A Class of ψ-Hilfer Fractional Pantograph Equations with Functional Boundary Data at Resonance

Abstract

In this paper, we explore the outcomes related to the existence of nonlocal functional boundary value problems associated with pantograph equations utilizing $\psi$-Hilfer fractional derivatives. The nonlinear term relies on unknown functions which contain a proportional delay term and their fractional derivatives in a higher order. We discuss various existence results for the different “smoothness” requirements of the unknown function by means of Mawhin’s coincidence theory at resonance. We wrap up by providing a detailed explanation accompanied by an illustration of one of the outcomes.

1. Introduction

Fractional differential equations (FDEs) play a crucial role in addressing non-standard dynamic behaviors intertwined with elements of short and long memory and hereditary influences, as highlighted in literature reviews of earlier research [1,2,3,4,5]. Recently, Vanterler and Capelas de Oliveira proposed a novel fractional derivative referred to as the $\psi$-Hilfer fractional derivative, which brings together various fractional definitions; a concise explanation can be found in the following referenced studies: [6,7,8]. Given that it comprises a large class of fractional derivatives, the reasonable inference is that it has a lot of advantages over classical derivatives and is better suited for practical applications. Recent improvements in FDEs using the $\psi$-Hilfer fractional derivative include those seen in [9,10,11,12,13,14,15,16,17,18].

One significant and unique topic in fractional calculus is the exploration of pantograph differential equations (PEs), which involve proportional delays. This concept arose from research on the electric current in the pantograph system of electric trains, conducted by Tayler and Ockendonare [19]. Over time, it has found applications in numerous fields of both pure and applied mathematics, such as quantum physics, electrodynamics, number theory, control systems, and probability theory. The pantograph, often utilized in electric trains and electric cells [20,21,22], is commonly seen as a device for measurement and graphing. As a result, there have been various efforts to develop pantograph-type fractional differential equations with a variety of specific boundary conditions [23,24,25,26,27,28,29,30,31,32]. Importantly, we can focus on problems with linear functional conditions [33,34,35,36,37,38], which greatly generalize and extend many specific boundary findings.

As far as we know, the pantograph equation boundary value problems that we talked about earlier with $\psi$-Hilfer fractional derivatives rarely mention resonance scenarios. Keeping the previous findings in perspective, we will introduce in this paper a set of $\psi$-Hilfer fractional functional boundary value problems related to pantograph equations:

$$\left\{\begin{array}{c}{}^{H}D_{0^{+}}^{\alpha ,\beta ;\psi }u\left(t\right)=f(t,u\left(t\right),u\left(\lambda t\right),D_{0^{+}}^{\gamma -2;\psi }u\left(t\right),D_{0^{+}}^{\gamma -1;\psi }u\left(t\right)),\lambda \in (0,1),t\in (0,1],\hfill \\ I_{0^{+}}^{n-\gamma ;\psi }u\left(0\right)=0,D_{0^{+}}^{\gamma -i;\psi }u\left(0\right)=0,i=3,4,\dots ,n-1,\hfill \\ B_1\left(u\right)=0,B_2\left(u\right)=0.\hfill \end{array}\right.$$

(1)

where ${}^{H}D_{0^{+}}^{\alpha ,\beta ;\psi }$ denotes the left $\psi$-Hilfer fractional derivative of order $\alpha (n-1<\alpha \le n,$$n>3)$ and type $\beta (0\le \beta \le 1)$, $f\in C([0,1]\times {\mathbb{R}}^4,\mathbb{R})$ and $\gamma =\alpha +\beta (n-\alpha )$. A boundary value problem is said to be at resonance if its corresponding homogeneous boundary value problem has a non-trivial solution; so, $B_1, B_2$ are continuous linear functionals with the resonance condition $\left(H\right)$:

$B_1\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)B_2\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right)=B_1\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right)B_2\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)$.

This research contains several key findings: To begin with, we examine functional boundary conditions in a highly general manner. Next, there is an effective integration of the $\psi$-Hilfer fractional derivative $\alpha (n-1<\alpha \le n, n>3)$. Additionally, the pantograph equation we explore has substantial practical applications, which we detail. Lastly, we address the resonance issue for this category of equations, which adds to the initial boundary value challenge for $\psi$-Hilfer fractional derivatives with pantograph equations.

The structure of this document is laid out as follows: Section 2 includes some introductions and essential concepts regarding linear operators, the coincidence degree continuation theorem, general fractional calculus, and supporting Banach spaces and lemmas. In Section 3, we present three existence findings concerning the various “smoothness” criteria of the unknown function related to resonance, relying on Mawhin’s coincidence theory, and we include a numerical example to illustrate our main findings.

Definition 1. 

We say that $u\in X$ (please refer to Section 2 for the definition of X) is a solution to the functional boundary value problem (FBVP) (Equation (1)) if u satisfies the equation and boundary conditions in (Equation (1)).

2. Basic Definitions and Preliminaries

We first restate the following definitions, theorem, and auxiliary lemmas related to fractional calculus theory; for details, see [6,7,8].

Definition 2 

([6,7,8]). The left-sided ψ-Riemann–Liouville fractional integral for an integrable function $y:[a,b]\to \mathbb{R}$ with respect to another function $\psi :[a,b]\to \mathbb{R},{\psi }^{\prime }\left(t\right)>0$, for all $t\in [a,b]$, is defined as follows:

$$I_{a^{+}}^{\alpha ;\psi }y\left(t\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_a^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}y\left(s\right)ds,t>a.$$

Definition 3 

([6,7,8]). The left-sided ψ-Riemann–Liouville fractional derivative of order $\alpha ,(n-1<\alpha <n)$ for a function $y\in C^n[a,b]$ with respect to another function $\psi \in C^n[a,b],{\psi }^{\prime }\left(t\right)>0$, for all $t\in [a,b]$, is defined as follows:

$$D_{a^{+}}^{\alpha ;\psi }y\left(t\right)={\displaystyle \frac{1}{\Gamma (n-\alpha )}}{\left({\displaystyle \frac{1}{{\psi }^{\prime }\left(t\right)}}{\displaystyle \frac{d}{dt}}\right)}^n{\int }_a^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{n-\alpha -1}y\left(s\right)ds,t>a.$$

Definition 4 

([6]). Let $[a,b]$ be an interval such that $-\infty \le a<b\le +\infty$ and $\psi \in C^n[a,b],$${\psi }^{\prime }\left(t\right)>0$ for all $t\in [a,b]$. The (left-sided) ψ-Hilfer fractional derivative of function $y\in C^n[a,b]$ of order α and type $0\le \beta \le 1$ is determined as follows:

$${}^{H}D_{a^{+}}^{\alpha ,\beta ;\psi }y\left(t\right)=I_{a^{+}}^{\beta (n-\alpha )}{\left({\displaystyle \frac{1}{{\psi }^{\prime }\left(t\right)}}{\displaystyle \frac{d}{dt}}\right)}^n{I}_{a^{+}}^{(1-\beta )(n-\alpha )}y\left(t\right)=I_{a^{+}}^{\gamma -\alpha ;\psi }{D}_{a^{+}}^{\gamma ;\psi }y\left(t\right),$$

where $n-1<\alpha <n, 0\le \beta \le 1, \gamma =\alpha +\beta (n-\alpha ), n=\left[\alpha \right]+1.$

Lemma 1 

([6,7,8]). Assume $f\in C^1[0,1], \alpha >0$ and $0\le \beta \le 1$; then, ${}^{H}D_{a^{+}}^{\alpha ,\beta ;\psi }{I}_{a^{+}}^{\alpha ;\psi }f\left(t\right)=f\left(t\right).$

Lemma 2 

([6,7,8]).

(1)

If $f\in C^n[a,b],n-1<\alpha <n,0\le \beta \le 1$ and $\gamma =\alpha +\beta (n-\alpha )$, then

$${\displaystyle I_{a^{+}}^{\alpha ;\psi }{}^{H}D_{a^{+}}^{\alpha ,\beta ;\psi }f\left(t\right)=f\left(t\right)-\sum _{k=1}^n{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(a\right))}^{\gamma -k}}{\Gamma (\gamma -k+1)}}{f}_{\psi }^{[n-k]}{I}_{a^{+}}^{(1-\beta )(n-\alpha );\psi }f\left(a\right),}$$

where $f_{\psi }^{[n-k]}f\left(t\right):={\left({\displaystyle \frac{1}{{\psi }^{\prime }\left(t\right)}}{\displaystyle \frac{d}{dt}}\right)}^{n-k}f\left(t\right).$

(2)

Assume $\alpha >0$ and $\delta >0$; then,

$$D_{0^{+}}^{\alpha ;\psi }{(\psi \left(x\right)-\psi \left(a\right))}^{\delta -1}={\displaystyle \frac{\Gamma \left(\delta \right)}{\Gamma (\delta -\alpha )}}{(\psi \left(x\right)-\psi \left(a\right))}^{\delta -\alpha -1},$$

and

$$I_{0^{+}}^{\alpha ;\psi }{(\psi \left(x\right)-\psi \left(a\right))}^{\delta -1}={\displaystyle \frac{\Gamma \left(\delta \right)}{\Gamma (\delta +\alpha )}}{(\psi \left(x\right)-\psi \left(a\right))}^{\delta +\alpha -1}.$$

Example 1. 

$D_{0^{+}}^{\alpha ;\psi }{\left(\psi \right(x)-\psi (a\left)\right)}^{\alpha -i}=0,i=1,2,\dots ,n.$

Definition 5 

([39,40]). Let X, Z be real Banach spaces and $L:domL\subset X\to Z$ be a linear operator. X is said to be the Fredholm operator of index zero provided that the following are true:

(i)

$\mathrm{Im}L$ is a closed subset of Z;

(ii)

$\dim \mathrm{Ker}L=\mathrm{codim}\mathrm{Im}L<+\infty$.

Let $P:X\to X$ and $Q:Z\to Z$ be continuous projectors such that $\mathrm{Im}P=\mathrm{Ker}L$, $\mathrm{Ker}Q=\mathrm{Im}L$, $X=\mathrm{Ker}L\oplus \mathrm{Ker}P$, and $Z=\mathrm{Im}L\oplus \mathrm{Im}Q$. It follows that ${L|}_{\mathrm{dom}L\cap \mathrm{Ker}P}:\mathrm{dom}L\cap \mathrm{Ker}P\to \mathrm{Im}L$ is reversible. We denote the inverse of the mapping as $K_P$ (the generalized inverse operator of L). If $\Omega$ is an open bounded subset of X such that $\mathrm{dom}L\cap \Omega \ne \varnothing$, the mapping $N:X\to Z$ will be called L-compact on $\overline{\Omega }$, if $QN\left(\overline{\Omega }\right)$ is bounded and $K_P(I-Q)N:\overline{\Omega }\to X$ is compact.

Theorem 1 

(see [39,40] for the Mawhin continuation theorem). Let $L:\mathrm{dom}L\subset X\to Z$ be a Fredholm operator of index zero and $N:X\to Z$ be L-compact on $\overline{\Omega }$. We assume that the following conditions are satisfied:

(i)

$Lu\ne \lambda Nu$ for every $(u,\lambda )\in \left[\right(\mathrm{dom}L\setminus \mathrm{Ker}L)\cap \partial \Omega ]\times (0,1)$;

(ii)

$Nu\notin \mathrm{Im}L$ for every $u\in \mathrm{Ker}L\cap \partial \Omega$;

(iii)

$\mathrm{deg}(QN{|}_{\mathrm{Ker}L},\Omega \cap \mathrm{Ker}L,0)\ne 0$, where $Q:Z\to Z$ is a continuous projection such that $\mathrm{Im}L=\mathrm{Ker}Q$.

