In this section, we discuss the differentiability of the solution relative to the control function for IFS (3). Before that, we make some preparations. For convenience, we consider the smooth optimal control problem, namely the allowable control set \(\mathscr{U}([0,T];\mathbb{R})\subset C([0,T];\mathbb{R})\). Suppose that \((\bar{x},\bar{u})\) is the solution of the smooth optimal control problem given by IFS (3), where the optimal trajectory \(\bar{x}(\cdot ;\bar{u}(\cdot ))\) is the solution of IFS (3) corresponding to the optimal control *ū*. For any \(\alpha \in \mathbb{R}\) and \(u(\cdot )\in \mathscr{U}([0,T];\mathbb{R})\), we have that \(u^{\alpha }(\cdot )=\bar{u}(\cdot )+\alpha u(\cdot )\in \mathscr{U}([0,T]; \mathbb{R})\). Furthermore, it can be seen from Lemma 2.1 that IFS (3) has a unique approximate solution \(u_{\alpha }(\cdot ;u^{\alpha })\). From IFS (2), we can claim that IFS may be not continuous dependence relative to the control function. Naturally, we cannot expect to the differentiability of the solution relative to the control function. Hence, we need to modify the classical definition of the variation.

### Definition 3.1

The solution \(x(\cdot ;u)\) of IFS (3) is said to be Gâteaux differentiable relative to \(u(\cdot )\) in the direction of the function \(u^{\alpha }(\cdot )\) if

$$ \lim_{\alpha \searrow 0} \frac{x_{\alpha }(t;u^{\alpha })-x(t;u)}{\alpha } $$

exists for all *t* such that \(x(t;u)\neq y_{i}\) for all \(i\in \mathscr{P}\). Furthermore, let

$$\begin{aligned} \varphi \bigl(t;u^{\alpha }\bigr)= \textstyle\begin{cases} \lim_{\alpha \searrow 0} \frac{x_{\alpha }(t;u^{\alpha })-x(t;u)}{\alpha },&x(t;u)\neq y_{i} \text{ for all }i\in \mathscr{P}, \\ \lim_{s\nearrow t}\lim_{\alpha \searrow 0} \frac{x_{\alpha }(t;u^{\alpha })-x(t;u)}{\alpha },&x(t)= y_{i} \text{ for some }i\in \mathscr{P}, \end{cases}\displaystyle \end{aligned}$$

(9)

then \(\varphi (\cdot ;u^{\alpha })\) is called the Gâteaux derivative of the solution \(x(\cdot ;u)\) relative to the control function \(u(\cdot )\) in the direction of \(u^{\alpha }(\cdot )\).

To obtain the differentiability of the solution relative to the control function for IFS (3), we need the following lemma. For this purpose, we first define several functions as follows:

$$\begin{aligned} \tilde{y}(\alpha )= \textstyle\begin{cases} y_{i}-\alpha ^{2},&\text{if }y_{i}>y_{j}, \\ y_{i}+\alpha ^{2},&\text{if }y_{i}< y_{j}, \end{cases}\displaystyle \end{aligned}$$

where \(i,j\in \mathscr{P}\),

$$\begin{aligned}& H(\alpha ,t)=x_{\alpha }\bigl(t;u^{\alpha } \bigr)-\tilde{y}(\alpha ), \\& h_{t}(\alpha ) \text{ denotes the solution of } H(\theta ,t)=0. \end{aligned}$$

(10)

Furthermore, we get the following lemma.

### Lemma 3.1

*If* \(f_{i}\in ([0,+\infty ),\mathbb{R})\) *and* \(f_{i}(t)+\bar{u}(t)\neq ay_{j}\) *for all* \(t\in [0,+\infty )\) *and* \(i,j\in \mathscr{P}\), *then there is* \(\delta >0\) *such that* \(h_{t}\) *is differentiable on* \([0,\delta ]\) *and its derivative is given by*

$$ \dot{h}_{t_{m}}(0)= \textstyle\begin{cases} - \frac{\int _{0}^{t_{1}}e^{a(t_{1}-\tau )}u(\tau )\,d\tau }{f_{1}(t_{1})+ay_{2}+\bar{u}(t_{1})},&m=1 ,\\ - \frac{\int _{0}^{t_{m}}e^{a(t_{m}-\tau )}u(\tau )\,d\tau +\sum_{i=1}^{m-1}(-1)^{i+1}\dot{h}_{t_{i}}(0)e^{a(t_{m}-t_{i})}(f_{1}(t_{i})-f_{2}(t_{i}))-\dot{\tilde{y}}(\alpha )}{f_{i}(t_{m})+ay_{j}+\bar{u}(t_{m})}, &m=2,3,\ldots ,k, \end{cases} $$

*where* \(i=2\), \(j=1\) *if* *m* *is even and* \(i=1\), \(j=2\) *if* *m* *is odd*.

