D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA

Σχετικά έγγραφα
Example Sheet 3 Solutions

Matrices and Determinants

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

2 Composition. Invertible Mappings

Uniform Convergence of Fourier Series Michael Taylor

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Other Test Constructions: Likelihood Ratio & Bayes Tests

C.S. 430 Assignment 6, Sample Solutions

Memoirs on Differential Equations and Mathematical Physics

Congruence Classes of Invertible Matrices of Order 3 over F 2

Areas and Lengths in Polar Coordinates

Homomorphism in Intuitionistic Fuzzy Automata

Homework 3 Solutions

Areas and Lengths in Polar Coordinates

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

SCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018

Section 8.3 Trigonometric Equations

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Statistical Inference I Locally most powerful tests

Coefficient Inequalities for a New Subclass of K-uniformly Convex Functions

Concrete Mathematics Exercises from 30 September 2016

A Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering

Every set of first-order formulas is equivalent to an independent set

Second Order Partial Differential Equations

Partial Differential Equations in Biology The boundary element method. March 26, 2013

EE512: Error Control Coding

D Alembert s Solution to the Wave Equation

Solution Series 9. i=1 x i and i=1 x i.

ORDINAL ARITHMETIC JULIAN J. SCHLÖDER

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

ST5224: Advanced Statistical Theory II

derivation of the Laplacian from rectangular to spherical coordinates

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

2. Let H 1 and H 2 be Hilbert spaces and let T : H 1 H 2 be a bounded linear operator. Prove that [T (H 1 )] = N (T ). (6p)

6.3 Forecasting ARMA processes

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

Lecture 13 - Root Space Decomposition II

Reminders: linear functions

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

Second Order RLC Filters

Lecture 34 Bootstrap confidence intervals

Approximation of distance between locations on earth given by latitude and longitude

Commutative Monoids in Intuitionistic Fuzzy Sets

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

The Simply Typed Lambda Calculus

SOME PROPERTIES OF FUZZY REAL NUMBERS

On a four-dimensional hyperbolic manifold with finite volume

PROPERTIES OF CERTAIN INTEGRAL OPERATORS. a n z n (1.1)

( y) Partial Differential Equations

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

CRASH COURSE IN PRECALCULUS

Research Article Existence of Positive Solutions for m-point Boundary Value Problems on Time Scales

Homework 8 Model Solution Section

Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.

Tridiagonal matrices. Gérard MEURANT. October, 2008

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω

4.6 Autoregressive Moving Average Model ARMA(1,1)

MINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS

A General Note on δ-quasi Monotone and Increasing Sequence

Inverse trigonometric functions & General Solution of Trigonometric Equations

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Finite Field Problems: Solutions

Exercises to Statistics of Material Fatigue No. 5

POSITIVE SOLUTIONS FOR A FUNCTIONAL DELAY SECOND-ORDER THREE-POINT BOUNDARY-VALUE PROBLEM

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Section 7.6 Double and Half Angle Formulas

A Note on Intuitionistic Fuzzy. Equivalence Relation

Solutions to Exercise Sheet 5

Approximation of Entire Functions of Slow Growth 1

Math221: HW# 1 solutions

Spherical Coordinates

The Pohozaev identity for the fractional Laplacian

5. Choice under Uncertainty

Optimal Parameter in Hermitian and Skew-Hermitian Splitting Method for Certain Two-by-Two Block Matrices

The k-α-exponential Function

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

w o = R 1 p. (1) R = p =. = 1

Fractional Colorings and Zykov Products of graphs

n=2 In the present paper, we introduce and investigate the following two more generalized

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

ΗΜΥ 220: ΣΗΜΑΤΑ ΚΑΙ ΣΥΣΤΗΜΑΤΑ Ι Ακαδημαϊκό έτος Εαρινό Εξάμηνο Κατ οίκον εργασία αρ. 2

The Negative Neumann Eigenvalues of Second Order Differential Equation with Two Turning Points

Srednicki Chapter 55

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

Depth versus Rigidity in the Design of International Trade Agreements. Leslie Johns

New bounds for spherical two-distance sets and equiangular lines

Online Appendix I. 1 1+r ]}, Bψ = {ψ : Y E A S S}, B W = +(1 s)[1 m (1,0) (b, e, a, ψ (0,a ) (e, a, s); q, ψ, W )]}, (29) exp( U(d,a ) (i, x; q)

