On the existence of solutions of some non-linear functional integral equations in Banach algebra with applications

Abstract

In this article, we establish some results for the existence of solution of nonlinear functional integral equations by using Darbo’s fixed point theorem in Banach algebra. As an application, we discuss some examples of nonlinear functional integral equations and existence of solutions.

Keywords:

• Functional integral equations
• fixed point theorem
• Banach algebra
• measure of non-compactness

MSC 2020:

• 47H10

1. Introduction

Integral equations are an important branch of mathematical analysis, science and equations of such types are applicable in many physical problems such as in the vehicular traffic, the biology, control theory and mathematical physics(see Abdou, 2003; Argyros, 1985; Corduneanu, 1990; Deimling, 1985). Recently the theory of FIE and various kind of functional differential equations are developed effectively and emerged in the field of analysis, engineering, applied mathematics, and nonlinear functional analysis (see Aghajani et al., 2014; Arab, 2016; Cabada et al., 2018; Deepmala, 2013a; Roshan, 2017; Tunç, 2010a;Tunç 2010b; Tunç, 2016; Tunç, 2020; Tunç and Golmankhaneh, 2020; Tunç and Tunç, 2018a; Tunç and Tunç, 2018b; Tunç and Tunç, 2018c; Tunç and Tunç, 2019; Tunç, 2020 and references therein). In this article, we prove the existence of solution of the following generalized FIE: (1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) for $\zeta \in [0,b].$

The FIE (1(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) ) consists of many special types of FIEs, those are very useful in real-world problems of physics, biology, differential equations, etc. Here, our aim to examine the difficulty of the existence of the solutions of FIE (1(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) ) using the techniques of MNC and Darbo’s fixed point theorem in [0, b]. Many authors have taken out some successful attempts to solve many FIE by utilizing Darbo’s condition which is an important tool to study these equations (Aghajani et al., 2014; Banaś and Sadarangani, 2003; Deepmala, 2013a; Deepmala, 2013b; Maleknejad et al., 2008; Maleknejad et al., 2009a).

Recently, there are some developments in the field of the travelling wave solutions as well as its applications, so it will enhance the readability and comprehension of the manuscript (for details, see (Chen et al., 2019; Dai et al., 2019; Dai et al., 2020; Wang et al., 2018) and references therein).

2. Preliminaries

Throughout this entire paper, we use the following assumptions:

• M: Real Banach space;

• $\left|\right|.\left|\right|:$ norm on a Banach space;

• $B(y_0,\sigma ):$ closed ball having y₀ as a center with radius $\sigma ;$

• coZ: convex hull of a set Z;

• $co\overline{Z}:$ closed convex hull of a set Z;

• E_M: set of all bounded subsets of a space M;

• N_M: set of all relatively compact subsets of a space M;

Definition 2.1

(Banas and Goebel, 1980). Assume $Y\in E_M$ and $$\mu (Y)=\inf \left\{ϵ>0:Y=\underset{j=1}{\overset{n}{\cup }}{Y}_j \text{with diam}{\mathrm{Y}}_{\mathrm{j}}\le ϵ,\mathrm{j}=1,2,\dots ,\mathrm{n}\right\},$$ where $$\text{diam} Y=\sup \left\{\right||\alpha -\beta ||:\alpha ,\beta \in Y\}.$$

Hence, $0\le \mu (Y)<\infty ,$ where $\mu (Y)$ is called the Kuratowski MNC.

Theorem 2.1.

Assume $Y,Z\in E_M$ and $\lambda \in \mathbb{R}$. Then,

1. $\mu (Y)=0$ if and only if $Y\in N_M;$

2. $Y\subseteq Z$ $\Rightarrow$ $\mu (Y)\le \mu (Z);$

3. $\mu (\overline{Y})=\mu (\mathit{C}\mathit{\text{onv}}Y)=\mu (Y);$

4. $\mu (Y\cup Z)=\max \{\mu (Y),\mu (Z)\};$

5. $\mu (\lambda Y)=\left|\lambda \right|\mu (Y),$ where $\lambda Y=\{\lambda y:y\in Y\};$

6. $\mu (Y+Z)\le \mu (Y)+\mu (Z),$ where $Y+Z=\{y+z:y\in Y,z\in Z\};$

7. $|\mu (Y)-\mu (Z)|\le 2d(Y,Z),$ where $d(Y,Z)$ denotes the Hausdorff metric of Y and Z, i.e. $$d(Y,Z)=\max \left\{\underset{z\in Z}{\sup }d(z,Y),\underset{y\in Y}{\sup }d(y,Z)\right\},$$

here $d(.)$ is the distance from an element M to a set of M.

Theorem 2.2

(Banas and Goebel, 1980). Let G be a nonempty, bounded, closed and convex subset of M and let $D:G\to G$ be continuous mapping such that there exists a constant $\tilde{k}\in (0,1)$, with $$\mu (DY)\le \tilde{k}\mu (Y)$$ for any subset of Y of G. Then D has a fixed point in G.

