$$ \operatorname{det}(I-Q)=0. $$
To investigate the existence of solutions of system (2), the following auxiliary equation is often considered:
$$\begin{aligned}& \begin{aligned} &\dot{x}=\lambda f(t,x), \quad t\neq t_{k} , t\in R, \\ &\Delta x=\lambda I _{k}(x), \quad t=t_{k},k \in Z. \end{aligned} \end{aligned}$$
(10)
Then we give the following existence theorem for (Q,T)-affine-periodic solutions by using the topological degree theory [6, 7, 9–11].
Theorem 4.1
Let
\(D\subset R^{n}\)
be a bounded open set. Assume that the following hypotheses hold for system (10):
-
(H1)
For each
\(\lambda\in(0,1]\), every Q-affine-periodic solution
\(x(t)\)
of system (10) satisfies
$$ x(t)\notin\partial D \quad \textit{for all }t; $$
-
(H2)
the Brouwer degree,
$$ \operatorname{deg}\bigl(g,D\cap \operatorname{Ker}(I-Q),0 \bigr)\neq0 \quad \textit{if } \operatorname{Ker}(I-Q) \neq{0}, $$
where
$$ g(a)=\frac{1}{T} \biggl[ \int_{0}^{T}Pf(s,a)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t _{k}) \bigr) \biggr], $$
with an orthogonal projection
\(P:R^{n}\rightarrow \operatorname{Ker}(I-Q)\).
Then system (2) has at least one Q-affine-periodic solution
\(x_{*}(t)\in D\)
for all t.
Proof
Consider the auxiliary equation (10) with the boundary value condition \(x(T)=Qx(t)\), where \(\lambda\in(0,1]\). Let \(x(t)\) be any solution of (10) with \(x(T)=Qx(0)\). Then
$$\begin{aligned}& (I-Q)x_{0} \\& \quad =-\lambda \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I _{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$
(11)
In this case, \((I-Q)^{-1}\) does not exist. By coordinate transformation, without loss of generality, we can just let
$$ Q= \left ( \begin{matrix} I & 0 \\ 0 & Q_{1} \end{matrix} \right ) , $$
(12)
where \((I-Q_{1})^{-1}\) exists. Here \(Q=Q_{1}\oplus I\).
Let \(P:R^{n}\rightarrow{ \operatorname{Ker}(I-Q)}\) be the orthogonal projection. Then
$$ \begin{aligned}[b] (I-Q)x_{0}={} &(I-Q) \bigl(x^{0}_{\ker }+x^{0}_{\bot} \bigr) \\ ={}&{-}\lambda \biggl[ \int _{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ ={}&{-}\lambda \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}- \lambda \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t _{k}) \bigr) \biggr], \end{aligned} $$
(13)
where \(x^{0}_{\ker }\in \operatorname{Ker}(I-Q)\), \(x^{0}_{\bot}\in \operatorname{Im}(I-Q)\) and \(x_{0}=x^{0}_{\ker }+x^{0}_{\bot}\).
Let \(L_{p}=(I-Q)|_{\operatorname{Im}(I-Q)}\). It is easy to see that \(L^{-1}_{p}\) exists. Thus equation (13) is equivalent to
$$\begin{aligned}& (I-Q)x^{0}_{\ker }=-\lambda \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t_{k}) \bigr) \biggr]=0, \\& (I-Q)x^{0}_{\bot}=-\lambda \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$
Thus we have
$$ x^{0}_{\bot}=\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. $$
For \(x\in\textrm{X}\) such that \(x(t)\in\overline{D}\) for all \(t\in[0,T]\), we define the operator \(\mathtt{T}(x^{0}_{\ker },x, \lambda)\) by
$$ \mathtt{T} \bigl(x^{0}_{\ker },x,\lambda \bigr) = \left ( \begin{matrix} {x^{0}_{\ker }+ \frac{1}{T} [ \int_{0}^{T}Pf (s,x(s) )\,ds+\sum_{0\leq t _{k}< T}PI_{k} (x(t_{k}) ) ]} \\ {x^{0}_{\ker }-\lambda L^{-1}_{p} [ \int_{0}^{T}(I-P)f (s,x(s) )\,ds+\sum_{0\leq t_{k}< T}(I-P)I_{k} (x(t_{k}) ) ]} \\ {{}+\lambda [ \int_{0}^{t}f (s,x(s) )\,ds+\sum_{0\leq t_{k}< t}I_{k} (x(t _{k}) ) ]} \end{matrix} \right ), $$
(14)
where \(\lambda\in[0,1]\). We claim that each fixed point x of T in X is a solution of (10) with \(x(T)=Qx(0)\).
