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Article

Analytical Solution of Urysohn Integral Equations by Fixed Point Technique in Complex Valued Metric Spaces

by
Hasanen A. Hammad
1 and
Manuel De la Sen
2,*
1
Department of Mathematics, Faculty of Science, Sohag University, Sohag 82524, Egypt
2
Institute of Research and Development of Processes University of the Basque Country, 48940 Leioa, Bizkaia, Spain
*
Author to whom correspondence should be addressed.
Submission received: 18 July 2019 / Revised: 8 September 2019 / Accepted: 10 September 2019 / Published: 15 September 2019
(This article belongs to the Special Issue Fixed Point Theory and Related Nonlinear Problems with Applications)

Abstract

:
The purpose of this article is to introduce a fixed point result for a general contractive condition in the context of complex valued metric spaces. Also, some important corollaries under this contractive condition are obtained. As an application, we find a unique solution for Urysohn integral equations, and some illustrative examples are given to support our obtaining results. Our results extend and generalize some other known results in the literature.

1. Introduction

The fixed point theorem, generally known as the Banach contraction principle, appeared in explicit form in Banach thesis in 1922. Fixed point theory is very famous because of its variety of applications in numerous areas, such as computer science, engineering, economics, etc. The contractive type conditions play an important role in the fixed point theory. Many researchers have extended and generalized this principle because it is the heart of fixed point theory (see, for example, the works of the authors of [1,2,3,4,5,6,7]).
The complex valued metric spaces is more general than ordinary metric spaces. According to this concept, a number of articles related to fixed point theory and it’s application are presented (see, for example, [8,9,10,11,12,13,14,15,16,17,18,19,20,21,22]).
The aim of this paper is to prove a fixed point theorem in complex valued metric spaces under contractive condition for single-valued mappings. Moreover, we give a result of existence and uniqueness for solutions of a nonlinear system of integral equations. Finally, we give some explained examples to strengthen our results.

2. Preliminaries and Known Results

This section is prepared to discuss some known notations and definitions that will be used later. Suppose that C is the set of complex numbers and z 1 , z 2 C . Define a partial order ⪯ on C :
z 1 z 2 iff R e ( z 1 ) R e ( z 2 ) and I m ( z 1 ) I m ( z 2 ) .
So, z 1 z 2 if one of the following conditions hold.
( i ) I m ( z 1 ) < I m ( z 2 ) and R e ( z 1 ) = R e ( z 2 ) , ( i i ) I m ( z 1 ) = I m ( z 2 ) and R e ( z 1 ) < R e ( z 2 ) , ( i i i ) I m ( z 1 ) < I m ( z 2 ) and R e ( z 1 ) < R e ( z 2 ) , ( i v ) I m ( z 1 ) = I m ( z 2 ) and R e ( z 1 ) = R e ( z 2 ) .
Here, we write z 1 z 2 if z 1 z 2 and one of conditions ( i ) , ( i i ) , and ( i i i ) is satisfied and if only ( i i i ) is satisfied, we write z 1 z 2 .
Definition 1
([8]). Let Ω . A mapping ω : Ω × Ω C is said to a complex valued metric on Ω if the following conditions holds for all κ , μ , τ Ω ,
  • ( C M 1 ) 0 ω ( κ , μ ) and ω ( κ , μ ) = 0 κ = μ ;
  • ( C M 2 ) ω ( κ , μ ) = ω ( μ , κ ) ;
  • ( C M 3 ) ω ( κ , μ ) ω ( κ , τ ) + ω ( τ , μ ) .
Then Ω is called a complex valued metric on Ω and ( Ω , ω ) is called a complex valued metric space.
For some examples of complex valued metric spaces see the works by the authors of [8,9,14,17].
Now, we state two examples not mentioned above.
Example 1.
Let Ω = C be a set of complex number. Define a distance ω : Ω × Ω Ω , by
ω z 1 , z 2 = ω ( κ 1 , κ 2 ) + i ω ( μ 1 , μ 2 ) ,
for all z 1 , z 2 Ω , where z 1 = ( κ 1 , μ 1 ) and z 2 = ( κ 2 , μ 2 ) . Then, ( Ω , ω ) is a complex valued metric space provided that ( Ω , ω ) is too.
Example 2.
Let Ω = C be a set of complex number. Define the distance ω : Ω × Ω Ω by
ω z 1 , z 2 = ( κ 1 κ 2 ) 2 + i ( μ 1 μ 2 ) 2 1 2 ,
where z 1 = κ 1 + i μ 1 and z 2 = κ 2 + i μ 2 . Then, ( Ω , ω ) is a complex valued metric space.
Definition 2
([8]). Let ( Ω , ω ) be a complex valued metric space. Then:
(i) A sequence κ n in Ω is said to be converge to κ Ω if for every 0 ε C there exists N N such that ω ( κ n , κ ) ε n > N . We denote this by lim n κ n = κ or κ n κ as n .
(ii) If for every 0 ε C there exists N N such that ω ( κ n , κ n + m ) ε for all n > N , m N , then a sequence κ n is called a Cauchy sequence in Ω.
(iii) If every Cauchy sequence is convergent, then ( Ω , ω ) is called a complete complex valued metric space.
Lemma 1
([8]). Let κ n be a sequence in ( Ω , ω ) . It is said that κ n converges to κ iff ω ( κ n , κ ) 0 as n .
Lemma 2
([8]). Let κ n be a sequence in ( Ω , ω ) . It is said that κ n in Ω is a Cauchy sequence if Ω ( κ n , κ n + m ) 0 as n , where m N .