Now, we discuss on $C[0,b],$ which contains set of all real continuous functions defined on the interval $[0,b]$ with the standard norm $$\left|\right|y\left|\right|=\sup \left\{\right|y(\zeta )|:\zeta \in [0,b\left]\right\}.$$

Clearly, $C[0,b]$ has also the structure of Banach algebra. Now, we will focus on a regular MNC defined in Banas and Lecko (2002). We fix a set $Y\in M_{C[0,b]}.$ For $y\in Y$ and given $ϵ>0$ denote by $\omega (y,ϵ)$ the modulus of continuity of y, i.e., $$\omega (y,ϵ)=\sup \left\{\right|y(\zeta )-y({\zeta }_1)|:\zeta ,{\zeta }_1\in [0,b],|\zeta -{\zeta }_1|\le ϵ\}.$$

Further, $$\omega (Y,ϵ)=\sup \{\omega (y,ϵ):y\in Y\},$$ $${\omega }_0(Y)=\underset{ϵ\to 0}{\lim }\omega (Y,ϵ).$$

Thus ${\omega }_0(Y)$ is a regular MNC in $C[0,b].$

Theorem 2.3

(Banas and Lecko, 2002). Suppose that G is a bounded, convex and closed subset of $C[0,b]$ and Q, H be the operators which transform continuously the set G into $C[0,b]$ such that Q(G) and H(G) are bounded. Again, the operator $D=Q.H$ transforms G into itself. If the operators Q and H satisfy the Darbo’s condition on the set G with the constants D₁ and D₂, respectively, then the operator D satisfies the Darbo’s condition on G with the constant $$\left|\right|Q(G)\left|\right|D_2+\left|\right|H(G)\left|\right|D_1.$$

If $\left|\right|Q(G)\left|\right|D_2+\left|\right|H(G)\left|\right|D_1<1,$ then D will be called contraction with respect to the measure ${\omega }_0$ and has a fixed point in the set G.

3. Main result

Now, we will analyze the solvability of the FIE (1(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) ) under the following assumptions:

$(B_1)$ $f:[0,b]\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ and $F,L:[0,b]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ are continuous functions and $\exists$ the constants $\phi$ and $\psi \ge 0$ such that $$|f(\zeta ,0,0)|\le \phi ,$$ $$|F(\zeta ,0,0,0)|\le \psi ,$$ $$|L(\zeta ,0,0,0)|\le \psi .$$

$(B_2)$ There exists the continuous functions $d_i(i=1,2,..,8):[0,b]\to [0,b]$ such that $$|f(\zeta ,y_1,y_2)-f(\zeta ,z_1,z_2)|\le d_1(\zeta )|y_1-z_1|+d_2(\zeta )|y_2-z_2|,$$ $$|F(\zeta ,y_1,z_1,y_2)-F(\zeta ,y_3,z_2,y_4)|\le d_3(\zeta )|y_1-y_3|+d_4(\zeta )|z_1-z_2|+d_5(\zeta )|y_2-y_4|,$$ $$|L(\zeta ,y_1,z_1,y_2)-L(\zeta ,y_3,z_2,y_4)|\le d_6(\zeta )|y_1-y_3|+d_7(\zeta )|z_1-z_2|+d_8(\zeta )|y_2-y_4|,$$ for all $\zeta \in [0,b]$ and $y_1,y_2,y_3,y_4,z_1,z_2\in \mathbb{R}.$

$(B_3)$ $r=r(\zeta ,s,y(\varphi (s)),u=u(\zeta ,s,y(a(s)),p=p(\zeta ,s,y(c(s)),q=q(\zeta ,s,y(\xi (s)):[0,b]\times [0,b]\times \mathbb{R}\to \mathbb{R}.$ Moreover, the functions $\varphi ,a,c,\xi ,d$ and η convert continuously the interval $[0,b]$ into itself.

$(B_4)$ There is a non-negative constant K such that $$\max \{d_i(\xi )\}\le K,\text{for}\xi \in [0,b],$$ for $i=1,2\dots 8.$

$(B_5)$ There are non-negative constants l and m such that $$\begin{array}{c}|p(\zeta ,s,y(c(s))|\le l+m|y|,\\ |q(\zeta ,s,y(\xi (s))|\le l+m|y|,\\ |r(\zeta ,s,y(\varphi (s))|\le l+m|y|,\\ |s(\zeta ,s,y(a(s))|\le l+m|y|,\end{array}$$ for all $\zeta , s\in [0,b]$ and $y\in \mathbb{R}.$

$(B_6)$ $4\gamma \rho <1$ for $\gamma =3K+2Kbm$ and $\rho =\phi +2Kbl+\psi .$

Theorem 3.1.

