Take the Banach space \((X,\| \cdot \|{_{\infty }})\),
$$ X = C [ {a,b} ]\quad \text{with the norm } \Vert x \Vert {_{\infty }} = \max_{t \in [a,b]} \bigl\vert x(t) \bigr\vert . $$
Lemma 3.1
Let
\(0 < {\alpha _{1}}\), \({\alpha _{2}} \leqslant 1\), \(1 < {\alpha _{1}} + {\alpha _{2}} \leqslant 2\), \(0 \leqslant {\beta _{1}}\), \({\beta _{2}} \leqslant 1\). Assume that
\((A_{1})\)
holds. Then, for
\(y\in X\), the function
\(x\in X\)
is a solution of the following BVP:
$$\begin{aligned}& D_{a+}^{{\alpha _{1}},{\beta _{1}}}D_{a+}^{{\alpha _{2}},{\beta _{2}}}x(t) +y(t)=0, \quad t\in (a,b), \end{aligned}$$
(3.1)
$$\begin{aligned}& x(a)=0,\qquad x(b)=\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}}),} \end{aligned}$$
(3.2)
if and only if
x
satisfies the integral equation
$$\begin{aligned} x(t) = \int _{a}^{b} {K(t,s)y(s)\,ds}+\frac{{{{(t - a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}K({\xi _{i}},s)y(s)\,ds} } ,\quad t \in [a,b], \end{aligned}$$
(3.3)
where
$$ K(t,s)=\frac{{{{(b-a)}^{(1-{\alpha _{1}})(1-{\beta _{1}})-{\alpha _{2}}}}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})}}\textstyle\begin{cases} {k_{1}}(t,s),\quad a \leqslant s \leqslant t \leqslant b, \\ {k_{2}}(t,s),\quad a \leqslant t \leqslant s \leqslant b, \end{cases} $$
and
$$\begin{aligned}& {k_{1}}(t,s) = {(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-s)^{{\alpha _{1}}+{\alpha _{2}}-1}}-{(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} {(t- s)^{{\alpha _{1}}+{\alpha _{2}}-1}}, \\& {k_{2}}(t,s) = {(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-s)^{ {\alpha _{1}}+{\alpha _{2}}-1}}. \end{aligned}$$
Proof
By using Lemma 2.3 twice and combining Lemma 2.2, we get that x is a solution of (3.1) if and only if
$$ x(t)=-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+{c_{1}} \frac{{{{(t-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{\varGamma ({\alpha _{2}}+1-(1-{\alpha _{1}})(1-{\beta _{1}}))}}+{c_{2}}\frac{{{{(t-a)}^{-(1- {\alpha _{2}})(1-{\beta _{2}})}}}}{{\varGamma (1-(1-{\alpha _{2}})(1-{\beta _{2}}))}}, $$
where \({c_{1}},{c_{2}} \in \mathbb{R}\). Considering the boundary conditions \(x(a) = 0\), \(x(b)=\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}\), we get
$$ {c_{2}} = 0, \qquad {c_{1}}=\frac{{\varGamma [{\alpha _{2}}+1-(1-{\alpha _{1}})(1-{\beta _{1}})]}}{ {{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \Biggl[ {I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t){|_{t=b}}+\sum _{i=1} ^{m-2} {{\sigma _{i}}x({\xi _{i}})} } \Biggr]. $$
Thus,
$$\begin{aligned} x(t) = &-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+\frac{{{{(t -a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} [ {I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)| _{t = b} } ]}}{{{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \\ &{} +\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {{{(b-a)}^{{\alpha _{2}} -(1-{\alpha _{1}})(1-{\beta _{1}})}}}}\sum_{i=1}^{m-2} {{ \sigma _{i}}x({\xi _{i}})} \\ =& \int _{a}^{b} {K(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}}}}{{{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \sum_{i = 1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}. \end{aligned}$$
(3.4)
Therefore,
$$ \sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}= \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)y(s)\,ds}+ \frac{ { \sum_{i=1}^{m-2} {{\sigma _{i}}{{({\xi _{i}}-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}} }}{{{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}. $$
That is,
$$\begin{aligned} \sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}=\frac{{{{(b-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)y(s)\,ds}. \end{aligned}$$
(3.5)
By substituting (3.5) into (3.4), we obtain
$$ x(t)= \int _{a}^{b} {K(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)y(s)\,ds} . $$
The proof is completed. □
Lemma 3.2
Let
\(0 < {\alpha _{1}}, {\alpha _{2}} \leqslant 1\), \(1 < {\alpha _{1}} + {\alpha _{2}} \leqslant 2\), \(0 \leqslant {\beta _{1}}, {\beta _{2}} \leqslant 1\). Assume that
\((A_{2})\)
holds. Then, for
\(y \in X\), the function
\(x\in X\)
is a solution of the following BVP:
$$\begin{aligned}& D_{a+}^{{\alpha _{1}},{\beta _{1}}}D_{a+}^{{\alpha _{2}},{\beta _{2}}}x(t) +y(t)=0,\quad t\in (a,b), \\ \end{aligned}$$
(3.6)
$$\begin{aligned}& x(a)=0,\qquad x'(b)=\sum _{i=1}^{m-2} {{\delta _{i}}} x({\eta _{i}}), \end{aligned}$$
(3.7)
if and only if
x
satisfies the integral equation
$$\begin{aligned} x(t)= \int _{a}^{b} {H(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{2}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}} H({\eta _{i}},s)y(s)\,ds},\quad t \in [a,b], \end{aligned}$$
(3.8)
where
$$ H(t,s) = \frac{{{{(b - s)}^{{\alpha _{1}} + {\alpha _{2}} - 2}}}}{ {\varGamma ({\alpha _{1}} + {\alpha _{2}})[{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})]}}{H_{1}}(t,s), $$
and
