## 1 Introduction

Fractional calculus is a generalization of ordinary differentiation and integration on an arbitrary order that can be noninteger. This subject, as old as the problem of ordinary differential calculus, can go back to the times when Leibniz and Newton invented differential calculus. As is known to all, the problem for fractional derivative was originally raised by Leibniz in a letter, dated September 30, 1695.

In recent years, the fractional differential equations have received more and more attention. The fractional derivative has been occurring in many physical applications such as a non-Markovian diffusion process with memory [1], charge transport in amorphous semiconductors [2], propagations of mechanical waves in viscoelastic media [3], etc. Phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry, and material science are also described by differential equations of fractional order (see [49]).

Recently, boundary value problems (BVPs for short) for fractional differential equations at nonresonance have been studied in many papers (see [1016]). Moreover, Kosmatov studied the BVPs for fractional differential equations at resonance (see [17]). Motivated by the work above, in this paper, we consider the following BVP of fractional equation at resonance

$\left\{\begin{array}{cc}\hfill {D}_{{0}^{+}}^{\alpha }x\left(t\right)=f\left(t,x\left(t\right),{x}^{\prime }\left(t\right),{x}^{″}\left(t\right)\right),\hfill & \hfill t\in \left[0,1\right],\hfill \\ \hfill x\left(0\right)=x\left(1\right),\hfill & \hfill {x}^{\prime }\left(0\right)={x}^{″}\left(0\right)=0,\hfill \end{array}\right\$
(1.1)

where ${D}_{{0}^{+}}^{\alpha }$ denotes the Caputo fractional differential operator of order α, 2 < α ≤ 3. f : [0, 1] × ℝ3 → ×ℝ is continuous.

The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions for BVP (1.1) under nonlinear growth restriction of f, basing on the coincidence degree theory due to Mawhin (see [18]). Finally, in Section 4, an example is given to illustrate the main result.

## 2 Preliminaries

In this section, we will introduce notations, definitions, and preliminary facts that are used throughout this paper.

Let X and Y be real Banach spaces and let L : domLXY be a Fredholm operator with index zero, and P : XX, Q : YY be projectors such that

$\begin{array}{c}\mathsf{\text{Im}}P=\mathsf{\text{Ker}}L,\phantom{\rule{1em}{0ex}}\mathsf{\text{Ker}}Q=\mathsf{\text{Im}}L,\\ X=\mathsf{\text{Ker}}L\oplus \mathsf{\text{Ker}}P,\phantom{\rule{1em}{0ex}}Y=\mathsf{\text{Im}}L\oplus \mathsf{\text{Im}}Q.\end{array}$

It follows that

$L{|}_{\mathsf{\text{dom}}L\cap \mathsf{\text{Ker}}P}\phantom{\rule{2.77695pt}{0ex}}:\mathsf{\text{dom}}L\cap \mathsf{\text{Ker}}P\to \mathsf{\text{Im}}L$

is invertible. We denote the inverse by K P .

If Ω is an open bounded subset of X, and $\mathsf{\text{dom}}L\cap \stackrel{̄}{\Omega }\ne \varnothing$, the map N : XY will be called L-compact on $\overline{\Omega }$ if $QN\left(\overline{\Omega }\right)$ is bounded and ${K}_{P}\left(I-Q\right)N:\overline{\Omega }\to X$ is compact. Where I is identity operator.

Lemma 2.1. ([18]) If Ω is an open bounded set, let L : domLXY be a Fredholm operator of index zero and N : XY L-compact on $\overline{\Omega }$. Assume that the following conditions are satisfied

1. (1)

LxλNx for every (x, λ) ∈ [(domL\KerL)] ∩ ∂Ω × (0, 1);

2. (2)

Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω;

3. (3)

deg(QN|KerL, KerL ∩ Ω, 0) ≠ 0, where Q : YY is a projection such that ImL = KerQ.

Then the equation Lx = Nx has at least one solution in $\mathsf{\text{dom}}L\cap \overline{\Omega }$.

