- Email: [email protected]
Positive solutions for three-point one-dimensional p-Laplacian boundary value problems with advanced arguments
Applied Mathematics and Computation 215 (2009) 125–131
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Positive solutions for three-point one-dimensional p-Laplacian boundary value problems with advanced arguments Tadeusz Jankowski ´ sk, Poland Gdansk University of Technology, Department of Differential Equations, 11/12 G. Narutowicz Str., 80-952 Gdan
a r t i c l e
i n f o
a b s t r a c t This paper considers the existence of positive solutions for advanced differential equations with one-dimensional p-Laplacian. To obtain the existence of at least three positive solutions we use a fixed point theorem due to Avery and Peterson. Ó 2009 Elsevier Inc. All rights reserved.
Keywords: Multiple positive solutions Differential equations with advanced arguments One-dimensional p-Laplacian Boundary value problems
1. Introduction Put J ¼ ½0; 1; J 0 ¼ ð0; 1Þ; Rþ ¼ ½0; 1Þ. Let us consider the following problem:
(
ð/p ðx0 ðtÞÞÞ0 þ hðtÞf ðt; xðaðtÞÞÞ ¼ 0; xð0Þ ¼ 0;
t 2 J;
bxðgÞ ¼ xð1Þ;
where /ðsÞ ¼ jsjp2 s; p > 1; /1 p ¼ /q ;
1 p
ð1Þ
þ 1q ¼ 1 and
H1 : f 2 CðJ Rþ ; Rþ Þ; a 2 CðJ; J 0 Þ. H2 : h 2 CðJ; Rþ Þ and h does not vanish identically on any subinterval. H3 : b; g 2 ð0; 1Þ. By a positive solution of problem (1) we mean a function x 2 C 2 ðJÞ which is positive on ð0; 1Þ. We have many fixed point theorems including corresponding theorems in a cone. Recently, there has been much attention on the existence of positive solutions for ordinary differential equations with boundary conditions including one-dimensional p-Laplacian. We have a vast literature devoted to the applications of fixed point theorems to obtain positive solutions of corresponding boundary value problems. Note that fixed point theorems in cones can also be applied to boundary value problems of differential equations with deviating arguments, but we have only a few such papers, see for example [5,6,11]. Indeed, papers [2–4,7,9,10] have also a word ‘‘delay” in the title but the corresponding functions f appearing on the right-hand-sides depend on xðt sÞ, where initial functions x are given on the initial set, for example ½s; 0. It means that problems from papers [1–4,7,9,10] have no advanced deviating arguments. To obtain positive solutions to problem (1) we use a fixed theorem due to Avery and Peterson [1]. Note that my paper is a first one when this fixed point theorem is applied to one-dimensional p-Laplacian boundary value problems with advanced arguments.
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2009.04.045
126
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
2. Some lemmas Lemma 1. ([see [8]]) Suppose u 2 C 1 ðJÞ with ð/ðu0 ðÞÞÞ0 2 CðJÞ satisfies
(
ð/p ðu0 ðtÞÞÞ0 P 0; t 2 J 0 ; uð0Þ ¼ 0; uð1Þ ¼ buðgÞ; b; g 2 J 0 :
Then u is concave and uðtÞ P 0; t 2 J. Moreover, there exists n 2 J 0 such that u0 ðnÞ ¼ 0. Let us consider the following problem:
(
ð/ðu0 ðtÞÞÞ0 þ yðtÞ ¼ 0;
t 2 J0 ;
ð2Þ
uð0Þ ¼ 0; buðgÞ ¼ uð1Þ:
ð3Þ
Lemma 2. Let Assumption H3 hold and y 2 CðJ; Rþ Þ. Assume that n 2 J 0 is a solution of the equation with respect to r
Z
Z
r
r
/q 0
Z yðsÞds ds ¼ b
Z
g
0
s
r
/q
Z yðsÞds ds þ
s
Z
1
s
/q
r
yðsÞds ds;
r
if g 6 n, and n 2 J 0 is a solution of
Z
Z
r
/q 0
r
yðsÞds ds ¼
s
1 1b
Z
Z
1
/q r
s
Z yðsÞds ds b
r
Z
g
r
/q
s
yðsÞds ds ;
r
if n 6 g. Then problem (2), (3) has the unique solution given by the following formula:
8R R > < 0t /q sn yðsÞds ds if t 2 ½0; n; R uðtÞ ¼ R > : uðnÞ nt /q ns yðsÞds ds if t 2 ½n; 1
ð4Þ
with
8 R R R R > < b 0g /q sn yðsÞds ds þ n1 /q ns yðsÞds ds if g 6 n; hR R i uðnÞ ¼ R g R s 1 s > 1 : 1b n /q n yðsÞds ds b n /q n yðsÞds ds if n 6 g:
ð5Þ
Proof. First, integrating the equation in (2) from n to 1, we have
/p ðu0 ðtÞÞ ¼ /p ðu0 ðnÞÞ
Z
t
yðsÞds:
n
Hence,
u0 ðtÞ ¼ /q
Z
yðsÞds ;
t n
0
because u ðnÞ ¼ 0, by Lemma 1. Integrating the last equation on ðn; 1Þ we obtain
uðtÞ ¼ uðnÞ
Z
Z
t
yðsÞds ds:
s
/q n
n
Now integrating the equation in (2) on ð0; nÞ we have
/p ðu0 ðtÞÞ ¼ /p ðu0 ð0ÞÞ
Z 0
and hence
/p ðu0 ð0ÞÞ ¼
Z
n
yðsÞds;
0
because u0 ðnÞ ¼ 0, so
u0 ðtÞ ¼ /q
Z t
n
yðsÞds :
t
yðsÞds
ð6Þ
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
127
Integrating the last equation on ð0; nÞ we obtain
uðtÞ ¼
Z
Z
t
n
/q 0
yðsÞds ds:
ð7Þ
s
It means that combining (6) and (7), we have (4). Now we need to eliminate uðnÞ using the boundary condition (3). Then, we have
uðnÞ
Z
Z
1
s
/q n
n
8 R R > < b 0g /q sn yðsÞds ds if g 6 n; yðsÞds ds ¼ R g R s > : b uðnÞ n /q n yðsÞds ds if n 6 g:
It yields (5). This ends the proof.
