Existence and Stability Results on Hadamard Type Fractional Time-Delay Semilinear Differential Equations

Abstract

A delayed perturbation of the Mittag-Leffler type matrix function with logarithm is proposed. This combines the classic Mittag–Leffler type matrix function with a logarithm and delayed Mittag–Leffler type matrix function. With the help of this introduced delayed perturbation of the Mittag–Leffler type matrix function with a logarithm, we provide an explicit form for solutions to non-homogeneous Hadamard-type fractional time-delay linear differential equations. We also examine the existence, uniqueness, and Ulam–Hyers stability of Hadamard-type fractional time-delay nonlinear equations.

1. Preliminaries

Let $0<a<b<\infty$ and $C\left[a,b\right]$ be the Banach space of all continuous functions $y:[a,b]\to {\mathbb{R}}^n$ with the norm ${∥y∥}_C:=max\left\{∥y\left(t\right)∥:t\in \left[a,b\right]\right\}$. For $0\le \gamma <1$, we denote the space $C_{\gamma ,ln}\left(a,b\right]$ by the weighted Banach space of the continuous function $y:[a,b]\to {\mathbb{R}}^n$, which is given by

$$C_{\gamma ,ln}\left(a,b\right]:=\left\{y\left(t\right):{\left(ln\frac{t}{a}\right)}^{\gamma }y\left(t\right)\in C\left[a,b\right]\right\},$$

endowed with the norm ${∥y∥}_{\gamma }:=sup\left\{{\left(ln\frac{t}{a}\right)}^{\gamma }∥y\left(t\right)∥:t\in \left(a,b\right]\right\}$.

The following definitions and lemmas will be used in this paper.

Definition 1.

[24] Hadamard fractional integral of order $\alpha \in {\mathbb{R}}^{+}$ of function $y\left(t\right)$ is defined by

$$\left({}^H{I}_{a^{+}}^{\alpha }y\right)\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^t{\left(ln\frac{t}{s}\right)}^{\alpha -1}y\left(s\right)\frac{ds}{s},1\le a<t\le b,$$

where Γ is the Gamma function.

Definition 2.

[24] Hadamard fractional derivative of order $\alpha \in \left[n-1,n\right)$, $n\in {\mathbb{Z}}^{+}$ of function $y\left(t\right)$ is defined by

$$\left({}^H{D}_{a^{+}}^{\alpha }y\right)\left(t\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\left(t\frac{d}{dt}\right)}^n{\int }_a^t{\left(ln\frac{t}{s}\right)}^{n-\alpha +1}y\left(s\right)\frac{ds}{s},1\le a<t\le b.$$

Lemma 1.

[24] If $a,\gamma ,\beta >0$ then

• $\left({}^H{I}_{a^{+}}^{\gamma }{\left(ln\frac{t}{a}\right)}^{\beta -1}\right)\left(t\right)={\displaystyle \frac{\Gamma \left(\beta \right)}{\Gamma \left(\beta +\gamma \right)}}{\left(ln\frac{t}{a}\right)}^{\beta +\gamma -1}.$
• $\left({}^H{D}_{a^{+}}^{\gamma }{\left(ln\frac{t}{a}\right)}^{\beta -1}\right)\left(t\right)={\displaystyle \frac{\Gamma \left(\beta \right)}{\Gamma \left(\beta -\gamma \right)}}{\left(ln\frac{t}{a}\right)}^{\beta -\gamma -1}.$
• For $0<\beta <1$, $\left({}^H{D}_{a^{+}}^{\beta }{\left(ln\frac{t}{a}\right)}^{\beta -1}\right)\left(t\right)=0.$

Definition 3.

M-L type matrix function with two parameters $e_{\alpha ,\beta }\left(\mathbb{A};t\right):\mathbb{R}\to {\mathbb{R}}^{n\times n}$ is defined by

$$e_{\alpha ,\beta }\left(\mathbb{A};t\right):=t^{\beta -1}{E}_{\alpha ,\beta }\left(\mathbb{A};t\right):=t^{\beta -1}{\displaystyle \sum _{k=0}^{\infty }}\frac{{\mathbb{A}}^k{t}^{\alpha k}}{\Gamma \left(k\alpha +\beta \right)},\alpha ,\beta >0,t\in \mathbb{R}.$$

Next, we introduce a definition of delayed M-L type matrix function $E_{h,\alpha ,\beta }^{\mathbb{B}}\left(lnt\right):{\mathbb{R}}^{+}\to {\mathbb{R}}^{n\times n}$ with logarithm generated by $\mathbb{B}$.

Definition 4.

Two parameters delayed M-L type matrix function $E_{h,\alpha ,\beta }^{\mathbb{B}}\left(lnt\right):{\mathbb{R}}^{+}\to {\mathbb{R}}^{n\times n}$ with logarithm generated by $\mathbb{B}$ is defined by

$$E_{h,\alpha ,\beta }^{\mathbb{B}}\left(lnt\right):=\left\{\begin{array}{ll}\Theta ,& 0<t\le \frac{1}{h},\\ I\frac{{\left(lnt+lnh\right)}^{\beta -1}}{\Gamma \left(\beta \right)},& \frac{1}{h}<t\le 1,\\ I\frac{{\left(lnt+lnh\right)}^{\beta -1}}{\Gamma \left(\beta \right)}+\mathbb{B}\frac{{\left(lnt\right)}^{\alpha +\beta -1}}{\Gamma \left(\alpha +\beta \right)}+\dots +{\mathbb{B}}^p\frac{{\left(lnt-\left(p-1\right)lnh\right)}^{p\alpha +\beta -1}}{\Gamma \left(p\alpha +\beta \right)},& h^{p-1}<t\le h^p.\end{array}\right.$$

(4)

Our definition of the two-parameter delayed M-L type matrix function with logarithm differs substantially from the definition given in [33].

In order to give a definition of delayed perturbation of the M-L type matrix functions with logarithm, we introduce the following matrices $Y_{\alpha ,\beta ,k},$ $k=0,1,2,\dots$

$$\begin{array}{cc}\hfill Y_{\alpha ,\beta ,0}\left(t,s\right)& ={\left(ln\frac{t}{s}\right)}^{\beta -1}{E}_{\alpha ,\beta }\left(\mathbb{A};ln\frac{t}{s}\right),\hfill \\ \hfill Y_{\alpha ,\beta ,1}\left(t,sh\right)& ={\int }_{sh}^t{e}_{\alpha ,\alpha }\left(\mathbb{A};ln\frac{t}{r}\right)\mathbb{B}{Y}_{\alpha ,\beta ,0}\left(\frac{r}{h},s\right)\frac{dr}{r},\hfill \\ \hfill Y_{\alpha ,\beta ,k}\left(t,s{h}^k\right)& ={\int }_{s{h}^k}^t{e}_{\alpha ,\alpha }\left(\mathbb{A};ln\frac{t}{r}\right)\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{r}{h},s{h}^{k-1}\right)\frac{dr}{r}.\hfill \end{array}$$

(5)

Definition 5.

Let $\mathbb{A},\mathbb{B}\in {\mathbb{R}}^{n\times n}$ be fixed matrices and $k\in \mathbb{N}\cup \left\{0\right\}$. Delayed perturbation of M-L type function $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(·,·\right):\mathbb{R}\times \mathbb{R}\to {\mathbb{R}}^n$ with logarithm generated by $\mathbb{A},\mathbb{B}$ is defined by

$$\begin{array}{ll}{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)& :={\displaystyle \sum _{j=0}^{\infty }}{Y}_{\alpha ,\beta ,j}\left(t,s{h}^j\right)H\left(t-s{h}^j\right)\\ & =\left\{\begin{array}{ll}\Theta ,& 0<t<s,\\ I,& t=s,\\ Y_{\alpha ,\beta ,0}\left(t,s\right)+Y_{\alpha ,\beta ,1}\left(t,sh\right)+\dots +Y_{\alpha ,\beta ,k}\left(t,s{h}^k\right),& s{h}^k<t\le s{h}^{k+1},\end{array}\right.\end{array}$$

(6)

where $H\left(t\right)$ is a Heaviside function: $H\left(t\right)=\left\{\begin{array}{ll}1,& t>0,\\ 0,& t\le 0.\end{array}\right.$.

