Existence Results for Resonant Functional Boundary Value Problems with Generalized Weighted Fractional Derivatives

Abstract

In this article, we deduce the existence of a solution to the weighted fractional differential equation with functional boundary data involving an $\omega$-weighted fractional derivative with Riemann–Liouville settings, ${\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }$ of order $\alpha \in ]n-1,n[$, on certain weighted Banach spaces when the nonlinear term contains the proportional delay term and fractional derivatives of order $(0,1)$. After carefully defining a few weighted spaces and building a few weighted projection operators, we use Mawhin’s coincidence theory to derive a number of existence results at resonance. Furthermore, our method generalizes some prior results because numerous fractional differential operators are specific instances of the operator ${\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }$ and represent functional boundary conditions in a highly generic way. Lastly, we illustrate and support our theoretical results with an example.

1. Introduction

Differential equations of fractional order are used in many disciplines, such as fluid mechanics [1,2], physics [3,4,5], engineering [6], medicine [7], and polymer science [8,9]. Therefore, fractional differential equations have been undergoing intensive development recently.

Riemann–Liouville and Caputo fractional derivatives are found in the majority of fractional operators (with singular kernels). Other types of fractional operators, however, have helped researchers better understand the world we live in. These include Hadamard-type fractional operators [10,11,12], generalized fractional integral and derivatives [13,14,15,16,17], weighted fractional operators [18,19], and Hilfer-Katugampola fractional derivatives [20]. As a result, this procedure has looked at numerous definitions of fractional differential operators, some of which are special cases of other definitions. Because these operators have advantages over Riemann–Liouville and Caputo derivative operators, authors [21,22,23,24] and others are still enthusiastic about utilizing them despite their many drawbacks and challenges. Since fractional differential equations encompass a wide range of fractional differential operators, it is useful to consider fractional differential equations containing these operators.

In [25,26], the author discussed the weighted fractional operators associated with the Caputo–Fabrizio [27] and Atangana–Baleanu [28] fractional operators. Jarad et al. [17] proposed a new class of weighted fractional operators on certain spaces, encompassing many well-known fractional operators (Riemann–Liouville and Caputo sense [29,30,31], Caputo–Hadamard [32,33], Erdélyi–Kober [34], Katugampola [35], and Hilfer–Katugampola derivative [36]). The author presented weighted fractional integrals bounded in Lebesgue measurable function spaces and showed that the weighted fractional derivatives of functions defined on certain spaces exist everywhere and listed some of their properties.

However, despite extensive research on the existence of solutions to fractional differential equations under specific boundary conditions, there is little mention in the literature of the existence of solutions to functional boundary value problems involving the weighted fractional derivative of one function with respect to another function, even though functional boundary value problems greatly generalize and extend many specific boundary data. The presence of solutions for integer-order and fractional-order functional boundary value issues has already been covered in [37,38,39], and research on weighted fractional-order operator equations has been examined in [40,41,42,43].

Furthermore, to our knowledge, the boundary value problem using $\omega$-weighted fractional derivatives discussed above has hardly mentioned resonance conditions. We shall examine the functional boundary value problem of the $\omega$-weighted fractional derivatives of a function with respect to another function while keeping in mind the earlier findings:

$$\left\{\begin{array}{c}{\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }u(x)=f(x,u(x),u(\lambda x),{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }u(x)),\lambda \in (0,1),x\in (0,1),\hfill \\ (\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -i,\psi ,\omega }u(x)){|}_{x=0}=0,i=1,2,\dots ,n-2,\hfill \\ {\mathfrak{B}}_1(u)=0,{\mathfrak{B}}_2(u)=0.\hfill \end{array}\right.$$

(1)

where ${\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }$ is the $\omega$-weighted fractional derivative of order $\alpha (n-1<\alpha \le n,n>3)$ with respect to another function $\psi (x)$, which is a strictly increasing continuously differentiable function where ${\psi }^{\prime }(x)\ne 0$, $\omega :[0,1]\to ]0,\infty [,$ is a weighted function, ${\omega }^{-1}(x)={\displaystyle \frac{1}{\omega (x)}}$. ${\mathfrak{B}}_1,{\mathfrak{B}}_2$ are continuous linear functionals. We will always suppose $f:[0,1]\times {\mathbb{R}}^3\to \mathbb{R}$ satisfies the following conditions:

(1)

$\omega (·){\psi }^{\prime }(·)f(·,u)$ is measurable for each fixed $u\in {\mathbb{R}}^3$, $\omega (x){\psi }^{\prime }(x)f(x,·)$ is continuous for a.e., $x\in [0,1]$.

(2)

$sup\{|\omega (x){\psi }^{\prime }(x)f(x,u)|:u\in D_0\}<+\infty$, for any compact set $D_0\in {\mathbb{R}}^3$.

Through our research on this fascinating specific topic, we have made novel and noteworthy discoveries. The following summarizes the focal contributions of our work: First, numerous well-known fractional order operators can be covered by the high-order weighted fractional order operator that we explored, such as Reimann-Liouville [29,44], Caputo, Hadamard [30,31,32,33], and Erdelyi-Kober fractional derivatives [34]. Second, the functional boundary value problem we studied has good generality and can include many homogeneous boundary value conditions. Additionally, it is worth noting that the nonlinear terms of the equation include proportional delay terms and weighted derivatives of unknown functions with ${\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }$ of order $\alpha \in ]n-1,n[$. In fact, functional differential equations with proportional delays are usually referred to as pantograph equations [45], which are widely applied in electric trains and electric cells [46,47,48], and a pantograph is commonly seen as a device for measurement and graphing. Another point is that incorporating derivative terms of unknown functions into nonlinear terms can more accurately describe and analyze complex nonlinear phenomena in practical applications. Subsequently, we discussed two results based on the potential requirements for the unknown function itself and the need for weighted fractional derivatives in the nonlinear terms in real-world applications. During the discussion, we carefully defined weighted Banach spaces and weighted projection operators, thereby providing solutions for more complex resonance scenarios. Finally, we provided an example to illustrate one of the theoretical cases.

The paper is divided into the following sections. The Section 2 includes some introductions and essential concepts regarding some symbols, weighted fractional calculus, the coincidence degree continuation theorem, and supporting Banach spaces and operators. We present two existence discoveries in Section 3 on the various needs of the unknown function related to resonance. Finally, to demonstrate our major findings, we provide a numerical example.

Definition 1.

We say that $u\in X$ is a solution to the functional boundary value problem (FBVP) (1), which means that u satisfies the equation and boundary conditions in (1).

2. Basic Definitions and Preliminaries

We begin this part by recalling several symbols which will be needed later.

Definition 2

([17]). Let $\omega \ne 0$ be a function defined on $[0,1]$, ψ is a differentiable strictly increasing function on $[0,1]$. The space $L^{p,\psi ,\omega }(0,1),1\le p\le \infty$ is the space of all Lebesgue measurable functions f defined on $[0,1]$ for which ${\parallel f\parallel }_{L^{p,\psi ,\omega }}$, where

$${\parallel f\parallel }_{L^{p,\psi ,\omega }}:={({\int }_0^1{\left|\omega (x)f(x)\right|}^p{\psi }^{\prime }(x)dx)}^{{\displaystyle \frac{1}{p}}},1\le p<\infty ,$$

and ${\parallel f\parallel }_{L^{\infty ,\psi ,\omega }}:=esssu{p}_{0\le x\le 1}\left|\omega (x)f(x)\right|<\infty .$

Remark 1.

It is worth noting that $f\in L^{p,\psi ,\omega }(0,1)\iff$ $\omega (x)f(x){({\psi }^{\prime }(x))}^{{\displaystyle \frac{1}{p}}}\in L^p(0,1)$ for $1\le p<\infty$ and $f\in L^{\infty ,\psi ,\omega }\iff \omega (x)f(x)\in L_{\infty }(0,1).$

Definition 3

([17]). Let $\alpha >0$. The ω-weighted Riemann–Liouville fractional integral of order α where the lower limit at 0 of a function $f\in L^{p,\psi ,\omega }(0,1)$ in regard to ψ is given by:

$$({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }f)(x)={\displaystyle \frac{{\omega }^{-1}(x)}{\Gamma (\alpha )}}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)f(t)dt.$$

Lemma 1

([17]). Assume $f\in L^{p,\psi ,\omega }(0,1)$, $1\le p\le \infty ,\alpha >0,$ and $\beta >0$. Then,

$$({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{\beta ,\psi ,\omega })f={\mathfrak{I}}_{0^{+}}^{\alpha +\beta ,\psi ,\omega }f.$$

Definition 4

([17]). Let $n\in \mathbb{N}$ and $\alpha \in ]n-1,n[.$ The ω-weighted Riemann–Liouville fractional derivative whose order α where the lower limit at 0 of a function $f:[0,1]\to \mathbb{R}$ in regard to ψ is given by:

$$({\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }f)(x)={\displaystyle \frac{{\omega }^{-1}(x)}{\Gamma (n-\alpha )}}{\mathfrak{D}}^{n,\psi ,\omega }\left({\int }_0^x{(\psi (x)-\psi (t))}^{n-\alpha -1}{\psi }^{\prime }(t)\omega (t)f(t)dt\right),x>0,\alpha >0,$$

assuming that the right-hand is well defined.

Remark 2.

The following notation is introduced:

$$({\mathfrak{D}}^{1,\psi ,\omega }f)(x):={\omega }^{-1}(x){\displaystyle \frac{1}{{\psi }^{\prime }(x)}}{\displaystyle \frac{d}{dx}}(\omega (x)f(x)),$$

$$({\mathfrak{D}}^{n,\psi ,\omega }f)(x):={\mathfrak{D}}^{1,\psi ,\omega }{D}^{n-1,\psi ,\omega }f(x)={\omega }^{-1}(x){[{\displaystyle \frac{1}{{\psi }^{\prime }(x)}}{\displaystyle \frac{d}{dx}}]}^n(\omega (x)f(x)),$$

and

$$f_n(x):=\omega (x){\mathfrak{D}}^{n,\psi ,\omega }f(x)={[{\displaystyle \frac{1}{{\psi }^{\prime }(x)}}{\displaystyle \frac{d}{dx}}]}^n(\omega (x)f(x)),n\in \mathbb{N}.$$

Lemma 2

([17]).

(1)

If $\alpha >0,$ and $\beta >0,$ then

$${\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }({\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\beta -1})(x)={\displaystyle \frac{\Gamma (\beta )}{\Gamma (\beta +\alpha )}}{\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\beta +\alpha -1};\forall x\in [0,1].$$

(2)

For $\alpha <\beta$ and $\beta >0$, we have

$${\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }({\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\beta -1})(x)={\displaystyle \frac{\Gamma (\beta )}{\Gamma (\beta -\alpha )}}{\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\beta -\alpha -1};\forall x\in [0,1].$$

The subsequent discussion will demonstrate the combination of weighted fractional derivatives and weighted fractional integrals.

Lemma 3

([17]).