Under the assumptions $(B_1)-(B_6)$ FIE(1) has at least one solution in $M=C[0,b].$

Proof.

Let the operators Q and H be defined on M by the formula: $$(Qy)(\zeta )=f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F\left(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }s(\zeta ,s,y(a(s))ds,y(d(\zeta ))\right),(Hy)(\zeta )=L\left(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))\right),$$

for $\zeta \in [0,b].$

From $(B_1)$ and $(B_3),$ we see that Q and H transform on M into itself. Now, we put $$Dy=(Qy)(Hy).$$

Clearly, D transforms M into itself. Now, fix $y\in M.$ Then, $$\begin{array}{c}|(Dy)(\zeta )|=|(Qy)(\zeta )|.|(Hy)(\zeta )|\\ =(|f(\zeta ,y(\zeta ),y(\theta (\zeta )))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s)))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s)))ds,y(d(\zeta )))|)\\ \times |L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s)))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s)))ds,y(\eta (\zeta )))|\\ \le (|f(\zeta ,y(\zeta ),y(\theta (\zeta )))-f(\zeta ,0,0)|+|f(\zeta ,0,0)|+|F(\zeta ,0,0,0)|\\ +|F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s)))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s)))ds,y(d(\zeta )))-F(\zeta ,0,0,0)|)\\ \times (|L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s)))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s)))ds,y(\eta (\zeta )))-L(\zeta ,0,0,0)|\\ +|L(\zeta ,0,0,0)|)\\ \le (d_1(\zeta )|y(\zeta )|+d_2(\zeta )|y(\theta (\zeta ))|+\phi +d_3(\zeta ){\int }_0^{\zeta }|r(\zeta ,s,y(\varphi (s)))|ds\\ d_4(\zeta ){\int }_0^{\zeta }|u(\zeta ,s,y(a(s)))|ds+d_5|(y(d(\zeta ))|+\psi )\\ \times (d_6(\zeta ){\int }_0^b|p(\zeta ,s,y(c(s)))|ds+d_7(\zeta ){\int }_0^b|q(\zeta ,s,y(\xi (s)))|ds+b_8(\zeta )|y(\eta (\zeta ))|+\psi )\\ \le (3K|\left|y\right||+\phi +2Kb(l+m\left|\right|y\left|\right|)+\psi )\\ \times (K|\left|y\right||+2Kb(l+m\left|\right|y\left|\right|)+\psi )\\ \le {((3K+2Kbm)|\left|y\right||+\phi +2Kbl+\psi )}^2.\end{array}$$

Taking $\gamma =3K+2Kbm$ and $\rho =\phi +2Kbl+\psi ,$ then we have (2) $$\left|\right|Qy\left|\right|\le \gamma \left|\right|y\left|\right|+\rho ,$$(2) (3) $$\left|\right|Hy\left|\right|\le \gamma \left|\right|y\left|\right|+\rho ,$$(3) (4) $$\left|\right|Dy\left|\right|\le {(\gamma |\left|y\right||+\rho )}^2,$$(4) for $y\in M.$

From (4(4) $$\left|\right|Dy\left|\right|\le {(\gamma |\left|y\right||+\rho )}^2,$$(4) ), we reduce the operator D maps the ball $B_{\sigma }\subset M$ into itself for ${\sigma }_1\le \sigma \le {\sigma }_2,$ where $${\sigma }_1=\frac{(1-2\gamma \rho )-\sqrt{1-4\gamma \rho }}{2{\gamma }^2},$$ $${\sigma }_2=\frac{(1-2\gamma \rho )+\sqrt{1-4\gamma \rho }}{2{\gamma }^2}.$$

Also, from the estimates (2(2) $$\left|\right|Qy\left|\right|\le \gamma \left|\right|y\left|\right|+\rho ,$$(2) ) and (3(3) $$\left|\right|Hy\left|\right|\le \gamma \left|\right|y\left|\right|+\rho ,$$(3) ), it follows that (5) $$\left|\right|Q{B}_{\sigma }\left|\right|\le \gamma \sigma +\rho ,$$(5) (6) $$\left|\right|H{B}_{\sigma }\left|\right|\le \gamma \sigma +\rho .$$(6)

