Semi-Hyers–Ulam–Rassias Stability via Laplace Transform, for an Integro-Differential Equation of the Second Order

Abstract

The Laplace transform method is applied to study the semi-Hyers–Ulam–Rassias stability of a Volterra integro-differential equation of the second order. A general equation is formulated first; then, some particular cases for the function from the kernel are considered.

1. Introduction

Hyers–Ulam stability has been the concern of many mathematicians since 1940 when Ulam posed an open problem regarding the stability of homomorphisms (see the book in reference [1]). The first result in this direction was given in 1941 by Hyers [2]. Since then, the research of the problem continued both by extending the classes of equations that are considered and by generalizing the notion of stability (Hyers–Ulam–Rassias stability, generalized Hyers–Ulam–Rassias stability, semi-Hyers–Ulam–Rassias stability, and Mittag–Leffler–Hyers–Ulam stability). The development of this field so far can be consulted, for instance, in the books of Brzdek, Popa, Rasa, Xu [3], and Tripathy [4], which provide a systematic approach to the subject.

Different kinds of stability results for linear ordinary differential equations were obtained in numerous papers, among which we mention those of Obloza [5], Alsina and Ger [6] (the first on differential equations), and references [7,8,9,10]. Systems of differential equations were considered too, see for instance reference [11] and the references therein. The Hyers–Ulam stability of integral or integro-differential equations was investigated in references [12,13,14,15,16]. Partial differential equations were also investigated using the point of view of their Hyers–Ulam stability, starting with Prastaro and Rassias [17], followed by many studies, such as the ones in references [18,19,20,21,22,23]. Some new types of stability can be found in references [24,25].

Various methods can be employed to establish the stability of the Hyers–Ulam type, such as the direct method, the substitution method, the fixed point method, the integral transform method, and the Gronwall inequality method. In this paper, we will use an integral transform method, more precisely the Laplace transform. In the context of Hyers–Ulam stability, this appeared first in the paper of Rezaei, Jung, and Rassias [26] for linear differential equations. Afterwards, the Laplace transform was used to prove stability results in several other papers, among which we mention reference [27] for linear differential equations, reference [28] for the Mittag–Leffler–Hyers–Ulam stability of a linear differential equation of the first order, reference [29] for the Laguerre differential equation and Bessel differential equation, reference [30] for fractional differential equations, and reference [31] for the convection partial differential equation. In reference [32], the discrete z-transform was used to investigate the stability of some linear difference equations with constant coefficients.

In the following, inspired by the method used in reference [26], we will study a Volterra integro-differential equation of the second order with a convolution-type kernel:

$$y^{″}\left(t\right)+{\int }_0^{t}y\left(u\right)g\left(t-u\right)du-f\left(t\right)=0,t\in \left(0,\infty \right),$$

(1)

where $y\in C^2(0,\infty ).$

This paper is a continuation of the work in reference [33], where we considered an integro-differential equation of the first order. The differences between the results for the first-order equations and those for the second-order equations are best seen when some particular instances of the function g are taken.

We recall (see reference [34]) that the equation

$$\begin{array}{cc}\hfill & a_2\left(t\right)y^{″}\left(t\right)+a_1\left(t\right)y^{\prime }\left(t\right)+a_0\left(t\right)y\left(t\right)=f\left(t\right)+\lambda {\int }_0^{t}g\left(t-u\right)y\left(u\right)du,\hfill \end{array}$$

is called the Volterra integro-differential equation of the convolution type of the second order. Volterra integro-differential equations have many applications, for instance in the diffusion process, the growth of cells, in heat and mass transfer, and the motion of satellites.

The outline of the paper is the following: In Section 2, we present the stability notion and properties of the Laplace transform, and prove several auxiliary results about the solutions of some algebraic equations that will appear. The main results (Theorems 1–4) are given in the next section and concern the semi-stability of the integro-differential Equation (1). The first theorem is proved for the general form and then some particular cases of the function g are treated. In Theorem 2, the function $g:\left(0,\infty \right)\to \mathbb{F}$, $g\left(t\right)=t^n$, with $n\in \mathbb{N}$ is considered. In Theorems 3 and 4, the function $g:\left(0,\infty \right)\to \mathbb{F}$, $g\left(t\right)=e^{\alpha t}$, $\alpha \in \mathbb{R}\backslash \{0,\frac{3\sqrt[3]{2}}{2}\}$ and the function $g:\left(0,\infty \right)\to \mathbb{F}$, $g\left(t\right)=cos\omega t$, $\omega \in {\mathbb{R}}^{*},$ were taken, respectively. Other functions g that satisfy the conditions of the Theorem 1 can be considered. In order to obtain Theorems 1–4, Lemmas 1–3 were used.

