# Existence of Unique Solution of Urysohn and Fredholm Integral Equations in Complex Double Controlled Metric Type Spaces

Authors: Sahibzada Waseem Ahmad; Muhammad Sarwar; Gul Rahmat; Fahd Jarad
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

Complex Urysohn integral equations and complex Fredholm integral equation of the second kind have intensified the attention of appreciable researchers to their solution due to their comprehensive applications. This study is devoted to the existence and uniqueness of solution to these integral equations in the setting of complete complex double controlled metric spaces via fixed point theory. For this motive, a fixed point result together with a numerical example for the convergence behavior of operator to the fixed point is analyzed in the context of the quoted metric spaces.


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## Body

## 1. Introduction and Preliminaries

A self-mapping℘:Ξ⟶Ξ, where Ξ≠∅, is called a contraction if ∃κ∈0,1, such that(1)∂℘a,℘b≤κ∂a,b,∀a,b∈Ξ,where Ξ,∂ is a metric space. Furthermore, if the metric space Ξ,∂ is complete and satisfies the inequality (1), then in such case, ℘ possesses a unique fixed point. Inequality (1) is often used to define contractivity, which implies the continuity of ℘ and is known as the Banach contraction principle. The generalization of Banach contraction principle stated by means of contraction mapping is one of the active areas of research for the practitioners of nonlinear phenomenon. Among these generalizations, weakly contractive mappings are one of the eminent results from scholars, which can be found in the articles [1, 2]. Moreover, Alqahtani et al. [3] have proved a result which is a generalization of Das and Gupta [4] and Jaggi [5], in which the rational type contractions were considered. Those considered rational type contractions are themselves, in fact, generalizations of Banach contraction principle. For more studies, refer to [6–11].Apart from the generalization of contractive conditions, researchers have dedicated plentiful time to the solution of diversified amount of equations via the fixed point method. Such generalizations include solving linear system of equations, differential equations, fractional differential and integral equations, etc. Abdeljawad et al. generalized the double controlled metric spaces [12] to a complex double controlled metric spaces [13], in which a contraction theorem was presented. Furthermore, the solutions to complex Riemann–Liouville fractional operator, Atangana–Baleanu fractional integral operator, and nonlinear telegraph equation were obtained via the fixed point method. For further study, we encourage readers to study [14].Integral equations play a key role in radioactive heat transfer problems, electricity and magnetism, kinetic theory of gases, mathematical economics, mathematical problems of radioactive equilibrium, quantum mechanics, fluid mechanics, optimization, optimal controlled systems, communication theory, radiation, population genetics, potential theory, queuing theory, geophysics, medicine, the particle transport problems of astrophysics, nuclear reactor theory, renewal theory, acoustics, fluid mechanics, hereditary phenomena in physics and biology, steady-state heat conduction, fracture mechanics, and continuum mechanics. Therefore, researchers have dedicated a lot of attention towards solving these equations (for instance, see [15–19] and references therein). Urysohn integral equations occupy comparable priority in the diversified area of research in multiplex subjects like mathematics, biological sciences, physical sciences, and so on. The Urysohn integral equations are defined as(2)δθ=ρθ+∫xyΔ1θ,ε,δεdε,ζθ=ρθ+∫xyΔ2θ,ε,ζεdε,where ρθ is a continuous function on the interval x,y, Δ1θ,ε,δε and Δ2θ,ε,ζε are kernels defined on the complex region, and Rx,y=θ,ε:x≤θ≤y,x≤ε≤y.The solution of Urysohn integral equations has been handled via distinctive approaches such as [20, 21] and references therein. One of the approaches is the fixed point approach (several articles can be found on such topic, for instance, see [22, 23] and references therein).Likewise, Fredholm integral equation is one of the significant integral equations and is given by(3)Λς=ℏς+α∫cdΩς,t,Λtdt,where ℏς is a complex continuous function on the given interval α≠0 and is a complex parameter, the kernel Ως,t,Λt=∑j=1ncjςbjt¯ is a known continuous separable kernel function on a complex plane, and Sc,d=ς,t:c≤ς≤d,c≤t≤d. It is