Analysis of the Sign of the Solution for Certain Second-Order Periodic Boundary Value Problems with Piecewise Constant Arguments

Abstract

We find sufficient conditions for the unique solution of certain second-order boundary value problems to have a constant sign. To this purpose, we use the expression in terms of a Green’s function of the unique solution for impulsive linear periodic boundary value problems associated with second-order differential equations with a functional dependence, which is a piecewise constant function. Our analysis lies in the study of the sign of the Green’s function.

1. Introduction

Delay differential equations find interesting applications in fields like biology, fisiology, physics, etc. Some interesting and recent applications of this type of equations appear, for instance, when proposing a mathematical model for electrohydraulic servomechanisms [1], absorption complexities in pharmacokinetics [2], the approximation of Fitzhugh–Nagumo and Hodgkin–Huxley models for action potential generation in excitable cells [3], tick population with diapause [4], or thermal models for variable pipe flow [5].

The study of the solutions’ properties to such kind of models is a relevant area of study [6]. In particular, the existence of solutions with a constant sign is important from the point of view of the interpretation of the biological models. Indeed, the non-negativity of the solutions [7] is essential for the coherence between mathematical expressions and the magnitudes that, in this context, normally refer to positive values such as the number of individuals in a certain population or the concentration of a certain substance. Recent work [8] deepens the problem of the existence of non-negative solutions to linear autonomous functional differential equations. See also [9] for order preserving evolution operators of functional differential equations.

This paper is devoted to delay differential equations with piecewise constant delay given by the greatest integer function. The appearance of such kind of delays can be interpreted as the dependence of the rate of change of the variable of interest not only on the present state but also on some previous (equidistributed with respect to time) memorized values of the state (see [10], where a Liénard-type differential equation with piecewise constant delays is considered).

The study of second-order differential equations independent of the first-order derivative $x^{\prime }$ with a functional dependence given by a piecewise constant argument goes back to references [11,12,13,14,15]. Here, the functional dependence considered is the greatest integer function. In [16], the authors consider boundary value problems for second-order differential equations with deviating arguments and, in [17], the uniqueness issue is addressed for second-order linear functional differential equations.

In particular, the periodic boundary value problem:

$$\left\{\begin{array}{c}-x^{″}\left(t\right)=f(t,x\left(t\right),x\left(\left[t\right]\right)),t\in J=[0,T],\\ x\left(0\right)=x\left(T\right),x^{\prime }\left(0\right)=x^{\prime }\left(T\right),\end{array}\right.$$

is studied in [15], where $T>0$, and $\left[t\right]$ denotes the floor of t, that is, the greatest integer less than or equal to t. The references [18,19] are devoted to similar problems for first-order differential equations. In order to consider a problem where a delay term also affects the derivative of the function, the existence of solutions for second-order functional differential equations with piecewise constant arguments of the form:

$$\left\{\begin{array}{c}{x}^{″}\left(t\right)+a{x}^{\prime }\left(t\right)+bx\left(t\right)+c{x}^{\prime }\left(\left[t\right]\right)+dx\left(\left[t\right]\right)=\sigma \left(t\right),t\in J=[0,T],\\ x\left(0\right)=x\left(T\right),\\ x^{\prime }\left(0\right)=x^{\prime }\left(T\right)+\lambda ,\end{array}\right.$$

(1)

is considered in [20] by studying the existence of solutions for the impulsive linear periodic boundary value problems:

$$\left\{\begin{array}{c}{x}^{″}\left(t\right)+a{x}^{\prime }\left(t\right)+bx\left(t\right)+c{x}^{\prime }\left(\left[t\right]\right)+dx\left(\left[t\right]\right)=0,t\in \mathbb{R},\\ x\left(0\right)=x\left(T\right),\\ x^{\prime }\left(0^{-}\right)=x^{\prime }\left(T^{+}\right),\\ x^{\prime }\left(s^{+}\right)=x^{\prime }\left(s^{-}\right)+1,\end{array}\right.$$

(2)

where $s\in J$, $a,b,c,d,\lambda \in \mathbb{R}$, $T>0$, and $x^{\prime }\left(z^{-}\right)$, $x^{\prime }\left(z^{+}\right)$ denote, respectively, the one-sided left and right limits of $x^{\prime }$ at the point $z\in J$.

Although the expression of the solution to (1) can be obtained directly from the solution of (2), which represents the Green’s function, it is quite tedious to find sufficient conditions for the Green’s function to have a constant sign. In this paper, we prove the existence of non-negative (resp. non-positive) solutions for (1) under the appropriate conditions.

Some other references dealing with similar problems are [21], where the existence and stability of periodic solutions for a quasi-linear differential equation with arguments which are piecewise constant functions is studied, or [22], where the stability of the null solution for the corresponding quasilinear equation is analyzed. In [23], a similar study is made to obtain the Green’s function to express the unique solution to a second-order nonautonomous differential equation with piecewise constant arguments.

The objectives of the main results in the paper are summarized as follows:

• Theorems 1–3 give sufficient conditions for the unique solution to problem (2) with $s=0$ or $s=T$ to have a constant sign. The technical Lemmas A2–A6 allow to interpret two of the sufficient conditions directly in terms of the coefficients of the linear problem $a,b,c,d$ (see Remarks 1 and 2). The other condition is algebraically manipulated in Lemma 1, Remark 3, Lemma 2, and Remark 4;
• An analogous study can be made for $0<s<T$ with $s\in {\mathbb{Z}}^{+}$, see Theorem 4, Lemma 3, Theorem 5, and Remark 5;
• To improve significantly these first results, it is important to determine what planar regions of the type:

  $$S:=\{(x,y)\in {\mathbb{R}}^2:x\ge 0,y\ge -Mx\},$$

  for a certain $M>0$, are such that their elements are mapped into the interval $[0,+\infty )$ when scalar multiplied by a certain vector $(h_1\left(z\right),h_2\left(z\right))$, which will be defined in relation with the differential problem. Lemma 4 is helpful to give an answer, therefore, some of the results are extended in the sense that the conditions required are reduced to some properties on the sign of functions $h_1$ and $h_2$, and the fact that certain iterates remain in $S$ (see Theorems 6 and 7);
• In order to study the sign of the solution to (1), we need to obtain similar results for problem (2) when $0<s<T$, $s\in (n,n+1)$, for some $n\in {\mathbb{Z}}^{+}$ (see Theorem 8). Remarks 6–8 provide some translations of the sufficient conditions given by using the technical lemmas in the appendices or direct manipulation;
• Finally, Theorems 9 and 10 provide sufficient conditions for the unique solution to problem (1) to have a constant sign on J.

Our study allows the establishment of some comparison results for boundary value problems related to second-order functional differential equations, allowing the application of some techniques to study the existence of solution to second-order nonlinear problems.