(1)

If $f\in L^{p,\psi ,\omega }(0,1)$ and $\alpha >\beta >0,$ then ${\mathfrak{D}}_{0^{+}}^{\beta ,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }f(x)={\mathfrak{I}}_{0^{+}}^{\alpha -\beta ,\psi ,\omega }f(x),\forall x\in [0,1].$

(2)

If $f\in L^{p,\psi ,\omega }(0,1),$ then

$${\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }f(x)=f(x),\forall x\in [0,1].$$

(3)

If $\alpha \in ]n-1,n[$ and $f\in L^{p,\psi ,\omega }(0,1)$ such that ${\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }f\in A{C}^{n,\psi ,\omega }[0,1],$ then for any $x\in [0,1]$

$$\begin{array}{cc}\hfill & {\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }({\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega })(x)\hfill \\ \hfill & =f(x)-{\omega }^{-1}(x)\sum _{k=1}^n{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -k}}{\Gamma (\alpha -k+1)}}{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }f)}_{n-k}(0),\hfill \end{array}$$

where ${({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }f)}_k(0)=\underset{x\to 0}{lim}{({\displaystyle \frac{1}{{\psi }^{\prime }(x)}}{\displaystyle \frac{d}{dx}})}^k\left(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }f(x)\right).$

Definition 5

([49,50]). Let X, Y be real Banach spaces, $\mathfrak{L}:dom\mathfrak{L}\subset X\to Y$ be a linear operator. X is said to be the Fredholm operator of index zero provided that:

(i)

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

(ii)

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

Let $\mathfrak{P}:X\to X$, $\mathfrak{Q}:Y\to Y$ are continuous projectors such that $\mathrm{Im}\mathfrak{P}=\mathrm{Ker}\mathfrak{L}$, $\mathrm{Ker}\mathfrak{Q}=\mathrm{Im}\mathfrak{L}$, $X=\mathrm{Ker}\mathfrak{L}\oplus \mathrm{Ker}\mathfrak{P}$ and $Y=\mathrm{Im}\mathfrak{L}\oplus \mathrm{Im}\mathfrak{Q}$.

It follows that ${\mathfrak{L}|}_{\mathrm{dom}\mathfrak{L}\cap \mathrm{Ker}\mathfrak{P}}:\mathrm{dom}\mathfrak{L}\cap \mathrm{Ker}\mathfrak{P}\to \mathrm{Im}\mathfrak{L}$ is reversible. We denote the inverse of the mapping by $K_{\mathfrak{P}}$ (generalized inverse operator of $\mathfrak{L}$).

If $\Omega$ is an open bounded subset of X such that $\mathrm{dom}\mathfrak{L}\cap \Omega \ne \varnothing$, the mapping $\mathfrak{N}:X\to Y$ will be called $\mathfrak{L}-compact$ on $\overline{\Omega }$, if $\mathfrak{QN}(\overline{\Omega })$ and $K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}:\overline{\Omega }\to X$ are continuous and compact.

Theorem 1

(see [49,50] Mawhin continuation theorem). Let $\mathfrak{L}:\mathrm{dom}\mathfrak{L}\subset X\to Y$ be a Fredholm operator of index zero and $\mathfrak{N}:X\to Y$ is $\mathfrak{L}$-compact on $\overline{\Omega }$. Assume that the following conditions are satisfied:

(i)

$\mathfrak{L}u\ne \mu \mathfrak{N}u$ for every $(u,\mu )\in [(\mathrm{dom}\mathfrak{L}\setminus \mathrm{Ker}\mathfrak{L})\cap \partial \Omega ]\times (0,1)$;

(ii)

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

(iii)

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

Then the equation $\mathfrak{L}u=\mathfrak{N}u$ has at least one solution in $\mathrm{dom}\mathfrak{L}\cap \overline{\Omega }$.

Let us consider the following Banach spaces:

$C_{\omega }[0,1]:=\{x\in C(0,1]:\omega x\in C[0,1]\}$, where ${\parallel x\parallel }_{C_{\omega }}=\underset{t\in [0,1]}{max}\left|\omega (t)x(t)\right|.$

$C_{n-\alpha ,\psi ,\omega }[0,1]:=\{x\in C(0,1]:{(\psi (·)-\psi (0))}^{n-\alpha }x(·)\in C_{\omega }[0,1]\},$ where ${\parallel x\parallel }_{C_{n-\alpha ,\psi ,\omega }}=\underset{t\in [0,1]}{max}$${|(\psi (t)-\psi (0))}^{n-\alpha }\omega (t)x(t)|$.

$A{C}^{1,\psi ,\omega }[0,1]:=\{x:[0,1]\to \mathbb{R},x\omega \in AC[0,1]\},$ where $AC[0,1]$ is the set of absolutely continuous functions on the interval $[0,1]$, and ${\parallel x\parallel }_{A{C}^{1,\psi ,\omega }}={\parallel x\omega \parallel }_{AC[0,1]}$.

$A{C}^{n,\psi ,\omega }[0,1]:=\{x:[0,1]\to \mathbb{R},{[{\displaystyle \frac{1}{{\psi }^{\prime }(t)}}{\displaystyle \frac{d}{dt}}]}^{n-1}(\omega (t)f(t))\in AC[0,1]\},$ and ${\parallel x\parallel }_{A{C}^{n,\psi ,\omega }}={\parallel x_{n-1}\parallel }_{AC[0,1]}$, where $x_{n-1}={[{\displaystyle \frac{1}{{\psi }^{\prime }(t)}}{\displaystyle \frac{d}{dt}}]}^{n-1}(\omega (t)f(t))$.

Based on the above spaces, we will further introduce the Banach spaces and their norms that are closely related to the narrative of this paper.

$X=\{u:[0,1]\to \mathbb{R}|u\in C_{n-\alpha ,\psi ,\omega }[0,1],{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }u\in C_{\omega }[0,1]\}$ with norm ${\parallel u\parallel }_X={\parallel u\parallel }_{C_{n-\gamma ,\psi ,\omega }}+{\parallel {\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }u\parallel }_{C_{\omega }}$ and $Y=L^{1,\psi ,\omega }(0,1)$ with norm ${\displaystyle {\parallel y\parallel }_{L^{1,\psi ,\omega }}={\int }_0^1\left|\omega (x)f(x)\right|{\psi }^{\prime }(x)dx}$.

Define the linear operator $\mathfrak{L}:X\cap \mathrm{dom}\mathfrak{L}\subset X\to Y$, and the nonlinear operator $\mathfrak{N}:X\to Y$ by

$$\mathfrak{L}u={\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }u(x),u\in \mathrm{dom}\mathfrak{L},$$

and

$$\mathfrak{N}u=\mu f(x,u(x),u(\lambda x),{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }u(x)),\mu ,\lambda \in (0,1),x\in [0,1],$$

where $\mathrm{dom}\mathfrak{L}=\left\{u\in X,{\mathfrak{D}}_{0^{+}}^{\alpha -1,\psi ,\omega }u\in A{C}^{1,\psi ,\omega }[0,1],\left(\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -i,\psi ,\omega }u\right)(0)=0,i=1,2,\dots ,n-2,{\mathfrak{B}}_j(u)=0,j=1,2\right\}$.

The operator equation $\mathfrak{L}u=\mathfrak{N}u,u\in \mathrm{dom}\mathfrak{L}$ is then equal to the $FBVP$ (1).

In order to better narrate the outcomes we want, we give the following hypotheses.

Hypothesis 1

($H_1$). The linear functionals ${\mathfrak{B}}_j:X\to \mathbb{R}$ satisfy ${\mathfrak{B}}_1({\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\alpha -n+1})=\varrho b,{\mathfrak{B}}_1({\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\alpha -n})=\varrho a,{\mathfrak{B}}_2({\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\alpha -n+1})=b,$ ${\mathfrak{B}}_2({\omega }^{-1}(x){(\psi (x)-\psi (0))}^{\alpha -n})=a,$ where $a^2+b^2\ne 0,\varrho ,a,b\in \mathbb{R}$.

Hypothesis 2

($H_2$). The functionals ${\mathfrak{B}}_1,{\mathfrak{B}}_2:X\to \mathbb{R}$ are continuous with the respective norms ${\beta }_1,{\beta }_2,$ that is, $|{\mathfrak{B}}_i(u)|\le {\beta }_i{\parallel u\parallel }_X$.

Hypothesis 3

($H_3$). The equation

$${\displaystyle ({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)(\frac{1}{\omega (x)\Gamma (\alpha )}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)h(t)dt)=1}$$

is valid for a function $h(x)\in L^{p,\psi ,\omega }(0,1).$

The next statement shows that there is indeed a function $h(x)$ for which the hypothesis $(H_3)$ is true.

Lemma 4.

Assume that $(H_1)$ holds. Then there exists $h(x)\in L^{p,\psi ,\omega }(0,1)$ such that

$${\displaystyle ({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)(\frac{1}{\omega (x)\Gamma (\alpha )}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)h(t)dt)\ne 0.}$$

Proof. 

For ease of use, set $\mathfrak{B}={\mathfrak{B}}_1-\sigma {\mathfrak{B}}_2$, and $h_n(x)={\omega }^{-1}(x){(\psi (x)-\psi (0))}^n.$ Then there must exist $n\in \mathbb{N}$ such that ${\displaystyle \mathfrak{B}\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{h}_n(x)\right)\ne 0}$.

If not, $\mathfrak{B}({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{h}_n(x))=0$, that is, $\mathfrak{B}\left({\omega }^{-1}(x){(\psi (x)-\psi (0))}^{n+\alpha }\right)=0$. Thus, $\mathfrak{B}(p)=0$ for every ’polynomial’ ${\displaystyle p(x)={\omega }^{-1}(x)\sum _{i=0}^n\left({(\psi (x)-\psi (0))}^{i+\alpha }\right)}$. It is easy to verify that ${\{{\omega }^{-1}(x)\left({(\psi (x)-\psi (0))}^{i+\alpha }\right)\}}_{i\in \mathbb{N}}$ is dense in X, so, we have $\mathfrak{B}(u)\equiv 0,u\in X$. Given that ${\mathfrak{B}}_1$ and ${\mathfrak{B}}_2$ are linearly independent on X, this is contradictory.

There must be some $i,i\in \mathbb{N}$ such that $\mathfrak{B}({\omega }^{-1}(x)\left({(\psi (x)-\psi (0))}^{i+\alpha }\right))\ne 0$. For these i, we simply take $h(x)={\displaystyle \frac{h_i(x)}{\mathfrak{B}({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{h}_i(x))}}.$ Thus, there exists $h\in L^{p,\psi ,\omega }(0,1)$ satisfying $(H_3)$. □

3. Main Results

Lemma 5.

In the event where hypotheses $(H)$, $(H_1)$, and $(H_3)$ are true. The mapping $\mathfrak{L}:\mathrm{dom}\mathfrak{L}\subset X\to Y$ is a Fredholm mapping of index zero.

Proof. 

