In order to prove the solvability of BVP (1.1), some notations are introduced.
In this paper, we define \(X = {C^{n - 1}} [ {0,1} ]\) with the norm \({ \Vert x \Vert _{X}} = \max \{ {{ \Vert x \Vert }_{\infty}},{{ \Vert {x'} \Vert }_{\infty}}, \ldots , {{ \Vert {{x^{ ( {n - 1} )}}} \Vert }_{\infty}} \}\) and \(Y = C [ {0,1} ]\) with the norm \({ \Vert y \Vert _{Y}} = { \Vert y \Vert _{\infty}}\), where \({ \Vert x \Vert _{\infty}} = \max_{t \in [ {0,1} ]} \vert {x ( t )} \vert \). It is clear that \(( {X,{{ \Vert \cdot \Vert }_{X}}} )\) and \(( {Y,{{ \Vert \cdot \Vert }_{Y}}} )\) are Banach space. Furthermore, we consider Banach space \(\overline{X} = X \times X\) with the norm \({ \Vert { ( {u,v} )} \Vert _{\overline{X} }} = \max \{ {{{ \Vert u \Vert }_{X}},{{ \Vert v \Vert }_{X}}} \}\) and \(\overline{Y}= Y \times Y\) with the norm \({ \Vert { ( {x,y} )} \Vert _{\overline{Y} }} = \max \{ {{{ \Vert x \Vert }_{Y}},{{ \Vert y \Vert }_{Y}}} \}\).
Define the linear operator \({L_{1}}:\operatorname{dom}{L_{1}} \subset X \to Y\) as
$${L_{1}}u = D_{0 + }^{\alpha}u, $$
where
$$\begin{aligned} \operatorname{dom} {L_{1}} =& \bigl\{ {u \in X|D_{0 + }^{\alpha}u ( t ) \in Y,u ( 0 ) = u' ( 0 ) = \cdots = {u^{ ( {n - 3} )}} ( 0 ) = 0,} \\ &{{u^{ ( {n - 2} )}} ( 0 ) = {\gamma _{1}} {u^{ ( {n - 1} )}} ( {{\xi _{1}}} ),{u^{ ( {n - 1} )}} ( 1 ) = {\delta _{1}} {u^{ ( {n - 2} )}} ( {{\eta _{1}}} )} \bigr\} . \end{aligned}$$
Define the linear operator \({L_{2}}:\operatorname{dom}{L_{2}} \subset X \to Y\) as
$${L_{2}}v = D_{0 + }^{\beta}v, $$
where
$$\begin{aligned} \operatorname{dom} {L_{2}} =& \bigl\{ v \in X|D_{0 + }^{\beta}v ( t ) \in Y,v ( 0 ) = v' ( 0 ) = \cdots = {v^{ ( {n - 3} )}} ( 0 ) = 0, \\ & {{v^{ ( {n - 2} )}} ( 0 ) = {\gamma _{2}} {v^{ ( {n - 1} )}} ( {{\xi _{2}}} ),{v^{ ( {n - 1} )}} ( 1 ) = {\delta _{2}} {v^{ ( {n - 2} )}} ( {{\eta _{2}}} )} \bigr\} . \end{aligned}$$
Define the operator \(L:\operatorname{dom}L \subset \overline{X} \to \overline{Y} \) as
$$ L ( {u,v} ) = ( {{L_{1}}u,{L_{2}}v} ), $$
(3.1)
where \(\operatorname{dom}L = \{ { ( {u,v} ) \in \overline{X} |u \in \operatorname{dom}{L_{1}},v \in \operatorname{dom}{L_{2}}} \}\), and we define \(N:\overline{X} \to \overline{Y} \) by setting
$$N ( {u,v} ) = ( {{N_{1}}v,{N_{2}}u} ), $$
where \({N_{1}}:Y \to X\) is defined as
$${N_{1}}v ( t ) = f \bigl( {t,v ( t ),v' ( t ), \ldots ,{v^{ ( {n - 1} )}} ( t )} \bigr), $$
and \({N_{2}}:Y \to X\) is defined as
$${N_{2}}u ( t ) = g \bigl( {t,u ( t ),u' ( t ), \ldots ,{u^{ ( {n - 1} )}} ( t )} \bigr). $$
Then BVP (1.1) is equivalent to the following operator equation:
$$L ( {u,v} ) = N ( {u,v} ), \quad ( {u,v} ) \in \operatorname{dom}L. $$
Next we establish the existence results for BVP (1.1) in the following cases:
- Case (i):
-
\({\gamma _{1}} = {\gamma _{2}} = 0\), \({\delta _{1}}{\eta _{1}} = {\delta _{2}}{\eta _{2}} = 1\);
- Case (ii):
-
\(( {{\gamma _{1}} + {\eta _{1}}} ){\delta _{1}} = ( {{\gamma _{2}} + {\eta _{2}}} ){\delta _{2}} = 1\).
Firstly, the main conclusions of Case (i) are given as follows.
Theorem 3.1
For Case (i), assume that the following conditions hold.
-
\((\mathrm{H}_{1})\)
:
-
If the functions
\(f,g: [ {0,1} ] \times {\mathbb{R}^{n}} \to \mathbb{R}\)
satisfy the Carathéodary condition, and there exist nonnegative functions
\({a_{i}},{d_{i}},{b_{1}},{b_{2}},{r_{1}},{r_{2}} \in Y\)
and constant
\({\theta _{1}},{\theta _{2}} \in [ {0,1} )\), \(i = \overline {0,n - 1} \), for
\(\forall ( {{x_{0}},{x_{1}}, \ldots ,{x_{n - 1}}} ) \in {\mathbb{R}^{n}}\), \(t \in [ {0,1} ]\), the following inequalities hold:
$$\begin{aligned} & \bigl\vert {f ( {t,{x_{0}},{x_{1}},{x_{2}}, \ldots ,{x_{n - 1}}} )} \bigr\vert \le \sum_{i = 0}^{n - 1} {{a_{i}} ( t )} \vert {{x_{i}}} \vert + {b_{1}} ( t )\sum_{i = 0}^{n - 1} {{ \vert {{x_{i}}} \vert }^{{\theta _{1}}}} + {r_{1}} ( t ), \\ & \bigl\vert {g ( {t,{x_{0}},{x_{1}},{x_{2}}, \ldots ,{x_{n - 1}}} )} \bigr\vert \le \sum_{i = 0}^{n - 1} {{d_{i}} ( t )} \vert {{x_{i}}} \vert + {b_{2}} ( t )\sum_{i = 0}^{n - 1} {{{ \vert {{x_{i}}} \vert }^{{\theta _{2}}}} + } {r_{2}} ( t ). \end{aligned}$$
-
\((\mathrm{H}_{2})\)
:
-
There exists a constant
\(M>0\)
such that for
\(\forall t \in [ {0,1} ]\), if
\(\vert {{u^{ ( {n- 1} )}} ( t )} \vert > M\)
and
\(\vert {{v^{ ( {n - 1} )}} ( t )} \vert > M\), then
\(QN ( {u,v} ) \ne ( {0,0} )\).
-
\((\mathrm{H}_{3})\)
:
-
There exists a constant
\(M^{*}>0\)
such that for
\(\forall {c_{1}},{c_{2}} \in \mathbb{R}\)
satisfying
\(\min \{ { \vert {{c_{1}}} \vert , \vert {{c_{2}}} \vert } \} > {M^{*}}\), one has either
$${c_{1}} {N_{1}} \bigl( {{c_{2}} {t^{n - 1}}} \bigr) > 0,\qquad {c_{2}} {N_{2}} \bigl( {{c_{1}} {t^{n - 1}}} \bigr) > 0, $$
or
$${c_{1}} {N_{1}} \bigl( {{c_{2}} {t^{n - 1}}} \bigr) < 0,\qquad {c_{2}} {N_{2}} \bigl( {{c_{1}} {t^{n - 1}}} \bigr) < 0. $$
-
\((\mathrm{H}_{4})\)
:
-
\(\max \{ 2{a_{1}}\sum_{i = 0}^{n - 1} { \Vert {{a_{i}}} \Vert }_{\infty}, {a_{1}}\sum_{i = 0}^{n - 1} { \Vert {{a_{i}}} \Vert }_{\infty}+ {a_{2}}\sum_{i = 0}^{n - 1} { \Vert {{d_{i}}} \Vert }_{\infty}, 2{a_{2}}\sum_{i = 0}^{n - 1} { \Vert {{d_{i}}} \Vert }_{\infty}\} < 1\), where
\({a_{1}} = \frac{1}{{\Gamma ( {\alpha - n + 2} )}}\), \({a_{2}} = \frac{1}{{\Gamma ( {\beta - n + 2} )}}\).