h
Lemma 3. Let Assumption H3 hold. Assume that y 2 CðJ; Rþ Þ with the maximum norm in the space CðJ; Rþ Þ. Then the unique solution u of problem (2), (3) satisfies the condition
min uðtÞ P Ckuk; ½g;1
where
C ¼ b min
1g ;g : 1 bg
Proof. Note that uð1Þ ¼ buðgÞ 6 uðgÞ. It means that
min uðtÞ ¼ uð1Þ: ½g;1
Put uðt Þ ¼ kuk. If t 6 g, then
uðt Þ 6 uðgÞ þ
uðgÞ uð1Þ uðgÞ uð1Þ 1 bg ðg t Þ 6 uðgÞ þ g¼ uð1Þ: 1g 1g bð1 gÞ
It yields
min uðtÞ P ½g;1
bð1 gÞ kuk: 1 bg
If g < t , then
uðt Þ 6 uðgÞ þ
uðgÞ uð0Þ uðgÞ 1 ð1 gÞ ¼ uð1Þ: ðt gÞ 6 uðgÞ þ bg g0 g
It yields
min uðtÞ P bgkuk: ½g;1
This ends the proof. h
3. Main results Now, we present the necessary definitions from the theory of cones in Banach spaces. Definition 1. Let E be a real Banach space. A nonempty convex set P E is said to be a cone provided that (i) ku 2 P for all u 2 P and all k P 0, and (ii) u; u 2 P implies u ¼ 0. Note that every cone P E induces an ordering in E given by x 6 y if y x 2 P. Definition 2. A map K is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if K : P ! Rþ is continuous and
Kðtx þ ð1 tÞyÞ P t KðxÞ þ ð1 tÞKðyÞ for all x; y 2 P and t 2 ½0; 1.
128
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
Similarly, we say the map u is a nonnegative continuous convex functional on a cone P of a real Banach space E if
u : P ! Rþ is continuous and
uðtx þ ð1 tÞyÞ 6 tuðxÞ þ ð1 tÞuðyÞ for all x; y 2 P and t 2 ½0; 1. Definition 3. An operator is called completely continuous if it is continuous and maps bounded sets into pre-compact sets. Let u and H be nonnegative continuous convex functionals on P, K be a nonnegative continuous concave functional on P, and W be a nonnegative continuous functional on P. Then for positive numbers a; b; c and d, we define the following sets:
Pðu; dÞ ¼ fx 2 P : uðxÞ < dg; Pðu; K; b; dÞ ¼ fx 2 P : b 6 KðxÞ; uðxÞ 6 dg; Pðu; H; K; b; c; dÞ ¼ fx 2 P : b 6 KðxÞ; HðxÞ 6 c; uðxÞ 6 dg and
Rðu; W; a; dÞ ¼ fx 2 P : a 6 WðxÞ; uðxÞ 6 dg: We will use the following fixed point theorem of Avery and Peterson to establish multiple positive solutions to problem (1). Theorem 1. ([see [1]]) Let P be a cone in a real Banach space E. Let u and H be nonnegative continuous convex functionals on P, K be a nonnegative continuous concave functional on P, and W be a nonnegative continuous functional on P satisfying WðkxÞ 6 kWðxÞ for 0 6 k 6 1, such that for some positive numbers M and d,
KðxÞ 6 WðxÞ and kxk 6 MuðxÞ for all x 2 Pðu; dÞ. Suppose
T : Pðu; dÞ ! Pðu; dÞ is completely continuous and there exist positive numbers a; b and c with a < b such that (S1): fx 2 Pðu; H; K; b; c; dÞ : KðxÞ > bg – ; and KðTxÞ > b for x 2 Pðu; H; K; b; c; dÞ. (S2): KðTxÞ > b for x 2 Pðu; K; b; dÞ with HðTxÞ > c. (S3): 0 R Rðu; W; a; dÞ and WðTxÞ < a for x 2 Rðu; W; a; dÞ with WðxÞ ¼ a. Then T has at least three fixed points x1 ; x2 ; x3 2 Pðu; dÞ, such that
uðxi Þ 6 d; for i ¼ 1; 2; 3; b < Kðx1 Þ;
a < Wðx2 Þ;
with Kðx2 Þ < b
and
Wðx3 Þ < a: Let X ¼ CðJ; RÞ be our Banach space with the maximum norm kxk ¼ maxt2J jxðtÞj. Let
P¼
x 2 X : x is nonnegative; concave on J and min xðtÞ P Ckxk ; ½g;1
Pr ¼ fx 2 P : kxk 6 rg; where C is defined as in Lemma 3. We define the nonnegative continuous concave functional K on P by
KðxÞ ¼ min jxðtÞj: ½g;1
Note that KðxÞ 6 kxk. Put WðxÞ ¼ HðxÞ ¼ kxk. Theorem 2. Let Assumptions H1 –H3 hold. In addition, we assume that there exist positive constants a; b; c; d; a < b and such that
l¼
Z
1
hðsÞds; Z 1 Z s Z g Z g b 0 < L 6 min /q hðsÞds ds; b /q hðsÞds ds 1b g g 0 s 0
129
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
and ðA1 Þ f ðt; uÞ 6 l1 /p ðdÞ for ðt; uÞ 2 J ½0; d.
