With the help of fixed point procedure we check the existence of fractal fractional to SE(Is)(Ih)AR epidemic model (1). We have
$$ \textstyle\begin{cases} S(t)-S(0) \\ \quad =\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}(b_{1} -[b_{2}+ \beta (I_{s}+\beta _{hr}I_{h}+ \beta _{ar}A )+K_{v}]S+\eta R)\,ds \\ \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}(b_{1} -[b_{2}+ \beta (I_{s}+ \beta _{hr}I_{h}+ \beta _{ar}A )+K_{v}]S+\eta R), \\ E(t)-E (0) \\ \quad =\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}(-(b_{2}+ \gamma )E+\beta (I_{s}+\beta _{hr}I_{h}+\beta _{ar}A)S)\,ds \\ \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}(-(b_{2}+\gamma )E+\beta (I_{s}+ \beta _{hr}I_{h}+\beta _{ar}A)S), \\ I_{s}(t)-I_{s} (0)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}(-(b_{2}+ \tau _{0})I_{s}+\gamma P_{s} E)\,ds \\ \hphantom{I_{s}(t)-I_{s} (0)=}{}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}(-(b_{2}+ \tau _{0})I_{s}+ \gamma P_{s} E), \\ I_{h}(t)-I_{h} (0)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}(-(b_{2}+\alpha + \tau _{0}+K_{T}) I_{h}+ \gamma P_{h}E)\,ds \\ \hphantom{I_{h}(t)-I_{h} (0)=}{}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}(-(b_{2}+\alpha + \tau _{0}+K_{T}) I_{h}+\gamma P_{h}E), \\ A(t)-A (0)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}(-(b_{2}+ \tau _{0})A + \gamma (1-P_{s}-P_{h})E)\,ds \\ \hphantom{A(t)-A (0)=}{}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}(-(b_{2}+ \tau _{0})A + \gamma (1-P_{s}-P_{h})E), \\ R(t)-R (0) \\ \quad =\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}(-(b_{2}+ \eta )R+ \tau _{0}(I_{s}+I_{h}+A)+K_{T} I_{h}+K_{v}S)\,ds \\ \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}(-(b_{2}+\eta )R+ \tau _{0}(I_{s}+I_{h}+A)+K_{T} I_{h}+K_{v}S). \end{cases} $$
(4)
Now, we define some functions \(Q_{i}\) and some constants \(\eta _{i}\), \(i\epsilon N_{1}^{6}\) as follows:
$$ \textstyle\begin{cases} Q_{1}(t,S)= b_{1} -[b_{2}+ \beta (I_{s}+\beta _{hr}I_{h}+ \beta _{ar}A )+K_{v}]S+\eta R, \\ Q_{2}(t,E)=-(b_{2}+\gamma )E+\beta (I_{s}+\beta _{hr}I_{h}+\beta _{ar}A)S, \\ Q_{3}(t,I_{s})=-(b_{2}+ \tau _{0})I_{s}+\gamma P_{s} E, \\ Q_{4}(t,I_{h})==-(b_{2}+\alpha + \tau _{0}+K_{T}) I_{h}+\gamma P_{h}E, \\ Q_{5}(t,A)=-(b_{2}+ \tau _{0})A + \gamma (1-P_{s}-P_{h})E, \\ Q_{6}(t,R)=-(b_{2}+\eta )R+ \tau _{0}(I_{s}+I_{h}+A)+K_{T} I_{h}+K_{v}S. \end{cases} $$
(5)
- (G*)::
-
For proving our results, we assume the following assumptions: The continuous functions \(S(t)\), \(E(t)\), \(I_{s}(t)\), \(I_{h}(t)\), \(A(t)\), \(R(t)\) and \(S^{*}(t)\), \(E^{*}(t)\), \(I_{s}^{*}(t)\), \(I_{h}^{*}(t)\), \(A^{*}(t)\), \(R ^{*}(t)\) all belong to \(L[0,1]\) such that \(\|I_{s}\|\leq \psi _{1}\), \(\|I_{h}\|\leq \psi _{2}\), \(\|A\|\leq \psi _{3}\) for \(\psi _{1}\), \(\phi _{2}\), \(\psi _{3}> 0\) and constants.
Theorem 2.1
The kernels \(Q_{i}\) for \(i=1,2,3,\ldots,6\) satisfy Lipschitz conditions if the assumption (G*) holds and satisfies \(\phi _{i}< 1\) for \(i\in N_{1}^{6}\).
