first-order impulsive ordinary differential equations with anti-periodic and nonlinear boundary...

Nonlinear Analysis 42 (2000) 163–173www.elsevier.nl/locate/na

First-order impulsive ordinary di�erentialequations with anti-periodic and nonlinear

boundary conditions(

Daniel Franco, Juan J. Nieto ∗

Departamento de An�alisis Matem�atico, Facultad de Matem�aticas, Universidad de Santiago deCompostela, 15706 Spain

Received 27 April 1998; accepted 1 June 1998

Keywords: Impulsive di�erential equation; Anti-periodic boundary conditions; Nonlinear bound-ary conditions; Monotone iterative technique

1. Introduction

The monotone iterative technique is a powerful method that have been used toapproximate the extremal solutions of several problems [1, 5].On the other hand, impulsive di�erential equations are a basic tool to study some

problems of biology, medicine, engineering, and physics [2].In this paper we deal with the following nonlinear boundary value problem for

�rst-order di�erential equation with impulses at �xed points:

u′(t)=f(t; u(t)); t ∈ J ′= J\{t1; t2; : : : ; tp}; (1.1)

u(t+k )= Ik(u(tk)); k =1; 2; : : : ; p; (1.2)

g(u(0); u(T ))= 0; (1.3)

where J = [0; T ], 0= t0¡t1¡t2¡ · · ·¡tp¡tp+1 =T , f : J ×R→R, Ik :R→R fork =1; 2; : : : ; p and g :R2→R.Problem (1:1) and (1:2) was considered in several papers with di�erent types of

boundary conditions. In [4], it is considered problem (1:1)–(1:3) when f is a

( Research partially supported by D.G.I.C.Y.T. (Spain), project PB94-0610.∗ Corresponding author. Tel.: 0034-81-563100/13177; fax.: 0034-81-597054.E-mail address: [email protected] (J.J. Nieto)

0362-546X/00/$ - see front matter ? 2000 Elsevier Science Ltd. All rights reserved.PII: S0362 -546X(98)00337 -X

164 D. Franco, J.J. Nieto / Nonlinear Analysis 42 (2000) 163–173

Carath�eodory function and g is continuous. There, the author proves the validity ofthe upper and lower solution method and the monotone iterative technique when gsatis�es some monotonicity conditions depending on the relative order of the lowerand upper solutions. Those results are applicable in some important cases such as theinitial or the periodic case. However, they are not valid, for example, for anti-periodicboundary conditions, i.e.

u(0)=−u(T ): (1.4)

This condition corresponds to g(x; y)= x+ y. Note that in this case g is increasing inthe second variable, and hence the results are inapplicable.Indeed, note that for M ≥ 0 we have that

u′(t) +Mu(t)≥ 0; t ∈ J (1.5)

and

u(0)− u(T )≥ 0

imply that u(t)≥ 0, t ∈ J . However, Eq. (1.5) and

u(0) + u(T )≥ 0

do not imply that u(t)≥ 0 for every t ∈ J . In consequence, we do not dispose of amaximum principle for anti-periodic boundary conditions and the abstract results of[5] are not applicable.The anti-periodic boundary value problem without impulses, i.e. problem (1:1)–(1:4)

is studied in [6] when �≤ �, and in [6] when �≤ �.In this work we study, via the monotone iterative method, the impulsive problem

(1:1)–(1:3) for f continuous, and either g(x; y)= x+ h(y), or g(x; y)= h(x)+ y, withh :R→R continuous.Hence, we consider the problems

u′(t)=f(t; u(t)); t ∈ J ′;u(t+k )= Ik(u(tk)); k =1; 2; : : : ; p;

u(0) + h(u(T ))= 0

(1.6)

and

u′(t)=f(t; u(t)); t ∈ J ′;u(t+k )= Ik(u(tk)); k =1; 2; : : : ; p;

u(T ) + h(u(0))= 0:

(1.7)

Our results include the anti-periodic boundary conditions and extend previous works,in particular we extend some results of [4, 6, 7].

