Existence of solutions for integral boundary value problems of mixed fractional differential equations under resonance

Abstract

In this paper, we concerned the existence of solutions of the following nonlinear mixed fractional differential equation with the integral boundary value problem:

$$\left \{ \textstyle\begin{array}{l} {}^{C}D^{\alpha}_{1-} D^{\beta}_{0+}u(t)=f(t,u(t),D^{\beta +1}_{0+}u(t),D^{\beta}_{0+}u(t)),\quad 0< t< 1,\\ u(0)=u'(0)=0,\qquad u(1)=\int^{1}_{0}u(t)\,dA(t), \end{array}\displaystyle \right . $$

where ${}^{C}D^{\alpha}_{1-}$ is the left Caputo fractional derivative of order $\alpha\in(1,2]$, and $D^{\beta}_{0+}$ is the right Riemann–Liouville fractional derivative of order $\beta\in(0,1]$. The coincidence degree theory is the main theoretical basis to prove the existence of solutions of such problems.

1 Introduction

In this paper, we study the following integral boundary value problems of the mixed fractional differential equations under resonance:

$$ \left \{ \textstyle\begin{array}{l} {}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}u(t)=f(t,u(t),D^{\beta +1}_{0+}u(t),D^{\beta}_{0+}u(t)),\quad 0< t< 1,\\ u(0)=u'(0)=0,\qquad u(1)=\int^{1}_{0}u(t)\,dA(t), \end{array}\displaystyle \right . $$

(1.1)

where ${}^{C}D^{\alpha}_{1-}$ and $D^{\beta}_{0+}$ are the left Caputo fractional derivative of order $\alpha\in(1,2]$ and the right Riemann–Liouville fractional derivative of order $\beta\in (0,1]$, respectively, $f\in C([0,1]\times\mathbb{R}^{3}, \mathbb{R})$, $A(t)$ is a bounded-variation function, $\int^{1}_{0}x(t)\,dA(t)$ is the Riemann–Stieltjes integral of x with respect to A. From the Lemma 2.3 we know that problem (1.1) is resonance if $\int^{1}_{0}t^{\beta+1}\,dA(t)=1$.

Due to the existence of solutions for boundary value problems of fractional differential equations widely used in applied science and technological science [1–5], they have become a popular research field. At present, many researchers study the existence of solutions of fractional differential equations such as the Riemann–Liouville fractional derivative problem at nonresonance [6–16], the Riemann–Liouville fractional derivative problem at resonance [17–23], the Caputo fractional boundary value problem [6, 24, 25], the Hadamard fractional boundary value problem [26–28], conformable fractional boundary value problems [29–32], impulsive problems [33–35], boundary value problems [8, 36–43], and variational structure problems [44, 45].

For example, Tang et al. [24] investigated the existence of solutions for the four-point boundary value problems of fractional differential equations

$$\left \{ \textstyle\begin{array}{l} D^{\alpha}_{0+}u(t)=f(t,u(t),u'(t)),\quad 0\leq t\leq1,\\ u'(0)-\beta u(\xi)=0,\qquad u'(1)+\gamma u(\eta)=0, \end{array}\displaystyle \right . $$

where $D^{\alpha}_{0+}$ denotes the Caputo fractional derivative with $1<\alpha\leq2$.

Zou and He [23] investigated the integral boundary value problem for resonant fractional differential equation

$$\left \{ \textstyle\begin{array}{l} -D^{p}_{0+}x(t)=f (t,x(t),D^{p-1}_{0+}x(t),D^{p-2}_{0+}x(t) ),\quad 0< t< 1,\\ x(0)=x'(0)=0,\\ x(1)=\int^{1}_{0}x(t)\,dA(t),\quad 2< p< 3, \end{array}\displaystyle \right . $$

where $D^{p}_{0+}$ is the standard Riemann–Liouville differentiation. Using Mawhin’s coincidence degree theory, they proved the existence of solutions.

In recent paper [9], the existence and uniqueness results for integral boundary value problem of two-term fractional differential equations

$$\left \{ \textstyle\begin{array}{l} D^{\delta}x(t)+f(t,x(t))= D^{\tau}g(t,x(t)),\quad t\in(0,1),\\ x(0)=0,\qquad x(1)=\frac{1}{\varGamma(\delta-\tau)}\int^{1}_{0}(1-s)^{\delta -\tau-1}g(s,x(s))\,ds \end{array}\displaystyle \right . $$

were considered by the Schauder fixed point theorem and the Banach contraction mapping principle.