$$\begin{aligned}& {H_{1}}(t,s) = \textstyle\begin{cases} {h_{1}}(t,s),\quad a \leqslant s \leqslant t \leqslant b, \\ {h_{2}}(t,s),\quad a \leqslant t \leqslant s \leqslant b, \end{cases}\displaystyle \\& {h_{1}}(t,s) = ({\alpha _{1}} + {\alpha _{2}} - 1){(b - a)^{2 - ({\alpha _{1}} + {\alpha _{2}}) - {\beta _{1}}(1 - {\alpha _{1}})}} {(t - a)^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} \\& \hphantom{{h_{1}}(t,s) =}{}- \bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr]\frac{{{{(t - s)}^{{\alpha _{1}} + {\alpha _{2}} - 1}}}}{{{{(b - s)}^{{\alpha _{1}} + {\alpha _{2}} - 2}}}}, \\& {h_{2}}(t,s) = ({\alpha _{1}} + {\alpha _{2}} - 1){(b - a)^{2 - ( {\alpha _{1}} + {\alpha _{2}}) - {\beta _{1}}(1 - {\alpha _{1}})}} {(t - a)^{ {\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}}. \end{aligned}$$
Proof
By a similar method used in Lemma 3.1, we obtain
$$ x(t)=-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+{c_{1}}\frac{{{{(t-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{\varGamma [{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1-{\alpha _{1}})]}}, $$
where \({c_{1}} \in \mathbb{R}\). Then, taking derivative to the both sides of the above equality, we have
$$\begin{aligned} x'(t) = &- \frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{\varGamma ({\alpha _{1}}+ {\alpha _{2}})}} \int _{a}^{t} {{{(t-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}} y(s)\,ds \\ &{}+{c_{1}}\frac{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(t-a)}^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {\varGamma [{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1 - {\alpha _{1}})]}}. \end{aligned}$$
Using the boundary condition \(x'(b) = \sum_{i = 1}^{m - 2} {{\delta _{i}}x({\eta _{i}})}\), we get
$$\begin{aligned} {c_{1}} = & \frac{{\varGamma [{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1- {\alpha _{1}})]}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(b-a)} ^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1- {\beta _{1}})}}}} \\ &{} \times \Biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{\varGamma ({\alpha _{1}}+ {\alpha _{2}})}} \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}} y(s)\,ds+ \sum_{i = 1}^{m-2} {{\delta _{i}}x({\eta _{i}})} } \Biggr]. \end{aligned}$$
Hence,
$$\begin{aligned} x(t) = &-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+ \frac{{{{(t-a)}^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}({\alpha _{1}}+{\alpha _{2}}-1) \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+ {\alpha _{2}}-2}}} y(s)\,ds}}{ {[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(b-a)}^{{\alpha _{2}}-1-(1- {\alpha _{1}})(1-{\beta _{1}})}}\varGamma ({\alpha _{1}}+{\alpha _{2}})}} \\ &{}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(b-a)}^{{\alpha _{2}}-1-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}\sum_{i=1}^{m-2} {{ \delta _{i}}x( {\eta _{i}})} \\ =& \int _{a}^{b} {H(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{[{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})]{{(b-a)}^{{\alpha _{2}}-1-(1- {\alpha _{1}})(1- {\beta _{1}})}}}} \\ &{}\times \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})}. \end{aligned}$$
(3.9)
Therefore,
$$ \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})}= \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} y(s)\,ds}+ \frac{ { \sum_{i=1}^{m-2} {{\delta _{i}}{{({\eta _{i}}-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}} \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})} }}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})] {{(b -a)}^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}. $$
It follows that
$$\begin{aligned} \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})}=\frac{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}{{{\Delta _{2}}{{(b-a)}^{1- {\alpha _{2}}+(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} y(s)\,ds}. \end{aligned}$$
(3.10)
If we plug (3.10) back into (3.9), we obtain
$$ x(t)= \int _{a}^{b} {H(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{2}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} y(s)\,ds} , $$
which completes the proof. □
Lemma 3.3
(See [14])
If
\(1<\omega <2\), then
$$ \frac{{2-\omega }}{{{{(\omega -1)}^{{{(\omega -1)} / {(\omega -2)}}}}}} \leqslant \frac{ {{{(\omega -1)}^{\omega -1}}}}{{{\omega ^{\omega }}}}. $$
Lemma 3.4
The functions
\(K(t,s)\)
and
\({H_{1}}(t,s)\)
defined in (3.3) and (3.8) satisfy the following properties:
-
(i)
\(K(t,s)\)
and
\({H_{1}}(t,s)\)
are two continuous functions for any
\((t,s) \in [a,b] \times [a,b]\);
-
(ii)
\(|K(t,s)| \leqslant \frac{{{{[(b-a)({\alpha _{1}}+{\alpha _{2}}-1)]}^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{\varGamma ({\alpha _{1}}+{\alpha _{2}}){{[2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})]} ^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}}\)
for all
\((t,s) \in [a,b] \times [a,b]\);
-
(iii)
\(|{H_{1}}(t,s)| \leqslant (b-a)\max \{ {\alpha _{1}}+ {\alpha _{2}}-1,{\beta _{1}}(1-{\alpha _{1}})\}\)
for every
\((t,s) \in [a,b] \times [a,b]\).