Definition 2.1. The Riemann-Liouville fractional integral operator of order α > 0 of a function x is given by

${I}_{{0}^{+}}^{\alpha }x\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -1}x\left(s\right)\mathsf{\text{d}}s,$

provided that the right side integral is pointwise defined on (0, +∞).

Definition 2.2. The Caputo fractional derivative of order α > 0 of a continuous function x is given by

${D}_{{0}^{+}}^{\alpha }x\left(t\right)={I}_{{0}^{+}}^{n-\alpha }\frac{{\mathsf{\text{d}}}^{n}x\left(t\right)}{\mathsf{\text{d}}{t}^{n}}=\frac{1}{\Gamma \left(n-\alpha \right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{n-\alpha -1}{x}^{\left(n\right)}\left(s\right)\mathsf{\text{d}}s,$

where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on (0, +∞).

Lemma 2.2. ([19]) For α > 0, the general solution of the Caputo fractional differential equation

${D}_{{0}^{+}}^{\alpha }x\left(t\right)=0$

is given by

$x\left(t\right)={c}_{0}+{c}_{1}t+{c}_{2}{t}^{2}+\cdots +{c}_{n-1}{t}^{n-1},$

where c i ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.

Lemma 2.3. ([19]) Assume that xC(0, 1) ∩ L(0, 1) with a Caputo fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L(0, 1). Then,

${I}_{{0}^{+}}^{\alpha }{D}_{{0}^{+}}^{\alpha }x\left(t\right)=x\left(t\right)+{c}_{0}+{c}_{1}t+{c}_{2}{t}^{2}+\cdots +{c}_{n-1}{t}^{n-1}$

where c i ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.

In this paper, we denote X = C2[0, 1] with the norm ||x|| X = max{||x||, ||x'||, ||x"||} and Y = C[0, 1] with the norm ||y|| Y = ||y||, where ||x|| = max t ∈[0, 1] |x(t)|. Obviously, both X and Y are Banach spaces.

Define the operator L : domLXY by

$Lx={D}_{{0}^{+}}^{\alpha }x,$
(2.1)

where

$\mathsf{\text{dom}}L=\left\{x\in X|{D}_{{0}^{+}}^{\alpha }x\left(t\right)\in Y,\phantom{\rule{2.77695pt}{0ex}}x\left(0\right)=x\left(1\right),\phantom{\rule{2.77695pt}{0ex}}{x}^{\prime }\left(0\right)={x}^{″}\left(0\right)=0\right\}.$

Let N : XY be the Nemytski operator

$Nx\left(t\right)=f\left(t,x\left(t\right),{x}^{\prime }\left(t\right),{x}^{″}\left(t\right)\right),\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right].$

Then, BVP (1.1) is equivalent to the operator equation

$Lx=Nx,\phantom{\rule{1em}{0ex}}x\in \mathsf{\text{dom}}L.$

## 3 Main result

In this section, a theorem on existence of solutions for BVP (1.1) will be given.

Theorem 3.1. Let f : [0, 1] × ℝ3 → ℝ be continuous. Assume that

(H1) there exist nonnegative functions p, q, r, sC[0, 1] with Γ(α - 1) - q1 - r1 - s1 > 0 such that

$|f\left(t,u,v,w\right)|\le p\left(t\right)+q\left(t\right)|u|+r\left(t\right)|v|+s\left(t\right)|w|,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{2.77695pt}{0ex}}\left(u,v,w\right)\in {ℝ}^{3},$

where p1 = ||p||, q1 = ||q||, r1 = ||r||, s1 = ||s||.

(H2) there exists a constant B > 0 such that for all u ∈ ℝ with |u| > B either

$uf\left(t,u,v,w\right)>0,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{2.77695pt}{0ex}}\left(v,w\right)\in {ℝ}^{2}$

or

$uf\left(t,u,v,w\right)<0,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{2.77695pt}{0ex}}\left(v,w\right)\in {ℝ}^{2}.$

Then, BVP (1.1) has at leat one solution in X.

Now, we begin with some lemmas below.