Among several types of fractional differential equations found in the literature, the Caputo and Riemann–Liouville derivatives are studied separately in many cases. However, the study of resonant boundary value problems involving mixed fractional-order derivatives have not been extensively studied (see [26, 46]). Motivated by the literature mentioned, we consider the existence of solutions for the resonant integral boundary value problem (1.1) involving the left Caputo and right Riemann–Liouville fractional derivatives by using the Mawhin’s coincidence degree theory.

In this paper, we always suppose that the following condition is satisfied:

$(H1)$:

$\int^{1}_{0}t^{\beta+1}\,dA(t)=1$, $\int^{1}_{0}t^{\beta}\,dA(t)-1\neq0$.

2 Preliminaries

In this paper, we first need the following necessary basic definitions.

Definition 2.1

([2])

The left and right Riemann–Liouville fractional integrals of order $\alpha>0$ of a function $g:(0,\infty)\rightarrow R$ are respectively given by

$$I^{\alpha}_{0+}g(t)= \int^{t}_{0}\frac{(t-s)^{\alpha-1}}{\varGamma(\alpha)}g(s)\,ds $$

and

$$I^{\alpha}_{1-}g(t)= \int^{1}_{t}\frac{(s-t)^{\alpha-1}}{\varGamma(\alpha )}g(s)\,ds, $$

where the right-hand sides are pointwise defined on $(0,\infty)$, and Γ is the gamma function.

Definition 2.2

([2])

The left Riemann-Liouville fractional derivative and the right Caputo fractional derivative of order $\alpha>0$ of a function $g\in C^{n}((0,\infty),R)$ are given by

$$D^{\alpha}_{0+}g(t)=\frac{d^{n}}{dt^{n}}\bigl(I^{n-\alpha}_{0+}g \bigr) (t) $$

and

$${}^{C}D^{\alpha}_{1-}g(t)=(-1)^{n}I^{n-\alpha}_{1-}g^{(n)}(t),\quad n-1< \alpha< n, $$

respectively.

Lemma 2.1

Let$\alpha\in(1,2]$and$\beta\in(0,1]$. For$y \in C[0,1]$, the fractional differential equation

$$ {}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}u(t)=y(t) $$

(2.1)

has the general solution

$$ u(t)=I^{\beta}_{0+} I^{\alpha}_{1-}y(t)+c_{0} \frac{t^{\beta}}{\varGamma (\beta+1)} +c_{1}\frac{t^{\beta+1}}{\varGamma(\beta+2)}+c_{2}t^{\beta-1}. $$

(2.2)

Proof

Applying the right fractional integral $I^{\alpha}_{1-}$ to (2.1) and using the properties of Caputo fractional derivatives, we can obtain that

$$D^{\beta}_{0+}u(t)=I^{\alpha}_{1-}y(t)+c_{0}+c_{1}t,\quad c_{0}, c_{1}\in\mathbb{R}. $$

Applying the left fractional integral $I^{\beta}_{0+}$ to this equation and using the properties of Riemann–Liouville fractional derivatives, we have

$$\begin{aligned} u(t)&= I^{\beta}_{0+} I^{\alpha}_{1-}y(t)+I^{\beta}_{0+} (c_{0}+c_{1}t)+c_{2}t^{\beta-1} \\ &= I^{\beta}_{0+} I^{\alpha}_{1-}y(t)+c_{0} \frac{t^{\beta}}{\varGamma (\beta+1)} +c_{1}\frac{t^{\beta+1}}{\varGamma(\beta+2)}+c_{2}t^{\beta-1},\quad c_{2}\in \mathbb{R}. \end{aligned} $$

 □

Lemma 2.2

Let$\alpha\in(1,2]$and$\beta\in(0,1]$. If$y \in C[0,1]$, thenuis a solution of the fractional differential equation

$$\left \{ \textstyle\begin{array}{l} {}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}u(t)=y(t),\quad 0< t< 1,\\ u(0)=u'(0)=u(1)=0, \end{array}\displaystyle \right . $$

if and only if

$$u(t)=I^{\beta}_{0+} I^{\alpha}_{1-}y(t)-t^{\beta+1} I^{\beta}_{0+} I^{\alpha}_{1-}y(1),\quad t\in[0,1]. $$

