By making use of Lemmas 2.2, 2.3 and 2.4, we can obtain the following existence theorem for FBVPs (1.1) at \(\dim \operatorname{Ker}L=3\).
Theorem 3.1
Assume
\((A_{1})\), \((H)\)
and the following conditions hold:
\((D_{1})\). There exist constants
\(M_{1}>0,M_{2}>0\)
such that, for
\((x,y)\in \operatorname{dom}L\), if
\(|x(t)|+|x'(t)|>M_{1}\), for
\(t\in[0,1]\), then
$$(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0, $$
if
\(|y(t)|+|y'(t)|>M_{2}\), for
\(t\in[0,1]\),
$$\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0, $$
or
$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0. $$
\((D_{2})\). There exist nonnegative functions
\(a_{i},b_{i},e_{i},d_{i},\rho_{i}\in L^{1}[0,1],i=1,2\)
such that
$$\begin{aligned}& \bigl\vert f(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < \rho _{1}(t)+a_{1}(t) \vert x_{1} \vert +b_{1}(t) \vert x_{2} \vert +e_{1}(t) \vert y_{1} \vert +d_{1}(t) \vert y_{2} \vert , \\& \bigl\vert g(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert \\& \quad < \rho _{2}(t)+a_{2}(t) \vert x_{1} \vert +b_{2}(t) \vert x_{2} \vert +e_{2}(t) \vert y_{1} \vert +d_{2}(t) \vert y_{2} \vert , \quad t\in [0,1],x_{i},y_{i}\in\mathbb{R},i=1,2. \end{aligned}$$
\((D_{3})\). There exist constants
\(E_{i}>0,i=1,2,3\), such that either for each
\((c_{1},b_{3},b_{4})\in\mathbb{R}^{3}\):
\(|c_{1}|>E_{1}\), then
$$ c_{1}(\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)>0, $$
(3.1)
\(|b_{3}|>E_{2}\), then
$$ b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)>0, $$
(3.2)
\(|b_{4}|>E_{3}\), then
$$ b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)>0, $$
(3.3)
or
\((c_{1},b_{3},b_{4})\in\mathbb{R}^{3}:|c_{1}|>E_{1}\), then
$$ c_{1}(\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0, $$
(3.4)
\(|b_{3}|>E_{2}\), then
$$ b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0, $$
(3.5)
\(|b_{4}|>E_{3}\), then
$$ b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0. $$
(3.6)
Then FBVPs (1.1) has at least one solution in
\(C^{1}[0,1]\times C^{1}[0,1]\)
provided that
$$B_{1}+\frac{C_{1}B_{2}}{1-C_{2}}< 1, \qquad C_{2}+\frac{C_{1}B_{2}}{1-B_{1}}< 1, $$
where
\(B_{1}= \Vert a_{1} \Vert _{1}+ \Vert e_{1} \Vert _{1}, B_{2}= \Vert a_{2} \Vert _{1}+ \Vert e_{2} \Vert _{1}, C_{1}= \Vert b_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1},C_{2}= \Vert b_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\).
The proof of Theorem 3.1 will be based on the next two lemmas.
Lemma 3.1
Assume that
\((A_{1}),(H),(D_{1}),(D_{2})\)
and
\((D_{3})\)
hold. Then
$$\Omega_{1}=\bigl\{ (x,y)\in \operatorname{dom}L\setminus \operatorname{Ker}L:L(x,y)=\lambda N(x,y), \textit{ for some } \lambda\in[0,1] \bigr\} , $$
and
$$\Omega_{2}=\bigl\{ (x,y)\in \operatorname{Ker}L:N(x,y)\in \operatorname{Im}L\bigr\} $$
are bounded.
Proof
For \((x,y)\in\Omega_{1}\), we have \((x,y)\notin \operatorname{Ker}L, \lambda\neq0\) and \(N(x,y)\in \operatorname{Im}L\).
So
$$\begin{aligned}& (\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)= 0, \\& \Gamma_{3} \biggl( \int _{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)=0, \end{aligned}$$
and
$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)= 0. $$
By \((D_{1})\), there exist constants \(t_{i}\in[0,1],i=1,2\) such that \(|x(t_{1})|\leq M_{1},|x'(t_{1})|\leq M_{1},|y(t_{2})|\leq M_{2},|y'(t_{2})|\leq M_{2}\).
Since \(x(t)=x(t_{1})+ \int _{t_{1}}^{t}x'(s)\,ds\), \(y(t)=y(t_{2})+ \int_{t_{2}}^{t}y'(s)\,ds\), we get
$$ \bigl\vert x(t) \bigr\vert \leq \bigl\Vert x' \bigr\Vert _{\infty}+M_{1}, \qquad \bigl\vert y(t) \bigr\vert \leq \bigl\Vert y' \bigr\Vert _{\infty}+M_{2},\quad t\in[0,1]. $$
(3.7)
Thus,
$$ \bigl\Vert (x,y) \bigr\Vert \leq\max\bigl\{ \bigl\Vert x' \bigr\Vert _{\infty}, \bigl\Vert y' \bigr\Vert _{\infty}\bigr\} +\max\{M_{1},M_{2}\}. $$
(3.8)
By \(L(x,y)=\lambda N(x,y)\), we obtain
$$\begin{aligned} \bigl(x'(t),y'(t) \bigr) =& \biggl(\lambda \int_{t_{1}}^{t}f \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds+x'(t_{1}),\\ &{}\lambda \int_{t_{2}}^{t}g \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds+y'(t_{2}) \biggr), \end{aligned}$$
thus, \(|x'(t)|< \Vert N_{1}x \Vert _{1}+M_{1},|y'(t)|< \Vert N_{2}y \Vert _{1}+M_{2}\), where \(N(x,y)=(N_{1}x,N_{2}y)\),
$$N_{1}x=f \bigl(s,x(s),x'(s),y(s),y'(s) \bigr), \qquad N_{2}y= g \bigl(s,x(s),x'(s),y(s),y'(s) \bigr). $$
That is, \(\max\{ \Vert x' \Vert _{\infty}, \Vert y' \Vert _{\infty}\}< \Vert N(x,y) \Vert _{1}+\max\{M_{1},M_{2}\}\).