If $u\in \mathrm{dom}\mathfrak{L}$, and $\mathfrak{L}u=0,$ we can deduce $u(x)={\omega }^{-1}(x)\left(c{(\psi (x)-\psi (0))}^{\alpha -n+1}+d{(\psi (x)-\psi (0))}^{\alpha -n}\right)$, ${\mathfrak{B}}_1({\omega }^{-1}(x)\left(c{(\psi (x)-\psi (0))}^{\alpha -n+1}+d{(\psi (x)-\psi (0))}^{\alpha -n})\right)=\varrho ca+\varrho db=0,$ and

${\mathfrak{B}}_2({\omega }^{-1}(x)\left(c{(\psi (x)-\psi (0))}^{\alpha -n+1}+d{(\psi (x)-\psi (0))}^{\alpha -n})\right)=ca+db=0,$ which means that

$$\mathrm{Ker}\mathfrak{L}=\{c{\omega }^{-1}(x)(a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}):a^2+b^2\ne 0,c\in \mathbb{R}\},\mathrm{dimKer}\mathfrak{L}=1.$$

We now demonstrate

$$\begin{array}{c}{\displaystyle \mathrm{Im}\mathfrak{L}=\{y\in Y:({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\displaystyle \frac{{\omega }^{-1}(x)}{\Gamma (\alpha )}}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)y(t)dt)=0\}.}\hfill \end{array}$$

(2)

Assuming $y=\mathrm{Im}\mathfrak{L},y=\mathfrak{L}u$, where $u\in \mathrm{dom}\mathfrak{L}$ exists. Consequently,

$u(x)={\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y(x)+{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)\Gamma (\alpha -n+2)}}{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)+{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)\Gamma (\alpha -n+1)}}({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0).$

With ${\mathfrak{B}}_1(u)={\mathfrak{B}}_2(u)=0,$ then results in

$$\begin{array}{cc}\hfill 0={\mathfrak{B}}_i(u)=& {\mathfrak{B}}_i\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+{\displaystyle \frac{{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)}{\Gamma (\alpha -n+2)}}{\mathfrak{B}}_i({\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)}})+{\displaystyle \frac{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+1)}}\hfill \\ \hfill & \times {\mathfrak{B}}_i({\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}),i=1,2.\hfill \end{array}$$

Then, we can deduce from $(H_1)$ that

$$\begin{array}{cc}\hfill & {\mathfrak{B}}_1\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+{\displaystyle \frac{{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)}{\Gamma (\gamma -n+2)}}{\mathfrak{B}}_1({\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)}})+{\displaystyle \frac{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+1)}}{\mathfrak{B}}_1({\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})\hfill \\ \hfill & ={\mathfrak{B}}_1\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+\varrho b{\displaystyle \frac{{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)}{\Gamma (\alpha -n+2)}}+\varrho a{\displaystyle \frac{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+1)}}=0,\hfill \\ \hfill & {\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+{\displaystyle \frac{{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)}{\Gamma (\gamma -n+2)}}{\mathfrak{B}}_2({\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)}})+{\displaystyle \frac{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+1)}}{\mathfrak{B}}_2({\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})\hfill \\ \hfill & ={\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+b{\displaystyle \frac{{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)}{\Gamma (\alpha -n+2)}}+a{\displaystyle \frac{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+1)}}=0.\hfill \end{array}$$

Therefore, ${\displaystyle y\in \{y\in Y:({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\displaystyle \frac{{\omega }^{-1}(x)}{\Gamma (\alpha )}}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)y(t)dt)=0\},}$ i.e.,

$${\displaystyle \mathrm{Im}\mathfrak{L}\subseteq \{y\in Y:({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\displaystyle \frac{{\omega }^{-1}(x)}{\Gamma (\alpha )}}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)y(t)dt)=0\}.}$$

If ${\displaystyle y\in \{y\in Y:({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\displaystyle \frac{{\omega }^{-1}(x)}{\Gamma (\alpha )}}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)y(t)dt)=0\},}$ take

$$u(t)=-{\displaystyle \frac{b{(\psi (x)-\psi (0))}^{\alpha -n+1}+a{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)(a^2+b^2)}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+{\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y(x).$$

It is simple to locate that $\mathfrak{L}u=y$,$\left(\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -i,\psi ,\omega }u\right)(0)=0,i=1,2,\dots ,n-2,$ and ${\mathfrak{B}}_1(u)={\mathfrak{B}}_2(u)=0,$ which suggests that

$${\displaystyle \{y\in Y:({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\displaystyle \frac{1}{\Gamma (\alpha )}}{\int }_0^t{\psi }^{\prime }(s){(\psi (t)-\psi (0))}^{\alpha -1}y(s)ds)=0\}\subseteq \mathrm{Im}\mathfrak{L}.}$$

As a result, we get $(2).$

Define $\mathfrak{Q}:Y\to Y$ as follows:

$${\displaystyle \mathfrak{Q}y=({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\displaystyle \frac{{\omega }^{-1}(x)}{\Gamma (\alpha )}}{\int }_0^x{(\psi (x)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)y(t)dt)h(x),}$$

where $h(x)$ is introduced in condition $(H_3)$.

With $(H_3)$, we have ${\mathfrak{Q}}^{2}y=\mathfrak{Q}y$ according to the property of $h(x)$. Thus, $\mathfrak{Q}:Y\to Y$ is a continuous liner projector such that $\mathrm{Im}\mathfrak{L}=\mathrm{Ker}\mathfrak{Q}$ and $\mathrm{Im}\mathfrak{Q}=\{ch(x)|c\in \mathbb{R}\}.$$Y=\mathrm{Im}\mathfrak{L}\oplus \mathrm{Im}\mathfrak{Q}$ and $\mathrm{dimKer}\mathfrak{L}=\mathrm{codimIm}\mathfrak{L}=1$ are evident. That is, $\mathfrak{L}$ is a Fredholm mapping of index zero. □

$\mathfrak{P}:X\to X$ is taken by

$\mathfrak{P}u(t)={\displaystyle \frac{a{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)-b(\alpha -n+1)(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+2)(a^2+b^2)}}{\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n})}{\omega (x)}}.$

For $u\in X$, set $u=u-\mathfrak{P}u+\mathfrak{P}u$. It is easy to verify that

$$\begin{array}{cc}\hfill & {\mathfrak{P}}^{2}u=\mathfrak{P}(\mathfrak{P}u)\hfill \\ \hfill =& {\displaystyle \frac{a{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }\mathfrak{P}u)}_1(0)-b(\alpha -n+1)(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }\mathfrak{P}u)(0)}{\Gamma (\alpha -n+2)(a^2+b^2)}}{\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n})}{\omega (x)}}\hfill \\ \hfill =& {\displaystyle \frac{a{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)-b(\alpha -n+1)(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+2)(a^2+b^2)}}(a^2\Gamma (\alpha -n+2)+b^2(\alpha -n+1)\Gamma (\alpha -n+1))\hfill \\ \hfill & \times {\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n})}{\omega (x)}}=\mathfrak{P}u,u\in X,\hfill \end{array}$$

since

$$\begin{array}{cc}\hfill {\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)}}& ={\displaystyle \frac{\Gamma (\alpha -n+2)}{\omega (x)}}(\psi (x)-\psi (0)),\hfill \\ \hfill {\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}& ={\displaystyle \frac{\Gamma (\alpha -n+1)}{\omega (x)}},\hfill \end{array}$$

${({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)}})}_1=\Gamma (\alpha -n+2),$ and ${({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})}_1=0.$

At the same time, it is convenient to verify that $\mathrm{Im}\mathfrak{P}=\mathrm{Ker}\mathfrak{L}$, moreover, take

$\tilde{u}=c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})\in \mathrm{Im}\mathfrak{P},c\in \mathbb{R}.$ If, in addition, $\tilde{u}\in \mathrm{Ker}\mathfrak{P}$, then

$$\begin{array}{cc}\hfill 0& =a{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }\tilde{u})}_1(0)-b(\alpha -n+1)(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }\tilde{u})(0)\hfill \\ \hfill & =c\Gamma (\alpha -n+2)(a^2+b^2).\hfill \end{array}$$

We have to have $c=0$ by $\Gamma (\alpha -n+2)\ne 0,a^2+b^2\ne 0,$ i.e., $\mathrm{Im}\mathfrak{P}\bigcap \mathrm{Ker}\mathfrak{P}=\{0\}$. So, $X=\mathrm{Ker}\mathfrak{L}\oplus \mathrm{Ker}\mathfrak{P}$.

Lemma 6.

The mapping $K_{\mathfrak{P}}:Y\to \mathrm{dom}\mathfrak{L}\bigcap \mathrm{Ker}\mathfrak{P}$ defined by

$$\begin{array}{c}\hfill K_{\mathfrak{P}}y(x)=-{\displaystyle \frac{b{(\psi (x)-\psi (0))}^{\alpha -n+1}+a{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)(a^2+b^2)}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+{\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y(x)\end{array}$$

is the generalized inverse operator of $\mathfrak{L}$.

Proof. 

Considering Lemma 3 (3) and Definition 4, one may demonstrate $\mathfrak{L}{K}_{\mathfrak{P}}y={\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }(K_{\mathfrak{P}}y)(x)={\mathfrak{D}}^{n,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }(K_{\mathfrak{P}}y)(x)={\mathfrak{D}}^{n,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y(x)=y(x)$ for all $y\in Y$. By the fact that $({\mathfrak{D}}_{0^{+}}^{\alpha -i,\psi ,\omega }{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)}})(0)=0$ and $({\mathfrak{D}}_{0^{+}}^{\alpha -i,\psi ,\omega }{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})(0)=0,i=1,2,\dots ,n-2$ hold, it is immediately possible to obtain the following equation

$$\begin{array}{cc}\hfill & (\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -i,\psi ,\omega }(K_{\mathfrak{P}}y))(0)\hfill \\ \hfill & =(\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -i,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y)(0)\hfill \\ \hfill & =(\omega (x){\mathfrak{I}}_{0^{+}}^{i,\psi ,\omega }y)(0)=0,i=1,2,\dots ,n-2.\hfill \end{array}$$

and ${\mathfrak{D}}_{0^{+}}^{\alpha -1,\psi ,\omega }(K_{\mathfrak{P}}y)={\mathfrak{I}}_{0^{+}}^{1,\psi ,\omega }y\in A{C}^{1,\psi ,\omega }[0,1].$

From Hypothesis 1$(H1)$, we directly obtain ${\mathfrak{B}}_j(K_{\mathfrak{P}}y)=0,j=1,2$.

$$\begin{array}{cc}\hfill \mathfrak{P}(K_{\mathfrak{P}}y)(x)=& {\displaystyle \frac{a{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }{K}_{\mathfrak{P}}y)}_1(0)-b(\alpha -n+1)(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }{K}_{P}y)(0)}{\Gamma (\alpha -n+2)(a^2+b^2)}}\hfill \\ \hfill & \times {\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}\hfill \\ \hfill =& {\displaystyle \frac{\frac{-ab(\alpha -n+1)\Gamma (\alpha -n+1)}{a^2+b^2}{\mathfrak{B}}_2({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y)+\frac{ab\Gamma (\alpha -n+2)}{a^2+b^2}{\mathfrak{B}}_2({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y)}{\Gamma (\alpha -n+2)(a^2+b^2)}}\hfill \\ \hfill & \times {\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}=0.\hfill \end{array}$$

In light of the aforementioned, $K_{\mathfrak{P}}y\in \mathrm{dom}\mathfrak{L}\bigcap \mathrm{Ker}\mathfrak{P}$.