Proof

Conditions $u(0)=u'(0)=0$ in (2.2) yield $c_{0}=c_{2}=0$. Consequently, (2.2) reduces to

$$u(t)=I^{\beta}_{0+} I^{\alpha}_{1-}y(t)+c_{1} \frac{t^{\beta+1}}{\varGamma (\beta+2)},\quad t\in[0,1]. $$

By the boundary condition $u(1)=0$ we have

$$c_{1}=-\varGamma(\beta+2) I^{\beta}_{0+} I^{\alpha}_{1-}y(1). $$

Therefore

$$u(t)=I^{\beta}_{0+} I^{\alpha}_{1-}y(t)-t^{\beta+1} I^{\beta}_{0+} I^{\alpha}_{1-}y(1),\quad t\in[0,1]. $$

This process is reversible. □

Let $L:\operatorname{Dom}L\subset X\rightarrow Y$ be a Fredholm operator of index zero, where X and Y are two real Banach spaces, and let $N:X\rightarrow Y$ be a nonlinear continuous map. If $P:X\rightarrow X$ and $Q:Y\rightarrow Y$ are continuous projectors such that $\operatorname{Im}P=\operatorname{Ker}L$, $\operatorname {Ker}Q=\operatorname{Im}L$, $X=\operatorname{Ker}L\oplus\operatorname {Ker}P$, and $Y=\operatorname{Im}L\oplus\operatorname{Im}Q$, then $L_{P}=L|_{\operatorname{Dom}L\cap\operatorname {Ker}P}:\operatorname{Dom}L\cap\operatorname{Ker}P\rightarrow \operatorname{Im}L$ is invertible. By $K_{P}$ we denote the inverse of the operator $L_{P}$.

Let Ω is an open bounded subset of X with $\operatorname {Dom}L\cap\varOmega\neq\varnothing$. If $QN(\overline{\varOmega})$ is bounded and $K_{P}(I-Q)N:\overline{\varOmega}\rightarrow X$ is compact, then we call the mapping $N:X\rightarrow Y$L-compact on Ω̅.

Theorem 2.1

([47])

LetLbe a Fredholm operator of index zero, and letNbe L-compact onΩ̅. Assume that the following conditions are satisfied:

• (i) $Lu\neq\lambda Nu$for every$(u,\lambda)\in[(\operatorname {dom}L\setminus\operatorname{Ker}L)\cap\partial\varOmega]\times(0,1)$;

• (ii) $Nu\notin\operatorname{Im}L$for every$u\in\operatorname {Ker}L \cap\partial\varOmega$;

• (iii) $\operatorname{deg} (JQN|_{\operatorname{Ker}L},\varOmega \cap\operatorname{Ker}L,0 )\neq0$, where$J:\operatorname{Im} Q\rightarrow\operatorname{Ker} L$is an isomorphism.

Then the equation$Lu=Nu$has at least one solution in$\operatorname {dom}L\cap\overline{\varOmega}$.

We use the classical Banach space $Y=C[0,1]$ with the norm $\|u\| _{\infty}=\max_{t\in[0,1]}|u(t)|$ and the Banach space $X=\{u:[0,1]\rightarrow\mathbb{R}\mid u,D^{\beta+1}_{0+}u,D^{\beta }_{0+}u\in C[0,1]\}$ with the norm $\|x\|_{X}=\max\{\|u\|_{\infty},\|D^{\beta+1}_{0+}u\| _{\infty},\|D^{\beta}_{0+}u\|_{\infty}\}$ (see [22, 23]).

After further discussion for problems (1.1), we define two operators L and N as follows:

$$ \begin{gathered} (Lu) (t)={}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}u(t),\quad u\in\operatorname {Dom} L, \\ (Nu) (t)=f\bigl(t,u(t),D^{\beta+1}_{0+}u(t),D^{\beta}_{0+}u(t) \bigr),\quad u\in X,\end{gathered} $$

(2.3)

where

$$\operatorname{Dom}L= \biggl\{ u\in X\Bigm| {}^{C}D^{\alpha}_{1-}D^{\beta }_{0+}u \in Y, u(0)=u'(0)=0, u(1)= \int^{1}_{0}u(t)\,dA(t) \biggr\} , $$

then we can write problem (1.1) as $Lx=Nx$.

Next, the following lemmas play an important role in proving the existence of solutions to (1.1).

