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Existence results for impulsive second-order periodic problems
Nonlinear Analysis 59 (2004) 133 – 146 www.elsevier.com/locate/na
Existence results for impulsive second-order periodic problems Irena Rach˚unkováa,1 , Milan Tvrdýb,∗,2 a Department of Mathematics, Palacký University, 779 00 Olo-Mouc, Tomkova 40, Czech Republic b Mathematical Institute, Academy of Sciences of the Czech Republic, Zitna 25, 115 67 Praha, Czech Republic
Received 28 January 2004; accepted 15 July 2004
Abstract This paper provides existence results for the nonlinear impulsive periodic boundary value problem u = f (t, u, u ), u(ti +) = Ji (u(ti )), u(0) = u(T ),
u (ti +) = Mi (u (ti )), i = 1, 2, . . . , m,
u (0) = u (T ),
where f ∈ Car([0, T ]× R2 ) and Ji , Mi ∈ C(R). The basic assumption is the existence of lower/upper functions 1 /2 associated with the problem. Here, we generalize and extend the existence results of our previous papers. 䉷 2004 Elsevier Ltd. All rights reserved. MSC: 34B37; 34B15; 34C25 Keywords: Second-order nonlinear ordinary differential equation with impulses; Periodic solutions; Lower and upper functions; Leray–Schauder topological degree; A priori estimates
∗ Corresponding author. Tel.: +420-222090762; fax: +420-222211638.
E-mail addresses: [email protected] (I. Rach˚unková), [email protected] (M. Tvrdý). 1 Supported by the Grant no. 201/04/1077 of the Grant Agency of the Czech Republic and by the Council of
Czech Government J14/98:153100011. 2 Supported by the Grant no. 201/01/1077 of the Grant Agency of the Czech Republic.
0362-546X/$ - see front matter 䉷 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.07.006
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1. Introduction This paper deals with the solvability of the nonlinear impulsive boundary value problem (2.1)–(2.3) and provides conditions for f, Ji and Mi , i = 1, 2, . . . , m, which guarantee the existence of at least one solution. Boundary value problems of this kind have received a considerable attention, see e.g. [1–7,10,12]. The results of these papers rely on the existence of a well-ordered pair 1 2 of lower/upper functions associated with the problem under consideration. In [11], we extended these results to the case that 1 /2 are not well-ordered, i.e.
1 () > 2 () for some ∈ [0, T ].
(1.1)
The goal of this paper is to generalize the main existence results of [11], where we restricted our attention to impulsive functions Mi , i = 1, 2, . . . , m, fulfilling the conditions yMi (y)0
for y ∈ R, i = 1, 2, . . . , m.
(1.2)
Here we prove existence criteria without restriction (1.2). Throughout the paper we keep the following notation and conventions: real-valued function u defined a.e. on [0, T ], we put T u ∞ = sup ess |u(t)| and u 1 = |u(s)| ds. t∈[0,T ]
For
a
0
For a given interval J ⊂ R, by C(J ) we denote the set of real-valued functions which are continuous on J. Furthermore, C1 (J ) is the set of functions having continuous first derivatives on J and L(J ) is the set of functions which are Lebesgue integrable on J. Let m ∈ N and let 0=t0 < t1 < t2 < · · · < tm < tm+1 =T be a division of the interval [0, T ]. We denote D = {t1 , t2 , . . . , tm } and define C1D [0, T ] as the set of functions u : [0, T ] → R of the form u (t) if t ∈ [0, t1 ], [0] u[1] (t) if t ∈ [t1 , t2 ], u(t) = . ... .. u[m] (t) if t ∈ [tm , T ], where u[i] ∈ C1 [ti , ti+1 ] for i = 0, 1, . . . , m. Moreover, AC1D [0, T ] stands for the set of functions u ∈ C1D [0, T ] having first derivatives absolutely continuous on each subinterval (ti , ti+1 ), i = 0, 1, . . . , m. For u ∈ C1D [0, T ] and i = 1, 2, . . . , m + 1 we define u (ti ) = u (ti −) = lim u (t), t→ti −
u (0) = u (0+) = lim u (t) t→0+
(1.3)
and u D = u ∞ + u ∞ . Note that the set C1D [0, T ] becomes a Banach space when equipped with the norm . D and with the usual algebraic operations. We say that f : [0, T ] × R2 → R satisfies the Carathéodory conditions on [0, T ] × R2 if (i) for each x ∈ R and y ∈ R the function f (., x, y) is measurable on [0, T ]; (ii) for almost every t ∈ [0, T ] the function f (t, ., .) is continuous on R2 ; (iii) for each compact set K ⊂ R2 there is a function mK (t) ∈ L[0, T ] such that |f (t, x, y)|mK (t) holds for a.e.
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135
t ∈ [0, T ] and all (x, y) ∈ K. The set of functions satisfying the Carathéodory conditions on [0, T ] × R2 will be denoted by Car([0, T ] × R2 ). Given a Banach space X and its subset M, let cl(M) and *M denote the closure and the boundary of M, respectively. Let be an open bounded subset of X. Assume that the operator F : cl() → X is completely continuous and Fu = u for all u ∈ * . Then deg (I − F, ) denotes the Leray–Schauder topological degree of I − F with respect to , where I is the identity operator on X. For the definition and properties of the degree see e.g. [8]. 2. Formulation of the problem and main assumptions Here we study the existence of solutions to the problem u = f (t, u, u ),
(2.1)
u(ti +) = Ji (u(ti )), u(0) = u(T ),
u (ti +) = Mi (u (ti )),
i = 1, 2, . . . , m,
u (0) = u (T ),
(2.2) (2.3)
where u (ti ) are understood in the sense of (1.3), f ∈ Car([0, T ] × R2 ), Ji ∈ C(R) and Mi ∈ C(R). Definition 2.1. A solution of problem (2.1)–(2.3) is a function u ∈ AC1D [0, T ] which satisfies impulsive conditions (2.2), periodic conditions (2.3) and for a.e. t ∈ [0, T ] fulfils Eq. (2.1). Definition 2.2. A function 1 ∈ AC1D [0, T ] is called a lower function of problem (2.1)–(2.3) if
1 (t)f (t, 1 (t), 1 (t)) for a.e. t ∈ [0, T ], 1 (ti +) = Ji (1 (ti )), 1 (0) = 1 (T ),
1 (ti +)Mi (1 (ti )), i = 1, 2, . . . , m,
1 (0)1 (T ).
(2.4) (2.5) (2.6)
Similarly, a function 2 ∈ AC1D [0, T ] is an upper function of problem (2.1)–(2.3) if
2 (t)f (t, 2 (t), 2 (t)) for a.e. t ∈ [0, T ], 2 (ti +) = Ji (2 (ti )), 2 (0) = 2 (T ),
2 (ti +)Mi (2 (ti )), i = 1, 2, . . . , m,
2 (0)2 (T ).
Assumptions 2.3. In the paper we work with the following assumptions: 0 = t0 < t1 < · · · < tm < tm+1 = T < ∞, D = {t1 , t2 , . . . , tm }, f ∈ Car([0, T ] × R2 ), Ji ∈ C(R), Mi ∈ C(R), i = 1, 2, . . . , m;
1 and 2 are, respectively, lower and upper functions of (2.1)–(2.3);
(2.7) (2.8) (2.9)
(2.10) (2.11)
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x > 1 (ti ) ⇒ Ji (x) > Ji (1 (ti )), x < 2 (ti ) ⇒ Ji (x) < Ji (2 (ti )),
i = 1, 2, . . . , m;
(2.12)
i = 1, 2, . . . , m.
(2.13)
y 1 (ti ) ⇒ Mi (y)Mi (1 (ti )), y 2 (ti ) ⇒ Mi (y)Mi (2 (ti )),
Operator reformulation of (2.1)–(2.3). By G(t, s) we denote the Green function of the Dirichlet boundary value problem u = 0, u(0) = u(T ) = 0, i.e., t (s − T ) if 0t s T , (2.14) G(t, s) = s (t T− T ) if 0s < t T . T Furthermore, we define the operator F : C1D [0, T ] → C1D [0, T ] by T (Fx)(t)=x(0) + x (0) − x (T ) + G(t, s) f (s, x(s), x (s)) ds −
m *G i=1
*s
0
m (t, ti )(Ji (x(ti ))−x(ti ))+ G(t, ti )(Mi (x (ti ))−x (ti )). (2.15) i=1
As in [9, Lemma 3.1], where m = 1, we can prove (see Proposition 2.6 below) that F is completely continuous and that a function u is a solution of (2.1)–(2.3) if and only if u is a fixed point of F. To this aim we need the following lemma which extends Lemma 2.1 from [9]. Lemma 2.5. For each h ∈ L[0, T ], c, di , ei ∈ R, i =1, 2, . . . , m, there is a unique function x ∈ AC1D [0, T ] fulfilling x (t) = h(t) a.e. on [0, T ], (2.16) x(ti +) − x(ti ) = di , x (ti +) − x (ti ) = ei , i = 1, 2, . . . , m, x(0) = x(T ) = c.