Lemma 2.

Let $a,b>-1$. For $s{h}^k<t\le s{h}^{k+1},$ $k\in \mathbb{N}\cup \left\{0\right\}$, one has

$$\begin{array}{cc}\hfill {\int }_r^t{\left(ln\frac{t}{s}\right)}^a{\left(ln\frac{s}{r}\right)}^b\frac{ds}{s}& ={\left(ln\frac{t}{r}\right)}^{a+b+1}\mathcal{B}[a+1,b+1],\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill \frac{1}{\Gamma \left(1-\alpha \right)}{\int }_{s{h}^k}^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{Y}_{\alpha ,\beta ,k}\left(r,s{h}^k\right)\frac{dr}{r}& ={\int }_{s{h}^k}^t{E}_{\alpha ,1}\left(\mathbb{A};ln\frac{t}{r}\right)\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{r}{h},s{h}^{k-1}\right)\frac{dr}{r}.\hfill \end{array}$$

(8)

Proof. 

Let $ln{\displaystyle \frac{s}{r}}=\tau ln{\displaystyle \frac{t}{r}}.$ Then $s=r{\left({\displaystyle \frac{t}{r}}\right)}^{\tau }$, $ds=r{\left({\displaystyle \frac{t}{r}}\right)}^{\tau }ln{\displaystyle \frac{t}{r}}d\tau$. So we have

$$ln\frac{t}{s}=ln{\left({\displaystyle \frac{t}{r}}\right)}^{1-\tau }=\left(1-\tau \right)ln\left({\displaystyle \frac{t}{r}}\right),$$

and

$$\begin{array}{cc}\hfill {\int }_r^t{\left(ln\frac{t}{s}\right)}^a{\left(ln\frac{s}{r}\right)}^b\frac{ds}{s}& ={\int }_0^1{\left(ln\frac{t}{r}\right)}^a{\left(1-\tau \right)}^a{\left(ln\frac{t}{r}\right)}^b{\tau }^{b}ln{\displaystyle \frac{t}{r}}d\tau \hfill \\ \hfill & ={\left(ln\frac{t}{r}\right)}^{a+b+1}\mathcal{B}\left[a+1,b+1\right].\hfill \end{array}$$

To prove (8), firstly using (7) we calculate it for $k=0$:

$$\begin{array}{ll}\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_s^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{Y}_{\alpha ,\beta ,0}\left(r,s\right)\frac{dr}{r}& =\frac{1}{\Gamma \left(1-\alpha \right)}\frac{1}{\Gamma \left(\beta \right)}{\int }_s^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{\left(ln\frac{r}{s}\right)}^{\beta -1}\frac{dr}{r}\\ & =\frac{{\left(lnt-lns\right)}^{-\alpha +\beta }}{\Gamma \left(-\alpha +\beta +1\right)}.\end{array}$$

(9)

Similarly, for any $k\in \mathbb{N}$, we have

$$\begin{array}{l}\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_{s{h}^k}^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{Y}_{\alpha ,\beta ,k}\left(r,s{h}^k\right)\frac{dr}{r}\\ =\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_{s{h}^k}^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{\int }_{s{h}^k}^r{e}_{\alpha ,\alpha }\left(\mathbb{A};ln\frac{r}{r_1}\right)\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{r_1}{h},s{h}^{k-1}\right)H\left(r-s{h}^k\right)\frac{d{r}_1}{r_1}\frac{dr}{r}\\ =\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_{s{h}^k}^t{\int }_{r_1}^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{e}_{\alpha ,\alpha }\left(\mathbb{A};ln\frac{r}{r_1}\right)\frac{dr}{r}\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{r_1}{h},s{h}^{k-1}\right)H\left(r-s{h}^k\right)\frac{d{r}_1}{r_1}\\ ={\int }_{s{h}^k}^t{E}_{\alpha ,1}\left(\mathbb{A};ln\frac{t}{r_1}\right)\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{r_1}{h},s{h}^{k-1}\right)\frac{d{r}_1}{r_1}.\end{array}$$

 □

Lemma 3.

If $\mathbb{A}$ and $\mathbb{B}$ are commutative matrices, then

$$Y_{\alpha ,\beta ,k}\left(t,s{h}^k\right)={\mathbb{B}}^k{\displaystyle \sum _{n=0}^{\infty }}\left(\begin{array}{c}n+k\\ k\end{array}\right){\mathbb{A}}^n\frac{{\left(lnt-lns{h}^k\right)}^{n\alpha +k\alpha +\beta -1}}{\Gamma \left(n\alpha +k\alpha +\beta \right)}.$$

(10)

Proof. 

The proof is based on the equality (7). Using Lemma 2, for $k=1,$ we have

$$\begin{array}{ll}{Y}_{\alpha ,\beta ,1}\left(t,sh\right)& ={\int }_{sh}^t{e}_{\alpha ,\alpha }\left(\mathbb{A};ln\frac{t}{r}\right)\mathbb{B}{Y}_{\alpha ,\beta ,0}\left(\frac{r}{h},s\right)\frac{dr}{r}\\ & =\mathbb{B}{\int }_{sh}^t{\displaystyle \sum _{n=0}^{\infty }}{\mathbb{A}}^n{\left(ln\frac{t}{r}\right)}^{\alpha n+\alpha -1}\frac{1}{\Gamma \left(\alpha n+\alpha \right)}{\displaystyle \sum _{k=0}^{\infty }}{\mathbb{A}}^k{\left(ln\frac{r}{sh}\right)}^{\alpha n+\beta -1}\frac{1}{\Gamma \left(\alpha k+\beta \right)}\frac{dr}{r}\\ & =\mathbb{B}{\int }_{sh}^t{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{k=0}^k}{\mathbb{A}}^k{\left(ln\frac{t}{r}\right)}^{\alpha k+\alpha -1}\frac{1}{\Gamma \left(\alpha k+\alpha \right)}{\mathbb{A}}^{n-k}{\left(ln\frac{r}{sh}\right)}^{\alpha \left(n-k\right)+\beta -1}\frac{1}{\Gamma \left(\alpha \left(n-k\right)+\beta \right)}\frac{dr}{r}\\ & =\mathbb{B}{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{k=0}^k}{\mathbb{A}}^n\frac{1}{\Gamma \left(\alpha k+\alpha \right)\Gamma \left(\alpha \left(n-k\right)+\beta \right)}{\int }_{sh}^t{\left(ln\frac{t}{r}\right)}^{\alpha k+\alpha -1}{\left(ln\frac{r}{sh}\right)}^{\alpha \left(n-k\right)+\beta -1}\frac{dr}{r}\\ & =\mathbb{B}{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{k=0}^k}{\mathbb{A}}^n\frac{{\left(lnt-lnsh\right)}^{\alpha n+\alpha +\beta -1}}{\Gamma \left(\alpha k+\alpha \right)\Gamma \left(\alpha \left(n-k\right)+\beta \right)}\mathcal{B}(\alpha k+\alpha ,\alpha \left(n-k\right)+\beta )\\ & =\mathbb{B}{\displaystyle \sum _{n=0}^{\infty }}\left(\begin{array}{c}n+1\\ 1\end{array}\right){\mathbb{A}}^n\frac{{\left(lnt-lnsh\right)}^{\alpha n+\alpha +\beta -1}}{\Gamma \left(\alpha n+\alpha +\beta \right)}.\end{array}$$