D. Franco, J.J. Nieto / Nonlinear Analysis 42 (2000) 163–173 165

2. Preliminaries and comparison results

We introduce some de�nitions in order to de�ne the concept of solution forEqs. (1.1)–(1.3)

PC(J ) = {u : J→R: u is continuous for any t ∈ J ′;u(0+); u(T−); u(t+k ); u(t

−k ) exist; and u(t

−k )= u(tk); k =1; : : : ; p}

and

PC1(J ) = {u∈PC(J ): u is continuously di�erentiable for anyt ∈ J ′; u′(0+); u′(T−); u′(t+k ); u

′(t−k ) exist; k =1; : : : ; p; }:

PC(J ) and PC1(J ) are Banach spaces with the norms

‖u‖PC(J ) = sup{|u(t)|: t ∈ J}

and

‖u‖PC1(J ) = ‖u‖PC(J ) + ‖u′‖PC(J ):

For u∈PC(J ), we consider the functions

uk : Jk = [tk−1; tk ]→R; k =1; 2; : : : ; p+ 1;

uk(t)= u(t) if t ∈ (tk−1; tk ];

uk(t)= u(t+k−1) if t= tk−1:

Then

‖u‖PC(J ) = sup{‖uk‖C(Jk ): k =1; 2; : : : ; p+ 1}

and, in this sense, PC(J ) is equivalent to∏p+1k=1 C(Jk).

We say that a function u is a solution for Eqs. (1:1)–(1:3) if u∈PC1(J ) and itsatis�es Eqs. (1:1)–(1:3).Now, we introduce three simple results:

Lemma 1. Let u∈PC1(J ), m; �∈PC(J ), and ck ; dk ∈R, k =1; 2; : : : ; p, such that

u′(t)=m(t)u(t) + �(t); t ∈ J ′

u(t+k )= cku(tk) + dk ; k =1; 2; : : : ; p;

u(0)= u0:


Then u can be expressed as

u(t) = u0∏

{k:0¡tk¡t}ckeM(t) +

∫ t

0

∏{k : s¡tk¡t}

ckeM(t)−M(s)�(s) ds

+∑

{k : 0¡tk¡t}

∏{i : tk¡ti¡t}

cieM(t)−M(tk )dk ; (2.1)

where M(t)=∫ t0 m(r) dr.

Proof. See Theorem 2.5.1 in [2].

Corollary 1. Let u∈PC1(J ) and m∈PC(J ) such thatu′(t)≤m(t)u(t); t ∈ J ′;u(t+k )≤ 0; k =1; 2; : : : ; p;

u(0)≤ 0:Then u(t)≤ 0 for each t ∈ J .

Proof. Apply Lemma 1 with �(t)≤ 0, t ∈ J , u0≤ 0, and for k =1; 2; : : : ; p, ck =0 anddk ≤ 0:

Lemma 2. Assume that u∈PC1(J ), m∈PC(J ), and ck ∈R for k =1; 2; : : : ; p verifyu′(t)=m(t)u(t); t ∈ J ′;u(t+k )= cku(tk); k =1; 2; : : : ; p;

au(0)= bu(T )

with a; b∈R, and

a− b ·p∏k=1

ck exp(∫ T

0m(s) ds

)6=0:

Then u≡ 0 on J .

Proof. By Lemma 1 we have, for any t ∈ J , that

u(t)= u(0) ·∏

{k : tk¡t}ck exp

(∫ t

0m(s) ds

): (2.2)

In particular, for t=T ,

u(T )

[a− b ·

p∏k=1

ck exp(∫ T

0m(s) ds

)]=0:


Hence u(T )= 0. We have now two cases:(i) a 6=0, then u(0)= (b=a)u(T )= 0 and by Eq. (2.2) u≡ 0 on J .(ii) a=0, then by hypotheses b 6=0 and ck 6=0, k =1; 2; : : : ; p. Therefore using again

Eq. (2.2) for t=T we have that u(0)= 0, which completes the proof.

3. Lower and upper solutions

De�nition 1. We say that �∈PC1(J ) and �∈PC1(J ) are a pair of lower and upperrelated solutions for Eq. (1.6) if they satisfy

�′(t)≤f(t; �(t)); t ∈ J ′;�(t+k )≤ Ik(�(tk)); k =1; 2; : : : ; p;

�(0) + h(�(T ))≤ 0and

�′(t)≥f(t; �(t)); t ∈ J ′;�(t+k )≥ Ik(�(tk)); k =1; 2; : : : ; p;

�(0) + h(�(T ))≥ 0:The following result for problem (1:6) generalises Lemma 5 in [7].

Lemma 3. Suppose that h∈C1(R) is nondecreasing and that there exists N ≥ 0 andck ≥ 0, k =1; 2; : : : ; p such that for x≥y

f(t; x)− f(t; y)≤N (x − y)

and

0≤ Ik(x)− Ik(y)≤ ck(x − y):

Then Eq: (1:6) has at most one solution.

