Letα < β. Consider the system of impulsive differential equations (1.5).
Definition 1.1. The functionϕ: (α, β) is said to be asolutionof (1.5) if:
1. (t, ϕ(t))∈R×Ωfort∈(α, β).
2. For t ∈ (α, β), t = τk(ϕ(t)), k=±1, ±2, . . . the function ϕ(t) is differentiable and dϕ(t)
dt =f(t, ϕ(t)).
3. The functionϕ(t) is continuous from the left in (α, β) and ift=τk(ϕ(t)), t=β, thenϕ(t+) =ϕ(t) +Ik(ϕ(t)), and for eachk=±1,±2, . . .and some δ >0, s=τk(ϕ(s)) fort < s < t+δ.
Definition 1.2. Each solutionϕ(t) of (1.5) which is defined in an interval of the form (t0, β) and satisfies the conditionϕ(t+0) =x0is said to be asolution of the initial value problem(1.5), (1.4).
We note that, instead of the initial conditionx(t0) =x0, we have imposed the limiting conditionx(t+0) =x0which, in general, is natural for (1.5) since (t0, x0) may be such thatt0=τk(x0) for somek. Whenevert0=τk(x0), for allk, we shall understand the initial conditionx(t+0) =x0in the usual sense, that is,x(t0) =x0.
It is clear that if t0 = τk(x0), k = ±1,±2, . . ., then the existence and uniqueness of the solution of the initial value problem (1.5), (1.4) depends only on the properties of the function f(t, x). Thus, for instance, if the function f(t, x) is continuous in a neighborhood of the point (t0, x0), then there exists a solution of the initial problem (1.5), (1.4) and this solution is unique iff(t, x) is Lipschitz continuous in this neighborhood.
If, however, t0 = τk(x0) for some k, that is, (t0, x0) belongs to the hypersurface σk ≡ t = τk(x), then it is possible that the solution x(t) of the initial value problem
˙
x=f(t, x), x(t0) =x0 (1.9) lies entirely inσk.
Consequently, we need some additional conditions onf(t, x) andτk(x) to guarantee the existence of solutionx(t) of the initial value problem (1.9) in some interval [t0, β), and the validity of the condition
t=τk(x(t)) fort∈(t0, β) andk=±1,±2, . . . . Conditions of this type are imposed in the next theorem.
Theorem 1.2 ([15]). Let the following conditions hold:
1. The functionf :R×Ω→Rn is continuous int=τk(x), k=±1,±2, . . ..
2. For any(t, x)∈R×Ω there exists a locally integrable functionl(t) such that in a small neighborhood of(t, x)
||f(s, y)|| ≤l(s).
3. For each k =±1,±2, . . . the condition t1 =τk(x1) implies the existence of δ >0 such that
t=τk(x) for all 0< t−t1< δ and||x−x1||< δ.
Then for each (t0, x0)∈R×Ω there exists a solutionx: [t0, β)→Rn of the initial value problem (1.5), (1.4) for some β > t0.
1.1 Impulsive Differential Equations 7
Remark 1.2. Condition 2 of Theorem1.2can be replaced by the condition:
2’.For anyk =±1,±2, . . .and (t, x)∈ σk there exists the finite limit of f(s, y)as(s, y)→(t, x), s > τk(y).
Remark 1.3. The solution x(t) of the initial value problem (1.5), (1.4) is unique, if the function f(t, x) is such that the solution of the initial value problem (1.9) is unique. This requirement is met if, for instance, f(t, x) is (locally) Lipschitz continuous with respect toxin a neighborhood of (t0, x0).
Now, we shall consider in more detail the system with fixed moments of impulsive effect:
x˙ =f(t, x), t=tk,
Δx(tk) =Ik(x(tk)), k=±1,±2, . . . , (1.10) wheretk < tk+1, k=±1,±2, . . .and lim
k→±∞tk=±∞. Let us first establish some theorems.
Theorem 1.3 ([15]). Let the following conditions hold:
1. The functionf :R×Ω→Rn is continuous in the sets(tk, tk+1]×Ω, k=
±1,±2, . . ..
2. For eachk=±1,±2, . . . and x∈Ω there exists the finite limit of f(t, y) as(t, y)→(tk, x), t > tk.
Then for each (t0, x0) ∈ R×Ω there exists β > t0 and a solution x: [t0, β)→Rn of the initial value problem (1.10), (1.4).
If, moreover, the function f(t, x) is locally Lipschitz continuous with respect to x∈Ωthen this solution is unique.
Let us consider the problem of the continuability to the right of a given solutionϕ(t)of system (1.10).
Theorem 1.4 ([15]). Let the following conditions hold:
1. The functionf :R×Ω→Rn is continuous in the sets(tk, tk+1]×Ω, k=
±1,±2, . . ..
2. For eachk=±1,±2, . . . andx∈Ω there exists the finite limit off(t, y) as(t, y)→(tk, x), t > tk.
3. The function ϕ: (α, β)→Rn is a solution of (1.10).
Then the solution ϕ(t) is continuable to the right of β if and only if there exists the limit
lim
t→β−ϕ(t) =η and one of the following conditions hold:
(a) β=tk for eachk=±1,±2, . . .andη∈Ω.