always assumed that functions cjς and djt are complex-valued. In addition, the sets cjς and djt are linearly independent on the interval c,d, and c,d are the real limits of the integral. Similarly, the applications of Fredholm integral equation can be found in the aforementioned fields of the research, which highlight the dominance of integral (3) in the area of research. Therefore, the authors conceive to find the solution of (3) via variant strategies. Similarly, the solution of Fredholm integral equation is managed in distinctive methods. An illustration of some of those methods has been analyzed in [24–26]. One of the methods to solve integral (3) is the B-spline wavelet method by Maleknejad and Nosrati Sahlan [27]. Similarly, He proposed a variational iteration method [28–30] to solve integral (3). The Adomian decomposition method and various numerical methods have also been applied to obtain the solution to (3). Fixed point theory is an alternative supporting approach to solve (3). Numerous researchers have followed this method. For further study, one can see [31–33].As mentioned earlier, there are diverse methodologies to solve an integral equation. The fixed point technique is among the various other techniques. In the current study, we have extended the result proven in [3] to common fixed point weak contraction in the setting of complete complex double controlled metric type spaces with a numerical experiment. Moreover, we apply our proven result to obtain sufficient conditions for the existence and uniqueness of complex-valued Urysohn integral equations and Fredholm integral equation.Theorem 1 (see [3]).
LetΞ,∂ be a complete extended b-metric space and ℘:Ξ⟶Ξ be a continuous self-mapping such that for all a≠b∈Ξ,(4)∂℘a,℘b≤κMa,b,where κ∈0,1 and(5)Ma,b=max∂a,b,∂b,℘b1+∂a,℘a1+∂a,b,∂a,℘a1+∂b,℘b1+∂a,b,∂b,℘b∂a,℘a∂a,b.Consider that for eacha∈Ξ and limm,n⟶∞ϑan,am<1/κ, an=℘na0, where a0∈Ξ for all n∈ℕ. Then, ℘ has a fixed point say h. In addition, for a∈Ξ, we have ℘na⟶h.In [13], Panda et al. proposed a more generalized structure of double controlled metric spaces, i.e., complex double controlled metric spaces, where a couple of fixed point results are proved. In addition, the solutions to a couple of fractional operators and telegraph equation have been investigated via the fixed point technique. To achieve the main objectives of this work, we need to revisit some basic concepts from literature.Definition 1.
Consider the set of complex numbersℂ and suppose ζ1,ζ2,ζ3∈ℂ, are complex numbers. A partial order ≺¯ on ℂ can be defined as(6)ζ1<¯ζ2⇔Reζ1≤Reζ2andImζ1≤Imζ2.
Therefore, it can be inferred thatζ1≺¯ζ2, if one of the given property holds.(1)
Reζ1=Reζ2 and Imζ1=Imζ2.(2)
Reζ1=Reζ2 and Imζ1=Imζ2.(3)
Reζ1=Reζ2 and Imζ1=Imζ2.(4)
Reζ1=Reζ2 and Imζ1=Imζ2.
The notation can be listed asζ1⋩ζ2, if ζ1≠ζ2, along with property (4) to be satisfied by considering the equivalent fashion, i.e.,(1)
0≺¯ζ1⋩ζ2⇒ζ1<ζ2.(2)
ζ1≺¯ζ2 and ζ2≺¯ζ3⇒ζ1≤ζ3.(3)
0≺¯ζ1≺¯ζ2⇒ζ1≤ζ2.(4)
If0≤a≤b∈ℝ and ζ1≺¯ζ2⇒aζ1≺¯bζ2∀ζ1,ζ2∈ℂ.Definition 2.
LetΞ be a nonempty set and ∂:Ξ×Ξ⟶ℂ be a distance mapping, where ℂ is a complex-valued set. Moreover, consider the functions μ,ν:Ξ×Ξ⟶1,∞. A complex-valued double controlled metric can be defined as(1)
∂η,ζ=0; ⇔η=ζ.(2)
∂η,ζ=∂ζ,η.(3)
∂η,ς≺¯μη,ζ∂η,ζ+νζ,ς∂ζ,ς.
∀η,ζ,ς∈Ξ. The pair Ξ,∂ is called a complex-valued double controlled metric space.Table 1
Distance function.
∂η,ζ01/3100ι2+3ι1/3ι01+ι12+3ι1+ι0Example 1.
Consider a nonempty setΞ=0,1/3,1, and we define distances as follows (Table 1).
Lets define the functions,μ,ν:Ξ⟶1,∞ by (Table 2).
From the above data, it can be analyzed that(7)2+3ι=∂η,ζ=∂0,1≻μ0,13∂0,13+μ13,1∂13,1=32ι+541+ι=54+2.75ι,which shows that Ξ,∂ is not a complex controlled metric space and hence nor a complex extended b-metric space. On the other hand,(8)2+3ι=∂η,ζ=∂0,1≺μη,ζ∂η,ζ+νζ,ς∂ζ+ς=32ι+41+ι=4+5.5ι.
Similarly, all the remaining cases can be verified, which shows thatΞ,∂ is a complex double controlled metric space.Table 2
Controlled functions.
μη,ζ01/31013/221/33/217/6127/61υη,ζ01/3101231/32141341Definition 3.
(see [13]). Let Ξ,∂ be a complex double controlled metric type space and let ζn∈ℕ∈Ξ be a sequence and ζ∈Ξ.
For anyκ∈ℂ, there exists N∈ℕ such that ∀n>N,(1)
A sequenceζn∈ℕ is said to be convergent to ζ∈Ξ if ∂ζn,ζ≺κ; then, ζnn∈ℕ⟶ζ as n⟶∞. Mathematically, it can be expressed as limn⟶∞ζn=ζ.(2)
A sequenceζnn∈ℕ⟶ζ, is said to be Cauchy if for m>n>N, ∂ζn,ζ≺κ.(3)
The metric spaceΞ,∂ is said to be complete if every Cauchy sequence is convergent in Ξ.