2. Preliminary Notions and Results

In the rest of the paper, we denote by $\mathbb{F}$ the real field $\mathbb{R}$ or the complex field $\mathbb{C}$, and by $\Re \left(s\right)$ and $\Im \left(s\right)$ the real part and the imaginary part of a complex number s, respectively. Throughout the work we assume that the functions $f,g,y:\left(0,\infty \right)\to \mathbb{F}$ are continuous and of exponential order.

The Laplace transform of the function f is denoted by $\mathcal{L}\left(f\right)$ and defined by

$$\mathcal{L}\left(f\right)\left(s\right)=F\left(s\right)={\int }_0^{\infty }f\left(t\right)e^{-st}dt,$$

on the open half plane $\{s\in \mathbb{C}|\Re \left(s\right)>{\sigma }_f\}$, where ${\sigma }_f$ is the abscissa of convergence. The inverse Laplace transform of a function F is denoted by ${\mathcal{L}}^{-1}\left(F\right)$. The following properties of the Laplace transform, which will be used in the paper, are well known:

$$\mathcal{L}\left(e^{\alpha t}\right)=\frac{1}{s-\alpha },s>\alpha \iff {\mathcal{L}}^{-1}\left(\frac{1}{s-\alpha }\right)=e^{\alpha t},$$

$$\mathcal{L}\left(t^n\right)\left(s\right)=\frac{n!}{s^{n+1}},s>0,n\in \mathbb{N},$$

$$\mathcal{L}\left(cos\omega t\right)=\frac{s}{s^2+{\omega }^2},\omega \in \mathbb{R},$$

$$\mathcal{L}\left(y^{″}\right)=s^2\mathcal{L}\left(y\right)-sy\left(0^{+}\right)-y^{\prime }\left(0^{+}\right),$$

$$\mathcal{L}(f\ast g)\left(s\right)=\mathcal{L}\left(f\right)\left(s\right)·\mathcal{L}\left(g\right)\left(s\right),$$

where $(f\ast g)\left(t\right)={\int }_0^{t}f(t-u)g\left(u\right)du$ is the convolution product of f and g. In the rest of the paper we write $y\left(0\right),y^{\prime }\left(0\right),$ instead of the lateral limits $y\left(0^{+}\right),y^{\prime }\left(0^{+}\right),$ respectively.

Let $\epsilon >0$ and consider the inequality

$$\left|y^{″}\left(t\right)+{\int }_0^{t}y\left(u\right)g\left(t-u\right)du-f\left(t\right)\right|\le \epsilon ,t\in \left(0,\infty \right).$$

(2)

We say, as in reference [35], that Equation (1) is semi-Hyers–Ulam–Rassias stable if there exists a function $k:\left(0,\infty \right)\to \left(0,\infty \right)$, such that for each solution y of the inequality (2), there exists a solution $y_0$ of Equation (1) with

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le k\left(t\right),\forall t\in \left(0,\infty \right).$$

(3)

A function $y:\left(0,\infty \right)\to \mathbb{F}$ is a solution of (2) if and only if there exists a function $h:\left(0,\infty \right)\to \mathbb{F}$ such that

(1)

$\left|h\left(t\right)\right|\le \epsilon ,\forall t\in \left(0,\infty \right),$

(2)

$y^{″}\left(t\right)+{\int }_0^{t}y\left(u\right)g\left(t-u\right)du-f\left(t\right)=h\left(t\right),\forall t\in \left(0,\infty \right).$

The following Lemmas will be used in the paper.

Lemma 1.

Let $m\in \mathbb{N}$, $m\ge 3$, and $a_j,b_j\in \mathbb{C}$. Let $s_1,s_2,\dots ,s_m\in \mathbb{C}$ be the distinct solutions of the equation $s^m+a_1{s}^{m-1}+\cdots +a_m=0$ and $A_1,A_2,\dots ,A_m\in \mathbb{F}$, such that

$$\frac{s^{m-2}+b_1{s}^{m-3}+\cdots +b_{m-2}}{s^m+a_1{s}^{m-1}+\cdots +a_m}=\sum _{k=1}^m\frac{A_k}{s-s_k}.$$

Then

$$\sum _{k=1}^m{A}_k=0$$

and

$$\sum _{k=1}^m{A}_k{s}_k=1.$$

Proof. 

