In this section, we show the existence of solutions for the problem (1.1) by applying some fixed point theorems.
In relation to the problem (1.1), we define the fixed point problem
where the operator \(\Psi:C(J,\mathbb{R} )\rightarrow C(J,\mathbb{R})\) is defined by
$$\begin{aligned} \Psi x(t) =&\sum_{k=0}^{n-1} \frac{(t-\theta)^{k}}{k!} \biggl( b_{k}+ \int _{t_{0}}^{\theta }g_{k} \bigl(s,x(s) \bigr)- \int _{t_{0}}^{\theta}\frac{ ( \theta -s ) ^{\alpha-k-1}}{\Gamma ( \alpha -k ) }f \bigl(s,x(s) \bigr)\,ds \biggr) \\ &{}+ \int _{t_{0}}^{t}\frac{ ( t-s ) ^{\alpha -1}}{\Gamma ( \alpha ) }f \bigl(s,x(s) \bigr)\,ds. \end{aligned}$$
(4.2)
Observe that the problem (1.1) has solutions if the operator equation (4.1) has fixed points.
Lemma 4.1
The operator
\(\Psi:C(J,\mathbb{R} )\rightarrow C(J,\mathbb{R} )\)
given by (4.2) is completely continuous.
Proof
Obviously continuity of the operator Ψ follows from the continuity of the functions f and \(g_{k}\), \(k=0,1,\ldots,n-1\). Let Ω be a bounded subset of \(C(J,\mathbb{R} )\), then for any \(t\in J\), and \(x\in\Omega\), there exist positive constants \(L_{f}\), and \(L_{k}\), \(k=0,1,\ldots,n-1\), such that \(\vert f(t,x(t))\vert \leq L_{f}\), and \(\vert g_{k}(t,x(t))\vert \leq L_{k}\). Then we have
$$\begin{aligned} &\bigl|(\Psi x) (t)\bigr| \\ &\quad\leq\sum_{k=0}^{n-1}\frac{\vert {t-\theta} \vert ^{k}}{k!} \biggl( \vert b_{k}\vert + \int _{t_{0}}^{\theta }\bigl|g_{k} \bigl(s,x(s) \bigr)\bigr| \,ds+ \int _{t_{0}}^{\theta}\frac{ ( \theta-s ) ^{\alpha-k-1}}{\Gamma ( \alpha-k ) } \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \biggr) \\ &\qquad{}+ \int _{t_{0}}^{t}\frac{ ( t-s ) ^{\alpha -1}}{\Gamma ( \alpha ) } \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \\ &\quad\leq\sum_{k=0}^{n-1}\frac{\vert {t-\theta} \vert ^{k}}{k!} \biggl( \vert b_{k}\vert + ( \theta-t_{0} ) L_{k}+\frac{ ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha -k+1 ) }L_{f} \biggr) + \frac{ ( t-t_{0} ) ^{\alpha}}{\Gamma ( \alpha +1 ) }L_{f} \\ &\quad=: L, \end{aligned}$$
which implies that \(\Vert (\Psi x)\Vert \leq L\). Furthermore,