Next, we show that Q is continuous on the ball $B_{\sigma }.$ To do this, fix $ϵ>0$ and arbitrary $y,z\in B_{\sigma }$ such that $\left|\right|y-z\left|\right|\le ϵ.$ Then, for $\zeta \in [0,b],$ we get

$$\begin{array}{c}|(Qy)(\zeta )-(Qz)(\zeta )|=|f(\zeta ,y(\zeta ),y(\theta (\zeta )))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta )))\\ -f(\zeta ,z(\zeta ),z(\theta (\zeta ))-F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,z(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,z(a(s))ds,z(d(\zeta )))|\\ \le d_1(\zeta )|y(\zeta )-z(\zeta )|+d_2(\zeta )|y(\theta (\zeta ))-z(\theta (\zeta ))|\\ +|F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta )))\\ -F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,z(a(s))ds,z(d(\zeta )))\\ +F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,z(a(s))ds,z(d(\zeta )))\\ -F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,z(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,z(a(s))ds,z(d(\zeta )))|\\ \le d_1(\zeta )|y(\zeta )-z(\zeta )|+d_2(\zeta )|y(\theta (\zeta )-z(\theta (\zeta )|\\ +d_4(\zeta ){\int }_0^{\zeta }|u(\zeta ,s,y(a(s))-u(\zeta ,s,z(a(s))|ds+d_5(\zeta )|y(d(\zeta ))-z(d(\zeta ))|\\ +d_3(\zeta )|{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds-{\int }_0^{\zeta }r(\zeta ,s,z(\varphi (s))ds|\\ \le 3K\left|\right|y-z\left|\right|+Kb\omega (u,ϵ)+Kb\omega (r,ϵ)\\ \le 3Kϵ+Kb(\omega (u,ϵ)+\omega (r,ϵ)),\end{array}$$ where$$\omega (u,ϵ)=\mathit{sup}\left\{\right|u(\zeta ,s,y)-u(\zeta ,s,z)|:\zeta ,s\in [0,b];y,z\in [-\sigma ,\sigma ];||y-z||\le ϵ\},$$$$\omega (r,ϵ)=\mathit{sup}\left\{\right|r(\zeta ,s,y)-r(\zeta ,s,z)|:\zeta ,s\in [0,b];y,z\in [-\sigma ,\sigma ];||y-z||\le ϵ\}.$$

The function $r=r(\zeta ,s,y)$ and $u=u(\zeta ,s,y)$ are uniform continuous on the bounded subset $[0,b]\times [0,b]\times [-\sigma ,\sigma ],$ then $\omega (r,ϵ)$ and $\omega (u,ϵ)\to 0$ as $ϵ\to 0.$ Thus, Q is continuous on $B_{\sigma }.$ Similarly, H is also continuous on $B_{\sigma }.$ Hence, D is a continuous operator on $B_{\sigma }.$

Now, we will show that the Q and H satisfy the Darbo’s condition in the ball $B_{\sigma }.$ Assume that a non empty subset Z of $B_{\sigma }$ and $y\in Z.$ Let $ϵ>0$ be fixed and ${\zeta }_1,{\zeta }_2\in [0,b]$ such that ${\zeta }_1\le {\zeta }_2$ and ${\zeta }_1-{\zeta }_2\le ϵ.$ Then, we obtain $$\begin{array}{c}|(Qy)({\zeta }_2)-(Qy)({\zeta }_1)|=|f({\zeta }_2,y({\zeta }_2),y(\theta ({\zeta }_2))\\ +F\left({\zeta }_2,{\int }_0^{{\zeta }_2}r({\zeta }_2,s,y(\varphi (s))ds,{\int }_0^{{\zeta }_2}u({\zeta }_2,s,y(a(s))ds,y(d({\zeta }_2))\right)\\ -f({\zeta }_1,y({\zeta }_1),y(\theta ({\zeta }_1))\\ -F\left({\zeta }_1,{\int }_0^{{\zeta }_1}r({\zeta }_1,s,y(\varphi (s))ds,{\int }_0^{{\zeta }_1}u({\zeta }_1,s,y(a(s))ds,y(d({\zeta }_1))\right)|\\ \le |f({\zeta }_2,y({\zeta }_2),y(\theta ({\zeta }_2))-f({\zeta }_2,y({\zeta }_1),y(\theta ({\zeta }_1))|\\ +|f({\zeta }_2,y({\zeta }_1),y(\theta ({\zeta }_1))-f({\zeta }_1,y({\zeta }_1),y(\theta ({\zeta }_1))|\\ +|F\left({\xi }_2,{\int }_0^{{\xi }_2}r({\xi }_2,s,y(\varphi (s))ds,{\int }_0^{{\zeta }_2}u({\zeta }_2,s,y(a(s))ds,y(d({\zeta }_2))\right)\\ -F\left({\zeta }_2,{\int }_0^{{\zeta }_1}r({\zeta }_1,s,y(\varphi (s))ds{\int }_0^{{\zeta }_1}u({\zeta }_1,s,y(a(s))ds,y(d({\zeta }_1))\right)|\end{array}$$ $$\begin{array}{c}+|F\left({\zeta }_2,{\int }_0^{{\zeta }_1}r({\zeta }_1,s,y(\varphi (s))ds,{\int }_0^{{\zeta }_1}u({\zeta }_1,s,y(a(s))ds,y(d({\zeta }_1))\right)\\ -F\left({\zeta }_1,{\int }_0^{{\zeta }_1}r({\zeta }_1,s,y(\varphi (s))ds,{\int }_0^{{\zeta }_1}u({\zeta }_1,s,y(a(s))ds,y(d({\zeta }_1))\right)|\\ \le d_1(\zeta )|y({\zeta }_2)-y({\zeta }_1)|+d_2(\zeta )|y(\theta ({\zeta }_2))-y(\theta ({\zeta }_1))|+\omega (f,ϵ)\\ +d_3(\zeta )|{\int }_0^{{\zeta }_2}r({\zeta }_2,s,y(\varphi (s))ds-r({\zeta }_1,s,y(\varphi (s))ds|\\ +d_4(\zeta )|{\int }_0^{{\zeta }_2}u({\zeta }_2,s,y(a(s))ds-{\int }_0^{{\zeta }_1}u({\zeta }_1,s,y(a(s))ds|\\ d_5(\zeta )|y(d({\zeta }_2)-y(d({\zeta }_1))|+{\omega }_F(ϵ)\\ \le K\omega (y,ϵ)+K\omega (y,\omega (\theta ,ϵ))+{\omega }_f(ϵ)\\ +K\omega (y,\omega (d,ϵ))+{\omega }_F(ϵ)+K\{{\int }_0^{{\zeta }_2}|r({\zeta }_2,s,y(\varphi (s))-r({\zeta }_1,s,y(\varphi (s))|ds\\ +{\int }_{{\zeta }_1}^{{\zeta }_2}|r({\zeta }_2,s,y(\varphi (s))|ds\}+K\{{\int }_0^{{\zeta }_2}|u({\zeta }_2,s,y(a(s))-u({\zeta }_1,s,y(a(s))|ds\\ +{\int }_{{\zeta }_1}^{{\zeta }_2}|u({\zeta }_2,s,y(a(s))|ds\}\end{array}$$