By \((D_{2})\) and (3.7), we have
$$\begin{aligned} \bigl\vert x'(t) \bigr\vert < & \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1} \Vert x \Vert _{\infty}+ \Vert b_{1} \Vert _{1} \Vert y \Vert _{\infty}+ \Vert e_{1} \Vert _{1} \bigl\Vert x' \bigr\Vert _{\infty}+ \Vert d_{1} \Vert _{1} \bigl\Vert y' \bigr\Vert _{\infty}+M_{1} \\ < & \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1}M_{1}+ \Vert b_{1} \Vert _{1}M_{2}+\bigl( \Vert a_{1} \Vert _{1}+ \Vert e_{1} \Vert _{1}\bigr) \bigl\Vert x' \bigr\Vert _{\infty} \\ &{}+\bigl( \Vert b_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1}\bigr) \bigl\Vert y' \bigr\Vert _{\infty}+M_{1}, \end{aligned}$$
(3.9)
$$\begin{aligned} \bigl\vert y'(t) \bigr\vert < & \Vert \rho_{2} \Vert _{1}+ \Vert a_{2} \Vert _{1}M_{1}+ \Vert b_{2} \Vert _{1}M_{2}+\bigl( \Vert a_{2} \Vert _{1}+ \Vert e_{2} \Vert _{1}\bigr) \bigl\Vert x' \bigr\Vert _{\infty} \\ &{}+\bigl( \Vert b_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\bigr) \bigl\Vert y' \bigr\Vert _{\infty}+M_{2}, \end{aligned}$$
(3.10)
for the sake of brevity, let \(A_{1}= \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1}M_{1}+ \Vert b_{1} \Vert _{1}M_{2}+M_{1}, A_{2}= \Vert \rho_{2} \Vert _{1}+ \Vert a_{2} \Vert _{1}M_{1}+ \Vert b_{2} \Vert _{1}M_{2}+M_{2}\), then by (3.10) and (3.9), we have \(\Vert y' \Vert _{\infty}<\frac{A_{2}+B_{2} \Vert x' \Vert _{\infty}}{1-C_{2}}\), \(\Vert x' \Vert _{\infty}<\frac{A_{1}+\frac{C_{1}A_{2}}{1-C_{2}}}{1-B_{1}-\frac {C_{1}B_{2}}{1-C_{2}}}\).
Similarly, \(\Vert y' \Vert _{\infty}<\frac{A_{2}+\frac {B_{2}A_{1}}{1-B_{1}}}{1-C_{2}-\frac{C_{1}B_{2}}{1-B_{1}}}\).
By (3.8), \(\Vert (x,y) \Vert <\infty\). Therefore \(\Omega_{1}\) is bounded.
For \((x,y)\in \Omega_{2},(x,y)= (c_{1}(at-b),b_{3}t+b_{4} ),c_{1},b_{3},b_{4}\in\mathbb{R}\) and \(N(x,y)\in \operatorname{Im}L\). So,
$$(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)=0 $$
and
$$b_{j}\Gamma_{j} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)=0,\quad j=3,4. $$
Considering \((D_{3}),|c_{1}|\leq E_{1},|b_{3}|\leq E_{2},|b_{4}|\leq E_{3}\), we have \(\Vert x \Vert \leq E_{1} \Vert at-b \Vert \), \(\Vert y \Vert \leq E_{2}+E_{3}\). Therefore \(\Omega_{2}\) is bounded. □
Lemma 3.2
Assume that
\((A_{1})\), \((H)\)
and
\((D_{3})\)
hold. Then
$$\Omega_{3}=\bigl\{ (x,y)\in \operatorname{Ker}L:\lambda J(x,y)+(1- \lambda)QN(x,y)=0,\lambda\in [0,1]\bigr\} $$
is bounded, where
\(J:\operatorname{Ker}L\rightarrow \operatorname{Im}Q\)
is homeomorphous: \((x,y)= (c_{1}(at-b),b_{3}+b_{4}t ),c_{1},b_{3}, b_{4}\in\mathbb{R}\),
$$\begin{aligned}& J(x,y)\\& \hspace{6pt}= \biggl(c_{1}h_{1},\\& \hspace{12pt}\frac{n_{2}(n_{2}-1)[\Gamma_{4}(t^{m_{2}})b_{3} -\Gamma_{3}(t^{m_{2}})b_{4}]t^{n_{2}-2}-m_{2}({m_{2}}-1)[\Gamma_{4}(t^{n_{2}})b_{3}-\Gamma _{3}(t^{n_{2}})b_{4}]t^{m_{2}-2}}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma _{3}(t^{m_{2}})\Gamma_{4}(t^{n_{2}})} \biggr). \end{aligned}$$
Proof
For \((x,y)\in\Omega_{3}\), \(\lambda J(x,y)+(1-\lambda )QN(x,y)=0\). If \(\lambda=1\), then \(c_{1}=0,b_{3}=0,b_{4}=0\). That is, \((x,y)=0\). If \(\lambda\neq1\), we can have
$$\begin{aligned}& \lambda c_{1}h_{1}=-(1-\lambda) ( \Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)h_{1}, \\& \Gamma_{4}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr) \\& \quad{}-\Gamma_{3}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr)=0, \end{aligned}$$
(3.11)
and
$$\begin{aligned}& \Gamma_{4}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr) \\& \quad {}-\Gamma_{3}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma _{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr)=0. \end{aligned}$$
From Lemma 2.3,
$$ \begin{vmatrix} \Gamma_{4}\bigl(t^{m_{2}}\bigr) & \Gamma_{3}\bigl(t^{m_{2}}\bigr) \\ \Gamma_{4}\bigl(t^{n_{2}}\bigr) & \Gamma_{3} \bigl(t^{n_{2}}\bigr) \end{vmatrix} \neq0, $$
it yields
$$\textstyle\begin{cases} \lambda b_{3} +(1-\lambda)\Gamma_{3} (\int_{0}^{t}(t-s)g (s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} )\,ds )=0,\\ \lambda b_{4} +(1-\lambda)\Gamma_{4} (\int_{0}^{t}(t-s)g (s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} )\,ds )=0, \end{cases} $$
if \(|c_{1}|>E_{1},|b_{3}|>E_{2},|b_{4}|>E_{3}\), considering above equalities, (3.11) and (3.1)-(3.3), we have
$$\begin{aligned}& \lambda c^{2}_{1}h_{1} =-(1-\lambda)c_{1}( \Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0, \\& \lambda b^{2}_{3} =-(1-\lambda)b_{3} \Gamma_{3} \biggl( \int _{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0, \\& \lambda b^{2}_{4} =-(1-\lambda)b_{4} \Gamma_{4} \biggl( \int _{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0. \end{aligned}$$
Thus \(|c_{1}|\leq E_{1},|b_{3}|\leq E_{2},|b_{4}|\leq E_{3}\). So, \(\Omega_{3}\) is bounded.