From $\frac{s^{m-2}+b_1{s}^{m-3}+\cdots +b_{m-2}}{s^m+a_1{s}^{m-1}+\cdots +a_m}={\displaystyle \sum _{k=1}^m}\frac{A_k}{s-s_k}$, we have

$$\begin{array}{cc}\hfill & s^{m-2}+b_1{s}^{m-3}+\cdots +b_{m-2}=A_1\left(s-s_2\right)\cdots \left(s-s_m\right)\hfill \\ \hfill & +\sum _{k=2}^{m-1}{A}_k\left(s-s_1\right)\cdots \left(s-s_{k-1}\right)\left(s-s_{k+1}\right)\left(s-s_m\right)\hfill \\ \hfill & +A_m\left(s-s_1\right)\cdots \left(s-s_{m-1}\right).\hfill \end{array}$$

For $k\in \{2,\dots ,m-1\}$, denote by

$$\begin{array}{cc}\hfill & S_{k,1}=s_1+\cdots +s_{k-1}+s_{k+1}+\cdots +s_m,\hfill \\ \hfill & \cdots \hfill \\ \hfill & S_{k,m-1}=s_1{s}_2\cdots s_{k-1}{s}_{k+1}\cdots s_m,\hfill \end{array}$$

the sums corresponding to the Viète relations of the following polynomial of degree $m-1$:

$$\left(s-s_1\right)\left(s-s_2\right)\cdots \left(s-s_{k-1}\right)\left(s-s_{k+1}\right)\cdots \left(s-s_m\right).$$

For $k=1$, we denote

$$\begin{array}{cc}\hfill & S_{1,1}=s_2+s_3+\cdots s_m,\hfill \\ \hfill & \cdots \hfill \\ \hfill & S_{1,m-1}=s_2{s}_3\cdots s_m.\hfill \end{array}$$

For $k=m$, we denote

$$\begin{array}{cc}\hfill & S_{m,1}=s_1+s_2+\cdots s_{m-1},\hfill \\ \hfill & \cdots \hfill \\ \hfill & S_{m,m-1}=s_1{s}_2\cdots s_{m-1}.\hfill \end{array}$$

We obtain

$$s^{m-2}+b_1{s}^{m-3}+\cdots +b_{m-2}=\sum _{k=1}^m{A}_k\left(s^{m-1}-S_{k,1}{s}^{m-2}+\cdots +{\left(-1\right)}^{m-1}{S}_{k,m-1}\right).$$

Identifying the coefficients of $s^{m-1}$ and $s^{m-2}$, we obtain

$$\sum _{k=1}^m{A}_k=0\mathrm{and}\sum _{k=1}^m{A}_k(-S_{k,1})=1.$$

On the other hand, we have $s_1+s_2+\cdots +s_m=-a_1$, so $S_{k,1}=-a_1-s_k$, and it follows that

$$1=\sum _{k=1}^m{A}_k(a_1+s_k)=a_1\sum _{k=1}^m{A}_k+\sum _{k=1}^m{A}_k{s}_k=\sum _{k=1}^m{A}_k{s}_k.$$

□

Lemma 2.

Let ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{n+3}$ be the solutions of the equation $s^{n+3}+n!=0,n\in \mathbb{N},$ and $B_1,B_2,\dots ,B_{n+3}\in \mathbb{F}$, such that

$$\frac{s^{n+1}}{s^{n+3}+n!}=\sum _{k=1}^{n+3}\frac{B_k}{s-{\sigma }_k}.$$

Then

$$\left|B_k\right|=\frac{1}{(n+3)\sqrt[n+3]{n!}},\forall k\in \{1,\dots ,n+3\}.$$

Proof. 