and (7) $$\begin{array}{c}\omega (Qy,ϵ)\le K\omega (y,ϵ)+K\omega (y,\omega (\theta ,ϵ))+{\omega }_f(ϵ)+K\omega (y,\omega (d,ϵ))+{\omega }_F(ϵ)\\ +K\{{\omega }_r(ϵ)b+K_1ϵ\}+K\{{\omega }_u(ϵ)b+K_1ϵ\},\end{array}$$(7) where $$\begin{array}{c}{\omega }_f(ϵ,..)=\mathit{sup}\left\{\right|f(\zeta ,y_1,y_2)-f({\zeta }_1,y_1,y_2)|:\zeta ,{\zeta }_1\in [0,b];|\zeta -{\zeta }_1|\le ϵ:y_1,y_2\in [-\sigma ,\sigma \left]\right\},\\ {\omega }_r(ϵ,..)=\mathit{sup}\left\{\right|r(\zeta ,s,y)-r({\zeta }_1,s,y)|:\zeta ,{\zeta }_1\in [0,b];|\zeta -{\zeta }_1|\le ϵ:y\in [-\sigma ,\sigma \left]\right\},\\ {\omega }_u(ϵ,..)=\mathit{sup}\left\{\right|u(\zeta ,s,y)-u({\zeta }_1,s,y)|:\zeta ,{\zeta }_1\in [0,b];|\zeta -{\zeta }_1|\le ϵ:y\in [-\sigma ,\sigma \left]\right\},\\ {\omega }_F(ϵ,..)=\mathit{sup}\left\{\right|F(\zeta ,y_1,z_1,y_2)-F({\zeta }_1,y_1,z_1,y_2)|:\zeta ,{\zeta }_1\in [0,b],\\ |\zeta -{\zeta }_1|\le ϵ:y_1,y_2\in [-\sigma ,\sigma ];z_1\in [-K_{1}b,K_{1}b]\},\\ K_1=\mathit{sup}\left\{\right|r(\zeta ,s,y)|,|u(\zeta ,s,y)|:\zeta ,s\in [0,b];\in [-\sigma ,\sigma \left]\right\}.\end{array}$$

In view of our assumptions we deduce that the functions $f=f(\zeta ,y_1,y_2)$ and $F=F(\zeta ,y_1,z_1,y_2)$ are uniform continuous on $[0,b]\times \mathbb{R}\times \mathbb{R}$ and $[0,b]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R},$ respectively and the functions $r=r(\zeta ,s,y)$ and $u=u(\zeta ,s,y)$ are uniform continuous on $[0,b]\times [0,b]\times \mathbb{R}.$ Hence, we infer that ${\omega }_f(ϵ\dots )\to 0,{\omega }_r(ϵ\dots )\to 0,{\omega }_u(ϵ\dots )\to 0$ and ${\omega }_F(ϵ\dots )\to 0$ as $ϵ\to 0.$ Thus, we get (8) $${\omega }_0(QY)\le 3K{\omega }_0(Y).$$(8)

Similarly, it is obtained that (9) $${\omega }_0(HY)\le K{\omega }_0(H).$$(9)