### Proof

We can deduce from Theorem 2.1 that there exists \(\delta >0\) such that \(h_{t}:[0,\delta ]\rightarrow O(t)\) is a function for all \(\alpha \in [0,\delta ]\) and \(h_{t}(0)=t\), if \(x(t)=y_{i}\) for some \(i\in \mathscr{P}\), where \(O(t)\) denotes some neighborhood of *t*. Furthermore, we have that

$$ x_{\alpha }\bigl(h_{t_{m}}(\alpha );u^{\alpha }\bigr)= \tilde{y}(\alpha ) \quad \text{for all } \alpha \in [0,\delta ], m=1,2,\ldots ,k, $$

which implies

$$ H\bigl(\alpha ,h_{t_{m}}(\alpha )\bigr)=0 \quad \text{for all } \alpha \in [0, \delta ], m=1,2,\ldots ,k. $$

Since

$$ \frac{\partial H(\alpha ,t)}{\partial t}= \frac{\partial [x_{\alpha }(t;u^{\alpha })-\tilde{y}(\alpha )]}{\partial t}=ax_{\alpha }\bigl(t;u^{\alpha } \bigr)+u^{\alpha }(t)+f_{i}(t)\quad \text{for all } \alpha \in [0, \delta ], $$

then it can be seen from \(f_{i}(\cdot )\in C([0,T];\mathbb{R})\) and \(u^{\alpha }(\cdot )\in C([0,T];\mathbb{R})\) that \(\frac{\partial H(\alpha ,t)}{\partial t}\) is continuous on \([0,\delta ]\times [h_{t_{m}},h_{t_{m+1}}]\). Combined with \(f_{i}(t)+\bar{u}(t)\neq -ay_{i}\) for all \(i\in \mathscr{P}\), we have that

$$\begin{aligned} \frac{\partial H(\alpha ,t)}{\partial t}\bigg|_{t=h_{t_{m}}(\alpha )} =& \frac{\partial [x_{\alpha }(t;u^{\alpha })-\tilde{y}(\alpha )]}{\partial t}\bigg|_{t=h_{t_{m}}( \alpha )} \\ =&f_{i}\bigl(h_{t_{m}}(\alpha ) \bigr)+a\tilde{y}(\alpha )+u^{\alpha }\bigl(h_{t_{1}}( \alpha )\bigr) \neq 0, \end{aligned}$$

(11)

where \(m=1,2,\ldots ,k\), \(i\in \mathscr{P}\). Otherwise, for the time interval \([0,h_{t_{1}}(\alpha ))\), we obtain that

$$\begin{aligned}& H_{\alpha }(\alpha ,t) \\& \quad = \lim_{\xi \rightarrow 0} \frac{H(\alpha +\xi ,t)-H(\alpha ,t)}{\xi } \\& \quad = \lim_{\xi \rightarrow 0} \frac{x_{\alpha +\xi }(t;u^{\alpha +\xi })-x_{\alpha }(t;u^{\alpha })}{\xi }- \dot{\tilde{y}}(\alpha ) \\& \quad = \lim_{\xi \rightarrow 0} \frac{e^{at}y_{1}+\int _{0}^{t}e^{a(t-\tau )}(u^{\alpha +\xi }(\tau )+f_{1}(\tau ))\,d\tau -[e^{at}y_{1}+\int _{0}^{t}e^{a(t-\tau )}(u^{\alpha }(\tau )+f_{1}(\tau ))\,d\tau ]}{\xi } \\& \qquad {} -\dot{\tilde{y}}(\alpha ) \\& \quad = \int _{0}^{t}e^{a(t-\tau )}u(\tau )\,d\tau -\dot{ \tilde{y}}(\alpha ). \end{aligned}$$

(12)

The above implies that the continuous partial derivative of \(H(\alpha ,t)\) with respect to *α* exists on \([0,\delta ]\times [0,h_{t_{1}}(\alpha ))\). Furthermore, we induce from (11) and (12) that

$$ \dot{h}_{t_{1}}(\alpha )=- \frac{\frac{\partial H(\alpha ,t)}{\partial \alpha }}{\frac{\partial H(\alpha ,t)}{\partial t}} \bigg|_{t=h_{t_{1}}(\alpha )}=- \frac{\int _{0}^{h_{t_{1}}(\alpha )}e^{a(h_{t_{1}}(\alpha )-\tau )}u(\tau )\,d\tau -\dot{\tilde{y}}(\alpha )}{f_{1}(h_{t_{1}}(\alpha ))+a\tilde{y}(\alpha )+u^{\alpha }(h_{t_{1}}(\alpha ))} $$

and

$$ \dot{h}_{t_{1}}(0)=- \frac{\int _{0}^{t_{1}}e^{a(t_{1}-\tau )}u(\tau )\,d\tau }{f_{1}(t_{1})+ay_{2}+\bar{u}(t_{1})}. $$