For $k=2,$ we get

$$\begin{array}{ll}{Y}_{\alpha ,\beta ,2}\left(t,s{h}^2\right)& ={\int }_{s{h}^2}^t{e}_{\alpha ,\alpha }\left(\mathbb{A};ln\frac{t}{r}\right)\mathbb{B}{Y}_{\alpha ,\beta ,1}\left(\frac{r}{h},sh\right)dr\\ & =\mathbb{B}{\int }_{s{h}^2}^t{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{k=0}^n}{\mathbb{A}}^k{\left(ln\frac{t}{r}\right)}^{k\alpha +\alpha -1}\frac{1}{\Gamma \left(k\alpha +\alpha \right)}\mathbb{B}\left(\begin{array}{c}n+1-k\\ 1\end{array}\right)\\ & \times {\mathbb{A}}^{n-k}{\left(ln\frac{r}{s{h}^2}\right)}^{n\alpha -k\alpha +\alpha +\beta -1}\frac{1}{\Gamma \left(n\alpha -k\alpha +\alpha +\beta \right)}\\ & ={\mathbb{B}}^2{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{k=0}^n}\left(\begin{array}{c}n+1-k\\ 1\end{array}\right){\mathbb{A}}^n\frac{1}{\Gamma \left(k\alpha +\beta \right)\Gamma \left(n\alpha -k\alpha +\alpha +\beta \right)}{\int }_{s{h}^2}^t{\left(ln\frac{t}{r}\right)}^{k\alpha +\alpha -1}{\left(ln\frac{r}{s{h}^2}\right)}^{n\alpha -k\alpha +\alpha +\beta -1}dr\\ & ={\mathbb{B}}^2{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \sum _{k=0}^n}\left(\begin{array}{c}n+1-k\\ 1\end{array}\right)\right){\mathbb{A}}^n{\left(ln\frac{r}{s{h}^2}\right)}^{n\alpha +2\alpha +\beta -1}\frac{1}{\Gamma \left(n\alpha +2\alpha +\beta \right)}\\ & ={\mathbb{B}}^2{\displaystyle \sum _{n=0}^{\infty }}\left(\begin{array}{c}n+2\\ 2\end{array}\right){\mathbb{A}}^n\frac{{\left(lnt-lns{h}^2\right)}^{n\alpha +2\alpha +\beta -1}}{\Gamma \left(n\alpha +2\alpha +\beta \right)}.\end{array}$$

Using the Mathematical Induction in a similar manner we can get (10). □

According to Lemma 3 in the case $\mathbb{AB}=\mathbb{BA}$ delayed M-L type function $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)$ has a simple form:

$$Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right):=\left\{\begin{array}{ll}\Theta ,& 0<t<s,\\ I,& t=s,\\ {\displaystyle \sum _{i=0}^{\infty }}{\mathbb{A}}^i{\displaystyle \frac{{\left(lnt-lns\right)}^{i\alpha +\beta -1}}{\Gamma \left(i\alpha +\beta \right)}}+{\displaystyle \sum _{i=1}^{\infty }}\left(\begin{array}{c}i\\ 1\end{array}\right){\mathbb{A}}^{i-1}\mathbb{B}{\displaystyle \frac{{\left(lnt-lnsh\right)}^{i\alpha +\beta -1}}{\Gamma \left(i\alpha +\beta \right)}}& \\ +\dots +{\displaystyle \sum _{i=p}^{\infty }}\left(\begin{array}{c}i\\ p\end{array}\right){\mathbb{A}}^{i-p}{\mathbb{B}}^p{\displaystyle \frac{{\left(lnt-lns{h}^p\right)}^{i\alpha +\beta -1}}{\Gamma \left(i\alpha +\beta \right)}},& s{h}^p<t\le s{h}^{p+1}.\end{array}\right.$$

(11)

Next lemma shows some special cases of the delayed M-L type function.

Lemma 4.

Let $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)$ be defined by (6). Then the following holds true:

(i) 

if $\mathbb{A}=\Theta$ then $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,1\right)=E_{h,\alpha ,\beta }^{\mathbb{B}}\left(ln\frac{t}{h}\right),h^{k-1}<\frac{t}{h}\le h^k$,

(ii) 

if $\mathbb{B}=\Theta$ then $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)={\left(ln\frac{t}{s}\right)}^{\beta -1}{E}_{\alpha ,\beta }\left(\mathbb{A}{\left(ln\frac{t}{s}\right)}^{\alpha }\right)=e_{\alpha ,\beta }\left(\mathbb{A};ln\frac{t}{s}\right),$

(iii) 

if $\alpha =\beta =1$ and $\mathbb{AB}=\mathbb{BA}$ then $Y_{h,1,1}^{\mathbb{A},\mathbb{B}}\left(t,s\right)=e^{\mathbb{A}\left(lnt-lns\right)}{e}_h^{\mathbb{B}\left(lnt-lnh\right)}$, ${\mathbb{B}}_1=\mathbb{B}{e}^{-\mathbb{A}lnh},$ $s{h}^k<t\le s{h}^{k+1}$.

Proof. 

(i) If $\mathbb{A}=\Theta ,$ then the formula (5)

$$\begin{array}{ll}{Y}_{\alpha ,\beta ,0}\left(t,s\right)& =e_{\alpha ,\beta }\left(\Theta ,ln\frac{t}{s}\right)=I\frac{{\left(lnt-lns\right)}^{\beta -1}}{\Gamma \left(\beta \right)},\\ Y_{\alpha ,\beta ,1}\left(t,sh\right)& ={\int }_{sh}^t{e}_{\alpha ,\alpha }\left(\Theta ,ln\frac{t}{r}\right)\mathbb{B}{Y}_0\left(\frac{r}{h},s\right)\frac{dr}{r}=\frac{1}{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}\mathbb{B}{\int }_{sh}^t{\left(ln\frac{t}{r}\right)}^{\alpha -1}{\left(ln\frac{r}{sh}\right)}^{\beta -1}\frac{dr}{r}\\ & =\frac{1}{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}\mathbb{B}{\left(ln\frac{t}{sh}\right)}^{\alpha +\beta -1}\mathcal{B}\left[\alpha ,\beta \right]=\frac{1}{\Gamma \left(\alpha +\beta \right)}\mathbb{B}{\left(ln\frac{t}{sh}\right)}^{\alpha +\beta -1},\\ Y_{\alpha ,\beta ,2}\left(t,s{h}^2\right)& ={\int }_{s{h}^2}^t{e}_{\alpha ,\alpha }\left(\Theta ,ln\frac{t}{r}\right)\mathbb{B}{Y}_1\left(\frac{r}{h},sh\right)dr=\frac{1}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta \right)}{\mathbb{B}}^2{\int }_{s{h}^2}^t{\left(ln\frac{t}{r}\right)}^{\alpha -1}{\left(ln\frac{r}{s{h}^2}\right)}^{\alpha +\beta -1}\frac{dr}{r}\\ & =\frac{1}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta \right)}{\mathbb{B}}^2{\left(ln\frac{t}{s{h}^2}\right)}^{2\alpha +\beta -1}\mathcal{B}\left[\alpha ,\alpha +\beta \right]=\frac{1}{\Gamma \left(2\alpha +\beta \right)}{\mathbb{B}}^2{\left(ln\frac{t}{s{h}^2}\right)}^{2\alpha +\beta -1},\\ Y_{\alpha ,\beta ,k}\left(t,s{h}^k\right)& ={\mathbb{B}}^k{\left(ln\frac{t}{s{h}^k}\right)}^{k\alpha +\beta -1}\frac{1}{\Gamma \left(k\alpha +\beta \right)},k\ge 0.\end{array}$$

