We first look at [8, Example 1] with appropriate modifications.
Example 5.1
Consider
$$ \begin{aligned} &\bigl({}_{0.25}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,1}x \bigr) (t) =- \biggl(24+\frac{1}{1+t} \biggr) \bigl(4x(t)+x(t)e^{-x^{2}(t)-t^{2}}\bigr) \\ &\hphantom{\bigl({}_{0.25}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,1}x \bigr) (t) ={}}{}-3x(t)-\sin \bigl(x(t)\bigr)+\frac{\sin (x(t))}{1+e^{t}} \\ &\hphantom{\bigl({}_{0.25}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,1}x \bigr) (t) ={}}{}+ \int _{t-0.25}^{t}e^{s-t} \frac{\sin (x(s))}{1+e^{s^{2}}}\,\mathrm{d}s,\quad t>0.25, \\ &x(0.25+\theta )=\phi (\theta ), \quad \theta \in [-0.25,0]. \end{aligned} $$
(5.1)
Problem (5.1) is a special case of (2.1) with
$$\begin{aligned}& n=1, \qquad \rho =1,\qquad t_{0}=0.25,\qquad g_{1}(t)\equiv \tau _{1}=0.25, \\& C_{1}(t,s)=e^{s-t},\qquad f_{1}(t,x)= \frac{\sin (x)}{1+e^{t^{2}}},\qquad \psi _{1}(t,x,y)=\frac{\sin (x)}{1+e^{t}}, \\& F(t,x)= \biggl(24+\frac{1}{1+t} \biggr) \bigl(4x+xe^{-x^{2}-t^{2}} \bigr)+3x+\sin (x). \end{aligned}$$
It is easy to check that
$$\begin{aligned}& F(t,0)=0, \qquad xF(t,x)\geq \biggl(98+\frac{4}{1+t} \biggr)x^{2},\qquad b(t)=98+\frac{4}{1+t}, \\& \psi _{1}(t,0,0)=0, \qquad \bigl\vert \psi _{1}(t,x,y) \bigr\vert \leq \frac{ \vert x \vert }{1+e^{t}}, \qquad A_{1}(t)=\frac{1}{1+e^{t}},\qquad B_{1}(t)\equiv 0, \\& f_{1}(t,0)=0, \qquad \bigl\vert f_{1}(t,x) \bigr\vert \leq \frac{ \vert x \vert }{1+e^{t^{2}}}\leq \frac{ \vert x \vert }{2},\qquad \ell _{1}(t)= \frac{1}{2}. \end{aligned}$$
Therefore, \((\mathrm{C}_{1})\)–\((\mathrm{C}_{3})\) are satisfied. Moreover, we have (recall that \(B_{1}(t)\equiv 0\))
$$\begin{aligned} A_{1}(t)+ \int _{t-g_{1}(t)}^{t} \bigl\vert C_{1}(t,s) \bigr\vert \ell _{1}(s) \,\mathrm{d}s =&\frac{1}{1+e^{t}}+ \frac{1}{2} \int _{t-0.25}^{t}e^{s-t} \,\mathrm{d}s \\ =&\frac{1}{1+e^{t}}+\frac{1}{2 (1-e^{-0.25} )}< 3.3< 98< 98+ \frac{4}{1+t}=b(t), \end{aligned}$$
so that (4.4) is fulfilled. According to Corollary 4.4, the zero solution is stable.