If (3.4)-(3.6) hold, then let
$$\Omega_{3}=\bigl\{ (x,y)\in \operatorname{Ker}L:-\lambda J(x,y)+(1- \lambda)QN(x,y)=0,\lambda \in[0,1]\bigr\} . $$
By the same method we can also see that \(\Omega_{3}\) is bounded. □
Proof of Theorem 3.1
Let Ω be a bounded open subset of Y such that \(\bigcup_{j=1}^{3}\overline{\Omega}_{j} \subset\Omega\). The compactness of \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow Y\) and \(QN(\overline{\Omega}) \) will follow from the Arzela-Ascoli theorem and the Kolmogorov-Riesz criterion, respectively. Thus N is L-compact on Ω̅.
Then from above arguments, we have
-
(i)
\(L(x,y) \neq\lambda N(x,y)\), for every \(((x,y),\lambda )\in[(\operatorname{dom} L\setminus\operatorname{Ker} L) \cap\partial\Omega]\times(0,1)\);
-
(ii)
\(N(x,y) \notin\operatorname{Im} L\), for every \((x,y)\in\operatorname{Ker} L \cap \partial\Omega\).
At last we will prove that (iii) of Lemma 2.2. is satisfied.
Let \(H ((x,y),\lambda )=\pm\lambda J(x,y)+(1-\lambda)QN(x,y)=0\), noting that \(\Omega_{3}\subset\Omega\), we know \(H ((x,y), \lambda )\neq0\) for every \(((x,y),\lambda)\in\partial\Omega\cap \operatorname{Ker} L\). Thus, by the homotopic property of degree
$$\begin{aligned} \operatorname{deg}(QN|_{\operatorname{Ker}L},\Omega\cap \operatorname{Ker}L,0) =& \operatorname{deg} \bigl(H(x,y,0),\Omega\cap \operatorname{Ker}L,0 \bigr)\\ =& \operatorname{deg} \bigl(H(x,y,1),\Omega\cap \operatorname{Ker}L,0 \bigr) = \operatorname{deg}(\pm J,\Omega\cap \operatorname{Ker}L,0)\neq0. \end{aligned}$$
Then by Lemma 2.2, \(L(x,y)=N(x,y)\) has at least one solution in \(\operatorname{dom}L\cap\overline{\Omega}\). The proof of Theorem 3.1 is completed. □
Theorem 3.2
Assume
\((A_{3}),(D_{2}),(H)\)
and the following conditions hold:
\((D_{4})\). There exist constants
\(M_{3}>0,M_{4}>0\)
such that, for
\((x,y)\in \operatorname{dom}L\), if
\(|x(t)|+|x'(t)|>M_{3}\), for
\(t\in[0,1]\), then
$$\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0, $$
or
$$\Gamma_{2} \biggl( \int_{0}^{t}(t-s)f \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0, $$
if
\(|y(t)|+|y'(t)|>M_{4}\), for
\(t\in[0,1]\), then
$$\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0, $$
or
$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0. $$
\((D_{5})\). There exist constants
\(E_{i}>0,i=4,5\), such that either for each
\((a_{1},a_{2},b_{3},b_{4})\in\mathbb{R}^{4}\):
\(\vert a_{1} \vert >E_{4}\), then
$$ a_{1}\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0, $$
(3.12)
\(|a_{2}|>E_{5}\), then
$$ a_{2}\Gamma_{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0, $$
(3.13)
\(|b_{3}|>E_{6}\), then
$$ b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0, $$
(3.14)
\(|b_{4}|>E_{7}\), then
$$ b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0, $$
(3.15)
or for each
\((a_{1},a_{2},b_{3},b_{4})\in\mathbb{R}^{4}\):
\(\vert a_{1} \vert >E_{4}\), then
$$ a_{1}\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0, $$
(3.16)
\(|a_{2}|>E_{5}\), then
$$ a_{2}\Gamma_{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0, $$
(3.17)
\(|b_{3}|>E_{6}\), then
$$ b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0, $$
(3.18)
\(|b_{4}|>E_{7}\), then
$$ b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0. $$
(3.19)
Then FBVP (1.1) has at least one solution in
\(C^{1}[0,1]\times C^{1}[0,1]\)
provided that
$$B_{1}+\frac{C_{1}B_{2}}{1-C_{2}}< 1,\qquad C_{2}+\frac{C_{1}B_{2}}{1-B_{1}}< 1, $$
where
\(B_{1}= \Vert a_{1} \Vert _{1}+ \Vert e_{1} \Vert _{1}, B_{2}= \Vert a_{2} \Vert _{1}+ \Vert e_{2} \Vert _{1},C_{1}= \Vert b_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1},C_{2}= \Vert b_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\).
The proof of Theorem 3.2 will also be based on the next two lemmas.