So $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,1\right)$ coincides with $E_{h,\alpha ,\beta }^{\mathbb{B}}\left(lnt-lnh\right):$

$$\begin{array}{ll}{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,1\right)& ={\displaystyle \sum _{i=0}^k}{\mathbb{B}}^i{\left(ln\frac{t}{h^i}\right)}^{i\alpha +\beta -1}{\displaystyle \frac{1}{\Gamma \left(i\alpha +\beta \right)}}=I\frac{{\left(lnt\right)}^{\beta -1}}{\Gamma \left(\beta \right)}+\mathbb{B}\frac{{\left(lnt-lnh\right)}^{\alpha +\beta -1}}{\Gamma \left(\alpha +\beta \right)}+\dots +{\mathbb{B}}^k\frac{{\left(lnt-klnh\right)}^{k\alpha +\beta -1}}{\Gamma \left(k\alpha +\beta \right)}\\ & =E_{h,\alpha ,\beta }^{\mathbb{B}}\left(lnt-lnh\right),h^{k-1}<\frac{t}{h}\le h^k.\end{array}$$

(ii) Trivially, from definition of $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)$ we have: if $\mathbb{B}=\Theta$, then

$$Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)={\left(ln\frac{t}{s}\right)}^{\beta -1}{E}_{\alpha ,\beta }\left(\mathbb{A}{\left(ln\frac{t}{s}\right)}^{\alpha }\right).$$

(iii) By (11) for the case $\alpha =\beta =1$ and $\mathbb{AB}=\mathbb{BA}$, we have

$$\begin{array}{ll}{Y}_{h,1,1}^{\mathbb{A},\mathbb{B}}\left(t,s\right)& ={\displaystyle \sum _{i=0}^{\infty }}{\mathbb{A}}^i{\displaystyle \frac{{\left(lnt-lns\right)}^i}{\Gamma \left(i+1\right)}}+{\displaystyle \sum _{i=0}^{\infty }}\left(\begin{array}{c}i+1\\ 1\end{array}\right){\mathbb{A}}^i\mathbb{B}{\displaystyle \frac{{\left(lnt-lnsh\right)}^{i+1}}{\Gamma \left(i+1\right)}}\\ & +\dots +{\displaystyle \sum _{i=0}^{\infty }}\left(\begin{array}{c}i+k\\ k\end{array}\right){\mathbb{A}}^i{\mathbb{B}}^k{\displaystyle \frac{{\left(lnt-lns{h}^k\right)}^{i+k}}{\Gamma \left(i+k+1\right)}}\\ & =e^{\mathbb{A}\left(lnt-lns\right)}+e^{\mathbb{A}\left(lnt-lnsh\right)}\mathbb{B}\left(lnt-lnsh\right)+\dots +e^{\mathbb{A}\left(lnt-lns{h}^k\right)}{\mathbb{B}}^k\frac{1}{k!}{\left(lnt-klnh\right)}^k\\ & =e^{\mathbb{A}\left(lnt-lns\right)}{e}_h^{{\mathbb{B}}_1\left(lnt-lnh\right)}.\end{array}$$

 □

Lemma 5.

For any $s\in \mathbb{R}$ the function $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(·,s\right):\left(1,\infty \right)\to {\mathbb{R}}^{n\times n}$ is continuous.

Proof. 

The proof is similar to that of [34] and is omitted. □

It turns out that $Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)$ is a delayed perturbation of the Cauchy matrix with logarithm of the homogeneous Equation (2) with $f=0.$

Lemma 6.

$Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}:\mathbb{R}\times \mathbb{R}\to {\mathbb{R}}^{n\times n}$ is a solution of

$${}^H{D}_{1^{+}}^{\alpha }{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)={\left(ln\frac{t}{s}\right)}^{-\alpha +\beta -1}\frac{1}{\Gamma \left(-\alpha +\beta \right)}+\mathbb{A}{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)+\mathbb{B}{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(\frac{t}{h},s\right).$$

(12)

Proof. 

According to (9) we have

$$\begin{array}{ll}\left({}^H{D}_{1^{+}}^{\alpha }{Y}_{\alpha ,\beta ,0}\left(t,s\right)\right)\left(t\right)& =\frac{1}{\Gamma \left(1-\alpha \right)}\left(t\frac{d}{dt}\right){\int }_1^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{Y}_{\alpha ,\beta ,0}\left(r,s\right)\frac{dr}{r}\\ & =\frac{1}{\Gamma \left(1-\alpha \right)}\left(t\frac{d}{dt}\right){\int }_1^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{e}_{\alpha ,\beta }\left(\mathbb{A};ln\frac{r}{s}\right)\frac{dr}{r}\\ & =\left(t\frac{d}{dt}\right)e_{\alpha ,1-\alpha +\beta }\left(\mathbb{A};ln\frac{t}{s}\right)\\ & ={\left(ln\frac{t}{s}\right)}^{-\alpha +\beta -1}\frac{1}{\Gamma \left(-\alpha +\beta \right)}+\mathbb{A}{e}_{\alpha ,\beta }\left(\mathbb{A};ln\frac{t}{s}\right)\\ & ={\left(ln\frac{t}{s}\right)}^{-\alpha +\beta -1}\frac{1}{\Gamma \left(-\alpha +\beta \right)}+\mathbb{A}{Y}_{\alpha ,\beta ,0}\left(t,s\right).\end{array}$$

(13)

On the other hand for any $k\in \mathbb{N}$:

$$\begin{array}{ll}\left({}^H{D}_{1^{+}}^{\alpha }{Y}_{\alpha ,\beta ,k}\left(t,s{h}^k\right)\right)\left(t\right)& =\frac{1}{\Gamma \left(1-\alpha \right)}\left(t\frac{d}{dt}\right){\int }_1^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{Y}_{\alpha ,\beta ,k}\left(r,s\right)\frac{dr}{r}\\ & =\left(t\frac{d}{dt}\right){\int }_{s{h}^k}^t{E}_{\alpha ,1}\left(\mathbb{A};ln\frac{t}{r}\right)\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{r}{h},s{h}^{k-1}\right)\frac{dr}{r}\\ & ={\int }_{s{h}^k}^t\left(t\frac{d}{dt}\right)E_{\alpha ,1}\left(\mathbb{A};ln\frac{t}{r}\right)\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{r}{h},s{h}^{k-1}\right)\frac{dr}{r}\\ & +\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{t}{h},s{h}^{k-1}\right)\\ & =\mathbb{A}{Y}_{\alpha ,\beta ,k}\left(t,s{h}^k\right)+\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{t}{h},s{h}^{k-1}\right).\end{array}$$

(14)

From (13) and (14) it follows that for $s{h}^k<t\le s{h}^{k+1}$

$$\begin{array}{ll}{}^H{D}_{1^{+}}^{\alpha }{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)& ={}^H{D}_{1^{+}}^{\alpha }{Y}_{\alpha ,\beta ,0}\left(t,s\right)+{}^H{D}_{1^{+}}^{\alpha }{Y}_{\alpha ,\beta ,1}\left(t,sh\right)+\dots +{}^H{D}_{1^{+}}^{\alpha }{Y}_{\alpha ,\beta ,k}\left(t,s{h}^k\right)\\ & ={\left(ln\frac{t}{s}\right)}^{-\alpha +\beta -1}\frac{1}{\Gamma \left(-\alpha +\beta \right)}+\mathbb{A}{Y}_{\alpha ,\beta ,0}\left(t,s\right)+\mathbb{A}{Y}_{\alpha ,\beta ,1}\left(t,sh\right)+\mathbb{B}{Y}_{\alpha ,\beta ,0}\left(\frac{t}{h},s\right)\\ & +\dots +\mathbb{A}{Y}_{\alpha ,\beta ,k}\left(t,s{h}^k\right)+\mathbb{B}{Y}_{\alpha ,\beta ,k-1}\left(\frac{t}{h},s{h}^{k-1}\right)\\ & ={\left(ln\frac{t}{s}\right)}^{-\alpha +\beta -1}\frac{1}{\Gamma \left(-\alpha +\beta \right)}+\mathbb{A}{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)+\mathbb{B}{Y}_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(\frac{t}{h},s\right).\end{array}$$

The proof is complete. □

Theorem 1.