Then BVP (1.1) has at least one solution.
To prove the above theorem, we begin with the following lemmas.
Lemma 3.1
Let
L
be defined by (3.1), then
$$ \begin{aligned} &\operatorname{Ker}L = ( \operatorname{Ker} {L_{1}}, \operatorname{Ker} {L_{2}} ) = \bigl\{ { ( {u,v} ) \in \overline{X} | ( {u,v} ) = \bigl( {{c_{1}} {t^{n - 1}},{c_{2}} {t^{n - 1}}} \bigr),{c_{1}}, {c_{2}} \in \mathbb{R}} \bigr\} , \\ &\operatorname{Im} L = ( \operatorname{Im} {L_{1}},\operatorname{Im} {L_{2}} ) = \bigl\{ { ( {x,y} ) \in \overline{Y}|{T_{1}}x = 0, {T_{2}}y = 0} \bigr\} , \end{aligned} $$
where
$$\begin{aligned} &{T_{1}}x = { \int_{0}^{1} { ( {1 - s} )} ^{\alpha - n}}x ( s ) \,ds - \frac{{{\delta _{1}}}}{{\alpha - n + 1}}{ \int_{0}^{{\eta _{1}}} { ( {{\eta _{1}} - s} )} ^{\alpha - n + 1}}x ( s )\,ds , \\ &{T_{2}}y = { \int_{0}^{1} { ( {1 - s} )} ^{\beta - n}}y ( s ) \,ds - \frac{{{\delta _{2}}}}{{\beta - n + 1}}{ \int_{0}^{{\eta _{2}}} { ( {{\eta _{2}} - s} )} ^{\beta - n + 1}}y ( s )\,ds . \end{aligned}$$
Proof
According to Lemma 2.2, \({L_{1}}u = D_{0 + }^{\alpha}u ( t ) = 0\) has the solution
$$u ( t ) = {c_{0}} + {c_{1}}t + {c_{2}} {t^{2}} + \cdots + {c_{n - 1}} {t^{n - 1}},\quad {c_{i}} \in \mathbb{R}, i = \overline {0,n - 1} . $$
By the definition of \(\operatorname{dom}{L_{1}}\), we have \({c_{i}} = 0\), \(i = \overline {0,n - 2} \), thus
$$\operatorname{Ker} {L_{1}} = \bigl\{ {u \in X|u ( t ) = {c_{1}} {t^{n - 1}}, \forall t \in [ {0,1} ], {c_{1}} \in \mathbb{R}} \bigr\} . $$
For \(x \in \operatorname{Im} {L_{1}}\), there exists \({u \in \operatorname{dom}{L_{1}}}\) such that \(x = {L_{1}}u \in Y\). From Lemma 2.1, we have
$$u ( t ) = I_{0 + }^{\alpha}x ( t ) + {c_{0}} + {c_{1}}t + {c_{2}} {t^{2}} + \cdots + {c_{n - 1}} {t^{n - 1}}. $$
Then by the definition of \({\operatorname{dom}{L_{1}}}\) we have \({c_{i}} = 0,i = \overline {0,n - 2} \). Hence
$$u ( t ) = I_{0 + }^{\alpha}x ( t ) + {c_{n - 1}} {t^{n - 1}}. $$
According to Lemma 2.4, we obtain
$$\begin{aligned} &{u^{ ( {n - 1} )}} ( t ) = I_{0 + }^{\alpha - n + 1}x ( t ) + {c_{n - 1}} ( {n - 1} )!, \\ &{u^{ ( {n - 2} )}} ( t ) = I_{0 + }^{\alpha - n + 2}x ( t ) + {c_{n - 1}} ( {n - 1} )!t. \end{aligned}$$
Taking into account the boundary condition \({u^{ ( {n - 1} )}} ( 1 ) = {\delta _{1}}{u^{ ( {n - 2} )}} ( {{\eta _{1}}} )\) and \({\delta _{1}}{\eta _{1}} = 1\) of Case (i), we see that x satisfies
$${T_{1}}x = { \int_{0}^{1} { ( {1 - s} )} ^{\alpha - n}}x ( s ) \,ds - \frac{{{\delta _{1}}}}{{\alpha - n + 1}}{ \int_{0}^{{\eta _{1}}} { ( {{\eta _{1}} - s} )} ^{\alpha - n + 1}}x ( s )\,ds = 0. $$
On the other hand, assume that \(x \in Y\) satisfies the equation \({T_{1}}x = 0\). Let \(u ( t ) = I_{0 + }^{\alpha}x ( t )\), then \(u \in \operatorname{dom}{L_{1}}\). By Lemma 2.3, we have \(D_{0 + }^{\alpha}u ( t ) = x ( t )\), so \(x \in \operatorname{Im} {L_{1}}\). Then we get
$$\operatorname{Im} {L_{1}} = \{ {x \in Y|{T_{1}}x = 0} \}. $$
Similarly, we have
$$ \begin{aligned} &\operatorname{Ker} {L_{2}} = \bigl\{ {v \in X|v ( t ) = {c_{2}} {t^{n - 1}},\forall t \in [ {0,1} ],{c_{2}} \in \mathbb{R}} \bigr\} , \\ &\operatorname{Im} {L_{2}} = \{ {y \in Y|{T_{2}}y = 0} \}. \end{aligned} $$
Then, the proof is complete. □
Lemma 3.2
Let
L
be defined by (3.1), then
L
is a Fredholm operator of index zero. The linear continuous projector operators
\(P:\overline{X} \to \overline{X}\)
and
\(Q:\overline{Y} \to \overline{Y}\)
can be defined as
$$\begin{aligned} &P ( {u,v} ) = ( {{P_{1}}u,{P_{2}}v} ) = \biggl( { \frac{{{u^{ ( {n - 1} )}} ( 0 )}}{{ ( {n - 1} )!}}{t^{n - 1}},\frac{{{v^{ ( {n - 1} )}} ( 0 )}}{{ ( {n - 1} )!}}{t^{n - 1}}} \biggr), \\ &Q ( {x,y} ) = ( {{Q_{1}}x,{Q_{2}}y} ) = \biggl( { \frac{{\alpha - n + 1}}{{1 - \frac{{{\eta _{1}}^{\alpha - n + 1}}}{{\alpha - n + 2}}}}{T_{1}}x,\frac{{\beta - n + 1}}{{1 - \frac{{{\eta _{2}}^{\beta - n + 1}}}{{\beta - n + 2}}}}{T_{2}}y} \biggr), \end{aligned}$$
and
\({K_{P}}:\operatorname{Im} L \to \operatorname{dom}L \cap \operatorname{Ker}P\)
by
$${K_{P}} ( {x,y} ) = \bigl( {I_{0 + }^{\alpha}x ( t ),I_{0 + }^{\beta}y ( t )} \bigr). $$
Proof
Obviously \(\operatorname{Im} P = \operatorname{Ker}L\) and \({P^{2}} ( {u,v} ) = P ( {u,v} )\). Since \(( {u,v} ) = ( { ( {u,v} ) - P ( {u,v} )} ) + P ( {u,v} )\), it is clear that \(\overline{X} = \operatorname{Ker}P + \operatorname{Ker}L\). By calculation, we get \(\operatorname{Ker}L \cap \operatorname{Ker}P = \{ { ( {0,0} )} \}\). Thus, we obtain
$$\overline{X} = \operatorname{Ker}L \oplus \operatorname{Ker}P. $$
For every \(( {u,v} ) \in \overline{X}\), we have
$$ { \bigl\Vert {P ( {u,v} )} \bigr\Vert _{\overline{X}}} \le \max \bigl\{ { \bigl\vert {{u^{ ( {n - 1} )}} ( 0 )} \bigr\vert , \bigl\vert {{v^{ ( {n - 1} )}} ( 0 )} \bigr\vert } \bigr\} . $$
(3.2)
Taking \(( {x,y} ) \in \overline{Y}\), one has
$${Q^{2}} ( {x,y} ) = Q ( {{Q_{1}}x,{Q_{2}}y} ) = \bigl( {{Q_{1}}^{2}x,{Q_{2}}^{2}y} \bigr). $$
By the definition of \({{Q_{1}}}\), we obtain
$$Q_{1}^{2}x = {Q_{1}}x \cdot \frac{{\alpha - n + 1}}{{1 - \frac{{{\eta _{1}}^{\alpha - n + 1}}}{{\alpha - n + 2}}}} \biggl( \int_{0}^{1} ( {1 - s} )^{\alpha - n} \,ds - \frac{\delta _{1}}{\alpha - n + 1} \int_{0}^{\eta _{1}} ( {{\eta _{1}} - s} ) ^{\alpha - n + 1} \,ds \biggr) = {Q_{1}}x, $$
where for the denominator \(1 - \frac{{{\eta _{1}}^{\alpha - n + 1}}}{{\alpha - n + 2}} > 0\) can be verified.