Proof
Obviously, (i) is true. To prove (ii), for \((t,s) \in [a,b] \times [a,b]\), it is straightforward to show that
$$ 0 \leqslant {k_{2}}(t,s) \leqslant {k_{2}}(s,s). $$
Differentiating \({k_{1}}(t,s)\) with respect to s, we have
$$\begin{aligned} \frac{{\partial {k_{1}}(t,s)}}{{\partial s}} = &-({\alpha _{1}}+{\alpha _{2}}-1){(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-s)^{ {\alpha _{1}}+{\alpha _{2}}-2}} \\ &{}+({\alpha _{1}}+{\alpha _{2}}-1) {(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(t-s)^{{\alpha _{1}}+{\alpha _{2}}-2}} \\ =&({\alpha _{1}}+{\alpha _{2}}-1){(b-a)^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(t-s)^{{\alpha _{1}}+ {\alpha _{2}}-2}} \\ &{}\times \biggl[ {1-{{ \biggl( {\frac{{t-s}}{ {b-s}}} \biggr)}^{2-({\alpha _{1}} + {\alpha _{2}})}} {{ \biggl( {\frac{ {t-a}}{{b-a}}} \biggr)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}} \biggr] \\ \geqslant & 0, \end{aligned}$$
which shows \({k_{1}}(t,s)\) is increasing with respect to \(s \in [a,t]\). Thus,
$$ {k_{1}}(t,a) \leqslant {k_{1}}(t,s) \leqslant {k_{1}}(t,t). $$
Since
$$\begin{aligned} {k_{1}}(t,a) = &{(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}}-{(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} {(t-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} \\ =&{(t-a)^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-a)^{{\alpha _{1}}+ {\alpha _{2}}-1}} \biggl[ {1-{{ \biggl( { \frac{{b-a}}{{t-a}}} \biggr)} ^{{\beta _{1}}(1-{\alpha _{1}})}}} \biggr] \leqslant 0. \end{aligned}$$
Thus,
$$ \bigl\vert {k_{1}}(t,s) \bigr\vert \leqslant {\mathrm{{max}}} \Bigl\{ {\max_{t \in [a,b]} {k_{1}}(t,t),\max _{t \in [a,b]} \bigl(-{k_{1}}(t,a)\bigr)} \Bigr\} . $$
We consider the functions
$$\begin{aligned}& f(t) = {k_{1}}(t,t) = {(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} {(b-t)^{{\alpha _{1}}+{\alpha _{2}}-1}},\quad t \in [a,b], \\& \tilde{f}(t) = -{k_{1}}(t,a) \\& \hphantom{\tilde{f}(t)}={(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} {(t-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} \bigl[ {{{(b-a)}^{ {\beta _{1}}(1-{\alpha _{1}})}}-{{(t-a)}^{{\beta _{1}}(1- {\alpha _{1}})}}} \bigr],\quad t \in [a,b]. \end{aligned}$$
Differentiating \(f(t)\) on \((a,b)\), we have
$$\begin{aligned} f'(t) = & \bigl[{\alpha _{2}} - (1 - { \alpha _{1}}) (1 - {\beta _{1}})\bigr]{(t - a)^{ {\alpha _{2}} - 1 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{ {\alpha _{1}} + {\alpha _{2}} - 1}} \\ &{}- ({\alpha _{1}} + {\alpha _{2}} - 1){(t - a)^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 2}} \\ =& {(t - a)^{{\alpha _{2}} - 1 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 2}} \\ &{} \times \bigl\{ {\bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr](b - t) - ({\alpha _{1}} + {\alpha _{2}} - 1) (t - a)} \bigr\} , \end{aligned}$$
and
$$\begin{aligned} f''(t) = & \bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr] \bigl[{ \alpha _{2}} - 1 - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr] \\ &{} \times {(t - a)^{{\alpha _{2}} - 2 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 1}} \\ &{} - 2({\alpha _{1}} + {\alpha _{2}} - 1)\bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}}) \bigr] \\ &{} \times {(t - a)^{{\alpha _{2}} - 1 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 2}} \\ &{} + ({\alpha _{1}} + {\alpha _{2}} - 1) ({\alpha _{1}} + {\alpha _{2}} - 2){(t - a)^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 3}}. \end{aligned}$$
By calculating, we get \(f'(t)=0\) has a unique zero in \((a,b)\) as follows:
$$\begin{aligned} t ={t^{*}} = & a + \frac{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a) \\ =& b-\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a) \in (a,b). \end{aligned}$$
(3.11)
Because
$$\begin{aligned}& {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}}) \geqslant 0,\qquad {\alpha _{2}}-1-(1-{\alpha _{1}}) (1-{\beta _{1}}) \leqslant 0, \\& {\alpha _{1}}+{\alpha _{2}}-1>0, \qquad {\alpha _{1}}+{\alpha _{2}}-2 \leqslant 0, \end{aligned}$$
it is easy to verify that
$$ f''\bigl({t^{*}}\bigr) \leqslant 0. $$
Hence, we obtain