## 2. Fixed Point Result

In this section, a fixed point result is established which generalizes the result in [3] to a common weak contraction in the sense of complete complex double controlled metric space.Theorem 2.
LetΞ,∂ be a complete complex double controlled metric space. Furthermore, let ℘,ℵ:Ξ⟶Ξ be two self-operators. Now for λ∈0,1 and L>0, such that for all distinct a,b∈Ξ, we have(9)∂℘a,ℵb≺¯λMa,b−LNa,b,where(10)Ma,b=max∂a,b,∂b,ℵb1+∂a,℘a1+∂a,b,∂a,℘a1+∂b,ℵb1+∂a,b,∂b,ℵb∂a,℘a∂a,b,−b±b2−4ac2a,Na,b=min∂a,℘a,∂b,ℵb,∂b,℘a.Moreover, forλ∈0,1 and for any ai∈Ξ,(11)supm≥1limi⟶∞μai+1,ai+2νai+1,a2m+1μai,ai+1<1λ.Additionally, for eacha∈Ξ, suppose that(12)limn⟶∞μa,a2n+1andlimn⟶∞μa2n+1,aexist and are finite.Then,℘,ℵ have a common fixed point.Proof.
Leta0∈Ξ be an arbitrary point. Suppose the sequence an=℘na0 satisfies the conditions of the theorem and is defined as(13)a2n+1=℘a2n,a2n+2=ℵa2n+1.
Consider(14)∂a2n+1,a2n+2=∂℘a2n,ℵa2n+1≺¯λMa2n,a2n+1−LNa2n,a2n+1,together with(15)Ma2n,a2n+1=max∂a2n,a2n+1,∂a2n+1,ℵa2n+11+∂a2n,℘a2n1+∂a2n,a2n+1,∂a2n,℘a2n1+∂a2n+1,ℵa2n+11+∂a2n,a2n+1,∂a2n+1,ℵa2n+1∂a2n,℘a2n∂a2n,a2n+1,=max∂a2n,a2n+1,∂a2n+1,a2n+21+∂a2n,a2n+11+∂a2n,a2n+1,∂a2n,a2n+11+∂a2n+1,a2n+21+∂a2n,a2n+1,∂a2n+1,a2n+2∂a2n,a2n+1∂a2n,a2n+1,from which it is concluded that(16)Ma2n,a2n+1=max∂a2n,a2n+1,∂a2n+1,a2n+2,∂a2n,a2n+11+∂a2n+1,a2n+21+∂a2n,a2n+1.
Now(17)Na2n,a2n+1=min∂a2n,℘a2n,∂a2n+1,ℵa2n+1,∂a2n+1,℘a2n,=min∂a2n,a2n+1,∂a2n+1,a2n+2,∂a2n+1,a2n+1.
From (17), we get that Na2n,a2n+1=∂a2n+1,a2n+1=0.
In order to proceed, consider the following cases from (16).Case 1.
Case1:If Ma2n,a2n+1=∂a2n+1,a2n+2, using (9) implies ∂a2n+1,a2n+2≺¯λ∂a2n+1,a2n+2−LNa2n,a2n+1; as L>0 and Na2n,a2n+1=0, we get ∂a2n+1,a2n+2≺¯λ∂a2n+1,a2n+2, which is an obvious contradiction as λ∈0,1.Case2:.
IfMa2n,a2n+1=∂a2n,a2n+11+∂a2n+1,a2n+2/1+∂a2n,a2n+1, then(18)max∂a2n,a2n+1,∂a2n+1,a2n+2≺¯∂a2n,a2n+11+∂a2n+1,a2n+21+∂a2n,a2n+1.
In order to conclude, consider the following subcases.
Case2i: if max∂a2n,a2n+1,∂a2n+1,a2n+2=∂a2n,a2n+1, which follows(19)∂a2n+1,a2n+2≺∂a2n,a2n+2,while from (18), we have(20)∂a2n,a2n+1≺¯∂a2n,a2n+11+∂a2n+1,a2n+21+∂a2n,a2n+1.
By simple calculation from (20), we get(21)∂a2n,a2n+1≺¯∂a2n+1,a2n+2,which is a contradiction to inequality (19).