Proof
First, we prove that \(Q_{1}(t, S)\) satisfies the Lipschitz condition. Using \(S(t)\), \(S^{*}(t)\), we have
$$\begin{aligned} \bigl\Vert Q_{1}(t, S)-Q_{1}\bigl(t, S^{*} \bigr) \bigr\Vert =& \bigl\Vert \bigl(b_{1} -\bigl[b_{2}+ \beta (I_{s}+ \beta _{hr}I_{h}+ \beta _{ar}A )+K_{v}\bigr]S+\eta R\bigr) \\ &{}-\bigl(b_{1} -\bigl[b_{2}+ \beta (I_{s}+\beta _{hr}I_{h}+ \beta _{ar}A )+K_{v} \bigr]S+ \eta R\bigr) \bigr\Vert \\ =& \bigl\Vert (b_{2}+\beta I_{s}+ \beta _{hr} I_{h}+\beta _{ar} A+K_{v}) \bigl(S^{*}-S \bigr) \bigr\Vert \\ \leq & \bigl(b_{2}+\beta \Vert I_{s} \Vert + \beta _{hr} \Vert I_{h} \Vert +\beta _{ar} \Vert A \Vert +K_{v}\bigr) \bigl\Vert \bigl(S^{*}-S\bigr) \bigr\Vert \\ \leq &(b_{2}+\beta \psi _{1}+\beta _{hr} \psi _{2})+\beta _{ar} \psi _{3}+K_{v} \bigl\Vert (S-S*) \bigr\Vert \\ \leq &\phi _{1} \bigl\Vert \bigl(S-S^{*}\bigr) \bigr\Vert . \end{aligned}$$
Hence \(Q_{1}\) satisfies the Lipschitz condition and \(\phi _{1}< 1\). Next we prove that \(Q_{2}(t, E)\) satisfies the Lipschitz condition. Now, using \(E(t)\), \(E^{*}(t)\), we have
$$\begin{aligned} \bigl\Vert Q_{2}(t, E)-Q_{2}\bigl(t, E^{*} \bigr) \bigr\Vert =& \bigl\Vert \bigl(-(b_{2}+\gamma )E+\beta (I_{s}+ \beta _{hr}I_{h}+\beta _{ar}A)S \bigr) \\ &{}-\bigl(-(b_{2}+\gamma )E^{*}+\beta (I_{s}+ \beta _{hr}I_{h}+\beta _{ar}A)S\bigr) \bigr\Vert \\ =& \bigl\Vert (b_{2}+\lambda ) \bigl(E^{*}-E\bigr) \bigr\Vert \\ \leq &(b_{2}+\lambda ) \bigl\Vert \bigl(E^{*}-E\bigr) \bigr\Vert \\ \leq &\phi _{2} \bigl\Vert E-E^{*} \bigr\Vert . \end{aligned}$$
Hence \(Q_{2}\) satisfies the Lipschitz condition and \(\phi _{2}< 1\). Next we prove that \(Q_{3}(t, I_{s})\) satisfies the Lipschitz condition. Using \(I_{s}(t)\), \(I_{s}^{*}(t)\), we have
$$\begin{aligned} \bigl\Vert Q_{3}(t, I_{s})-Q_{3}\bigl(t, I_{s}^{*}\bigr) \bigr\Vert =& \bigl\Vert \bigl(-(b_{2}+ \tau _{0})I_{s}+ \gamma P_{s} E\bigr)-\bigl(-(b_{2}+ \tau _{0})I_{s}+ \gamma P_{s} E\bigr) \bigr\Vert \\ =& \bigl\Vert (b_{2}+\tau _{0}) \bigl(I_{s}^{*}-I_{s} \bigr) \bigr\Vert \\ \leq &(b_{2}+\tau _{0}) \bigl\Vert \bigl(I_{s}^{*}-I_{s}\bigr) \bigr\Vert \\ \leq &\phi _{3} \bigl\Vert I_{s}-I_{s}^{*} \bigr\Vert . \end{aligned}$$
Hence \(Q_{3}\) satisfies the Lipschitz condition and \(\phi _{3}< 1\). Next we prove that \(Q_{4}(t, I_{h})\) satisfies the Lipschitz condition. Using \(I_{h}(t)\), \(I_{h}^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert Q_{4}(t, I_{h})-Q_{4}\bigl(t, I_{h}^{*}\bigr) \bigr\Vert \\& \quad = \bigl\Vert \bigl(-(b_{2}+\alpha + \tau _{0}+K_{T}) I_{h}+\gamma P_{h}E\bigr) \\& \qquad {}-\bigl(-(b_{2}+\alpha + \tau _{0}+K_{T}) I_{h}^{*}+ \gamma P_{h}E\bigr) \bigr\Vert \\& \quad = \|((b_{2}+\alpha + \tau _{0}+K_{T}) \bigl(I_{h}^{*}-I_{h}\bigr)\| \\& \quad \leq ((b_{2}+\alpha + \tau _{0}+K_{T}) \bigl\Vert \bigl(I_{h}^{*}-I_{h}\bigr) \bigr\Vert \\& \quad \leq \phi _{4} \bigl\Vert I_{h}-I_{h}^{*} \bigr\Vert . \end{aligned}$$
Hence \(Q_{4}\) satisfies the Lipschitz condition and \(\phi _{4}< 1\). Next we prove that \(Q_{5}(t, A)\) satisfies the Lipschitz condition. Using \(A(t)\), \(A^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert Q_{5}(t, A)-Q_{5}\bigl(t, A^{*} \bigr) \bigr\Vert \\& \quad = \bigl\Vert \bigl(-(b_{2}+ \tau _{0})A + \gamma (1-P_{s}-P_{h})E\bigr) \\& \qquad {}- \bigl(-(b_{2}+ \tau _{0})A^{*} + \gamma (1-P_{s}-P_{h})E\bigr) \bigr\Vert \\& \quad = \bigl\Vert (b_{2}+ \tau _{0}) \bigl(A^{*}-A \bigr) \bigr\Vert \\& \quad \leq \|(b_{2}+ \tau _{0} \bigl\Vert \bigl(A^{*}-A\bigr) \bigr\Vert \\& \quad \leq \phi _{5} \bigl\Vert A-A^{*} \bigr\Vert . \end{aligned}$$
Hence \(Q_{5}\) satisfies the Lipschitz condition and \(\phi _{5}< 1\). Next we prove that \(Q_{6}(t, R)\) satisfies the Lipschitz condition. Using \(R(t)\), \(R^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert Q_{6}(t, E_{M})-Q_{6}\bigl(t, R^{*}\bigr) \bigr\Vert \\& \quad = \bigl\Vert (\beta S_{M} I -vE_{M}-\mu _{M}E_{M})-\bigl( \beta S_{M} I -vE_{M}^{*}-\mu _{M}E_{M}^{*} \bigr) \bigr\Vert \\& \quad = \bigl\Vert (b_{2}+\eta ) \bigl(R^{*}-R\bigr) \bigr\Vert \\& \quad \leq (b_{2}+\eta ) \bigl\Vert \bigl(R^{*}-R\bigr) \bigr\Vert \\& \quad \leq \phi _{6} \bigl\Vert R-R^{*} \bigr\Vert . \end{aligned}$$
Hence \(Q_{6}\) satisfies the Lipschitz condition and \(\phi _{6}< 1\). Ultimately all the functions satisfy Lipschitz conditions and are contractions with \(\phi _{i}< 1\) for \(i\in N_{1}^{6}\). Hence this completes the proof. □
We rewrite the system of equations (4) in the following form by using the kernels \(Q_{i}\), \(i\in N_{1}^{6}\) and the initial conditions \(S(0)=E(0)=I_{s}(0)=I_{h}(0)=A(0)=R(0)=0 \), we have