For every root ${\sigma }_k$,

$$s^{n+3}+n!=(s-{\sigma }_k)(s^{n+2}+{\sigma }_k{s}^{n+1}+{\sigma }_k^2{s}^n+\cdots +{\sigma }_k^{n+2}).$$

From $\frac{s^{n+1}}{s^{n+3}+n!}={\displaystyle \sum _{k=1}^{n+3}}\frac{B_k}{s-{\sigma }_k}$, we have

$$s^{n+1}=\sum _{k=1}^{n+3}{B}_k(s^{n+2}+{\sigma }_k{s}^{n+1}+{\sigma }_k^2{s}^n+\cdots +{\sigma }_k^{n+2}).$$

(4)

For $l\in \left\{1,2,\dots ,n+3\right\}$ fixed, let $s={\sigma }_l$ in (4) to obtain:

$${\sigma }_l^{n+1}=B_l({\sigma }_l^{n+2}+{\sigma }_l{\sigma }_l^{n+1}+\cdots +{\sigma }_l^{n+2})=(n+3)B_l{\sigma }_l^{n+2},$$

which gives $B_l=\frac{1}{(n+3){\sigma }_l}$. Writing $-n!$ in polar form, the equation becomes $s^{n+3}=n!(cos\pi +isin\pi )$, so $\left|{\sigma }_l\right|=\sqrt[n+3]{n!}$. Finally, $\left|B_l\right|=\frac{1}{(n+3)\sqrt[n+3]{n!}},\forall l\in \{1,\cdots ,n+3\}$. □

Lemma 3.

Let ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{n+3}$ be the solutions of the equation $s^{n+3}+n!=0$, $n\in \mathbb{N}$.

(i)

If $n=4q$, $n=4q+1$ or $n=4q+2$, with $q\in \mathbb{N}$, then $\Re \left({\sigma }_k\right)\ne 0$, for all $k\in \{1,\dots ,n+3\}$.

(ii)

If $n=4q+3$, with $q\in \mathbb{N}$, then there exist exactly two roots ${\sigma }_1$, ${\sigma }_2$, such that $\Re \left({\sigma }_1\right)=\Re \left({\sigma }_2\right)=0$.

Proof. 

Let ${\sigma }_k$ be a solution of the equation $s^{n+3}+n!=0$ and suppose that $\Re \left({\sigma }_k\right)=0$, that is ${\sigma }_k=i\beta$, where $\beta \in \mathbb{R}$. The equation becomes $i^{n+3}{\beta }^{n+3}=-n!$.

If $n=4q$, $q\in \mathbb{N}$, we have $i{\beta }^{4q+3}=-\left(4q\right)!$, which is impossible for $\beta \in \mathbb{R}$.

If $n=4q+1$, $q\in \mathbb{N}$, we have ${\beta }^{4q+4}=-(4q+1)!$, which is impossible for $\beta \in \mathbb{R}$.

If $n=4q+2$, $q\in \mathbb{N}$, we have $i{\beta }^{4q+5}=-(4q+2)!$, which is impossible for $\beta \in \mathbb{R}$.

If $n=4q+3$, $q\in \mathbb{N}$, we have ${\beta }^{4q+6}=(4q+3)!$, an equation which admits exactly two real solutions. □

3. Main Stability Results

The main result of the paper is given by the following Theorem 1. Some special cases are further studied.

Theorem 1.

Let $g:\left(0,\infty \right)\to \mathbb{F}$ be a continuous function and of exponential order, such that the inverse ${\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)$ exists, ${\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(0\right)=0$, and ${\left({\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right)}^{\prime }\left(0\right)=1.$

If a function $y:\left(0,\infty \right)\to \mathbb{F}$ satisfies the inequality (2) and $y\left(0\right)=0$, $y^{\prime }\left(0\right)=1,$ then there exists a solution $y_0:\left(0,\infty \right)\to \mathbb{F}$ of (1), such that

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon {\int }_0^t\left|{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)\right|du,\forall t\in \left(0,\infty \right),$$

that is, Equation (1) is semi-Hyers–Ulam–Rassias stable.

Proof. 

Let $h:\left(0,\infty \right)\to \mathbb{F}$,

$$h\left(t\right)=y^{″}\left(t\right)+{\int }_0^{t}y\left(u\right)g\left(t-u\right)du-f\left(t\right),t\in \left(0,\infty \right).$$

(5)

We have

$$\mathcal{L}\left(h\right)=s^2\mathcal{L}\left(y\right)-sy\left(0\right)-y^{\prime }\left(0\right)+\mathcal{L}\left(y\right)·\mathcal{L}\left(g\right)-\mathcal{L}\left(f\right),$$

hence

$$\mathcal{L}\left(y\right)=\frac{\mathcal{L}\left(h\right)}{s^2+\mathcal{L}\left(g\right)}+\frac{1+\mathcal{L}\left(f\right)}{s^2+\mathcal{L}\left(g\right)}.$$

