Generalized Fixed-Point Principles for Integral Equations and Their Quadrature Approximations

1. Introduction

Integral equations arise naturally in many branches of mathematics, physics, engineering, astronomy, biology, economics, and applied sciences. Numerous boundary value problems, differential equations, radiative transfer models, population dynamics models, and transport phenomena can be reformulated as integral equations. Consequently, the development of effective theoretical and numerical techniques for establishing the existence, uniqueness, and approximation of their solutions has remained an active area of research.

One of the most powerful tools in the study of nonlinear integral equations is fixed-point theory. In particular, the Banach contraction principle and its numerous extensions provide a unified framework for proving the existence and uniqueness of solutions for a wide range of functional, differential, and integral equations. The fundamental idea is to reformulate the original problem as a fixed-point equation $Tx=x,$ and then establish suitable contractive properties of the associated operator T. Classical results typically require the operator to satisfy a global contraction condition on a complete metric space. Although highly successful, these assumptions are often restrictive in applications, especially for nonlinear integral operators that fail to be contractions on the entire space.

To overcome these limitations, various generalizations of contraction mappings have been developed. Such extensions have considerably enlarged the class of operators for which fixed-point methods can be applied.

Motivated by these observations, this paper develops a fixed-point framework based on a class of generalized comparison functions ${\Psi }_0\left(\epsilon \right)$, where $\epsilon \in (0,+\infty ]$. This class extends the classical notion of comparison functions by requiring only the convergence of iterates ${\psi }^n\left(t\right)\to 0,t<\epsilon ,$ rather than on the whole positive half-line. This localization leads naturally to the notions of d-$\psi$-$\epsilon$-contractions and weak d-$\psi$-$\epsilon$-contractions, which generalize several classical fixed-point principles based on comparison functions.

A distinctive feature of the proposed approach is the separation between the metric governing the contraction and the metric governing completeness. More precisely, the contractive condition is formulated with respect to a metric d, whereas convergence is established in a complete metric space $(X^{*},d^{*})$. This separation enables the treatment of situations in which the natural contractive metric is not complete or is not suitable for directly proving convergence. As a consequence, the resulting fixed-point theorem provides a flexible framework that is particularly well adapted to nonlinear integral equations.

The abstract theory is subsequently applied to nonlinear Fredholm-type integral equations. By constructing suitable invariant subsets and introducing a $\psi$-$\epsilon$-max inequality, existence, uniqueness, localization, and convergence results are obtained under assumptions that differ substantially from classical Lipschitz-type conditions. The developed framework is further illustrated through Chandrasekhar-type equations and several nonlinear weighted integral equations.

Besides the theoretical analysis, numerical approximation of the obtained solutions is also investigated. The integral operators are replaced by suitable quadrature formulas, including Newton–Cotes, Gauss–Legendre, and Gauss–Jacobi rules, leading to quadrature integral equations. Combining these discretizations with Picard iteration produces numerical schemes whose convergence is supported by the underlying fixed-point theory. In this way, the paper establishes a direct connection between analytical existence theory and practical computational methods.

The principal contributions of this paper can be summarized as follows:

1.

Introduction of the generalized class ${\Psi }_0\left(\epsilon \right)$ together with the notions of d-$\psi$-$\epsilon$-contractions and weak d-$\psi$-$\epsilon$-contractions.

2.

Establishment of a fixed-point theorem in which the contraction metric and the completeness metric are allowed to be different.

3.

Development of existence, uniqueness, localization, and convergence results for nonlinear integral equations under a $\psi$-$\epsilon$-max inequality.

4.

Extension of the theory to quadrature integral equations generated by numerical integration rules.

5.

Construction and analysis of Picard-type numerical schemes based on Newton–Cotes and Gaussian quadrature formulas.

6.

Applications to Chandrasekhar-type integral equations and several nonlinear weighted integral equations.

The remainder of the paper is organized as follows. Section 2 contains the necessary preliminaries and auxiliary results which yields generalized fixed-point principles. Section 3 and Section 4 develops the abstract fixed-point framework and establishes the main existence and convergence theorem for nonlinear integral equations and quadrature integral equations, respectively. Section 4 introduces numerical schemes based on Picard iteration and quadrature rules and establishes convergence result and local error estimation of numerical scheme for general quadrature integral equations. Numerical experiments and applications are presented in Section 5. Finally, conclusions and directions for future research are discussed in the last section.

2. A new and significant fixed point theorem

This section presents a new and significant theorem that allows us to resolve integral equations. To prove it, some preliminaries are needed.

Definition 1. 

If d and $d^{*}$ are two metrics on a space X, we say that $d^{*}$ is Cauchy weaker than d, denoted $d^{*}⪯d$, if every d-Cauchy sequence in $(X,d)$ is also a $d^{*}$-Cauchy sequence in $(X,d^{*})$. For example, if there exist constants $\mathcal{M}>0$ and $\alpha >0$ such that $d^{*}(x,y)\le \mathcal{M}{\left(d(x,y)\right)}^{\alpha }$ for all $x,y\in X$, then $d^{*}⪯d$. In the case $\alpha =1$, we say that $d^{*}$ is equivalent from above to d.

Example 1. 

Suppose $0\ne f\in C[a,b]$ is fixed. Let us introduce a metric on $C[a,b]$ by

$$\begin{array}{c}\hfill d_f(x,y)=\left\{\begin{array}{cc}max\{\parallel x-f{\parallel }_u,\parallel y-f{\parallel }_u\},\hfill & x\ne y\in C[a,b],\hfill \\ 0,\hfill & \mathit{otherwise},\hfill \end{array}\right.\end{array}$$

(1)

which enables us to find the solution of the integral equations in the subsequent sections.

A sequence $x_n$ in the metric space $(C[a,b],d_f)$ is Cauchy if

$$\forall \epsilon >0,\exists N\in \mathbb{N}\mathrm{such}\mathrm{that}{d}_f(x_n,x_m)=max\{\parallel x_n{-f\parallel }_u,\parallel x_m-f{\parallel }_u\}<\epsilon \mathrm{for}\mathrm{all}m,n\ge N.$$

Hence, a Cauchy sequence must eventually lie arbitrarily close to f. In particular, this implies $x_n\stackrel{d_f}{\to }f$. Thus, $(C[a,b],d_f)$ is a complete metric space, and every Cauchy sequence in the topology ${\tau }_{d_f}$ converges to f.

The following definition extends several previously introduced notions, see [1,13,18,22].

Definition 2. 

Let $\epsilon \in (0,+\infty ]$. Denote by ${\Psi }_0\left(\epsilon \right)$ the family of all non-decreasing functions $\psi :[0,\infty )\to [0,\infty )$ satisfying $\psi \left(t\right)=0$ if and only if $t=0$, and

$$\underset{n\to \infty }{lim}{\psi }^n\left(t\right)=0\mathrm{for}\mathrm{every}t<\epsilon ,$$

where ${\psi }^n$ denotes the n-fold composition of ψ. If $\epsilon =+\infty$, then every $\psi \in {\Psi }_0(+\infty )$ is called a comparison function (see [1]).

Now let $(X,d)$ be a metric space and let $T:X\to X$ be a self-map.

• If there exist $\epsilon >0$ and a function $\psi \in {\Psi }_0\left(\epsilon \right)$ such that

  $$\begin{array}{c}\hfill d(Tu,Tv)\le \psi \left(d(u,v)\right)\mathrm{for}\mathrm{all}u,v\in X\mathrm{with}d(u,v)<\epsilon ,\end{array}$$

  (2)
  then T is called a d-ψ-ε-contraction.
• If there exist $\epsilon >0$ and a function $\psi \in {\Psi }_0\left(\epsilon \right)$ such that

  $$\begin{array}{c}\hfill d(Tu,T^{2}u)\le \psi \left(d(u,Tu)\right)\mathrm{for}\mathrm{all}u\in X\mathrm{with}d(u,Tu)<\epsilon ,\end{array}$$

  (3)
  then T is called a weak d-ψ-ε-contraction.

A point $x_0\in X$ is called a d-ε-initial value for T if

$$d(x_0,T{x}_0)<\epsilon .$$

In the case $\epsilon =+\infty$, the restrictions $d(u,v)<\epsilon$ and $d(u,Tu)<\epsilon$ are unnecessary in (2) and (3), respectively, and every point $x_0\in X$ is a d-ε-initial value.

Clearly, every d-ψ-ε-contraction is also a weak d-ψ-ε-contraction.

The following lemma shows that every d-$\psi$-$\epsilon$-contraction T on X is continuous.

Lemma 1 

(see also [3]).

1.

If $\psi \in {\Psi }_0\left(\epsilon \right)$ for some $0<\epsilon \le +\infty$, then $\psi \left(t\right)<t$ for all $0<t<\epsilon$.

Assume that $(X,d)$ is a metric space.

2.

Every d-ψ-ε-contraction self-map T on X is continuous in topology ${\tau }_d$.

3.

Let T be a continuous self-map on X. Then the limit of every convergent Picard sequence in X is a fixed point of T.

Proof. 

1.

Assume that there exists $0<t<\epsilon$ such that $\psi \left(t\right)\ge t$. Since $\psi$ is nondecreasing, an induction argument yields ${\psi }^n\left(t\right)\ge t>0,\forall n\in \mathbb{N},$ which contradicts ${lim}_{n\to \infty }{\psi }^n\left(t\right)=0.$ Therefore, $\psi \left(t\right)<t$ for all $0<t<\epsilon$.

2.

Suppose that $d(x_n,x)\to 0$. Then there exists $N\in \mathbb{N}$ such that $d(x_n,x)<\epsilon ,\forall n\ge N.$ Hence, $d(T{x}_n,Tx)\le \psi \left(d(x_n,x)\right),\forall n\ge N.$ Since $\psi \left(d(x_n,x)\right)\to 0$, we conclude that $d(T{x}_n,Tx)\to 0.$ Thus, T is continuous.

3.

Let $x_n$ be a convergent Picard sequence such that $x_n\to x^{*}$. Since T is continuous, $T{x}_n\to T{x}^{*}.$ On the other hand, since $x_{n+1}=T{x}_n$ and $x_{n+1}\to x^{*}$, it follows that $T{x}^{*}=x^{*}.$ Therefore, $x^{*}$ is a fixed point of T.

□

Some elementary properties related to Definition 2 are given below:

Lemma 2. 

1.

If $0<\epsilon <{\epsilon }^{\prime }$, then ${\Psi }_0\left({\epsilon }^{\prime }\right)\subseteq {\Psi }_0\left(\epsilon \right)$. Moreover, $\psi \left(t\right)={\displaystyle \frac{1}{\epsilon }}{t}^2\in {\Psi }_0\left(\epsilon \right)\setminus {\Psi }_0\left({\epsilon }^{\prime }\right)$; hence ${\Psi }_0\left({\epsilon }^{\prime }\right)⊊{\Psi }_0\left(\epsilon \right)$.

2.

If $\psi \in {\Psi }_0\left(\epsilon \right)$ for some $\epsilon >0$, then $L\psi \in {\Psi }_0\left(\epsilon \right)$ for all $0<L\le 1$.

3.

For ${\psi }_{\lambda }\left(t\right)=\lambda t$ ($t\ge 0$) with $0<\lambda <1$, we have ${\psi }_{\lambda }\in {\Psi }_0\left(\epsilon \right)$ for every $\epsilon >0$. If $\lambda \ge 1$, then Lemma 1-1 implies that ${\psi }_{\lambda }\notin {\Psi }_0\left(\epsilon \right)$ for any $\epsilon >0$.

4.

If $\psi \left(t\right)=\lambda t^2$ ($t\ge 0$) with $\lambda >0$, then $\psi \in {\Psi }_0\left(\epsilon \right)$ for $\epsilon ={\lambda }^{-1}$. The given ε is the greatest value for which $\psi \in {\Psi }_0\left(\epsilon \right)$.

5.

If $\psi \left(t\right)=\lambda t^{\alpha }$ ($t\ge 0$) with $\lambda >0$ and $\alpha >1$, then $\psi \in {\Psi }_0\left(\epsilon \right)$ for $\epsilon ={\lambda }^{-{\displaystyle \frac{1}{\alpha -1}}}$. The given ε is the greatest value for which $\psi \in {\Psi }_0\left(\epsilon \right)$.

6.

Let $\lambda >0$ and $0<L\le 1$. If $\psi \in {\Psi }_0\left(\epsilon \right)$ for some $\epsilon >0$, then ${\psi }^{\prime }\left(t\right)=L\psi \left(\lambda t\right)$ ($t\ge 0$) belongs to ${\Psi }_0\left({\epsilon }^{\prime }\right)$ with ${\epsilon }^{\prime }={\displaystyle \frac{\epsilon }{\lambda }}$.

7.

Let $\lambda ,L>0$ and $\alpha >1$. For $\psi \left(t\right)={\left(\lambda t\right)}^{\alpha }$, define ${\psi }^{\prime }\left(t\right)=L\psi \left(\lambda t\right)={\left(L^{{\displaystyle \frac{1}{\alpha }}}\lambda t\right)}^{\alpha }$. Then ${\psi }^{\prime }\in {\Psi }_0\left(\epsilon \right)$ with $\epsilon ={\left(L^{-{\displaystyle \frac{1}{\alpha -1}}}\lambda \right)}^{-{\displaystyle \frac{\alpha }{\alpha -1}}}$. The given ε is the greatest value for which $\psi \in {\Psi }_0\left(\epsilon \right)$.

8.

If $\psi \left(t\right)={\left(\lambda t\right)}^{\alpha }$ with $\lambda >0$ and $\alpha >1$, then $\psi \in {\Psi }_0\left(\epsilon \right)$ for $\epsilon ={\lambda }^{-{\displaystyle \frac{\alpha }{\alpha -1}}}$. Moreover, if $(X,d)$ is a metric space such that $d(Tu,Tv)\le {\lambda }^{\alpha }{d}^{\alpha }(u,v),\forall u,v\in X,$ then T is a d-ψ-ε-contraction.

9.

If $\psi \left(t\right)=ln(t+1)$ ($t\ge 0$), then $\psi \in {\Psi }_0\left(\epsilon \right)$, for all $0<\epsilon \le +\infty$.

10.

If $\psi \left(t\right)=t-ln(t+1)$ ($t\ge 0$), then $\psi \in {\Psi }_0\left(\epsilon \right)$, for all $0<\epsilon \le +\infty$.

Proof. 

Items 1–3 are straightforward. We prove item 5. Items 4–8 are either immediate consequences or can be proved similarly.

The fixed points of $\psi$ are 0 and $\epsilon :={\lambda }^{-{\displaystyle \frac{1}{\alpha -1}}}$. Suppose that $0<t<\epsilon$. Then $\psi \left(\psi \left(t\right)\right)=\lambda {\left(\psi \left(t\right)\right)}^{\alpha }=\lambda {\left(\lambda t^{\alpha }\right)}^{\alpha }={\lambda }^{1+\alpha }{t}^{{\alpha }^2}.$ The inequality $\psi \left(\psi \right(t\left)\right)<\psi \left(t\right)$ is equivalent to ${\lambda }^{1+\alpha }{t}^{{\alpha }^2}<\lambda t^{\alpha },$ which holds if and only if $t<{\lambda }^{-{\displaystyle \frac{1}{\alpha -1}}}=\epsilon .$

By induction, ${\psi }^{n+1}\left(t\right)<{\psi }^n\left(t\right)$ for all $n\in \mathbb{N}$ and every $0<t<\epsilon$. Hence ${\psi }^n\left(t\right)$ is decreasing and bounded below by 0, and therefore converges to some limit L.

Since $\psi$ is continuous, Lemma 1-3 implies that L is a fixed point of $\psi$ on $[0,\infty )$. Therefore $L=0$ for all $0<t<\epsilon$, and consequently $\psi \in {\Psi }_0\left(\epsilon \right)$.

If $t_0\ge \epsilon$, then ${\psi }^n\left(t_0\right)\ge t_0$ for all $n\in \mathbb{N}$. Hence ${lim}_{n\to \infty }{\psi }^n\left(t_0\right)\ge t_0>0.$ This proves that $\epsilon$ is the greatest value such that $\psi \in {\Psi }_0\left(\epsilon \right)$.

Items 9 and 10:

Fix $t>0$ and define the iterates $t_0=t$ and $t_{n+1}=\psi \left(t_n\right)=ln(1+t_n)$. Since $0<\psi \left(t\right)<t$ for all $t>0$, and $\psi$ is continuous, the sequence $t_n$ is strictly decreasing and converges to the unique fixed point 0. Thus, $\psi \in {\Psi }_0(+\infty )$. Similarly, 10 holds.

□

In the case of Lemma 2-3, a mapping T satisfying (2) is called a contraction on $(X,d)$.

The following fundamental theorem provides the principal framework for the analysis of the integral equations considered in the subsequent sections. Its main novelty is the decoupling of the contraction structure, described by a metric d on a possibly non-complete space X, from the completeness structure, described by a metric $d^{*}$ on a complete subspace $X^{*}\subseteq X$, together with the extension of the theory to the class ${\Psi }_0\left(\epsilon \right)$, where $\epsilon \in (0,+\infty ]$.

Theorem 1. 

Let $(X,d)$ be a (not necessarily complete) metric space, and $\epsilon \in (0,+\infty ]$.

(I)

Assume that T is a weak d-ψ-ε-contraction on X, where $\psi \in {\Psi }_0\left(\epsilon \right)$. Then the Picard sequence $x_n=T^n\left(x_0\right)$ is asymptotically regular for every d-ε-initial value $x_0\in X$.

(II)

Assume that T is a d-ψ-ε-contraction on X, where $\psi \in {\Psi }_0\left(\epsilon \right)$. Then the Picard sequence $x_n=T^n\left(x_0\right)$ is a d-Cauchy sequence for every d-ε-initial value $x_0\in X$.