Finally, it follows that D satisfies the Darbo’s condition on $B_{\sigma }$ with respect to the measure ω₀ with constant $(\gamma \sigma +\rho )3K+(\gamma \sigma +\rho )K.$ Now, we have $$\begin{array}{c}(\gamma \sigma +\rho )3K+(\gamma \sigma +\rho )K=4K(\gamma \sigma +\rho )\\ =4K(\gamma {\sigma }_1+\rho )\\ =4K\left(\gamma \left(\frac{(1-2\gamma \rho )-\sqrt{1-4\gamma \rho }}{2{\gamma }^2}\right)+\rho \right)\\ =4K\left(\frac{1-\sqrt{1-4\gamma \rho }}{2\gamma }\right)\\ <1.\end{array}$$

Hence, D is a contraction on $B_{\sigma }$ with respect to ${\omega }_0.$ Consequently, we conclude that the nonlinear FIE (1(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) ) has at least one solution in ball $B_{\sigma }.$ □

4. Applications

Our proposed functional integral equation contains several integral equations, considered by several authors as a special case.

• If $f(\zeta ,y_1,y_2)=f(\zeta ,y_1),F(\zeta ,y_1,z,y_2)=F(\zeta ,z,y_2)$ and $L(\zeta ,y_1,z,y_2)=L(\zeta ,z,y_2),$ then equation reduces to the following FIE, which was studied in Deepmala and Pathak (2013a). $$\begin{array}{c}y(\zeta )=\left(f(\zeta ,y(\zeta ))+F\left(\zeta ,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))\right)\right)\\ \times L\left(\zeta ,{\int }_0^{b}q(\zeta ,\vartheta ,y(\xi (\vartheta )),y(\eta (\zeta ))\right).\end{array}$$

• Taking $f(\zeta ,y_1,y_2)=f(\zeta ,y_1),F(\zeta ,y_1,z,y_2)=F(\zeta ,z,y_2)$ and $L(\zeta ,y_1,z,y_2)=1,$ then FIE(1(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) ) is converted into the following form which has been studied in Maleknejad et al. (2009a). $$y(\zeta )=f(\zeta ,y(\zeta ))+F\left(\zeta ,{\int }_0^{\zeta }u(\zeta ,s,y(s))ds,y(d(\zeta ))\right).$$

• On putting $f(\zeta ,y_1,y_2)=0,F(\zeta ,y_1,z,y_2)=F(\zeta ,z,y_2)$ and $L(\zeta ,y_1,z,y_2)=L(\zeta ,z,y_2),$ we obtain the following FIE studied in Banaś and Sadarangani (2003) and Maleknejad et al. (2008). $$\begin{array}{c}y(\zeta )=F\left(\zeta ,{\int }_0^{\zeta }u(\zeta ,s,y(s))ds,y(d(\zeta ))\right)\\ \times L\left(\zeta ,{\int }_0^{b}q(\zeta ,s,y(s))ds,y(\eta (\zeta ))\right).\end{array}$$

• If $f(\zeta ,y_1,y_2)=f(\zeta ,y_1),F(\zeta ,y_1,z,y_2)=z$ and $L(\zeta ,y_1,z,y_2)=1,$ then we get the following FIE studied in Banaś and Rzepka (2003). $$y(\zeta )=f(\zeta ,y(\zeta ))+{\int }_0^{\zeta }u(\zeta ,s,y(s))ds.$$

• Taking $f(\zeta ,y_1,y_2)=0,L(\zeta ,y_1,z,y_2)=1$ and $F(\zeta ,y_1,z,y_2)=f(\zeta ,y)z,$ then FIE (1(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) ) has the following form studied in Maleknejad et al. (2009b)). $$y(\zeta )=f(\zeta ,y(\zeta )){\int }_0^{\zeta }u(\zeta ,s,y(s))ds.$$

• On putting $f(\zeta ,y_1,y_2)=0,L(\zeta ,y_1,z,y_2)=1$ and $F(\zeta ,y_1,z,y_2)=b(\zeta )+z,$ then, we get following non-linear Volterra integral equation (Corduneanu, 1990). $$y(\zeta )=b(\zeta )+{\int }_0^{\zeta }u(\zeta ,s,y(s))ds.$$

• Taking $f(\zeta ,y_1,y_2)=0,F(\zeta ,y_1,z,y_2)=1$ and $L(\zeta ,y_1,z,y_2)=v(\zeta )+z,$ then we obtain Urysohn integral equation (Corduneanu, 1990). $$y(\zeta )=v(\zeta )+{\int }_0^{b}q(\zeta ,s,y(s))ds.$$

• If $f(\zeta ,y_1,y_2)=0,F(\zeta ,y_1,z,y_2)=1,L(\zeta ,y_1,z,y_2)=1+yz$ and $q(\zeta ,y)=\frac{\zeta }{\zeta +s}\varphi (\zeta )y,\beta (\zeta )=\zeta ,$ then FIE(1(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) ) has the following form $$y(\zeta )=1+y(\zeta ){\int }_0^b\frac{\zeta }{\zeta +s}\varphi (s)y(s)ds.$$

The above integral equation is the famous quadratic integral equation of Chandrasekhar type (Chandrasekhar, 1950).