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

As is usual, we shall use the classical Banach space $Y=C[0,1]$ with norm ${\parallel y\parallel }_{C[0,1]}=\underset{t\in [0,1]}{max}\left|y\left(t\right)\right|$ and the Banach space $C_{n-\gamma ;\psi }(0,1]:=\left\{u\right|{(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }u\left(t\right)\in C[0,1]\}$ with norm ${\parallel u\parallel }_{C_{n-\gamma ;\psi }}$${=\parallel (\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{u\parallel }_{C[0,1]}=\underset{t\in [0,1]}{max}{\left\{\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }u\left(t\right)\left|\right\}$. Then, we can define and easily verify that the space $X=\left\{u\right|u\in C_{n-\gamma ;\psi }(0,1],$$D_{0^{+}}^{\gamma -1;\psi }u\in C[0,1],D_{0^{+}}^{\gamma -2;\psi }u\in C[0,1]\}$ with norm ${\parallel u\parallel }_X{=\parallel (\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{u\parallel }_{C[0,1]}+\parallel D_{0^{+}}^{\gamma -1;\psi }{u\parallel }_{C[0,1]}+{\parallel D_{0^{+}}^{\gamma -2;\psi }u\parallel }_{C[0,1]}$ is the Banach space.

We define the linear operator as $L:X\cap \mathrm{dom}L\subset X\to Y$, the nonlinear operator $N:X\to Y$ as

$$Lu{=}^H{D}_{0^{+}}^{\alpha ,\beta ;\psi }u\left(t\right),u\in \mathrm{dom}L,$$

and

$$Nu=f(t,u\left(t\right),u\left(\lambda t\right),D_{0^{+}}^{\gamma -2;\psi }u\left(t\right),D_{0^{+}}^{\gamma -1;\psi }u\left(t\right)),\lambda \in (0,1),t\in [0,1],$$

where $\mathrm{dom}L=\{u\in X{,}^H{D}_{0^{+}}^{\alpha ,\beta ;\psi }u\in C[0,1],I_{0^{+}}^{n-\gamma ;\psi }u\left(0\right)=0,D_{0^{+}}^{\gamma -i;\psi }u\left(0\right)=0,$$i=3,4,\dots ,n-1,B_i\left(u\right)=0,i=1,2\}$.

Then, the $FBVP$(Equation (1)) is equivalent to the operator equation $Lu=Nu,u\in \mathrm{dom}L.$

To achieve the outcomes we want, we need to use the following hypotheses.

Hypothesis 1 (H₁).

The linear functionals $B_i:X\to \mathbb{R}$ satisfy $B_1\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)=\sigma d,B_1\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right)=\sigma c,B_2\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)=d,$  $B_2\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right)=c,$ where $c^2+d^2\ne 0,\sigma ,c,d\in \mathbb{R}$.

Hypothesis 2 (H₂).

The functionals $B_1,B_2:X\to \mathbb{R}$ are continuous and linearly independent, with the respective norms ${\beta }_1,{\beta }_2,$; that is, $|B_i\left(u\right)|\le {\beta }_i{\parallel u\parallel }_X$.

Hypothesis 3 (H₃).

There exists a function $h\left(t\right)\in C[0,1]$ such that the equation

$${\displaystyle (B_1-\sigma B_2)\left(\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}h\left(s\right)ds\right)=1.}$$

The next statement shows that there is indeed a function $h\left(t\right)$ for which the assumption in $\left(H_3\right)$ is true.

Lemma 3. 

Assume that $\left(H_1\right)$ holds. Then, there exists $h\left(t\right)\in C[0,1]$ such that

$${\displaystyle (B_1-\sigma B_2)\left(\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}h\left(s\right)ds\right)\ne 0.}$$

Proof. 

For convenience, we set $B=B_1-\sigma B_2$, and $h_n\left(t\right)={(\psi \left(t\right)-\psi \left(0\right))}^{n-2+\gamma -\alpha },$$n\ge 2,\psi \left(t\right)\in C^n[0,1].$ Then, there must exist $n\in \mathbb{N}$ such that ${\displaystyle B\left(I_{0^{+}}^{\alpha ;\psi }{h}_n\left(t\right)\right)\ne 0}$.

If not, $B\left(I_{0^{+}}^{\alpha ;\psi }{h}_n\left(t\right)\right)=0$; that is, $B\left({(\psi \left(t\right)-\psi \left(0\right))}^{n-2+\gamma }\right)=0$. From $\left(H_1\right),B\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)=B\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right)=0$. Thus, $B\left(p\right)=0$ for every ‘polynomial’ $p\left(t\right)={\displaystyle \sum _{i=0}^n}{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma +i-2}$. It is easy to verify that ${\left\{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma +n-2}\right\}}_{n\in \mathbb{N}}$ is dense in X, thus we obtain $B\left(u\right)\equiv 0,u\in X$. This is a contradiction (as $B_1$ and $B_2$ are linearly independent on X).

There must be some $i\ge 2,i\in \mathbb{N}$ such that $B\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma +i-2}\right)\ne 0$. For these i, we simply take $h\left(t\right)={\displaystyle \frac{h_i\left(t\right)}{B\left(I_{0^{+}}^{\alpha ;\psi }{h}_i\left(t\right)\right)}}.$ Thus, there exists an $h\in Y$ satisfying $\left(H_3\right)$. □

3. Main Results

Lemma 4. 

If assumptions $\left(H\right)$, $\left(H_1\right)$, and $\left(H_3\right)$ hold, then the mapping $L:\mathrm{dom}L\subset X\to Y$ is a Fredholm mapping of index zero.

Proof. 

If $u\in \mathrm{dom}L$, and $Lu=0,$ we can deduce that $u=c_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}$, $B_1(c_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})=\sigma d{c}_1+\sigma c{c}_2=0,$ and $B_2(c_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})=d{c}_1+c{c}_2=0,$ which means that

$$\mathrm{Ker}L=\left\{c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}):c^2+d^2\ne 0,c_0\in \mathbb{R}\right\},\mathrm{dimKer}L=1.$$

Now, we can show that

$$\begin{array}{c}\mathrm{Im}L=\left\{{\displaystyle y\in Y:(B_1-\sigma B_2)\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}y\left(s\right)ds\right)=0}\right\}.\hfill \end{array}$$

(2)

Let $y\in \mathrm{Im}L$; then, there exists $u\in \mathrm{dom}L$ such that $y=Lu,$ and it follows that

$$\begin{array}{c}u\left(t\right)={\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}}{\Gamma \left(\gamma \right)}}{f}_{\psi }^{[n-1]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)+{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{\Gamma (\gamma -1)}}{f}_{\psi }^{[n-2]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)\hfill \\ {\displaystyle +{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}y\left(s\right)ds,}\hfill \end{array}$$

which, with $B_1\left(u\right)=B_2\left(u\right)=0,$ leads to

$$\begin{array}{c}0=B_i\left(u\right)=B_i\left(I_{0^{+}}^{\alpha ;\psi }y\right)+{\displaystyle \frac{f_{\psi }^{[n-1]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma \left(\gamma \right)}}{B}_i\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)+{\displaystyle \frac{f_{\psi }^{[n-2]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma (\gamma -1)}}·\hfill \\ B_i\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right),i=1,2.\hfill \end{array}$$

Then, we can deduce from $\left(H_1\right)$ that

$$\begin{array}{cc}\hfill & B_1\left(I_{0^{+}}^{\alpha ;\psi }y\right)+{\displaystyle \frac{f_{\psi }^{[n-1]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma \left(\gamma \right)}}{B}_1\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)+{\displaystyle \frac{f_{\psi }^{[n-2]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma (\gamma -1)}}·B_1\left(\right(\psi \left(t\right)-\hfill \\ \hfill & {\psi \left(0\right))}^{\gamma -2})=B_1\left(I_{0^{+}}^{\alpha ;\psi }y\right)+\sigma d{\displaystyle \frac{f_{\psi }^{[n-1]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma \left(\gamma \right)}}+\sigma c{\displaystyle \frac{f_{\psi }^{[n-2]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma (\gamma -1)}}=0,\hfill \\ \hfill & B_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)+{\displaystyle \frac{f_{\psi }^{[n-1]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma \left(\gamma \right)}}{B}_2\left({(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}\right)+{\displaystyle \frac{f_{\psi }^{[n-2]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma (\gamma -1)}}{B}_2\left(\right(\psi \left(t\right)-\hfill \\ \hfill & {\psi \left(0\right))}^{\gamma -2})=B_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)+d{\displaystyle \frac{f_{\psi }^{[n-1]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma \left(\gamma \right)}}+c{\displaystyle \frac{f_{\psi }^{[n-2]}{I}_{0^{+}}^{(1-\beta )(n-\alpha );\psi }u\left(0\right)}{\Gamma (\gamma -1)}}=0.\hfill \end{array}$$

Therefore, ${\displaystyle y\in \left\{{\displaystyle y\in Y:(B_1-\sigma B_2)\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}y\left(s\right)ds\right)=0}\right\}.}$ That is,

$${\displaystyle \mathrm{Im}L\subseteq \{y\in Y:(B_1-\sigma B_2)\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }{\left(s\right)(\psi \left(t\right)-\psi \left(s\right)}^{\alpha -1}y\left(s\right)ds\right)=0\}.}$$

If ${\displaystyle y\in \left\{{\displaystyle y\in Y:(B_1-\sigma B_2)\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}y\left(s\right)ds\right)=0}\right\},}$ then

$$u\left(t\right)=-{\displaystyle \frac{d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)+I_{0^{+}}^{\alpha ;\psi }y\left(t\right).$$

It is easy to find that $Lu=y$ and $B_1\left(u\right)=B_2\left(u\right)=0,$ which implies that

$${\displaystyle \left\{{\displaystyle y\in Y:(B_1-\sigma B_2)\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(0\right))}^{\alpha -1}y\left(s\right)ds\right)=0}\right\}\subseteq \mathrm{Im}L.}$$

Hence, we obtain Equation (2).

We define $Q:Y\to Y$ as follows:

$${\displaystyle Qy=(B_1-\sigma B_2)\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(s\right))}^{\alpha -1}y\left(s\right)ds\right)h\left(t\right),}$$

where $h\left(t\right)$ is introduced in condition $\left(H_3\right)$.