(III)

In addition to part (II), suppose that there exists a subset $X^{*}\subseteq X$ such that $(X^{*},d^{*})$ is a complete metric space, $T\left(X^{*}\right)\subseteq X^{*}$, $d^{*}⪯d$ on $X^{*}$, and T is continuous with respect to $d^{*}$. Then, for every d-ε-initial value $x_0\in X^{*}$, the Picard sequence $x_n=T^n\left(x_0\right)$ converges to a fixed point $u^{*}\in X^{*}$ of T with respect to the topology ${\tau }_{d^{*}}$. Moreover, the fixed point of T is unique in the set

$${\mathcal{B}}_{d,\epsilon }\left(u^{*}\right):=\{v\in X^{*}:d(u^{*},v)<\epsilon \}.$$

(IV)

In addition to part (III), suppose that $d^{*}$ is equivalent from above to d, that is, there exists a constant $\mathcal{M}>0$ such that $d^{*}\le \mathcal{M}d$, and that $\psi \in {\Psi }_0\left(\epsilon \right)$ satisfies the condition

$${\displaystyle \frac{\psi \left(t\right)}{t}}\mathrm{is}\mathrm{non}-\mathrm{decreasing}\mathrm{on}(0,\epsilon ).$$

(4)

Then the following a posteriori and a priori estimates hold, respectively, for every Picard sequence $x_n=T^n\left(x_0\right)$ generated by a d-ε-initial value $x_0\in X^{*}$ satisfying $x_0\ne T{x}_0$:

$$\begin{array}{cc}\hfill d^{*}(x_n,u^{*})& \le \mathcal{M}{\displaystyle \frac{d(x_0,T{x}_0)}{d(x_0,T{x}_0)-\psi \left(d(x_0,T{x}_0)\right)}}d(x_n,x_{n+1}),\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill d^{*}(x_n,u^{*})& \le \mathcal{M}{\displaystyle \frac{d(x_0,T{x}_0)}{d(x_0,T{x}_0)-\psi \left(d(x_0,T{x}_0)\right)}}{\psi }^n\left(d(x_0,T{x}_0)\right),\forall n\in \mathbb{N}.\hfill \end{array}$$

(6)

Proof. 

(I)

Consider the Picard sequence $x_n=T^n\left(x_0\right)$ for $x_0\in X$ with $d(x_0,T{x}_0)<\epsilon$. From (3) and the non-decreasing property of $\psi$ we obtain

$$\begin{array}{c}\hfill d(x_n,x_{n+1})=d(T^n{x}_0,T^{n+1}{x}_0)\le \psi \left(d(T^{n-1}{x}_0,T^n{x}_0)\right)\le \dots \le {\psi }^n\left(d(x_0,T{x}_0)\right)\end{array}$$

(7)

for all $n\in \mathbb{N}$. Hence

$$\underset{n\to \infty }{lim}d(x_n,x_{n+1})\le \underset{n\to \infty }{lim}{\psi }^n\left(d(x_0,T{x}_0)\right)=0,$$

because $\psi \in {\Psi }_0\left(\epsilon \right)$ and $d(x_0,T{x}_0)<\epsilon$. Thus $x_n$ is an asymptotically regular sequence.

(II)

Assume that T is a d-$\psi$-$\epsilon$-contraction for some $\psi \in {\Psi }_0\left(\epsilon \right)$ with $\epsilon >0$ (possibly $\epsilon =+\infty$). Let $x_0\in X$ be a d-$\epsilon$-initial value, i.e. $d(x_0,T{x}_0)<\epsilon$, and define $x_n=T^n{x}_0$.

From part (I) we know ${lim}_{n\to \infty }d(x_n,x_{n+1})=0$.

Fix an arbitrary $\eta >0$. Choose a number $\delta$ such that

$$0<\delta <min\{\epsilon ,\eta /2\}.$$

Since $\delta <\epsilon$, Lemma 1-1 gives $\psi \left(\delta \right)<\delta$; thus $\alpha :=\delta -\psi \left(\delta \right)>0$. Because $d(x_n,x_{n+1})\to 0$, there exists $n_0\in \mathbb{N}$ such that

$$d(x_{i+1},x_i)<\alpha \mathrm{for}\mathrm{all}i\ge n_0.$$

We claim that $d(x_{n_0+k},x_{n_0})<\delta$ for every $k\in \mathbb{N}$. For $k=1$ this follows directly from the choice of $n_0$. Assume it holds for some $k\ge 1$. Then, using the contraction property (note that $d(x_{n_0+k},x_{n_0})<\delta <\epsilon$),

$$\begin{array}{cc}\hfill d(x_{n_0+k+1},x_{n_0})& \le d(x_{n_0+k+1},x_{n_0+1})+d(x_{n_0+1},x_{n_0})\hfill \\ & \le \psi \left(d(x_{n_0+k},x_{n_0})\right)+d(x_{n_0+1},x_{n_0})\hfill \\ & <\psi \left(\delta \right)+\alpha =\delta .\hfill \end{array}$$

Thus the claim holds for all k by induction.

Consequently, for all $m,n\ge n_0$,

$$d(x_m,x_n)\le d(x_m,x_{n_0})+d(x_n,x_{n_0})<2\delta \le \eta .$$

Since $\eta >0$ was arbitrary, $x_n$ is a d-Cauchy sequence.

(III)

Since $d^{*}⪯d$, part (II) implies that the Picard sequence $x_n:=T^n\left(x_0\right)$ is also a $d^{*}$-Cauchy sequence for every d-$\epsilon$-initial value $x_0\in X^{*}$. Since $(X^{*},d^{*})$ is complete, there exists $u^{*}\in X^{*}$ such that

$$T^n\left(x_0\right)\stackrel{d^{*}}{\to }{u}^{*},\mathrm{i}.\mathrm{e}.,\underset{n\to \infty }{lim}{d}^{*}(T^n\left(x_0\right),u^{*})=0.$$

By the continuity of T, it follows that $u^{*}$ is a fixed point of T in $X^{*}$; see Lemma 1.

To prove uniqueness, suppose that $u^{*}$ and $v^{*}$ are fixed points of T satisfying $0<d(u^{*},v^{*})<\epsilon$. Then, by (2) and Lemma 1-1,

$$0<d(u^{*},v^{*})=d(T{u}^{*},T{v}^{*})\le \psi \left(d(u^{*},v^{*})\right)<d(u^{*},v^{*}),$$

which is impossible. Therefore, $u^{*}$ is the unique fixed point of T in the ball

$${\mathcal{B}}_{d,\epsilon }\left(u^{*}\right):=\{v\in X^{*}:d(u^{*},v)<\epsilon \}.$$

(IV)

Assume $0<d(x_n,x_{n+1})$ for all $n\in \mathbb{N}$. By part (I), $d(x_n,x_{n+1})\to 0$. Induction using (3) and Lemma 1-1 gives

$$d(x_n,x_{n+1})\le \psi \left(d(x_{n-1},x_n)\right)<d(x_{n-1},x_n)<\dots <d(x_0,T{x}_0)<\epsilon$$

for all n. From the monotonicity condition (4) on $\psi \left(t\right)/t$ we deduce

$${\displaystyle \frac{d(x_1,x_0)-\psi \left(d(x_1,x_0)\right)}{d(x_1,x_0)}}\le {\displaystyle \frac{d(x_n,x_{n+1})-\psi \left(d(x_n,x_{n+1})\right)}{d(x_n,x_{n+1})}}\le {\displaystyle \frac{d(x_n,x_{n+1})-d(x_{n+1},x_{n+2})}{d(x_n,x_{n+1})}}$$

for every $n\in \mathbb{N}$. Using the right-hand inequality and the fact that $d^{*}\le \mathcal{M}d$, we estimate for any $m\ge n$:

$$\begin{array}{cc}\hfill d^{*}(x_n,x_m)& \le {\displaystyle \sum _{i=n}^{m-1}}{d}^{*}(x_i,x_{i+1})\le \mathcal{M}{\displaystyle \sum _{i=n}^{m-1}}d(x_i,x_{i+1})\hfill \\ & \le \mathcal{M}{\displaystyle \frac{d(x_1,x_0)}{d(x_1,x_0)-\psi \left(d(x_1,x_0)\right)}}{\displaystyle \sum _{i=n}^{\infty }}\left[d(x_i,x_{i+1})-d(x_{i+1},x_{i+2})\right]\hfill \\ & =\mathcal{M}{\displaystyle \frac{d(x_1,x_0)}{d(x_1,x_0)-\psi \left(d(x_1,x_0)\right)}}\left(d(x_n,x_{n+1})-\underset{i\to \infty }{lim}d(x_{i+1},x_{i+2})\right)\hfill \\ & =\mathcal{M}{\displaystyle \frac{d(x_1,x_0)}{d(x_1,x_0)-\psi \left(d(x_1,x_0)\right)}}d(x_n,x_{n+1}).\hfill \end{array}$$

Letting $m\to \infty$ we obtain the a posteriori estimate (5).

If it is not true that $d(x_n,x_{n+1})>0$ for all n, the non-increasing nature of this sequence guarantees the existence of some N such that $d(x_n,x_{n+1})>0$ for $n\le N$ and $d(x_n,x_{n+1})=0$ for $n>N$. Consequently $x_n=u^{*}$ for all $n>N$, and the estimate (5) holds trivially for $n>N$; for $n\le N$ the same reasoning as above applies.

Finally, the a priori estimate (6) follows from the a posteriori estimate (5) together with inequality (7), valid for every d-$\epsilon$-initial value $x_0\in X^{*}$.

□

Remark 1. 

1.

The first proof of Theorem 1-(II) was given by Matkowski [18] in the case $\epsilon =+\infty$; see also [2]. The preceding proof generalizes this result and establishes the theorem for every $\epsilon \in (0,+\infty ]$.

2.

In the preceding theorem we need the following conditions such that all the conclusion of parts (I) to (IV) hold true:

(a)

The space $(X,d)$ is a (not necessarily complete) metric space and T be a self map on X.

(b)

The space $(X^{*},d^{*})$ is a complete metric space for a subset $X^{*}\subseteq X$.

(c)

T be invariant under $X^{*}$, continuous with the metric $d^{*}$, and T is a d-ψ-ε-contraction on X or $X^{*}$, where $\psi \in {\Psi }_0\left(\epsilon \right)$, for some $0<\epsilon$.

(d)

ψ satisfies (4) and $d^{*}\le \mathcal{M}d$, where $0<\mathcal{M}$.

In the above circumstances we say the pairs $(X^{*},d^{*})⪯(X,d)$ satisfies properties (P) of Theorem 1 for T. We will show that the properties (P) enable us to obtain the solutions of (non-linear) integral equations properly.

3.

In the preceding theorem it can be $X=X^{*}$. From Lemma 2-3, ${\psi }_{\lambda }$ belongs to ${\Psi }_0\left(\epsilon \right)$, for all $0<\epsilon$, where ${\psi }_{\lambda }\left(t\right)=\lambda t,0<\lambda <1$, when $d^{*}=d$ be a metric on X, Theorem 1 reduces to Banach contraction principle on the complete metric space $(X,d)$ with error estimations (5) and (6), where $\mathcal{M}=1$.

4.

The roles of d and $d^{*}⪯d$ are creation Cauchy sequences and convergence of Picard sequences (in a complete metric space $(X^{*},d^{*})$), respectively. In the next sections we see that separating roles of d and $d^{*}$ gives a great significant advantage over Banach contraction principle.

5.

The ψ in the items 3 to 10 of Lemma 2 except 9 are satisfied in condition (4).

3. Applications to the Existence of Solutions of Integral Equations

The next theorem is a straightforward application of Theorem 1 to the existence of solutions of nonlinear integral equations. In the sequel, we denote by $(C\left(I\right),\parallel ·{\parallel }_u)$ the space of continuous functions endowed with the uniform norm, and by ${\mathcal{B}}_R\left(x_0\right)=\left\{x\in C\left(I\right);\parallel x-x_0{\parallel }_u<R\right\},$ the open ball centered at $x_0\in C\left(I\right)$ with radius $R>0$. We also use the uniform metric $d_u(x,y)={\parallel x-y\parallel }_u,x,y\in X,$ where X is a closed subset of $(C\left(I\right),\parallel ·{\parallel }_u)$.

Theorem 2. 

Let $f\in \mathcal{C}\left(I\right)$ and $K\in C(I\times I\times \mathbb{R})$. Consider the integral equation

$$\begin{array}{c}\hfill x\left(t\right)=Tx\left(t\right):=f\left(t\right)+{\int }_a^{b}K(t,s,x\left(s\right))ds,t\in I.\end{array}$$

(8)

(I)

Assume

(i)

There exists $0<r_0\le R_0$ such that the following

$$\begin{array}{c}\hfill r_0\le {\parallel Tx-f\parallel }_u\le R_0⟹r_0\le {\parallel T^{2}x-f\parallel }_u\le R_0.\end{array}$$

(9)

hold, for all $x\in C\left(I\right)$.

(ii)

Put

$$\begin{array}{cc}\hfill X^{*}:& =\{x\in C(I);\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill & {\parallel x\parallel }_u\le w_0{\parallel x-f\parallel }_u,\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill & r_0\le {\parallel Tx-f\parallel }_u\le R_0\},\hfill \end{array}$$

(12)

in which $w_0$ is a constant greater than ${\displaystyle \frac{{\parallel f\parallel }_u+R_0}{r_0}}$, then $X^{*}$ is non-empty and complete subset of $(C\left(I\right),\parallel ·{\parallel }_u)$, and T is invariant on $X^{*}$, that is, $T\left(X^{*}\right)\subseteq X^{*}$.

(iii)

If K satisfies ψ-ε-max-inequality on $I\times I\times \mathbb{R}$, that is,

$$\underset{t\in I}{sup}{\int }_a^{b}max\left\{\right|K(t,s,u)|,|K(t,s,v)\left|\right\}ds\le max\left\{\psi \right(\left|u\right|),\psi (\left|v\right|\left)\right\},\forall u,v\in \mathbb{R},$$

(13)

where $\psi \in {\Psi }_0\left(\epsilon \right)$, for some $0<\epsilon \le +\infty$,

then the Picard sequence $T^n\left(x_0\right)$ is a $d_f$-Cauchy sequence in $X^{*}$, for all $d_f$-${\epsilon }_0$-initial value $x_0\in X^{*}$, where $d_f$ is given by (1).

(II)

If T is continuous with respect to the uniform metric then integral equation (8) has a solution $u^{*}\in X^{*}$ and it is unique in the set ${\mathcal{B}}_{d_f,{\epsilon }_0}\left(u^{*}\right):=\{v\in X^{*},d_f(u^{*},v)<{\epsilon }_0\}$. The Picard sequence $x_n=T^n\left(x_0\right)$ converge to $u^{*}\in X^{*}$ in the topology $d_u$, for all $d_f$-${\epsilon }_0$-initial value $x_0\in X^{*}$.

(III)

Moreover, suppose ψ satisfies condition (4) then the following posteriori and the priori estimates

$$\begin{array}{cc}\hfill \parallel x_n-u^{*}{\parallel }_u& \le 2{\displaystyle \frac{d_f(x_0,x_1)}{d_f(x_0,x_1)-\psi \left(d_f(x_0,x_1)\right)}}{d}_f(x_n,x_{n+1}),\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill & \le 2{\displaystyle \frac{d_f(x_0,x_1)}{d_f(x_0,x_1)-\psi \left(d_f(x_0,x_1)\right)}}{\psi }^n\left(d_f(x_0,x_1)\right),\mathrm{for}\mathrm{all}n\in \mathbb{N},\hfill \end{array}$$

(15)

hold true, for all $d_f$-${\epsilon }_0$-initial value $x_0\in X^{*}$, respectively, where $x_n=T^n\left(x_0\right)$ and $x_1\ne x_0$.

Proof. 

We shall show the pairs $(X^{*},d_u)⪯(X^{*},d_f)$ satisfies the properties (P) of Theorem 1 with the uniform metric $d_u$ and the metric $d_f$ given by (1).

(I)

Put $W:=\{x\in C(I),Tx\mathrm{satisfies}(12\left)\right\}$, clearly $W\ne \varnothing$. Let $x\in W$ then we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\parallel Tx\parallel }_u& =\underset{t\in I}{sup}\left|f\left(t\right)+{\int }_a^{b}K(t,s,x\left(s\right))ds\right|\hfill \\ \hfill & \le {\parallel f\parallel }_u+R_0\hfill \\ \hfill & \le {\displaystyle \frac{{\parallel f\parallel }_u+R_0}{r_0}}{\parallel Tx-f\parallel }_u\hfill \\ \hfill & =w_0{\parallel Tx-f\parallel }_u,\hfill \end{array}\end{array}$$

(16)

for all $x\in C\left(I\right)$. Thus, $Tx$ satisfies (11), which together with (9) implies $\varnothing \ne T\left(W\right)\subseteq X^{*}$, that is, $X^{*}$ is a non-empty set. Moreover, from (16) together with (9) we get $T\left(X^{*}\right)\subseteq X^{*}$. Since $X^{*}$ is a closed subset of $(C\left(I\right),\parallel ·{\parallel }_u)$, we get $(X^{*},d_u)$ is a complete metric space. The inequality

$$\begin{array}{c}\hfill d_u(x,y)={\parallel x-y\parallel }_u\le {\parallel x-f\parallel }_u{+\parallel f-y\parallel }_u\le {2max\{\parallel x-f\parallel }_u{,\parallel f-y\parallel }_u\}=2d_f(x,y),\end{array}$$

for all $x,y\in X^{*}$, implies the uniform metric $d_u$ is equivalent from above to the metric $d_f$ ($d_u\le \mathcal{M}{d}_f,\mathcal{M}=2$). Finally, from conditions (12) and (13) we show T is a $d_f$-${\psi }^{\prime }$-${\epsilon }^{\prime }$-contraction, where ${\psi }^{\prime }\in {\Psi }_0\left({\epsilon }^{\prime }\right)$ defined as follows ${\psi }^{\prime }\left(t\right)=\psi \left(w_{0}t\right)$, and ${\epsilon }^{\prime }:={\displaystyle \frac{\epsilon }{w_0}}$: For all $x,y\in X^{*}$, from (11) to (13) and non-decreasing condition of $\psi$ we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill d_f(Tx,Ty)& =max\left\{{\parallel Tx-f\parallel }_u,{\parallel Ty-f\parallel }_u\right\}\hfill \\ \hfill & =max\left\{\underset{t\in I}{sup}\left|{\int }_a^{b}K(t,s,x\left(s\right))ds\right|,\underset{t\in I}{sup}\left|{\int }_a^{b}K(t,s,y\left(s\right))ds\right|\right\}\hfill \\ \hfill & =\underset{t\in I}{sup}max\left\{\left|{\int }_a^{b}K(t,s,x\left(s\right))ds\right|,\left|{\int }_a^{b}K(t,s,y\left(s\right))ds\right|\right\}\hfill \\ \hfill & \le \underset{t\in I}{sup}{\int }_a^{b}max\left\{\left|K(t,s,x(s\left)\right)\right|,\left|K(t,s,y(s\left)\right)\right|\right\}ds\hfill \\ \hfill & \le max\left\{{\psi (\parallel x\parallel }_u{),\psi (\parallel y\parallel }_u)\right\}\hfill \\ \hfill & \le max\left\{\psi (w_0{\parallel x-f\parallel }_u),\psi (w_0{\parallel y-f\parallel }_u)\right\}\hfill \\ \hfill & =\psi \left(w_0{d}_f(x,y)\right)\hfill \\ \hfill & ={\psi }^{\prime }\left(d_f(x,y)\right).\hfill \end{array}\end{array}$$

(17)

Thus, T is a $d_f$-${\psi }^{\prime }$-${\epsilon }^{\prime }$-contraction, see Example 2-6. From parts (I)-(II) of Theorem 1, the part (I) obtains, and from parts (I) to (III) the part (II) obtains. The posteriori and priori estimates (14) and (15) obtain from corresponding estimates (5) and (6), where $\mathcal{M}=2$, which prove part (III).