Secondly, when \(t\in (h_{t_{1}}(\alpha ),h_{t_{2}}(\alpha ))\), we can reduce that

$$\begin{aligned}& H_{\alpha }(\alpha ,t) \\& \quad =\frac{\partial }{\partial \alpha } \biggl(e^{at}y_{1}+ \int _{0}^{t}e^{a(t-\tau )}u^{\alpha }(\tau )\,d\tau + \int _{0}^{h_{t_{1}( \alpha )}}e^{a(t-\tau )}f_{1}(\tau )\,d\tau + \int _{h_{t_{1}}(\alpha )}^{t}e^{a(t- \tau )}f_{2}(\tau )\,d\tau \biggr) \\& \qquad {} -\dot{\tilde{y}}(\alpha ) \\& \quad = \int _{0}^{t}e^{a(t-\tau )}u(\tau )\,d\tau + \dot{h}_{t_{1}}(\alpha )e^{a(t-h_{t_{1}}( \alpha ))}[f_{1} \bigl(h_{t_{1}}(\alpha )\bigr)-f_{2}\bigl(h_{t_{1}}(\alpha )\bigr)- \dot{\tilde{y}}(\alpha ), \end{aligned}$$

(13)

which implies that the continuous partial derivative of \(H(\alpha ,t)\) with respect to *α* exists on \([0,\delta ]\times [h_{t_{1}}(\alpha ),h_{t_{2}}(\alpha ))\). Furthermore, we induce from (11) and (13) that

$$ \dot{h}_{t_{2}}(\alpha )=- \frac{\int _{0}^{h_{t_{2}}(\alpha )}e^{a(h_{t_{2}}(\alpha )-\tau )}u(\tau )\,d\tau +\dot{h}_{t_{1}}(\alpha )e^{a(h_{t_{2}}(\alpha )-h_{t_{1}}(\alpha ))}(f_{1}(h_{t_{1}}(\alpha ))-f_{2}(h_{t_{1}}(\alpha )))-\dot{\tilde{y}}(\alpha )}{f_{2}(h_{t_{2}}(\alpha ))+a\tilde{y}(\alpha )+u^{\alpha }(h_{t_{2}}(\alpha ))} $$

and

$$ \dot{h}_{t_{2}}(0)=- \frac{\int _{0}^{t_{2}}e^{a(t_{2}-\tau )}u(\tau )\,d\tau +\dot{h}_{t_{1}}(0)e^{a(t_{2}-t_{1})}(f_{1}(t_{1})-f_{2}(t_{1}))}{f_{2}(h_{t_{2}}(\alpha ))+ay_{1}+\bar{u}(t_{2})}. $$

In general, for any \(t\in (h_{t_{m}}(\alpha ),h_{t_{m+1}}(\alpha ))\) (\(m=1,2,\ldots ,k-1\)) or \(t\in (h_{t_{k}}(\alpha ),T]\), we have

$$\begin{aligned}& H_{\alpha }(\alpha ,t) \\& \quad =\frac{\partial }{\partial \alpha } \Biggl(e^{at}y_{1}+ \int _{0}^{t}e^{a(t-\tau )}u^{\alpha }(\tau )\,d\tau +\sum_{i=1}^{m} \int _{h_{t_{i-1}}(\alpha )}^{h_{t_{i}}(\alpha )}e^{a(t-\tau )}f_{p}( \tau )\,d\tau \\& \qquad {} + \int _{h_{t_{m}}(\alpha )}^{t}e^{a(t-\tau )}f_{j}(\tau )\,d\tau \Biggr) \\& \qquad {} -\dot{\tilde{y}}(\alpha ) \\& \quad = \int _{0}^{t}e^{a(t-\tau )}u(\tau )\,d\tau +\sum _{i=1}^{m}(-1)^{i+1} \dot{h}_{t_{i}}(\alpha )e^{a(t-h_{t_{i}}(\alpha ))}(f_{1} \bigl(h_{t_{i}}( \alpha )\bigr)-f_{2}\bigl(h_{t_{i}}( \alpha )\bigr)-\dot{\tilde{y}}(\alpha ), \end{aligned}$$