Example 5.2
Let \(\alpha \in (0,1)\) and let \(\rho \in (0.0247292,1)\). Consider (5.1) with the generalized Caputo proportional fractional derivative instead of the Caputo fractional derivative, i.e., consider
$$ \begin{aligned} &\bigl({}_{0.25}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,\rho }x \bigr) (t) =- \biggl(24+\frac{1}{1+t} \biggr) \bigl(4x(t)+x(t)e^{-x^{2}(t)-t^{2}}\bigr) \\ &\hphantom{\bigl({}_{0.25}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,\rho }x \bigr) (t) ={}}{}-3x(t)-\sin \bigl(x(t)\bigr)+\frac{\sin (x(t))}{1+e^{t}} \\ &\hphantom{\bigl({}_{0.25}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,\rho }x \bigr) (t) ={}}{}+ \int _{t-0.25}^{t}e^{s-t} \frac{\sin (x(s))}{1+e^{s^{2}}}\,\mathrm{d}s,\quad t>0.25, \\ &x(0.25+\theta )=\phi (\theta ), \quad \theta \in [-0.25,0]. \end{aligned} $$
(5.2)
As in Example 5.1, \((\mathrm{C}_{1})\)–\((\mathrm{C}_{3})\) are satisfied. Let us define
$$ \kappa :=\frac{1}{4\rho }-\frac{1}{2}\leq 9.6095062. $$
We have (recall that \(B_{1}(t)\equiv 0\))
$$\begin{aligned} &A_{1}(t)+ \int _{t-g_{1}(t)}^{t} \frac{1+e^{\frac{1-\rho }{\rho }(t-s)}}{2} \bigl\vert C_{1}(t,s) \bigr\vert \ell _{1}(s)\,\mathrm{d}s \\ &\quad =\frac{1}{1+e^{t}}+ \int _{t-0.25}^{t} \frac{1+e^{\frac{1-\rho }{\rho }(t-s)}}{2}e^{s-t} \frac{\mathrm{d}s}{2} \\ &\quad < 1+\frac{1}{4} \int _{t-0.25}^{t} \bigl(e^{s-t}+e^{-4\kappa (s-t)} \bigr)\,\mathrm{d}s \\ &\quad =1+\frac{1}{4} \bigl(1-e^{-0.25} \bigr)+ \frac{1}{4} \int _{t-0.25}^{t}e^{-4 \kappa (s-t)}\,\mathrm{d}s \\ &\quad =1+\frac{1}{4} \bigl(1-e^{-0.25} \bigr) + \frac{1}{16} \textstyle\begin{cases} 1& \text{if } \kappa =0, \\ \frac{e^{\kappa }-1}{\kappa }& \text{otherwise} \end{cases}\displaystyle \\ &\quad < 97.995812< 98< 98+\frac{4}{1+t}=b(t), \end{aligned}$$
and the first inequality in the last line is due to the fact that the function f defined by \(f(x)=(e^{x}-1)/x\) for \(x\in \mathbb{R}\setminus \{0\}\) and \(f(0)=1\) is strictly increasing. Thus, the condition \((\mathrm{C}_{4})\) is fulfilled. According to Corollary 4.5, the zero solution of (5.2) is exponentially stable.
Example 5.3
Consider
$$ \begin{aligned} &\bigl({}_{0}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,1}x \bigr) (t) =- \frac{2+t}{1+t}x(t)+ \frac{x(t)}{1+t^{2}} + \int _{t-1}^{t}(t-s)x(s)\,\mathrm{d}s,\quad t>0, \\ &x(\theta )=\phi (\theta ),\quad \theta \in [-1,0]. \end{aligned} $$
(5.3)
Problem (5.3) is a special case of (2.1) with
$$\begin{aligned}& n=1, \qquad \rho =1, \qquad t_{0}=0,\qquad g_{1}(t)\equiv \tau _{1}=1, \\& C_{1}(t,s)=t-s,\qquad f_{1}(t,x)=x,\qquad \psi _{1}(t,x,y)= \frac{x}{1+t^{2}}, \\& F(t,x)=\frac{2+t}{1+t}x. \end{aligned}$$
It is easy to check that
$$\begin{aligned}& F(t,0)=0,\qquad xF(t,x)=\frac{2+t}{1+t}x^{2}, \qquad b(t)= \frac{2+t}{1+t}, \\& \psi _{1}(t,0,0)=0, \qquad \bigl\vert \psi _{1}(t,x,y) \bigr\vert \leq \frac{ \vert x \vert }{1+t^{2}},\qquad A_{1}(t)=\frac{1}{1+t^{2}},\qquad B_{1}(t)\equiv 0, \\& f_{1}(t,0)=0, \qquad \bigl\vert f_{1}(t,x) \bigr\vert = \vert x \vert , \qquad \ell _{1}(t)=1. \end{aligned}$$
Therefore, \((\mathrm{C}_{1})\)–\((\mathrm{C}_{3})\) are satisfied. Moreover, we have (recall that \(B_{1}(t)\equiv 0\))
$$\begin{aligned} A_{1}(t)+ \int _{t-g_{1}(t)}^{t} \bigl\vert C_{1}(t,s) \bigr\vert \ell _{1}(s) \,\mathrm{d}s&=\frac{1}{1+t^{2}}+ \int _{t-1}^{t} \vert t-s \vert \,\mathrm{d}s \\ &=\frac{1}{1+t^{2}}+\frac{1}{2}< \frac{2+t}{1+t}=b(t), \end{aligned}$$
so that (4.4) is fulfilled. According to Corollary 4.4, the zero solution is stable.
