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Multiple positive solutions for first order nonlinear impulsive integro-differential equations in a Banach space
Applied Mathematics and Computation 143 (2003) 233–249 www.elsevier.com/locate/amc
Multiple positive solutions for first order nonlinear impulsive integro-differential equations in a Banach space Dajun Guo Department of Mathematics, Shandong University, Jinan, Shandong 250100, PR China
Abstract By using the fixed point index theory, the author obtains the existence of two positive solutions for a boundary value problem of first order nonlinear impulsive integro-differential equations on an infinite interval with an infinite number of impulsive times in a real Banach space. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Impulsive integro-differential equation; Boundary value problem; Ordered Banach space; Fixed point index
1. Introduction In a recent paper [1], we have discussed the maximal and minimal solutions for an initial value problem of second order nonlinear impulsive integro-differential equations on an infinite interval with an infinite number of impulsive times in a real Banach space by means of establishing a comparison result and using the monotone iterative technique. Now, in this paper, we shall investigate the multiple positive solutions of a boundary value problem (BVP) for first order such equations by means of completely different method, that is, by using the fixed point index theory.
E-mail address: [email protected] (D. Guo). 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. doi:10.1016/S0096-3003(02)00356-9
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D. Guo / Appl. Math. Comput. 143 (2003) 233–249
Let E be a real Banach space and P be a cone in E which defined a partial ordering in E by x 6 y if and only if y x 2 P . P is said to be normal if there exists a positive constant N such that h 6 x 6 y implies kxk 6 N kyk, where h denotes the zero element of E, and the smallest N is called the normal constant of P (it is clear, N P 1). P is called solid if its interior P is nonempty. If x 6 y and x 6¼ y, we write x < y. If P is solid and y x 2 P , we write x y. For details on cone theory, see [2]. Consider the BVP for first order nonlinear impulsive integro-differential equation in E: 8 0 < u ðtÞ ¼ f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ; 8t 2 J 0 ; Dujt¼tk ¼ Ik ðuðtk ÞÞ ðk ¼ 1; 2; 3; . . .Þ; ð1Þ : uð1Þ ¼ buð0Þ; where J ¼ ½0; 1Þ, f 2 C½J P P P ; P , Ik 2 C½P ; P (k ¼ 1; 2; 3; . . .), 0 < t1 < < tk < , tk ! 1, J 0 ¼ J n ft1 ; . . . ; tk ; . . .g, b > 1, uð1Þ ¼ limt!1 uðtÞ and Z t Z 1 ðTuÞðtÞ ¼ Kðt; sÞuðsÞ ds; ðSuÞðtÞ ¼ H ðt; sÞuðsÞ ds; ð2Þ 0
0
K 2 C½D; Rþ , D ¼ fðt; sÞ 2 J J : t P sg, H 2 C½J J ; Rþ and Rþ is the set of all nonnegative numbers. Dujt¼tk denotes the jump of uðtÞ at t ¼ tk , i.e. Dujt¼tk ¼ uðtkþ Þ uðtk Þ; where uðtkþ Þ and uðtk Þ represent the right and left limits of uðtÞ at t ¼ tk , respectively. Let PC½J ; E ¼ fu: u is a map from J into E such that uðtÞ is continuous at t 6¼ tk , left continuous at t ¼ tk , and uðtkþ Þ exists for k ¼ 1; 2; 3; . . .g and BPC½J ; E ¼ fu 2 PC½J ; E: supt2J kuðtÞk < 1g, PC½J ; P ¼ fu 2 PC½J ; E: uðtÞ P h, 8t 2 J g, BPC½J ; P ¼ fu 2 BPC½J ; E: uðtÞ P h, 8t 2 J g. It is clear, BPC½J ; E is a Banach space with norm kukB ¼ sup kuðtÞk t2J
and BPC½J ; P is a cone in BPC½J ; E. u 2 BPC½J ; P \ C 1 ½J 0 ; E is called a positive solution of BVP(1) if uðtÞ satisfies (1) and uðtÞ 6 h. In what follows, let J0 ¼ ½0; t1 , Jk ¼ ðtk ; tkþ1 (k ¼ 1; 2; 3; . . .).
2. Several lemmas Let us list some conditions. Rt R1 ðH1 Þ k ¼ supt2J 0 Kðt; sÞ ds < 1, h ¼ supt2J 0 H ðt; sÞ ds < 1 and
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
Z lim 0 t !t
235
1
jH ðt0 ; sÞ H ðt; sÞj ds ¼ 0;
8t 2 J :
0
ðH2 Þ There exist a; b 2 C½J ; Rþ \ L½J ; Rþ and g 2 C½Rþ Rþ Rþ ; Rþ such that kf ðt; u; v; wÞk 6 aðtÞ þ bðtÞgðkuk; kvk; kwkÞ;
8t 2 J ; u; v; w 2 P :
ðH3 Þ There exist ck P 0 (k ¼ 1; 2; 3; . . .) and F 2 C½Rþ ; Rþ such that kIk ðuÞk 6 ck F ðkukÞ; 8u 2 P ðk ¼ 1; 2; 3; . . .Þ P1 and c ¼ k¼1 ck < 1. ðH4 Þ For any t 2 J and r > 0, f ðt; Pr ; Pr ; Pr Þ ¼ ff ðt; u; v; wÞ: u; v; w 2 Pr g and Ik ðPr Þ ¼ fIk ðuÞ: u 2 Pr g (k ¼ 1; 2; 3; . . .) are relatively compact in E, where Pr ¼ fu 2 P : kuk 6 rg. Remark 1. It is clear, condition ðH4 Þ is satisfied automatically when E is finite dimensional. Lemma 1. If condition ðH1 Þ is satisfied, then the operators T and S defined by (2) are bounded linear operators from BPC½J ; E into BPC½J ; E and kT k 6 k , kSk 6 h . Moreover, T ðBPC½J ; P Þ BC½J ; P , SðBPC½J ; P Þ BC½J ; P , where BC½J ; P ¼ fu 2 C½J ; E: supt2J kuðtÞk < 1; uðtÞ P h, 8t 2 J g. The proof is obvious. We shall reduce BVP(1) to an impulsive integral equation in E. To this end, we first consider operator A defined by ! Z 1 1 X 1 ðAuÞðtÞ ¼ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ b1 0 k¼1 Z t X þ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ; 8t 2 J : 0
0
ð3Þ Lemma 2. If conditions ðH1 Þ–ðH4 Þ are satisfied, then operator A defined by (3) is a completely continuous (i.e. continuous and compact) operator from BPC½J ; P into BPC½J ; P . Proof. Let r > 0 be given and Mr ¼ maxfgðx; y; zÞ: 0 6 x; y; z 6 r g;
Nr ¼ maxfF ðxÞ: 0 6 x 6 rg;
ð4Þ
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D. Guo / Appl. Math. Comput. 143 (2003) 233–249
where r ¼ maxfr; k r; h rg:
ð5Þ
For u 2 BPC½J ; P , kukB 6 r, we see that by virtue of Lemma 1, condition ðH2 Þ and (4), (5), kf ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞk 6 aðtÞ þ Mr bðtÞ;
8t 2 J ;
which implies the convergence of the infinite integral Z 1 f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ dt
ð6Þ
ð7Þ
0
and
Z
f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ dt
0 Z 1 6 kf ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞk dt 6 a þ Mr b ; 1
ð8Þ
0
where a ¼
Z
1
aðtÞ dt < 1;
0
b ¼
Z
1
bðtÞ dt < 1:
ð9Þ
0
On the other hand, condition ðH3 Þ implies the convergence of the series 1 X
Ik ðuðtk ÞÞ
ð10Þ
k¼1
and
X
X 1 1 1 X
Ik ðuðtk ÞÞ 6 kIk ðuðtk ÞÞk 6 Nr ck ¼ Nr c < 1;
k¼1
k¼1 k¼1
ð11Þ
where Nr is given by (4). Hence, we have proved that A maps BPC½J ; P into BPC½J ; P and, by (3), (8) and (11), kAukB 6
b ða þ Mr b þ Nr c Þ; b1
8u 2 BPC½J ; P ; kukB 6 r:
ð12Þ
Now, we are going to show that A is continuous. Let un ; u 2 BPC½J ; P , kun ukB ! 0 ðn ! 1Þ. Then r ¼ supn kun kB < 1 and kukB 6 r. By (3), we have
