Existence Results for an m-Point Mixed Fractional-Order Problem at Resonance on the Half-Line

Abstract

This work considers the existence of solutions for a mixed fractional-order boundary value problem at resonance on the half-line. The Mawhin’s coincidence degree theory will be used to prove existence results when the dimension of the kernel of the linear fractional differential operator is equal to two. An example is given to demonstrate the main result obtained.

1. Introduction

Fractional calculus has become increasingly popular lately as a result of some interesting properties of the fractional derivative. For instance, the fractional derivative has a memory property that enables its future state to be determined by the current state and all the previous states. This makes fractional differential equations applicable in various fields of science and engineering [1,2,3].

When the corresponding homogeneous equation of a fractional boundary value problem (FBVP) has a trivial solution then the FBVP is a non-resonance problem and its solution can be obtained using fixed point theorems, see [4,5,6,7] and the references cited therein. When the homogeneous equation of a FBVP has a non-trivial solution then the problem is a resonance problem and the solution can be obtained using topological degree methods [8,9,10,11,12,13,14,15].

In [16], the authors consider a higher-order fractional boundary value problem involving mixed fractional derivatives:

$${(-1)}^m{}^C{D}_{1-}^{\alpha }{D}_{1+}^{\beta }+f(t,u\left(t\right))=0,0\le t\le 1,$$

$$u\left(0\right)=u^{\left(i\right)}\left(0\right)=0,i=1,\cdots ,m+n-2,D_{0+}^{\beta +m-1}u\left(1\right)=0,$$

where ${}^C{D}_{1-}^{\alpha }$ is the left Caputo fractional derivative of order $\alpha \in (m-1,m)$ and $D_{1+}^{\beta }$ is the right Caputo fractional derivative of order $\beta \in (n-1,n)$, where $m,n\ge 2$ are integers.

Guezane Lakoud et al. [17] obtained existence results for a fractional boundary value problem at resonance on the half-line:

$${-}^C{D}_{0-}^{\alpha }{D}_{0+}^{\beta }x\left(t\right)+f(t,x\left(t\right))=0,t\in [0,1],$$

$$u\left(0\right)=u^{\prime }\left(0\right)=u\left(1\right)=0,$$

where ${-}^C{D}_{0-}^{\alpha }$ is the left Caputo fractional derivative of order $\alpha \in (0,1]$, and $D_{0+}^{\beta }$ is the right Caputo fractional derivative of order $\beta \in (1,2]$.

Zhang and Liu [15] considered the following FBVP

$$D_{0+}^{\alpha }x\left(t\right)=f(t,x\left(t\right),D_{0+}^{\alpha -2}x\left(t\right),D_{0+}^{\alpha -1}x\left(t\right)),t\in (0,1),$$

$$x\left(0\right)=0,D_{0+}^{\alpha -1}x\left(0\right)=\sum _{i=1}^{+\infty }{\alpha }_i{D}_{0+}^{\alpha -1}x\left({\xi }_i\right),D_{0+}^{\alpha -1}x\left(1\right)=\sum _{i=1}^{+\infty }{\alpha }_i{D}_{0+}^{\alpha -1}x\left({\gamma }_i\right),$$

where $2<\alpha \le 3$, $D_{0+}^{\alpha }$ is the Riemann–Liouville derivative of order $\alpha$, $f\in [0,1]\times {\mathbb{R}}^3\to \mathbb{R}$ is a Caratheodory function, ${\xi }_i,{\gamma }_i\in (0,1)$ and ${\left\{{\xi }_i\right\}}_{i=1}^{+\infty },{\left\{{\gamma }_i\right\}}_{i=1}^{+\infty }$ are two monotonic sequences with ${lim}_{i\to +\infty }{\xi }_i=a,li{m}_{i\to +\infty }{\gamma }_i=b$, $a,b\in (0,1)$, ${\alpha }_i,{\beta }_{i}f\in \mathbb{R}$.

Imaga et al. [18] obtained existence results for the following fractional-order boundary value problem at resonance on the half-line with integral boundary conditions:

$$D_{-}^a{\varphi }_p\left(D_{0+}^{b}u\left(t\right)\right)+e^{-t}w(t,u\left(t\right),D_{0^{+}}^{b}u\left(t\right))=0,t\in (0,\infty ),$$

(1)

$$I_{0^{+}}^{1-b}u\left(0\right)=0,{\varphi }_p\left(D_{0^{+}}^{b}u(+\infty )\right)={\varphi }_p\left(D_{0+}^{b}u\left(0\right)\right),$$

(2)

where $D_{-}^a$ is the left Caputo fractional derivative on the half line and $D_{0+}^b$ the right Riemann–Louville fractional derivative on the half-line, $0<a,b\le 1$, $1<a+b\le 2$, ${\varphi }_p\left(r\right)={\left|r\right|}^{p-2},p>1$, with ${\varphi }_q={\varphi }_p^{-1}$ and $1/q+1/p=1$. $w:[0,+\infty )\times {\mathbb{R}}^2\to \mathbb{R}$ is a continuous function.

Chen and Tang [9] established existence of positive solutions for a FBVP at resonance in an unbounded domain:

$$D_{0+}^{\alpha }u\left(t\right)=f(t,u\left(t\right)),t\in [0,+\infty ),$$

$$u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(0\right)=0,D_{0+}^{\alpha -1}u\left(0\right)=\underset{t\to +\infty }{lim}{D}_{0+}^{\alpha -1}u\left(t\right),$$

where $D_{0+}^{\alpha }$ is Riemann–Liouville fractional derivative, $3<\alpha <4$ and $f:[0,+\infty )\times \mathbb{R}\to \mathbb{R}$ is continuous.

Motivated by the results above, we will use the Mawhin coincidence degree theory [19] to study the solvability of the following mixed fractional-order m-point boundary value problem at resonance on the half-line:

$${}^C{D}_{0^{+}}^a{D}_{0^{+}}^{b}u\left(t\right)=f(t,u\left(t\right),D_{0^{+}}^{b-1}u\left(t\right),D_{0^{+}}^{b}u\left(t\right)),t\in [0,+\infty )$$

(3)

$$I_{0^{+}}^{2-b}u\left(0\right)=0,D_{0+}^{b-1}u\left(0\right)=\sum _{j=1}^m{\alpha }_j{D}_{0^{+}}^{b-1}u\left({\xi }_j\right),D_{0^{+}}^{b}u(+\infty )=\sum _{k=1}^n{\beta }_k{D}_{0^{+}}^{b}u\left({\eta }_k\right)$$

(4)

where $f:[0,+\infty )\times {\mathbb{R}}^3\to \mathbb{R}$ is a continuous function, ${}^C{D}_{0^{+}}^a$ is the Caputo fractional derivative, $D_{0^{+}}^b$ is the Riemann–Liouville fractional derivative, $0<a\le 1$, $1<b\le 2$, $0<a+b\le 3$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_m<+\infty$, $0<{\eta }_1<{\eta }_2<\cdots <{\xi }_m<+\infty$, ${\alpha }_j\in \mathbb{R},j=1,2,\cdots ,m$ and ${\beta }_k\in \mathbb{R},k=1,2,\cdots ,n$. The resonant conditions are ${\sum }_{k=1}^n{\beta }_k={\sum }_{j=1}^m{\alpha }_j=1$ and ${\sum }_{k=1}^n{\beta }_k{\eta }_k^{-1}={\sum }_{j=1}^m{\alpha }_j{\xi }_j^{-1}=0$.