Lemma 3.1. Let L be defined by (2.1), then

$\mathsf{\text{Ker}}L=\left\{x\in X|x\left(t\right)={c}_{0},\phantom{\rule{2.77695pt}{0ex}}{c}_{0}\in ℝ,\phantom{\rule{2.77695pt}{0ex}}\forall t\in \left[0,1\right]\right\},$
(3.1)
$\mathsf{\text{Im}}L=\left\{y\in Y|\underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}y\left(s\right)\mathsf{\text{d}}s=0\right\}.$
(3.2)

Proof. By Lemma 2.2, ${D}_{{0}^{+}}^{\alpha }x\left(t\right)=0$ has solution

$x\left(t\right)={c}_{0}+{c}_{1}t+{c}_{2}{t}^{2},\phantom{\rule{1em}{0ex}}{c}_{0},{c}_{1},{c}_{2}\in ℝ.$

Combining with the boundary value condition of BVP (1.1), one has (3.1) hold.

For y ∈ ImL, there exists x ∈ domL such that y = LxY. By Lemma 2.3, we have

$x\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -1}y\left(s\right)\mathsf{\text{d}}s+{c}_{0}+{c}_{1}t+{c}_{2}{t}^{2}.$

Then, we have

${x}^{\prime }\left(t\right)=\frac{1}{\Gamma \left(\alpha -1\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -2}y\left(s\right)\mathsf{\text{d}}s+{c}_{1}+2{c}_{2}t$

and

${x}^{″}\left(t\right)=\frac{1}{\Gamma \left(\alpha -2\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -3}y\left(s\right)\mathsf{\text{d}}s+2{c}_{2}.$

By conditions of BVP (1.1), we can get that y satisfies

$\underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}y\left(s\right)\mathsf{\text{d}}s=0.$

Thus, we get (3.2). On the other hand, suppose yY and satisfies ${\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}y\left(s\right)\mathsf{\text{d}}s=0$. Let $x\left(t\right)={I}_{{0}^{+}}^{\alpha }y\left(t\right)$, then x ∈ domL and ${D}_{{0}^{+}}^{\alpha }x\left(t\right)=y\left(t\right)$. So that, y ∈ ImL. The proof is complete.

Lemma 3.2. Let L be defined by (2.1), then L is a Fredholm operator of index zero, and the linear continuous projector operators P : XX and Q : YY can be defined as

$\begin{array}{lll}\hfill Px\left(t\right)& =x\left(0\right),\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill Qy\left(t\right)& =\alpha \underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}y\left(s\right)\mathsf{\text{d}}s,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right].\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

Furthermore, the operator K P : ImL → domL ∩ KerP can be written by

${K}_{P}y\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -1}y\left(s\right)\mathsf{\text{d}}s,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right].$

Proof. Obviously, ImP = KerL and P2x = Px. It follows from x = (x - Px) + Px that X = KerP + KerL. By simple calculation, we can get that KerL ∩ KerP = {0}. Then, we get

$X=\mathsf{\text{Ker}}L\oplus \mathsf{\text{Ker}}P.$

For yY, we have

${Q}^{2}y=Q\left(Qy\right)=Qy\cdot \alpha \underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}\mathsf{\text{d}}s=Qy.$

Let y = (y - Qy) + Qy, where y - Qy ∈ KerQ = ImL, Qy ∈ ImQ. It follows from KerQ = ImL and Q2y = Qy that ImQ ∩ ImL = {0}. Then, we have

$Y=\mathsf{\text{Im}}L\oplus \mathsf{\text{Im}}Q.$

Thus,

$\mathsf{\text{dim}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{Ker}}L=\mathsf{\text{dim}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{Im}}Q=\mathsf{\text{codim}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{Im}}L=1.$

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

From the definitions of P, K P , it is easy to see that the generalized inverse of L is K P . In fact, for y ∈ ImL, we have

$L{K}_{P}y={D}_{{0}^{+}}^{\alpha }{I}_{{0}^{+}}^{\alpha }y=y.$
(3.3)

Moreover, for x ∈ domL ∩ KerP, we get x(0) = x'(0) = x"(0) = 0. By Lemma 2.3, we obtain that

${I}_{{0}^{+}}^{\alpha }Lx\left(t\right)={I}_{{0}^{+}}^{\alpha }{D}_{{0}^{+}}^{\alpha }x\left(t\right)=x\left(t\right)+{c}_{0}+{c}_{1}t+{c}_{2}{t}^{2},\phantom{\rule{1em}{0ex}}{c}_{0},\phantom{\rule{2.77695pt}{0ex}}{c}_{1},\phantom{\rule{2.77695pt}{0ex}}{c}_{2}\in ℝ,$

which together with x(0) = x'(0) = x"(0) = 0 yields that

${K}_{P}Lx=x.$
(3.4)

Combining (3.3) with (3.4), we know that K P = (L|domL∩KerP)-1. The proof is complete.

