsÞ ds > R;
P C1 NR R
t 2 ½a þ d; b d ;
aþd
which implies that kTxk > R ¼ kxk
for all x 2 K \ oXR :
Now, by the first part of Lemma 2, T has a fixed point x 2 K \ ðXR n Xq Þ. Similarly we can prove that there exists 0 < q1 < R1 such that T has a fixed point x1 2 K \ ðXR ; nXq1 Þ if f is sublinear. Since a fixed point of T in K is a positive solution of (1), in both cases, we have shown that (1) has at least one positive solution. Remark 1. A solution xðtÞ of (1) is said to be a positive solution of (1) if x 2 C n ða; bÞ \ C n1 ½a; b and xðtÞ > 0 for t 2 ða; bÞ. Theorem 2 also implies that aðtÞ may be singular at t ¼ a or t ¼ b. Theorem 3. Suppose conditions (i), (ii) of Theorem 1 are replaced by f0 ¼ f1 ¼ þ 1 and there exists q > 0 such that for 0 6 x 6 q; 0 6 f ðxÞ < M 1 q, or
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X. Yang / Appl. Math. Comput. 136 (2003) 379–393
equivalently f : ½0; p ! ½0; M 1 qÞ. Then the boundary value problem (1) has at least two positive solutions. Proof. Similar to the proof of Theorem 2, we need only to prove that the mapping T has at least two fixed point is K. From the proof of Theorem 2, TK K and T is completely continuous. We examine now conditions (a)–(d) of Lemma 3. Since f0 ¼ 1 there exists e > 0 such that for 0 6 x 6 e; f ðxÞ > 2MC11 , where C1 ¼
min
aþd 6 t 6 bd
pðtÞ
and Z
bd
O< aþd
1 1 qðsÞaðsÞ ds ¼ M ¼ 2 2
Z
b
qðsÞaðsÞ ds < 1: a
Let x 2 K; kxk ¼ e and Tx ¼ kx, for k > 0. Then Z b kTxk ¼ max Gðt; sÞaðsÞf ðxðsÞÞ ds a6t6b
P P
a
Z
bd
max
aþd 6 t 6 bd
dþa Z bd
max
aþd 6 t 6 bd
Z
Gðt; sÞaðsÞf ðxðsÞÞ ds pðtÞqðsÞaðsÞf ðxðsÞÞ ds
dþa
bd
qðsÞaðsÞ ds > e:
P C1 Me dþa
This implies that k > 1. Similarly, we can prove that there exists R > q > e such that for x 2 K with kxk ¼ R we have kTxk > kxk ¼ R. Hence for Tx ¼ lx; kxk ¼ R; x 2 K, we have l > 1. Now let x 2 K; kxk ¼ q, then f ðxÞ < M 1 p and Z b Z b kTxk 6 qðsÞaðsÞf ðxðsÞÞ ds < qðsÞaðsÞ ds M 1 q ¼ p: a
a
This implies if Tx ¼ kx, then k < 1. Let r1 ¼ e < r2 ¼ q < r3 ¼ R; D ¼ fx 2 K j e 6 kxk 6 Rg. Then the above discussion shows that conditions (a)–(d) of Lemma 3 are fulfilled. Therefore Lemma 3 implies that there exist two positive solutions x1 ðtÞ, x2 ðtÞ such that 0 < kx1 k < q < kx2 k:
Let K be a cone in Banach space E.