The paper is organized as follows. In Section 2, the main results on the existence of non-negative (resp. non-positive) solutions to problems (2) and (1) are given. These results are based on some preliminary and technical results that are placed on the appendices for simplicity. In Section 3, some examples are shown to illustrate the applicability of the main results. At the end, it is possible to find Appendix A, Appendix B and Appendix C. The first one is devoted to recall some preliminary results from [20], which are essential to the procedure since they provide the expression of the solution to the problems of interest; the second and third ones focus on the properties of the sign of three auxiliary functions, namely $h_1$, $h_2$, and g, and their derivatives, which is crucial to decide the sign of the Green’s function to problem (1). The proofs of these auxiliary lemmas have been written in appendices in order to improve the readability of the work, focusing the interest in the main results.

2. Main Results

We start by introducing some additional notation that is essential to our procedure. We denote ${\mathbb{Z}}^{+}:=\mathbb{N}\cup \left\{0\right\}$. Define the functions $h_1$, $h_2$, and g by the respective expressions:

$$h_1\left(s\right)=\left\{\begin{array}{c}{\displaystyle 1-\frac{d}{a}s+\frac{d}{a^2}(1-e^{-as}),\mathrm{if} b=0,a\ne 0,}\hfill \\ \\ {\displaystyle 1-\frac{d}{2}{s}^2,\mathrm{if} b=0,a=0,}\hfill \\ \\ {\displaystyle \left({\displaystyle 1+\frac{d}{b}}\right)\left({\displaystyle 1+\frac{a}{2}s}\right)e^{-\frac{a}{2}s}-\frac{d}{b},\mathrm{if} b\ne 0,a^2=4b,}\hfill \\ \\ {\displaystyle \left({\displaystyle 1+\frac{d}{b}}\right)\frac{\beta e^{\alpha s}-\alpha e^{\beta s}}{\beta -\alpha }-\frac{d}{b},\mathrm{if} b\ne 0,a^2>4b,}\hfill \\ \\ {\displaystyle \left({\displaystyle 1+\frac{d}{b}}\right)e^{-\frac{a}{2}s}\left\{cos\sqrt{b-\frac{a^2}{4}}s+\frac{a}{2\sqrt{b-\frac{a^2}{4}}}sin\sqrt{b-\frac{a^2}{4}}s\right\}-\frac{d}{b},\mathrm{if} b\ne 0,a^2<4b,}\hfill \end{array}\right.$$

$$h_2\left(s\right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{a}\left(1-e^{-as}-cs+\frac{c}{a}(1-e^{-as})\right),\mathrm{if} b=0,a\ne 0,}\hfill \\ \\ {\displaystyle s-\frac{c}{2}{s}^2,\mathrm{if} b=0,a=0,}\hfill \\ \\ {\displaystyle e^{-\frac{a}{2}s}\left[{\displaystyle \frac{c}{b}\left(1+\frac{a}{2}s\right)+s}\right]-\frac{c}{b},\mathrm{if} b\ne 0,a^2=4b,}\hfill \\ \\ {\displaystyle \frac{\left(\frac{\beta c}{b}-1\right)e^{\alpha s}+\left(1-\frac{\alpha c}{b}\right)e^{\beta s}}{\beta -\alpha }-\frac{c}{b},\mathrm{if} b\ne 0,a^2>4b,}\hfill \\ \\ {\displaystyle e^{-\frac{a}{2}s}\left\{{\displaystyle \frac{c}{b}cos\sqrt{b-\frac{a^2}{4}}s+\frac{1+\frac{ac}{2b}}{\sqrt{b-\frac{a^2}{4}}}sin\sqrt{b-\frac{a^2}{4}}s}\right\}-\frac{c}{b},\mathrm{if} b\ne 0,a^2<4b,}\hfill \end{array}\right.$$

$$g\left(z\right)=\left\{\begin{array}{cc}\frac{1-e^{-az}}{a},\hfill & \mathrm{if} b=0,a\ne 0,\hfill \\ z,\hfill & \mathrm{if} b=0,a=0,\hfill \\ z{e}^{-\frac{a}{2}z},\hfill & \mathrm{if} b\ne 0,a^2=4b,\hfill \\ \frac{e^{\beta z}-e^{\alpha z}}{\beta -\alpha },\hfill & \mathrm{if} b\ne 0,a^2>4b,\hfill \\ e^{-\frac{a}{2}z}\frac{sin\sqrt{b-\frac{a^2}{4}}z}{\sqrt{b-\frac{a^2}{4}}},\hfill & \mathrm{if} b\ne 0,a^2<4b.\hfill \end{array}\right.$$

(3)

In the previous definitions, we denote, for $b\ne 0,a^2>4b$,

$$\alpha :=-\frac{a}{2}+\sqrt{{\left(\frac{a}{2}\right)}^2-b},\beta :=-\frac{a}{2}-\sqrt{{\left(\frac{a}{2}\right)}^2-b}.$$

For $s\in J$, we consider $E_s$ the class of functions $y:\mathbb{R}⟶\mathbb{R}$ which satisfy the conditions:

(i)

y continuous for $t\in \mathbb{R}$,

(ii)

$y^{\prime }$ continuous for $t\in \mathbb{R}\backslash \left\{s\right\}$, and there exist $y^{\prime }\left(s^{-}\right)$, $y^{\prime }\left(s^{+}\right)\in \mathbb{R}$,

(iii)

$y^{″}$ exists and is continuous for $t\in [n,n+1)\backslash \left\{s\right\}$, $n\in \mathbb{Z}$, and there exist $y^{″}\left(s^{-}\right)$, $y^{″}\left(s^{+}\right)$, $y^{″}\left(n^{-}\right)\in \mathbb{R}$, $\forall n\in \mathbb{Z}$.

In condition (iii), $y^{″}\left(z^{-}\right)$, $y^{″}\left(z^{+}\right)$ denote, respectively, the one-sided left and right limits of $y^{″}$ at the point $z\in J$.

Definition 1

(c.f. Definition 1, [20]). For $s\in J$, $y:\mathbb{R}⟶\mathbb{R}$ is a solution to problem (2) if $y\in E_s$ and satisfies conditions in (2), taking $y^{\prime }\left(s\right)=y^{\prime }\left(s^{+}\right)$ and $y^{″}\left(t\right)=y^{″}\left(t^{+}\right)$, for all $t\in \mathbb{Z}\cup \left\{s\right\}$.

Recall that $\left[T\right]$ is the integer part of T, i.e., the greatest integer k, with $k\le T.$

Definition 2

(c.f. Definition 2, [20]). Let:

$$\begin{array}{c}\Lambda :=\{y:J⟶\mathbb{R}:ycontinuous in J\backslash \{1,2,\dots ,\left[T\right]\},\hfill \\ \hfill and there exist y\left(n^{-}\right)\in \mathbb{R},y\left(n^{+}\right)=y\left(n\right),\forall n\in \{1,2,\dots ,\left[T\right]\}\},\end{array}$$

and

$$E:=\{x:J⟶\mathbb{R}:x, x^{\prime } are continuous and x^{″}\in \Lambda \}.$$

Definition 3

(c.f. Definition 3, [20]). A function x is a solution to (1) if $x\in E$ and satisfies conditions in (1), taking $x^{″}\left(n\right)=x^{″}\left(n^{+}\right)$, $\forall n\in \{0,1,2,\dots ,[T\left]\right\}$.