$$\begin{aligned} & \bigl\vert (\Psi x) (t_{2})-(\Psi x) (t_{1}) \bigr\vert \\ &\quad\leq\sum_{k=0}^{n-1} \frac{\vert (t_{2}-\theta)^{k}-(t_{1}-\theta )^{k}\vert }{k!}\\ &\qquad{}\times \biggl( \vert b_{k}\vert + \int _{t_{0}}^{\theta }\bigl|g_{k} \bigl(s,x(s) \bigr)\bigr| \,ds+ \int _{t_{0}}^{\theta}\frac{ ( \theta-s ) ^{\alpha-k-1}}{\Gamma ( \alpha-k ) } \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \biggr) \\ &\qquad{}+ \int _{t_{0}}^{t_{1}}\frac{\vert ( t_{2}-s ) ^{\alpha-1}- ( t_{1}-s ) ^{\alpha-1}\vert }{\Gamma ( \alpha )} \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds + \int _{t_{1}}^{t_{2}}\frac{ ( t_{2}-s )^{\alpha -1}}{\Gamma ( \alpha )} \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \\ &\quad\leq\sum_{k=0}^{n-1} \frac{\vert (t_{2}-\theta)^{k}-(t_{1}-\theta )^{k}\vert }{k!} \biggl( \vert b_{k}\vert + ( \theta-t_{0} ) L_{k}+\frac{ ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha -k+1 ) }L_{f} \biggr) \\ &\qquad{}+\frac{L_{f}}{\Gamma(\alpha+1)} \bigl( 2\vert {t_{2}-t_{1}}\vert ^{\alpha}+ \bigl\vert (t_{2}-t_{0})^{\alpha }-(t_{1}-t_{0})^{\alpha} \bigr\vert \bigr), \end{aligned}$$
which tends to zero independent of x as \(t_{2}\rightarrow t_{1}\). This implies that Ψ is equicontinuous on J. In consequence, it follows by the Arzela-Ascoli theorem that the operator Ψ is completely continuous. □
Our first existence result is based on Krasnoselskii’s fixed point theorem (4.2).
Theorem 4.2
([33])
Let
M
be a closed convex and nonempty subset of a Banach space
X. Let
A, B
be the operators such that
-
(i)
\(Ax+By\in M\)
whenever
\(x,y\in M\);
-
(ii)
A
is compact and continuous;
-
(iii)
B
is a contraction.
Then there exists
\(z\in M\)
such that
\(z=Az+Bz\).
Theorem 4.3
Assume that
- (A1):
-
For any
\(t\in J\)
and
\(k=0,1,\ldots,n-1\), there exist positive constants
\(C_{f}\)
and
\(C_{k}\)
such that
$$\bigl\vert f(t,x)-f(t,y) \bigr\vert \leq C_{f}\vert x-y\vert ,\qquad \bigl\vert g_{k}(t,x)-g_{k}(t,y) \bigr\vert \leq C_{k}\vert x-y\vert , $$
for all
\(x, y \in\mathbb{R}\), and we can find
\({\mu}_{f}, \mu_{k}\in C(J, \mathbb{R}^{+})\)
such that
$$ \bigl\vert f(t,x) \bigr\vert \leq { \mu}_{f}(t), \qquad \bigl\vert g_{k}(t,x) \bigr\vert \leq { \mu}_{k}(t),\quad \textit{for all } x \in \mathbb{R}; $$
- (A2):
-
\(\eta<1\), where
$$ \eta=\sum_{k=0}^{n-1}\frac{(T-\theta)^{k}}{k!} \biggl[ C_{k} ( \theta-t_{0} ) +\frac{C_{f} ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) } \biggr]. $$
Then the problem (1.1) has at least one solution on
J.