Lemma 3.3. Assume Ω ⊂ X is an open bounded subset such that $\mathsf{\text{dom}}L\cap \stackrel{̄}{\Omega }\ne \varnothing$, then N is L-compact on $\overline{\Omega }$.

Proof. By the continuity of f, we can get that $QN\left(\overline{\Omega }\right)$ and ${K}_{P}\left(I-Q\right)N\left(\overline{\Omega }\right)$ are bounded. So, in view of the Arzelà -Ascoli theorem, we need only prove that ${K}_{P}\left(I-Q\right)N\left(\overline{\Omega }\right)\subset X$ is equicontinuous.

From the continuity of f, there exists constant A > 0 such that |(I - Q)Nx| ≤ A, $\forall x\in \overline{\Omega }$, t ∈ [0, 1]. Furthermore, denote K P,Q = K P (I - Q)N and for 0 ≤ t1 < t2 ≤ 1, $x\in \overline{\Omega }$, we have

$\begin{array}{c}\left|\left({K}_{P,Q}x\right)\left({t}_{2}\right)-\left({K}_{P,Q}x\right)\left({t}_{1}\right)\right|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\Gamma \left(\alpha \right)}\left|\underset{0}{\overset{{t}_{2}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -1}\left(I-Q\right)Nx\left(s\right)\mathsf{\text{d}}s-\underset{0}{\overset{{t}_{1}}{\int }}{\left({t}_{1}-s\right)}^{\alpha -1}\left(I-Q\right)Nx\left(s\right)\mathsf{\text{d}}s\right|\\ \phantom{\rule{1em}{0ex}}\le \frac{A}{\Gamma \left(\alpha \right)}\left[\underset{0}{\overset{{t}_{1}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -1}-{\left({t}_{1}-s\right)}^{\alpha -1}\mathsf{\text{d}}s+\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -1}\mathsf{\text{d}}s\right]\\ \phantom{\rule{1em}{0ex}}=\frac{A}{\Gamma \left(\alpha +1\right)}\left({t}_{2}^{\alpha }-{t}_{1}^{\alpha }\right),\\ |{\left({K}_{P,Q}x\right)}^{\prime }\left({t}_{2}\right)-{\left({K}_{P,Q}x\right)}^{\prime }\left({t}_{1}\right)|\\ \phantom{\rule{1em}{0ex}}=\frac{\alpha -1}{\Gamma \left(\alpha \right)}\left|\underset{0}{\overset{{t}_{2}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -2}\left(I-Q\right)Nx\left(s\right)\mathsf{\text{d}}s-\underset{0}{\overset{{t}_{1}}{\int }}{\left({t}_{1}-s\right)}^{\alpha -2}\left(I-Q\right)Nx\left(s\right)\mathsf{\text{d}}s\right|\\ \phantom{\rule{1em}{0ex}}\le \frac{A}{\Gamma \left(\alpha -1\right)}\left[\underset{0}{\overset{{t}_{1}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -2}-{\left({t}_{1}-s\right)}^{\alpha -2}\mathsf{\text{d}}s+\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -2}\mathsf{\text{d}}s\right]\\ \phantom{\rule{1em}{0ex}}\le \frac{A}{\Gamma \left(\alpha \right)}\left({t}_{2}^{\alpha -1}-{t}_{1}^{\alpha -1}\right)\end{array}$

