Some Stability and Boundedness Criteria for a Class of Volterra Integro-differential Systems

Jito Vanualailai ^{∗}

Department of Mathematics and Computing Science

University of the South Pacific, Suva, FIJI. †

Abstract

Using Lyapunov and Lyapunov-like functionals, we study the stability and boundedness of the solutions of a system of Volterra integrodifferential equations. Our results, also extending some of the more well-known criteria, give new sufficient conditions for stability of the zero solution of the nonperturbed system, and prove that the same conditions for the perturbed system yield boundedness when the perturbation is L^{2}.

Key words and phrases: Lyapunov Stability, Volterra Equations 2000 Mathematics Subject Classification: 34D20, 34K20

1 Introduction

We consider the stability and boundedness of solutions of systems of Volterra integro-differential equations, with forcing functions, of the form

\frac{d}{dt}[x(t)] = A(t)f(x(t)) + ∫_{0}^{t} B(t, s)g(x(s))ds + h(t) ,  (1)

∗Supported partially by the Japanese Government JSPS Grant-in-Aid for Scientific Research (# 13640213), distributed by Kobe University, Japan, which the author visited as a Visiting Researcher in November, 2001, and the University of the South Pacific URC Grant # 6C011-1341-70762-00.

†e-mail: vanualailai@usp.ac.fj

in which A(t) is an n × n matrix function continuous on [0, ∞), B(t, s) is an n × n matrix continuous for 0 ≤ s ≤ t < ∞, f and g are n × 1 vector functions continuous on (−∞, ∞) and h is an n × 1 vector function defined almost everywhere on [0, ∞). Here, h(t) represent the forcing functions or external disturbances or inputs into system (1).

The qualitative behaviour of the solutions of systems of Volterra integrodifferential equations, especially the case where f(x) = g(x) = x and h(t) = 0, has been thoroughly analyzed by many researchers. Among the contributions in the 1980s, those of Burton are worthy of mention. His work ([1], [2]) laid the foundation for a systematic treatment of the basic structure and stability properties of Volterra integro-differential equations, mainly, via the direct method of Lyapunov. This paper essentially looks into some of the many interesting results established by Burton and proposes ways of utilizing the form of the Lyapunov functionals proposed by Burton to construct new or similar ones for system (1).

Now, if f(0) = g(0) = 0 and h(t) = 0, then system (1) reduces to

\frac{d}{dt}[x(t)] = A(t)f(x(t)) + ∫_{0}^{t} B(t, s)g(x(s))ds ,  (2)

so that x(t) ≡ 0 is a solution of (2) called the zero solution. The initial conditions for integral equations such as (1) or (2) involve continuous initial functions on an initial interval, say, x(t) = ϕ(t) for 0 ≤ t ≤ t_{0}. Hence, x(t; t_{0}, ϕ), t ≥ t_{0} ≥ 0 denotes the solution of (1) or (2), with the initial function ϕ : [0, t_{0}] → R^{n} assumed to be bounded and continuous on [0, t_{0}].

The definitions of the stability and the boundedness of solutions of (1) are given in Burton [1]. It is assumed that the functions in (1) are well-behaved, that continuous initial functions generate solutions, and that solutions which remain bounded can be continued.

2 The Scalar Equation

2.1 Nonperturbed Case

Consider the scalar equation

x′(t) = A(t)f(x(t)) + ∫_{0}^{t} B(t, s)g(x(s))ds .  (3)

We suppose that

A(t) is continuous for 0 ≤ t < ∞;  (4)
B(t, s) is continuous for 0 ≤ s ≤ t < ∞;  (5)
∫_{0}^{t} |B(u, s)|du is defined and continuous for 0 ≤ s ≤ t < ∞; (6)
f(x) and g(x) are continuous on (−∞, ∞);  (7)
xf(x) > 0 ∀x ≠ 0 , and f(0) = g(0) = 0 .  (8)

For comparison sake, we first state Burton’s theorem regarding the stability of the zero solution of (3).

Theorem 1 (Burton [3]). Let (4)–(8) hold and suppose there are constants m > 0 and M > 0 such that g^{2}(x) ≤ m^{2} f^{2}(x) if | x |≤ M. Define

β(t, k) = A(t) + k ∫_{t}^{∞} |B(u, t)|du + \frac{1}{2} ∫_{0}^{t} |B(t, s)|ds

If there exists k > 0 with m^{2} < 2k and β(t, k) ≤ 0 for t ≥ 0, then the zero solution of (3) is stable.

We next state an extension of Theorem 1, which Burton proved via the Lyapunov functional

V_{1}(t, x(·)) = ∫_{0}^{x} f(s)ds + k ∫_{0}^{t} ∫_{t}^{∞} |B(u, s)|duf^{2}(x(s))ds ,  (9)

the time-derivative along a trajectory of (3) of which is,

V′_{1(3)} ≤ β(t, k)f^{2}(x) − (2k − m^{2}) ∫_{0}^{t} |B(t, s)|f^{2}(x(s))ds ≤ β(t, k)f^{2}(x) .