□

Remark 2. 

(1)

Theorem 2, without any modification, can also be applied to the Volterra integral equation

$$\begin{array}{c}\hfill x\left(t\right)=Tx\left(t\right):=f\left(t\right)+{\int }_a^{t}K(t,s,x\left(s\right))ds,t\in I.\end{array}$$

(2)

The subset $X^{*}\subset C\left(I\right)$ is closed with respect to the metric $d_u$. Hence, $(X^{*},d_u)$ is a complete metric space. However, this is not the case for the metric $d_f$ defined by (1), since every $d_f$-Cauchy sequence converges to $f\ne 0$ (see Example 1), while $f\notin X^{*}$ because f does not satisfy condition (11). Therefore, $(X^{*},d_f)$ is not a complete metric space.

(3)

The metric $d_f$ plays the role of ensuring the Cauchy property of the Picard sequence through the $d_f$-ψ-ε-contraction property, whereas $d_u$ is used to guarantee the convergence of the Picard sequence $T^n\left(x_0\right)$, since the space $X^{*}$ is complete with respect to $d_u$. The existence of these two different metrics provides greater flexibility in the study of integral equations and extends the applicability of the Banach contraction framework.

(4)

Condition (13), called the ψ-ε-max inequality, is new and, as will be seen in the subsequent sections, provides a suitable alternative to Lipschitz-type conditions.

(5)

Other or additional conditions may be added to (10)–(12) in such a way that $X^{*}$ remains invariant under T. Suppose there exists $x_0\in X$ such that

$$\begin{array}{c}\hfill x_0\le x\in C\left(I\right)\Rightarrow x_0\le Tx.\end{array}$$

(18)

Define

$$\begin{array}{c}\hfill X^{**}:=X^{*}\cap \{x\in C\left(I\right):x_0\le x\}.\end{array}$$

(19)

Clearly, $X^{**}$ is a closed subset of $(X^{*},d_u)$, $T\left(X^{**}\right)\subseteq X^{**}$, and all arguments in the proof of Theorem 2 remain valid on $X^{**}$.

For example,

• If $0\le x\in C\left(I\right)\Rightarrow 0\le Tx$, then condition (10) can be replaced by $0\le x\in C\left(I\right)$, which yields a positive solution.
• If $0\ge x\in C\left(I\right)\Rightarrow 0\ge Tx$, then condition (10) can be replaced by $0\ge x\in C\left(I\right)$, which yields a negative solution.
• Suppose that T is a non-decreasing operator, that is, $x\le y\Rightarrow Tx\le Ty,\forall x,y\in C\left(I\right),$ and assume that $x_0\le T{x}_0$ for some $x_0\in C\left(I\right)$. Then condition (18) holds. Indeed, if $x_0\le x$, then, by the monotonicity of T, $x_0\le T{x}_0\le Tx.$

(6)

In practice, the continuity of T with respect to the uniform metric $d_u$, required in Theorem 2-(II), is often guaranteed by assuming that K is uniformly continuous (or Lipschitz-type conditions) on $I\times I\times [-M,M]$ for every $M>0$, and that the Picard sequence $T^n\left(x_0\right)$ is uniformly bounded.

Such boundedness can be ensured by an appropriate choice of $X^{*}$, for example, by imposing an explicit uniform bound on ${\parallel x\parallel }_u$.

Theorem 2 can naturally be extended to weighted kernels of the form $\overline{K}(t,s,x\left(s\right))=w(t,s)K(t,s,x\left(s\right))$, where usually K is a smooth function and w is an integrable (possibly singular) function with respect to s, such as those considered in Table 1. For instance, we have the following result.

Theorem 3. 

Under the same assumptions as in Theorem 2, all the conclusions remain valid for the fractional integral equation in the Riemann–Liouville sense (see [16]), where the operator $T:C\left(I\right)\to C\left(I\right)$ is defined by

$$\begin{array}{c}\hfill x\left(t\right)=Tx\left(t\right)=f\left(t\right)+{\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^t{(t-s)}^{\alpha -1}K(t,s,x\left(s\right))ds,t\in I:=[0,b],0<\alpha \le 1,\end{array}$$

and $\mathsf{\Gamma }\left(\alpha \right)$ denotes the Gamma function.

Proof. 

The proof is similar to that of Theorem 2. The only difference is in verifying the $d_f$-${\psi }^{\prime }$-${\epsilon }^{\prime }$-contraction condition corresponding to (17), which now becomes

$$\begin{array}{cc}\hfill d_f(Tx,Ty)& =max\left\{{\parallel Tx-f\parallel }_u,{\parallel Ty-f\parallel }_u\right\}\hfill \\ \hfill & ={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}max\left\{\underset{t\in I}{sup}\left|{\int }_0^t{(t-s)}^{\alpha -1}K(t,s,x\left(s\right))ds\right|,\underset{t\in I}{sup}\left|{\int }_0^t{(t-s)}^{\alpha -1}K(t,s,y\left(s\right))ds\right|\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\underset{t\in I}{sup}{\int }_0^t{(t-s)}^{\alpha -1}max\left\{\left|K\right(t,s,x\left(s\right)\left)\right|,\left|K\right(t,s,y\left(s\right)\left)\right|\right\}ds\hfill \\ \hfill & \le {\displaystyle \frac{{sup}_{t\in I}{\displaystyle \frac{t^{\alpha }}{\alpha }}}{\mathsf{\Gamma }\left(\alpha \right)}}max\left\{{\psi (\parallel x\parallel }_u{),\psi (\parallel y\parallel }_u)\right\}\hfill \\ \hfill & \le {\displaystyle \frac{b^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}}\psi \left(w_{0}max\left\{{\parallel x-f\parallel }_u,{\parallel y-f\parallel }_u\right\}\right)\hfill \\ \hfill & ={\displaystyle \frac{b^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}}\psi \left(w_0{d}_f(x,y)\right)\hfill \\ \hfill & ={\psi }^{\prime }\left(d_f(x,y)\right),\hfill \end{array}$$

where ${\psi }^{\prime }\left(t\right)={\displaystyle \frac{b^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}}\psi \left(w_{0}t\right),t\ge 0.$

By a suitable change of variable, the interval $[0,b]$ can be transformed into an interval $[0,b^{\prime }]$ such that $b^{\prime }\le \mathsf{\Gamma }{(\alpha +1)}^{{\displaystyle \frac{1}{\alpha }}},$ equivalently, ${\displaystyle \frac{{b^{\prime }}^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}}\le 1.$ Hence, by Lemma 2-6, ${\psi }^{\prime }\in {\Psi }_0\left({\epsilon }^{\prime }\right)$, where ${\epsilon }^{\prime }={\displaystyle \frac{\epsilon }{w_0}}.$ Consequently, T is a $d_f$-${\psi }^{\prime }$-${\epsilon }^{\prime }$-contraction. □

Theorem 4. 

Under the same assumptions as in Theorem 2, all the conclusions remain valid for the $s^{-\alpha }$-fractional integral equation $T:C\left(I\right)\to C\left(I\right)$ defined by

$$\begin{array}{c}\hfill x\left(t\right)=Tx\left(t\right)=f\left(t\right)+{\int }_0^b{s}^{-\alpha }K(t,s,x\left(s\right))ds,t\in I:=[0,b],0\le \alpha <1.\end{array}$$

Proof. 

The proof is similar to those of Theorems 2 and 3. □

3.1. Some Tests and Examples for the Existence of Solutions of Integral Equations

In this section, several examples are investigated in order to illustrate and validate the existence results established in Theorems 1 and 2.

Example 2 

([24, Chapter 7, Example 2]).

Consider the nonlinear Fredholm integral equation on $(C[0,1],\parallel ·{\parallel }_u)$ given by

$$\begin{array}{c}\hfill x\left(t\right)=T\left(x\right)\left(t\right)={\displaystyle \frac{7}{8}}t+{\displaystyle \frac{1}{2}}{\int }_0^{1}ts{x}^2\left(s\right)ds,t\in [0,1],\end{array}$$

(20)

where $f\left(t\right)={\displaystyle \frac{7}{8}}t$ and $K(t,s,u)=ts{u}^2$.

Wazwaz [24] solved equation (20) analytically (by substituting $x\left(t\right)=at$) and showed that its exact solutions are $x_1^{*}\left(t\right)=t$ and $x_2^{*}\left(t\right)=7t$.

Suppose that $x\in C\left(I\right)$ and $r\le {\parallel Tx-f\parallel }_u={\displaystyle \frac{1}{2}}{\int }_0^{1}s{x}^2\left(s\right)ds\le R$. Then

$$\begin{array}{cc}\hfill \parallel T^2{x-f\parallel }_u& =\underset{t\in I}{sup}{\displaystyle \frac{1}{2}}t{\int }_0^{1}s{\left(Tx\left(s\right)\right)}^{2}ds\hfill \\ \hfill & =\underset{t\in I}{sup}{\displaystyle \frac{1}{2}}t{\int }_0^{1}s{\left({\displaystyle \frac{7}{8}}s+{\displaystyle \frac{s}{2}}{\int }_0^1{s}_{1}x{\left(s_1\right)}^{2}d{s}_1\right)}^{2}ds\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\int }_0^{1}s{\left({\displaystyle \frac{7}{8}}s+s{\parallel Tx-f\parallel }_u\right)}^{2}ds\hfill \\ \hfill & ⟶\left\{\begin{array}{c}\begin{array}{cc}\hfill \le & {\displaystyle \frac{1}{2}}{\int }_0^{1}s{\left({\displaystyle \frac{7}{8}}s+sR\right)}^{2}ds={\displaystyle \frac{{(8R+7)}^2}{512}},\hfill \end{array}\hfill \\ \begin{array}{cc}\hfill \ge & {\displaystyle \frac{1}{2}}{\int }_0^{1}s{\left({\displaystyle \frac{7}{8}}s+sr\right)}^{2}ds={\displaystyle \frac{{(8r+7)}^2}{512}}.\hfill \end{array}\hfill \end{array}\right.\hfill \end{array}$$

The inequalities ${\displaystyle \frac{{(8R+7)}^2}{512}}\le R$ and ${\displaystyle \frac{{(8r+7)}^2}{512}}\ge r$ hold whenever $r\le {\displaystyle \frac{1}{8}}$ and ${\displaystyle \frac{1}{8}}\le R\le {\displaystyle \frac{49}{8}}.$

Choose $r_0=R_0={\displaystyle \frac{1}{8}}$ (note that any $0<r\le {\displaystyle \frac{1}{8}}\le R\le {\displaystyle \frac{49}{8}}$ may also be selected), and define $w_0={\displaystyle \frac{{\parallel f\parallel }_u+R_0}{r_0}}={\displaystyle \frac{\frac{7}{8}+\frac{1}{8}}{\frac{1}{8}}}=8.$

Therefore, according to (16), the operator T is invariant on

$$X^{*}=\left\{x\in {C\left(I\right):\parallel x\parallel }_u\le {8\parallel x-f\parallel }_u,{\parallel Tx-f\parallel }_u={\displaystyle \frac{1}{2}}{\int }_0^{1}s{x}^2\left(s\right)ds={\displaystyle \frac{1}{8}}\right\}.$$

Next, we verify the max-inequality condition. For all $u,v\in \mathbb{R}$,

$$\begin{array}{cc}\hfill \underset{t\in I}{sup}{\int }_0^{1}max\left\{\right|K(t,s,u)|,|K(t,s,v)\left|\right\}ds& =\underset{t\in I}{sup}{\int }_0^{1}max\left\{\right|ts{u}^2|,|ts{v}^2\left|\right\}ds\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\underset{t\in I}{sup}max\{t{u}^2,t{v}^2\}\hfill \\ \hfill & =max\left\{\psi \right(\left|u\right|),\psi (\left|v\right|\left)\right\},\hfill \end{array}$$

where $\psi \left(t\right)=\lambda t^2,t\ge 0,\lambda ={\displaystyle \frac{1}{2}}.$

Since $\psi \in {\Psi }_0\left(2\right)$, we obtain ${\psi }^{\prime }\left(t\right)=\psi \left(w_{0}t\right)=\lambda w_0^2{t}^2=32t^2,$ and hence $\epsilon ={\epsilon }_0={\displaystyle \frac{1}{\lambda w_0^2}}={\displaystyle \frac{1}{32}},$ see Theorem 2-(iii).

Therefore, for every $d_f$-ε-initial value $x_0\in X^{*}$ satisfying

$$d_f(x_0,T{x}_0)=max\{\parallel x_0{-f\parallel }_u,\parallel T{x}_0{-f\parallel }_u\}=\parallel x_0{-f\parallel }_u<{\displaystyle \frac{1}{32}},$$

(the equality follows from $\parallel T{x}_0{-f\parallel }_u={\displaystyle \frac{1}{8}}$), the Picard sequence $T^n\left(x_0\right)$ converges in the topology induced by $d_u$ to a fixed point $u^{*}\in X^{*}$. Moreover, this fixed point is unique in the ball

$${\mathcal{B}}_{d_f,{\displaystyle \frac{1}{32}}}\left(u^{*}\right):=\left\{v\in X^{*}:d_f(v,u^{*})=max\{\parallel u^{*}{-f\parallel }_u{,\parallel v-f\parallel }_u\}<{\displaystyle \frac{1}{32}}\right\}.$$

Notice that the maximal admissible value of $R_0$ for defining $X^{*}$ is ${\displaystyle \frac{49}{8}}$. However,

$$\begin{array}{cc}\hfill \parallel T{x}_2^{*}{-f\parallel }_u& =\underset{t\in I}{sup}\left|{\displaystyle \frac{7}{8}}t+{\displaystyle \frac{1}{2}}{\int }_0^{1}ts{\left(7s\right)}^{2}ds\right|\hfill \\ \hfill & ={\displaystyle \frac{56}{8}}>{\displaystyle \frac{49}{8}}.\hfill \end{array}$$

Hence condition (12) is not satisfied for $x=x_2^{*}=7t$, that is, $x_2^{*}\notin X^{*}$. Consequently, $T^n\left(x_0\right)\to x_1^{*},$ for every ε-initial value $x_0\in X^{*}$.

Finally, observe that larger values of $R_0$ together with smaller values of $r_0$ may enlarge the set of admissible initial values.

The following show that using Theorem 2, instead of Banach contraction principle gains better results even for a linear integral equation. Also, this example as mentioned in Remark 2-(5) shows inequality (11) in Theorem 2, which is used to define $X^{*}$, may be changed to a more appropriate condition, see also Section 3.2.

Example 3 

([11]). Consider the Fredholm linear integral equation T on $\left(C\right[0,1],\parallel ·\parallel )$ as follows

$$\begin{array}{c}\hfill x\left(t\right)=T\left(x\right)\left(t\right)=2t+\rho t{\int }_0^1{e}^{-s}x\left(s\right)ds,t\in [0,1],\rho \in \mathbb{R}\setminus \left\{0\right\},\end{array}$$

(21)

where $f\left(t\right)=2t$ and $K(t,s,u)=\rho t{e}^{-s}u$.

Equation $T_1$Hernández-Verón and Romero [11] demonstrated that T (denoted by $T_1$ in their paper) is a contraction on $C\left(I\right)$ if and only if $0<\left|\rho \right|<{\displaystyle \frac{e}{e-1}}\simeq 1.5819,$ which follows from Banach contraction inequality.

To achieve better results we change inequality (11) and use Theorem 1 directly. Suppose $x\in C\left(I\right)$ and $r\le {\parallel Tx-f\parallel }_u=\left|\rho \right|{\int }_0^1{e}^{-s}x\left(s\right)ds\le R.$ Then

$$\begin{array}{cc}\hfill \parallel T^2{x-f\parallel }_u& =\left|\rho \right|{\int }_0^1{e}^{-s}Tx\left(s\right)ds\hfill \\ \hfill & =\left|\rho \right|{\int }_0^1{e}^{-s}\left(2s+\left|\rho \right|s{\int }_0^1{e}^{-s_1}x\left(s_1\right)d{s}_1\right)ds\hfill \\ \hfill & =\left|\rho \right|{\int }_0^1{e}^{-s}\left({2s+s\parallel Tx-f\parallel }_u\right)ds\hfill \\ \hfill & ⟶\left\{\begin{array}{c}\begin{array}{cc}\hfill \le & \left|\rho \right|{\int }_0^1{e}^{-s}s\left(2+R\right)ds=\left|\rho \right|\left(2+R\right)(1-2e^{-1}),\hfill \end{array}\hfill \\ \begin{array}{cc}\hfill \ge & \left|\rho \right|{\int }_0^1{e}^{-s}s\left(2+r\right)ds=\left|\rho \right|\left(2+r\right)(1-2e^{-1}).\hfill \end{array}\hfill \end{array}\right.\hfill \end{array}$$

Thus, the inequalities

$$\{\begin{array}{c}\left|\rho \right|(2+R)\frac{e-2}{e}\le R,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(22\right)\\ \left|\rho \right|(2+r)\frac{e-2}{e}\ge r,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(23\right)\end{array}$$

must have positive solutions.

The inequalities (22) and (23) hold if $\left|\rho \right|<{\displaystyle \frac{e}{e-2}},$ with solutions

$$r\le {\displaystyle \frac{2\left|\rho \right|\frac{e-2}{e}}{1-\left|\rho \right|\frac{e-2}{e}}},R\ge {\displaystyle \frac{2\left|\rho \right|\frac{e-2}{e}}{1-\left|\rho \right|\frac{e-2}{e}}}.$$

Let us choose

$$r_0=R_0={\displaystyle \frac{2\left|\rho \right|(e-2)}{e-\left|\rho \right|(e-2)}}$$

(notice that every $0<r\le {\displaystyle \frac{2\left|\rho \right|(e-2)}{e-\left|\rho \right|(e-2)}}\le R$ is allowed).