The solution $y\left(t\right)$ of (2) with zero initial condition has the form

$$y\left(t\right)={\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)f\left(s\right)\frac{ds}{s},t\ge 0.$$

Proof. 

Assume that any solution of a nonhomogeneous system $y\left(t\right)$ has the form

$$y\left(t\right)={\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)h\left(s\right)\frac{ds}{s},t\ge 0,$$

(15)

where $h\left(s\right),$ $1\le s\le t\le T$ is an unknown continuous vector function and $y\left(1\right)=0$. Having Hadamard fractional differentiation on both sides of (15), for $1<t\le h$ we have

$$\begin{array}{ll}\left({}^H{D}_{1^{+}}^{\alpha }y\right)\left(t\right)& =\mathbb{A}y\left(t\right)+\mathbb{B}y\left(\frac{t}{h}\right)+f\left(t\right)\\ & =\mathbb{A}{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)h\left(s\right)\frac{ds}{s}+\mathbb{B}{\int }_1^{t/h}{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(\frac{t}{h},s\right)h\left(s\right)\frac{ds}{s}+f\left(t\right)\\ & =\mathbb{A}{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)h\left(s\right)\frac{ds}{s}+f\left(t\right).\end{array}$$

On the other hand, according to Lemma 2, we have

$$\begin{array}{ll}\left({}^H{D}_{1^{+}}^{\alpha }y\right)\left(t\right)& =\frac{1}{\Gamma \left(1-\alpha \right)}\left(t\frac{d}{dt}\right){\int }_1^t{\left(ln\frac{t}{r}\right)}^{-\alpha }\left({\int }_1^{\mathbb{R}}{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(r,s\right)h\left(s\right)\frac{ds}{s}\right)\frac{dr}{r}\\ & =\frac{1}{\Gamma \left(1-\alpha \right)}\left(t\frac{d}{dt}\right){\int }_1^t{\int }_s^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(r,s\right)h\left(s\right)\frac{dr}{r}\frac{ds}{s}\\ & =c\left(t\right)+\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_1^t\left(t\frac{d}{dt}\right){\int }_s^t{\left(ln\frac{t}{r}\right)}^{-\alpha }{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(r,s\right)h\left(s\right)\frac{dr}{r}\frac{ds}{s}\\ & =h\left(t\right)+\mathbb{A}{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)h\left(s\right)\frac{ds}{s}.\end{array}$$

Therefore, $h\left(t\right)\equiv f\left(t\right)$. The proof is complete. □

Theorem 2.

The solution $y\in C\left(\left[1,T\right],{\mathbb{R}}^n\right)$ of (2) with $f=0$ has a form

$$y\left(t\right)=Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)a+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)\mathbb{B}\phi \left(\frac{s}{h}\right)\frac{ds}{s}.$$

Proof. 

We are looking for a solution which depends on an unknown constant $c,$ and a vector function $g\left(t\right),$ of the form

$$y\left(t\right)=Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)c+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)g\left(s\right)\frac{ds}{s},$$

Moreover, $y\left(t\right)$ satisfies initial conditions

$$\begin{array}{cc}\hfill y\left(t\right)& =Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)c+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)g\left(s\right)\frac{ds}{s},1<t\le h,\hfill \\ \hfill \left({}^H{I}_{1^{+}}^{1-\alpha }y\right)\left(1^{+}\right)& =a.\hfill \end{array}$$

We have

$$\begin{array}{ll}a& =\left({}^H{I}_{1^{+}}^{1-\alpha }y\right)\left(1^{+}\right)={\displaystyle \underset{t\to 1^{+}}{lim}}\left({}^H{I}_{1^{+}}^{1-\alpha }y\right)\left(t\right)\\ & ={\displaystyle \underset{t\to 1^{+}}{lim}}\left(\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_1^t{\left(lnt-lns\right)}^{-\alpha }{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(s,1\right)c\frac{ds}{s}\right)\\ & ={\displaystyle \underset{t\to 1^{+}}{lim}}\left(\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_1^t{\left(lnt-lns\right)}^{-\alpha }{e}_{\alpha ,\alpha }\left(\mathbb{A},lns\right)\frac{ds}{s}c\right)=c.\end{array}$$

Thus $c=a$. Since $1<t\le h$, we obtain that

$$Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)=\left\{\begin{array}{ll}{\left(ln\frac{t}{s}\right)}^{\alpha -1}{E}_{\alpha ,\alpha }\left(\mathbb{A}{\left(ln\frac{t}{s}\right)}^{\alpha }\right),& 1\le s<t\le h,\\ \Theta ,& t<s\le h.\end{array}\right.$$

Consequently, on interval $1<t\le h$, we can easily derive

$$\begin{array}{ll}y\left(t\right)& =Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)a+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)g\left(s\right)\frac{ds}{s}\\ & =Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)a+{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)g\left(s\right)\frac{ds}{s}+{\int }_t^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)g\left(s\right)\frac{ds}{s}\\ & ={\left(lnt\right)}^{\alpha -1}{E}_{\alpha ,\alpha }\left(\mathbb{A}{\left(lnt\right)}^{\alpha }\right)a+{\int }_1^t{\left(ln\frac{t}{s}\right)}^{\alpha -1}{E}_{\alpha ,\alpha }\left(\mathbb{A}{\left(ln\frac{t}{s}\right)}^{\alpha }\right)g\left(s\right)\frac{ds}{s}.\end{array}$$

(16)

Having differentiated (16) in the Hadamard sense, we obtain

$$\begin{array}{ll}\left({}^H{D}_{1^{+}}^{\alpha }y\right)\left(t\right)& =\mathbb{A}{\left(lnt\right)}^{\alpha -1}{E}_{\alpha ,\alpha }\left(\mathbb{A}{\left(lnt\right)}^{\alpha }\right)a+\mathbb{A}{\int }_1^t{\left(ln\frac{t}{s}\right)}^{\alpha -1}{E}_{\alpha ,\alpha }\left(\mathbb{A}{\left(ln\frac{t}{s}\right)}^{\alpha }\right)g\left(s\right)\frac{ds}{s}+g\left(t\right)\\ & =\mathbb{A}y\left(t\right)+g\left(t\right).\end{array}$$

Therefore, $g\left(t\right)=\left({}^H{D}_{1^{+}}^{\alpha }y\right)\left(t\right)-\mathbb{A}y\left(t\right)=\mathbb{B}\phi \left(\frac{t}{h}\right)$ and the desired formula holds. □

Combining Theorems 1 and 2 together we get the following result.

Corollary 1.

A solution $y\in C\left(\left[1,T\right],{\mathbb{R}}^n\right)$ of (2) has a form

$$\begin{array}{ll}y\left(t\right)& =Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)a+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)\mathbb{B}\phi \left(\frac{s}{h}\right)\frac{ds}{s}\\ & +{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)f\left(s\right)\frac{ds}{s}.\end{array}$$

2. Existence Uniqueness and Stability

In this section, we consider the following equivalent integral form of the nonlinear Cauchy problem for fractional time-delay differential equations with Hadamard derivative (3):

$$\begin{array}{ll}y\left(t\right)& =Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)a+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)\mathbb{B}\phi \left(\frac{s}{h}\right)\frac{ds}{s}\\ & +{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)f\left(s,y\left(s\right)\right)\frac{ds}{s}.\end{array}$$

(17)

Let us introduce the conditions under which existence and uniqueness of the integral Equation (17) will be investigated.