ðA2 Þ f ðt; uÞ P /p ðbLÞ for ðt; uÞ 2 ½g; 1 b; Cb . 1 ðA3 Þ f ðt; uÞ 6 l /p ðaÞ for ðt; uÞ 2 J ½0; a. Then, problem (1) has at least three positive solutions x1 ; x2 ; x3 satisfying kxi k 6 d; i ¼ 1; 2; 3 and
b 6 Kðx1 Þ;
a < kx2 k with Kðx2 Þ < b
and kx3 k < a. Proof. Put ðFxÞðsÞ ¼ hðsÞf ðs; xðaðsÞÞÞ. Now we define an operator T by
8R R > < 0t /q sn ðFxÞðsÞds ds if t 2 ½0; n; R ðTxÞðtÞ ¼ R t s > : ux ðnÞ n /q n ðFxÞðsÞds ds if t 2 ½n; 1
ð8Þ
with
8 R R R R > < b 0g /q sn ðFxÞðsÞds ds þ n1 /q ns ðFxÞðsÞds ds if g 6 n; hR R R i ux ðnÞ ¼ R 1 s g s > 1 : 1b /q n ðFxÞðsÞds ds b n /q n ðFxÞðsÞds ds if n 6 g; n where n is a solution of the equation with respect to r
Z
Z
r
r
/q 0
Z ðFxÞðsÞds ds ¼ b
Z
g
0
s
r
/q
Z ðFxÞðsÞds ds þ
s
Z
1
s
/q r
ðFxÞðsÞds ds;
r
if g 6 n or n is a solution of the equation
Z
Z
r
/q 0
r
ðFxÞðsÞds ds ¼
1 1b
s
Z
Z
1
/q r
s
Z ðFxÞðsÞds ds b
r
n
Z
g
/q
s
ðFxÞðsÞds ds ;
r
if n 6 g. Indeed, T : X ! X. Problem (1) has a solution x if and only if x solves the operator equation x ¼ Tx. Note that
8 R > < /q tn ðTxÞðsÞds if t 2 ½0; n; 0 R ðTxÞ ðtÞ ¼ t > : /q n ðTxÞðsÞds if t 2 ½n; 1: We see that Tx is nondecreasing on ½0; n, nonincreasing on ½n; 1 and ðTxÞ0 ðnÞ ¼ 0. Hence ðTxÞðnÞ is the maximum value of Tx on J. Moreover, ðTxÞð0Þ ¼ 0,
ðTxÞð1Þ bðTxÞðgÞ ¼ b
Z
Z
g
n
/q s
0
b
Z ðFxÞðsÞds ds þ
Z
Z
g
s
/q
n
n
/q 0
Z
1
Z ðFxÞðsÞds ds
n
ðFxÞðsÞds ds ¼ 0 if g 6 n;
Z
1
ðFxÞðsÞds ds
s
/q
n
n
s
and
ðTxÞð1Þ bðTxÞðgÞ ¼ b
Z
Z
g
n
þb
s
/q Z
n
Z
g
/q n
Z ðFxÞðsÞds ds þ n
n
s
ð/p ððTxÞ0 ÞðtÞÞ0 ¼ hðtÞf ðt; xðaðtÞÞÞ 6 0; so Tx is concave and ðTxÞðtÞ P 0; t 2 J, by Lemma 1. Consequently, in view of Lemma 3, we have ½g;1
so TP P.