In fact, if x is a fixed point of T, we have
$$ \left ( \begin{matrix} {x^{0}_{\ker}} \\ {x(t)} \end{matrix} \right ) = \left ( \begin{matrix} {x^{0}_{\ker}+\frac{1}{T} [ \int_{0}^{T}Pf (s,x(s) )\,ds+\sum_{0\leq t_{k}< T}PI _{k} (x(t_{k}) ) ]} \\ {x^{0}_{\ker}-\lambda L^{-1}_{p} [ \int_{0}^{T}(I-P)f (s,x(s) )\,ds+ \sum_{0\leq t_{k}< T}(I-P)I_{k} (x(t_{k}) ) ]} \\ {{}+\lambda [ \int_{0}^{t}f (s,x(s) )\,ds+ \sum_{0\leq t_{k}< t}I_{k} (x(t_{k}) ) ]} \end{matrix} \right ) . $$
Thus
$$\begin{aligned}& \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t _{k}) \bigr) \biggr]=0, \end{aligned}$$
(15)
$$\begin{aligned}& x(t)= x^{0}_{\ker } -\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \hphantom{x(t)=}{}+\lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< t}I_{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$
(16)
By equation (16) we know that
$$ x_{0}=x^{0}_{\ker }-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. $$
According to \((I-Q)x^{0}_{\ker }=0\), we have
$$ \begin{aligned} Qx_{0} &=Qx^{0}_{\ker }- \lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &=x^{0}_{\ker }-\lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned} $$
Since equation (15) holds, we have
$$\begin{aligned}& (I-Q)L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad = \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad = \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \qquad {} + \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad = \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t_{k}) \bigr). \end{aligned}$$
Thus
$$\begin{aligned}& \lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad =\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \qquad {}-\lambda \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}I_{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$
Then
$$\begin{aligned} Qx_{0} = &x^{0}_{\ker }- \lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ =&x^{0}_{\ker }-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}+\lambda \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t _{k}< T}I_{k} \bigl(x(t_{k}) \bigr) \biggr] =x(T). \end{aligned}$$
(17)
By equations (16) and (17), equation (11) holds. Thus,
$$ x^{0}_{\bot}=-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. $$
Then,
$$\begin{aligned} x(t) = &x^{0}_{\ker }-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}+\lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< t}I \bigl(x(t_{k}) \bigr) \biggr] \\ =&x^{0}_{\ker }+x^{0}_{\bot}+ \lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ =&x_{0}+\lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I_{k} \bigl(x(t _{k}) \bigr) \biggr]. \end{aligned}$$
This means that the fixed point x is a solution of (10) with \(x(T)=Qx(0)\).
Now, we need to prove the existence of the fixed point of T. Take a constant M such that \(M> \sup_{t\in[0,T],x\in\overline{D}}|f(t,x)|\), and let
$$ \textrm{X}_{\lambda}= \biggl\{ x\in\textrm{X}: \biggl\vert \frac{x(t)-x(r)}{t-r} \biggr\vert \leq \lambda M \text{ for all } t,r \in(t_{k},t_{k+1}],t\neq r \biggr\} . $$
Then, it is easy to make a retraction \(\alpha_{\lambda}:\textrm{X} \rightarrow\textrm{X}_{\lambda}\).