In the process, and motivated by the work of Miyagi et. al in the construction of generalized Lyapunov functions for power systems [4] and single-machine systems [5], we propose a new Lyapunov functional. As a simple example will show, the new stability criterion may be used in situations where Theorem 1 cannot be applied.

Theorem 2. Let (4)–(8) hold, with A(t) < 0, and suppose there are constants

m > 0 and M > 0 such that g^{2}(x) ≤ m^{2} f^{2}(x) if |x| ≤ M ,  (10)
α > 4 and N > 0 such that 4x^{2} ≤ (α − 4) f^{2}(x) if |x| ≤ N , and(11)
J ≥ 1 such that − \frac{1}{4A(t)} ∫_{0}^{t} |B(t, s)|ds < \frac{1}{J} for every t ≥ 0 .(12)

Suppose further there is some constant k > 0 such that

\frac{(1 + α)m^{2}}{J} < k ,  (13)

and

A(t) + k ∫_{t}^{∞} |B(u, t)|du ≤ 0  (14)

for t ≥ 0. Then the zero solution of (3) is stable.

Proof. Consider the functional

V_{2}(t, x(·)) = \frac{1}{2}x^{2} + \sqrt{α} ∫_{0}^{x} \sqrt{uf(u)} du + \frac{1}{2}α ∫_{0}^{x} f(u)du
+ k ∫_{0}^{t} ∫_{t}^{∞} |B(u, s)|duf^{2}(x(s))ds .

We have, along a trajectory of (3),

V′_{2(3)} = x x′ + \sqrt{α} \sqrt{xf(x)} x′ + \frac{1}{2}αf(x) x′

+ \frac{d}{dt} [k ∫_{0}^{t} ∫_{t}^{∞} |B(u, s)|duf^{2}(x(s))ds] .

Recalling that A(t) < 0 for all t ≥ 0 and noting that the Schwarz inequality yields,

(∫_{0}^{t} B(t, s)g(x(s))ds)^{2} ≤ ∫_{0}^{t} |B(t, s)|ds ∫_{0}^{t} |B(t, s)|g^{2}(x(s))ds,

we have,

x x′ = A(t)xf(x) + x ∫_{0}^{t} B(t, s)g(x(s))ds

= A(t)xf(x) − (\sqrt{−A(t)} x − \frac{1}{2\sqrt{−A(t)}} ∫_{0}^{t} B(t, s)g(x(s))ds)^{2}

−A(t)x^{2} − \frac{1}{4A(t)} (∫_{0}^{t} B(t, s)g(x(s))ds)^{2}

≤ A(t)xf(x) − \frac{1}{4}(α − 4)A(t)f^{2}(x)

− \frac{1}{4A(t)} ∫_{0}^{t} |B(t, s)|ds ∫_{0}^{t} |B(t, s)|g^{2}(x(s))ds

≤ A(t)xf(x) − \frac{1}{4}(α − 4)A(t)f^{2}(x) + \frac{m^{2}}{J} ∫_{0}^{t} |B(t, s)|f^{2}(x(s))ds
= A(t)xf(x) − \frac{1}{4}αA(t)f^{2}(x) + A(t)f^{2}(x) + \frac{m^{2}}{J} ∫_{0}^{t} |B(t, s)|f^{2}(x(s))ds ,

and

\sqrt{α} \sqrt{xf(x)} x′ = − (\frac{\sqrt{α}}{2\sqrt{−A(t)}} x′ − \sqrt{−A(t)} \sqrt{xf(x)})^{2}

− A(t)xf(x) − \frac{α}{4A(t)} (x′)^{2}

≤ −A(t)xf(x) − \frac{1}{4}αA(t)f^{2}(x) − \frac{1}{2}αf(x) ∫_{0}^{t} B(t, s)g(x(s))ds

− \frac{α}{4A(t)} (∫_{0}^{t} B(t, s)g(x(s))ds)^{2}

≤ −A(t)xf(x) − \frac{1}{4}αA(t)f^{2}(x) − \frac{1}{2}αf(x) ∫_{0}^{t} B(t, s)g(x(s))ds

+ \frac{m^{2}α}{J} ∫_{0}^{t} |B(t, s)|f^{2}(x(s))ds .

The third and fourth terms of V′_{2(3)} yield

\frac{1}{2}αf(x) x′ = \frac{1}{2}αA(t)f^{2}(x) + \frac{1}{2}αf(x) ∫_{0}^{t} B(t, s)g(x(s))ds ,

and

\frac{d}{dt} [k ∫_{0}^{t} ∫_{t}^{∞} |B(u, s)|duf^{2}(x(s))ds] = k ∫_{t}^{∞} |B(u, t)|duf^{2}(x)
− k ∫_{0}^{t} |B(t, s)|f^{2}(x(s))ds ,

respectively.

