For system (1.1), we have
$$\begin{aligned} &\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0, \ldots,\Delta_{m} ( u(t_{m}^{ -} ) ) \to 0} \left \{ \textstyle\begin{array}{@{}l} {}_{C - H}D_{a^{ +}}^{q}u(t) = f(t,u(t)),\\ \quad t \in (a,T] \mbox{ and } t \ne t_{k}\ (k = 1,2,\ldots,m), \\ \Delta u |_{t = t_{k}} = u(t_{k}^{ +} ) - u(t_{k}^{ -} ) = \Delta_{k} ( u(t_{k}^{ -} ) ) \in \mathbb{C},\quad k = 1,2,\ldots,m, \\ u(a^{ +} ) = u_{a},\quad u_{a} \in \mathbb{C} \end{array}\displaystyle \right . \\ &\quad\to \left \{ \textstyle\begin{array}{@{}l} {}_{C - H}D_{a^{ +}}^{q}u(t) = f(t,u(t)),\quad t \in (a,T], \\ u(a^{ +} ) = u_{a},\quad u_{a} \in \mathbb{C}. \end{array}\displaystyle \right . \end{aligned}$$
(3.1)
That is,
$$\begin{aligned} &\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0, \ldots,\Delta_{m} ( u(t_{m}^{ -} ) ) \to 0} \bigl\{ \mbox{the solution of system } (1.1) \bigr\} \\ &\quad= \bigl\{ \mbox{the solution of system } (3.1) \bigr\} . \end{aligned}$$
(3.2)
Thus, the definition of solution for system (1.1) is provided.
Definition 3.1
A function \(z(t):[a,T] \to \mathbb{C}\) is said to be a solution of the fractional Cauchy problem (1.1) if \(z(a) = u_{a}\), the equation condition \({}_{C - H}D_{a^{ +}}^{q}z(t) = f(t,z(t))\) for each \(t \in (a, T]\) is verified, the impulsive conditions \(\Delta z|_{t = t_{k}} = \Delta_{k} ( z(t_{k}^{ -} ) )\) (here \(k = 1, 2, \ldots, m\)) are satisfied, the restriction of \(z( \cdot )\) to the interval \((t_{k},t_{k + 1}]\) (here \(k = 0,1, 2, \ldots, m\)) is continuous, and condition (3.2) holds.
A piecewise function is defined by
$$\begin{aligned} &\tilde{u}(t) = u\bigl(t_{k}^{ +} \bigr) + \frac{1}{\Gamma (q)} \int_{t_{k}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s}, \\ &\quad\mbox{for } t \in (t_{k},t_{k + 1}]\ (\mbox{where } k = 0,1,2,\ldots,m). \end{aligned}$$
By Theorem 2.4, we have
$$\begin{aligned} &\bigl[ {}_{C - H}D_{a^{ +}}^{q}\tilde{u}(t) \bigr]_{t \in (t_{k},t_{k + 1}]} \\ &\quad= \biggl\{ \frac{1}{\Gamma (1 - q)}\int_{a}^{t} \biggl( \ln \frac{t}{s} \biggr)^{1 - q - 1} \biggl[ s\frac{d}{ds} \biggl( u\bigl(t_{k}^{ +} \bigr) \\ &\qquad{}+ \frac{1}{\Gamma (q)}\int _{t_{k}}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f\bigl(\eta,u(\eta )\bigr)\frac{d\eta}{\eta} \biggr) \biggr] \frac{ds}{s} \biggr\} _{t \in (t_{k},t_{k + 1}]} \\ &\quad= \biggl\{ \frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{t_{k}}^{t} \biggl( \ln \frac{t}{s} \biggr)^{1 - q - 1} \biggl[ s\frac{d}{ds} \biggl( \int_{t_{k}}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f\bigl(\eta,u(\eta )\bigr)\frac{d\eta}{\eta} \biggr) \biggr] \frac{ds}{s} \biggr\} _{t \in (t_{k},t_{k + 1}]} \\ &\quad= \biggl\{ \frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{t_{k}}^{t} \biggl( \ln \frac{t}{s} \biggr)^{1 - q - 1} \biggl( s\frac{ - 1}{q} \frac{d}{ds}\int_{t_{k}}^{s} f\bigl(\eta,u(\eta )\bigr)\,d\biggl(\ln \frac{s}{\eta} \biggr)^{q} \biggr) \frac{ds}{s} \biggr\} _{t \in (t_{k},t_{k + 1}]} \\ &\quad= \biggl\{ \frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{t_{k}}^{t} \biggl( \ln \frac{t}{s} \biggr)^{1 - q - 1} \biggl( s\frac{ - 1}{q} \frac{d}{ds} \biggl[ \biggl(\ln \frac{s}{\eta} \biggr)^{q}f\bigl(\eta,u(\eta )\bigr) \Big| _{t_{k}}^{s}\\ &\qquad{}- \int_{t_{k}}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q}f'\bigl(\eta,u(\eta )\bigr)\,d\eta \biggr] \biggr)\frac{ds}{s} \biggr\} _{t \in (t_{k},t_{k + 1}]} \\ &\quad= f\bigl(t,u(t)\bigr) |_{t \in (t_{k},t_{k + 1}]}. \end{aligned}$$
It shows that the piecewise function \(\tilde{u}(t)\) satisfies the condition of fractional derivative in system (1.1). Thus, we assume that the piecewise function \(\tilde{u}(t)\) is an approximate solution of (1.1).