Define

$$\begin{array}{c}\hfill X^{*}=\left\{x\in {C\left(I\right);\parallel x\parallel }_u\le w_0{\parallel x-f\parallel }_u^2{,\parallel Tx-f\parallel }_u=\left|\rho \right|{\int }_0^1{e}^{-s}x\left(s\right)ds=R_0\right\},\end{array}$$

where

$$w_0={\displaystyle \frac{{\parallel f\parallel }_u+R_0}{r_0^2}}={\displaystyle \frac{2+R_0}{R_0^2}}.$$

Notice that if $x\in X^{*}$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\parallel Tx\parallel }_u& =\underset{t\in I}{sup}\left|f\left(t\right)+\rho t{\int }_0^1{e}^{-s}x\left(s\right)ds\right|\hfill \\ \hfill & \le {\parallel f\parallel }_u+R_0\hfill \\ \hfill & ={\displaystyle \frac{{\parallel f\parallel }_u+R_0}{R_0^2}}{\parallel Tx-f\parallel }_u^2\hfill \\ \hfill & =w_0{\parallel Tx-f\parallel }_u^2.\hfill \end{array}\end{array}$$

(24)

Therefore, from (22)–(24) we obtain $T\left(X^{*}\right)\subseteq X^{*}.$

Also, T is a $d_f$-ψ-ε-contraction on $X^{*}$, where $\psi \left(t\right)=\lambda t^2,t\ge 0,$ with $\lambda =\left|\rho \right|(1-e^{-1})w_0.$

Indeed, for all $x,y\in X^{*}$,

$$\begin{array}{cc}\hfill d_f(Tx,Ty)& =max\left\{\parallel Tx-f\parallel ,\parallel Ty-f\parallel \right\}\hfill \\ \hfill & =max\left\{\left|\rho \underset{t\in I}{sup}t{\int }_0^1{e}^{-s}x\left(s\right)\right|ds,\left|\rho \underset{t\in I}{sup}t{\int }_0^1{e}^{-s}y\left(s\right)\right|ds\right\}\hfill \\ \hfill & \le \left|\rho \right|{\int }_0^1{e}^{-s}max\left\{\right|x\left(s\right)|,|y\left(s\right)\left|\right\}ds\hfill \\ \hfill & \le \left(\left|\rho \right|{\int }_0^1{e}^{-s}ds\right)max\{\parallel x\parallel ,\parallel y\parallel \}\hfill \\ \hfill & \le \left|\rho \right|(1-e^{-1})max\{w_0{\parallel x-f\parallel }_u^2,w_0{\parallel y-f\parallel }_u^2\}\hfill \\ \hfill & =\left|\rho \right|(1-e^{-1})w_0{d}_f^2(x,y)\hfill \\ \hfill & =\psi \left(d_f(x,y)\right).\hfill \end{array}$$

Hence, $\psi \in {\Psi }_0\left(\epsilon \right),$ where $\epsilon ={\displaystyle \frac{1}{\lambda }},$ see Lemma 2-4.

Therefore, for every $d_f$-ε-initial value $x_0\in X^{*}$, namely

$$d_f(x_0,T{x}_0)=max\{\parallel x_0{-f\parallel }_u,\parallel T{x}_0-f{\parallel }_u\}<\epsilon ,$$

the Picard sequence $T^n\left(x_0\right)$ converges in the topology $d_u$ to a fixed point $u^{*}\in X^{*}$, and it is unique in the ball

$${\mathcal{B}}_{d_f,\epsilon }\left(u^{*}\right):=\left\{v\in X^{*};d_f(v,u^{*})=max\{\parallel u^{*}{-f\parallel }_u{,\parallel v-f\parallel }_u\}<\epsilon \right\}.$$

Thus, for all $0<\left|\rho \right|<{\displaystyle \frac{e}{e-2}}\simeq 3.7844,$ the assumptions of Theorem 2 hold on $X^{*}$, which shows that the result of Theorem 1 is stronger than Banach contraction principle (compare with [11]). Numerical experiments also verify this result; see Example 11, where $\rho =-{\displaystyle \frac{7}{2}}.$

Note 3. 

1.

The previous example illustrates that condition (9) is more important than condition (11) in calculation of the upper bound for ρ such that Picard sequences converges. In other words, when applying Theorem 1, condition (9) plays the effective role in determining the admissible upper bound of ρ through an appropriate adjustment of the subset $X^{*}\subseteq C\left(I\right)$, whereas in Banach contraction principle this role is arises entirely from the contraction condition itself.

2.

By selecting larger admissible values for $R_0$ and smaller admissible values for $r_0$, the subset $X^{*}$ can be enlarged in such a way that $T\left(X^{*}\right)\subseteq X^{*}$ holds in all examples considered in this paper.

Example 4 

([11]). Consider the Fredholm nonlinear integral equation T on $\left(C\right[0,1],\parallel ·\parallel )$ given by

$$T\left(x\right)\left(t\right)=t+\rho t{\int }_0^1{e}^{-s}x{\left(s\right)}^{2}ds,t\in [0,1],\rho \in \mathbb{R}\setminus \left\{0\right\},$$

where $f\left(t\right)=t$ and $K(t,s,u)=\rho t{e}^{-s}{u}^2$.

Equations $T_2$ and $T_3$Hernández-Verón and Romero [11] considered two cases, denoted by $T_2$ and $T_3$, where $\rho ={\displaystyle \frac{1}{5}}$ and $\rho ={\displaystyle \frac{1}{2}},$ respectively. They demonstrated that $T_2$ is a contraction on the closed ball $\overline{{\mathcal{B}}_2\left(0\right)}$, while for $T_3$ it is impossible to find a ball $\overline{{\mathcal{B}}_r\left(0\right)}\subset C[0,1]$, with $r>0$, such that $T_3$ is invariant on $\overline{{\mathcal{B}}_r\left(0\right)}$, since the inequality ${\displaystyle \frac{1-e^{-1}}{2}}{r}^2-r+1\le 0$ has no solution.

We show that T satisfies the assumptions of Theorem 2 whenever $0<\left|\rho \right|\le {\displaystyle \frac{1}{4e^{-1}(2e-5)}}\simeq 1.5566.$

Suppose that $x\in C\left(I\right)$ and $r\le {\parallel Tx-f\parallel }_u\le R,$ where ${\parallel Tx-f\parallel }_u=\left|\rho \right|{\int }_0^1{e}^{-s}x{\left(s\right)}^{2}ds.$ Then

$$\begin{array}{cc}\hfill \parallel T^2{x-f\parallel }_u& =\left|\rho \right|{\int }_0^1{e}^{-s}{\left(Tx\left(s\right)\right)}^{2}ds\hfill \\ \hfill & =\left|\rho \right|{\int }_0^1{e}^{-s}{\left(s+\left|\rho \right|s{\int }_0^1{e}^{-s_1}x{\left(s_1\right)}^{2}d{s}_1\right)}^{2}ds\hfill \\ \hfill & =\left|\rho \right|{\int }_0^1{e}^{-s}{\left({s+s\parallel Tx-f\parallel }_u\right)}^{2}ds\hfill \\ \hfill & ⟶\left\{\begin{array}{c}\begin{array}{cc}\hfill \le & \left|\rho \right|{\int }_0^1{e}^{-s}{s}^2{\left(1+R\right)}^{2}ds=\left|\rho \right|{\left(1+R\right)}^2{e}^{-1}(2e-5),\hfill \end{array}\hfill \\ \begin{array}{cc}\hfill \ge & \left|\rho \right|{\int }_0^1{e}^{-s}{s}^2{\left(1+r\right)}^{2}ds=\left|\rho \right|{\left(1+r\right)}^2{e}^{-1}(2e-5).\hfill \end{array}\hfill \end{array}\right.\hfill \end{array}$$

Therefore, the inequalities

$$\{\begin{array}{c}\left|\rho \right|{(1+R)}^2{e}^{-1}(2e-5)\le R,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(25\right)\\ \left|\rho \right|{(1+r)}^2{e}^{-1}(2e-5)\ge r,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(26\right)\end{array}$$

must admit positive solutions.

The inequality (25) has a solution whenever $\Delta ={\left(2\left|\rho \right|e^{-1}(2e-5)-1\right)}^2-4{\left(\left|\rho \right|e^{-1}(2e-5)\right)}^2=1-4e^{-1}(2e-5)\left|\rho \right|\ge 0,$ that is, $0<\left|\rho \right|\le {\displaystyle \frac{1}{4e^{-1}(2e-5)}}\simeq 1.5566.$

It is sufficient to choose

$$0<r\le {\displaystyle \frac{1-2e^{-1}(2e-5)\left|\rho \right|-\sqrt{\Delta }}{4\left|\rho \right|e^{-1}(2e-5)}}\le R\le {\displaystyle \frac{1-2e^{-1}(2e-5)\left|\rho \right|+\sqrt{\Delta }}{4\left|\rho \right|e^{-1}(2e-5)}},$$

so that inequalities (25) and (26) hold.

To further illustrate the proof of Theorem 2, we verify directly that T is a $d_f$-ψ-ε-contraction, without using the max-inequality (13). For all $x,y\in X^{*}$,

$$\begin{array}{cc}\hfill d_f(Tx,Ty)& =max\{\parallel Tx-f\parallel ,\parallel Ty-f\parallel \}\hfill \\ \hfill & =max\left\{\left|\rho \right|\underset{t\in I}{sup}t{\int }_0^1{e}^{-s}{x}^2\left(s\right)ds,\left|\rho \right|\underset{t\in I}{sup}t{\int }_0^1{e}^{-s}{y}^2\left(s\right)ds\right\}\hfill \\ \hfill & =\left|\rho \right|\underset{t\in I}{sup}t{\int }_0^1{e}^{-s}max\{x^2\left(s\right),y^2\left(s\right)\}ds\hfill \\ \hfill & \le \left(\left|\rho \right|{\int }_0^1{e}^{-s}ds\right){max\{\parallel x\parallel }^2{,\parallel y\parallel }^2\}\hfill \\ \hfill & \le \left|\rho \right|(1-e^{-1})max\{w_0^2{\parallel x-f\parallel }_u^2,w_0^2{\parallel y-f\parallel }_u^2\}\hfill \\ \hfill & =\left|\rho \right|(1-e^{-1})w_0^2{d}_f^2(x,y)\hfill \\ \hfill & =\psi \left(d_f(x,y)\right),\hfill \end{array}$$

where $\psi \left(t\right)=\lambda t^2,t\ge 0,$ with $\lambda =\left|\rho \right|(1-e^{-1})w_0^2,\epsilon ={\displaystyle \frac{1}{\lambda }}.$ Hence, $\psi \in {\Psi }_0\left(\epsilon \right),$ see Lemma 2-4.

For example, let $\rho ={\displaystyle \frac{1}{2}}$. Then we may choose $r_0=R_0=2.0965$ (which lies in the admissible interval), where every $0<r\le 2.0965\le R\le 12.3565$ is admissible.

We obtain

$$w_0={\displaystyle \frac{{\parallel f\parallel }_u+R_0}{r_0}}\simeq {\displaystyle \frac{1+2.0965}{2.0965}}\simeq 1.4769,$$

and therefore $\epsilon ={\displaystyle \frac{1}{\lambda }}\simeq 1.4503.$

Consequently, according to (16), T is invariant on

$$X^{*}=\left\{x\in {C\left(I\right);\parallel x\parallel }_u\le {1.4769\parallel x-f\parallel }_u,{\parallel Tx-f\parallel }_u={\displaystyle \frac{1}{2}}{\int }_0^1{e}^{-s}{x}^2\left(s\right)ds=2.0965\right\}.$$

Hence, for every $d_f$-ε-initial value $x_0\in X^{*}$ satisfying

$$d_f(x_0,T{x}_0)=max\{\parallel x_0{-f\parallel }_u,\parallel T{x}_0-f{\parallel }_u\}<1.4503,$$

the Picard sequence $T^n\left(x_0\right)$ converges in the topology $d_u$ to a fixed point $u^{*}\in X^{*}$, and this fixed point is unique in the ball ${\mathcal{B}}_{d_f,\epsilon }\left(u^{*}\right)=\left\{v\in X^{*};max\{\parallel u^{*}{-f\parallel }_u{,\parallel v-f\parallel }_u\}<1.4503\right\}.$

Note 4. 

Hernández-Verón and Romero [11] proved that, for $T_3$ (corresponding to $\rho =0.5$), the Krasnoselskij iterative method with the ${\epsilon }^{\prime }$-initial value $\parallel x_0-T_3{x}_0\parallel \le {\epsilon }^{\prime }:={\displaystyle \frac{2e-5}{5e}}\simeq 0.0321,$ converges to the unique fixed point $u^{*}\in \overline{{\mathcal{B}}_{0.13}\left(f\right)}.$ Theorem 1 further shows that the Picard iteration also converges to the same fixed point.

Example 5 

([6,11]). Consider the Fredholm non-linear integral equation T on $(C[0,1],\parallel ·{\parallel }_u)$ defined by

$$x\left(t\right)=T\left(x\right)\left(t\right)=sin\left(\pi t\right)+\rho cos\left(\pi t\right){\int }_0^{1}sin\left(\pi s\right)x{\left(s\right)}^{3}ds,t\in [0,1],\rho \in \mathbb{R}\setminus \left\{0\right\},$$

where $f\left(t\right)=sin\left(\pi t\right)$ and $K(t,s,u)=\rho cos\left(\pi t\right)sin\left(\pi s\right)u^3$.

Suppose that $x\in C\left(I\right)$ and

$$r\le {\parallel Tx-f\parallel }_u=\underset{t\in I}{sup}\left|\rho \right|\left|cos\left(\pi t\right){\int }_0^{1}sin\left(\pi s\right)x^3\left(s\right)ds\right|=\left|\rho \right|\left|{\int }_0^{1}sin\left(\pi s\right)x^3\left(s\right)ds\right|\le R.$$

Then

$$\begin{array}{cc}\hfill \parallel T^2{x-f\parallel }_u& =\left|\rho \right|\left|{\int }_0^{1}sin\left(\pi s\right){\left(Tx\left(s\right)\right)}^{3}ds\right|\hfill \\ \hfill & =\left|\rho \right|{\int }_0^{1}sin\left(\pi s\right){\left(sin\left(\pi s\right)+\left|\rho \right|\left|cos\left(\pi s\right){\int }_0^{1}sin\left(\pi s_1\right)x^3\left(s_1\right)d{s}_1\right|\right)}^{3}ds\hfill \\ \hfill & =\left|\rho \right|{\int }_0^{1}sin\left(\pi s\right){\left({sin\left(\pi s\right)+|cos\left(\pi s\right)|\parallel Tx-f\parallel }_u\right)}^{3}ds\hfill \\ \hfill & ⟶\left\{\begin{array}{c}\begin{array}{cc}\hfill \le & \left|\rho \right|{\int }_0^{1}sin\left(\pi s\right){\left(sin\left(\pi s\right)+|cos(\pi s\left)\right|R\right)}^{3}ds={\displaystyle \frac{\left|\rho \right|}{8\pi }}(4R^3+3\pi R^2+12R+3\pi )\hfill \end{array}\hfill \\ \begin{array}{cc}\hfill \ge & \left|\rho \right|{\int }_0^{1}sin\left(\pi s\right){\left(sin\left(\pi s\right)+|cos(\pi s\left)\right|r\right)}^{3}ds={\displaystyle \frac{\left|\rho \right|}{8\pi }}(4r^3+3\pi r^2+12r+3\pi ).\hfill \end{array}\hfill \end{array}\right.\hfill \end{array}$$

Therefore, the inequalities

$$\{\begin{array}{c}{\displaystyle \frac{\left|\rho \right|}{8\pi }}(4R^3+3\pi R^2+12R+3\pi )\le R,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(27\right)\\ {\displaystyle \frac{\left|\rho \right|}{8\pi }}(4r^3+3\pi r^2+12r+3\pi )\ge r\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(28\right)\end{array}$$

must admit positive solutions.

Since $4r^3+3\pi r^2+12r+3\pi$ is increasing on $[0,+\infty )$, it is sufficient to consider

$$r_0\le {\displaystyle \frac{3}{8}}\left|\rho \right|$$

in order for inequality (28) to hold.

Moreover, inequality (27) holds provided that

$$\left|\rho \right|\le \underset{R\in [0,+\infty )}{sup}{\displaystyle \frac{8\pi R}{4R^3+3\pi R^2+12R+3\pi }}\simeq 0.7420,$$

where the supremum is attained at $R_0\simeq 0.7763$.

Thus,

$$w_0\ge {\displaystyle \frac{1+0.7763}{\frac{3}{8}\left|\rho \right|}}$$

is obtained for $\left|\rho \right|\le 0.7420$, where inequalities (27) and (28) admit the solutions

$$r_0={\displaystyle \frac{3}{8}}\left|\rho \right|,R_0=0.7763.$$

Next, let us investigate the max-inequality condition (13). For all $u,v\in \mathbb{R}$,

$$\begin{array}{cc}\hfill \underset{t\in I}{sup}{\int }_0^{1}max\left\{\right|K(t,s,u)|,|K(t,s,v)\left|\right\}ds& \le \underset{t\in I}{sup}{\int }_0^{1}max\left\{\right|\rho cos\left(\pi t\right)sin\left(\pi s\right)u^3|,|\rho cos\left(\pi t\right)sin\left(\pi s\right)v^3\left|\right\}ds\hfill \\ \hfill & ={\displaystyle \frac{2\left|\rho \right|}{\pi }}{max\left\{\right|u|}^3{,\left|v\right|}^3\}\hfill \\ \hfill & =max\left\{\psi \right(\left|u\right|),\psi (\left|v\right|\left)\right\},\hfill \end{array}$$

where $\psi \left(t\right)=\lambda t^3,\lambda ={\displaystyle \frac{2\left|\rho \right|}{\pi }},t\ge 0.$ By Lemma 2-5, $\psi \in {\Psi }_0\left(\epsilon \right),\epsilon ={\lambda }^{-1/2}=\sqrt{{\displaystyle \frac{\pi }{2\left|\rho \right|}}}.$

Therefore, for every $d_f$-ε-initial value $x_0\in X^{*}$ satisfying

$$d_f(x_0,T{x}_0)=max\{\parallel x_0{-f\parallel }_u,\parallel T{x}_0-f{\parallel }_u\}<\sqrt{{\displaystyle \frac{\pi }{2\left|\rho \right|}}},$$

the Picard sequence $T^n\left(x_0\right)$ converges in the topology induced by $d_u$ to a fixed point $u^{*}\in X^{*}$, and this fixed point is unique in the ball

$${\mathcal{B}}_{d_f,\epsilon }\left(u^{*}\right):=\left\{v\in X^{*}:max\{\parallel u^{*}{-f\parallel }_u{,\parallel v-f\parallel }_u\}<\sqrt{{\displaystyle \frac{\pi }{2\left|\rho \right|}}}\right\}.$$

Two questions may be asked if condition (I)-(i) of Theorem 2 does not hold: whether Eq. (8) may still have a solution or may fail to have one. The next two examples give positive answers to both questions, respectively.

Example 6. 

Consider Example 3. Suppose $u^{*}$ is a fixed point of T and put $a={\int }_0^1{e}^{-s}{u}^{*}\left(s\right)ds$. Then from (21) we have $u^{*}\left(t\right)=(2+\rho a)t$. Thus, the solution is of the form $x\left(t\right)=At$, $A\in \mathbb{R}$. Substituting $x\left(t\right)=At$ into (21), the analytical solution is obtained with

$$A={\displaystyle \frac{2e}{e-\rho (e-2)}},\rho \ne {\displaystyle \frac{e}{e-2}}.$$

For $\rho ={\displaystyle \frac{e}{e-2}}$, straightforward calculations show that $u\left(t\right)=(2-e)t$ is a fixed point.