Thus,

V′_{2(3)} ≤ [A(t) + k ∫_{t}^{∞} |B(u, t)|du] f^{2}(x)

− [k − \frac{m^{2}(1 + α)}{J}] ∫_{0}^{t} |B(t, s)|f^{2}(x(s))ds ,

which will be nonpositive if equations (13) and (14) are satisfied. Finally, to prove the positive definiteness of V_{2}, we see that if we define

r(u) = \{ \begin{matrix}(\sqrt{u} + \frac{1}{2} \sqrt{α}\sqrt{f(u)} )^{2} , & u ≥ 0 , \\ − (\sqrt{−u} − \frac{1}{2} \sqrt{α}\sqrt{−f(u)} )^{2} , & u < 0 , \end{matrix}

then we can rewrite V_{2} as

V_{2}(t, x(·)) = ∫_{0}^{x} r(u)du + \frac{1}{4}α ∫_{0}^{x} f(u)du
+ k ∫_{0}^{t} ∫_{t}^{∞} |B(u, s)|duf^{2}(x(s))ds ,

which is clearly positive definite given that ur(u) > 0 for u ≠ 0.

This completes the proof of Theorem 2.

Thus, we have proposed an alternate stability criterion for the scalar equation (3), and the criterion may be considered for cases where Burton’s Theorem 1, though simpler, cannot be applied.

Example 1. For the equation

x′ = −x + ∫_{0}^{t} \frac{1}{(1 + t − s)^{2}} [x^{2}(s) + \frac{1}{2}x(s)] ds ,

both Theorems 1 and 2 establish the stability of the zero solution. To see this, Theorem 1 yields M = m − 1/2 so that |x| ≤ m − 1/2. Also, we have β(t, k) < −1 + k + 1/2 ≤ 0 so that 0 < k ≤ 1/2. From m^{2} < 2k, we have, if we choose k = 1/2, the inequality 0 < m < 1. Thus, we may choose 1/2 < m < 1 to satisfy all conditions of Theorem 1. Theorem 2 yields, from (10), M = m − 1/2. From (11), α ≥ 8 and N = ∞. From (12), 4 > J. From (13), 0 < 9m^{2}/4 < k if we pick α = 8. From (14), 0 < k ≤ 1. Choose k = 1. Then 0 < m < 2/3. Thus, we may choose 1/2 < m < 2/3 to satisfy all conditions of Theorem 2.

□

Example 2. Analysis via Theorem 2 shows that the zero solution of

x′ = −x + ∫_{0}^{t} e^{−(t−s)/2} [x^{2}(s) + \frac{1}{4}x(s)] ds ,

is stable. That is, Theorem 2 yields, from (10), M = m − 1/4. From (11), α ≥ 8 and N = ∞. From (12), 2 > J. From (13), 0 < 9m^{2}/2 < k if we pick α = 8. From (14), 0 < k ≤ 1/2. Choose k = 1/2. Then 0 < m < 1/3. Thus, we may choose 1/4 < m < 1/3 to satisfy all conditions of the theorem.

Theorem 1 is not applicable.

□

2.2 Perturbed Case

The next two results, which extend Theorem 1 and Theorem 2, give a class of forcing functions that maintains the boundedness of the solutions of the equation

x′(t) = A(t)f(x(t)) + ∫_{0}^{t} B(t, s)g(x(s))ds + h(t) ,  (15)

where h : [0, ∞) → R is defined almost everywhere on [0, ∞).

Theorem 3. Let (4)–(8) hold and suppose there is a constant m > 0 such that g^{2}(x) ≤ m^{2} f^{2}(x) for all x ∈ R. Define

β(t, k) = A(t) + k ∫_{t}^{∞} |B(u, t)|du + \frac{1}{2} ∫_{0}^{t} |B(t, s)|ds

and let there be constants ρ > 0 and k > 0 such that m^{2} < 2k and β(t, k) ≤ −ρ for t ≥ 0. If

∫_{0}^{x} f(x)dx → ∞ as |x| → ∞ ,  (16)

and

h(·) ∈ L^{2}[0, ∞) ,

then all solutions of (15) are bounded.

Proof. Let ϵ > 0 and consider the functional

V_{3}(t, x(·)) = V_{1}(t, x(·)) + \frac{1}{4ϵ} ∫_{t}^{∞} h^{2}(u)du .

Since

∫_{0}^{∞}[h(u)]^{2}du < ∞ ,

we have,

\frac{d}{dt} [∫_{t}^{∞} h^{2}(u)du] = \frac{d}{dt} [∫_{0}^{∞} h^{2}(u)du − ∫_{0}^{t} h^{2}(u)du] = −h^{2}(t) ,

implying, therefore, the differentiability and hence the existence on [0, ∞) of the second term of the functional V_{3}. Thus, we have

V_{3}′_{(15)} ≤ β(t, k)f^{2}(x) + f(x)h(t) − \frac{1}{4ϵ}h^{2}(t)
≤ −ρf^{2}(x) + ϵf^{2}(x) + \frac{1}{4ϵ}h^{2}(t) − \frac{1}{4ϵ}h^{2}(t)
= −(ρ − ϵ)f^{2}(x) .

This completes the proof of Theorem 3 since we can always find some ϵ > 0 small enough such that (ρ − ϵ) > 0. Note that (16) ensures the radial unboundedness of V_{3}.

In the same fashion, we prove the following extension of Theorem 2.