Theorem 3.2
Let
\(0 < \Re (q) < 1\)
and
ħ
be a constant. A function
\(u(t):[a,T] \to \mathbb{C}\)
is a general solution of system (1.1) if and only if
\(u(t)\)
satisfies the fraction integral equation
$$ u(t) = \left \{ \textstyle\begin{array}{@{}l@{\quad}l} u_{a} + \frac{1}{\Gamma (q)}\int_{a}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} \quad \textit{for } t \in (a,t_{1}],\\ u_{a} + \sum_{i = 1}^{k} \Delta_{i} ( u(t_{i}^{ -} ) ) + \frac{1}{\Gamma (q)}\int_{a}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} \\ \quad{}+ \sum_{i = 1}^{k} \{ \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} [ \int_{a}^{t_{i}} (\ln \frac{t_{i}}{s})^{q - 1}f(s,u(s))\frac{ds}{s} \\ \quad{}+ \int_{t_{i}}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} - \int_{a}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} ] \} \\ \quad\textit{for } t \in (t_{k},t_{k + 1}],k = 1,2,\ldots,m, \end{array}\displaystyle \right . $$
(3.3)
provided that the integral in (3.3) exists.
Proof
‘Necessity’, it will be verified that Eq. (3.3) satisfies the conditions of system (1.1).
Taking the Caputo-Hadamard fractional derivative to the both sides of Eq. (3.3), for \(t \in (a,t_{1}]\), we have
$$\begin{aligned} &{}_{C - H}D_{a^{ +}}^{q}u(t) |_{t \in (a,t_{1}]} \\ &\quad= {}_{C - H}D_{a^{ +}}^{q} \biggl( u_{a} + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr) \\ &\quad= {}_{C - H}D_{a^{ +}}^{q} \biggl( \frac{1}{\Gamma (q)} \int_{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr) \\ &\quad= \frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1} \biggl[ s\frac{d}{ds} \biggl( \int _{a}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f\bigl(\eta,u(\eta )\bigr)\frac{d\eta}{\eta} \biggr) \biggr] \frac{ds}{s} \\ &\quad=\frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1} \biggl[ s\frac{ - 1}{q} \frac{d}{ds} \biggl( \int_{a}^{s} f\bigl( \eta,u(\eta )\bigr)\,d\biggl(\ln \frac{s}{\eta} \biggr)^{q} \biggr) \biggr]\frac{ds}{s} \\ &\quad= \frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1} \biggl[ s\frac{ - 1}{q} \frac{d}{ds} \biggl( \biggl(\ln \frac{s}{\eta} \biggr)^{q}f\bigl(\eta,u(\eta )\bigr) \Big| _{a}^{s}\\ &\qquad{}- \int_{a}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q}f'\bigl(\eta,u(\eta )\bigr)\,d\eta \biggr) \biggr]\frac{ds}{s} \\ &\quad= \frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1} \biggl[ s\frac{ - 1}{q} \frac{d}{ds} \biggl( - \biggl(\ln \frac{s}{a}\biggr)^{q}f \bigl(a,u(a)\bigr)\\ &\qquad{} - \int_{a}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q}f'\bigl(\eta,u(\eta )\bigr)\,d\eta \biggr) \biggr]\frac{ds}{s} \\ &\quad=\frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1} \biggl[ \biggl(\ln \frac{s}{a} \biggr)^{q - 1}f\bigl(a,u(a)\bigr)\\ &\qquad{} + \int_{a}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f'\bigl(\eta,u( \eta )\bigr)\,d\eta \biggr]\frac{ds}{s} \\ &\quad= f\bigl(a,u(a)\bigr) + \frac{1}{\Gamma (1 - q)\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1} \biggl[ \int _{a}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f'\bigl(\eta,u(\eta )\bigr)\,d\eta \biggr] \frac{ds}{s} \\ &\quad= f\bigl(t,u(t)\bigr). \end{aligned}$$