If $u\in \mathrm{dom}\mathfrak{L}\bigcap \mathrm{Ker}\mathfrak{P}$, considering

$$\begin{array}{cc}\hfill & {({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_i(0)\hfill \\ \hfill =& (\omega (x){\mathfrak{D}}_{0^{+}}^{i,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)\hfill \\ \hfill =& (\omega (x){\mathfrak{D}}_{0^{+}}^{i,\psi ,\omega }{\mathfrak{I}}_{0^{+}}^{i-(\alpha -n+i),\psi ,\omega }u)(0)\hfill \\ \hfill =& (\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+i,\psi ,\omega }u)(0)=0,i=2,3,\dots ,n-1,\hfill \end{array}$$

by $u\in \mathrm{dom}\mathfrak{L},$ One can infer that

$$\begin{array}{cc}\hfill & K_{\mathfrak{P}}\mathfrak{L}u(x)=K_{\mathfrak{P}}({\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }u)(x)\hfill \\ \hfill =& -{\displaystyle \frac{b{(\psi (x)-\psi (0))}^{\alpha -n+1}+a{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)(a^2+b^2)}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }u\right)+{\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{\mathfrak{D}}_{0^{+}}^{\alpha ,\psi ,\omega }u(x)\hfill \\ \hfill =& -{\displaystyle \frac{b{(\psi (x)-\psi (0))}^{\alpha -n+1}+a{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)(a^2+b^2)}}{\mathfrak{B}}_2(u(x)-{\omega }^{-1}(x)\sum _{k=1}^n{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -k}}{\Gamma (\alpha -k+1)}}\hfill \\ \hfill & \times {({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_{n-k}(0))+u(x)-{\omega }^{-1}(x)\sum _{k=1}^n{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -k}}{\Gamma (\alpha -k+1)}}{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_{n-k}(0)\hfill \\ \hfill =& {\displaystyle \frac{b{(\psi (x)-\psi (0))}^{\alpha -n+1}+a{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)(a^2+b^2)}}{\mathfrak{B}}_2({\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)\Gamma (\alpha -n+1)}}(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)\hfill \\ \hfill & +{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)\Gamma (\alpha -n+2)}}{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0))+u(x)-{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)\Gamma (\alpha -n+1)}}(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)\hfill \\ \hfill & -{\displaystyle \frac{{(\psi (x)-\psi (0))}^{\alpha -n+1}}{\omega (x)\Gamma (\alpha -n+2)}}{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)\hfill \\ \hfill =& u(x)-{\displaystyle \frac{a{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)-b(\alpha -n+1)(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+2)(a^2+b^2)}}\hfill \\ \hfill & \times {\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n})}{\omega (x)}}=u(x)-\mathfrak{P}u(x)=u(x).\hfill \end{array}$$

Thus,

$$K_{\mathfrak{P}}={(\mathfrak{L}{|}_{\mathrm{dom}\mathfrak{L}\bigcap \mathrm{Ker}\mathfrak{P}})}^{-1}.$$

□

The norm estimations required for the following findings are given in the following lemma.

Lemma 7.

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

$${∥K_{\mathfrak{P}}y∥}_{C_{n-\alpha ,\psi ,\omega }}\le B·{\parallel y\parallel }_{L^{1,\psi ,\omega }},$$

(3)

$${∥{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }\left(K_{\mathfrak{P}}y\right)∥}_{C_{\omega }}\le C·{\parallel y\parallel }_{L^{1,\psi ,\omega }},$$

(4)

and

$$\parallel K_{\mathfrak{P}}{y\parallel }_X\le ∥K_{\mathfrak{P}}∥·{\parallel y\parallel }_{L^{1,\psi ,\omega }},$$

(5)

$$∥K_{\mathfrak{P}}∥=B+C.$$

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

Proof. 

To compute the norm of operator $K_{\mathfrak{P}}y$, we need to first establish the following estimates. For $y\in Y$, we can deduce

$$\begin{array}{cc}{|(\psi (x)-\psi (0))}^{n-\alpha }\omega (x){\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y(x)|\le \hfill & {\displaystyle \frac{{(\psi (1)-\psi (0))}^{n-1}}{\Gamma (\alpha )}}{\parallel y\parallel }_{L^{1,\psi ,\omega }},\hfill \\ and|\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y)(x)|\le \hfill & {\displaystyle \frac{{(\psi (1)-\psi (0))}^{n-2}}{\Gamma (n-1)}}{\parallel y\parallel }_{L^{1,\psi ,\omega }},\hfill \end{array}$$

which means that

$$\begin{array}{cc}\hfill \parallel {\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{y\parallel }_X& \le \left({\displaystyle \frac{{(\psi (1)-\psi (0))}^{n-1}}{\Gamma (\alpha )}}+{\displaystyle \frac{{(\psi (1)-\psi (0))}^{n-2}}{\Gamma (n-1)}}\right){\parallel y\parallel }_{L^{1,\psi ,\omega }}:=A·{\parallel y\parallel }_{L^{1,\psi ,\omega }}.\hfill \end{array}$$

By the definition of the norm on space X, we compute in two steps:

• first, we provide the following estimate, in light of Hypothesis 2 $(H_2)$, we can determine

$$\begin{array}{cc}\hfill & {|(\psi (x)-\psi (0))}^{n-\alpha }\omega (x)\left(K_{\mathfrak{P}}y\right)(x)|\hfill \\ \hfill =& |-{\displaystyle \frac{b(\psi (t)-\psi (0))+a}{a^2+b^2}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha \psi ,\omega }y\right)+{(\psi (x)-\psi (0))}^{n-\alpha }\omega (x){\mathfrak{I}}_{0^{+}}^{\alpha \psi ,\omega }y(x)|\hfill \\ \hfill \le & {\displaystyle \frac{\left|b\right|(\psi (1)-\psi (0))+\left|a\right|}{a^2+b^2}}{\beta }_2·\parallel {\mathfrak{I}}_{0^{+}}^{\alpha ;\psi }{y\parallel }_X+{\displaystyle \frac{{(\psi (1)-\psi (0))}^{\alpha -1}}{\Gamma (\alpha )}}{\parallel y\parallel }_{L^{1,\psi ,\omega }}\hfill \\ \hfill \le & \left({\displaystyle \frac{\left|b\right|(\psi (1)-\psi (0))+\left|a\right|}{a^2+b^2}}{\beta }_{2}A+{\displaystyle \frac{{(\psi (1)-\psi (0))}^{\alpha -1}}{\Gamma (\alpha )}}\right)·{\parallel y\parallel }_{L^{1,\psi ,\omega }}:=B·{\parallel y\parallel }_{L^{1,\psi ,\omega }},\hfill \end{array}$$

and second, we provide the following estimate

$$\begin{array}{cc}\hfill |\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }\left(K_{\mathfrak{P}}y\right)(x)|& =|{\displaystyle \frac{b\Gamma (\alpha -n+2)}{a^2+b^2}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }y\right)+\omega (x){\mathfrak{I}}_{0^{+}}^{n-1,\psi ,\omega }y(x)|\hfill \\ \hfill & \le \left({\displaystyle \frac{\left|b\right|\Gamma (\alpha -n+2)}{a^2+b^2}}{\beta }_{2}A+{\displaystyle \frac{{(\psi (1)-\psi (0))}^{n-2}}{\Gamma (n-1)}}\right)·{\parallel y\parallel }_{L^{1,\psi ,\omega }}\hfill \\ \hfill & :=C·{\parallel y\parallel }_{L^{1,\psi ,\omega }}.\hfill \end{array}$$

Thus, we can deduce that

$$\begin{array}{cc}\hfill {∥K_{\mathfrak{P}}y∥}_X=& {∥K_{\mathfrak{P}}y∥}_{C_{n-\alpha ,\psi ,\omega }}+{∥{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }\left(K_{\mathfrak{P}}y\right)∥}_{C_{\omega }}\hfill \\ \hfill \le & (B+C)·{\parallel y\parallel }_{L^{1,\psi ,\omega }}:=\parallel K_{\mathfrak{P}}{\parallel ·\parallel y\parallel }_{L^{1,\psi ,\omega }},\hfill \end{array}$$

which produces the intended outcome. □

Lemma 8.

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

Proof. 

Given that $\Omega$ is a bounded set and that the condition $(H)$ holds, there is a constant $r>0$ and a function $g_r\in Y$ such that ${\parallel u\parallel }_X\le r,$ and ${\displaystyle {\int }_0^1\left|\omega (x)\mathfrak{N}u(x)\right|{\psi }^{\prime }(x)dx\le {\parallel g_r\parallel }_{L^{1,\psi ,\omega }}}$ hold while $u\in \overline{\Omega }$. The function $h\in Y$ and the hypothesis $(H_2)$ hold when $\parallel {\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{y\parallel }_X\le A·{\parallel y\parallel }_{L^{1,\psi ,\omega }}$ is invoked. It is easy to see that

$$\begin{array}{cc}\hfill {\parallel \mathfrak{QN}u\parallel }_{L^{1,\psi ,\omega }}& {\displaystyle ={\int }_0^1\left|\omega (x)({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }Nu\right)h(x)\right|{\psi }^{\prime }(x)dx}\hfill \\ \hfill & \le |({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }Nu\right){|·\parallel h\parallel }_{L^{1,\psi ,\omega }}\hfill \\ \hfill & \le ({\beta }_1+\left|\varrho \right|{\beta }_2)\parallel {\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }{Nu\parallel }_X·{\parallel h\parallel }_{L^{1,\psi ,\omega }}\le A({\beta }_1+\left|\varrho \right|{\beta }_2)\parallel g_r{\parallel }_{L^{1,\psi ,\omega }}·{\parallel h\parallel }_{L^{1,\psi ,\omega }}.\hfill \end{array}$$

The boundedness of $\mathfrak{QN}(\overline{\Omega })$ follows.

We shall now demonstrate the compactness of $K_{\mathfrak{P}}(\mathfrak{I}-\mathfrak{Q})\mathfrak{N}:\overline{\Omega }\to X$. Below we will proceed in two steps:

Step 1: Hypothesis $(H_2)$ and $\parallel K_{\mathfrak{P}}{y\parallel }_X\le \parallel K_{\mathfrak{P}}{\parallel ·\parallel y\parallel }_{L^{1,\psi ,\omega }}$ have led to

$$\begin{array}{cc}\hfill \parallel K_{\mathfrak{P}}{(\mathfrak{I}-\mathfrak{Q})\mathfrak{N}u\parallel }_X& \le \parallel K_{\mathfrak{P}}{\parallel ·\parallel (\mathfrak{I}-\mathfrak{Q})\mathfrak{N}u\parallel }_X\le \parallel K_{\mathfrak{P}}{\parallel ·(\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }}+{\parallel \mathfrak{QN}u\parallel }_{L^{1,\psi ,\omega }})\hfill \\ \hfill & < \parallel K_{\mathfrak{P}}\parallel ·\left(1+A({\beta }_1+\left|\varrho \right|{\beta }_2{)\parallel h\parallel }_{L^{1,\psi ,\omega }}\right){\parallel g_r\parallel }_{L^{1,\psi ,\omega }}<\infty .\hfill \end{array}$$

i.e., $K_{\mathfrak{P}}(\mathfrak{I}-\mathfrak{Q})\mathfrak{N}:\overline{\Omega }\to X$ is unformed bounded.