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
391
Definition 1. A map a : K ! ½0; 1Þ is said to be a nonnegative continuous concave functional if a is continuous and aðkx þ ð1 kÞyÞ P kaðxÞ þ ð1 kÞaðyÞ
8x; y 2 K and 0 6 k 6 1:
Definition 2. Let 0 < a < b be positive numbers and a be a nonnegative continuous functional on K, define convex sets Kr and Sða; a; bÞ, respectively, by Kr ¼ fx 2 K j kxk < rg and Sða; a; bÞ ¼ fx 2 k j 6 aðxÞ; kxk 6 bg: Lemma 4 (Legget–Williams fixed point theorem [8]). Let T : K C ! K C be a completely continuous and a be a nonnegative continuous concave functional on K such that aðxÞ 6 kxk for all x 2 K C . Suppose there exists 0 < u < b < d 6 C such that (C1) fx 2 Sða; b; dÞ j aðxÞ > bg 6¼ ; and aðTxÞ > b for x 2 Sða; b; dÞ, (C2) kTxk < a for kxk 6 a, and (C3) aðTxÞ > b for x 2 Sða; b; CÞ with kTxk > d. Then T has at least three fixed points x1 , x2 and x3 satisfying kx1 k < a;
b < aðx2 Þ and
kx3 k > a
with aðx3 Þ < b:
Theorem 4. Consider the ðk; n kÞ conjugate boundary value problem ð1Þnk xðnÞ ðtÞ ¼ bðtÞf ðxðtÞÞ;
a < t < b;
ð27Þ 0 6 i 6 k 1; 0 6 j 6 n k 1; Rb where b 2 Cð½a; b ; ½0; 1ÞÞ, f 2 Cð½0; 1Þ; ½0; 1ÞÞ. If 0 < a qðsÞbðsÞ ds ¼ M < 1, and there exist positive numbers 0 < A < B < C < D such that (i) f ½0; A ½0; M 1 AÞ; (ii) f ½0; D ½0; M 1 D ; (iii) f ½B; C ð2BM 1 C11 ; M 1 D . Then problem (27) has at least three positive solutions x1 ðtÞ, x2 ðtÞ and x3 ðtÞ such that ðiÞ
ðjÞ
x ðaÞ ¼ x ðbÞ ¼ 0;
kx1 k < A;
B<
min
aþd 6 t 6 bd
x2 ðtÞ
and kx3 k > A with minaþd 6 t 6 bd x3 ðtÞ < B; where d 2 ða; ðða þ bÞ=2ÞÞ is such that Z bd 12 and C1 ¼ min pðtÞ; qðsÞbðsÞ ds ¼ aþd 6 t 6 bd M aþd and pðsÞ; qðsÞ are given by Lemma 1.
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X. Yang / Appl. Math. Comput. 136 (2003) 379–393
Proof. Let K ¼ fx 2 C½a; b ; xðtÞ P 0 and minaþd 6 t 6 bd xðtÞ P C1 kxk, where C1 ¼ mind 6 t 6 bd pðtÞ 2 ð0; 1Þ; pðtÞ is given by Lemma 1}. Then from Lemma 1, we have Z b TxðtÞ Gðt; sÞbðsÞf ðxðsÞÞ ds a
and min
aþd 6 t 6 bd
Z b min pðtÞ qðsÞbðsÞf ðxðsÞÞ ds aþd 6 t 6 bd a Z b qðsÞbðsÞf ðxðsÞÞ ds ¼ C1 a Z b Gðt; sÞbðsÞf ðxðsÞÞ ds P C1 ¼ C1 kxk:
TxðtÞ P
a
This implies that TK K. Let KD ¼ fx 2 K; kxk < Dg. We show that T K D K D . In fact, by Lemma 1, Gðt; sÞ 6 qðsÞ and by (ii), f ð½0; DÞ ½0; M 1 D , we have therefore Z b kTxk ¼ max TxðtÞ 6 qðsÞbðsÞf ðxðsÞÞ ds a6t6b a Z b qðsÞbðsÞ ds DM 1 ¼ D: 6 a
If x 2 KD and kxk 6 A, then from (i) f ðxÞ 6 M 1 A and Z b kTxk < qðsÞbðsÞ ds M 1 A ¼ A; a
which implies (C2) of Lemma 4. Since xðtÞ ððB þ CÞ=2Þ 2 Sða; B; CÞ, where aðxÞ ¼ minaþd 6 t 6 bd xðtÞ, then a is a nonnegative continuous concave functional on K and we have fx 2 Sða; B; CÞ j aðxÞ > Bg 6¼ ;. For x 2 Sða; B; CÞ, we have by (iii), aðTxÞ ¼
min
TxðtÞ Z b ¼ min Gðt; sÞbðsÞf ðxðsÞÞ ds aþd 6 t 6 bd a Z bd Z bd P C1 qðsÞf ðxðsÞÞ ds > C1 qðsÞbðsÞ ds 2C11 M 1 B ¼ B: aþd 6 t 6 bd
aþd
This proves (C1) of Lemma 4.
aþd
X. Yang / Appl. Math. Comput. 136 (2003) 379–393
393
Now let x 2 Sða; B; DÞ, then minaþd 6 t 6 bd xðtÞ P B and Z bd aðTxÞ ¼ min TxðtÞ P C1 qðsÞbðsÞf ðxðsÞÞ ds aþd 6 t 6 bd
Z
aþd
bd
> C1
qðsÞbðsÞ ds2M 1 C11 B ¼ B:
aþd
This implies that (C3) of Lemma 4 is fulfilled. Combining the above results, Theorem 4 is proved.
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