(14)

where \(p=1\) if *i* is odd and \(p=2\) if *i* is even \(j\in \mathscr{P}\) and \(j\neq p\), which implies that the continuous partial derivative of \(H(\epsilon ,t)\) with respect to *ϵ* exists on \([0,\delta ]\times (h_{t_{m}}(\epsilon ),h_{t_{m+1}}(\epsilon ))\). Furthermore, from the implicit function theory, (11), and (14), we discover that

$$\begin{aligned}& \dot{h}_{t_{m+1}}(\alpha ) \\& \quad =-\frac{\int _{0}^{h_{t_{m+1}}(\alpha )}e^{a(h_{t_{m+1}}(\alpha )-\tau )}u(\tau )\,d\tau +\sum_{i=1}^{m}(-1)^{i+1}\dot{h}_{t_{i}}(\alpha )e^{a(h_{t_{m+1}}(\alpha )-h_{t_{i}}(\alpha ))}(f_{1}(h_{t_{i}}(\alpha ))-f_{2}(h_{t_{i}}(\alpha )))-\dot{\tilde{y}}(\alpha )}{f_{i}(h_{t_{m+1}}(\alpha ))+a\tilde{y}(\alpha )+u^{\alpha }(h_{t_{m+1}}(\alpha ))} \end{aligned}$$

and

$$ \dot{h}_{t_{m+1}}(0)=- \frac{\int _{0}^{t_{m+1}}e^{a(t_{m+1}-\tau )}u(\tau )\,d\tau +\sum_{i=1}^{m}(-1)^{i+1}\dot{h}_{t_{i}}(0)e^{a(t_{m+1}-t_{i})}(f_{1}(t_{i})-f_{2}(t_{i}))-\dot{\tilde{y}}(\alpha )}{f_{i}(t_{m+1})+ay_{j}+\bar{u}(t_{m+1})}, $$

where \(i=1\), \(j=2\) if *m* is even and \(i=2\), \(j=1\) if *m* is odd. This completes the proof of Lemma 3.1. □

Based on Lemma 3.1 and Definition 3.1, we will give our final result.

### Theorem 3.1

*Suppose that* \(f_{i}\in ([0,+\infty ),\mathbb{R})\) *and* \(f_{i}(t)+\bar{u}(t)\neq ay_{j}\) *for all* \(t\in [0,+\infty )\) *and* \(i,j\in \mathscr{P}\), *then for any* \(u(\cdot )\in \mathscr{U}([0,T];\mathbb{R})\), *the solution* \(\bar{x}(\cdot ;\bar{u})\) *of IFS* (3) *is Gâteaux differentiable at* \(\bar{u}(\cdot )\) *in the direction of* \(u(\cdot )\) *in the sense of Definition *3.1. *Moreover*, *its derivative* *φ* *satisfies the following impulsive differential equation*:

$$ \textstyle\begin{cases} \overset{.}{\varphi }(t;u^{\alpha })=a\varphi (t;u^{\alpha })+u(t),&t\in (t_{0},T],x(t) \neq y_{p}~(p\in \mathscr{P}). \\ \varphi (t+;u^{\alpha })=\varphi (t;u^{\alpha })+\dot{h}_{t}(0)[f_{j}(t)-f_{i}(t)], &x(t)=y_{i}~(i,j \in \mathscr{P},i\neq j), \\ \varphi (0;u^{\alpha })=0. \end{cases} $$

(15)

### Proof

For the number of the irregular points on \([t_{0},T]\), there are only two possibilities which are Case (1): \(\bar{x}(\cdot ;\bar{u})\) has no irregular point on \([t_{0},T]\) and Case (2): \(\bar{x}(\cdot ;\bar{u})\) has at least one irregular point on \([t_{0},T]\). We will prove Theorem 3.1 from the two aspects.

Case (1), one can directly check that \(x(\cdot )\) is Gâteaux differentiable and its Gâteaux derivative *φ* (see Definition 3.1) is a solution of the following differential equation:

$$ \textstyle\begin{cases} \overset{.}{\varphi }(t;u^{\alpha })=a\varphi (t;u^{\alpha })+u(t),&t\in (0,T], \\ \varphi (0;u^{\alpha })=0. \end{cases} $$

Case (2), from Definition 3.1, we first have that

$$\begin{aligned} \varphi \bigl(t;u^{\alpha }\bigr) =&\lim _{\alpha \rightarrow 0} \frac{x_{\alpha }(t;u^{\alpha })-\bar{x}(t;\bar{u})}{\alpha } \\ =&\lim_{\alpha \rightarrow 0} \frac{e^{at}y_{1}+\int _{0}^{t}e^{a(t-\tau )}(f_{1}(\tau )+u^{\alpha }(\tau ))\,d\tau -[e^{at}y_{1}+\int _{0}^{t}e^{a(t-\tau )}(f_{1}(\tau )+\bar{u}(\tau ))\,d\tau ]}{\alpha } \\ =& \int _{0}^{t}e^{a(t-\tau )}u(\tau )\,d\tau , \quad \forall t\in (0,t_{1}). \end{aligned}$$