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
kAun A ukB 6
b b1
Z
237
1
kf ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ 0
f ðs; uðsÞ; ðT uÞðsÞ; ðSuÞðsÞÞk ds þ
kIk ðun ðtk ÞÞ
k¼1
! uðtk ÞÞk ; Ik ð
1 X
ðn ¼ 1; 2; 3; . . .Þ:
ð13Þ
It is clear, uðtÞ; ðT uÞðtÞ; ðSuÞðtÞÞ f ðt; un ðtÞ; ðTun ÞðtÞ; ðSun ÞðtÞÞ ! f ðt; as n ! 1; 8t 2 J :
ð14Þ
Moreover, we see from (6) that kf ðt; un ðtÞ; ðTun ÞðtÞ; ðSun ÞðtÞÞ f ðt; uðtÞ; ðT uÞðtÞ; ðSuÞðtÞÞk 6 2aðtÞ þ 2Mr bðtÞ ¼ rðtÞ;
8t 2 J ðn ¼ 1; 2; 3; . . .Þ;
r 2 L½J ; Rþ : ð15Þ
It follows from (14) and (15) and the dominated convergence theorem that Z 1 kf ðt; un ðtÞ; ðTun ÞðtÞ; ðSun ÞðtÞÞ f ðt; uðtÞ; ðT uÞðtÞ; ðSuÞðtÞÞk dt ! 0 0
as n ! 1:
ð16Þ
On the other hand, for any > 0, we can choose a positive integer m such that 1 X
Nr
ð17Þ
ck < :
k¼mþ1
And then, choose an integer n0 such that m X
kIk ðun ðtk ÞÞ Ik ð uðtk ÞÞk < ;
8n > n0 ðk ¼ 1; 2; . . . ; mÞ:
ð18Þ
k¼1
From (17) and (18) we see that 1 X
kIk ðun ðtk ÞÞ Ik ð uðtk ÞÞk < þ 2Nr
k¼1
1 X
ck < 3;
8n > n0 ;
k¼mþ1
i.e. lim
n!1
1 X k¼1
kIk ðun ðtk ÞÞ Ik ð uðtk ÞÞk ¼ 0:
ð19Þ
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D. Guo / Appl. Math. Comput. 143 (2003) 233–249
It follows from (13), (16) and (19) that kAun AukB ! 0 ðn ! 1Þ, and the continuity of A is proved. Finally, we prove that A is compact. Let V ¼ fun g BPC½J ; P be bounded and kun kB 6 r (n ¼ 1; 2; 3; . . .). We have, by (3), (6) and condition ðH3 Þ,
0
kðAun Þðt Þ ðAun ÞðtÞk 6
Z
t0
kf ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞk ds t
þ
X
kIk ðun ðtk ÞÞk 6
Z
ðaðsÞ þ Mr bðsÞÞ ds t
t 6 tk
þ Nr
t0
X
8t; t0 2 J ; t0 > t ðn ¼ 1; 2; 3; . . .Þ;
ck ;
ð20Þ
t 6 tk
which implies that fðAun ÞðtÞg (n ¼ 1; 2; 3; . . .) are equicontinuous on each Jk (k ¼ 0; 1; 2; . . .). On the other hand, for any > 0, choose a sufficiently large s > 0 and a sufficiently large positive integer m such that Z
1
ðaðsÞ þ Mr bðsÞÞ ds < ;
s
Nr
1 X
ck < :
ð21Þ
k¼mþ1
We have, by (3), (6) and (21),
ðAun ÞðtÞ ¼
1 b1
Z
s
f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds þ 0
m 1 X Ik ðun ðtk ÞÞ b 1 k¼1
Z 1 1 f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds þ b1 s Z t 1 X 1 þ Ik ðun ðtk ÞÞ þ f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds b 1 k¼mþ1 0 X þ Ik ðun ðtk ÞÞ; 8t 2 J ðn ¼ 1; 2; 3; . . .Þ ð22Þ 0
and
Z 1
f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds
< ;
s
X 1
Ik ðun ðtk ÞÞ < ðn ¼ 1; 2; 3; . . .Þ:
k¼mþ1
ð23Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
It follows from (22), (23), (6) and [3, Theorem 1.2.3] that Z s 2 aððAV ÞðtÞÞ 6 aðf ðs; V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞÞÞ ds b1 0 m 1 X 2 2 þ þ aðIk ðV ðtk ÞÞÞ þ b 1 k¼1 b1 b1 Z t þ2 aðf ðs; V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞÞÞ ds X0 aðIk ðV ðtk ÞÞÞ; 8t 2 J ; þ
239
ð24Þ
0
where ðAV ÞðtÞ ¼ fðAun ÞðtÞ: n ¼ 1; 2; 3; . . .g, V ðsÞ ¼ fun ðsÞ: n ¼ 1; 2; 3; . . .g, ðTV ÞðsÞ ¼ fðTun ÞðsÞ: n ¼ 1; 2; 3; . . .g, ðSV ÞðsÞ ¼ fðSun ÞðsÞ: n ¼ 1; 2; 3; . . .g and aðU Þ denotes the Kuratowski measure of noncompactness of bounded set U E (see [3, Section 1.2]). Since V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞ Pr for s 2 J , where r is given by (5), we see that, by condition ðH4 Þ, aðf ðs; V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞÞÞ ¼ 0;
8s 2 J ;
aðIk ðV ðtk ÞÞÞ ¼ 0 ðk ¼ 1; 2; 3; . . .Þ:
ð25Þ
It follows from (24) and (25) that aððAV ÞðtÞÞ 6
4 ; b1
8t 2 J ;
which implies by virtue of the arbitrariness of that aððAV ÞðtÞÞ ¼ 0 for t 2 J . By Ascoli–Arzela theorem (see [3, Theorem 1.2.5]), we conclude that AV is relatively compact in each C½Jk ; E (k ¼ 0; 1; 2; . . .). Hence, by using the diagonal method, we see that there exists a subsequence funi g of fun g such that fðAuni ÞðtÞg converges to some vðtÞ uniformly on each Jk (k ¼ 0; 1; 2; . . .). Consequently, v 2 PC½J ; P . We have, by (12), kAuni kB 6
b ða þ Mr b þ Nr c Þ b1
ði ¼ 1; 2; 3; . . .Þ;
which implies v 2 BPC½J ; P and kvkB 6
b ða þ Mr b þ Nr c Þ: b1
From (20) we get 0
kvðt Þ vðtÞk 6
Z t
t0
ðaðsÞ þ Mr bðsÞÞ ds þ Nr
X
ck ;
8t; t0 2 J ; t0 > t:
t 6 tk
ð26Þ Let > 0 be arbitrarily given. Choose s > 0 and a positive integer m such that (21) holds. And then, choose a positive integer i0 such that
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D. Guo / Appl. Math. Comput. 143 (2003) 233–249
kðAuni ÞðtÞ vðtÞk < ;
80 6 t 6 s ; i P i0 :
ð27Þ
where s ¼ maxfs; tmþ1 g. For t > s , we have by virtue of (20), (21), (26) and (27), kðAuni ÞðtÞ vðtÞk 6 kðAuni ÞðtÞ ðAuni Þðs Þk þ kðAuni Þðs Þ vðs Þk þ kvðs Þ vðtÞk < 2 þ þ 2 ¼ 5; ði ¼ 1; 2; 3; . . .Þ:
ð28Þ
It follows from (27) and (28) that kAuni vkB 6 5 for i P i0 . Hence kAuni vkB ! 0 as i ! 1, and the compactness of A is proved. Lemma 3 ([1] Lemma 1(a)). If u 2 PC½J ; E \ C 1 ½J 0 ; E, then Z t X uðtÞ ¼ uð0Þ þ u0 ðsÞ ds þ ½uðtkþ Þ uðtk Þ; 8t 2 J : 0
ð29Þ
0
Lemma 4. Let conditions ðH1 Þ–ðH4 Þ be satisfied. Then, u 2 BPC½J ; P \ C 1 ½J 0 ; E is a solution of BVP (1) if and only if u 2 BPC½J ; P is a solution of the following impulsive integral equation in E: ! Z 1 1 X 1 uðtÞ ¼ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ b1 0 k¼1 Z t X þ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ; 8t 2 J : ð30Þ 0
0
Proof. Let u 2 BPC½J ; P \ C 1 ½J 0 ; E be a solution of BVP(1). We have by formula (29), Z t X uðtÞ ¼ uð0Þ þ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ; 8t 2 J : 0
0
ð31Þ By virtue of Lemma 2, the infinite integral and the series in the right-hand side of (3) are convergent, so, taking limits as t ! 1 in (31), we get Z 1 1 X f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ: ð32Þ uð1Þ ¼ uð0Þ þ 0
k¼1
It follows from (32) and the relation uð1Þ ¼ buð0Þ that ! Z 1 1 X 1 uð0Þ ¼ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ ; b1 0 k¼1 and, substituting (33) into (31), we see that uðtÞ satisfies (30).
ð33Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
241
Conversely, assume that u 2 BPC½J ; P is a solution of Eq. (30). Taking limits as t ! 1 in (30), we see that uð1Þ exists and ! Z 1 1 X b f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ ¼ buð0Þ: uð1Þ ¼ b1 0 k¼1 On the other hand, direct differentiation of (30) gives u0 ðtÞ ¼ f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ;
8t 2 J 0 ;
and, it is clear by (30), Dujt¼tk ¼ Ik ðuðtk ÞÞ ðk ¼ 1; 2; 3; . . .Þ: Hence u 2 C 1 ½J 0 ; E and uðtÞ satisfies (1).
