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Positive solutions for first order periodic boundary value problem
Applied Mathematics and Computation 158 (2004) 345–351 www.elsevier.com/locate/amc
Positive solutions for first order periodic boundary value problem Shiguo Peng Department of Applied Mathematics, Guangdong University of Technology, Guangzhou 510090, PeopleÕs Republic of China
Abstract This paper discusses the existence and multiplicity of positive solutions for the following first order periodic boundary value problem 0 x ðtÞ þ f ðt; xðtÞÞ ¼ 0; 0 6 t 6 x; xð0Þ ¼ xðxÞ: By utilizing a fixed point theorem on cone, some existence and multiplicity results of positive solutions are derived. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Positive solutions; Periodic boundary value problem; Fixed point theorem; Cone
1. Introduction In this paper, we are concerned with the existence and multiplicity of positive solutions for the following periodic boundary value problem (PBVP for short) 0 x ðtÞ þ f ðt; xðtÞÞ ¼ 0; t 2 J ¼ ½0; x; ð1Þ xð0Þ ¼ xðxÞ: where f 2 CðJ R; RÞ, x > 0. We always assume that the following hypothesis holds:
E-mail address: [email protected] (S. Peng). 0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.08.090
346
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
(H1) There exists a positive number M > 0 such that Mx f ðt; xÞ P 0 for x P 0, t 2 J . In recent years, the method of upper and lower solutions coupled with monotone iterative techniques has been widely used to study PBVPs for differential equations (see [1–6]). Some excellent results of PBVPs have been obtained. It is easy to see that all the proofs which the monotone iterative method has been used are constructive. Meanwhile, in order to apply these results, finding upper and lower solutions is very important. Sometimes, it is difficult to find upper and lower solutions. In this paper, by utilizing a fixed point theorem on cone, some brief conditions on the existence and multiplicity of positive solutions for PBVP (1) are obtained. Some examples are also worked out to illustrate our results.
2. Preliminaries Consider the following first order linear PBVP: 0 x ðtÞ þ MxðtÞ ¼ rðtÞ; t 2 J ; xð0Þ ¼ xðxÞ;
ð2Þ
where M is a positive number, r 2 CðJ ; RÞ. Let rðtÞ ¼
expð MtÞ ; 1 expð MxÞ
t 2 J;
ð3Þ
we can see r 2 C 1 ðJ ; RÞ, rðtÞ > 0, and rðtÞ satisfies the following inequality: expð MxÞ 1 6 rðtÞ 6 : 1 expð MxÞ 1 expð MxÞ
ð4Þ
We can easily claim that the following lemma holds. Lemma 2.1. Suppose that M > 0. Then 8r 2 CðJ ; RÞ, PBVP (2) has a unique solution Z t Z x xðtÞ ¼ rðt sÞrðsÞ ds þ rðx þ t sÞrðsÞ ds: ð5Þ 0
t
Clearly, let rðtÞ ¼ 1, we have Z x Z Z t rðt sÞ ds þ rðx þ t sÞ ds ¼ 0
t
x
rðsÞ ds ¼ 0
1 : M
ð6Þ
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
347
Let Cx ¼ CðJ ; RÞ, for x 2 Cx , kxk ¼ supt2J jxðtÞj, then Cx is a Banach space. For any u 2 Cx , we consider the following PBVP: 0 x ðtÞ þ MxðtÞ ¼ MuðtÞ f ðt; uðtÞÞ; t 2 J ; ð7Þ xð0Þ ¼ xðxÞ: From Lemma 2.1, we shall know that PBVP (7) has a unique solution Z t Z x xu ðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds; ð8Þ 0
t
where ru ðsÞ ¼ MuðsÞ f ðs; uðsÞÞ. Define an operator T : Cx ! Cx : Z t Z T ðuÞðtÞ ¼ rðt sÞru ðsÞ ds þ 0
x
rðx þ t sÞru ðsÞ ds;
u 2 Cx ; t 2 J :
t
ð9Þ It is obvious that if the operator T has a fixed point u 2 Cx , then uðtÞ is a solution of PBVP (1). Let d ¼ expð MxÞ, we can see 0 < d < 1. Define a cone in Cx as follows: Kd ¼ fx 2 Cx : xðtÞ P dkxk; t 2 J g: Lemma 2.2. T ðKd Þ Kd . Proof. For any t 2 J and u 2 Kd , uðtÞ P 0, by (H1) and (4), we have Z t Z x T ðuÞðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds 0 t Z x 1 6 ru ðsÞ ds; 1 expð MxÞ 0 Hence T ðuÞðtÞ P
expð MxÞ 1 expð MxÞ
Z
x
ru ðsÞ ds P dkT ðuÞk:
0
Lemma 2.3. Let g be a positive number, X ¼ fx 2 Cx : kxk < gg. Then T : Kd \ X ! Kd is completely continuous. By the continuity of f ðt; xÞ and Arzela–Ascoli theorem, we can easily give the proof of Lemma 2.3, we omit it here.