Step 2: We will show $\{K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}u:u\in \overline{\Omega }\}$ is equicontinuous on $[0,1]$.

For each $u\in \overline{\Omega },0\le x_1<x_2\le 1,$ one can show

$$\begin{array}{cc}\hfill & |{(\psi (x_2)-\psi (0))}^{n-\alpha }\omega (x_2)K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}u(x_2)-{(\psi (x_1)-\psi (0))}^{n-\alpha }\omega (x_1)K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}u(x_1)|\hfill \\ \hfill =& |-{\displaystyle \frac{b(\psi (x_2)-\psi (0))+a}{a^2+b^2}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u\right)+{\displaystyle \frac{b(\psi (x_1)-\psi (0))+a}{a^2+b^2}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u\right)\hfill \\ \hfill +& {(\psi (x_2)-\psi (0))}^{n-\alpha }\psi (x_2){\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u(x_2)-{(\psi (x_1)-\psi (0))}^{n-\alpha }\psi (x_1){\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u(x_1)|.\hfill \end{array}$$

To begin with, one has

$$\begin{array}{cc}\hfill & |-{\displaystyle \frac{b(\psi (x_2)-\psi (0))+a}{a^2+b^2}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u\right)+{\displaystyle \frac{b(\psi (x_1)-\psi (0))+a}{a^2+b^2}}{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u\right)|\hfill \\ \hfill & =|-{\displaystyle \frac{b(\psi (x_2)-\psi (x_1))}{a^2+b^2}}\left|\right|{\mathfrak{B}}_2\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u\right)|\hfill \\ \hfill & \le {\displaystyle \frac{\left|b\right|(\psi (x_2)-\psi (x_1))}{a^2+b^2}}{\beta }_2{\parallel {\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u\parallel }_X\hfill \\ \hfill & \le {\displaystyle \frac{{\beta }_2\left|b\right|(\psi (x_2)-\psi (x_1))}{a^2+b^2}}A·\left(1+A({\beta }_1+\left|\varrho \right|{\beta }_2{)\parallel h\parallel }_{L^{1,\psi ,\omega }}\right){\parallel g_r\parallel }_{L^{1,\psi ,\omega }},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & |{(\psi (x_2)-\psi (0))}^{n-\alpha }\psi (x_2){\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u(x_2)-{(\psi (x_1)-\psi (0))}^{n-\alpha }\psi (x_1){\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u(x_2)|\hfill \\ \hfill =& {\displaystyle |{\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}{\int }_0^{x_2}{(\psi (x_2)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt}\hfill \\ \hfill & {\displaystyle -{\displaystyle \frac{{(\psi (x_1)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}{\int }_0^{x_1}{(\psi (x_1)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt|}\hfill \\ \hfill =& {\displaystyle |{\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}{\int }_0^{x_1}{(\psi (x_2)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt+{\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}}\hfill \\ \hfill & {\int }_{x_1}^{x_2}{(\psi (x_2)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt\hfill \\ \hfill +& {\displaystyle |{\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}{\int }_0^{x_1}{(\psi (x_1)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt}\hfill \\ \hfill -& {\displaystyle |{\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}{\int }_0^{x_1}{(\psi (x_1)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt}\hfill \\ \hfill -& {\displaystyle |{\displaystyle \frac{{(\psi (x_1)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}{\int }_0^{x_1}{(\psi (x_1)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt|}\hfill \\ \hfill \le & {\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}|{\int }_0^{x_1}[{(\psi (x_2)-\psi (t))}^{\alpha -1}-{(\psi (x_1)-\psi (t))}^{\alpha -1}]{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt|\hfill \\ \hfill & +{\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }-{(\psi (x_1)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}|{\int }_0^{x_1}{(\psi (x_1)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt|\hfill \\ \hfill & +{\displaystyle \frac{{(\psi (x_2)-\psi (0))}^{n-\alpha }}{\Gamma (\alpha )}}|{\int }_{x_1}^{x_2}{(\psi (x_2)-\psi (t))}^{\alpha -1}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt|,\hfill \end{array}$$

Both ${(\psi (x)-\psi (0))}^{n-\alpha }$ and ${(\psi (x)-\psi (0))}^{\alpha -1}$ are continuous on $x\in [0,1]$; thus, given the absolute continuity of the integral, as $x_1\to x_2$, we obtain

$$\begin{array}{c}\hfill |{(\psi (x_2)-\psi (0))}^{n-\alpha }\omega (x_2)K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}u(x_2)-{(\psi (x_1)-\psi (0))}^{n-\alpha }{K}_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}u(x_1)|\to 0.\end{array}$$

(6)

Next, using the definition of $K_P$, we can determine

$$\begin{array}{cc}\hfill & |\omega (x_2){\mathfrak{D}}_{0+}^{\alpha -n+1,\psi ,\omega }(K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}u)(x_2)-\omega (x_1){\mathfrak{D}}_{0+}^{\alpha -n+1,\psi ,\omega }(K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}u)(x_1)|\hfill \\ \hfill =& |{\mathfrak{I}}_{0^{+}}^{n-1,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u(x_2)-{\mathfrak{I}}_{0^{+}}^{n-1,\psi ,\omega }(I-\mathfrak{Q})\mathfrak{N}u(x_1)|\hfill \\ \hfill \le & {\displaystyle \frac{1}{\Gamma (n-1)}}{\int }_0^{x_2}({(\psi (x_2)-\psi (t))}^{n-2}){\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt\hfill \\ \hfill & -{\displaystyle \frac{1}{\Gamma (n-1)}}{\int }_0^{x_1}({(\psi (x_1)-\psi (t))}^{n-2}){\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt\hfill \\ \hfill =& {\displaystyle \frac{1}{\Gamma (n-1)}}{\int }_0^{x_1}[{(\psi (x_2)-\psi (t))}^{n-2}-{(\psi (x_1)-\psi (t))}^{n-2}]{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt\hfill \\ \hfill & +{\displaystyle \frac{1}{\Gamma (n-1)}}{\int }_{x_1}^{x_2}{(\psi (x_2)-\psi (t))}^{n-2}{\psi }^{\prime }(t)\omega (t)(I-\mathfrak{Q})\mathfrak{N}u(t)dt\hfill \end{array}$$

Similar to $x_1\to x_2$, we can hold that the above equation likewise goes to zero by using the continuity of ${(\psi (x)-\psi (t))}^{n-2}$ for $x\in [0,1]$ and the absolute continuity of the integral. The equicontinuity of $\{K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}:\overline{\Omega }\}$ on $[0,1]$ is thus verified.

Therefore, by the Arzéla-Ascoli theorem, it holds that the operator $K_{\mathfrak{P}}(I-\mathfrak{Q})\mathfrak{N}:\overline{\Omega }\to X$ is compact and $\mathfrak{QN}(\overline{\Omega })$ is bounded, i.e., $\mathfrak{N}$ is $\mathfrak{L}$-compact. □

Lemma 9.

Assume $a\ne 0$ and the following conditions hold:

Hypothesis 4

($H_4$). There exists a constant $M>0$ such that if $|\omega (x)D_{0^{+}}^{\alpha -n+1,\psi ,\omega }u(x)|>M$, for all $x\in [0,1]$, then

$$({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }\mathfrak{N}u)\ne 0.$$

Hypothesis 5

($H_5$). There exist some positive functions $a_i(x),i=0,1,2,3$ with $a_0(x),$${\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}},{\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}$ and ${\displaystyle \frac{a_3(x)}{\omega (x)}}\in Y$ such that for all $(u,v,w)\in {\mathbb{R}}^3,x\in [0,1],$

$$\left|f(x,u,v,w)\right|\le a_0(x)+a_1(x)\left|u\right|+a_2(x)\left|v\right|+a_3(x)\left|w\right|,$$

provided that $(\parallel K_{\mathfrak{P}}\parallel +E)(\parallel {\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+{\parallel {\displaystyle \frac{a_3(x)}{\omega (x)}}\parallel }_{L^{1,\psi ,\omega }})<1,$ where $E:=C(1+{\displaystyle \frac{\psi (1)-\psi (0)}{\Gamma (\alpha -n+2)}}+{\displaystyle \frac{\left|b\right|}{\left|a\right|\Gamma (\alpha -n+2)}}).$ Then the set ${\Omega }_1=\{u\in \mathrm{dom}\mathfrak{L}\setminus \mathrm{Ker}\mathfrak{L}:\mathfrak{L}u=\mu \mathfrak{N}u$, for some $\mu \in (0,1)\}$ is bounded.

Proof. 

$\mathfrak{N}u\in \mathrm{Im}\mathfrak{L},$ if $u\in {\Omega }_1,$ thus we have $({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }\mathfrak{N}u)=0.$ This, together with $(H_4)$, implies that there exists $x_0\in [0,1]$ such that $|(\omega (x)D_{0^{+}}^{\alpha -n+1,\psi ,\omega }u)(x_0)|\le M$.

Write $u(x)=u_1(x)+u_2(x),$ where $u_1=(I-\mathfrak{P})u\in \mathrm{dom}\mathfrak{L}\cap \mathrm{Ker}\mathfrak{P}$, and $u_2=\mathfrak{P}u\in \mathrm{Im}\mathfrak{P}.$ Thus, $u_1=K_{\mathfrak{P}}\mathfrak{L}{u}_1=K_{\mathfrak{P}}\mathfrak{L}(I-\mathfrak{P})u=K_{\mathfrak{P}}\mathfrak{L}u=\mu K_{\mathfrak{P}}\mathfrak{N}u.$ In light of (5) in Lemma 7, one obtains

$$\begin{array}{c}\hfill \parallel u_1{\parallel }_X \le \parallel K_{\mathfrak{P}}{\parallel ·\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }}.\end{array}$$

(7)

Now, $u_2=u-u_1,$ so, ${\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_2={\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }u-{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_1$ and

$$\begin{array}{cc}& |(\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_2)(x_0)| \le (\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }u)(x_0)| +(\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_1)(x_0)|\\ & \le M+C·{\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }},\end{array}$$

(8)

where C from (4) of Lemma 7.

Recall that $u_2(x)=\mathfrak{P}u(x)=c(u){\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}$, where

$$c(u)={\displaystyle \frac{a{({\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)}_1(0)-b(\alpha -n+1)(\omega (x){\mathfrak{I}}_{0^{+}}^{n-\alpha ,\psi ,\omega }u)(0)}{\Gamma (\alpha -n+2)(a^2+b^2)}}$$

is used for the sake of brevity. Hence, it follows from $\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_2(x)=c(u)a\Gamma (\alpha -n+2)$ that $\left|c(u)\right|\le {\displaystyle \frac{M+C·{\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }}}{\left|a\right|\Gamma (\alpha -n+2)}}$.