3. Main theorems Let us list more conditions. ðH5 Þ There exists c 2 C½J ; Rþ \ L½J ; Rþ such that kf ðt; u; v; wÞk !0 cðtÞðkuk þ kvk þ kwkÞ
as u; v; w 2 P ; kuk þ kvk þ kwk ! 1
uniformly for t 2 J , and F ðxÞ !0 x
as x ! 1;
where F is the function in condition ðH3 Þ. ðH6 Þ There exists d 2 C½J ; Rþ \ L½J ; Rþ such that kf ðt; u; v; wÞk !0 dðtÞðkuk þ kvk þ kwkÞ
as u; v; w 2 P ; kuk þ kvk þ kwk ! 0
uniformly for t 2 J , and F ðxÞ !0 x
as x ! 0þ ;
where F is the function in condition ðH3 Þ. ðH7 Þ P is normal and solid, and there exist u0 h, 0 6 t1 < t2 < 1 and r 2 C½I; Rþ such that I ¼ ½t1 ; t2 Jk for some k and f ðt; u; v; wÞ P rðtÞu0 ;
8t 2 I; u P u0 ; v P h; w P h
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D. Guo / Appl. Math. Comput. 143 (2003) 233–249
and Z
t2
rðtÞ dt > b 1:
t1
ðH8 Þ There ½t1 ; t2 Jk for
exist u0 > h, 0 6 t1 < t2 < 1 and r 2 C½I; Rþ such that I ¼ some k and
f ðt; u; v; wÞ P rðtÞu0 ;
8t 2 I; u P u0 ; v P h; w P h
and Z
t2
rðtÞ dt P b 1:
t1
Remark 2. It is clear, condition ðH8 Þ is weaker than condition ðH7 Þ. Theorem 1. If conditions ðH1 Þ–ðH7 Þ are satisfied, then BVP (1) has at least two positive solutions u1 ; u2 2 BPC½J ; P \ C 1 ½J 0 ; E such that u1 ðtÞ u0 for t 2 J . Proof. By Lemma 2, operator A defined by (3) is completely continuous from BPC½J ; P into BPC½J ; P and, by Lemma 4, we need only to show that A has two positive fixed points u1 and u2 in BPC½J ; P such that u1 ðtÞ u0 for t 2 J . By virtue of condition ðH5 Þ, there exists r > 0 such that kf ðt; u; v; wÞk 6 0 cðtÞðkuk þ kvk þ kwkÞ;
8t 2 J ; u; v; w 2 P ;
kuk þ kvk þ kwk > r;
ð34Þ
and F ðxÞ 6 0 x;
8x > r;
ð35Þ
where b1 ; 0 ¼ 2b½c ð1 þ h þ k Þ þ c
c ¼
Z
1
cðtÞ dt:
ð36Þ
0
It is clear, condition ðH2 Þ implies kf ðt; u; v; wÞk 6 sðtÞ;
8t 2 J ; u; v; w 2 P ; kuk þ kvk þ kwk 6 r;
where sðtÞ ¼ aðtÞ þ MbðtÞ;
M ¼ maxfgðx; y; zÞ: 0 6 x; y; z 6 rg; Z 1 sðtÞ dt < 1: s 2 C½J ; Rþ \ L½J ; Rþ ; s ¼ 0
ð37Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
243
It follows from (34) and (37) that kf ðt; u; v; wÞk 6 0 cðtÞðkuk þ kvk þ kwkÞ þ sðtÞ;
8t 2 J ; u; v; w 2 P : ð38Þ
On the other hand, (35) implies F ðxÞ 6 0 x þ M ;
8x P 0;
ð39Þ
where M ¼ maxfF ðxÞ: 0 6 x 6 rg. Let u 2 BPC½J ; P . From (3), (38), (39) and condition ðH3 Þ, we have b kðAuÞðtÞk 6 b1 þ
(Z
1
ð0 cðsÞðkuðsÞk þ kðTuÞðsÞk þ kðSuÞðsÞkÞ þ sðsÞÞ ds 0
1 X
)
ck ð0 kuðtk Þk þ M Þ 6
k¼1
b f0 ½c ð1 þ k þ h Þ b1
1 b þ c kukB þ s þ c M g ¼ kukB þ ðs þ c M Þ; 2 b1
8t 2 J ;
which implies kAukB 6
1 b kukB þ ðs þ c M Þ; 2 b1
8u 2 BPC½J ; P :
ð40Þ
Choose
2b ðs þ c M Þ ; R > max 2ku0 k; b1
ð41Þ
where u0 h is given in condition ðH7 Þ, and let U1 ¼ fu 2 BPC½J ; P : kukB < Rg. Then U 1 ¼ fu 2 BPC½J ; P : kukB 6 Rg and, by (40) and (41), we have AðU 1 Þ U1 :
ð42Þ
Similarly, by condition ðH6 Þ, there exists r1 > 0 such that kf ðt; u; v; wÞk 6 1 dðtÞðkuk þ kvk þ kwkÞ; kuk þ kvk þ kwk 6 r1
8t 2 J ; u; v; w 2 P ;
and F ðxÞ 6 1 x;
80 6 x 6 r1 ;
where 1 ¼
b1 ; 2b½d ð1 þ k þ h Þ þ c
d ¼
Z
1
dðtÞ dt; 0
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D. Guo / Appl. Math. Comput. 143 (2003) 233–249
and therefore, it is easy to see that kAukB 6
1 kukB ; 2
8u 2 BPC½J ; P ;
kukB 6 r2 ¼
r1 : 1 þ k þ h
ð43Þ
Choose 0 < r < minfN 1 ku0 k; r2 g;
ð44Þ
where N denotes the normal constant of P, and let U2 ¼ fu 2 BPC½J ; P : kukB < rg. Then, by (43), AðU 2 Þ U2 :
ð45Þ
Let U3 ¼ fu 2 BPC½J ; P : kukB < R, uðtÞ u0 , 8t 2 I ¼ ½t1 ; t2 g, and we are going to show that U3 is an open set of BPC½J ; P . It is clear, we need only to show: for any u 2 U3 , there exists g > 0 such that u 2 BPC½J ; P , ku ukB < g implies that uðtÞ u0 for t 2 I. We have uðtÞ u0 for t 2 I. So, for any t0 2 I, 0 0 0 there exists a ¼ ðt Þ > 0 such that uðt0 Þ P ð1 þ 30 Þu0 :
ð46Þ
uðtÞ is continuous on I, we can find an open interval Since u0 h and Iðt0 ; d0 Þ ¼ ðt0 d0 ; t0 þ d0 Þðd0 > 0Þ such that 0 u0 þ ½ uðtÞ uðt0 Þ P h;
8t 2 Iðt0 ; d0 Þ;
which implies by virtue of (46) that uðtÞ P ð1 þ 20 Þu0 ;
8t 2 Iðt0 ; d0 Þ:
Since I is compact, there is a finite collection of such intervals fIðti0 ; d0i Þg (i ¼ 1; 2; . . . ; m) which covers I, and uðtÞ P ð1 þ 20i Þu0 ;
8t 2 Iðti0 ; d0i Þ ði ¼ 1; 2; . . . ; mÞ;
where 0i > 0 (i ¼ 1; 2; . . . ; m). Consequently, uðtÞ P ð1 þ 2 Þu0 ;
8t 2 I;
ð47Þ
where ¼ minf01 ; . . . ; 0m g > 0. Since u0 h, there exists g > 0 such that u0 þ ½uðtÞ uðtÞ P h;
8t 2 J
ð48Þ
whenever u 2 BPC½J ; P satisfying ku ukB < g. It follows from (47) and (48) that uðtÞ P ð1 þ Þu0 u0 ;
8t 2 I
whenever u 2 BPC½J ; P satisfying ku ukB < g. Thus, we have proved that U3 is open in BPC½J ; P . U3 6¼ ; because 2u0 2 U3 on account of (41). It is clear, U 3 fu 2 BPC½J ; P : kukB 6 R, uðtÞ P u0 , 8t 2 Ig. Let u 2 U 3 . By (42), we have kAukB < R. On the other hand, (3) and condition ðH7 Þ imply
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
ðAuÞðtÞ P
1 b1
1 P b1
Z
t2
245
f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds
t1
Z
t2
! rðsÞ ds u0 u0 ;
8t 2 J :
ð49Þ
t1
Hence AðU 3 Þ U3 :
ð50Þ
Since U1 , U2 and U3 are nonempty bounded convex open sets of BPC½J ; P , we see that (42), (45) and (50) imply by virtue of [3, Corollary 1.2.3] the fixed point indices iðA; Ui ; BPC½J ; P Þ ¼ 1
ði ¼ 1; 2; 3Þ:
ð51Þ
On the other hand, by (41) and (44) we find U2 U1 ;
U3 U1 ;
U2 \ U3 ¼ ;:
ð52Þ
It follows from (51) and (52) that iðA; U1 n ðU2 [ U3 Þ; BPC½J ; P Þ ¼ iðA; U1 ; BPC½J ; P Þ iðA; U2 ; BPC½J ; P Þ iðA; U3 ; BPC½J ; P Þ ¼ 1: ð53Þ Finally, (51)–(53) imply that A has two fixed points u1 2 U3 and u2 2 U1 n ðU2 [ U3 Þ. We have, by (49), u1 ðtÞ u0 for t 2 J and, it is clear, ku2 kB > r, hence u1 ðtÞ 6 h and u2 ðtÞ 6 h. The proof is complete. Remark 3. Condition ðH6 Þ and the continuity of f imply that f ðt; h; h; hÞ ¼ h for t 2 J . Hence, under conditions of Theorem 1, BVP(1) has the trivial solution uðtÞ h besides two positive solutions u1 and u2 . Theorem 2. If conditions ðH1 Þ–ðH5 Þ and ðH8 Þ are satisfied, then BVP (1) has at least one positive solution u1 2 BPC½J ; P \ C 1 ½J 0 ; E such that u1 ðtÞ P u0 for t 2 J. Proof. By Lemma 2, operator A defined by (3) is completely continuous from BPC½J ; P into BPC½J ; P and, by Lemma 4, we need only to show that A has one positive fixed point u1 in BPC½J ; P such that u1 ðtÞ P u0 for t 2 J . As in the proof of Theorem 1, (40) holds. Choose R satisfying (41) and let W ¼ fu 2 BPC½J ; P : kukB 6 R, uðtÞ P u0 , 8t 2 ½t1 ; t2 g, where u0 > h is given by condition ðH8 Þ. It is clear, W is a nonempty bounded closed convex set in space BPC½J ; E (W 6¼ ; because 2u0 2 W ). Let u 2 W . By (40) and (41), we have kAukB < R. On the other hand, as in the proof of Theorem 1, (3) and condition ðH8 Þ imply
246
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
ðAuÞðtÞ P
1 b1
1 P b1
Z
t2
f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds
t1
Z
!