Theorem 4. Let (4)–(8) hold, with A(t) < 0, and suppose there are constants

m > 0 such that g^{2}(x) ≤ m^{2} f^{2}(x) for all x ∈ R ,

α > 4 such that 4x^{2} ≤ (α − 4) f^{2}(x) for all x ∈ R , and

J ≥ 1 such that − \frac{1}{4A(t)} ∫_{0}^{t} |B(t, s)|ds < \frac{1}{J} for every t ≥ 0 .

Further, suppose there are constants k > 0 and ρ > 0 such that

\frac{(1 + α)m^{2}}{J} < k ,

and

A(t) + k ∫_{t}^{∞} |B(u, t)|du ≤ −ρ ,

for all t ≥ 0. If h(·) ∈ L^{2}[0, ∞), then all solutions of (15) are bounded. Proof. For ϵ > 0, the functional

V_{4}(t, x(·)) = V_{2}(t, x(·)) + \frac{1}{4ϵ} ∫_{t}^{∞} h^{2}(u)du ,

yields, given the definition,

τ = (\sqrt{\frac{\sqrt{α − 4}}{2}} + \sqrt{α})^{2} ,

the time-derivative,

V_{4}′_{(15)} ≤ [A(t) + k ∫_{t}^{∞} |B(u, t)|du] f^{2}(x)

+ [x + \sqrt{α}\sqrt{xf(x)} + \frac{1}{2}αf(x)] h(t) − \frac{1}{4ϵ}h^{2}(t)

≤ −ρf^{2}(x) + [\sqrt{|x|} + \sqrt{α}\sqrt{|f(x)|}]^{2} h(t) − \frac{1}{4ϵ}h^{2}(t)
≤ −ρf^{2}(x) + [\sqrt{\frac{\sqrt{α − 4}}{2}}\sqrt{|f(x)|} + \sqrt{α}\sqrt{|f(x)|}]^{2} h(t) − \frac{1}{4ϵ}h^{2}(t)
= −ρf^{2}(x) + τ|f(x)|h(t) − \frac{1}{4ϵ}h^{2}(t)
≤ −ρf^{2}(x) + τ^{2}ϵf^{2}(x) + \frac{1}{4ϵ}h^{2}(t) − \frac{1}{4ϵ}h^{2}(t)
= − [ρ − τ^{2}ϵ] f^{2}(x) .

This completes the proof of Theorem 4 since we can always find some ϵ > 0 small enough such that (ρ − τ^{2}ϵ) > 0. We note that V_{4} → ∞ if |x| → ∞.

3 The Vector Equation

3.1 Nonperturbed Case

First, we consider the nonperturbed system (2). If we suppose that f, g ∈ C^{1}[R^{n}, R^{n}], then we can define

D(x) = [d_{ij}(x)]_{n×n} with d_{ij}(x) = ∫_{0}^{1} \frac{∂f_{i}(ux)}{∂(ux_{j})}du ,

and

E(x) = [e_{ij}(x)]_{n×n} with e_{ij}(x) = ∫_{0}^{1} \frac{∂g_{i}(ux)}{∂(ux_{j})}du ,

which are defined for all x ∈ R^{n}. Hence, assuming f(0) = g(0) = 0, system (2) can be written as

x′(t) = A(t)D(x(t))x(t) + ∫_{0}^{t} B(t, s)E(x(s))x(s) ds ,  (17)

the i-th component of which is

x′_{i}(t) = a_{ii}(t) [d_{ii}(x)x_{i} + ∑_{{j=1 \atop j≠i}}^{n} d_{ij}(x)x_{j}]

+ ∑_{{j=1 \atop j≠i}}^{n} a_{ij}(t) [d_{ji}(x)x_{i} + ∑_{{k=1 \atop k≠i}}^{n} d_{jk}(x)x_{k}]

+ ∑_{k=1}^{n} ∫_{0}^{t} [b_{ii}(t, s)e_{ik}(x(s)) + ∑_{{j=1 \atop j≠i}}^{n} b_{ij}(t, s)e_{jk}(x(s))] x_{k}(s)ds ,

noting that in the above equation d_{ij}(x)x_{j} and e_{ij}(x)x_{j}, for i, j = 1, …, n, are continuously differentiable with respect to x ∈ R^{n} simply for the reason that D(x)x = f(x) and E(x)x = g(x) with f, g ∈ C^{1}[R^{n}, R^{n}].

The next result is new.

Theorem 5. Let f, g ∈ C^{1}[R^{n}, R^{n}] , f(0) = g(0) = 0 and

β_{i}(t, x) = a_{ii}(t)d_{ii}(x) + ∑_{{j=1 \atop j≠i}}^{n} a_{ij}(t)d_{ji}(x)

+ \frac{1}{2} ∑_{{j=1 \atop j≠i}}^{n} [|a_{ii}(t)d_{ij}(x)| + |a_{jj}(t)d_{ji}(x)| + |a_{ji}(t)d_{ii}(x)| + |a_{ij}(t)d_{jj}(x)|]

+ \frac{1}{2} ∑_{{j=1 \atop j≠i}}^{n} ∑_{{k=1 \atop k≠i} ; k≠j}^{n} [|a_{ik}(t)d_{kj}(x)| + |a_{jk}(t)d_{ki}(x)|]

+ \frac{1}{2} ∑_{k=1}^{n} ∫_{t}^{∞} [|b_{kk}(u, t)|e_{ki}^{2}(x) + ∑_{{j=1 \atop j≠i}}^{n} |b_{kj}(u, t)|e_{ji}^{2}(x)] du

+ \frac{n}{2} ∫_{0}^{t} [|b_{ii}(t, s)| + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(t, s)|] ds .

