SOME OPIAL-TYPE INEQUALITIES APPLICABLE TO DIFFERENTIAL EQUATIONS INVOLVING IMPULSES

Abstract

The purpose of this paper is to obtain Opial-type inequalities that are useful to study various qualitative properties of certain differential equations involving impulses. After we obtain some Opial-type inequalities, we apply our results to certain differential equations involving impulses.

1. INTRODUCTION

Opial-type inequalities are very useful to study various qualitative properties of differential equations. For a good reference of the work on such inequalities together with various applications, we recommend the monograph [1]. In this paper we obtain some Opial-type inequalities that involve Stieltjes derivatives which are applicable to differential equations with impulses. Differential equations involving impulses arise in various real world phenomena, we refer to the monograph [8].

 

2. PRELIMINARIES

To obtain our results in this paper we need some preliminaries.

Let R be the set of all real numbers. Assume that [a, b] ⊂ R is a bounded interval. A function f : [a, b] → R is called regulated on [a, b] if both

exist for every point s ∈ [a, b), t ∈ (a, b], respectively. Let G([a, b]) be the set of all regulated functions on [a, b]. For f ∈ G([a, b]) we define f(a−) = f(a); f(b+) = f(b). For convenience we define

Remark 2.1. Let f ∈ G([a, b]). Since both f(s+) and f(s−) exist for every s ∈ [a, b] it is obvious that f is bounded on [a, b], and since f is the uniform limit of step functions, f is Borel measurable (see [3, Theorem 3.1.]).

For a closed interval I = [c, d], we define f(I) = f(d) − f(c).

A tagged interval (τ, [c, d]) in [a, b] consists of an interval [c, d] ⊂ [a, b] and a point τ ⊂ [c, d].

Let Ii = [ci, di] ⊂ [a, b], i = 1, …, m. We say that the intervals Ii are pairwise non-overlapping if

for i ≠ j where int(I) denotes the interior of an interval I.

A finite collection {(τi, Ii) : i = 1, 2, …, m} of pairwise non-overlapping tagged intervals is called a tagged partition of [a, b] if . A positive function δ on [a, b] is called a gauge on [a, b].

From now on we use notation .

Definition 2.2 ([6, 9]). Let δ be a gauge on [a, b]. A tagged partition of [a, b] is said to be δ−fine if for every we have

Moreover if a δ−fine partition P satisfies the implications

then it is called a δ*−fine partition of [a, b].

The following lemma implies that for a gauge δ on [a, b] there exists a δ*−fine partition of [a, b]. This also implies the existence of a δ− partition of [a, b].

Lemma 2.3 ([6]). Let δ be a gauge on [a, b] and a dense subset ­ Ω⊂ (a, b) be given. Then there exists a δ*−fine partition of [a, b] such that ti ∈ Ω for

We now give a formal definition of two types of the Kurzweil integrals.

Definition 2.4 ([6, 9]). Assume that f, g : [a, b] → R are given. We say that f dg is Kurzweil integrable (or shortly, K-integrable) on [a, b] and v ∈ R is its integral if for every ε > 0 there exists a gauge δ on [a, b] such that

provided is a δ−fine tagged partition of [a, b]. In this case we define (or, shortly, ).

If, in the above definition, δ−fine is replaced by δ*−fine, then we say that f dg is Kurzweil* integrable(or, shortly, K*-integrable) on [a, b] and we define .

Remark 2.5. By the above definition it is obvious that K-integrability implies K*-integrability.

The following results are needed in this paper. For other properties of the K-integrals, see, e.g., [2, 7, 9, 10].

In this paper BV ([a, b]) denotes the set of all functions that are of bounded variation on [a, b].

Theorem 2.6 ([11, 2.15. Theorem]). Assume that f ∈ G([a, b]) and g ∈ BV ([a, b]). Then both f dg and g df are K-integrable on [a, b] and

Remark 2.7. In the above theorem, the sum Σt∈[a,b][Δ−f(t)Δ−g(t)−Δ+f(t)Δ+g(t)] is actually a countable sum because every regulated function has only countable dis-continuities.

Theorem 2.8 ([10, p. 40, 4.25. Theorem]). Let h ∈ BV ([a, b]), g : [a, b] → R and f: [a, b] → R. If the integral g dh exists and f is bounded on[a, b], then the integral exists if and only if the integral exists and in this case we have

Theorem 2.9 ([10, p. 34, 4.13. Corollary]). Assume that f ∈ G([a, b]) and g ∈ BV ([a, b]). Then we have for every t ∈ [a, b]

The following result is the Hölder’s inequality for K-integral. In this paper we frequently use this inequality.