t2
rðsÞ ds u0 P u0 ;
8t 2 J :
ð54Þ
t1
Hence Au 2 W , and therefore AðW Þ W . Thus, the Schauder fixed point theorem implies that A has a fixed point u1 in W, and, by (54), u1 ðtÞ P u0 for t 2 J . The theorem is proved. Example 1. Consider the BVP for scalar first order impulsive integro-differential equation 1=2 Z t Z 1 2 0 t ðtþ1Þs s u ðtÞ ¼ 2e 4uðtÞ þ 3 e uðsÞ ds þ 2 e sin ðt sÞuðsÞ ds 0 0 Z t Z 1 2 ðtþ1Þs s ln 1 þ uðtÞ þ e uðsÞ ds þ e sin ðt sÞuðsÞ ds ; 0
0
80 6 t < 1; t 6¼ k ðk ¼ 1; 2; 3; . . .Þ; 1=3
Dujt¼k ¼ k 2 ðuðkÞÞ uð1Þ ¼ 2uð0Þ:
lnð1 þ uðkÞÞ;
ðk ¼ 1; 2; 3; . . .Þ; ð55Þ
Evidently, uðtÞ 0 is the trivial solution of BVP (55). Conclusion BVP(55) has at least two positive solutions u1 and u2 such that u1 ðtÞ > 1 for 0 6 t < 1. Proof. Let E ¼ R1 and P ¼ Rþ . Then P is a normal and solid cone in E, and BVP(55) is a BVP of form (1) in E. In this situation, Kðt; sÞ ¼ eðtþ1Þs , H ðt; sÞ ¼ es sin2 ðt sÞ, b ¼ 2 and f ðt; u; v; wÞ ¼ 2et ð4u þ 3v þ 2wÞ
1=2
lnð1 þ u þ v þ wÞ;
8t 2 J ¼ ½0; 1Þ; u; v; w P 0; 2 1=3
Ik ðuÞ ¼ k u
lnð1 þ uÞ;
ð56Þ
8u P 0 ðk ¼ 1; 2; 3; . . .Þ:
ð57Þ
It is clear, conditions ðH1 Þ–ðH4 Þ are satisfied with k < 1, h < 1 and F ðxÞ ¼ x1=3 lnð1 þ xÞ:
ð58Þ
On the other hand, we have, by (56), 0 6 f ðt; u; v; wÞ 6 4et ðu þ v þ wÞ 8t 2 J ; u; v; w P 0;
1=2
lnð1 þ u þ v þ wÞ; ð59Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
247
It follows from (59) and (58) that conditions ðH5 Þ and ðH6 Þ are satisfied for cðtÞ ¼ dðtÞ ¼ 4et . Finally, we check condition ðH7 Þ. For 0 6 t 6 1 and u P 1, v P 0, w P 0, we have, by (56), Z 1 et dt ¼ 4 ln 2ð1 e1 Þ > 1; f ðt; u; v; wÞ P 4et ln 2; 4 ln 2 0
which implies that ðH7 Þ is satisfied for u0 ¼ 1, t1 ¼ 0, t2 ¼ 1 and rðtÞ ¼ ð4 ln 2Þet . Thus, our conclusion follows from Theorem 1. Example 2. Consider the infinite system of scalar first order impulsive integrodifferential equations !1=3 Z t 4 1 unþ1 ðsÞ ds 1=2 0 un ðtÞ ¼ pffiffiffi ð1 þ un cos u2n Þ þ 2 nð1 þ tÞ2 nð1 þ tÞ3 0 ð1 þ t þ sÞ !1=5 Z 1 1 u3n ðsÞ ds þ ; 80 6 t < 1 t 6¼ k n2 ð1 þ tÞ4 ð1 þ t þ sÞ3 0 ðk ¼ 1; 2; 3; . . . ; n ¼ 1; 2; 3; . . .Þ; 1 2=3 Dun jt¼k ¼ 3 ðunþ2 ðkÞÞ ; ðk ¼ 1; 2; 3; . . . ; n ¼ 1; 2; 3; . . .Þ; nk 2un ð1Þ ¼ 3un ð0Þ ðn ¼ 1; 2; 3; . . .Þ: ð60Þ Evidently, un ðtÞ 0 (n ¼ 1; 2; 3; . . .) is the trivial solution of infinite system (60). Conclusion Infinite system (60) has at least one positive solution fun ðtÞg satisfying un ðtÞ P 1n for 0 6 t < 1 (n ¼ 1; 2; 3; . . .). Proof. Let E ¼ c0 ¼ fu ¼ ðu1 ; . . . ; un ; . . .Þ: un ! 0g with norm kuk ¼ supn jun j and P ¼ fu ¼ ðu1 ; . . . ; un ; . . .Þ 2 c0 : un P 0, n ¼ 1; 2; 3; . . .g. Then P is a normal cone in E (but P is not solid), and infinite system (60) can be regarded as a BVP of form (1) in E. In this situation, u ¼ ðu1 ; . . . ; un ; . . .Þ, v ¼ ðv1 ; . . . ; vn ; . . .Þ, 2 3 w ¼ ðw1 ; . . . ; wn ; . . .Þ, Kðt; sÞ ¼ ð1 þ t þ sÞ , H ðt; sÞ ¼ ð1 þ t þ sÞ , b ¼ 32, f ¼ ðf1 ; . . . ; fn ; . . .Þ and Ik ¼ ðIk1 ; . . . ; Ikn ; . . .Þ, in which 4 1 1=2 1=3 ð1 þ un cos u2n Þ þ v fn ðt; u; v; wÞ ¼ pffiffiffi 2 3 nþ1 nð1 þ tÞ nð1 þ tÞ 1 1=5 þ w3n ; 8t 2 J ¼ ½0; 1Þ; u; v; w 2 P 2 n ð1 þ tÞ4
ð61Þ
248
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
and 1 2=3 u ; 8u 2 P : ð62Þ nk 3 nþ2 Obviously, f 2 C½J P P P ; P ðJ ¼ ½0; 1ÞÞ, Ik 2 C½P ; P (k ¼ 1; 2; 3; . . .) and condition ðH1 Þ is satisfied for k < 1 and h < 12. We have, by (61) and (62), n o 4 1=2 1=3 1=5 ð2 þ kukÞ þ kvk þ kw j kf ðt; u; v; wÞk 6 ; 2 ð1 þ tÞ Ikn ðuÞ ¼
8t 2 J ; u; v; w 2 P
ð63Þ
and 1 2=3 kuk ; 8u 2 P ðk ¼ 1; 2; 3; . . .Þ: ð64Þ k3 It follows from (63) and (64) that condition ðH2 Þ and ðH3 Þ are satisfied for 2 1=2 aðtÞ ¼ 0, bðtÞ ¼ 4ð1 þ tÞ , gðx; y; zÞ ¼ ð2 þ xÞ þ y 1=3 þ z1=5 , ck ¼ k 3 and kIk ðuÞk 6
F ðxÞ ¼ x2=3 ;
8x P 0:
ð65Þ
Now, we are going to show that condition ðH4 Þ is satisfied. Let t 2 J and r > 0 ðmÞ be fixed and fzðmÞ g be any sequence in f ðt; Pr ; Pr ; Pr Þ, where zðmÞ ¼ ðz1 ; . . . ; zðmÞ n ; . . .Þ. Then, by (61), we have 4 r1=3 1=2 0 6 zðmÞ ð2 þ rÞ þ n 6 pffiffiffi 2 3 nð1 þ tÞ nð1 þ tÞ þ
r1=5 n2 ð1 þ tÞ4
ðn; m ¼ 1; 2; 3; . . .Þ:
ð66Þ
So, fzðmÞ n g is bounded, and, by diagonal method, we can choose a subsequence fmi g fmg such that znðmi Þ ! zn
as i ! 1 ðn ¼ 1; 2; 3; . . .Þ;
ð67Þ
which implies by virtue of (66) that 4 r1=3 1=2 ð2 þ rÞ þ 0 6 zn 6 pffiffiffi nð1 þ tÞ3 nð1 þ tÞ2 þ
r1=5 n2 ð1 þ tÞ4
ðn ¼ 1; 2; 3; . . .Þ:
ð68Þ
Hence z ¼ ðz1 ; . . . ; zn ; . . .Þ 2 c0 . It is easy to see from (66)–(68) that iÞ kzðmi Þ zk ¼ sup jzðm zn j ! 0 n
as i ! 1:
n
Thus, we have proved that f ðt; Pr ; Pr ; Pr Þ is relatively compact in E. Similarly, by virtue of (62) we can show that Ik ðPr Þ (k ¼ 1; 2; 3; . . .) are relatively compact in E. Consequently, condition ðH4 Þ is satisfied. By (63), we have
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
kf ðt; u; v; wÞk 6
4 ð1 þ tÞ
2
249
n 1=2 ð2 þ kuk þ kvk þ kwkÞ
þ ðkuk þ kvk þ kwkÞ
1=3
þ ðkuk þ kvk þ kwkÞ
8t 2 J ; u; v; w 2 P ;
1=5
o
; ð69Þ
It follows from (69) and (65) that condition ðH5 Þ is satisfied for cðtÞ ¼ 2 4ð1 þ tÞ . Finally, let u0 ¼ ð1; . . . ; 1n ; . . .Þ. Then u0 > h. For 12 6 t 6 1, u P u0 , v P h and w P h, we have, by (61), fn ðt; u; v; wÞ P
4 nð1 þ tÞ
2
ðn ¼ 1; 2; 3; . . .Þ
and Z 1 2
1
4 dt ð1 þ tÞ
2
¼
2 1 > ¼ b 1; 3 2 2
so, condition ðH8 Þ is satisfied for t1 ¼ 12, t2 ¼ 1 and rðtÞ ¼ 4ð1 þ tÞ . Thus, our conclusion follows from Theorem 2.
Acknowledgements Research supported by the National Nature Science Foundation of China and the Nature Science Foundation of Shandong Province of China.
References [1] D. Guo, A class of second order impulsive integro-differential equations on unbounded domain in a Banach space, Appl. Math. Comput. 125 (2002) 59–77. [2] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Boston, MA, 1988. [3] D. Guo, V. Lakshmikantham, X.Z. Liu, Nonlinear Integral Equations in Abstract Spaces, Kluwer Academic Publishers, Dordrecht, 1996.