Define an operator \(\widehat{\mathtt{T}}(x^{0}_{\ker },x,\lambda)\) by
$$ \begin{aligned} &\widehat{\mathtt{T}} \bigl(x^{0}_{\ker },x, \lambda \bigr) \\ &\quad = \left ( \begin{matrix} {x^{0}_{\ker} +\frac{1}{T} [ \int_{0}^{T}Pf (s,\alpha_{\lambda}\circ x(s) )\,ds+ \sum_{0\leq t_{k}< T}PI_{k} ( \alpha_{\lambda}\circ x(t_{k}) ) ]} \\ {\alpha_{\lambda}\circ x^{0}_{\ker }-\lambda L^{-1}_{p} [ \int_{0} ^{T}(I-P)f (s,\alpha_{\lambda} \circ x(s) )\,ds+\sum_{0\leq t_{k}< T}(I-P)I _{k} (\alpha_{\lambda}\circ x(t_{k}) ) ]} \\ {{}+\lambda [ \int_{0}^{t}f (s, \alpha_{\lambda}\circ x(s) )\,ds+\sum_{0\leq t_{k}< t}I_{k} ( \alpha_{ \lambda}\circ x(t_{k}) ) ]} \end{matrix} \right ). \end{aligned} $$
(18)
Since \(P:R^{n}\rightarrow \operatorname{Ker}(I-Q)\), it is easy to see that
$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s, x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t _{k}) \bigr) \biggr] \in \operatorname{Ker}(I-Q). $$
Also,
$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,\alpha_{\lambda}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< T}PI_{k} \bigl( \alpha_{\lambda}\circ x(t_{k}) \bigr) \biggr]\in \operatorname{Ker}(I-Q). $$
Let us consider the homotopy
$$\begin{aligned}& H \bigl(x^{0}_{\ker },x,\lambda \bigr)= \widehat{ \mathtt{T}} \bigl(x^{0}_{\ker },x, \lambda \bigr), \end{aligned}$$
(19)
$$\begin{aligned}& \bigl(x^{0}_{\ker },x,\lambda \bigr)\in \bigl(D \cap \operatorname{Ker}(I-Q)\times\widetilde{D} \times[0,1] \bigr), \end{aligned}$$
(20)
where \(\widetilde{D}=\{x\in X:x(t)\in D \text{ for all } t \in[0,T]\}\).
We claim that
$$ 0\notin(id-H) (\partial \bigl( \bigl(D\cap \operatorname{Ker}(I-Q)\times \widetilde{D} \bigr)\times[0,1] \bigr). $$
(21)
Suppose, on the contrary, that there exists \((\widehat{x}^{0}_{\ker }, \widehat{x},\widehat{\lambda})\in\partial((D\cap \operatorname{Ker}(I-Q)\times \widetilde{D})\times[0,1]\) such that \((id-H)(\widehat{x}^{0}_{\ker }, \widehat{x},\widehat{\lambda})=0\). Since \(\widehat{x}^{0}_{\ker } \in\partial D\) is contradictory to (\(H_{1}\)) and since \(\partial(D \cap \operatorname{Ker}(I-Q))\subset\partial D\), we have that \(\widehat{x}^{0}_{ \ker }\notin\partial(D\cap \operatorname{Ker}(I-Q))\). In other words, \(\widehat{x}\in\partial D\). Then (21) can be proved as follows.