$$\begin{aligned} \max_{t \in [a,b]} f(t) =&f\bigl({t^{*}}\bigr) \\ =& \biggl[ {\frac{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}{ {2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a)} \biggr]^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})} \\ &{}\times \biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a)} \biggr]^{{\alpha _{1}}+{\alpha _{2}}-1} \\ =&\frac{{}}{{}}{ \bigl[ {{\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})} \bigr]^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {({\alpha _{1}}+ {\alpha _{2}}-1)^{{\alpha _{1}}+ {\alpha _{2}}-1}} \\ &{}\times { \biggl[ {\frac{{b-a}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})}}} \biggr]^{{\alpha _{1}}+2{\alpha _{2}}-1-(1-{\alpha _{1}})(1- {\beta _{1}})}}. \end{aligned}$$
We now prove that \(\max_{t \in [a,b]} \tilde{f}(t) \leqslant \max_{t \in [a,b]} f(t)\). In fact, if \({\beta _{1}}(1-{\alpha _{1}})=0\), then \(\tilde{f}(t) \equiv 0\), and the conclusion is obvious. If \({\beta _{1}}(1-{\alpha _{1}}) \ne 0\), differentiating \(\tilde{f}(t)\) on \((a,b)\), we have
$$\begin{aligned} \tilde{f}'(t) = & ({\alpha _{1}}+{\alpha _{2}}-1){(b-a)^{{\alpha _{2}}-(1- {\beta _{1}})(1-{\alpha _{1}})}} {(t-a)^{{\alpha _{1}}+{\alpha _{2}}-2}} \\ &{}-\bigl[{\alpha _{2}}-(1-{\beta _{1}}) (1-{\alpha _{1}})\bigr]{(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} {(t-a)^{{\alpha _{2}}-1-(1-{\beta _{1}})(1-{\alpha _{1}})}}, \end{aligned}$$
and
$$\begin{aligned} \tilde{f}''(t) = & ({\alpha _{1}}+{\alpha _{2}}-1) ({\alpha _{1}}+{\alpha _{2}}-2){(b-a)^{{\alpha _{2}}-(1-{\beta _{1}})(1-{\alpha _{1}})}} {(t-a)^{ {\alpha _{1}}+{\alpha _{2}}-3}} \\ &{} -\bigl[{\alpha _{2}}-(1-{\beta _{1}}) (1- {\alpha _{1}})\bigr] \bigl[{\alpha _{2}}-1-(1-{\beta _{1}}) (1-{\alpha _{1}})\bigr] \\ & {} \times {(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} {(t-a)^{{\alpha _{2}}-2-(1- {\beta _{1}})(1-{\alpha _{1}})}}. \end{aligned}$$
(3.12)
By calculating, we get \(\tilde{f}'(t) = 0\) has a unique zero in \((a,b)\) as follows:
$$ t={t_{*}}=a+{ \biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}} \biggr]^{{1 / {{\beta _{1}}(1-{\alpha _{1}})}}}}(b-a) \in (a,b). $$
(3.13)
Submitting (3.13) into (3.12), we have
$$\begin{aligned} \tilde{f}''({t_{*}}) = & - \bigl[ {{\alpha _{2}}-(1-{\alpha _{1}}) (1- {\beta _{1}})} \bigr]{\beta _{1}}(1-{\alpha _{1}}){(b-a)^{-2(1-{\alpha _{2}})-(2-{\beta _{1}})(1-{\alpha _{1}})}} \\ &{}\times { \biggl[ {\frac{ {{\alpha _{1}}+{\alpha _{2}}-1}}{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}} \biggr]^{{{[{\alpha _{2}}-2-(1-{\beta _{1}})(1-{\alpha _{1}})]} / {{\beta _{1}}(1-{\alpha _{1}})}}}} \\ \leqslant & 0. \end{aligned}$$
Thus,
$$\begin{aligned} \max_{t \in [a,b]} \tilde{f}(t) =&\max _{t \in [a,b]} \tilde{f}({t_{*}}) \\ =& \frac{{{\beta _{1}}(1-{\alpha _{1}}){{(b-a)}^{2({\alpha _{1}}+{\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}}}}{{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} \\ &{}\times { \biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}} \biggr]^{{{({\alpha _{1}}+ {\alpha _{2}}-1)} / {{\beta _{1}}(1-{\alpha _{1}})}}}}. \end{aligned}$$
Take \(\omega = \frac{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}{{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}\), then \(1< \omega <2\). It follows from Lemma 3.3 that
$$\begin{aligned} \max_{t \in [a,b]} \tilde{f}(t) = & {(b-a)^{2({\alpha _{1}}+{\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}}(2- \omega ){( \omega -1)^{{{(\omega -1)} / {(2-\omega )}}}} \\ \leqslant & {(b-a)^{2( {\alpha _{1}}+{\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}}\frac{{{{( \omega -1)}^{(\omega -1)}}}}{{{\omega ^{\omega }}}} \\ = & { \biggl\{ {\frac{{{{({\alpha _{1}}+{\alpha _{2}}-1)}^{({\alpha _{1}}+ {\alpha _{2}}-1)}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}}}}{{{{[2{\alpha _{2}}-(1- {\alpha _{1}})(2-{\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})}}}}} \biggr\} ^{\frac{1}{{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}}}} \\ &{} \times {(b-a)^{2({\alpha _{1}}+{\alpha _{2}}-1)+ {\beta _{1}}(1-{\alpha _{1}})}} \\ \leqslant & {(b-a)^{2({\alpha _{1}}+ {\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}} {{({\alpha _{1}}+{\alpha _{2}}-1)}^{({\alpha _{1}}+{\alpha _{2}}-1)}} \\ &{} \times \frac{{{{[ {\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})]}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{{[2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}} \\ =& \max_{t \in [a,b]} f(t). \end{aligned}$$
From the above we get
$$\begin{aligned} \bigl\vert {k_{1}}(t,s) \bigr\vert \leqslant & \max _{t \in [a,b]} {k_{1}}(t,t)=\max_{s \in [a,b]} {k_{2}}(s,s) \\ =& { \bigl[ {{\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})} \bigr]^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {({\alpha _{1}}+{\alpha _{2}}-1)^{{\alpha _{1}}+{\alpha _{2}} -1}} \\ &{} \times { \biggl[ {\frac{ {b-a}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}} \biggr]^{2 {\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}. \end{aligned}$$