/q s
n
Z ðFxÞðsÞds ds
ðFxÞðsÞds ds ¼ 0 if n 6 g:
Indeed,
minðTxÞðtÞ P CkTxk;
Z
1
n
Z
1
/q n
s
ðFxÞðsÞds ds
130
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
Now we prove that the operator T : P ! P is completely continuous. Let x 2 P r . Then jxj 6 r. Note that h and f are continuous so h is bounded on J and f is bounded on J ½r; r. It means that there exists a constant K > 0 such kTxk 6 K. This proves that TP is uniformly bounded. On the other hand for t 1 ; t2 2 J there exists a constant L1 > 0 such that
jðTxÞðt1 Þ ðTxÞðt2 Þj 6 L1 jt1 t 2 j: This shows that TP is equicontinuous on J, so T is completely continuous. Let x 2 Pðu; dÞ, so 0 6 xðtÞ 6 d; t 2 J, and kxk 6 d. Note that also 0 6 xðaðtÞÞ 6 d; t 2 J because 0 6 t 6 aðtÞ 6 1 on J. By Assumption ðA1 Þ, we see that
uðTxÞ ¼ kTxk ¼ max jðTxÞðtÞj ¼ maxðTxÞðtÞ ¼ ðTxÞðnÞ ¼ t2J
6d
Z
1
/q
1
t2J
Z
l
0
1
Z
/q 0
hðsÞds ds ¼ d:
Z
n
n
Z ðFxÞðsÞds ds 6
s
n
/q
0
1
l
/p ðdÞ
Z
n
hðsÞds ds
s
0
It proves that T : Pðu; dÞ ! Pðu; dÞ. Now we need to show that condition ðS1 Þ is satisfied. Take
xðtÞ ¼
1 b bþ ; 2 C
kxk ¼
bðC þ 1Þ b 6 2c C
t 2 J:
Then
so KðxÞ ¼ min xðtÞ ¼ ½g;1
bðC þ 1Þ b > b ¼ C P Ckxk: 2C C
It proves that
b x 2 P u; H; K; b; ; d : b < KðxÞ – ;:
C
Let b 6 uðtÞ 6 Cb for t 2 ½g; 1. Then g 6 t 6 aðtÞ 6 1 on ½g; 1. It yields b 6 uðaðtÞÞ 6 Cb on ½g; 1. Note that
minðTxÞðtÞ ¼ minððTxÞðgÞ; ðTxÞð1ÞÞ ¼ min ½g;1
1 ðTxÞð1Þ; ðTxÞð1Þ ¼ ðTxÞð1Þ: b
Hence, in view of Assumption ðA2 Þ, we have
Z g Z n Z g Z n b KðTxÞ ¼ minðTxÞðtÞ ¼ ðTxÞð1Þ ¼ b /q ðFxÞðsÞds ds P b /q /p hðsÞds ds ½g;1 L s s 0 0 Z g Z g b / hðsÞds ds P b if g 6 n Pb L 0 q s
and
Z 1 Z s Z 1 Z s b b b /q ðFxÞðsÞds ds P /q /p hðsÞds ds ½g;1 1b g 1b g L n n Z s Z b b 1 / hðsÞds ds P b if n 6 g: P 1b L g q g
KðTxÞ ¼ minðTxÞðtÞ ¼ ðTxÞð1Þ ¼
Consequently, KðTxÞ > b, so condition ðS1 Þ holds.
Now we need to prove that condition ðS2 Þ is satisfied. Take x 2 P u; K; b; Cb and kTuk > Cb ¼ d. Then
KðTxÞ ¼ minðTxÞðtÞ P CkTxk > C ½g;1
b
C
¼ b;
so condition ðS2 Þ holds. Indeed, uð0Þ ¼ 0 < a, so 0 R Rðu; W; a; dÞ. Suppose that x 2 Rðu; W; a; dÞ with WðxÞ ¼ kxk ¼ a. Then
WðTxÞ ¼ kTxk ¼ maxðTxÞðtÞ ¼ ðTxÞðnÞ ¼ t2J
6a
Z
1
/q 0
1
l
Z
1
hðsÞds ds ¼ a:
Z
Z
n
/q 0
s
n
Z ðFxÞðsÞds ds 6
n
/q 0
1
l
/p ðaÞ
Z
n
hðsÞds ds
s
0
It shows that condition ðS3 Þ is satisfied. By Theorem 1, there exist at least three positive solutions x1 ; x2 ; x3 of problem (1) such that kxi k 6 d for i ¼ 1; 2; 3,
b 6 min½g;1 x1 ðtÞ;
a < kx2 k with min x2 ðtÞ < b
and kx3 k < a. This ends the proof.
½g;1
h
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
131
Remark 1. Note that we can also discuss the situation when b ¼ 1, so when the boundary condition in problem (1) has the form xð1Þ ¼ xðgÞ but in this case we have to assume that g 6 n. References [1] R.I. Avery, A.C. Peterson, Three positive fixed points of nonlinear operators on ordered Banach spaces, Comput. Math. Appl. 42 (2001) 313–322. [2] D. Bai, Y. Xu, Positive solutions of second-order two-delay differential systems with twin parameter, Nonlinear Anal. 63 (2005) 601–617. [3] D. Bai, Y. Xu, Existence of positive solutions for boundary value problem of second-order delay differential equations, Appl. Math. Lett. 18 (2005) 621– 630. [4] B. Du, X. Hu, W. Ge, Positive solutions to a type of multi-point boundary value problem with delay and one-dimensional p-Laplacian, Appl. Math. Comput. 208 (2009) 501–510. [5] T. Jankowski, Positive solutions of three-point boundary value problems for second order impulsive differential equations with advanced arguments, Appl. Math. Comput. 197 (2008) 179–189. [6] T. Jankowski, Three positive solutions to second-order three-point impulsive differential equations with deviating arguments, Int. J. Comput. Math., in press. [7] D. Jiang, Multiple positive solutions for boundary value problems of second-order delay differential equations, Appl. Math. Lett. 15 (2002) 575–583. [8] B. Sun, Y. Qu, W. Ge, Existence and iteration of positive solutions for a multipoint one-dimensional p-Laplacian boundary value problem, Appl. Math. Comput. 197 (2008) 389–398. [9] W. Wang, J. Sheng, Positive solutions to a multi-point boundary value problem with delay, Appl. Math. Comput. 188 (2007) 96–102. [10] Y. Wang, W. Zhao, W. Ge, Multiple positive solutions for boundary value problems of second order delay differential equations with one-dimensional p-Laplacian, J. Math. Anal. Appl. 326 (2007) 641–654. [11] C. Yang, C. Zhai, J. Yan, Positive solutions of the three-point boundary value problem for second order differential equations with an advanced argument, Nonlinear Anal. 65 (2006) 2013–2023.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Positive solutions for three-point one-dimensional p-Laplacian boundary value problems with advanced arguments Tadeusz Jankowski ´ sk, Poland Gdansk University of Technology, Department of Differential Equations, 11/12 G. Narutowicz Str., 80-952 Gdan
a r t i c l e
i n f o
a b s t r a c t This paper considers the existence of positive solutions for advanced differential equations with one-dimensional p-Laplacian. To obtain the existence of at least three positive solutions we use a fixed point theorem due to Avery and Peterson. Ó 2009 Elsevier Inc. All rights reserved.
