Nonlocal problems of fractional systems involving left and right fractional derivatives at resonance

1.   Introduction

In the recent decades, fractional differential equation has received extensive attention in mathematical theory and application research, see ^[1,2,3,4,5] and the references therein. A great deal of research results have been obtained in the theory and application of fractional differential equations, see ^{[6,7,8,9,10,11,12,13,14,15,16,17,18]} and the references therein. Meanwhile, the differential equations with left and right fractional derivatives are also playing an important role in many different applications. For example, in ^[20,19], this type of differential equations is used to describe the temperature distribution of building walls while in ^[21], it is used to simulate the movement of particulate matter in the process of silo emptying. The theoretical research of this kind of problem has also attracted lots of attention, see ^{[22,23,24,25,26,27,28,29,30,31]}.

In this paper, we study the following fractional systems which involve both left and right fractional derivatives

with the nonlocal boundary conditions

where $1 < \alpha, \beta \leq 2$, $_{0}^C D_{t}^\alpha$ and $_{t}^C D_{1}^\beta$ represent the left and right Caputo fractional derivative operator, respectively. $f_i\in C([0, 1]\times \mathbb{R}^2, \mathbb{R})$ may be nonlinear functions, and $\omega\in C([0, 1], [0, +\infty))$ is a given function, $r$ is a real.

The purpose of this paper is to obtain the existence results for solutions of boundary value problem (1.1) and (1.2) under the resonant condition

As an application of our results, we deal with the existence result of the solution to fractional differential equation under the resonant conditions which involves both left and right fractional derivatives

which satisfies certain nonlocal boundary conditions.

2.   Preliminaries

In this section, we show some basic definitions for the fractional calculus and related lemmas which are used to establish the main results.

Definition 2.1. (See ^[1,3]) Suppose $\gamma > 0$, then the order $\gamma$ Riemann-Liouville left fractional integral and Caputo left fractional derivative of function $y:[0, 1]\rightarrow \mathbb{R}$ are defined by

respectively, provided the right sides exist. And the order $\gamma$ Riemann-Liouville right fractional integral and Caputo right fractional derivative of $y$ are given by

respectively, provided the right-side integral converges, where $n$ is an integer with $n-1 < \gamma < n$.

Lemma 2.1. (See ^[1,3]) For $n-1 < \gamma < n$, $n$ is a positive integer, then the general solution of fractional differential equation $_{0}^C D_{t}^{\gamma } y(t) = 0$ is given by

and the general solution of fractional differential equation $_{t}^C D_{1}^{\gamma } y(t) = 0$ is given by

where $c_{j}, \, d_{j}\in \mathbb{R}, \ j = 0, 1, \cdots, n$.

Definition 2.2. (see ^[32], P39) Let $\mathcal{X}$, $\mathcal{Y}$ be normed vector spaces, $L :\operatorname{Dom} L \subset \mathcal{X} \rightarrow\mathcal{Y}$ a linear mapping. The mapping $L$ will be called a Fredholm mapping of index zero if

(a) $\dim \operatorname{Ker} L = \operatorname{codim} \mathrm{Im} L < +\infty$;

(b) $\mathrm{Im} L$ is closed in $\mathcal{Y}$.

Lemma 2.2. (Generalized Krasnosel'skii theorem, see ^[32], P32) Let $\mathcal{X}$ and $\mathcal{Y}$ be Banach spaces. Let $L:\, \operatorname{dom} L\subset \mathcal{X}\to \mathcal{Y}$ be a Fredholm mapping of index zero, $N:\, \mathcal{X}\to \mathcal{Y}$ be an $L$-compact mapping in $\overline\Omega$ with $\Omega$ open, bounded, symmetric with respect to the origin and containing it. If

for every $x\in \operatorname{Dom} L\cap\partial\Omega$ and every $\lambda\in[0, 1]$, where $\partial\Omega$ is the boundary of $\Omega$ with respect to $\mathcal{X}$, then equation $Lx = Nx$ has at least one solution in $\Omega$.

3.   The existence of solutions for the systems

In this section, we present the existence results of the solutions of boundary value problem (1.1) and (1.2).

Let

be endowed with the norm

Then $(\mathcal{X}, \|\cdot\|)$ and $(\mathcal{Y}, \|\cdot\|)$ are Banach spaces.

Denote vector functions

and an operator

Let $L:\, \operatorname{Dom} L\subseteq \mathcal{X}\rightarrow \mathcal{Y}$ by

where

Define $N:\, \mathcal{X}\rightarrow \mathcal{Y}$ by

It is clear that boundary value problem (1.1) and (1.2) is equivalent to the following operator equation

Lemma 3.1. Let $L$ be defined by (3.1). Then $L$ is a Fredholm operator of index zero.