In Section 2 of this work the required lemmas, theorem, and definitions will be given, while Section 3 is dedicated to stating and proving the main existence results. An example will be given in Section 4.

2. Materials and Methods

In this section, we will give some definitions and lemmas that will be used in this work.

Let U, Z be normed spaces, $L:domL\subset U\to Z$ a Fredholm mapping of zero index and $A:U\to U,B:Z\to Z$ projectors that are continuous, such that:

$$ImA=kerL,kerB=ImL,U=kerL\oplus kerA,Z=ImL\oplus ImB.$$

Then,

$${L|}_{domL\cap kerA}:domL\cap kerA\to ImL$$

is invertible. The inverse of the mapping L will be denoted by $K_A:ImL\to domL\cap kerA$ while the generalized inverse, $K_{A,B}:Z\to domL\cap kerA$ is defined as $K_{A,B}=K_A(I-B)$.

Definition 1.

Let $L:domL\subset X\to Z$ be a Fredholm mapping, E a metric space and $N:E\to Z$ a non-linear mapping. N is said to be L-compact on E if $BN:E\to Z$ and $K_{A,B}N:E\to X$ are continuous and compact on E. Additionally, N is L-completely continuous if it is L-compact on every bounded $E\subset U$.

Theorem 1.

([19]). Let L be a Fredholm map of index zero and let N be L-compact on $\overline{\mathrm{\Omega }}$ where $\mathrm{\Omega }\subset U$ is an open and bounded. Assume that the following conditions are satisfied:

(i) 

$Lx\ne \lambda Nx$ for every $(x,\lambda )\in \left[\right(domLkerL\cap \partial \mathrm{\Omega }]\times (0,1)$;

(ii) 

$Nx\notin ImL$ for every $x\in kerL\cap \partial \mathrm{\Omega }$;

(iii) 

$deg(BN{|}_{kerL},kerL,0)\ne 0$, where $B:Z\to Z$ is a projection with $ImL=kerB$.

Then, the abstract equation $Lu=Nu$ has at least one solution in $domL\cap \overline{\mathrm{\Omega }}$.

Definition 2.

([20]). Let $\alpha >0$, the Caputo and Riemann–Liouville fractional integral of a function x on $(0,+\infty )$ is defined by:

$$I_{0+}^{\alpha }x\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^t\frac{x\left(r\right)}{{(r-t)}^{1-\alpha }}dr,t\in [0,1]$$

Definition 3.

([20]). Let $\alpha >0$, the Caputo $\left({}^C{D}_{0+}^{\alpha }x\left(t\right)\right)$ and Riemann–Liouville $\left(D_{0+}^{\alpha }x\left(t\right)\right)$ fractional derivative of a function x on $(0,+\infty )$ is defined by:

$${}^C{D}_{0+}^{\alpha }x\left(t\right)=D_{0+}^{\alpha }x\left(t\right)=\frac{1}{\mathrm{\Gamma }(n-\alpha )}\frac{d^n}{d{t}^n}{\int }_0^t\frac{x\left(r\right)}{{(t-r)}^{\alpha -n+1}}dr,t\in (0,+\infty )$$

where $n=\left[a\right]+1$.

Lemma 1.

([21]). Let $a\in (0,+\infty )$. The general solution of the Riemman–Liouville fractional differential equation:

$$D_{0^{+}}^{a}g\left(t\right)=0$$

is $g\left(t\right)=b_1{t}^{a-1}+b_2{t}^{a-2}+\cdots +b_n{t}^{a-n}$, where $b_j\in \mathbb{R},j=1,2\cdots ,n$ while, the general solution of the Caputo fractional differential equation:

$$D_{0^{+}}^{a}g\left(t\right)=0$$

is $g\left(t\right)=d_0+d_{1}t+\cdots +d_n{t}^n$, where $d_i\in \mathbb{R},i=0,1,\cdots ,n$ and $n=\left[a\right]+1$ is the smallest integer greater than or equal to a.

Lemma 2.

([21]). Let $a\in (0,+\infty )$ and $i=1,2,\cdots ,n$, $n=\left[a\right]+1$ then

$$\left(I_{0^{+}}^a{D}_{0^{+}}^{a}g\right)\left(t\right)=g\left(t\right)+d_1{t}^{a-1}+d_2{t}^{a-2}+\cdots +d_n{t}^{a-n}$$

holds almost everywhere on $[0,+\infty )$ for some $d_i\in \mathbb{R}$. Similarly,

$$\left(I_{0^{+}}^a{}^C{D}_{0^{+}}^{a}g\right)\left(t\right)=g\left(t\right)+d_0+d_1{t}^1+d_2{t}^2+\cdots +d_n{t}^n$$

holds almost everywhere on $[0,+\infty )$ for some $d_i\in \mathbb{R}$, $i=0,1,\cdots ,n$.

Lemma 3.

([21]). Let $a>0,\rho >-1,t>0$, $g\left(t\right)\in C[0,+\infty )$, then:

(i) 

$I_{0^{+}}^a{t}^{\rho }=\frac{\mathrm{\Gamma }(\rho +1)}{\mathrm{\Gamma }(\rho +1+a)}{t}^{a+\rho }$;

(ii) 

$D_{0^{+}}^a{t}^{\rho }=\frac{\mathrm{\Gamma }(\rho +1)}{\mathrm{\Gamma }(\rho +1-a)}{t}^{a-\rho }$, for $\rho >-1$, in particular for $D_{0^{+}}^a{t}^{a-k}=0$, $k=1,2,\cdots ,N$, where N is the smallest integer greater than or equal to a;

(iii) 

$D_{0^{+}}^a{I}_{0+}^{a}g\left(t\right)=g\left(t\right)$, $g\left(t\right)\in C[0,+\infty )$;

(iv) 

$I_{0+}^a{I}_{0+}^{b}g\left(t\right)=I_{0+}^{a+b}g\left(t\right)$.