(2.17)
This function is given by T m m *G (t, ti ) di + G(t, s) h(s) ds − G(t, ti ) ei x(t)=c + *s 0 i=1 i=1 for t ∈ [0, T ],
(2.18)
where G(t, s) is defined by (2.14). Proof. It is easy to check that x ∈ AC1D [0, T ] fulfils (2.16) together with x(0) = c if and only if there is c ∈ R such that x(t)=c + t c+
t
+ 0
m i=1
(ti ,T ] (t) di +
(t − s) h(s) ds
m i=1
(ti ,T ] (t) (t − ti ) ei
for t ∈ [0, T ],
(2.19)
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where (ti ,T ] (t) = 1 if t ∈ (ti , T ] and (ti ,T ] (t) = 0 if t ∈ R\(ti , T ]. Furthermore, x(T ) = c if and only if T m m T −s di T − t i h(s) ds. (2.20) ei − − c=− T T T 0 i=1
i=1
Inserting (2.20) into (2.19), we get x(t)=
ti (t − T ) t (ti − T ) (t − T ) t ei + ei − di − di T T T T ti
Hence, taking into account (2.14), we conclude that the function x given by (2.18) is the unique solution of (2.16), (2.17) in AC1D [0, T ]. Proposition 2.6. Assume that (2.10) holds. Let the operator F : C1D [0, T ] → C1D [0, T ] be defined by (2.14) and (2.15). Then F is completely continuous and a function u is a solution of (2.1)–(2.3) if and only if u = Fu. Proof. Choose an arbitrary y ∈ C1D [0, T ] and put h(t) = f (t, y(t), y (t)) for a.e. t ∈ [0, T ], di = Ji (y(ti )) − y(ti ), ei = Mi (y (ti )) − y (ti ), i = 1, 2, = . . . , m, c = y(0) + y (0) − y (T ).
(2.21)
Then h ∈ L[0, T ], c, di , ei ∈ R i = 1, 2, . . . , m. By Lemma 2.5, there is a unique x ∈ AC1D [0, T ] fulfilling (2.16), (2.17) and it is given by (2.18). Due to (2.21), we have x(t) = (Fy)(t) for t ∈ [0, T ]. Therefore, u ∈ C1D [0, T ] is a solution to (2.1)–(2.3) if and only if uF u. Define an operator F1 : C1D [0, T ] → C1D [0, T ] by T (F1 y)(t) = G(t, s) f (s, y(s), y (s)) ds, t ∈ [0, T ]. 0
As F1 is a composition of the Green type operator for the Dirichlet problem u = 0, u(0) = u(T ) = 0, and of the superposition operator generated by f ∈ Car([0, T ] × R2 ), making use of the Lebesgue Dominated Convergence Theorem and the Arzelà–Ascoli Theorem, we get in a standard way that F1 is completely continuous. Since Ji , Mi , i = 1, 2, . . . , m, are continuous, the operator F2 = F − F1 is continuous, as well. Having in mind that F2 maps bounded sets onto bounded sets and its values are contained in a (2m + 1)-dimensional subspace of C1D [0, T ], we conclude that the operators F2 and F = F1 + F2 are completely continuous. In the proof of our main result we will need the next proposition which concerns the case of well-ordered lower/upper functions and which follows from [10, Corollary 3.5].
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Proposition 2.7. Assume that (2.10) holds and let and be respectively lower and upper functions of (2.1)–(2.3) such that
(t) < (t) f or t ∈ [0, T ] and
(+) < (+) f or ∈ D,
(ti ) < x < (ti ) ⇒ Ji ((ti )) < Ji (x) < Ji ((ti )), and
y (ti ) ⇒ Mi (y)Mi ( (ti )), y (ti ) ⇒ Mi (y)Mi ( (ti )),
i = 1, 2, . . . , m
(2.22) (2.23)
(2.24)
i = 1, 2, . . . , m.
Further, let h ∈ L[0, T ] be such that |f (t, x, y)|h(t)
f or a.e. t ∈ [0, T ] and all (x, y) ∈ [(t), (t)] × R
(2.25)
and let the operator F be defined by (2.15). Finally, for ∈ (0, ∞) denote
(, , )={u ∈ C1D [0, T ] : (t) < u(t) < (t) f or t ∈ [0, T ], (+) < u(+) < (+) f or ∈ D, u ∞ < }.
(2.26)
Then deg (I − F, (, , )) = 1 whenever Fu = u on *(, , ) and
> h 1 +
∞ + ∞ ,
where =
min
i=1,2,...,m+1
(ti − ti−1 ).
(2.27)
Proof. Using the mean value theorem, we can show that u ∞ h 1 +
∞ + ∞
(2.28)
holds for each u ∈ C1D [0, T ] fulfilling (t) < u(t) < (t) for t ∈ [0, T ] and (+) < u(+) < (+) for ∈ D. Thus, if we denote by c the right-hand side of (2.28), we can follow the proof of [10, Corollary 3.5]. 3. A priori estimates In Section 3, we will need a priori estimates which are contained in Lemmas 3.1–3.3. Lemma 3.1. Let 1 ∈ (0, ∞), h ∈ L[0, T ], Mi ∈ C(R), i = 1, 2, . . . , m. Then there exists d ∈ ( 1 , ∞) such that the estimate u ∞ < d
(3.1)
i ∈ C(R), i = 1, 2, . . . , m, satisfying (2.3), is valid for each u ∈ AC1D [0, T ] and each M |u ( u )| < 1
f or some u ∈ [0, T ],
i (u (ti )), u (ti +) = M
i = 1, 2, . . . , m,
|u (t)| < h(t) f or a.e. t ∈ [0, T ]
(3.2) (3.3) (3.4)
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139
and i (y)| : |y| < a} < b sup {|Mi (y)| : |y| < a}
(3.5)
i ∈ C(R), i = 1, 2, . . . , m, satisfy (2.3) and Proof. Suppose that u ∈ AC1D [0, T ] and M (3.2)–(3.5). Due to (2.3), we can assume that u ∈ (0, T ], i.e. there is j ∈ {1, 2, ..., m + 1} such that u ∈ (tj −1 , tj ]. We will distinguish 3 cases: either j = 1 or j = m + 1 or 1 < j < m + 1. Let j = 1. Then, using (3.2) and (3.4), we obtain |u (t)| < a1
on [0, t1 ],
(3.6)
where a1 = 1 + h 1 . Since M1 ∈ C(R), we can find b1 (a1 ) ∈ (a1 , ∞) such that |M1 (y)| < b1 (a1 ) for all y ∈ (−a1 , a1 ). Hence, in view of (3.3) and (3.5), we have h 1 for |u (t1 +)| < b1 (a1 ), wherefrom, using (3.4), we deduce that |u (t)| < b1 (a1 ) + t ∈ (t1 , t2 ]. Continuing by induction, we get bi (ai ) ∈ (ai , ∞) such that |u (t)| < ai+1 = bi (ai ) + h 1 on (ti , ti+1 ] for i = 2, . . . , m, i.e. u ∞ < d := max{ai : i = 1, 2, . . . , m + 1}.
(3.7)
Assume that j = m + 1. Then, using (3.2) and (3.4), we obtain |u (t)| < am+1
on (tm , T ],
(3.8)
where am+1 = 1 + h 1 . Furthermore, due to (2.3), we have |u (0)| < am+1 which together h 1 . Now, proceeding as in the case with (3.4) yields that (3.6) is true with a1 = am+1 + j = 1, we show that (3.7) is true also in the case j = m + 1. Assume that 1 < j < m + 1. Then (3.2) and (3.4) yield |u (t)| < aj +1 = 1 + h 1 h 1 on (tj +1 , tj +2 ], where on (tj , tj +1 ]. If j < m, then |u (t)| < aj +2 = bj +1 (aj +1 ) + bj +1 (aj +1 ) > aj +1 . Proceeding by induction we get (3.8) with am+1 = bm (am ) + h 1 and bm (am ) > am , wherefrom (3.7) again follows as in the previous case. Lemma 3.2. Let 0 , d, q ∈ (0, ∞) and Ji ∈ C(R), i = 1, 2, . . . , m. Then there exists c ∈ ( 0 , ∞) such that the estimate u ∞ < c
(3.9)
is valid for each u ∈ C1D [0, T ] and each Ji ∈ C(R), i = 1, 2, . . . , m, satisfying (2.3), (3.1), u(ti +) = Ji (u(ti )), |u(u )| < 0
i = 1, 2, . . . , m,
f or some u ∈ [0, T ]
(3.10) (3.11)
and sup {|Ji (x)|:|x| < a}
(3.12)
Proof. We will argue similarly as in the proof of Lemma 3.1. Suppose that u ∈ C1D [0, T ] satisfies (2.3), (3.1), (3.10), (3.11) and that Ji ∈ C(R), i = 1, 2, . . . , m, satisfy (3.12).