Additionally, the following inequalities can be easily deduced:

$$\begin{array}{cc}\hfill {|(\psi (x)-\psi (0))}^{n-\alpha }\omega (x)u_2(x)|& =|c(u)|·|a(\psi (x)-\psi (0))-b|\hfill \\ \hfill & \le |c(u)|·(|a|(\psi (1)-\psi (0))+|b|)\hfill \\ \hfill & ={(M+C·\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }})({\displaystyle \frac{\psi (1)-\psi (0)}{\Gamma (\alpha -n+2)}}+{\displaystyle \frac{\left|b\right|}{\left|a\right|\Gamma (\alpha -n+2)}}),\hfill \end{array}$$

which leads to

$$\begin{array}{cc}\hfill \parallel u_2{\parallel }_X =& \parallel u_2{\parallel }_{C_{n-\alpha ,\psi ,\omega }}+ {\parallel {\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_2\parallel }_{C_{\omega }}\hfill \\ \hfill \le & {(M+C·\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }})\left({\displaystyle \frac{\psi (1)-\psi (0)}{\Gamma (\alpha -n+2)}}+{\displaystyle \frac{\left|b\right|}{\left|a\right|\Gamma (\alpha -n+2)}}\right)+M+C·{\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }}\hfill \\ \hfill =& M\left(1+{\displaystyle \frac{\psi (1)-\psi (0)}{\Gamma (\alpha -n+2)}}+{\displaystyle \frac{\left|b\right|}{\left|a\right|\Gamma (\alpha -n+2)}}\right)+(1+{\displaystyle \frac{\psi (1)-\psi (0)}{\Gamma (\alpha -n+2)}}\hfill \\ \hfill & +{\displaystyle \frac{\left|b\right|}{\left|a\right|\Gamma (\alpha -n+2)}})C·{\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }}:=D+E·{\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }},\hfill \end{array}$$

where $D:=M\left(1+{\displaystyle \frac{\psi (1)-\psi (0)}{\Gamma (\alpha -n+2)}}+{\displaystyle \frac{\left|b\right|}{\left|a\right|\Gamma (\alpha -n+2)}}\right)$ and $E:=C·\left(1+{\displaystyle \frac{\psi (1)-\psi (0)}{\Gamma (\alpha -n+2)}}+{\displaystyle \frac{\left|b\right|}{\left|a\right|\Gamma (\alpha -n+2)}}\right),$ which together with (7) and $(H_5)$, yields

$$\begin{array}{cc}\hfill {\parallel u\parallel }_X\le & \parallel u_1{\parallel }_X+\parallel u_2{\parallel }_X\le (\parallel K_{\mathfrak{P}}{\parallel +E)\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }}+D\hfill \\ \hfill \le & D+(\parallel K_{\mathfrak{P}}\parallel +E)(\parallel a_0{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}·{\parallel u\parallel }_{C_{n-\alpha ,\psi ,\omega }}\hfill \\ \hfill & + \parallel {\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}·{\parallel u(\lambda x)\parallel }_{C_{n-\alpha ,\psi ,\omega }}+\parallel {\displaystyle \frac{a_3(x)}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}·{\parallel {\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }u\parallel }_{C_{\omega }})\hfill \\ \hfill \le & D+(\parallel K_{\mathfrak{P}}\parallel +E)[\parallel a_0{\parallel }_{L^{1,\psi ,\omega }}+({\parallel {\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}\parallel }_{L^{1,\psi ,\omega }}\hfill \\ \hfill & + \parallel {\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_3(x)}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}){\parallel u\parallel }_X]\hfill \\ \hfill \le & {\displaystyle \frac{D+(\parallel K_{\mathfrak{P}}\parallel +E)\parallel a_0{\parallel }_{L^{1,\psi ,\omega }}}{1-(\parallel K_{\mathfrak{P}}\parallel +E)\left(\parallel \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}{\parallel }_{L^{1,\psi ,\omega }}+\parallel \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}{\parallel }_{L^{1,\psi ,\omega }}+{\parallel \frac{a_3(x)}{\omega (x)}\parallel }_{L^{1,\psi ,\omega }}\right)}}.\hfill \end{array}$$

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

Given that $\mathfrak{L}$ is a Fredholem mapping of index zero, $J:\mathrm{Im}\mathfrak{Q}\to \mathrm{Ker}\mathfrak{L}$ is an isomorphism. The operator J for this is defined by

$$Jg(x)=({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }g)({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}).$$

Obviously, $J:Y\to \mathrm{Ker}\mathfrak{L}.$ If $g\in \mathrm{Im}\mathfrak{Q},$ then $g=ch$, where h is introduced in $(H_3)$, and

$$\begin{array}{cc}\hfill Jg(x)& =J(ch)(x)\hfill \\ \hfill & =({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }ch\right)({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})\hfill \\ \hfill & =c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}).\hfill \end{array}$$

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

Lemma 10.

Assume $(H_1)$,$(H_2)$ and the following hypothesis holds:

Hypothesis 6

($H_6$). There exists a constant $M_1>0$ such that either for each $c\in \mathbb{R}$:

$$\begin{array}{c}\hfill c({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }f(x,u_c(x),u_c(\lambda x),{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_c(x))\right)>0,for\left|c\right|>M_1,\end{array}$$

(9)

or

$$\begin{array}{c}\hfill c({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }f(x,u_c(x),u_c(\lambda x),{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_c(x))\right)<0,for\left|c\right|>M_1,\end{array}$$

(10)

where $u_c(x)=c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})$. Then sets ${\Omega }_2=\{u\in \mathrm{Ker}\mathfrak{L}:\mathfrak{N}u\in \mathrm{Im}\mathfrak{L}\}$.

Proof. 

It is simple to identify that $u_c(x)=c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}$ if $u_c\in {\Omega }_2$. as well as $\mathfrak{QN}{u}_c=0$. This suggests that $({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }\mathfrak{N}{u}_c\right)h(x)=0$, $\left|c\right|\le M_1.$

Let us now estimate the norm of $u_c$ in X.

$$\begin{array}{cc}\hfill {|(\psi (x)-\psi (0))}^{n-\alpha }\omega (x)u_c|& =\left|c(a(\psi (x)-\psi (0))-b)\right|\le M_1(|a|(\psi (1)-\psi (0))+\left|b\right|),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill |\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_c|& =|ca\Gamma (\alpha -n+2)|\le M_1\left|a\right|\Gamma (\alpha -n+2).\hfill \end{array}$$

So, $\parallel u_c{\parallel }_X\le M_1\left(\left|a\right|(\Gamma (\alpha -n+2)+\psi (1)-\psi (0))+\left|b\right|\right)<\infty ,$ that is, ${\Omega }_2$ is bounded. □

Lemma 11.

Assume $(H_1)$,$(H_2)$, and $(H_6)$ hold. Then ${\Omega }_3=\{u\in \mathrm{Ker}\mathfrak{L}:\rho \delta Iu+(1-\delta )J\mathfrak{QN}u=0,\delta \in [0,1]\}$ are bounded, where $J:Im\mathfrak{Q}\to Ker\mathfrak{L}$ is a homeomorphism with

$$J(g)=J(ch)=c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}),$$

$$\rho =\left\{\begin{array}{c}1,if(9)holds;\hfill \\ -1.if(10)holds.\hfill \end{array}\right.$$

(11)

Proof. 

For $u\in {\Omega }_3,u(x)=c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})$ and $\rho \delta Iu+(1-\delta )J\mathfrak{QN}u=0.$

$u\equiv 0$, if $\delta =1.$$J\mathfrak{QN}u=0\iff \mathfrak{QN}u=0,$ if $\delta =0$. This is because the proof of boundedness of ${\Omega }_2$ shows that ${\parallel u\parallel }_X<\infty .$

$\rho \delta c+(1-\delta )({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }f(x,u_c(x),u_c(\lambda x),{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_c(x))\right)=0$, if $\delta \in (0,1)$. After that, if $\left|c\right|>M_1,$ we get

$$\rho \delta c^2=-(1-\delta )c({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)\left({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }f(x,u_c(x),u_c(\lambda x),{\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_c(x))\right),$$

which contradicts itself. As a result, $\left|c\right|\le M_1$. Thus, ${\Omega }_3$’s boundedness follows

$$\begin{array}{cc}\hfill {\parallel u\parallel }_X\le & M_1{\parallel {\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}\parallel }_X\hfill \\ \hfill \le & M_1\left[\left|a\right|(\psi (1)-\psi (0)+|b|)+\left|a\right|\Gamma (\alpha -n+2)\right].\hfill \end{array}$$

□

Theorem 2.

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

Proof. 

Let $\Omega$ be open and bounded such that $\Omega \supset \overline{{\Omega }_1}\cup \overline{{\Omega }_2}\cup \overline{{\Omega }_3}\cup \{\varnothing \}$. We can derive $\mathfrak{L}u\ne \mu \mathfrak{N}u$ for $u\in \partial \Omega \cap (\mathrm{dom}\mathfrak{L}\setminus Ker\mathfrak{L}),\mu \in (0,1)$ and $\mathfrak{N}u\ne 0,u\in \mathrm{Ker}\mathfrak{L}\cap \partial \Omega$ given the boundedness of ${\Omega }_1$ and ${\Omega }_2$ in Lemmas 9 and 10. So, $(i)$ and $(ii)$ of Theorem 1 hold.

Let $H(u,\delta )=\rho \delta Iu+(1-\delta )J\mathfrak{QN}u,\delta \in [0,1],u\in \mathrm{Ker}\mathfrak{L}\cap \partial \Omega ,$ noticing the boundedness of ${\Omega }_3$ in Lemma 10 and ${\Omega }_3\subset \Omega$, we know $H(u,\delta )\ne 0,u\in \mathrm{Ker}\mathfrak{L}\cap \partial \Omega ,\delta \in [0,1]$.

For $u\in \mathrm{Ker}\mathfrak{L}\cap \partial \Omega$, we have $u(x)=c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})\ne 0,$ and $H(u,1)=\rho c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})\ne 0.$ For this aim, via Lemma 10, we know that $H(u,0)=J\mathfrak{QN}(c({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}))\ne 0.$

Thus, by invariance of degree under a homotopy, we get that

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

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

Example 1.