Let

$$\begin{array}{c}\hfill U=\left\{u\in C[0,+\infty ):\underset{t\to +\infty }{lim}\frac{\left|u\right(t\left)\right|}{1+t^{a+b}},\underset{t\to +\infty }{lim}\frac{|D_{0^{+}}^{b-1}u\left(t\right)|}{1+t^{a+1}}\mathrm{and}\underset{t\to +\infty }{lim}\frac{|D_{0^{+}}^{b}u\left(t\right)|}{1+t^a}\mathrm{exists}\right\}\end{array}$$

with the norm ${\parallel u\parallel }_U={max\{\parallel u\parallel }_0,\parallel D_{0^{+}}^{b-1}{u\parallel }_1,\parallel D_{0^{+}}^{b}u{\parallel }_2\}$ defined on U where:

$${\parallel u\parallel }_0=\underset{t\in [0,+\infty ]}{sup}\frac{\left|u\right(t\left)\right|}{1+t^{a+b}},\parallel D_{0+}^{b-1}{u\parallel }_1=\underset{t\in [0,+\infty ]}{sup}\frac{|D_{0+}^{b-1}u\left(t\right)|}{1+t^{a+1}}\mathrm{and}{\parallel D_{0+}^{b}u\parallel }_2=\underset{t\in [0,+\infty ]}{sup}\frac{|D_{0+}^{b}u\left(t\right)|}{1+t^a}.$$

Let $Z=\{z:C[0,+\infty ):{sup}_{t\in [0,+\infty )}|z\left(t\right)|<+\infty \}$ equipped with the norm ${\parallel z\parallel }_Z={sup}_{t\in [0,+\infty )}\left|z\left(t\right)\right|$. The spaces $(U,\parallel ·{\parallel }_U)$ and $(Z,\parallel ·{\parallel }_Z)$ can be shown to be Banach Spaces. Additionally, define $Lu={}^C{D}_{0+}^a{D}_{0^{+}}^{b}u\left(t\right)$, with domain

$$\begin{array}{cc}\hfill \mathrm{dom}L=\{& u\in U:{}^C{D}_{0+}^a{D}_{0^{+}}^{b}u\left(t\right)\in Z,\mathrm{boundary}\mathrm{conditions}\left(4\right)\mathrm{is}\mathrm{satisfied}\mathrm{by}u\},\hfill \end{array}$$

and the non-linear operator $N:U\to Z$ will be defined by

$$\left(Nu\right)t=f(t,u\left(t\right),D_{0^{+}}^{b-1}u\left(t\right),D_{0^{+}}^{b}u\left(t\right)),t\in [0,+\infty ),$$

hence, Equations (3) and (4) may be written as

$$Lu=Nu.$$

Definition 4.

The set $Y\subset U$ is said to be relatively compact if

$$Y_1=\left\{\frac{u\left(t\right)}{1+t^{a+b}}:u\in Y\right\},Y_2=\left\{\frac{D_{0+}^{b-1}u\left(t\right)}{1+t^{a+1}}:u\in Y\right\},Y_3=\left\{\frac{D_{0+}^{b}u\left(t\right)}{1+t^a}:u\in Y\right\}$$

are uniformly bounded; equicontinuous on any compact subinterval of $[0,+\infty )$ and equiconvergent at: $+\infty$.

Definition 5.

The set $Y\subset U$ is said to be equiconvergent at $+\infty$ if given $ϵ>0$ there exists a $\tau \left(ϵ\right)>0$, such that:

$$\left|\frac{u\left(t_1\right)}{1+t_1^{a+b}}-\frac{u\left(t_2\right)}{1+t_2^{a+b}}\right|<ϵ,\left|\frac{D_{0+}^{b-1}u\left(t_1\right)}{1+t_1^{a+1}}-\frac{D_{0+}^{b-1}u\left(t_2\right)}{1+t_2^{a+1}}\right|<ϵand\left|\frac{D_{0+}^{b}u\left(t_1\right)}{1+t_1^a}-\frac{D_{0+}^{b}u\left(t_2\right)}{1+t_2^a}\right|<ϵ$$

where $t_1,t_2>\tau$.

Lemma 4.

$kerL=\{c_1{t}^b+c_2{t}^{b-1}:c_1,c_2\in \mathbb{R},t\in [0,+\infty )\}$ and $\mathrm{Im}L=\{z\in Z:B_{1}z=B_{2}z=0\}$

where ${\displaystyle B_{1}z=\sum _{k=1}^n{\beta }_k{\int }_0^{{\eta }_k}{({\eta }_k-r)}^{a-1}z\left(r\right)dr}$ and ${\displaystyle B_{2}z=\sum _{j=1}^m{\alpha }_j{\int }_0^{{\xi }_j}{({\xi }_j-r)}^{a}z\left(r\right)dr}$.

Proof. 

Consider ${}^C{D}_{0+}^a{D}_{0^{+}}^{b}u\left(t\right)=0$ for $u\in kerL$, then by Lemma 1

$$u\left(t\right)=c_1{t}^b+c_2{t}^{b-1}+c_3{t}^{b-2},c_1,c_2,c_3\in \mathbb{R}.$$

Applying the boundary condition $I_{0^{+}}^{2-b}u\left(0\right)=0$, gives $c_3=0$. Thus, $u\left(t\right)=c_1{t}^b+c_2{t}^{b-1}$. Next, consider ${}^C{D}_{0+}^a{D}_{0^{+}}^{b}u\left(t\right)=z\left(t\right)$ for $z\left(t\right)\in \mathrm{Im}L$ and $u\in \mathrm{dom}L$, then

$$u\left(t\right)=I_{0+}^{a+b}z\left(t\right)+c_1{t}^b+c_2{t}^{b-1}+c_3{t}^{b-2}.$$

From $I_{0+}^{2-b}u\left(0\right)=0$ we obtain $c_3=0$. Therefore,

$$D_{0+}^{b}u\left(t\right)=I_{0+}^{a}z\left(t\right)+c_1+c_2{t}^{-1}$$

(5)

By boundary condition $D_{0+}^{b}u(+\infty )={\sum }_{k=1}^n{\beta }_k{D}_{0+}^{b}u\left({\eta }_k\right)$ and the conditions ${\sum }_{k=1}^n{\beta }_k=1$, ${\sum }_{k=1}^n{\beta }_k{\eta }_k^{-1}=0$, (5) gives

$$B_{1}z=\sum _{k=1}^n{\beta }_k{\int }_0^{{\eta }_k}{({\eta }_k-r)}^{a-1}z\left(r\right)dr=0,$$

Similarly,

$$D_{0+}^{b-1}u\left(t\right)=I_{0+}^{a+1}z\left(t\right)+c_{1}t+c_2,$$

(6)

by boundary condition $D_{0+}^{b-1}u\left(0\right)={\sum }_{j=1}^m{\alpha }_j{D}_{0^{+}}^{b-1}u\left({\xi }_j\right)$ and resonant conditions ${\sum }_{j=1}^m{\alpha }_j=1$ and ${\sum }_{j=1}^m{\alpha }_j{\xi }_j^{-1}=0$, (6) gives

$$B_{2}z=\sum _{j=1}^m{\alpha }_j{\int }_0^{{\xi }_j}{({\xi }_j-r)}^{a}z\left(r\right)dr.$$

□

Let $\Delta =(B_1{t}^{b-1}{e}^{-t}·B_2{t}^b{e}^{-t})-(B_2{t}^{b-1}{e}^{-t}·B_1{t}^b{e}^{-t}):=(g_{11}·g_{22})-(g_{21}·g_{12})\ne 0$. Let the operator $B:Z\to Z$ be defined as

$$Bz=\left({\Delta }_{1}z\right)+\left({\Delta }_{2}z\right)·t^b$$

where

$${\Delta }_{1}z=\frac{1}{\Delta }({\delta }_{11}{B}_{1}z+{\delta }_{12}{B}_{2}z)e^{-t},{\Delta }_{2}z=\frac{1}{\Delta }({\delta }_{21}{B}_{1}z+{\delta }_{22}{B}_{2}z)e^{-t},$$

and ${\delta }_{ij}$ is the algebraic cofactor of $g_{ij}$.