Assume $\rho \ge {\displaystyle \frac{e}{e-2}}$. Similar to Example 7, put

$$L={\parallel Tx-f\parallel }_u=\rho {\int }_0^1{e}^{-s}x\left(s\right)ds,$$

where $0\le x\in C\left(I\right)$. Since $K\ge 0$ and $f\ge 0$, we obtain $Tx-f,T^{2}x-f\ge 0$, and

$$\begin{array}{cc}\hfill \parallel T^2{x-f\parallel }_u& =\rho {\int }_0^1{e}^{-s}Tx\left(s\right)ds\hfill \\ \hfill & =\rho {\int }_0^1{e}^{-s}\left(2s+\rho s{\int }_0^1{e}^{-s_1}x\left(s_1\right)d{s}_1\right)ds\hfill \\ \hfill & =\rho {\int }_0^1{e}^{-s}\left({2s+s\parallel Tx-f\parallel }_u\right)ds\hfill \\ \hfill & =\rho (2+L){\displaystyle \frac{e-2}{e}}.\hfill \end{array}$$

The inequality

$$\rho (2+L){\displaystyle \frac{e-2}{e}}\le L$$

has no positive solution. Thus, there exists no $R>0$ such that inequalities (9) hold.

Example 7. 

Consider the following Fredholm integral equation, which has no solution in $C\left(I\right)$ (see [19] and [20]):

$$\begin{array}{c}\hfill x\left(t\right)=T\left(x\right)\left(t\right)=t+{\int }_0^{1}k(t,s)x\left(s\right)ds,\end{array}$$

where

$$k(t,s)=\left\{\begin{array}{cc}{\pi }^{2}s(1-t),\hfill & s\le t,\hfill \\ {\pi }^{2}t(1-s),\hfill & t\le s.\hfill \end{array}\right.$$

Suppose $0\le x\in C\left(I\right)$ and put

$$L={\parallel Tx-f\parallel }_u=\underset{t\in I}{sup}{\int }_0^{1}k(t,s)x\left(s\right)ds.$$

Since $k\ge 0$ and $f\ge 0$, we obtain $Tx-f,T^{2}x-f\ge 0$, and

$$\begin{array}{cc}\hfill \parallel T^2{x-f\parallel }_u& =\underset{t\in I}{sup}{\int }_0^{1}k(t,s)Tx\left(s\right)ds\hfill \\ \hfill & =\underset{t\in I}{sup}{\int }_0^{1}k(t,s)\left(s+{\int }_0^{1}k(s,s_1)x\left(s_1\right)d{s}_1\right)ds\hfill \\ \hfill & =\underset{t\in I}{sup}{\int }_0^{1}k(t,s)\left({s+\parallel Tx-f\parallel }_u\right)ds\hfill \\ \hfill & =\underset{t\in I}{sup}{\int }_0^{1}k(t,s)(s+L)ds\hfill \\ \hfill & =\underset{t\in I}{sup}\left({\int }_0^t{\pi }^{2}s(1-t)(s+L)ds+{\int }_t^1{\pi }^{2}t(1-s)(s+L)ds\right)\hfill \\ \hfill & =\underset{t\in I}{sup}{\pi }^2\left({\displaystyle \frac{1}{6}}(1-t)t^2(3L+2t)+{\displaystyle \frac{1}{6}}t{(t-1)}^2(2t+3L+1)\right)\hfill \\ \hfill & =\underset{t\in I}{sup}{\displaystyle \frac{{\pi }^2}{6}}\left(3Lt(1-t)+t(t+1)(1-t)\right).\hfill \end{array}$$

The function $t(t+1)(1-t)=t-t^3$ attains its maximum at $t=1/\sqrt{3}$, and its maximum value is ${\displaystyle \frac{2\sqrt{3}}{9}}$. Hence,

$$\parallel T^2{x-f\parallel }_u\le {\displaystyle \frac{{\pi }^2}{6}}\left({\displaystyle \frac{3}{4}}L+{\displaystyle \frac{2\sqrt{3}}{9}}\right).$$

The inequality

$${\displaystyle \frac{{\pi }^2}{6}}\left({\displaystyle \frac{3}{4}}L+{\displaystyle \frac{2\sqrt{3}}{9}}\right)\le L$$

has no positive solution because ${\displaystyle \frac{{\pi }^2}{8}}>1$. Thus, there exists no $R>0$ such that inequalities (9) hold.

3.2. Chandrasekhar problem

The Chandrasekhar problem [7] remains unresolved by analytical methods and therefore provides an important example for the application and investigation of Theorem 1.

Consider the Fredholm non-linear integral equation T on $\left(C\right[0,1],\parallel ·\parallel )$ given by

$$x\left(t\right)=Tx\left(t\right)=1+tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds,t\in [0,1],$$

(29)

where $0\ne \psi \in C[0,1]$.

The kernel is not of the form required in Theorem 2, so we apply Theorem 1 directly. Nevertheless, similarly to Theorem 2, it may be possible to establish a more general theorem for kernels of the form

$$K(t,s,u,v)={\displaystyle \frac{tu}{t+s}}\psi \left(s\right)v,K\in C(I\times I\times \mathbb{R}\times \mathbb{R}).$$

Let us consider two cases:

• Case 1:

Suppose $\psi$ is a positive function (the negative case is similar). We can benefit from Remark 2-(5), since T is a non-decreasing functional and

$$T1\left(t\right)=1+t{\int }_0^1{\displaystyle \frac{\psi \left(s\right)}{s+t}}ds\ge 1,t\in [0,1].$$

Put

$$X^{**}=\left\{x\in C\left(I\right):1\le {x,\parallel x\parallel }_u\le {w\parallel x-1\parallel }_u,r\le {\parallel Tx-1\parallel }_u\le R\right\},$$

and similarly to the previous examples calculate the parameters R, r, and w.

Suppose $1\le x\in C\left(I\right)$ and

$$r\le {\parallel Tx-1\parallel }_u=\underset{t\in I}{sup}tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds\le R.$$

Let us calculate r and R. Put

$$I:={\int }_0^1{\displaystyle \frac{\psi \left(s\right)}{s+1}}ds.$$

Then

$$\begin{array}{cc}\hfill \parallel T^{2}x-1\parallel & =\underset{t\in I}{sup}tTx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)Tx\left(s\right)}{s+t}}ds\hfill \\ \hfill & =\underset{t\in I}{sup}t\left(1+tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)}{s+t}}\left(1+sx\left(s\right){\int }_0^1{\displaystyle \frac{\psi \left(s_1\right)x\left(s_1\right)}{s_1+s}}d{s}_1\right)ds\hfill \\ \hfill & =\left({1+\parallel Tx-1\parallel }_u\right)\underset{t\in I}{sup}t{\int }_0^1{\displaystyle \frac{\psi \left(s\right)}{s+t}}\left(1+sx\left(s\right){\int }_0^1{\displaystyle \frac{\psi \left(s_1\right)x\left(s_1\right)}{s_1+s}}d{s}_1\right)ds\hfill \\ \hfill & ⟶\left\{\begin{array}{c}\le {(1+R)}^{2}I,\hfill \\ \ge I.\hfill \end{array}\right.\hfill \end{array}$$

Thus, the inequalities

$$\left\{\begin{array}{c}{(1+R)}^{2}I\le R,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(30\right)\hfill \\ I\ge r,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\left(31\right)\hfill \end{array}\right.$$

must have positive solutions.

The inequalities (30) and (31) hold provided that

$$I={\int }_0^1{\displaystyle \frac{\psi \left(s\right)}{s+1}}ds\le \underset{R\in [0,+\infty )}{sup}{\displaystyle \frac{R}{{(1+R)}^2}}={\displaystyle \frac{1}{4}},$$

(32)

where the supremum occurs at $R=1$.

Moreover, the inequalities admit solutions

$$0<r\le I\le {\displaystyle \frac{1-2I-\sqrt{1-4I}}{2I}}\le R\le {\displaystyle \frac{1-2I+\sqrt{1-4I}}{2I}}.$$

One may choose

$$r_0=I,R_0={\displaystyle \frac{1-2I-\sqrt{1-4I}}{2I}},$$

noting that every choice satisfying

$$0<r\le I\le {\displaystyle \frac{1-2I-\sqrt{1-4I}}{2I}}\le R\le {\displaystyle \frac{1-2I+\sqrt{1-4I}}{2I}}$$

is admissible.

Put

$$w_0={\displaystyle \frac{1+R_0}{r_0}}.$$

Then for all $x,y\in X^{**}$ we obtain

$$\begin{array}{cc}\hfill d_f(Tx,Ty)& =max\{\parallel Tx-1{\parallel }_u,\parallel Ty-1{\parallel }_u\}\hfill \\ \hfill & \le max\left\{\underset{t\in I}{sup}t\left|x\left(t\right)\right|{\int }_0^1\left|{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}\right|ds,\underset{t\in I}{sup}t\left|y\left(t\right)\right|{\int }_0^1\left|{\displaystyle \frac{\psi \left(s\right)y\left(s\right)}{s+t}}\right|ds\right\}\hfill \\ \hfill & \le \underset{t\in I}{sup}t{\int }_0^1\left|{\displaystyle \frac{\psi \left(s\right)}{s+t}}\right|{dsmax\{\parallel x\parallel }^2{,\parallel y\parallel }^2\}\hfill \\ \hfill & \le {\int }_0^1{\displaystyle \frac{\left|\psi \right(s\left)\right|}{s+1}}ds{w}_0^2{max\{\parallel x-1\parallel }^2{,\parallel y-1\parallel }^2\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}{w}_0^2{d}_f^2(x,y).\hfill \end{array}$$

Thus, T is a $d_f$-$\psi$-$\epsilon$-contraction, where $\psi \left(t\right)={\displaystyle \frac{1}{4}}{w}_0^2{t}^2\in {\Psi }_0\left(\epsilon \right),\epsilon ={\displaystyle \frac{4}{w_0^2}},$ see Lemma 2-4.

• Case 2:

In physical applications certain restrictions such as ${\int }_0^1\psi \left(s\right)ds\le {\displaystyle \frac{1}{2}}$ are necessary (see [9]), but from the viewpoint of pure mathematics such assumptions are not essential. Suppose $\psi$ changes sign, i.e.

$${\psi }^{(+)}=max\{\psi ,0\},{\psi }^{(-)}=max\{-\psi ,0\}\ge 0,$$

and both are nonzero. Then $\psi ={\psi }^{(+)}-{\psi }^{(-)}$. Put

$$I^{(+)}:={\int }_0^1{\displaystyle \frac{{\psi }^{(+)}\left(s\right)}{s+1}}ds,I^{(-)}:={\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)}{s+1}}ds>0.$$

Define

$$\begin{array}{c}\hfill X^{**}={\{x\in [a,A];\parallel x\parallel }_u\le w_0{\parallel x-1\parallel }_u,r_0\le {\parallel Tx-1\parallel }_u\le R_0\},\end{array}$$

(33)

where

$$[a,A]=\{x\in C(I);a\le x\le A\}.$$

Theorem 1 applies provided $0\le a\le A$ and $0<r\le R$ are chosen so that $T\left(X^{**}\right)\subseteq X^{**}$.

Let us first estimate A. If $x\le A$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill Tx\left(t\right)& =1+tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds\hfill \\ \hfill & =1+tx\left(t\right){\int }_0^1{\displaystyle \frac{{\psi }^{(+)}\left(s\right)x\left(s\right)}{s+t}}ds-tx\left(t\right){\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)x\left(s\right)}{s+t}}ds\hfill \\ \hfill & \le 1+A^2{I}^{(+)}.\hfill \end{array}\end{array}$$

(34)

Hence $Tx\le A$ admits a solution whenever

$$1\le {\displaystyle \frac{1-\sqrt{1-4I^{(+)}}}{2I^{(+)}}}\le A\le {\displaystyle \frac{1+\sqrt{1-4I^{(+)}}}{2I^{(+)}}},$$

which holds under

$$\begin{array}{c}\hfill 0<I^{(+)}={\int }_0^1{\displaystyle \frac{{\psi }^{(+)}\left(s\right)}{s+1}}ds\le {\displaystyle \frac{1}{4}}.\end{array}$$

(35)

Next, if $a\le x$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill Tx\left(t\right)& =1+tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds\hfill \\ \hfill & =1+tx\left(t\right){\int }_0^1{\displaystyle \frac{{\psi }^{(+)}\left(s\right)x\left(s\right)}{s+t}}ds-tx\left(t\right){\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)x\left(s\right)}{s+t}}ds\hfill \\ \hfill & \ge 1-A^2{I}^{(-)}.\hfill \end{array}\end{array}$$

(36)

Therefore $Tx\ge a$ has a solution whenever

$$0\le a\le 1-A^2{I}^{(-)}<1,$$

provided

$$\begin{array}{c}\hfill 0<I^{(-)}={\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)}{s+1}}ds<{\displaystyle \frac{1}{A^2}}.\end{array}$$

(37)

Thus one may choose

$$0<a\le 1-A^2{I}^{(-)}<1,$$

and consequently

$$\begin{array}{c}\hfill T\left(\right[a,A\left]\right)\subseteq [a,A].\end{array}$$

(38)

Now let us determine $r_0,R_0$. Suppose $x\in [a,A]$ and

$$r\le {\parallel Tx-1\parallel }_u=\underset{t\in I}{sup}\left|tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds\right|\le R.$$

Using (38),

$$\begin{array}{cc}\hfill \parallel T^2{x-1\parallel }_u& =\underset{t\in I}{sup}\left|tTx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)Tx\left(s\right)}{s+t}}ds\right|\hfill \\ \hfill & \le \underset{t\in I}{sup}tA\left|{\int }_0^1{\displaystyle \frac{{\psi }^{(+)}\left(s\right)}{s+t}}Tx\left(s\right)ds-{\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)}{s+t}}Tx\left(s\right)ds\right|\hfill \\ \hfill & \le \underset{t\in I}{sup}t{A}^{2}max\left\{{\int }_0^1{\displaystyle \frac{{\psi }^{(+)}\left(s\right)}{s+t}}Ads,{\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)}{s+t}}Ads\right\}\hfill \\ \hfill & \le A^{2}max\{I^{(+)},I^{(-)}\}.\hfill \end{array}$$

Hence every

$$A^{2}max\{I^{(+)},I^{(-)}\}\le R$$

is admissible.

On the other hand, from (36) and (38), we have $Tx\left(t\right)\ge a$, and therefore

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel T^2{x-1\parallel }_u& =\underset{t\in I}{sup}\left|tTx\left(t\right)\times {\int }_0^1{\displaystyle \frac{\psi \left(s\right)Tx\left(s\right)}{s+t}}ds\right|\hfill \\ \hfill & \ge a\underset{t\in I}{sup}\left|t{\int }_0^1{\displaystyle \frac{\psi \left(s\right)Tx\left(s\right)}{s+t}}ds\right|\hfill \\ \hfill & =a\underset{t\in I}{sup}t\times \left|{\int }_0^1{\displaystyle \frac{{\psi }^{(+)}\left(s\right)}{s+t}}Tx\left(s\right)ds-{\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)}{s+t}}Tx\left(s\right)ds\right|\hfill \\ \hfill & \ge \underset{t\in I}{sup}ta\times \left|{\int }_0^{1}a{\displaystyle \frac{{\psi }^{(+)}\left(s\right)}{s+t}}ds-A{\int }_0^1{\displaystyle \frac{{\psi }^{(-)}\left(s\right)}{s+t}}ds\right|\hfill \\ \hfill & \ge a\left|a{I}^{(+)}-A{I}^{(-)}\right|,\hfill \end{array}\end{array}$$

(39)

Now choose a and A such that $a{I}^{(+)}-A{I}^{(-)}>0$ (this is possible by taking a sufficiently large or A sufficiently small while respecting the earlier constraints). Then, for $t=1$, the integrand is non-negative and we obtain the lower bound

$$\parallel T^2{x-1\parallel }_u\ge a\left(a{I}^{(+)}-A{I}^{(-)}\right).$$

Thus we may set

$$r_0:=a\left(a{I}^{(+)}-A{I}^{(-)}\right).$$

Hence every $0<r\le r_0$ is admissible. Therefore (39) shows that the left-hand side of (9) has solutions whenever (35) and (37) hold.

Finally, choose

$$w_0\ge {\displaystyle \frac{1+R_0}{r_0}}.$$

Then for all $x\in X^{**}$,

$$\begin{array}{cc}\hfill {\parallel Tx\parallel }_u& =\underset{t\in I}{sup}\left|1+tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds\right|\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill & \le 1+R_0\le {\displaystyle \frac{1+R_0}{r_0}}\underset{t\in I}{sup}\left|tx\left(t\right){\int }_0^1{\displaystyle \frac{\psi \left(s\right)x\left(s\right)}{s+t}}ds\right|\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill & =w_0{\parallel Tx-1\parallel }_u.\hfill \end{array}$$

(42)

Thus (34)–(40) imply

$$T\left(X^{**}\right)\subseteq X^{**},$$

provided (35) and (37) hold.

The proof that T is a $d_f$-$\psi$-$\epsilon$-contraction is analogous to Case 1.

Note 5. 

1.

Observe that in Case 2 the condition (9) is obtained after calculating a and A, and the values of $r_0$ and $R_0$ depend on a and A, respectively. This method cannot be applied in general. For example, there is no $A\ge 0$ such that $x\le A\Rightarrow Tx\le A$ in Example 4, since the inequality

$${\displaystyle \frac{1-e^{-1}}{2}}{A}^2-A+1\le 0$$

has no solution.

2.

If $0\le \psi$ and ${\parallel \psi \parallel }_u\le {\displaystyle \frac{1}{4ln2}}\simeq 0.3606$, see [5], then condition (32) holds since

$${\int }_0^1{\displaystyle \frac{\psi \left(s\right)}{s+1}}ds\le {\displaystyle \frac{1}{4ln2}}{\int }_0^1{\displaystyle \frac{1}{s+1}}ds={\displaystyle \frac{1}{4}}.$$

3.

If $0\le \psi$ and

$${\parallel \psi \parallel }_2={\left({\int }_0^1\psi {\left(s\right)}^{2}ds\right)}^{1/2}\le {\displaystyle \frac{\sqrt{2}}{4}}\simeq 0.3535,$$

then Cauchy’s inequality implies

$${\int }_0^1{\displaystyle \frac{\psi \left(s\right)}{s+1}}ds\le {\parallel \psi \parallel }_2{\left({\int }_0^1{\displaystyle \frac{1}{{(s+1)}^2}}ds\right)}^{1/2}\le {\displaystyle \frac{\sqrt{2}}{4}}·{\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{1}{4}},$$

which shows that condition (32) differs from the previous case. For example, if $\psi \left(t\right)=2t^{16}$, then

$${\parallel \psi \parallel }_2=\sqrt{{\displaystyle \frac{4}{33}}}\approx 0.348<0.3535,{\parallel \psi \parallel }_u=2.$$

Substitution: Examples (ψ₂) and (ψ₄) below are investigated numerically in SubSection 5.1, although the implemented program can also be applied to other cases.