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Due to (2.3) we can assume that u ∈ (0, T ], i.e. there is j ∈ {1, 2, . . . , m + 1} such that u ∈ (tj −1 , tj ]. We will consider three cases: j = 1, j = m + 1, 1 < j < m + 1. If j = 1, then (3.1) and (3.11) yield |u(t)| < a1 = 0 + d T on [0, t1 ]. In particular, |u(t1 )| < a1 . Since J1 ∈ C(R), we can find b1 (a1 ) ∈ (a1 +q, ∞) such that |J1 (x)| < b1 (a1 ) for all x ∈ (−a1 , a1 ) and consequently, by (3.12), also | J1 (x)| < b1 (a1 )for all x ∈ (−a1 , a1 ). Therefore, by (3.1), |u(t)| < |u(t1 +)| + d T = | J1 (u(t1 ))| + d T < a2 = b1 (a1 ) + d T on (t1 , t2 ]. Proceeding by induction we get bi (ai ) ∈ (ai + q, ∞) such that |u(t)| < ai+1 = bi (ai ) + d T for t ∈ (ti , ti+1 ] and i = 2, . . . , m. As a result, (3.9) is true with c = max{ai : i = 1, 2, . . . , m + 1}. Analogously we would proceed in the remaining cases j = m + 1 or 1 < j < m + 1. Finally, we will need two estimates for functions u satisfying one of the following conditions: u(su ) < 1 (su ) and u1 on [0, T ] and u2 on [0, T ]
and
u(tu ) > 2 (tu ) for some su , tu ∈ [0, T ],
(3.13)
inf
|u(t) − 1 (t)| = 0,
(3.14)
inf
|u(t) − 2 (t)| = 0.
(3.15)
t∈[0,T ] t∈[0,T ]
i ∈ C(R), i = 1, 2, . . . , m, Lemma 3.3. Assume that 1 , 2 ∈ AC1D [0, T ], Ji , Mi , Ji , M satisfy (2.12), (2.13) and Ji (1 (ti )) = Ji (1 (ti )), x > 1 (ti ) ⇒ Ji (x) > (3.16) x < 2 (ti ) ⇒ Ji (x) < Ji (2 (ti )) = Ji (2 (ti )), i = 1, 2, . . . , m and
i (y)Mi ( (ti )), y 1 (ti ) ⇒ M 1 i (y)Mi ( (ti )), y 2 (ti ) ⇒ M 2
i = 1, 2, . . . , m.
(3.17)
Define B={u ∈ C1D [0, T ] : u satisf ies (2.3), (3.10), (3.3) and one of the conditions (3.13), (3.14), (3.15)}. Then each function u ∈ B satisfies |u ( u )| < 1 f or some u ∈ [0, T ], where 2 1 = ( 1 ∞ + 2 ∞ ) + 1 ∞ + 2 ∞ + 1. t1
(3.18)
(3.19)
Proof. This lemma can be proved by the same arguments as Lemma 2.3 in [11] with the i (u (ti )) in place of Mi (u (ti )) and that we use (3.17) instead only difference that we write M of (3.13). 4. Main results Our main result consists in a generalization of [11, Theorem 3.1]. Particularly, we remove condition (1.2) which was assumed in [11] and prove the following theorem.
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Theorem 4.1. Assume that (2.10)–(2.13) and (1.1) hold and let h ∈ L[0, T ] be such that |f (t, x, y)|h(t)
f or a.e. t ∈ [0, T ] and all (x, y) ∈ R2 .
(4.1)
Then problem (2.1)–(2.3) has a solution u satisfying one of the conditions (3.13)–(3.15). Proof. Step 1 : We construct a proper auxiliary problem. Let 1 and 2 be, respectively, lower and upper functions of (2.1)–(2.3) and let 1 be associated with them as in (3.19). Put h(t) = 2 h(t) + 1 for a.e. t ∈ [0, T ] and = 1 +
m
(|Mi (1 (ti ))| + |Mi (2 (ti ))|).
i=1
, ∞) satisfying (3.1). Furthermore, put 0 = 1 ∞ + 2 ∞ +1 By Lemma 3.1, find d ∈ ( and
m T (4.2) max max |Mi (y)|, d + 1 q= |y|d+1 m i=1
and, by Lemma 3.2, find c ∈ ( 0 + q, ∞) fulfilling (3.9). In particular, we have c > 1 ∞ + 2 ∞ + q + 1,
d > 1 ∞ + 2 ∞ + 1.
Finally, for a.e. t ∈ [0, T ] and all x, y ∈ R and i = 1, 2, . . . , m, define functions f (t, x, y) − h(t) − 1 if x − c − 1, f (t, x, y) + (x + c) (h(t) + 1) if − c − 1 < x < − c, f(t, x, y) = f (t, x, y) if − cx c, f (t, x, y) + (x − c) (h(t) + 1) if c < x < c + 1, f (t, x, y) + h(t) + 1 if x c + 1, x+q if x − c − 1, Ji (−c) (c + 1 + x) − (x + q) (x + c) if − c − 1 < x < − c, Ji (x) = Ji (x) if − cx c, J (c) (c + 1 − x) + (x − q) (x − c) if c < x < c + 1, i x−q if x c + 1, y if y − d − 1, (−d) (d + 1 + y) − y (y + d) if − d − 1 < y < − d, M i i (y) = Mi (y) M if − d y d, if d < y < d + 1, Mi (d) (d + 1 − y) + y (y − d) y if y d + 1
(4.3)
(4.4)
(4.5)
(4.6)
and consider the auxiliary problem u = f(t, u, u ),
(3.10),
(3.3),
(2.3).
(4.7)
i ∈ C(R) for i = 1, 2, . . . , m. According to Ji , M Due to (2.10), f ∈ Car([0, T ] × R) and (4.3)–(4.6) the functions 1 and 2 are, respectively, lower and upper functions of (4.7). By (4.1) we have h(t) for a.e. t ∈ [0, T ] and all (x, y) ∈ R2 |f(t, x, y)|
(4.8)
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and
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f(t, x, y) < 0 f(t, x, y) > 0
for a.e. t ∈ [0, T ] and all (x, y) ∈ (−∞, −c − 1] × R, for a.e. t ∈ [0, T ] and all (x, y) ∈ [c + 1, ∞) × R.
(4.9)
i satisfy the assumptions of Lemmas 3.1–3.3. Ji and M Step 2 : We show that Choose an arbitrary i ∈ {1, 2, . . . , m}. (i) Condition (3.5): Let a ∈ (0, ∞), b ∈ (a, ∞) and Mi∗ = sup{|Mi (y)| : |y| < a} < b. i (y)| : |y| < a} max{a, M ∗ } < b. Then, by (4.6), we have sup{|M i (ii) Condition (3.12): Let a ∈ (0, ∞), b ∈ (a +q, ∞) and Ji∗ =sup{|Ji (x)| : |x| < a} < b. Then, by (4.5), we have sup{| Ji (x)| : |x| < a} max{a + q, Ji∗ } < b. (iii) Condition (3.16): Due to (2.12), (4.3) and (4.5), we see that (3.16) holds if |x|c. Assume that x > c. Then x > max{1 (ti ), 2 (ti )} which means that the second condition in (3.16) need not be considered in this case. Since |1 (ti )| < c, we have Ji (1 (ti )) = Ji (1 (ti )). Furthermore, due to (4.3), x − q > 1 ∞ + 2 ∞ + 1. If x c + 1, then Ji (x) = x − q > 1 (ti +) = Ji (1 (ti )). Finally, if x ∈ (c, c + 1), then Ji (x) = Ji (c) (c + 1 − x) + (x − q) (x − c) > Ji (1 (ti ))because Ji (c) > Ji (1 (ti )) by (2.12). For x < (∞, −c) we can argue similarly. (iv) Condition (3.17): Due to (2.13), (4.3) and (4.6), we see that (3.17) holds for |y| < d. Assume that y > d. Then y > max{1 (ti ), 2 (ti )} which means that the first condition in i (y) = (3.17) need not be considered in this case. Since d > > Mi (2 (ti )), we have M i (y) = Mi (d) (d + 1 − y) + y (y − d) > Mi ( (ti )) if y > Mi (2 (ti )) if y > d + 1 and M 2 y ∈ (d, d + 1). Hence the second condition in (3.17) is satisfied for y ∈ (d, ∞). Similarly we can verify the first condition in (3.17) for y ∈ (−∞, −d). Step 3 : We construct a well-ordered pair of lower/upper functions for (4.7). Put A∗ = q +
m i=1
and
max | Ji (x)|
|x|c+1
4 (0) = A∗ + m q, 4 (t) = A∗ + (m − i) q + mTq t for t ∈ (ti , ti+1 ], i = 0, 1, . . . , m, 3 (t) = −4 (t) for t ∈ [0, T ].
(4.10)
(4.11)
Then 3 , 4 ∈ AC1D [0, T ] and, by (4.5) and (4.10),
3 (t) < − A∗ < − c − 1,
4 (t) > A∗ > c + 1 for t ∈ [0, T ].
(4.12)
In view of (4.2), mq mq − (d + 1) and 4 (t) = d + 1 for t ∈ [0, T ]. (4.13) T T Now, we prove that 4 is an upper function of (4.7): By (4.9) and (4.12), we have 0 = 4 (t) < f(t, 4 (t), 4 (t)) for a.e. t ∈ [0, T ]. Furthermore, by (4.5), mq ti = 4 (ti ) − q = 4 (ti +) = A∗ + (m − i) q + Ji (4 (ti )). T
3 (t) = −
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143
By virtue of (4.2) and (4.6), we get
4 (ti +) =
mq i (4 (ti )) for i = 1, 2, . . . , m. = 4 (ti ) = M T
Finally, 4 (0) = A∗ + m q = 4 (T ) and 4 (0) = m q/T = 4 (T ), i.e. 4 is an upper function of (4.7). Since 3 = −4 , we can see that 3 is a lower function of (4.7). Clearly,
3 < 4 on [0, T ] and 3 (+) < 4 (+) for ∈ D.