Similarly, \(Q_{2}^{2}y = {Q_{2}}y\). This gives \({Q^{2}} ( {x,y} ) = Q ( {x,y} )\). Let \(( {x,y} ) = ( { ( {x,y} ) - Q ( {x,y} )} ) + Q ( {x,y} )\), where \(( {x,y} ) - Q ( {x,y} ) \in \operatorname{Ker}Q = \operatorname{Im} L\), \(Q ( {x,y} ) \in \operatorname{Im} Q\). It follows from \(\operatorname{Ker}Q = \operatorname{Im} L\) and \({Q^{2}} ( {x,y} ) = Q ( {x,y} )\) that \(\operatorname{Im} Q \cap \operatorname{Im} L = \{ { ( {0,0} )} \}\). Then, we have
$$\overline{Y} = \operatorname{Im} L \oplus \operatorname{Im} Q. $$
Thus
$$\dim \operatorname{Ker} L = \dim \operatorname{Im} Q = \operatorname{codim} \operatorname{Im} L. $$
This means that L is a Fredholm operator of index zero.
Now we prove that \({K_{P}}\) is the inverse operator of \(L|_{\operatorname{dom}L \cap \operatorname{Ker}P}\). By Lemma 2.3, for \(( {x,y} ) \in \operatorname{Im} L\), we obtain
$$L{K_{P}} ( {x,y} ) = \bigl( {D_{0 + }^{\alpha}\bigl( {I_{0 + }^{\alpha}x} \bigr),D_{0 + }^{\beta}\bigl( {I_{0 + }^{\beta}y} \bigr)} \bigr) = ( {x,y} ). $$
Moreover, for \(( {u,v} ) \in \operatorname{dom}L \cap \operatorname{Ker}P\), we have \({u^{ ( {n - 1} )}} ( 0 ) = {v^{ ( {n - 1} )}} ( 0 ) = 0\). Together with the boundary condition, we get
$${K_{P}}L ( {u,v} ) = \bigl( {I_{0 + }^{\alpha}D_{0 + }^{\alpha}u ( t ),I_{0 + }^{\beta}D_{0 + }^{\beta}v ( t )} \bigr) = ( {u,v} ). $$
To summarize, \({K_{P}} = { ( L|_{\operatorname{dom}L \cap \operatorname{Ker}P} )^{ - 1}}\).
Hence, for each \(( {x,y} ) \in \operatorname{Im} L\), by the definition of \({ \Vert \cdot \Vert _{\overline{X}}}\) we have
$$\begin{aligned} { \bigl\Vert {{K_{P}} ( {x,y} )} \bigr\Vert _{\overline{X}}} &\le \max \biggl\{ {\frac{1}{{\Gamma ( {\alpha - n + 2} )}}{{ \Vert x \Vert }_{\infty}}, \frac{1}{{\Gamma ( {\beta - n + 2} )}}{{ \Vert y \Vert }_{\infty}}} \biggr\} \\ &: = \max \bigl\{ {{a_{1}} {{ \Vert x \Vert }_{\infty}},{a_{2}} {{ \Vert y \Vert }_{\infty}}} \bigr\} , \end{aligned}$$
(3.3)
where \({a_{1}} = \frac{1}{{\Gamma ( {\alpha - n + 2} )}}\), \({a_{2}} = \frac{1}{{\Gamma ( {\beta - n + 2} )}}\). The proof is complete. □
The main proof of Theorem 3.1 is given by the following three steps.
Proof of Theorem 3.1
Step 1 Let
$${\Omega _{1}} = \bigl\{ { ( {u,v} ) \in \operatorname{dom}L\backslash \operatorname{Ker} L \vert {L ( {u,v} ) = \lambda N ( {u,v} )} , \lambda \in ( {0,1} )} \bigr\} . $$
For \(( {u,v} ) \in {\Omega _{1}}\), we have \(L ( {u,v} ) = \lambda N ( {u,v} ) \in \operatorname{Im} L = \operatorname{Ker}Q\), thus \(QN ( {u,v} ) = ( {0,0} )\), i.e.
\({Q_{1}}{N_{1}}v ( t ) = 0\), \({Q_{2}}{N_{2}}u ( t ) = 0\). From \((\mathrm{H}_{2})\), we know there exists \({t_{0}},{t_{1}} \in ( {0,1} )\), such that \(\vert {{v^{ ( {n - 1} )}} ( {{t_{0}}} )} \vert \le M\) and \(\vert {{u^{ ( {n - 1} )}} ( {{t_{1}}} )} \vert \le M\). It is easy to check that \({ \Vert u \Vert _{X}} = { \Vert {{u^{ ( {n - 1} )}}} \Vert _{\infty}},{ \Vert v \Vert _{X}} = { \Vert {{v^{ ( {n - 1} )}}} \Vert _{\infty}}\) for all \(u \in \operatorname{dom}{L_{1}}\), \(v \in \operatorname{dom}{L_{2}}\). Again for \(( {u,v} ) \in {\Omega _{1}}\), then \(( {I - P} ) ( {u,v} ) \in \operatorname{dom}L \cap \operatorname{Ker} P\) and \(LP ( {u,v} ) = ( {0,0} )\). Hence, from (3.3), we get
$$ \begin{aligned}[b] { \bigl\Vert { ( {I - P} ) ( {u,v} )} \bigr\Vert _{\overline{X}}}& = { \bigl\Vert {{K_{p}}L ( {I - P} ) ( {u,v} )} \bigr\Vert _{\overline{X}}} = { \bigl\Vert {{K_{p}}L ( {u,v} )} \bigr\Vert _{\overline{X}}} = { \bigl\Vert {{K_{p}} ( {{L_{1}}u,{L_{2}}v} )} \bigr\Vert _{\overline{X}}} \\ &\le \max \bigl\{ {{a_{1}} {{ \Vert {{N_{1}}v} \Vert }_{\infty}},{a_{2}} {{ \Vert {{N_{2}}u} \Vert }_{\infty}}} \bigr\} . \end{aligned} $$
(3.4)
By \(L_{1} u = \lambda N_{1} u\) and \(u \in \operatorname{dom} L_{1}\), we have
$$ \begin{aligned} u ( t ) ={} &\frac{\lambda }{{\Gamma ( \alpha )}} \int_{0}^{t} {{{ ( {t - s} )}^{\alpha - 1}}f \bigl( {s,v ( s ),v' ( s ), \ldots ,{v^{ ( {n - 1} )}} ( s )} \bigr)} \,ds \\ &{}- u ( 0 ) - u' ( 0 )t - \cdots - \frac{{{u^{ ( {n - 1} )}} ( 0 )}}{{ ( {n - 1} )!}}{t^{n - 1}}. \end{aligned} $$
Furthermore, we obtain
$${u^{ ( {n - 1} )}} ( t ) = \frac{\lambda }{{\Gamma ( {\alpha - n + 1} )}} \int_{0}^{t} {{{ ( {t - s} )}^{\alpha - n}}f \bigl( {s,v ( s ),v' ( s ), \ldots ,{v^{ ( {n - 1} )}} ( s )} \bigr)} \,ds - {u^{ ( {n - 1} )}} ( 0 ), $$
then substituting \(t=t_{1}\) into the above equation, we get
$${u^{ ( {n - 1} )}} ( {{t_{1}}} ) = \frac{\lambda }{{\Gamma ( {\alpha - n + 1} )}} \int_{0}^{{t_{1}}} {{{ ( {{t_{1}} - s} )}^{\alpha - n}}f \bigl( {s,v ( s ),v' ( s ), \ldots ,{v^{ ( {n - 1} )}} ( s )} \bigr)} \,ds - {u^{ ( {n - 1} )}} ( 0 ). $$
Together with \(\vert {{u^{ ( {n - 1} )}} ( {{t_{1}}} )} \vert \le M\) and \((\mathrm{H}_{1})\), we get