(A1)

$f:\left[1,T\right]\times {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a function such that $f\left(t,y\right)\in C_{\gamma ,ln}\left[1,T\right]$ with $\gamma <\alpha$ for any $y\in {\mathbb{R}}^n;$

(A2)

There exists a positive constant $L_f>0$ such that

$$∥f\left(t,y_1\right)-f\left(t,y_2\right)∥\le L_f∥y_1-y_2∥,$$

for each $\left(t,y_1\right),\left(t,y_2\right)\in \left[1,T\right]\times {\mathbb{R}}^n$.

From (A1) and (A2), it follows that

$$∥f\left(t,y\right)∥\le L_f∥y∥+L_2\mathrm{for}\mathrm{some}{L}_2>0.$$

To prove existence uniqueness and stability of (17) we use the following properties of $Y_{\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right).$

Lemma 7.

We have for $s{h}^p<t\le s{h}^{p+1},$ $p=0,1,\dots ,$

$$∥Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)\left(t,s\right)∥\le Y_{h,\alpha ,\beta }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)\le Y_{1,\alpha ,\beta }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,1\right).$$

Proof. 

Indeed,

$$\begin{array}{ll}∥Y_{h,\alpha ,\beta }^{\mathbb{A},\mathbb{B}}\left(t,s\right)∥& \le Y_{h,\alpha ,\beta }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)={\displaystyle \sum _{k=0}^p}{\displaystyle \sum _{n=0}^{\infty }}\left(\begin{array}{c}n+k\\ k\end{array}\right){∥\mathbb{B}∥}^k{∥\mathbb{A}∥}^n\frac{{\left(lnt-lns{h}^k\right)}^{n\alpha +k\alpha +\beta -1}}{\Gamma \left(n\alpha +k\alpha +\beta \right)}\\ & \le {\displaystyle \sum _{k=0}^p}{\displaystyle \sum _{n=0}^{\infty }}\left(\begin{array}{c}n+k\\ k\end{array}\right){∥\mathbb{B}∥}^k{∥\mathbb{A}∥}^n\frac{{\left(lnt\right)}^{n\alpha +k\alpha +\beta -1}}{\Gamma \left(n\alpha +k\alpha +\beta \right)}\\ & ={\displaystyle \sum _{k=0}^p}{\displaystyle \sum _{n=0}^{\infty }}\left(\begin{array}{c}n+k\\ k\end{array}\right){∥\mathbb{B}∥}^k{∥\mathbb{A}∥}^n\frac{{\left(lnt\right)}^{n\alpha +k\alpha +\beta -1}}{\Gamma \left(n\alpha +k\alpha +\beta \right)}\\ & =Y_{1,\alpha ,\beta }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,1\right).\end{array}$$

 □

Our first result on existence and uniqueness of (17) is based on the Banach contraction principle.

Theorem 3.

Assume that (A1), (A2) hold. If

$$L_f\left(lnT\right)Y_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)<1,$$

then the Cauchy problem (3) has a unique solution on $[1,T]$.

Proof. 

We define an operator $\Theta$ on ${\mathcal{B}}_r:=\left\{y\in C_{\gamma ,ln}\left[1,T\right]:{∥y∥}_{\gamma }\le r\right\}$ as follows

$$\begin{array}{ll}\left(\Theta y\right)\left(t\right)& =Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)a+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)\mathbb{B}\phi \left(\frac{s}{h}\right)\frac{ds}{s}\\ & +{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)f\left(s,y\left(s\right)\right)\frac{ds}{s},\end{array}$$

where $r\ge {\displaystyle \frac{M_2}{1-M_1}}$,

$$\begin{array}{ll}{M}_2& :={\left(lnt\right)}^{\gamma }{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,1\right)∥a∥+{\left(lnh\right)}^{-\gamma +1}{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥}_{\gamma ,ln}\\ & +L_2{\left(lnt\right)}^{\gamma +1}{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right),\\ M_1& :=L_f{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥y∥}_{\gamma ,ln}.\end{array}$$

It is obvious that $\Theta$ is well-defined due to (A1). Therefore, the existence of a solution of the Cauchy problem (3) is equivalent to that of the operator $\Theta$ has a fixed point on ${\mathcal{B}}_r$. We will use the Banach contraction principle to prove that $\Theta$ has a fixed point. The proof is divided into two steps.

Step 1. $\Theta y\in {\mathcal{B}}_r$ for any $y\in {\mathcal{B}}_r$.

Indeed, for any $y\in {\mathcal{B}}_r$ and any $\delta >0,$ by (A3)

$$\begin{array}{ll}∥{\left(lnt\right)}^{\gamma }\left(\Theta y\right)\left(t\right)∥& \le {\left(lnt\right)}^{\gamma }{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,1\right)∥a∥+{\left(lnt\right)}^{\gamma }{\int }_1^h{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥\frac{ds}{s}\\ & +{\left(lnt\right)}^{\gamma }{\int }_1^t{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)∥f\left(s,y\left(s\right)\right)∥\frac{ds}{s}\end{array}$$

(18)

Firstly, we estimate the first integral:

$$\begin{array}{l}{\left(lnt\right)}^{\gamma }{\int }_1^h{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥\frac{ds}{s}\\ \le {\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){\int }_1^h{\left(lns\right)}^{-\gamma }\frac{ds}{s}{∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥}_{\gamma ,ln}\\ \le {\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){\left(lnh\right)}^{-\gamma +1}{∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥}_{\gamma ,ln}\\ ={\left(lnh\right)}^{-\gamma +1}{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥}_{\gamma ,ln}.\end{array}$$

(19)

Similarly,

$$\begin{array}{l}{\left(lnt\right)}^{\gamma }{\int }_1^t{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)∥f\left(s,y\left(s\right)\right)∥\frac{ds}{s}\\ \le {\left(lnt\right)}^{\gamma }{\int }_1^t{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)\left(L_1∥y\left(s\right)∥+L_2\right)\frac{ds}{s}\\ \le L_f{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥y∥}_{\gamma ,ln}+L_2{\left(lnt\right)}^{\gamma +1}{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right).\end{array}$$

(20)

Inserting (19) and (20) into (18) we get

$$\begin{array}{l}∥{\left(lnt\right)}^{\gamma }\left(\Theta y\right)\left(t\right)∥\\ \le {\left(lnt\right)}^{\gamma }{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,1\right)∥a∥+{\left(lnh\right)}^{-\gamma +1}{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥}_{\gamma ,ln}\\ +L_f{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥y∥}_{\gamma ,ln}+L_2{\left(lnt\right)}^{\gamma +1}{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)\\ \le M_2+M_1{∥y∥}_{\gamma ,ln}\le M_2+M_{1}r\le r.\end{array}$$

Step 2. Let $y,z\in C_{\gamma ,ln}\left[1,T\right]$. Then similar to the estimation (20) we get

$$\begin{array}{l}∥{\left(lnt\right)}^{\gamma }\left(\left(\Theta y\right)\left(t\right)-\left(\Theta z\right)\left(t\right)\right)∥\le {\left(lnt\right)}^{\gamma }{\int }_1^t{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,s\right)∥f\left(s,y\left(s\right)\right)-f\left(s,z\left(s\right)\right)∥\frac{ds}{s}\\ \le L_f{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){\int }_1^t{\left(lns\right)}^{-\gamma }{\left(lns\right)}^{\gamma }∥y\left(s\right)-z\left(s\right)∥\frac{ds}{s}\\ \le L_f{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){\int }_1^t{\left(lns\right)}^{-\gamma }\frac{ds}{s}{∥y-z∥}_{\gamma ,ln}\\ \le L_{f}lnt{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥y-z∥}_{\gamma ,ln}\end{array}$$

which implies that

$${∥\Theta y-\Theta z∥}_{\gamma ,ln}\le L_f\left(lnT\right)Y_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥y-z∥}_{\gamma ,ln}.$$

(21)

Hence, the operator $\Theta$ is contraction on ${\mathcal{B}}_r$ and the proof is competed by using the Banach fixed point theorem. □

Secondly, we discuss the Ulam–Hyers stability for the problems (3) by means of integral operator given by

$$y\left(t\right)=\left(\Theta y\right)\left(t\right),$$

where $\Theta$ is defined by (17).