3. Main Result

We introduce our first result.
Theorem 1.
Let A and B be self-mappings on a complete complex valued metric space ( Ω , ω ) , such that
ω ( A κ , B μ ) α Θ ( κ , μ ) ,
κ , μ Ω , where α ( 0 , 1 ) ,
Θ ( κ , μ ) = max ω ( κ , y ) , ω ( κ , A κ ) ω ( μ , B μ ) 1 + ω ( κ , μ ) , ω ( κ , B μ ) ω ( μ , A κ ) 1 + ω ( κ , μ ) .
Then there exists a unique common fixed point of the pair mappings ( A , B ) .
Proof. 
Let κ Ω be an arbitrary point. Define a sequence { κ n } as follows.
κ 2 n + 1 = A κ 2 n and κ 2 n + 2 = B κ 2 n + 1 , n = 0 , 1 , 2 ,
Then, by Equations (1) and (2), we get
ω ( κ 2 n + 1 , κ 2 n + 2 ) = ω ( A κ 2 n , B κ 2 n + 1 ) α max ω ( κ 2 n , κ 2 n + 1 ) , ω ( κ 2 n , A κ 2 n ) ω ( κ 2 n + 1 , B κ 2 n + 1 ) 1 + ω ( κ 2 n , κ 2 n + 1 ) , ω ( κ 2 n , B κ 2 n + 1 ) ω ( κ 2 n + 1 , A κ 2 n ) 1 + ω ( κ 2 n , κ 2 n + 1 ) α max ω ( κ 2 n , κ 2 n + 1 ) , ω ( κ 2 n , κ 2 n + 1 ) ω ( κ 2 n + 1 , κ 2 n + 2 ) 1 + ω ( κ 2 n , κ 2 n + 1 ) , ω ( κ 2 n , κ 2 n + 2 ) ω ( κ 2 n + 1 , κ 2 n + 1 ) 1 + ω ( κ 2 n , κ 2 n + 1 ) α max ω ( κ 2 n , κ 2 n + 1 ) , ω ( κ 2 n + 1 , κ 2 n + 2 ) .
If max ω ( κ 2 n , κ 2 n + 1 ) , ω ( κ 2 n + 1 , κ 2 n + 2 ) = ω ( κ 2 n + 1 , κ 2 n + 2 ) , then
ω ( κ 2 n + 1 , κ 2 n + 2 ) α ω ( κ 2 n + 1 , κ 2 n + 2 ) ,
Since α < 1 and both distances in the left and right-hand sides are identical this equation is not possible unless both distances are zero. So
ω ( κ 2 n + 1 , κ 2 n + 2 ) α ω ( κ 2 n , κ 2 n + 1 ) .
Similarly, we can obtain that
ω ( κ 2 n + 2 , κ 2 n + 3 ) α ω ( κ 2 n + 1 , κ 2 n + 2 ) .
From Equations (3) and (4) for all n = 0 , 1 , 2 , , we can write
ω ( κ n + 1 , κ n + 2 ) α ω ( κ n , κ n + 1 ) α n + 1 ω ( κ , κ 1 ) .
So, for m > n ,
ω ( κ n , κ m ) ω ( κ n , κ n + 1 ) + ω ( κ n + 1 , κ n + 2 ) + + ω ( κ m 1 , κ m ) α n + α n + 1 + . . . + α m 1 ω ( κ , κ 1 ) α n 1 α ω ( κ , κ 1 ) .
So,
ω ( κ n , κ m ) α n 1 α ω ( κ , κ 1 ) 0 as n .
Therefore, { κ n } is a Cauchy sequence in Ω . Since Ω is complete, then there exists u Ω , such that κ n u . If A and B are not continuous, then u = A u , unlike that ω ( u , A u ) = z > 0 and one gets
z ω ( u , κ 2 l + 2 ) + ω ( A u , κ 2 l + 2 ) ω ( u , κ 2 l + 2 ) + ω ( A u , B κ 2 l + 1 ) ω ( u , κ 2 l + 2 ) + α max ω ( u , κ 2 l + 1 ) , ω ( u , A u ) ω ( κ 2 l + 1 , B κ 2 l + 1 ) 1 + d ( u , κ 2 l + 1 ) , ω ( u , B κ 2 l + 1 ) ω ( κ 2 l + 1 , A u ) 1 + ω ( u , κ 2 l + 1 ) ω ( u , x 2 l + 2 ) + α max ω ( u , κ 2 l + 1 ) , ω ( u , A u ) ω ( κ 2 l + 1 , κ 2 l + 2 ) 1 + ω ( u , κ 2 l + 1 ) , ω ( u , κ 2 l + 2 ) ω ( κ 2 l + 1 , A u ) 1 + ω ( u , κ 2 l + 1 ) ω ( u , κ 2 l + 2 ) + α max 0 , 0 , z ω ( u , x 2 l + 2 ) + α z .