(18)

Suppose β_{i}(t, x) ≤ 0 for i = 1, …, n, t ≥ 0 and x ∈ R^{n}. Then the zero solution of system (2) is stable. Proof. Consider the functional

V_{5}(t, x(·)) = \frac{1}{2} ∑_{i=1}^{n} x_{i}^{2}(t)

+ \frac{1}{2} ∑_{i=1}^{n} ∑_{k=1}^{n} ∫_{0}^{t} ∫_{t}^{∞} [|b_{ii}(u, s)|e_{ik}^{2}(x(s)) + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(u, s)|e_{jk}^{2}(x(s))] dux_{k}^{2}(s)ds .

Now,

\frac{1}{2} ∑_{i=1}^{n} \frac{d}{dt} [x_{i}^{2}/2]_{(2)} = ∑_{i=1}^{n} x_{i}x′_{i}

= ∑_{i=1}^{n} x_{i} \{a_{ii}(t) [d_{ii}(x)x_{i} + ∑_{{j=1 \atop j≠i}}^{n} d_{ij}(x)x_{j}]

+ ∑_{{j=1 \atop j≠i}}^{n} a_{ij}(t) [d_{ji}(x)x_{i} + ∑_{{k=1 \atop k≠i}}^{n} d_{jk}(x)x_{k}]

+ ∑_{k=1}^{n} ∫_{0}^{t} [b_{ii}(t, s)e_{ik}(x(s)) + ∑_{{j=1 \atop j≠i}}^{n} b_{ij}(t, s)e_{jk}(x(s))] x_{k}(s)ds\}

≤ ∑_{i=1}^{n} \{a_{ii}(t)d_{ii}(x) + ∑_{{j=1 \atop j≠i}}^{n} a_{ij}(t)d_{ji}(x)\} x_{i}^{2} + ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} a_{ii}(t)d_{ij}(x)x_{j}x_{i}

+ ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} ∑_{{k=1 \atop k≠i}}^{n} a_{ij}(t)d_{jk}(x)x_{k}x_{i}

+ \frac{1}{2} ∑_{i=1}^{n} ∑_{k=1}^{n} ∫_{0}^{t} |b_{ii}(t, s)|[e_{ik}^{2}(x(s))x_{k}^{2}(s) + x_{i}^{2}]ds

+ \frac{1}{2} ∑_{i=1}^{n} ∑_{k=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} ∫_{0}^{t} |b_{ij}(t, s)|[e_{jk}^{2}(x(s))x_{k}^{2}(s) + x_{i}^{2}]ds

= ∑_{i=1}^{n} \{a_{ii}(t)d_{ii}(x) + ∑_{{j=1 \atop j≠i}}^{n} a_{ij}(t)d_{ji}(x) + \frac{n}{2} ∫_{0}^{t} [|b_{ii}(t, s)| + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(t, s)|] ds\} x_{i}^{2}

+ ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} a_{ii}(t)d_{ij}(x)x_{j}x_{i} + ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} ∑_{{k=1 \atop k≠i}}^{n} a_{ij}(t)d_{jk}(x)x_{k}x_{i}

+ \frac{1}{2} ∑_{i=1}^{n} ∑_{k=1}^{n} ∫_{0}^{t} [|b_{ii}(t, s)|e_{ik}^{2}(x(s)) + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(t, s)|e_{jk}^{2}(x(s))] x_{k}^{2}(s)ds ,

where

∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} a_{ii}(t)d_{ij}(x)x_{j}x_{i} ≤ \frac{1}{2} ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} [|a_{ii}(t)d_{ij}(x)| + |a_{jj}(t)d_{ji}(x)|] x_{i}^{2} ,

and

∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} ∑_{{k=1 \atop k≠i}}^{n} a_{ij}(t)d_{jk}(x)x_{k}x_{i}

= ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} a_{ji}(t)d_{ii}(x)x_{j}x_{i} + ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} ∑_{{k=1 \atop k≠i} ; k≠j}^{n} a_{ik}(t)d_{kj}(x)x_{j}x_{i}

≤ \frac{1}{2} ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} [|a_{ji}(t)d_{ii}(x)| + |a_{ij}(t)d_{jj}(x)|] x_{i}^{2}

+ \frac{1}{2} ∑_{i=1}^{n} ∑_{{j=1 \atop j≠i}}^{n} ∑_{{k=1 \atop k≠i} ; k≠j}^{n} [|a_{ik}(t)d_{kj}(x)| + |a_{jk}(t)d_{ki}(x)|] x_{i}^{2} .