Lemma 5.

The following holds:

(i) 

$L:\mathrm{dom}L\subset U$ is a Fredholm operator of index zero;

(ii) 

the generalized inverse $K_A:\mathrm{Im}L\to \mathrm{dom}L\cap kerA$ may be written as

$$K_{A}z=I_{0+}^{a+b}z\left(t\right).$$

Additionally,

$$\parallel K_A{z\parallel =\parallel z\parallel }_Z.$$

Proof. 

(i) For $z\in Z$, it is easily be seen that ${\Delta }_1\left(\left({\Delta }_{1}z\right)\right)=\left({\Delta }_{1}z\right)$, ${\Delta }_1\left(\left({\Delta }_{2}z\right)t^b\right)=0$, ${\Delta }_2\left(\left({\Delta }_{1}z\right)\right)=0$, and ${\Delta }_2\left(\left({\Delta }_{2}z\right)t^b\right)=\left({\Delta }_{2}y\right)$. Hence, $B^{2}z=Bz$, thus $Bz$ is a projector.

We now prove that $kerB=\mathrm{Im}L$. Let $z\in kerB$, since $Bz=0$ then $z\in \mathrm{Im}L$. Conversely, if $z\in \mathrm{Im}L$, then by $Bz=0$, $z\in kerB$. Therefore, $kerB=\mathrm{Im}L$.

Let $z\in Z$, then $z\in \mathrm{Im}L$ and $z\in kerB$, hence, $Z=\mathrm{Im}L+kerB$. Assuming $z=c_1{t}^{b-1}+c_2{t}^b$, then since $z\in \mathrm{Im}L$, then from equation

$$\{\begin{array}{c}{\Delta }_1{c}_1{t}^{b-1}{e}^{-t}+{\Delta }_2{c}_2{t}^{b-1}{e}^{-t}=0,\hfill \\ {\Delta }_1{c}_1{t}^b{e}^{-t}+{\Delta }_2{c}_2{t}^b{e}^{-t}=0.\hfill \end{array}$$

(7)

gives $c_1=c_2=0$, since $\Delta \ne 0$. Therefore $\mathrm{Im}L\cap \mathrm{Im}B=\left\{0\right\}$ and $A=\mathrm{Im}L\oplus \mathrm{Im}B$. Thus $dimkerL=\mathrm{codim}\mathrm{Im}L=2$ implying L is a Fredholm mapping of index zero.

(ii) Let $A:U\to U$ a continuous projector be defined as:

$$Au\left(t\right)=\frac{D_{0+}^{b}u\left(0\right)}{\mathrm{\Gamma }\left(b\right)}{t}^{b-1}+\frac{D_{0+}^{b}u\left(0\right)}{\mathrm{\Gamma }(b+1)}{t}^b$$

For $z\in \mathrm{Im}L$, we have

$$\left(L{K}_A\right)z\left(t\right)={}^C{D}_{0+}^a{D}_{0+}^b\left(K_{A}z\right)={}^C{D}_{0+}^a{D}_{0+}^b{I}_{0+}^b{I}_{0+}^{a}z\left(t\right)=z\left(t\right).$$

Similarly, for $u\in \mathrm{dom}L\cap kerA$, we have

$$\begin{array}{cc}\hfill \left(K_{A}L\right)u\left(t\right)& =\left(K_A\right){}^C{D}_{0+}^a{D}_{0+}^{b}u\left(t\right)\hfill \\ \hfill & =I_{0+}^b{I}_{0+}^a{}^C{D}_{0+}^a{D}_{0+}^{b}u\left(t\right)\hfill \\ & =I_{0+}^b(D_{0+}^{b}u\left(t\right)+d_1)\hfill \\ \hfill & =u\left(t\right)-\frac{D_{0+}^{b-1}u\left(0\right)}{\mathrm{\Gamma }\left(b\right)}{t}^{b-1}-\frac{I_{0+}^{2-b}u\left(0\right)}{\mathrm{\Gamma }(b-1)}{t}^{b-2}-\frac{D_{0+}^{b}u\left(0\right)}{\mathrm{\Gamma }(b+1)}{t}^b.\hfill \end{array}$$

Since $u\in \mathrm{dom}L\cap kerA$, $Au\left(t\right)=0$ and $I_{0+}^{2-b}u\left(0\right)=0$, then $\left(K_{A}L\right)u\left(t\right)=u\left(t\right).$ Therefore, $K_A={\left(L{|}_{\mathrm{dom}L\cap kerA}\right)}^{-1}$. Furthermore,

$$\begin{array}{cc}\hfill \parallel K_A{z\parallel }_0& =\underset{t\in [0,+\infty )}{sup}\frac{|I_{0+}^{a+b}z\left(t\right)|}{1+t^{a+b}}=\underset{t\in [0,+\infty )}{sup}\frac{1}{1+t^{a+b}}\left|\frac{1}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\int }_0^t{(t-r)}^{a+b-1}z\left(r\right)dr\right|\hfill \\ \hfill & \le \frac{1}{(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\parallel z\parallel }_Z\le {\parallel z\parallel }_Z,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel D_{0+}^{b-1}{K}_P{z\parallel }_1& =\underset{t\in [0,+\infty )}{sup}\frac{|I_{0+}^{a+1}z\left(t\right)|}{1+t^{a+1}}=\underset{t\in [0,+\infty )}{sup}\frac{1}{1+t^{a+1}}\left|\frac{1}{\mathrm{\Gamma }(a+1)}{\int }_0^t{(t-r)}^{a}z\left(r\right)dr\right|\hfill \\ \hfill & \le \frac{1}{(a+1)\mathrm{\Gamma }(a+1)}{\parallel z\parallel }_Z\le {\parallel z\parallel }_Z\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \parallel D_{0+}^b{K}_A{z\parallel }_2& =\underset{t\in [0,+\infty )}{sup}\frac{|I_{0+}^{a}z\left(t\right)|}{1+t^a}=\underset{t\in [0,+\infty )}{sup}\frac{t^a}{1+t^a}\frac{{\parallel z\parallel }_Z}{\mathrm{\Gamma }(a+1)}\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }(a+1)}{\parallel z\parallel }_Z\le {\parallel z\parallel }_Z.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill \parallel K_{A}z\parallel & =max\{\parallel K_A{z\parallel }_0,\parallel D_{0+}^{b-1}{K}_A{z\parallel }_1,\parallel D_{0+}^b{K}_A{z\parallel }_2{\}\le \parallel z\parallel }_Z.\hfill \end{array}$$

Proof of Lemma 5 is complete. □

Lemma 6.