For $z\in [0,1]$, we denote by $H\left(z\right)$ the matrix:

$$H\left(z\right):=\left(\begin{array}{cc}{h}_1\left(z\right)& h_2\left(z\right)\\ h_1^{\prime }\left(z\right)& h_2^{\prime }\left(z\right)\end{array}\right),$$

and $C:=H\left(1\right)$, so that:

$$C=H\left(1\right)=\left(\begin{array}{cc}{C}_1& C_2\\ C_1^{\prime }& C_2^{\prime }\end{array}\right),H(T-\left[T\right])=\left(\begin{array}{cc}{h}_1(T-\left[T\right])& h_2(T-\left[T\right])\\ h_1^{\prime }(T-\left[T\right])& h_2^{\prime }(T-\left[T\right])\end{array}\right).$$

Consider also I the identity matrix $\left(\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right)$.

Theorem 1.

Assume that condition:

$$det\left(I-H(T-\left[T\right])C^{\left[T\right]}\right)\ne 0$$

(4)

holds, that is, the matrix:

$$I-H(T-\left[T\right])C^{\left[T\right]} is invertible.$$

For $s=0$ or $s=T$, the unique solution to problem (2) is non-negative if the following conditions hold:

(i) 

$h_1\ge 0$, $h_2\ge 0$ on $(0,1)$;

(ii) 

$h_1^{\prime }\ge 0$, $h_2^{\prime }\ge 0$ on $(0,1)$;

(iii) 

The elements in the second column of the matrix ${\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}$ are non-negative.

Proof. 

It is obvious from the expression of the unique solution given in Theorem 3.1 [20]:

$$K(t,s)=\left(\begin{array}{cc}{h}_1(t-n)\hfill & h_2(t-n)\hfill \end{array}\right)C^n{V}_0,t\in [n,n+1),n\in {\mathbb{Z}}^{+},$$

where $V_0={\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right).$ □

Theorem 2.

Under the hypotheses of Theorem 1, and replacing Condition (iii) by:

$(\widetilde{iii})$

The elements in the second column of the matrix ${\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}$ are non-positive,

the unique solution of problem (2) (with $s=0$ or $s=T$) is non-positive.

Remark 1.

According to Lemmas A2–A6, conditions (i) and (ii) in Theorem 1 are fulfilled if the following conditions hold:

• If $b=0$, $a\ne 0$:
  $d\le 0$ and one of the following hypotheses are satisfied:

  *

  $c=0$; or

  *

  $c\ne 0$, $a+c\le 0$; or

  *

  $c\ne 0$, $a+c>0$, $\frac{(a+c)e^{-a}-c}{a}\ge 0$;

• If $a=b=0$:
  $d\le 0$ and $c\le 1$;
• If $b\ne 0,a^2=4b$:
  $b+d\le 0$, and $\frac{a}{2}+c\le 1$;
• If $b\ne 0$, $a^2>4b$:
  $b+d\le 0$, and $a+2c\le \sqrt{a^2-4b}\left(1+\frac{2}{e^{\sqrt{a^2-4b}}-1}\right)$;
• If $b\ne 0$, $a^2<4b$:
  One of the following conditions holds:

  *

  $b+d=0$; or

  *

  $b+d<0$ and $\sqrt{b-\frac{a^2}{4}}\le \pi$,

  and one of the following conditions holds:

  *

  $\sqrt{b-\frac{a^2}{4}}\le \frac{\pi }{2}$ and $\frac{a}{2}+c\le 0$; or

  *

  $\sqrt{b-\frac{a^2}{4}}<\frac{\pi }{2}$ and $0<\frac{a}{2}+c\le \sqrt{b-\frac{a^2}{4}}cot\left(\sqrt{b-\frac{a^2}{4}}\right)$; or

  *

  $\frac{\pi }{2}<\sqrt{b-\frac{a^2}{4}}<\pi$, and $\frac{a}{2}+c\le \sqrt{b-\frac{a^2}{4}}cot\left(\sqrt{b-\frac{a^2}{4}}\right)<0.$

Theorem 3.

Theorems 1 and 2 are still valid, if we replace condition (ii) by the more general condition:

$(\widetilde{ii})$

$h_1^{\prime }\left(1\right)\ge 0$, $h_2^{\prime }\left(1\right)\ge 0$.

Remark 2.

Taking into account Theorem 3, it is possible to improve the conditions given in Remark 1, in the last case. According to Lemma A6, conditions (i) and $(\widetilde{ii})$ are fulfilled for the case $b\ne 0$, $a^2<4b$, if:

one of the following conditions holds:

• $b+d=0$; or
• $b+d<0$ and $\sqrt{b-\frac{a^2}{4}}\le \pi$,

and one of the following conditions holds:

• $\sqrt{b-\frac{a^2}{4}}\le \frac{\pi }{2}$ and $\frac{a}{2}+c\le 0$; or
• $\sqrt{b-\frac{a^2}{4}}<\frac{\pi }{2}$ and $0<\frac{a}{2}+c\le \sqrt{b-\frac{a^2}{4}}cot\left(\sqrt{b-\frac{a^2}{4}}\right)$; or
• $\frac{\pi }{2}<\sqrt{b-\frac{a^2}{4}}<\pi$, and $\frac{a}{2}+c\le \sqrt{b-\frac{a^2}{4}}cot\left(\sqrt{b-\frac{a^2}{4}}\right)<0$; or
• $\sqrt{b-\frac{a^2}{4}}\in (\pi ,\frac{3\pi }{2})$, and $\frac{a}{2}+c>\sqrt{b-\frac{a^2}{4}}cot\left(\sqrt{b-\frac{a^2}{4}}\right)>0$, provided that $B\ge 0$; or
• $\sqrt{b-\frac{a^2}{4}}=\frac{3\pi }{2}$, and $\frac{a}{2}+c>0$, provided that $B\ge 0$.