$$ \textstyle\begin{cases} S(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1}(s,S(s))\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1}(t,S(t)), \\ E(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2}(s,E(s))\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2}(t,E(t)), \\ I_{s}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3}(s,I_{s}(s))\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3}(t,I_{s}(t)), \\ I_{h}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4}(s,I_{h}(s))\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4}(t,I_{h}(t)), \\ A(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5}(s,A(s))\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5}(t,A(t)), \\ R(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6}(s,R(s))\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6}(t,R(t)). \end{cases} $$
(6)
Now we define the following recursive formulas:
$$\begin{aligned}& S_{n}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma (u_{1})} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S_{n-1}(s)\bigr)\,ds \\& \hphantom{S_{n}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S_{n-1}(t)\bigr), \\& E_{n}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E_{n-1}(s)\bigr)\,ds \\& \hphantom{E_{n}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E_{n-1}(t)\bigr), \\& I_{s_{n}}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s_{n-1}}(s)\bigr)\,ds \\& \hphantom{I_{s_{n}}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s_{n-1}}(t)\bigr), \\& I_{h_{n}}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h_{n-1}}(s)\bigr)\,ds \\& \hphantom{I_{h_{n}}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h_{n-1}}(t)\bigr), \\& A_{n}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A_{n-1}(s)\bigr)\,ds \\& \hphantom{A_{n}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A_{n-1}(t)\bigr), \\& R_{n}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R_{n-1}(s)\bigr)\,ds \\& \hphantom{R_{n}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R_{n-1}(t)\bigr). \end{aligned}$$
Now we consider the following differences:
$$\begin{aligned}& D S_{n+1}(t) = S_{n+1}-S_{n} \\& \hphantom{D S_{n+1}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S_{n}(t)\bigr) \\& \hphantom{D S_{n+1}(t)=} {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S_{n-1}(s)\bigr)\,ds \\& \hphantom{D S_{n+1}(t)=} {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1}\bigl(t,S_{n-1}(t)\bigr) \biggr) \\& \hphantom{D S_{n+1}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl(Q_{1}\bigl(s,S_{n}(s)\bigr)-Q_{1} \bigl(s,S_{n-1}(s)\bigr) \bigr)\,ds \\& \hphantom{D S_{n+1}(t)=} {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl(Q_{1}\bigl(t,S_{n}(t) \bigr)-Q_{1}\bigl(t,S_{n-1}(t)\bigr) \bigr), \\& D E_{n+1}(t) = E_{n+1}-E_{n} \\& \hphantom{D E_{n+1}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E_{n}(t)\bigr) \\& \hphantom{D E_{n+1}(t)=} {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E_{n-1}(s)\bigr)\,ds \\& \hphantom{D E_{n+1}(t)=} {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E_{n-1}(t)\bigr) \biggr) \\& \hphantom{D E_{n+1}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl(Q_{2}\bigl(s,E_{n}(s)\bigr)-Q_{1} \bigl(s,E_{n-1}(s)\bigr) \bigr)\,ds \\& \hphantom{D E_{n+1}(t)=}{} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl(Q_{2}\bigl(t,E_{n}(t) \bigr)-Q_{2}\bigl(t,E_{n-1}(t)\bigr) \bigr), \\& D I_{s_{n+1}}(t) = I_{s_{n+1}}-I_{H_{n}} \\& \hphantom{D I_{s_{n+1}}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s_{n}}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s_{n}}(t)\bigr) \\& \hphantom{D I_{s_{n+1}}(t)=} {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s_{n-1}}(s)\bigr)\,ds \\& \hphantom{D I_{s_{n+1}}(t)=} {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s_{n-1}}(t)\bigr) \biggr) \\& \hphantom{D I_{s_{n+1}}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl(Q_{3}\bigl(s,I_{s_{n}}(s)\bigr)-Q_{3} \bigl(s,I_{s_{n-1}}(s)\bigr) \bigr)\,ds \\& \hphantom{D I_{s_{n+1}}(t)=}{} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl(Q_{3}\bigl(t,I_{s_{n}}(t) \bigr)-Q_{3}\bigl(t,I_{s_{n-1}}(t)\bigr) \bigr), \\& D I_{h_{n+1}}(t) = I_{h_{n+1}}-I_{h_{n}} \\& \hphantom{D I_{h_{n+1}}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h_{n}}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h_{n}}(t)\bigr) \\& \hphantom{D I_{h_{n+1}}(t)=} {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h_{n-1}}(s)\bigr)\,ds \\& \hphantom{D I_{h_{n+1}}(t)=} {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h_{n-1}}(t)\bigr) \biggr) \\& \hphantom{D I_{h_{n+1}}(t)}= \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl(Q_{4}\bigl(s,I_{h_{n}}(s)\bigr)-Q_{4} \bigl(s,I_{h_{n-1}}(s)\bigr) \bigr)\,ds \\& \hphantom{D I_{h_{n+1}}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl(Q_{4}\bigl(t,I_{h_{n}}(t) \bigr)-Q_{4}\bigl(t,I_{h_{n-1}}(t)\bigr) \bigr), \\& D A_{n+1}(t) = A_{n+1}-A_{n} \\& \hphantom{D A_{n+1}(t)}= \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A_{n}(t)\bigr) \\& \hphantom{D