Multiple nonnegative solutions for second order impulsive differential equations

Applied Mathematics and Computation 114 (2000) 51±59 www.elsevier.nl/locate/amc

Multiple nonnegative solutions for second order impulsive di€erential equations Ravi P. Agarwal a

a,*

, Donal O'Regan

b,1

Department of Mathematics, National University of Singapore, 10 Kent Ridge Crescent, Singapore 119260, Singapore b Department of Mathematics, National University of Ireland, Galway, Ireland

Abstract We study the existence of single and multiple solutions to a second order impulsive equation with ®xed moments. Our analysis relies on combining a nonlinear alternative of Leray±Schauder type with Krasnoselskii's ®xed point theorem in a cone. Ó 2000 Elsevier Science Inc. All rights reserved.

1. Introduction Let 0 ˆ t0 < t1 <


< tm < tm‡1 ˆ 1 be given. In this paper we examine the second order impulsive equation y 00 …t† ‡ /…t†f …t; y…t†† ˆ 0 My…tk † ˆ

Ik …y…tkÿ ††;

My 0 …tk † ˆ Jk …y…tkÿ ††;

for t 2 …0; 1† n ft1 ; . . . ; tm g;

k ˆ 1; . . . ; m; k ˆ 1; . . . ; m;

…1:1†

y…0† ˆ y…1† ˆ 0: Here My…tk † ˆ y…tk‡ † ÿ y…tkÿ † where y…tk‡ † (respectively y…tkÿ †) denote the right limit (respectively left limit) of y…t† at t ˆ tk . Also My 0 …tk † ˆ y 0 …tk‡ † ÿ y 0 …tkÿ †. We de®ne the Banach space (r ˆ 0 or 2 in this paper), *

Corresponding author. E-mail address: [email protected] (R.P. Agarwal). 1 INTAS±96±0915. 0096-3003/00/$ - see front matter Ó 2000 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 9 9 ) 0 0 0 7 4 - 0

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R.P. Agarwal, D. O'Regan / Appl. Math. Comput. 114 (2000) 51±59

PC r ‰0; 1Š ˆ fy: ‰0; 1Š  R: for all j ˆ 0; . . . ; m there exists yj 2 C r ‰tj ; tj‡1 Š such that y ˆ yj on …tj ; tj‡1 Š; y…0† ˆ y0 …0†g; with the norm

 kykPCr ˆ max jyjH ; jy 0 jH ; . . . . . . ; jy …r† jH ;

here

 jyjH ˆ sup fjy…t†j: t 2 ‰0; 1Š n ft1 ; . . . ; tm gg ˆ max jyjk : k ˆ 0; . . . ; m ; jyjk ˆ sup jy…t†j: t2‰tk ;tk‡1 Š

By a solution to Eq. (1.1) we mean a function y 2 PC 2 ‰0; 1Š which satis®es Eq. (1.1). This paper has one main section, Section 2. In Section 2 we begin by presenting a new result (via a nonlinear alternative of Leray±Schauder type) which guarantees the existence of a solution to Eq. (1.1). This result will be used together with Theorem 1.2 and Theorem 1.3 below to guarantee the existence of twin nonnegative solutions to Eq. (1.1). In addition we also present a new multiplicity result using only Theorem 1.2 and Theorem 1.3 below. The existence theory presented in Section 2 below is very general and it extends and complements many results in the literature [1,3±9]. For the remainder of this section we present some results which will be needed in Section 2. First we use a nonlinear alternative of Leray±Schauder type [10] to establish an existence principle for Eq. (1.1). Theorem 1.1. Suppose f : ‰0; 1Š  R ! R is continuous;

…1:2†

/ 2 C…0; 1† with / > 0 on …0; 1† and / 2 L1 ‰0; 1Š

…1:3†

and Ik ; Jk : R ! R are continuous for k ˆ 1; . . . ; m hold. In addition assume there is a constant M0 , independent of k, with

…1:4†

jyjH 6ˆ M0 for any solution y to y 00 …t† ‡ k/…t†f …t; y…t†† ˆ 0 for t 2 …0; 1† n ft1 ; . . . ; tm g; My…tk † ˆ kIk …y…tkÿ ††; 0

My …tk † ˆ

kJk …y…tkÿ ††;

k ˆ 1; . . . ; m; k ˆ 1; . . . ; m;

…1:5k †

y…0† ˆ y…1† ˆ 0; for each k 2 …0; 1†. Then Eq. (1.1) has at least one solution y 2 PC 2 ‰0; 1Š with jyjH 6 M0 .