(4.14)
Having G from (2.15), define an operator F : C1D [0, T ] → C1D [0, T ] by T (Fu)(t)=u(0) + u (0) − u (T ) + G(t, s) f(s, u(s), u (s)) ds −
m *G i=1
+
m
*s
0
(t, ti )( Ji (u(ti )) − u(ti ))
i (u (ti )) − u (ti )), G(t, ti )(M
t ∈ [0, T ].
(4.15)
i=1
By Proposition 2.6, F is completely continuous and u is a solution of (4.7) whenever Fu = u. Step 4 : We prove the first a priori estimate for solutions of (4.7). Define
0 ={u ∈ C1D [0, T ] : u ∞ < C ∗ , 3 (+) < u(+) < 4 (+)
3 < u < 4 on [0, T ], for ∈ D},
(4.16)
where C ∗ = 1 + h 1 +
3 ∞ + 4 ∞
(4.17)
and is defined in (2.27). We are going to prove that for each solution u of (4.7) the estimate u ∈ cl(0 ) ⇒ u ∈ 0
(4.18)
is true. To this aim, suppose that u is a solution of (4.7) and u ∈ cl(0 ), i.e. u ∞ C ∗ and
3 u4
on [0, T ].
(4.19)
By the Mean Value Theorem, there are i ∈ (ti , ti+1 ), i = 1, 2, . . . , m, such that |u ( i )|( 3 ∞ + 4 ∞ )/. Hence, by (4.8), we get u ∞ < C ∗ ,
(4.20)
where C ∗ is defined in (4.17). It remains to show that 3 < u < 4 on [0, T ] and 3 (+) < u(+) < 4 (+) for ∈ D. Assume the contrary. Then there exists k ∈ {3, 4} such that u( ) = k ( )
for some ∈ [0, T ]
(4.21)
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or u(ti +) = k (ti +)
for some ti ∈ D.
(4.22)
Case A: Let (4.21) hold for k = 4. (i) If = 0, then u(0) = 4 (0) = 4 (T ) = u(T ) = A∗ + q m which gives, in view of (2.3), (4.13) and (4.19), u (0) = u (T ) =
mq = 4 (t) for t ∈ [0, T ]. T
Further, due to (4.9) and (4.12), we can find > 0 such that u > c + 1 on [0, ] and t f(s, u(s), u (s)) ds > 0 for t ∈ [0, ]. u (t) − u (0) = 0
Hence u (t) > u (0)= 4 (t) on (0, ] which implies that u > 4 on (0, ], contrary to (4.19). (ii) If ∈ (ti , ti+1 ) for some ti ∈ D, then u ( ) = 4 ( ) = m q/T = 4 (t) for t ∈ [0, T ] and we reach a contradiction as above. (iii) If = ti ∈ D, then u(ti ) = 4 (ti ) and, by (4.5) and (4.12), u(ti +) = 4 (ti +) = 4 (ti ) − q > c + 1 − q > 1 ∞ + 2 ∞ . By virtue of (4.19) we have u (ti +)4 (ti +) and u (ti )4 (ti ). Now, since the last inequality together with (4.6) and (4.13) yield u (ti +)4 (ti +), we get u (ti +) = 4 (ti +) = m q/T = 4 (t)for t ∈ [0, T ]. Similarly as above, this leads again to a contradiction. Case B: Let (4.22) hold for k = 4, i.e. u(ti +) = 4 (ti +). By (4.5) and (4.12), Ji (u(ti )) = 4 (ti +) = 4 (ti ) − q > A∗ − q, wherefrom, with respect to (4.10), we get u(ti ) > c + 1 and hence Ji (u(ti )) = u(ti ) − q. Therefore u(ti ) = 4 (ti ) and we can continue as in Case A (iii). If (4.21) or (4.22) hold for k = 3, then we use analogical arguments as in Case A or Case B. Step 5 : We prove the second a priori estimate for solutions of (4.7). Define sets
1 ={u ∈ 0 : u(t) > 1 (t) for t ∈ [0, T ], u(+) > 1 (+) for ∈ D}, 2 ={u ∈ 0 : u(t) < 2 (t) for t ∈ [0, T ], u(+) < 2 (+) for ∈ D} and = 0 \cl(1 ∪ 2 ). Then, by (1.1), 1 ∩ 2 = ∅ and = {u ∈ 0 : u satisfies (3.13)}.
(4.23)
Furthermore, with respect to (2.26), (4.16) and (4.11) we have 0 = (3 , 4 , C ∗ ), 1 = (1 , 4 , C ∗ ) and 2 = (3 , 2 , C ∗ ). Consider c from Step 1. We are going to prove that the estimates
) ⇒ u ∞ < c, u ∈ cl(
u ∞ < d
(4.24)
). are valid for each solution u of (4.7). So, assume that u is a solution of (4.7) and u ∈ cl( Then, due to (4.18), u fulfils one of the conditions (3.13), (3.14), (3.15) and so, by (3.18), u ∈ B. Since we have already proved that (3.16) and (3.17) hold, we can use Lemma 3.3
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145
i , i = 1, 2, . . . , m, fulfil (3.5) and get u ∈ [0, T ] such that (3.19) is true. Further, since M and since (2.3), (3.3) and (4.8) are valid, we can apply Lemma 3.1 to show that u satisfies the estimate (3.1). Finally, by [11, Lemma 2.4], u satisfies (3.11) with 0 defined in Step 1. Moreover, let us recall that Ji , i = 1, 2, . . . , m, verify the condition (3.12). Hence, by Lemma 3.2, we have (3.9), i.e. each solution u of (4.7) satisfies (4.24). Step 6. We prove the existence of a solution to problem (2.1)–(2.3). Consider the operator F defined by (4.15). We distinguish two cases: either F has a fixed point in * . or it has no fixed point in * Assume that F u = u for some u ∈ * . Then u is a solution of (4.7) and, with respect to (4.24), we have u ∞ < c, u ∞ < d, which means, by (4.4)–(4.6), that u is a solution of (2.1)–(2.3). Furthermore, due to (4.18), u satisfies (3.14) or (3.15). . Then Fu = u for all u ∈ *0 ∪ *1 ∪ *2 . Now, assume that F u = u for all u ∈ * i , If we replace f, h, Ji , Mi , i = 1, 2, . . . , m, , and , respectively, by f, h, Ji , M i = 1, 2, . . . , m, 3 , 4 and C ∗ in Proposition 2.7, we see that the assumptions (2.22)–(2.25) and (2.27) are satisfied. Thus, by Proposition 2.7, we obtain that F, 0 ) = 1. deg (I − F, (3 , 4 , C ∗ )) = deg (I −
(4.25)
Similarly, we can apply Proposition 2.7 to show that F, 1 ) = 1 deg (I − F, (1 , 4 , C ∗ )) = deg (I −
(4.26)
F, 2 ) = 1. deg (I − F, (3 , 2 , C ∗ )) = deg (I −
(4.27)
and
Using the additivity property of the Leray-Schauder topological degree we derive from (4.25)–(4.27) that F, 1 ) − deg (I − F, 2 ) = −1. ) = deg (I − F, 0 ) − deg (I − deg (I − F,
. By (4.24) we have u ∞ < c and u ∞ < d. This Therefore, F has a fixed point u ∈ together with (4.4)–(4.6) and (4.23) yields that u is a solution to (2.1)–(2.3) fulfilling (3.13). References [1] D. Bainov, P. Simeonov, Impulsive Differential Equations: Periodic Solutions and Applications, Longman Science and Technology, Harlow, 1993. [2] A. Cabada, J.J. Nieto, D. Franco, S.I. Trofimchuk, A generalization of the monotone method for second order periodic boundary value problem with impulses at fixed points, Dynam. Contin. Discrete Impuls. Systems 7 (2000) 145–158. [3] Dong Yujun, Periodic solutions for second order impulsive differential systems, Nonlinear Anal. 27 (1996) 811–820. [4] L.H. Erbe, Liu Xinzhi, Existence results for boundary value problems of second order impulsive differential equations, J. Math. Anal. Appl. 149 (1990) 56–69. [5] Hu Shouchuan, V. Laksmikantham, Periodic boundary value problems for second order impulsive differential systems, Nonlinear Anal. 13 (1989) 75–85. [6] E. Liz, J.J. Nieto, Periodic solutions of discontinuous impulsive differential systems, J. Math. Anal. Appl. 161 (1991) 388–394.