From the property of $h\left(t\right)$ in $\left(H_3\right)$, we know that $Q^{2}y=Qy.$ So, $Q:Y\to Y$ is a continuous linear projector such that $\mathrm{Im}L=\mathrm{Ker}Q$ and $\mathrm{Im}Q=\left\{ch\right(t\left)\right|c\in \mathbb{R}\}.$ It is clear that $Y=\mathrm{Im}L\oplus \mathrm{Im}Q$ and $\mathrm{dimKer}L=\mathrm{codimIm}L=1$; that is, L is a Fredholm mapping of index zero. □

Take $P:X\to X$, where

$$Pu\left(t\right)={\displaystyle \frac{c{D}_{0^{+}}^{\gamma -1;\psi }u\left(0\right)-d(\gamma -1)D_{0^{+}}^{\gamma -2;\psi }u\left(0\right)}{\Gamma \left(\gamma \right)c^2+d^2}}(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}).$$

For $u\in X$, set $u=u-Pu+Pu$. It is not hard to check that $P^{2}u=Pu,u\in X$, since $D_{0^{+}}^{\gamma -1;\psi }{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}=\Gamma \left(\gamma \right),D_{0^{+}}^{\gamma -1;\psi }{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}=0,D_{0^{+}}^{\gamma -2;\psi }{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}=\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))$ and $D_{0^{+}}^{\gamma -2;\psi }{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}=\Gamma (\gamma -1).$

It is quite simple to verify that $\mathrm{Im}P=\mathrm{Ker}L$; moreover, we take $u_0=c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\in \mathrm{Im}P.$ If, in addition, $c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\in \mathrm{Ker}P$, then $0=c{D}_{0^{+}}^{\gamma -1;\psi }{u}_0\left(0\right)-d(\gamma -1)D_{0^{+}}^{\gamma -2;\psi }{u}_0\left(0\right)=c_0\Gamma \left(\gamma \right)(c^2+d^2).$ We must obtain $c_0=0$ by setting $\Gamma \left(\gamma \right)\ne 0,c^2+d^2\ne 0,$ i.e., $\mathrm{Im}P\cap \mathrm{Ker}P=\left\{0\right\}$. So, $X=\mathrm{Ker}L\oplus \mathrm{Ker}P$.

Lemma 5. 

The mapping $K_P:Y\to \mathrm{dom}L\cap \mathrm{Ker}P$ is defined by

$$\begin{array}{c}\hfill K_{P}y\left(t\right)=-{\displaystyle \frac{d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)+I_{0^{+}}^{\alpha ;\psi }y\left(t\right)\end{array}$$

and is the generalized inverse operator of L.

Proof. 

In view of Definition 4 and Lemma 2, one can show that $L{K}_{P}y{=}^H{D}_{0^{+}}^{\alpha ,\beta ;\psi }\left(K_{P}y\left(t\right)\right)$$=I_{0^{+}}^{\gamma -\alpha ;\psi }{D}_{0^{+}}^{\gamma ;\psi }\left(K_{P}y\left(t\right)\right)=I_{0^{+}}^{\gamma -\alpha ;\psi }{D}_{0^{+}}^{\gamma ;\psi }{I}_{0^{+}}^{\alpha ;\psi }y\left(t\right)=y\left(t\right)$ for all $y\in Y$. Meanwhile, it is suffices to show that $I_{0^{+}}^{n-\gamma ;\psi }\left(K_{P}y\right)\left(0\right)=0, D_{0^{+}}^{\gamma -i;\psi }\left(K_{P}y\right)\left(0\right)=0, i=3,4,\dots ,n-1, B_i\left(K_{P}y\right)=0,$$i=1,2$ and

$$\begin{array}{cc}\hfill P\left(K_{P}y\right)\left(t\right)& ={\displaystyle \frac{c{D}_{0^{+}}^{\gamma -1;\psi }\left(K_{P}y\right)\left(0\right)-d(\gamma -1)D_{0^{+}}^{\gamma -2;\psi }\left(K_{P}y\right)\left(0\right)}{\Gamma \left(\gamma \right)c^2+d^2}}(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\hfill \\ \hfill & ={\displaystyle \frac{\frac{-cd\Gamma \left(\gamma \right)}{c^2+d^2}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)+\frac{cd(\gamma -1)\Gamma (\gamma -1)}{c^2+d^2}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)}{\Gamma \left(\gamma \right)(c^2+d^2)}}(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\hfill \\ \hfill & =0.\hfill \end{array}$$

In accordance with the equations above, $K_{P}y\in \mathrm{dom}L\cap \mathrm{Ker}P$. If $u\in \mathrm{dom}L\cap \mathrm{Ker}P$, from Lemma 2(1) and $\left(H_1\right)$ it can be deduced that

$$\begin{array}{cc}\hfill K_{P}Lu\left(t\right)=& K_P{(}^H{D}_{0^{+}}^{\alpha ,\beta ;\psi }u)\left(t\right)\hfill \\ \hfill =& -{\displaystyle \frac{d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }{}^{H}D_{0^{+}}^{\alpha ,\beta ;\psi }u\right)+I_{0^{+}}^{\alpha ;\psi }{}^{H}D_{0^{+}}^{\alpha ,\beta ;\psi }u\left(t\right)\hfill \\ \hfill =& -{\displaystyle \frac{d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{c^2+d^2}}{B}_2(u\left(t\right)-\sum _{k=1}^{n-1}{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -k}}{\Gamma (\gamma -k+1)}}{D}_{0^{+}}^{\gamma -k;\psi }u\left(0\right)\hfill \\ \hfill & -{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}}{\Gamma (\gamma -n+1)}}{I}_{0^{+}}^{n-\gamma ;\psi }u\left(0\right))+u\left(t\right)-\sum _{k=1}^{n-1}{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -k}}{\Gamma (\gamma -k+1)}}{D}_{0^{+}}^{\gamma -k;\psi }u\left(0\right)\hfill \\ \hfill & -{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}}{\Gamma (\gamma -n+1)}}{I}_{0^{+}}^{n-\gamma ;\psi }u\left(0\right)\hfill \\ \hfill =& -{\displaystyle \frac{d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}+c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{c^2+d^2}}{B}_2(u\left(t\right)-{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}}{\Gamma \left(\gamma \right)}}{D}_{0^{+}}^{\gamma -1;\psi }u\left(0\right)\hfill \\ \hfill & -{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{\Gamma (\gamma -1)}}{D}_{0^{+}}^{\gamma -2;\psi }u\left(0\right))+u\left(t\right)-{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}}{\Gamma \left(\gamma \right)}}{D}_{0^{+}}^{\gamma -1;\psi }u\left(0\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}}{\Gamma (\gamma -1)}}{D}_{0^{+}}^{\gamma -2;\psi }u\left(0\right)\hfill \\ \hfill =& u\left(t\right)-{\displaystyle \frac{c{D}_{0^{+}}^{\gamma -1;\psi }u\left(0\right)-d(\gamma -1)D_{0^{+}}^{\gamma -2;\psi }u\left(0\right)}{\Gamma \left(\gamma \right)c^2+d^2}}(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\hfill \\ \hfill =& u\left(t\right)-Pu\left(t\right)=u\left(t\right).\hfill \end{array}$$

Thus,

$$K_P={\left(L{|}_{\mathrm{dom}L\cap \mathrm{Ker}P}\right)}^{-1}.$$

□

The next lemma provides the norm estimates needed for the following results.

Lemma 6. 

For $y\in Y$, we now show that

$${∥{(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }\left(K_{P}y\right)∥}_{C[0,1]}\le B·{\parallel y\parallel }_{C[0,1]},$$

(3)

$${∥D_{0^{+}}^{\gamma -1;\psi }\left(K_{P}y\right)∥}_{C[0,1]}\le C·{\parallel y\parallel }_{C[0,1]},$$

(4)

$$\begin{array}{cc}\hfill B& ={\displaystyle \frac{{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-1}+\left|c\right|{(\psi \left(t\right)-\psi \left(0\right))}^{n-2}}{c^2+d^2}}{\beta }_{2}A+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +n}}{\Gamma (\alpha +1)}},\hfill \\ \hfill C& =\left|d\right|\Gamma \left(\gamma \right){\beta }_{2}A+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +1}}{\Gamma (\alpha -\gamma +2)}},\hfill \\ \hfill A& ={\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +n}}{\Gamma (\alpha +1)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +1}}{\Gamma (\alpha -\gamma +2)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +2}}{\Gamma (\alpha -\gamma +3)}},\hfill \end{array}$$

and

$${∥D_{0^{+}}^{\gamma -2;\psi }\left(K_{P}y\right)∥}_{C[0,1]}\le D·{\parallel y\parallel }_{C[0,1]},$$

(5)

$$D={\displaystyle \frac{\left|d\right|\Gamma \left(\gamma \right)+\left|c\right|\Gamma (\gamma -1)}{c^2+d^2}}{\beta }_{2}A+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +2}}{\Gamma (\alpha -\gamma +3)}}.$$

Moreover,

$$\parallel K_P{y\parallel }_X\le ∥K_P∥·{\parallel y\parallel }_{C[0,1]},$$

(6)

$$∥K_P∥=B+C+D.$$

Proof. 

For $y\in Y$, we can deduce that

$$\begin{array}{cc}\hfill {|(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }y\left(t\right)|\le & {\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +n}}{\Gamma (\alpha +1)}}{\parallel y\parallel }_{C[0,1]},\hfill \\ \hfill |D_{0^{+}}^{\gamma -1;\psi }{I}_{0^{+}}^{\alpha ;\psi }y\left(t\right)|\le & {\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +1}}{\Gamma (\alpha -\gamma +2)}}{\parallel y\parallel }_{C[0,1]},\hfill \\ \hfill and|D_{0^{+}}^{\gamma -2;\psi }{I}_{0^{+}}^{\alpha ;\psi }y\left(t\right)|\le & {\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +2}}{\Gamma (\alpha -\gamma +3)}}{\parallel y\parallel }_{C[0,1]},\hfill \end{array}$$

which means that

$$\begin{array}{cc}\hfill \parallel I_{0^{+}}^{\alpha ;\psi }{y\parallel }_X& \le \left({\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +n}}{\Gamma (\alpha +1)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +1}}{\Gamma (\alpha -\gamma +2)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +2}}{\Gamma (\alpha -\gamma +3)}}\right){\parallel y\parallel }_{C[0,1]}\hfill \\ \hfill & :=A·{\parallel y\parallel }_{C[0,1]}.\hfill \end{array}$$

Based on the assumption of $\left(H_2\right)$, we can derive the following:

$$\begin{array}{cc}\hfill {|(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }\left(K_{P}y\right)\left(t\right)|=& |-{\displaystyle \frac{d{(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+c{(\psi \left(t\right)-\psi \left(0\right))}^{n-2}}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)\hfill \\ \hfill & +{(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }y\left(t\right)|\hfill \\ \hfill \le & {\displaystyle \frac{{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-1}+\left|c\right|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}}{c^2+d^2}}{\beta }_2·{\parallel I_{0^{+}}^{\alpha ;\psi }y\parallel }_X\hfill \\ \hfill & +{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +n}}{\Gamma (\alpha +1)}}{\parallel y\parallel }_{C[0,1]}\hfill \\ \hfill \le & ({\displaystyle \frac{{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-1}+\left|c\right|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}}{c^2+d^2}}{\beta }_{2}A\hfill \\ \hfill & +{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +n}}{\Gamma (\alpha +1)}})·{\parallel y\parallel }_{C[0,1]}:=B·{\parallel y\parallel }_{C[0,1]},\hfill \end{array}$$

$$\begin{array}{cc}\hfill |D_{0^{+}}^{\gamma -1;\psi }\left(K_{P}y\right)\left(t\right)|& =|d\Gamma \left(\gamma \right)B_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)+D_{0^{+}}^{\gamma -1;\psi }{I}_{0^{+}}^{\alpha ;\psi }y\left(t\right)|\hfill \\ \hfill & \le \left(\left|d\right|\Gamma \left(\gamma \right){\beta }_{2}A+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +1}}{\Gamma (\alpha -\gamma +2)}}\right)·{\parallel y\parallel }_{C[0,1]}:=C·{\parallel y\parallel }_{C[0,1]},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill |D_{0^{+}}^{\gamma -2;\psi }\left(K_{P}y\right)\left(t\right)|& =|{\displaystyle \frac{d\Gamma \left(\gamma \right)+c\Gamma (\gamma -1)}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }y\right)+D_{0^{+}}^{\gamma -2;\psi }{I}_{0^{+}}^{\alpha ;\psi }y\left(t\right)|\hfill \\ \hfill & \le \left({\displaystyle \frac{\left|d\right|\Gamma \left(\gamma \right)+\left|c\right|\Gamma (\gamma -1)}{c^2+d^2}}{\beta }_{2}A+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{\alpha -\gamma +2}}{\Gamma (\alpha -\gamma +3)}}\right)·{\parallel y\parallel }_{C[0,1]}:=D·{\parallel y\parallel }_{C[0,1]}.\hfill \end{array}$$