Let

$$y_0\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right)+{\mathcal{L}}^{-1}\left(\frac{\mathcal{L}\left(f\right)}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right),\forall t\in \left(0,\infty \right),$$

that is

$$y_0\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right)+{\int }_0^{t}f\left(u\right){\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)du.$$

We have

$$\begin{array}{cc}\hfill y_0^{\prime }\left(t\right)& ={\left({\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right)}^{\prime }\left(t\right)+f\left(t\right){\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(0\right)\hfill \\ \hfill & +{\int }_0^{t}f\left(u\right)\frac{\partial }{\partial t}\left({\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right)\left(t-u\right),\forall t\in \left(0,\infty \right).\hfill \end{array}$$

We remark that $y_0\left(0\right)=y\left(0\right)=0$ and $y_0^{\prime }\left(0\right)=y^{\prime }\left(0\right)=1$.

Hence, we obtain

$$\begin{array}{cc}\hfill & \mathcal{L}\left[y_0^{″}\left(t\right)+{\int }_0^t{y}_0\left(u\right)g\left(t-u\right)du-f\left(t\right)\right]\hfill \\ \hfill & =s^2\mathcal{L}\left(y_0\right)-s{y}_0\left(0\right)-y_0^{\prime }\left(0\right)+\mathcal{L}\left(y_0\right)·\mathcal{L}\left(g\right)-\mathcal{L}\left(f\right)\hfill \\ \hfill & =\left[s^2+\mathcal{L}\left(g\right)\right]\mathcal{L}\left(y_0\right)-1-\mathcal{L}\left(f\right)\hfill \\ \hfill & =\left[s^2+\mathcal{L}\left(g\right)\right]\left[\frac{1}{s^2+\mathcal{L}\left(g\right)}+\frac{\mathcal{L}\left(f\right)}{s^2+\mathcal{L}\left(g\right)}\right]-1-\mathcal{L}\left(f\right)=0.\hfill \end{array}$$

Since $\mathcal{L}$ is one-to-one, it follows that

$$y_0^{″}\left(t\right)+{\int }_0^t{y}_0\left(u\right)g\left(t-u\right)du-f\left(t\right)=0,$$

that is, $y_0$ is a solution of (1).

We have

$$\mathcal{L}\left(y\right)-\mathcal{L}\left(y_0\right)=\frac{\mathcal{L}\left(h\right)}{s^2+\mathcal{L}\left(g\right)},$$

hence

$$\begin{array}{cc}\hfill & \left|y\left(t\right)-y_0\left(t\right)\right|=\left|{\mathcal{L}}^{-1}\left(\frac{\mathcal{L}\left(h\right)}{s^2+\mathcal{L}\left(g\right)}\right)\right|=\left|{\mathcal{L}}^{-1}\left(\mathcal{L}\left(h\right)\right)\ast {\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right|\hfill \\ \hfill & =\left|h\ast {\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right|=\left|{\int }_0^{t}h\left(u\right)·{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)du\right|\hfill \\ \hfill & \le {\int }_0^t\left|h\left(u\right)\right|·\left|{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)\right|du\le \epsilon {\int }_0^t\left|{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)\right|du.\hfill \end{array}$$

□

We will consider next some particular cases for the function g.

• Let $g:\left(0,\infty \right)\to \mathbb{F}$, $g\left(t\right)=t^n$, with $n\in \mathbb{N}$.

Using Lemmas 1 and 3 and Theorem 1 we obtain the following stability result in this case.

Theorem 2.

Let $g:\left(0,\infty \right)\to \mathbb{F},g\left(t\right)=t^n,n\in \mathbb{N}.$ Let ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{n+3}$ be the roots of the equation $s^{n+3}+n!=0$.

If a function $y:\left(0,\infty \right)\to \mathbb{F}$ satisfies the inequality (2) and $y\left(0\right)=0$, $y^{\prime }\left(0\right)=1,$ then there exists a solution $y_0:\left(0,\infty \right)\to \mathbb{F}$ of (1), such that

$$\begin{array}{cc}\hfill & \left|y\left(t\right)-y_0\left(t\right)\right|\hfill \\ \hfill & \le \left\{\begin{array}{c}\frac{\epsilon }{(n+3)\sqrt[n+3]{n!}}{\displaystyle \sum _{k=1}^{n+3}}\frac{1}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right),\mathrm{if}n\in \{4q,4q+1,4q+2|q\in \mathbb{N}\},\hfill \\ \frac{\epsilon }{(n+3)\sqrt[n+3]{n!}}\left(2t+{\displaystyle \sum _{k=3}^{n+3}}\frac{1}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right)\right),\mathrm{if}n=4q+3,q\in \mathbb{N}.\hfill \end{array}\right.\hfill \end{array}$$

Proof. 