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[7] E. Liz, J.J. Nieto, The monotone iterative technique for periodic boundary value problems of second order impulsive differential equations, Comment. Math. Univ. Carolin. 34 (1993) 405–411. [8] J. Mawhin, Topological degree and boundary value problems for nonlinear differential equations, in: M. Furi, P. Zecca (Eds.), Topological Methods for Ordinary Differential Equations, Lecture Notes in Mathematics, vol. 1537, Springer, Berlin, 1993, pp. 73–142. [9] I. Rach˚unková, M. Tvrdý, Impulsive periodic boundary value problem and topological degree, Funct. Differ. Equ. 9 (3–4) (2002) 471–498. [10] I. Rach˚unková, M. Tvrdý, Nonmonotone impulse effects in second order periodic boundary value problems, Abstr. Anal. Appl. 2004 (2004) 577–590. [11] I. Rach˚unková, M. Tvrdý, Non-ordered lower and upper functions in second order impulsive periodic problems, Dynam. Contin. Discrete Impuls. Systems, to appear. [12] Zang Zhitao, Existence of solutions for second order impulsive differential equations, Appl. Math. Ser. B (Engl. Ed.) 12 (1997) 307–320.
Existence results for impulsive second-order periodic problems Irena Rach˚unkováa,1 , Milan Tvrdýb,∗,2 a Department of Mathematics, Palacký University, 779 00 Olo-Mouc, Tomkova 40, Czech Republic b Mathematical Institute, Academy of Sciences of the Czech Republic, Zitna 25, 115 67 Praha, Czech Republic
Received 28 January 2004; accepted 15 July 2004
Abstract This paper provides existence results for the nonlinear impulsive periodic boundary value problem u = f (t, u, u ), u(ti +) = Ji (u(ti )), u(0) = u(T ),
u (ti +) = Mi (u (ti )), i = 1, 2, . . . , m,
u (0) = u (T ),
where f ∈ Car([0, T ]× R2 ) and Ji , Mi ∈ C(R). The basic assumption is the existence of lower/upper functions 1 /2 associated with the problem. Here, we generalize and extend the existence results of our previous papers. 䉷 2004 Elsevier Ltd. All rights reserved. MSC: 34B37; 34B15; 34C25 Keywords: Second-order nonlinear ordinary differential equation with impulses; Periodic solutions; Lower and upper functions; Leray–Schauder topological degree; A priori estimates
∗ Corresponding author. Tel.: +420-222090762; fax: +420-222211638.
E-mail addresses: [email protected] (I. Rach˚unková), [email protected] (M. Tvrdý). 1 Supported by the Grant no. 201/04/1077 of the Grant Agency of the Czech Republic and by the Council of
Czech Government J14/98:153100011. 2 Supported by the Grant no. 201/01/1077 of the Grant Agency of the Czech Republic.
0362-546X/$ - see front matter 䉷 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.07.006
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1. Introduction This paper deals with the solvability of the nonlinear impulsive boundary value problem (2.1)–(2.3) and provides conditions for f, Ji and Mi , i = 1, 2, . . . , m, which guarantee the existence of at least one solution. Boundary value problems of this kind have received a considerable attention, see e.g. [1–7,10,12]. The results of these papers rely on the existence of a well-ordered pair 1 2 of lower/upper functions associated with the problem under consideration. In [11], we extended these results to the case that 1 /2 are not well-ordered, i.e.
1 () > 2 () for some ∈ [0, T ].
(1.1)
The goal of this paper is to generalize the main existence results of [11], where we restricted our attention to impulsive functions Mi , i = 1, 2, . . . , m, fulfilling the conditions yMi (y)0
for y ∈ R, i = 1, 2, . . . , m.
(1.2)
Here we prove existence criteria without restriction (1.2). Throughout the paper we keep the following notation and conventions: real-valued function u defined a.e. on [0, T ], we put T u ∞ = sup ess |u(t)| and u 1 = |u(s)| ds. t∈[0,T ]
For
a
0
For a given interval J ⊂ R, by C(J ) we denote the set of real-valued functions which are continuous on J. Furthermore, C1 (J ) is the set of functions having continuous first derivatives on J and L(J ) is the set of functions which are Lebesgue integrable on J. Let m ∈ N and let 0=t0 < t1 < t2 < · · · < tm < tm+1 =T be a division of the interval [0, T ]. We denote D = {t1 , t2 , . . . , tm } and define C1D [0, T ] as the set of functions u : [0, T ] → R of the form u (t) if t ∈ [0, t1 ], [0] u[1] (t) if t ∈ [t1 , t2 ], u(t) = . ... .. u[m] (t) if t ∈ [tm , T ], where u[i] ∈ C1 [ti , ti+1 ] for i = 0, 1, . . . , m. Moreover, AC1D [0, T ] stands for the set of functions u ∈ C1D [0, T ] having first derivatives absolutely continuous on each subinterval (ti , ti+1 ), i = 0, 1, . . . , m. For u ∈ C1D [0, T ] and i = 1, 2, . . . , m + 1 we define u (ti ) = u (ti −) = lim u (t), t→ti −
u (0) = u (0+) = lim u (t) t→0+
(1.3)
and u D = u ∞ + u ∞ . Note that the set C1D [0, T ] becomes a Banach space when equipped with the norm . D and with the usual algebraic operations. We say that f : [0, T ] × R2 → R satisfies the Carathéodory conditions on [0, T ] × R2 if (i) for each x ∈ R and y ∈ R the function f (., x, y) is measurable on [0, T ]; (ii) for almost every t ∈ [0, T ] the function f (t, ., .) is continuous on R2 ; (iii) for each compact set K ⊂ R2 there is a function mK (t) ∈ L[0, T ] such that |f (t, x, y)|mK (t) holds for a.e.
I. Rach˚unková, M. Tvrdý / Nonlinear Analysis 59 (2004) 133 – 146
135
t ∈ [0, T ] and all (x, y) ∈ K. The set of functions satisfying the Carathéodory conditions on [0, T ] × R2 will be denoted by Car([0, T ] × R2 ). Given a Banach space X and its subset M, let cl(M) and *M denote the closure and the boundary of M, respectively. Let be an open bounded subset of X. Assume that the operator F : cl() → X is completely continuous and Fu = u for all u ∈ * . Then deg (I − F, ) denotes the Leray–Schauder topological degree of I − F with respect to , where I is the identity operator on X. For the definition and properties of the degree see e.g. [8]. 2. Formulation of the problem and main assumptions Here we study the existence of solutions to the problem u = f (t, u, u ),
(2.1)
u(ti +) = Ji (u(ti )), u(0) = u(T ),
u (ti +) = Mi (u (ti )),
i = 1, 2, . . . , m,
u (0) = u (T ),
(2.2) (2.3)
where u (ti ) are understood in the sense of (1.3), f ∈ Car([0, T ] × R2 ), Ji ∈ C(R) and Mi ∈ C(R). Definition 2.1. A solution of problem (2.1)–(2.3) is a function u ∈ AC1D [0, T ] which satisfies impulsive conditions (2.2), periodic conditions (2.3) and for a.e. t ∈ [0, T ] fulfils Eq. (2.1). Definition 2.2. A function 1 ∈ AC1D [0, T ] is called a lower function of problem (2.1)–(2.3) if
1 (t)f (t, 1 (t), 1 (t)) for a.e. t ∈ [0, T ], 1 (ti +) = Ji (1 (ti )), 1 (0) = 1 (T ),
1 (ti +)Mi (1 (ti )), i = 1, 2, . . . , m,
1 (0)1 (T ).
(2.4) (2.5) (2.6)
Similarly, a function 2 ∈ AC1D [0, T ] is an upper function of problem (2.1)–(2.3) if
2 (t)f (t, 2 (t), 2 (t)) for a.e. t ∈ [0, T ], 2 (ti +) = Ji (2 (ti )), 2 (0) = 2 (T ),
2 (ti +)Mi (2 (ti )), i = 1, 2, . . . , m,
2 (0)2 (T ).
Assumptions 2.3. In the paper we work with the following assumptions: 0 = t0 < t1 < · · · < tm < tm+1 = T < ∞, D = {t1 , t2 , . . . , tm }, f ∈ Car([0, T ] × R2 ), Ji ∈ C(R), Mi ∈ C(R), i = 1, 2, . . . , m;
1 and 2 are, respectively, lower and upper functions of (2.1)–(2.3);
(2.7) (2.8) (2.9)
(2.10) (2.11)
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x > 1 (ti ) ⇒ Ji (x) > Ji (1 (ti )), x < 2 (ti ) ⇒ Ji (x) < Ji (2 (ti )),
i = 1, 2, . . . , m;
(2.12)
i = 1, 2, . . . , m.
(2.13)
y 1 (ti ) ⇒ Mi (y)Mi (1 (ti )), y 2 (ti ) ⇒ Mi (y)Mi (2 (ti )),
Operator reformulation of (2.1)–(2.3). By G(t, s) we denote the Green function of the Dirichlet boundary value problem u = 0, u(0) = u(T ) = 0, i.e., t (s − T ) if 0t s T , (2.14) G(t, s) = s (t T− T ) if 0s < t T . T Furthermore, we define the operator F : C1D [0, T ] → C1D [0, T ] by T (Fx)(t)=x(0) + x (0) − x (T ) + G(t, s) f (s, x(s), x (s)) ds −
m *G i=1
*s
0
m (t, ti )(Ji (x(ti ))−x(ti ))+ G(t, ti )(Mi (x (ti ))−x (ti )). (2.15) i=1
As in [9, Lemma 3.1], where m = 1, we can prove (see Proposition 2.6 below) that F is completely continuous and that a function u is a solution of (2.1)–(2.3) if and only if u is a fixed point of F. To this aim we need the following lemma which extends Lemma 2.1 from [9]. Lemma 2.5. For each h ∈ L[0, T ], c, di , ei ∈ R, i =1, 2, . . . , m, there is a unique function x ∈ AC1D [0, T ] fulfilling x (t) = h(t) a.e. on [0, T ], (2.16) x(ti +) − x(ti ) = di , x (ti +) − x (ti ) = ei , i = 1, 2, . . . , m, x(0) = x(T ) = c.