Hence, we can conclude that

$$\begin{array}{cc}\hfill {∥K_{P}y∥}_X=& {\parallel (\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{K}_P{y\parallel }_{C[0,1]}+\parallel D_{0^{+}}^{\gamma -1;\psi }\left(K_{P}y\right){\parallel }_{C[0,1]}+{\parallel D_{0^{+}}^{\gamma -2;\psi }\left(K_{P}y\right)\parallel }_{C[0,1]}\hfill \\ \hfill \le & (B+C+D)·{\parallel y\parallel }_{C[0,1]}:=\parallel K_P{\parallel ·\parallel y\parallel }_{C[0,1]},\hfill \end{array}$$

which yields our desired result. □

Lemma 7. 

N is L-compact on $\overline{\Omega }$ if $\mathrm{dom}L\cap \overline{\Omega }\ne \varnothing ,$ where Ω is an open and bounded subset of X.

Proof. 

Since $\Omega$ is a bounded set and functions $h\left(t\right)$ and $f(t,u\left(t\right),u\left(\lambda t\right),D_{0^{+}}^{\gamma -2;\psi }u\left(t\right),$$D_{0^{+}}^{\gamma -1;\psi }u\left(t\right))$ are continuous for $t\in [0,1]$, there exist constants $M_1,M_2>0$ such that ${\parallel Nu\parallel }_{C[0,1]}\le M_1$ and $\underset{t\in [0,1]}{sup}\left|h\left(t\right)\right|\le M_2$ hold and $u\in \overline{\Omega }$. Invoking $\parallel I_{0^{+}}^{\alpha ;\psi }{y\parallel }_X\le A·{\parallel y\parallel }_{C[0,1]}$ and Hypothesis $\left(H_2\right)$, it is easy to see that

$$\begin{array}{cc}\hfill \left|QNu\right|& {\displaystyle =|(B_1-\sigma B_2)\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{\psi }^{\prime }\left(s\right){(\psi \left(t\right)-\psi \left(0\right))}^{\alpha -1}Nu\left(s\right)ds\right)h\left(t\right)|}\hfill \\ \hfill & \le ({\beta }_1+\left|\sigma \right|{\beta }_2)\parallel I_{0^{+}}^{\alpha ;\psi }{f\parallel }_X·\left|h\left(t\right)\right|\le A{M}_1{M}_2({\beta }_1+\left|\sigma \right|{\beta }_2),\hfill \end{array}$$

hence, $QN\left(\overline{\Omega }\right)$ is bounded.

Now, we will prove that $K_P(I-Q)N:\overline{\Omega }\to X$ is compact. The proof process is divided into two steps:

• Step 1  $K_P(I-Q)N:\overline{\Omega }\to X$ is uniformly bounded. Based on $\parallel K_P{y\parallel }_{C[0,1]}\le \parallel K_P{\parallel ·\parallel y\parallel }_{C[0,1]}$ and $\left(H_2\right)$, we obtain

  $$\begin{array}{cc}\hfill \parallel K_P{(I-Q)Nu\parallel }_X& \le \parallel K_P{\parallel ·\parallel (I-Q)Nu\parallel }_{C[0,1]}\le \parallel K_P{\parallel ·(\parallel Nu\parallel }_{C[0,1]}+{\parallel QNu\parallel }_{C[0,1]})\hfill \\ \hfill & <\parallel K_P\parallel ·(M_1+A{M}_1{M}_2({\beta }_1+\left|\sigma \right|{\beta }_2\left)\right)<\infty .\hfill \end{array}$$
• Step 2  $\{K_P(I-Q)Nu:u\in \overline{\Omega }\}$ is equicontinuous on $[0,1]$.

For each $u\in \overline{\Omega },0\le t_1<t_2\le 1,$ one can show that

$$\begin{array}{cc}\hfill & |{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }{K}_P(I-Q)Nu\left(t_2\right)-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }{K}_P(I-Q)Nu\left(t_1\right)|\hfill \\ \hfill =& |-{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }\left(d{(\psi \left(t_2\right)-\psi \left(0\right))}^{\gamma -1}+c{(\psi \left(t_2\right)-\psi \left(0\right))}^{\gamma -2}\right)}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\right)\hfill \\ \hfill & +{\displaystyle \frac{{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }\left(d{(\psi \left(t_1\right)-\psi \left(0\right))}^{\gamma -1}+c{(\psi \left(t_1\right)-\psi \left(0\right))}^{\gamma -2}\right)}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\right)\hfill \\ \hfill & +{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\left(t_2\right)-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\left(t_1\right)|\hfill \\ \hfill =& |-{\displaystyle \frac{\left(d{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-1}+c{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-2}\right)}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\right)\hfill \\ \hfill & +{\displaystyle \frac{\left(d{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-1}+c{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-2}\right)}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\right)\hfill \\ \hfill & +{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\left(t_2\right)-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\left(t_1\right)|.\hfill \end{array}$$

To begin with, one knows that

$$\begin{array}{cc}\hfill & |-{\displaystyle \frac{\left(d{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-1}+c{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-2}\right)}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\right)\hfill \\ \hfill & +{\displaystyle \frac{\left(d{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-1}+c{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-2}\right)}{c^2+d^2}}{B}_2\left(I_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\right)|\hfill \\ \hfill & \le {\displaystyle \frac{\left|d\right|[{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-1}-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-1}]+\left|c\right|[{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-2}-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-2}]}{c^2+d^2}}\hfill \\ \hfill & ·A{M}_1{\beta }_2(1+A{M}_2({\beta }_1+\left|\sigma \right|{\beta }_2\left)\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & |{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\left(t_2\right)-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }{I}_{0^{+}}^{\alpha ;\psi }(I-Q)Nu\left(t_1\right)|\hfill \\ \hfill =& {\displaystyle |{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}{\int }_0^{t_2}{\psi }^{\prime }\left(s\right){(\psi \left(t_2\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds-{\displaystyle \frac{{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}}\hfill \\ \hfill & {\int }_0^{t_1}{\psi }^{\prime }\left(s\right){(\psi \left(t_1\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds|\hfill \\ \hfill =& {\displaystyle |{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}{\int }_0^{t_1}{\psi }^{\prime }\left(s\right){(\psi \left(t_2\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds+{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}}\hfill \\ \hfill & {\int }_{t_1}^{t_2}{\psi }^{\prime }\left(s\right){(\psi \left(t_2\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds\hfill \\ \hfill +& {\displaystyle |{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}{\int }_0^{t_1}{\psi }^{\prime }\left(s\right){(\psi \left(t_1\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds}\hfill \\ \hfill -& {\displaystyle |{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}{\int }_0^{t_1}{\psi }^{\prime }\left(s\right){(\psi \left(t_2\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds}\hfill \\ \hfill -& {\displaystyle |{\displaystyle \frac{{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}{\int }_0^{t_1}{\psi }^{\prime }\left(s\right){(\psi \left(t_1\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds|}\hfill \\ \hfill \le & {\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}\left|{\int }_0^{t_1}{\psi }^{\prime }\left(s\right)[{(\psi \left(t_2\right)-\psi \left(s\right))}^{\alpha -1}-{(\psi \left(t_1\right)-\psi \left(s\right))}^{\alpha -1}](I-Q)Nu\left(s\right)ds\right|\hfill \\ \hfill & +{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}\left|{\int }_0^{t_1}{\psi }^{\prime }\left(s\right){(\psi \left(t_1\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds\right|\hfill \\ \hfill & +{\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }}{\Gamma \left(\alpha \right)}}\left|{\int }_{t_1}^{t_2}{\psi }^{\prime }\left(s\right){(\psi \left(t_2\right)-\psi \left(s\right))}^{\alpha -1}(I-Q)Nu\left(s\right)ds\right|\hfill \\ \hfill \le & M_1(1+A{M}_2({\beta }_1+\left|\sigma \right|{\beta }_2\left)\right)[{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma +\alpha }-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma +\alpha }].\hfill \end{array}$$

As $t_1\to t_2$, it is easy to find that

$$\begin{array}{c}\hfill |{(\psi \left(t_2\right)-\psi \left(0\right))}^{n-\gamma }{K}_P(I-Q)Nu\left(t_2\right)-{(\psi \left(t_1\right)-\psi \left(0\right))}^{n-\gamma }{K}_P(I-Q)Nu\left(t_1\right)|\to 0.\end{array}$$

(7)

Next, from the definition of $K_P$, we can derive the following:

$$\begin{array}{cc}& |D_{0+}^{\gamma -1;\psi }\left(K_P(I-Q)Nu\right)\left(t_2\right)-D_{0+}^{\gamma -1;\psi }\left(K_P(I-Q)Nu\right)\left(t_1\right)|\\ & =|I_{0^{+}}^{\alpha -\gamma +1;\psi }(I-Q)Nu\left(t_2\right)-I_{0^{+}}^{\alpha -\gamma +1;\psi }(I-Q)Nu\left(t_1\right)|\\ & \le M_1(1+A{M}_2({\beta }_1+\left|\sigma \right|{\beta }_2\left)\right){\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{\alpha -\gamma +1}-{(\psi \left(t_1\right)-\psi \left(0\right))}^{\alpha -\gamma +1}}{\Gamma (\alpha -\gamma +2)}}.\end{array}$$

(8)

Furthermore,

$$\begin{array}{cc}& |D_{0+}^{\gamma -2;\psi }\left(K_P(I-Q)Nu\right)\left(t_2\right)-D_{0+}^{\gamma -2;\psi }\left(K_P(I-Q)Nu\right)\left(t_1\right)|\\ & =|I_{0^{+}}^{\alpha -\gamma +2;\psi }(I-Q)Nu\left(t_2\right)-I_{0^{+}}^{\alpha -\gamma +2;\psi }(I-Q)Nu\left(t_1\right)|\\ & \le M_1(1+A{M}_2({\beta }_1+\left|\sigma \right|{\beta }_2\left)\right){\displaystyle \frac{{(\psi \left(t_2\right)-\psi \left(0\right))}^{\alpha -\gamma +2}-{(\psi \left(t_1\right)-\psi \left(0\right))}^{\alpha -\gamma +2}}{\Gamma (\alpha -\gamma +3)}}.\end{array}$$

(9)

Taking $t_1\to t_2$, from the continuity of ${(\psi \left(t\right)-\psi \left(0\right))}^{\alpha -\gamma +1}$ and ${(\psi \left(t\right)-\psi \left(0\right))}^{\alpha -\gamma +2}$ for $t\in [0,1]$, we can deduce that Equation (8) $\to 0$ and Equation (9) $\to 0$.