We notice that the roots ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{n+3}$ are distinct and ${\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(0\right)=0,$${\left({\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right)}^{\prime }\left(0\right)=1.$ Indeed, with $B_1,B_2,\dots ,B_{n+3}\in \mathbb{F}$, such that

$$\frac{s^{n+1}}{s^{n+3}+n!}=\sum _{k=1}^{n+3}\frac{B_k}{s-{\sigma }_k},$$

$$\begin{array}{c}\hfill {\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s^2+\frac{n!}{s^{n+1}}}\right)\left(t\right)={\mathcal{L}}^{-1}\left(\frac{s^{n+1}}{s^{n+3}+n!}\right)\left(t\right)=\sum _{k=1}^{n+3}{B}_k{e}^{{\sigma }_{k}t},\end{array}$$

hence, by Lemma 1, with $m=n+3$,

$$\begin{array}{c}\hfill {\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(0\right)=\sum _{k=1}^{n+3}{B}_k=0.\end{array}$$

We also have

$${\left({\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right)}^{\prime }\left(t\right)=\sum _{k=1}^{n+3}{B}_k{\sigma }_k{e}^{{\sigma }_{k}t},$$

hence

$${\left({\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\right)}^{\prime }\left(0\right)=\sum _{k=1}^{n+3}{B}_k{\sigma }_k=1.$$

If n has one of the forms $4q$, $4q+1$, or $4q+2$, with $q\in \mathbb{N}$, then, by Lemma 3, it follows that $\Re \left({\sigma }_k\right)\ne 0$, for all $k=\overline{1,n+3}$. Applying Theorem 1, we obtain

$$\begin{array}{cc}\hfill & \left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon {\int }_0^t\left|{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)\right|du\le \epsilon {\int }_0^t\left|\stackrel{n+3}{\sum _{k=1}}{B}_k{e}^{{\sigma }_k\left(t-u\right)}\right|du\hfill \\ \hfill & \le \epsilon \sum _{k=1}^{n+3}{\int }_0^t\left|B_k{e}^{{\sigma }_k\left(t-u\right)}\right|du=\epsilon \sum _{k=1}^{n+3}\left|B_k\right|e^{\Re \left({\sigma }_k\right)t}{\int }_0^t{e}^{-\Re \left({\sigma }_k\right)u}du=\hfill \\ \hfill & =\frac{\epsilon }{(n+3)\sqrt[n+3]{n!}}\sum _{k=1}^{n+3}\frac{1}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right).\hfill \end{array}$$

If $n=4q+3$, $q\in \mathbb{N}$, then exactly two roots have the real parts zero, for instance $\Re \left({\sigma }_1\right)=\Re \left({\sigma }_2\right)=0$. We apply Theorem 1 and obtain

$$\begin{array}{cc}\hfill & \left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon {\int }_0^t\left|{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)\right|du\le \epsilon {\int }_0^t\left|\sum _{k=1}^{n+3}{B}_k{e}^{{\sigma }_k\left(t-u\right)}\right|du\hfill \\ \hfill & =\epsilon \sum _{k=1}^{n+3}\left|B_k\right|e^{\Re \left({\sigma }_k\right)t}{\int }_0^t{e}^{-\Re \left({\sigma }_k\right)u}du\hfill \\ \hfill & =\epsilon \sum _{k=1}^2\left|B_k\right|{\int }_0^{t}du+\epsilon \sum _{k=3}^{n+3}\left|B_k\right|e^{\Re \left({\sigma }_k\right)t}{\int }_0^t{e}^{-\Re \left({\sigma }_k\right)u}du\hfill \\ \hfill & =\epsilon \sum _{k=1}^2\left|B_k\right|t+\epsilon \sum _{k=3}^{n+3}\frac{\left|B_k\right|}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right)\hfill \\ \hfill & =\frac{\epsilon }{(n+3)\sqrt[n+3]{n!}}\left(2t+\sum _{k=3}^{n+3}\frac{1}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right)\right).\hfill \end{array}$$

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Example 1.