Theorem 2.10. (Hölder’s inequality) Assume that f, g ∈ G([a, b]) and h is a nondecreasing function defined on [a, b]. Let . Then we have

Proof. The proof of this theorem is very similar to the proof of the classical Hölder’s inequality. So we omit the proof.   ☐

 

3. STIELTJES DERIVATIVES

In this section we state the results in [4, 5] that are essential to obtain our main results.

Throughout this section, we assume that f ∈ G([a, b]) and g is a nondecreasing function on [a, b].

We say that the function g is not locally constant at t ∈ (a, b) if there exists η> 0 such that g is not constant on (t − ε, t + ε) for every 0 < ε < η. We also say that the function g is not locally constant at a and b, respectively if there exist η, η* > 0 such that g is not constant on [a, a + ε), (b − ε*, b] respectively, for every ε ∈ (0, η), ε*∈ (0, η*).

Definition 3.1 ([4]). If g is not locally constant at t ∈ (a, b), we define

provided that the limit exists.

If g is not locally constant at t = a and t = b respectively, we define

respectively, provided that the limits exist. Frequently we use instead of .

If both f and g are constant on some neighborhood of t, then we define .

Remark 3.2. It is obvious that if g is not continuous at t then exists. Thus if does not exist then g is continuous at t. is called a Stieltjes derivative of f with respect to g.

Theorem 3.3 ([4]). Assume that if g is not locally constant at t ∈ [a, b]. If f is continuous at t or g is not continuous at t; then we have

K*-integrals recover Stieltjes derivatives.

Theorem 3.4 ([4]). Assume that if g is constant on some neighborhood of t then there is a neighborhood of t where both f and g are constant. Suppose that exists at every t ∈ [a, b] − {c1, c2, …}, where f is continuous at every t ∈ {c1, c2, …}. Then we have

Lemma 3.5 ([4]). Assume that if g is constant on some neighborhood of t then there is a neighborhood of t such that both f1 and f2 are constant there. If both and exist and f1, f2 ∈ G([a, b]), then we have

Similarly to the Riemann integral we have the following integration by parts formula.

Theorem 3.6. (Integration by Parts) Assume that functions f, g, h ∈ G([a, b]) are all left-continuous and h is nondecreasing. Suppose that both and exist for every t ∈ [a, b] and . Then we have

Proof. By Theorem 2.8 and Theorem 3.4 we have

So by Theorem 2.6 we get

This completes the proof.   ☐

Let

The Heaviside function Hτ : R → {0, 1} is defined by

Using the Heaviside function Hτ , we define function ϕ : [a, b] → R by

Remark 3.7. It is obvious that the function ϕ is strictly increasing and of bounded variation on [a, b], and left-continuous on [a, b].

Lemma 3.8 ([5]). Assume that f ∈ G([a, b]) and f'(t) exists for t ≠ tk, Then we have

 

4. OPIAL-TYPE INTEGRAL INEQUALITIES INVOLVING STIELTJES DERIVATIVES

In this section we obtain some Opial-type integral inequalities involving Stieltjes derivatives. The Opial-type inequalities have many interesting applications in the theory of differential equations(see, e.g., [1]).

Throughout this paper we always assume that

and that a function α : [a, b] → R is strictly increasing on [a, b], and continuous at t ≠ tk, and Δα(tk) ≠ 0; for every .

Remark 4.1. Note that strictly increasing implies nondecreasing, and a nondecreasing function is regulated.

Let PC([a, b]) = {u ∈ G([a, b]) : u is continuous at every t ≠ tk, }.

From now on we always assume that u, , and we define

The following result is an Opial-type inequality with Stieltjes derivatives.

Theorem 4.2. Assume that u(a) = u(b) = 0. If both u and α are left-continuous on [a, b], then we have

where Kα = inf h∈[a, b] max{α(h) − α(a), α(b) − α(h)}.

Proof. Let for t∈[a, b],

By Theorem 2.9, the functions, y and z are left-continuous on [a, b]. Also by Theorem 3.3, we have

and we have by Theorem 3.4 and u(a) = u(b) = 0

for t ∈ [a, b]. So by Theorem 3.4, Lemma 3.5, and using Hölder’s inequality, we get

and similarly we obtain

So we have

The proof is complete.   ☐

A slightly more general result is as follows.