By taking the prior estimation (Steps 1–2) into account and applying the Arzelà–Ascoli theorem, it holds that the operator $K_P(I-Q)N:\overline{\Omega }\to X$ is compact and $QN\left(\overline{\Omega }\right)$ is bounded, i.e., N is L-compact. □

Lemma 8. 

Assume $c\ne 0$ and the following hypotheses hold:

Hypothesis 4 (H₄).

There exists a constant $M_0>0$ such that if $|D_{0^{+}}^{\gamma -1;\psi }u\left(t\right)|>M_0$, for all $t\in [0,1]$, then

$$(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }Nu\right)\ne 0.$$

Hypothesis 5 (H₅).

There exist some positive functions $a_i\left(t\right)i=0,1,2,3,4$, with $a_0\left(t\right),a_1\left(t\right)$${(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n},a_2\left(t\right)$${(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n},a_3\left(t\right)$, and $a_4\left(t\right)\in Y$, such that for all $(u_1,u_2,u_3,u_4)\in {\mathbb{R}}^4,t\in [0,1]$

$$|f(t,u_1,u_2,u_3,u_4)|\le a_0\left(t\right)+a_1\left(t\right)|u_1|+a_2\left(t\right)|u_2|+a_3\left(t\right)|u_3|+a_4\left(t\right)\left|u_4\right|,$$

provided that $(\parallel K_P\parallel +F\left)\right(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+{\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}\parallel }_{C[0,1]}+$$\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})<1.$ Then, the set ${\Omega }_1=\{u\in \mathrm{dom}L\setminus \mathrm{Ker}L:Lu=\lambda Nu$, for some $\lambda \in (0,1)\}$ is bounded.

Proof. 

Take $u\in {\Omega }_1$; then, $Nu\in \mathrm{Im}L,$ thus we obtain $(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }Nu\right)=0.$ This, together with $\left(H_4\right)$, means that there exists $t_0\in [0,1]$ such that $|D_{0^{+}}^{\gamma -1;\psi }u\left(t_0\right)|\le M_0$.

Write $u\left(t\right)=u_1\left(t\right)+u_2\left(t\right),$ where $u_1=(I-P)u\in \mathrm{dom}L\cap \mathrm{Ker}P$ and $u_2=Pu\in \mathrm{Im}P.$ Thus, $u_1=K_{P}L{u}_1=K_{P}L(I-P)u=K_{P}Lu=\lambda K_{P}Nu.$ In light of (Equation (5)) in Lemma 7, one obtains

$$\begin{array}{c}\hfill \parallel u_1{\parallel }_X\le \parallel K_P{\parallel ·\parallel Nu\parallel }_{C[0,1]}.\end{array}$$

(10)

Now, $u_2=u-u_1,$ so $D_{0^{+}}^{\gamma -1;\psi }{u}_2=D_{0^{+}}^{\gamma -1;\psi }u-D_{0^{+}}^{\gamma -1;\psi }{u}_1$ and

$$|D_{0^{+}}^{\gamma -1;\psi }{u}_2\left(t_0\right)|\le |D_{0^{+}}^{\gamma -1;\psi }{u}_2\left(t_0\right)|+|D_{0^{+}}^{\gamma -1;\psi }{u}_1\left(t_0\right)|\le M_0+C·{\parallel Nu\parallel }_{C[0,1]},$$

(11)

where C is the same as in (Lemma 6). Recall that $u_2\left(t\right)=Pu\left(t\right)=c\left(u\right)(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})$, where

$$c\left(u\right)={\displaystyle \frac{c{D}_{0^{+}}^{\gamma -1;\psi }u\left(0\right)-d(\gamma -1)D_{0^{+}}^{\gamma -2;\psi }u\left(0\right)}{\Gamma \left(\gamma \right)(c^2+d^2)}}$$

is introduced for the sake of brevity. Hence, it follows from Equation (11) that $\left|c\left(u\right)\right|\le {\displaystyle \frac{M_0+C·{\parallel Nu\parallel }_{C[0,1]}}{\left|c\right|\Gamma \left(\gamma \right)}}$.

Furthermore, it is straightforward to derive that the following inequalities:

$$\begin{array}{cc}\hfill |u_2\left(t\right){(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }|& \le {\left|c\left(u\right)\right|·|c(\psi \left(t\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{n-2}|\hfill \\ \hfill & \le (M_0+C·{\parallel Nu\parallel }_{C[0,1]})({\displaystyle \frac{{(\psi \left(t\right)-\psi \left(0\right))}^{n-1}}{\Gamma \left(\gamma \right)}}+{\displaystyle \frac{{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}}{\left|c\right|\Gamma \left(\gamma \right)}}),\hfill \\ \hfill |D_{0^{+}}^{\gamma -1;\psi }{u}_2\left(t\right)|& =|c\left(u\right)·c\Gamma \left(\gamma \right)|\le M_0+C·{\parallel Nu\parallel }_{C[0,1]},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill |D_{0^{+}}^{\gamma -2;\psi }{u}_2\left(t\right)|& =|c\left(u\right)·\left(c\Gamma \left(\gamma \right)\left(\psi \right(t)-\psi (0\left)\right)-d\Gamma (\gamma -1)\right)|\hfill \\ \hfill & \le (M_0+C·{\parallel Nu\parallel }_{C[0,1]})((\psi \left(1\right)-\psi \left(0\right))+{\displaystyle \frac{\left|d\right|}{\left|c\right|(\gamma -1)}}),\hfill \end{array}$$

which leads to

$$\begin{array}{cc}\hfill \parallel u_2{\parallel }_X& \le M_0(1+\psi \left(1\right)-\psi \left(0\right)+{\displaystyle \frac{\left|d\right|}{\left|c\right|(\gamma -1)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}}{\Gamma \left(\gamma \right)}}+{\displaystyle \frac{{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-2}}{\left|c\right|\Gamma \left(\gamma \right)}})\hfill \\ \hfill & +C·{\parallel Nu\parallel }_{C[0,1]}\left(1+\psi \left(1\right)-\psi \left(0\right)+{\displaystyle \frac{\left|d\right|}{\left|c\right|(\gamma -1)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}}{\Gamma \left(\gamma \right)}}+{\displaystyle \frac{{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-2}}{\left|c\right|\Gamma \left(\gamma \right)}}\right)\hfill \\ \hfill & :=E+F·{\parallel Nu\parallel }_{C[0,1]},\hfill \end{array}$$

where $E:=M_0(1+\psi \left(1\right)-\psi \left(0\right)+{\displaystyle \frac{\left|d\right|}{\left|c\right|(\gamma -1)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}}{\Gamma \left(\gamma \right)}}+{\displaystyle \frac{{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-2}}{\left|c\right|\Gamma \left(\gamma \right)}})$ and $F:=C·\left(1+\psi \left(1\right)-\psi \left(0\right)+{\displaystyle \frac{\left|d\right|}{\left|c\right|(\gamma -1)}}+{\displaystyle \frac{{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}}{\Gamma \left(\gamma \right)}}+{\displaystyle \frac{{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-2}}{\left|c\right|\Gamma \left(\gamma \right)}}\right),$ which together with Equation (10) and $\left(H_5\right)$, yield

$$\begin{array}{cc}\hfill & {\parallel u\parallel }_X\le \parallel u_1{\parallel }_X+\parallel u_2{\parallel }_X\le (\parallel K_P{\parallel +F)\parallel Nu\parallel }_{C[0,1]}+E\hfill \\ \hfill & \le E+(\parallel K_P\parallel +F\left)\right(\parallel a_0{\parallel }_{C[0,1]}+(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+{\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}\parallel }_{C[0,1]}\hfill \\ \hfill & +\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})·{\parallel u\parallel }_X)\hfill \\ \hfill & \le {\displaystyle \frac{E+(\parallel K_P\parallel +F)\parallel a_0{\parallel }_{C[0,1]}}{1-(\parallel K_P\parallel +F\left)\right(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})}}.\hfill \end{array}$$

As a result, ${\Omega }_1$ is bounded. This proof is thus complete. □

Since L is a Fredholem mapping of index zero, there exists an isomorphism $J:\mathrm{Im}Q\to \mathrm{Ker}L$. For this, the operator J is defined by

$$Jg\left(t\right)=(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }g\right)(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}).$$

Obviously, $J:Y\to \mathrm{Ker}L.$ If $g\in \mathrm{Im}Q,$ then $g=c_{0}h$, where h the same as in $\left(H_3\right)$, and

$$\begin{array}{cc}\hfill Jg\left(t\right)& =J\left(c_{0}h\right)\left(t\right)\hfill \\ \hfill & =(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }{c}_{0}h\right)(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\hfill \\ \hfill & =c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}).\hfill \end{array}$$

Thus, $J:\mathrm{Im}Q\to \mathrm{Ker}L$ is an isomorphism.

Lemma 9. 

Assume $\left(H_1\right)$ and $\left(H_2\right)$ and the following hypothesis holds:

Hypothesis 6 (H₆).

There exists a constant $M_3>0$ such that either for each $c_0\in \mathbb{R}$

$$\begin{array}{c}\hfill c_0(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }f(t,u_{c_0}\left(t\right),u_{c_0}\left(\lambda t\right),D_{0^{+}}^{\gamma -2;\psi }{u}_{c_0}\left(t\right),D_{0^{+}}^{\gamma -1;\psi }{u}_{c_0}\left(t\right))\right)>0, for\left|c_0\right|>M_3,\end{array}$$

(12)

or

$$\begin{array}{c}\hfill c_0(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }f(t,u_{c_0}\left(t\right),u_{c_0}\left(\lambda t\right),D_{0^{+}}^{\gamma -2;\psi }{u}_{c_0}\left(t\right),D_{0^{+}}^{\gamma -1;\psi }{u}_{c_0}\left(t\right))\right)<0, for\left|c_0\right|>M_3,\end{array}$$

(13)

where $u_{c_0}\left(t\right)=c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})$. Then, sets ${\Omega }_2=\{u\in \mathrm{Ker}L:Nu\in \mathrm{Im}L\}$ and ${\Omega }_3=\{u\in \mathrm{Ker}L:\rho \delta Iu+(1-\delta )JQNu=0,\delta \in [0,1]\}$ are bounded, where $J:ImQ\to KerL$ is a homeomorphism with $J\left(g\right)=c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})$:

$$\rho =\left\{\begin{array}{c}1,if\left(12\right)holds;\hfill \\ -1.if\left(13\right)holds.\hfill \end{array}\right.$$

(14)

Proof. 