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S. Peng / Appl. Math. Comput. 158 (2004) 345–351
Lemma 2.4 [7]. Let K be a cone in a Banach space E and X1 , X2 be two bounded open sets in E such that 0 2 X1 and X1 X2 . Let T : K \ ðX2 n X1 Þ ! K be completely continuous operator. If (C1) There exists u0 2 K n f0g, such that u Tu 6¼ ku0 , u 2 K \ oX2 , k P 0; Tu 6¼ lu, u 2 K \ oX1 , l P 1, or (C2) There exists u0 2 K n f0g, such that u Tu 6¼ ku0 , u 2 K \ oX1 , k P 0; Tu 6¼ lu, u 2 K \ oX2 , l P 1. Then T has at least one fixed point in K \ ðX2 n X1 Þ.
3. Existence of positive solutions In this section, we denote f ðt; xÞ f ðt; xÞ ; f 0 ¼ lim inf ; f0 ¼ lim sup max min þ t2J t2J x!0 þ x x x!0 f ðt; xÞ f ðt; xÞ ; f 1 ¼ lim inf min : f1 ¼ lim sup max x!þ1 t2J t2J x x x!þ1 Theorem 3.1. Assume that (H1) holds. If (H2) f 0 > 0, f1 < 0; or (H3) f 1 > 0, f0 < 0. Then PBVP (1) has at least one positive solution. Proof. At first, assume that (H2) holds. We can choose e > 0, r1 > 0, R0 > 0 (e and r1 are small enough, R0 is large enough), such that f ðt; xÞ P ex;
ð10Þ
0 < x 6 r1 ;
f ðt; xÞ 6 ex;
ð11Þ
x P R0 :
Let X1 ¼ fx 2 Cx : kxk < r1 g, then for any u 2 Kd \ oX1 , from (10), we have Z t Z x T ðuÞðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds 0
t
M e kuk < kuk: 6 M Hence T ðuÞ 6¼ lu;
8u 2 Kd \ oX1 ; 8l P 1:
ð12Þ
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
349
On the other hand, let r2 ¼ R0 =d, X2 ¼ fx 2 Cx : kxk < r2 g, then for u 2 Kd \ oX2 , we have uðtÞ P dkuk ¼ R0 . We choose u0 ¼ 1, then u0 2 Kd n f0g. Suppose that there exists u 2 Kd \ oX2 , such that u T ðuÞ ¼ k0 u0 for some k0 P 0. Let n ¼ mint2J uðtÞ, then n P R0 , from (11), we have Z t Z x M þe n þ k0 ; uðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds þ k0 P M 0 t ð13Þ where ru ðsÞ ¼ M uðsÞ f ðs; uðsÞÞ. Thus, n P tradiction. So we derive that u T ðuÞ 6¼ ku0 ;
Mþe n M
8u 2 Kd \ oX2 ; k P 0:
þ k0 > n, which is a conð14Þ
2 n X1 Þ ! Kd is completely continuous. So all conBy Lemma 2.3, T : Kd \ ðX 2 n X1 Þ such that ditions of Lemma 2.4 are satisfied, there exists u1 2 Kd \ ðX T ðu1 ÞðtÞ ¼ u1 ðtÞ, u1 ðtÞ P dku1 k P dr1 > 0, u1 ðtÞ is a positive solution of PBVP (1). Next, suppose that (H3) holds. We can choose e1 > 0, r3 > 0, R1 > 0 (e1 and r3 are small enough, R1 is large enough), such that f ðt; xÞ 6 e1 x; f ðt; xÞ P e1 x;