For $n=0$ in Theorem 2, we have $g:\left(0,\infty \right)\to \mathbb{F}$, $g\left(t\right)=1.$ Then

$$\begin{array}{cc}\hfill & {\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s^2+\frac{1}{s}}\right)\left(t\right)={\mathcal{L}}^{-1}\left(\frac{s}{s^3+1}\right)\left(t\right)\hfill \\ \hfill & ={\mathcal{L}}^{-1}\left(\frac{1}{3}\left(\frac{1}{s+1}+\frac{1}{s-s_1}+\frac{1}{s-s_2}\right)\right)=\frac{1}{3}\left(e^{-t}+e^{s_{1}t}+e^{s_{2}t}\right),\hfill \end{array}$$

where $s_{1,2}=\frac{1\pm i\sqrt{3}}{2}$. We apply Theorem 2 and obtain

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le \frac{\epsilon }{3}\left(-e^{-t}+4e^{\frac{t}{2}}-3\right).$$

II.

Let $g:\left(0,\infty \right)\to \mathbb{F}$, $g\left(t\right)=e^{\alpha t}$, $\alpha \in \mathbb{R}$.

Using Lemma 1 and Theorem 1 we can state the following stability result, in the case when g is an exponential function.

Theorem 3.

Let $\alpha \in \mathbb{R}\setminus \{0,\frac{3\sqrt[3]{2}}{2}\}$ and $g:\left(0,\infty \right)\to \mathbb{F}$, defined by $g\left(t\right)=e^{\alpha t}$.

If a function $y:\left(0,\infty \right)\to \mathbb{F}$ satisfies the inequality (2), then there exists a solution $y_0:\left(0,\infty \right)\to \mathbb{F}$ of (1), such that

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon \sum _{k=1}^3\frac{\left|C_k\right|}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right),\forall t\in \left(0,\infty \right),$$

where ${\sigma }_k$, $k\in \{1,2,3\}$ are the roots of the equation $s^3-\alpha s^2+1=0$ and

$$\frac{s-\alpha }{s^3-\alpha s^2+1}=\sum _{k=1}^3\frac{C_k}{s-{\sigma }_k}.$$

(6)

Proof. 

From Theorem 1, we have

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon {\int }_0^t\left|{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)\right|du,\forall t\in \left(0,\infty \right).$$

For $g\left(t\right)=e^{\alpha t}$, we obtain

$${\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right)={\mathcal{L}}^{-1}\left(\frac{s-\alpha }{s^3-\alpha s^2+1}\right)\left(t\right).$$

For $\alpha \in \mathbb{R}\backslash \{0,\frac{3\sqrt[3]{2}}{2}\}$, the equation $s^3-\alpha s^2+1=0$ admits three distinct roots, denoted by ${\sigma }_k$. Indeed, if we suppose that $\sigma$ is a double root, it verifies ${\sigma }^3-\alpha {\sigma }^2+1=0$ and also $3{\sigma }^2-2\alpha \sigma =0$. Solving this system, we obtain $\sigma =\sqrt[3]{2}$ and $\alpha =\frac{3\sqrt[3]{2}}{2}$, which is excluded. It follows that

$${\mathcal{L}}^{-1}\left(\frac{s-\alpha }{s^3-\alpha s^2+1}\right)\left(t\right)=\sum _{k=1}^3{C}_k{e}^{{\sigma }_{k}t},$$

with the coefficients $C_k$ defined by (6).

From this equality, by Lemma 1, we obtain that ${\mathcal{L}}^{-1}\left(\frac{s-\alpha }{s^3-\alpha s^2+1}\right)\left(0\right)=0$ and ${\left({\mathcal{L}}^{-1}\left(\frac{s-\alpha }{s^3-\alpha s^2+1}\right)\right)}^{\prime }\left(0\right)=1$, so Theorem 1 can be applied. Next,

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon {\int }_0^t\left|\sum _{k=1}^3{C}_k{e}^{{\sigma }_k(t-u)}\right|du\le \epsilon \sum _{k=1}^3\left|C_k\right|e^{\Re \left({\sigma }_k\right)t}{\int }_0^t{e}^{-\Re \left({\sigma }_k\right)u}du.$$

Since $\alpha \in \mathbb{R}$ and $\alpha \ne 0$, it follows that $\mathcal{R}\left({\sigma }_k\right)\ne 0$, too. Indeed, if ${\sigma }_k=i\beta$, with $\beta \in \mathbb{R}$, the equation $s^3-\alpha s^2+1=0$ becomes $-i{\beta }^3+\alpha {\beta }^2+1=0$, which is impossible. Further on,

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon \sum _{k=1}^3\left|C_k\right|e^{\Re \left({\sigma }_k\right)t}\frac{1-e^{-\Re \left({\sigma }_k\right)t}}{\Re \left({\sigma }_k\right)}=\epsilon \sum _{k=1}^3\frac{\left|C_k\right|}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right).$$

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III.