Example 4.1.

Consider the following non-linear FIE: (10) $$\begin{array}{cc}\hfill & y(\xi )=[\frac{\zeta }{8(1+{\zeta }^2)}ln(1+|y(1-{\zeta }^2)|)+\frac{\zeta }{17+{\zeta }^2}a\mathit{rctan}(|y(\zeta )|)\hfill \\ \hfill & +\frac{1}{13}{\int }_0^{\zeta }(\frac{\zeta }{3+\zeta }\mathit{\text{sin}}\hspace{0.17em}(1+y(s))+(1+\zeta )\mathit{arctan}\frac{|y(s)|}{2+|y(s)|}ds)\hfill \\ \hfill & +\frac{1}{15}{\int }_0^{\zeta }(\frac{\zeta }{2(1+{\zeta }^2)}\mathit{\text{sin}}\hspace{0.17em}\frac{y{s}^{\frac{1}{2}}}{1+y{s}^{\frac{1}{2}}}+3ln(1+|y{s}^{\frac{1}{2}}|))ds]\hfill \\ \hfill & \times [\frac{1}{18}{\int }_0^1(\frac{1}{2(1+{\zeta }^3}\mathit{\text{sin}}\hspace{0.17em}\frac{y^2(t)}{1+y^2(s)}+3\xi \mathit{arctan}{y}^2(s))ds\hfill \\ \hfill & +\frac{1}{17}{\int }_0^1(\frac{2\zeta }{1+{\zeta }^2}\mathit{\text{sin}}\hspace{0.17em}(|1+y(s^2)|)+(2+\zeta )\mathit{arctan}\frac{s^2(y(s^2)}{1+y(s^2)})ds],\hfill \end{array}$$(10) where $\xi \in [0,1].$

The FIE (10(10) $$\begin{array}{cc}\hfill & y(\xi )=[\frac{\zeta }{8(1+{\zeta }^2)}ln(1+|y(1-{\zeta }^2)|)+\frac{\zeta }{17+{\zeta }^2}a\mathit{rctan}(|y(\zeta )|)\hfill \\ \hfill & +\frac{1}{13}{\int }_0^{\zeta }(\frac{\zeta }{3+\zeta }\mathit{\text{sin}}\hspace{0.17em}(1+y(s))+(1+\zeta )\mathit{arctan}\frac{|y(s)|}{2+|y(s)|}ds)\hfill \\ \hfill & +\frac{1}{15}{\int }_0^{\zeta }(\frac{\zeta }{2(1+{\zeta }^2)}\mathit{\text{sin}}\hspace{0.17em}\frac{y{s}^{\frac{1}{2}}}{1+y{s}^{\frac{1}{2}}}+3ln(1+|y{s}^{\frac{1}{2}}|))ds]\hfill \\ \hfill & \times [\frac{1}{18}{\int }_0^1(\frac{1}{2(1+{\zeta }^3}\mathit{\text{sin}}\hspace{0.17em}\frac{y^2(t)}{1+y^2(s)}+3\xi \mathit{arctan}{y}^2(s))ds\hfill \\ \hfill & +\frac{1}{17}{\int }_0^1(\frac{2\zeta }{1+{\zeta }^2}\mathit{\text{sin}}\hspace{0.17em}(|1+y(s^2)|)+(2+\zeta )\mathit{arctan}\frac{s^2(y(s^2)}{1+y(s^2)})ds],\hfill \end{array}$$(10) ) is a special case of equation (1)(1) $$\begin{array}{c}y(\zeta )=(f(\zeta ,y(\zeta ),y(\theta (\zeta ))+F(\zeta ,{\int }_0^{\zeta }r(\zeta ,s,y(\varphi (s))ds,{\int }_0^{\zeta }u(\zeta ,s,y(a(s))ds,y(d(\zeta ))))\times L(\zeta ,{\int }_0^{b}p(\zeta ,s,y(c(s))ds,{\int }_0^{b}q(\zeta ,s,y(\xi (s))ds,y(\eta (\zeta ))),\end{array}$$(1) . Here $$\begin{array}{c}f(\zeta ,y_1,y_2)=\frac{\zeta }{8(1+{\zeta }^2)}ln(1+|y(1-{\zeta }^2)|)+\frac{\zeta }{17+{\zeta }^2}a\mathit{rctan}(|y(\zeta )|),\\ F(\zeta ,y_1,z{,}_1,y_2)=\frac{1}{13}{y}_1+\frac{1}{15}{z}_1,L(t,y_1,z_1,y_2)=\frac{1}{18}{y}_1+\frac{1}{17}{z}_1,\\ r(\zeta ,s,y)=\frac{\zeta }{3+\zeta }\mathit{\text{sin}}\hspace{0.17em}(1+y(s))+(1+s)\mathit{arctan}\frac{|y(s)|}{2+|y(s)|},\\ u(\zeta ,s,y)=\frac{\zeta }{2(1+{\zeta }^2)}\mathit{\text{sin}}\hspace{0.17em}\frac{y{t}^{\frac{1}{2}}}{1+y{s}^{\frac{1}{2}}}+3ln(1+|y{s}^{\frac{1}{2}}\left|\right),\\ p(\zeta ,s,y)=\frac{1}{2(1+{\zeta }^3}\mathit{\text{sin}}\hspace{0.17em}\frac{y^2(s)}{1+y^2(s)}+3\zeta \mathit{arctan}{y}^2(s),\\ q(\zeta ,s,y)=\frac{2\zeta }{1+{\zeta }^2}\mathit{\text{sin}}\hspace{0.17em}(|1+y(s^2)|)+(2+\zeta )\mathit{arctan}\frac{s^2(y(s^2)}{1+y(s^2)}.\end{array}$$