Proof
Let us define a set \(B_{r}=\{x\in C(J,\mathbb{R} ):\Vert x\Vert \leq r\}\), where r is a positive constant satisfying the inequality
$$ r\geq\frac{\vert T{-t}_{0}\vert ^{\alpha}}{\Gamma(\alpha+1 )}{\Vert { \mu}_{f}\Vert }+\sum _{k=0}^{n-1}\frac{\vert T{-\theta} \vert ^{k}}{k!} \biggl( \vert b_{k}\vert +{\Vert {\mu}_{k} \Vert }+{\Vert {\mu}_{f} \Vert }\frac{ ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) } \biggr). $$
Introduce the operators Θ and Φ on \(B_{r}\) as
$$\begin{aligned}& ( \Theta x ) ( t ) = \int _{t_{0}}^{t}{\frac {{(t-s)}^{\alpha-1}}{\Gamma(\alpha)}}f \bigl( s,x ( s ) \bigr)\,ds, \\& (\Phi x) (t)=\sum_{k=0}^{n-1} \frac{(t-\theta)^{k}}{k!} \biggl( b_{k}+ \int _{t_{0}}^{\theta }g_{k} \bigl(s,x(s) \bigr) \,ds- \int _{t_{0}}^{\theta}\frac{ ( \theta -s ) ^{\alpha-k-1}}{\Gamma ( \alpha-k ) }f \bigl(s,x(s) \bigr)\,ds \biggr) . \end{aligned}$$
For \(x,y\in B_{r}\), \(t\in J\), using assumption (A1), we find that
$$\begin{aligned}[b] \bigl\vert \Theta x(t)+\Phi y(t) \bigr\vert \leq{}&\frac{ ( T{-t}_{0} ) ^{\alpha}}{\Gamma(\alpha+1 )}{\Vert {\mu}_{f}\Vert } \\ &{}+\sum_{k=0}^{n-1}\frac{\vert T-\theta \vert ^{k}}{k!} \biggl( \vert b_{k}\vert +{\Vert { \mu}_{k}\Vert }+{\Vert { \mu}_{f}\Vert }\frac{ ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) } \biggr) \le r. \end{aligned} $$
Thus, \(\Theta x+\Phi y\in B_{r}\). By assumption (A1), for \(x,y\in B_{r}\), \(t\in J\), we have
$$\begin{aligned} \bigl\vert ( \Phi x ) ( t ) - ( \Phi y ) ( t ) \bigr\vert \leq &\sum _{k=0}^{n-1}\frac{(T-\theta)^{k}}{k!} \biggl[ C_{k} ( \theta-t_{0} ) +\frac{C_{f} ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) } \biggr] \Vert x-y\Vert \\ \le&\eta \Vert x-y\Vert , \end{aligned}$$
that is, \(\Vert ( \Phi x ) - ( \Phi y ) \Vert \leq\eta \Vert x-y\Vert \). Since \(\eta<1\) by (A2), Φ is a contraction.
Continuity of f implies that the operator Θ is continuous. Also, Θ is uniformly bounded on \(B_{r} \) as
$$\Vert \Theta x\Vert \leq\frac{ ( T{-t}_{0} ) ^{\alpha}}{\Gamma(\alpha+1 )}{\Vert {\mu}_{f}\Vert }. $$
Now we show the compactness of the operator Θ. In view of assumption (A1), we define
$$ {\sup_{(t,x)\in J\times Br}} \bigl\vert {f(t,x)} \bigr\vert =f_{\max}< \infty. $$
Then, for \(t_{1},t_{2}\in J\), we have
$$\begin{aligned} \bigl| ( \Theta x ) ( t_{2} ) - ( \Theta x ) ( t_{1} )\bigr| =& \biggl\vert \int _{t_{0}}^{t_{1}}\frac{ [{(t_{2}-s)}^{\alpha -1}-{(t_{1}-s)}^{\alpha-1} ]}{\Gamma(\alpha )}{f \bigl( s,x ( s ) \bigr)\,ds} \\ &{} + \int _{t_{1}}^{t_{2}}{\frac{{(t_{2}-s)}^{q-1}}{\Gamma(\alpha )}f \bigl( s,x ( s ) \bigr)\,ds} \biggr\vert \\ \leq&\frac{f_{\max}}{\Gamma(\alpha+1)} \bigl( 2\vert {t_{2}-t_{1}}\vert ^{\alpha}+ \bigl\vert (t_{2}-t_{0})^{\alpha }-(t_{1}-t_{0})^{\alpha} \bigr\vert \bigr) , \end{aligned}$$
which is independent of x and tends to zero as \(t_{2}\rightarrow t_{1}\). So Θ is relatively compact on \(B_{r}\). Hence, by the Arzela-Ascoli theorem, Θ is compact on \(B_{r}\). Thus all the assumptions of Theorem 4.2 are satisfied. Therefore, the problem (1.1) has at least one solution on J. This completes the proof. □
Our next result deals with the uniqueness of solutions for the problem (1.1) and is based on the contraction mapping principle due to Banach.