R.P. Agarwal, D. O'Regan / Appl. Math. Comput. 114 (2000) 51±59

53

Proof. Solving …1:5†k is equivalent (see [8]) to ®nding a solution y 2 PC‰0; 1Š which satis®es ! Z 1 m X G…t; s†/…s†f …s; y…s††ds ‡ Wk …t; y† for t 2 ‰0; 1Š; …1:6† y…t† ˆ k 0

here

( G…t; s† ˆ

kˆ1

…1 ÿ t†s;

0 6 s 6 t;

…1 ÿ s†t;

t 6 s 6 1;

and for k ˆ 1; . . . ; m, ( t‰ ÿ Ik …y…tk ††Š ÿ …1 ÿ tk †Jk …y…tk †; Wk …t; y† ˆ …1 ÿ t†‰Ik …y…tk †† ÿ tk Jk …y…tk †Š;

0 6 t 6 tk ; tk < t 6 1:

Let F : PC‰0; 1Š ! PC‰0; 1Š be de®ned by Z 1 m X G…t; s†/…s†f …s; y…s††ds ‡ Wk …t; y†: Fy…t† ˆ 0

kˆ1

A standard argument implies F : PC‰0; 1Š ! PC‰0; 1Š is continuous and completely continuous. Let U ˆ fu 2 PC‰0; 1Š: jujH < M0 g and apply [10 Theorem 2.5 (notice …A2† cannot occur)] to deduce the result.



Remark. Notice that Eq. (1.2) could be replaced by: f : J  R ! R is continuous and limt!tk f …t; y† exists (here J ˆ ‰0; 1Š n ft1 ; . . . ; tm g). Next we state for completeness Krasnoselskii's Fixed Point Theorem in cones. This will be used throughout Section 2. Theorem 1.2. Let E ˆ …E; k:k† be a Banach space and let K  E be a cone in E. Assumeÿ X1 and
X2 are open subsets of E with 0 2 X1 and X1  X2 and let A: K \ X2 n X1 ! K be continuous and completely continuous. In addition suppose either kAuk 6 kuk for u 2 K \ oX1

and

kAuk P kuk for u 2 K \ oX2

or and kAuk 6 kuk for u 2 K \ oX2 ÿ
hold. Then A has a ®xed point in K \ X2 n X1 . kAuk P kuk for u 2 K \ oX1

In applications to show A : K ! K in Theorem 2.1 most papers in the literature use the following well known result [2,3,5].

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R.P. Agarwal, D. O'Regan / Appl. Math. Comput. 114 (2000) 51±59

Theorem 1.3. Let 0 6 a < b; y 2 C 2 ‰a; bŠ with y 00 6 0 on ‰a; bŠ; y…a† P 0 and y…b† P 0. Then min y…t† P

t2‰c;dŠ

…b ÿ a† jyj1 ; 4

here jyj1 ˆ sup jy…t†j; c ˆ t2‰a;bŠ

3a ‡ b 4

and

dˆ

a ‡ 3b : 4

2. Existence In this section we discuss the existence and multiplicity of solutions of Eq. (1.1). For our ®rst result we will assume the following conditions are satis®ed: / 2 C…0; 1† with / > 0 on …0; 1† and / 2 L1 ‰0; 1Š;

…2:1†

f : ‰0; 1Š  R ! ‰0; 1† is continuous with

…2:2†

f …t; u† > 0 for …t; u† 2 ‰0; 1Š  …0; 1†; Ik ; Jk : R ! R are continuous for k ˆ 1; . . . ; m;

…2:3†

Ik …v† P tk Jk …v† for v 2 R and k ˆ 1; . . . ; m;

…2:4†

tIk …v† 6 …tk ÿ 1†Jk …v† for v 2 R; t 2 ‰0; tk Š and k ˆ 1; . . . ; m;

…2:5†

f …t; u† 6 w…u† on ‰0; 1Š  ‰0; 1† with w P 0

…2:6†

continuous and nondecreasing on ‰0; 1†; Wk …t; u† 6 X…u…tk †† on ‰0; 1Š  ‰0; 1† with X P 0 continuous and nondecreasing on ‰0; 1† and k ˆ 1; . . . ; m;

…2:7†

and there exists r > 0 with

w…r† sup

t2‰0;1Š

here

 G…t; s† ˆ

…1 ÿ t†s; …1 ÿ s†t;

0 6 s 6 t; t 6 s 6 1;

R1 0

r G…t; s†/…s†ds ‡ mX…r†

> 1;

…2:8†

R.P. Agarwal, D. O'Regan / Appl. Math. Comput. 114 (2000) 51±59

and for k ˆ 1; . . . ; m,  t‰ ÿ Ik …y…tk ††Š ÿ …1 ÿ tk †Jk …y…tk †; Wk …t; y† ˆ …1 ÿ t†‰Ik …y…tk †† ÿ tk Jk …y…tk †Š;

55

0 6 t 6 tk ; tk < t 6 1:

Theorem 2.1. Suppose Eqs. (2.1)±(2.8) hold. Then Eq. (1.1) has a solution y 2 PC 2 ‰0; 1Š with jyjH < r and y…t† P 0 for t 2 ‰0; 1Š. Proof. To show Eq. (1.1) has a solution we will use Theorem 1.1. Let y 2 PC 2 ‰0; 1Š be any solution of …1:5†k . Then ! Z 1 m X G…t; s†/…s†f …s; y…s††ds ‡ Wk …t; y† for t 2 ‰0; 1Š: …2:9† y…t† ˆ k 0

kˆ1

Now Eqs. (2.2), (2.4), (2.5) imply y…t† P 0 for t 2 ‰0; 1Š. In addition Eqs. (2.6), (2.7) and (2.9) imply Z 1 m X G…t; s†/…s†ds ‡ X…y…tk †† jy…t†j 6 w…jyjH † 0

kˆ1

Z

6 w…jyjH † sup

t2‰0;1Š

1 0

G…t; s†/…s†ds ‡ mX…jyjH †

for t 2 ‰0; 1Š. Consequently w…jyjH † sup

t2‰0;1Š

R1 0

jyjH G…t; s†/…s†ds ‡ mX…jyjH †

6 1:

…2:10†

Now Eq. (2.8) together with Eq. (2.10) implies jyjH 6ˆ r. Thus Theorem 1.1 implies that …1:1† has a solution y with jyjH < r (note jyjH 6 r by Theorem 1.1 and jyjH 6ˆ r by an argument similar to the one above).  Remark. Notice in Eq. (2.2) that f : ‰0; 1Š  R ! ‰0; 1† is continuous could be replaced by: f : J  R ! ‰0; 1† is continuous and limt!tk f …t; y† exists (here J ˆ ‰0; 1Š n ft1 ; . . . ; tm g). Remark. It is easy to replace f : ‰0; 1Š  R ! ‰0; 1† in Eq. (2.2) by f : ‰0; 1Š  ‰0; 1† ! ‰0; 1†. Next we will use the above result together with Krasnoselskii's ®xed point theorem in a cone to establish the existence of two solutions to Eq. (1.1). Theorem 2.2. Suppose Eqs. (2.1)±(2.8) hold. In addition assume the following two conditions are satis®ed:

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R.P. Agarwal, D. O'Regan / Appl. Math. Comput. 114 (2000) 51±59

for each k 2 f0; 1; . . . ; mg there exists sk 2 C‰ak ; bk Š with sk > 0 on ‰ak ; bk Š and with /…t† f …t; u† P sk …t†w…u† on ‰ak ; bk Š  …0; 1†; 3tk ‡ tk‡1 tk ‡ 3tk‡1 and bk ˆ with t0 ˆ 0 and tm‡1 ˆ 1 here ak ˆ 4 4 and

for each k 2 f0; 1; . . . ; mg there exists R > r with R   6 1; Rb ‰tk‡1 ÿtk Š R sup akk G…t; s†sk …s†ds w 4

…2:11†

…2:12†

t2‰0;1Š

here ak and bk are as in …2:11†: Then Eq. (1.1) has two solutions y1 ; y2 2 PC 2 ‰0; 1Š with y1 P 0; y2 P 0 on ‰0; 1Š and with 0 6 jy1 jH < r < jy2 jH 6 R. Proof. From Theorem 2.1 we have immediately the existence of y1 . To show the existence of y2 we will use Theorem 1.2. Let E ˆ …PC‰0; 1Š; j:jH † and  K ˆ u 2 PC‰0; 1Š: u…t† P 0 for t 2 ‰0; 1Š and  ‰tk‡1 ÿ tk Š jujk for k ˆ 0; 1; . . . ; m : min u…t† P t2‰ak ;bk Š 4 …2:13† Clearly K is a cone of E. Let A : K ! PC‰0; 1Š be de®ned by Z 1 m X G…t; s†/…s†f …s; y…s††ds ‡ Wk …t; y†: Ay…t† ˆ 0

…2:14†

kˆ1

A standard argument implies A : K ! PC‰0; 1Š is continuous and completely continuous. Also a standard argument [3,5,7] immediately implies A : K ! K. For completeness we supply the details here. Let y 2 A. Then Eqs. (2.2), (2.4), (2.5) imply Ay…t† P 0 for t 2 ‰0; 1Š. Next ®x k 2 f0; 1; . . . ; mg. Now Theorem 1.3 implies that ‰tk‡1 ÿ tk Š jAyjk : min Ay…t† P t2‰ak ;bk Š 4 Thus A : K ! K. Let X1 ˆ fu 2 PC‰0; 1Š: jujH < rg

and

X2 ˆ fu 2 PC‰0; 1Š: jujH < Rg:

We ®rst show jAyjH 6 jyjH

for y 2 K \ oX1 :

…2:15†

To see this let y 2 K \ oX1 ; so jyjH ˆ r. Now Eq. (2.14) together with Eqs. (2.6)±(2.8) implies (see the proof of Theorem 2.1),