Let $g:\left(0,\infty \right)\to \mathbb{F}$, $g\left(t\right)=cos\omega t$, with $\omega \in {\mathbb{R}}^{*}$.

Theorem 4.

Let $\omega \in {\mathbb{R}}^{*}$ and $g:\left(0,\infty \right)\to \mathbb{F}$, defined by $g\left(t\right)=cos\omega t$.

If a function $y:\left(0,\infty \right)\to \mathbb{F}$ satisfies the inequality (2), then there exists a solution $y_0:\left(0,\infty \right)\to \mathbb{F}$ of (1), such that

$$\left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon \left({\omega }^{2}t+\sum _{k=1}^3\frac{\left|D_k\right|}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right)\right),\forall t\in \left(0,\infty \right),$$

where ${\sigma }_k$, $k\in \{1,2,3\}$ are the roots of the equation $s^3+{\omega }^{2}s+1=0$ and $\frac{s^2+{\omega }^2}{s^4+{\omega }^2{s}^2+s}=\frac{{\omega }^2}{s}+{\displaystyle \sum _{k=1}^3}\frac{D_k}{s-{\sigma }_k}.$

Proof. 

For $g\left(t\right)=cos\omega t$, we obtain

$${\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right)={\mathcal{L}}^{-1}\left(\frac{s^2+{\omega }^2}{s^4+{\omega }^2{s}^2+s}\right)\left(t\right).$$

Let ${\sigma }_k$, $k\in \{1,2,3\}$ be the distinct roots of the equation $s^3+{\omega }^{2}s+1=0$ and $D_k$, such that

$$\frac{s^2+{\omega }^2}{s^4+{\omega }^2{s}^2+s}=\frac{D_0}{s}+\sum _{k=1}^3\frac{D_k}{s-{\sigma }_k}.$$

It is easy to see that $D_0={\omega }^2$, so ${\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t\right)={\omega }^2+{\sum }_{k=1}^3{D}_k{e}^{{\sigma }_{k}t}$. From this and Lemma 1, it follows that ${\mathcal{L}}^{-1}\left(\frac{s^2+{\omega }^2}{s^4+{\omega }^2{s}^2+s}\right)\left(0\right)={\omega }^2+{\sum }_{k=1}^3{D}_k=0$ and ${\left({\mathcal{L}}^{-1}\left(\frac{s^2+{\omega }^2}{s^4+{\omega }^2{s}^2+s}\right)\right)}^{\prime }\left(0\right)={\sum }_{k=1}^3{\sigma }_k{D}_k=1$.

Like in Theorem 3, it can be proved that $\Re \left({\sigma }_k\right)\ne 0$ for every $k\in \{1,2,3\}$. By Theorem 1, we obtain

$$\begin{array}{cc}\hfill & \left|y\left(t\right)-y_0\left(t\right)\right|\le \epsilon {\int }_0^t\left|{\mathcal{L}}^{-1}\left(\frac{1}{s^2+\mathcal{L}\left(g\right)}\right)\left(t-u\right)\right|du\hfill \\ \hfill & \le \epsilon {\int }_0^t\left({\omega }^2+\sum _{k=1}^3\left|D_k\right|\left|e^{{\sigma }_k(t-u)}\right|\right)du=\epsilon \left({\omega }^{2}t+\sum _{k=1}^3\left|D_k\right|e^{\Re \left({\sigma }_k\right)t}{\int }_0^t{e}^{-\Re \left({\sigma }_k\right)u}du\right)\hfill \\ \hfill & =\epsilon \left({\omega }^{2}t+\sum _{k=1}^3\frac{\left|D_k\right|}{\Re \left({\sigma }_k\right)}\left(e^{\Re \left({\sigma }_k\right)t}-1\right)\right).\hfill \end{array}$$

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4. Conclusions

In this paper, we studied the semi-Hyers–Ulam–Rassias stability of Equation (1) using the Laplace transform. We first studied the general case and then considered the various functions g that appear in the equation. Some examples were given. The results continue those from the paper in reference [33]. We intend to study further the general case, when the equation has order n.