Example 5.4
Let \(\alpha \in (0,1)\) and let \(\rho \in (0.3420471,1)\). Consider (5.3) with the generalized Caputo proportional fractional derivative instead of the Caputo fractional derivative, i.e., consider
$$ \begin{aligned} &\bigl({}_{0}^{\mathrm{C}} \!{\mathcalligra{D}}^{\alpha ,\rho }x \bigr) (t) =- \frac{2+t}{1+t}x(t)+ \frac{x(t)}{1+t^{2}} + \int _{t-1}^{t}(t-s)x(s)\,\mathrm{d}s, \quad t>0, \\ &x(\theta )=\phi (\theta ), \quad \theta \in [-1,0]. \end{aligned} $$
(5.4)
As in Example 5.3, \((\mathrm{C}_{1})\)–\((\mathrm{C}_{3})\) are satisfied. Let us define
$$\begin{aligned}& \kappa :=\frac{1}{\rho }-1\leq 1.9235739,\qquad g(t):= \frac{1}{1+t^{2}}- \frac{2+t}{1+t}, \\& \mu :=\max_{t>0}g(t)=-1-2 \biggl(\frac{1}{2}+ \frac{\sqrt{5}}{2}-\sqrt{ \frac{1+\sqrt{5}}{2}} \biggr)^{2}. \end{aligned}$$
We have for \(t>0\) (recall that \(B_{1}(t)\equiv 0\))
$$\begin{aligned} &A_{1}(t)+ \int _{t-g_{1}(t)}^{t} \frac{1+e^{\frac{1-\rho }{\rho }(t-s)}}{2} \bigl\vert C_{1}(t,s) \bigr\vert \ell _{1}(s)\,\mathrm{d}s \\ &\quad =\frac{1}{1+t^{2}}+ \int _{t-1}^{t}\frac{1+e^{\kappa (t-s)}}{2}(t-s) \, \mathrm{d}s \\ &\quad =\frac{1}{1+t^{2}}+\frac{1}{2} \int _{0}^{1} \bigl(v+ve^{\kappa v} \bigr) \,\mathrm{d}v \\ &\quad =\frac{1}{1+t^{2}}+\frac{1}{2} \biggl\{ \frac{1}{2}+ \frac{(\kappa -1)e^{\kappa }+1}{\kappa ^{2}} \biggr\} \\ &\quad =g(t)+b(t)+\frac{1}{2} \biggl\{ \frac{1}{2}+ \frac{(\kappa -1)e^{\kappa }+1}{\kappa ^{2}} \biggr\} \\ &\quad \leq \mu +\frac{1}{2} \biggl\{ \frac{1}{2}+ \frac{(\kappa -1)e^{\kappa }+1}{\kappa ^{2}} \biggr\} +b(t) \\ &\quad < b(t), \end{aligned}$$
and the last inequality is due to the fact that the function f defined by \(f(x)= ((x-1)e^{x}+1)/x^{2}\) for \(x>0\) is strictly increasing. Thus, the condition \((\mathrm{C}_{4})\) is fulfilled. According to Corollary 4.5, the zero solution of (5.4) is exponentially stable.