A_{n+1}(t)=} {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A_{n-1}(s)\bigr)\,ds \\& \hphantom{D A_{n+1}(t)=} {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A_{n-1}(t)\bigr) \biggr) \\& \hphantom{D A_{n+1}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl(Q_{5}\bigl(s,A_{n}(s)\bigr)-Q_{5} \bigl(s,A_{n-1}(s)\bigr) \bigr)\,ds \\& \hphantom{D A_{n+1}(t)=}{} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl(Q_{5}\bigl(t,A_{n}(t) \bigr)-Q_{5}\bigl(t,A_{n-1}(t)\bigr) \bigr), \\& D R_{n+1}(t) = R_{n+1}-R_{n} \\& \hphantom{D R_{n+1}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R_{n}(t)\bigr) \\& \hphantom{D R_{n+1}(t)=} {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R_{n-1}(s)\bigr)\,ds \\& \hphantom{D R_{n+1}(t)=}{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R_{n-1}(t)\bigr) \biggr) \\& \hphantom{D R_{n+1}(t)} = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl(Q_{6}\bigl(s,R_{n}(s)\bigr)-Q_{6} \bigl(s,R_{n-1}(s)\bigr) \bigr)\,ds \\& \hphantom{D R_{n+1}(t)=}{} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl(Q_{6}\bigl(t,R_{n}(t) \bigr)-Q_{6}\bigl(t,R_{n-1}(t)\bigr) \bigr). \end{aligned}$$
Taking norm of the above differences, we have
$$\begin{aligned}& \bigl\Vert D S_{n+1}(t) \bigr\Vert \\& \quad = \Vert S_{n+1}-S_{n} \Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S_{n}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S_{n-1}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S_{n-1}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert Q_{1}\bigl(s,S_{n}(s)\bigr)-Q_{1} \bigl(s,S_{n-1}(s)\bigr) \bigr\Vert \,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{1}\bigl(t,S_{n}(t) \bigr)-Q_{1}\bigl(t,S_{n-1}(t)\bigr) \bigr\Vert , \\& \bigl\Vert D E_{n+1}(t) \bigr\Vert \\& \quad = \Vert E_{n+1}-E_{n} \Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E_{n}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E_{n-1}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E_{n-1}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert \bigl(Q_{2}\bigl(s,E_{n}(s) \bigr)-Q_{1}\bigl(s,E_{n-1}(s)\bigr)\bigr) \bigr\Vert \,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}\|(Q_{2}\bigl(t,E_{n}(t) \bigr)-Q_{2}\bigl(t,E_{n-1}(t)\bigr) \|, \\& \bigl\Vert D I_{s_{n+1}}(t) \bigr\Vert \\& \quad = \Vert I_{s_{n+1}}-I_{s_{n}} \Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s_{n}}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s_{n}}(t)\bigr) \\& \qquad {} {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s_{n-1}}(s)\bigr)\,ds \\& \qquad {} +\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s_{n-1}}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \| \bigl(Q_{3}\bigl(s,I_{s_{n}}(s)\bigr)-Q_{3} \bigl(s,I_{s_{n-1}}(s)\bigr)\|\bigr)\,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{3}\bigl(t,I_{s_{n}}(t) \bigr)-Q_{3}\bigl(t,I_{s_{n-1}}(t)\bigr) \bigr\Vert , \\& \bigl\Vert D I_{h_{n+1}}(t) \bigr\Vert \\& \quad = \Vert I_{h_{n+1}}-I_{h_{n}} \Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h_{n}}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h_{n}}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h_{n-1}}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h_{n-1}}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert Q_{4}\bigl(s,I_{h_{n}}(s)\bigr)-Q_{4} \bigl(s,I_{h_{n-1}}(s)\bigr) \bigr\Vert \,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{4}\bigl(t,I_{h_{n}}(t) \bigr)-Q_{4}\bigl(t,I_{h_{n-1}}(t)\bigr) \bigr\Vert , \\& \bigl\Vert D A_{n+1}(t) \bigr\Vert \\& \quad = \Vert A_{n+1}-A_{n} \Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A_{n}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A_{n-1}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A_{n-1}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert Q_{5}\bigl(s,A_{n}(s)\bigr)-Q_{5} \bigl(s,A_{n-1}(s)\bigr) \bigr\Vert \,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{5}\bigl(t,A_{n}(t) \bigr)-Q_{5}\bigl(t,A_{n-1}(t)\bigr) \bigr\Vert , \\& \bigl\Vert D R_{n+1}(t) \bigr\Vert \\& \quad = \Vert R_{n+1}-R_{n} \Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R_{n}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R_{n-1}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R_{n-1}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert Q_{6}\bigl(s,R_{n}(s)\bigr)-Q_{6} \bigl(s,R_{n-1}(s)\bigr) \bigr\Vert \,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{6}\bigl(t,R_{n}(t) \bigr)-Q_{6}\bigl(t,R_{n-1}(t)\bigr) \bigr\Vert . \end{aligned}$$
Theorem 2.2
The fractal fractional of diffusion model SE(Is)(Ih)AR epidemic has a solution if the following holds true:
$$\sigma =\max \{\phi _{1},\phi _{2},\ldots, \phi _{6}\}< 1. $$
Proof
Let us define the following functions:
$$ \textstyle\begin{cases} G_{1}n(t)=S_{n+1}(t)-S(t), \\ G_{2}n(t)=E_{n+1}(t)-E(t), \\ G_{3}n(t)=I_{s_{n+1}}(t)-I_{s}(t), \\ G_{4}n(t)=I_{h_{n+1}}(t)-I_{h}(t), \\ G_{5}n(t)=A_{n+1}(t)-A(t), \\ G_{6}n(t)=R_{n+1}(t)-R(t). \end{cases} $$
(7)
Taking norm of the above system, we have
$$\begin{aligned}& \bigl\Vert G_{1}n(t) \bigr\Vert \\& \quad = \bigl\Vert S_{n+1}(t)-S(t) \bigr\Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S_{n}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1}(s,S) (s)\biggr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1}(t,S) (t)) ) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert Q_{1}\bigl(s,S_{n}(s)\bigr)-Q_{1}(s,S) (s) \bigr\Vert \,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{1}\bigl(t,S_{n}(t) \bigr)-Q_{1}(t,S) (t) \bigr\Vert \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{ AB(u_{1})} \biggr)\phi _{1} \Vert S_{n}-S \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{1} \Vert S_{n}-S \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)^{n} \sigma ^{n} \Vert S_{1}-S \Vert , \end{aligned}$$
where \(\sigma < 1\) and as \(n \rightarrow \infty \) so \(S_{n}\rightarrow S\), and using the formula \(B(u, v)=(b-a)^{-u+v+1}\int _{a}^{b}(s-a)^{u-1}(b-s)^{v-1}\,ds\) and as \(t\in [0, 1]\) so \(t^{-1-u_{1}+u_{2}}\leq 1\) and \(t^{u_{2}}\leq 1\),
$$\begin{aligned} \bigl\Vert G_{2}n(t) \bigr\Vert =& \bigl\Vert E_{n+1}(t)-E(t) \bigr\Vert \\ =& \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E_{n}(s)\bigr)\,ds \\ &{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E_{n}(t)\bigr) \\ &{}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2}(s,E) (s)\biggr)\,ds \\ &{}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2}\bigl(t,E(t)\bigr) ) \biggr\Vert \\ =&\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \big\| Q_{2}\bigl(s,E_{n}(s)\bigr)-Q_{2}(s,E) (s)) \big\Vert \,ds \\ &{}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \big\Vert Q_{2}\bigl(t,E_{n}(t) \bigr)-Q_{2}(t,E) (t)) \big\| \\ \leq & \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{2} \Vert E_{n}-E \Vert \\ \leq & \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{2} \Vert E_{n}-E \Vert \\ \leq & \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)^{n} \sigma ^{n} \Vert E_{1}-E \Vert , \end{aligned}$$
where \(\sigma < 1\) and as \(n \rightarrow \infty \) so \(E_{H_{n}}\rightarrow E\).
$$\begin{aligned}& \bigl\Vert G_{3}n(t) \bigr\Vert \\& \quad = \bigl\Vert I_{s_{n+1}}(t)-I_{s}(t) \bigr\Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s_{n}}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s_{n}}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert Q_{3}\bigl(s,I_{s_{n}}(s)\bigr)-Q_{3} \bigl(s,I_{s}(s)\bigr) \bigr\Vert \,ds \\& \qquad {} + \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{3}\bigl(t,I_{s_{n}}(t) \bigr)-Q_{3}\bigl(t,I_{s}(t)\bigr) \bigr\Vert \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{3} \Vert I_{s_{n}}-I_{s} \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{3} \Vert I_{s_{n}}-I_{s} \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)^{n} \sigma ^{n} \Vert I_{s_{1}}-I_{s} \Vert , \end{aligned}$$
where \(\sigma < 1\) and as \(n \rightarrow \infty \) so \(I_{s_{n}}\rightarrow I_{s}\).
$$\begin{aligned}& \bigl\Vert G_{4}n(t) \bigr\Vert \\& \quad = \bigl\Vert I_{h_{n+1}}(t)-I_{h}(t) \bigr\Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h_{n}}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h_{n}}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\Vert Q_{4}\bigl(s,I_{h_{n}}(s)\bigr)-Q_{4} \bigl(s,I_{h}(s)\bigr) \bigr\Vert \,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\Vert Q_{4}\bigl(t,I_{h_{n}}(t) \bigr)-Q_{4}\bigl(t,I_{h}(t)\bigr) \bigr\Vert \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{4} \Vert I_{h_{n}}-I_{h} \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{4} \Vert I_{h_{n}}-I_{h} \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)^{n} \sigma ^{n} \Vert I_{h_{1}}-I_{h} \Vert , \end{aligned}$$
where \(\sigma < 1\) and as \(n \rightarrow \infty \) so \(I_{h_{n}}\rightarrow I_{h}\).
$$\begin{aligned}& \bigl\Vert G_{5}n(t) \bigr\Vert \\& \quad = \bigl\Vert A_{n+1}(t)-A(t) \bigr\Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A_{n}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5}(s,A) (s)\biggr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5}(t,A) (t)) ) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{5}\bigl(s,A_{n}(s)\bigr)-Q_{5}(s,A) (s)) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{5}\bigl(t,A_{n}(t) \bigr)-Q_{5}(t,A) (t)) \| \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{5} \Vert A_{n}-A \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{5} \Vert A_{n}-A \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)^{n} \sigma ^{n} \Vert A_{1}-A \Vert , \end{aligned}$$
where \(\sigma < 1\) and as \(n \rightarrow \infty \) so \(A_{n}\rightarrow A\).