and

$\begin{array}{l}|\left({K}_{P,Q}x{\right)}^{″}\left({t}_{2}\right)-\left({K}_{P,Q}x{\right)}^{″}\left({t}_{1}\right)|\\ \phantom{\rule{0.25em}{0ex}}=\frac{\left(\alpha -2\right)\left(\alpha -1\right)}{\Gamma \left(\alpha \right)}|\underset{0}{\overset{{t}_{2}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -3}\left(I-Q\right)Nx\left(s\right)\text{d}s-\underset{0}{\overset{{t}_{1}}{\int }}{\left({t}_{1}-s\right)}^{\alpha -3}\left(I-Q\right)Nx\left(s\right)\text{d}s|\\ \phantom{\rule{0.25em}{0ex}}\le \frac{A}{\Gamma \left(\alpha -2\right)}\left[\underset{0}{\overset{{t}_{1}}{\int }}{\left({t}_{1}-s\right)}^{\alpha -3}-{\left({t}_{2}-s\right)}^{\alpha -3}\text{d}s+\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}{\left({t}_{2}-s\right)}^{\alpha -3}\text{d}s\right]\\ \phantom{\rule{0.25em}{0ex}}\le \frac{A}{\Gamma \left(\alpha -1\right)}\left[{t}_{1}^{\alpha -2}-{t}_{2}^{\alpha -2}+2\left({t}_{2}-{t}_{1}{\right)}^{\alpha -2}\right].\end{array}$

Since tα , tα-1 and tα-2 are uniformly continuous on [0, 1], we can get that ${K}_{P,Q}\left(\overline{\Omega }\right)\subset C\left[0,1\right]$, ${\left({K}_{P,Q}\right)}^{\prime }\left(\overline{\Omega }\right)\subset C\left[0,1\right]$ and ${\left({K}_{P,Q}\right)}^{″}\left(\overline{\Omega }\right)\subset C\left[0,1\right]$ are equicontinuous. Thus, we get that ${K}_{P,Q}:\overline{\Omega }\to X$ is compact. The proof is completed.

Lemma 3.4. Suppose (H1), (H2) hold, then the set

${\Omega }_{1}=\left\{x\in \mathsf{\text{dom}}L\\mathsf{\text{Ker}}L|Lx=\lambda Nx,\phantom{\rule{1em}{0ex}}\lambda \in \left(0,1\right)\right\}$

is bounded.

Proof. Take x ∈ Ω1, then Nx ∈ ImL. By (3.2), we have

$\underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}f\left(s,x\left(s\right),{x}^{\prime }\left(s\right),{x}^{″}\left(s\right)\right)\mathsf{\text{d}}s=0.$

Then, by the integral mean value theorem, there exists a constant ξ ∈ (0, 1) such that f(ξ, x(ξ), x'(ξ), x"(ξ)) = 0. Then from (H2), we have |x(ξ)| ≤ B.

Then, we have

$|x\left(t\right)|\phantom{\rule{2.77695pt}{0ex}}=\left|x\left(\xi \right)+\underset{\xi }{\overset{t}{\int }}{x}^{\prime }\left(s\right)\mathsf{\text{d}}s\right|\le B+\parallel {x}^{\prime }{\parallel }_{\infty }.$

That is

$\parallel x{\parallel }_{\infty }\le B+\parallel {x}^{\prime }{\parallel }_{\infty }.$
(3.5)

From x ∈ domL, we get x'(0) = 0. Therefore,

$|{x}^{\prime }\left(t\right)|=\left|{x}^{\prime }\left(0\right)+\underset{0}{\overset{t}{\int }}{x}^{″}\left(s\right)\mathsf{\text{d}}s\right|\le \parallel {x}^{″}{\parallel }_{\infty }.$

That is

$\parallel {x}^{\prime }{\parallel }_{\infty }\le \parallel {x}^{″}{\parallel }_{\infty }.$
(3.6)

By Lx = λNx and x ∈ domL, we have

$x\left(t\right)=\frac{\lambda }{\Gamma \left(\alpha \right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -1}f\left(s,x\left(s\right),{x}^{\prime }\left(s\right),{x}^{″}\left(s\right)\right)\mathsf{\text{d}}s+x\left(0\right).$