(Case 2)ii: if max∂a2n,a2n+1,∂a2n+1,a2n+2=∂a2n+1,a2n+2, then(22)∂a2n,a2n+1≺∂a2n+1,a2n+2.
Moreover, from (18), it follows that(23)∂a2n+1,a2n+2≺∂a2n,a2n+11+∂a2n+1,a2n+21+∂a2n,a2n+1.
From inequality (23), it can be deduced that(24)∂a2n+1,a2n+2≺¯∂a2n,a2n+1,which is a contradiction to (22). Consequently, Case 2 cannot be true.Case3:.
IfMa2n,a2n+1=∂a2n,a2n+1, then using (9), we have(25)∂a2n+1,a2n+2≺¯λMa2n,a2n+1−LNa2n,a2n+1,∂a2n+1,a2n+2≺¯λ∂a2n,a2n+1−LNa2n,a2n+1.
SinceL>0 and Nan,an+1=0, then from (25),(26)∂a2n+1,a2n+2≺¯λ∂a2n,a2n+1,≺¯λn∂a0,a1.
Sinceλ∈0,1, letting n⟶∞, it is obvious that(27)limn⟶∞∂a2n+1,a2n+2=0.
Further, we prove that the given sequence is a Cauchy sequence. Letn,m,N∈ℕ be integers such that m>n>N, and we have(28)∂a2n+1,a2n+2≺∂a2n,a2n+2.
Hereμa,b>1 is used. Define(29)Sp=∑i=0p∏j=0iνaj,a2m+1μai,ai+1λi.
Hence,(30)∂a2n+1,a2m+1≺¯∂a0,a1λnμa2n+1,a2n+2+S2m−S2n+1.
The ratio test together with (10) implies that the limit of real number sequence Snn=1∞ exists, and so Snn=1∞ is a Cauchy sequence. In fact, the ratio test is applied to the term xi=∏j=0iνaj,a2m+1μaj,ai+1 and using (30), we have(31)limm,n⟶∞∂a2m+1,a2n+1=0,∀i,j≥1,which follows that the sequence ann=1∞ is a Cauchy sequence. Now, it is known that the double controlled metric space is a complete space. So, the sequence ann=1∞ converges to some point say f∈Ξ. We claim that f is the common fixed point for ℘,ℵ. Using triangle inequality, it follows that(32)∂f,a2n+2≺¯μf,a2n+1∂f,a2n+1+νa2n+1,a2n+2∂a2n+1,a2n+2.
Now it is claimed that℘f=f. Using triangle inequality, continuity of ℘, and (9), it follows that(33)∂f,℘f≺¯μf,a2n+1∂f,a2n+1+νa2n+1,℘f∂a2n+1,℘f,=μf,a2n+1∂f,a2n+1+νa2n+1,℘f∂℘a2n+1,℘f,≺¯μf,a2n+1∂f,a2n+1+νa2n+1,℘f∂℘a2n+1,℘f,≺¯μf,a2n+1∂f,a2n+1+νa2n+1,℘fλMa2n,f−LNa2n,f.
From (27) and (32) and letting n⟶∞, we get ∂f,℘f=0 which implies f=℘f. Similarly, it can be checked that f=ℵf. Thus, ℘,ℵ have a common fixed point say f. Now we show that the given point is unique; for this, consider f≠g∈Ξ, such that ℘f=f and ℘g=g. From (4), it follows that(34)∂℘f,℘g=∂f,g,≺¯λMf,g−LNf,g,≺¯λn∂f,g.
It can be deduced thatf=g. Similarly, the uniqueness of fixed point for ℵ can also be checked. Therefore, ℘,ℵ have a unique common fixed point.
This completes the proof.
If in Theorem2, Na,b=0. Then, we get the following corollary.Corollary 1.