R.P. Agarwal, D. O'Regan / Appl. Math. Comput. 114 (2000) 51±59

Z jAy…t†j 6 w…jyjH † sup

t2‰0;1Š

Z

6 w…r† sup

t2‰0;1Š

1 0

1

57

G…t; s†/…s† ds ‡ mX…jyjH †

G…t; s†/…s† ds ‡ mX…r† < r ˆ jyjH ;

0

for t 2 ‰0; 1Š. Thus jAyjH 6 jyjH for y 2 K \ oX1 so (2.15) is true. Next we show jAyjH P jyjH

for y 2 K \ oX2 :

…2:16†

To see this let y 2 K \ oX2 so jyjH ˆ R. Also there exists p 2 f0; 1; . . . ; mg with R ˆ jyjH ˆ jyjp . Notice also that min y…t† P

t2‰ap ;bp Š

and so

‰tp‡1 ÿ tp Š ‰tp‡1 ÿ tp Š jyjH ˆ R 4 4



‰tp‡1 ÿ tp Š R; R y…t† 2 4

 for t 2 ‰ap ; bp Š:

Now Eq. (2.14) together with Eqs. (2.11) and (2.12) imply ! Z 1 m X G…t; s†/…s†f …s; y…s†† ds ‡ Wk …t; y† sup Ay…t† ˆ sup t2‰0;1Š

t2‰0;1Š

0

Z P sup

t2‰0;1Š

ap

Z P sup

t2‰0;1Š

bp

bp ap

!

kˆ1

G…t; s†/…s†f …s; y…s†† ds ! G…t; s†sp …s†w…y…s†† ds

  Z bp ‰tp‡1 ÿ tp Š R sup G…t; s†sp …s† ds Pw 4 t2‰0;1Š ap P R ˆ jyjH : Thus jAyjH P jyjH for y 2 K \ oX2 ÿso Eq. (2.16) is true. Now Theorem 1.2
implies A has a ®xed point y2 2 K \ X2 n X1 i.e. r 6 jy2 jH 6 R. In fact r < jy2 jH (argue as in the ®rst part of Theorem 2.1).  In Theorem 2.2 it is possible for jy1 jH to be zero in some applications. Our next theorem guarantees the existence of two solutions y1 , y2 2 PC 2 ‰0; 1Š with jy1 jH 6ˆ 0 and jy2 jH 6ˆ 0. We will assume the following conditions are satis®ed: f : ‰0; 1Š  ‰0; 1† ! ‰0; 1† is continuous with f …t; u† > 0 for …t; u† 2 ‰0; 1Š  …0; 1†; Ik ; Jk : ‰0; 1† ! R are continuous for k ˆ 1; . . . ; m;

…2:17† …2:18†

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R.P. Agarwal, D. O'Regan / Appl. Math. Comput. 114 (2000) 51±59

Ik …v† P tk Jk …v† for v P 0 and k ˆ 1; . . . ; m;

…2:19†

tIk …v† 6 …tk ÿ 1†Jk …v† for v P 0; t 2 ‰0; tk Š and k ˆ 1; . . . ; m:

…2:20†

and

Theorem 2.3. Suppose Eqs. (2.1), (2.6)±(2.8), (2.11), (2.12), (2.17)±(2.20) hold. In addition assume the following condition is satis®ed: for each k 2 f0; 1; . . . ; mg there exists L; 0 < L < r; with L   6 1; Rb ‰tk‡1 ÿtk Š L sup akk G…t; s†sk …s†ds w 4

…2:21†

t2‰0;1Š

here ak ; bk and sk are as in …2:11†: Then Eq. (1.1) has two solutions y1 ; y2 2 PC 2 ‰0; 1Š with y1 P 0, y2 P 0 on ‰0; 1Š and with L 6 jy1 jH < r < jy2 jH 6 R. Proof. The existence of y2 follows immediately from Theorem 2.2. To guarantee the existence of y1 we will use Theorem 1.2. Let E; K and A be as in Theorem 2.2 and as in Theorem 2.2 we have A : K ! K. Let   X3 ˆ u 2 PC‰0; 1Š: jujH < L and X4 ˆ u 2 PC‰0; 1Š: jujH < r : As in Theorem 2.2 (see Eq. (2.15)) we have jAyjH 6 jyjH for y 2 K \ oX4 :

…2:22†

In addition (follow the reasoning used to prove Eq. (2.16) using Eq. (2.21) now) we have jAyjH P jyjH for y 2 K \ oX3 :

…2:23†
Now Theorem 1.2 implies A has a ®xed point y1 2 K \ X4 n X3 i.e. L 6 jy1 jH 6 r. In fact jy1 jH < r (argue as in the ®rst part of Theorem 2.1).  ÿ

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