Proof. Obviously $L$ is a linear operator. Next, we consider the kernel of the linear operator $L$.

which implies that

Then

Take the boundary conditions $u_1'(0) = 0$ and $u_2'(1) = 0$ into account, and we can get $c_1 = d_1 = 0$.

In view of $u_1(0) = r u_2(1)$ and $u_2(0) = \int^1_0\omega(t)u_1(t)\mathrm{d}t$ and the resonant condition $r\int^1_0\omega(t)\mathrm{d}t = 1$, we can show that $c_0 = rd_0$. Therefore,

which implies that $\mathrm{dimKer} L = 1$.

Following, we denote

And for ${\bf{y}} = \left({\begin{array}{c} y_1(t)\\ y_2(t) \end{array}} \right)\in \mathcal{Y}$, we denote

We prove that

Since

for any ${\bf{y}} = \left({\begin{array}{c} y_1(t)\\y_2(t)\end{array}} \right)\in \mathrm{Im} L$, there exists ${\bf{u}} = \left({\begin{array}{c} u_1(t)\\u_2(t) \end{array}} \right)\in \mathrm{Dom} L$ such that

Then

And

By the boundary conditions $u_1'(0) = u_2'(1) = 0$, we get $c_1 = d_1 = 0$. And by $u_1(0) = r u_2(1)$, $u_2(0) = \int^1_0\omega(t)u_1(t)\mathrm{d}t$, we have

and

Then

It follows

from the resonant condition $r\int^1_0\omega(s)\mathrm{d}s = 1$. That is

Then, Eq. (3.11) is equivalent to

and

On the other hand, for every ${\bf{y}} = \left({\begin{array}{c} y_1(t)\\ y_2(t) \end{array}} \right)\in\Big\{{\bf{y}}\in \mathcal{Y}:\; \left < {\bf{\Lambda}}, \; {\bf{y}}\right > = 0\Big\}$, let

then $u\in \mathrm{Dom}L$ and $L{\bf{u}} = {\bf{y}}$. So

Hence (3.6) holds.

Following we prove that $\mathrm{Im} L$ is closed.

Assume that ${\bf{y}}_k = \left({\begin{array}{c} y_{k, 1}(t)\\ y_{k, 2}(t) \end{array}} \right)\in \mathrm{Im} L,$ and ${\lim\limits_{k\rightarrow \infty}{\bf{y}}_k} = {\bf{y}}_0 = \left({\begin{array}{c} y_{0, 1}(t)\\ y_{0, 2}(t) \end{array}} \right)$.

Since $\lim\limits_{k\rightarrow \infty}\|{\bf{y}}_k-{\bf{y}}_0\| = 0$, then $\lim\limits_{k\rightarrow \infty}|y_{k, i}-y_{0, i}| = 0$ for $i = 1, 2$.

Because ${\bf{y}}_k\in \mathrm{Im} L$ which implies $y_{k, i}\in C[0, 1]$ for $k = 1, 2, \cdots$ and $i = 1, 2$, then $y_{0, i}\in C[0, 1]$ for $i = 1, 2$, which implies ${\bf{y}}_0\in \mathcal{Y}$.

By (3.6), we can get that

Then

which implies that ${\bf{y}}_0\in\mathrm{Im} L$ and $\mathrm{Im} L$ is a closed in $\mathcal{Y}$.

Because ${\bf{\Lambda}}$ is a fixed vector function, then $\left < {\bf{\Lambda}}, \; {\bf{y}}\right > \in \mathbb{R}$, for any ${\bf{y}}\in\mathcal{Y}$, which implies $\mathrm{dim} (\mathcal{Y}/\mathrm{Im} L) = 1$.

So

Therefore, we get that the linear operator $L$ is a Fredholm operator with index zero. For the definition of Fredholm operator with index zero, see Definition 2.2.

Define $P :\, \mathcal{X}\to \mathcal{X}$ by

Then $P$ is a linear continuous projector operator. We can easily check that $\mathrm{Im} P = \mathrm{Ker} L$ and $\mathcal{X} = \mathrm{Ker} P\oplus\mathrm{Ker} L$.

So the operator $L|_{\operatorname{Dom} L \cap \mathrm{Ker} P}\; :\; \operatorname{Dom} L \cap \mathrm{Ker} P \longrightarrow \mathrm{Im} L$ is reversible.

For every $y\in \mathrm{Im} L$, there exists $u\in \operatorname{Dom} L \cap \mathrm{Ker} P$ such that

Then

Combining the boundary conditions and noticing $u\in \operatorname{Dom} L \cap \mathrm{Ker} P$, and we can get that $c_0 = c_1 = d_0 = d_1 = 0$. Then

where $L_{P}^{-1}$ is the inverse of $L|_{\operatorname{Dom} L \cap \mathrm{Ker} P}$.