Theorem 4.4
Assume that (A1) holds and that
\(\beta<1\), where
$$ \beta=\sum_{k=0}^{n-1} \frac{ ( T-\theta ) ^{k}}{k!} \biggl( C_{k}(\theta-t_{0})+ \frac{ ( \theta -t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) }C_{f} \biggr) +\frac{ ( T-t_{0} ) ^{\alpha}}{\Gamma ( \alpha+1 ) }C_{f}. $$
(4.3)
Then there exists a unique solution for the problem (1.1) on
J.
Proof
Setting \(\sup_{t\in J}|f(t,0)|=A_{f}\), \(\sup_{t\in J}|g_{k}(t,0)|=A_{k}\), and
$$ r\geq ( 1-\beta ) ^{-1} \Biggl[ \sum_{k=0}^{n-1} \frac{ ( T-\theta ) ^{k}}{k!} \biggl( \vert b_{k}\vert +A_{k}( \theta-t_{0})+\frac{ ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha -k+1 ) }A_{f} \biggr) + \frac{ ( T-t_{0} ) ^{\alpha }}{\Gamma ( \alpha+1 ) }A_{f} \Biggr], $$
we show that \(\Psi B_{r}\subset B_{r}\), where \(B_{r}=\{x\in C(J,\mathbb{R} ):\Vert x\Vert \leq r\}\). For \(x\in B_{r}\), we have
$$\begin{aligned} \bigl\vert \Psi x(t) \bigr\vert \leq &\sum_{k=0}^{n-1} \frac{\vert t-\theta \vert ^{k}}{k!} \biggl( \vert b_{k}\vert + \int _{t_{0}}^{\theta} \bigl\vert g_{k} \bigl(s,x(s) \bigr) \bigr\vert \,ds + \int _{t_{0}}^{\theta}\frac{ ( \theta-s ) ^{\alpha -k-1}}{\Gamma ( \alpha-k ) } \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \biggr) \\ &{}+ \int _{t_{0}}^{t}\frac{ ( t-s ) ^{\alpha -1}}{\Gamma ( \alpha ) } \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \\ \leq&\sum_{k=0}^{n-1}\frac{\vert t-\theta \vert ^{k}}{k!} \biggl( \vert b_{k}\vert + \int _{t_{0}}^{\theta} \bigl( \bigl\vert g_{k} \bigl(s,x(s) \bigr)-g_{k}(s,0) \bigr\vert + \bigl\vert g_{k}(s,0) \bigr\vert \bigr)\,ds \\ &{} + \int _{t_{0}}^{\theta}\frac{ ( \theta -s ) ^{\alpha-k-1}}{\Gamma ( \alpha-k ) } \bigl(\bigl|f \bigl(s,x(s) \bigr)-f(s,0)\bigr|+\bigl|f(s,0)\bigr| \bigr)\,ds \biggr) \\ &{}+ \int _{t_{0}}^{t}\frac{ ( t-s ) ^{\alpha -1}}{\Gamma ( \alpha ) } \bigl(\bigl|f \bigl(s,x(s) \bigr)-f(s,0)\bigr|+\bigl|f(s,0)\bigr| \bigr)\,ds \\ \leq&\sum_{k=0}^{n-1}\frac{\vert t-\theta \vert ^{k}}{k!} \biggl(\vert b_{k}\vert +(\theta-t_{0}) \bigl( C_{k}\Vert x\Vert +A_{k} \bigr) \\ &{}+\frac{ ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) } \bigl( C_{f}\Vert x\Vert +A_{f} \bigr) \biggr) +\frac{ ( t-t_{0} ) ^{\alpha}}{\Gamma ( \alpha+1 ) } \bigl( C_{f}\Vert x\Vert +A_{f} \bigr) \\ \le&\sum_{k=0}^{n-1}\frac{\vert T-\theta \vert ^{k}}{k!