Multiple positive solutions for first order nonlinear impulsive integro-differential equations in a Banach space Dajun Guo Department of Mathematics, Shandong University, Jinan, Shandong 250100, PR China
Abstract By using the fixed point index theory, the author obtains the existence of two positive solutions for a boundary value problem of first order nonlinear impulsive integro-differential equations on an infinite interval with an infinite number of impulsive times in a real Banach space. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Impulsive integro-differential equation; Boundary value problem; Ordered Banach space; Fixed point index
1. Introduction In a recent paper [1], we have discussed the maximal and minimal solutions for an initial value problem of second order nonlinear impulsive integro-differential equations on an infinite interval with an infinite number of impulsive times in a real Banach space by means of establishing a comparison result and using the monotone iterative technique. Now, in this paper, we shall investigate the multiple positive solutions of a boundary value problem (BVP) for first order such equations by means of completely different method, that is, by using the fixed point index theory.
E-mail address: [email protected] (D. Guo). 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. doi:10.1016/S0096-3003(02)00356-9
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D. Guo / Appl. Math. Comput. 143 (2003) 233–249
Let E be a real Banach space and P be a cone in E which defined a partial ordering in E by x 6 y if and only if y x 2 P . P is said to be normal if there exists a positive constant N such that h 6 x 6 y implies kxk 6 N kyk, where h denotes the zero element of E, and the smallest N is called the normal constant of P (it is clear, N P 1). P is called solid if its interior P is nonempty. If x 6 y and x 6¼ y, we write x < y. If P is solid and y x 2 P , we write x y. For details on cone theory, see [2]. Consider the BVP for first order nonlinear impulsive integro-differential equation in E: 8 0 < u ðtÞ ¼ f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ; 8t 2 J 0 ; Dujt¼tk ¼ Ik ðuðtk ÞÞ ðk ¼ 1; 2; 3; . . .Þ; ð1Þ : uð1Þ ¼ buð0Þ; where J ¼ ½0; 1Þ, f 2 C½J P P P ; P , Ik 2 C½P ; P (k ¼ 1; 2; 3; . . .), 0 < t1 < < tk < , tk ! 1, J 0 ¼ J n ft1 ; . . . ; tk ; . . .g, b > 1, uð1Þ ¼ limt!1 uðtÞ and Z t Z 1 ðTuÞðtÞ ¼ Kðt; sÞuðsÞ ds; ðSuÞðtÞ ¼ H ðt; sÞuðsÞ ds; ð2Þ 0
0
K 2 C½D; Rþ , D ¼ fðt; sÞ 2 J J : t P sg, H 2 C½J J ; Rþ and Rþ is the set of all nonnegative numbers. Dujt¼tk denotes the jump of uðtÞ at t ¼ tk , i.e. Dujt¼tk ¼ uðtkþ Þ uðtk Þ; where uðtkþ Þ and uðtk Þ represent the right and left limits of uðtÞ at t ¼ tk , respectively. Let PC½J ; E ¼ fu: u is a map from J into E such that uðtÞ is continuous at t 6¼ tk , left continuous at t ¼ tk , and uðtkþ Þ exists for k ¼ 1; 2; 3; . . .g and BPC½J ; E ¼ fu 2 PC½J ; E: supt2J kuðtÞk < 1g, PC½J ; P ¼ fu 2 PC½J ; E: uðtÞ P h, 8t 2 J g, BPC½J ; P ¼ fu 2 BPC½J ; E: uðtÞ P h, 8t 2 J g. It is clear, BPC½J ; E is a Banach space with norm kukB ¼ sup kuðtÞk t2J
and BPC½J ; P is a cone in BPC½J ; E. u 2 BPC½J ; P \ C 1 ½J 0 ; E is called a positive solution of BVP(1) if uðtÞ satisfies (1) and uðtÞ 6 h. In what follows, let J0 ¼ ½0; t1 , Jk ¼ ðtk ; tkþ1 (k ¼ 1; 2; 3; . . .).
2. Several lemmas Let us list some conditions. Rt R1 ðH1 Þ k ¼ supt2J 0 Kðt; sÞ ds < 1, h ¼ supt2J 0 H ðt; sÞ ds < 1 and
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
Z lim 0 t !t
235
1
jH ðt0 ; sÞ H ðt; sÞj ds ¼ 0;
8t 2 J :
0
ðH2 Þ There exist a; b 2 C½J ; Rþ \ L½J ; Rþ and g 2 C½Rþ Rþ Rþ ; Rþ such that kf ðt; u; v; wÞk 6 aðtÞ þ bðtÞgðkuk; kvk; kwkÞ;
8t 2 J ; u; v; w 2 P :
ðH3 Þ There exist ck P 0 (k ¼ 1; 2; 3; . . .) and F 2 C½Rþ ; Rþ such that kIk ðuÞk 6 ck F ðkukÞ; 8u 2 P ðk ¼ 1; 2; 3; . . .Þ P1 and c ¼ k¼1 ck < 1. ðH4 Þ For any t 2 J and r > 0, f ðt; Pr ; Pr ; Pr Þ ¼ ff ðt; u; v; wÞ: u; v; w 2 Pr g and Ik ðPr Þ ¼ fIk ðuÞ: u 2 Pr g (k ¼ 1; 2; 3; . . .) are relatively compact in E, where Pr ¼ fu 2 P : kuk 6 rg. Remark 1. It is clear, condition ðH4 Þ is satisfied automatically when E is finite dimensional. Lemma 1. If condition ðH1 Þ is satisfied, then the operators T and S defined by (2) are bounded linear operators from BPC½J ; E into BPC½J ; E and kT k 6 k , kSk 6 h . Moreover, T ðBPC½J ; P Þ BC½J ; P , SðBPC½J ; P Þ BC½J ; P , where BC½J ; P ¼ fu 2 C½J ; E: supt2J kuðtÞk < 1; uðtÞ P h, 8t 2 J g. The proof is obvious. We shall reduce BVP(1) to an impulsive integral equation in E. To this end, we first consider operator A defined by ! Z 1 1 X 1 ðAuÞðtÞ ¼ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ b1 0 k¼1 Z t X þ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ; 8t 2 J : 0
0
ð3Þ Lemma 2. If conditions ðH1 Þ–ðH4 Þ are satisfied, then operator A defined by (3) is a completely continuous (i.e. continuous and compact) operator from BPC½J ; P into BPC½J ; P . Proof. Let r > 0 be given and Mr ¼ maxfgðx; y; zÞ: 0 6 x; y; z 6 r g;
Nr ¼ maxfF ðxÞ: 0 6 x 6 rg;
ð4Þ
236
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
where r ¼ maxfr; k r; h rg:
ð5Þ
For u 2 BPC½J ; P , kukB 6 r, we see that by virtue of Lemma 1, condition ðH2 Þ and (4), (5), kf ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞk 6 aðtÞ þ Mr bðtÞ;
8t 2 J ;
which implies the convergence of the infinite integral Z 1 f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ dt
ð6Þ
ð7Þ
0
and
Z
f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ dt
0 Z 1 6 kf ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞk dt 6 a þ Mr b ; 1
ð8Þ
0
where a ¼
Z
1
aðtÞ dt < 1;
0
b ¼
Z
1
bðtÞ dt < 1:
ð9Þ
0
On the other hand, condition ðH3 Þ implies the convergence of the series 1 X
Ik ðuðtk ÞÞ
ð10Þ
k¼1
and
X
X 1 1 1 X
Ik ðuðtk ÞÞ 6 kIk ðuðtk ÞÞk 6 Nr ck ¼ Nr c < 1;
k¼1
k¼1 k¼1
ð11Þ
where Nr is given by (4). Hence, we have proved that A maps BPC½J ; P into BPC½J ; P and, by (3), (8) and (11), kAukB 6
b ða þ Mr b þ Nr c Þ; b1
8u 2 BPC½J ; P ; kukB 6 r:
ð12Þ
Now, we are going to show that A is continuous. Let un ; u 2 BPC½J ; P , kun ukB ! 0 ðn ! 1Þ. Then r ¼ supn kun kB < 1 and kukB 6 r. By (3), we have
