A new stability test for linear neutral differential equations

TL;DR

This paper introduces new explicit exponential stability criteria for linear scalar neutral differential equations with two bounded delays, utilizing the Bohl-Perron theorem and a transformation approach, with applications to neutral logistic equations.

Contribution

The paper presents novel explicit stability conditions for neutral differential equations with delays, expanding the theoretical understanding and practical criteria for stability analysis.

Findings

Derived new exponential stability conditions for neutral equations

Applied results to neutral logistic equations

Provided explicit criteria using Bohl-Perron theorem

Abstract

We obtain new explicit exponential stability conditions for the linear scalar neutral equation with two bounded delays $ \dot{x}(t)-a(t)\dot{x}(g(t))+b(t)x(h(t))=0, $ where $ 0\leq a(t)\leq A_0<1$, $0<b_0\leq b(t)\leq B$, using the Bohl-Perron theorem and a transformation of the neutral equation into a differential equation with an infinite number of delays. The results are applied to the neutral logistic equation.

Equations

$$\dot{x}(t)-a(t)\dot{x}(g(t))=-b(t)x(h(t))$$

$$\dot{x}(t)=r(t)x(t)\left(1-\frac{x(h(t))-\rho \dot{x}(g(t))}{K}\right),$$

$$\dot{x}(t)=r_{0}x(t)\left(1-\frac{x(t-\tau )-\rho \dot{x}(t-\tau )}{K}\right)$$

$$\dot{x}(t)-a(t)\dot{x}(g(t))+b(t)x(h(t))=f(t),t\ge t_0,x(t)=\phi (t),t\le t_0,\dot{x}(t)=\psi (t),t<t_0,$$

$$\dot{x}(t)-a(t)\dot{x}(g(t))+b(t)x(h(t))=0,x(t)=0,\dot{x}(t)=0,t<s,x(s)=1$$

$$x(t)=X(t,t_0)x_0+{\int }_{t_0}^{t}X(t,s)[(I-S{)}^{-1}f](s)ds+{\int }_{t_0}^{t}X(t,s)[(I-S{)}^{-1}F](s)ds,$$

$$\dot{x}(t)-a(t)\dot{x}(g(t))+b(t)x(h(t))=f(t),t\ge t_0,x(t)=0,t\le t_0,\dot{x}(t)=0,t<t_0$$

$$\dot{x}(t)+B(t)x(h_0(t))=0,B(t)\ge 0,0\le t-h_0(t)\le {\tau }_0.$$

$$\Vert (I-S{)}^{-1}{\Vert }_{L_{\infty }[t_0,\infty )\to L_{\infty }[t_0,\infty )}\le \frac{1}{1-A_0}.$$

$$|\dot{x}{|}_J\le {\Vert (I-S{)}^{-1}\Vert }_{L_{\infty }(J)\to L_{\infty }(J)}[B_0|x{|}_J+|f{|}_J]\le \frac{B_0}{1-A_0}|x{|}_J+M_1,$$

$$\dot{x}(t)=-b(t)x(h(t))-\sum _{j=1}^{\infty }\prod _{k=0}^{j-1}a\left(g^{[k]}(t)\right)b\left(g^{[j]}(t)\right)x\left(h(g^{[j]}(t))\right)+f_1(t),$$

$$B(t)=b(t)+\sum _{j=1}^{\infty }\prod _{k=1}^{j-1}a\left(g^{[k]}(t)\right)b\left(g^{[j]}(t)\right).$$

$$\dot{x}(t)+B(t)x(t)=b(t){\int }_{h(t)}^t\dot{x}(\xi )d\xi +\sum _{j=1}^{\infty }\prod _{k=0}^{j-1}a\left(g^{[k]}(t)\right)b\left(g^{[j]}(t)\right){\int }_{h(g^{[j]}(t))}^t\dot{x}(\xi )d\xi +f_1(t),$$

$$x(t)=\underset{t_0}{\overset{t}{\int }}{e}^{-\underset{s}{\overset{t}{\int }}B(\xi )d\xi }B(s)\left(\frac{1}{B(s)}[b(s)\underset{h(s)}{\overset{s}{\int }}\dot{x}(\xi )d\xi +\sum _{j=1}^{\infty }\prod _{k=0}^{j-1}a\left(g^{[k]}(s)\right)b\left(g^{[j]}(s)\right)\underset{h(g^{[j]}(s))}{\overset{s}{\int }}\dot{x}(\xi )d\xi ]\right)ds+f_2(t),$$