For \(t \in (t_{k},t_{k + 1}]\) (where \(k = 1, 2,\ldots, m\)), we get
$$\begin{aligned} &{}_{C - H}D_{a^{ +}}^{q} \Biggl( \sum _{i = 1}^{k} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int _{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} + \int _{t_{i}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &\qquad{}- \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr) \Biggr) \bigg|_{t \in (t_{k},t_{k + 1}]} \\ &\quad= \sum_{i = 1}^{k} \biggl[ \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)}{}_{C - H}D_{a^{ +}}^{q} \biggl( \int _{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} + \int _{t_{i}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &\qquad{}- \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr) \biggr]\Big|_{t \in (t_{k},t_{k + 1}]} \\ &\quad= \sum_{i = 1}^{k} \biggl[ \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (1 - q)\Gamma (q)}\int_{t_{0}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1} \biggl[ s\frac{d}{ds} \biggl( \int _{t_{i}}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f\bigl(\eta,u(\eta )\bigr)\frac{d\eta}{\eta} \\ &\qquad{}- \int _{a}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f\bigl(\eta,u(\eta )\bigr)\frac{d\eta}{\eta} \biggr) \biggr] \frac{ds}{s} \biggr] \Big|_{t \in (t_{k},t_{k + 1}]} \\ &\quad= \sum_{i = 1}^{k} \biggl[ \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (1 - q)\Gamma (q)} \biggl( \int_{t_{i}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1}s\frac{d}{ds} \biggl( \int _{t_{i}}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f\bigl(\eta,u(\eta )\bigr)\frac{d\eta}{\eta} \biggr) \frac{ds}{s} \\ &\qquad{}- \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{1 - q - 1}s\frac{d}{ds} \biggl( \int _{a}^{s} \biggl(\ln \frac{s}{\eta} \biggr)^{q - 1}f\bigl(\eta,u(\eta )\bigr)\frac{d\eta}{\eta} \biggr) \frac{ds}{s} \biggr) \biggr] \Big|_{t \in (t_{k},t_{k + 1}]} \\ &\quad= \sum_{i = 1}^{k} \bigl[ \hbar \Delta_{i} \bigl( u(t_{i}^{ -} ) \bigr) \bigl( f\bigl(t,u(t)\bigr) |_{t \ge t_{i}} - f \bigl(t,u(t)\bigr) \vert _{t \ge a} \bigr) \bigr]|_{t \in (t_{k},t_{k + 1}]} = 0. \end{aligned}$$
Thus,
$$\begin{aligned} &{}_{C - H}D_{a^{ +}}^{q}u(t) |_{t \in (t_{k},t_{k + 1}]} \\ &\quad= {}_{C - H}D_{a^{ +}}^{q}\Biggl\{ u_{a} + \sum _{i = 1}^{k} \Delta_{i} \bigl( u \bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &\qquad{}+ \sum_{i = 1}^{k} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &\qquad{}+ \int_{t_{i}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr) \Biggr\} _{t \in (t_{k},t_{k + 1}]} \\ &\quad= \biggl\{ {}_{C - H}D_{a^{ +}}^{q} \biggl( \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr) \biggr\} _{t \in (t_{k},t_{k + 1}]} \\ &\quad= \bigl\{ f\bigl(t,u(t)\bigr) |_{t \ge a} \bigr\} _{t \in (t_{k},t_{k + 1}]} = f\bigl(t,u(t)\bigr)_{t \in (t_{k},t_{k + 1}]}. \end{aligned}$$
So, Eq. (3.3) satisfies the condition of fractional derivative in system (1.1).