The operator N is L-compact on $\overline{\mathrm{\Omega }}$, where $\mathrm{\Omega }\subset U$ is open and bounded with $\mathrm{dom}$$L\cap \overline{\mathrm{\Omega }}\ne \varnothing$.

Proof. 

Let $u\in \overline{\mathrm{\Omega }}$ then

$${\parallel Nu\parallel }_Z=\underset{t\in [0,+\infty )}{sup}\left|f(t,u\left(t\right),D_{0^{+}}^{b-1}u\left(t\right),D_{0^{+}}^{b}u\left(t\right))\right|<+\infty ,t\in [0,+\infty ).$$

(8)

It follows that

$$|B_{1}Nu|=|\sum _{k=1}^n{\beta }_k{\int }_0^{{\eta }_k}{({\eta }_k-r)}^{a-1}Nu\left(r\right)dr|\le \frac{{\parallel Nu\parallel }_Z}{a}\sum _{k=1}^n\left|{\beta }_k\right|{\eta }_k^a<+\infty$$

(9)

and

$$|B_{2}Nu|=\left|\sum _{j=1}^m{\alpha }_j{\int }_0^{{\xi }_j}{({\xi }_j-r)}^{a}Nu\left(r\right)drds\right|\le \frac{{\parallel Nu\parallel }_Z}{(a+1)}\sum _{j=1}^m\left|{\alpha }_j\right|{\xi }_j^{a+1}<+\infty .$$

(10)

Then,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\parallel BNu\parallel }_Z& =\underset{t\in [0,+\infty )}{sup}|\left({\Delta }_{1}Nu\left(t\right)\right)+\left({\Delta }_{2}Nu\left(t\right)\right)|\hfill \\ \hfill & \le \frac{{\parallel Nu\parallel }_Z}{|\Delta |}\left[\left(\right|{\delta }_{11}|+|{\delta }_{21}\left|\right)\frac{1}{a}\sum _{k=1}^n|{\beta }_k|{\eta }_k^a+\left(\right|{\delta }_{12}|+|{\delta }_{22}\left|\right)\frac{1}{(a+1)}\sum _{j=1}^m\left|{\alpha }_j\right|{\xi }_j^{a+1}\right]<+\infty .\hfill \end{array}\end{array}$$

(11)

Therefore, $BN\left(\overline{\mathrm{\Omega }}\right)$ is bounded. In addition, ${\parallel Nu\parallel }_Z+{\parallel BNu\parallel }_Z<+\infty$. In the following steps, we show that $K_A(I-B)N\left(\overline{\mathrm{\Omega }}\right)$ is compact. Let $u\in \overline{\mathrm{\Omega }}$ and $m\left(t\right)=(I-B)Nu\left(t\right)$, then:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \frac{|K_A(I-B)Nu\left(t\right)|}{1+t^{a+b}}& =\frac{|I_{0+}^{a+b}m\left(t\right)|}{1+t^{a+b}}\le \underset{t\in [0,+\infty )}{sup}\frac{t^{a+b}}{1+t^{a+b}}\frac{{\parallel m\parallel }_Z}{(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}\hfill \\ \hfill & \le \frac{1}{(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\parallel m\parallel }_Z,\hfill \end{array}\end{array}$$

(12)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \frac{|D_{0+}^{b-1}{K}_A(I-B)Nu\left(t\right)|}{1+t^{a+1}}& =\frac{|I_{0+}^{a+1}m\left(t\right)|}{1+t^{a+1}}\underset{t\in [0,+\infty )}{sup}\frac{t^{a+1}}{1+t^{a+1}}\frac{{\parallel m\parallel }_Z}{(a+1)\mathrm{\Gamma }(a+1)}\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }(a+2)}{\parallel m\parallel }_Z\hfill \end{array}\end{array}$$

(13)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \frac{|D_{0+}^b{K}_A(I-B)Nu\left(t\right)|}{1+t^a}& =\frac{|I_{0+}^{a}m\left(t\right)|}{1+t^a}\le \underset{t\in [0,+\infty )}{sup}\frac{t^a}{1+t^a}\frac{{\parallel m\parallel }_Z}{\mathrm{\Gamma }(a+1)}\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }(a+1)}{\parallel m\parallel }_Z.\hfill \end{array}\end{array}$$

(14)

From (8), (11)–(14), we see that $K_A(I-B)N\left(\overline{\mathrm{\Omega }}\right)$ is bounded. Next, the equi-continuity of $K_A(I-B)N\left(\overline{\mathrm{\Omega }}\right)$ will be proved. For $u\in \overline{\mathrm{\Omega }}$, $t_1,t_2\in [0,M]$ with $t_1<t_2$ and $M\in (0,+\infty )$, then:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |& \frac{K_A(I-B)Nu\left(t_1\right)}{1+t_1^{a+b}}-\frac{K_A(I-B)Nu\left(t_2\right)}{1+t_2^{a+b}}|\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }(a+b)}\left[|{\int }_0^{t_1}\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}m\left(r\right)dr-{\int }_0^{t_1}\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}m\left(r\right)dr|\right]\hfill \\ \hfill & \le \frac{{\parallel m\parallel }_Z}{\mathrm{\Gamma }(a+b)}\left[{\int }_0^{t_1}\left|\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}-\frac{{(t_2-r)}^{a+b-1}}{1+t_2^{a+b}}\right|dr+\frac{1}{a+b}\frac{{(t_2-t_1)}^{a+b}}{1+t_2^{a+b}}\right]\hfill \\ \hfill & \to 0\mathrm{as}{t}_1\to t_2,\hfill \end{array}\end{array}$$

(15)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |& \frac{D_{0+}^{b-1}\left(K_A(I-B)Nu\right)\left(t_1\right)}{1+t_1^{a+1}}-\frac{D_{0+}^{b-1}\left(K_A(I-B)Nu\right)\left(t_2\right)}{1+t_2^{a+1}}|\hfill \\ \hfill & \le \frac{{\parallel m\parallel }_Z}{\mathrm{\Gamma }(a+1)}\left[{\int }_0^{t_1}\left|\frac{{(t_1-r)}^a}{1+t_1^{a+1}}-\frac{{(t_2-r)}^a}{1+t_2^{a+1}}\right|dr+\frac{1}{a+1}\frac{{(t_2-t_1)}^{a+1}}{1+t_2^{a-1}}\right]\hfill \\ \hfill & \to 0\mathrm{as}{t}_1\to t_2,\hfill \end{array}\end{array}$$