} \biggl(\vert b_{k}\vert +A_{k}(\theta -t_{0})+\frac{ ( \theta-t_{0} )^{\alpha-k}}{\Gamma (\alpha-k+1 )}A_{f} \biggr) + \frac{ ( T-t_{0} ) ^{\alpha}}{\Gamma ( \alpha+1 ) }A_{f} \\ &{}+ \Biggl( \sum_{k=0}^{n-1} \frac{\vert T-\theta \vert ^{k}}{k!} \biggl( C_{k}(\theta-t_{0})+ \frac{ ( \theta-t_{0} ) ^{\alpha -k}}{\Gamma ( \alpha-k+1 ) }C_{f} \biggr) +\frac{ ( T-t_{0} ) ^{\alpha}}{\Gamma ( \alpha+1 ) }C_{f} \Biggr) \Vert x\Vert \\ \leq& ( 1-\beta ) r+\beta r=r. \end{aligned}$$
Now, for \(x, y \in C(J, \mathbb{R})\) and for each \(t \in J\), we obtain
$$\begin{aligned} & \bigl\vert (\Psi x) (t)-(\Psi y) (t) \bigr\vert \\ &\quad\leq\sum_{k=0}^{n-1}\frac{\vert t-\theta \vert ^{k}}{k!} \biggl( C_{k}(\theta-t_{0})\Vert x-y\Vert + \frac{ ( \theta -t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) }C_{f}\Vert x-y\Vert \biggr) \\ &\qquad{}+\frac{ ( t-t_{0} ) ^{\alpha}}{\Gamma ( \alpha+1 ) }C_{f}\Vert x-y\Vert \\ &\quad\leq \Biggl( \sum_{k=0}^{n-1} \frac{\vert T-\theta \vert ^{k}}{k!} \biggl( C_{k}(\theta-t_{0})+ \frac{ ( \theta-t_{0} ) ^{\alpha -k}}{\Gamma ( \alpha-k+1 ) }C_{f} \biggr) +\frac{ ( T-t_{0} ) ^{\alpha}}{\Gamma ( \alpha+1 ) }C_{f} \Biggr) \Vert x-y\Vert \\ &\quad\leq{\beta} \Vert x-y\Vert , \end{aligned}$$
where β is given by (4.3). As \(\beta<1\), Ψ is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle. This completes the proof. □
Theorem 4.5
Let
\(f, g_{k}:J\times\mathbb{R} \rightarrow\mathbb{R}\) (\(k=0,1,\ldots,n-1\)) be continuous functions and let there exist positive constants
\(D_{f}\), \(E_{f}\), \(D_{k}\), \(E_{k}\), M, and
N
such that
$$\begin{aligned}& \bigl\vert f(t,x) \bigr\vert \leq D_{f}\vert x\vert +E_{f}, \qquad\bigl\vert g_{k}(t,x) \bigr\vert \leq D_{k}\vert x\vert +E_{k}, \quad\forall t \in J, x \in \mathbb{R}, \\& M=\frac{E_{f} ( T-t_{0} ) ^{\alpha}}{\Gamma ( \alpha +1 ) }+\sum_{k=0}^{n-1} \frac{\vert {T-\theta} \vert ^{k}}{k!} \biggl( \vert b_{k}\vert + ( \theta-t_{0} ) E_{k}+\frac{E_{f} ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha -k+1 ) } \biggr) >0, \\& N=\frac{D_{f} ( T-t_{0} ) ^{\alpha}}{\Gamma ( \alpha +1 ) }+\sum_{k=0}^{n-1} \frac{\vert T-\theta \vert ^{k}}{k!} \biggl( ( \theta-t_{0} ) D_{k}+\frac{D_{f} ( \theta -t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) } \biggr) < 1. \end{aligned}$$
Then the problem (1.1) has at least one solution on
J.