Namely, \(\varphi (\cdot )\) satisfies the following equation:

$$ \textstyle\begin{cases} \overset{.}{\varphi }(t;u^{\alpha })=a\varphi (t;u^{\alpha })+u(t),&t\in (0,t_{1}), \\ \varphi (0;u^{\alpha })=0. \end{cases} $$

(16)

Meanwhile, when \(h_{t_{1}}(\alpha )>t_{1}\), combining equation (3) with (9), we discover that

$$\begin{aligned}& \lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(t_{1};\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-x_{\alpha }(t_{1};u^{\alpha })}{\alpha }+ \lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{1};u^{\alpha })-\bar{x}(t_{1};\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-x_{\alpha }(t_{1};u^{\alpha })}{h_{t_{1}}(\alpha )-t_{1}} \frac{h_{t_{1}}(\alpha )-t_{1}}{\alpha }+\varphi \bigl(t_{1};u^{\alpha }\bigr) \\& \quad =\dot{h}_{t_{1}}(0) \bigl(ay_{2}+f_{1}(t_{1})+ \bar{u}(t_{1})\bigr)+\varphi \bigl(t_{1};u^{\alpha } \bigr). \end{aligned}$$

Furthermore, we can reduce that

$$\begin{aligned} \begin{aligned} \varphi \bigl(t_{1}+;u^{\alpha } \bigr)&=\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(h_{t_{1}}(\alpha );\bar{u})}{\alpha } \\ &=\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(t_{1};\bar{u})}{\alpha }+ \lim_{\alpha \rightarrow 0} \frac{\bar{x}(t_{1};\bar{u})-\bar{x}(h_{t_{1}}(\alpha );\bar{u})}{\alpha } \\ &=\dot{h}_{t_{1}}(0) \bigl(f_{1}(t_{1})-f_{2}(t_{1}) \bigr)+\varphi \bigl(t_{1};u^{\alpha }\bigr). \end{aligned} \end{aligned}$$

Similarly, when \(h_{t_{1}}(\alpha )< t_{1}\), we have that

$$\begin{aligned}& \lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(t_{1};\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(h_{t_{1}}(\alpha );\bar{u})}{\alpha }+ \lim_{\alpha \rightarrow 0} \frac{\bar{x}(h_{t_{1}}(\alpha );\bar{u})-\bar{x}(t_{1};\bar{u})}{\alpha } \\& \quad =\varphi \bigl(t_{1};u^{\alpha }\bigr)+\lim_{\alpha \rightarrow 0} \frac{\bar{x}(h_{t_{1}}(\alpha );\bar{u})-\bar{x}(t_{1};\bar{u})}{h_{t_{1}}(\alpha )-t_{1}} \frac{h_{t_{1}}(\alpha )-t_{1}}{\alpha } \\& \quad =\varphi \bigl(t_{1};u^{\alpha }\bigr)+\dot{h}_{t_{1}}(0) \bigl(ay_{2}+f_{1}(t_{1})+ \bar{u}(t_{1}) \bigr), \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \varphi \bigl(t_{1}+;u^{\alpha } \bigr)&=\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{1};u^{\alpha })-\bar{x}(t_{1};\bar{u})}{\alpha } \\ &=\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{1};u^{\alpha })-x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })}{\alpha }+ \lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(t_{1};\bar{u})}{\alpha } \\ &=\dot{h}_{t_{1}}(0) \bigl(f_{1}(t_{1})-f_{2}(t_{1}) \bigr)+\varphi \bigl(t_{1};u^{\alpha }\bigr). \end{aligned} \end{aligned}$$

In conclusion, we can claim that

$$ \lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(t_{1};\bar{u})}{\alpha }= \varphi (t_{1})+ \dot{h}_{t_{1}}(0) \bigl(ay_{2}+f_{1}(t_{1})+ \bar{u}(t_{1})\bigr) $$

and

$$\begin{aligned} \varphi \bigl(t_{1}+;u^{\alpha }\bigr)= \dot{h}_{t_{1}}(0) \bigl(f_{1}(t_{1})-f_{2}(t_{1}) \bigr)+ \varphi \bigl(t_{1};u^{\alpha }\bigr). \end{aligned}$$

(17)