Theorem 4.3. Assume that u(b) = 0. If both u and α are left-continuous on [a, b], then we have

Proof. From (4.2) we have

So we get

This gives (4.3). The proof is complete.   ☐

More generally we have the following result.

Theorem 4.4. Let p ≥ 0, q ≥ 1, r ≥ 0, m ≥ 1 be real numbers and let f ∈ PC([a, b]) be a positive function on [a, b] with infs∈[a, b] f(s) > 0. Assume that both functions u and α are left-continuous on [a, b]. If u(b) = 0, then we have

where , for m ≠ 1, and for m = 1.

Proof. Let for t ∈ [a, b],

Then by Theorem 3.4, |u(t)| ≤ z(t) and by Theorem 2.9, z is left-continuous, and non-increasing on [a, b].

If t ≠ tk, , Then by Theorem 3.3, exists, and by Theorem 2.9, z is continuous at t. Using the Mean Value Theorem and by the definition of the Stieltjes derivatives, if z is not locally constant at t, then we have,

If z is constant on some neighborhood of t, then since , the above equality is also true. If t = tk, , since z is non-increasing on [a, b], and and by the Mean Value Theorem, and by the definition of the Stieltjes derivatives, we have,

where z(tk+) ≤ ω ≤ z(tk) = z(tk−): Thus we have

Let . Then by hypotheses, β is strictly increasing on [a, b].

Since

we have by Theorem 3.4 and (4.5), since z(b) = 0 and ,

Using Hölder’s inequality with indices m,, we have

Integrating (4.6) on [a, b] and using Hölder’s inequality with indices q, , and considering by Theorem 2.8, we get

If , then

is obviously true, otherwise, dividing both sides of (4.7) by and then taking the qth power on both sides of the resulting inequality we get also (4.8).

Using the Hölder’s inequality with indice , , we have, by (4.8),

Using Hölder’s inequality with indices , , we get by (4.9)

If , then the inequality

is obviously true, otherwise, dividing both sides of (4.10) by and then taking the power on both sides of the resulting inequality we get also (4.11). Since |u| ≤ z and we have

This gives (4.4). The proof is complete.   ☐

 

5. SOME APPLICATIONS TO CERTAIN DIFFERENTIAL EQUATIONS INVOLVING IMPULSES

In this section we always assume that both functions u and u' are left- continuous on [a, b], and that α = ϕ (see (3.2)). Consider the following impulsive differential equation: for ,

where q1 ∈ PC([a, b]): Now we define

Since by Lemma 3.8 for

the equation (5.1) implies the following equation:

where

We need the following result.

Lemma 5.1. If the function u satisfies the equation (5.1) and c ∈ [a, b], then we have

Proof. In the proof, we frequently use Lemma 3.8, . , and , .

This gives (5.4). And

This gives (5.5). And

This gives (5.6). Also

This gives (5.7). The proof is complete.   ☐

Theorem 5.2. Assume that u satisfies the equation (5.1) and u'(a) = 0, u(a) ≠ 0. If we have

where , then u(t) ≠ 0 for every t ∈ [a, b].

Proof. Assume that there is a number c ∈ (a, b] with u(c) = 0. Then multiplying both sides of (5.2) by u and integrating we have

Using Theorem 3.3, Lemma 3.5 and Theorem 3.6, and u(c) = Q(a) = 0, we get, since, by Theorem 2.9 and Remark 3.7, Q is left-continuous on [a, b]; and Δα(tk) = Δ+ α(tk) = 1, q(tk) = 0, ,

Since both u and u' are left-continuous

By Lemma 3.8 and Lemma 5.1, we get, since u(c) = u'(a) = 0,

By (5.9), (5.10) and (5.11), we have

Hence by Theorem 4.3 and Lemma 5.1, we get

If

then, since , and u' α(tk) = u(tk+) − u(tk) = 0. This implies that u is a constant on [a, c]. So u(c) = u(a) ≠0. But this is a contradiction to u(c) = 0. Hence we conclude that .

In (5.12), canceling , we get a contradiction to (5.8). This completes the proof.   ☐

In the following result we apply Theorem 4.4.

Theorem 5.3. Let q ∈ PC([a, b]) and let α = ϕ (see (3.2)). If u ∈ PC([a, b]) is left-continuous and a nontrivial solution of the following equation:

then we have

Proof. Substituting f ≡ 1, p = 0, q = 1, r = 0 into Theorem 4.4, then we have

So we have

Canceling , we get (5.13).   ☐