(2.17)
This function is given by T m m *G (t, ti ) di + G(t, s) h(s) ds − G(t, ti ) ei x(t)=c + *s 0 i=1 i=1 for t ∈ [0, T ],
(2.18)
where G(t, s) is defined by (2.14). Proof. It is easy to check that x ∈ AC1D [0, T ] fulfils (2.16) together with x(0) = c if and only if there is c ∈ R such that x(t)=c + t c+
t
+ 0
m i=1
(ti ,T ] (t) di +
(t − s) h(s) ds
m i=1
(ti ,T ] (t) (t − ti ) ei
for t ∈ [0, T ],
(2.19)
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137
where (ti ,T ] (t) = 1 if t ∈ (ti , T ] and (ti ,T ] (t) = 0 if t ∈ R\(ti , T ]. Furthermore, x(T ) = c if and only if T m m T −s di T − t i h(s) ds. (2.20) ei − − c=− T T T 0 i=1
i=1
Inserting (2.20) into (2.19), we get x(t)=
ti (t − T ) t (ti − T ) (t − T ) t ei + ei − di − di T T T T ti
Hence, taking into account (2.14), we conclude that the function x given by (2.18) is the unique solution of (2.16), (2.17) in AC1D [0, T ]. Proposition 2.6. Assume that (2.10) holds. Let the operator F : C1D [0, T ] → C1D [0, T ] be defined by (2.14) and (2.15). Then F is completely continuous and a function u is a solution of (2.1)–(2.3) if and only if u = Fu. Proof. Choose an arbitrary y ∈ C1D [0, T ] and put h(t) = f (t, y(t), y (t)) for a.e. t ∈ [0, T ], di = Ji (y(ti )) − y(ti ), ei = Mi (y (ti )) − y (ti ), i = 1, 2, = . . . , m, c = y(0) + y (0) − y (T ).
(2.21)
Then h ∈ L[0, T ], c, di , ei ∈ R i = 1, 2, . . . , m. By Lemma 2.5, there is a unique x ∈ AC1D [0, T ] fulfilling (2.16), (2.17) and it is given by (2.18). Due to (2.21), we have x(t) = (Fy)(t) for t ∈ [0, T ]. Therefore, u ∈ C1D [0, T ] is a solution to (2.1)–(2.3) if and only if uF u. Define an operator F1 : C1D [0, T ] → C1D [0, T ] by T (F1 y)(t) = G(t, s) f (s, y(s), y (s)) ds, t ∈ [0, T ]. 0
As F1 is a composition of the Green type operator for the Dirichlet problem u = 0, u(0) = u(T ) = 0, and of the superposition operator generated by f ∈ Car([0, T ] × R2 ), making use of the Lebesgue Dominated Convergence Theorem and the Arzelà–Ascoli Theorem, we get in a standard way that F1 is completely continuous. Since Ji , Mi , i = 1, 2, . . . , m, are continuous, the operator F2 = F − F1 is continuous, as well. Having in mind that F2 maps bounded sets onto bounded sets and its values are contained in a (2m + 1)-dimensional subspace of C1D [0, T ], we conclude that the operators F2 and F = F1 + F2 are completely continuous. In the proof of our main result we will need the next proposition which concerns the case of well-ordered lower/upper functions and which follows from [10, Corollary 3.5].
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Proposition 2.7. Assume that (2.10) holds and let and be respectively lower and upper functions of (2.1)–(2.3) such that
(t) < (t) f or t ∈ [0, T ] and
(+) < (+) f or ∈ D,
(ti ) < x < (ti ) ⇒ Ji ((ti )) < Ji (x) < Ji ((ti )), and
y (ti ) ⇒ Mi (y)Mi ( (ti )), y (ti ) ⇒ Mi (y)Mi ( (ti )),
i = 1, 2, . . . , m
(2.22) (2.23)
(2.24)
i = 1, 2, . . . , m.
Further, let h ∈ L[0, T ] be such that |f (t, x, y)|h(t)
f or a.e. t ∈ [0, T ] and all (x, y) ∈ [(t), (t)] × R
(2.25)
and let the operator F be defined by (2.15). Finally, for ∈ (0, ∞) denote
(, , )={u ∈ C1D [0, T ] : (t) < u(t) < (t) f or t ∈ [0, T ], (+) < u(+) < (+) f or ∈ D, u ∞ < }.
(2.26)
Then deg (I − F, (, , )) = 1 whenever Fu = u on *(, , ) and
> h 1 +
∞ + ∞ ,
where =
min
i=1,2,...,m+1
(ti − ti−1 ).
(2.27)
Proof. Using the mean value theorem, we can show that u ∞ h 1 +
∞ + ∞
(2.28)
holds for each u ∈ C1D [0, T ] fulfilling (t) < u(t) < (t) for t ∈ [0, T ] and (+) < u(+) < (+) for ∈ D. Thus, if we denote by c the right-hand side of (2.28), we can follow the proof of [10, Corollary 3.5]. 3. A priori estimates In Section 3, we will need a priori estimates which are contained in Lemmas 3.1–3.3. Lemma 3.1. Let 1 ∈ (0, ∞), h ∈ L[0, T ], Mi ∈ C(R), i = 1, 2, . . . , m. Then there exists d ∈ ( 1 , ∞) such that the estimate u ∞ < d
(3.1)
i ∈ C(R), i = 1, 2, . . . , m, satisfying (2.3), is valid for each u ∈ AC1D [0, T ] and each M |u ( u )| < 1
f or some u ∈ [0, T ],
i (u (ti )), u (ti +) = M
i = 1, 2, . . . , m,
|u (t)| < h(t) f or a.e. t ∈ [0, T ]
(3.2) (3.3) (3.4)
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139
and i (y)| : |y| < a} < b sup {|Mi (y)| : |y| < a}
(3.5)
i ∈ C(R), i = 1, 2, . . . , m, satisfy (2.3) and Proof. Suppose that u ∈ AC1D [0, T ] and M (3.2)–(3.5). Due to (2.3), we can assume that u ∈ (0, T ], i.e. there is j ∈ {1, 2, ..., m + 1} such that u ∈ (tj −1 , tj ]. We will distinguish 3 cases: either j = 1 or j = m + 1 or 1 < j < m + 1. Let j = 1. Then, using (3.2) and (3.4), we obtain |u (t)| < a1
on [0, t1 ],
(3.6)
where a1 = 1 + h 1 . Since M1 ∈ C(R), we can find b1 (a1 ) ∈ (a1 , ∞) such that |M1 (y)| < b1 (a1 ) for all y ∈ (−a1 , a1 ). Hence, in view of (3.3) and (3.5), we have h 1 for |u (t1 +)| < b1 (a1 ), wherefrom, using (3.4), we deduce that |u (t)| < b1 (a1 ) + t ∈ (t1 , t2 ]. Continuing by induction, we get bi (ai ) ∈ (ai , ∞) such that |u (t)| < ai+1 = bi (ai ) + h 1 on (ti , ti+1 ] for i = 2, . . . , m, i.e. u ∞ < d := max{ai : i = 1, 2, . . . , m + 1}.
(3.7)
Assume that j = m + 1. Then, using (3.2) and (3.4), we obtain |u (t)| < am+1
on (tm , T ],
(3.8)
where am+1 = 1 + h 1 . Furthermore, due to (2.3), we have |u (0)| < am+1 which together h 1 . Now, proceeding as in the case with (3.4) yields that (3.6) is true with a1 = am+1 + j = 1, we show that (3.7) is true also in the case j = m + 1. Assume that 1 < j < m + 1. Then (3.2) and (3.4) yield |u (t)| < aj +1 = 1 + h 1 h 1 on (tj +1 , tj +2 ], where on (tj , tj +1 ]. If j < m, then |u (t)| < aj +2 = bj +1 (aj +1 ) + bj +1 (aj +1 ) > aj +1 . Proceeding by induction we get (3.8) with am+1 = bm (am ) + h 1 and bm (am ) > am , wherefrom (3.7) again follows as in the previous case. Lemma 3.2. Let 0 , d, q ∈ (0, ∞) and Ji ∈ C(R), i = 1, 2, . . . , m. Then there exists c ∈ ( 0 , ∞) such that the estimate u ∞ < c
(3.9)
is valid for each u ∈ C1D [0, T ] and each Ji ∈ C(R), i = 1, 2, . . . , m, satisfying (2.3), (3.1), u(ti +) = Ji (u(ti )), |u(u )| < 0
i = 1, 2, . . . , m,
f or some u ∈ [0, T ]
(3.10) (3.11)
and sup {|Ji (x)|:|x| < a}
(3.12)
Proof. We will argue similarly as in the proof of Lemma 3.1. Suppose that u ∈ C1D [0, T ] satisfies (2.3), (3.1), (3.10), (3.11) and that Ji ∈ C(R), i = 1, 2, . . . , m, satisfy (3.12).