As an application of the previous result, we introduce an example. Considering the non-local boundary value problems of the following pantograph equation:

$$\left\{\begin{array}{c}{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x)={\displaystyle \frac{A_0}{x+1}}+{\displaystyle \frac{3A_0}{2}}xsinu(x)+{\displaystyle \frac{3A_0}{2}}xsinu(\lambda x)+A_0{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x),x\in (0,1],\lambda \in [0,1],\hfill \\ (\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x)){|}_{x=0}=0,\hfill \\ {\mathfrak{B}}_1(u)={\displaystyle \frac{\sqrt{6}-1}{2}}{\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(0)+{\displaystyle \frac{\sqrt{6\pi }}{2}}u(1)=0,\hfill \\ {\mathfrak{B}}_2(u)={\displaystyle \frac{\sqrt{6\pi }}{2}}(\sqrt{{\displaystyle \frac{3}{2}}x}(1+x)u){|}_{x=1}-{\displaystyle \frac{3}{2}}{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(0)=0,\hfill \end{array}\right.$$

where $A_0={\displaystyle \frac{9}{1000}}$. It is easy to see that $a_0(x)={\displaystyle \frac{A_0}{1+x}},a_1(x)=a_2(x)={\displaystyle \frac{3A_0}{2}}x,a_3(x)=A_0.$$B_1({\displaystyle \frac{{(\frac{3}{2}x)}^{\frac{1}{2}}}{1+x}})={\displaystyle \frac{3}{4}}\Gamma ({\displaystyle \frac{1}{2}}),B_2({\displaystyle \frac{{(\frac{3}{2}x)}^{\frac{1}{2}}}{1+x}})={\displaystyle \frac{3}{4}}\Gamma ({\displaystyle \frac{1}{2}}),B_1({\displaystyle \frac{{(\frac{3}{2}x)}^{-\frac{1}{2}}}{1+x}})={\displaystyle \frac{\sqrt{6}}{2}}\Gamma ({\displaystyle \frac{1}{2}}),B_2({\displaystyle \frac{{(\frac{3}{2}x)}^{-\frac{1}{2}}}{1+x}})={\displaystyle \frac{\sqrt{6}}{2}}\Gamma ({\displaystyle \frac{1}{2}}).$ The problem is at resonance and $a={\displaystyle \frac{\sqrt{6}}{2}}\Gamma ({\displaystyle \frac{1}{2}}),b={\displaystyle \frac{3}{4}}\Gamma ({\displaystyle \frac{1}{2}}),\varrho =1$. At this point, we can take $h(x)=-{\displaystyle \frac{64}{3\sqrt{\pi }}}{\displaystyle \frac{x}{x+1}}$ to ensure that $({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}h(x))=1$, i.e., $(H_3)$ holds.

Also,

$$\begin{array}{cc}\hfill |B_2(u)|& = |{\displaystyle \frac{\sqrt{6\pi }}{2}}(\sqrt{{\displaystyle \frac{3}{2}}x}(1+x)u){|}_{x=1}-{\displaystyle \frac{3}{2}}{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(0)|\hfill \\ \hfill & \le ({\displaystyle \frac{3}{2}}+{\displaystyle \frac{\sqrt{6\pi }}{2}})·{\parallel u\parallel }_X={\displaystyle \frac{3+\sqrt{6\pi }}{2}}{\parallel u\parallel }_X:={\beta }_2{\parallel u\parallel }_X.\hfill \end{array}$$

Then we can directly calculate

$$\begin{array}{cc}\hfill (\parallel K_{\mathfrak{P}}\parallel +E)& (\parallel {\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+{\parallel {\displaystyle \frac{a_3(x)}{\omega (x)}}\parallel }_{L^{1,\psi ,\omega }})\hfill \\ \hfill & ={\displaystyle \frac{3+2\sqrt{6}}{2}}{A}_0(\parallel K_{\mathfrak{P}}\parallel +E)\approx 0.8826<1.\hfill \end{array}$$

Hence, $(H_5)$ holds.

If $\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x)>8,$ then ${\psi }^{\prime }(x)\omega (x)f(x,u(x),u(\lambda x),{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x))>0,$ and

if $\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x)<-8$, then ${\psi }^{\prime }(x)\omega (x)f(x,u(x),u(\lambda x),{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x))<0$.

Hence,

$$\begin{array}{cc}\hfill & ({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}Nu)\hfill \\ \hfill & ={\displaystyle \frac{\sqrt{6}}{2}}\Gamma ({\displaystyle \frac{1}{2}})({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}Nu(1)-({({\displaystyle \frac{3}{2}}x)}^{{\displaystyle \frac{1}{2}}}(1+x){\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}Nu(x)){|}_{x=1})\hfill \\ \hfill & ={\displaystyle \frac{12-3\sqrt{6}}{8}}{\int }_0^1{(1-t)}^{{\displaystyle \frac{3}{2}}}{\psi }^{\prime }(t){\omega }^{\prime }(t)Nu(t)dt\ne 0\hfill \end{array}$$

provided $u\in \mathrm{dom}L\setminus \mathrm{Ker}L$ satisfies $|\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x)|>M=8.$ So, the condition $(H_4)$ holds.

Last, for $u\in \mathrm{Ker}L$, $u_c(t)=c({\displaystyle \frac{a{(\frac{3}{2}x)}^{\frac{1}{2}}-b{(\frac{3}{2}x)}^{-\frac{1}{2}}}{1+x}})$, one can choose $M_1=4>0$ such that $c({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}N{u}_c(x))>0$, which shows that $(H_6)$ is confirmed, since

$$\begin{array}{cc}\hfill c{\psi }^{\prime }(x)\omega (x)N{u}_c(x)& ={\displaystyle \frac{3c}{2}}{A}_0\left(1+{\displaystyle \frac{3}{2}}(x^2+x)sin{u}_c(x)+{\displaystyle \frac{3}{2}}(x^2+x)sin{u}_c(\lambda x)+(x+1){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}x,x+1}{u}_c(x)\right)\hfill \\ \hfill & \ge {\displaystyle \frac{3}{2}}{A}_0\left(-\left|c\right|-3\left|c\right|-3\left|c\right|+{\displaystyle \frac{\sqrt{6}\pi c^2}{4}}\right)\hfill \\ \hfill & ={\displaystyle \frac{3}{2}}{A}_0\left|c\right|\left({\displaystyle \frac{\sqrt{6}\pi \left|c\right|}{4}}-7\right)>0,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & c({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}N{u}_c(x))\hfill \\ \hfill & =c{\displaystyle \frac{\sqrt{6}}{2}}\Gamma ({\displaystyle \frac{1}{2}})({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}Nu(1)-({({\displaystyle \frac{3}{2}}x)}^{{\displaystyle \frac{1}{2}}}(1+x){\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},{\displaystyle \frac{3}{2}}x,x+1}Nu(x)){|}_{x=1})\hfill \\ \hfill & ={\displaystyle \frac{12-3\sqrt{6}}{8}}{\int }_0^1{(1-t)}^{{\displaystyle \frac{3}{2}}}c{\psi }^{\prime }(t)\omega (t)N{u}_c(t)dt>0,\left|c\right|>4.\hfill \end{array}$$

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

Example 2.

The following considers another boundary value problem, namely the Hadamard-type fractional-order pantograph equation, here, $\psi (x)=ln(x+1),\omega (x)={(x+1)}^{\mu }$.

$$\left\{\begin{array}{c}{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{5}{2}},ln(x+1),{(x+1)}^{\mu }}u(x)={\displaystyle \frac{B_0}{{(x+1)}^{\mu }}}+B_0{\displaystyle \frac{u(x)+1}{|u(x)|+1}}+B_{0}sinu(\lambda x)+B_0{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u(x),\hfill \\ x\in (0,1],\lambda \in [0,1],\mu \in [i,+\infty ),\hfill \\ (\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{2}}x,x+1}u(x)){|}_{x=0}=0,\hfill \\ {\mathfrak{B}}_1(u)=({\displaystyle \frac{1}{\sqrt{ln(x+1)}}}{\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u){|}_{x=1}=0,\hfill \\ {\mathfrak{B}}_2(u)={\displaystyle \frac{\sqrt{\pi }{2}^{-\mu }}{{(ln(2))}^{\frac{1}{2}}}}(\sqrt{ln(x+1)}{(x+1)}^{\mu }u){|}_{x=1}-2^{-\mu }{(ln(2))}^{{\displaystyle \frac{1}{2}}}{\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u(0)=0,\hfill \end{array}\right.$$

where $B_0={\displaystyle \frac{1}{18}}$. Clearly, $a_0(x)={\displaystyle \frac{B_0}{{(1+x)}^{\mu }}},a_1(x)=a_2(x)=a_3(x)=B_0,$$B_1({\displaystyle \frac{ln{(x+1)}^{\frac{1}{2}}}{{(x+1)}^{\mu }}})=2^{-\mu }\Gamma ({\displaystyle \frac{3}{2}}){(ln2)}^{{\displaystyle \frac{1}{2}}},B_2({\displaystyle \frac{ln{(x+1)}^{\frac{1}{2}}}{{(x+1)}^{\mu }}})=2^{-\mu }\Gamma ({\displaystyle \frac{3}{2}}){(ln2)}^{{\displaystyle \frac{1}{2}}},B_1({\displaystyle \frac{1}{{(x+1)}^{\mu }ln{(x+1)}^{\frac{1}{2}}}})=2^{-\mu }\Gamma ({\displaystyle \frac{1}{2}}){(ln2)}^{-{\displaystyle \frac{1}{2}}},$ and $B_2({\displaystyle \frac{1}{{(x+1)}^{\mu }ln{(x+1)}^{\frac{1}{2}}}})=2^{-\mu }\Gamma ({\displaystyle \frac{1}{2}}){(ln2)}^{-{\displaystyle \frac{1}{2}}}.$ The problem is at resonance, with $a=2^{-\mu }\Gamma ({\displaystyle \frac{1}{2}}){(ln2)}^{-{\displaystyle \frac{1}{2}}},b=2^{-\mu }\Gamma ({\displaystyle \frac{3}{2}}){(ln2)}^{{\displaystyle \frac{1}{2}}},\varrho =1$, and ${\beta }_2={\displaystyle \frac{\sqrt{\pi }+ln2}{2^{\mu }{(ln2)}^2}}$. Now, we may use $h(x)={\displaystyle \frac{2^{\mu }ln(x+1)}{(\frac{4}{3\pi }-1)ln2\sqrt{\pi }}}$ to make sure that $({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},ln(x+1),{(x+1)}^{\mu }}h(x))=1$, $i.e.,(H_3)$ holds.

After that, we can compute

$$\begin{array}{cc}\hfill (\parallel K_{\mathfrak{P}}\parallel +E)& (\parallel {\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+{\parallel {\displaystyle \frac{a_3(x)}{\omega (x)}}\parallel }_{L^{1,\psi ,\omega }})\hfill \\ \hfill & =B_0(ln2+2{(ln2)}^{{\displaystyle \frac{1}{2}}})(\parallel K_{\mathfrak{P}}\parallel +E)\approx 0.95298<1.\hfill \end{array}$$

Therefore, $(H_5)$ is true.

If $\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u(x)>2^{\mu +1}-1,$ then ${\psi }^{\prime }(x)\omega (x)f(x,u(x),u(\lambda x)$, ${\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u(x))>0,$ and if $\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u(x)<1-2^{\mu +1}$, then ${\psi }^{\prime }(x)\omega (x)f(x,u(x),u(\lambda x)$, ${\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u(x))<0$.

Therefore,

$$\begin{array}{cc}\hfill & ({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},ln(x+1),{(x+1)}^{\mu }}Nu)\hfill \\ \hfill & ={(ln2)}^{-{\displaystyle \frac{1}{2}}}{\mathfrak{I}}_{0^{+}}^{3,ln(x+1),{(x+1)}^{\mu }}Nu(1)-\sqrt{\pi }{\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},ln(x+1),{(x+1)}^{\mu }}Nu(1)\hfill \\ \hfill & >{\displaystyle \frac{1}{3·2^{\mu -2}}}{\int }_0^1[{(ln2-ln(1+t))}^2-{(ln2-ln(1+t))}^{{\displaystyle \frac{3}{2}}}]{\psi }^{\prime }(t)\omega (t)Nu(t)dt\ne 0\hfill \end{array}$$

provided $u\in \mathrm{dom}L\setminus \mathrm{Ker}L$ fulfills $|\omega (x){\mathfrak{D}}_{0^{+}}^{{\displaystyle \frac{1}{2}},ln(x+1),{(x+1)}^{\mu }}u(x)|>2^{\mu +1}-1.$ Thus, the condition $(H_4)$ is true.