Then we get

${x}^{\prime }\left(t\right)=\frac{\lambda }{\Gamma \left(\alpha -1\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -2}f\left(s,x\left(s\right),{x}^{\prime }\left(s\right),{x}^{″}\left(s\right)\right)\mathsf{\text{d}}s$

and

${x}^{″}\left(t\right)=\frac{\lambda }{\Gamma \left(\alpha -2\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -3}f\left(s,x\left(s\right),{x}^{\prime }\left(s\right),{x}^{″}\left(s\right)\right)\mathsf{\text{d}}s.$

From (3.5),(3.6), and (H1), we have

$\begin{array}{lll}\hfill {∥{x}^{″}∥}_{\infty }& \le \frac{1}{\Gamma \left(\alpha -2\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -3}|f\left(s,x\left(s\right),{x}^{\prime }\left(s\right),{x}^{″}\left(s\right)\right)|\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ \le \frac{1}{\Gamma \left(\alpha -2\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -3}\left[p\left(s\right)+q\left(s\right)|x\left(s\right)|+r\left(s\right)|{x}^{\prime }\left(s\right)|+s\left(s\right)|{x}^{″}\left(s\right)|\right]\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ \le \frac{1}{\Gamma \left(\alpha -2\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -3}\left({p}_{1}+{q}_{1}\parallel x{\parallel }_{\infty }+{r}_{1}\parallel {x}^{\prime }{\parallel }_{\infty }+{s}_{1}\parallel {x}^{″}{\parallel }_{\infty }\right)\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ \le \frac{1}{\Gamma \left(\alpha -2\right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -3}\left[{p}_{1}+{q}_{1}B+\left({q}_{1}+{r}_{1}+{s}_{1}\right)\parallel {x}^{″}{\parallel }_{\infty }\right]\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ \le \frac{1}{\Gamma \left(\alpha -1\right)}\left[{p}_{1}+{q}_{1}B+\left({q}_{1}+{r}_{1}+{s}_{1}\right)\parallel {x}^{″}{\parallel }_{\infty }\right].\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

Thus, from Γ(α - 1) - q1 - r1 - s1 > 0, we obtain that

$\parallel {x}^{″}{\parallel }_{\infty }\le \frac{{p}_{1}+{q}_{1}B}{\Gamma \left(\alpha -1\right)-{q}_{1}-{r}_{1}-{s}_{1}}:={M}_{1}.$

Thus, we get

$\parallel {x}^{\prime }{\parallel }_{\infty }\le \parallel {x}^{″}{\parallel }_{\infty }\le {M}_{1}$

and

$\parallel x{\parallel }_{\infty }\le B+\parallel {x}^{\prime }{\parallel }_{\infty }\le B+{M}_{1}.$

Therefore,

$\parallel x{\parallel }_{X}\le max\left\{{M}_{1},B+{M}_{1}\right\}.$

So Ω1 is bounded. The proof is complete.

Lemma 3.5. Suppose (H2) holds, then the set

${\Omega }_{2}=\left\{x|x\in \mathsf{\text{Ker}}L,Nx\in \mathsf{\text{Im}}L\right\}$

is bounded.

Proof. For x ∈ Ω2, we have x(t) = c, c ∈ ℝ, and Nx ∈ ImL. Then, we get

$\underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}f\left(s,c,0,0\right)\mathsf{\text{d}}s=0,$

which together with (H2) implies |c| ≤ B. Thus, we have

$\parallel x{\parallel }_{X}\le B.$

Hence, Ω2 is bounded. The proof is complete.

Lemma 3.6. Suppose the first part of (H2) holds, then the set

${\Omega }_{3}=\left\{x|x\in \mathsf{\text{Ker}}L,\lambda x+\left(1-\lambda \right)QNx=0,\phantom{\rule{1em}{0ex}}\lambda \in \left[0,1\right]\right\}$

is bounded.