$$\begin{aligned} \bigl\vert {{u^{ ( {n - 1} )}} ( 0 )} \bigr\vert \le {}&\biggl\vert {\frac{\lambda }{{\Gamma ( {\alpha - n + 1} )}} \int_{0}^{{t_{1}}} {{{ ( {{t_{1}} - s} )}^{\alpha - n}}f \bigl( {s,v ( s ),v' ( s ), \ldots ,{v^{ ( {n - 1} )}} ( s )} \bigr)} \,ds} \biggr\vert + \bigl\vert {{u^{ ( {n - 1} )}} ( {{t_{1}}} )} \bigr\vert \\ \le{}& \frac{1}{{\Gamma ( {\alpha - n + 1} )}} \int_{0}^{{t_{1}}} {{{ ( {{t_{1}} - s} )}^{\alpha - n}} \bigl\vert {f \bigl( {s,v ( s ),v' ( s ), \ldots ,{v^{ ( {n - 1} )}} ( s )} \bigr)} \bigr\vert } \,ds + M \\ \le {}&\frac{1}{{\Gamma ( {\alpha - n + 1} )}} \int_{0}^{{t_{1}}} {{{ ( {{t_{1}} - s} )}^{\alpha - n}} \Biggl( {\sum_{i = 0}^{n - 1} {{a_{i}} ( t )} \bigl\vert {{v^{ ( i )}}} \bigr\vert + {b_{1}} ( t )\sum_{i = 0}^{n - 1} {{ \bigl\vert {{v^{ ( i )}}} \bigr\vert }^{{\theta _{1}}}} + {r_{1}} ( t )} \Biggr)} \,ds \\ &{} + M \\ \le {}&\frac{1}{{\Gamma ( {\alpha - n + 1} )}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}+ {{ \Vert {{b_{1}}} \Vert }_{\infty}}\sum_{i = 0}^{n - 1} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + {{ \Vert {{r_{1}}} \Vert }_{\infty}}} } \Biggr) \\ &{}\times\int_{0}^{{t_{1}}} {{{ ( {{t_{1}} - s} )}^{\alpha - n}}} \,ds + M \\ \le{}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}+ {{ \Vert {{b_{1}}} \Vert }_{\infty}} \sum_{i = 0}^{n - 1} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + {{ \Vert {{r_{1}}} \Vert }_{\infty}}}} \Biggr) + M, \end{aligned}$$
(3.5)
where \({a_{1}} = \frac{1}{{\Gamma ( {\alpha - n + 2} )}}\). Similarly, we obtain
$$ \bigl\vert {v^{ ( {n - 1} )}} ( 0 ) \bigr\vert \le {a_{2}} \Biggl( \sum_{i = 0}^{n - 1} \Vert {{d_{i}}} \Vert _{\infty}\bigl\Vert {u^{ ( i )}} \bigr\Vert _{\infty}+ \Vert {{b_{2}}} \Vert _{\infty}\sum _{i = 0}^{n - 1} \bigl\Vert {u^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{2}}} + \Vert {{r_{2}}} \Vert _{\infty}\Biggr) + M, $$
(3.6)
where \({a_{2}} = \frac{1}{{\Gamma ( {\beta - n + 2} )}}\). Combined with (3.2) and (3.4), we get
$$ \begin{aligned} { \bigl\Vert { ( {u,v} )} \bigr\Vert _{\overline{X}}}= {}& { \bigl\Vert {P ( {u,v} ) + ( {I - P} ) ( {u,v} )} \bigr\Vert _{\overline{X}}} \\ \le{}& { \bigl\Vert {P ( {u,v} )} \bigr\Vert _{\overline{X}}} + { \bigl\Vert { ( {I - P} ) ( {u,v} )} \bigr\Vert _{\overline{X}}} \\ \le {}&\max \bigl\{ { \bigl\vert {{u^{ ( {n - 1} )}} ( 0 )} \bigr\vert + {a_{1}} {{ \Vert {{N_{1}}v} \Vert }_{\infty}}, \bigl\vert {{u^{ ( {n - 1} )}} ( 0 )} \bigr\vert + {a_{2}} {{ \Vert {{N_{2}}u} \Vert }_{\infty}},} \\ & { \bigl\vert {{v^{ ( {n - 1} )}} ( 0 )} \bigr\vert + {a_{1}} {{ \Vert {{N_{1}}v} \Vert }_{\infty}}, \bigl\vert {{v^{ ( {n - 1} )}} ( 0 )} \bigr\vert + {a_{2}} {{ \Vert {{N_{2}}u} \Vert }_{\infty}}} \bigr\} . \end{aligned} $$
Next, we will prove this conclusion in four cases.
Case 1
\({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le \vert {{u^{ ( {n - 1} )}} ( 0 )} \vert + {a_{1}}{ \Vert {{N_{1}}v} \Vert _{\infty}}\).
By (3.5) and \((\mathrm{H}_{1})\), we get
$$ \begin{aligned} { \bigl\Vert { ( {u,v} )} \bigr\Vert _{\overline{X}}} \le{}& \bigl\vert {{u^{ ( {n - 1} )}} ( 0 )} \bigr\vert + {a_{1}} { \Vert {{N_{1}}v} \Vert _{\infty}} \\ \le{}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}} + {{ \Vert {{b_{1}}} \Vert }_{\infty}}\sum_{i = 0}^{n - 1} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + {{ \Vert {{r_{1}}} \Vert }_{\infty}}} } \Biggr) + M \\ &{} + {a_{1}} { \bigl\Vert {f \bigl( {t,v ( t ),v' ( t ), \ldots ,{v^{ ( {n - 1} )}} ( t )} \bigr)} \bigr\Vert _{\infty}} \\ \le {}& 2{a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}} + {{ \Vert {{b_{1}}} \Vert }_{\infty}}\sum_{i = 0}^{n - 1} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}}+ {{ \Vert {{r_{1}}} \Vert }_{\infty}}} \Biggr) + M \\ \le{}& 2{a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}+ n{{ \Vert {{b_{1}}} \Vert }_{\infty}} \bigl\Vert {v^{ ( n-1 )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + {{ \Vert {{r_{1}}} \Vert }_{\infty}}} \Biggr) + M \\ \le {} & 2{a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} {{ \Vert v \Vert }_{X}} + n{{ \Vert {{b_{1}}} \Vert }_{\infty}} \Vert v \Vert _{X}^{{\theta _{1}}} + {{ \Vert {{r_{1}}} \Vert }_{\infty}}} \Biggr) + M. \end{aligned} $$
According to \((\mathrm{H}_{4})\) and the definition of \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}}\), from the above inequality, we can derive that \({ \Vert v \Vert _{X}}\) is bounded. Therefore \({\Omega _{1}}\) is bounded.
Case 2
\({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le \vert {{v^{ ( {n - 1} )}} ( 0 )} \vert + {a_{2}}{ \Vert {{N_{2}}u} \Vert _{\infty}}\). The proof is similar to Case 1. Here, we omit it.
Case 3
\({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le \vert {{u^{ ( {n - 1} )}} ( 0 )} \vert + {a_{2}}{ \Vert {{N_{2}}u} \Vert _{\infty}}\).