This yields,
z ω ( u , κ 2 l + 2 ) + α z .
That is, z = 0 , is a contradiction, and hence u = A u . It follows, similarly, that u = B u .
If A and B are continuous, i.e., the continuity of A, yields
u = lim n κ 2 n + 2 = lim n A κ 2 n + 1 = A lim n κ 2 n + 1 = A u .
Similarly, u = B u . Hence the pair ( A , B ) has a common fixed point.
Uniqueness. Suppose that q Ω is a another common fixed point of the nonlinear self-mappings A and B. Then,
ω ( u , q ) = ω ( A u , B q ) α max ω ( u , q ) , ω ( u , A u ) ω ( q , B q ) 1 + ω ( u , q ) , ω ( u , B q ) ω ( q , A u ) 1 + ω ( u , q ) α d ( u , q ) .
This implies that u = q , this completes the proof. ☐
The following example support Theorem 1.
Example 3.
Let Ω = [ 0 , ) define the distance ω : Ω × Ω C by
ω ( κ , μ ) = i κ μ .
It is clear that ( Ω , ω ) is a complete complex valued metric space. We define the two self-mappings A and B as
A κ = 2 κ 2 1 and B κ = ( 2 κ ) 2 .
Then the contractive condition Equation (1) is satisfied, indeed for κ = 1 3 and μ = 3 , we can write by the simple calculations of
ω ( A κ , B μ ) = 16 9 i .
and
Θ ( κ , μ ) = max 8 3 i , 40 3 + 8 i , 68 9 ( 3 + 8 i ) = 8 3 i .
So,
16 i 9 α 8 i 3 .
Therefore, the conditions of Theorem 1 are verified with α = 2 3 < 1 , and 1 Ω is a unique common fixed point of A and B.
If we take A = B in the above theorem we have the following immediate consequences.
Corollary 1.
Suppose that A is a self-mapping on a complete complex valued metric space ( Ω , ω ) , such that
ω ( A κ , A μ ) α Θ ( κ , μ ) ,
for all κ , μ Ω , where 0 < α < 1 and
Θ ( κ , μ ) = max ω ( κ , μ ) , ω ( κ , A κ ) ω ( μ , A μ ) 1 + ω ( κ , μ ) , ω ( κ , A μ ) ω ( μ , A κ ) 1 + ω ( κ , μ ) .
Then, in Ω , a mapping A has a unique fixed point.
To justify the requirements of Corollary 1, we present the following example.
Example 4.
Let Ω = [ 0 , ) and ω : Ω × Ω C be a mapping defined by
ω ( κ , μ ) = κ μ + i κ μ .
Clearly ( Ω , ω ) is a complete complex valued metric space. Define a self-mapping A by
A κ = 2 π sin 1 κ .
To verify the contractive condition of Corollary 1, we take κ = 1 2 and μ = 3 2 , one can write by the simple calculations of
ω ( A κ , A μ ) 0.1667 ( 1 + i ) .
and
Θ ( κ , μ ) max 0.3660 ( 1 + i ) , 0.0483 i , 0.1301 i 0.3660 ( 1 + i ) .
So,
0.1667 ( 1 + i ) α 0.3660 ( 1 + i ) .
Therefore, all conditions of Corollary 1 are satisfied with α 0.4555 < 1 and A has a unique fixed point 1 Ω .
Corollary 2.