ð15Þ
0 < x 6 r3 ;
ð16Þ
x P R1 :
Let X3 ¼ fx 2 Cx : kxk < r3 g, then for any u 2 Kd \ oX3 , uðtÞ P dkuk ¼ dr3 > 0. It is similar to the proof of (14), we have u T ðuÞ 6¼ ku0 ;
8u 2 Kd \ oX3 ; k P 0:
ð17Þ
Let r4 ¼ R1 =d, X4 ¼ fx 2 Cx : kxk < r4 g, then for u 2 Kd \ oX4 , we have uðtÞ P dkuk ¼ R1 , from (16), we can easily obtain T ðuÞ 6¼ lu;
8u 2 Kd \ oX4 ; l P 1:
ð18Þ
By Lemma 2.4, T has a fixed point u2 2 Kd \ ðX4 n X3 Þ, u2 ðtÞ P dr3 , u2 is positive solution of PBVP (1). h Example 3.1. Consider the PBVP: 0 x þ arctan x ð1 þ tÞx3 ¼ 0; xð0Þ ¼ xð1Þ;
0 6 t 6 1;
ð19Þ
where f ðt; xÞ ¼ arctan x ð1 þ tÞx3 , x ¼ 1. We choose M ¼ 1, d ¼ 1=e, and compute that f 0 P 1, f1 ¼ 1. It is easy to check that (H1) and (H2) are satisfied. Thus, by Theorem 3.1, PBVP (19) has at least one positive solution.
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S. Peng / Appl. Math. Comput. 158 (2004) 345–351
4. Multiplicity of positive solutions Theorem 4.1. Assume that (H1) holds. If (H4) f0 < 0, f1 < 0, and (H5) there exists q > 0 such that minff ðt; xÞjt 2 J ; dq 6 x 6 qg > 0: Then PBVP (1) has at least two positive solutions. Proof. By condition (H4), from the proof of Theorem 3.1, we should know that there exist r5 > 0, r6 > 0 (r5 is small enough, and r6 is large enough, r5 < q < r6 ), such that u T ðuÞ 6¼ ku0 ;
8u 2 Kd \ oX5 ; k P 0;
ð20Þ
u T ðuÞ 6¼ ku0 ;
8u 2 Kd \ oX6 ; k P 0;
ð21Þ
where X5 ¼ fx 2 Cx : kxk < r5 g, X6 ¼ fx 2 Cx : kxk < r6 g. From (H5), we can choose e > 0 (e is small enough), such that f ðt; xÞ P ex;
t 2 J ; dq 6 x 6 q:
ð22Þ
Let X7 ¼ fx 2 Cx : kxk < qg, for any u 2 Kd \ oX7 , dq 6 uðtÞ 6 q, from (22), it is similar to the proof of (12), we have T ðuÞ 6¼ lu;
8l P 1; u 2 Kd \ oX7 :
ð23Þ
It is obvious that X5 X7 X6 , by Lemma 2.4, we conclude that T has two 6 n X7 Þ, u3 ðtÞ and u4 ðtÞ are fixed points u3 2 Kd \ ðX7 n X5 Þ and u4 2 Kd \ ðX positive solutions of PBVP (1). h It is similar to the proof of Theorem 4.1, we have the following Theorem 4.2. Theorem 4.2. Assume that (H1) holds. If (H6) f 0 > 0; f 1 > 0, and (H7) there exists q > 0 such that maxff ðt; xÞjt 2 J ; dq 6 x 6 qg < 0: Then PBVP (1) has at least two positive solutions.