Now, we show that all the assumptions of Theorem 3.1 are satisfied. It is obvious that $d_1=\frac{1}{16},d_2=\frac{1}{18},d_3=\frac{1}{13},d_4=\frac{1}{15}{d}_6=\frac{1}{18},d_7=\frac{1}{17},d_5=d_8=0.$ In this case, $K=\mathit{max}\{\frac{1}{16},\frac{1}{18},\frac{1}{13},\frac{1}{15},\frac{1}{17},0\}=\frac{1}{13}.$

Moreover, it is obvious that $$|f(\zeta ,0,0)|=0,|F(\zeta ,0,0,0)|=0,|L(\zeta ,0,0,0)|=0,|r(\zeta ,s,y)|=\frac{1}{4}+3\left|y\right|,|u(\zeta ,s,y)|=\frac{1}{4}+3\left|y\right|,$$ $$|p(\zeta ,s,y)|=\frac{1}{4}+3\left|y\right|,|q(\zeta ,s,y)|=\frac{1}{4}+3\left|y\right|.$$

We also have $s=\psi =0,$ $l=\frac{1}{4},m=3$ and b = 1.

Finally, we see that $$4\rho \gamma =4(3K+2Kbm)(\phi +2Kbl+\psi )<1.$$

Hence, all assumptions $(B_1)$ to $(B_6)$ are satisfied. The FIE (10(10) $$\begin{array}{cc}\hfill & y(\xi )=[\frac{\zeta }{8(1+{\zeta }^2)}ln(1+|y(1-{\zeta }^2)|)+\frac{\zeta }{17+{\zeta }^2}a\mathit{rctan}(|y(\zeta )|)\hfill \\ \hfill & +\frac{1}{13}{\int }_0^{\zeta }(\frac{\zeta }{3+\zeta }\mathit{\text{sin}}\hspace{0.17em}(1+y(s))+(1+\zeta )\mathit{arctan}\frac{|y(s)|}{2+|y(s)|}ds)\hfill \\ \hfill & +\frac{1}{15}{\int }_0^{\zeta }(\frac{\zeta }{2(1+{\zeta }^2)}\mathit{\text{sin}}\hspace{0.17em}\frac{y{s}^{\frac{1}{2}}}{1+y{s}^{\frac{1}{2}}}+3ln(1+|y{s}^{\frac{1}{2}}|))ds]\hfill \\ \hfill & \times [\frac{1}{18}{\int }_0^1(\frac{1}{2(1+{\zeta }^3}\mathit{\text{sin}}\hspace{0.17em}\frac{y^2(t)}{1+y^2(s)}+3\xi \mathit{arctan}{y}^2(s))ds\hfill \\ \hfill & +\frac{1}{17}{\int }_0^1(\frac{2\zeta }{1+{\zeta }^2}\mathit{\text{sin}}\hspace{0.17em}(|1+y(s^2)|)+(2+\zeta )\mathit{arctan}\frac{s^2(y(s^2)}{1+y(s^2)})ds],\hfill \end{array}$$(10) ) has at least one solution in $C[0,1].$

5. Conclusion

Integral equations represent an important field in the area of applied mathematics and a powerful tool for modeling diverse problems arising in all areas of scientific research. Our result contains outcome of several research papers as a particular case. These result may be further extended for the developments in the field of the traveling wave solutions as well as its applications (Dhage, 1994; Hu et al., 1989; Kelly, 1982).

Disclosure statement

No potential conflict of interest was reported by the authors.

Additional information

Funding

The research work of the first author is thankful to the CSIR JRF Fellowship under the Government of India, Program No. 09/1174(0003)/2017-EMR-1, CSIR New Delhi.