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
kAun A ukB 6
b b1
Z
237
1
kf ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ 0
f ðs; uðsÞ; ðT uÞðsÞ; ðSuÞðsÞÞk ds þ
kIk ðun ðtk ÞÞ
k¼1
! uðtk ÞÞk ; Ik ð
1 X
ðn ¼ 1; 2; 3; . . .Þ:
ð13Þ
It is clear, uðtÞ; ðT uÞðtÞ; ðSuÞðtÞÞ f ðt; un ðtÞ; ðTun ÞðtÞ; ðSun ÞðtÞÞ ! f ðt; as n ! 1; 8t 2 J :
ð14Þ
Moreover, we see from (6) that kf ðt; un ðtÞ; ðTun ÞðtÞ; ðSun ÞðtÞÞ f ðt; uðtÞ; ðT uÞðtÞ; ðSuÞðtÞÞk 6 2aðtÞ þ 2Mr bðtÞ ¼ rðtÞ;
8t 2 J ðn ¼ 1; 2; 3; . . .Þ;
r 2 L½J ; Rþ : ð15Þ
It follows from (14) and (15) and the dominated convergence theorem that Z 1 kf ðt; un ðtÞ; ðTun ÞðtÞ; ðSun ÞðtÞÞ f ðt; uðtÞ; ðT uÞðtÞ; ðSuÞðtÞÞk dt ! 0 0
as n ! 1:
ð16Þ
On the other hand, for any > 0, we can choose a positive integer m such that 1 X
Nr
ð17Þ
ck < :
k¼mþ1
And then, choose an integer n0 such that m X
kIk ðun ðtk ÞÞ Ik ð uðtk ÞÞk < ;
8n > n0 ðk ¼ 1; 2; . . . ; mÞ:
ð18Þ
k¼1
From (17) and (18) we see that 1 X
kIk ðun ðtk ÞÞ Ik ð uðtk ÞÞk < þ 2Nr
k¼1
1 X
ck < 3;
8n > n0 ;
k¼mþ1
i.e. lim
n!1
1 X k¼1
kIk ðun ðtk ÞÞ Ik ð uðtk ÞÞk ¼ 0:
ð19Þ
238
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
It follows from (13), (16) and (19) that kAun AukB ! 0 ðn ! 1Þ, and the continuity of A is proved. Finally, we prove that A is compact. Let V ¼ fun g BPC½J ; P be bounded and kun kB 6 r (n ¼ 1; 2; 3; . . .). We have, by (3), (6) and condition ðH3 Þ,
0
kðAun Þðt Þ ðAun ÞðtÞk 6
Z
t0
kf ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞk ds t
þ
X
kIk ðun ðtk ÞÞk 6
Z
ðaðsÞ þ Mr bðsÞÞ ds t
t 6 tk
þ Nr
t0
X
8t; t0 2 J ; t0 > t ðn ¼ 1; 2; 3; . . .Þ;
ck ;
ð20Þ
t 6 tk
which implies that fðAun ÞðtÞg (n ¼ 1; 2; 3; . . .) are equicontinuous on each Jk (k ¼ 0; 1; 2; . . .). On the other hand, for any > 0, choose a sufficiently large s > 0 and a sufficiently large positive integer m such that Z
1
ðaðsÞ þ Mr bðsÞÞ ds < ;
s
Nr
1 X
ck < :
ð21Þ
k¼mþ1
We have, by (3), (6) and (21),
ðAun ÞðtÞ ¼
1 b1
Z
s
f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds þ 0
m 1 X Ik ðun ðtk ÞÞ b 1 k¼1
Z 1 1 f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds þ b1 s Z t 1 X 1 þ Ik ðun ðtk ÞÞ þ f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds b 1 k¼mþ1 0 X þ Ik ðun ðtk ÞÞ; 8t 2 J ðn ¼ 1; 2; 3; . . .Þ ð22Þ 0
and
Z 1
f ðs; un ðsÞ; ðTun ÞðsÞ; ðSun ÞðsÞÞ ds
< ;
s
X 1
Ik ðun ðtk ÞÞ < ðn ¼ 1; 2; 3; . . .Þ:
k¼mþ1
ð23Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
It follows from (22), (23), (6) and [3, Theorem 1.2.3] that Z s 2 aððAV ÞðtÞÞ 6 aðf ðs; V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞÞÞ ds b1 0 m 1 X 2 2 þ þ aðIk ðV ðtk ÞÞÞ þ b 1 k¼1 b1 b1 Z t þ2 aðf ðs; V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞÞÞ ds X0 aðIk ðV ðtk ÞÞÞ; 8t 2 J ; þ
239
ð24Þ
0
where ðAV ÞðtÞ ¼ fðAun ÞðtÞ: n ¼ 1; 2; 3; . . .g, V ðsÞ ¼ fun ðsÞ: n ¼ 1; 2; 3; . . .g, ðTV ÞðsÞ ¼ fðTun ÞðsÞ: n ¼ 1; 2; 3; . . .g, ðSV ÞðsÞ ¼ fðSun ÞðsÞ: n ¼ 1; 2; 3; . . .g and aðU Þ denotes the Kuratowski measure of noncompactness of bounded set U E (see [3, Section 1.2]). Since V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞ Pr for s 2 J , where r is given by (5), we see that, by condition ðH4 Þ, aðf ðs; V ðsÞ; ðTV ÞðsÞ; ðSV ÞðsÞÞÞ ¼ 0;
8s 2 J ;
aðIk ðV ðtk ÞÞÞ ¼ 0 ðk ¼ 1; 2; 3; . . .Þ:
ð25Þ
It follows from (24) and (25) that aððAV ÞðtÞÞ 6
4 ; b1
8t 2 J ;
which implies by virtue of the arbitrariness of that aððAV ÞðtÞÞ ¼ 0 for t 2 J . By Ascoli–Arzela theorem (see [3, Theorem 1.2.5]), we conclude that AV is relatively compact in each C½Jk ; E (k ¼ 0; 1; 2; . . .). Hence, by using the diagonal method, we see that there exists a subsequence funi g of fun g such that fðAuni ÞðtÞg converges to some vðtÞ uniformly on each Jk (k ¼ 0; 1; 2; . . .). Consequently, v 2 PC½J ; P . We have, by (12), kAuni kB 6
b ða þ Mr b þ Nr c Þ b1
ði ¼ 1; 2; 3; . . .Þ;
which implies v 2 BPC½J ; P and kvkB 6
b ða þ Mr b þ Nr c Þ: b1
From (20) we get 0
kvðt Þ vðtÞk 6
Z t
t0
ðaðsÞ þ Mr bðsÞÞ ds þ Nr
X
ck ;
8t; t0 2 J ; t0 > t:
t 6 tk
ð26Þ Let > 0 be arbitrarily given. Choose s > 0 and a positive integer m such that (21) holds. And then, choose a positive integer i0 such that
240
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
kðAuni ÞðtÞ vðtÞk < ;
80 6 t 6 s ; i P i0 :
ð27Þ
where s ¼ maxfs; tmþ1 g. For t > s , we have by virtue of (20), (21), (26) and (27), kðAuni ÞðtÞ vðtÞk 6 kðAuni ÞðtÞ ðAuni Þðs Þk þ kðAuni Þðs Þ vðs Þk þ kvðs Þ vðtÞk < 2 þ þ 2 ¼ 5; ði ¼ 1; 2; 3; . . .Þ:
ð28Þ
It follows from (27) and (28) that kAuni vkB 6 5 for i P i0 . Hence kAuni vkB ! 0 as i ! 1, and the compactness of A is proved. Lemma 3 ([1] Lemma 1(a)). If u 2 PC½J ; E \ C 1 ½J 0 ; E, then Z t X uðtÞ ¼ uð0Þ þ u0 ðsÞ ds þ ½uðtkþ Þ uðtk Þ; 8t 2 J : 0
ð29Þ
0
Lemma 4. Let conditions ðH1 Þ–ðH4 Þ be satisfied. Then, u 2 BPC½J ; P \ C 1 ½J 0 ; E is a solution of BVP (1) if and only if u 2 BPC½J ; P is a solution of the following impulsive integral equation in E: ! Z 1 1 X 1 uðtÞ ¼ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ b1 0 k¼1 Z t X þ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ; 8t 2 J : ð30Þ 0
0
Proof. Let u 2 BPC½J ; P \ C 1 ½J 0 ; E be a solution of BVP(1). We have by formula (29), Z t X uðtÞ ¼ uð0Þ þ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ; 8t 2 J : 0
0
ð31Þ By virtue of Lemma 2, the infinite integral and the series in the right-hand side of (3) are convergent, so, taking limits as t ! 1 in (31), we get Z 1 1 X f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ: ð32Þ uð1Þ ¼ uð0Þ þ 0
k¼1
It follows from (32) and the relation uð1Þ ¼ buð0Þ that ! Z 1 1 X 1 uð0Þ ¼ f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ ; b1 0 k¼1 and, substituting (33) into (31), we see that uðtÞ satisfies (30).
ð33Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
241
Conversely, assume that u 2 BPC½J ; P is a solution of Eq. (30). Taking limits as t ! 1 in (30), we see that uð1Þ exists and ! Z 1 1 X b f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds þ Ik ðuðtk ÞÞ ¼ buð0Þ: uð1Þ ¼ b1 0 k¼1 On the other hand, direct differentiation of (30) gives u0 ðtÞ ¼ f ðt; uðtÞ; ðTuÞðtÞ; ðSuÞðtÞÞ;
8t 2 J 0 ;
and, it is clear by (30), Dujt¼tk ¼ Ik ðuðtk ÞÞ ðk ¼ 1; 2; 3; . . .Þ: Hence u 2 C 1 ½J 0 ; E and uðtÞ satisfies (1).