Therefore,
$$\begin{aligned} \bigl\vert K(t,s) \bigr\vert \leqslant & \frac{{{\max_{t \in [a,b]}}{k_{1}}(t,t)}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \\ =& \frac{{{{[(b-a)({\alpha _{1}}+{\alpha _{2}}-1)]}^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){{[2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}}. \end{aligned}$$
To prove (iii), for \((t,s) \in [a,b] \times [a,b]\), obviously, we have that the following inequalities hold:
$$ 0 \leqslant {h_{2}}(t,s) \leqslant {h_{2}}(s,s)={h_{1}}(s,s). $$
Differentiating \({h_{1}}(t,s)\) with respect to t, we have
$$\begin{aligned} \frac{{\partial {h_{1}}(t,s)}}{{\partial t}} = & ({\alpha _{1}}+{\alpha _{2}}-1)\bigl[{\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr] \\ &{} \times {(b-a)^{2-({\alpha _{1}}+{\alpha _{2}})-{\beta _{1}}(1-{\alpha _{1}})}} {(t-a)^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1-{\beta _{1}})}} \\ &{}- \bigl[ {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr]({\alpha _{1}}+{\alpha _{2}}-1) \frac{{{{(t-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}}}{{{{(b-s)} ^{{\alpha _{1}}+{\alpha _{2}}-2}}}} \\ =& ({\alpha _{1}}+{\alpha _{2}}-1)\bigl[ {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr] \\ &{}\times \biggl[ {-{{ \biggl( {\frac{{b-s}}{{t-s}}} \biggr)}^{2-({\alpha _{1}}+{\alpha _{2}})}}+{{ \biggl( {\frac{{b-a}}{{t-a}}} \biggr)}^{1-{\alpha _{2}}+(1- {\beta _{1}})(1- {\alpha _{1}})}}} \biggr] \\ \leqslant & 0. \end{aligned}$$
Hence, \({h_{1}}(t,s)\) is a decreasing function of \(t \in [s,b]\), which implies that
$$\begin{aligned}& \begin{aligned} &{h_{1}}(b,s)\leqslant {h_{1}}(t,s) \leqslant {h_{1}}(s,s)= {h_{2}}(s,s), \\ &{h_{1}}(s,s)=({\alpha _{1}}+{\alpha _{2}}-1){(b-a)^{2-( {\alpha _{1}}+{\alpha _{2}})-{\beta _{1}}(1-{\alpha _{1}})}} {(s-a)^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}} \\ &\hphantom{{h_{1}}(s,s)}\leqslant ({\alpha _{1}}+{\alpha _{2}}-1) (b-a), \\ &{h_{1}}(b,s)= ({\alpha _{1}}+{\alpha _{2}}-1) (b-a)-\bigl[ {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr](b-s). \end{aligned} \end{aligned}$$
(3.14)
It is trivial to show that \({h_{1}}(t,s)\) is an increasing function with respect to \(s \in [a,b]\). Thus,
$$ {h_{1}}(b,a) \leqslant {h_{1}}(b,s) \leqslant {h_{1}}(b,b). $$
Note that
$$\begin{aligned}& {h_{1}}(b,b)= ({\alpha _{1}}+{\alpha _{2}}-1) (b-a)> 0, \\& {h_{1}}(b,a) =-{\beta _{1}}(1-{\alpha _{1}}) (b-a)< 0. \end{aligned}$$
We get
$$ \bigl\vert {{h_{1}}(b,s)} \bigr\vert \leqslant (b-a)\max \bigl\{ {{\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1-{\alpha _{1}})} \bigr\} . $$
(3.15)
Combining (3.14) and (3.15), we obtain
$$ \bigl\vert {{h_{1}}(t,s)} \bigr\vert \leqslant (b-a)\max \bigl\{ {{\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1-{ \alpha _{1}})} \bigr\} . $$
Therefore,
$$ \bigl\vert {H_{1}}(t,s) \bigr\vert \leqslant (b-a)\max \bigl\{ { \alpha _{1}}+{\alpha _{2}}-1, {\beta _{1}}(1-{ \alpha _{1}})\bigr\} . $$
Then we complete the proof of Lemma 3.4. □
Remark 3.1
From the proof of Lemma 3.4, we have the following conclusions:
-
(i)
\(K(t,s)\) has a unique maximum, given by
$$\begin{aligned}& \max_{(t,s) \in {{[a,b]}^{2}}} \bigl\vert K(t,s) \bigr\vert \\& \quad = K\bigl({t ^{*}},{t^{*}}\bigr) \\& \quad = \frac{{{{[(b-a)({\alpha _{1}}+{\alpha _{2}}-1)]} ^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){{[2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}}, \end{aligned}$$
where \(t^{*}\) is defined by (3.11);
-
(ii)
\(\max_{(t,s) \in {{[a,b]}^{2}}} | {H_{1}}(t,s)| = (b-a)\max \{ {\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1- {\alpha _{1}})\}\) and
$$ \bigl\vert {H_{1}}(t,s) \bigr\vert = \textstyle\begin{cases} (b-a)({\alpha _{1}}+{\alpha _{2}}-1),\quad \text{if and only if } t=s=b, \\ (b-a){\beta _{1}}(1-{\alpha _{1}}),\quad \text{if and only if } t=b, s=a. \end{cases} $$
Theorem 3.1
Assume that
\((\mathrm{{A}_{1})}\)
holds. If the fractional BVP (1.14), (1.15) has a nontrivial continuous solution for a real-valued continuous function
q, then