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
Example 4.1. Consider the following PBVP: 0 1 x þ 2p x arctanð1 þ xÞ ð1 þ tÞx2 expð 12 xÞ ¼ 0; xð0Þ ¼ xð1Þ;
0 6 t 6 1;
351
ð24Þ
1 x arctanð1 þ xÞ ð1 þ tÞx2 expð 12 xÞ, x ¼ 1. We choose where f ðt; xÞ ¼ 2p 1
0:25 M ¼ 4, d ¼ e , q ¼ 2, we can compute that f 0 P 18, f 1 P 14, maxff ðt; xÞjt 2 J ; 2d 6 x 6 2g < 0:3 < 0. By Theorem 4.2, PBVP (24) has two positive solutions.
5. Remark We can also use the methods of this paper to deal with the following PBVP: 0
x ðtÞ þ f ðt; xðtÞÞ ¼ 0; t 2 J ; ð25Þ xð0Þ ¼ xðxÞ; where f 2 CðJ R; RÞ, and (H1) holds. For PBVP (25), the function rðtÞ in (3) is replaced by rðtÞ ¼
expðMtÞ ; expðMxÞ 1
t 2 J:
ð26Þ
Let d ¼ expð MxÞ, we can also verify that Theorem 3.1 and Theorems 4.1 and 4.2 are valid for PBVP (25). In addition, if the function f ðt; xÞ is periodic in its first variable, the existence and multiplicity of positive periodic solutions can also be discussed by using our method, we omit the details here.
References [1] V. Lakshmikantham, S. Leela, Existence and monotone method for periodic solutions of firstorder differential equations, J. Math. Anal. Appl. 91 (1983) 237–243. [2] V. Lakshmikantham, S. Leela, Remarks on first and second order periodic boundary value problems, Nonlinear Anal. 8 (1984) 281–287. [3] V. Lakshmikantham, Periodic boundary value problems of first and second order differential equations, J. Appl. Math. Simulat. 2 (1989) 131–138. [4] S. Leela, M.N. Oguzt€ oreli, Periodic boundary value problem for differential equations with delay and monotone iterative method, J. Math. Anal. Appl. 122 (1987) 301–307. [5] J.R. Haddock, M.N. Nkashama, Periodic boundary value problems and monotone iterative methods for functional differential equations, Nonlinear Anal. 22 (1994) 267–276. [6] Eduardo Liz, J.J. Nieto, Periodic boundary value problems for a class of functional differential equations, J. Math. Anal. Appl. 200 (1996) 680–686. [7] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, New York, 1988.
Positive solutions for first order periodic boundary value problem Shiguo Peng Department of Applied Mathematics, Guangdong University of Technology, Guangzhou 510090, PeopleÕs Republic of China
Abstract This paper discusses the existence and multiplicity of positive solutions for the following first order periodic boundary value problem 0 x ðtÞ þ f ðt; xðtÞÞ ¼ 0; 0 6 t 6 x; xð0Þ ¼ xðxÞ: By utilizing a fixed point theorem on cone, some existence and multiplicity results of positive solutions are derived. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Positive solutions; Periodic boundary value problem; Fixed point theorem; Cone
1. Introduction In this paper, we are concerned with the existence and multiplicity of positive solutions for the following periodic boundary value problem (PBVP for short) 0 x ðtÞ þ f ðt; xðtÞÞ ¼ 0; t 2 J ¼ ½0; x; ð1Þ xð0Þ ¼ xðxÞ: where f 2 CðJ R; RÞ, x > 0. We always assume that the following hypothesis holds:
E-mail address: [email protected] (S. Peng). 0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.08.090
346
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
(H1) There exists a positive number M > 0 such that Mx f ðt; xÞ P 0 for x P 0, t 2 J . In recent years, the method of upper and lower solutions coupled with monotone iterative techniques has been widely used to study PBVPs for differential equations (see [1–6]). Some excellent results of PBVPs have been obtained. It is easy to see that all the proofs which the monotone iterative method has been used are constructive. Meanwhile, in order to apply these results, finding upper and lower solutions is very important. Sometimes, it is difficult to find upper and lower solutions. In this paper, by utilizing a fixed point theorem on cone, some brief conditions on the existence and multiplicity of positive solutions for PBVP (1) are obtained. Some examples are also worked out to illustrate our results.