(ψ₁)

Consider $\psi \left(t\right)={\displaystyle \frac{3}{8}}{t}^2(1-t^2)$, see [10, Example 2]. Then

$${\parallel \psi \parallel }_u={\displaystyle \frac{3}{32}},R_0=1,r_0={\displaystyle \frac{3}{8}}{\int }_0^1{\displaystyle \frac{s^2(1-s^2)}{s+1}}ds={\displaystyle \frac{1}{32}}=0.03125.$$

Hence, $w_0={\displaystyle \frac{2}{1/32}}=64,\epsilon ={\displaystyle \frac{4}{{64}^2}}={\displaystyle \frac{1}{1024}}.$ Therefore, for every $d_f$-$\epsilon$-initial value $x_0\in X^{*}$ satisfying

$$d_f(x_0,T{x}_0)=max\{\parallel x_0{-1\parallel }_u,\parallel T{x}_0-1{\parallel }_u\}<\epsilon ,$$

the Picard sequence $T^n\left(x_0\right)$ converges in the topology $d_u$ to a fixed point $u^{*}\in X^{*}$, unique in

$${\mathcal{B}}_{d,\epsilon }\left(u^{*}\right)=\{v\in X^{*};max\{\parallel u^{*}{-1\parallel }_u{,\parallel v-1\parallel }_u\}<\epsilon \}.$$

(ψ₂)

Consider $\psi \left(t\right)={\displaystyle \frac{3}{32}}{(1-t^2)}^2$, see [7] and [10, Example 3]. Then ${\parallel \psi \parallel }_u={\displaystyle \frac{3}{32}}.$

(ψ₃)

Consider $\psi \left(t\right)=\rho \left({\displaystyle \frac{1}{2}}-t\right)\left({\displaystyle \frac{1}{2}}+t\right)$, $\rho \ne 0$. Then ${\parallel \psi \parallel }_u={\displaystyle \frac{3}{4}}\left|\rho \right|.$ If $\left|\rho \right|\le {\displaystyle \frac{1}{3ln2}}\simeq 0.4808,$ then the condition of Note 5-2 holds. However, conditions (35) and (37) provide a better estimate. Indeed,

$${\int }_{[0,{\displaystyle \frac{1}{2}}]}{\displaystyle \frac{(\frac{1}{2}-s)(\frac{1}{2}+s)}{s+1}}ds\simeq 0.0709,$$

hence

$$I^{(+)}=\left|\rho \right|{\int }_{[0,{\displaystyle \frac{1}{2}}]}{\displaystyle \frac{(\frac{1}{2}-s)(\frac{1}{2}+s)}{s+1}}ds\simeq 0.0709\left|\rho \right|\le {\displaystyle \frac{1}{4}},$$

which yields $\left|\rho \right|⪅3.5260.$ Moreover,

$$1\le {\displaystyle \frac{1-\sqrt{1-4I^{(+)}}}{2I^{(+)}}}\le A.$$

Choosing the lowest admissible value $A=1$, we obtain

$$I^{(-)}=\left|\rho \right|{\int }_{[{\displaystyle \frac{1}{2}},1]}-{\displaystyle \frac{(\frac{1}{2}-s)(\frac{1}{2}+s)}{s+1}}ds\simeq 0.0907\left|\rho \right|<{\displaystyle \frac{1}{A^2}}=1,$$

that is, $\left|\rho \right|⪅11.0253.$ Therefore, Case 2 predicts convergence of the Picard sequence for $\left|\rho \right|⪅3.5260.$ However, the numerical experiments in SubSection 5.1 show that the Picard sequence associated with the quadrature equation (46), obtained by Newton–Cotes rules, converges for $-5⪅\rho ⪅10.5,$ which is considerably larger than the theoretical prediction.

(ψ₄)

Consider $\psi \left(t\right)=\rho (1-2t)(1+t)$, $\rho \ne 0$. Then ${\parallel \psi \parallel }_u=2\left|\rho \right|.$ Hence, the condition of Note 5-2 holds whenever $\left|\rho \right|\le {\displaystyle \frac{1}{8ln2}}\simeq 0.1803.$ Again, conditions (35) and (37) yield a better estimate. Since

$${\int }_{[0,{\displaystyle \frac{1}{2}}]}{\displaystyle \frac{(1-2s)(1+s)}{s+1}}ds={\displaystyle \frac{1}{4}},$$

we obtain

$$I^{(+)}={\displaystyle \frac{\left|\rho \right|}{4}}\le {\displaystyle \frac{1}{4}}⟹\left|\rho \right|\le 1,$$

and

$$1\le {\displaystyle \frac{1-\sqrt{1-4I^{(+)}}}{2I^{(+)}}}\le A.$$

Taking $A=1$, we get

$$I^{(-)}=\left|\rho \right|{\int }_{[{\displaystyle \frac{1}{2}},1]}-{\displaystyle \frac{(\frac{1}{2}-s)(\frac{1}{2}+s)}{s+1}}ds={\displaystyle \frac{\left|\rho \right|}{4}}<1,$$

hence $\left|\rho \right|<4.$ Therefore, Case 2 predicts convergence for $\left|\rho \right|\le 1.$ Nevertheless, the numerical computations in SubSection 5.1 show that the Picard sequence of the quadrature equation (46), generated by Newton–Cotes rules, converges for $-2⪅\rho ⪅3,$ which again exceeds the theoretical estimate.

4. Application to the Existence of Solutions of Quadrature Integral Equations

Suppose that the integral in (8) can be approximated by a quadrature rule of the form

$$I\left(x\right):={\int }_a^{b}w\left(s\right)x\left(s\right),ds={\overline{I}}_k(x,[a,b])+E(x,k),{\overline{I}}_k(x,[a,b]):={\displaystyle \sum _{j=1}^k}{\omega }_{i}x\left(s_j\right),$$

(43)

where $w\ge 0$ is a weight function chosen to incorporate possible singularities, so that x becomes sufficiently smooth and can be accurately approximated by polynomials. Notice that, in Newton–Cotes quadrature formulas, the nodes are indexed by $j=0,\dots ,k$. Consequently, in (43) the index j starts at 0 and throughout the subsequent formulas, every occurrence of ${\sum }_{j=1}^k$ should be replaced by ${\sum }_{j=0}^k.$

The degree of exactness of a quadrature rule is the largest integer $p\in \mathbb{N}$ for which (43) is exact for all polynomials $1,s,\dots ,s^p$. The remaining parameters are as follows: $kh=b-a$, ${\omega }_i\in \mathbb{R}$, and $a\le s_1<\dots <s_k\le b,$ where the nodes are equally spaced for Newton–Cotes formulas (the first index start at $s_0=a$) and generally nonuniform for Gaussian-type quadratures. The integer k is called the order of the quadrature rule. The weights ${\omega }_i$, nodes $s_i$, and error term $E(x,k)$ can be computed in advance so that a prescribed degree of exactness is achieved; see [8,12,15].

Quadrature methods include the open and closed Newton–Cotes formulas and Gaussian quadratures such as Gauss–Jacobi, Gauss–Chebyshev, Gauss–Legendre, Gauss–Laguerre, Gauss–Radau, Gauss–Lobatto, Gauss–Log, and others. These methods can be applied to appropriately smooth, weakly singular, logarithmic, and fractional-type integrands; see [8,12,15]. Their main advantage is that the associated error terms are explicitly known and can therefore be incorporated into rigorous estimates.

It is also possible to employ composite (repeated) quadrature formulas, analogous to the composite Newton–Cotes rules; see [15, Subsection 3.2]. Let $m\in \mathbb{N}$ be a multiple of a fixed integer $k\in \mathbb{N}$, and use the notation $i=0:k:m\mathrm{for}i=0,k,2k,\dots ,m,$ with ${\sum }_{i=0:k}^m{a}_i=a_0+a_k+\dots +a_m.$ Then the composite quadrature formula is

$${\overline{I}}_{m,k}\left(x\right)={\displaystyle \sum _{i=0:k}^{m-k}}{\overline{I}}_k(x,[s_i,s_{i+k}])={\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_{j}x\left(s_{i+j}\right).$$

(44)

Note 6. 

1.

Table 1 is not exhaustive. It is well known that Gauss quadrature rules have strictly positive weights [12], whereas Newton–Cotes and Gauss–Log rules do not generally guarantee positivity of weights [4]. Asymptotic error estimates of these rules can be found in standard references on numerical integration. Moreover, different quadrature rules may be combined in repeated formulations. For example, to compute

$${\int }_0^1{(1-s)}^{\alpha }ds,\alpha >-1,$$

one may apply Gauss–Legendre quadrature on $[0,{\displaystyle \frac{1}{2}}]$, where the integrand is smooth, and Gauss–Jacobi quadrature with parameters $(\alpha ,0)$ on $[{\displaystyle \frac{1}{2}},1]$, where the endpoint singularity occurs.

2.

If a repeated formulation uses $i={\displaystyle \frac{m}{k}}$ subintervals and the corresponding quadrature rules have errors

$$O\left(f_1\left(h_1\right)\right),O\left(f_2\left(h_2\right)\right),\dots ,O\left(f_i\left(h_i\right)\right),$$

then the overall asymptotic error is

$$O\left(\underset{1\le j\le i}{max}{f}_j\left(h_j\right)\right).$$

4.1. Iterative Schemes Based on Picard Iteration and Quadrature Rules

In this section, we study the quadrature integral equation (46) associated with the weighted integral equation $T:C\left(I\right)\to C\left(I\right)$ of the form

$$\begin{array}{c}\hfill x\left(t\right)=Tx\left(t\right):=f\left(t\right)+{\int }_a^{b}w\left(s\right)K(t,s,x\left(s\right))ds,t\in I,\end{array}$$

(45)

where $f\in C\left(I\right)$, $K\in C(I\times I\times \mathbb{R})$, and $w\ge 0$ is a weight function such as those listed in Table 1. We assume that the integral in (45) can be approximated by a suitable quadrature rule from Table 1.

Using the notation introduced in Section 4, we define the quadrature integral equation

$$\begin{array}{c}\hfill x\left(t\right)=T_{m}x\left(t\right):=f\left(t\right)+{\overline{I}}_{m,k}(K(t,·,x(·)),I)=f\left(t\right)+{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_{j}K(t,s_{i+j},x\left(s_{i+j}\right)),t\in I,\end{array}$$

(46)

which is obtained from the weighted integral equation (45) by applying a quadrature rule of order k, where m is a multiple of k.

We shall show that the Picard sequence

$$\begin{array}{c}\hfill x_r\left(t_i\right)=T_m{x}_{r-1}\left(t_i\right)=T_m^r{x}_0\left(t_i\right),t_i\in I_m,\end{array}$$

(47)

where

$$I_m:=\{s_{i+j}:i=0,k,\dots ,m-k,j=0\mathrm{or}1,\dots ,k\},$$

denotes the set of quadrature nodes, provides an effective numerical approximation of the quadrature integral equation (46), and consequently of the integral equation (45).

Theorem 5. 

Replace the weighted integral operator T defined by (45) with the quadrature operator $T_m$ defined by (46) throughout the statements of Theorem 2. In particular, condition (9) becomes

$$\begin{array}{c}\hfill r_0\le \parallel T_m{x-f\parallel }_u\le R_0⟹r_0\le {\parallel T_m^{2}x-f\parallel }_u\le R_0.\end{array}$$

(48)

Moreover, replace the ψ-ε-max inequality by

$$\underset{t,s\in I}{sup}max\left\{\right|K(t,s,u)|,|K(t,s,v)\left|\right\}\le max\left\{\psi \right(\left|u\right|),\psi (\left|v\right|\left)\right\},\forall u,v\in \mathbb{R},$$

(49)

where $\psi \in {\Psi }_0\left(\epsilon \right)$ for some $\epsilon \in (0,+\infty ]$.

If ${\omega }_j\ge 0$ for all $j=0\mathrm{or}1,\dots ,k$, then all conclusions of Theorem 2 remain valid for the quadrature integral equation (46). In particular, $T_m$ admits a solution

$$u_m^{*}\in X^{*}:=\left\{x\in {C\left(I\right):\parallel x\parallel }_u\le w_0{\parallel x-f\parallel }_u,r_0\le {\parallel T_{m}x-f\parallel }_u\le R_0\right\},$$

which is unique in

$${\mathcal{B}}_{d,{\epsilon }_0}\left(u_m^{*}\right):=\{v\in X^{*}:d(u_m^{*},v)<{\epsilon }_0\}.$$

Furthermore, for every $d_f$-${\epsilon }_0$-initial value $x_0\in X^{*}$, the Picard sequence $T_m^n\left(x_0\right)$ converges to $u_m^{*}$ in the topology induced by $d_u$.

Proof. 

By a change of variables, we may assume that $b-a\le 1$. Since the quadrature rule is exact for constants, setting $f\equiv 1$ in (44) gives

$${\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j=b-a\le 1.$$

The proof is analogous to that of Theorem 2. For instance, using (49) and ${\omega }_j\ge 0$, we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill d_f(T_{m}x,T_{m}y)& =max\{\parallel T_{m}x-f{\parallel }_u,\parallel T_{m}y-f{\parallel }_u\}\hfill \\ \hfill & \le {\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j\underset{t,s\in I}{sup}max\left\{\right|K(t,s,x\left(s\right))|,|K(t,s,y\left(s\right))\left|\right\}\hfill \\ \hfill & \le \left({\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j\right){max\left\{\psi \right(\parallel x\parallel }_u{),\psi (\parallel y\parallel }_u\left)\right\}\hfill \\ \hfill & \le \psi \left(w_0{max\{\parallel x-f\parallel }_u{,\parallel y-f\parallel }_u\}\right)\hfill \\ \hfill & =\psi \left(w_0{d}_f(x,y)\right)=:{\psi }^{\prime }\left(d_f(x,y)\right).\hfill \end{array}\end{array}$$

(50)

where ${\psi }^{\prime }\in \Psi \left({\epsilon }^{\prime }\right)$ defined as follows ${\psi }^{\prime }\left(t\right)=\psi \left(w_{0}t\right)$, and ${\epsilon }^{\prime }:={\displaystyle \frac{\epsilon }{w_0}}$, see Lemma 2-6. The remaining assertions follow exactly as in the proof of Theorem 2. □

Note 7. 

1.

Observe that the ψ-ε-max condition (13) is weaker than (49); indeed, (49)$⟹$ (13).

2.

If $\psi \left(t\right)={\left(\lambda t\right)}^{\alpha }$ ($\alpha >1,\lambda >0$) as in Lemma 2-7, the positivity condition on ${\omega }_j$ can be dropped. Using (50) and ${\parallel x\parallel }_u\le w_0{\parallel x-f\parallel }_u$, we get

$$d_f(T_{m}x,T_{m}y)\le L\psi \left(w_0{d}_f(x,y)\right),L:={\displaystyle \sum _{j=1}^k}\left|{\omega }_j\right|.$$

Define ${\psi }^{\prime }\left(t\right):=L\psi \left(w_{0}t\right)=L{\left(\lambda w_{0}t\right)}^{\alpha }={\left(\left(L^{1/\alpha }\lambda w_0\right)t\right)}^{\alpha }$. Then ${\psi }^{\prime }\in {\Psi }_0\left(\epsilon \right)$ with

$$\epsilon ={\displaystyle \frac{1}{{\left(L^{1/\alpha }\lambda w_0\right)}^{\frac{\alpha }{\alpha -1}}}}={\displaystyle \frac{1}{{\left(L{\lambda }^{\alpha }{w}_0^{\alpha }\right)}^{\frac{1}{\alpha -1}}}}.$$

Thus $T_m$ is a $d_f$-${\psi }^{\prime }$-ε-contraction.

4.2. Tests and Examples Concerning the Existence of Solutions for Quadrature Integral Equations

In this section, we present two examples to demonstrate the applicability of Theorems 1 and 5 and to validate the corresponding existence results.

Example 8. 

Consider the crosspounding quadrature integral equations of Example 2:

$$\begin{array}{c}\hfill x\left(t\right)=T_{m}x\left(t\right):={\displaystyle \frac{7}{8}}t+{\displaystyle \frac{t}{2}}{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{s}_{i+j}{x}^2\left(s_{i+j}\right),\forall t\in I,\end{array}$$

(51)

where $s_r\in I_m,r=0\mathrm{or}1,\dots ,m$. The similar calculations apply to the quadrature version (Example 2) by replacing the continuous integrals with quadrature sums and using the corresponding discrete bounds. To investigate condition (48) in Theorem 5 two cases may be occur:

• Case 1 (Quadrature with positive weights):

Suppose that all quadrature weights ${\omega }_j$ are nonnegative. Suppose

$$r\le {∥T_{m}x-{\displaystyle \frac{7}{8}}t∥}_u={\displaystyle \frac{1}{2}}{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{s}_{i+j}{x}^2\left(s_{i+j}\right)\le R.$$

Then

$${∥T_m^{2}x-{\displaystyle \frac{7}{8}}t∥}_u={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{7}{8}}+{\parallel T_{m}x-f\parallel }_u\right)}^2{\displaystyle \sum _{i=0:k}^m}{\displaystyle \sum _{j=1}^k}{\omega }_j{s}_{i+j}^3,$$

and therefore

$${\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{7}{8}}+r\right)}^2{\displaystyle \sum _{i=0:k}^m}{\displaystyle \sum _{j=1}^k}{\omega }_j{s}_{i+j}^3\le {∥T_m^{2}x-{\displaystyle \frac{7}{8}}t∥}_u\le {\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{7}{8}}+R\right)}^2{\displaystyle \sum _{i=0:k}^m}{\displaystyle \sum _{j=1}^k}{\omega }_j{s}_{i+j}^3.$$

Since

$$\underset{m\to \infty }{lim}{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{s}_{i+j}^3={\int }_0^1{s}^{3}ds={\displaystyle \frac{1}{4}},$$

there exists $M_0\in \mathbb{N}$ such that for all $m\ge M_0$,

$${\displaystyle \frac{1}{5}}<{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{s}_{i+j}^3<{\displaystyle \frac{2}{5}}.$$

Consequently, the inequalities

$${\displaystyle \frac{1}{10}}{\left({\displaystyle \frac{7}{8}}+r\right)}^2\ge r,{\displaystyle \frac{1}{5}}{\left({\displaystyle \frac{7}{8}}+R\right)}^2\le R$$

must hold. Solving them yields

$$r_0\le {\displaystyle \frac{33}{8}}-{\displaystyle \frac{\sqrt{65}}{2}}\approx 0.09385,{\displaystyle \frac{13}{8}}-{\displaystyle \frac{\sqrt{7.5}}{2}}\le R_0\le {\displaystyle \frac{13}{8}}+{\displaystyle \frac{\sqrt{7.5}}{2}}.$$

Thus condition (48) holds for both T and $T_m$ whenever $m\ge M_0$. Choosing

$$r_0={\displaystyle \frac{33}{8}}-{\displaystyle \frac{\sqrt{65}}{2}},R_0={\displaystyle \frac{13}{8}}-{\displaystyle \frac{\sqrt{7.5}}{2}},w_0={\displaystyle \frac{{\parallel f\parallel }_u+R_0}{r_0}}\approx {\displaystyle \frac{0.875+0.2556}{0.09385}}\approx 12.05,$$

we conclude, as in Example 2, that T and $T_m$ are invariant on

$$X^{**}=\left\{x\in {C\left(I\right):\parallel x\parallel }_u\le w_0{\parallel x-f\parallel }_u,r_0\le {\displaystyle \frac{1}{2}}{\int }_0^{1}s{x}^2\left(s\right)ds\le R_0\right\}.$$

• Case 2 (Quadrature rules with possibly negative weights).