Analogously, we consider \((t_{1},t_{2})\). From the following equation

$$\begin{aligned}& \varphi \bigl(t;u^{\alpha }\bigr) \\& \quad =\lim _{\alpha \rightarrow 0} \frac{x(t;u^{\alpha })-\bar{x}(t;\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{e^{a(t-h_{t_{1}}(\alpha ))}\tilde{y}(\alpha )+\int _{h_{t_{1}}(\alpha )}^{t}e^{a(t-\tau )}(f_{2}(\tau )+u^{\alpha }(\tau ))\,d\tau -[e^{a(t-t_{1})}y_{2}+\int _{t_{1}}^{t}e^{a(t-\tau )}(f_{2}(\tau )+\bar{u}(\tau ))\,d\tau ]}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{(e^{a(t-h_{t_{1}}(\alpha ))}-e^{a(t-t_{1})})\tilde{y}(\alpha )}{h_{t_{1}}(\alpha )-t_{1}} \frac{h_{t_{1}}(\alpha )-t_{1}}{\alpha }+\lim _{\alpha \rightarrow 0}\frac{e^{a(t-t_{1})}(\tilde{y}(\alpha )-y_{2})}{\alpha } \\& \qquad {} +\lim_{\alpha \rightarrow 0} \frac{\int _{h_{t_{1}}(\alpha )}^{t_{1}}e^{a(t-\tau )}(f_{2}(\tau )+\bar{u}(\tau ))\,d\tau }{h_{t_{1}}(\alpha )-t_{1}} \frac{h_{t_{1}}(\alpha )-t_{1}}{\alpha }+\lim _{\alpha \rightarrow 0} \int _{h_{t_{1}}(\alpha )}^{t}e^{a(t-\tau )}u(\tau )\,d\tau \\& \quad =\lim_{\alpha \rightarrow 0} \frac{(e^{a(t-h_{t_{1}}(\alpha ))}-e^{a(t-t_{1})})\tilde{y}(\alpha )}{h_{t_{1}}(\alpha )-t_{1}} \frac{h_{t_{1}}(\alpha )-t_{1}}{\alpha }+\lim _{\alpha \rightarrow 0} \frac{e^{a(t-t_{1})}(x_{\alpha }(h_{t_{1}}(\alpha );u^{\alpha })-\bar{x}(t_{1};\bar{u}))}{\alpha } \\& \qquad {} +\lim_{\alpha \rightarrow 0} \frac{\int _{h_{t_{1}}(\alpha )}^{t_{1}}e^{a(t-\tau )}(f_{2}(\tau )+\bar{u}(\tau ))\,d\tau }{h_{t_{1}}(\alpha )-t_{1}} \frac{h_{t_{1}}(\alpha )-t_{1}}{\alpha }+\lim _{\alpha \rightarrow 0} \int _{h_{t_{1}}(\alpha )}^{t}e^{a(t-\tau )}u(\tau )\,d\tau \\& \quad = \int _{t_{1}}^{t}e^{a(t_{1}-\tau )}u(\tau )\,d\tau + \dot{h}_{t_{1}}(0) \bigl(f_{1}(t_{1})-f_{2}(t_{1}) \bigr)+ \varphi \bigl(t_{1};u^{\alpha }\bigr) \\& \quad = \int _{t_{1}}^{t}e^{a(t_{1}-\tau )}u(\tau )\,d\tau +\varphi \bigl(t_{1}+;u^{\alpha }\bigr), \quad \forall t\in (t_{1},t_{2}), \end{aligned}$$

we know from (17) that \(\varphi (\cdot ;u^{\alpha })\) satisfies the following equation:

$$ \textstyle\begin{cases} \overset{.}{\varphi }(t;u^{\alpha })=a\varphi (t;u^{\alpha })+u(t),\quad t\in (t_{1},t_{2}), \\ \varphi (t_{1}+;u^{\alpha })=\dot{h}_{t_{1}}(0)(f_{1}(t_{1})-f_{2}(t_{1}))+ \varphi (t_{1};u^{\alpha }). \end{cases} $$

(18)