Lemma 3.3
Assume that
\((A_{3}),(B_{2}),(H),(D_{2}),(D_{4})\)
and
\((D_{5})\)
hold. Then
$$\Omega_{1}=\bigl\{ (x,y)\in \operatorname{dom}L\setminus \operatorname{Ker}L:L(x,y)=\lambda N(x,y), \textit{ for some }\lambda\in[0,1]\bigr\} $$
and
$$\Omega_{2}=\bigl\{ (x,y)\in \operatorname{Ker}L:N(x,y)\in \operatorname{Im}L\bigr\} $$
are bounded.
Proof
For \((x,y)\in\Omega_{1}\), we have \((x,y)\notin \operatorname{Ker}L,\lambda\neq0\) and \(N(x,y)\in \operatorname{Im}L\).
So
$$\begin{aligned}& \Gamma_{1} \biggl( \int_{0}^{t}(t-s)f \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)=0,\\& \Gamma_{2} \biggl( \int _{0}^{t}(t-s)f \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)=0, \\& \Gamma_{3} \biggl( \int_{0}^{t}(t-s)g \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)=0, \\& \Gamma_{4} \biggl( \int _{0}^{t}(t-s)g \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)= 0. \end{aligned}$$
By \((D_{4})\), there exist constants \(t_{i}\in[0,1],i=3,4\) such that
$$\bigl\vert x(t_{3}) \bigr\vert \leq M_{3},\qquad \bigl\vert x'(t_{3}) \bigr\vert \leq M_{3}, \qquad \bigl\vert y(t_{4}) \bigr\vert \leq M_{4},\qquad \bigl\vert y'(t_{4}) \bigr\vert \leq M_{4}. $$
Since
$$x(t)=x(t_{3})+ \int _{t_{3}}^{t}x'(s)\,ds,\qquad y(t)=y(t_{4})+ \int_{t_{4}}^{t}y'(s)\,ds, $$
we get
$$ \bigl\vert x(t) \bigr\vert \leq \bigl\Vert x' \bigr\Vert _{\infty}+M_{3},\qquad \bigl\vert y(t) \bigr\vert \leq \bigl\Vert y' \bigr\Vert _{\infty}+M_{4}, \quad t\in [0,1]. $$
(3.20)
Thus,
$$ \bigl\Vert (x,y) \bigr\Vert \leq\max\bigl\{ \bigl\Vert x' \bigr\Vert _{\infty}, \bigl\Vert y' \bigr\Vert _{\infty}\bigr\} +\max\{M_{3},M_{4}\}. $$
(3.21)
By the proof of method in Lemma 3.1, we obtain \(\Vert x' \Vert _{\infty}<\frac{A_{3}+\frac{C_{1}A_{4}}{1-C_{2}}}{1-B_{1}-\frac {C_{1}B_{2}}{1-C_{2}}}\), \(\Vert y' \Vert _{\infty}<\frac{A_{4}+\frac {B_{2}A_{3}}{1-B_{1}}}{1-C_{2}-\frac{C_{1}B_{2}}{1-B_{1}}}\), where \(A_{3}= \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1}M_{3}+ \Vert b_{1} \Vert _{1}M_{4}+M_{3}, A_{4}= \Vert \rho_{2} \Vert _{1}+ \Vert a_{2} \Vert _{1}M_{3}+ \Vert b_{2} \Vert _{1}M_{4}+M_{4}\), by (3.21), \(\Vert (x,y) \Vert <\infty\). Therefore \(\Omega_{1}\) is bounded.
For \((x,y)\in\Omega_{2}, (x,y)(t)=(a_{1}+a_{2}t,b_{3}+b_{4}t),a_{i},b_{j}\in\mathbb {R},i=1,2,j=3,4, t\in[0,1]\) and \(N(x,y)\in \operatorname{Im}L\).
So
$$\begin{aligned}& \Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)= 0, \\& \Gamma_{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)= 0, \\& \Gamma_{3} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)=0, \end{aligned}$$
and
$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)= 0. $$
Considering \((D_{5})\), \(|a_{1}|\leq E_{4},|a_{2}|\leq E_{5},|b_{3}|\leq E_{6},|b_{4}|\leq E_{7}\), so \(\Vert x \Vert \leq E_{4}+E_{5}\), \(\Vert y \Vert \leq E_{6}+E_{7}\).