$$\begin{aligned}& \bigl\Vert G_{6}n(t) \bigr\Vert \\& \quad = \bigl\Vert R_{n+1}(t)-R(t) \bigr\Vert \\& \quad = \biggl\Vert \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R_{n}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R_{n}(t)\bigr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6}(s,R) (s)\biggr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6}(t,R) (t)) ) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{6}\bigl(s,R_{n}(s)\bigr)-Q_{6}(s,R) (s)) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{6}\bigl(t,R_{n}(t) \bigr)-Q_{6}(t,R) (t)) \| \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{6} \Vert R_{n}-R \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{6} \Vert R_{n}-R \Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)^{n} \sigma ^{n} \Vert R_{1}-R \Vert , \end{aligned}$$
where \(\sigma < 1\) and as \(n \rightarrow \infty \) so \(R_{n}\rightarrow R\). Thus we find that \(G_{i}n(t)\rightarrow 0\) as \(n\rightarrow \infty \) for \(i\in N_{1}^{6}\) AND \(\sigma < 1\). Hence this completes the proof. □
2.1 Uniqueness of the solution
Theorem 2.3
The fractal fractional model (1) has a unique solution if the following inequalities hold true:
$$ \biggl(\frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2} (1-u_{1})}{AB(u_{1})} \biggr)\phi _{i}\leq 1,\quad i\in N_{1}^{6}.$$
(8)
Proof
Let us consider the contradiction that there exists another solution of fractal fractional model (1) such that \(S^{*}\), \(E^{*}\), \(I_{s}^{*}\), \(I_{h}^{*}\), \(A^{*}\), \(R^{*}\) satisfying the given model. We have
$$\begin{aligned}& S^{*}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S^{*}(t)\bigr), \\& E^{*}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E^{*}(t)\bigr), \\& I_{s}^{*}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}^{*}(t)\bigr), \\& I_{h}^{*}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}^{*}(t)\bigr), \\& A^{*}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A^{*}(t)\bigr), \\& R^{*}(t)=\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R^{*}(t)\bigr). \end{aligned}$$
Now, taking norm of the difference of \(S(t)\), \(S^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert S(t)-S^{*}(t) \bigr\Vert \\& \quad = \biggl\Vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1}\bigl(t,S(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S^{*}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{1}(s, S(s)-Q_{1}\bigl(s, S^{*}(s)\bigr) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{1}(t, S(t)-Q_{1} \bigl(t, S^{*}(t)\bigr)\|\,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{1} \bigl\Vert S-S^{*} \bigr\Vert \\& \qquad {}\times \biggl[1- \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{1} \biggr] \bigl\Vert S-S^{*} \bigr\Vert \leq 0. \end{aligned}$$
The above inequality is true if \(\|S-S^{*}\|=0\), which implies \(S=S^{*}\). Similarly, taking norm of the difference of \(E(t)\), \(E^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert E(t)-E^{*}(t) \bigr\Vert \\& \quad = \biggl\Vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2}\bigl(t,E(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E^{*}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{2}(s, E(s)-Q_{2}\bigl(s, E^{*}(s)\bigr) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{2}(t, E(t)-Q_{2} \bigl(t, E^{*}(t)\bigr)\|\,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{2} \bigl\Vert E-E^{*} \bigr\Vert \\& \qquad {}\times \biggl[1- \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{2} \biggr] \bigl\Vert E-E^{*} \bigr\Vert \leq 0. \end{aligned}$$
The above inequality is true if \(\|E-E^{*}\|=0\), which implies \(E=E^{*}\). Similarly, taking norm of the difference of \(I_{s}(t)\), \(I_{s}^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert I_{s}(t)-I_{s}^{*}(t) \bigr\Vert \\& \quad = \biggl\Vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}^{*}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{3}(s, I_{s}(s)-Q_{3}\bigl(s, I_{s}^{*}(s)\bigr) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{3}(t, I_{s}(t)-Q_{3} \bigl(t, I_{s}^{*}(t)\bigr)\|\,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{3} \bigl\Vert I_{s}-I_{s}^{*} \bigr\Vert \\& \qquad {}\times \biggl[1- \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{3} \biggr] \bigl\Vert I_{s}-I_{s}^{*} \bigr\Vert \leq 0. \end{aligned}$$
The above inequality is true if \(\|I_{s}-I_{s}^{*}\|=0\), which implies \(I_{s}=I_{s}^{*}\). Similarly, taking norm of the difference of \(I_{h}(t)\), \(I_{h}^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert I_{h}(t)-I_{h}^{*}(t) \bigr\Vert \\& \quad = \biggl\Vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}^{*}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{4}(s, I_{h}(s)-Q_{4}\bigl(s, I_{h}^{*}(s)\bigr) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{4}(t, I_{h}(t)-Q_{4} \bigl(t, I_{h}^{*}(t)\bigr)\|\,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{4} \bigl\Vert I_{h}-I_{h}^{*} \bigr\Vert \\& \qquad {}\times \biggl[1- \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{4} \biggr] \bigl\Vert I_{h}-I_{h}^{*} \bigr\Vert \leq 0. \end{aligned}$$
The above inequality is true if \(\|I_{h}-I_{h}^{*}\|=0\), which implies \(I_{h}=I_{h}^{*}\). Similarly, taking norm of the difference of \(A(t)\), \(A^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert A(t)-A^{*}(t) \bigr\Vert \\& \quad = \biggl\Vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5}\bigl(t,A(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A^{*}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A^{*}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{5}(s, A(s)-Q_{5}\bigl(s, A^{*}(s)\bigr) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{5}(t, A(t)-Q_{5} \bigl(t, A^{*}(t)\bigr)\|\,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{5} \bigl\Vert A-A^{*} \bigr\Vert \\& \qquad {}\times \biggl[1- \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{5} \biggr] \bigl\Vert A-A^{*} \bigr\Vert \leq 0. \end{aligned}$$