In the last two options, we consider:

$${\displaystyle B=h_2\left(s_1\right)=e^{-\frac{a}{2}{s}_1}\left\{{\displaystyle \frac{c}{b}cosR{s}_1+\frac{1+\frac{ac}{2b}}{R}sinR{s}_1}\right\}-\frac{c}{b},}$$

where,

$$R=\sqrt{b-\frac{a^2}{4}},and s_1=\frac{1}{R}\left(arctan\left(\frac{R}{\frac{a}{2}+c}\right)+\pi \right).$$

Note that:

$${\displaystyle B=-e^{-\frac{a}{2}{s}_1}\left\{{\displaystyle \frac{c}{b}cos\left(arctan\frac{R}{\frac{a}{2}+c}\right)+\frac{1+\frac{ac}{2b}}{R}sin\left(arctan\frac{R}{\frac{a}{2}+c}\right)}\right\}-\frac{c}{b}.}$$

In Figure 1, we illustrate how the last cases included the consideration of different situations not covered by the estimates in Remark 1. Indeed, taking $R=\frac{3\pi }{2}$, $a=7$, $b=R^2+\frac{a^2}{4}$, and $c=-2.5$, it is satisfied that $b\ne 0$, $a^2<4b$, $\frac{a}{2}+c>0$, and $B\ge 0$, and function $h_2$ is non-negative on $(0,1)$, such that $h_2^{\prime }\left(1\right)>0$.

Lemma 1.

Condition (iii) in Theorem 1 is fulfilled if the following conditions hold:

$$\frac{\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)}{det(I-H(T-\left[T\right])C^{\left[T\right]})}\ge 0,$$

(5)

$$\frac{1-\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}1\hfill \\ 0\hfill \end{array}\right)}{det(I-H(T-\left[T\right])C^{\left[T\right]})}\ge 0.$$

(6)

Proof. 

We use that, for $A=\left(\begin{array}{cc}{r}_1& r_2\\ r_3& r_4\end{array}\right)$,

$$A^{-1}=\frac{1}{det\left(A\right)}\left(\begin{array}{cc}{r}_4& -r_2\\ -r_3& r_1\end{array}\right),$$

in consequence, taking $A=(I-H(T-\left[T\right])C^{\left[T\right]})$, the first row of A is:

$$\left(\begin{array}{cc}1\hfill & 0\hfill \end{array}\right)-\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]},$$

and the second row of A is:

$$\left(\begin{array}{cc}0\hfill & 1\hfill \end{array}\right)-\left(\begin{array}{cc}{h}_1^{\prime }(T-\left[T\right])\hfill & h_2^{\prime }(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}.$$

Thus, the elements in the second column of $A^{-1}$ are given by:

$$\frac{\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)}{det(I-H(T-\left[T\right])C^{\left[T\right]})}$$

and

$$\frac{1-\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}1\hfill \\ 0\hfill \end{array}\right)}{det(I-H(T-\left[T\right])C^{\left[T\right]})}.$$

□

Remark 3.

If $det(I-H(T-\left[T\right])C^{\left[T\right]})>0$, conditions (5) and (6) hold if:

$$\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}$$

(7)

has the first component less than or equal to 1 and the second component is non-negative. However, in the context of Theorem 1, due to the validity of (i) and (ii), (or, more generally, (i) and $(\widetilde{ii})$), it is clear that the part above in (5) is non-negative, thus, if $det(I-H(T-\left[T\right])C^{\left[T\right]})>0$, and the first component in (7) is less than or equal to 1, then (iii) is satisfied.

Similarly, if $det(I-H(T-\left[T\right])C^{\left[T\right]})<0$, conditions (5) and (6) hold if (7) has the first component greater than or equal to 1 and the second component non-positive (null, by the restrictions).

Lemma 2.

The reverse inequalities in Lemma 1 imply the validity of $(\widetilde{iii})$ in Theorem 2.

Remark 4.

In relation with Theorem 2, under the hypotheses (i) and $(\widetilde{ii})$, it is possible to assure that, if $det(I-H(T-\left[T\right])C^{\left[T\right]})<0$, then the opposite to (5) holds. Assuming also that the first component in (7) is less than or equal to 1, then the opposite to (6) holds. Thus, by Lemma 2, these two conditions imply the validity of $(\widetilde{iii})$.

On the other hand, if $det(I-H(T-\left[T\right])C^{\left[T\right]})>0$, $(\widetilde{iii})$ holds if the first component in (7) is greater than or equal to 1 and the second component in (7) is null.

Next, we study the sign of the solution to the problem (2), for $0<s<T$ with $s\in {\mathbb{Z}}^{+}$.

Theorem 4.

Assume that condition (4) holds. For $0<s<T$, $s\in {\mathbb{Z}}^{+}$, the unique solution to problem (2) is non-negative if the following conditions hold: (i) in Theorem 1, $(\widetilde{ii})$ in Theorem 3 (see Remarks 1 and 2), and

$(\widetilde{iii})$

The elements in the second column of the matrix:

$${\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}H(T-\left[T\right])C^{\left[T\right]-s}$$

are non-negative.

Proof. 

It is deduced from the expression of the unique solution to (2) given in Theorem A4 (Theorem 3.1 [20]):

$$K(t,s)=\left(\begin{array}{cc}{h}_1(t-n)\hfill & h_2(t-n)\hfill \end{array}\right)C^n{V}_0,t\in [n,n+1),n\in \{0,1,\dots ,s-1\},$$

and

$$K(t,s)=\left(\begin{array}{cc}{h}_1(t-n)\hfill & h_2(t-n)\hfill \end{array}\right)C^{n-s}\left[C^s{V}_0+\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)\right],t\in [n,n+1),n\ge s,$$

where

$$V_0={\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}H(T-\left[T\right])C^{\left[T\right]-s}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right).$$

Note that $V_0\ge 0$ (non-negative components) implies that $C^s{V}_0+\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)\ge 0$, that is, this vector has also non-negative components. □

Lemma 3.

Assume one of the following hypotheses:

• Conditions (i), (ii) in Theorem 1, that is, $h_1$, $h_2$, $h_1^{\prime }$ and $h_2^{\prime }$ are non-negative on $(0,1)$; or more generally
• Condition (i) in Theorem 1, and $h_1^{\prime }\left(1\right)\ge 0$, $h_2^{\prime }\left(1\right)\ge 0$, $h_1^{\prime }(T-\left[T\right])\ge 0$, $h_2^{\prime }(T-\left[T\right])\ge 0$.

Then, condition ($\widehat{iii}$) in Theorem 4 is satisfied if ${\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}$ has non-negative components. In particular, this last condition is valid if $det\left(I-H(T-\left[T\right])C^{\left[T\right]}\right)>0$ and the following numbers are non-negative:

$$1-\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}1\hfill \\ 0\hfill \end{array}\right)\ge 0,$$

$$1-\left(\begin{array}{cc}{h}_1^{\prime }(T-\left[T\right])\hfill & h_2^{\prime }(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)\ge 0.$$

Proof. 

Under these particular conditions, and using the same procedure as in the proof of Lemma 1, we prove that ${[I-H(T-\left[T\right])C^{\left[T\right]}]}^{-1}$ has non-negative components, since, obviously,

$$\left(\begin{array}{cc}{h}_1(T-\left[T\right])\hfill & h_2(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)\ge 0,$$

and

$$\left(\begin{array}{cc}{h}_1^{\prime }(T-\left[T\right])\hfill & h_2^{\prime }(T-\left[T\right])\hfill \end{array}\right){\left(\begin{array}{cc}{c}_1\hfill & c_2\hfill \\ c_1^{\prime }\hfill & c_2^{\prime }\hfill \end{array}\right)}^{\left[T\right]}\left(\begin{array}{c}1\hfill \\ 0\hfill \end{array}\right)\ge 0.$$

□

Theorem 5.