Define the following nonlinear operator $Q:C_{\gamma ,ln}([1,T],{\mathbb{R}}^n)\to C_{\gamma ,ln}([1,T],{\mathbb{R}}^n)$:

$$Q\left(y\right)\left(t\right):=\left({}^H{D}_{1^{+}}^{\alpha }y\right)\left(t\right)-\mathbb{A}y\left(t\right)-\mathbb{B}y\left(\frac{t}{h}\right)-f\left(t,y\left(t\right)\right).$$

For some $\epsilon >0$, we look at the following inequality:

$${∥Q\left(y\right)∥}_{\gamma ,ln}\le \epsilon .$$

(22)

Definition 6.

We say that the Equation (17) is Ulam–Hyers stable, if there exist $V>0$ such that for every solution $y^{*}\in C_{\gamma ,ln}([\frac{1}{h},T],{\mathbb{R}}^n)$ of the inequality (22), there exists a unique solution $y\in C_{\gamma ,ln}([\frac{1}{h},T],{\mathbb{R}}^n)$ of problem (17) with

$${∥y-y^{*}∥}_{\gamma ,ln}\le V\epsilon .$$

(23)

Theorem 4.

Under the assumptions of Theorem 3, the problem (17) is stable in Ulam–Hyers sense.

Proof. 

Let $y\in C_{\gamma ,ln}([\frac{1}{h},T],{\mathbb{R}}^n)$ be the solution of the problem (17). Let $y^{*}$ be any solution satisfying (22):

$$\left({}^H{D}_{1^{+}}^{\alpha }{y}^{*}\right)\left(t\right)=\mathbb{A}{y}^{*}\left(t\right)+\mathbb{B}{y}^{*}\left(\frac{t}{h}\right)+f\left(t,y^{*}\left(t\right)\right)+Q\left(y^{*}\right)\left(t\right).$$

So

$$y^{*}\left(t\right)=\Theta \left(y^{*}\right)\left(t\right)+{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)Q\left(y^{*}\right)\left(s\right)\frac{ds}{s}.$$

It follows that

$$\begin{array}{l}{\left(lnt\right)}^{\gamma }∥\Theta \left(y^{*}\right)\left(t\right)-y^{*}\left(t\right)∥\le {\left(lnt\right)}^{\gamma }{\int }_1^t∥Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)∥∥Q\left(y^{*}\right)\left(s\right)∥\frac{ds}{s}\\ \le {\left(lnT\right)}^{\gamma +1}{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)\epsilon .\end{array}$$

Therefore, we deduce by the fixed-point property (21) of the operator $\Theta ,$ that

$$\begin{array}{ll}{\left(lnt\right)}^{\gamma }∥y\left(t\right)-y^{*}\left(t\right)∥& \le {\left(lnt\right)}^{\gamma }∥\Theta \left(y\right)\left(t\right)-\Theta \left(y^{*}\right)\left(t\right)∥+{\left(lnt\right)}^{\gamma }∥\Theta \left(y^{*}\right)\left(t\right)-y^{*}\left(t\right)∥\\ & \le L_f\left(lnT\right)Y_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥y-y^{*}∥}_{\gamma ,ln}+{\left(lnT\right)}^{\gamma +1}{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)\epsilon ,\end{array}$$

(24)

and

$${∥y-y^{*}∥}_{\gamma ,ln}\le \frac{{\left(lnT\right)}^{\gamma +1}{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)}{1-L_f\left(lnT\right)Y_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)}\epsilon .$$

Thus, the problem (3) is Ulam–Hyers stable with

$$V=\frac{{\left(lnT\right)}^{\gamma +1}{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)}{1-L_f\left(lnT\right)Y_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)}.$$

 □

3. Existence Result

Our second existence result is based on the well known Schaefer’s fixed point theorem. We use the following linear growth condition to replace (A₂):

(A₃) There exists $M_f>0$ such that

$$\begin{array}{cc}\hfill ∥f\left(t,y\right)∥& \le M_f∥y∥,\mathrm{for}\mathrm{each}t\in \left[1,T\right],y\in {\mathbb{R}}^n,\hfill \\ \hfill M_f{\left(lnT\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)& <1.\hfill \end{array}$$

Theorem 5.

Assume that (A₁) and (A₃) hold. Then the Cauchy problem (3) has at least one solution on $C_{\gamma ,ln}\left[1,T\right]$.

Proof. 

Consider the operator $\Theta :C_{\gamma ,ln}\left[1,T\right]\to C_{\gamma ,ln}\left[1,T\right]$ defined as follows

$$\begin{array}{ll}\left(\Theta y\right)\left(t\right)& =Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,1\right)a+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)\mathbb{A}\phi \left(\frac{s}{h}\right)\frac{ds}{s}\\ & +{\int }_1^t{Y}_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)f\left(s,y\left(s\right)\right)\frac{ds}{s}.\end{array}$$

For the sake of convenience, we will split the proof into several steps.

Step 1. $\Theta$ is continuous.

Let $\left\{y_n\right\}\subset C_{\gamma ,ln}\left[1,T\right]$ be a sequence converging to $y\in C_{\gamma ,ln}\left[1,T\right]$. Then for each $t\in \left[1,T\right]$, we have

$$\begin{array}{l}∥{\left(lnt\right)}^{\gamma }\left(\Theta y_n\right)\left(t\right)-\left(\Theta y\right)\left(t\right)∥\\ \le {\left(lnt\right)}^{\gamma }{\int }_1^t∥Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t,s\right)∥∥f\left(s,y_n\left(s\right)\right)-f\left(s,y\left(s\right)\right)∥\frac{ds}{s}\\ \le L_f\left(lnT\right)Y_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥f\left(·,y_n\left(·\right)\right)-f\left(·,y\left(·\right)\right)∥}_{\gamma ,ln}.\end{array}$$

Since $f\in C_{\gamma ,ln}\left[1,T\right]$, we have

$$\begin{array}{l}{∥\Theta y_n-\Theta y∥}_{\gamma ,ln}\\ \le L_f\left(lnT\right)Y_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥f\left(·,y_n\left(·\right)\right)-f\left(·,y\left(·\right)\right)∥}_{\gamma ,ln}\to 0\end{array}$$

as $n\to \infty$.

Step 2. $\Theta$ maps bounded sets into bounded sets in $C_{\gamma ,ln}\left[1,T\right]$.

Step 3. $\Theta$ is equicontinuous of $C_{\gamma ,ln}\left[1,T\right]$.