Also, we have

\frac{1}{2} ∑_{i=1}^{n} ∑_{k=1}^{n} \frac{d}{dt} [∫_{0}^{t} ∫_{t}^{∞} (|b_{ii}(u, s)|e_{ik}^{2}(x(s)) + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(u, s)|e_{jk}^{2}(x(s))) dux_{k}^{2}(s)ds]

= \frac{1}{2} ∑_{i=1}^{n} ∑_{k=1}^{n} \{∫_{t}^{∞} [|b_{ii}(u, t)|e_{ik}^{2}(x(t)) + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(u, t)|e_{jk}^{2}(x(t))] dux_{k}^{2}(t)

− ∫_{0}^{t} [|b_{ii}(t, s)|e_{ik}^{2}(x(s)) + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(t, s)|e_{jk}^{2}(x(s))] x_{k}^{2}(s)ds\}

= \frac{1}{2} ∑_{i=1}^{n} ∑_{k=1}^{n} \{∫_{t}^{∞} [|b_{kk}(u, t)|e_{ki}^{2}(x(t)) + ∑_{{j=1 \atop j≠i}}^{n} |b_{kj}(u, t)|e_{ji}^{2}(x(t))] dux_{i}^{2}(t)

− ∫_{0}^{t} [|b_{ii}(t, s)|e_{ik}^{2}(x(s)) + ∑_{{j=1 \atop j≠i}}^{n} |b_{ij}(t, s)|e_{jk}^{2}(x(s))] x_{k}^{2}(s)ds\} .

Thus,

\frac{d}{dt}[V_{5}]_{(2)} ≤ ∑_{i=1}^{n} β_{i}(t, x)x_{i}^{2} ≤ 0 .

Moreover, V_{5} is clearly positive definite, given that

V_{5}(t, x(·)) ≥ \frac{1}{2} ∑_{i=1}^{n} x_{i}^{2}(t) .

Hence, we obtain the conclusion of Theorem 5.

Putting n = 1 in Theorem 5 yields a new stability criterion for the scalar case (3),

x′(t) = A(t)f(x(t)) + ∫_{0}^{t} B(t, s)g(x(s))ds ,

rewritten as

x′(t) = A(t)D(x(t))x(t) + ∫_{0}^{t} B(t, s)E(x(s))x(s)ds,

on the assumption that f, g ∈ C^{1}[R, R] and f(0) = g(0) = 0, and on letting

D(x) = \{ \begin{matrix}\frac{f(x)}{x}, & x ≠ 0 , \\ f′(0), & x = 0 , \end{matrix} and E(x) = \{ \begin{matrix}\frac{g(x)}{x}, & x ≠ 0 , \\ g′(0), & x = 0 . \end{matrix}

Now, if n = 1, then, from (18),

β_{1}(t, x_{1}) = a_{11}(t)d_{11}(x_{1}) + \frac{1}{2} ∫_{t}^{∞} |b_{11}(u, t)|due_{11}^{2}(x_{1}) + \frac{1}{2} ∫_{0}^{t} |b_{11}(t, s)|ds .

Putting x_{1} = x, a_{11}(t) = A(t), d_{11}(x_{1}) = D(x), e_{11}(x_{1}) = E(x) and b_{11}(t, s) = B(t, s), we have the following result:

Corollary 1. Let f, g ∈ C^{1}[R, R], f(0) = g(0) = 0 and

β(t, x) = A(t)D(x) + \frac{1}{2}E^{2}(x) ∫_{t}^{∞} |B(u, t)| du + \frac{1}{2} ∫_{0}^{t} |B(t, s)| ds .

If β(t, x) ≤ 0 for t ≥ 0 and x ∈ R, then the zero solution of (3) is stable.

Proof. Using V_{5}, with n = 1, so that

V_{5}(t, x(·)) = \frac{1}{2}x^{2}(t) + \frac{1}{2} ∫_{0}^{t} ∫_{t}^{∞} |B(u, s)|duE^{2}(x(s))x^{2}(s)ds ,

we get

\frac{d}{dt}[V_{5}]_{(3)} ≤ β(t, x)x^{2} ≤ 0 .

Corollary 1 is thus proved. Example 3. For the equation

x′ = −e^{t} (x + \frac{sin x}{2}) + k ∫_{0}^{t} e^{−(t−s)}[1 − cos x(s)] ds , k > 0 ,  (19)

Corollary 1 is easier than either Theorem 1 or Theorem 2 to apply. Thus, we have, for all t ≥ 0 and x ≠ 0,

β(t, x) = −e^{t} (1 + \frac{1}{2} \frac{sin x}{x}) + \frac{1}{2} [k (\frac{1 − cos x}{x})^{2} + k (1 − e^{−t})]
≤ −e^{t} + \frac{1}{2}e^{t} |\frac{sin x}{x}| + \frac{k}{2} (\frac{1 − cos x}{x})^{2} + \frac{k}{2} |1 − e^{−t}|
≤ −e^{t} + \frac{1}{2}e^{t} + k ≤ −\frac{1}{2} + k ,

so that β(t, x) ≤ 0 if 0 < k ≤ 1/2. Moreover, for these values of k,

β(t, 0) = −\frac{3}{2}e^{t} + \frac{k}{2}|1 − e^{−t}| ≤ −\frac{3}{2} + \frac{k}{2} < 0 .