Assume that condition (4) holds. For $0<s<T$, $s\in {\mathbb{Z}}^{+}$, the unique solution to problem (2) is non-positive if the following conditions hold: (i) in Theorem 1, ($(\widetilde{ii})$ in Theorem 3 (see Remarks 1 and 2), and

(iv) 

The elements in $V_0$, the second column of the matrix:

$${\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}H(T-\left[T\right])C^{\left[T\right]-s},$$

are non-positive, and the elements of $C^s{V}_0+\left(\begin{array}{c}0\\ 1\end{array}\right)$ are non-positive.

Remark 5.

The obvious changes in Lemma 3 lead to $V_0\le 0$, which joint to Conditions (i), ($\widetilde{ii}$) implies that $K(t,s)\le 0$ for $0<s<T$, $s\in {\mathbb{Z}}^{+}$, and $t\in [n,n+1),n\in \{0,1,\dots ,s-1\}$. To obtain the non-positiveness of $K(t,s)$ on $[n,n+1)$, for $n\ge s$, we have to guarantee that $C^s{V}_0+\left(\begin{array}{c}0\\ 1\end{array}\right)\le 0$.

Next, we give some extensions of Theorems 1, and 2 by virtue of the following Lemma.

Lemma 4.

Suppose that $h_1\left(1\right)>0$, and that $h_2>0$ on $(0,1)$. Define the set:

$$S:=\{(x,y)\in {\mathbb{R}}^2:x\ge 0,y\ge -Mx\},$$

with

$$M=\underset{z\in (0,1)}{inf}\frac{h_1\left(z\right)}{h_2\left(z\right)}.$$

If $(x,y)\in S$, then $h_1\left(z\right)x+h_2\left(z\right)y\ge 0$ for every $z\in [0,1]$.

Proof. 

Let $(x,y)\in S$. Then:

• If $z=0$, by the properties of $h_1$ and $h_2$, $h_1\left(0\right)x+h_2\left(0\right)y=x\ge 0$ by the definition of $S$;
• On the other hand, for $z\in (0,1)$, $h_2\left(z\right)>0$ and, since $x\ge 0$,

  $$y\ge -\underset{\xi \in (0,1)}{inf}\frac{h_1\left(\xi \right)}{h_2\left(\xi \right)}x\ge -\frac{h_1\left(z\right)}{h_2\left(z\right)}x,$$

  thus the non-negativity of $h_1\left(z\right)x+h_2\left(z\right)y$ is derived;
• Finally, if $z=1$, we have $h_1\left(1\right)x+h_2\left(1\right)y$ and we distinguish two cases. If $h_2\left(1\right)=0$, it is reduced to $h_1\left(1\right)x$, which is non-negative, and if $h_2\left(1\right)>0$, then, by the continuity of $h_1$ and $h_2$,

  $$M=\underset{\xi \in (0,1)}{inf}\frac{h_1\left(\xi \right)}{h_2\left(\xi \right)}=\underset{\xi \in (0,1]}{inf}\frac{h_1\left(\xi \right)}{h_2\left(\xi \right)},$$

  and we conclude. □

Theorem 6.

Assume that condition (4) holds. For $s=0$ or $s=T$, the unique solution to problem (2) is non-negative if the following conditions hold:

(I) 

$h_1\left(1\right)>0$, and $h_2>0$ on $(0,1)$;

(II) 

The vector $V_0$ given by:

$$V_0={\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)$$

(8)

satisfies that $C^k{V}_0\in S$ for every $k=0,1,\dots ,\left[T\right]$.

Theorem 7.

Assume that condition (4) holds. For $s=0$ or $s=T$, the unique solution to problem (2) is non-positive if the following conditions hold:

(I) 

$h_1\left(1\right)>0$, and $h_2>0$ on $(0,1)$;

(II*) 

The vector $V_0$ given by (8) satisfies that $(-1)C^k{V}_0\in S$ for every $k=0,1,\dots ,\left[T\right]$.

Similar considerations can be made for Theorems 4 and 5. We omit them since they are not required for the study of the sign of the nonhomogeneus Equation (1).

Theorem 8.

Assume that condition (4) holds. If $0<s<T$, $s\in (n,n+1)$, for some $n\in {\mathbb{Z}}^{+}$, then the unique solution to problem (2) is non-negative if the following conditions hold:

(I) 

$h_1\left(1\right)>0$, and $h_2>0$ on $(0,1)$;

(III) 

The function g given by (3) is non-negative on $(0,1)$;

(IV) 

The vector $V_0$ given, for $T<n+1$, by:

$$V_0={\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}\left(\begin{array}{c}g(T-s)\hfill \\ g^{\prime }(T-s)\hfill \end{array}\right),$$

and, for $n+1\le T$, by

$$V_0={\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}H(T-\left[T\right])C^{\left[T\right]-n-1}\left(\begin{array}{c}g(n+1-s)\hfill \\ g^{\prime }(n+1-s)\hfill \end{array}\right),$$

satisfies that $C^k{V}_0\in S$, for every $k=0,1,\dots ,\left[T\right]$;

(V) 

If $T\ge n+1$, assume also that the vector:

$$V_1=\left[C^{n+1}{V}_0+\left(\begin{array}{c}g(n+1-s)\hfill \\ g^{\prime }(n+1-s)\hfill \end{array}\right)\right],\left(with{V}_{0}giveninIV\right),$$

satisfies that $C^k{V}_1\in S$, for every $k=0,1,\dots ,\left[T\right]-n-1$.

Proof. 

In Theorem 3.1 [20] (see Theorem A4), it is proved that, under these assumptions, the expression of the unique solution to problem (2), for $0<s<T$, $s\in (n,n+1)$, for some $n\in {\mathbb{Z}}^{+}$, is given by:

$$K(t,s)=\left(\begin{array}{cc}{h}_1(t-k)\hfill & h_2(t-k)\hfill \end{array}\right)C^k{V}_0,t\in [k,k+1)\cap [0,s),k\in \{0,1,\dots ,n\},$$

$$K(t,s)=\left(\begin{array}{cc}{h}_1(t-n)\hfill & h_2(t-n)\hfill \end{array}\right)C^n{V}_0+g(t-s),t\in [s,n+1)\cap [0,T],$$

$$\begin{array}{c}K(t,s)=\left(\begin{array}{cc}{h}_1(t-n-1-k)\hfill & h_2(t-n-1-k)\hfill \end{array}\right)C^{n+1+k}{V}_0\hfill \\ \hfill +\left(\begin{array}{cc}{h}_1(t-n-1-k)\hfill & h_2(t-n-1-k)\hfill \end{array}\right)C^k\left(\begin{array}{c}g(n+1-s)\hfill \\ g^{\prime }(n+1-s)\hfill \end{array}\right),\end{array}$$

for $t\in [n+1+k,n+2+k)\cap [0,T]$, $k\in {\mathbb{Z}}^{+}$, where $V_0$ is given in IV) and g is given by (3). Hence, the proof is concluded by using Lemma 4. □

Remark 6.