Therefore, \(\Omega _{2}\) is bounded. □
Lemma 3.4
Assume that
\((A_{3}),(B_{2}),(H)\)
and
\((D_{5})\)
hold. Then
$$\Omega_{3}=\bigl\{ (x,y)\in \operatorname{Ker}L:\lambda J(x,y)+(1- \lambda)QN(x,y)=0,\lambda\in [0,1]\bigr\} $$
is bounded, where
\(J:\operatorname{Ker}L\rightarrow \operatorname{Im}Q\)
is homeomorphous: \((x,y)(t)=(a_{1}+a_{2}t,b_{3}+b_{4}t)\), \(a_{1},a_{2},b_{3},b_{4}\in\mathbb{R}\),
$$\begin{aligned}& J(x,y) (t)\\& \hspace{6pt}= \biggl(\frac{n_{1}(n_{1}-1)[\Gamma _{2}(t^{m_{1}})a_{1}-\Gamma_{1}(t^{m_{1}})a_{2}]t^{n_{1}-2}-{m_{1}}({m_{1}}-1)[\Gamma _{2}(t^{n_{1}})a_{1}-\Gamma_{1}(t^{n_{1}})a_{2}]t^{m_{1}-2}}{\Gamma_{1}(t^{n_{1}})\Gamma _{2}(t^{m_{1}})-\Gamma_{1}(t^{m_{1}})\Gamma_{2}(t^{n_{1}})} , \\& \hspace{12pt}{}\frac{n_{2}(n_{2}-1)[\Gamma _{4}(t^{m_{2}})b_{3}-\Gamma_{3}(t^{m_{2}})b_{4}]t^{n_{2}-2}-{m_{2}}({m_{2}}-1)[\Gamma _{4}(t^{n_{2}})b_{3}-\Gamma_{3}(t^{n_{2}})b_{4}]t^{m_{2}-2}}{\Gamma_{3}(t^{n_{2}})\Gamma _{4}(t^{m_{2}}) -\Gamma_{3}(t^{m_{2}})\Gamma_{4}(t^{n_{2}})} \biggr). \end{aligned}$$
Proof
For every \((x,y)\in\Omega_{3}\), \(\lambda J(x,y)+(1-\lambda )QN(x,y)=0\). If \(\lambda=1\), then \(a_{1}=0,a_{2}=0,b_{3}=0,b_{4}=0\). That is, \((x,y)=0\). If \(\lambda\neq1\), we can have
$$\begin{aligned}& \Gamma_{2}\bigl(t^{m_{1}}\bigr) \biggl(\lambda a_{1} +(1-\lambda)\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad{}-\Gamma _{1}\bigl(t^{m_{1}}\bigr) \biggl(\lambda a_{2} +(1-\lambda)\Gamma_{2} \biggl( \int _{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0, \\& \Gamma_{2}\bigl(t^{n_{1}}\bigr) \biggl(\lambda a_{1} +(1-\lambda)\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad {}-\Gamma_{1}\bigl(t^{n_{1}}\bigr) \biggl(\lambda a_{2} +(1-\lambda)\Gamma _{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0, \\& \Gamma_{4}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad{}-\Gamma_{3}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma_{4} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0, \end{aligned}$$
and
$$\begin{aligned}& \Gamma_{4}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad{}-\Gamma_{3}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma_{4} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0. \end{aligned}$$
From Lemma 2.3,
$$ \begin{vmatrix} \Gamma_{2} \bigl(t^{m_{1}}\bigr) & \Gamma_{1}\bigl(t^{m_{1}}\bigr) \\ \Gamma_{2}\bigl(t^{n_{1}}\bigr) & \Gamma_{1} \bigl(t^{n_{1}}\bigr) \end{vmatrix} \neq0 \quad \mbox{and} \quad \begin{vmatrix} \Gamma_{4} \bigl(t^{m_{2}}\bigr) & \Gamma_{3}\bigl(t^{m_{2}}\bigr) \\ \Gamma_{4}\bigl(t^{n_{2}}\bigr) & \Gamma_{3} \bigl(t^{n_{2}}\bigr) \end{vmatrix} \neq0, $$
it yields
$$\textstyle\begin{cases} \lambda a^{2}_{1} +(1-\lambda)a_{1}\Gamma_{1} (\int _{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0,\\ \lambda a^{2}_{2} +(1-\lambda)a_{2}\Gamma_{2} (\int _{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0,\\ \lambda b^{2}_{3} +(1-\lambda)b_{3}\Gamma_{3} (\int _{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0,\\ \lambda b^{2}_{4} +(1-\lambda)b_{4}\Gamma_{4} (\int _{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0, \end{cases} $$
if \(|a_{1}|>E_{4},|a_{2}|>E_{5},|b_{3}|>E_{6},|b_{4}|>E_{7}\), considering the above equalities and (3.12)-(3.15), we have \(\Vert x \Vert \leq E_{4}+E_{5}\), \(\Vert y \Vert \leq E_{6}+E_{7}\). So, \(\Omega_{3}\) is bounded.
If (3.16)-(3.19) hold, then let \(\Omega_{3}=\{(x,y)\in \operatorname{Ker}L:-\lambda J(x,y)+(1-\lambda)QN(x,y)=0,\lambda\in[0,1]\}\). Similar to the above arguments, we can show that \(\Omega_{3}\) is bounded, too. □
Proof of Theorem 3.2
Let Ω be a bounded open subset of Y such that \(\bigcup_{j=1}^{3}\overline{\Omega}_{j} \subset\Omega\). The compactness of \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow Y\) and \(QN(\overline{\Omega}) \) will follow from the Arzela-Ascoli theorem and the Kolmogorov-Riesz criterion, respectively. Thus N is L-compact on Ω̅. Then from the above arguments, we have
-
(i)
\(L(x,y) \neq\lambda N(x,y)\), for every \(((x,y),\lambda )\in[(\operatorname{dom} L\setminus\operatorname{Ker} L) \cap\partial\Omega]\times(0,1)\);
-
(ii)
\(N(x,y) \notin\operatorname{Im} L\), for every \((x,y)\in\operatorname{Ker} L \cap \partial\Omega\).
At last we will prove that (iii) of Lemma 2.2 is satisfied.
Let \(H ((x,y),\lambda )=\pm\lambda J(x,y)+(1-\lambda )QN(x,y)=0\), noting that \(\Omega_{3}\subset\Omega\), we know \(H ((x,y), \lambda )\neq0\) for every \(((x,y),\lambda)\in\partial\Omega \cap \operatorname{Ker} L\). Thus, by the homotopic property of degree
$$\begin{aligned} \operatorname{deg}(QN|_{\operatorname{Ker}L},\Omega\cap \operatorname{Ker}L,0) =& \operatorname{deg} \bigl(H(x,y,0),\Omega\cap \operatorname{Ker}L,0 \bigr)\\ =& \operatorname{deg} \bigl(H(x,y,1),\Omega\cap \operatorname{Ker}L,0 \bigr) = \operatorname{deg}(\pm J,\Omega\cap \operatorname{Ker}L,0)\neq0. \end{aligned}$$
Then by Lemma 2.2, \(L(x,y)=N(x,y)\) has at least one solution in \(\operatorname{dom}L\cap\overline{\Omega}\). The proof of Theorem 3.2 is completed. □
The next lemma provides norm estimates needed for the following result.