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Due to (2.3) we can assume that u ∈ (0, T ], i.e. there is j ∈ {1, 2, . . . , m + 1} such that u ∈ (tj −1 , tj ]. We will consider three cases: j = 1, j = m + 1, 1 < j < m + 1. If j = 1, then (3.1) and (3.11) yield |u(t)| < a1 = 0 + d T on [0, t1 ]. In particular, |u(t1 )| < a1 . Since J1 ∈ C(R), we can find b1 (a1 ) ∈ (a1 +q, ∞) such that |J1 (x)| < b1 (a1 ) for all x ∈ (−a1 , a1 ) and consequently, by (3.12), also | J1 (x)| < b1 (a1 )for all x ∈ (−a1 , a1 ). Therefore, by (3.1), |u(t)| < |u(t1 +)| + d T = | J1 (u(t1 ))| + d T < a2 = b1 (a1 ) + d T on (t1 , t2 ]. Proceeding by induction we get bi (ai ) ∈ (ai + q, ∞) such that |u(t)| < ai+1 = bi (ai ) + d T for t ∈ (ti , ti+1 ] and i = 2, . . . , m. As a result, (3.9) is true with c = max{ai : i = 1, 2, . . . , m + 1}. Analogously we would proceed in the remaining cases j = m + 1 or 1 < j < m + 1. Finally, we will need two estimates for functions u satisfying one of the following conditions: u(su ) < 1 (su ) and u1 on [0, T ] and u2 on [0, T ]
and
u(tu ) > 2 (tu ) for some su , tu ∈ [0, T ],
(3.13)
inf
|u(t) − 1 (t)| = 0,
(3.14)
inf
|u(t) − 2 (t)| = 0.
(3.15)
t∈[0,T ] t∈[0,T ]
i ∈ C(R), i = 1, 2, . . . , m, Lemma 3.3. Assume that 1 , 2 ∈ AC1D [0, T ], Ji , Mi , Ji , M satisfy (2.12), (2.13) and Ji (1 (ti )) = Ji (1 (ti )), x > 1 (ti ) ⇒ Ji (x) > (3.16) x < 2 (ti ) ⇒ Ji (x) < Ji (2 (ti )) = Ji (2 (ti )), i = 1, 2, . . . , m and
i (y)Mi ( (ti )), y 1 (ti ) ⇒ M 1 i (y)Mi ( (ti )), y 2 (ti ) ⇒ M 2
i = 1, 2, . . . , m.
(3.17)
Define B={u ∈ C1D [0, T ] : u satisf ies (2.3), (3.10), (3.3) and one of the conditions (3.13), (3.14), (3.15)}. Then each function u ∈ B satisfies |u ( u )| < 1 f or some u ∈ [0, T ], where 2 1 = ( 1 ∞ + 2 ∞ ) + 1 ∞ + 2 ∞ + 1. t1
(3.18)
(3.19)
Proof. This lemma can be proved by the same arguments as Lemma 2.3 in [11] with the i (u (ti )) in place of Mi (u (ti )) and that we use (3.17) instead only difference that we write M of (3.13). 4. Main results Our main result consists in a generalization of [11, Theorem 3.1]. Particularly, we remove condition (1.2) which was assumed in [11] and prove the following theorem.
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141
Theorem 4.1. Assume that (2.10)–(2.13) and (1.1) hold and let h ∈ L[0, T ] be such that |f (t, x, y)|h(t)
f or a.e. t ∈ [0, T ] and all (x, y) ∈ R2 .
(4.1)
Then problem (2.1)–(2.3) has a solution u satisfying one of the conditions (3.13)–(3.15). Proof. Step 1 : We construct a proper auxiliary problem. Let 1 and 2 be, respectively, lower and upper functions of (2.1)–(2.3) and let 1 be associated with them as in (3.19). Put h(t) = 2 h(t) + 1 for a.e. t ∈ [0, T ] and = 1 +
m
(|Mi (1 (ti ))| + |Mi (2 (ti ))|).
i=1
, ∞) satisfying (3.1). Furthermore, put 0 = 1 ∞ + 2 ∞ +1 By Lemma 3.1, find d ∈ ( and
m T (4.2) max max |Mi (y)|, d + 1 q= |y|d+1 m i=1
and, by Lemma 3.2, find c ∈ ( 0 + q, ∞) fulfilling (3.9). In particular, we have c > 1 ∞ + 2 ∞ + q + 1,
d > 1 ∞ + 2 ∞ + 1.
Finally, for a.e. t ∈ [0, T ] and all x, y ∈ R and i = 1, 2, . . . , m, define functions f (t, x, y) − h(t) − 1 if x − c − 1, f (t, x, y) + (x + c) (h(t) + 1) if − c − 1 < x < − c, f(t, x, y) = f (t, x, y) if − cx c, f (t, x, y) + (x − c) (h(t) + 1) if c < x < c + 1, f (t, x, y) + h(t) + 1 if x c + 1, x+q if x − c − 1, Ji (−c) (c + 1 + x) − (x + q) (x + c) if − c − 1 < x < − c, Ji (x) = Ji (x) if − cx c, J (c) (c + 1 − x) + (x − q) (x − c) if c < x < c + 1, i x−q if x c + 1, y if y − d − 1, (−d) (d + 1 + y) − y (y + d) if − d − 1 < y < − d, M i i (y) = Mi (y) M if − d y d, if d < y < d + 1, Mi (d) (d + 1 − y) + y (y − d) y if y d + 1
(4.3)
(4.4)
(4.5)
(4.6)
and consider the auxiliary problem u = f(t, u, u ),
(3.10),
(3.3),
(2.3).
(4.7)
i ∈ C(R) for i = 1, 2, . . . , m. According to Ji , M Due to (2.10), f ∈ Car([0, T ] × R) and (4.3)–(4.6) the functions 1 and 2 are, respectively, lower and upper functions of (4.7). By (4.1) we have h(t) for a.e. t ∈ [0, T ] and all (x, y) ∈ R2 |f(t, x, y)|
(4.8)
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and
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f(t, x, y) < 0 f(t, x, y) > 0
for a.e. t ∈ [0, T ] and all (x, y) ∈ (−∞, −c − 1] × R, for a.e. t ∈ [0, T ] and all (x, y) ∈ [c + 1, ∞) × R.
(4.9)
i satisfy the assumptions of Lemmas 3.1–3.3. Ji and M Step 2 : We show that Choose an arbitrary i ∈ {1, 2, . . . , m}. (i) Condition (3.5): Let a ∈ (0, ∞), b ∈ (a, ∞) and Mi∗ = sup{|Mi (y)| : |y| < a} < b. i (y)| : |y| < a} max{a, M ∗ } < b. Then, by (4.6), we have sup{|M i (ii) Condition (3.12): Let a ∈ (0, ∞), b ∈ (a +q, ∞) and Ji∗ =sup{|Ji (x)| : |x| < a} < b. Then, by (4.5), we have sup{| Ji (x)| : |x| < a} max{a + q, Ji∗ } < b. (iii) Condition (3.16): Due to (2.12), (4.3) and (4.5), we see that (3.16) holds if |x|c. Assume that x > c. Then x > max{1 (ti ), 2 (ti )} which means that the second condition in (3.16) need not be considered in this case. Since |1 (ti )| < c, we have Ji (1 (ti )) = Ji (1 (ti )). Furthermore, due to (4.3), x − q > 1 ∞ + 2 ∞ + 1. If x c + 1, then Ji (x) = x − q > 1 (ti +) = Ji (1 (ti )). Finally, if x ∈ (c, c + 1), then Ji (x) = Ji (c) (c + 1 − x) + (x − q) (x − c) > Ji (1 (ti ))because Ji (c) > Ji (1 (ti )) by (2.12). For x < (∞, −c) we can argue similarly. (iv) Condition (3.17): Due to (2.13), (4.3) and (4.6), we see that (3.17) holds for |y| < d. Assume that y > d. Then y > max{1 (ti ), 2 (ti )} which means that the first condition in i (y) = (3.17) need not be considered in this case. Since d > > Mi (2 (ti )), we have M i (y) = Mi (d) (d + 1 − y) + y (y − d) > Mi ( (ti )) if y > Mi (2 (ti )) if y > d + 1 and M 2 y ∈ (d, d + 1). Hence the second condition in (3.17) is satisfied for y ∈ (d, ∞). Similarly we can verify the first condition in (3.17) for y ∈ (−∞, −d). Step 3 : We construct a well-ordered pair of lower/upper functions for (4.7). Put A∗ = q +
m i=1
and
max | Ji (x)|
|x|c+1
4 (0) = A∗ + m q, 4 (t) = A∗ + (m − i) q + mTq t for t ∈ (ti , ti+1 ], i = 0, 1, . . . , m, 3 (t) = −4 (t) for t ∈ [0, T ].
(4.10)
(4.11)
Then 3 , 4 ∈ AC1D [0, T ] and, by (4.5) and (4.10),
3 (t) < − A∗ < − c − 1,
4 (t) > A∗ > c + 1 for t ∈ [0, T ].