(16)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |& \frac{D_{0+}^b\left(K_A(I-B)Nu\right)\left(t_1\right)}{1+t_1^a}-\frac{D_{0+}^b\left(K_A(I-B)Nu\right)\left(t_2\right)}{1+t_2^a}|\hfill \\ \hfill & \le \frac{{\parallel m\parallel }_Z}{\mathrm{\Gamma }\left(a\right)}\left[{\int }_0^{t_1}\left|\frac{{(t_1-r)}^{a-1}}{1+t_1^a}-\frac{{(t_2-r)}^{a-1}}{1+t_2^a}\right|dr+\frac{1}{a}\frac{{(t_2-t_1)}^a}{1+t_2^a}\right]\to 0\mathrm{as}{t}_1\to t_2.\hfill \end{array}\end{array}$$

(17)

Thus, (15)–(17) shows that $K_A(I-B)Nu\left(\overline{\mathrm{\Omega }}\right)$ is equi-continuous on the compact set $[0,M]$. Finally, we show equi-convergence at $+\infty$. Let $\tau >0$ be a constant such that

$$\left|g\left(r\right)\right|=\left|(I-B)Nu\left(r\right)\right|\le r,u\in \overline{\mathrm{\Omega }}.$$

In addition, since ${lim}_{t\to +\infty }\frac{t^{a+b-1}}{1+t^{a+b}}={lim}_{t\to +\infty }\frac{t^a}{1+t^{a+1}}={lim}_{t\to +\infty }\frac{t^{a-1}}{1+t^a}=0$, then for same $ϵ>0$, there exist $M>0$, such that for $M<t_1<t_2$, we have

$$\left|\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}-\frac{{(t_2-r)}^{a+b-1}}{1+t_2^{a+b}}\right|\le \frac{t_1^{a+b-1}}{1+t_1^{a+b}}-\frac{t_2^{a+b-1}}{1+t_2^{a+b}}<ϵ,$$

$$\left|\frac{{(t_1-r)}^a}{1+t_1^{a+1}}-\frac{{(t_2-r)}^a}{1+t_2^{a+1}}\right|\le \frac{t_1^a}{1+t_1^{a+1}}-\frac{t_2^a}{1+t_2^{a+1}}<ϵ,$$

and

$$\left|\frac{{(t_1-r)}^{a-1}}{1+t_1^a}-\frac{{(t_2-r)}^{a-1}}{1+t_2^a}\right|\le \frac{t_1^{a-1}}{1+t_1^a}-\frac{t_2^{a-1}}{1+t_2^a}<ϵ,$$

Hence,

$$\begin{array}{cc}\hfill |& \frac{K_A(I-B)Nu\left(t_1\right)}{1+t_1^{a+b}}-\frac{K_A(I-B)Nu\left(t_2\right)}{1+t_2^{a+b}}|\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}\left[|{\int }_0^{t_1}\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}g\left(r\right)dr-{\int }_0^{t_1}\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}g\left(r\right)dr|\right]\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\int }_0^M\left|\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}-\frac{{(t_2-r)}^{a+b-1}}{1+t_2^{a+b}}\right|\left|g\left(r\right)\right|dr\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\int }_M^{t_1}\frac{{(t_1-r)}^{a+b-1}}{1+t_1^{a+b}}\left|g\left(r\right)\right|dr+\frac{1}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\int }_M^{t_2}\frac{{(t_2-r)}^{a+b-1}}{1+t_2^{a+b}}\left|g\left(r\right)\right|dr\hfill \\ \hfill & \le \frac{M\tau ϵ}{(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}+\frac{2\tau ϵ}{(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)},\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill |& \frac{D_{0+}^{b-1}\left(K_A(I-B)Nu\right)\left(t_1\right)}{1+t_1^{a+1}}-\frac{D_{0+}^{b-1}\left(K_A(I-B)Nu\right)\left(t_2\right)}{1+t_2^{a+1}}|\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }(a+1)}[{\int }_0^M\left|\frac{{(t_1-r)}^a}{1+t_1^{a+1}}-\frac{{(t_2-r)}^a}{1+t_2^{a+1}}\right|\left|g\left(r\right)\right|dr\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(a+1)}\left[{\int }_M^{t_1}\frac{{(t_1-r)}^a}{1+t_1^{a+1}}g\left(r\right)dr+{\int }_M^{t_2}\frac{{(t_2-r)}^a}{1+t_2^{a+1}}g\left(r\right)dr\right]\hfill \\ \hfill & \le \frac{M\tau ϵ}{\mathrm{\Gamma }(a+1)}+\frac{2\tau ϵ}{\mathrm{\Gamma }(a+2)}\hfill \end{array}$$

(19)

and

$$\begin{array}{cc}\hfill |& \frac{D_{0+}^b\left(K_A(I-B)Nu\right)\left(t_1\right)}{1+t_1^a}-\frac{D_{0+}^b\left(K_A(I-B)Nu\right)\left(t_2\right)}{1+t_2^a}|\hfill \\ \hfill & \le \frac{1}{\mathrm{\Gamma }\left(a\right)}[{\int }_0^M\left|\frac{{(t_1-r)}^{a-1}}{1+t_1^a}-\frac{{(t_2-r)}^{a-1}}{1+t_2^a}\right|\left|g\left(r\right)\right|dr\hfill \\ \hfill & +\frac{1}{a\mathrm{\Gamma }\left(a\right)}\left[{\int }_M^{t_1}\frac{{(t_1-r)}^{a-1}}{1+t_1^a}g\left(r\right)dr+{\int }_M^{t_2}\frac{{(t_2-r)}^{a-1}}{1+t_2^a}g\left(r\right)dr\right]\hfill \\ \hfill & \le \frac{M\tau ϵ}{\mathrm{\Gamma }\left(a\right)}+\frac{2\tau ϵ}{\mathrm{\Gamma }(a+1)}.\hfill \end{array}$$

(20)

Hence, $K_A(I-B)Nu\left(\overline{\mathrm{\Omega }}\right)$ is equi-convergent at $+\infty$. Therefore, by Definition 1, $K_A(I-B)Nu\left(\overline{\mathrm{\Omega }}\right)$ is compact, therefore, the non-linear operator N is L-compact on $\overline{\mathrm{\Omega }}$. This concludes proof of Lemma 6. □

3. Results and Discussion

Here, the conditions for the existence of solutions to problem (1.1) subject to (1.2) is proved.

Theorem 2.

Let f be a continuous function. If (${\varphi }_1$) and (${\varphi }_1$) holds, then, the following conditions also hold:

($H_1$)

There exists functions $\rho \left(t\right),\mu \left(t\right),\nu \left(t\right),\sigma \in C[0,+\infty )$, such that for all $(j,k,l)\in {\mathbb{R}}^3$ and $t\in [0,+\infty )$,

$$|f(t,u\left(t\right),D_{0+}^{b-1}u\left(t\right),D_{0+}^{b}u\left(t\right))|\le \rho \left(t\right)\frac{\left|j\right|}{1+t^{a+b}}+\mu \left(t\right)\frac{\left|k\right|}{1+t^{a+1}}+\nu \left(t\right)\frac{\left|l\right|}{1+t^a}+\sigma \left(t\right).$$

(21)

($H_2$)

There exist constants $M>0$, such that for $u\in \mathrm{dom}L$ if $|D_{0+}^{b}u\left(t\right)|>M$ for $t\in [0,+\infty )$, then either

$$B_{1}Nu\left(t\right)\ne 0\mathrm{or}{B}_{2}Nu\left(t\right)\ne 0.$$

($H_3$)

There exists a constant $C>0$, such that if $|c_1|>C$ or $|c_2|>C$, then either

$$B_{1}N(c_1{t}^{b-1}+c_2{t}^b)+B_{2}N(c_1{t}^{b-1}+c_2{t}^b)<0$$

(22)

or

$$B_{1}N(c_1{t}^{b-1}+c_2{t}^b)+B_{2}N(c_1{t}^{b-1}+c_2{t}^b)>0$$

(23)

where $c_1,c_2\in \mathbb{R}$ satisfying $c_1^2+c_2^2>0$.