LetΞ,∂ be a complete complex double controlled metric space. Furthermore, suppose that ℘,ℵ:Ξ⟶Ξ are two self-operators, such that ∀ distinct a,b∈Ξ,(35)∂℘a,ℵb≺¯κMa,b,where Ma,b=max∂a,b,∂b,ℵb1+a,℘a/1+∂a,b,∂a,℘a1+∂b,ℵb/1+∂a,b,∂b,ℵb∂a,℘a/∂a,b and 0<κ<1. With the additional assumption for κ∈0,1 and any ai∈Ξ,(36)supm≥1limi⟶∞μai+1,ai+2νai+1,a2m+1μai,ai+1<1κ.Moreover, for eacha∈Ξ, suppose that(37)limn⟶∞μa,a2n+1andlimn⟶∞μa2n+1,aexist and are finite.Then,℘,ℵ have a common unique fixed point.Corollary 2.
LetΞ,∂ be a complete complex double controlled metric space. Furthermore, suppose that ℘,ℵ:Ξ⟶Ξ are two self-operators, such that ∀ distinct a,b∈Ξ, we have(38)∂℘a,ℵb≺¯∇a,b,where ∇a,b=κ1∂a,b+κ2∂b,ℵb1+∂a,℘a/1+∂a,b+κ3∂a,℘a1+∂b,ℵb/1+∂a,b+κ4∂b,ℵb∂a,℘a/∂a,b such that each κn>0 for n=1,2,3,4 such that ∑n=14κn<1. With the additional assumption for κ∈0,1 and for any ai∈Ξ,(39)supm≥1limi⟶∞μai+1,ai+2νai+1,a2m+1μai,ai+1<1κ.Moreover, for eacha∈Ξ, suppose that(40)limn⟶∞μa,a2n+1andlimn⟶∞μa2n+1,aexist and are finite.Then,℘,ℵ have a common unique fixed point.IfMa,b=∂a,b and ℘=ℵ, then Corollary 1 reduces to the following corollary.Corollary 3.
LetΞ,∂ be a complete complex double controlled metric space. Furthermore, suppose that ℘:Ξ⟶Ξ is a self-operator, such that(41)∂℘a,℘b≺¯κ∂a,b.With the additional assumption forκ∈0,1 and for any ai∈E,(42)supm≥1limi⟶∞μai+1,ai+2νai+1,amμai,ai+1<1κ.Moreover, for eacha∈Ξ, suppose that(43)limn⟶∞μa,anandlimn⟶∞νan,aexist and are finite.Then,℘ has a unique fixed point.Example 2.
LetΞ=0,1 and define a distance function ∂:Ξ×Ξ⟶ℂ by ∂a,b=a−b1+b2eιcot−1b, where b=1 and a≠b∈Ξ. Consider the functions μ,ν:Ξ×Ξ⟶1,∞ by μa,b=2a+1/3b+1 and νa,b=a+4b+3, respectively. Then, it is easy to verify that Ξ,∂ is a complete complex double controlled type metric space. Now consider the operators ℘,ℵ:Ξ⟶Ξ by ℘a=1/75a, and ℵa=2/91a∀a∈0,1. Then, the left hand side of the inequality (9) implies that(44)∂℘a,ℵb=175a−291b2eιcot−11.
Now to calculate right hand side of the inequality (9), consider(45)Ma,b=maxa−b2eιcot−11,b−2/91b2eιcot−111+a−1/75a2eιcot−111+a−b2eιcot−11,a−1/75a2eιcot−111+b−2/91b2eιcot−111+a−b2eιcot−11,a−1/75a2eιcot−11b−2/91b2eιcot−11a−b2eιcot−11,Na,b=a−175a2eιcot−11,b−2/91b2eιcot−11,b−175a2eιcot−11.