The above inequality is true if \(\|A-A^{*}\|=0\), which implies \(A=A^{*}\). Similarly, taking norm of the difference of \(R(t)\), \(R^{*}(t)\), we have
$$\begin{aligned}& \bigl\Vert R(t)-R^{*}(t) \bigr\Vert \\& \quad = \biggl\Vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6}\bigl(t,R(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R^{*}(t)\bigr) \biggr) \biggr\Vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \|Q_{6}(s, R(s)-Q_{2}\bigl(s, R^{*}(s)\bigr) \biggl\Vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\Vert Q_{6}(t, R(t)-Q_{6} \bigl(t, R^{*}(t)\bigr)\|\,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{6} \bigl\Vert R-R^{*} \bigr\Vert \\& \qquad {}\times \biggl[1- \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{6} \biggr] \bigl\Vert R-R^{*} \bigr\Vert \leq 0. \end{aligned}$$
The above inequality is true if \(\|R-R^{*}\|=0\), which implies \(R(t)\), \(R^{*}(t)\). Hence we see that \(S=S^{*}\), \(E=E^{*}\), \(I_{s}=I_{s}^{*}\), \(rI_{h}=I_{h}^{*}\), \(A=A^{*}\), \(R=R^{*}\), so our supposition is wrong and the theorem has a unique solution. □
Hyers–Ulam stability
Definition 2.4
The fractal fractional integrals (6) are said to be Hyers–Ulam stable if there exist constants \(\alpha _{i} > 0\), \(i \in N_{1}^{6}\) satisfying, for every \(\beta _{i} > 0\), \(i \in N_{1}^{6}\), the following:
$$\begin{aligned}& \biggl\vert S(t)-\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1}\bigl(t,S(t)\bigr) \biggr\vert \leq \beta _{1}, \\& \biggl\vert E(t)-\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2}\bigl(t,E(t)\bigr) \biggr\vert \leq \beta _{2}, \\& \biggl\vert I_{s}(t)-\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}(t)\bigr) \biggr\vert \\& \quad \leq \beta _{3}, \\& \biggl\vert I_{h}(t)-\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}(t)\bigr) \biggr\vert \\& \quad \leq \beta _{4}, \\& \biggl\vert A(t)-\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5}\bigl(t,A(t)\bigr) \biggr\vert \\& \quad \leq \beta _{5}, \\& \biggl\vert R(t)-\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6}\bigl(t,R(t)\bigr) \biggr\vert \leq \beta _{6}. \end{aligned}$$
There exists an approximate solution of model (1) \(S^{*}(t)\), \(E^{*}(t)\), \(I_{s}^{*}(t)\), \(I_{h}^{*}(t)\), \(A^{*}(t)\), \(R^{*}(t)\) that satisfies the given model, such that
$$\begin{aligned}& \bigl\vert S(t)-S^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1}\bigl(t,S(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{1}(s, S(s)-Q_{1}\bigl(s, S^{*}(s) \bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{1}(t, S(t)-Q_{1} \bigl(t, S^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{1} \bigl\Vert S-S^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{1} \bigl\Vert S-S^{*} \bigr\Vert . \end{aligned}$$
Let \(\zeta _{1}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|S-S^{*}\|\), \(\eta _{1}=\phi _{1}\),so the above inequality becomes \(|S-S^{*}|\leq \zeta _{1} \eta _{1}\).
$$\begin{aligned}& \bigl\vert E(t)-E^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2}\bigl(t,E(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{2}(s, E(s)-Q_{2}\bigl(s, E^{*}(s) \bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{2}(t, E(t)-Q_{2} \bigl(t, E^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{2} \bigl\Vert E-E^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{2} \bigl\Vert E-E^{*} \bigr\Vert . \end{aligned}$$
Let \(\zeta _{2}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|E-E^{*}\|\), \(\eta _{2}=\phi _{2}\),so the above inequality becomes \(|E-E^{*}|\leq \zeta _{2} \eta _{2}\).
$$\begin{aligned}& \bigl\vert I_{s}(t)-I_{s}^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{3}(s, I_{s}(s)-Q_{3}\bigl(s, I_{s}^{*}(s)\bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{3}(t, I_{s}(t)-Q_{3} \bigl(t, I_{s}^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{3} \bigl\Vert I_{s}-I_{s}^{*} \bigr\Vert . \end{aligned}$$
Let \(\zeta _{3}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|I_{s}-I_{s}^{*}\|\), \(\eta _{3}=\phi _{3}\), so the above inequality becomes \(|I_{s}-I_{s}^{*}|\leq \zeta _{3} \eta _{3}\).
$$\begin{aligned}& \bigl\vert I_{h}(t)-I_{h}^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{4}(s, I_{h}(s)-Q_{4}\bigl(s, I_{h}^{*}(s)\bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{4}(t, I_{h}(t)-Q_{4} \bigl(t, I_{h}^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{4} \bigl\Vert I_{h}-I_{h}^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{4} \bigl\Vert I_{h}-I_{h}^{*} \bigr\Vert . \end{aligned}$$
Let \(\zeta _{4}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|I_{h}-I_{h}^{*}\|\), \(\eta _{4}=\phi _{4}\), so the above inequality becomes \(|I_{h}-I_{h}^{*}|\leq \zeta _{4} \eta _{4}\).
$$\begin{aligned}& \bigl\vert A(t)-A^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5}\bigl(t,A(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{5}(s, A(s)-Q_{5}\bigl(s, A^{*}(s) \bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{5}(t, A(t)-Q_{5} \bigl(t, A^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{5} \bigl\Vert A-A^{*} \bigr\Vert . \end{aligned}$$
Let \(\zeta _{5}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|A-A^{*}\|\), \(\eta _{5}=\phi _{5}\), so the above inequality becomes \(|A-A^{*}|\leq \zeta _{5} \eta _{5}\).
$$\begin{aligned}& \bigl\vert R(t)-R^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6}\bigl(t,R(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}|Q_{6}(s, R(s)-Q_{6}\bigl(s, R^{*}(s)\bigr) \biggl\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\vert Q_{6}(t, R(t)-Q_{6} \bigl(t, R^{*}(t)\bigr)|\,ds \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{6} \bigl\Vert R-R^{*} \bigr\Vert . \end{aligned}$$
Let \(\zeta _{6}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|R-R^{*}\|\), \(\eta _{6}=\phi _{6}\), so the above inequality becomes \(|R-R^{*}|\leq \zeta _{6} \eta _{6}\).
Theorem 2.5
With assumption (G*), the fractal fractional model (1) is Hyers–Ulam stable.