Hence, by Corollary 1, the zero solution of (19) is stable if k ∈ (0, 1/2].

Example 4. The system

[ \begin{matrix}x′_{1}(t) \\ x′_{2}(t) \end{matrix} ] = [ \begin{matrix}a_{11}(t) & a_{12}(t) \\ a_{21}(t) & a_{22}(t) \end{matrix} ] [ \begin{matrix}f_{1}(x_{1}(t), x_{2}(t)) \\ f_{2}(x_{1}(t), x_{2}(t)) \end{matrix} ]

+ ∫_{0}^{t} [ \begin{matrix}b_{11}(t, s) & b_{12}(t, s) \\ b_{21}(t, s) & b_{22}(t, s) \end{matrix} ] [ \begin{matrix}g_{1}(x_{1}(s), x_{2}(s)) \\ g_{2}(x_{1}(s), x_{2}(s)) \end{matrix} ] ds ,

is stable if f, g ∈ C^{1}[R^{2}, R^{2}], with f(0) = g(0) = 0, and if for c_{1}, c_{2} > 0, t ≥ 0 and x ∈ R^{2}, we have, using (18) and condition (b) of Theorem 5,

β_{1}(t, x)x_{1}^{2} = {a_{11}(t)d_{11}(x) + a_{12}(t)d_{21}(x)

+ \frac{1}{2} [|a_{11}(t)d_{12}(x)| + |a_{22}(t)d_{21}(x)| + |a_{21}(t)d_{11}(x)| + |a_{12}(t)d_{22}(x)|]

+ \frac{1}{2} ∫_{t}^{∞} [|b_{11}(u, t)e_{11}^{2}(x)| + |b_{12}(u, t)e_{21}^{2}(x)| + 2|b_{22}(u, t)e_{21}^{2}(x)|] du

+ ∫_{0}^{t} [|b_{11}(t, s)| + |b_{12}(t, s)|] ds}x_{1}^{2} ≤ −c_{1}x_{1}^{2} ,

and

β_{2}(t, x)x_{2}^{2} = {a_{22}(t)d_{22}(x) + a_{21}(t)d_{12}(x)

+ \frac{1}{2} [|a_{22}(t)d_{21}(x)| + |a_{11}(t)d_{12}(x)| + |a_{12}(t)d_{22}(x)| + |a_{21}(t)d_{11}(x)|]

+ \frac{1}{2} ∫_{t}^{∞} [2|b_{11}(u, t)e_{12}^{2}(x)| + |b_{22}(u, t)e_{22}^{2}(x)| + |b_{21}(u, t)e_{12}^{2}(x)|] du

+ ∫_{0}^{t} [|b_{22}(t, s)| + |b_{21}(t, s)|] ds}x_{2}^{2} ≤ −c_{2}x_{2}^{2} .

The following simple, but illustrative, case is one such stable system:

[ \begin{matrix}x′_{1}(t) \\ x′_{2}(t) \end{matrix} ] = [ \begin{matrix}\frac{10}{t + 1} & −20 \\ −20 & \frac{10}{t + 1} \end{matrix} ] [ \begin{matrix}x_{2}(t) − \frac{1}{20}x_{1} tanh(x_{1}(t)) \\ x_{1}(t) + \frac{1}{20}x_{2} tanh(x_{2}(t)) \end{matrix} ]

+ ∫_{0}^{t} [ \begin{matrix}\frac{1}{(1 + t − s)^{2}} & 0 \\ 0 & \frac{1}{4[(t − s)^{2} + 1]} \end{matrix} ] [ \begin{matrix}x_{1}(s) + x_{2}(s) \\ x_{1}(s) + x_{2}(s) \end{matrix} ] ds ,

(20) or, in the form of (17),

[ \begin{matrix}x′_{1}(t) \\ x′_{2}(t) \end{matrix} ] = [ \begin{matrix}\frac{10}{t + 1} & −20 \\ −20 & \frac{10}{t + 1} \end{matrix} ] [ \begin{matrix}−\frac{1}{20} tanh(x_{1}(t)) & 1 \\ 1 & \frac{1}{20} tanh(x_{2}(t)) \end{matrix} ] [ \begin{matrix}x_{1}(t) \\ x_{2}(t) \end{matrix} ]

+ ∫_{0}^{t} [ \begin{matrix}\frac{1}{(1 + t − s)^{2}} & 0 \\ 0 & \frac{1}{4[(t − s)^{2} + 1]} \end{matrix} ] [ \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} ] [ \begin{matrix}x_{1}(s) \\ x_{2}(s) \end{matrix} ] ds .