Keywords: Multiple positive solutions Differential equations with advanced arguments One-dimensional p-Laplacian Boundary value problems
1. Introduction Put J ¼ ½0; 1; J 0 ¼ ð0; 1Þ; Rþ ¼ ½0; 1Þ. Let us consider the following problem:
(
ð/p ðx0 ðtÞÞÞ0 þ hðtÞf ðt; xðaðtÞÞÞ ¼ 0; xð0Þ ¼ 0;
t 2 J;
bxðgÞ ¼ xð1Þ;
where /ðsÞ ¼ jsjp2 s; p > 1; /1 p ¼ /q ;
1 p
ð1Þ
þ 1q ¼ 1 and
H1 : f 2 CðJ Rþ ; Rþ Þ; a 2 CðJ; J 0 Þ. H2 : h 2 CðJ; Rþ Þ and h does not vanish identically on any subinterval. H3 : b; g 2 ð0; 1Þ. By a positive solution of problem (1) we mean a function x 2 C 2 ðJÞ which is positive on ð0; 1Þ. We have many fixed point theorems including corresponding theorems in a cone. Recently, there has been much attention on the existence of positive solutions for ordinary differential equations with boundary conditions including one-dimensional p-Laplacian. We have a vast literature devoted to the applications of fixed point theorems to obtain positive solutions of corresponding boundary value problems. Note that fixed point theorems in cones can also be applied to boundary value problems of differential equations with deviating arguments, but we have only a few such papers, see for example [5,6,11]. Indeed, papers [2–4,7,9,10] have also a word ‘‘delay” in the title but the corresponding functions f appearing on the right-hand-sides depend on xðt sÞ, where initial functions x are given on the initial set, for example ½s; 0. It means that problems from papers [1–4,7,9,10] have no advanced deviating arguments. To obtain positive solutions to problem (1) we use a fixed theorem due to Avery and Peterson [1]. Note that my paper is a first one when this fixed point theorem is applied to one-dimensional p-Laplacian boundary value problems with advanced arguments.
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2009.04.045
126
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
2. Some lemmas Lemma 1. ([see [8]]) Suppose u 2 C 1 ðJÞ with ð/ðu0 ðÞÞÞ0 2 CðJÞ satisfies
(
ð/p ðu0 ðtÞÞÞ0 P 0; t 2 J 0 ; uð0Þ ¼ 0; uð1Þ ¼ buðgÞ; b; g 2 J 0 :
Then u is concave and uðtÞ P 0; t 2 J. Moreover, there exists n 2 J 0 such that u0 ðnÞ ¼ 0. Let us consider the following problem:
(
ð/ðu0 ðtÞÞÞ0 þ yðtÞ ¼ 0;
t 2 J0 ;
ð2Þ
uð0Þ ¼ 0; buðgÞ ¼ uð1Þ:
ð3Þ
Lemma 2. Let Assumption H3 hold and y 2 CðJ; Rþ Þ. Assume that n 2 J 0 is a solution of the equation with respect to r
Z
Z
r
r
/q 0
Z yðsÞds ds ¼ b
Z
g
0
s
r
/q
Z yðsÞds ds þ
s
Z
1
s
/q
r
yðsÞds ds;
r
if g 6 n, and n 2 J 0 is a solution of
Z
Z
r
/q 0
r
yðsÞds ds ¼
s
1 1b
Z
Z
1
/q r
s
Z yðsÞds ds b
r
Z
g
r
/q
s
yðsÞds ds ;
r
if n 6 g. Then problem (2), (3) has the unique solution given by the following formula:
8R R > < 0t /q sn yðsÞds ds if t 2 ½0; n; R uðtÞ ¼ R > : uðnÞ nt /q ns yðsÞds ds if t 2 ½n; 1
ð4Þ
with
8 R R R R > < b 0g /q sn yðsÞds ds þ n1 /q ns yðsÞds ds if g 6 n; hR R i uðnÞ ¼ R g R s 1 s > 1 : 1b n /q n yðsÞds ds b n /q n yðsÞds ds if n 6 g:
ð5Þ
Proof. First, integrating the equation in (2) from n to 1, we have
/p ðu0 ðtÞÞ ¼ /p ðu0 ðnÞÞ
Z
t
yðsÞds:
n
Hence,
u0 ðtÞ ¼ /q
Z
yðsÞds ;
t n
0
because u ðnÞ ¼ 0, by Lemma 1. Integrating the last equation on ðn; 1Þ we obtain
uðtÞ ¼ uðnÞ
Z
Z
t
yðsÞds ds:
s
/q n
n
Now integrating the equation in (2) on ð0; nÞ we have
/p ðu0 ðtÞÞ ¼ /p ðu0 ð0ÞÞ
Z 0
and hence
/p ðu0 ð0ÞÞ ¼
Z
n
yðsÞds;
0
because u0 ðnÞ ¼ 0, so
u0 ðtÞ ¼ /q
Z t
n
yðsÞds :
t
yðsÞds
ð6Þ
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
127
Integrating the last equation on ð0; nÞ we obtain
uðtÞ ¼
Z
Z
t
n
/q 0
yðsÞds ds:
ð7Þ
s
It means that combining (6) and (7), we have (4). Now we need to eliminate uðnÞ using the boundary condition (3). Then, we have
uðnÞ
Z
Z
1
s
/q n
n
8 R R > < b 0g /q sn yðsÞds ds if g 6 n; yðsÞds ds ¼ R g R s > : b uðnÞ n /q n yðsÞds ds if n 6 g:
It yields (5). This ends the proof.