Proof. We �rst note that the growth condition of f implies that for any t0 ∈ J , u0 ∈R,the initial value problem

u′(t)=f(t; u(t)); t ∈ [t0; T ]; u(t0)= 0

has a unique solution de�ned on [t0; T ].Let u1; u2 be two solutions of Eq. (1.6). Consider the set

D= {t ∈ J : u1(t)= u2(t)}⊂ J:

We distinguish two cases:


Case 1: D 6= ∅. If 0∈D, then by the uniqueness of the solution for the initial valueproblem

u′=f(t; u); t ∈ J ′ u(t+k )= Ik(u(tk)); k =1; 2; : : : ; p; u(0)= u0;

we have that D= J and u1≡ u2 on J . Now, if T ∈D, then 0∈D since

u1(0)=−h(u1(T ))=−h(u2(T ))= u2(0)and u1≡ u2 on J . If t ∈D, then by the uniqueness of solution we have that [ t; T ]⊂D.Hence 0∈D and, in consequence, D= J .Case 2: D= ∅. Since Ik , k =1; 2; : : : ; p are nondecreasing either u1¡u2 or u2¡u1.

We assume, without lost of generality, that u1¡u2. Let v= u2 − u1, we havev′(t)=f(t; u2(t))− f(t; u1(t))≤Nv(t); t ∈ J ′;v(t+k )= Ik(u2(tk))− Ik(u1(tk))≤ ckv(tk); k =1; 2; : : : ; p;

v(0)= h(u1(T ))− h(u2(T )):Using Lemma 1 with m(t)=N , t ∈ J we see that

v(T )≤ v(0)p∏k=1

ckeNT :

On the other hand, v(0)=−h′(�)v(T ) for some �∈ (u1(T ); u2(T )) by the mean valuetheorem. Thus,

v(T )

(1 + h′(�) ·

p∏k=1

ckeNT)≤ 0

and v(T )= u2(T ) − u1(T ) ≤ 0, which is a contradiction with D= ∅. In consequenceD 6= ∅ and D= J:

Let � and � be a pair of lower and upper related solutions of Eq. (1.6) such that

�≤ � on J: (3.1)

We de�ne the set

= {(t; u): t ∈ J; �(t)≤ u≤ �(t)}and the sector

[�; �] = {u∈PC(J ): �(t)≤ u(t)≤ �(t); t ∈ J}:

Corollary 2. Let � and � be a pair of lower and upper related solutions of Eq: (1:6)satisfying Eq. (3.1). Suppose that h∈C1([�(T ); �(T )]) is nondecreasing and that thereexists N ≥ 0 and ck ≥ 0, k =1; 2; : : : ; p such that for

f(t; x)− f(t; y)≤N (x − y); �(t)≤y≤ x≤ �(t)


and

Ik(x)− Ik(y)≤ ck(x − y); �(tk)≤y≤ x≤ �(tk):Then Eq: (1:6) has at most one solution on [�; �].

Proof. Use Theorem 1.4.1 of [3] and the proof of the Lemma 3.

The following theorem is the most important result of this section

Theorem 1. Let � and � be a pair of lower and upper related solutions of Eq: (1:6)satisfying Eq: (3:1). Suppose that f, h and Ik , k =1; 2; : : : ; p verify the followingconditions:(H1) f∈C(); @f=@u∈C() and there exists M¿0 such that

f(t; x)− f(t; y)≥ −M (x − y); t ∈ J; �(t)≤y≤ x≤ �(t):(H2) For each k =1; 2; : : : ; p, Ik ∈C1(R) is nondecreasing.(H3) h∈C1(R) is nondecreasing.(H4) For k ∈ (�(tk); �(tk)), k =1; 2; : : : ; p, �∈ (�(T ); �(T )) and � : J→R, (t; �(t))∈

h′(�) ·p∏k=1

I ′k( k) exp(∫ T

0

@f@u(s; �(s)) ds

)6=1:

Then, problem (1:6) has an unique solution u∈ [�; �]. Moreover, there exist mono-tone sequences {�n} and {�n} such that {�n} ↗ u and {�n} ↘ u uniformly on J .