Next, by (3.3), for each \(t_{k}\) (here \(k = 1, 2,\ldots, m\)), we have
$$\begin{aligned} u\bigl(t_{k}^{ +} \bigr) - u\bigl(t_{k}^{ -} \bigr) ={}& \lim_{t \to t_{k}^{ +}} u(t) - u(t_{k}) \\ ={}& u_{a} + \sum_{i = 1}^{k} \Delta_{i} \bigl( u\bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t_{k}} \biggl(\ln \frac{t_{k}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \sum_{i = 1}^{k} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \int_{t_{i}}^{t_{k}} \biggl(\ln \frac{t_{k}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{a}^{t_{k}} \biggl(\ln \frac{t_{k}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr) \\ &{}- u_{a} - \sum_{i = 1}^{k - 1} \Delta_{i} \bigl( u\bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t_{k}} \biggl(\ln \frac{t_{k}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \sum_{i = 1}^{k - 1} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \int_{t_{i}}^{t_{k}} \biggl(\ln \frac{t_{k}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{a}^{t_{k}} \biggl(\ln \frac{t_{k}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr) \\ ={}& \Delta_{k} \bigl( u\bigl(t_{k}^{ -} \bigr) \bigr). \end{aligned}$$
Therefore, Eq. (3.3) satisfies the impulsive condition of (1.1).
Finally, it can be easily verified that Eq. (3.3) satisfies condition (3.2).
‘Sufficiency’, we will prove that the solutions of system (1.1) satisfy Eq. (3.3) by using the inductive method. For \(t \in (a,t_{1}]\), we obtain the following equation by Lemma 2.6:
$$ u(t) = u_{a} + \frac{1}{\Gamma (q)}\int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \quad\mbox{for } t \in (a,t_{1}]. $$
(3.4)
Using (3.4), we have
$$\begin{aligned} u\bigl(t_{1}^{ +} \bigr) &= u\bigl(t_{1}^{ -} \bigr) + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) \\ &= u_{a} + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s}. \end{aligned}$$
Then the approximate solution is given by
$$\begin{aligned} \tilde{u}(t) ={}& u\bigl(t_{1}^{ +} \bigr) + \frac{1}{\Gamma (q)}\int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ ={}& u_{a} + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\ &{}+ \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \quad\mbox{for } t \in (t_{1},t_{2}]. \end{aligned}$$
(3.5)
Let \(e_{1}(t) = u(t) - \tilde{u}(t)\), for \(t \in (t_{1},t_{2}]\). Due to
$$\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}u(t) = u_{a} + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} $$
for \(t \in (t_{1},t_{2}]\), we get
$$\begin{aligned} \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}e_{1}(t) ={}& \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0} \bigl\{ u(t) - \tilde{u}(t) \bigr\} \\ ={}& \frac{1}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} - \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr]. \end{aligned}$$
Then we assume
$$\begin{aligned} e_{1}(t) ={}& \sigma \bigl( \Delta_{1}\bigl(u \bigl(t_{1}^{ -} \bigr)\bigr) \bigr)\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}e_{1}(t) \\ ={}& \frac{\sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} - \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int _{t_{1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr], \end{aligned}$$
where the function \(\sigma ( \cdot )\) is an undetermined function with \(\sigma (0) = 1\). Therefore,
$$\begin{aligned} u(t) ={}& \tilde{u}(t) + e_{1}(t) = u_{a} + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\ &{}+ \frac{1 - \sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}-\int_{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \quad\mbox{for } t \in (t_{1},t_{2}]. \end{aligned}$$
(3.6)
By (3.6), we get
$$\begin{aligned} u\bigl(t_{2}^{ +} \bigr) ={}& u\bigl(t_{2}^{ -} \bigr) + \Delta_{2} \bigl( u\bigl(t_{2}^{ -} \bigr) \bigr) \\ ={}& u_{a} + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) + \Delta_{2} \bigl( u\bigl(t_{2}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\ &{}+ \frac{1 - \sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{1}}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr]. \end{aligned}$$
Therefore, the approximate solution is provided by
$$\begin{aligned} \tilde{u}(t) ={}& u\bigl(t_{2}^{ +} \bigr) + \frac{1}{\Gamma (q)}\int_{t_{2}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ ={}& u_{a} + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) + \Delta_{2} \bigl( u\bigl(t_{2}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)} \biggl[ \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\ &{}+ \int_{t_{2}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\ &{}+ \frac{1 - \sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{1}}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \quad\mbox{for } t \in (t_{2},t_{3}]. \end{aligned}$$