(i) When \(\widehat{\lambda}=0\), by the definition of the set \(\textrm{X}_{\lambda}\) we have
$$ \textrm{X}_{0}= \biggl\{ x\in X: \biggl\vert \frac{x(t)-x(r)}{t-r} \biggr\vert \leq0 \text{ for all } t,r\in(t_{k},t_{k+1}],t \neq r \biggr\} . $$
Hence \(\alpha_{0}\circ x(t)\equiv\alpha_{0}\circ x(t_{k+1})\) for all \(t\in(t_{k},t_{k+1}]\). Since \((id-H)(\widehat{x}^{0}_{\ker}, \widehat{x},0)=0\), we have
$$ \left ( \begin{matrix} {\widehat{x}^{0}_{\ker }} \\ {\widehat{x}(t)} \end{matrix} \right ) = \left ( \begin{matrix} { \widehat{x}^{0}_{\ker } +\frac{1}{T} [ \int_{0}^{T}Pf (s,\alpha_{ \lambda}\circ x(s) )\,ds+\sum_{0\leq t_{k}< T}PI_{k} ( \alpha_{\lambda} \circ x(t_{k}) ) ]} \\ {\alpha_{0} \circ\widehat{x}^{0}_{\ker }} \end{matrix} \right ) . $$
(22)
This means that \(\widehat{x}(t)\equiv\widehat{x}(0)\) for all \(t\in[0,T]\). Taking \(\widehat{x}(t)=p\), we have \(\alpha_{0}\circ \widehat{x}^{0}_{\ker }=\widehat{x}(t)=p\). Consequently,
$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,\alpha_{\lambda}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< T}PI_{k} \bigl( \alpha_{\lambda}\circ x(t_{k}) \bigr) \biggr]=0, $$
and this is equivalent to \(g(p)=0\) by the definition of \(g(a)\). Notice that \(\widehat{x}\in\partial\widetilde{D}\) and \(\widetilde{D}=\{x \in D \text{ for all } t\in[0,T]\}\). Then there exists \(t_{0}\in[0,T]\) such that \(\widehat{x(t)}_{0}\in\partial D\). Since \(\widehat{x}(t)\equiv p\) for all \(t\in[0,T]\), we obtain that \(p\in\partial D\). Thus, we have \(p\in\partial D\) and \(g(p)=0\). It is contradictory to (\(H_{2}\)) because the Brouwer degree \(\operatorname{deg}(g,D,0) \neq0\).
(ii) When \(\widehat{\lambda}\in(0,1]\), as \(0=(id-H)(\widehat{x}^{0} _{\ker },\widehat{x},\widehat{\lambda})\), we have
$$\begin{aligned}& \left ( \begin{matrix} {\widehat{x}^{0}_{\ker }} \\ {\widehat{x}(t)} \end{matrix} \right ) \\& \quad = \left ( \begin{matrix} {\widehat{x}^{0}_{\ker } +\frac{1}{T} [ \int_{0}^{T}Pf (s, \alpha_{\widehat{\lambda}}\circ x(s) )\,ds+\sum_{0\leq t_{k}< T}PI_{k} ( \alpha_{\widehat{\lambda}}\circ x(t_{k}) ) ]} \\ {\alpha_{\widehat{\lambda}}\circ x^{0}_{\ker }-\widehat{\lambda} L ^{-1}_{p} [ \int_{0}^{T}(I-P)f (s,\alpha_{\widehat{\lambda}} \circ x(s) )\,ds+ \sum_{0\leq t_{k}< T}(I-P)I_{k} ( \alpha_{\widehat{\lambda}}\circ x(t _{k}) ) ]} \\ {{}+\widehat{\lambda} [ \int_{0}^{t}f (s,\alpha_{ \widehat{\lambda}}\circ x(s) )\,ds+\sum_{0\leq t_{k}< t}I_{k} ( \alpha_{\widehat{\lambda}}\circ x(t_{k}) ) ]} \end{matrix} \right ) . \end{aligned}$$
Thus
$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,\alpha_{\widehat{\lambda}}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< T}PI_{k} \bigl( \alpha_{\widehat{\lambda}}\circ x(t_{k}) \bigr) \biggr]=0 $$
and