Lemma 2.3

LetLbe defined as in (2.3). Then

$$\begin{aligned}& \operatorname{ker} L=\bigl\{ u\in X\mid u(t)=ct^{\beta+1}, c\in\mathbb{R}, t\in[0,1]\bigr\} , \end{aligned}$$

(2.4)

$$\begin{aligned}& \operatorname{Im}L= \biggl\{ y\in Y\Bigm| \int^{1}_{0}I^{\beta }_{0+}I^{\alpha}_{1-}y(t) \,dA(t)-I^{\beta}_{0+}I^{\alpha}_{1-}y(t)\big| _{t=1}=0 \biggr\} . \end{aligned}$$

(2.5)

Proof

By Lemma 2.1${}^{C}D^{\alpha}_{1-}D^{\beta }_{0+}u(t)=0$ has the solution

$$ u(t)=c_{0}\frac{t^{\beta}}{\varGamma(\beta+1)}+c_{1}\frac{t^{\beta +1}}{\varGamma(\beta+2)}+c_{2}t^{\beta-1}. $$

(2.6)

By the boundary value condition $u(0)=u'(0)=0$ we can infer that $c_{0}=c_{2}=0$. Consequently, (2.6) reduces to

$$u(t)=c_{1}\frac{t^{\beta+1}}{\varGamma(\beta+2)}. $$

Then, combining with the boundary value condition $u(1)=\int ^{1}_{0}u(t)\,dA(t)$, we have that (2.4) holds.

If $y\in\operatorname{Im}L$, then there exists $u\in\operatorname {dom}L$ such that $y(t)={}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}u(t)$. It follows from Lemma 2.1 and the boundary value condition $u(0)=u'(0)=0$ that

$$u(t)=I^{\beta}_{0+}I^{\alpha}_{1-}y(t)+c_{1} \frac{t^{\beta+1}}{\varGamma (\beta+2)}. $$

Thus we have

$$u(1)=I^{\beta}_{0+}I^{\alpha}_{1-}y(t) \big|_{t=1}+c_{1}\frac{1}{\varGamma (\beta+2)} $$

and

$$\begin{aligned} \int_{0}^{1}u(t)\,dA(t)&= \int_{0}^{1}I^{\beta}_{0+}I^{\alpha }_{1-}y(t) \,dA(t)+c_{1}\frac{ \int_{0}^{1}t^{\beta+1}\,dA(t)}{\varGamma(\beta +2)} \\ &= \int_{0}^{1}I^{\beta}_{0+}I^{\alpha}_{1-}y(t) \,dA(t)+c_{1}\frac {1}{\varGamma(\beta+2)}. \end{aligned} $$

Using the condition $u(1)=\int^{1}_{0}u(t)\,dA(t)$, we obtain that

$$\int^{1}_{0}I^{\beta}_{0+}I^{\alpha}_{1-}y(t) \,dA(t)-I^{\beta }_{0+}I^{\alpha}_{1-}y(t) \big|_{t=1}=0, $$

so that $\operatorname{Im}L\subset\{y\in Y\mid\int^{1}_{0}I^{\beta }_{0+}I^{\alpha}_{1-}y(t)\,dA(t)-I^{\beta}_{0+}I^{\alpha}_{1-}y(t)| _{t=1}=0\}$.

On the other hand, suppose $y\in Y$ satisfies

$$\int^{1}_{0}I^{\beta}_{0+}I^{\alpha}_{1-}y(t) \,dA(t)-I^{\beta }_{0+}I^{\alpha}_{1-}y(t) \big|_{t=1}=0. $$

Let

$$u(t)=I^{\beta}_{0+}I^{\alpha}_{1-}y(t)+t^{\beta+1}. $$

Then ${}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}u(t)=y(t)$, $u(0)=u'(0)=0$, and $u(1)=\int^{1}_{0}u(t)\,dA(t)$. So we obtain that $y\in\operatorname{Im}L$.

Thus the proof of

$$\operatorname{Im}L= \biggl\{ y\in Y\Bigm| \int^{1}_{0}I^{\beta}_{0+}I^{\alpha }_{1-}y(t) \,dA(t)-I^{\beta}_{0+}I^{\alpha}_{1-}y(t) \big|_{t=1}=0 \biggr\} $$

is completed. □

Lemma 2.4

Assume that$(H_{1})$is satisfied. Then the operatorLis a Fredholm operator with index zero, and two linear continuous projectors$P:X\rightarrow X$and$Q: Y\rightarrow Y$are respectively defined by