3. Main theorems Let us list more conditions. ðH5 Þ There exists c 2 C½J ; Rþ \ L½J ; Rþ such that kf ðt; u; v; wÞk !0 cðtÞðkuk þ kvk þ kwkÞ
as u; v; w 2 P ; kuk þ kvk þ kwk ! 1
uniformly for t 2 J , and F ðxÞ !0 x
as x ! 1;
where F is the function in condition ðH3 Þ. ðH6 Þ There exists d 2 C½J ; Rþ \ L½J ; Rþ such that kf ðt; u; v; wÞk !0 dðtÞðkuk þ kvk þ kwkÞ
as u; v; w 2 P ; kuk þ kvk þ kwk ! 0
uniformly for t 2 J , and F ðxÞ !0 x
as x ! 0þ ;
where F is the function in condition ðH3 Þ. ðH7 Þ P is normal and solid, and there exist u0 h, 0 6 t1 < t2 < 1 and r 2 C½I; Rþ such that I ¼ ½t1 ; t2 Jk for some k and f ðt; u; v; wÞ P rðtÞu0 ;
8t 2 I; u P u0 ; v P h; w P h
242
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
and Z
t2
rðtÞ dt > b 1:
t1
ðH8 Þ There ½t1 ; t2 Jk for
exist u0 > h, 0 6 t1 < t2 < 1 and r 2 C½I; Rþ such that I ¼ some k and
f ðt; u; v; wÞ P rðtÞu0 ;
8t 2 I; u P u0 ; v P h; w P h
and Z
t2
rðtÞ dt P b 1:
t1
Remark 2. It is clear, condition ðH8 Þ is weaker than condition ðH7 Þ. Theorem 1. If conditions ðH1 Þ–ðH7 Þ are satisfied, then BVP (1) has at least two positive solutions u1 ; u2 2 BPC½J ; P \ C 1 ½J 0 ; E such that u1 ðtÞ u0 for t 2 J . Proof. By Lemma 2, operator A defined by (3) is completely continuous from BPC½J ; P into BPC½J ; P and, by Lemma 4, we need only to show that A has two positive fixed points u1 and u2 in BPC½J ; P such that u1 ðtÞ u0 for t 2 J . By virtue of condition ðH5 Þ, there exists r > 0 such that kf ðt; u; v; wÞk 6 0 cðtÞðkuk þ kvk þ kwkÞ;
8t 2 J ; u; v; w 2 P ;
kuk þ kvk þ kwk > r;
ð34Þ
and F ðxÞ 6 0 x;
8x > r;
ð35Þ
where b1 ; 0 ¼ 2b½c ð1 þ h þ k Þ þ c
c ¼
Z
1
cðtÞ dt:
ð36Þ
0
It is clear, condition ðH2 Þ implies kf ðt; u; v; wÞk 6 sðtÞ;
8t 2 J ; u; v; w 2 P ; kuk þ kvk þ kwk 6 r;
where sðtÞ ¼ aðtÞ þ MbðtÞ;
M ¼ maxfgðx; y; zÞ: 0 6 x; y; z 6 rg; Z 1 sðtÞ dt < 1: s 2 C½J ; Rþ \ L½J ; Rþ ; s ¼ 0
ð37Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
243
It follows from (34) and (37) that kf ðt; u; v; wÞk 6 0 cðtÞðkuk þ kvk þ kwkÞ þ sðtÞ;
8t 2 J ; u; v; w 2 P : ð38Þ
On the other hand, (35) implies F ðxÞ 6 0 x þ M ;
8x P 0;
ð39Þ
where M ¼ maxfF ðxÞ: 0 6 x 6 rg. Let u 2 BPC½J ; P . From (3), (38), (39) and condition ðH3 Þ, we have b kðAuÞðtÞk 6 b1 þ
(Z
1
ð0 cðsÞðkuðsÞk þ kðTuÞðsÞk þ kðSuÞðsÞkÞ þ sðsÞÞ ds 0
1 X
)
ck ð0 kuðtk Þk þ M Þ 6
k¼1
b f0 ½c ð1 þ k þ h Þ b1
1 b þ c kukB þ s þ c M g ¼ kukB þ ðs þ c M Þ; 2 b1
8t 2 J ;
which implies kAukB 6
1 b kukB þ ðs þ c M Þ; 2 b1
8u 2 BPC½J ; P :
ð40Þ
Choose
2b ðs þ c M Þ ; R > max 2ku0 k; b1
ð41Þ
where u0 h is given in condition ðH7 Þ, and let U1 ¼ fu 2 BPC½J ; P : kukB < Rg. Then U 1 ¼ fu 2 BPC½J ; P : kukB 6 Rg and, by (40) and (41), we have AðU 1 Þ U1 :
ð42Þ
Similarly, by condition ðH6 Þ, there exists r1 > 0 such that kf ðt; u; v; wÞk 6 1 dðtÞðkuk þ kvk þ kwkÞ; kuk þ kvk þ kwk 6 r1
8t 2 J ; u; v; w 2 P ;
and F ðxÞ 6 1 x;
80 6 x 6 r1 ;
where 1 ¼
b1 ; 2b½d ð1 þ k þ h Þ þ c
d ¼
Z
1
dðtÞ dt; 0
244
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
and therefore, it is easy to see that kAukB 6
1 kukB ; 2
8u 2 BPC½J ; P ;
kukB 6 r2 ¼
r1 : 1 þ k þ h
ð43Þ
Choose 0 < r < minfN 1 ku0 k; r2 g;
ð44Þ
where N denotes the normal constant of P, and let U2 ¼ fu 2 BPC½J ; P : kukB < rg. Then, by (43), AðU 2 Þ U2 :
ð45Þ
Let U3 ¼ fu 2 BPC½J ; P : kukB < R, uðtÞ u0 , 8t 2 I ¼ ½t1 ; t2 g, and we are going to show that U3 is an open set of BPC½J ; P . It is clear, we need only to show: for any u 2 U3 , there exists g > 0 such that u 2 BPC½J ; P , ku ukB < g implies that uðtÞ u0 for t 2 I. We have uðtÞ u0 for t 2 I. So, for any t0 2 I, 0 0 0 there exists a ¼ ðt Þ > 0 such that uðt0 Þ P ð1 þ 30 Þu0 :
ð46Þ
uðtÞ is continuous on I, we can find an open interval Since u0 h and Iðt0 ; d0 Þ ¼ ðt0 d0 ; t0 þ d0 Þðd0 > 0Þ such that 0 u0 þ ½ uðtÞ uðt0 Þ P h;
8t 2 Iðt0 ; d0 Þ;
which implies by virtue of (46) that uðtÞ P ð1 þ 20 Þu0 ;
8t 2 Iðt0 ; d0 Þ:
Since I is compact, there is a finite collection of such intervals fIðti0 ; d0i Þg (i ¼ 1; 2; . . . ; m) which covers I, and uðtÞ P ð1 þ 20i Þu0 ;
8t 2 Iðti0 ; d0i Þ ði ¼ 1; 2; . . . ; mÞ;
where 0i > 0 (i ¼ 1; 2; . . . ; m). Consequently, uðtÞ P ð1 þ 2 Þu0 ;
8t 2 I;
ð47Þ
where ¼ minf01 ; . . . ; 0m g > 0. Since u0 h, there exists g > 0 such that u0 þ ½uðtÞ uðtÞ P h;
8t 2 J
ð48Þ
whenever u 2 BPC½J ; P satisfying ku ukB < g. It follows from (47) and (48) that uðtÞ P ð1 þ Þu0 u0 ;
8t 2 I
whenever u 2 BPC½J ; P satisfying ku ukB < g. Thus, we have proved that U3 is open in BPC½J ; P . U3 6¼ ; because 2u0 2 U3 on account of (41). It is clear, U 3 fu 2 BPC½J ; P : kukB 6 R, uðtÞ P u0 , 8t 2 Ig. Let u 2 U 3 . By (42), we have kAukB < R. On the other hand, (3) and condition ðH7 Þ imply
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
ðAuÞðtÞ P
1 b1
1 P b1
Z
t2
245
f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds
t1
Z
t2
! rðsÞ ds u0 u0 ;
8t 2 J :
ð49Þ
t1
Hence AðU 3 Þ U3 :
ð50Þ
Since U1 , U2 and U3 are nonempty bounded convex open sets of BPC½J ; P , we see that (42), (45) and (50) imply by virtue of [3, Corollary 1.2.3] the fixed point indices iðA; Ui ; BPC½J ; P Þ ¼ 1
ði ¼ 1; 2; 3Þ:
ð51Þ
On the other hand, by (41) and (44) we find U2 U1 ;
U3 U1 ;
U2 \ U3 ¼ ;:
ð52Þ
It follows from (51) and (52) that iðA; U1 n ðU2 [ U3 Þ; BPC½J ; P Þ ¼ iðA; U1 ; BPC½J ; P Þ iðA; U2 ; BPC½J ; P Þ iðA; U3 ; BPC½J ; P Þ ¼ 1: ð53Þ Finally, (51)–(53) imply that A has two fixed points u1 2 U3 and u2 2 U1 n ðU2 [ U3 Þ. We have, by (49), u1 ðtÞ u0 for t 2 J and, it is clear, ku2 kB > r, hence u1 ðtÞ 6 h and u2 ðtÞ 6 h. The proof is complete. Remark 3. Condition ðH6 Þ and the continuity of f imply that f ðt; h; h; hÞ ¼ h for t 2 J . Hence, under conditions of Theorem 1, BVP(1) has the trivial solution uðtÞ h besides two positive solutions u1 and u2 . Theorem 2. If conditions ðH1 Þ–ðH5 Þ and ðH8 Þ are satisfied, then BVP (1) has at least one positive solution u1 2 BPC½J ; P \ C 1 ½J 0 ; E such that u1 ðtÞ P u0 for t 2 J. Proof. By Lemma 2, operator A defined by (3) is completely continuous from BPC½J ; P into BPC½J ; P and, by Lemma 4, we need only to show that A has one positive fixed point u1 in BPC½J ; P such that u1 ðtÞ P u0 for t 2 J . As in the proof of Theorem 1, (40) holds. Choose R satisfying (41) and let W ¼ fu 2 BPC½J ; P : kukB 6 R, uðtÞ P u0 , 8t 2 ½t1 ; t2 g, where u0 > h is given by condition ðH8 Þ. It is clear, W is a nonempty bounded closed convex set in space BPC½J ; E (W 6¼ ; because 2u0 2 W ). Let u 2 W . By (40) and (41), we have kAukB < R. On the other hand, as in the proof of Theorem 1, (3) and condition ðH8 Þ imply
246
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
ðAuÞðtÞ P
1 b1
1 P b1
Z
t2
f ðs; uðsÞ; ðTuÞðsÞ; ðSuÞðsÞÞ ds
t1
Z
!