(3.7)
Let \(e_{2}(t) = u(t) - \tilde{u}(t)\), for \(t \in (t_{2},t_{3}]\). For the exact solution \(u(t)\) of system (1.1), we have
$$\begin{aligned}& \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0,\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0}u(t) = u_{a} + \frac{1}{\Gamma (q)}\int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \quad\mbox{for } t \in (t_{2},t_{3}], \\& \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}u(t) = u_{a} + \Delta_{2} \bigl( u\bigl(t_{2}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\& \hphantom{ \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}u(t) =}{}+ \frac{1 - \sigma ( \Delta_{2}(u(t_{2}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\& \hphantom{ \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}u(t) =}{}+ \int_{t_{2}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int_{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\& \hphantom{\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}u(t) =} {}\mbox{for } t \in (t_{2},t_{3}], \\& \begin{aligned}[b] \lim_{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0}u(t) ={}& u_{a} + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \frac{1 - \sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\ &{}+ \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int_{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\ &{}\mbox{for } t \in (t_{2},t_{3}]. \end{aligned} \end{aligned}$$
Thus,
$$\begin{aligned}& \begin{aligned}[b] \mathop{\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0, }}_{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0}e_{2}(t) &= \mathop{\lim_{ \Delta_{1} ( u(t_{1}^{ -} ) ) \to 0, }}_{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0} \bigl\{ u(t) - \tilde{u}(t) \bigr\} \\ &= \frac{1}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} - \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}-\int_{t_{2}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr], \end{aligned} \end{aligned}$$
(3.8)
$$\begin{aligned}& \lim_{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0}e_{2}(t) = \lim _{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0} \bigl\{ u(t) - \tilde{u}(t) \bigr\} = \frac{\sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\& \hphantom{\lim_{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0}e_{2}(t) =}{}- \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{2}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\& \hphantom{\lim_{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0}e_{2}(t) =}{}+ \frac{1 - \sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\& \hphantom{\lim_{\Delta_{2} ( u(t_{2}^{ -} ) ) \to 0}e_{2}(t) =}{}- \int_{t_{1}}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{2}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr], \end{aligned}$$
(3.9)
$$\begin{aligned}& \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}e_{2}(t) = \lim _{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0} \bigl\{ u(t) - \tilde{u}(t) \bigr\} \\& \hphantom{\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}e_{2}(t)}= \frac{\sigma ( \Delta_{2}(u(t_{2}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\& \hphantom{\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0}e_{2}(t) =}{}- \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{2}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr]. \end{aligned}$$
(3.10)
By (3.8)-(3.10), we obtain
$$\begin{aligned} e_{2}(t) ={}& \frac{\sigma ( \Delta_{1}(u(t_{1}^{ -} )) ) + \sigma ( \Delta_{2}(u(t_{2}^{ -} )) ) - 1}{\Gamma (q)} \biggl[ \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int _{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{2}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\ &{}+ \frac{1 - \sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} - \int_{t_{1}}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int _{t_{2}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr]. \end{aligned}$$
(3.11)
Therefore,