According to Lemmas A2–A6, Condition (I) in Theorems 6–8 is satisfied under the following circumstances:

• $b=0,a\ne 0$, ${\displaystyle 1-\frac{d}{a}+\frac{d}{a^2}(1-e^{-a})>0}$, and one of the following conditions hold:

  *

  $c=0$; or

  *

  $c\ne 0$, and $a+c\le 0$; or

  *

  $c\ne 0$, $a+c>0$, and $\frac{(a+c)e^{-a}-c}{a}\ge 0$; or

  *

  $c\ne 0$, $a+c>0$, $\frac{(a+c)e^{-a}-c}{a}<0$, and ${\displaystyle \frac{1}{a}\left(1-e^{-a}-c+\frac{c}{a}(1-e^{-a})\right)\ge 0}$.

• $b=0,a=0$, ${\displaystyle 1-\frac{d}{2}>0}$, and one of the following conditions hold:

  *

  $c\le 1$; or

  *

  $c>1$, and $c\le 2$.

• $b\ne 0,a^2=4b$, ${\displaystyle \left({\displaystyle 1+\frac{d}{b}}\right)\left({\displaystyle 1+\frac{a}{2}}\right)e^{-\frac{a}{2}}-\frac{d}{b}>0}$, and one of the following conditions hold:

  *

  $\frac{a}{2}+c\le 1$; or

  *

  $\frac{a}{2}+c>1$, and ${\displaystyle e^{-\frac{a}{2}}\left[{\displaystyle \frac{c}{b}\left(1+\frac{a}{2}\right)+s}\right]-\frac{c}{b}\ge 0}$.

• $b\ne 0,a^2>4b$, ${\displaystyle \left({\displaystyle 1+\frac{d}{b}}\right)\frac{\beta e^{\alpha }-\alpha e^{\beta }}{\beta -\alpha }-\frac{d}{b}>0}$, and one of the following conditions hold:

  *

  $a+2c\le \sqrt{a^2-4b}\left(1+\frac{2}{e^{\sqrt{a^2-4b}}-1}\right)$; or

  *

  $a+2c>\sqrt{a^2-4b}\left(1+\frac{2}{e^{\sqrt{a^2-4b}}-1}\right)$, and ${\displaystyle \frac{\left(\frac{\beta c}{b}-1\right)e^{\alpha }+\left(1-\frac{\alpha c}{b}\right)e^{\beta }}{\beta -\alpha }-\frac{c}{b}\ge 0}$.

• $b\ne 0,a^2<4b$, ${\displaystyle \left({\displaystyle 1+\frac{d}{b}}\right)e^{-\frac{a}{2}}\left\{cosR+\frac{a}{2R}sinR\right\}-\frac{d}{b}>0}$, where $R:=\sqrt{b-\frac{a^2}{4}}$, and one of the following conditions hold:

  *

  $R\le \frac{\pi }{2}$ and $\frac{a}{2}+c\le 0$; or

  *

  $R<\frac{\pi }{2}$, and $0<\frac{a}{2}+c\le Rcot\left(R\right)$; or

  *

  $\frac{\pi }{2}<R<\pi$, and $\frac{a}{2}+c\le Rcot\left(R\right)<0$; or

  *

  ${\displaystyle e^{-\frac{a}{2}}\left\{{\displaystyle \frac{c}{b}cosR+\frac{1+\frac{ac}{2b}}{R}sinR}\right\}-\frac{c}{b}\ge 0}$, and one of the following restrictions hold:

  ⋆

  $R=\frac{\pi }{2}$ and $\frac{a}{2}+c>0$; or

  ⋆

  $R<\frac{\pi }{2}$ and $\frac{a}{2}+c>Rcot\left(R\right)$; or

  ⋆

  $R\in (\frac{\pi }{2},\pi ]$, and $\frac{a}{2}+c>0$; or

  ⋆

  $R\in (\pi ,\frac{3\pi }{2})$, and $0<\frac{a}{2}+c\le Rcot\left(R\right)$; or

  ⋆

  $\frac{3\pi }{2}\ge R\ge \pi$, and $\frac{a}{2}+c<0$; or

  ⋆

  $\frac{\pi }{2}<R<\pi$, and $0>\frac{a}{2}+c>Rcot\left(R\right).$

  *

  ${\displaystyle e^{-\frac{a}{2}{s}_1}\left\{{\displaystyle \frac{c}{b}cosR{s}_1+\frac{1+\frac{ac}{2b}}{R}sinR{s}_1}\right\}-\frac{c}{b}>0}$ and one of the following restrictions hold:

  ⋆

  $R\in (\pi ,\frac{3\pi }{2})$, and $\frac{a}{2}+c>Rcot\left(R\right)>0$; or

  ⋆

  $R=\frac{3\pi }{2}$, and $\frac{a}{2}+c>0$.

  *

  $\frac{a}{2}+c=0$, and $R\le \frac{\pi }{2}$.

Remark 7.

According to Lemmas A7–A11, Condition (III) in Theorem 8 is satisfied if one of the following conditions holds:

• $b=0$;
• $b\ne 0,a^2\ge 4b$;
• $b\ne 0,a^2<4b$, and $R\le \pi$.

Remark 8.

If $T\in (0,1)$, Conditions IV) and V) in Theorem 8 are reduced to $V_0\in S$. If $T\ge 1$, then several successive iterations by C of the vectors in IV) and V) should belong to the set $S$.