If $u_{c_0}\in {\Omega }_2$, it is easy to find that $u_{c_0}\left(t\right)=c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})$ and $QN{u}_{c_0}=0,$ which represents $(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }N{u}_{c_0}\right)h\left(t\right)=0$, $|c_0|\le M_3.$ Now, let us estimate the norm of $u_{c_0}$ in X.

$$\begin{array}{cc}\hfill {\parallel (\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{u}_{c_0}{\parallel }_{C[0,1]}& =\underset{t\in [0,1]}{max}\left\{c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{n-2})\right\}\hfill \\ \hfill & \le M_3{\left(\right|c|(\psi \left(1\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-2}),\hfill \end{array}$$

$$\begin{array}{c}\hfill \parallel D_{0^{+}}^{\gamma -1;\psi }{u}_{c_0}{\parallel }_{C[0,1]}=|c_{0}c\Gamma \left(\gamma \right)|\le M_{3}c\Gamma \left(\gamma \right),\end{array}$$

and

$$\begin{array}{cc}\hfill \parallel D_{0^{+}}^{\gamma -2;\psi }{u}_{c_0}{\parallel }_{C[0,1]}& =\underset{t\in [0,1]}{max}\left\{c_0\left(c\left(\psi \right(t)-\psi (0\left)\right)-d\Gamma (\gamma -1)\right)\right\}\hfill \\ \hfill & \le M_3\left(\left|c\right|\left(\psi \right(1)-\psi (0\left)\right)+\left|d\right|\Gamma (\gamma -1)\right).\hfill \end{array}$$

By using the above estimates, we obtain $\parallel u_{c_0}{\parallel }_X\le M_3\left(\left|c\right|(\Gamma \left(\gamma \right)+\psi \left(1\right)-\psi \left(0\right)+{(\psi \left(t\right)-\psi \left(0\right))}^{n-1})+\left|d\right|(\Gamma (\gamma -1)+{(\psi \left(t\right)-\psi \left(0\right))}^{n-2})\right)<\infty$; that is, ${\Omega }_2$ is bounded.

For $u\in {\Omega }_3,$$u\left(t\right)=c_0\left(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right)$ and $\rho \delta Iu+(1-\delta )JQNu=0.$

If $\delta =1,$ then $u\equiv 0.$ If $\delta =0,$ then $QNu=0,$ i.e., $JQNu=JQN[c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}]=0,$ which follows from the proof of the boundedness of ${\Omega }_2$ as ${\parallel u\parallel }_X<\infty .$

If $\delta \in (0,1)$, we claim that $\rho \delta c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})+(1-\delta )JQN\left[c_0\left(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\right)\right]=0,$ and then, taking $|c_0|>M_3,$ we obtain

$$\rho \delta c_0+(1-\delta )(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }f(t,u_{c_0}\left(t\right),u_{c_0}\left(\lambda t\right),D_{0^{+}}^{\gamma -2;\psi }{u}_{c_0}\left(t\right),D_{0^{+}}^{\gamma -1;\psi }{u}_{c_0}\left(t\right))\right)=0,$$

which, with multiplying by $c_0$, leads to

$$\rho \delta c_0^2=-(1-\delta )c_0(B_1-\sigma B_2)(I_{0^{+}}^{\alpha ;\psi }f(t,u_{c_0}\left(t\right),u_{c_0}\left(\lambda t\right),D_{0^{+}}^{\gamma -1;\psi }{u}_{c_0}\left(t\right),D_{0^{+}}^{\gamma -2;\psi }{u}_{c_0}\left(t\right)),$$

which is impossible! Thus, we have $|c_0|\le M_3$. So, the boundedness of ${\Omega }_3$ follows ${\parallel u\parallel }_X\le M_3{\parallel c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2}\parallel }_X$. □

Theorem 2. 

Assume (H) and $(H_1$– $H_6)$ hold. Then, the functional boundary value problem (Equation (1))has at least one solution in X.

Proof. 

Let $\Omega$ be open and bounded such that $\Omega \supset \overline{{\Omega }_1}\cup \overline{{\Omega }_2}\cup \overline{{\Omega }_3}\cup \{\varnothing \}$. In light of the boundedness of ${\Omega }_1$ and ${\Omega }_2$ in Lemmas 8 and 9, we can derive $Lu\ne N_{\lambda }u$ for $u\in \partial \Omega \cap \left(\mathrm{dom}LKerL\right), \lambda \in (0,1)$ and $Nu\ne 0, u\in \mathrm{Ker}L\cap \partial \Omega .$ Thus, $\left(i\right)$ and $\left(ii\right)$ of Theorem 1 hold.

Let $H(u,\delta )=\rho \delta Iu+(1-\delta )JQNu,\delta \in [0,1],u\in \mathrm{Ker}L\cap \partial \Omega$; noticing the boundedness of ${\Omega }_3$ in Lemma 9 and ${\Omega }_3\subset \Omega$, we know that $H(u,\delta )\ne 0,u\in \mathrm{Ker}L\cap \partial \Omega ,\delta \in [0,1]$.

For $u\in \mathrm{Ker}L\cap \partial \Omega$, we find that $u\left(t\right)=c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\ne 0,$ and $H(u,1)=\rho c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\ne 0.$ To achieve this, via Lemma 9, we know that $H(u,0)=JQN\left(c_0(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})\right)\ne 0.$

Thus, by calculating the invariance of degree under a homotopy, we find that

$$\begin{array}{cc}\hfill deg(JQN,\Omega \cap \mathrm{Ker}L,0)& =deg\left(H(·,0),\Omega \cap \mathrm{Ker}L,0\right)\hfill \\ \hfill & =deg\left(H(·,1),\Omega \cap \mathrm{Ker}L,0\right)\hfill \\ \hfill & =deg\left(\rho I,\Omega \cap \mathrm{Ker}L,0\right)=\pm 1\ne 0,\hfill \end{array}$$

therefore, the condition $\left(iii\right)$ of Theorem 1 holds, and the existence result for $FBVP$(Equation (1)) is provided in $\overline{\Omega }$. □

Example 2. 

We introduce an example in order to demonstrate the application of the previous result. Consider the nonlocal boundary value problems of the following pantograph equation:

$$\left\{\begin{array}{c}{}^{H}D_{0^{+}}^{{\displaystyle \frac{7}{2}},{\displaystyle \frac{1}{2}};{\displaystyle \frac{5t}{2}}}u\left(t\right)=t+{\displaystyle \frac{{\left(\frac{5t}{2}\right)}^{\frac{1}{4}}sinu\left(t\right)}{{10}^4}}+{\displaystyle \frac{{\left(\frac{5t}{2}\right)}^{\frac{1}{4}}sinu\left(\frac{t}{2}\right)}{{10}^4}}+{\displaystyle \frac{1}{{10}^4}}{\displaystyle \frac{D_{0^{+}}^{\frac{7}{4};\frac{5t}{2}}u\left(t\right)}{|D_{0^{+}}^{\frac{7}{4};\frac{5t}{2}}u\left(t\right)|+1}}+{\displaystyle \frac{D_{0^{+}}^{\frac{11}{4};\frac{5t}{2}}u\left(t\right)}{{10}^4}},t\in (0,1],\hfill \\ I_{0^{+}}^{{\displaystyle \frac{1}{4}};{\displaystyle \frac{5t}{2}}}u\left(0\right)=0,D_{0^{+}}^{{\displaystyle \frac{3}{4}};{\displaystyle \frac{5t}{2}}}u\left(0\right)=0,\hfill \\ B_1\left(u\right)=D_{0^{+}}^{{\displaystyle \frac{7}{4}};{\displaystyle \frac{5t}{2}}}u\left(1\right)-{\displaystyle \frac{5}{3}}{D}_{0^{+}}^{{\displaystyle \frac{11}{4}};{\displaystyle \frac{5t}{2}}}u\left(0\right)=0,\hfill \\ B_2\left(u\right)=u\left(0\right)+{\displaystyle \frac{8}{25}}{I}_{0^{+}}^{{\displaystyle \frac{1}{4}};{\displaystyle \frac{5t}{2}}}u\left(1\right)=0.\hfill \end{array}\right.$$

It is easy to see that $a_0\left(t\right)=1, a_1\left(t\right)=a_2\left(t\right)={\displaystyle \frac{{\left(\frac{5t}{2}\right)}^{\frac{1}{4}}}{{10}^4}}, a_3\left(t\right)=a_4\left(t\right)={\displaystyle \frac{1}{{10}^4}},$$B_1\left({\left({\displaystyle \frac{5t}{2}}\right)}^{{\displaystyle \frac{11}{4}}}\right)={\displaystyle \frac{5}{6}}\Gamma \left({\displaystyle \frac{15}{4}}\right), B_1\left({\left({\displaystyle \frac{5t}{2}}\right)}^{{\displaystyle \frac{7}{4}}}\right)=\Gamma \left({\displaystyle \frac{11}{4}}\right), B_2\left({\left({\displaystyle \frac{5t}{2}}\right)}^{{\displaystyle \frac{11}{4}}}\right)={\displaystyle \frac{5}{6}}\Gamma \left({\displaystyle \frac{15}{4}}\right),$ and $B_2\left({\left({\displaystyle \frac{5t}{2}}\right)}^{{\displaystyle \frac{7}{4}}}\right)=\Gamma \left({\displaystyle \frac{11}{4}}\right).$ The problem is seen at resonance and when $c=\Gamma \left({\displaystyle \frac{11}{4}}\right), d={\displaystyle \frac{5}{6}}\Gamma \left({\displaystyle \frac{15}{4}}\right),$ and $\sigma =1$. At this point, we can take $h\left(t\right)={\displaystyle \frac{48}{125}}{\displaystyle \frac{{\left(\frac{5t}{2}\right)}^{\frac{1}{4}}}{\Gamma \left(\frac{19}{4}\right)}}$ to ensure that $(B_1-\sigma B_2)\left(I_{0^{+}}^{{\displaystyle \frac{7}{2}};{\displaystyle \frac{5t}{2}}}h\left(t\right)\right)=1$, i.e., $\left(H_3\right)$ holds.

Also,

$$\begin{array}{cc}\hfill |B_2\left(u\right)|& =|u\left(0\right)+{\displaystyle \frac{8}{25}}{I}_{0^{+}}^{{\displaystyle \frac{1}{4}};{\displaystyle \frac{5t}{2}}}u\left(1\right)|\hfill \\ \hfill & \le \left|u\left(0\right)\right|+{\displaystyle \frac{8}{25}}{I}_{0^{+}}^{{\displaystyle \frac{1}{4}};{\displaystyle \frac{5t}{2}}}\left|u\left(1\right)\right|\hfill \\ \hfill & \le ({\displaystyle \frac{8}{25}}·{\left({\displaystyle \frac{5}{2}}\right)}^{{\displaystyle \frac{1}{4}}}+1)·{\parallel u\parallel }_X\le {\displaystyle \frac{3}{2}}{\parallel u\parallel }_X,\hfill \end{array}$$

which represents ${\beta }_2={\displaystyle \frac{3}{2}}$. Then, we can directly calculate

$$\begin{array}{cc}\hfill (\parallel K_P\parallel +F)& (\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})\hfill \\ \hfill & \approx 0.9764<1.\hfill \end{array}$$

Hence, $\left(H_5\right)$ holds.

If $D_{0^{+}}^{{\displaystyle \frac{11}{4}};{\displaystyle \frac{5t}{2}}}u\left(t\right)>10,004,$ then $f(t,u\left(t\right),D_{0^{+}}^{{\displaystyle \frac{1}{2}}}u\left(t\right))>0$.