$$t-g(t)\le \sigma ,t-g^{[2]}(t)=t-g(t)+(g(t)-g(g(t)))\le 2\sigma ,\dots ,t-g^{[n]}(t)\le n\sigma ,$$

$$t-h(t)\le \tau ,t-h(g(t))=t-g(t)+(g(t)-h(g(t)))\le \sigma +\tau ,\dots ,t-h(g^{[n]}(t))\le n\sigma +\tau .$$

$$\frac{1}{B(s)}\left[b(s){\int }_{h(s)}^s\dot{x}(\xi )d\xi +\sum _{j=1}^{\infty }\prod _{k=0}^{j-1}a\left(g^{[k]}(s)\right)b\left(g^{[j]}(s)\right){\int }_{h(g^{[j]}(s))}^s\dot{x}(\xi )d\xi \right]$$

$$\le$$

$$\dot{x}(t)+B(t)x\left(t-\frac{1-A_0}{B_{0}e}\right)=0.$$

$$\tau \ge t-h(t)\ge \frac{1-A_0}{B_{0}e},\tau +\sigma \ge t-h(g(t))\ge \frac{1-A_0}{B_{0}e},\tau +n\sigma \ge t-h(g^{[n]}(t))\ge \frac{1-A_0}{B_{0}e}.$$

$$x(t)={\int }_{t_0}^t{X}_0(t,s)B(s)(\frac{1}{B(s)}\left[b(s)\underset{h(s)}{\overset{s-\frac{1-A_0}{B_{0}e}}{\int }}\dot{x}(\xi )\right)d\xi +\sum _{j=1}^{\infty }\prod _{k=0}^{j-1}a\left(g^{[k]}(s)\right)b\left(g^{[j]}(s)\right)\underset{h(g^{[j]}(s))}{\overset{s-\frac{1-A_0}{B_{0}e}}{\int }}\dot{x}(\xi )d\xi ]ds+f_3(s),$$

$$|x{|}_J\le \left(\left[\tau +\frac{\sigma A_0{B}_0(1-a_0)}{b_0(1-A_0{)}^2}\right]\frac{B_0}{1-A_0}-\frac{1}{e}\right)|x{|}_J+M_4,$$

$$\dot{x}(t)-a\dot{x}(g(t))=-bx(h(t)),$$

$$\dot{x}(t)-a(t)\dot{x}(g(t))=-b(t)x(t).$$

$$\dot{x}(t)-0.6\dot{x}(t-0.1)=-r(1+0.1\sin t)x(h(t)),0.9\le t-h(t)\le 1,$$

$$\dot{x}(t)-0.6\dot{x}(t-0.1)=-r(1+0.1\sin t)x(t-1).$$

$$\underset{t\to \infty }{\mathrm{limsup}}{\int }_{t-(3\tau +(N-1)\sigma )}^{t}b(s)ds<\frac{3}{2}-2a\left(1-\frac{1}{4}a\right)=0.48$$

$$\underset{t\to \infty }{\mathrm{limsup}}{\int }_{t-(\tau +(N-1)\sigma )}^{t}b(s)ds<\frac{3-4a^N}{2(1-a^N)}(1-a)\approx 0.544898$$

Full text

††journal: Applied Mathematics Letters

1 Introduction

Many applied problems lead to neutral differential equations as their mathematical models, for example, a model of a controlled motion of a rigid body, a distributed network (a long line with tunnel diodes), models of infection diseases, a price model in economic dynamics, see, for example, [3, 14, 15]. Though neutral delay differential equations describe important applied models, from mechanics to disease spread in epidemiology, compared to other classes of equations, stability theory for neutral equations with variable coefficients and delays is not sufficiently developed. In particular, there are no explicit stability results for general linear equations but only for particular classes of neutral equations, see [7, 8, 10, 11, 16] and references therein.

The aim of the present paper is to obtain stability conditions for the equation

[TABLE]

which depend on both delays. To this end, we transform (1.1) into a linear delay differential equation with an infinite number of delays. This method has not been applied before to stability problems, but used to study oscillation in [4].