$$\begin{aligned}& \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} \\& \quad > \frac{{\varGamma ({\alpha _{1}}+{\alpha _{2}}) {{[2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})]}^{2{\alpha _{2}}-(1- {\alpha _{1}})(2-{\beta _{1}})}}{\Delta _{1}}}}{{{{(b-a)}^{{\alpha _{1}} +{\alpha _{2}}-1}}{{({\alpha _{1}}+{\alpha _{2}}-1)}^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {{\tilde{\Delta }}_{1}}}}, \end{aligned}$$
(3.16)
where
$$ {\tilde{\Delta }_{1}}: = {\Delta _{1}}+\sum _{i=1}^{m-2} {{\sigma _{i}}} {(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}. $$
Proof
Assume \(x(t)\) is a nontrivial solution of BVP (1.14), (1.15), then
$$ x(t)= \int _{a}^{b} {K(t,s)q(s)x(s)\,ds}+ \frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)q(s)x(s)\,ds}, $$
and
$$\begin{aligned} \bigl\vert x(t) \bigr\vert \leqslant & \int _{a}^{b} { \bigl\vert K(t,s) \bigr\vert \cdot \bigl\vert q(s) \bigr\vert \cdot \bigl\vert x(s) \bigr\vert \,ds} \\ &{} + \frac{{{{(t - a)}^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i = 1}^{m - 2} {{\sigma _{i}}} \bigl\vert K({\xi _{i}},s) \bigr\vert \cdot \bigl\vert q(s) \bigr\vert \cdot \bigl\vert x(s) \bigr\vert \,ds}, \quad t \in [a,b]. \end{aligned}$$
Since x is a nontrivial solution, which will require \(q(s)\not \equiv 0\) on \([a,b]\). Moreover, from \(q(s) \in C[a,b]\), we obtain that there exists an interval \([{a_{1}},{b_{1}}] \subset [a,b]\) such that \(|q(s)| > 0\) on \([{a_{1}},{b_{1}}]\). Then, by Lemma 3.4 and Remark 3.1, we have
$$\begin{aligned} \Vert x \Vert {_{\infty }} < & \frac{{{{[(b - a)({\alpha _{1}} + {\alpha _{2}} - 1)]}^{{\alpha _{1}} + {\alpha _{2}} - 1}}{{[{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})]}^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}}{{\tilde{\Delta }}_{1}}}}{{\varGamma ( {\alpha _{1}} + {\alpha _{2}}){{[2{\alpha _{2}} - (1 - {\alpha _{1}})(2 - {\beta _{1}})]}^{2{\alpha _{2}} - (1 - {\alpha _{1}})(2 - {\beta _{1}})}}{\Delta _{1}}}} \\ &{}\times \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} \Vert x \Vert {_{\infty }}. \end{aligned}$$
Thus, inequality (3.16) holds. This completes the proof of Theorem 3.1. □
Theorem 3.2
Assume that
\((\mathrm{{A}_{2})}\)
holds. If the fractional BVP (1.14), (1.16) has a nontrivial continuous solution for a real-valued continuous function
q, then
$$ \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds} > \frac{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]{\Delta _{2}}}}{{\max \{ {\alpha _{1}}+{\alpha _{2}}-1, {\beta _{1}}(1-{\alpha _{1}})\} {{\tilde{\Delta }}_{2}}}}, $$
(3.17)
where
$$ {\tilde{\Delta }_{2}}:={\Delta _{2}}(b-a)+\sum _{i=1}^{m-2} {{\delta _{i}}} {(b-a)^{{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1-{\alpha _{1}})}}. $$
Proof
Assume that \(x(t)\) is a nontrivial solution of BVP (1.14), (1.16), then
$$\begin{aligned} x(t) = & \int _{a}^{b}{H(t,s)q(s)x(s)\,ds}+ \frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{2}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} q(s)x(s)\,ds} \\ =& \frac{ { \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}{H_{1}}(t,s)q(s)x(s)\,ds}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}} \\ &{}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} \int _{a}^{b}{\sum_{i=1}^{m-2} {{\delta _{i}}{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}- 2}}{H_{1}}({\eta _{i}},s)q(s)x(s)\,ds}}}}{{{\Delta _{2}}\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}}, \end{aligned}$$
and
$$\begin{aligned} \bigl\vert x(t) \bigr\vert \leqslant & \frac{{ \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \vert {H_{1}}(t,s) \vert \cdot \vert q(s) \vert \cdot \vert x(s) \vert \,ds}}}{{\varGamma ({\alpha _{1}}+{\alpha _{2}})[ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]}} \\ &{}+ \frac{{{{(t-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}} \vert {H_{1}}({\eta _{i}},s) \vert \cdot \vert q(s) \vert \cdot \vert x(s) \vert \,ds}}}{ {{\Delta _{2}}\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}}. \end{aligned}$$
An argument similar to the one used in Theorem 3.1 shows that there exists an interval \([{a_{2}},{b_{2}}] \subset [a,b]\) such that \(|q(s)| > 0\) on \([{a_{2}},{b_{2}}]\). Now, applying Lemma 3.4 and Remark 3.1, we have
$$ \Vert x \Vert {_{\infty }} < \frac{{\max \{{\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1-{\alpha _{1}})\} {{\tilde{\Delta }}_{2}}}}{{\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})] {\Delta _{2}}}} \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds} \Vert x \Vert {_{\infty }}, $$
from which inequality in (3.17) follows. The proof is completed. □
Theorem 3.1 gives the following corollaries.