Proof
We know that the fractal fractional model (1) has a unique solution. Let there exist an approximate solution of model (1) \(S^{*}(t)\), \(E^{*}(t)\), \(I_{s}^{*}(t)\), \(I_{h}^{*}(t)\), \(A^{*}(t)\), \(R^{*}(t)\) that satisfies the given model, such that
$$\begin{aligned}& \bigl\vert S(t)-S^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1}\bigl(t,S(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{1} \bigl(s,S^{*}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{1} \bigl(t,S^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{1}(s, S(s)-Q_{1}\bigl(s, S^{*}(s) \bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{1}(t, S(t)-Q_{1} \bigl(t, S^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{1} \bigl\Vert S-S^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{1} \bigl\Vert S-S^{*} \bigr\Vert . \end{aligned}$$
Let \(\alpha _{1}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|S-S^{*}\|\), \(\Delta _{1}=\phi _{1}\), so the above inequality becomes \(|S-S^{*}|\leq \alpha _{1} \Delta _{1}\).
$$\begin{aligned}& \bigl\vert E(t)-E^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2}\bigl(t,E(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{2} \bigl(s,E^{*}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{2} \bigl(t,E^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{2}(s, E(s)-Q_{2}\bigl(s, E^{*}(s) \bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{2}(t, E(t)-Q_{2} \bigl(t, E^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{2} \bigl\Vert E-E^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{2} \bigl\Vert E-E^{*} \bigr\Vert . \end{aligned}$$
Let \(\alpha _{2}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|E-E^{*}\|\), \(\Delta _{2}=\phi _{2}\), so the above inequality becomes \(|E-E^{*}|\leq \alpha _{2} \Delta _{2}\).
$$\begin{aligned}& \bigl\vert I_{s}(t)-I_{s}^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{3} \bigl(s,I_{s}^{*}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{3} \bigl(t,I_{s}^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{3}(s, I_{s}(s)-Q_{3}\bigl(s, I_{s}^{*}(s)\bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{3}(t, I_{s}(t)-Q_{3} \bigl(t, I_{s}^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{3} \bigl\Vert I_{s}-I_{s}^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{3} \bigl\Vert I_{s}-I_{s}^{*} \bigr\Vert . \end{aligned}$$
Let \(\alpha _{3}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|I_{s}-I_{s}^{*}\|\), \(\Delta _{3}=\phi _{3}\), so the above inequality becomes \(|I_{s}-I_{s}^{*}|\leq \alpha _{3} \Delta _{3}\).
$$\begin{aligned}& \bigl\vert I_{h}(t)-I_{h}^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{4} \bigl(s,I_{h}^{*}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{4} \bigl(t,I_{h}^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}|Q_{4}(s, I_{h}(s)-Q_{4}\bigl(s, I_{h}^{*}(s) \bigr) \biggl\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr\vert Q_{4}(t, I_{h}(t)-Q_{4} \bigl(t, I_{h}^{*}(t)\bigr)|\,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{4} \bigl\Vert I_{h}-I_{h}^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{4} \bigl\Vert I_{h}-I_{h}^{*} \bigr\Vert . \end{aligned}$$
Let \(\alpha _{4}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|I_{h}-I_{h}^{*}\|\), \(\Delta _{4}=\phi _{4}\), so the above inequality becomes \(|I_{h}-I_{h}^{*}|\leq \alpha _{4} \Delta _{4}\).
$$\begin{aligned}& \bigl\vert A(t)-A^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A(s)\bigr)\,ds+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5}\bigl(t,A(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{5} \bigl(s,A^{*}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{5} \bigl(t,A^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{5}(s, A(s)-Q_{5}\bigl(s, A^{*}(s) \bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{5}(t, A(t)-Q_{5} \bigl(t, A^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{5} \bigl\Vert A-A^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{5} \bigl\Vert A-A^{*} \bigr\Vert . \end{aligned}$$
Let \(\alpha _{5}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|A-A^{*}\|\), \(\Delta _{5}=\phi _{5}\), so the above inequality becomes \(|A-A^{*}|\leq \alpha _{5} \Delta _{5}\).
$$\begin{aligned}& \bigl\vert R(t)-R^{*}(t) \bigr\vert \\& \quad = \biggl\vert \biggl( \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6}\bigl(t,R(t)\bigr) \biggr) \\& \qquad {}- \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}Q_{6} \bigl(s,R^{*}(s)\bigr)\,ds \\& \qquad {}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})}Q_{6} \bigl(t,R^{*}(t)\bigr) \biggr) \biggr\vert \\& \quad = \frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1} \bigl\vert Q_{6}(s, R(s)-Q_{6}\bigl(s, R^{*}(s) \bigr) \bigr\vert \,ds \\& \qquad {}+\frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \bigl\vert Q_{6}(t, R(t)-Q_{6} \bigl(t, R^{*}(t)\bigr) \bigr\vert \,ds \\& \quad \leq \biggl(\frac{u_{1} u_{2}}{AB(u_{1})\Gamma u_{1}} \int _{0}^{t} s^{u_{2}-1}(t-s)^{u_{1}-1}+ \frac{u_{2}(1-u_{1})t^{u_{2}-1}}{AB(u_{1})} \biggr)\phi _{6} \bigl\Vert R-R^{*} \bigr\Vert \\& \quad \leq \biggl( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} \biggr)\phi _{6} \bigl\Vert R-R^{*} \bigr\Vert . \end{aligned}$$
Let \(\alpha _{6}= ( \frac{u_{1} u_{2}\Gamma u_{2}}{AB(u_{1}\Gamma (u_{1}+u_{2}))}+ \frac{u_{2}(1-u_{1})}{AB(u_{1})} )\|R-R^{*}\|\), \(\Delta _{6}=\phi _{6}\), so the above inequality becomes \(|R-R^{*}|\leq \alpha _{6} \Delta _{6}\). Consequently, by definition the fractal fractional model (1) is Hyers–Ulam stable. This completes the proof. □