From (3.5) and \((\mathrm{H}_{1})\), we get
$$ \begin{aligned} \bigl\Vert (u,v) \bigr\Vert _{\overline{X}} \le{}& \bigl\vert {u^{(n - 1)}}(0) \bigr\vert + {a_{2}} \Vert {N_{2}}u \Vert _{\infty}\\ \le{}& {a_{1}} \Biggl( \sum_{i = 0}^{n - 1} \Vert {a_{i}} \Vert _{\infty}\bigl\Vert {v^{(i)}} \bigr\Vert _{\infty}+ \Vert {b_{1}} \Vert _{\infty}\sum_{i = 0}^{n - 1} { \bigl\Vert {v^{(i)}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} \Biggr) + M \\ &{} + {a_{2}} \bigl\Vert g \bigl(t,u(t),u'(t), \ldots ,{u^{(n - 1)}}(t) \bigr) \bigr\Vert _{\infty}\\ \le{}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} \Vert {a_{i}} \Vert _{\infty}\bigl\Vert {v^{(i)}} \bigr\Vert _{\infty}+ \Vert {b_{1}} \Vert _{\infty}\sum_{i = 0}^{n - 1} \bigl\Vert {v^{(i)}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} \Biggr) + M \\ &{} + {a_{2}} \Biggl( {\sum_{i = 0}^{n - 1} \Vert {d_{i}} \Vert _{\infty}\bigl\Vert {u^{(i)}} \bigr\Vert _{\infty}+ \Vert {b_{2}} \Vert _{\infty}\sum_{i = 0}^{n - 1} \bigl\Vert {u^{(i)}} \bigr\Vert _{\infty}^{{\theta _{2}}} + \Vert {r_{2}} \Vert _{\infty}} \Biggr) \\ \le{}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}} \bigl\Vert {v^{ ( {n - 1} )}} \bigr\Vert _{\infty}+ n \Vert {b_{1}} \Vert _{\infty}\bigl\Vert {v^{ ( {n - 1} )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} \Biggr) + M \\ &{}+ {a_{2}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {d_{i}} \Vert _{\infty}} \bigl\Vert {u^{ ( {n - 1} )}} \bigr\Vert _{\infty}+ n \Vert {b_{2}} \Vert _{\infty}\bigl\Vert {u^{ ( {n - 1} )}} \bigr\Vert _{\infty}^{{\theta _{2}}} + \Vert {r_{2}} \Vert _{\infty}} \Biggr) \\ = {}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} {{ \Vert v \Vert }_{X}} + n{{ \Vert {{b_{1}}} \Vert }_{\infty}} \Vert v \Vert _{X}^{{\theta _{1}}} + {{ \Vert {{r_{1}}} \Vert }_{\infty}}} \Biggr) \\ &{} + {a_{2}} \Biggl( {\sum_{i = 0}^{n - 1} {{{ \Vert {{d_{i}}} \Vert }_{\infty}}} {{ \Vert u \Vert }_{X}} + n{{ \Vert {{b_{2}}} \Vert }_{\infty}} \Vert u \Vert _{X}^{{\theta _{2}}} + {{ \Vert {{r_{2}}} \Vert }_{\infty}}} \Biggr) + M. \end{aligned} $$
By \((\mathrm{H}_{4})\), from the above inequality, we see that \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}}\) is bounded. Therefore \({\Omega _{1}}\) is bounded.
Case 4
\({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le \vert {{v^{ ( {n - 1} )}} ( 0 )} \vert + {a_{1}}{ \Vert {{N_{1}}v} \Vert _{\infty}}\). The proof is similar to Case 3. Here, we omit it.
According to the above arguments, we prove that \({\Omega _{1}}\) is bounded.
Step 2 Let
$${\Omega _{2}} = \bigl\{ { ( {u,v} ) \vert { ( {u,v} ) \in \operatorname{Ker} L,N ( {u,v} )} \in \operatorname{Im} L} \bigr\} . $$
For \(( {u,v} ) \in {\Omega _{2}}\), we have \({ ( {u,v} ) = ( {{c_{1}}{t^{n - 1}},{c_{2}}{t^{n - 1}}} ),{c_{1}},{c_{2}} \in \mathbb{R}}\). In view of \(N ( {u,v} ) = ( {{N_{1}}v,{N_{2}}u} ) \in \operatorname{Im} L = \operatorname{Ker}Q\), we have \(QN ( {u,v} ) = ( {0,0} )\), then \({Q_{1}}{N_{1}}v ( t ) = 0\), \({Q_{2}}{N_{2}}u ( t ) = 0\). Together with \((\mathrm{H}_{2})\), there exist \({t_{0}},{t_{1}} \in ( {0,1} )\) such that \(\vert {{v^{ ( {n - 1} )}} ( {{t_{0}}} )} \vert \le M\), \(\vert {{u^{ ( {n - 1} )}} ( {{t_{1}}} )} \vert \le M\), which imply \(\vert {{c_{i}}} \vert \le \frac{M}{{ ( {n - 1} )!}}\), \(i = 1,2\). Thus, we get
$${ \bigl\Vert { ( {u,v} )} \bigr\Vert _{\overline{X}}} \le M. $$
Hence, \({\Omega _{2}}\) is bounded.
Step 3 Let
$${\Omega _{3}} = \bigl\{ { ( {u,v} ) \in \operatorname{Ker} L \vert { \lambda ( {u,v} ) + ( {1 - \lambda } )QN ( {u,v} )} = ( {0,0} ), \lambda \in [ {0,1} ]} \bigr\} , $$
for \(( {u,v} ) \in {\Omega _{3}}\), we get \(( {u,v} ) = ( {{c_{1}}{t^{n - 1}},{c_{2}}{t^{n - 1}}} ),{c_{1}},{c_{2}} \in \mathbb{R}\), and
$$\begin{aligned} &\lambda {c_{1}} {t^{n - 1}} + ( {1 - \lambda } ){Q_{1}} {N_{1}} ( v ) = 0, \\ &\lambda {c_{2}} {t^{n - 1}} + ( {1 - \lambda } ){Q_{2}} {N_{2}} ( u ) = 0, \end{aligned}$$
that is to say,
$$\begin{aligned} &\begin{aligned} - \lambda {c_{1}}^{2}{t^{n - 1}} ={}& ( {1 - \lambda } ){c_{1}} {Q_{1}} {N_{1}} ( v ) \\ = {}&( {1 - \lambda } ){c_{1}}\frac{{\alpha - n + 1}}{{1 - \frac{{{\eta _{1}}^{\alpha - n + 1}}}{{\alpha - n + 2}}}} \biggl( \int_{0}^{1} ( {1 - s} ) ^{\alpha - n} f \bigl( {s,{c_{2}} {s^{n - 1}},{c_{2}} ( {n - 1} ){s^{n - 2}}, \ldots ,{c_{2}} ( {n - 1} )!} \bigr)\,ds \\ & {} - \frac{{{\delta _{1}}}}{{\alpha - n + 1}} \int_{0}^{{\eta _{1}}} ( {{\eta _{1}} - s} )^{\alpha - n + 1} f \bigl( {s,{c_{2}} {s^{n - 1}},{c_{2}} ( {n - 1} ){s^{n - 2}}, \ldots ,{c_{2}} ( {n - 1} )!} \bigr) \,ds \biggr), \end{aligned} \\ &\begin{aligned} - \lambda {c_{2}}^{2}{t^{n - 1}} ={}& ( {1 - \lambda } ){c_{2}} {Q_{2}} {N_{2}} ( u ) \\ ={}& ( {1 - \lambda } ){c_{2}}\frac{{\beta - n + 1}}{{1 - \frac{{{\eta _{2}}^{\beta - n + 1}}}{{\beta - n + 2}}}} \biggl( \int_{0}^{1} ( {1 - s} ) ^{\beta - n} g \bigl( {s,{c_{1}} {s^{n - 1}},{c_{1}} ( {n - 1} ){s^{n - 2}}, \ldots ,{c_{1}} ( {n - 1} )!} \bigr) \,ds \\ &{}- \frac{{{\delta _{2}}}}{{\beta - n + 1}} \int_{0}^{{\eta _{2}}} {( {{\eta _{2}} - s} )^{\alpha - n + 1} g \bigl( {s,{c_{1}} {s^{n - 1}},{c_{1}} ( {n - 1} ){s^{n - 2}}, \ldots ,{c_{1}} ( {n - 1} )!} \bigr)\,ds } \biggr). \end{aligned} \end{aligned}$$
If \(\lambda = 0\), then \({Q_{1}}{N_{1}} ( v ) = {Q_{2}}{N_{2}} ( u ) = 0\), together with \((\mathrm{H}_{2})\), we have \(\vert {{u^{ ( {n - 1} )}} ( t )} \vert \le M\), \(\vert {{v^{ ( {n - 1} )}} ( t )} \vert \le M\), which imply \(\vert {{c_{i}}} \vert \le \frac{M}{{ ( {n - 1} )!}}\), \(i = 1,2\). If \(\lambda \in ( {0,1} ]\), then \(\vert {{c_{i}}} \vert \le \frac{M}{{ ( {n - 1} )!}}\), \(i = 1,2\). Otherwise, if \(\vert {{c_{i}}} \vert > \frac{M}{{ ( {n - 1} )!}}\), \(i = 1,2\), in view of the first part of \((\mathrm{H}_{3})\), the left of the above two equations is less than 0, while the right is greater than 0, which is apparently contradictory. Thus, \({\Omega _{3}}\) is bounded.