Corollary 3.1
The necessary condition for the existence of a nontrivial solution for BVP (1.1) is
$$ \int _{a}^{b}{ \bigl\vert q(s) \bigr\vert \,ds}> \frac{4}{{(b-a)}}. $$
(3.18)
Proof
Apply Theorem 3.1 for \({\alpha _{1}}= {\alpha _{2}}=1\), \({\sigma _{i}}=0\) (\(i = 1,2, \ldots ,m - 2\)), then (3.18) holds. Obviously, (3.18) coincides with the classical Lyapunov inequality, i.e., inequality (1.2). □
Corollary 3.2
The necessary condition for the existence of a nontrivial solution for BVP (1.5) of case (i) is
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} > \frac{{\varGamma ({\alpha }+{\beta }){2^{2( {\alpha }+{\beta }-1)}}}}{{{{(b-a)}^{{\alpha }+{\beta }-1}}}}. $$
(3.19)
Proof
Apply Theorem 3.1 for \({\alpha _{1}}= \alpha \), \({\alpha _{2}}=\beta \), \({\beta _{1}}={\beta _{2}}=0\), \({\sigma _{i}}=0\) (\(i=1,2,\ldots ,m-2\)), then (3.19) holds. Obviously, (3.19) coincides with inequality (1.6). □
Corollary 3.3
The necessary condition for the existence of a nontrivial solution for BVP (1.3) is
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\varGamma (\alpha ){ \biggl({\frac{4}{{b-a}}} \biggr) ^{\alpha -1}}. $$
(3.20)
Proof
Apply Theorem 3.1 for \({\alpha _{1}}=1\), \({\beta _{2}}=0\), \(\alpha =1+{\alpha _{2}}\), \({\sigma _{i}}=0\) (\(i=1,2, \ldots ,m-2\)), then (3.20) holds. Obviously, (3.20) coincides with inequality (1.4). □
Corollary 3.4
The necessary condition for the existence of a nontrivial solution for BVP (1.5) of case (ii) is
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma ({\alpha }+{\beta }){{({\alpha }+2 {\beta }-1)}^{{\alpha }+2{\beta }-1}}}}{{{{(b-a)}^{{\alpha }+{\beta }- 1}}{{({\alpha }+{\beta }-1)}^{{\alpha }+{\beta }-1}}\beta ^{{\beta }}}}. $$
(3.21)
Proof
Apply Theorem 3.1 for \({\alpha _{1}} =\alpha , {\alpha _{2}}=\beta , {\beta _{1}}={\beta _{2}}=1, {\sigma _{i}}=0\) (\(i=1,2,\ldots ,m-2\)), then (3.21) holds, which coincides with inequality (1.7). □
Corollary 3.5
Consider the following fractional BVP:
$$ \textstyle\begin{cases} {}^{C}D_{a+}^{\alpha }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x(b)=0, \end{cases} $$
(3.22)
where
\(q\in C([a,b], \mathbb{R})\), \({}^{C}D_{a+}^{\alpha }\)
is the left Caputo fractional derivative. If (3.22) has a nontrivial continuous solution in
\([a,b]\), then
$$ \int _{a}^{b}{ \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha ){\alpha ^{\alpha }}}}{ {{{[(b-a)(\alpha -1)]}^{\alpha -1}}}}. $$
(3.23)
Proof
Apply Theorem 3.1 for \({\alpha _{2}}=1\), \({\beta _{1}}=1\), \(\alpha ={\alpha _{1}}+1\), \({\sigma _{i}}=0\) (\(i=1,2, \ldots ,m-2T\), then (3.23) holds. Corollary 3.5 coincides with [14, Theorem 1]. □
Corollary 3.6
The necessary condition for the existence of a nontrivial solution for BVP (1.8), (1.9) is
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha ){{[\alpha -(2-\alpha )(1-\beta )]}^{\alpha -(2-\alpha )(1-\beta )}}}}{{{{(b-a)}^{\alpha -1}} {{(\alpha -1)}^{\alpha - 1}}{{[\alpha -1+\beta (2-\alpha )]}^{\alpha -1+\beta (2-\alpha )}}}}. $$
(3.24)
Proof
Apply Theorem 3.1 for \({\alpha _{2}} =1\), \(\alpha ={\alpha _{1}}+1\), \(\beta ={\beta _{1}}\), \({\sigma _{i}} =0\) (\(i=1,2,\ldots ,m-2\)), then (3.24) holds, which coincides with inequality (1.11). By Remark 3.1, we show that the non-strict inequality (1.11) can be replaced by strict inequality (3.24). □
Corollary 3.7
Assume that the following boundary value problem
$$ \textstyle\begin{cases} D_{a+}^{\alpha ,\beta }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=0, \qquad x(b)=\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}}),} \end{cases} $$
where
\(q\in C([a,b], \mathbb{R})\), \(D_{a+}^{\alpha ,\beta }\)
is the Hilfer fractional derivative of order
α
and type
\(\beta \in [0,1]\), has a nontrivial continuous solution in
\([a,b]\), then
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha ){{[2-(2-\alpha )(2- \beta )]}^{2-(2-\alpha )(2-\beta )}}{\Delta _{11}}}}{{{{(b-a)}^{\alpha -1}}{{(\alpha -1)}^{\alpha -1}}{{[1- (2-\alpha )(1-\beta )]}^{1-(2- \alpha )(1-\beta )}}{{\tilde{\Delta }}_{11}}}}, $$
(3.25)
where
$$\begin{aligned}& {\Delta _{11}}:= {(b-a)^{1-(2-\alpha )(1-\beta )}}-\sum _{i=1} ^{m-2} {{\sigma _{i}}} {({\xi _{i}}-a)^{1-(2-\alpha )(1-\beta )}}, \\& {\tilde{\Delta }_{11}}:= {\Delta _{11}}+\sum _{i=1}^{m-2} {{\sigma _{i}}} {(b-a)^{1-(2-\alpha )(1-\beta )}}. \end{aligned}$$
Proof
Apply Theorem 3.1 for \({\alpha _{2}} =1\), \(\alpha ={\alpha _{1}}+1\), \(\beta ={\beta _{1}}\), then (3.25) holds, which coincides with [30, Theorem 3.1]. □
Theorem 3.2 gives the following corollaries.