To show the first iteration, suppose that $h_1\left(1\right)>0$, $h_2>0$ on $(0,1)$, and $h_2^{\prime }\left(1\right)+M{h}_2\left(1\right)>0$. Let $V=(x,y)\in S$. Then V is such that $CV\in S$ if and only if:

$$h_1\left(1\right)x+h_2\left(1\right)y\ge 0,h_1^{\prime }\left(1\right)x+h_2^{\prime }\left(1\right)y\ge -M(h_1\left(1\right)x+h_2\left(1\right)y).$$

(9)

Since $(x,y)\in S$, $x\ge 0$ and $y\ge -{inf}_{z\in (0,1)}\frac{h_1\left(z\right)}{h_2\left(z\right)}x$. If $h_2\left(1\right)=0$, then $h_1\left(1\right)x\ge 0$ is trivially valid by hypotheses and, if $h_2\left(1\right)>0$,

$$y\ge -\underset{z\in (0,1)}{inf}\frac{h_1\left(z\right)}{h_2\left(z\right)}x=-\underset{z\in (0,1]}{inf}\frac{h_1\left(z\right)}{h_2\left(z\right)}x\ge -\frac{h_1\left(1\right)}{h_2\left(1\right)}x,$$

so that the first inequality in (9) holds. On the other hand, if

$$y\ge -\frac{h_1^{\prime }\left(1\right)+M{h}_1\left(1\right)}{h_2^{\prime }\left(1\right)+M{h}_2\left(1\right)}x,$$

the second inequality in (9) holds. Hence, the interesting set for one iteration is:

$$S_1:=\left\{(x,y)\in {\mathbb{R}}^2:x\ge 0,y\ge -Mx,y\ge -\frac{h_1^{\prime }\left(1\right)+M{h}_1\left(1\right)}{h_2^{\prime }\left(1\right)+M{h}_2\left(1\right)}x\right\}.$$

If we define the (finite) sequence (assuming that the denominators are positive):

$$M_0:=M,M_n=\frac{h_1^{\prime }\left(1\right)+M_{n-1}{h}_1\left(1\right)}{h_2^{\prime }\left(1\right)+M_{n-1}{h}_2\left(1\right)},n=1,\dots ,l,$$

the interesting set for iterations until order l is:

$$S_l:=\left\{(x,y)\in {\mathbb{R}}^2:x\ge 0,y\ge -\left(\underset{k=0,1,\dots ,l}{min}{M}_k\right)x\right\}.$$

If we are interested in analyzing the sign of the solution to the second-order functional-differential equation with piecewise constant arguments (1), which exists and is unique under the appropriate hypothesis, we need to determine the sign of $K(t,0)$ and $K(t,s)$ on $[0,T]\times [0,T]$ except at most in a set of measure zero. This is a consequence of the fact that the expression of $K(t,s)$, the solution to problem (2), represents the Green’s function for the solution to problem (1), as established in Theorem 3.2 [20].

Theorem 9.

Suppose that hypothesis (4) holds. If $\sigma \in \Lambda$ is non-negative on J, $\lambda \ge 0$, conditions (I), (II) in Theorem 6 hold and conditions (III), (IV), and (V) in Theorem 8 hold [it is assumed that (IV) and (V) will be fulfilled for all $0<s<T$, with $s\in (n,n+1)$, for some $n\in {\mathbb{Z}}^{+}$], then the unique solution to problem (1) is non-negative on J.

Proof. 

This result is obtained by applying Theorems 6, and 8, and using that the unique solution to problem (1) is, according to Theorem A5 (Theorem 3.2 [20]), given by:

$$x\left(t\right)={\int }_0^{T}K(t,s)\sigma \left(s\right)ds+\lambda K(t,0),t\in J,$$

where, for all $s\in J$, $K(·,s)$ is the unique solution to problem (2). □

Theorem 10.

Suppose that the hypothesis (4) holds. Assume also that the conditions (I), (II) in Theorem 6 hold and conditions (III), (IV), and (V) in Theorem 8 hold [it is assumed (IV) and (V) are to be fulfilled for all $0<s<T$, with $s\in (n,n+1)$, for some $n\in {\mathbb{Z}}^{+}$], if $\sigma \in \Lambda$ is non-positive on J and $\lambda \le 0$, then the unique solution to problem (1) is non-positive on J.

Remark 9.

Note that we only need to impose Condition (V) in Theorems 9 and 10 if $T>n+1$, since the value of $K(T,s)$ is not required.

3. Examples

Finally, we present some examples to illustrate the main results.

Example 1.

Consider $b=0,a\ne 0$, and $0<T<1$. We check that conditions in Theorem 1 hold. Indeed, according to Remark 1, conditions (i), (ii) are true for $d\le 0$ and one of the following hypotheses:

• $c=0$; or
• $c\ne 0$, $a+c\le 0$; or
• $c\ne 0$, $a+c>0$, $\frac{(a+c)e^{-a}-c}{a}\ge 0$.

For condition (iii), using that $\left[T\right]=0$, then the matrix:

$$\begin{array}{lll}I\hfill & -H& (T-\left[T\right])C^{\left[T\right]}=I-H(T-\left[T\right])=I-H\left(T\right)\\ & & =I-\left(\begin{array}{cc}1-\frac{d}{a}T+\frac{d}{a^2}(1-e^{-aT})& \frac{1}{a}(1-e^{-aT}-cT+\frac{c}{a}(1-e^{-aT})\\ \frac{d}{a}(e^{-aT}-1)& \frac{1}{a}(a{e}^{-aT}-c+c{e}^{-aT})\end{array}\right)\hfill \\ & & =\left(\begin{array}{cc}\frac{d}{a}T-\frac{d}{a^2}(1-e^{-aT})& -\frac{1}{a}(1-e^{-aT}-cT+\frac{c}{a}(1-e^{-aT}))\\ -\frac{d}{a}(e^{-aT}-1)& 1-\frac{1}{a}(a{e}^{-aT}-c+c{e}^{-aT})\end{array}\right)\hfill \end{array}$$

Note that $det(I-H(T\left)\right)=\frac{dT}{a}(1-e^{-aT})\ne 0$ if and only if $d\ne 0$, due to $a\ne 0$.

Suppose that $d\ne 0$. According to Lemma 1, for the validity of (5) and (6), we only have to check that:

$$\frac{h_2\left(T\right)}{det(I-H(T\left)\right)}\ge 0,$$

(10)

and

$$\frac{1-h_1\left(T\right)}{det(I-H(T\left)\right)}\ge 0.$$

(11)

Indeed, we have to impose that:

$$\begin{array}{c}\frac{h_2\left(T\right)}{det(I-H(T\left)\right)}=\frac{\frac{1}{a}((1-e^{-aT})(1+\frac{c}{a})-cT)}{\frac{dT}{a}(1-e^{-aT})}\hfill \\ \hfill =\frac{(1-e^{-aT})(1+\frac{c}{a})-cT}{dT(1-e^{-aT})}=\frac{(1-e^{-aT})(a+c)-acT}{daT(1-e^{-aT})}\ge 0,\end{array}$$

and

$$\frac{1-h_1\left(T\right)}{det(I-H(T\left)\right)}=\frac{\frac{d}{a^2}(aT-1+e^{-aT})}{\frac{dT}{a}(1-e^{-aT})}=\frac{aT-1+e^{-aT}}{aT(1-e^{-aT})}\ge 0.$$

(12)

We take into account that function $\nu \left(z\right)=z-(1-e^{-z})$ is positive for $z\ne 0$, because $\nu \left(0\right)=0$ and ${\nu }^{\prime }\left(z\right)=1-e^{-z}>0$ if $z>0$ and ${\nu }^{\prime }\left(z\right)<0$ if $z<0$. Then $\nu \left(aT\right)>0$, since $a\ne 0$, and hence the condition we must assume for the validity of (12) is:

$$aT(1-e^{-aT})>0,$$

which is trivially true independently of the value of $a\ne 0$. Thus, condition (12) is trivially satisfied.