Let Ω is a bounded open set of X̅, such that \(\bigcup _{i = 1}^{3} \overline {{\Omega _{i}}} \subset \Omega \). It follows from Lemma 3.2 that L is a Fredholm operator of index zero. Based on the Arzela-Ascoli theorem, we obtain the result that N is L-compact on Ω̅. By Step 1 and Step 2, we see that the following two conditions hold:
-
\(( {{a_{1}}} )\)
:
-
\(L ( {u,v} ) \ne \lambda N ( {u,v} )\), \(( { ( {u,v} ),\lambda } ) \in [ { ( {{\operatorname{dom}}L\backslash \operatorname{Ker} L} ) \cap \partial \Omega } ] \times ( {0,1} )\),
-
\(( {{a_{2}}} )\)
:
-
\(Nx \notin \operatorname{Im}L\), \(( {u,v} ) \in \operatorname{Ker} L \cap \partial \Omega \).
Let
$$H \bigl( { ( {u,v} ),\lambda } \bigr) = \lambda ( {u,v} ) + ( {1 - \lambda } )QN( {u,v} ). $$
According to Step 3, we get \(H ( { ( {u,v} ),\lambda } ) \ne 0\) for \(( {u,v} ) \in \operatorname{Ker} L \cap \partial \Omega \). Therefore,
$$\begin{aligned} \deg \bigl( QN|_{{\operatorname{Ker}}L},\Omega \cap \operatorname{Ker}L, ( {0,0} ) \bigr)& = \deg \bigl( {H ( { \cdot ,0} ),\Omega \cap \operatorname{Ker}L, ( {0,0} )} \bigr) \\ & = \deg \bigl( {H ( { \cdot ,1} ),\Omega \cap \operatorname{Ker}L, ( {0,0} )} \bigr) \\ & = \deg \bigl( {I, \Omega \cap \operatorname{Ker}L, ( {0,0} )} \bigr) \\ & \ne 0. \end{aligned}$$
Hence, the condition \(( {{a_{3}}} )\) of Theorem 2.1 is satisfied. By Theorem 2.1, we see that \(L ( {u,v} ) = N ( {u,v} )\) has at least one set of fixed points in \(\operatorname{dom}L \cap \overline{\Omega}\), so BVP (1.1) has at least one set of solutions. The proof is complete. □
Remark 3.1
If the second part of \((\mathrm{H}_{3})\) is satisfied, then the set
$$\Omega _{3}^{\prime}= \bigl\{ { ( {u,v} ) \in \operatorname{Ker}L \vert { - \lambda ( {u,v} ) + ( {1 - \lambda } )QN ( {u,v} )} = ( {0,0} ), \lambda \in [ {0,1} ]} \bigr\} $$
is bounded.
Now we consider BVP (1.1) in the Case (ii); the main conclusion is given as follows.
Theorem 3.2
For Case (ii), assume that the following conditions hold.
-
\({ ( {{{\mathrm{H}}_{1}}} )^{\prime}}\)
:
-
If the functions
\(f,g \in [ {0,1} ] \times \mathbb{R}^{n} \to {\mathbb{R}}\)
satisfy the Carathéodary condition, and there exist nonnegative functions
\({a_{i}},{d_{i}},{b_{1}},{b_{2}},{r_{1}},{r_{2}} \in Y\)
and constant
\({\theta _{1}},{\theta _{2}} \in [ {0,1} )\), \(i = \overline {0,n - 1} \), for
\(\forall ( {{x_{0}},{x_{1}}, \ldots ,{x_{n - 1}}} ) \in {{\mathbb{R}}^{n}}\), \(t \in [ {0,1} ]\), the following inequalities hold:
$$\begin{aligned} &\bigl\vert {f ( {t,{x_{0}},{x_{1}}, \ldots ,{x_{n - 1}}} )} \bigr\vert \le \sum_{i = 0}^{n - 1} {{a_{i}} ( t ) \vert {{x_{i}}} \vert } + {b_{1}} ( t )\sum_{i = 0}^{n - 1} {{{ \vert {{x_{i}}} \vert }^{{\theta _{1}}}}} + {r_{1}} ( t ), \\ &\vert {g ( {t,{x_{0}},{x_{1}}, \ldots ,{x_{n - 1}}} )} \vert \le \sum_{i = 0}^{n - 1} {{d_{i}} ( t ) \vert {{x_{i}}} \vert } + {b_{2}} ( t )\sum_{i = 0}^{n - 1} {{{ \vert {{x_{i}}} \vert }^{{\theta _{2}}}}} + {r_{2}} ( t ). \end{aligned}$$
-
\({ ( {{{\mathrm{H}}_{2}}} )^{\prime}}\)
:
-
There exists a constant
\(M > 0\), such that, for
\(\forall t \in [ {0,1} ]\), if
\(\vert {{u^{ ( {n - 1} )}} ( t )} \vert > M\)
and
\(\vert {{v^{ ( {n - 1} )}} ( t )} \vert > M\), then
\(QN ( {u,v} ) \ne ( {0,0} )\).
-
\({ ( {{{\mathrm{H}}_{3}}} )^{\prime}}\)
:
-
There exists a constant
\({M^{*}} > 0\)
such that, for every
\({c_{1}},{c_{2}} \in {\mathbb{R}}\)
satisfying
\(\min \{ { \vert {{c_{1}}} \vert , \vert {{c_{2}}} \vert } \} > {M^{*}}\), one has either
$${c_{1}}{N_{1}} ( {{c_{2}} ( {{t^{n - 1}} + ( {n - 1} ){\gamma _{2}}{t^{n - 2}}} )} ) > 0,\qquad {c_{2}}{N_{2}} ( {{c_{1}} ( {{t^{n - 1}} + ( {n - 1} ){\gamma _{1}}{t^{n - 2}}} )} ) > 0, $$
or
$${c_{1}}{N_{1}} ( {{c_{2}} ( {{t^{n - 1}} + ( {n - 1} ){\gamma _{2}}{t^{n - 2}}} )} ) < 0,\qquad {c_{2}}{N_{2}} ( {{c_{1}} ( {{t^{n - 1}} + ( {n - 1} ){\gamma _{1}}{t^{n - 2}}} )} ) < 0. $$
-
\({ ( {{{\mathrm{H}}_{4}}} )^{\prime}}\)
:
-
$$\begin{aligned} &\max \Biggl\{ ( {2 + n{\gamma _{1}}} ){a_{1}}\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} ,\bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr){a_{1}}\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} + ( {1 + {\gamma _{2}}} ){a_{2}}\sum_{i = 0}^{n - 1} {{{ \Vert {{d_{i}}} \Vert }_{\infty}}} ,\\ &\quad { ( {1 + {\gamma _{1}}} ){a_{1}}\sum_{i = 0}^{n - 1} {{{ \Vert {{a_{i}}} \Vert }_{\infty}}} + \bigl( {1 + ( {n - 1} ){\gamma _{2}}} \bigr){a_{2}}\sum_{i = 0}^{n - 1} {{{ \Vert {{d_{i}}} \Vert }_{\infty}}} , ( {2 + n{\gamma _{2}}} ){a_{2}}\sum_{i = 0}^{n - 1} {{{ \Vert {{d_{i}}} \Vert }_{\infty}}} } \Biggr\} < 1, \end{aligned}$$
where
\({a_{1}} = \frac{1}{{\Gamma ( {\alpha - n + 2} )}}\), \({a_{2}} = \frac{1}{{\Gamma ( {\beta - n + 2} )}}\).
Then BVP (1.1) has at least one solution.
To prove the above theorem, we have the following lemma, whose proof is similar to that of Lemma 3.1, Lemma 3.2 and is omitted.