$$\begin{gathered} (Pu) (t)=u(1)t^{\beta+1},\quad u\in X, \\ Qy=\frac{1}{\theta(\int^{1}_{0} \,dA(t)-1)} \biggl( \int^{1}_{0}I^{\beta }_{0+}I^{\alpha}_{1-}y(t) \,dA(t)-I^{\beta}_{0+}I^{\alpha}_{1-}y(1) \biggr),\quad y\in Y,\end{gathered} $$

where$\theta=I^{\beta}_{0+}I^{\alpha}_{1-}1=\frac{1}{(\alpha+\beta )\varGamma(\alpha+1)\varGamma(\beta)}$. Furthermore, let$K_{P}:\operatorname{Im}L\rightarrow\operatorname{Dom}L\cap \operatorname{Ker}P$be a linear operator defined by

$$\begin{aligned} K_{P}y(t)&= I^{\beta}_{0+}I^{\alpha}_{1-}y(t)-t^{\beta+1}I^{\beta }_{0+}I^{\alpha}_{1-}y(1) \\ &= \int^{t}_{0}\frac{(t-s)^{\beta-1}}{\varGamma(\beta)} \int ^{1}_{s}\frac{(\tau-s)^{\alpha-1}}{\varGamma(\alpha)}y(\tau)\,d\tau \,ds \\ &\quad -t^{\beta+1} \int^{1}_{0}\frac{(1-s)^{\beta-1}}{\varGamma(\beta)} \int ^{1}_{s}\frac{(\tau-s)^{\alpha-1}}{\varGamma(\alpha)}y(\tau)\,d\tau \,ds. \end{aligned} $$

Then$K_{P}$is the inverse of$L_{P}=L|_{\operatorname{Dom}L\cap \operatorname{Ker}P}$.

Proof

For $u\in X$, we have

$$\bigl(P^{2}u\bigr) (t)=P(Pu) (t)=t^{\beta+1} \bigl[t^{\beta+1}u(1)\bigr]\big|_{t=1}=(Pu) (t). $$

So $P:X\rightarrow X$ is a linear continuous projector operator with $\operatorname{Ker} P=\operatorname{Im} L$.

Since $u=u-Pu+Pu$, it is easy to see that $u-Pu\in\operatorname{Ker} P$ and $Pu\in\operatorname{Ker} L$. Thus $X= \operatorname{Ker} P+\operatorname{Ker} L$. If $u\in\operatorname{Ker} P\cap \operatorname{Ker} L$ and so $u(t)=ct^{\beta+1}$, then we can conclude that $(Pu)(t)=ct^{\beta +1}=0$, and so $c=0$. Then

$$X=\operatorname{Ker} P\oplus\operatorname{Ker} L. $$

Take $z(t)\equiv1$ for $t\in[0,1]$. For $y\in Y$, we have

$$Q^{2}y(t)=\frac{Qy(t)}{\theta(\int^{1}_{0}\,dA(t)-1)} \biggl( \int ^{1}_{0}I^{\beta}_{0+}I^{\alpha}_{1-}z(t) \,dA(t)-I^{\beta}_{0+}I^{\alpha }_{1-}z(1) \biggr)=Qy(t), $$

which implies that $Q^{2}=Q$ and $\operatorname{Ker} Q=\operatorname{Im} L$.

For $y\in Y$, $y=y-Qy+Qy$, we have $Y= \operatorname{Im} L +\operatorname{Im} Q$. Moreover, by direct computation we get $\operatorname{Im} L \cap\operatorname{Im} Q=\{0\}$. Thus $Y= \operatorname{Im} L\oplus\operatorname{Im} Q$. Therefore

$$\operatorname{dim} \operatorname{Ker}L=\operatorname{dim} \operatorname {Im}Q = \operatorname{codim} \operatorname{Im}L = 1. $$

This shows that L is a Fredholm operator of index zero.

Next, we will prove that $K_{P}:\operatorname{Im}L\rightarrow \operatorname{Dom}L\cap\operatorname{Ker}P$ is the inverse of $L_{P}=L|_{\operatorname{Dom}L\cap\operatorname{Ker}P}$.

In fact, for $y\in \operatorname{Im}L$, we have

$$L_{P}K_{P}y={}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}I^{\beta}_{0+}I^{\alpha}_{1-}y=y, $$

and for $u\in \operatorname{dom}L\cap\ker P$, we know that there exists $y\in Y$ such that

$$\left \{ \textstyle\begin{array}{l} {}^{C}D^{\alpha}_{1-}D^{\beta}_{0+}u(t)=y(t),\quad 0< t< 1,\\ u(0)=u'(0)=u(1)=0. \end{array}\displaystyle \right . $$

In view of Lemma 2.2, we get

$$(K_{p}L)u(t)=(K_{p}y) (t)=u(t), $$

which shows that $K_{P}=(L|_{ \operatorname{dom} L \cap\ker P})^{-1}$.