Finally, for $u\in \mathrm{Ker}L$, $u_c(t)=c({\displaystyle \frac{a{(ln(x+1))}^{\frac{1}{2}}-b{(ln(x+1))}^{-\frac{1}{2}}}{{(1+x)}^{\mu }}})$, one can pick $M_1=2^{\mu +1}-1>0$, similar to the proof for condition $(H_4)$, it is easy to demonstrate $c({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{{\displaystyle \frac{5}{2}},ln(x+1),{(x+1)}^{\mu }}N{u}_c(x))>0$, indicating that $(H_6)$ is confirmed. Theorem 2 implies that X must include at least one solution.

Theorem 3.

Assume that $(H),$$(H_1-H_3)$ with ${\displaystyle \frac{d}{c}}\notin [0,\psi (1)-\psi (0)]$, and $(H_6)(of$ Lemma 10) and the following hypothesis hold:

Hypothesis 7

($H_7$). There exists a constant $M_2>0$ such that if $\left|\omega (x)u(t)\right|>M_2$, for all $x\in [0,1]$, then$({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }\mathfrak{N}u)\ne 0.$

Hypothesis 8

($H_8$). There exist some positive functions $a_i(x),i=0,1,2,3$ with $a_0(x),$${\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}},{\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}$ and ${\displaystyle \frac{a_3(x)}{\omega (x)}}\in Y$ such that for all $(u,v,w)\in {\mathbb{R}}^3,x\in [0,1],$

$$\left|f(x,u,v,w)\right|\le a_0(x)+a_1(x)\left|u\right|+a_2(x)\left|v\right|+a_3(x)\left|w\right|,$$

provided that $(\parallel K_P\parallel +E^{\prime })(\parallel {\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_3(x)}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }})<1,$ where $E^{\prime }:={\displaystyle \frac{B(|a|\left(\Gamma (\alpha -n+2)+\psi (1)-\psi (0)\right)+|b|)}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}.$ Then the functional boundary value problem (1) has at least one solution in X.

Proof. 

As in the proof of Lemma 9, $u\in {\Omega }_1$ means $({\mathfrak{B}}_1-\varrho {\mathfrak{B}}_2)({\mathfrak{I}}_{0^{+}}^{\alpha ,\psi ,\omega }\mathfrak{N}u)=0.$ By $(H_7)$, there exists a constant $x_0\in [0,1]$ such that $|\omega (x_0)u(x_0)|\le M_2$.

Remark 3.

Note that $|\omega (x_0)D_{0^{+}}^{\alpha -n+1,\psi ,\omega }u(x_0)|\le M_2$, which follows immediately from $(H_4)$, is not readily available to us.

Similarly, $u(x)=u_1(x)+u_2(x),$ where $u_1=(I-\mathfrak{P})u\in \mathrm{dom}\mathfrak{L}\cap \mathrm{Ker}\mathfrak{P}$, and $u_2=\mathfrak{P}u\in \mathrm{Im}\mathfrak{P}.$ Thus, $u_1=K_{\mathfrak{P}}\mathfrak{L}{u}_1=K_{\mathfrak{P}}\mathfrak{L}(I-\mathfrak{P})u=K_{\mathfrak{P}}\mathfrak{L}u=\mu K_{\mathfrak{P}}\mathfrak{N}u.$ As in the proof of Lemma 9,

$$\parallel u_1{\parallel }_X\le \parallel K_{\mathfrak{P}}{\parallel ·\parallel \mathfrak{N}u\parallel }_{L^{1,\psi ,\omega }}.$$

Now, $u_2=u-u_1,$ invoke (3) to show

$$\begin{array}{c}\hfill |\left({(\psi (x)-\psi (0))}^{n-\alpha }\omega (x)u_2\right)(x_0)|\le |\left({(\psi (x)-\psi (0))}^{n-\alpha }\omega (x)u\right)(x_0)|+|\left({(\psi (x)-\psi (0))}^{n-\alpha }\omega (x)u_1\right)(x_0)|\\ \hfill \le M_2·{(\psi (1)-\psi (0))}^{n-\alpha }+B·{\parallel Nu\parallel }_{L^{1,\psi ,\omega }}.\end{array}$$

(12)

Similar to Lemma 9’s data, $u_2(x)=\mathfrak{P}u(x)=c(u)({\displaystyle \frac{a{(\psi (x)-\psi (0))}^{\alpha -n+1}-b{(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}})$. By (12), one obtains

$$\begin{array}{cc}\hfill |c(u)|& \le {\displaystyle \frac{M_2·{(\psi (1)-\psi (0))}^{n-\alpha }+B·{\parallel Nu\parallel }_{L^{1,\psi ,\omega }}}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}\hfill \end{array}$$

by ${\displaystyle \frac{d}{c}}\notin [0,\psi (1)-\psi (0)]$. Then,

$$\begin{array}{cc}\hfill & {|(\psi (x)-\psi (0))}^{n-\alpha }\omega (x)u_2(x)|\hfill \\ \hfill & =|c(u)|·|a(\psi (x)-\psi (0))-b|\le |c(u)|·(|a|(\psi (1)-\psi (0))+|b|)\hfill \\ \hfill & \le {\displaystyle \frac{M_2·{(\psi (1)-\psi (0))}^{n-\alpha }+B·{\parallel Nu\parallel }_{L^{1,\psi ,\omega }}}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}(|a|(\psi (1)-\psi (0))+\left|b\right|),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & |\omega (x){\mathfrak{D}}_{0^{+}}^{\alpha -n+1,\psi ,\omega }{u}_2(x)|=|c(u)·a\Gamma (\alpha -n+2)|\hfill \\ \hfill & \le {\displaystyle \frac{M_2·{(\psi (1)-\psi (0))}^{n-\alpha }+B·{\parallel Nu\parallel }_{L^{1,\psi ,\omega }}}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}(|a|\Gamma (\alpha -n+2)),\hfill \end{array}$$

which allows us to acquire

$$\begin{array}{cc}\hfill \parallel u_2{\parallel }_X\le & {\displaystyle \frac{(|a|\left(\Gamma (\alpha -n+2)+\psi (1)-\psi (0)\right)+\left|b\right|)(M_2·{(\psi (1)-\psi (0))}^{n-\alpha }+B·{\parallel Nu\parallel }_{L^{1,\psi ,\omega }})}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}\hfill \\ \hfill \le & {\displaystyle \frac{M_2·{(\psi (1)-\psi (0))}^{n-\alpha }(|a|\left(\Gamma (\alpha -n+2)+\psi (1)-\psi (0)\right)+\left|b\right|)}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}\hfill \\ \hfill & +{\displaystyle \frac{(|a|\left(\Gamma (\alpha -n+2)+\psi (1)-\psi (0)\right)+\left|b\right|)B·{\parallel Nu\parallel }_{L^{1,\psi ,\omega }}}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}\hfill \\ \hfill & :=D^{\prime }+E^{\prime }·{\parallel Nu\parallel }_{L^{1,\psi ,\omega }},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill D^{\prime }:=& {\displaystyle \frac{M_2·{(\psi (1)-\psi (0))}^{n-\alpha }(|a|\left(\Gamma (\alpha -n+2)+\psi (1)-\psi (0)\right)+\left|b\right|)}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}\hfill \end{array}$$

and

$$\begin{array}{c}\hfill E^{\prime }:={\displaystyle \frac{B(|a|\left(\Gamma (\alpha -n+2)+\psi (1)-\psi (0)\right)+|b|)}{\underset{x\in [0,1]}{min}\{|a(\psi (x)-\psi (0))-b|\}}}.\end{array}$$

Additionally, it maintains that

$$\begin{array}{cc}\hfill {\parallel u\parallel }_X\le & \parallel u_1{\parallel }_X+\parallel u_2{\parallel }_X\le (\parallel K_{\mathfrak{P}}\parallel +E^{\prime }{)\parallel Nu\parallel }_{L^{1,\psi ,\omega }}+D^{\prime }\hfill \\ \hfill \le & D^{\prime }+(\parallel K_{\mathfrak{P}}\parallel +E^{\prime })[\parallel a_0{\parallel }_{L^{1,\psi ,\omega }}+({\parallel {\displaystyle \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}\parallel }_{L^{1,\psi ,\omega }}\hfill \\ \hfill & +\parallel {\displaystyle \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}+\parallel {\displaystyle \frac{a_3(x)}{\omega (x)}}{\parallel }_{L^{1,\psi ,\omega }}){\parallel u\parallel }_X]\hfill \\ \hfill \le & {\displaystyle \frac{D^{\prime }+(\parallel K_{\mathfrak{P}}\parallel +E^{\prime })\parallel a_0{\parallel }_{L^{1,\psi ,\omega }}}{1-(\parallel K_P\parallel +E^{\prime })(\parallel \frac{a_1(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}{\parallel }_{L^{1,\psi ,\omega }}+\parallel \frac{a_2(x){(\psi (x)-\psi (0))}^{\alpha -n}}{\omega (x)}{\parallel }_{L^{1,\psi ,\omega }}+\parallel \frac{a_3(x)}{\omega (x)}{\parallel }_{L^{1,\psi ,\omega }})}}.\hfill \end{array}$$

By $(H_8)$, ${\Omega }_1$ is bounded. The steps in the remaining portion of the proof are identical to those in Theorem 2. □

4. Conclusions

Examining a class of weighted fractional operators with certain choices of $\omega$ and $\psi$ equations under functional boundary value conditions at resonance was the main goal of this paper. It is worth noting that these operators cover many well-known fractional operators such as the Reimann–Liouville fractional integrals, the Reimann–Liouville fractional derivatives and the Caputo fractional derivative (when $\omega =1$ and $\psi (x)=x$), as well as other types of fractional operators such as the Hadamard fractional operators, the Erdelyi–Kober fractional operators, and the Katugampola fractional operators, which are special weighted fractional operators with different $\omega$ and $\psi$ functions.

First, the functional boundary conditions we study can generalize many homogeneous boundary value conditions, such as two-point, multi-point, integral, and integral–differential boundary conditions. Second, the nonlinear terms we study include proportional delay terms (applied in settings like electric trains and electric cells) and weighted fractional-order operators (applied in scenarios where the properties of unknown functions are required), which undoubtedly increase the application prospects of this equation. As a result, we present diverse existence solutions for two different needs of the unknown function using Mawhin’s coincidence theory. Finally, we can further extend the established weighted fractional-order derivative theoretical framework to broader fractional-order differential equation systems (such as coupled systems and delayed systems) by analyzing the existence and uniqueness of the solutions. We also explore combining Mawhin’s coincidence degree theory with other tools in nonlinear analysis (such as critical point theory and bifurcation theory) to investigate solution multiplicity, stability, and the bifurcation phenomena.