For the quadrature integral equation (51) associated with Example 2, the positivity assumption on the quadrature weights is not required. Indeed, by Note 7-2, it is sufficient to consider $\psi \left(t\right)=t^2,$ which corresponds to the family $\psi \left(t\right)={\left(\lambda t\right)}^{\alpha }$ with $\lambda =1$ and $\alpha =2$. Arguing as in (50), we obtain

$$d_f(T_{m}x,T_{m}y)\le L\psi \left(w_0{d}_f(x,y)\right),L={\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=0}^k}\left|{\omega }_j\right|.$$

Consequently, ${\psi }^{\prime }\left(t\right)=L\psi \left(w_{0}t\right)=L{\left(w_{0}t\right)}^2$ belongs to ${\Psi }_0\left(\epsilon \right)$, where ε is determined by Lemma 2-7. Therefore, $T_m$ is a $d_f$-${\psi }^{\prime }$-ε-contraction and all conclusions of Theorem 5 remain valid.

Example 9 (Quadrature Chandrasekhar problem)

Consider the crosspounding quadrature integral equations of Chandrasekhar problem (29) as follows

$$\begin{array}{c}\hfill x\left(t\right)=T_{m}x\left(t\right):=1+tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}},\forall t\in I,\end{array}$$

(52)

where $w_i\ge 0$. Let us consider the Case 2: Suppose ψ be a both positive and negative function and the restrictions (35) and (37) hold. Put $I_m^{(+)}:={\sum }_{i=0:k}^{m-k}{\sum }_{j=1}^k{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)}{1+s_{i+j}}},I_m^{(-)}:={\sum }_{i=0:k}^{m-k}{\sum }_{j=1}^k{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)}{1+s_{i+j}}}$. Treats similar as Chandrasekhar problem Case 2 and define $X_m^{**}$ as (33), i.e.,

$$\begin{array}{c}\hfill X_m^{**}=\left\{x\in [a_m,A_m]{,\parallel x\parallel }_u\le w_0\left(m\right){\parallel x-1\parallel }_u,r_0\left(m\right)\le {\parallel T_{m}x-1\parallel }_u\le R_0\left(m\right)\right\}.\end{array}$$

(53)

It can be apply Theorem 1 if real numbers $0\le a_m\le A_m$ and $0<R_0\left(m\right),r_0\left(m\right),w_0\left(m\right)$ are found such that $T_m$ become invariant on $X^{**}$. Let us calculate $A_m\ge 0$: If $0\le x\le A_m$, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill Tx\left(t\right)& =1+tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\hfill \\ \hfill & =1+tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}-tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\hfill \\ \hfill & \le 1+A_m^2{I}_m^{(+)},\hfill \end{array}\end{array}$$

(54)

for all $t>0$. The inequality $Tx\le A_m$ has a solution $1\le {\displaystyle \frac{1-\sqrt{1-4I_m^{(+)}}}{2I_m^{(+)}}}\le A_m\le {\displaystyle \frac{1+\sqrt{1-4I_m^{(+)}}}{2I_m^{(+)}}}$ if the restriction

$$\begin{array}{c}\hfill 0<I_m^{(+)}={\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)}{1+s_{i+j}}}\le {\displaystyle \frac{1}{4}},\end{array}$$

(55)

holds. Let us calculate $a_m\ge 0$: If $a_m\le x$, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill T_{m}x\left(t\right)& =1+tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\hfill \\ \hfill & =1+tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}-tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\hfill \\ \hfill & \ge 1-A_m^2{I}_m^{(-)},\hfill \end{array}\end{array}$$

(56)

for all $t\in I$. The inequality $T_{m}x\ge a_m$ has a solution $0\le a_m\le 1-A_m^2{I}_m^{(-)}<1$ if the restriction

$$\begin{array}{c}\hfill 0<I_m^{(-)}={\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)}{1+s_{i+j}}}<{\displaystyle \frac{1}{A_m^2}},\end{array}$$

(57)

holds. If (57) holds then we can select $0<a_m\le 1-A_m^2{I}^{(-)}<1$. So far we have shown

$$\begin{array}{c}\hfill T_m\left([a_m,A_m]\right)\subseteq T_m\left([a_m,A_m]\right).\end{array}$$

(58)

Let us calculate $0<r_0\left(m\right)\le R_0\left(m\right)$: Suppose

$$\begin{array}{c}\hfill r\left(m\right)\le {∥T_{m}x-1∥}_u=\underset{t\in I}{sup}\left|tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\le R\left(m\right).\end{array}$$

(59)

From (56) to (58) we get $T_{m}x\left(t\right)\ge a_m,$ for all $t\in I$, and From (59) we get

$$\begin{array}{cc}\hfill \parallel T_m^2{x-1\parallel }_u& =\underset{t\in I}{sup}\left|t{T}_{m}x\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)T_{m}x\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\hfill \\ \hfill & \le \underset{t\in I}{sup}t{A}_m\left|x\left(t\right)\left({\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}-{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\right)T_{m}x\left(s_{i+j}\right)\right|\hfill \\ \hfill & \le A_m^{2}max\left\{{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}},{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\right\}\hfill \\ \hfill & =A_m^{2}max\{I_m^{(+)},I_m^{(-)}\}.\hfill \end{array}$$

Thus, the right hand side of condition (48) has solutions has a solution $max\{I_m^{(+)},I_m^{(-)}\}{A}_m\le R_0\left(m\right)$.

Also, from (56) to (58) we get $T_{m}x\left(t\right)\ge a$ for all $t\in I$, and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel T_m^2{x-1\parallel }_u& =\underset{t\in I}{sup}\left|t{T}_{m}x\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)T_{m}x\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\hfill \\ \hfill & \ge a_m\underset{t\in I}{sup}\left|t{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)T_{m}x\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\hfill \\ \hfill & =a_m\underset{t\in I}{sup}t\left|{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)T_{m}x\left(s_{i+j}\right)}{t+s_{i+j}}}-{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)T_{m}x\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\hfill \\ \hfill & \ge a_m\underset{t\in I}{sup}t\left|a_m{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(+)}\left(s_{i+j}\right)}{t+s_{i+j}}}-A_m{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{{\psi }^{(-)}\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\hfill \\ \hfill & \ge \left|a_m{I}_m^{(+)}-A_m{I}_m^{(-)}\right|.\hfill \end{array}\end{array}$$

(60)

Put $r_0\left(m\right):=a_m\left|a_m{I}_m^{(+)}-A_m{I}_m^{(-)}\right|\ge 0$, if $a_m{I}_m^{(+)}=A_m{I}_m^{(-)}$ then by choosing another allowed choice of $a_m$ or $A_m$ we get $r_0\left(m\right)>0$. Thus, (60) shows the left hand side of condition (48) has solutions. Then similar to proof of Theorem 2 consider $w_0\left(m\right)\ge {\displaystyle \frac{1+R_0\left(m\right)}{r_0\left(m\right)}}$, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel T_m{x\parallel }_u& =\underset{t\in I}{sup}\left|1+tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\hfill \\ \hfill & \le (1+R_0\left(m\right))\le {\displaystyle \frac{1+R_0\left(m\right)}{r_0\left(m\right)}}\underset{t\in I}{sup}\left|tx\left(t\right){\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j{\displaystyle \frac{\psi \left(s_{i+j}\right)x\left(s_{i+j}\right)}{t+s_{i+j}}}\right|\hfill \\ \hfill & =w_0\left(m\right){\parallel T_{m}x-1\parallel }_u,\hfill \end{array}\end{array}$$

(61)

for all $x\in X_m^{**}$. The relations (54) to (61) show that $T\left(X_m^{**}\right)\subseteq X_m^{**}$, where the restrictions (55) and (57) hold.

To showing T is a $d_f$-ψ-ε-contraction treat as Chandrasekhar problem.

5. Numerical Solution of Integral Equations via the Picard Scheme

The existence and convergence results established in the preceding sections not only guarantee the solvability of a broad class of nonlinear Fredholm integral equations, but also provide a rigorous basis for their numerical treatment. In many important problems, including equations for which no explicit analytical solution is known, quadrature methods offer a natural way to approximate the underlying integral operator.

In this section, we combine Picard iteration with quadrature rules to construct convergent numerical schemes for nonlinear integral equations. Theorems 6–8 establish conditions ensuring that the discrete Picard sequence converges to a solution of the corresponding quadrature integral equation and approximates a solution of the original continuous problem. The results provide a direct link between the fixed-point theory developed earlier and practical computational methods.

Theorem 6. 

Let $T,T_m:X\to X$, for all $m\ge M_0\in \mathbb{N}$, be self-mappings. Suppose the pair $(X^{*},d^{*})$ satisfies properties (P) of Theorem 1 for each T and $T_m$ (with the same $\psi \in {\Psi }_0\left(\epsilon \right)$ and the same ε). In particular, whenever condition (4) is satisfied, the a priori and a posteriori error estimates of Theorem 1 are available.

Let $u^{*}$ and $u_m^{*}$ be the unique fixed points of T and $T_m$ in $X^{*}$, respectively (their existence follows from Theorem 1). If the sequence $T_m$ converges point-wise to T on $(X^{*},d^{*})$, i.e.

$$\underset{m\to \infty }{lim}{d}^{*}(T_{m}x,Tx)=0\forall x\in X^{*},$$

then $u_m^{*}\stackrel{d^{*}}{\to }{u}^{*}$. Moreover, for every d-ε-initial value $x_0\in X^{*}$ (with $x_0\ne T{x}_0$), the following a posteriori and a priori estimates hold:

$$\begin{array}{cc}\hfill d^{*}(x_{r,m},u^{*})& \le \mathcal{M}{\displaystyle \frac{d(x_0,T{x}_0)}{d(x_0,T{x}_0)-\psi \left(d(x_0,T{x}_0)\right)}}d(x_r,x_{r+1})\hfill \\ \hfill & +\mathcal{M}{\displaystyle \frac{d(x_0,T_m{x}_0)}{d(x_0,T_m{x}_0)-\psi \left(d(x_0,T_m{x}_0)\right)}}d(x_{r,m},x_{r+1,m})\hfill \\ \hfill & +d^{*}(T_m^r{x}_0,T^r{x}_0),\hfill \end{array}$$

(62)

$$\begin{array}{cc}\hfill d^{*}(x_{r,m},u^{*})& \le \mathcal{M}{\displaystyle \frac{d(x_0,T{x}_0)}{d(x_0,T{x}_0)-\psi \left(d(x_0,T{x}_0)\right)}}{\psi }^r\left(d(x_0,T{x}_1)\right)\hfill \\ \hfill & +\mathcal{M}{\displaystyle \frac{d(x_0,T_m{x}_0)}{d(x_0,T_m{x}_0)-\psi \left(d(x_0,T_m{x}_0)\right)}}{\psi }^r\left(d(x_0,T_m{x}_0)\right)\hfill \\ \hfill & +d^{*}(T_m^r{x}_0,T^r{x}_0),\hfill \end{array}$$

(63)

where $x_r=T^r{x}_0$, $x_{r,m}=T_m^r{x}_0$, and $d^{*}\le \mathcal{M}d$.

Proof. 

We first prove by induction on r that

$$\begin{array}{c}\hfill \underset{m\to \infty }{lim}{d}^{*}(T_m^{r}x,T^{r}x)=0\forall x\in X^{*},r\in \mathbb{N}.\end{array}$$

(64)

For $r=1$ this is exactly the point-wise convergence of $T_m$ to T. Assume (64) holds for some $r\ge 1$. Then

$$\begin{array}{cc}\hfill d^{*}(T_m^{r+1}x,T^{r+1}x)& \le d^{*}(T_m^{r+1}x,T_m^r\left(Tx\right))+d^{*}(T_m^r\left(Tx\right),T^{r+1}x)\hfill \\ & \le \mathcal{M}d(T_m^{r+1}x,T_m^r\left(Tx\right))+d^{*}(T_m^r\left(Tx\right),T^r\left(Tx\right))\hfill \\ & \le \mathcal{M}\psi \left(d(T_m^{r}x,T^{r}x)\right)+d^{*}(T_m^r\left(Tx\right),T^r\left(Tx\right)).\hfill \end{array}$$

Since $\psi \left(t\right)<t$ and $d(T_m^{r}x,T^{r}x)\to 0$, the first term tends to 0. The second term tends to 0 by the induction hypothesis applied to the point $Tx$. Thus (64) holds for $r+1$, completing the induction.

Now let $x_0\in X^{*}$ be a d-$\epsilon$-initial value. Using the a priori estimate from Theorem 1-(IV) we have

$$d^{*}(T_m^r{x}_0,u_m^{*})\le \mathcal{M}{\displaystyle \frac{d(x_0,T_m{x}_0)}{d(x_0,T_m{x}_0)-\psi \left(d(x_0,T_m{x}_0)\right)}}{\psi }^r\left(d(x_0,T_m{x}_0)\right),$$

and similarly for T and $u^{*}$. Then

$$\begin{array}{cc}\hfill d^{*}(u_m^{*},u^{*})& \le d^{*}(T_m^r{x}_0,u_m^{*})+d^{*}(T_m^r{x}_0,T^r{x}_0)+d^{*}(T^r{x}_0,u^{*})\hfill \\ & \le \mathcal{M}{\displaystyle \frac{d(x_0,T_m{x}_0)}{d(x_0,T_m{x}_0)-\psi \left(d(x_0,T_m{x}_0)\right)}}{\psi }^r\left(d(x_0,T_m{x}_0)\right)\hfill \\ & +d^{*}(T_m^r{x}_0,T^r{x}_0)\hfill \\ & +\mathcal{M}{\displaystyle \frac{d(x_0,T{x}_0)}{d(x_0,T{x}_0)-\psi \left(d(x_0,T{x}_0)\right)}}{\psi }^r\left(d(x_0,T{x}_0)\right).\hfill \end{array}$$

Because $\psi \in {\Psi }_0\left(\epsilon \right)$, we have ${\psi }^r\left(d(x_0,T{x}_0)\right)\to 0$ and ${\psi }^r\left(d(x_0,T_m{x}_0)\right)\to 0$ as $r\to \infty$, uniformly in m (since $d(x_0,T_m{x}_0)<\epsilon$). Moreover, by (56) with r fixed, $d^{*}(T_m^r{x}_0,T^r{x}_0)\to 0$ as $m\to \infty$. Therefore, given any $ϵ>0$, we can first choose r so large that the two ${\psi }^r$ terms are less than $ϵ/3$, and then choose m sufficiently large so that the middle term is also less than $ϵ/3$. This yields $d^{*}(u_m^{*},u^{*})<ϵ$ for all large m, i.e. $u_m^{*}\stackrel{d^{*}}{\to }{u}^{*}$.

The posteriori estimate (62) follows from the triangle inequality and the corresponding estimates for T and $T_m$:

$$\begin{array}{cc}\hfill d^{*}(x_{r,m},u^{*})& \le d^{*}(x_{r,m},u_m^{*})+d^{*}(u_m^{*},u^{*})+d^{*}(u^{*},x_r)\hfill \\ & \le \mathcal{M}{\displaystyle \frac{d(x_0,T_m{x}_0)}{d(x_0,T_m{x}_0)-\psi \left(d(x_0,T_m{x}_0)\right)}}d(x_{r,m},x_{r+1,m})\hfill \\ & +d^{*}(T_m^r{x}_0,T^r{x}_0)\hfill \\ & +\mathcal{M}{\displaystyle \frac{d(x_0,T{x}_0)}{d(x_0,T{x}_0)-\psi \left(d(x_0,T{x}_0)\right)}}d(x_r,x_{r+1}).\hfill \end{array}$$

The a priori estimate (63) is obtained by replacing $d(x_{r,m},x_{r+1,m})$ and $d(x_r,x_{r+1})$ by the bounds ${\psi }^r\left(d(x_0,T_m{x}_0)\right)$ and ${\psi }^r\left(d(x_0,T{x}_0)\right)$ (see inequality (7) of Theorem 1). This completes the proof. □

Theorem 7. 

Let $T,T_m:X\to X$, for all $m\ge M_0\in \mathbb{N}$, be self-mappings given by (45) and (46), respectively. Suppose $w_i\ge 0$ for all $i=0,\dots ,k$. Then $T_m$ converges pointwise to T on $(C\left(I\right),\parallel ·{\parallel }_u)$. Consequently, all conclusions of Theorem 6 hold.

Proof. 