$$\begin{aligned} u(t) ={}& \tilde{u}(t) + e_{2}(t) \\ ={}& u_{a} + \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr) + \Delta_{2} \bigl( u\bigl(t_{2}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\ &{}+ \frac{1 - \sigma ( \Delta_{1}(u(t_{1}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int_{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\ &{}+ \frac{1 - \sigma ( \Delta_{2}(u(t_{2}^{ -} )) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{2}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int_{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr]. \end{aligned}$$
(3.12)
Letting \(t_{2} \to t_{1}\), we have
$$\begin{aligned}& \lim_{t_{2} \to t_{1}} \left \{ \textstyle\begin{array}{@{}l} {}_{C - H}D_{a^{ +}}^{q}u(t) = f(t,u(t)),\quad t \in (a,t_{3}] \mbox{ and } t \ne t_{1} \mbox{ and } t \ne t_{2}, \\ \Delta u \vert _{t = t_{k}} = u(t_{k}^{ +} ) - u(t_{k}^{ -} ) = \Delta_{k} ( u(t_{k}^{ -} ) ) \in \mathbb{C},\quad k = 1,2, \\ u(a^{ +} ) = u_{a},\quad u_{a} \in \mathbb{C} \end{array}\displaystyle \right . \end{aligned}$$
(3.13)
$$\begin{aligned}& \quad\to \left \{ \textstyle\begin{array}{@{}l} {}_{C - H}D_{a^{ +}}^{q}u(t) = f(t,u(t)),\quad t \in (a,t_{3}] \mbox{ and } t \ne t_{1}, \\ \Delta u \vert _{t = t_{1}} = u(t_{1}^{ +} ) - u(t_{1}^{ -} ) + u(t_{2}^{ +} ) - u(t_{2}^{ -} ) = \Delta_{1} ( u(t_{1}^{ -} ) ) + \Delta_{2} ( u(t_{2}^{ -} ) ), \\ u(a^{ +} ) = u_{a},\quad u_{a} \in \mathbb{C}. \end{array}\displaystyle \right . \end{aligned}$$
(3.14)
Using (3.6) and (3.12) to systems (3.14) and (3.13), respectively, we get
$$\begin{aligned} &1 - \sigma \bigl( \Delta_{1}\bigl(u\bigl(t_{1}^{ -} \bigr)\bigr) + \Delta_{2}\bigl(u\bigl(t_{2}^{ -} \bigr)\bigr) \bigr) = 1 - \sigma \bigl( \Delta_{1}\bigl(u \bigl(t_{1}^{ -} \bigr)\bigr) \bigr) + 1 - \sigma \bigl( \Delta_{2}\bigl(u\bigl(t_{2}^{ -} \bigr)\bigr) \bigr), \\ &\quad\forall \Delta_{1} \bigl( u\bigl(t_{1}^{ -} \bigr) \bigr),\Delta_{2} \bigl( u\bigl(t_{2}^{ -} \bigr) \bigr) \in \mathbb{C}. \end{aligned}$$
Letting \(\rho (z) = 1 - \sigma (z)\) (\(\forall z \in \mathbb{C}\)), we have \(\rho (z + w) = \rho (z) + \rho (w)\) (\(\forall z,w \in \mathbb{C}\)). Therefore \(\rho (z) = \hbar z\), where ħ is a constant. So, we obtain the following two equations:
$$\begin{aligned} u(t) ={}& u_{a} + \Delta_{1} \bigl( u \bigl(t_{1}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \frac{\hbar \Delta_{1} ( u(t_{1}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \quad\mbox{for } t \in (t_{1},t_{2}] \end{aligned}$$
(3.15)
and
$$\begin{aligned} u(t) ={}& u_{a} + \Delta_{1} \bigl( u \bigl(t_{1}^{ -} \bigr) \bigr) + \Delta_{2} \bigl( u\bigl(t_{2}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \frac{\hbar \Delta_{1} ( u(t_{1}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{1}} \biggl(\ln \frac{t_{1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\ &{}+ \frac{\hbar \Delta_{2} ( u(t_{2}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{2}} \biggl(\ln \frac{t_{2}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} + \int_{t_{2}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{} - \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \quad\mbox{for } t \in (t_{2},t_{3}]. \end{aligned}$$
(3.16)
For \(t \in (t_{n},t_{n + 1}]\), suppose
$$\begin{aligned} u(t) ={}& u_{a} + \sum_{i = 1}^{n} \Delta_{i} \bigl( u\bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \sum_{i = 1}^{n} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} + \int_{t_{i}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{} - \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr) \quad\mbox{for } t \in (t_{n},t_{n + 1}]. \end{aligned}$$
(3.17)
By (3.17), we have
$$\begin{aligned} u\bigl(t_{n + 1}^{ +} \bigr) ={}& u\bigl(t_{n + 1}^{ -} \bigr) + \Delta_{n + 1} \bigl( u\bigl(t_{n + 1}^{ -} \bigr) \bigr) \\ ={}& u_{a} + \sum_{i = 1}^{n + 1} \Delta_{i} \bigl( u\bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \sum_{i = 1}^{n} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \int_{t_{i}}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{a}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr). \end{aligned}$$
So, the approximate solution is provided by
$$\begin{aligned} \tilde{u}(t) ={}& u\bigl(t_{n + 1}^{ +} \bigr) + \frac{1}{\Gamma (q)}\int_{t_{n + 1}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ ={}& u_{a} + \sum_{i = 1}^{n + 1} \Delta_{i} \bigl( u\bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)} \biggl[ \int_{a}^{t_{n + 1}} \biggl( \ln \frac{t_{n + 1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \int_{t_{n + 1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] + \sum_{i = 1}^{n} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \int_{t_{i}}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{a}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr) \\ &{}\quad\mbox{for } t \in (t_{n + 1},t_{n + 2}]. \end{aligned}$$