On the Galois Group of Linear Difference-Differential Equations

Parametrized Surfaces

Lecture 15 - Root System Axiomatics

Takeaki Yamazaki (Toyo Univ.) 山崎丈明 ( 東洋大学 ) Oct. 24, RIMS

Trigonometry 1.TRIGONOMETRIC RATIOS

MA 342N Assignment 1 Due 24 February 2016

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

Transcript:

GEORGIAN MATHEMATICAL JOURNAL: Vol. 3, No. 1, 1996, 11-26 PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS WITH SUPREMUM D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA Abstract. Partial averaging for impulsive differential equations with supremum is justified. The proposed averaging schemes allow one to simplify considerably the equations considered. 1. Introduction For mathematical simulation in various fields of science and technology impulsive differential equations with supremum are successfully used. The investigation of these equations is rather difficult due to the discontinuous character of their solutions and the presence of the supremum of the unknown function, hence the need for approximate methods for solving them. In the present paper three schemes for partial averaging of impulsive differential equations with supremum are considered and justified. 2. Preliminary Notes We note, for the sake of definiteness, that by the value of a piecewise continuous function at a point of discontinuity we mean the limit from the left of the function (provided it exists at this point. By the symbol <τ k <T we denote summation over all values of k for which the inequality < τ k < T is satisfied, and by the symbol the Euclidean norm in R n. In the proof of the main result we shall use the following two lemmas. Lemma 1. Let u(t be a piecewise continuous function for t a with points of discontinuity of the first kind τ k > a, k = 1, 2,..., for which τ k < τ k+1 for k = 1, 2,..., and lim τ k =. Then, if for t a the k 1991 Mathematics Subject Classification. 34A37, 34C29. Key words and phrases. Impulsive differential equations with supremum, partial averaging schemes. 11 172-947X/96/1-11$9.5/ c 1996 Plenum Publishing Corporation

12 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA inequality u(t c 1 + c 2 a u(τ dτ + c 3 a<τ k <t u(τ k holds, where c 2 is a positive constant and c 1, c 3 are nonnegative constants, then for t a the estimate u(t c 1 (1 + c 3 i(a,t exp{c 2 (t a} is valid, where i(a, t is the number of points τ k which belong to the interval a, t. Lemma 1 is proved by means of the Gronwall Bellman inequality and by induction. Lemma 1 is a particular case of the theorems on integral inequalities obtained in 1] and 2]. Lemma 2. Let the sequence τ 1,..., τ k... be such that for any k N the inequality τ k τ k 1 holds, where is a positive constant and τ =. Then for any T and t the inequality is valid. t τ k <t +T (τ k t < 3T 2 Proof. Let n 2 and τ j < t τ j+1 < < τ j+n < t + T. Then for T we have t τ k <t +T 1 j+n 1 k=j+1 τ j+n (τ k t = j+n 1 k=j+1 2 (τ k t + (τ j+n t (τ k t (τ k+1 τ k + (τ j+n t 1 (t t dt + (τ j+n t < τ j+1 T 2 2 + T 3T 2 2. Let τ j < t τ j+1 < t + T τ j+2. Then for T we have t τ k <t +T (τ k t = τ j+1 t < T 3T 2 2.

PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS 13 3. Main Results Let the impulsive system of differential equations with supremum have the form ż(t = εz(t, z(t, z(t, z(t, t >, t τ k, z(t = κ(t, ż(t = κ(t, h t, z(t z(t + z(t = εl k (z(t, t = τ k, k N, where z(t = (z 1 (t,..., z n (t, h is a positive constant, z(t = (z 1 (t,..., z n (t, z(t = ( z 1 (t,..., z n (t, z(t + =(z 1 (t +,..., z n (t +, z(t =(z 1 (t,..., z n (t, z i (t = sup{z i (s : s t h, t]}, z i (t = sup{ż i (s : s t h, t]}, z i (t + = lim z i(s, z i (t = lim z i(s, i = 1,..., n, s t, s>t s t, s<t κ(t = (κ 1 (t,..., κ n (t is an initial function, =τ <τ 1 < <τ k <, lim k τ k = and ε> is a small parameter. Let the functions Z(t, z and L k (z exist for which (1 and 1 T ] lim Z(t, z, z, Z(t, z dt = (2 T T 1 lim T T <τ k <T Lk (z L k (z ] dt =, k N. (3 Then with the initial-value problem (1 we associate the averaged impulsive system of differential equations ζ(t = εz(t, ζ(t, t >, t τ k, ζ(t ζ(t + ζ(t = εl k (ζ(t, t = τ k, k N, ζ( = κ(. (4 We shall prove a theorem on nearness of the solutions of the initial-value problems (1 and (4. Theorem 1. Let the following conditions hold: 1. The functions Z(t, z, u, v, Z(t, z, L k (z, and L k (z, k N, are continuous in the respective projections of the domain {t, z, u G, v H}, where G and H are open domains in R n. The functions κ(t and κ(t are continuous in the interval h, ], h = const >, κ i (t and κ i (t

14 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA (i = 1,..., n have a finite number of extremums in the interval h, ], and κ(t G, κ(t H for t h, ]. 2. There exist positive constants λ and M such that for all t, z, u, z, u G, v, v H, k N, the following inequalities are valid: Z(t, z, u, v + Z(t, z + L k (z + L k (z M, Z(t, z, u, v Z(t, z, u, v λ ( z z + u u + v v, Z(t, z Z(t, z + L k (z L k (z + L k (z L k (z λ z z and for t h, ] the estimate κ(t + κ(t M is valid. 3. For each z G there exist the limits (2 and (3. 4. There exists a positive constant such that for k N the inequality τ k τ k 1 holds, where τ =. 5. For any ε (, ε ], ε = const >, the initial-value problem (1 has a unique solution z(t which is defined for t ; z i (t and ż i (t (i = 1,..., n have a finite number of extremums in each interval of length h; z(t and ż(t satisfy respectively the conditions z( + = κ( and ż( + = κ(. 6. For any ε (, ε ] the initial-value problem (4 has a unique solution ζ(t which is defined for t and for t belongs to the compact Q G together with some ρ-neighborhood of it (ρ = const >. Then for any η > and L > there exists ε (, ε (ε = ε (η, L such that for ε (, ε ] and t, Lε 1 ] the inequality z(t ζ(t < η Proof. From the conditions of Theorem 1 it follows that for the compact Q G there exists a continuous function α(t which monotonically tends to zero as T and is such that for z Q the inequalities T Z(t, z, z, Z(t, z ] dt T α(t (5 and Lk (z L k (z ] T α(t (6 <τ k <T hold. For t T ε =, Lε 1 ] we have z(t = z( + ε Z(τ, z(τ, z(τ, z(τdτ + ε <τ k <t L k (z(τ k, (7

PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS 15 ζ(t = ζ( + ε Z(τ, ζ(τdτ + ε <τ k <t where z(t = κ(t and ż(t = κ(t for t h, ]. Subtracting (8 from (7 for t we obtain z(t ζ(t ε + ε L k (ζ(τ k, (8 Z(τ, z(τ, z(τ, z(τ Z(τ, ζ(τ, ζ(τ, dτ + <τ k <t L k (z(τ k L k (ζ(τ k + t ] + ε Z(τ, ζ(τ, ζ(τ, Z(τ, ζ(τ dτ + + ε <τ k <t Lk (ζ(τ k L k (ζ(τ k ]. (9 Denote successively by β(t, γ(t, δ(t, and σ(t the summands on the right-hand side of (9. Without loss of generality, for the estimation of the function β(t we shall assume that h < t, z i (τ k = max(z i (τ k, z i (τ k +, and ż i (τ k = max(ż i (τ k, ż i (τ k + for i = 1,..., n and k N. From the last assumption it follows that the supremums in z i (t and z i (t (i = 1,..., n are achieved. Denote by s i (t (resp., s i (t the leftmost point of the interval t h, t], t, at which z i (s (resp., ż i (s takes its greatest value in this interval. Then z i (t = z i (s i (t and z i (t = ż i ( s i (t (i = 1,..., n. Using the conditions of Theorem 1, we obtain β(t = ε 2ελ Z(τ, z(τ, z(τ, z(τ Z(τ, ζ(τ, ζ(τ, dτ z(τ ζ(τ dτ + ελ z(τ z(τ + z(τ ] dτ. (1 Denote by β (t the second summand on the right-hand side of (1. We obtain the following estimate: ] β (t = ελ z(τ z(τ + z(τ dτ