Proof
Define a suitable ball \(B_{R}\subset C(J,\mathbb{R} )\) with radius \(R>0\) as \(B_{R}=\{x\in C(J,\mathbb{R} ): \Vert x\Vert < R\}\), where R will be fixed later. Then it is sufficient to show that \(\Psi:\overline{B}_{R}\rightarrow C(J,\mathbb{R} )\) satisfies
$$ 0\notin(I-\lambda\Psi) ( \partial B_{R} ) , $$
(4.4)
for any \(x\in\partial B_{R}\), and \(\lambda\in{}[0,1]\). Define the homotopy
$$ h_{\lambda}(x)=H(\lambda,x)=x-\lambda\Psi x,\quad x\in C(J,\mathbb{R} ), \lambda\in [0,1]. $$
Then, by the Arzela-Ascoli theorem, \(h_{\lambda}\) is completely continuous. If (4.4) is true, then the Leray-Schauder degrees are well defined. Let I denote the identity operator. Then the homotopy invariance and normalization properties of topological degrees imply that
$$\begin{aligned} \operatorname{deg}(h_{\lambda},B_{R},0) =&\operatorname{deg} \bigl( ( I- \lambda\Psi ) ,B_{R},0 \bigr)=\operatorname{deg}(h_{1},B_{R},0) \\ =&\operatorname{deg}(h_{0},B_{R},0)=\operatorname{deg}(I,B_{R},0)=1, \end{aligned}$$
since \(0\in B_{R}\). By the nonzero property of the Leray-Schauder degree, \(h_{1}(x)=x-\Psi x=0\) for at least one \(x\in B_{R}\). In order to find R, we assume that \(x(t)=\lambda\Psi x(t)\) for some \(\lambda\in{}[0,1]\) and for all \(t\in J\). Then
$$\begin{aligned} & \bigl\vert x(t) \bigr\vert \\ &\quad= \bigl\vert \lambda\Psi x(t) \bigr\vert \leq \int _{t_{0}}^{t}\frac{ ( t-s ) ^{\alpha-1}}{\Gamma ( \alpha ) } \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \\ &\qquad{}+\sum_{k=0}^{n-1}\frac{(t-\theta)^{k}}{k!} \biggl( \vert b_{k}\vert + \int _{t_{0}}^{\theta}\bigl|g_{k} \bigl(s,x(s) \bigr)\bigr| \,ds + \int _{t_{0}}^{\theta}\frac{ ( \theta-s ) ^{\alpha-k-1}}{\Gamma ( \alpha-k ) } \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \biggr) \\ &\quad\leq\sum_{k=0}^{n-1}\frac{\vert {t-\theta} \vert ^{k}}{k!} \biggl( \vert b_{k}\vert + ( \theta-t_{0} ) \bigl( D_{k}\Vert x\Vert +E_{k} \bigr) + \frac{ ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha -k+1 ) } \bigl( D_{f}\Vert x\Vert +E_{f} \bigr) \biggr) \\ &\qquad{}+\frac{ ( t-t_{0} ) ^{\alpha}}{\Gamma ( \alpha+1 ) } \bigl( D_{f}\Vert x\Vert +E_{f} \bigr) \\ &\quad\leq\frac{E_{f} ( t-t_{0} ) ^{\alpha}}{\Gamma ( \alpha +1 ) }+\sum_{k=0}^{n-1} \frac{\vert {t-\theta} \vert ^{k}}{k!} \biggl( \vert b_{k}\vert + ( \theta-t_{0} ) E_{k}+\frac{E_{f} ( \theta-t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha -k+1 ) } \biggr) \\ &\qquad{}+ \Biggl( \frac{D_{f} ( t-t_{0} ) ^{\alpha}}{\Gamma ( \alpha +1 ) }+\sum_{k=0}^{n-1} \frac{\vert {t-\theta} \vert ^{k}}{k!} \biggl( ( \theta-t_{0} ) D_{k}+\frac{D_{f} ( \theta -t_{0} ) ^{\alpha-k}}{\Gamma ( \alpha-k+1 ) } \biggr) \Biggr) \Vert x\Vert \\ &\quad\leq M+N\Vert x\Vert , \end{aligned}$$
which, after taking the supremum norm and solving for \(\Vert x\Vert \), yields
$$ \Vert x\Vert \leq\frac{M}{1-N}. $$
Letting \(R=\frac{M-N+1}{1-N}\), (4.4) holds. This completes the proof. □
Remark 4.6
Following the method of proof employed in this section, we can obtain the existence results for nonlinear variants of problems (3.8) and (3.9).