t2
rðsÞ ds u0 P u0 ;
8t 2 J :
ð54Þ
t1
Hence Au 2 W , and therefore AðW Þ W . Thus, the Schauder fixed point theorem implies that A has a fixed point u1 in W, and, by (54), u1 ðtÞ P u0 for t 2 J . The theorem is proved. Example 1. Consider the BVP for scalar first order impulsive integro-differential equation 1=2 Z t Z 1 2 0 t ðtþ1Þs s u ðtÞ ¼ 2e 4uðtÞ þ 3 e uðsÞ ds þ 2 e sin ðt sÞuðsÞ ds 0 0 Z t Z 1 2 ðtþ1Þs s ln 1 þ uðtÞ þ e uðsÞ ds þ e sin ðt sÞuðsÞ ds ; 0
0
80 6 t < 1; t 6¼ k ðk ¼ 1; 2; 3; . . .Þ; 1=3
Dujt¼k ¼ k 2 ðuðkÞÞ uð1Þ ¼ 2uð0Þ:
lnð1 þ uðkÞÞ;
ðk ¼ 1; 2; 3; . . .Þ; ð55Þ
Evidently, uðtÞ 0 is the trivial solution of BVP (55). Conclusion BVP(55) has at least two positive solutions u1 and u2 such that u1 ðtÞ > 1 for 0 6 t < 1. Proof. Let E ¼ R1 and P ¼ Rþ . Then P is a normal and solid cone in E, and BVP(55) is a BVP of form (1) in E. In this situation, Kðt; sÞ ¼ eðtþ1Þs , H ðt; sÞ ¼ es sin2 ðt sÞ, b ¼ 2 and f ðt; u; v; wÞ ¼ 2et ð4u þ 3v þ 2wÞ
1=2
lnð1 þ u þ v þ wÞ;
8t 2 J ¼ ½0; 1Þ; u; v; w P 0; 2 1=3
Ik ðuÞ ¼ k u
lnð1 þ uÞ;
ð56Þ
8u P 0 ðk ¼ 1; 2; 3; . . .Þ:
ð57Þ
It is clear, conditions ðH1 Þ–ðH4 Þ are satisfied with k < 1, h < 1 and F ðxÞ ¼ x1=3 lnð1 þ xÞ:
ð58Þ
On the other hand, we have, by (56), 0 6 f ðt; u; v; wÞ 6 4et ðu þ v þ wÞ 8t 2 J ; u; v; w P 0;
1=2
lnð1 þ u þ v þ wÞ; ð59Þ
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
247
It follows from (59) and (58) that conditions ðH5 Þ and ðH6 Þ are satisfied for cðtÞ ¼ dðtÞ ¼ 4et . Finally, we check condition ðH7 Þ. For 0 6 t 6 1 and u P 1, v P 0, w P 0, we have, by (56), Z 1 et dt ¼ 4 ln 2ð1 e1 Þ > 1; f ðt; u; v; wÞ P 4et ln 2; 4 ln 2 0
which implies that ðH7 Þ is satisfied for u0 ¼ 1, t1 ¼ 0, t2 ¼ 1 and rðtÞ ¼ ð4 ln 2Þet . Thus, our conclusion follows from Theorem 1. Example 2. Consider the infinite system of scalar first order impulsive integrodifferential equations !1=3 Z t 4 1 unþ1 ðsÞ ds 1=2 0 un ðtÞ ¼ pffiffiffi ð1 þ un cos u2n Þ þ 2 nð1 þ tÞ2 nð1 þ tÞ3 0 ð1 þ t þ sÞ !1=5 Z 1 1 u3n ðsÞ ds þ ; 80 6 t < 1 t 6¼ k n2 ð1 þ tÞ4 ð1 þ t þ sÞ3 0 ðk ¼ 1; 2; 3; . . . ; n ¼ 1; 2; 3; . . .Þ; 1 2=3 Dun jt¼k ¼ 3 ðunþ2 ðkÞÞ ; ðk ¼ 1; 2; 3; . . . ; n ¼ 1; 2; 3; . . .Þ; nk 2un ð1Þ ¼ 3un ð0Þ ðn ¼ 1; 2; 3; . . .Þ: ð60Þ Evidently, un ðtÞ 0 (n ¼ 1; 2; 3; . . .) is the trivial solution of infinite system (60). Conclusion Infinite system (60) has at least one positive solution fun ðtÞg satisfying un ðtÞ P 1n for 0 6 t < 1 (n ¼ 1; 2; 3; . . .). Proof. Let E ¼ c0 ¼ fu ¼ ðu1 ; . . . ; un ; . . .Þ: un ! 0g with norm kuk ¼ supn jun j and P ¼ fu ¼ ðu1 ; . . . ; un ; . . .Þ 2 c0 : un P 0, n ¼ 1; 2; 3; . . .g. Then P is a normal cone in E (but P is not solid), and infinite system (60) can be regarded as a BVP of form (1) in E. In this situation, u ¼ ðu1 ; . . . ; un ; . . .Þ, v ¼ ðv1 ; . . . ; vn ; . . .Þ, 2 3 w ¼ ðw1 ; . . . ; wn ; . . .Þ, Kðt; sÞ ¼ ð1 þ t þ sÞ , H ðt; sÞ ¼ ð1 þ t þ sÞ , b ¼ 32, f ¼ ðf1 ; . . . ; fn ; . . .Þ and Ik ¼ ðIk1 ; . . . ; Ikn ; . . .Þ, in which 4 1 1=2 1=3 ð1 þ un cos u2n Þ þ v fn ðt; u; v; wÞ ¼ pffiffiffi 2 3 nþ1 nð1 þ tÞ nð1 þ tÞ 1 1=5 þ w3n ; 8t 2 J ¼ ½0; 1Þ; u; v; w 2 P 2 n ð1 þ tÞ4
ð61Þ
248
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
and 1 2=3 u ; 8u 2 P : ð62Þ nk 3 nþ2 Obviously, f 2 C½J P P P ; P ðJ ¼ ½0; 1ÞÞ, Ik 2 C½P ; P (k ¼ 1; 2; 3; . . .) and condition ðH1 Þ is satisfied for k < 1 and h < 12. We have, by (61) and (62), n o 4 1=2 1=3 1=5 ð2 þ kukÞ þ kvk þ kw j kf ðt; u; v; wÞk 6 ; 2 ð1 þ tÞ Ikn ðuÞ ¼
8t 2 J ; u; v; w 2 P
ð63Þ
and 1 2=3 kuk ; 8u 2 P ðk ¼ 1; 2; 3; . . .Þ: ð64Þ k3 It follows from (63) and (64) that condition ðH2 Þ and ðH3 Þ are satisfied for 2 1=2 aðtÞ ¼ 0, bðtÞ ¼ 4ð1 þ tÞ , gðx; y; zÞ ¼ ð2 þ xÞ þ y 1=3 þ z1=5 , ck ¼ k 3 and kIk ðuÞk 6
F ðxÞ ¼ x2=3 ;
8x P 0:
ð65Þ
Now, we are going to show that condition ðH4 Þ is satisfied. Let t 2 J and r > 0 ðmÞ be fixed and fzðmÞ g be any sequence in f ðt; Pr ; Pr ; Pr Þ, where zðmÞ ¼ ðz1 ; . . . ; zðmÞ n ; . . .Þ. Then, by (61), we have 4 r1=3 1=2 0 6 zðmÞ ð2 þ rÞ þ n 6 pffiffiffi 2 3 nð1 þ tÞ nð1 þ tÞ þ
r1=5 n2 ð1 þ tÞ4
ðn; m ¼ 1; 2; 3; . . .Þ:
ð66Þ
So, fzðmÞ n g is bounded, and, by diagonal method, we can choose a subsequence fmi g fmg such that znðmi Þ ! zn
as i ! 1 ðn ¼ 1; 2; 3; . . .Þ;
ð67Þ
which implies by virtue of (66) that 4 r1=3 1=2 ð2 þ rÞ þ 0 6 zn 6 pffiffiffi nð1 þ tÞ3 nð1 þ tÞ2 þ
r1=5 n2 ð1 þ tÞ4
ðn ¼ 1; 2; 3; . . .Þ:
ð68Þ
Hence z ¼ ðz1 ; . . . ; zn ; . . .Þ 2 c0 . It is easy to see from (66)–(68) that iÞ kzðmi Þ zk ¼ sup jzðm zn j ! 0 n
as i ! 1:
n
Thus, we have proved that f ðt; Pr ; Pr ; Pr Þ is relatively compact in E. Similarly, by virtue of (62) we can show that Ik ðPr Þ (k ¼ 1; 2; 3; . . .) are relatively compact in E. Consequently, condition ðH4 Þ is satisfied. By (63), we have
D. Guo / Appl. Math. Comput. 143 (2003) 233–249
kf ðt; u; v; wÞk 6
4 ð1 þ tÞ
2
249
n 1=2 ð2 þ kuk þ kvk þ kwkÞ
þ ðkuk þ kvk þ kwkÞ
1=3
þ ðkuk þ kvk þ kwkÞ
8t 2 J ; u; v; w 2 P ;
1=5
o
; ð69Þ
It follows from (69) and (65) that condition ðH5 Þ is satisfied for cðtÞ ¼ 2 4ð1 þ tÞ . Finally, let u0 ¼ ð1; . . . ; 1n ; . . .Þ. Then u0 > h. For 12 6 t 6 1, u P u0 , v P h and w P h, we have, by (61), fn ðt; u; v; wÞ P
4 nð1 þ tÞ
2
ðn ¼ 1; 2; 3; . . .Þ
and Z 1 2
1
4 dt ð1 þ tÞ
2
¼
2 1 > ¼ b 1; 3 2 2
so, condition ðH8 Þ is satisfied for t1 ¼ 12, t2 ¼ 1 and rðtÞ ¼ 4ð1 þ tÞ . Thus, our conclusion follows from Theorem 2.
Acknowledgements Research supported by the National Nature Science Foundation of China and the Nature Science Foundation of Shandong Province of China.
References [1] D. Guo, A class of second order impulsive integro-differential equations on unbounded domain in a Banach space, Appl. Math. Comput. 125 (2002) 59–77. [2] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Boston, MA, 1988. [3] D. Guo, V. Lakshmikantham, X.Z. Liu, Nonlinear Integral Equations in Abstract Spaces, Kluwer Academic Publishers, Dordrecht, 1996.