As an application, we give local asymptotic stability tests for the logistic neutral equation

[TABLE]

where $\rho>0$ corresponds to higher resources consumption by a shrinking population. The model

[TABLE]

which is an autonomous version of (1.2), was studied in [9, 13, 17].

2 Preliminaries

We consider scalar delay differential equation (1.1) under the following conditions:

(a1) $a,b,g,h$ are Lebesgue measurable, $a$ and $b$ are essentially bounded on $[0,\infty)$ functions;

(a2) $0\leq a_{0}\leq a(t)\leq A_{0}<1,0<b_{0}\leq b(t)\leq B_{0}$ for all $t\geq t_{0}\geq 0$ and some fixed $t_{0}\geq 0$;

(a3) $mes~{}E=0\Longrightarrow mes~{}g^{-1}(E)=0$, where $mes~{}E$ is the Lebesgue measure of the set $E$;

(a4) $0\leq t-g(t)\leq\sigma$, $0\leq t-h(t)\leq\tau$ for $t\geq t_{0}$ and some $\sigma>0$, $\tau>0$ and $t_{0}\geq 0$.

Along with (1.1), we consider for each $t_{0}\geq 0$ an initial value problem

[TABLE]

where $f:[t_{0},\infty)\rightarrow{\mathbb{R}}$ is a Lebesgue measurable locally essentially bounded function, $\varphi:(-\infty,t_{0}]\rightarrow{\mathbb{R}}$ and $\psi:(-\infty,t_{0})\rightarrow{\mathbb{R}}$ are Borel measurable bounded functions.

Further, we assume that the above conditions hold without mentioning it.

Definition 1**.**

A locally absolutely continuous on $[t_{0},\infty)$ function $x:{\mathbb{R}}\rightarrow{\mathbb{R}}$ is called a solution of problem (2.1) if it satisfies the equation in (2.1) for almost all $t\in[t_{0},\infty)$ and the equalities in (2.1) for $t\leq t_{0}$. For each $s\geq t_{0}$ the solution $X(t,s)$ of the problem

[TABLE]

is called the fundamental function of equation (1.1). We assume $X(t,s)=0$ for $0\leq t<s$. We will say that equation (1.1) is uniformly exponentially stable if there exist $M>0$ and $\gamma>0$ such that the solution of problem (2.1) with $f\equiv 0$ has the estimate $\displaystyle|x(t)|\leq Me^{-\gamma(t-t_{0})}\sup_{t\in(-\infty,t_{0}]}(|\varphi(t)|+|\psi(t)|)$, $t\geq t_{0}$, where $M$ and $\gamma$ do not depend on $t_{0}\geq 0$, $\varphi$ and $\psi$. The fundamental function $X(t,s)$ of equation (1.1) has an exponential estimate if it satisfies $\displaystyle|X(t,s)|\leq M_{0}e^{-\gamma_{0}(t-s)}$ for some $t\geq s\geq t_{0}$, $M_{0}>0$ and $\gamma_{0}>0$.

For a fixed bounded interval $J=[t_{0},t_{1}]$, consider the space $L_{\infty}[t_{0},t_{1}]$ of all essentially bounded on $J$ functions with the norm $|y|_{J}=\operatorname*{ess\,sup}_{t\in J}|y(t)|$, denote $\|f\|_{[t_{0},\infty)}=\operatorname*{ess\,sup}_{t\geq t_{0}}|f(t)|$ for an unbounded interval, $I$ is the identity operator. Define an operator on the space $L_{\infty}[t_{0},t_{1}]$ as \displaystyle(Sy)(t)=\left\{\begin{array}[]{ll}a(t)y(g(t)),&g(t)\geq t_{0},\\ 0,&g(t)<t_{0}.\\ \end{array}\right. Note that there exists a unique solution of problem (2.1), and it can be presented as

[TABLE]

where $F(t)=a(t)\psi(g(t))-b(t)\varphi(h(t))$ and $\psi(g(t))=0$ for $g(t)\geq t_{0}$, $\varphi(h(t))=0$ for $h(t)\geq t_{0}$, see, for example, [2].