(b) β=tk for somek=±1,±2, . . . andη+Ik(η)∈Ω.
Theorem 1.5 ([15]). Let the following conditions hold:
1. Conditions 1 and 2 of Theorem 1.1.11 hold.
2. The function f is locally Lipschitz continuous with respect tox∈Ω.
3. η+Ik(η)∈Ω for eachk=±1,±2, . . .andη∈Ω.
Then for any (t0, x0) ∈ R×Ω there exists a unique solution of the initial value problem (1.10), (1.4) which is defined in an interval of the form[t0, ω)and is not continuable to the right ofω.
Let the conditions of Theorem 1.5 be satisfied and let (t0, x0) ∈ R×Ω.
Denote by J+=J+(t0, x0)the maximal interval of the form[t0, ω)in which the solutionx(t;t0, x0) is defined.
Theorem 1.6 ([15]). Let the following conditions hold:
1. Conditions 1, 2 and 3 of Theorem 1.1.12 are met.
2. ϕ(t)is a solution of the initial value problem (1.10), (1.4).
3. There exists a compact Q⊂Ω such that ϕ(t)∈Qfor t∈J+(t0, x0).
Then J+(t0, x0) = (t0,∞).
Let ϕ(t) : (α, ω) → Rn be a solution of system (1.10) and consider the question of the continuability of this solution to the left ofα.
Ifα =tk, k =±1,±2, . . . then the problem of continuability to the left ofαis solved in the same way as for ordinary differential equations without impulses [45]. In this case such an extension is possible if and only if there exists the limit
lim
t→σ+
ϕ(t) =η (1.11)
andη∈Ω.
If α = tk, for some k = ±1,±2, . . ., then the solution ϕ(t) will be continuable to the left of tk when there exists the limit (1.11), η ∈ Ω, and the equationx+Ik(x) =η has a unique solution xk ∈ Ω. In this case the extension ψ(t) of ϕ(t) for t ∈ (tk−1, tk] coincides with the solution of the initial value problem
ψ˙ =f(t, ψ), tk−1< t≤tk, ψ(tk) =xk.
If the solutionϕ(t) can be continued up totk−1, then the above procedure is repeated, and so on. Under the conditions of Theorem 1.5 for each (t0, x0) ∈ R×Ω there exists a unique solution x(t;t0, x0) of the initial value problem (1.10), (1.4) which is defined in an interval of the form (α, ω) and is not continuable to the right of ω and to the left of α. Denote by J(t0, x0) this maximal interval of existence of the solution x(t;t0, x0) and set J− =J−(t0, x0) = (α, t0]. A straightforward verification shows that the solutionx(t) =x(t;t0, x0) of the initial value problem (1.10), (1.4) satisfies the following integro-summary equation
1.1 Impulsive Differential Equations 9
x(t) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩ x0+
t t0
f(s, x(s))ds+
t0<tk<t
Ik(x(tk)), for t∈J+, x0+
t t0
f(s, x(s))ds−
t≤tk≤t0
Ik(x(tk)), for t∈J−.
(1.12)
Now, we consider the linear system impulsive equations x˙ =A(t)x, t=tk,
Δx(tk) =Bkx(tk), k=±1,±2, . . . , (1.13) under the assumption that the following conditions hold:
H1.1. tk< tk+1, k=±1,±2, . . .and lim
k→±∞tk =±∞. H1.2. A∈P C[R,Rn×n], Bk ∈Rn×n, k=±1,±2, . . ..
Theorem 1.7 ([15]). Let conditions H1.1 and H1.2 hold. Then for any (t0, x0)∈ R×Rn there exists a unique solution x(t) of system (1.13) with x(t+0) =x0 and this solution is defined fort≥t0.
If moreover,det(E+Bk)= 0, k=±1,±2, . . ., then this solution is defined for allt∈R.
Let Uk(t, s) (t, s ∈ (tk−1, tk]) be the Cauchy matrix [65] for the linear equation
˙
x(t) =A(t)x(t), tk−1< t≤tk, k=±1,±2, . . . .
Then by virtue of Theorem1.7the solution of the initial problem (1.13), (1.4) can be decomposed as:
x(t;t0, x0) =W(t, t+0)x0, (1.14) where
W(t, s) =
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
Uk(t, s) as t, s∈(tk−1, tk],
Uk+1(t, t+k)(E+Bk)Uk(tk, s) as tk−1< s≤tk< t≤tk+1, Uk(t, tk)(E+Bk)−1Uk+1(t+k, s) as tk−1< t≤tk< s≤tk+1, Uk+1(t, t+k)
i+1 j=k
(E+Bj)Uj(tj, t+j−1)(E+Bi)Ui(ti, s) as ti−1< s≤ti< tk < t≤tk+1,
Ui(t, ti)
k−1 j=i
(E+Bj)−1Uj+1(t+j, tj+1)(E+Bk)−1Uk+1(t+k, s) as ti−1< t≤ti< tk < s≤tk+1,
is the solving operator of the (1.13).