Define the operator $Q :\, \mathcal{Y} \to \mathcal{Y}/\mathrm{Im} L$ by

Let $\bf{y_0} = \left({\begin{array}{c}0\\1 \end{array}} \right),$ then

and $Q^{2} = Q$. That is, $Q$ is a linear continuous projector operator.

We can easily see that $\mathrm{Ker} Q = \mathrm{Im} L$ and $Y = \mathrm{Im} L\oplus\mathrm{Im} Q$.

Since ${\bf{f}}$ is continuous, it follows that Lemma 3.2 holds from (3.13) and (3.14).

Lemma 3.2. $N:\, \mathcal{X}\to \mathcal{Y}$ is an $L$-compact operator.

Denote

(H1) The functions $f_i\in C([0, 1]\times \mathbb{R}^2, \mathbb{R})$, there exist constants $\sigma_i\geq 0$ and functions $a_{i}, \, b_{i}\in C([0, 1], \, [0, \, +\infty))$, $i = 1, 2$, such that

for any $t\in [0, 1]$, $x_j, \, y_j\in \mathbb{R}$, $j = 1, 2$.

(H2) The functions $f_i\in C([0, 1]\times \mathbb{R}^2, \mathbb{R})$ and

where $R_1 = \frac{\Gamma(\alpha+1)}{2}$ and $R_2 = \frac{\Gamma(\beta+1)}{2}$.

For convenience, let

and

Theorem 3.1. Suppose (H1) holds and $0\leq \sigma < 1$. Then boundary value problem (1.1) and (1.2) has at least one solution.

Proof. If ${\bf{u}}\in \mathrm{Dom} L \cap \mathrm{Ker} P$ satisfies the following equation

then

and

We can get

and

where $m_0$ and $f_0$ are given by (3.15) and (3.16).

Similarly, we can show

As a result,

Since $0\leq \sigma < 1$, we take $M\geq \max\{(2m_0)^{\frac{1}{1-\sigma}}, 2f_0\}+1$ and

Then $\Omega$ is open, bounded, symmetric with respect to the origin and containing it.

If ${\bf{u}}\in \operatorname{Dom} L\cap\partial\Omega$ and satisfies (3.17), by (3.21), we can show that

which is a contradiction.

Therefore, we can obtain that

By Lemma 3.2, $N$ is an $L$-compact operator.

According to Lemma 2.2, we have the equation $L{\bf{u}} = N{\bf{u}}$ has at least one solution on $\operatorname{Dom} L\cap\overline\Omega$. Namely, boundary value problem (1.1) and (1.2) has at least one solution.

Theorem 3.2. Suppose (H1) holds. If $\sigma = 1$ and $m_0 < 1$, then boundary value problem (1.1) and (1.2) has at least one solution.

Proof. Since $m_0 < 1$, we take $M > \frac{f_0}{1-m_0}$ and

Then $\Omega$ is open, bounded, symmetric with respect to the origin and containing it.

If ${\bf{u}}\in \mathrm{Dom} L \cap \mathrm{Ker} P$ satisfies equation (3.17), similar to the proof of the theorem 1, we can get that

It follows

from (3.22). We can show that

By Lemma 3.2, $N:\, \mathcal{X}\to \mathcal{Y}$ is an $L$-compact operator.

According to Lemma 2.2, we have the equation $L{\bf{u}} = N{\bf{u}}$ has at least one solution on $\operatorname{Dom} L\cap\overline\Omega$. Namely, boundary value problem (1.1) and (1.2) has at least one solution.

Theorem 3.3. Suppose (H2) holds, then boundary value problem (1.1) and (1.2) has at least one solution.

Proof. For $i = 1, 2$, let $\epsilon_i = \frac{1}{2}\big(R_i-\limsup\limits_{|x|+|y|\to\infty} \sup_{t\in[0, 1]}\frac{|f_i(t, x, y)|}{|x|+|y|}\big) > 0$. By (H2), there exist constants $M_i$ such that

Since $f_i$ are continuous, there exist constants $R_{0i}$ such that

We have

Let $M > \max\{\frac{R_{01}}{\Gamma(\alpha+1)-2(R_{1}-\epsilon_1)}, \frac{R_{02}}{\Gamma(\beta+1)-2(R_{2}-\epsilon_1)}\}$ and

Then $\Omega$ is open, bounded, symmetric with respect to the origin and containing it.

If ${\bf{u}}\in \mathrm{Dom} L \cap \mathrm{Ker} P$ satisfies Eq. (3.17), we can get that

Hence, if ${\bf{u}}\in \operatorname{Dom} L\cap\partial\Omega$ and $\lambda\in[0, 1]$, then

Similarly, we can show

It follows

from (3.24) and (3.25), which is a contradiction. We show that

By Lemma 3.2, $N$ is an $L$-compact operator.