(3.18)
Let \(e_{n + 1}(t) = u(t) - \tilde{u}(t)\), for \(t \in (t_{n + 1},t_{n + 2}]\). For the exact solution \(u(t)\) of system (1.1), we have
$$\begin{aligned}& \lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0, \ldots,\Delta_{n + 1} ( u(t_{n + 1}^{ -} ) ) \to 0}u(t) = u_{a} + \frac{1}{\Gamma (q)}\int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \quad\mbox{for } t \in (t_{n + 1},t_{n + 2}], \\& \begin{aligned}[b] \lim_{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0}u(t) ={}& u_{a} + \mathop{\sum_{1 \le i \le n + 1 }}_{ \mathrm{and}\ i \ne j} \Delta_{i} \bigl( u\bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \mathop{\sum_{1 \le i \le n + 1 }}_{ \mathrm{and}\ i \ne j} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s}\\ &{} + \int_{t_{i}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr), \\ &{}\mbox{here } 1 \le j \le n + 1. \end{aligned} \end{aligned}$$
Then
$$\begin{aligned}& \begin{aligned}[b] &\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0, \ldots, \Delta_{n + 1} ( u(t_{n + 1}^{ -} ) ) \to 0}e_{n + 1}(t)\\ &\quad=\lim_{\Delta_{1} ( u(t_{1}^{ -} ) ) \to 0,\ldots, \Delta_{n + 1} ( u(t_{n + 1}^{ -} ) ) \to 0} \bigl\{ u(t) - \tilde{u}(t) \bigr\} \\ &\quad= \frac{1}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} - \int_{a}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &\qquad{}- \int_{t_{n + 1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr], \end{aligned} \end{aligned}$$
(3.19)
$$\begin{aligned}& \lim_{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0}e_{n + 1}(t) = \lim _{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0} \bigl\{ u(t) - \tilde{u}(t) \bigr\} \\& \hphantom{\lim_{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0}e_{n + 1}(t)}= \frac{1 - \hbar \mathop{\sum_{1 \le i \le n + 1 }}\limits_{ \hphantom{ss}\mathrm{and}\ i \ne j} \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\& \hphantom{\lim_{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0}e_{n + 1}(t) =}{}- \int_{a}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{n + 1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\& \hphantom{\lim_{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0}e_{n + 1}(t) =}{}+ \mathop{\sum_{1 \le i \le n }}_{\mathrm{and}\ i \ne j} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{t_{i}}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\& \hphantom{\lim_{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0}e_{n + 1}(t) =}{}- \int_{t_{i}}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{n + 1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr), \\& \hphantom{\lim_{\Delta_{j} ( u(t_{j}^{ -} ) ) \to 0}e_{n + 1}(t) =}{}\mbox{here } 1 \le j \le n + 1. \end{aligned}$$
(3.20)
So, by (3.19) and (3.20), we obtain
$$\begin{aligned} e_{n + 1}(t) ={}& \frac{1 - \hbar \sum_{1 \le i \le n + 1} \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr) \frac{ds}{s} \\ &{}- \int_{a}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{n + 1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \\ &{}+ \sum_{1 \le i \le n} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{t_{i}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int _{t_{i}}^{t_{n + 1}} \biggl(\ln \frac{t_{n + 1}}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} - \int _{t_{n + 1}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr). \end{aligned}$$
Thus,
$$\begin{aligned} u(t) ={}& \tilde{u}(t) + e_{n + 1}(t) \\ ={}& u_{a} + \sum_{i = 1}^{n + 1} \Delta_{i} \bigl( u\bigl(t_{i}^{ -} \bigr) \bigr) + \frac{1}{\Gamma (q)}\int_{a}^{t} \biggl(\ln \frac{t}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}+ \sum_{i = 1}^{n + 1} \biggl( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} \biggl[ \int_{a}^{t_{i}} \biggl(\ln \frac{t_{i}}{s}\biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} + \int_{t_{i}}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \\ &{}- \int _{a}^{t} \biggl(\ln \frac{t}{s} \biggr)^{q - 1}f\bigl(s,u(s)\bigr)\frac{ds}{s} \biggr] \biggr) \quad\mbox{for } t \in (t_{n + 1},t_{n + 2}]. \end{aligned}$$
(3.21)
So, the solution of system (1.1) satisfies Eq. (3.3).