(4.12)
In view of (4.2), mq mq − (d + 1) and 4 (t) = d + 1 for t ∈ [0, T ]. (4.13) T T Now, we prove that 4 is an upper function of (4.7): By (4.9) and (4.12), we have 0 = 4 (t) < f(t, 4 (t), 4 (t)) for a.e. t ∈ [0, T ]. Furthermore, by (4.5), mq ti = 4 (ti ) − q = 4 (ti +) = A∗ + (m − i) q + Ji (4 (ti )). T
3 (t) = −
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143
By virtue of (4.2) and (4.6), we get
4 (ti +) =
mq i (4 (ti )) for i = 1, 2, . . . , m. = 4 (ti ) = M T
Finally, 4 (0) = A∗ + m q = 4 (T ) and 4 (0) = m q/T = 4 (T ), i.e. 4 is an upper function of (4.7). Since 3 = −4 , we can see that 3 is a lower function of (4.7). Clearly,
3 < 4 on [0, T ] and 3 (+) < 4 (+) for ∈ D.
(4.14)
Having G from (2.15), define an operator F : C1D [0, T ] → C1D [0, T ] by T (Fu)(t)=u(0) + u (0) − u (T ) + G(t, s) f(s, u(s), u (s)) ds −
m *G i=1
+
m
*s
0
(t, ti )( Ji (u(ti )) − u(ti ))
i (u (ti )) − u (ti )), G(t, ti )(M
t ∈ [0, T ].
(4.15)
i=1
By Proposition 2.6, F is completely continuous and u is a solution of (4.7) whenever Fu = u. Step 4 : We prove the first a priori estimate for solutions of (4.7). Define
0 ={u ∈ C1D [0, T ] : u ∞ < C ∗ , 3 (+) < u(+) < 4 (+)
3 < u < 4 on [0, T ], for ∈ D},
(4.16)
where C ∗ = 1 + h 1 +
3 ∞ + 4 ∞
(4.17)
and is defined in (2.27). We are going to prove that for each solution u of (4.7) the estimate u ∈ cl(0 ) ⇒ u ∈ 0
(4.18)
is true. To this aim, suppose that u is a solution of (4.7) and u ∈ cl(0 ), i.e. u ∞ C ∗ and
3 u4
on [0, T ].
(4.19)
By the Mean Value Theorem, there are i ∈ (ti , ti+1 ), i = 1, 2, . . . , m, such that |u ( i )|( 3 ∞ + 4 ∞ )/. Hence, by (4.8), we get u ∞ < C ∗ ,
(4.20)
where C ∗ is defined in (4.17). It remains to show that 3 < u < 4 on [0, T ] and 3 (+) < u(+) < 4 (+) for ∈ D. Assume the contrary. Then there exists k ∈ {3, 4} such that u( ) = k ( )
for some ∈ [0, T ]
(4.21)
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or u(ti +) = k (ti +)
for some ti ∈ D.
(4.22)
Case A: Let (4.21) hold for k = 4. (i) If = 0, then u(0) = 4 (0) = 4 (T ) = u(T ) = A∗ + q m which gives, in view of (2.3), (4.13) and (4.19), u (0) = u (T ) =
mq = 4 (t) for t ∈ [0, T ]. T
Further, due to (4.9) and (4.12), we can find > 0 such that u > c + 1 on [0, ] and t f(s, u(s), u (s)) ds > 0 for t ∈ [0, ]. u (t) − u (0) = 0
Hence u (t) > u (0)= 4 (t) on (0, ] which implies that u > 4 on (0, ], contrary to (4.19). (ii) If ∈ (ti , ti+1 ) for some ti ∈ D, then u ( ) = 4 ( ) = m q/T = 4 (t) for t ∈ [0, T ] and we reach a contradiction as above. (iii) If = ti ∈ D, then u(ti ) = 4 (ti ) and, by (4.5) and (4.12), u(ti +) = 4 (ti +) = 4 (ti ) − q > c + 1 − q > 1 ∞ + 2 ∞ . By virtue of (4.19) we have u (ti +)4 (ti +) and u (ti )4 (ti ). Now, since the last inequality together with (4.6) and (4.13) yield u (ti +)4 (ti +), we get u (ti +) = 4 (ti +) = m q/T = 4 (t)for t ∈ [0, T ]. Similarly as above, this leads again to a contradiction. Case B: Let (4.22) hold for k = 4, i.e. u(ti +) = 4 (ti +). By (4.5) and (4.12), Ji (u(ti )) = 4 (ti +) = 4 (ti ) − q > A∗ − q, wherefrom, with respect to (4.10), we get u(ti ) > c + 1 and hence Ji (u(ti )) = u(ti ) − q. Therefore u(ti ) = 4 (ti ) and we can continue as in Case A (iii). If (4.21) or (4.22) hold for k = 3, then we use analogical arguments as in Case A or Case B. Step 5 : We prove the second a priori estimate for solutions of (4.7). Define sets
1 ={u ∈ 0 : u(t) > 1 (t) for t ∈ [0, T ], u(+) > 1 (+) for ∈ D}, 2 ={u ∈ 0 : u(t) < 2 (t) for t ∈ [0, T ], u(+) < 2 (+) for ∈ D} and = 0 \cl(1 ∪ 2 ). Then, by (1.1), 1 ∩ 2 = ∅ and = {u ∈ 0 : u satisfies (3.13)}.
(4.23)
Furthermore, with respect to (2.26), (4.16) and (4.11) we have 0 = (3 , 4 , C ∗ ), 1 = (1 , 4 , C ∗ ) and 2 = (3 , 2 , C ∗ ). Consider c from Step 1. We are going to prove that the estimates
) ⇒ u ∞ < c, u ∈ cl(
u ∞ < d
(4.24)
). are valid for each solution u of (4.7). So, assume that u is a solution of (4.7) and u ∈ cl( Then, due to (4.18), u fulfils one of the conditions (3.13), (3.14), (3.15) and so, by (3.18), u ∈ B. Since we have already proved that (3.16) and (3.17) hold, we can use Lemma 3.3
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i , i = 1, 2, . . . , m, fulfil (3.5) and get u ∈ [0, T ] such that (3.19) is true. Further, since M and since (2.3), (3.3) and (4.8) are valid, we can apply Lemma 3.1 to show that u satisfies the estimate (3.1). Finally, by [11, Lemma 2.4], u satisfies (3.11) with 0 defined in Step 1. Moreover, let us recall that Ji , i = 1, 2, . . . , m, verify the condition (3.12). Hence, by Lemma 3.2, we have (3.9), i.e. each solution u of (4.7) satisfies (4.24). Step 6. We prove the existence of a solution to problem (2.1)–(2.3). Consider the operator F defined by (4.15). We distinguish two cases: either F has a fixed point in * . or it has no fixed point in * Assume that F u = u for some u ∈ * . Then u is a solution of (4.7) and, with respect to (4.24), we have u ∞ < c, u ∞ < d, which means, by (4.4)–(4.6), that u is a solution of (2.1)–(2.3). Furthermore, due to (4.18), u satisfies (3.14) or (3.15). . Then Fu = u for all u ∈ *0 ∪ *1 ∪ *2 . Now, assume that F u = u for all u ∈ * i , If we replace f, h, Ji , Mi , i = 1, 2, . . . , m, , and , respectively, by f, h, Ji , M i = 1, 2, . . . , m, 3 , 4 and C ∗ in Proposition 2.7, we see that the assumptions (2.22)–(2.25) and (2.27) are satisfied. Thus, by Proposition 2.7, we obtain that F, 0 ) = 1. deg (I − F, (3 , 4 , C ∗ )) = deg (I −
(4.25)
Similarly, we can apply Proposition 2.7 to show that F, 1 ) = 1 deg (I − F, (1 , 4 , C ∗ )) = deg (I −
(4.26)
F, 2 ) = 1. deg (I − F, (3 , 2 , C ∗ )) = deg (I −
(4.27)
and
Using the additivity property of the Leray-Schauder topological degree we derive from (4.25)–(4.27) that F, 1 ) − deg (I − F, 2 ) = −1. ) = deg (I − F, 0 ) − deg (I − deg (I − F,
. By (4.24) we have u ∞ < c and u ∞ < d. This Therefore, F has a fixed point u ∈ together with (4.4)–(4.6) and (4.23) yields that u is a solution to (2.1)–(2.3) fulfilling (3.13). References [1] D. Bainov, P. Simeonov, Impulsive Differential Equations: Periodic Solutions and Applications, Longman Science and Technology, Harlow, 1993. [2] A. Cabada, J.J. Nieto, D. Franco, S.I. Trofimchuk, A generalization of the monotone method for second order periodic boundary value problem with impulses at fixed points, Dynam. Contin. Discrete Impuls. Systems 7 (2000) 145–158. [3] Dong Yujun, Periodic solutions for second order impulsive differential systems, Nonlinear Anal. 27 (1996) 811–820. [4] L.H. Erbe, Liu Xinzhi, Existence results for boundary value problems of second order impulsive differential equations, J. Math. Anal. Appl. 149 (1990) 56–69. [5] Hu Shouchuan, V. Laksmikantham, Periodic boundary value problems for second order impulsive differential systems, Nonlinear Anal. 13 (1989) 75–85. [6] E. Liz, J.J. Nieto, Periodic solutions of discontinuous impulsive differential systems, J. Math. Anal. Appl. 161 (1991) 388–394.
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