h
Lemma 3. Let Assumption H3 hold. Assume that y 2 CðJ; Rþ Þ with the maximum norm in the space CðJ; Rþ Þ. Then the unique solution u of problem (2), (3) satisfies the condition
min uðtÞ P Ckuk; ½g;1
where
C ¼ b min
1g ;g : 1 bg
Proof. Note that uð1Þ ¼ buðgÞ 6 uðgÞ. It means that
min uðtÞ ¼ uð1Þ: ½g;1
Put uðt Þ ¼ kuk. If t 6 g, then
uðt Þ 6 uðgÞ þ
uðgÞ uð1Þ uðgÞ uð1Þ 1 bg ðg t Þ 6 uðgÞ þ g¼ uð1Þ: 1g 1g bð1 gÞ
It yields
min uðtÞ P ½g;1
bð1 gÞ kuk: 1 bg
If g < t , then
uðt Þ 6 uðgÞ þ
uðgÞ uð0Þ uðgÞ 1 ð1 gÞ ¼ uð1Þ: ðt gÞ 6 uðgÞ þ bg g0 g
It yields
min uðtÞ P bgkuk: ½g;1
This ends the proof. h
3. Main results Now, we present the necessary definitions from the theory of cones in Banach spaces. Definition 1. Let E be a real Banach space. A nonempty convex set P E is said to be a cone provided that (i) ku 2 P for all u 2 P and all k P 0, and (ii) u; u 2 P implies u ¼ 0. Note that every cone P E induces an ordering in E given by x 6 y if y x 2 P. Definition 2. A map K is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if K : P ! Rþ is continuous and
Kðtx þ ð1 tÞyÞ P t KðxÞ þ ð1 tÞKðyÞ for all x; y 2 P and t 2 ½0; 1.
128
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
Similarly, we say the map u is a nonnegative continuous convex functional on a cone P of a real Banach space E if
u : P ! Rþ is continuous and
uðtx þ ð1 tÞyÞ 6 tuðxÞ þ ð1 tÞuðyÞ for all x; y 2 P and t 2 ½0; 1. Definition 3. An operator is called completely continuous if it is continuous and maps bounded sets into pre-compact sets. Let u and H be nonnegative continuous convex functionals on P, K be a nonnegative continuous concave functional on P, and W be a nonnegative continuous functional on P. Then for positive numbers a; b; c and d, we define the following sets:
Pðu; dÞ ¼ fx 2 P : uðxÞ < dg; Pðu; K; b; dÞ ¼ fx 2 P : b 6 KðxÞ; uðxÞ 6 dg; Pðu; H; K; b; c; dÞ ¼ fx 2 P : b 6 KðxÞ; HðxÞ 6 c; uðxÞ 6 dg and
Rðu; W; a; dÞ ¼ fx 2 P : a 6 WðxÞ; uðxÞ 6 dg: We will use the following fixed point theorem of Avery and Peterson to establish multiple positive solutions to problem (1). Theorem 1. ([see [1]]) Let P be a cone in a real Banach space E. Let u and H be nonnegative continuous convex functionals on P, K be a nonnegative continuous concave functional on P, and W be a nonnegative continuous functional on P satisfying WðkxÞ 6 kWðxÞ for 0 6 k 6 1, such that for some positive numbers M and d,
KðxÞ 6 WðxÞ and kxk 6 MuðxÞ for all x 2 Pðu; dÞ. Suppose
T : Pðu; dÞ ! Pðu; dÞ is completely continuous and there exist positive numbers a; b and c with a < b such that (S1): fx 2 Pðu; H; K; b; c; dÞ : KðxÞ > bg – ; and KðTxÞ > b for x 2 Pðu; H; K; b; c; dÞ. (S2): KðTxÞ > b for x 2 Pðu; K; b; dÞ with HðTxÞ > c. (S3): 0 R Rðu; W; a; dÞ and WðTxÞ < a for x 2 Rðu; W; a; dÞ with WðxÞ ¼ a. Then T has at least three fixed points x1 ; x2 ; x3 2 Pðu; dÞ, such that
uðxi Þ 6 d; for i ¼ 1; 2; 3; b < Kðx1 Þ;
a < Wðx2 Þ;
with Kðx2 Þ < b
and
Wðx3 Þ < a: Let X ¼ CðJ; RÞ be our Banach space with the maximum norm kxk ¼ maxt2J jxðtÞj. Let
P¼
x 2 X : x is nonnegative; concave on J and min xðtÞ P Ckxk ; ½g;1
Pr ¼ fx 2 P : kxk 6 rg; where C is defined as in Lemma 3. We define the nonnegative continuous concave functional K on P by
KðxÞ ¼ min jxðtÞj: ½g;1
Note that KðxÞ 6 kxk. Put WðxÞ ¼ HðxÞ ¼ kxk. Theorem 2. Let Assumptions H1 –H3 hold. In addition, we assume that there exist positive constants a; b; c; d; a < b and such that
l¼
Z
1
hðsÞds; Z 1 Z s Z g Z g b 0 < L 6 min /q hðsÞds ds; b /q hðsÞds ds 1b g g 0 s 0
129
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
and ðA1 Þ f ðt; uÞ 6 l1 /p ðdÞ for ðt; uÞ 2 J ½0; d.