Proof. We construct the sequences {�n} and {�n} by de�ning �1 = �, �1 = �, and forn¿1, �n and �n are the solutions of

�′n(t)=f(t; �n−1(t))−M [�n(t)− �n−1(t)]; t ∈ J ′;�n(t+k )= Ik(�n−1(tk)); k =1; 2; : : : ; p;

�n(0)=−h(�n−1(T ))(3.2)

and

�′n(t)=f(t; �n−1(t))−M [�n(t)− �n−1(t)]; t ∈ J ′;�n(t+k )= Ik(�n−1(tk)); k =1; 2; : : : ; p;

�n(0)=−h(�n−1(T )):(3.3)

This is an adequate de�nition since by general results on the initial value problemof impulsive di�erential equations [2] the existence and uniqueness of solution forEqs. (3.2) and (3.3) are guarantee.We prove that these sequences satisfy the property

�n≤ �n+1≤ �n+1≤ �n; n≥ 1: (3.4)


Indeed, if we consider v= �1 − �2, thenv′(t)≤M (�2(t)− �1(t))= −Mv(t); t ∈ J ′;v(t+k )≤ Ik(�1(tk))− Ik(�1(tk))= 0; k =1; 2; : : : ; p;

v(0)≤ − h(�1(T )) + h(�1(T ))= 0and by Corollary 1 we have v= �1 − �2≤ 0 on J . Analogously, �2≤ �1 on J .Now, let v= �2 − �2, using (H1)–(H3) we havev′(t)=f(t; �1(t))− f(t; �1(t)) +M [�1(t)− �1(t)− v(t)]≤ −Mv(t); t ∈ J ′;v(t+k )= Ik(�1(tk))− Ik(�1(tk))≤ 0; k =1; 2; : : : ; p;

v(0)= h(�1(T ))− h(�1(T ))≤ 0:Again by Corollary 1, v= �2 − �2≤ 0.Thus, Eq. (3.4) is valid for n=1. Assuming that it is valid for m≥ 1 we show that

it is valid for m+ 1. Let v= �m − �m+1, by (H1)–(H3),v′(t)=f(t; �m−1(t))− f(t; �m(t)) +M [�m(t)− �m−1(t)− v(t)]≤ −Mv(t); t ∈ J ′;v(t+k )= Ik(�m−1(tk))− Ik(�m(tk))≤ 0; k =1; 2; : : : ; p;

v(0)= h(�m−1(T ))− h(�m(T ))≤ 0:Therefore, v= �m − �m+1≤ 0 on J . Analogously, �m+1≤ �m on J . Finally, letv= �m+1 − �m+1,

v′(t)=f(t; �m(t))− f(t; �m(t)) +M [�m(t)− �m(t)− v(t)]≤ −Mv(t); t ∈ J ′;v(t+k )= Ik(�m(tk))− Ik(�m(tk))≤ 0; k =1; 2; : : : ; p;

v(0)= h(�m(T ))− h(�m(T ))≤ 0:In consequence v= �m+1− �m+1≤ 0 on J , and we have proved the validity ofEq. (3.4).We have two monotone sequences in PC(J ) that are bounded. By standard argu-

ments, there exist � and � with {�n} ↗ � and {�n} ↘ �. Moreover the convergenceis uniformly on J . Also, we obtain that the functions � and � satisfy

�′(t)=f(t; �(t)); t ∈ J ′;�(t+k )= Ik(�(tk)); k =1; 2; : : : ; p;

�(0) + h(�(T ))= 0

and

�′(t)=f(t; �(t)); t ∈ J ′;�(t+k )= Ik(�(tk)); k =1; 2; : : : ; p;

�(0) + h(�(T ))= 0:


If we show that �= �, then � is a solution of Eq. (1.6). It is clear that �≤ �. Weconsider two cases:Case 1: �¡� on J . Consider v= � − �, we take � : J→R with �(t)¡�(t)¡�(t),

t ∈ J and

v′(t)=f(t; �(t))− f(t; �(t))= @f@u(t; �(t))v(t); t ∈ [0; T ]

for k =1; 2; : : : ; p, k ∈ (�(tk); �(tk)) with

v(t+k )= Ik(�(tk))− Ik(�(tk))= I ′k( k)v(tk)

and �∈ (�(T ); �(T )) with

v(0)= h(�(T ))− h(�(T ))= h′(�)(�(T )− �(T ))= h′(�)v(T ):

By Lemma 2, �≡ � on J which is a contradiction with �¡� on J . And this case isnot possible.Case 2: There exists s∈ J with �(s)= �(s).If s=T , then

0= �(0) + h(�(T ))= �(0) + h(�(T ))

and � is a solution of problem (1:6). In the same way we have that � is solution ofEq. (1.6). Now, by the regularity of the functions f, h and Ik we have that Eq. (1.6)has at most one solution in [�; �] (Corollary 2), thus, �≡ �.On the other hand, if �(t)= �(t) for any t ∈ [0; T ), it is easy to see that �(T )= �(T )

(see the proof of Lemma 3). Hence, �≡ �:

In the particular case h(x)= x (anti-periodic boundary conditions) and Ik(x)= x,k =1; 2; : : : ; p (no impulses) we have the nonimpulsive anti-periodic boundary problem(1:1)–(1:4). In this case condition (H4) reads

∫ T

0

@f@u(s; �(s)) ds 6=0 with � : J→R; (t; �(t))∈ (3.5)

and our Theorem 1 is precisely Theorem 1 of [7].We note that Eq. (3.5) is always satis�ed if, for instance, f∈C1() and is strictly

monotone. This is a relevant di�erence with the periodic case.To conclude we consider problem (1:7).We say that �; �∈PC1(J ) are a pair of lower and upper related solutions for

Eq. (1.7) if they satisfy

�′(t)≤f(t; �(t)); t ∈ J ′;�(t+k )≤ Ik(�(tk)); k =1; 2; : : : ; p;

�(T ) + h(�(0))≤ 0


and

�′(t)≥f(t; �(t)); t ∈ J ′;�(t+k )≥ Ik(�(tk)); k =1; 2; : : : ; p;

�(T ) + h(�(0))≥ 0:We enunciate two dual results corresponding to Corollary 1 and Lemma 3. There

are some di�erences but the ideas and techniques are the same. Thus, we present theresults without proofs.

Lemma 4. Let u∈PC1(J ), m∈PC(J ), and ck ∈R, ck¿0, k =1; 2; : : : ; p, such thatu′(t)≥m(t)u(t); t ∈ J ′;u(t+k )≥ cku(tk); k =1; 2; : : : ; p;

u(T )≤ 0:Then u≤ 0 in J .

Lemma 5. Suppose that h∈C1(R) is nondecreasing and that there exists N ≥ 0 andck¿0, k =1; 2; : : : ; p such that for x≥y

f(t; x)− f(t; y)≥ − N (x − y)and

Ik(x)− Ik(y)≥ ck(x − y):Then Eq: (1:7) has at most one solution.

The following result show the validity of the monotone iterative technique forEq. (1.7) when �≤ � on J .

Theorem 2. Let � and � be a pair of lower and upper related solutions of Eq: (1:7)satisfying �≤ � on J . Suppose that f, h and Ik , k =1; 2; : : : ; p verify the followingconditions:(H1′) f∈C(), @f=@u∈C() and there exists M¿0 such that

f(t; x)− f(t; y)≤M (x − y); t ∈ J; �(t)≤y≤ x≤ �(t):(H2′) For each k =1; 2; : : : ; p, Ik ∈C1(R) and there exist dk¿0 such that

Ik(x)− Ik(y)≤dk(x − y) �(tk)≤y≤ x≤ �(tk):(H3′) h∈C1(R) is nondecreasing.(H4′) For k ∈ (�(tk); �(tk)), k =1; 2; : : : ; p, �∈ (�(0); �(0)) and � : J→R, (t; �(t))∈

h′(�) +p∏k=1

I ′k( k) exp(∫ T

0

@f@u(s; �(s)) ds

)6=0:


Then, problem (1:7) has an unique solution u∈ [�; �]. Moreover, there exist mono-tone sequences {�n} and {�n} such that {�n} ↘ u and {�n} ↗ u uniformly on J .

Proof. In this case we construct the sequences {�n} and {�n} by de�ning �1 = �,�1 = �, and for n¿1, �n and �n are the solutions of

�′n(t)=f(t; �n−1(t))−M [�n(t)− �n−1(t)]; t ∈ J ′;�n(t+k )= Ik(�n−1(tk))− dk(�n−1(tk)− �n(tk)); k =1; 2; : : : ; p;

�n(T )=−h(�n−1(0))(3.6)

and

�′n(t)=f(t; �n−1(t))−M [�n(t)− �n−1(t)]; t ∈ J ′;�n(t+k )= Ik(�n−1(tk))− dk(�n−1(tk)− �n(tk)); k =1; 2; : : : ; p;

�n(T )=−h(�n−1(0)):(3.7)

Combining Lemmas 4, 5, 2 and the proof of Theorem 1 it is easy to conclude theproof.

We note that Theorem 1 in [6] is a particular case of this last theorem.

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[3] V. Lakshmikantham, S. Leela, Di�erential and Integral Inequalities, vol. I, Academic Press, New York,1968.

[4] E. Liz, Existence and approximation of solutions for impulsive �rst order problems with nonlinearboundary conditions, Nonlinear Anal. 25 (1995) 1191.

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first-order impulsive ordinary differential equations with anti-periodic and nonlinear boundary...

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