Thus the proof that $K_{P}$ is the inverse of $L_{P}=L|_{\operatorname {Dom}L\cap\operatorname{Ker}P}$ is complete. □

By standard arguments we have the following lemma.

Lemma 2.5

$K_{P}(I-Q)N:Y\rightarrow Y$is completely continuous.

Lemma 2.6

For$y\in Y$, let

$$ (Ty) (t)=I^{\beta}_{0+} I^{\alpha}_{1-}y(t)= \int^{t}_{0}\frac{(t-s)^{\beta-1}}{\varGamma(\beta)} \int^{1}_{s}\frac {(\tau-s)^{\alpha-1}}{\varGamma(\alpha)}y(\tau)\,d\tau \,ds. $$

(2.7)

Then

$$\begin{gathered} \Vert Ty \Vert _{\infty}\leq \frac{1}{\varGamma(\beta+1)\varGamma(\alpha+1)} \Vert y \Vert _{\infty}, \\ \bigl\Vert D_{0+}^{\beta}(Ty) \bigr\Vert _{\infty}\leq \frac{1}{\varGamma(\alpha+1)} \Vert y \Vert _{\infty}, \\ \bigl\Vert D_{0+}^{\beta+1}(Ty) \bigr\Vert _{\infty} \leq \frac{1}{\varGamma(\alpha)} \Vert y \Vert _{\infty}.\end{gathered} $$

Moreover,

$$\Vert Ty \Vert _{X}\leq \Delta \Vert y \Vert _{\infty}, $$

where$\Delta=\max\{\frac{1}{\alpha\varGamma(\beta+1)}, 1\}\frac {1}{\varGamma(\alpha)}$.

Proof

Applying the left fractional derivative $D_{0+}^{\beta}$ and $D_{0+}^{\beta+1}$, respectively, and using the properties of Riemann–Liouville fractional derivatives, we get

$$D_{0+}^{\beta}(Ty) (t)=I^{\alpha}_{1-}y(t)= \int^{1}_{t}\frac{(s-t)^{\alpha-1}}{\varGamma(\alpha)}y(s)\,ds $$

and

$$D_{0+}^{\beta+1}(Ty) (t)= - \int^{1}_{t}\frac{(s-t)^{\alpha-2}}{\varGamma(\alpha-1)}y(s)\,ds. $$

Consequently,

$$\begin{aligned}& \begin{aligned} \bigl\vert (Ty) (t) \bigr\vert &\leq \biggl\vert \int^{t}_{0}\frac{(t-s)^{\beta-1}}{\varGamma(\beta)} \frac {(1-s)^{\alpha}}{\varGamma(\alpha+1)} \,ds \biggr\vert \Vert y \Vert _{\infty} \leq \biggl\vert \int^{t}_{0}\frac{(t-s)^{\beta-1}}{\varGamma(\beta)\varGamma (\alpha+1)}\,ds \biggr\vert \Vert y \Vert _{\infty} \\ &= \biggl\vert \frac{t^{\beta}}{\varGamma(\beta+1)\varGamma(\alpha+1)} \biggr\vert \Vert y \Vert _{\infty} \leq \frac{1}{\varGamma(\beta+1)\varGamma(\alpha+1)} \Vert y \Vert _{\infty}, \end{aligned} \\& \bigl\vert D_{0+}^{\beta}(Ty) (t) \bigr\vert \leq \biggl\vert \int^{1}_{t}\frac{(s-t)^{\alpha-1}}{\varGamma(\alpha)}\,ds \biggr\vert \Vert y \Vert _{\infty}\leq\frac{1}{\varGamma(\alpha+1)} \Vert y \Vert _{\infty}, \end{aligned}$$

and

$$\bigl\vert D_{0+}^{\beta+1}(Ty) (t) \bigr\vert \leq \biggl\vert \int^{1}_{t}\frac{(s-t)^{\alpha-2}}{\varGamma(\alpha-1)}\,ds \biggr\vert \Vert y \Vert _{\infty}\leq\frac{1}{\varGamma(\alpha)} \Vert y \Vert _{\infty}, $$

which, on taking the norm for $t\in[0,1]$, yields

$$\Vert Ty \Vert _{X}=\max\bigl\{ \Vert Ty \Vert _{\infty}, \bigl\Vert D_{0+}^{\beta}(Ty) \bigr\Vert _{\infty}, \bigl\Vert D_{0+}^{\beta+1}(Ty) \bigr\Vert _{\infty}\bigr\} \leq \Delta \Vert y \Vert _{\infty}. $$

 □

3 Main results

In this section, we use Theorem 2.1 to prove the existence of solutions to IBVP (1.1).