Existence of an exponential estimate for the fundamental function is equivalent [2] to the exponential stability for equations with bounded delays. The following result is usually referred to as the Bohl-Perron principle.

Lemma 1**.**

[2, Theorem 4.7.1]** Assume that the solution of the problem

[TABLE]

is bounded on $[t_{0},\infty)$ for any essentially bounded on $[t_{0},\infty)$ function $f$. Then equation (1.1) is uniformly exponentially stable.

In Lemma 1 we can consider boundedness of solutions not for all essentially bounded on $[t_{0},\infty)$ functions $f$ but only for essentially bounded on $[t_{0}+\delta,\infty)$ functions $f$ that vanish on $[t_{0},t_{0}+\delta)$ for any fixed $\delta>0$, see [5]. We further use this fact in the paper without an additional reference.

Denote by $X_{0}(t,s)$ the fundamental function of the equation with a single delay

[TABLE]

Lemma 2**.**

[5]** If $X_{0}(t,s)>0$ for $t\geq s\geq t_{0}$ then $\displaystyle\int_{t_{0}+\tau_{0}}^{t}X_{0}(t,s)B(s)\,ds\leq 1.$

Lemma 3**.**

[5, 12]** If for some $t_{0}\geq 0$, $\displaystyle\int_{h_{0}(t)}^{t}B(s)\,ds\leq\frac{1}{e},~{}t\geq t_{0}$ then $X_{0}(t,s)>0$ for $t\geq s\geq t_{0}$. If in addition $B(t)\geq b_{0}>0$ then equation (2.4) is exponentially stable.

Finally, the properties of the operator $S$ are outlined in the following lemma.

Lemma 4**.**

[1]** If $\|a\|_{[t_{0},\infty)}\leq A_{0}<1$ then $I-S$ is invertible in the space $L_{\infty}[t_{0},\infty)$,we have $\displaystyle((I-S)^{-1}y)(t)=y(t)+\sum_{j=1}^{\infty}\prod_{k=0}^{j-1}a\left(g^{[k]}(t)\right)y\left(g^{[j]}(t)\right),$ where $g^{[0]}(t)=t$, $g^{[1]}(t)=g(t)$, $g^{[n]}(t)=g((g^{[n-1]}(t)))$, and the operator norm satisfies

[TABLE]

3 Explicit Stability Conditions

Theorem 1**.**

*Assume that for $t\geq t_{0}$ at least one of the following conditions holds:

a) $\displaystyle\tau B_{0}+\frac{\sigma A_{0}B_{0}^{2}(1-a_{0})}{(1-A_{0})^{2}b_{0}}<1-A_{0};$

b) $\displaystyle t-h(t)\geq\frac{1-A_{0}}{eB_{0}}$    and    $\displaystyle\tau B_{0}+\frac{\sigma A_{0}B_{0}^{2}(1-a_{0})}{(1-A_{0})^{2}b_{0}}<\left(1+\frac{1}{e}\right)(1-A_{0}).$

Then equation (1.1) is uniformly exponentially stable.*

Proof.

Applying $(I-S)^{-1}$ to (2.3), using (2.5) on $J$ instead of $[t_{0},\infty)$ and (a2), we get

[TABLE]

where $\displaystyle M_{1}=\frac{\|f\|_{[t_{0},\infty)}}{1-A_{0}}$ and $\|f\|_{[t_{0},\infty)}<\infty$. By Lemma 4, (2.3) is equivalent to the equation with an infinite number of delays

[TABLE]

where $~{}f_{1}(t)=((I-S)^{-1}f)(t)$ and $\|f_{1}\|_{[t_{0},\infty)}<\infty$. Since $x(t)=0$ for $t\leq t_{0}$, we can assume that $b(t)=b_{0}$, $t\leq t_{0}$. Denote

[TABLE]

By Lemma 4, using the bounds for $a$ and $b$, we obtain $\displaystyle\frac{b_{0}}{1-a_{0}}\leq B(t)\leq\frac{B_{0}}{1-A_{0}}.$ Equation (3.2) can be rewritten in the form

[TABLE]

therefore

[TABLE]

where $B(t)\geq b_{0}>0$ implies $\displaystyle\|f_{2}\|_{[t_{0},\infty)}\leq\int_{t_{0}}^{t}e^{-\int_{s}^{t}B(\xi)d\xi}|f_{1}(s)|\,ds<\infty$. We have

[TABLE]

[TABLE]

Hence for $t\in J$,

[TABLE]

where the constant $M_{2}$ does not depend on $J$, and the last inequality is due to (3.1).