$$ \begin{aligned}[b] \widehat{x}(t)={} & \alpha_{\widehat{\lambda}} \circ x^{0}_{\ker }- \widehat{\lambda} L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s, \alpha_{\widehat{\lambda}} \circ x(s) \bigr)\,ds+\sum_{0\leq t_{k}< T}(I-P)I _{k} \bigl(\alpha_{\widehat{\lambda}}\circ x(t_{k}) \bigr) \biggr] \\ &{}+ \widehat{\lambda} \biggl[ \int_{0}^{t}f \bigl(s,\alpha_{\widehat{\lambda}}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< t}I_{k} \bigl( \alpha_{\widehat{\lambda}}\circ x(t_{k}) \bigr) \biggr]. \end{aligned} $$
(23)
Note that
$$\begin{aligned}& \biggl\vert \frac{x(t)-x(r)}{t-r} \biggr\vert \\& \quad = \frac{1}{ \vert t-r \vert } \biggl\vert \widehat{\lambda} \int _{0}^{t}f \bigl(s,\alpha_{\widehat{\lambda}}\circ \widehat{x}(s) \bigr)\,ds- \widehat{\lambda} \int_{0}^{r}f \bigl(s,\alpha_{\widehat{\lambda}}\circ \widehat{x}(s) \bigr)\,ds \biggr\vert \\& \quad =\frac{1}{ \vert t-r \vert } \biggl\vert \widehat{\lambda} \int _{r}^{t}f \bigl(s,\alpha_{\widehat{\lambda}}\circ \widehat{x}(s) \bigr)\,ds \biggr\vert \\& \quad \leq \lambda M. \end{aligned}$$
By the definition of \(\textrm{X}_{\lambda}\) we obtain \(\widehat{x} \in\textrm{X}_{\widehat{\lambda}}\), which means that \(\alpha_{\widehat{\lambda}}\circ\widehat{x}=\widehat{x}\). Now we can rewrite equation (23) as
$$\begin{aligned} \widehat{x}(t) =& x^{0}_{\ker } - \widehat{ \lambda} L^{-1}_{p} \biggl[ \int _{0}^{T}(I-P)f \bigl(s, x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}+\widehat{\lambda} \biggl[ \int_{0}^{t}f \bigl(s, x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I _{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$
By a similar discussion of equation (16) we can prove that \(\widehat{x}(t)\) is a solution of equation (10). By hypothesis (\(H_{1}\)) we know that \(\widehat{x}(t)\notin\partial\widetilde{D}\) for any \(t\in[0,T]\). This is a contradiction to \(\widehat{x}\in\partial \widetilde{D}\).
By (i) and (ii) we obtain that
$$ 0\notin(id-H) \bigl(\partial \bigl( \bigl(D\cap \operatorname{Ker}(I-Q) \bigr)\times \widetilde{D} \bigr)\times [0,1] \bigr). $$
Therefore, by the homotopy invariance and the theory of Brouwer degree we have
$$ \begin{aligned} &\operatorname{deg}\bigl(id-H \bigl(x^{0}_{\ker}, \cdot,1 \bigr), \bigl(D\cap \operatorname{Ker}(I-Q) \bigr)\times\widetilde{D},0 \bigr) \\ &\quad =\operatorname{deg}\bigl(id-H \bigl(x^{0}_{\ker},\cdot,0 \bigr), \bigl(D \cap \operatorname{Ker}(I-Q) \bigr)\times \widetilde{D},0 \bigr) \\ &\quad =\operatorname{deg}\bigl(g,D\cap \operatorname{Ker}(I-Q),0 \bigr)\neq0. \end{aligned} $$
This means that there exists \(\widehat{x}_{*}\in\widetilde{D}\) such that
$$ \left ( \begin{matrix} {\widehat{x}^{0}_{*\ker }} \\ {\widehat{x}_{*}(t)} \end{matrix} \right ) =\widehat{ T} \bigl(\widehat{x}^{0}_{*\ker }, \widehat{x} _{*}(t),1 \bigr). $$
(24)
Similarly to the proof in (ii), we get \(\widehat{x}_{*}\in\textrm{X} _{\lambda}\). Then
$$ { } \widehat{T} \bigl(\widehat{x}^{0}_{*\ker }, \widehat{x}_{*}(t),1 \bigr)= T \bigl(\widehat{x}^{0}_{*\ker }, \widehat{x}_{*}(t),1 \bigr). $$
(25)
By equations (24) and (25) we obtain that \(\widehat{x}_{*}\) is a fixed point of T in X. Thus, \(\widehat{x}_{*}\) is a solution of system (2) with boundary value condition \(x(T)=Qx(0)\). □