ðA2 Þ f ðt; uÞ P /p ðbLÞ for ðt; uÞ 2 ½g; 1 b; Cb . 1 ðA3 Þ f ðt; uÞ 6 l /p ðaÞ for ðt; uÞ 2 J ½0; a. Then, problem (1) has at least three positive solutions x1 ; x2 ; x3 satisfying kxi k 6 d; i ¼ 1; 2; 3 and
b 6 Kðx1 Þ;
a < kx2 k with Kðx2 Þ < b
and kx3 k < a. Proof. Put ðFxÞðsÞ ¼ hðsÞf ðs; xðaðsÞÞÞ. Now we define an operator T by
8R R > < 0t /q sn ðFxÞðsÞds ds if t 2 ½0; n; R ðTxÞðtÞ ¼ R t s > : ux ðnÞ n /q n ðFxÞðsÞds ds if t 2 ½n; 1
ð8Þ
with
8 R R R R > < b 0g /q sn ðFxÞðsÞds ds þ n1 /q ns ðFxÞðsÞds ds if g 6 n; hR R R i ux ðnÞ ¼ R 1 s g s > 1 : 1b /q n ðFxÞðsÞds ds b n /q n ðFxÞðsÞds ds if n 6 g; n where n is a solution of the equation with respect to r
Z
Z
r
r
/q 0
Z ðFxÞðsÞds ds ¼ b
Z
g
0
s
r
/q
Z ðFxÞðsÞds ds þ
s
Z
1
s
/q r
ðFxÞðsÞds ds;
r
if g 6 n or n is a solution of the equation
Z
Z
r
/q 0
r
ðFxÞðsÞds ds ¼
1 1b
s
Z
Z
1
/q r
s
Z ðFxÞðsÞds ds b
r
n
Z
g
/q
s
ðFxÞðsÞds ds ;
r
if n 6 g. Indeed, T : X ! X. Problem (1) has a solution x if and only if x solves the operator equation x ¼ Tx. Note that
8 R > < /q tn ðTxÞðsÞds if t 2 ½0; n; 0 R ðTxÞ ðtÞ ¼ t > : /q n ðTxÞðsÞds if t 2 ½n; 1: We see that Tx is nondecreasing on ½0; n, nonincreasing on ½n; 1 and ðTxÞ0 ðnÞ ¼ 0. Hence ðTxÞðnÞ is the maximum value of Tx on J. Moreover, ðTxÞð0Þ ¼ 0,
ðTxÞð1Þ bðTxÞðgÞ ¼ b
Z
Z
g
n
/q s
0
b
Z ðFxÞðsÞds ds þ
Z
Z
g
s
/q
n
n
/q 0
Z
1
Z ðFxÞðsÞds ds
n
ðFxÞðsÞds ds ¼ 0 if g 6 n;
Z
1
ðFxÞðsÞds ds
s
/q
n
n
s
and
ðTxÞð1Þ bðTxÞðgÞ ¼ b
Z
Z
g
n
þb
s
/q Z
n
Z
g
/q n
Z ðFxÞðsÞds ds þ n
n
s
ð/p ððTxÞ0 ÞðtÞÞ0 ¼ hðtÞf ðt; xðaðtÞÞÞ 6 0; so Tx is concave and ðTxÞðtÞ P 0; t 2 J, by Lemma 1. Consequently, in view of Lemma 3, we have ½g;1
so TP P.