Lemma 3.3
Let
L
be defined by (3.1), then
$$\begin{aligned} &\begin{aligned} \operatorname{Ker}L&= ( {\operatorname{Ker} {L_{1}}, \operatorname{Ker} {L_{2}}} ) \\ & = \bigl\{ ( {u,v} ) \in \overline{X}| ( {u,v} )= \bigl( {{c_{1}} \bigl( {{t^{n - 1}} + ( {n - 1} ){\gamma _{1}} {t^{n - 2}}} \bigr),{c_{2}} \bigl( {{t^{n - 1}} + ( {n - 1} ){\gamma _{2}} {t^{n - 2}}} \bigr)} \bigr),\\ &\quad {c_{1}},{c_{2}} \in \mathbb{R} \bigr\} , \end{aligned} \\ &\operatorname{Im}L = ( {\operatorname{Im} {L_{1}},\operatorname{Im} {L_{2}}} ) = \bigl\{ ( {x,y} ) \in \bar{Y} \vert {{T_{3}}x = 0,{T_{4}}y = 0} \bigr\} , \end{aligned}$$
where
$$\begin{aligned} &\begin{aligned} {T_{3}}x ={}& \int_{0}^{1} {{ ( {1 - s} )}^{\alpha - n}}x ( s )\,ds - \frac{{{\delta _{1}}}}{{\alpha - n + 1}} \int_{0}^{{\eta _{1}}} {{{ ( {{\eta _{1}} - s} )}^{\alpha - n + 1}}x ( s )\,ds } \\ &{}- {\gamma _{1}} {\delta _{1}} \int_{0}^{{\xi _{1}}} {{{ ( {{\xi _{1}} - s} )}^{\alpha - n}}x ( s )\,ds } , \end{aligned} \\ &\begin{aligned} {T_{4}}y = {}&\int_{0}^{1} {{{ ( {1 - s} )}^{\beta - n}}y ( s )\,ds } - \frac{{{\delta _{2}}}}{{\beta - n + 1}}\int_{0}^{{\eta _{2}}} {{{ ( {{\eta _{2}} - s} )}^{\beta - n + 1}}y ( s )\,ds } \\ &{}- {\gamma _{2}}{\delta _{2}}\int_{0}^{{\xi _{2}}} {{{ ( {{\xi _{2}} - s} )}^{\beta - n}}y ( s )\,ds } . \end{aligned} \end{aligned}$$
For
\(\forall t \in [ {0,1} ]\), the linear continuous projector operators
\(P:\overline{X} \to \overline{X}\)
and
\(Q:\overline{Y} \to \overline{Y} \)
can be defined as
$$ \begin{aligned} &P ( {u,v} ) = ( {{P_{1}}u,{P_{2}}v} ) = \biggl( {\frac{{{u^{ ( {n - 1} )}} ( 0 )}}{{ ( {n - 1} )!}} \bigl( {{t^{n - 1}} + ( {n - 1} ){\gamma _{1}} {t^{n - 2}}} \bigr), \frac{{{v^{ ( {n - 1} )}} ( 0 )}}{{ ( {n - 1} )!}} \bigl( {{t^{n - 1}} + ( {n - 1} ){\gamma _{2}} {t^{n - 2}}} \bigr)} \biggr), \\ &Q ( {x,y} ) = ( {{Q_{1}}x,{Q_{2}}y} ) = ( {{\Lambda _{1}} {T_{3}}x,{\Lambda _{2}} {T_{4}}y} ), \end{aligned} $$
where
$${\Lambda _{1}} = \frac{{\alpha - n + 1}}{{1 - \frac{{{\delta _{1}}}}{{\alpha - n + 2}}\eta _{1}^{\alpha - n + 2} - {\gamma _{1}}{\delta _{1}}\xi _{1}^{\alpha - n + 1}}},\qquad {\Lambda _{2}} = \frac{{\beta - n + 1}}{{1 - \frac{{{\delta _{2}}}}{{\beta - n + 2}}\eta _{2}^{\beta - n + 2} - {\gamma _{2}}{\delta _{2}}\xi _{2}^{\beta - n + 1}}}. $$
Define the operator
\({K_{P}}:\operatorname{Im} L \to \operatorname{dom}L \cap \operatorname{Ker}P\)
as
$$ \begin{aligned} {K_{P}} ( {x,y} ) ={}& \biggl( {\frac{1}{{\Gamma ( \alpha )}}} \int_{0}^{t} {{{ ( {t - s} )}^{\alpha - 1}}x ( s )\,ds } + \frac{{{\gamma _{1}}{t^{n - 2}}}}{{ ( {n - 2} )!\Gamma ( {\alpha - n + 1} )}} \int_{0}^{{\xi _{1}}} {{{ ( {{\xi _{1}} - s} )}^{\alpha - n}}x ( s )\,ds } , \\ &\frac{1}{{\Gamma ( \beta )}} \int_{0}^{t} {{{ ( {t - s} )}^{\beta - 1}}y ( s )\,ds } + \frac{{{\gamma _{2}}{t^{n - 2}}}}{{ ( {n - 2} )!\Gamma ( {\beta - n + 1} )}} \int_{0}^{{\xi _{2}}} {{{ ( {{\xi _{2}} - s} )}^{\beta - n}}y ( s )\,ds } \biggr). \end{aligned} $$
Next, we give the proof of Theorem 3.2 (similar to Theorem 3.1).
Proof
Firstly, it will be proved that the set
$${\Omega _{1}} = \bigl\{ { ( {u,v} ) \in \operatorname{dom} L\backslash \operatorname{Ker} L \vert {L ( {u,v} ) = \lambda N ( {u,v} )} , \lambda \in ( {0,1} )} \bigr\} $$
is bounded. If \(( {u,v} ) \in {\Omega _{1}}\), similar to Step 1 in the proof of Theorem 3.1, we get
$$ \bigl\vert {{u^{ ( {n - 1} )}} ( 0 )} \bigr\vert \le {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}\bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}} + \Vert {b_{1}} \Vert _{\infty}\sum _{i = 0}^{n - 1} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + } \Vert {r_{1}} \Vert _{\infty}} \Biggr) + M, $$
(3.7)
where \({a_{1}} = \frac{1}{{\Gamma ( {\alpha - n + 2} )}}\), and
$$ \bigl\vert {{v^{ ( {n - 1} )}} ( 0 )} \bigr\vert \le {a_{2}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {d_{i}} \Vert _{\infty}} \bigl\Vert {u^{ ( i )}} \bigr\Vert _{\infty}+ \Vert {b_{2}} \Vert _{\infty}\sum_{i = 0}^{n - 1} { \bigl\Vert {u^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{2}}} + } \Vert {r_{2}} \Vert _{\infty}} \Biggr) + M, $$
(3.8)
where \({a_{2}} = \frac{1}{{\Gamma ( {\beta - n + 2} )}}\).
So
$$ { \bigl\Vert {P ( {u,v} )} \bigr\Vert _{\overline{X}}} \le \max \bigl\{ { \bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr) \bigl\vert {{u^{ ( {n - 1} )}} ( 0 )} \bigr\vert , \bigl( {1 + ( {n - 1} ){\gamma _{2}}} \bigr) \bigl\vert {{v^{ ( {n - 1} )}} ( 0 )} \bigr\vert } \bigr\} . $$
(3.9)
On the other hand, for \(( {x,y} ) \in \operatorname{Im} L\), by the definition of \({ \Vert \cdot \Vert _{\overline{X}}}\) and \(K_{P}\), it is easy to see that
$$ { \bigl\Vert {{K_{P}} ( {x,y} )} \bigr\Vert _{\overline{X}}} \le \max \bigl\{ { ( {1 + {\gamma _{1}}} ){a_{1}} {{ \Vert x \Vert }_{\infty}}, ( {1 + {\gamma _{2}}} ){a_{2}} {{ \Vert y \Vert }_{\infty}}} \bigr\} . $$
(3.10)
Hence,
$$ \begin{aligned}[b] {{ \bigl\Vert { ( {I - P} ) ( {u,v} )} \bigr\Vert }_{\overline{X}}}& = {{ \bigl\Vert {{K_{P}}L ( {I - P} ) ( {u,v} )} \bigr\Vert }_{\overline{X}}} = {{ \bigl\Vert {{K_{P}}L ( {u,v} )} \bigr\Vert }_{\overline{X}}} = {{ \bigl\Vert {{K_{P}} ( {{L_{1}}u,{L_{2}}v} )} \bigr\Vert }_{\overline{X}}} \\ & \le \max \bigl\{ { ( {1 + {\gamma _{1}}} ){a_{1}} {{ \Vert {{N_{1}}v} \Vert }_{\infty}}, ( {1 + {\gamma _{2}}} ){a_{2}} {{ \Vert {{N_{2}}u} \Vert }_{\infty}}} \bigr\} . \end{aligned} $$
(3.11)
Thus,
$$ \begin{aligned} {{ \bigl\Vert { ( {u,v} )} \bigr\Vert }_{\overline{X}}} ={}& {{ \bigl\Vert {P ( {u,v} ) + ( {I - P} ) ( {u,v} )} \bigr\Vert }_{\overline{X}}} \\ \le{}& {{ \bigl\Vert {P ( {u,v} )} \bigr\Vert }_{\overline{X}}} + {{ \bigl\Vert { ( {I - P} ) ( {u,v} )} \bigr\Vert }_{\overline{X}}} \\ \le{}& \max \bigl\{ { \bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr) \bigl\vert {u^{ ( {n - 1} )}} ( 0 ) \bigr\vert + ( {1 + {\gamma _{1}}} ){a_{1}} {{ \Vert {{N_{1}}v} \Vert }_{\infty}},} \\ & \bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr) \bigl\vert {u^{ ( {n - 1} )}} ( 0 ) \bigr\vert + ( {1 + {\gamma _{2}}} ){a_{2}} { \Vert {{N_{2}}u} \Vert _{\infty}}, \\ & \bigl( {1 + ( {n - 1} ){\gamma _{2}}} \bigr) \bigl\vert {v^{ ( {n - 1} )}} ( 0 ) \bigr\vert + ( {1 + {\gamma _{1}}} ){a_{1}} { \Vert {{N_{1}}v} \Vert _{\infty}}, \\ &{ \bigl( {1 + ( {n - 1} ){\gamma _{2}}} \bigr) \bigl\vert {v^{ ( {n - 1} )}} ( 0 ) \bigr\vert + ( {1 + {\gamma _{2}}} ){a_{2}} {{ \Vert {{N_{2}}u} \Vert }_{\infty}}} \bigr\} . \end{aligned} $$
Next we will prove this conclusion in four cases.