Now, for all t ≥ 0 and for all x ∈ R^{2}, we have,

β_{1}(t, x)x_{1}^{2} = \{(\frac{10}{t + 1}) (−\frac{tanh x_{1}}{20} ) + (−20) · 1

+ \frac{1}{2} [|\frac{10}{t + 1} · 1| + |\frac{10}{t + 1} · 1| + |−20 (−\frac{tanh x_{1}}{20} )| + |−20 · \frac{tanh x_{2}}{20} |]

+\frac{1}{2} [∫_{t}^{∞} \frac{1}{(1 + u − t)^{2}} · 1 du + 2 ∫_{t}^{∞} \frac{1}{4[(u − t)^{2} + 1]} · 1 du]

+ ∫_{0}^{t} \frac{1}{(1 + t − s)^{2}} ds\} x_{1}^{2}

= \{−\frac{tanh x_{1}}{2(t + 1)} − 20 + \frac{10}{t + 1} + \frac{1}{2} [| tanh x_{1}| + | tanh x_{2}|] + \frac{1}{2} [1 + \frac{2π}{8} ]

+ [1 − \frac{1}{1 + t}]\} x_{1}^{2} ,

which clearly shows that β_{1}(t, x)x_{1}^{2} is continuous on [0, ∞) × R^{2}. Moreover

β_{1}(t, x)x_{1}^{2} ≤ − (6 − \frac{π}{8} ) x_{1}^{2} ≤ 0 .

Next, we have,

β_{2}(t, x)x_{2}^{2} = \{(\frac{10}{t + 1}) (\frac{tanh x_{2}}{20} ) + (−20) · 1

+ \frac{1}{2} [|\frac{10}{t + 1} · 1| + |\frac{10}{t + 1} · 1| + |−20 (\frac{tanh x_{2}}{20} )| + |−20 (−\frac{tanh x_{1}}{20} )|]

+\frac{1}{2} [2 ∫_{t}^{∞} \frac{1}{(1 + u − t)^{2}} · 1 du + ∫_{t}^{∞} \frac{1}{4[(u − t)^{2} + 1]} · 1 du]

+ ∫_{0}^{t} \frac{1}{4[(t − s)^{2} + 1]} ds\} x_{2}^{2}

= \{\frac{tanh x_{2}}{2(t + 1)} − 20 + \frac{10}{t + 1} + \frac{1}{2} [| tanh x_{1}| + | tanh x_{2}|] + \frac{1}{2} [2 + \frac{π}{8} ]

+\frac{tan^{−1} t}{4}\} x_{2}^{2} ,

which shows that β_{2}(t, x)x_{2}^{2} is continuous on [0, ∞) × R^{2}. Moreover

β_{2}(t, x)x_{2}^{2} ≤ − (7 − \frac{π}{4} ) x_{2}^{2} ≤ 0 .

Hence, we have shown that β_{i}(t, x)x_{i}^{2} ≤ 0 for i = 1, 2, t ≥ 0 and x ∈ R^{2}. The zero solution of system (20) is therefore stable by Theorem 5.

3.2 Perturbed Case

We finally consider system (1), where h(t) = (h_{1}(t), …, h_{n}(t))^{T}.

Theorem 6. Let the conditions of Theorem 5 hold, with the last condition replaced by the assumption that there are constants c_{i} > 0 such that β_{i}(t, x) ≤ − c_{i} for i = 1, …, n, t ≥ 0 and x ∈ R^{n}. If h_{i}(·) ∈ L^{2}[0, ∞) for i = 1, …, n, then all solutions of (1) are bounded.

Proof. Let ϵ > 0 and consider the functional

V_{6}(t, x(·)) = V_{5}(t, x(·)) + \frac{1}{4ϵ} ∑_{i=1}^{n} ∫_{t}^{∞} h_{i}^{2}(u)du ,

which is clearly radially unbounded. Now, we have, for c = − min{c_{1}, …, c_{n}}, t ≥ 0 and x ∈ R^{n},

\frac{d}{dt}[V_{6}]_{(1)} ≤ ∑_{i=1}^{n} β_{i}(t, x)x_{i}^{2} + ∑_{i=1}^{n} x_{i}h_{i}(t) − \frac{1}{4ϵ} ∑_{i=1}^{n} h_{i}^{2}(t)
≤ −(c − ϵ) ∑_{i=1}^{n} x_{i}^{2} .

We have thus proved the boundedness of solutions of (1), since we can always find ϵ > 0 such that (c − ϵ) ≥ 0.

The following corollary follows directly from Theorem 6 by putting n = 1.

Corollary 2. Let the conditions of Corollary 1 hold, with the last condition replaced by the assumption that there is a constant c > 0 such that β(t, x) ≤ − c for t ≥ 0 and x ∈ R. If h(·) ∈ L^{2}[0, ∞), then all solutions of the scalar equation (15) are bounded.

4 Acknowledgement

The author would like to thank foremost the anonymous referees for the detailed checking of the original version of this paper and the ensuing critical comments. Also the author is indepted to Professor Shin-ichi Nakagiri of Kobe University, Japan, and Professor Junhong Ha of Korea University of Technology and Education, South Korea, for their helpful suggestions.

References

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[3] T. A. Burton, Construction of Liapunov functionals for Volterra equations, Journal of Mathematical Analysis and Applications, 85 (1982), 90–105.

[4] H. Miyagi, and K. Yamashita, Construction of non-Luré-type Lyapunov function for multimachine power systems, IEE Proceedings, 134 (1988), 805–812.

[5] H. Miyagi, T. Ohshiro and K. Yamashita, Generalized Lyapunov function for Liénard-type non-linear systems, International Journal of Control, 48 (1987), 95–100.