16 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA ελ + ελ ελ + ελ + ελ h h h h h z(τ + z(τ + z(τ ] dτ + z(τ z(τ + z(τ ] dτ κ(τ h + κ(τ h + 2 z(τ + ż(τ ] dτ + n ( zi (s i (τ z i (τ ] 2 1/2 dτ + n 1/2dτ żi 2 ( s i (τ] ελh 3 + ε(1 + h + ε h ] M + ε 2 λ(t hm n + + ελ + ε h n s i(τ<τ k <τ ( ε τ s i(τ Z i (l, z(l, z(l, z(ldl + L ki (z(τ k 2 ] 1/2dτ. Setting A = λh3 + ε(1 + h + ε h ]M + λ(l εhm n and applying Minkowski s inequality, we obtain for t T ε β (t εa + ε 2 λ n + h s i(τ<τ k <τ εa + ε 2 λ h { n τ s i(τ Z i (l, z(l, z(l, z(ldl 2 ] 1/2 + L ki (z(τ k 2 ] 1/2 }dτ { n h 2 M 2] 1/2 n + εa + ελ(l εh 1 + hm n. ( h 2M 2 ] 1/2} dτ

PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS 17 For γ(t and t T ε by the conditions of Theorem 1 we get the estimate γ(t = ε <τ k <t L k (z(τ k L k (ζ(τ k ελ <τ k <t z(τ k ζ(τ k. In order to obtain estimates of the functions δ(t for t T ε we partition the interval T ε into q equal parts by the points t i = il εq, i =, 1,..., q. Let t be an arbitrary chosen and fixed number from the interval T ε, and let t (t s, t s+1 ], where s q 1. Then, using the conditions of Theorem 1 and inequality (5, we obtain s 1 δ(t ε t i+1 i= t i Z(τ, ζ(τ, ζ(τ, Z(τ, ζ(τ Z(τ, ζ(t i, ζ(t i, + Z(τ, ζ(t i ] dτ + s 1 + ε + ε t s s 1 ε t i+1 i= t i Z(τ, ζ(ti, ζ(t i, Z(τ, ζ(t i ] dτ + Z(τ, ζ(τ, ζ(τ, + Z(τ, ζ(τ ] dτ t i+1 i= t i Z(τ, ζ(τ, ζ(τ, Z(τ, ζ(ti, ζ(t i, + + Z(τ, ζ(τ Z(τ, ζ(t i ] t s 1 i+1 dτ +ε Z(τ, ζ(ti, ζ(t i, Z(τ, ζ(t i ] s 1 t i dτ +ε Z(τ, ζ(ti, ζ(t i, Z(τ, ζ(t i ] + dτ + ML q s 1 3ελ t i+1 i= t i s 1 + ε t i α(t i + ML q + 2ε s t i α(t i + ML q i= s 1 ζ(τ ζ(t i dτ + ε t i+1 α(t i+1 + 3ε 2 λ 1 + 3ε 2 λ 1 + 2 M s 1 i= i= t i+1 (τ t i dτ + t i ( L 2 + εq M s 1 i=

18 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA + 2s2 L q ( L α + ML (3λs(1 + L + 2qML = εq q 2q 2 + 2s2 L ( L α. q εq We pass to the estimation of σ(t. Using the conditions of Theorem 1 and inequality (6 for t T ε, we obtain σ(t = ε Lk (ζ(τ k L k (ζ(τ k ] <τ k <t εtα(t ( τ sup τα τ L ε = δ 1 (ε, δ 1 (ε, ε. From (9 and the estimates obtained for the functions β(t, γ(t, δ(t, and σ(t it follows that for t T ε the inequality holds, where z(t ζ(t b(q + c(q, ε + 2ελ b(q = + ελ <τ k <t (3λ(1 + L + 2ML, 2q c(q, ε = εa + ελ(l εh 1 + z(τ ζ(τ dτ + z(τ k ζ(τ k (11 hm ( L n + 2qLα + δ 1 (ε. εq Choose a sufficiently large q N so that the inequality b(q 1 2 exp { λl ( 2 + 1 } min(η, ρ Consider inequality (11 for q = q, ε (, ε ], t T ε and apply to it Lemma 1. Thus we obtain z(t ζ(t ( b(q + c(q, ε exp{2ελt}(1 + ελ i(,t ( b(q + c(q, ε exp { 2ελt + t ln(1 + ελ} ( b(q + c(q, ε exp { λl ( 2 + 1 }. Choose a sufficiently small ε (, ε ] so that for ε (, ε ] we have c(q, ε < 1 2 exp { λl ( 2 + 1 } min(η, ρ.

PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS 19 Then for any ε (, ε ] (ε = ε (η, L and t T ε the inequality z(t ζ(t < min(η, ρ Hence, for ε (, ε ] and t T ε, z(t belongs to the domain G, and the estimate z(t ζ(t < η is valid. Consider the initial-value problem (1, where ( ( x(t X(t, z(t, z(t, z(t z(t =, Z(t, z(t, z(t, z(t =, y(t Y (t, z(t, z(t, z(t ( ( ( (12 ϕ(t x(t Ik (z κ(t =, z(t =, L ψ(t y(t k (z =, k N, J k (z x(t and y(t are l- and m-dimensional vector-functions, and l + m = n. For problem (1, (12 various schemes for partial averaging are possible, which lead to averaged systems of differential equations not containing supremum. and First scheme for partial averaging. Let the following limits exist: 1 T lim X(t, z, z, dt = X(z (13 T T 1 lim T T <τ k <T I k (z = I(z. (14 Then with the initial problem (1, (12 we associate the partially averaged impulsive system of differential equations (4 with ( X(z Z(t, z =, (15 Y (t, z, z, ( I(z L k (z =, k N. (16 J k (z Theorem 2. Let the following conditions hold: 1. The functions X(t, z, u, v, Y (t, z, u, v, I k (z, J k (z, k N, are continuous in the respective projections of the domain {t, z, u G, v H}, where G and H are open domains in R n. The functions ϕ(t, ϕ(t, ψ(t, ψ(t are continuous in the interval h, ], h = const >, ϕ i (t, ϕ i (t (i = 1,..., l, ψ j (t, ψ j (t (j = 1,..., m have a finite number of extremums in the interval h, ], and κ(t = (ϕ(t, ψ(t G, κ(t = ( ϕ(t, ψ(t H for t h, ].

2 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA 2. There exist positive constants λ and M such that for all t, z, u, z, u G, v, v H, k N, the following inequalities are valid: X(t, z, u, v + Y (t, z, u, v + I k (z + J k (z M, X(t, z, u, v X(t, z, u, v λ ( z z + u u + v v, Y (t, z, u, v Y (t, z, u, v λ ( z z + u u + v v, I k (z I k (z + J k (z J k (z λ z z and for t h, ] the estimate ϕ(t + ϕ(t + ψ(t + ψ(t M is valid. 3. For each z G there exist the limits (13 and (14. 4. There exists a positive constant such that for k N the inequality τ k τ k 1 holds, where τ =. 5. For any ε (, ε ], ε = const > the initial-value problem (1, (12 has a unique solution z(t which is defined for t ; z i (t and ż i (t (i = 1,..., n have a finite number of extremums in each interval of length h; z(t and ż(t satisfy respectively the conditions z( + = κ( and ż( + = κ(. 6. For any ε (, ε ] the initial-value problem (4, (15, (16 has a unique solution ζ(t which is defined for t and for t belongs to the compact Q G together with some ρ-neighborhood of it (ρ = const >. Then for any η > and L > there exists ε (, ε ] (ε = ε (η, L such that for ε (, ε ] and t, Lε 1 ] the inequality z(t ζ(t < η Proof. From (13, (14 and the conditions of Theorem 2 it follows that in the domain G the functions X(z and I(z are bounded, continuous, and satisfy the Lipschitz condition. Hence for the functions Z(t, z, u, v, Z(t, z, L k (z, and L k (z, k N, of problem (1, (12, conditions 1 and 2 of Theorem 1 are met. Further on, the proof of Theorem 2 is analogous to the proof of Theorem 1. Second scheme for partial averaging. Let there exist the limit (13 Then with the initial-value problem (1, (12 we associate the partially averaged impulsive system of differential equations (4 with Z(t, z determined by (15 and L k (z = L k (z, k N. (17 Theorem 3. Let the following conditions hold: 1. Conditions 1, 2, 4, and 5 of Theorem 2 are met. 2. For each z G there exists the limit (13.

PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS 21 3. For any ε (, ε ] the initial-value problem (4, (15, (17 has a unique solution ζ(t which is defined for t and for t belongs to the compact Q G together with some ρ-neighborhood of it (ρ = const >. Then for any η > and L > there exists ε (, ε ] (ε = ε (η, L such that for ε (, ε ] and t, Lε 1 ] the inequality z(t ζ(t < η The proof of Theorem 3 is analogous to the proof of Theorem 1. Third scheme for partial averaging. Let there exist (13, (14 and 1 lim T T <τ k <T J k (z = J(z. (18 Then with the initial-value problem (1, (12 we associate the averaged system of ordinary differential equations with the initial condition χ(t = ε Z(t, χ(t + L(χ(t ] (19 χ( = κ(, (2 where Z(t, z is determined by (15 and ( I(z L(z =. (21 J(z Theorem 4. Let the following conditions hold: 1. Conditions 1, 2, 4, and 5 of Theorem 2 are met. 2. For each z G there exist the limits (13, (14, (16. 3. For any ε (, ε ] the initial-value problem (4, (15, (17 and the initial-value problem (19, (2, (15, (21 have unique solutions ζ(t and χ(t, respectively, which are defined for t and lie in the compact Q G, together with some of its ρ-neighborhood (ρ = const >. Then for any η > and L > there exists ε (, ε ] (ε = ε (η, L such that for ε (, ε ] and t, Lε 1 ] the inequality z(t χ(t < η Proof. From (13, (14, (16 and the conditions of Theorem 4 it follows that in the domain G the functions X(z, I(z, and J(z are bounded, continuous, and satisfy the Lipschitz condition, namely for all z, z G the inequalities X(z M, I(z M, J(z M, X(z X(z 2λ z z,

22 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA I(z I(z λ z z, J(z J(z λ z z are valid. By Theorem 3 for each η > and L > there exists a number ε (, ε ] ( ε = ε (η, L such that for ε (, ε ] and t, Lε 1 ] the inequality z(t ζ(t < 1 min(η, ρ (22 2 holds, where z(t is a solution of the initial-value problem (1, (12, and ζ(t is a solution of the initial-value problem (4, (15, (17. From the conditions of Theorem 4 it follows that for the compact Q G there exists a continuous function α(t which monotonically tends to zero as T and is such that for z Q the inequality <τ k <T L k (z L(z T T α(t (23 For t T ε =, Lε 1 ], from (4 and (19, (2 we have ζ(t = κ( + ε χ(t = κ( + ε Z(τ, ζ(τdτ + ε Subtracting (25 from (24 for t we obtain ζ(t χ(t ε + ε + ε <τ k <t L k (ζ(τ k, (24 Z(τ, χ(τ + L(χ(τ ] dτ. (25 Z(τ, ζ(τ Z(τ, χ(τ dτ + <τ k <t <τ k <t L k (ζ(τ k L k (χ(τ k + L k (χ(τ k L(χ(τdτ. (26 Denote by ω(t the third summand in the right-hand side of (26. In order to estimate ω(t for t T ε =, Lε 1 ], we partition the interval T ε into q equal parts by means of the points t i = il εq, i =, 1,..., q.

PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS 23 Let t be an arbitrary chosen and fixed number of the interval T ε and let t (t s, t s+1 ], where s N and s q 1. Then, using the conditions of Theorem 4 and inequality (23, we obtain ω(t ε L k (χ(τ k L k (χ( +ε L(χ(τ L(χ( dτ + <τ k <t 1 s 1 ( + ε L k (χ(τ k L k (χ(t i + t i τ k <t i+1 t i+1 + L(χ(τ L(χ(t i dτ + t i 1 + ε L k (χ( L(χ(dτ + <τ k <t 1 t s 1 i+1 + ε L k (χ(t i L(χ(t i dτ + t i τ k <t i+1 + ε ελ t s τ k <t L k (χ(t k + ε t s <τ k <t 1 χ(τ k χ( + ε 2λ t i 1 L(χ(τ dτ 1 χ(τ χ( dτ + s 1 ( + ελ χ(τ k χ(t i + 2 t i+1 χ(τ χ(t i dτ + t i τ k <t i+1 t i s 1 + εt 1 α(t 1 + ε L k (χ(t i t i+1 L(χ(t i + <τ k <t i+1 s 1 + ε Lk (χ(t i t i L(χ(t i ] + <τ k <t i s 1 + ε L k (χ(t i + 3LM q τ k =t i 2ε2 λm(1 + s 1 ( t i τ k <t i+1 (τ k t i + 2 t i+1 t i (τ t i dτ +

24 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA s 1 +εt 1 α(t 1 +ε 2ε2 λm(1 + + 2s2 L q s 1 t i+1 α(t i+1 +ε s 1 ( (τ k t i i= t i τ k <t i+1 ( L α + 2ε(s 1M + 3LM εq q. t i α(t i + 2ε(s 1M + 3LM q + 2sλL2 M(1 + 2 q 2 + From (26, the conditions of Theorem 4, and the estimate obtained for ω(t it follows that for t T ε the inequality holds, where ζ(t χ(t b(q, ε + c(q, ε + 4ελ b(q, ε = + ελ ( 3 + <τ k <t ζ(τ χ(τ dτ + ζ(τ k χ(τ k (27 2λL(1 + LM q + + 2ε2 λm(1 + m 1 i= ( L c(q, ε = 2qLα + 2εqM. εq ( (τ k t i, t i τ k <t i+1 Choose successively a sufficiently large q N and a sufficiently small ε 1 (, ε ] so that the inequalities ( 3 + 5λL(1 + LM q 1 4 exp { λl ( 4 + 1 } min(η, ρ, (28 ε 1 q L For q = q and ε (, ε 1 ] we apply Lemma 2 to the summand on the right-hand side of b(q, ε and obtain 2ε 2 λm(1 + q 1 i= ( (τ k t i t i τ k <t i+1 3λ2 L 2 M(1 + 2 q. (29

PARTIAL AVERAGING FOR IMPULSIVE DIFFERENTIAL EQUATIONS 25 From (27, (28, and (29 for q = q, ε (, ε 1 ], and t T ε there follows the inequality ζ(t χ(t b(q, ε + c(q, ε + 4ελ + ελ <τ k <t where b(q, ε 1 4 exp{ λl(4 + 1 } min(η, ρ. Applying Lemma 1 to (3, we get ζ(τ χ(τ dτ + ζ(τ k χ(τ k, (3 ζ(t χ(t ( b(q, ε + c(q, ε (1 + ελ i(,t exp{4ελt} ( b(q, ε + c(q, ε exp { 4ελt + t ln(1 + ελ} ( b(q, ε + c(q, ε exp { λl ( 4 + 1 }. Choose a sufficiently small ˇε (, ε 1 ] so that for ε (, ˇε ] we have c(q, ε < 1 4 exp { λl ( 4 + 1 } min(η, ρ. Then for ε (, ˇε ] (ˇε = ˇε (η, L and t T ε the inequality ζ(t χ(t < 1 2 min(η, ρ (31 From (22 and (31 it follows that for ε (, ε ], ε = min( ε, ˇε, and t T ε the inequality z(t χ(t z(t ζ(t + ζ(t χ(t min(η, ρ Hence for ε (, ε ] and t T ε, z(t lies in the domain G, and the estimate z(t χ(t < η is valid. Acknowledgement The present investigation was partially supported by the Bulgarian Ministry of Education and Science under grant MM-7.

26 D. BAINOV, YU. DOMSHLAK, AND S. MILUSHEVA References 1. D. D. Bainov and P. S. Simeonov, Inequalities and applications. Kluwer, 1992. 2. V. Lakshmikantham, D. D. Bainov, and P. S. Simeonov, Theory of impulsive differential equations. World Scientific, Singapore, 1989. Authors addresses: D. Bainov Academy of Medicine Sofia, Bulgaria (Received 28.3.1994 Yu. Domshlak Ben Gurion University of the Negev Israel S. Milusheva Technical University Sofia, Bulgaria