By Lemma 2, the solution of problem (2.3) satisfies $\displaystyle|x|_{J}\leq\left[\tau+\frac{\sigma A_{0}B_{0}(1-a_{0})}{b_{0}(1-A_{0})^{2}}\right]\frac{B_{0}}{1-A_{0}}|x|_{J}+M_{3},$ where $M_{3}$ is a constant not dependent on $J$.

Condition a) of the theorem implies $\displaystyle\left[\tau+\frac{\sigma A_{0}B_{0}(1-a_{0})}{b_{0}(1-A_{0})^{2}}\right]\frac{B_{0}}{1-A_{0}}<1.$ Hence $|x(t)|\leq M$ for $t\geq t_{0}$, for some constant $M$ which does not depend on $J$. By Lemma 1, equation (1.1) is uniformly exponentially stable.

Next, assume that the conditions in b) hold. Consider the following delay equation

[TABLE]

Since $B(t)\geq b_{0}$ and $\displaystyle\frac{B(t)(1-A_{0})}{B_{0}e}\leq\frac{1}{e}$, by Lemma 3 equation (3.3) is exponentially stable, and its fundamental function is positive: $X_{0}(t,s)>0$, $t\geq s\geq t_{0}$. We have

[TABLE]

Problem (2.3) is equivalent to (3.2) which has a solution

[TABLE]

where $f_{3}(t)=\int_{t_{0}}^{t}X_{0}(t,s)f_{1}(s)ds$, and $\|f_{3}\|_{[t_{0},\infty)}<\infty$, since (3.3) is exponentially stable.

By the same calculations as in a) we have

[TABLE]

where $M_{4}$ does not depend on the interval $J$.

By the second condition in b), we have $\displaystyle\left[\tau+\frac{\sigma A_{0}B_{0}(1-a_{0})}{b_{0}(1-A_{0})^{2}}\right]\frac{B_{0}}{1-A_{0}}-\frac{1}{e}<1.$ Hence $|x(t)|\leq M$ for $t\geq t_{0}$ for some constant $M$ which does not depend on $I$. By Lemma 1, equation (1.1) is exponentially stable. ∎

Consider now two partial cases of equation (1.1), one with constant coefficients

[TABLE]

where $a,b$ are positive constants, and another with a non-delayed term

[TABLE]

Corollary 1**.**

*If either a) $\displaystyle\tau b+\frac{\sigma ab}{1-a}<1-a;$ or b) $\displaystyle t-h(t)\geq\frac{1-a}{eb}$ and $\displaystyle\tau b+\frac{\sigma ab}{1-a}<\left(1+\frac{1}{e}\right)(1-a)$ then equation (3.4) is uniformly exponentially stable. *

Corollary 2**.**

If   $\displaystyle\frac{\sigma A_{0}B_{0}^{2}(1-a_{0})}{(1-A_{0})^{3}b_{0}}<1$   then equation (3.5) is uniformly exponentially stable.

4 Examples and Applications

First, we illustrate the results obtained in the paper with examples.

Example 1**.**

*Equation (3.4) with $g(t)\equiv t$ and variable $h(t)$, by Corollary 1, is uniformly exponentially stable if $\displaystyle\frac{b\tau}{1-a}<1+\frac{1}{e}\approx 1.37$. The well-known Myshkis test establishes stability for $\displaystyle\frac{b\tau}{1-a}<\frac{3}{2}$, under the assumption that the delay function is continuous. Corollary 1 gives a close estimate for a measurable delay. *

Example 2**.**

Consider an equation with a variable coefficient and time-dependent $h(t)$

[TABLE]

and its particular case with a constant delay

[TABLE]

We compare Theorem 1 with applicable results obtained in [16]. For both (4.1) and (4.2), we have $\tau=1$, $\sigma=0.1$, $b_{0}=0.9r$, $B_{0}=1.1r$, $a=0.6$. By Part a) of Theorem 1, $r<0.307$ implies exponential stability, while Part b) requires $r>0.149$ for (4.1), while $r>0.134$ for (4.2), and $r<r_{0}\approx 0.420347$.