If $D_{0^{+}}^{{\displaystyle \frac{11}{4}};{\displaystyle \frac{5t}{2}}}u\left(t\right)<-10,004$, then $f(t,u\left(t\right),D_{0^{+}}^{{\displaystyle \frac{1}{2}}}u\left(t\right))<0$.

Hence,

$$\begin{array}{cc}\hfill & (B_1-\sigma B_2)\left(I_{0^{+}}^{{\displaystyle \frac{7}{2}};{\displaystyle \frac{5t}{2}}}Nu\right)\hfill \\ \hfill & =I_{0^{+}}^{{\displaystyle \frac{7}{2}};{\displaystyle \frac{5t}{2}}}Nu\left(1\right)-{\displaystyle \frac{8}{25}}{I}_{0^{+}}^{{\displaystyle \frac{15}{4}};{\displaystyle \frac{5t}{2}}}Nu\left(1\right)\hfill \\ \hfill & ={\int }_0^1[{\displaystyle \frac{{\left(\frac{5}{2}\right)}^{\frac{7}{2}}}{\Gamma \left(\frac{7}{4}\right)}}{(1-s)}^{{\displaystyle \frac{5}{2}}}-{\displaystyle \frac{8}{25}}{\displaystyle \frac{{\left(\frac{5}{2}\right)}^{\frac{11}{4}}}{\Gamma \left(\frac{15}{4}\right)}}{(1-s)}^{{\displaystyle \frac{11}{4}}}]Nu\left(s\right)ds\ne 0\hfill \end{array}$$

provided that $u\in \mathrm{dom}L\mathrm{Ker}L$ satisfies $|D_{0^{+}}^{{\displaystyle \frac{11}{4}};{\displaystyle \frac{5t}{2}}}u\left(t\right)|>M_0=10,004,$ since ${\displaystyle \frac{{\left(\frac{5}{2}\right)}^{\frac{7}{2}}}{\Gamma \left(\frac{7}{4}\right)}}>{\displaystyle \frac{8}{25}}{\displaystyle \frac{{\left(\frac{5}{2}\right)}^{\frac{11}{4}}}{\Gamma \left(\frac{15}{4}\right)}}$ and ${(1-s)}^{{\displaystyle \frac{5}{2}}}>{(1-s)}^{{\displaystyle \frac{11}{4}}},s\in (0,1).$ So, Hypothesis $\left(H_4\right)$ holds.

Last, for $u\in \mathrm{Ker}L$, where $u_{c_0}\left(t\right)=c_0(\Gamma \left({\displaystyle \frac{11}{4}}\right){\left({\displaystyle \frac{5t}{2}}\right)}^{{\displaystyle \frac{11}{4}}}-{\displaystyle \frac{5}{6}}\Gamma \left({\displaystyle \frac{15}{4}}\right){\left({\displaystyle \frac{5t}{2}}\right)}^{{\displaystyle \frac{7}{4}}})$, one can choose $M_3=1407>0$ such that $c_0(B_1-\sigma B_2)\left(I_{0^{+}}^{{\displaystyle \frac{7}{2}};{\displaystyle \frac{5t}{2}}}N{u}_{c_0}\left(t\right)\right)>0$, which shows that $\left(H_6\right)$ is confirmed, since

$$\begin{array}{cc}\hfill c_{0}N{u}_{c_0}\left(t\right)& =c_{0}t+{\displaystyle \frac{c_0{\left(\frac{5t}{2}\right)}^{\frac{1}{4}}}{{10}^4}}sin\left(u_{c_0}\left(t\right)\right)+{\displaystyle \frac{c_0{\left(\frac{5t}{2}\right)}^{\frac{1}{4}}}{{10}^4}}sin\left(u_{c_0}\left({\displaystyle \frac{t}{2}}\right)\right)+{\displaystyle \frac{c_0}{{10}^4}}{\displaystyle \frac{D_{0^{+}}^{\frac{7}{4};\frac{5t}{2}}{u}_{c_0}\left(t\right)}{|D_{0^{+}}^{\frac{7}{4};\frac{5t}{2}}{u}_{c_0}\left(t\right)|+1}}+{\displaystyle \frac{c_0{D}_{0^{+}}^{\frac{11}{4};\frac{5t}{2}}u\left(t\right)}{{10}^4}}\hfill \\ \hfill & \ge -|c_0|-{\displaystyle \frac{2|c_0|{\left(\frac{5}{2}\right)}^{\frac{1}{4}}}{{10}^4}}-{\displaystyle \frac{|c_0|}{{10}^4}}+{\displaystyle \frac{c_0^2}{{10}^4}}\Gamma \left({\displaystyle \frac{11}{4}}\right)\Gamma \left({\displaystyle \frac{15}{4}}\right)\hfill \\ \hfill & =|c_0\left|\right(|c_0|{\displaystyle \frac{\Gamma \left(\frac{11}{4}\right)\Gamma \left(\frac{15}{4}\right)}{{10}^4}}-{\displaystyle \frac{1+2{\left(\frac{5}{2}\right)}^{\frac{1}{4}}+{10}^4}{{10}^4}})>0,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & c_0(B_1-\sigma B_2)\left(I_{0^{+}}^{{\displaystyle \frac{7}{2}};{\displaystyle \frac{5t}{2}}}N{u}_{c_0}\left(t\right)\right)\hfill \\ \hfill & =c_0(I_{0^{+}}^{{\displaystyle \frac{7}{2}};{\displaystyle \frac{5t}{2}}}N{u}_{c_0}\left(1\right)-{\displaystyle \frac{8}{25}}{I}_{0^{+}}^{{\displaystyle \frac{15}{4}};{\displaystyle \frac{5t}{2}}}N{u}_{c_0}\left(1\right))\hfill \\ \hfill & ={\int }_0^1[{\displaystyle \frac{{\left(\frac{5}{2}\right)}^{\frac{7}{2}}}{\Gamma \left(\frac{7}{4}\right)}}{(1-s)}^{{\displaystyle \frac{5}{2}}}-{\displaystyle \frac{8}{25}}{\displaystyle \frac{{\left(\frac{5}{2}\right)}^{\frac{11}{4}}}{\Gamma \left(\frac{15}{4}\right)}}{(1-s)}^{{\displaystyle \frac{11}{4}}}]c_{0}N{u}_{c_0}\left(s\right)ds>0,\left|c_0\right|>1407.\hfill \end{array}$$

It follows from Theorem 2 that there must be at least one solution in X.

Theorem 3. 

Assume that $\left(H\right),$$(H_1-H_3)$ with ${\displaystyle \frac{d}{c}}<0$ or ${\displaystyle \frac{d}{c}}>(\gamma -1)(\psi \left(1\right)-\psi \left(0\right))$, $(H_6$) (of Lemma 9), and the following conditions hold:

Hypothesis 4′ (H₄′).

There exists a constant $M_0>0$ such that if $|D_{0^{+}}^{\gamma -2;\psi }u\left(t\right)|>M_0$, for all $t\in [0,1]$, then

$$(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }Nu\right)\ne 0.$$

Hypothesis 5′ (H₅′).

There exist some positive functions $a_i\left(t\right),i=0,1,2,3,4$, with $a_0\left(t\right),a_1\left(t\right)$${(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n},a_2\left(t\right)$${(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n},a_3\left(t\right)$, and $a_4\left(t\right)\in Y$, such that for all $(u_1,u_2,u_3,u_4)\in {\mathbb{R}}^4,t\in [0,1],$

$$|f(t,u_1,u_2,u_3,u_4)|\le a_0\left(t\right)+a_1\left(t\right)|u_1|+a_2\left(t\right)|u_2|+a_3\left(t\right)|u_3|+a_4\left(t\right)\left|u_4\right|,$$

provided that $(\parallel K_P\parallel +F^{\prime }\left)\right(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+{\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}\parallel }_{C[0,1]}+$$\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})<1.$ Then, the functional boundary value problem (Equation (1)) has at least one solution in X.

Proof. 

As in the proof of Lemma 8, $u\in {\Omega }_1$ means $(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }Nu\right)=0.$ From (H₄′), we know that there exists a constant $t_0\in [0,1]$ such that $|D_{0^{+}}^{\gamma -2;\psi }u\left(t_0\right)|\le M_0$. □

Remark 1. 

Similarly to Lemma 8, in this case $|D_{0^{+}}^{\gamma -1;\psi }u\left(t_0\right)|\le M_0|$ does not come for free.

Similarly, $u\left(t\right)=u_1\left(t\right)+u_2\left(t\right),$ where $u_1=(I-P)u\in \mathrm{dom}L\cap \mathrm{Ker}P$ and $u_2=Pu\in \mathrm{Im}P.$ Thus, $u_1=K_{P}L{u}_1=K_{P}L(I-P)u=K_{P}Lu=\lambda K_{P}Nu.$ As in the proof of Lemma 8,

$$\parallel u_1{\parallel }_X\le \parallel K_P{\parallel ·\parallel Nu\parallel }_{C[0,1]}.$$

Now, $u_2=u-u_{1}0$. In view of (Equation (5)), it is easy to deduce that $D_{0^{+}}^{\gamma -2;\psi }{u}_2=D_{0^{+}}^{\gamma -2;\psi }u-D_{0^{+}}^{\gamma -2;\psi }{u}_1$ and

$$|D_{0^{+}}^{\gamma -2;\psi }{u}_2\left(t_0\right)|\le |D_{0^{+}}^{\gamma -2;\psi }u\left(t_0\right)|+|D_{0^{+}}^{\gamma -2;\psi }{u}_1\left(t_0\right)|\le M_0+D·{\parallel Nu\parallel }_{C[0,1]}.$$

(15)

As in the proof of Lemma 8, $u_2\left(t\right)=Pu\left(t\right)=c\left(u\right)(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})$. Hence, it follows from (15) that

$$\left|c\left(u\right)\right|\le {\displaystyle \frac{M_0+D·{\parallel Nu\parallel }_{C[0,1]}}{\underset{t\in [0,1]}{min}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}}={\displaystyle \frac{M_0+D·{\parallel Nu\parallel }_{C[0,1]}}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}},$$

by $c\Gamma \left(\gamma \right)(\psi \left(t_0\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\ne 0$ in $[0,1]$. Then,

$$\begin{array}{cc}\hfill |u_2\left(t\right){(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }|& \le {\left|c\left(u\right)\right|·|c(\psi \left(t\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{n-2}|\hfill \\ \hfill & \le {\displaystyle \frac{(M_0+D·{\parallel Nu\parallel }_{C[0,1]})\left({\left|c\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-1}+\left|d\right|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right)}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}},\hfill \\ \hfill |D_{0^{+}}^{\gamma -1;\psi }{u}_2\left(t\right)|& =|c\left(u\right)·c\Gamma \left(\gamma \right)|\le {\displaystyle \frac{\left|c\right|\Gamma \left(\gamma \right)(M_0+D·\parallel Nu{\parallel }_{C[0,1]})}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill |D_{0^{+}}^{\gamma -2;\psi }{u}_2\left(t\right)|& =\left|c\right(u)·(c\Gamma \left(\gamma \right)\left(\psi \right(t)-\psi (0\left)\right)-d\Gamma (\gamma -1)\left)\right|\hfill \\ \hfill & \le {\displaystyle \frac{(M_0+D·{\parallel Nu\parallel }_{C[0,1]})·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}.\hfill \end{array}$$