Analogous to \(\varphi (t_{1}+;u^{\alpha })\), from equations (3), (9), and Lemma 3.1, we have the following results about \(\varphi (t_{2}+;u^{\alpha })\): when \(h_{t_{2}}(\alpha )>t_{2}\),

$$\begin{aligned}& \varphi \bigl(t_{2}+;u^{\alpha } \bigr) \\& \quad =\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{2}}(\alpha );u^{\alpha })-\bar{x}(h_{t_{2}}(\alpha );\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{2}}(\alpha );u^{\alpha })-\bar{x}(t_{2};\bar{u})}{\alpha }+ \lim_{\alpha \rightarrow 0} \frac{\bar{x}(t_{2};\bar{u})-\bar{x}(h_{t_{2}}(\alpha );\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{2}}(\alpha );u^{\alpha })-x_{\alpha }(t_{2};u^{\alpha })}{\alpha }+ \lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{2};u^{\alpha })-\bar{x}(t_{2};\bar{u})}{\alpha } \\& \qquad {}+ \lim_{\alpha \rightarrow 0} \frac{\bar{x}(t_{2};\bar{u})-\bar{x}(h_{t_{2}}(\alpha );\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{2}}(\alpha );u^{\alpha })-x_{\alpha }(t_{2};u^{\alpha })}{h_{t_{2}}(\alpha )-t_{2}} \frac{h_{t_{2}}(\alpha )-t_{2}}{\alpha }+\lim _{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{2};u^{\alpha })-\bar{x}(t_{2};\bar{u})}{\alpha } \\& \qquad {} +\lim_{\alpha \rightarrow 0} \frac{\bar{x}(t_{2};\bar{u})-\bar{x}(h_{t_{2}}(\alpha );\bar{u})}{h_{t_{2}}(\alpha )-t_{2}} \frac{h_{t_{2}}(\alpha )-t_{2}}{\alpha } \\& \quad =\dot{h}_{t_{2}}(0) \bigl(f_{2}(t_{2})-f_{1}(t_{2}) \bigr)+\varphi \bigl(t_{2};u^{\alpha }\bigr) \end{aligned}$$

and when \(h_{t_{2}}(\alpha )< t_{2}\),

$$\begin{aligned} \varphi \bigl(t_{2}+;u^{\alpha } \bigr) =&\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{2};u^{\alpha })-\bar{x}(t_{2};\bar{u})}{\alpha } \\ =&\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{2};u^{\alpha })-x_{\alpha }(h_{t_{2}}(\alpha );u^{\alpha })}{\alpha }+ \lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(h_{t_{2}}(\alpha );u^{\alpha })-\bar{x}(h_{t_{2}}(\alpha );\bar{u})}{\alpha } \\ &{} +\lim_{\alpha \rightarrow 0} \frac{\bar{x}(h_{t_{2}}(\alpha );\bar{u})-\bar{x}(t_{2};\bar{u})}{\alpha } \\ =&\lim_{\alpha \rightarrow 0} \frac{x_{\alpha }(t_{2};u^{\alpha })-x_{\alpha } (h_{t_{2}}(\alpha );u^{\alpha })}{h_{t_{2}}(\alpha )-t_{2}} \frac{h_{t_{2}}(\alpha ) -t_{2}}{\alpha } \\ &{}+\lim _{\alpha \rightarrow 0} \frac{\bar{x}(h_{t_{2}}(\alpha );\bar{u})-\bar{x}(t_{2};\bar{u})}{h_{t_{2}}(\alpha )-t_{2}} \frac{h_{t_{2}}(\alpha )-t_{2}}{\alpha } \\ &{} +\varphi \bigl(t_{2};u^{\alpha }\bigr) \\ =&\dot{h}_{t_{2}}(0) \bigl(f_{2}(t_{2})-f_{1}(t_{2}) \bigr)+\varphi \bigl(t_{2};u^{\alpha }\bigr). \end{aligned}$$

The above equations yield that

$$ \varphi \bigl(t_{2}+;u^{\alpha }\bigr)=\dot{h}_{t_{2}}(0) \bigl(f_{2}(t_{2})-f_{1}(t_{2})\bigr)+ \varphi \bigl(t_{2};u^{\alpha }\bigr). $$

Repeat the above process, we consider the time interval \((t_{m},t_{m}+1)\) (\(m=1,2,\ldots ,k-1\)) or \((t_{k},T)\). Firstly, we have the following equation:

$$\begin{aligned}& \varphi \bigl(t;u^{\alpha }\bigr) \\& \quad =\lim _{\alpha \rightarrow 0} \frac{x(t;u^{\alpha })-\bar{x}(t;\bar{u})}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{e^{a(t-h_{t_{m}}(\alpha ))}\tilde{y}(\alpha )+\int _{h_{t_{m}}(\alpha )}^{t}e^{a(t-\tau )}(f_{i}(\tau )+u^{\alpha }(\tau ))\,d\tau -[e^{a(t-t_{m})}y_{i}+\int _{t_{m}}^{t}e^{a(t-\tau )}(f_{i}(\tau )+\bar{u}(\tau ))\,d\tau ]}{\alpha } \\& \quad =\lim_{\alpha \rightarrow 0} \frac{(e^{a(t-h_{t_{m}}(\alpha ))}-e^{a(t-t_{m})})\tilde{y}(\alpha )}{h_{t_{m}}(\alpha )-t_{m}} \frac{h_{t_{m}}(\alpha )-t_{m}}{\alpha }+\lim _{\alpha \rightarrow 0}\frac{e^{a(t-t_{m})}(\tilde{y}(\alpha )-y_{i})}{\alpha } \\& \qquad {} +\lim_{\alpha \rightarrow 0} \frac{\int _{h_{t_{m}}(\alpha )}^{t_{m}}e^{a(t-\tau )}(f_{i}(\tau )+\bar{u}(\tau ))\,d\tau }{h_{t_{m}}(\alpha )-t_{m}} \frac{h_{t_{m}}(\alpha )-t_{m}}{\alpha }+\lim _{\alpha \rightarrow 0} \int _{h_{t_{m}}(\alpha )}^{t}e^{a(t-\tau )}u(\tau )\,d\tau \\& \quad =\lim_{\alpha \rightarrow 0} \frac{(e^{a(t-h_{t_{m}}(\alpha ))}-e^{a(t-t_{m})})\tilde{y}(\alpha )}{h_{t_{m}}(\alpha )-t_{m}} \frac{h_{t_{m}}(\alpha )-t_{m}}{\alpha }+\lim _{\alpha \rightarrow 0} \frac{e^{a(t-t_{m})}(x_{\alpha }(h_{t_{m}}(\alpha );u^{\alpha })-\bar{x}(t_{m};\bar{u}))}{\alpha } \\& \qquad {} +\lim_{\alpha \rightarrow 0} \frac{\int _{h_{t_{m}}(\alpha )}^{t_{m}}e^{a(t-\tau )}(f_{i}(\tau )+\bar{u}(\tau ))\,d\tau }{h_{t_{m}}(\alpha )-t_{m}} \frac{h_{t_{m}}(\alpha )-t_{m}}{\alpha }+\lim _{\alpha \rightarrow 0} \int _{h_{t_{m}}(\alpha )}^{t}e^{a(t-\tau )}u(\tau )\,d\tau \\& \quad = \int _{t_{m}}^{t}e^{a(t_{m}-\tau )}u(\tau )\,d\tau + \dot{h}_{t_{m}}(0) \bigl(f_{j}(t_{m})-f_{i}(t_{1}) \bigr)+ \varphi \bigl(t_{m};u^{\alpha }\bigr) \\& \quad = \int _{t_{m}}^{t}e^{a(t_{m}-\tau )}u(\tau )\,d\tau +\varphi \bigl(t_{m}+;u^{\alpha }\bigr),\quad \forall t\in (t_{m},t_{m}+1) \cup (t_{k},T), \end{aligned}$$

where \(i=2\), \(j=1\) if *m* is even and \(i=1\), \(j=2\) if *m* is odd. The above equation implies that, for any \(t\in (t_{m},t_{m}+1)\cup (t_{k},T)\) (\(m=1,2,\ldots ,k-1\)), \(\varphi (t;u^{\alpha })\) satisfies the following differential equation:

$$ \overset{.}{\varphi }\bigl(t;u^{\alpha }\bigr)=a \varphi \bigl(t;u^{\alpha }\bigr)+u(t),\quad t\in (t_{m},t_{m}+1) \cup (t_{k},T). $$

(19)

On the other hand, analogous to \(\varphi (t_{1}+;u^{\alpha })\) and \(\varphi (t_{2}+;u^{\alpha })\), we obtain that

$$ \varphi \bigl(t_{m+1}+;u^{\alpha } \bigr)=\dot{h}_{t_{m+1}}(0) \bigl(f_{i}(t_{m+1})-f_{j}(t_{m+1}) \bigr)+ \varphi \bigl(t_{m+1};u^{\alpha }\bigr), $$

(20)

where \(i=2\), \(j=1\) if *m* is even and \(i=1\), \(j=2\) if *m* is odd.

In conclusion, we can deduce from (18), (19), and (20) that \(\varphi (\cdot ;u^{\alpha })\) satisfies the following impulsive differential equation:

$$ \textstyle\begin{cases} \overset{.}{\varphi }(t;u^{\alpha })=a\varphi (t;u^{\alpha })+u(t),&t\in (t_{0},T],x(t) \neq y_{p}~(p\in \mathscr{P}), \\ \varphi (t+;u^{\alpha })=\varphi (t;u^{\alpha })+\dot{h}_{t}(0)[f_{j}(t)-f_{i}(t)], &x(t)=y_{i}~(i,j \in \mathscr{P},i\neq j), \\ \varphi (0;u^{\alpha })=0. \end{cases} $$

The proof of Theorem 3.1 is complete. □