Let $1<t_1<t_2\le T$, $y\in {\mathcal{B}}_r$. Using the assumption (A₃), we have

$$\begin{array}{l}∥\left(\Theta y\right)\left(t_2\right)-\left(\Theta y\right)\left(t_1\right)∥\\ \le ∥Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t_2,1\right)a-Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t_1,1\right)a∥\\ +{\int }_1^h∥\left(Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t_2,s\right)-Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t_1,s\right)\right)∥∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥\frac{ds}{s}\\ +{\int }_1^{t_1}∥\left(Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t_1,s\right)-Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t_2,s\right)\right)∥∥f\left(s,y\left(s\right)\right)∥\frac{ds}{s}\\ +{\int }_{t_1}^{t_2}∥Y_{h,\alpha ,\alpha }^{\mathbb{A},\mathbb{B}}\left(t_2,s\right)∥∥f\left(s,y\left(s\right)\right)∥\frac{ds}{s}.\end{array}$$

The case $1=t_1<t_2\le T$ is similar. By Lemmas 5 and 7, as $t_2\to t_1$, the right hand side tends to zero, so $\Theta$ is equicontinuous.

Steps 1–3 imply that $\Theta$ is continuous and completely continuous.

Step 4. A priori bounds.

Now it remains to show that the set

$$W=\left\{y\in C_{\gamma ,ln}\left[1,T\right]:y=\lambda \Theta y,\mathrm{for}\mathrm{some}0<\lambda <1\right\}$$

is bounded. Assume that $y\in W$, then $y=\lambda \Theta y$ for some $0<\lambda <1$. Thus, for any $t\in \left[1,T\right]$, we have

$$\begin{array}{l}∥{\left(lnt\right)}^{\gamma }y\left(t\right)∥=\lambda ∥{\left(lnt\right)}^{\gamma }\left(\Theta y\right)\left(t\right)∥\\ \le {\left(lnt\right)}^{\gamma }{Y}_{1,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(t,1\right)∥a∥+{\left(lnh\right)}^{-\gamma +1}{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥\mathbb{B}\phi \left(\frac{s}{h}\right)∥}_{\gamma ,ln}\\ +M_f{\left(lnt\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right){∥y∥}_{\gamma ,ln}\end{array}$$

Since $M_f{\left(lnT\right)}^{\gamma }{Y}_{h,\alpha ,\alpha }^{∥\mathbb{A}∥,∥\mathbb{B}∥}\left(T,1\right)<1$, this shows that the set W is bounded. As a consequence of Schaefer’s fixed point theorem, we deduce that $\Theta$ has a fixed point which is a solution of the Cauchy problem (3). □

4. Example

In this section, we give an examples to illustrate the obtained theoretical result.

Example 1.

Let $\alpha =0.3$, $h=1.2,$ $k=4.$ Consider

$$\begin{array}{cc}\hfill \left({}^H{D}_{1^{+}}^{0.3}y\right)\left(t\right)& =\mathbb{B}y\left(\frac{t}{1.2}\right),t\in \left(1,2.0736\right],\hfill \\ \hfill y\left(t\right)& =\phi \left(t\right),\frac{1}{1.2}\le t\le 1,\hfill \\ \hfill \left({}^H{I}_{{\frac{1}{1.2}}^{+}}^{0.7}y\right)\left({\frac{1}{1.2}}^{+}\right)& =a\in {\mathbb{R}}^2,\hfill \end{array}$$

(25)

where

$$\mathbb{A}=\Theta ,\mathbb{B}=\left(\begin{array}{cc}\hfill 2& 1\\ \hfill 3& 5\end{array}\right),a=\left(\begin{array}{c}1\\ 2\end{array}\right).$$

The solution of (25) can be represented by $Y_{h,\alpha ,\alpha }^{\Theta ,\mathbb{B}}\left(t,s\right)$

$$y\left(t\right)=Y_{h,\alpha ,\alpha }^{\Theta ,\mathbb{B}}\left(t,1\right)a+{\int }_1^h{Y}_{h,\alpha ,\alpha }^{\Theta ,\mathbb{B}}\left(t,s\right)\mathbb{B}\phi \left(\frac{s}{h}\right)\frac{ds}{s},$$

where

$$Y_{h,\alpha ,\alpha }^{\Theta ,\mathbb{B}}\left(t,s\right)=\left\{\begin{array}{ll}0,& -\infty <t<s,\\ I,& t=s,\\ I{\displaystyle \frac{{\left(lnt-lns\right)}^{-0.7}}{\Gamma \left(0.3\right)}},& s<t\le 1.2s,\\ I{\displaystyle \frac{{\left(lnt-lns\right)}^{-0.7}}{\Gamma \left(0.3\right)}}+\mathbb{B}{\displaystyle \frac{{\left(\left(lnt-ln1.2s\right)\right)}^{-0.4}}{\Gamma \left(0.6\right)}},& 1.2s<t\le {\left(1.2\right)}^{2}s,\\ I{\displaystyle \frac{{\left(lnt-lns\right)}^{-0.7}}{\Gamma \left(0.3\right)}}+\mathbb{B}{\displaystyle \frac{{\left(\left(lnt-ln1.2s\right)\right)}^{-0.4}}{\Gamma \left(0.6\right)}}+{\mathbb{B}}^2{\displaystyle \frac{{\left(lnt-ln{\left(1.2\right)}^{2}s\right)}^{-0.1}}{\Gamma \left(0.9\right)}}& {\left(1.2\right)}^{2}s<t\le {\left(1.2\right)}^{3}s,\\ I{\displaystyle \frac{{\left(lnt-lns\right)}^{-0.7}}{\Gamma \left(0.3\right)}}+\mathbb{B}{\displaystyle \frac{{\left(\left(lnt-ln1.2s\right)\right)}^{-0.4}}{\Gamma \left(0.6\right)}}+{\mathbb{B}}^2{\displaystyle \frac{{\left(lnt-ln{\left(1.2\right)}^{2}s\right)}^{-0.1}}{\Gamma \left(0.9\right)}}& \\ +{\mathbb{B}}^3{\displaystyle \frac{{\left(lnt-ln{\left(1.2\right)}^{3}s\right)}^{0.2}}{\Gamma \left(1.2\right)}}& {\left(1.2\right)}^{3}s<t\le {\left(1.2\right)}^{4}s.\end{array}\right.$$

Example 2.

Consider

$$\begin{array}{cc}\hfill \left({}^H{D}_{1^{+}}^{0.3}y\right)\left(t\right)& =\mathbb{B}y\left(\frac{t}{1.2}\right)+L_f\left(\begin{array}{c}sin{y}_1\left(t\right)\\ tsin{y}_2\left(t\right)-y_1\left(t\right)cost\end{array}\right),t\in \left(1,2.0736\right],\hfill \\ \hfill y\left(t\right)& =\phi \left(t\right),\frac{1}{1.2}\le t\le 1,\hfill \\ \hfill \left({}^H{I}_{{\frac{1}{1.2}}^{+}}^{0.7}y\right)\left({\frac{1}{1.2}}^{+}\right)& =a\in {\mathbb{R}}^2.\hfill \end{array}$$

Clearly, the function

$$f\left(t,y\right)=L_f\left(\begin{array}{c}sin{y}_1\\ tsin{y}_2-y_{1}cost\end{array}\right)$$

is jointly continuous and Lipschitz continuous with respect to $y.$ We can choose $L_f>0$ so that the conditions of Theorems 3 and 4 are satisfied. Thus, the above problem has a unique solution which is Ulam–Hyers stable.

5. Conclusions

In this paper, we have introduced delayed perturbation of the M-L matrix exponential with logarithms, to get a representation formula for time-delay Hadamard type fractional differential equations with non-commutative linear part. Using this representation formula we have obtained several existence results for an initial value problem of time-delay Hadamard-type fractional differential equations. Furthermore, we have presented a sufficient condition for stability in the Ulam–Hyers sense. In our future work, we are planing to investigate the existence, stability and controllability of solutions to an initial value problem for time-delay fractional differential equations involving a combination of Caputo and Hadamard fractional derivatives.