This implies that

$$\begin{array}{cc}\hfill \parallel u_2{\parallel }_X\le & {\displaystyle \frac{M_0·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}\hfill \\ \hfill & +{\displaystyle \frac{M_0·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}\hfill \\ \hfill & +{\displaystyle \frac{D·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}{\parallel Nu\parallel }_{C[0,1]}\hfill \\ \hfill & +{\displaystyle \frac{D·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}{\parallel Nu\parallel }_{C[0,1]}\hfill \\ \hfill & =E^{\prime }+F^{\prime }{\parallel Nu\parallel }_{C[0,1]},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill E^{\prime }:=& {\displaystyle \frac{M_0·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}\hfill \\ \hfill +& {\displaystyle \frac{M_0·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill F^{\prime }:=& {\displaystyle \frac{D·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}}\hfill \\ \hfill & +{\displaystyle \frac{D·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c\Gamma \left(\gamma \right)\left(\psi \right(1)-\psi (0\left)\right)-d\Gamma (\gamma -1)|,|d\Gamma (\gamma -1)\left|\right\}}},\hfill \end{array}$$

which reveals that

$$\begin{array}{cc}\hfill & {\parallel u\parallel }_X\le \parallel u_1{\parallel }_X+\parallel u_2{\parallel }_X\le (\parallel K_P\parallel +F^{\prime }{)\parallel Nu\parallel }_{C[0,1]}+E^{\prime }\hfill \\ \hfill & \le E^{\prime }+(\parallel K_P\parallel +F^{\prime }\left)\right(\parallel a_0{\parallel }_{C[0,1]}+(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+{\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}\parallel }_{C[0,1]}\hfill \\ \hfill & +\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})·{\parallel u\parallel }_X)\hfill \\ \hfill & \le {\displaystyle \frac{E^{\prime }+(\parallel K_P\parallel +F^{\prime })\parallel a_0{\parallel }_{C[0,1]}}{1-(\parallel K_P\parallel +F^{\prime }\left)\right(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})}}\hfill \end{array}$$

From (H₅′), we know that ${\Omega }_1$ is bounded and the rest of the proof is identical to that of Theorem 2.

Theorem 4. 

Assume that (H), ( $H_1$– $H_3$) with ${\displaystyle \frac{d}{c}}<0$ or ${\displaystyle \frac{d}{c}}>(\psi \left(1\right)-\psi \left(0\right))$, $\left(H_6\right)$ (of Lemma 9), and the following conditions hold:

Hypothesis 4″ (H₄″).

There exists a constant $M_0>0$ such that if $\left|u\left(t\right)\right|>M_0$, for all $t\in [0,1]$, then

$$(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }Nu\right)\ne 0.$$

Hypothesis 5″ (H₅″).

There exist some positive functions $a_i\left(t\right),i=0,1,2,3,4$, with $a_0\left(t\right),a_1\left(t\right)$${(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n},a_2\left(t\right)$${(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n},a_3\left(t\right)$, and $a_4\left(t\right)\in Y$, such that for all $(u_1,u_2,u_3,u_4)\in {\mathbb{R}}^4,t\in [0,1]$

$$|f(t,u_1,u_2,u_3,u_4)|\le a_0\left(t\right)+a_1\left(t\right)|u_1|+a_2\left(t\right)|u_2|+a_3\left(t\right)|u_3|+a_4\left(t\right)\left|u_4\right|,$$

provided that $(\parallel K_P\parallel +F^{\prime \prime }\left)\right(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+{\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}\parallel }_{C[0,1]}+$$\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})<1.$ Then, the functional boundary value problem (Equation (1)) has at least one solution in X.

Proof. 

As in the proof of Lemma 8, $u\in {\Omega }_1$ means $(B_1-\sigma B_2)\left(I_{0^{+}}^{\alpha ;\psi }Nu\right)=0.$ From (H₄″), we know that there exists a constant $t_0\in [0,1]$ such that $|u\left(t_0\right)|\le M_0$. □

Remark 2. 

Note that we do not readily obtain $|D_{0^{+}}^{\gamma -1;\psi }u\left(t_0\right)|,|D_{0^{+}}^{\gamma -2;\psi }u\left(t_0\right)|\le M_0$, which follows directly from (H₄) and (H₄′).

Similarly, $u\left(t\right)=u_1\left(t\right)+u_2\left(t\right),$ where $u_1=(I-P)u\in \mathrm{dom}L\cap \mathrm{Ker}P$ and $u_2=Pu\in \mathrm{Im}P.$ Thus, $u_1=K_{P}L{u}_1=K_{P}L(I-P)u=K_{P}Lu=\lambda K_{P}Nu.$ As in the proof of Lemma 8,

$$\parallel u_1{\parallel }_X\le \parallel K_P{\parallel ·\parallel Nu\parallel }_{C[0,1]}.$$

Now, $u_2=u-u_1,$ and we can invoke Equation (3) to show that

$$\begin{array}{c}\hfill {|(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{u}_2\left(t_0\right)|\le |{(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }u\left(t_0\right)|+|{(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }{u}_1\left(t_0\right)|\\ \hfill \le M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }+B·{\parallel Nu\parallel }_{C[0,1]}.\end{array}$$

(16)

As in the proof of Lemma 8, $u_2\left(t\right)=Pu\left(t\right)=c\left(u\right)(c{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -2})$. From (16), one obtains

$$\begin{array}{cc}\hfill \left|c\right(u\left)\right|& \le {\displaystyle \frac{M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }+B·{\parallel Nu\parallel }_{C[0,1]}}{\underset{t\in [0,1]}{min}{\left\{\right|c(\psi \left(t\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{n-2}\left|\right\}}}\hfill \\ \hfill & ={\displaystyle \frac{M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }+B·{\parallel Nu\parallel }_{C[0,1]}}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}},\hfill \end{array}$$

by solving $c{(\psi \left(t_0\right)-\psi \left(0\right))}^{n-1}\ne d{(\psi \left(t_0\right)-\psi \left(0\right))}^{n-2}$ in $t\in [0,1]$. Then,

$$\begin{array}{cc}\hfill & |u_2\left(t\right){(\psi \left(t\right)-\psi \left(0\right))}^{n-\gamma }|\hfill \\ \hfill & \le {\left|c\left(u\right)\right|·|c(\psi \left(t\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(t\right)-\psi \left(0\right))}^{n-2}|\hfill \\ \hfill & \le {\displaystyle \frac{(M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }+B·{\parallel Nu\parallel }_{C[0,1]})\left({\left|c\right|(\psi \left(1\right)-\psi \left(0\right))}^{n-1}+\left|d\right|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right)}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}},\hfill \\ \hfill |D_{0^{+}}^{\gamma -1;\psi }{u}_2\left(t\right)|& =|c\left(u\right)·c\Gamma \left(\gamma \right)|\le {\displaystyle \frac{\left|c\right|\Gamma \left(\gamma \right)\left(M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }+B·{\parallel Nu\parallel }_{C[0,1]}\right)}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & |D_{0^{+}}^{\gamma -2;\psi }{u}_2\left(t\right)|\hfill \\ \hfill & =\left|c\right(u)·c\Gamma (\gamma \left)\right(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1\left)\right|\hfill \\ \hfill & \le {\displaystyle \frac{(M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }+B·{\parallel Nu\parallel }_{C[0,1]})·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}},\hfill \end{array}$$

which enables us to obtain

$$\begin{array}{cc}\hfill \parallel u_2{\parallel }_X\le & {\displaystyle \frac{M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}\hfill \\ \hfill & +{\displaystyle \frac{M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}\hfill \\ \hfill & +{\displaystyle \frac{B·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}{\parallel Nu\parallel }_{C[0,1]}\hfill \\ \hfill & +{\displaystyle \frac{B·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}{\parallel Nu\parallel }_{C[0,1]}\hfill \\ \hfill & =E^{\prime \prime }+F^{\prime \prime }{\parallel Nu\parallel }_{C[0,1]},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill E^{\prime \prime }:=& {\displaystyle \frac{M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}\hfill \\ \hfill +& {\displaystyle \frac{M_0·{(\psi \left(1\right)-\psi \left(0\right))}^{n-\gamma }·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill F^{\prime \prime }:=& {\displaystyle \frac{B·\left({\left|c\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-1}+{\left|d\right|(\psi \left(t\right)-\psi \left(0\right))}^{n-2}+\left|c\right|\Gamma \left(\gamma \right)\right)}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}\hfill \\ \hfill & +{\displaystyle \frac{B·\underset{t\in [0,1]}{max}\left\{\right|c\Gamma \left(\gamma \right)(\psi \left(t\right)-\psi \left(0\right))-d\Gamma (\gamma -1)\left|\right\}}{min\left\{\right|c{(\psi \left(1\right)-\psi \left(0\right))}^{n-1}-d{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}|,|d\left|{(\psi \left(1\right)-\psi \left(0\right))}^{n-2}\right\}}}.\hfill \end{array}$$

Furthermore, it holds that

$$\begin{array}{cc}\hfill & {\parallel u\parallel }_X\le \parallel u_1{\parallel }_X+\parallel u_2{\parallel }_X\le (\parallel K_P\parallel +F^{\prime \prime }{)\parallel Nu\parallel }_{C[0,1]}+E^{\prime \prime }\hfill \\ \hfill & \le E^{\prime \prime }+(\parallel K_P\parallel +F^{\prime \prime }\left)\right(\parallel a_0{\parallel }_{C[0,1]}+(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+{\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}\parallel }_{C[0,1]}\hfill \\ \hfill & +\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]}\left){\parallel u\parallel }_X\right)\hfill \\ \hfill & \le {\displaystyle \frac{E^{\prime \prime }+(\parallel K_P\parallel +F^{\prime \prime })\parallel a_0{\parallel }_{C[0,1]}}{1-(\parallel K_P\parallel +F^{\prime \prime }\left)\right(\parallel a_1{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_2{(\psi \left(t\right)-\psi \left(0\right))}^{\gamma -n}{\parallel }_{C[0,1]}+\parallel a_3{\parallel }_{C[0,1]}+\parallel a_4{\parallel }_{C[0,1]})}}.\hfill \end{array}$$

From (H₅″), we know that ${\Omega }_1$ is bounded. The remaining part of the proof follows the same steps as those in Theorem 2.

4. Conclusions

The primary aim of this paper was to explore the type of advanced $\psi$-Hilfer fractional pantograph equations that come with functional boundary value conditions at resonance. This indicates that the linear operator $Lu{=}^H{D}_{0^{+}}^{\alpha ,\beta ;\psi }u\left(t\right)$ related to Equation (1) possesses nontrivial solutions (further details are provided in Lemma 4).

First, an important aspect is that the conditions on boundaries can expand, as shown in recent studies related to multi-point and integral boundary conditions. Next, pantograph equations are applicable in settings like electric trains and electric cells, meaning that exploring these functional boundary value issues provides significant insights both theoretically and practically. Additionally, when looking at how applicable a certain type of equation is, the requirements for its solution, or the unknown function, may vary. Therefore, this article examined what might be needed for the unknown function to be “smooth” in real situations, and we presented various existence solutions for the unknown function while considering three different needs using Mawhin’s coincidence theory, showcasing our unique view on the topic. Lastly, our next paper will focus on a linked pantograph system with sequential fractional differential equations, pantograph systems involving piecewise fractional differential equations, and the stability of their solutions.