Then, the boundary value problem (3) and (4) has at least one solution provided:

$${\parallel \rho \parallel }_Z+{\parallel \mu \parallel }_Z+{\parallel \nu \parallel }_Z<\frac{\mathrm{\Gamma }(a+1)}{\mathrm{\Gamma }(a+1)+2}.$$

Proof. 

The proof will be completed in four stages.

Stage 1. We will establish that ${\mathrm{\Omega }}_1=\{u\in \mathrm{dom}L\setminus kerL:u=\lambda Nu,\mathrm{for}\lambda \in [0,1]\}$ is bounded. Let $u\in {\mathrm{\Omega }}_1$ then $u=(u-Au)+Au\in \mathrm{dom}L\setminus kerL$. This means that $(I-A)u\in \mathrm{dom}L\cap kerA$ and $Au\in kerA$, hence, $LAu=0$. By Lemma 5, we have

$$\begin{array}{cc}\hfill \parallel (I-A)u\parallel & =\parallel K_A{L(I-A)u\parallel \le \parallel L(I-A)u\parallel =\parallel Lu\parallel =\parallel Nu\parallel }_Z.\hfill \end{array}$$

(24)

Since $u\in {\mathrm{\Omega }}_1$, then $Lu=\lambda Nu$. Additionally, by ($H_2$) there exists $t_1\in [0,+\infty )$, such that $|D_{0+}^{b}u\left(t_1\right)|\le M$, therefore

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |D_{0+}^{b}u\left(0\right)|& \le |D_{0+}^{b}u\left(t_1\right)|+\frac{\lambda }{\mathrm{\Gamma }\left(a\right)}{\int }_0^{t_1}{(t_1-r)}^{a-1}\left|f(r,u\left(r\right),D_{0+}^{b-1}u\left(r\right),D_{0+}^{b}u\left(r\right))\right|dr\hfill \\ \hfill & \le M+\frac{1}{\mathrm{\Gamma }(a+1)}{\parallel Nu\parallel }_Z.\hfill \end{array}\end{array}$$

(25)

In addition,

$$\begin{array}{c}\hfill {\parallel Au\parallel }_0\le |D_{0+}^{b}u\left(0\right)|\left(\frac{1}{\mathrm{\Gamma }\left(b\right)}\underset{t\in [0,+\infty )}{sup}\frac{t^{b-1}}{1+t^{a+b}}+\frac{1}{\mathrm{\Gamma }(b+1)}\underset{t\in [0,+\infty )}{sup}\frac{t^b}{1+t^{a+b}}\right)\le 2\left|D_{0+}^{b}u\left(0\right)\right|,\end{array}$$

$$\begin{array}{c}\hfill \parallel D_{0+}^{b-1}{Au\parallel }_1\le |D_{0+}^{b}u\left(0\right)|\left(\frac{1}{\mathrm{\Gamma }\left(b\right)}\underset{t\in [0,+\infty )}{sup}\frac{1}{1+t^{a+1}}+\frac{1}{\mathrm{\Gamma }(b+1)}\underset{t\in [0,+\infty )}{sup}\frac{t}{1+t^{a+1}}\right)\le 2\left|D_{0+}^{b}u\left(0\right)\right|\end{array}$$

and

$$\begin{array}{c}\hfill \parallel D_{0+}^b{Au\parallel }_2\le |D_{0+}^{b}u\left(0\right)|\left(\frac{1}{\mathrm{\Gamma }\left(b\right)}\underset{t\in [0,+\infty )}{sup}\frac{t^{-1}}{1+t^a}+\frac{1}{\mathrm{\Gamma }(b+1)}\underset{t\in [0,+\infty )}{sup}\frac{1}{1+t^a}\right)\le 2\left|D_{0+}^{b}u\left(0\right)\right|.\end{array}$$

Therefore, from (25), we have

$$\begin{array}{cc}\hfill \parallel Au\parallel & \le {max\{\parallel u\parallel }_0,\parallel D_{0+}^{b-1}{u\parallel }_1,\parallel D_{0+}^b{u\parallel }_2\}\le 2|D_{0+}^{b}u\left(0\right)|\le 2M+\frac{2}{\mathrm{\Gamma }(a+1)}{\parallel Nu\parallel }_Z\hfill \end{array}$$

(26)

and from (24) and (26), we have

$$\begin{array}{cc}\hfill {\parallel u\parallel }_U& \le {\parallel Au\parallel }_U+{\parallel I-A\parallel }_U\hfill \\ \hfill & \le 2M+\left(1+\frac{2}{\mathrm{\Gamma }(a+1)}\right){\parallel Nu\parallel }_Z\hfill \\ \hfill & \le 2M+\left(1+\frac{2}{\mathrm{\Gamma }(a+1)}\right){\parallel u\parallel }_U{(\parallel \rho \parallel }_Z+{\parallel \mu \parallel }_Z+{\parallel \nu \parallel }_Z)+\left(1+\frac{2}{\mathrm{\Gamma }(a+1)}\right){\parallel \sigma \parallel }_Z.\hfill \end{array}$$

Hence,

$${\parallel u\parallel }_U\le \frac{2M+\left(1+\frac{2}{\mathrm{\Gamma }(a+1)}\right.∥\sigma {\parallel }_Z)}{1-\left(1+\frac{2}{\mathrm{\Gamma }(a+1)}\right){\parallel u\parallel }_U{(\parallel \rho \parallel }_Z+{\parallel \mu \parallel }_Z+{\parallel \nu \parallel }_Z)}.$$

Thus, ${\mathrm{\Omega }}_1$ is bounded.

Step 2. Let ${\mathrm{\Omega }}_2=\{u\in kerL:Nu\in \mathrm{Im}L\}$. For $u,Nu\in {\mathrm{\Omega }}_2$, then $u\left(t\right)=c_1{t}^{b-1}+c_2{t}^b$. and $BNu=0$. Thus, from ($H_3$), we have $|c_1|\le C$ and $|c_2|\le C$. Hence, ${\mathrm{\Omega }}_2$ is bounded.