By the proof of Sufficiency and Necessity, it is shown that system (1.1) is equivalent to the integral equation (3.3). The proof is now completed. □
Remark 3.1
Due to uncertainty of the constant ħ, the deviation caused by impulse for the fractional order nonlinear systems is undetermined.
Next, we provide an analysis of the connection between impulsive Caputo-Hadamard fractional differential equations and impulsive first-order differential equations. Letting \(q \to 1^{ -}\) for system (1.1), we have
$$\begin{aligned} &\lim_{q \to 1^{ -}} \left \{ \textstyle\begin{array}{@{}l} {}_{C - H}D_{a^{ +}}^{q}u(t) = f(t,u(t)),\quad t \in (a,T] \mbox{ and } t \ne t_{k}\ (k = 1,2,\ldots,m), \\ \Delta u \vert _{t = t_{k}} = u(t_{k}^{ +} ) - u(t_{k}^{ -} ) = \Delta_{k} ( u(t_{k}^{ -} ) ) \in \mathbb{C},\quad k = 1,2,\ldots,m, \\ u(a) = u_{a},\quad u_{a} \in \mathbb{C} \end{array}\displaystyle \right . \\ &\quad\to \left \{ \textstyle\begin{array}{@{}l} t\frac{dx(t)}{dt} = f(t,x(t)),\quad t \in (a,T] \mbox{ and } t \ne t_{k}\ (k = 1,2,\ldots,m), \\ \Delta u |_{t = t_{k}} = u(t_{k}^{ +} ) - u(t_{k}^{ -} ) = \Delta_{k}( u(t_{k}^{ -} ) ) \in \mathbb{C},\quad k = 1,2,\ldots,m, \\ u(a) = u_{a},\quad u_{a} \in \mathbb{C}. \end{array}\displaystyle \right . \end{aligned}$$
(3.22)
On the other hand, by Theorem 3.2, the general solution of system (1.1) satisfies
$$\begin{aligned} \lim_{q \to 1^{ -}} u(t) &= \left \{ \textstyle\begin{array}{@{}l} \lim_{q \to 1^{ -}} [ u_{a} + \frac{1}{\Gamma (q)}\int_{a}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} ] \quad\mbox{for } t \in (a,t_{1}], \\ \lim_{q \to 1^{ -}} \{ u_{a} + \sum_{i = 1}^{k} \Delta_{i} ( u(t_{i}^{ -} ) ) + \frac{1}{\Gamma (q)}\int_{a}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} \\ \quad{} + \sum_{i = 1}^{k} ( \frac{\hbar \Delta_{i} ( u(t_{i}^{ -} ) )}{\Gamma (q)} [ \int_{a}^{t_{i}} (\ln \frac{t_{i}}{s})^{q - 1}f(s,u(s))\frac{ds}{s} + \int_{t_{i}}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} \\ \quad{}- \int_{a}^{t} (\ln \frac{t}{s})^{q - 1}f(s,u(s))\frac{ds}{s} ] ) \} \quad\mbox{for } t \in (t_{k},t_{k + 1}], k = 1,2,\ldots,m \end{array}\displaystyle \right . \\ &= \left \{ \textstyle\begin{array}{@{}l} u_{a} + \int_{a}^{t} \frac{f(s,u(s))}{s}\,ds\quad \mbox{for } t \in (a,t_{1}], \\ u_{a} + \sum_{i = 1}^{k} \Delta_{i} ( u(t_{i}^{ -} ) ) + \int_{a}^{t} \frac{f(s,u(s))}{s}\,ds \\ \quad\mbox{for } t \in (t_{k},t_{k + 1}], k = 1,2,\ldots,m. \end{array}\displaystyle \right . \end{aligned}$$
(3.23)
Moreover, it is true that Eq. (3.23) is the solution of impulsive system (3.22) and indirectly verifies the formula of general solution for nonlinear systems with Caputo-Hadamard fractional derivatives and impulsive effect.