Lemma 3.5
For
\((u,v)\in Z, K_{P}(u,v)=(K_{P_{1}}u,K_{P_{2}}v)\), where
\(K_{P_{1}}u=-\frac{bt+a}{a^{2}+b^{2}}\Gamma_{2} (\int_{0}^{t}(t-s)u(s)\,ds )+\int_{0}^{t}(t-s)u(s)\,ds, K_{P_{2}}v=\int_{0}^{t}(t-s)v(s)\,ds\), then
-
(1)
\(\Vert K_{P_{1}}u \Vert \leq \Vert K_{P_{1}} \Vert \Vert u \Vert _{1}\),
-
(2)
\(\Vert K_{P_{2}}v \Vert \leq \Vert v \Vert _{1}\),
where
\(\Vert K_{P_{1}} \Vert =(\frac{ \Vert bt+a \Vert \beta_{2}}{a^{2}+b^{2}}+1)\).
Proof
Observe that due to \(|\Gamma_{2}(x)|\leq\beta_{2} \Vert x \Vert \),
$$\begin{aligned} \vert K_{P_{1}}u \vert =& \biggl\vert -\frac{bt+a}{a^{2}+b^{2}} \Gamma_{2}\biggl( \int_{0}^{t}(t-s)u(s)\,ds\biggr)+ \int _{0}^{t}(t-s)u(s)\,ds \biggr\vert \\ \leq & \frac{|bt+a|}{a^{2}+b^{2}}\beta_{2} \biggl\Vert \biggl( \int_{0}^{t}(t-s)u(s)\,ds\biggr) \biggr\Vert + \biggl\Vert \biggl( \int _{0}^{t}(t-s)u(s)\,ds\biggr) \biggr\Vert \\ \leq & \biggl(\frac{|bt+a|}{a^{2}+b^{2}}\beta_{2}+1\biggr) \Vert u \Vert _{1} \leq \biggl(\frac{ \Vert bt+a \Vert }{a^{2}+b^{2}}\beta_{2}+1\biggr) \Vert u \Vert _{1} \end{aligned}$$
and \(|(K_{P_{1}}u)'(t)|\leq(\frac{|b|}{a^{2}+b^{2}}\beta_{2}+1) \Vert u \Vert _{1}\); (1) follows from the above two inequalities. Similarly, we can obtain (2). □
Theorem 3.3
Assume
\((A_{1})\)
with
\(a\neq0\), \((H)\), \((D_{3})\) (of Theorem
3.1) and the following conditions hold:
\((D_{5})\). There exist constants
\(M_{1},M_{5},M_{6}>0\)
such that, for
\((x,y)\in \operatorname{dom}L\), if
\(|x'(t)|>M_{1}\), for
\(t\in[0,1]\), then
$$(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0, $$
if
\(|y'(t)|>M_{5}\),
$$\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0, $$
or if
\(|y(t)|>M_{6}\),
$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0. $$
\((D_{6})\). There exist nonnegative functions
\(a_{i},b_{i},e_{i},d_{i},\rho_{i}\in L^{1}[0,1],i=1,2\)
such that
$$\begin{aligned}& \bigl\vert f(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < \rho _{1}(t)+a_{1}(t) \vert x_{1} \vert +b_{1}(t) \vert x_{2} \vert +e_{1}(t) \vert y_{1} \vert +d_{1}(t) \vert y_{2} \vert , \\& \bigl\vert g(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < \rho _{2}(t)+a_{2}(t) \vert x_{1} \vert +b_{2}(t) \vert x_{2} \vert \\& \phantom{\bigl\vert g(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < }{}+e_{2}(t) \vert y_{1} \vert +d_{2}(t) \vert y_{2} \vert , \quad t\in [0,1],x_{i},y_{i}\in\mathbb{R},i=1,2, \end{aligned}$$
where
$$\begin{aligned}& \bigl( \Vert t-b/a \Vert +\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr) \bigl( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1}\bigr)+\bigl( \Vert t-b/a \Vert \\& \quad {}+\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr)\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})}< 1, \\& 6\bigl( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\bigr)+\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-( \Vert t-b/a \Vert +( \Vert t-b/a \Vert +1) \Vert K_{P_{1}} \Vert )( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}< 1. \end{aligned}$$
Then FBVP (1.1) has at least one solution in
\(C^{1}[0,1]\times C^{1}[0,1]\).
Proof
As in the proof of Lemma 3.1, by \((D_{5})\), there exist constants \(M_{i}>0, t_{i}\in[0,1],i=5,6,7\) such that \(|x'(t_{5})|\leq M_{1},|y'(t_{6})|\leq M_{5},|y(t_{7})|\leq M_{6}\). Since \(x'(t)=x'(t_{5})+ \int _{t_{5}}^{t}x''(s)\,ds\), \(y'(t)=y'(t_{6})+ \int_{t_{6}}^{t}y''(s)\,ds\), we get
$$ \bigl\vert x'(t) \bigr\vert \leq \Vert N_{1}x \Vert _{1}+M_{1},\qquad \bigl\vert y'(t) \bigr\vert \leq \Vert N_{2}y \Vert _{1}+M_{5},\quad t\in [0,1], $$
(3.22)
where \(N(x,y)=(N_{1}x,N_{2}y)\), \(N_{1}x= f (s,x(s),y(s),x'(s),y'(s) )\), and \(N_{2}y= g (s,x(s),y(s),x'(s), y'(s) )\). Write \((x,y)=(x_{1},y_{1})+(x_{2},y_{2})\), where \((x_{1},y_{1})=(I-P)(x,y)\in {\operatorname{dom}L\cap \operatorname{Ker}P}\) and \((x_{2},y_{2})=P(x,y)\in \operatorname{Im} P\).
Then since \((x_{1},y_{1})=(I-P)(x,y)\in{\operatorname{dom}L\cap \operatorname{Ker}P}\), \((x_{1},y_{1})=K_{P}L(x_{1},y_{1})=K_{P}L(I-P)(x,y)=\lambda K_{P}N(x,y)\).