On the other hand, for the validity of (10), we must assume that:

$$\frac{(1-e^{-aT})(a+c)-acT}{d}\ge 0.$$

(13)

This proves that, under condition (13), hypotheses (5) and (6) hold. Therefore, after analyzing the cases that lead to incompatible conditions, it is possible to affirm that, if $a<0$, $b=0$, $c=0$, $d<0$, and $0<T<1$, then the unique solution to problem (2) is non-negative, for $s=0$ or $s=T$.

Example 2.

Similarly to Example 1, consider $b=0,a\ne 0$, and $0<T<1$. We impose the following restrictions on the constants $c=0$, and $a>d>0$.

Condition (4) is satisfied due to $d\ne 0$ (see Example 1). Condition (I) is fulfilled (see Remark 6) since ${\displaystyle 1-\frac{d}{a}+\frac{d}{a^2}(1-e^{-a})>0}$ is valid by the choice of the constants. On the other hand, condition (III) is trivially satisfied by virtue of Remark 7. Since $T\in (0,1)$ and the choice of the constants,

$$\begin{array}{l}{\displaystyle {\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}}\hfill \\ ={\left(\begin{array}{cc}\frac{d}{a}T-\frac{d}{a^2}(1-e^{-aT})& -\frac{1}{a}(1-e^{-aT})\\ -\frac{d}{a}(e^{-aT}-1)& 1-e^{-aT}\end{array}\right)}^{-1}=\left(\begin{array}{cc}\frac{a}{dT}& \frac{1}{dT}\\ -\frac{1}{T}& \frac{aT-1+e^{-aT}}{aT(1-e^{-aT})}\end{array}\right).\hfill \end{array}$$

Note that it is not true that all the elements in the matrix ${\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}$ are non-negative, since its element $(2,1)$ is negative: $-\frac{1}{T}<0$. Next, we check condition II). Consider $V_0$ given by:

$$V_0={\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}\left(\begin{array}{c}0\hfill \\ 1\hfill \end{array}\right)=\left(\begin{array}{c}\frac{1}{dT}\\ \frac{aT-1+e^{-aT}}{aT(1-e^{-aT})}\end{array}\right),$$

and check that $V_0\in S$. In this case,

$$S:=\{(x,y)\in {\mathbb{R}}^2:x\ge 0,y\ge -Mx\},$$

with

$$M=\underset{z\in (0,1)}{inf}\frac{h_1\left(z\right)}{h_2\left(z\right)}=\underset{z\in (0,1)}{inf}\frac{a-dz+\frac{d}{a}(1-e^{-az})}{\left(1-e^{-az}\right)}=\underset{z\in (0,1)}{inf}\frac{a-dz}{1-e^{-az}}+\frac{d}{a}.$$

Let $\phi \left(z\right):=\frac{a-dz}{1-e^{-az}}$, $z\in (0,1)$. The sign of ${\phi }^{\prime }$ coincides with the sign of function ψ, defined on $[0,1]$ by $\psi \left(z\right)=d-a^2+adz-d{e}^{az}$. The derivative ${\psi }^{\prime }\left(z\right)=da(1-e^{az})$ is negative and $\psi \left(0\right)=-a^2<0$. Hence $\psi <0$ on $[0,1]$ and ${\phi }^{\prime }<0$ on $(0,1)$. In consequence,

$$M=\frac{a-d}{1-e^{-a}}+\frac{d}{a}>0.$$

Hence, $V_0\in S$, since $\frac{1}{dT}>0$ and

$$\frac{aT-1+e^{-aT}}{aT(1-e^{-aT})}\ge -\left(\frac{a-d}{1-e^{-a}}+\frac{d}{a}\right)\frac{1}{dT}.$$

Indeed, this inequality is equivalent to:

$$\frac{aT-1+e^{-aT}}{1-e^{-aT}}\ge -\left(\frac{a-d}{1-e^{-a}}+\frac{d}{a}\right)\frac{a}{d}=-\left(\frac{a(a-d)}{d(1-e^{-a})}+1\right),$$

and also to:

$$\frac{-aT}{1-e^{-aT}}=\frac{1-aT-e^{-aT}}{1-e^{-aT}}-1\le \frac{a(a-d)}{d(1-e^{-a})},$$

which is trivially valid since the left-hand side is negative and the right-hand side is positive.

On the other hand, we check IV), proving that $V_0$ given by:

$$\begin{array}{lll}{V}_0\hfill & =& {\left[I-H(T-\left[T\right])C^{\left[T\right]}\right]}^{-1}\left(\begin{array}{c}g(T-s)\\ g^{\prime }(T-s)\end{array}\right)\\ & =& \left(\begin{array}{cc}\frac{a}{dT}& \frac{1}{dT}\\ -\frac{1}{T}& \frac{aT-1+e^{-aT}}{aT(1-e^{-aT})}\end{array}\right)\left(\begin{array}{c}\frac{1-e^{-a(T-s)}}{a}\hfill \\ e^{-a(T-s)}\hfill \end{array}\right)\hfill \\ & =& \left(\begin{array}{c}\frac{1}{dT}\\ -\frac{1}{aT}\left(1-e^{-a(T-s)}+\frac{1-aT-e^{-aT}}{1-e^{-aT}}{e}^{-a(T-s)}\right)\end{array}\right)\hfill \end{array}$$

belongs to $S$, for every $s\in (0,T)$. Indeed, $V_0\in S$, since $\frac{1}{dT}>0$ and

$$-\frac{1}{aT}\left(1-e^{-a(T-s)}+\frac{1-aT-e^{-aT}}{1-e^{-aT}}{e}^{-a(T-s)}\right)\ge -\left(\frac{a-d}{1-e^{-a}}+\frac{d}{a}\right)\frac{1}{dT},$$

which is rewritten as:

$$1-e^{-a(T-s)}+\frac{1-aT-e^{-aT}}{1-e^{-aT}}{e}^{-a(T-s)}\le \left(\frac{a-d}{1-e^{-a}}+\frac{d}{a}\right)\frac{a}{d}=\frac{a(a-d)}{d(1-e^{-a})}+1,$$

or

$$e^{-a(T-s)}\left[\frac{1-aT-e^{-aT}}{1-e^{-aT}}-1\right]\le \frac{a(a-d)}{d(1-e^{-a})},$$

that is,

$$\frac{-aT{e}^{-aT}}{1-e^{-aT}}{e}^{as}\le \frac{a(a-d)}{d(1-e^{-a})},$$

which is trivially valid, since the left-hand side is negative and the right-hand side is positive.

Therefore, if $\sigma \in \Lambda$ is non-negative on J, and $\lambda \ge 0$, then the unique solution to problem (1) is non-negative on J.