Corollary 3.8
If a nontrivial continuous solution of the following boundary value problem
$$ \textstyle\begin{cases} x''(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$
exists, where
\(q\in C([a,b], \mathbb{R})\), then
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} > \frac{1}{{b-a}}. $$
(3.26)
Proof
Apply Theorem 3.2 for \({\alpha _{1}} ={\alpha _{2}}=1\), \({\delta _{i}}=0\) (\(i=1,2,\ldots ,m-2\)), then (3.26) holds, which coincides with [16, Corollary 5]. □
Corollary 3.9
Suppose that the following boundary value problem
$$ \textstyle\begin{cases} D_{a+}^{{\alpha _{1}}}D_{a+}^{{\alpha _{2}}}x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$
where
\(q\in C([a,b], \mathbb{R})\), \(D_{a+}^{(\cdot )}\)
is the left Riemann–Liouville fractional derivative, has a nontrivial continuous solution in
\([a,b]\), then
$$ \int _{a}^{b}{{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})}}{{b-a}}. $$
(3.27)
Proof
Apply Theorem 3.2 for \({\beta _{1}}={\beta _{2}}=0\), \({\delta _{i}}=0\) (\(i=1,2,\ldots ,m-2\)), then (3.27) holds. □
Corollary 3.10
Consider the following fractional BVP:
$$ \textstyle\begin{cases} D_{a+}^{\alpha }x(t)+q(t)x(t) = 0,\quad t \in (a,b), \\ x(a) = x'(b) = 0, \end{cases} $$
(3.28)
where
\(q\in C([a,b], \mathbb{R})\), \(D_{a+}^{\alpha }\)
is the Riemann–Liouville fractional derivative of fractional order
α. If (3.28) has a nontrivial continuous solution in
\([a,b]\), then
$$ \int _{a}^{b} {{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )}}{{b-a}}. $$
(3.29)
Proof
Apply Theorem 3.2 for \({\alpha _{1}}=1\), \({\beta _{2}} =0\), \(\alpha =1+{\alpha _{2}}\), \({\delta _{i}}=0\) (\(i=1,2,\ldots ,m -2\)), then (3.29) holds. □
Corollary 3.11
Consider the following fractional BVP:
$$ \textstyle\begin{cases} {}^{C}D_{a+}^{{\alpha _{1}}}{}^{C}D_{a+}^{{\alpha _{2}}}x(t)+q(t)x(t) =0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$
(3.30)
where
\(q\in C([a,b], \mathbb{R})\), \({}^{C}D_{a+}^{(\cdot )}\)
is the left Caputo fractional derivative. If (3.30) has a nontrivial continuous solution in
\([a,b]\), then
$$ \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){\alpha _{2}}}}{{(b-a)\max \{ {\alpha _{1}}+ {\alpha _{2}}-1,1-{\alpha _{1}}\} }}. $$
(3.31)
Proof
Apply Theorem 3.2 for \({\beta _{1}}={\beta _{2}}=1, {\delta _{i}}=0\) (\(i = 1,2,\ldots ,m-2\)), then (3.31) holds. □
Corollary 3.12
Consider the following fractional BVP:
$$ \textstyle\begin{cases} {}^{C}D_{a+}^{\alpha }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$
(3.32)
where
\(q\in C([a,b], \mathbb{R})\), \({}^{C}D_{a+}^{\alpha }\)
is the left Caputo fractional derivative of order
α. If (3.32) has a nontrivial continuous solution in
\([a,b]\), then
$$ \int _{a}^{b}{{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )}}{ {(b-a)\max \{\alpha -1,2-\alpha \}}}. $$
(3.33)
Proof
Apply Theorem 3.2 for \({\alpha _{2}} =1,{\beta _{1}}=1,\alpha ={\alpha _{1}}+1,{\delta _{i}}=0\) (\(i =1,2,\ldots ,m-2\)), then (3.33) holds, which coincides with [17] and [23, Theorem 3]. □
Corollary 3.13
The necessary condition for the existence of a nontrivial solution for BVP (1.8), (1.10) is
$$ \int _{a}^{b} {{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )[\alpha -1+\beta (2-\alpha )]}}{{(b-a)\max \{\alpha -1,\beta (2- \alpha )\}}}, $$
(3.34)
Proof
Apply Theorem 3.2 for \({\alpha _{2}} =1\), \(\alpha ={\alpha _{1}}+1\), \(\beta ={\beta _{1}}\), \({\delta _{i}} =0\) (\(i=1,2,\ldots ,m-2\)), then (3.34) holds, which coincides with inequality (1.12). By Remark 3.1, we show that the non-strict inequality (1.12) can be replaced by strict inequality (3.34). □
Corollary 3.14
If a nontrivial continuous solution of the following fractional boundary value problem
$$ \textstyle\begin{cases} D_{a+}^{\alpha ,\beta }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=0,\qquad x'(b)=\sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}}),} \end{cases} $$
exists, \(q\in C([a,b], \mathbb{R})\), \(D_{a+}^{\alpha ,\beta }\)
is the Hilfer fractional derivative of order
α
and type
\(\beta \in [0,1]\), then
$$ \int _{a}^{b} {{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )[\alpha -1+\beta (2-\alpha )]{\Delta _{22}}}}{{\max \{\alpha -1, \beta (2-\alpha )\} {{\tilde{\Delta }}_{22}}}}, $$
(3.35)
where
$$\begin{aligned}& {\Delta _{22}}:= \bigl[\alpha -1+\beta (2-\alpha ) \bigr]{(b-a)^{-(2-\alpha )(1- \beta )}}-\sum_{i=1}^{m-2}{{ \delta _{i}}} {({\eta _{i}}-a)^{ \alpha -1+\beta (2-\alpha )}}, \\& {{\tilde{\Delta }}_{22}}:={\Delta _{2}}(b-a)+\sum _{i=1}^{m-2} {{\delta _{i}}} {(b-a)^{\alpha + \beta (2-\alpha )}}. \end{aligned}$$
Proof
Apply Theorem 3.2 for \({\alpha _{2}}=1\), \(\alpha = {\alpha _{1}}+1\), \(\beta = {\beta _{1}}\), then (3.35) holds. □