With simple calculations, it can be analyzed thatMa,b=a−b2eιcot−11 and Na,b=a−1/75a2eιcot−11. Then, for λ=0.9 and L=0.001, inequality (9) is satisfied. Furthermore for each a∈Ξ, conditions (11) and (12) are also satisfied, respectively. Hence, by Theorem 2, ℘ and ℵ possess a common unique fixed point, i.e., 0. For visualizing the co-relation of left and right hand side of the inequality (9), consider Figure 1.
The real and imaginary parts can be separately visualized in Figures2 and 3, respectively.
It can be seen that Figures2 and 3 provide the same visuals. This is because of the equal values of real and complex parts as can be observed in Table 3. Furthermore, the numerical comparison of some values of left hand side and right hand side of inequality (9) is given in Table 3.
To visualize the convergence behavior of℘a and ℵb graphically, consider Figures 4 and 5, respectively.
For checking the convergence behavior of℘a and ℵb numerically, consider Tables 4 and 5, respectively.Figure 1
Comparison of left hand side and right hand side of (9).Figure 2
Real parts of (9).Figure 3
Imaginary parts of (9).Table 3
Numerical comparison of left hand side and right hand side of inequality (9).
ab∂℘a,ℵbλMa,b−ZNa,b where λ=0.9 and Z=0.001>00.00.10.0022+0.0022ι0.0699+0.1083ι0.00.50.0110+0.0110ι0.3343+0.5415ι0.00.90.0198+0.0198ι0.6018+0.9747ι0.11.00.0206+0.0206ι0.6018+0.9747ι0.10.90.0184+0.0184ι0.5349+0.8664ι0.10.50.0097+0.0097ι0.2675+0.4321ι0.51.00.0153+0.0153ι0.3343+0.5415ι0.50.90.0131+0.0131ι0.2675+0.4332ι0.50.10.0152+0.0153ι0.3343+0.5415ιFigure 4
Convergence behavior of℘a.Figure 5
Convergence behavior ofℵb.Table 4
Convergence behavior of℘a.
a0a0=0.1a0=0.4a0=0.75a0=0.99a10.001330.00530.01000.0132a20.0000170.0000700.0000130.00017a30.00000020.0000000090.00000730.0000023a40.0000000030.000000000010.0000000220.000000036a50.000000000040.000000000001200.00000000002660.000000000483a60.00000000000000.000000000000000.00000000000000000.0000000000000000⋮⋮⋮⋮⋮Table 5
Convergence behavior ofℵb.
b0b0=0.1b0=0.4b0=0.75b0=0.99b10.00220.00880.01650.0218b20.0000480.0001930.000360.00047b30.00000100.00000410.00000790.000010b40.00000002190.00000009000.00000017000.00000028b50.0000000004610.0000000019780.0000000039000.0000000061b60.00000000000000.00000000000000000.00000000000000000.0000000000000000⋮⋮⋮⋮⋮