Case 1′ \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le ( {1 + ( {n - 1} ){\gamma _{1}}} ) \vert {{u^{ ( {n - 1} )}} ( 0 )} \vert + ( {1 + {\gamma _{1}}} ){a_{1}}{ \Vert {{N_{1}}v} \Vert _{\infty}}\). By (3.7) and \({ ( {{{\mathrm{H}}_{1}}} )^{\prime}}\), we get
$$ \begin{aligned} { \bigl\Vert { ( {u,v} )} \bigr\Vert _{\overline{X}}} \le{}& \bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr){a_{1}} \Biggl( { \sum_{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}+ \Vert {b_{1}} \Vert _{\infty}\sum _{i = 0}^{n - 1} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert r \Vert _{\infty}} \Biggr) \\ &{} + \bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr)M + ( {1 + { \gamma _{1}}} ){a_{1}} \bigl\Vert f \bigl( {t,v ( t ),v' ( t ), \ldots ,{v^{ ( {n - 1} )}} ( t )} \bigr) \bigr\Vert _{\infty}\\ \le{}& ( {2 + n{\gamma _{1}}} ){a_{1}} \Biggl( {\sum _{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}+ \Vert {b_{1}} \Vert _{\infty}\sum _{i = 0}^{n - 1} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + } \Vert {r_{1}} \Vert _{\infty}} \Biggr)\\ &{} + \bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr)M \\ \le{}& ( {2 + n{\gamma _{1}}} ){a_{1}} \Biggl( {\sum _{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}} \bigl\Vert {v^{ ( {n - 1} )}} \bigr\Vert _{\infty}+ n \Vert {b_{1}} \Vert _{\infty}\bigl\Vert {v^{ ( {n - 1} )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} \Biggr)\\ &{} + \bigl( {1 + ( {n - 1} ){ \gamma _{1}}} \bigr)M \\ ={}& ( {2 + n{\gamma _{1}}} ){a_{1}} \Biggl( {\sum _{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}} \Vert v \Vert _{X} + n \Vert {b_{1}} \Vert _{\infty} \Vert v \Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} \Biggr) + \bigl( {1 + ( {n - 1} ){\gamma _{1}}} \bigr)M. \end{aligned} $$
According to \({ ( {{{\mathrm{H}}_{4}}} )^{\prime}}\) and the definition of \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}}\), we see that \({ \Vert v \Vert _{X}}\) is bounded, therefore \({\Omega _{1}}\) is bounded.
Case 2′ \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le \vert {{v^{ ( {n - 1} )}} ( 0 )} \vert + {a_{2}}{ \Vert {{N_{2}}u} \Vert _{\infty}}\). The proof is similar to Case 1′. Here, we omit it.
Case 3′ \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le \vert {{u^{ ( {n - 1} )}} ( 0 )} \vert + {a_{2}}{ \Vert {{N_{2}}u} \Vert _{\infty}}\). By (3.7) and \({ ( {{{\mathrm{H}}_{1}}} )^{\prime}}\), we get
$$\begin{aligned} \bigl\Vert ( {u,v} ) \bigr\Vert _{\overline{X}}\le{}& \bigl\vert {u^{ ( {n - 1} )}} ( 0 ) \bigr\vert + {a_{2}} \Vert {N_{2}}u \Vert _{\infty}\\ \le{}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}\bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}} + \Vert {b_{1}} \Vert _{\infty}\sum_{i = 0}^{n - 1} { \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} } \Biggr) + M \\ &{} + {a_{2}} \bigl\Vert g \bigl( {t,u ( t ),u' ( t ), \ldots ,{u^{ ( {n - 1} )}} ( t )} \bigr) \bigr\Vert _{\infty}\\ \le{}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} \Vert {a_{i}} \Vert _{\infty}\bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}+ \Vert {b_{1}} \Vert _{\infty}\sum_{i = 0}^{n - 1} \bigl\Vert {v^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} \Biggr) + M \\ &{} + {a_{2}} \Biggl( \sum_{i = 0}^{n - 1} \Vert {d_{i}} \Vert _{\infty}\bigl\Vert {u^{ ( i )}} \bigr\Vert _{\infty}+ \Vert {b_{2}} \Vert _{\infty}\sum_{i = 0}^{n - 1} \bigl\Vert {u^{ ( i )}} \bigr\Vert _{\infty}^{{\theta _{2}}} + \Vert {r_{2}} \Vert _{\infty}\Biggr) \\ \le{}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}\bigl\Vert {v^{ ( {n - 1} )}} \bigr\Vert _{\infty}} + n \Vert {b_{1}} \Vert _{\infty}\bigl\Vert {v^{ ( {n - 1} )}} \bigr\Vert _{\infty}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}} \Biggr) + M \\ &{} + {a_{2}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {d_{i}} \Vert _{\infty}} \bigl\Vert {u^{ ( {n - 1} )}} \bigr\Vert _{\infty}+ n \Vert {b_{2}} \Vert _{\infty}\bigl\Vert {u^{ ( {n - 1} )}} \bigr\Vert _{\infty}^{{\theta _{2}}} + \Vert {r_{2}} \Vert _{\infty}} \Biggr) \\ ={}& {a_{1}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {a_{i}} \Vert _{\infty}} \Vert v \Vert _{X}} + n \Vert {b_{1}} \Vert _{\infty} \Vert v \Vert _{X}^{{\theta _{1}}} + \Vert {r_{1}} \Vert _{\infty}\Biggr) \\ &{} + {a_{2}} \Biggl( {\sum_{i = 0}^{n - 1} { \Vert {d_{i}} \Vert _{\infty} \Vert u \Vert _{X}} + n \vert {b_{2}} \Vert _{\infty} \Vert u \Vert _{X}^{{\theta _{2}}} + \Vert {r_{2}} \Vert _{\infty}} \Biggr) + M. \end{aligned}$$
By \({ ( {{{\mathrm{H}}_{4}}} )^{\prime}}\), we get \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}}\) is bounded, therefore \({\Omega _{1}}\) is bounded.
Case 4′ \({ \Vert { ( {u,v} )} \Vert _{\overline{X}}} \le \vert {{v^{ ( {n - 1} )}} ( 0 )} \vert + {a_{1}}{ \Vert {{N_{1}}v} \Vert _{\infty}}\). The proof is similar to Case 3′. Here, we omit it.
In summary, we proved that \({\Omega _{1}}\) is bounded. The remainder of the proof is just similar to the proof of Theorem 3.1 and is omitted. □