To get our main result, we need the following conditions:

$(H2)$:

There exists a constant $B>0$ such that either for each $c\in \mathbb{R}:|c|>B$,

$$ cQN\bigl(ct^{\beta+1}\bigr)>0 $$

(3.1)

or for each $c\in\mathbb{R}:|c|>B$,

$$ cQN\bigl(ct^{\beta+1}\bigr)< 0. $$

(3.2)

$(H3)$:

There exist functions $\rho,\sigma,\tau,\gamma\in C[0,1]$ such that, for all $(u,v,w)\in\mathbb{R}^{3}$ and $t\in[0,1]$,

$$\bigl\vert f(t,u,v,w) \bigr\vert \leq\rho(t)+\sigma(t) \vert u \vert + \tau(t) \vert v \vert +\gamma(t) \vert w \vert . $$

$(H4)$:

There exists a constant $M>0$ such that if $|D^{\beta +1}_{0+}u(t)|>M$ for all $t\in[0,1]$, and then $QNu\neq0$.

Theorem 3.1

If$(H1)$, $(H2)$, $(H3)$, $(H4)$hold, then IBVP (1.1) has at least one solution inX, provided that

$$ \Vert \sigma \Vert _{\infty}+ \Vert \tau \Vert _{\infty}+ \Vert \gamma \Vert _{\infty}< \frac {\varGamma(\alpha)}{\varGamma(\alpha)+\Delta}. $$

(3.3)

Proof

Set

$$\varOmega_{1}=\bigl\{ u\in\operatorname{dom}L\setminus\operatorname {Ker}L:Lu=\lambda Nu \text{ for some } \lambda\in[0,1]\bigr\} . $$

For $u\in\varOmega_{1}$, since $Lu=\lambda Nu$ and so $\lambda\neq 0$, $Nu\in\operatorname{Im}L=\operatorname{Ker}Q$, and hence

$$QNu=0. $$

Thus, By $(H4)$ there exists $t_{0}\in[0,1]$ such that

$$\bigl\vert D^{\beta+1}_{0+}u(t_{0}) \bigr\vert \leq M. $$

It follows from Lemma 2.1 and $u(0)=u'(0)=0$ that there exists $c_{1}\in \mathbb{R}$ such that the function u satisfies

$$u(t)=\lambda I^{\beta}_{0+} I^{\alpha}_{1-}Nu(t)+c_{1} \frac{t^{\beta +1}}{\varGamma(\beta+2)}=\lambda T(Nu) (t)+c_{1}\frac{t^{\beta +1}}{\varGamma(\beta+2)}, $$

where the operator T is defined by (2.7). Applying the left fractional derivative $D_{0+}^{\beta+1}$ to this equation and using the properties of fractional derivative, we get

$$D_{0+}^{\beta+1}u(t)=\lambda D_{0+}^{\beta+1}I^{\beta}_{0+} I^{\alpha }_{1-}Nu(t)+c_{1}=-\lambda I^{\alpha-1}_{1-}Nu(t)+c_{1}. $$

Therefore

$$\vert c_{1} \vert \leq \bigl\vert D_{0+}^{\beta+1}u(t_{0}) \bigr\vert + \bigl\vert I^{\alpha-1}_{1-}Nu(t_{0}) \bigr\vert \leq M+\frac{1}{\varGamma(\alpha)} \Vert Nu \Vert _{\infty}. $$