## 3. Existence of a Unique Solution of Complex Urysohn Integral Equations

LetΞ=Cp,q,ℂ be the set of all continuous functions from the closed and bounded interval p,q to ℂ. Consider a distance function ∂:Ξ×Ξ⟶ℂ defined by(46)∂ϑ,σ=ϑ−σ+ιϑ−σ,where ϑ,σ∈Ξ. Furthermore, consider the functions μ,ν:Ξ×Ξ⟶1,∞ defined by μϑ,σ=ϑ+5σ+2 and νϑ,σ=ϑ+3σ+1, respectively. Then, Ξ,∂ is a complete complex double controlled metric space.In this section, we propose the existence and uniqueness of solution via the fixed point method for the complex-valued Urysohn integral equations.(47)ϑς=θς+∫pqΓ1ς,ϑ,ϑϑdϑ,σς=θς+∫pqΓ2ς,ϑ,σdϑ,where Γ1ς,,ϑ and Γ2ς,,σ are continuous complex-valued kernels on the given interval and θς is continuous and complex-valued on the interval p,q.Consider the operators℘,ℵ:Ξ⟶Ξ defined by(48)℘ϑς=θς+∫pqΓ1ς,,ϑdϑ,ℵσς=θς+∫pqΓ2ς,,σdϑ.Urysohn integral (47) would have a unique common solution if the following conditions:(49)|Γ1ς,,ϑ−Γ2ς,,σ|≺¯1p−qexpιpqMϑ,σ−LNϑ,σ,1expιpq≺¯κwhere0<κ<1,hold, where Mϑ,σ, Nϑ,σ, and L are defined in (9). Consider(50)℘ϑ−ℵσ+ι℘ϑ−ℵσ=θς+∫pqΓ1ς,,ϑd−θς+∫pqΓ2ς,,σd+ιθς+∫pqΓ1ς,,ϑd−θς+∫pqΓ2ς,,σd,≺¯∫pq|Γ1ς,,ϑ−Γ2ς,,σd+ι∫pqΓ1ς,,ϑ−Γ2ς,,σdϑ,≺∫pq1p−qexpιpqMϑ,σ−Nϑ,σdϑ,=1p−qexpιpqMϑ,σ−Nϑ,σ∫pqdϑ,=1expιpqMϑ,σ−Nϑ,σ.Ifexpιpq1/≺¯κ, then(51)∂℘ϑ,ℵσ≺¯κMϑ,σ−Nϑ,σ.It can be concluded that inequality (9) is satisfied. Additionally, for any ϑ∈Ξ, conditions (11) and (12) are also satisfied. Consequently, all hypotheses of Theorem 2 are satisfied. Therefore, system (47) possesses a common unique solution.


## 4. Existence of a Unique Solution of Complex Fredholm Integral Equation

LetΞ=Cp,q,ℂ be the set of all continuous functions from the closed and bounded interval p,q to ℂ. Consider a distance function ∂:Ξ×Ξ⟶ℂ defined by(52)∂ϑ,σ=ϑ−σ+ιϑ−σ,where ϑ,σ∈Ξ. Furthermore, consider the functions μ,ν:Ξ×Ξ⟶1,∞, defined by μϑ,σ=ϑ+2σ+5 and νϑ,σ=3ϑ+σ+1, respectively. Then, Ξ,∂ is a complete complex double controlled metric space.In this section, we inspect the existence and uniqueness of solution via the fixed point method for complex Fredholm integral equation.(53)ϑς=ℏς+α∫pqΩς,t,ϑtdt,where each of the component of (53) is defined in Section 1.Consider the operator℘:Ξ⟶Ξ defined by(54)℘ϑς=ℏς+α∫pqΩς,t,ϑtdt.The Fredholm integral equation would have a unique solution if the following condition:(55)Ως,t,ϑt−Ως,t,σt≤καq−pϑt−σtwhere0<κ<1,holds.Consider(56)℘ϑ−℘σ+ι℘ϑ−℘σ=ℏς+α∫pqΩς,tϑtdt−ℏς+α∫pqΩς,tσtdt+ιℏς+α∫pqΩς,tϑtdt−ℏς+α∫pqΩς,tσtdt,≺¯α∫pqΩς,t,ϑt−Ως,t,σtdt+ια∫pqΩς,t,ϑt−Ωσ,t,σtdt,≺¯κϑ−σ+ιϑ−σ,which implies(57)∂℘ϑ,℘σ≺¯κ∂ϑ,σ.Consequently, all hypotheses of the Corollary3 are satisfied. Therefore, integral (53) has a unique solution.


## 5. Conclusion

In this study, we have proved an extended rational weak contraction for common fixed point in the setting of complex double controlled type metric spaces. Our results are validated by a numerical example and existence and uniqueness for complex Urysohn integral equations and complex Fredholm integral equation. The existence and uniqueness of a solution for certain type of equation hold remarkable applications towards applied sciences, like applied mathematics, physics, biology, engineering, and so on. It helps the researchers to know about the nature, specific conditions involved, exact range, and domain of a solution to any under discussion problem and thus saves a lot of efforts, time, and energy. To contribute to the worldly problems, we have also considered complex Urysohn and Fredholm integral equations for existence and uniqueness of a solution. The Urysohn and Fredholm integral equations have remarkable applications towards geophysics, mechanics, quantum mechanics, population genetics, fluid mechanics, etc.



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*Source: 4791454-2022-05-29.xml*