Remark 4.7
(Special cases)
We obtain the existence results for an initial value problem with initial conditions: \(x^{(k)}(\theta )=b_{k}\), \(k=0,1,2,\ldots,n-1\) by taking \(\theta=t_{0}\) in the results of this paper. In this case, the operator given by (4.2) takes the form
$$ \Psi x(t)=\sum_{k=0}^{n-1} \frac{(t-t_{0})^{k}}{k!}b_{k} +\frac {1}{\Gamma ( \alpha ) }\int _{t_{0}}^{t} ( t-s ) ^{\alpha-1}f \bigl(s,x(s) \bigr)\,ds. $$
Further, our results correspond to the ones for a problem with classical nonlinear integral conditions:
$$x^{(k)}(T )=b_{k}+ \int _{t_{0}}^{T }g_{k} \bigl(s,x(s) \bigr) \,ds,\quad k=0,1,2,\ldots,n-1, $$
if we fix \(\theta=T\) in the obtained results.
Example 4.8
Consider the following nonlinear fractional boundary value problem:
$$ \left \{ \textstyle\begin{array}{@{}l} {}^{C}D_{0}^{\sqrt{10}}x(t)=\frac{t|x(t)|}{16(1+|x(t)|)}+\frac{7}{16},\quad t\in[0,1],\\ x^{(k)}(0.5)=1+\int _{0}^{0.5}\frac{s^{k}}{2(k+1)}\sin ( \frac{x(s)}{2} )\,ds,\quad k=0,1,2,3.\end{array}\displaystyle \right . $$
(4.5)
Here \(\alpha= \sqrt{10}\), \(\theta=0.5\), \(b_{k}=1\), \(f(t,x(t))=\frac {t|x(t)|}{16(1+|x(t)|)}+\frac{7}{16}\), and \(g_{k}(t,x(t))=\frac {t^{k}}{2(k+1)}\sin ( \frac{x(t)}{2} )\). With the given values, it is found that \({\mu}_{f}(t)=\frac{t}{16} +\frac{7}{16}\), \({\mu}_{k}(t)=\frac{t^{k}}{2(k+1)}\) with \(\Vert {\mu}_{f}\Vert =\frac{1}{2}\), \(\Vert {\mu}_{k}\Vert =\frac{1}{2(k+1)}\), \(k=0,1,2,3\), and
$$\begin{aligned} \eta =\sum_{k=0}^{3}\frac{1}{k! 2^{k}} \biggl[ \frac{1}{4(k+1)}+\frac{ ( \frac{1}{2} ) ^{\sqrt{10}-k}}{16 \Gamma (\sqrt{10}-k+1 ) } \biggr] \simeq0.333942< 1, \end{aligned}$$
that is, the assumption (A2) of Theorem 4.3 is satisfied. Thus, all the conditions of Theorem 4.3 are satisfied. Hence the problem (4.5) has a solution on \([0,1]\). Also β given by (4.3) is such that \(\beta\simeq0.342407<1\). This suggests that the problem (4.5) has a unique solution by the conclusion of Theorem 4.4.