2. Preliminaries Consider the following first order linear PBVP: 0 x ðtÞ þ MxðtÞ ¼ rðtÞ; t 2 J ; xð0Þ ¼ xðxÞ;
ð2Þ
where M is a positive number, r 2 CðJ ; RÞ. Let rðtÞ ¼
expð MtÞ ; 1 expð MxÞ
t 2 J;
ð3Þ
we can see r 2 C 1 ðJ ; RÞ, rðtÞ > 0, and rðtÞ satisfies the following inequality: expð MxÞ 1 6 rðtÞ 6 : 1 expð MxÞ 1 expð MxÞ
ð4Þ
We can easily claim that the following lemma holds. Lemma 2.1. Suppose that M > 0. Then 8r 2 CðJ ; RÞ, PBVP (2) has a unique solution Z t Z x xðtÞ ¼ rðt sÞrðsÞ ds þ rðx þ t sÞrðsÞ ds: ð5Þ 0
t
Clearly, let rðtÞ ¼ 1, we have Z x Z Z t rðt sÞ ds þ rðx þ t sÞ ds ¼ 0
t
x
rðsÞ ds ¼ 0
1 : M
ð6Þ
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
347
Let Cx ¼ CðJ ; RÞ, for x 2 Cx , kxk ¼ supt2J jxðtÞj, then Cx is a Banach space. For any u 2 Cx , we consider the following PBVP: 0 x ðtÞ þ MxðtÞ ¼ MuðtÞ f ðt; uðtÞÞ; t 2 J ; ð7Þ xð0Þ ¼ xðxÞ: From Lemma 2.1, we shall know that PBVP (7) has a unique solution Z t Z x xu ðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds; ð8Þ 0
t
where ru ðsÞ ¼ MuðsÞ f ðs; uðsÞÞ. Define an operator T : Cx ! Cx : Z t Z T ðuÞðtÞ ¼ rðt sÞru ðsÞ ds þ 0
x
rðx þ t sÞru ðsÞ ds;
u 2 Cx ; t 2 J :
t
ð9Þ It is obvious that if the operator T has a fixed point u 2 Cx , then uðtÞ is a solution of PBVP (1). Let d ¼ expð MxÞ, we can see 0 < d < 1. Define a cone in Cx as follows: Kd ¼ fx 2 Cx : xðtÞ P dkxk; t 2 J g: Lemma 2.2. T ðKd Þ Kd . Proof. For any t 2 J and u 2 Kd , uðtÞ P 0, by (H1) and (4), we have Z t Z x T ðuÞðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds 0 t Z x 1 6 ru ðsÞ ds; 1 expð MxÞ 0 Hence T ðuÞðtÞ P
expð MxÞ 1 expð MxÞ
Z
x
ru ðsÞ ds P dkT ðuÞk:
0
Lemma 2.3. Let g be a positive number, X ¼ fx 2 Cx : kxk < gg. Then T : Kd \ X ! Kd is completely continuous. By the continuity of f ðt; xÞ and Arzela–Ascoli theorem, we can easily give the proof of Lemma 2.3, we omit it here.
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S. Peng / Appl. Math. Comput. 158 (2004) 345–351
Lemma 2.4 [7]. Let K be a cone in a Banach space E and X1 , X2 be two bounded open sets in E such that 0 2 X1 and X1 X2 . Let T : K \ ðX2 n X1 Þ ! K be completely continuous operator. If (C1) There exists u0 2 K n f0g, such that u Tu 6¼ ku0 , u 2 K \ oX2 , k P 0; Tu 6¼ lu, u 2 K \ oX1 , l P 1, or (C2) There exists u0 2 K n f0g, such that u Tu 6¼ ku0 , u 2 K \ oX1 , k P 0; Tu 6¼ lu, u 2 K \ oX2 , l P 1. Then T has at least one fixed point in K \ ðX2 n X1 Þ.