In [16], a positive integer $N$ is introduced such that in (4.2), $a+\frac{3}{2}a^{N}=0.6+1.5\cdot 0.6^{N}\leq 1$; obviously, $N=3$. The first asymptotic stability condition for (4.2) from [16]

[TABLE]

is satisfied for $r<r_{1}\approx 0.14118$, while the second sufficient inequality from [16]

[TABLE]

holds for $r<r_{2}\approx 0.415025$. Note that in this case Theorem 1 gives a sharper estimate $\approx 0.420347$ for $r$; in addition, it provides a sufficient exponential stability condition for (4.1), while [16] for (4.2) only. To the best of our knowledge, other known conditions are also not applicable to (4.1).

Next, let us apply the results of Theorem 1 to logistic neutral equations (1.2) and (1.3), where $\rho>0$, $K>0$, $t-h(t)\leq\tau$, $t-g(t)\leq\tau$, $0<r_{0}\leq r(t)\leq R_{0}$, $r_{0}\rho\leq R_{0}\rho<1$, $r,g$ and $h$ are measurable functions. Equation (1.3) was studied in [9, 13, 17].

Proposition 1**.**

[17]** If $\displaystyle 2r_{0}|\rho|(2-r_{0}|\rho|)+r\tau<\frac{3}{2}$ then the positive equilibrium $K$ of equation (1.3) is locally asymptotically stable.

Note that the inequalities $2r_{0}|\rho|(2-r_{0}|\rho|)<\frac{3}{2}$ and $r_{0}\rho<1$ imply $r_{0}|\rho|<0.5$.

Theorem 2**.**

*If either a) $\displaystyle\tau R_{0}\rho+\frac{\sigma R_{0}^{2}\rho(1-r_{0})}{(1-R_{0})^{2}r_{0}}<1-R_{0},$ or

b) $\displaystyle t-h(t)\geq\frac{1-R_{0}}{eR_{0}\rho}$    and    $\displaystyle\tau R_{0}\rho+\frac{\sigma R_{0}^{2}\rho(1-r_{0})}{(1-R_{0})^{2}r_{0}}<\left(1+\frac{1}{e}\right)(1-R_{0})$

then the positive equilibrium $K$ of equation (1.2) is locally asymptotically stable.*

Proof.

Substituting $x=y-K$ in (1.2) leads to $\displaystyle\dot{y}(t)=-\frac{r(t)}{K}(y(t)+K)\left[y(h(t))-\rho\dot{y}(g(t))\right],$ its linearization about the zero equilibrium is $\displaystyle\dot{z}(t)=-r(t)\left[z(h(t))-\rho\dot{z}(g(t))\right]$. Applying Theorem 1 with $a_{0}=r_{0}$, $A_{0}=R_{0}$, $b_{0}=r_{0}\rho$, $B_{0}=R_{0}\rho$, we deduce that the linearization is exponentially stable, and thus $K$ is locally asymptotically stable. ∎

Remark 1**.**

The fact that exponential stability of the linearized equation implies local (and in some cases even global, see [6] and references therein) asymptotic stability of a nonlinear scalar equation was applied to conclude the proof of Theorem 2.

Corollary 3**.**

*If either $\displaystyle\tau r_{0}\rho<(1-r_{0})^{2}$ or $\displaystyle\frac{1-r_{0}}{e}<\tau r_{0}\rho<\left(1+\frac{1}{e}\right)(1-r_{0})^{2}$ then the positive equilibrium $K$ of equation (1.3) is locally asymptotically stable. *

Compared to Proposition 1, Theorem 2 is applicable to non-autonomous equations with different delays. Also, for $r_{0}=0.2$, $\rho=4$ and any $\displaystyle 0\leq\tau<0.8\left(1+\frac{1}{e}\right)$, Theorem 2 establishes local asymptotic stability of (1.3), while for these $r_{0}$ and $\rho$, Proposition 1 fails for any $\tau$.

Acknowledgment

The second author was partially supported by the NSERC research grant RGPIN-2015-05976.