As in the proof of Lemma 3.5,
$$ \Vert x_{1} \Vert \leq \Vert K_{P_{1}} \Vert \Vert N_{1}x \Vert _{1}, \qquad \Vert y_{1} \Vert \leq \Vert N_{2}y \Vert _{1}. $$
(3.23)
Now, \((x_{2},y_{2})=(x,y)-(x_{1},y_{1})\), so \(x_{2}'=x'-x'_{1},y_{2}'=y'-y'_{1}\) and \(|x'_{2}(t)|\leq|x'(t)|+|x'_{1}(t)|< M_{1}+( \Vert K_{P_{1}} \Vert +1) \Vert N_{1}x \Vert _{1}\), \(|y'_{2}(t)|\leq|y'(t)|+|y'_{1}(t)|< M_{5}+2 \Vert N_{2}y \Vert _{1}\) by (3.23). Recall that \((x_{2},y_{2})(t)=P(x,y)(t)=(c(x)(at-b),y'(0)t+y(0))\), where
$$c(x)=\frac{1}{a^{2}+b^{2}}\bigl(ax'(0)-bx(0)\bigr) $$
is introduced for the sake of brevity. Hence
$$\bigl\vert x'_{2}(t) \bigr\vert = \bigl\vert c(x)a \bigr\vert < M_{1}+\bigl( \Vert K_{P_{1}} \Vert +1\bigr) \Vert N_{1}x \Vert _{1}. $$
That is,
$$\bigl\vert c(x) \bigr\vert \leq\frac{1}{|a|}\bigl(M_{1}+ \bigl( \Vert K_{P_{1}} \Vert +1\bigr)' \Vert N_{1}x \Vert _{1}\bigr). $$
Thus,
$$ \Vert x_{2} \Vert = \bigl\vert c(x) \bigr\vert \Vert at-b \Vert < \Vert t-b/a \Vert \bigl(M_{1}+\bigl( \Vert K_{P_{1}} \Vert +1\bigr) \Vert N_{1}x \Vert _{1} \bigr). $$
(3.24)
Similarly, it is easy to obtain \(|y'(0)|< M_{5}+2 \Vert N_{2}y \Vert _{1}\). In addition, \(|y_{2}(t_{7})|\leq|y(t_{7})|+|y_{1}(t_{7})|\leq M_{6}+ \Vert N_{2}y \Vert _{1}\), so, \(|y_{2}(t_{7})|=|y'(0)t_{7}+y(0)|\leq M_{6}+ \Vert N_{2}y \Vert _{1}\) and \(|y(0)|\leq M_{5}+M_{6}+3 \Vert N_{2}y \Vert _{1}\), thus
$$ \Vert y_{2} \Vert \leq \bigl\Vert y'(0)t \bigr\Vert + \bigl\vert y(0) \bigr\vert \leq2M_{5}+M_{6}+5 \Vert N_{2}y \Vert _{1}. $$
(3.25)
By (3.23) and (3.24), \(\Vert x \Vert \leq \Vert x_{1} \Vert + \Vert x_{2} \Vert \leq C_{3}+C_{4} \Vert N_{1}x \Vert _{1}\), where
$$C_{3}= \Vert t-b/a \Vert M_{1},\qquad C_{4}= \Vert t-b/a \Vert +\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert . $$
\(\Vert y \Vert \leq \Vert y_{1} \Vert + \Vert y_{2} \Vert \leq2M_{5}+M_{6}+6 \Vert N_{2}y \Vert _{1}\) by (3.23) and (3.25). Finally, it follows from \((D_{6})\) that
$$\begin{aligned}& \Vert x \Vert \leq\frac{C_{3}+C_{4}( \Vert \rho_{1} \Vert +( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})\frac {2M_{5}+M_{6}+6 \Vert \rho_{2} \Vert _{1}}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})})}{1-C_{4}( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})-C_{4}\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})}}, \\& \Vert y \Vert \leq\frac{2M_{5}+M_{6}+6 \Vert \rho_{2} \Vert _{1}+\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})(C_{3}+C_{4} \Vert \rho_{1} \Vert _{1})}{1-C_{4}( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})-\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-C_{4}( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}}. \end{aligned}$$
Therefore \(\Omega_{1}\) is bounded. The rest of the proof repeats that of Theorem 3.1. □
We now provide an example that satisfies the assumptions of Theorem 3.3. Consider the kind of equation system
$$\textstyle\begin{cases} x''(t)=t-1+\frac{1}{32}\sin x(t)+\frac{1}{32}\sin y(t)+\frac {1}{32}x'(t)+\frac{1}{32}\sin y'(t),\\ y''(t)=g(t,x(t),y(t),x'(t),y'(t)),\\ \Gamma_{1}(x)=x'(0)+2x(\frac{1}{2})=0, \qquad \Gamma_{2}(x)=x(0)-2 \int_{0}^{1}x(s)\,ds=0,\\ \Gamma_{3}(y)= 2\int_{0}^{\frac{1}{2}}y(s)\,ds-y(\frac {1}{2})+\frac{1}{4}y'(\frac{1}{2})=0,\\ \Gamma_{4}(y)=y'(1)-y'(\frac{1}{2})=0, \end{cases} $$
where
$$\begin{aligned}& g\bigl(t,x(t),y(t),x'(t),y'(t)\bigr)\\& \quad= \textstyle\begin{cases} t+\frac{1}{32}\sin x(t)+\frac{1}{32} y(0)+\frac{1}{32}\sin x'(t)+\frac {1}{32}\sin y'(t), & t\in[0,\frac{1}{2}],\\ t+\frac{1}{32}\sin x(t)+\frac{1}{32}\sin y(t)+\frac{1}{32}\sin x'(t)+\frac{1}{32}y'(t), & t\in[\frac{1}{2},1]. \end{cases}\displaystyle \end{aligned}$$
It is easy to see that \(\Gamma_{1}(t)=2,\Gamma_{1}(1)=2,\Gamma _{2}(t)=-1,\Gamma_{2}(1)=-1,\Gamma_{3}(t)=\Gamma_{3}(1)=\Gamma_{4}(t)=\Gamma _{1}(1)=0\), so that \(\alpha_{1}=-2,a=b=-1\) and \({\operatorname{Ker}L}=\{ (c_{1}(t-1),b_{3}t+b_{4})|c_{1},b_{3},b_{4}\in\mathbb{R}\}\). It is not difficult to verify that \(h_{1}\equiv-\frac{12}{5}\) satisfies Lemma 1.1.