Proof. For x ∈ Ω3, we have x(t) = c, c ∈ ℝ, and

$\lambda c+\left(1-\lambda \right)\alpha \underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}f\left(s,c,0,0\right)\mathsf{\text{d}}s=0.$
(3.7)

If λ = 0, then |c| ≤ B because of the first part of (H2). If λ ∈ (0, 1], we can also obtain |c| ≤ B. Otherwise, if |c| > B, in view of the first part of (H2), one has

$\lambda {c}^{2}+\left(1-\lambda \right)\alpha \underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}cf\left(s,c,0,0\right)\mathsf{\text{d}}s>0,$

which contradicts to (3.7).

Therefore, Ω3 is bounded. The proof is complete.

Remark 3.1. Suppose the second part of (H2) hold, then the set

${\Omega }_{3}^{\prime }=\left\{x|x\in \mathsf{\text{Ker}}L,\phantom{\rule{2.77695pt}{0ex}}-\lambda x+\left(1-\lambda \right)QNx=0,\phantom{\rule{1em}{0ex}}\lambda \in \left[0,1\right]\right\}$

is bounded.

The proof of Theorem 3.1. Set Ω = {xX | ||x|| X < max{M1, B, B + M1} + 1}. It follows from Lemma 3.2 and 3.3 that L is a Fredholm operator of index zero and N is L-compact on $\overline{\Omega }$. By Lemma 3.4 and 3.5, we get that the following two conditions are satisfied

1. (1)

LxλNx for every (x, λ) ∈ [(domL\KerL) ∩ ∂Ω] × (0, 1);

2. (2)

Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω.

Take

$H\left(x,\lambda \right)=±\lambda x+\left(1-\lambda \right)QNx.$

According to Lemma 3.6 (or Remark 3.1), we know that H(x, λ) ≠ 0 for x ∈ KerL ∩ ∂Ω. Therefore,

$\begin{array}{lll}\hfill \mathsf{\text{deg}}\left({QN|}_{\mathsf{\text{Ker}}L},\Omega \cap \mathsf{\text{Ker}}L,0\right)& =\mathsf{\text{deg}}\left(H\left(\cdot ,0\right),\Omega \cap \mathsf{\text{Ker}}L,0\right)\phantom{\rule{2em}{0ex}}& \hfill \\ =\mathsf{\text{deg}}\left(H\left(\cdot ,1\right),\Omega \cap \mathsf{\text{Ker}}L,0\right)\phantom{\rule{2em}{0ex}}& \hfill \\ =\mathsf{\text{deg}}\left(±I,\Omega \cap \mathsf{\text{Ker}}L,0\right)\ne 0.\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

So that, the condition (3) of Lemma 2.1 is satisfied. By Lemma 2.1, we can get that Lx = Nx has at least one solution in $\mathsf{\text{dom}}L\cap \overline{\Omega }$. Therefore, BVP (1.1) has at least one solution. The proof is complete.

## 4 An example

Example 4.1. Consider the following BVP

$\left\{\begin{array}{ll}{D}_{{0}^{+}}^{\frac{5}{2}}x\left(t\right)=\frac{t}{16}\left(x-10\right)+\frac{{t}^{2}}{16}{\text{e}}^{-|{x}^{\prime }|}+\frac{{t}^{3}}{16}\mathrm{sin}\left[\left({x}^{″}{\right)}^{2}\right],\hfill & \phantom{\rule{0.25em}{0ex}}t\in \left[0,1\right]\hfill \\ x\left(0\right)=x\left(1\right),\hfill & \phantom{\rule{0.25em}{0ex}}{x}^{\prime }\left(0\right)={x}^{″}\left(0\right)=0.\hfill \end{array}$
(4.1)

where

$f\left(t,u,v,w\right)=\frac{t}{16}\left(u-10\right)+\frac{{t}^{2}}{16}{e}^{-|v|}+\frac{{t}^{3}}{16}sin\left({w}^{2}\right).$

Choose $p\left(t\right)=\frac{10t+2}{16}$, $q\left(t\right)=\frac{t}{16}$, r(t) = 0, s(t) = 0, B = 10. We can get that ${q}_{1}=\frac{1}{16}$, r1 = 0, s1 = 0 and

$\Gamma \left(\frac{5}{2}-1\right)-{q}_{1}-{r}_{1}-{s}_{1}>0.$

Then, all conditions of Theorem 3.1 hold, so BVP (4.1) has at least one solution.