This, together with Lemma 2.6, yields

$$\begin{aligned} \Vert u \Vert _{X}&= \max\bigl\{ \Vert u \Vert _{\infty}, \bigl\Vert D^{\beta}_{0+}u \bigr\Vert _{\infty}, \bigl\Vert D^{\beta+1}_{0+}u \bigr\Vert _{\infty}\bigr\} \\ &\leq \max\bigl\{ \bigl\Vert T(Nu) \bigr\Vert _{\infty}, \bigl\Vert D^{\beta}_{0+}T(Nu) \bigr\Vert _{\infty}, \bigl\Vert D^{\beta+1}_{0+}T(Nu) \bigr\Vert _{\infty}\bigr\} + \vert c_{1} \vert \\ &\leq M+\biggl(\frac{1}{\varGamma(\alpha)}+\Delta\biggr) \Vert Nu \Vert _{\infty} \\ &\leq M+\biggl(\frac{1}{\varGamma(\alpha)}+\Delta\biggr) \bigl( \Vert \sigma \Vert _{\infty}+ \Vert \tau \Vert _{\infty}+ \Vert \gamma \Vert _{\infty}\bigr) \Vert u \Vert _{X}. \end{aligned} $$

Thus from (3.3) we obtain that

$$\Vert u \Vert _{X}\leq\frac{M\varGamma(\alpha)}{\varGamma(\alpha)-(\varGamma (\alpha)+\Delta)( \Vert \sigma \Vert _{\infty}+ \Vert \tau \Vert _{\infty}+ \Vert \gamma \Vert _{\infty})}. $$

Therefore $\varOmega_{1}$ is bounded.

Now we denote $\varOmega_{2}=\{u\in\operatorname{Ker}L:Nu\in \operatorname{Im}L\}$. If $u\in\varOmega_{2}$, then $u=ct^{\beta+1}$, $c\in\mathbb{R}$, and it is easy to deduce that $QNu(t)=0$. By $(H2)$ we obtain $|c|\leq B$. Therefore $\varOmega_{2}$ is a bounded set.

Now we define the isomorphism $J: \operatorname{Im}Q\rightarrow \operatorname{Ker}L$ by

$$J(c)=ct^{\beta+1}. $$

If (3.1) holds, then let

$$\varOmega_{3}=\bigl\{ u\in\operatorname{Ker}L:\lambda u+(1-\lambda )JQNu=0,\lambda\in[0,1]\bigr\} . $$

For $u=ct^{\beta+1}\in\varOmega_{3}$, we have

$$\lambda ct^{\beta+1}=-(1-\lambda)t^{\beta+1}QN\bigl(ct^{\beta+1} \bigr). $$

So we get

$$\lambda c=-(1-\lambda)QN\bigl(ct^{\beta+1}\bigr). $$

If $\lambda=1$, then $c=0$. Otherwise, if $|c|>B$, in view of $(H2)$, we have

$$c(1-\lambda)QN\bigl(ct^{\beta+1}\bigr)>0, $$

which contradicts $\lambda c^{2}\geq0$. Thus $\varOmega_{3}$ is bounded.

If (3.2) holds, then define the set

$$\varOmega_{3}=\bigl\{ u\in\operatorname{Ker}L:-\lambda u+(1-\lambda )JQNu=0,\lambda\in[0,1]\bigr\} , $$

where J is as before. Similarly to the previous argument, we can show that $\varOmega_{3}$ also is bounded.

Next, we will prove that all the assumptions of Theorem 2.1 are satisfied. Let Ω be any bounded open subset of Y such that $\bigcup^{3}_{i=1}\overline{\varOmega_{i}}\subset\varOmega$. By Lemma 2.5$K_{P}(I-Q)N:\varOmega\rightarrow Y$ is compact, and thus N is L-compact on Ω̅.

Clearly, assumptions (i) and (ii) of Theorem 2.1 are fulfilled.

Finally, we will prove that (iii) of Theorem 2.1 is satisfied.

Let $F(u,\lambda)=\pm\lambda x+(1-\lambda)JQNu$. According to previous argument, we have

$$F(u,\lambda)\neq0 \quad\text{for } u\in\operatorname{Ker}L\cap \partial \varOmega. $$

Thus by the homotopy property of degree we have

$$\begin{aligned} \operatorname{deg}(JQN|_{\operatorname{Ker}L}, \operatorname{Ker}L\cap \varOmega,0)&=\operatorname{deg}\bigl(F(\cdot,0), \operatorname{Ker}L\cap \varOmega,0\bigr) \\ &=\operatorname{deg}\bigl(F(\cdot,1),\operatorname{Ker}L\cap\varOmega,0\bigr) \\ &=\operatorname{deg}(\pm I,\operatorname{Ker}L\cap\varOmega,0)\neq0. \end{aligned} $$

Then by Theorem 2.1$Lu=Nu$ has at least one solution in $\operatorname {dom}L\cap\overline{\varOmega}$, so that IBVP (1.1) has a solution. □