Also,
$$\bigl\vert \Gamma_{2}(x) \bigr\vert \leq \bigl\vert x(0) \bigr\vert +2 \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds \leq3 \Vert x \Vert , $$
that is, \(\beta_{2}=3,\rho_{1}=0,\rho_{2}=1, \Vert a_{1} \Vert _{1}=\frac{1}{32}, \Vert b_{1} \Vert _{1}=\frac{1}{32}, \Vert e_{1} \Vert _{1}=\frac{1}{32}, \Vert d_{1} \Vert _{1}=\frac{1}{32}, \Vert a_{2} \Vert _{1}= \Vert b_{2} \Vert _{1}= \Vert e_{2} \Vert _{1}= \Vert d_{2} \Vert _{1}=\frac{1}{32}, \Vert K_{P_{1}} \Vert =4, \Vert K_{P_{2}} \Vert =1, \Vert t-b/a \Vert =1\),
$$\begin{aligned}& \bigl( \Vert t-b/a \Vert +\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr) \bigl( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1}\bigr)+\bigl( \Vert t-b/a \Vert \\& \quad{}+\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr)\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})}=\frac{144}{160}< 1, \end{aligned}$$
and
$$6\bigl( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\bigr)+\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-( \Vert t-b/a \Vert +( \Vert t-b/a \Vert +1) \Vert K_{P_{1}} \Vert )( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}=\frac {3}{7}< 1. $$
Let \(M_{1}=36\). Since \(N(x,y)=(N_{1}x,N_{2}y)\), if \(x'(t)>36\), then \(N_{1}x(t)>-1-\frac{3}{32}+\frac{1}{32}M_{1}>0\), and if \(x'(t)<-36\), then \(N_{1}x(t)<\frac{3}{32}-\frac{1}{32}M_{1}<0\). Taking \(M_{5}=36, M_{6}=36\), if \(y'(t)>36\), then \(N_{2}y(t)>0\), and if \(y'(t)<-36\), then \(N_{2}y(t)<0\) for \(t\in[\frac{1}{2},1]\). And if \(y(t)>36\), then \(N_{2}y(t)>0\), and \(y(t)<-36\), then \(N_{2}y<0\) for \(t\in[0,\frac{1}{2}]\).
Observe that
$$\begin{aligned}& (\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)N_{1}x(s)\,ds\biggr)= \int _{0}^{1}\kappa(s)N_{1}(s)\,ds, \\& \Gamma_{3} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)= \int_{\frac{1}{2}}^{1} N_{2}y(s)\,ds,\\& \Gamma_{4} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)= \int _{0}^{\frac{1}{2}}s^{2}N_{2}y(s) \,ds, \end{aligned}$$
where
$$\kappa(s)= \textstyle\begin{cases} -1+2s-2s^{2}, & s\in[0,\frac{1}{2}],\\ -2+4s-2s^{2}, & s\in[\frac{1}{2},1]. \end{cases} $$
Obviously, \(\kappa(s)<0,s^{2}\geq0\) in\([0,1]\), therefore,
$$\begin{aligned}& (\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)N_{1}x(s)\,ds\biggr)\neq 0,\qquad\Gamma_{3} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr) \neq0,\\& \Gamma _{4} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)\neq0 \end{aligned}$$
provided \((x,y)\in \operatorname{dom} L \backslash \operatorname{Ker} L\) satisfies \(|x'(t)|>M_{1}=36,|y'(t)|>M_{5}=36,|y(t)|>M_{6}=36\). Hence \((D_{5})\) holds.
Finally, for\((x,y)\in{\operatorname{Ker} L},x_{c_{1}}(t)=c_{1}(t-1),y_{b}(t)=b_{3}t+b_{4}\).
Consequently,
$$c_{1}(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)N_{1}x(s)\,ds\biggr)= \int _{0}^{1}\kappa(s)c_{1}N_{1}x_{c_{1}}(s) \,ds>0, $$
since \(\kappa(s)<0\) in \([0,1]\) and
$$c_{1}N_{1}x_{c_{1}}(t)\leq\frac{1}{32}|c_{1}|+ \frac{1}{32}|c_{1}|+\frac {1}{32}|c_{1}|- \frac{1}{32}c_{1}^{2}< 0 $$
provided \(|c_{1}|>E_{1}=3\). When \(|b_{3}|>E_{2}=35,|b_{4}|>E_{3}=35\),
$$\begin{aligned}& b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)= \int_{\frac{1}{2}}^{1}s^{2}b_{3}N_{2}y_{b}(s) \,ds>0, \\& b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)N_{2}y_{b}(s) \,ds\biggr)= \int_{0}^{\frac{1}{2}}b_{4}N_{2}y_{b}(s) \,ds>0, \end{aligned}$$
since \(s^{2}>0\) in \([\frac{1}{2},1]\), and
$$\begin{aligned}& b_{3}N_{2}y_{b}(t)>- \vert b_{3} \vert -\frac{3}{32} \vert b_{3} \vert +\frac{1}{32}b_{3}^{2}>0, \quad t\in\biggl[\frac {1}{2},1\biggr], \\& b_{4}N_{2}y_{b}(t)>- \vert b_{4} \vert -\frac{3}{32} \vert b_{4} \vert + \frac{1}{32}b_{4}^{2}>0,\quad t\in\biggl[0, \frac{1}{2}\biggr], \end{aligned}$$
then condition \((D_{3})\) is satisfied. It follows from Theorem 3.3 that there must be at least one solution in \(C^{1}[0,1]\times C^{1}[0,1]\).