3. Existence of positive solutions In this section, we denote f ðt; xÞ f ðt; xÞ ; f 0 ¼ lim inf ; f0 ¼ lim sup max min þ t2J t2J x!0 þ x x x!0 f ðt; xÞ f ðt; xÞ ; f 1 ¼ lim inf min : f1 ¼ lim sup max x!þ1 t2J t2J x x x!þ1 Theorem 3.1. Assume that (H1) holds. If (H2) f 0 > 0, f1 < 0; or (H3) f 1 > 0, f0 < 0. Then PBVP (1) has at least one positive solution. Proof. At first, assume that (H2) holds. We can choose e > 0, r1 > 0, R0 > 0 (e and r1 are small enough, R0 is large enough), such that f ðt; xÞ P ex;
ð10Þ
0 < x 6 r1 ;
f ðt; xÞ 6 ex;
ð11Þ
x P R0 :
Let X1 ¼ fx 2 Cx : kxk < r1 g, then for any u 2 Kd \ oX1 , from (10), we have Z t Z x T ðuÞðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds 0
t
M e kuk < kuk: 6 M Hence T ðuÞ 6¼ lu;
8u 2 Kd \ oX1 ; 8l P 1:
ð12Þ
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
349
On the other hand, let r2 ¼ R0 =d, X2 ¼ fx 2 Cx : kxk < r2 g, then for u 2 Kd \ oX2 , we have uðtÞ P dkuk ¼ R0 . We choose u0 ¼ 1, then u0 2 Kd n f0g. Suppose that there exists u 2 Kd \ oX2 , such that u T ðuÞ ¼ k0 u0 for some k0 P 0. Let n ¼ mint2J uðtÞ, then n P R0 , from (11), we have Z t Z x M þe n þ k0 ; uðtÞ ¼ rðt sÞru ðsÞ ds þ rðx þ t sÞru ðsÞ ds þ k0 P M 0 t ð13Þ where ru ðsÞ ¼ M uðsÞ f ðs; uðsÞÞ. Thus, n P tradiction. So we derive that u T ðuÞ 6¼ ku0 ;
Mþe n M
8u 2 Kd \ oX2 ; k P 0:
þ k0 > n, which is a conð14Þ
2 n X1 Þ ! Kd is completely continuous. So all conBy Lemma 2.3, T : Kd \ ðX 2 n X1 Þ such that ditions of Lemma 2.4 are satisfied, there exists u1 2 Kd \ ðX T ðu1 ÞðtÞ ¼ u1 ðtÞ, u1 ðtÞ P dku1 k P dr1 > 0, u1 ðtÞ is a positive solution of PBVP (1). Next, suppose that (H3) holds. We can choose e1 > 0, r3 > 0, R1 > 0 (e1 and r3 are small enough, R1 is large enough), such that f ðt; xÞ 6 e1 x; f ðt; xÞ P e1 x;
ð15Þ
0 < x 6 r3 ;
ð16Þ
x P R1 :
Let X3 ¼ fx 2 Cx : kxk < r3 g, then for any u 2 Kd \ oX3 , uðtÞ P dkuk ¼ dr3 > 0. It is similar to the proof of (14), we have u T ðuÞ 6¼ ku0 ;
8u 2 Kd \ oX3 ; k P 0:
ð17Þ
Let r4 ¼ R1 =d, X4 ¼ fx 2 Cx : kxk < r4 g, then for u 2 Kd \ oX4 , we have uðtÞ P dkuk ¼ R1 , from (16), we can easily obtain T ðuÞ 6¼ lu;
8u 2 Kd \ oX4 ; l P 1:
ð18Þ
By Lemma 2.4, T has a fixed point u2 2 Kd \ ðX4 n X3 Þ, u2 ðtÞ P dr3 , u2 is positive solution of PBVP (1). h Example 3.1. Consider the PBVP: 0 x þ arctan x ð1 þ tÞx3 ¼ 0; xð0Þ ¼ xð1Þ;
0 6 t 6 1;
ð19Þ
where f ðt; xÞ ¼ arctan x ð1 þ tÞx3 , x ¼ 1. We choose M ¼ 1, d ¼ 1=e, and compute that f 0 P 1, f1 ¼ 1. It is easy to check that (H1) and (H2) are satisfied. Thus, by Theorem 3.1, PBVP (19) has at least one positive solution.