Set $f\equiv 1$ in (44); then ${\sum }_{i=0:k}^{m-k}{\sum }_{j=0}^k{w}_j=b-a$. Let $\delta =k{\displaystyle \frac{b-a}{m}}$, where k is fixed and m is a multiple of k (so we have ${\displaystyle \frac{m}{k}}$ repeated subintervals). For any $x\in C\left(I\right)$,

$$\begin{array}{cc}\hfill \parallel T_m{x-Tx\parallel }_u& =\underset{t\in I}{sup}\left|{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=0}^k}{\omega }_{j}K(t,s_{i+j},x\left(s_{i+j}\right))-{\int }_a^{b}K(t,s,x\left(s\right))ds\right|\hfill \\ & ={\displaystyle \frac{1}{b-a}}\underset{t\in I}{sup}\left|{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=0}^k}{\omega }_j{\int }_a^{b}K(t,s_{i+j},x\left(s_{i+j}\right))ds-{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=0}^k}{\omega }_j{\int }_a^{b}K(t,s,x\left(s\right))ds\right|\hfill \\ & \le {\displaystyle \frac{1}{b-a}}{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=0}^k}{\omega }_j\underset{t\in I}{sup}{\int }_a^b\left|K(t,s_{i+j},x\left(s_{i+j}\right))-K(t,s,x\left(s\right))\right|ds\hfill \\ & \le {\displaystyle \frac{1}{b-a}}\left({\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=0}^k}{\omega }_j\right){\omega }_{\delta }(K,I)\hfill \\ & ={\omega }_{\delta }(K,I),\hfill \end{array}$$

where

$${\omega }_{\delta }(K,I)=\underset{\begin{array}{c}t\in I,s,s^{\prime }\in I,|s-s^{\prime }|\le \delta \\ u,u^{\prime }\in [min\left(x\right),max\left(x\right)]\end{array}}{sup}\left|K(t,s,u)-K(t,s^{\prime },u^{\prime })\right|$$

is the modulus of continuity of K with respect to its second variable (and implicitly the third, because $u,u^{\prime }$ vary continuously as x is continuous). Here $min\left(x\right)={inf}_{t\in I}x\left(t\right)$ and $max\left(x\right)={sup}_{t\in I}x\left(t\right)$. As $m\to \infty$, $\delta \to 0$, and since K is uniformly continuous on the compact set $I\times I\times [min(x),max(x\left)\right]$, we obtain ${lim}_{\delta \to 0}{\omega }_{\delta }(K,I)=0$. Hence $\parallel T_m{x-Tx\parallel }_u\to 0$ for every $x\in C\left(I\right)$, i.e., $T_m\to T$ pointwise. □

Theorem 8 

(Local error estimation of numerical scheme (46) and (47)). With the same assumptions as in Theorem 5, suppose f and $K(t,·,·)$ are smooth functions on $I\times \mathbb{R}$ for all $t\in I_m$. If a repeated quadrature rule has asymptotic error of order $p>0$, i.e.

$$\left|{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_j\phi \left(s_{i+j}\right)-{\int }_a^{b}w\left(s\right)\phi \left(s\right)ds\right|=\mathcal{O}\left(h^p\right)$$

for every smooth function φ, where $h=(b-a)/m$ and p is the order of the quadrature rule, then the convergence of the Picard sequence $T_m^n\left(x_0\right)$ satisfies the following estimate

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |x_{r,m}\left(t\right)-u^{*}\left(t\right)|& \le 2{\displaystyle \frac{d_f(x_0,T{x}_0)}{d_f(x_0,T{x}_0)-\psi \left(d_f(x_0,T{x}_0)\right)}}{\psi }^r\left(d_f(x_0,T{x}_0)\right)\hfill \\ \hfill & +2{\displaystyle \frac{d_f(x_0,T_m{x}_0)}{d_f(x_0,T_m{x}_0)-\psi \left(d_f(x_0,T_m{x}_0)\right)}}{\psi }^r\left(d_f(x_0,T_m{x}_0)\right)\hfill \\ \hfill & +\mathcal{O}\left(h^p\right),\hfill \end{array}\end{array}$$

(65)

for all $d_f$-${\epsilon }_0$-initial values $x_0\in X^{*}$ and all $t\in I_m$. Here $u^{*}$ is a fixed point of T given by (45), $x_{r,m}=T_m^r{x}_0$, $r\in \mathbb{N}$, $m\ge M_0$, and $h={\displaystyle \frac{b-a}{m}}$.

Proof. 

Choose an initial value $x_0\in X^{*}$ that is smooth (such an element exists because smooth functions are dense in $C\left(I\right)$ and the set $X^{*}$ is non-empty). By the smoothness assumption on $K(t,·,·)$, each iterate $T_m^r\left(x_0\right)\left(t\right)$ is smooth on $I\times \mathbb{R}$. The local quadrature error at the first step is

$$E_{\mathrm{local}}^1\left(t\right)=|T_m\left(x_0\right)\left(t\right)-T\left(x_0\right)\left(t\right)|=\left|{\displaystyle \sum _{i=0:k}^{m-k}}{\displaystyle \sum _{j=1}^k}{\omega }_{j}K(t,s_{i+j},x_0\left(s_{i+j}\right))-{\int }_a^{b}w\left(s\right)K(t,s,x_0\left(s\right))ds\right|.$$

By the assumed order of the quadrature rule, this error is $\mathcal{O}\left(h^p\right)$ uniformly in $t\in I$. Inductively, for any $r\ge 1$,

$$E_{\mathrm{local}}^r\left(t\right)=|T_m^r\left(x_0\right)\left(t\right)-T^r\left(x_0\right)\left(t\right)|=\left|\sum _{i,j}{\omega }_{j}K(t,s_{i+j},T^{r-1}{x}_0\left(s_{i+j}\right))-{\int }_a^{b}w\left(s\right)K(t,s,T^{r-1}{x}_0\left(s\right))ds\right|$$

is also $\mathcal{O}\left(h^p\right)$, because $T^{r-1}{x}_0$ is smooth and the quadrature rule converges with the same order for all smooth integrands.

Now apply the a priori estimates from Theorem 1-(IV) (with $\mathcal{M}=2$ since $d_u\le 2d_f$). For the exact operator T,

$$|T^r{x}_0\left(t\right)-u^{*}\left(t\right)|\le d_u(T^r{x}_0,u^{*})\le 2{\displaystyle \frac{d_f(x_0,T{x}_0)}{d_f(x_0,T{x}_0)-\psi \left(d_f(x_0,T{x}_0)\right)}}{\psi }^r\left(d_f(x_0,T{x}_0)\right),$$

and analogously for $T_m$,

$$|T_m^r{x}_0\left(t\right)-u_m^{*}\left(t\right)|\le 2{\displaystyle \frac{d_f(x_0,T_m{x}_0)}{d_f(x_0,T_m{x}_0)-\psi \left(d_f(x_0,T_m{x}_0)\right)}}{\psi }^r\left(d_f(x_0,T_m{x}_0)\right).$$

Using the triangle inequality,

$$\begin{array}{cc}\hfill |x_{r,m}\left(t\right)-u^{*}\left(t\right)|& \le |T_m^r{x}_0\left(t\right)-u_m^{*}\left(t\right)|+|T_m^r{x}_0\left(t\right)-T^r{x}_0\left(t\right)|+|T^r{x}_0\left(t\right)-u^{*}\left(t\right)|\hfill \\ & \le 2{\displaystyle \frac{d_f(x_0,T_m{x}_0)}{d_f(x_0,T_m{x}_0)-\psi \left(d_f(x_0,T_m{x}_0)\right)}}{\psi }^r\left(d_f(x_0,T_m{x}_0)\right)+\mathcal{O}\left(h^p\right)\hfill \\ & +2{\displaystyle \frac{d_f(x_0,T{x}_0)}{d_f(x_0,T{x}_0)-\psi \left(d_f(x_0,T{x}_0)\right)}}{\psi }^r\left(d_f(x_0,T{x}_0)\right).\hfill \end{array}$$

This is exactly estimate (65). □

Note 8. 

The first two terms in (65) correspond to the convergence of the Picard sequences for T and $T_m$, respectively, while the third term is the quadrature error. For sufficiently large r, the first two terms become negligible (since ${\psi }^r\left(d_f(·,·)\right)\to 0$), so the overall error is dominated by the quadrature error $\mathcal{O}\left(h^p\right)$. Hence, in practice, the convergence rate is essentially the same as that of the underlying quadrature rule.

Example 10. 

Consider Chandrasekhar integral equation T given by (29) and its quadrature integral equation $T_m$ given by (52). If the conditions ${lim}_{m\to +\infty }{I}_m^{+}=I^{+}$ and ${lim}_{m\to +\infty }{I}_m^{-}=I^{-}$ hold then $R_m\to R$ and $r_m\to r$ and by choosing $R_m\simeq R$ and $r_m\simeq r$ it may assume $X^{**}=X_m^{**}$ given by (33) and (53), for all large enough $m\in \mathbb{N}$, and all the Theorems of this Section can be apply on $X^{**}$, see also Note 3-2. For instance, these conditions hold for all (ψ₁) to (ψ₄).

5.1. Numerical investigation of Picard sequence and quadrature integration rules

Let us examine some experimental examples to support the theoretical results of previous sections for Fredholm integral equations. The numerical solution is computed using the Picard sequence (47). We employed two termination criteria: ${max}_{t_i\in I_m}|x_{r+1}\left(t_i\right)-x_r\left(t_i\right)|<{10}^{-100}$ and a maximum of 200 iterations. In some examples the first criterion terminated the process, while in others the second one was used. The implementation codes are publicly available and can be accessed through the links provided in the Availability of data and materials section.

Example 11. 

We use two quadrature rules to validate the theoretical results for Example 3 with $\rho =-3.5$. The exact solution of integral equation (21) is obtained by substituting $x\left(t\right)=at$, which yields $a=2+a\rho (1-2e^{-1})$, hence $a={\displaystyle \frac{2e}{e-\rho (e-2)}}={\displaystyle \frac{4e}{9e-14}}\approx 1.0390$ (see Example 6). Since the problem is smooth, appropriate quadrature rules are Newton–Cotes and Gauss–Legendre.

Figure 1 and Table 2 compare the errors obtained for different values of m and quadrature orders k. The results for Newton–Cotes and Gauss–Legendre rules were obtained using the first and second termination criteria, respectively.

Note 9. 

In Example 11, numerical experiments based on the Newton–Cotes and Gauss–Legendre quadrature rules indicate that the Picard sequence converges only for $\left|\rho \right|⪅3.7$ and for $-45⪅\rho ⪅3.7$, respectively. In particular, the Gauss–Legendre discretization exhibits convergence for a significantly wider range of negative values of ρ than that guaranteed by Theorem 1; namely, convergence is observed throughout the interval $-45⪅\rho ⪅-3.7$, whereas the theorem predicts only the range $\left|\rho \right|\le {\displaystyle \frac{e}{e-2}}$ for the integral equation (21). However, no convergence is observed for $\rho ⪆3.7$.

Example 12. 

We apply the Gauss–Jacobi quadrature rule to solve the nonlinear $s^{\alpha }$-weighted integral equation

$$x\left(t\right)=Tx\left(t\right)=(1-\rho )t^{{\displaystyle \frac{1}{4}}}-\rho \left(cos(t+1)-cost\right)+\rho {\int }_0^1{s}^{-{\displaystyle \frac{1}{2}}}\left(sin(t+s)+2s{t}^{{\displaystyle \frac{1}{4}}}\right)x^2\left(s\right),ds,t\in I:=[0,1],$$

whose exact solution is $x\left(t\right)=t^{1/4}$.

Using the transformation $s={\displaystyle \frac{1-\xi }{2}}$, we obtain

$${\int }_0^1{s}^{\alpha }f\left(s\right),ds={\int }_{-1}^1{2}^{-(\alpha +1)}{(1-\xi )}^{\alpha }f\left({\displaystyle \frac{1-\xi }{2}}\right)d\xi .$$

Therefore, the Gauss–Jacobi rule can be applied by replacing the standard nodes and weights according to

$$s_i={\displaystyle \frac{1-{\xi }_i}{2}},{\omega }_i^{*}=2^{-(\alpha +1)}{\omega }_i.$$

In the implementation, the Gauss–Jacobi nodes and weights are computed by the eigenvalue method, which is both efficient and highly accurate.

The results are summarized in Figure 2 and Figure 3. In the first experiment, the Gauss–Jacobi rule with parameters $\alpha =-{\displaystyle \frac{1}{2}}$, $\beta =0$ and orders $k=10,100,1000$ is applied on the whole interval $[0,1]$. In the second experiment, the interval is divided into two subintervals. Since the integrand is singular only on the first subinterval, the Gauss–Jacobi rule is employed on $[0,{\displaystyle \frac{1}{2}}]$, whereas the Gauss–Legendre rule is used on $[{\displaystyle \frac{1}{2}},1]$. Because both quadrature formulas have comparable accuracy, the resulting errors are very similar, thereby confirming Note 6-2.

The same approach can be applied to the Riemann–Liouville-type integral equation

$$x\left(t\right)=Tx\left(t\right)={\displaystyle \frac{2-e}{2\mathsf{\Gamma }\left(\frac{2}{5}\right)}}{e}^{-t}+\sqrt{1-t}+{\displaystyle \frac{1}{2\mathsf{\Gamma }\left(\frac{2}{5}\right)}}{\int }_0^1{\displaystyle \frac{e^{,s-t}}{{(1-s)}^{\frac{2}{5}}}}{x}^{{\displaystyle \frac{14}{5}}}\left(s\right),ds,t\in I:=[0,1],$$

whose exact solution is $x\left(t\right)=\sqrt{1-t}$.

Using the transformation $s={\displaystyle \frac{\xi +1}{2}}$, we obtain

$${\int }_0^1{(1-s)}^{\alpha }f\left(s\right),ds={\int }_{-1}^1{2}^{-(\alpha +1)}{(1-\xi )}^{\alpha }f\left({\displaystyle \frac{\xi +1}{2}}\right)d\xi .$$

Hence, the Gauss–Jacobi rule can be implemented with the modified nodes and weights

$$s_i={\displaystyle \frac{1+{\xi }_i}{2}},{\omega }_i^{*}=2^{-(\alpha +1)}{\omega }_i.$$

Figure 4 presents the corresponding errors for $k=50,500,5000$. The computation with $k=5000$ required approximately 448 seconds of CPU time.

Example 13. 

We employ two quadrature rules to investigate Chandrasekhar’s problem corresponding to cases (ψ₂) and (ψ₄) of SubSection 3.2. Since the exact solution satisfies $x\left(0\right)=1$, the iterative scheme (47) is initialized with $x\left(0\right)=1$. Consequently, quadrature formulas whose first node is located at $t=0$, such as the Newton–Cotes rules, can be used without difficulty.

• Results for case (ψ₂).

An advantage of the Newton–Cotes family is that rules of different orders employ the same prescribed nodes, making their numerical solutions directly comparable. Table 3 presents the solutions obtained by Simpson’s rule, Boole’s rule, and the Newton–Cotes rule with $k=10$, together with the results reported in [7,10].

The agreement among all methods is excellent. To compare quadrature rules of different orders, let

$$e_m^{k,k^{\prime }}\left(s\right)=\left|{\mathrm{NS}}_k\left(s\right)-{\mathrm{NS}}_{k^{\prime }}\left(s\right)\right|,s\in I_m,$$

where ${\mathrm{NS}}_k$ denotes the numerical solution generated by (46)–(47) using a quadrature rule of order k. Figure 5 displays the differences between the solutions obtained with Simpson’s rule ($k=2$), Boole’s rule ($k=4$), and the Newton–Cotes rule ($k=10$). The largest discrepancies occur near the singular point $t=0$, while the differences become considerably smaller away from the singularity.

The Gauss–Legendre rule was also implemented with orders $k=4,5,$ and 10. Unlike Newton–Cotes formulas, Gauss–Legendre rules employ different quadrature nodes for different values of k. Therefore, their solutions cannot be compared pointwise in the same manner as in Figure 5. Instead, the computed solutions are plotted together in Figure 6 to illustrate their overall agreement.

• Results for case (ψ₄) with $\rho =-2$.

Figure 7 (left and right) shows the pointwise differences between the solutions obtained by Simpson’s rule and the Newton–Cotes rule ($k=10$), and between Boole’s rule and the Newton–Cotes rule ($k=10$), respectively. As in the previous case, the largest local errors occur near the singular point $t=0$.

Since Gauss–Legendre rules of different orders use distinct (non-equally spaced) quadrature nodes, a direct pointwise comparison is again impossible. Therefore, the corresponding numerical solutions are displayed together in Figure 8.

6. Conclusions and Future Research Directions

We established existence, uniqueness, localization, and convergence results for a broad class of nonlinear integral equations. The proposed framework allows the construction of invariant sets that are adapted to the structure of the problem and can be applied even in cases where classical contraction principles are not directly applicable. Several examples, including Chandrasekhar-type equations, illustrate the effectiveness of the obtained theoretical results.

The numerical experiments performed with Newton–Cotes, Gauss–Legendre, and Gauss–Jacobi rules demonstrate the practical applicability of the method and show good agreement with available exact or previously reported solutions.

The numerical investigations also reveal that, in several cases, the observed convergence region of the Picard sequence is significantly larger than that predicted by the theoretical estimates. This suggests that the sufficient conditions established in the present work are not optimal and that further refinements may be possible. Consequently, the proposed framework opens several directions for future research, including the study of alternative iterative schemes, sharper localization conditions, multiple-solution phenomena, and the extension of the theory to other classes of functional, fractional, and Volterra-type integral equations. The results presented in this paper open several directions for further investigation. The following list is not exhaustive, but it highlights some potential avenues for future research.

1.

Theorem 1 may have applications beyond integral equations, including matrix equations, equilibrium problems, optimization, and mathematical economics, in a manner analogous to the broad applicability of the Banach contraction principle, see [14].

2.

Theorem 1 may be extended in the spirit of classical fixed-point theory. In particular, it would be interesting to investigate whether analogues of celebrated results such as Caristi’s fixed-point theorem, Ekeland’s variational principle, and related equivalence theorems can be established within the present framework. Similar extensions may also be considered in generalized metric spaces and other metric-like structures.

3.

The Picard iteration method developed in this paper provides a mechanism for obtaining a fixed point of the operator T. However, as observed in Example 3, Theorem 2 guarantees the existence of solutions only for $\left|\rho \right|<{\displaystyle \frac{e}{e-2}},$ which already improves the corresponding result obtained via the Banach contraction principle in [11]. On the other hand, numerical experiments indicate that the Picard sequences associated with the quadrature integral equation (38) diverge when $\rho \gtrsim 3.7,$ for both Newton–Cotes and Gauss–Legendre quadrature rules.

This naturally leads to the following questions:

(a)

Can the Picard iteration be modified, for example by employing Krasnoselskiĭ-, Mann-, Ishikawa-, Newton-type, or other iterative schemes, to obtain convergence to additional solutions of the underlying functional equation?

(b)

Is condition (9) essential? More precisely, can alternative invariant-set conditions or modified forms of the $\psi$-$\epsilon$-max-inequality (13) be used to establish the existence of further solutions that are not captured by the present approach?

4.

In connection with the previous question, it remains unclear whether the Chandrasekhar problem admits multiple solutions. If multiple solutions exist, it would be important to develop analytical and numerical techniques capable of detecting and approximating them.

5.

The extension of the present theory to Volterra integral equations and related classes of functional equations deserves further investigation. In particular, the development of efficient numerical algorithms (e.g., [17,21,23]) and convergence analyses for such problems would be of significant interest.

6.

A systematic study of numerical integration schemes associated with the proposed framework would also be worthwhile. This includes a comparison of various quadrature rules, error estimates, computational efficiency, and stability properties for different classes of integral equations.

7.

Further applications to other nonlinear integral equations, integro-differential equations, fractional integral equations, and related problems may provide additional insight into the scope and limitations of the proposed fixed-point framework.