/q s
n
Z ðFxÞðsÞds ds
ðFxÞðsÞds ds ¼ 0 if n 6 g:
Indeed,
minðTxÞðtÞ P CkTxk;
Z
1
n
Z
1
/q n
s
ðFxÞðsÞds ds
130
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
Now we prove that the operator T : P ! P is completely continuous. Let x 2 P r . Then jxj 6 r. Note that h and f are continuous so h is bounded on J and f is bounded on J ½r; r. It means that there exists a constant K > 0 such kTxk 6 K. This proves that TP is uniformly bounded. On the other hand for t 1 ; t2 2 J there exists a constant L1 > 0 such that
jðTxÞðt1 Þ ðTxÞðt2 Þj 6 L1 jt1 t 2 j: This shows that TP is equicontinuous on J, so T is completely continuous. Let x 2 Pðu; dÞ, so 0 6 xðtÞ 6 d; t 2 J, and kxk 6 d. Note that also 0 6 xðaðtÞÞ 6 d; t 2 J because 0 6 t 6 aðtÞ 6 1 on J. By Assumption ðA1 Þ, we see that
uðTxÞ ¼ kTxk ¼ max jðTxÞðtÞj ¼ maxðTxÞðtÞ ¼ ðTxÞðnÞ ¼ t2J
6d
Z
1
/q
1
t2J
Z
l
0
1
Z
/q 0
hðsÞds ds ¼ d:
Z
n
n
Z ðFxÞðsÞds ds 6
s
n
/q
0
1
l
/p ðdÞ
Z
n
hðsÞds ds
s
0
It proves that T : Pðu; dÞ ! Pðu; dÞ. Now we need to show that condition ðS1 Þ is satisfied. Take
xðtÞ ¼
1 b bþ ; 2 C
kxk ¼
bðC þ 1Þ b 6 2c C
t 2 J:
Then
so KðxÞ ¼ min xðtÞ ¼ ½g;1
bðC þ 1Þ b > b ¼ C P Ckxk: 2C C
It proves that
b x 2 P u; H; K; b; ; d : b < KðxÞ – ;:
C
Let b 6 uðtÞ 6 Cb for t 2 ½g; 1. Then g 6 t 6 aðtÞ 6 1 on ½g; 1. It yields b 6 uðaðtÞÞ 6 Cb on ½g; 1. Note that
minðTxÞðtÞ ¼ minððTxÞðgÞ; ðTxÞð1ÞÞ ¼ min ½g;1
1 ðTxÞð1Þ; ðTxÞð1Þ ¼ ðTxÞð1Þ: b
Hence, in view of Assumption ðA2 Þ, we have
Z g Z n Z g Z n b KðTxÞ ¼ minðTxÞðtÞ ¼ ðTxÞð1Þ ¼ b /q ðFxÞðsÞds ds P b /q /p hðsÞds ds ½g;1 L s s 0 0 Z g Z g b / hðsÞds ds P b if g 6 n Pb L 0 q s
and
Z 1 Z s Z 1 Z s b b b /q ðFxÞðsÞds ds P /q /p hðsÞds ds ½g;1 1b g 1b g L n n Z s Z b b 1 / hðsÞds ds P b if n 6 g: P 1b L g q g
KðTxÞ ¼ minðTxÞðtÞ ¼ ðTxÞð1Þ ¼
Consequently, KðTxÞ > b, so condition ðS1 Þ holds.
Now we need to prove that condition ðS2 Þ is satisfied. Take x 2 P u; K; b; Cb and kTuk > Cb ¼ d. Then
KðTxÞ ¼ minðTxÞðtÞ P CkTxk > C ½g;1
b
C
¼ b;
so condition ðS2 Þ holds. Indeed, uð0Þ ¼ 0 < a, so 0 R Rðu; W; a; dÞ. Suppose that x 2 Rðu; W; a; dÞ with WðxÞ ¼ kxk ¼ a. Then
WðTxÞ ¼ kTxk ¼ maxðTxÞðtÞ ¼ ðTxÞðnÞ ¼ t2J
6a
Z
1
/q 0
1
l
Z
1
hðsÞds ds ¼ a:
Z
Z
n
/q 0
s
n
Z ðFxÞðsÞds ds 6
n
/q 0
1
l
/p ðaÞ
Z
n
hðsÞds ds
s
0
It shows that condition ðS3 Þ is satisfied. By Theorem 1, there exist at least three positive solutions x1 ; x2 ; x3 of problem (1) such that kxi k 6 d for i ¼ 1; 2; 3,
b 6 min½g;1 x1 ðtÞ;
a < kx2 k with min x2 ðtÞ < b
and kx3 k < a. This ends the proof.
½g;1
h
T. Jankowski / Applied Mathematics and Computation 215 (2009) 125–131
131
Remark 1. Note that we can also discuss the situation when b ¼ 1, so when the boundary condition in problem (1) has the form xð1Þ ¼ xðgÞ but in this case we have to assume that g 6 n. References [1] R.I. Avery, A.C. Peterson, Three positive fixed points of nonlinear operators on ordered Banach spaces, Comput. Math. Appl. 42 (2001) 313–322. [2] D. Bai, Y. Xu, Positive solutions of second-order two-delay differential systems with twin parameter, Nonlinear Anal. 63 (2005) 601–617. [3] D. Bai, Y. Xu, Existence of positive solutions for boundary value problem of second-order delay differential equations, Appl. Math. Lett. 18 (2005) 621– 630. [4] B. Du, X. Hu, W. Ge, Positive solutions to a type of multi-point boundary value problem with delay and one-dimensional p-Laplacian, Appl. Math. Comput. 208 (2009) 501–510. [5] T. Jankowski, Positive solutions of three-point boundary value problems for second order impulsive differential equations with advanced arguments, Appl. Math. Comput. 197 (2008) 179–189. [6] T. Jankowski, Three positive solutions to second-order three-point impulsive differential equations with deviating arguments, Int. J. Comput. Math., in press. [7] D. Jiang, Multiple positive solutions for boundary value problems of second-order delay differential equations, Appl. Math. Lett. 15 (2002) 575–583. [8] B. Sun, Y. Qu, W. Ge, Existence and iteration of positive solutions for a multipoint one-dimensional p-Laplacian boundary value problem, Appl. Math. Comput. 197 (2008) 389–398. [9] W. Wang, J. Sheng, Positive solutions to a multi-point boundary value problem with delay, Appl. Math. Comput. 188 (2007) 96–102. [10] Y. Wang, W. Zhao, W. Ge, Multiple positive solutions for boundary value problems of second order delay differential equations with one-dimensional p-Laplacian, J. Math. Anal. Appl. 326 (2007) 641–654. [11] C. Yang, C. Zhai, J. Yan, Positive solutions of the three-point boundary value problem for second order differential equations with an advanced argument, Nonlinear Anal. 65 (2006) 2013–2023.