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S. Peng / Appl. Math. Comput. 158 (2004) 345–351
4. Multiplicity of positive solutions Theorem 4.1. Assume that (H1) holds. If (H4) f0 < 0, f1 < 0, and (H5) there exists q > 0 such that minff ðt; xÞjt 2 J ; dq 6 x 6 qg > 0: Then PBVP (1) has at least two positive solutions. Proof. By condition (H4), from the proof of Theorem 3.1, we should know that there exist r5 > 0, r6 > 0 (r5 is small enough, and r6 is large enough, r5 < q < r6 ), such that u T ðuÞ 6¼ ku0 ;
8u 2 Kd \ oX5 ; k P 0;
ð20Þ
u T ðuÞ 6¼ ku0 ;
8u 2 Kd \ oX6 ; k P 0;
ð21Þ
where X5 ¼ fx 2 Cx : kxk < r5 g, X6 ¼ fx 2 Cx : kxk < r6 g. From (H5), we can choose e > 0 (e is small enough), such that f ðt; xÞ P ex;
t 2 J ; dq 6 x 6 q:
ð22Þ
Let X7 ¼ fx 2 Cx : kxk < qg, for any u 2 Kd \ oX7 , dq 6 uðtÞ 6 q, from (22), it is similar to the proof of (12), we have T ðuÞ 6¼ lu;
8l P 1; u 2 Kd \ oX7 :
ð23Þ
It is obvious that X5 X7 X6 , by Lemma 2.4, we conclude that T has two 6 n X7 Þ, u3 ðtÞ and u4 ðtÞ are fixed points u3 2 Kd \ ðX7 n X5 Þ and u4 2 Kd \ ðX positive solutions of PBVP (1). h It is similar to the proof of Theorem 4.1, we have the following Theorem 4.2. Theorem 4.2. Assume that (H1) holds. If (H6) f 0 > 0; f 1 > 0, and (H7) there exists q > 0 such that maxff ðt; xÞjt 2 J ; dq 6 x 6 qg < 0: Then PBVP (1) has at least two positive solutions.
S. Peng / Appl. Math. Comput. 158 (2004) 345–351
Example 4.1. Consider the following PBVP: 0 1 x þ 2p x arctanð1 þ xÞ ð1 þ tÞx2 expð 12 xÞ ¼ 0; xð0Þ ¼ xð1Þ;
0 6 t 6 1;
351
ð24Þ
1 x arctanð1 þ xÞ ð1 þ tÞx2 expð 12 xÞ, x ¼ 1. We choose where f ðt; xÞ ¼ 2p 1
0:25 M ¼ 4, d ¼ e , q ¼ 2, we can compute that f 0 P 18, f 1 P 14, maxff ðt; xÞjt 2 J ; 2d 6 x 6 2g < 0:3 < 0. By Theorem 4.2, PBVP (24) has two positive solutions.
5. Remark We can also use the methods of this paper to deal with the following PBVP: 0
x ðtÞ þ f ðt; xðtÞÞ ¼ 0; t 2 J ; ð25Þ xð0Þ ¼ xðxÞ; where f 2 CðJ R; RÞ, and (H1) holds. For PBVP (25), the function rðtÞ in (3) is replaced by rðtÞ ¼
expðMtÞ ; expðMxÞ 1
t 2 J:
ð26Þ
Let d ¼ expð MxÞ, we can also verify that Theorem 3.1 and Theorems 4.1 and 4.2 are valid for PBVP (25). In addition, if the function f ðt; xÞ is periodic in its first variable, the existence and multiplicity of positive periodic solutions can also be discussed by using our method, we omit the details here.
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