Step 3. For $c_1,c_2\in \mathbb{R}$, $t\in [0,+\infty )$, the isomorphism $J:kerL\to \mathrm{Im}B$ is as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill J(c_1{t}^{b-1}+c_2{t}^b)& =\frac{1}{\Delta }\left[({\delta }_{11}{c}_1+{\delta }_{12}{c}_2)+({\delta }_{21}{c}_1+{\delta }_{22}{c}_2)t\right]e^{-t}\hfill \end{array}\end{array}$$

(27)

Suppose (22) holds, let

$${\mathrm{\Omega }}_3=\{u\in kerL:\lambda Ju+(1-\lambda )BNu=0,\lambda \in [0,1]\}.$$

Let $u\in {\mathrm{\Omega }}_3$, then $u\left(t\right)=c_1{t}^{b-1}+c_2{t}^b$. Since $\Delta \ne 0$, then

$$\left\{\begin{array}{c}{c}_1\lambda +(1-\lambda )B_{1}N(c_1{t}^{b-1}+c_2{t}^b)=0,\\ c_2\lambda +(1-\lambda )B_{2}N(c_1{t}^{b-1}+c_2{t}^b)=0.\end{array}\right.$$

(28)

When $\lambda =1$, we obtain $c_1=c_2=0$. When $\lambda =0$, $B_{1}N(c_1{t}^{b-1}+c_2{t}^b)=B_{2}N(c_1{t}^{b-1}+c_2{t}^b)=0$, which contradicts (22) and (23). Hence, from ($H_3$), we obtain $|c_1|\le C$, and $|c_2|\le C$. For $\lambda \in (0,1)$, if $|c_1|>C$ or $|C_2|>A$ by (22) and (28), we have

$$\begin{array}{cc}\hfill \lambda (c_1^2+c_2^2)& =-(1-\lambda )[B_{1}N(c_1{t}^{b-1}+c_2{t}^b)+B_{2}N(c_1{t}^{b-1}+c_2{t}^b)]<0,\hfill \end{array}$$

which is a contradiction. Hence, ${\mathrm{\Omega }}_3$ is bounded.

Similarly, if (23) holds and ${\mathrm{\Omega }}_3=\{u\in kerL:\lambda Ju-(I-\lambda )BNu=0,\lambda \in [0,1]\}$, ${\mathrm{\Omega }}_3$ can be shown to be bounded by similar argument.

Step 4. Let $\mathrm{\Omega }\supset U_{i=1}^3{\overline{\mathrm{\Omega }}}_i$. Finally, we will show that a solution of (3) and (4) exists in $\mathrm{dom}L\cap \mathrm{\Omega }$. We have shown in Steps 1 and 2 that (i) and (ii) of Theorem 1 hold. Finally, we show that (iii) also holds. Let $H(u,\lambda )=\pm \lambda Ju+(1-\lambda )BNu$, then following the arguments of Step 3, it follows that for every $(u,\lambda )\in (kerL\cap \partial \mathrm{\Omega })\times [0,1]$, $H(u,\lambda )\ne 0$. Therefore, by the homotopy property of degree

$$\begin{array}{cc}\hfill deg(BN{|}_{kerL},\mathrm{\Omega }\cap kerL,0)& =deg(\pm J,\mathrm{\Omega }\cap kerL,0)\hfill \\ \hfill & =\pm 1\ne 0.\hfill \end{array}$$

Therefore, by Theorem 1 at least one solution of (3) and (4) exists in U. □

4. Conclusions

This work considered a mixed fractional-order boundary value problem at resonance on the half-line. The Mawhin’s coincidence degree theory was used to establish existence of solutions when the dimension of the kernel of the linear fractional differential operator is two. The result obtained is new and an example was used to demonstrate the result obtained.

5. Example

Example 1.

Consider the following boundary value problem:

$${}^C{D}_{0+}^{\frac{1}{2}}{D}_{0+}^{\frac{3}{2}}u\left(t\right)=\frac{e^{-5t}sin{D}_{0+}^{\frac{1}{2}}u\left(t\right)}{17(1+t^2)}+\frac{e^{-t}{D}_{0+}^{\frac{3}{2}}u\left(t\right)}{9(1+t^{\frac{3}{2}})}+\frac{e^{-2t}}{15(1+t^{\frac{1}{2}})},t\in [0,+\infty )$$

(29)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & I_{0+}^{\frac{1}{2}}u\left(0\right)=0,D_{0+}^{\frac{1}{2}}u\left(0\right)=\frac{2}{3}{D}_{0+}^{\frac{1}{2}}u\left(\frac{1}{4}\right)-\frac{1}{3}{D}_{0+}^{\frac{1}{2}}u\left(\frac{1}{2}\right),\hfill \\ \hfill & D_{0+}^{\frac{3}{2}}u(+\infty )=\frac{3}{4}{D}_{0+}^{\frac{1}{2}}u\left(\frac{1}{5}\right)+\frac{1}{4}{D}_{0+}^{\frac{1}{2}}u\left(\frac{3}{5}\right),.\hfill \end{array}\end{array}$$

(30)

Here $a=\frac{1}{2}$, $b=\frac{3}{2}$${\alpha }_1=\frac{2}{3}$, ${\alpha }_2=\frac{5}{2}$, ${\xi }_1=4$, ${\xi }_2=2$, ${\beta }_1=\frac{3}{4}$, ${\beta }_2=\frac{1}{4}$, ${\eta }_1=5$, ${\eta }_2=\frac{5}{3}$,

$n=m=2$. ${\sum }_{j=1}^2{\alpha }_j{\xi }_j^{-1}=0$, ${\sum }_{j=1}^2{\alpha }_j=1$, ${\sum }_{k=1}^2{\beta }_k{\eta }_k^{-1}=0$, ${\sum }_{k=1}^2{\beta }_k=1$.

${\parallel \rho \parallel }_Z=\frac{1}{17}{sup}_{t\in [0,+\infty )}\left|e^{-5t}\right|=\frac{1}{17}$, ${\parallel \mu \parallel }_Z=\frac{1}{9}{sup}_{t\in [0,+\infty )}\left|e^{-t}\right|=\frac{1}{9}$,

${\parallel \nu \parallel }_Z=\frac{1}{15}{sup}_{t\in [0,+\infty )}\left|e^{-6t}\right|=\frac{1}{15}$. Then, ${\parallel \rho \parallel }_Z+{\parallel \mu \parallel }_Z+{\parallel \nu \parallel }_Z=\frac{1}{17}+\frac{1}{9}+\frac{1}{15}=0.2367$

$\mathrm{\Gamma }(a+1)=\mathrm{\Gamma }(\frac{1}{2}+1)=1$. Then, $\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{2}\right)+2}=0.3071$. Hence,

$${\parallel \rho \parallel }_Z+{\parallel \mu \parallel }_Z+{\parallel \nu \parallel }_Z<\frac{\mathrm{\Gamma }(a+1)}{\mathrm{\Gamma }(a+1)+2}.$$

Finally, conditions ($H_1$) - ($H_3$) can also be shown to hold. Therefore (29) and (30) has at least one solution.