According to Lemma 2.2, we have that the equation $L{\bf{u}} = N{\bf{u}}$ has at least one solution on $\operatorname{Dom} L\cap\overline\Omega$. That is, boundary value problem (1.1) and (1.2) has at least one solution.

4.   The existence of solutions for the fractional equations

As an application of Theorem 3.2, in this section, we consider the existence of solutions for the following fractional differential equation which involves the left and right derivatives

with the following nonlocal boundary conditions

under the resonant condition $r\int_0^1\omega(s)ds = 1$.

Theorem 4.1. Assume $r\int_0^1\omega(s)ds = 1$, there exist $p, \, q\in C([0, 1], \, [0, +\infty))$ such that

for any $t\in [0, 1]$, $x_j, \, y_j\in \mathbb{R}$, $j = 1, 2$. If

then boundary value problem (4.1) and (4.2) has at least one solution.

Proof. Let $u_1(t) = u(t)$, $u_2(t) = {_{0}^C D_{t}^\alpha} u(t)$, $f_1(t, x, y) = y$ and $f_2(t, x, y) = g(t, x, y)$. Then Eq. (4.1) is equivalent to the following the system

and boundary conditions (4.2) is equivalent to (1.2).

We can easily check that all conditions in Theorem 3.2 are satisfied for the Eq. (4.1) with the boundary conditions (4.2).

By Theorem 3.2, we can get that the conclusion of Theorem 4.1 holds.

5.   Examples

In this section, we give out some examples to illustrate our main results.

Example 5.1 We consider the following fractional integral boundary value problems of the nonlinear fractional differential system

where $\alpha = \frac{3}{2}$, $\beta = \frac{5}{4}$, $r = \frac{3\sqrt{\pi}}{4}$. And the condition $u_2(0) = {_{0}I_{t}^\frac{3}{2}}u_1(1)$ is equivalent to

where $\omega(t) = \frac{2}{\sqrt{\pi}}(1-t)^{\frac{1}{2}}$. And

So boundary value problem (5.1) is a resonance problem.

Let $f_1(t, x, y) = \frac{1}{3} \arctan\big(2 (1-t)^{\frac{1}{2}} x+3 t^{\frac{1}{4}} y\big{)}+e^{t}$ and $f_2(t, x, y) = \frac{1}{4} \arctan\big(3(1-t)^{\frac{1}{2}} x+2t^{\frac{1}{4}} y\big{)}+e^{-t}$, $a_1(t) = 2 (1-t)^{\frac{1}{2}}$, $a_2(t) = 3(1-t)^{\frac{1}{2}}$, $b_1(t) = 3 t^{\frac{1}{4}}$, $b_2(t) = 2t^{\frac{1}{4}}$, then we can easily check $f_1$ and $f_2$ satisfy the condition (H1) with $\sigma = 1$. And we can get that

and

As a result, all conditions in Theorem 3.2 hold. By Theorem 3.2, boundary value problem (5.1) has at least one solution.

Example 5.2 We consider the following boundary value problem

Let $f_1(t, x, y) = \frac{2}{3} (1-t)^{\frac{1}{2}} x^\frac{3}{2}+\frac{4}{5} t^{\frac{1}{4}} y^\frac{5}{4}+e^{t}$ and $f_2(t, x, y) = \frac{4}{5} t^{\frac{1}{4}} x^\frac{5}{4}+\frac{2}{3} (1-t)^{\frac{1}{2}} y^\frac{3}{2}+e^{-t}$, $a_1(t) = 2(1-t)^{\frac{1}{2}}$, $a_2(t) = 2(1-t)^{\frac{1}{2}}$, $b_1(t) = 3 t^{\frac{1}{4}}$, $b_2(t) = 3t^{\frac{1}{4}}$. Other parameters are same as the ones in Example 5.1. Then we can easily check $f_i$, $i = 1, 2$, satisfy the conditions of (H1), where $\sigma = \frac{1}{2} < 1$.

Then the conditions in Theorem 3.1 hold. By Theorem 3.1, boundary value problem (5.3) has at least one solution.

Example 5.3 We consider the integral boundary value problem of fractional differential equation as following

Similar to Example 5.1, we can check that the conditions in Theorem 4.1 hold. As a result, boundary value problem (5.4) has at least one solution by Theorem 4.1.

Acknowledgments

This research was supported by the Natural Science Foundation of China (No. 111171220). Authors are grateful to the reviewers and editors for their suggestions and comments to improve the manuscript.

Conflict of interest

All authors declare no conflicts of interest in this paper.