Example 5.5
Consider
$$ \begin{aligned} &\bigl({}_{0}^{\mathrm{C}} \!{\mathcalligra{D}}^{0.3,0.8}x \bigr) (t) =- \frac{2+t}{1+t}x(t)+0.03x(t-1) \cos ^{2}t \\ &\hphantom{\bigl({}_{0}^{\mathrm{C}} \!{\mathcalligra{D}}^{0.3,0.8}x \bigr) (t) ={}}{}+\frac{x(t-1-\sin t)}{10+t^{2}} \\ &\hphantom{\bigl({}_{0}^{\mathrm{C}} \!{\mathcalligra{D}}^{0.3,0.8}x \bigr) (t) ={}}{}+ \int _{t-1}^{t}(t-s)x(s)\,\mathrm{d}s, \quad t>0, \\ &x(\theta )=\phi (\theta ),\quad \theta \in [-2,0]. \end{aligned} $$
(5.5)
Problem (5.5) is a special case of (2.1) with
$$\begin{aligned}& n=2, \qquad \rho =0.8, \qquad \alpha =0.3,\qquad t_{0}=0, \\& g_{1}(t)\equiv 1, \qquad g_{2}(t)=1+\sin t,\qquad \tau _{1}=1,\qquad \tau _{2}=2,\qquad \tau =2, \\& C_{1}(t,s)=t-s, \qquad C_{2}(t,s)\equiv 0, \qquad f_{1}(t,x)=f_{2}(t,x)=x, \\& \psi _{1}(t,x,y)=0.03y\cos ^{2}t,\qquad \psi _{2}(t,x,y)= \frac{y}{10+t^{2}},\qquad F(t,x)=\frac{2+t}{1+t}x. \end{aligned}$$
If we put
$$\begin{aligned}& b(t)=\frac{2+t}{1+t},\qquad \ell _{1}(t)=\ell _{2}(t) \equiv 1, \\& A_{1}(t)=A_{2}(t)\equiv 0,\qquad B_{1}(t)=0.03,\qquad B_{2}(t)=0.1, \end{aligned}$$
then \((\mathrm{C}_{1})\)–\((\mathrm{C}_{3})\) are satisfied. We put \(\kappa :=(1-\rho )/\rho =0.25\) and estimate for \(t>0\) (recall that \(A_{1}(t)=A_{2}(t)=C_{2}(t)\equiv 0\))
$$\begin{aligned} &\frac{1+e^{\kappa g_{1}(t)}}{2}B_{1}(t) + \frac{1+e^{\kappa g_{2}(t)}}{2}B_{2}(t) + \int _{t-g_{1}(t)}^{t} \frac{1+e^{\kappa (t-s)}}{2} \bigl\vert C_{1}(t,s) \bigr\vert \ell _{1}(s) \,\mathrm{d}s \\ &\quad =\frac{1+e^{\kappa }}{2}0.03+\frac{1+e^{\kappa (1+\sin t)}}{2}0.1 + \int _{t-1}^{t}\frac{1+e^{\kappa (t-s)}}{2}(t-s)\, \mathrm{d}s \\ &\quad \leq \frac{1+e^{\kappa }}{2}0.03+\frac{1+e^{2\kappa }}{2}0.1 + \frac{1}{2} \int _{0}^{1} \bigl(v+ve^{\kappa v} \bigr) \,\mathrm{d}v \\ &\quad =\frac{1+e^{\kappa }}{2}0.03+\frac{1+e^{2\kappa }}{2}0.1 +\frac{1}{2} \biggl\{ \frac{1}{2}+\frac{(\kappa -1)e^{\kappa }+1}{\kappa ^{2}} \biggr\} \\ &\quad < 0.712544< 1< 1+\frac{1}{1+t}=b(t), \end{aligned}$$
and thus \((\mathrm{C}_{4})\) is fulfilled. From Corollary 4.5, follows the exponential stability and therefore the asymptotic stability of the zero solution of (5.5), and according to (4.5), any solution of (5.5) satisfies
$$ \bigl\vert x(s) \bigr\vert \leq \sqrt{\max_{\theta \in [-2,0]}\phi ^{2}(\theta )}e^{-0.125s}\quad \text{for all } s\geq 0. $$