Consider ( Ω , ω ) is a complete complex valued metric space and A : Ω Ω , there exists n N such that
ω ( A n κ , A n μ ) α Θ ( κ , μ )
κ , μ Ω , where 0 < α < 1 ,
Θ ( κ , μ ) = max ω ( κ , μ ) , ω ( κ , A n κ ) ω ( μ , S n μ ) 1 + ω ( κ , μ ) , ω ( κ , A n μ ) ω ( μ , A n κ ) 1 + ω ( κ , μ ) .
Then A possesses a unique fixed point.
Proof. 
By Corollary 1, we obtain v Ω such that
A n v = v .
So,
ω ( A v , v ) = ω ( A A n v , A n v ) = ω ( A n A v , A n v ) α max ω ( A v , v ) , ω ( A v , A n A v ) ω ( v , A n v ) 1 + ω ( A v , v ) , ω ( A v , A n v ) ω ( v , A n A v ) 1 + ω ( A v , v ) α max ω ( A v , v ) , ω ( A v , A A n v ) ω ( v , A n v ) 1 + ω ( A v , v ) , ω ( A v , A n v ) ω ( v , A A n v ) 1 + ω ( A v , v ) = α ω ( A v , v ) .
Therefore, the result is follows. ☐
Now we shall give a numerical example to show the validity of Corollary 2.
Example 5.
Let Ω = C ( [ 0 , 2 ] , R ) , b > 0 ,
N κ μ = max t [ 0 , 2 ] κ ( t ) μ ( t ) , ω ( κ , μ ) = N κ μ 1 + b 3 3 e i   cot 1 b .
Define A : Ω Ω by
A κ ( t ) = 1 + 3 0 t u 2 κ ( u ) d u , t [ 0 , 2 ] .
For every κ , μ Ω , we have
ω ( A κ , A μ ) = N A κ A μ 1 + b 3 3 e i   cot 1 b = max t [ 0 , 2 ] A κ ( t ) A μ ( t ) 1 + b 3 3 e i   cot 1 b 3 0 2 max t [ 0 , 2 ] κ ( u ) μ ( u ) u 2 1 + b 3 3 e i   cot 1 b d u 8 ω ( κ , μ ) .
Similarly,
ω ( A n κ , A n μ ) 8 n n ! ω ( κ , μ ) 8 n n ! Θ ( κ , μ ) ,
where,
8 n n ! 295.894 I f n = 10 26.906 I f n = 15 1.185 I f n = 19 0.474 I f n = 20
Thus, for α 0.474 < 1 , n = 20 , all conditions of Corollary 2 are verified, so A has a unique fixed point, which is a unique solution of the nonlinear integral equation
κ ( t ) = 1 + 3 0 t u 2 κ ( u ) d u , t [ 0 , 2 ] ,
or the differential equation (initial value problem)
κ ( t ) 3 κ 2 t = 0 , t [ 0 , 2 ] , t ( 0 ) = 1 .

4. An Application to Urysohn Integral Type Equations

This section is the main result of the paper, here we apply Theorem 1 to find a unique solution of the Urysohn integral type equations
κ ( t ) = h ( t ) + a b χ 1 ( t , s , κ ( s ) ) d s μ ( t ) = h ( t ) + a b χ 2 ( t , s , μ ( s ) ) d s .
where,
(i)
κ ( t ) and μ ( t ) are unknown variables for each t [ a , b ] , a > 0 ,
(ii)
h ( t ) is the deterministic free term defined for t [ a , b ] ,
(iii)
χ 1 ( t , s ) and χ 2 ( t , s ) are deterministic kernels defined for t , s [ a , b ] .
Let Ω = ( C [ a , b ] , R n ) , a > 0 and ω : Ω × Ω R n defined by
ω ( κ , μ ) = sup t [ a , b ] κ ( t ) μ ( t ) 1 + b 3 3 e i   cot 1 b ,
for all κ , μ Ω , i = 1 C .
It’s obvious that ( C [ a , b ] , R n , . ) is a complete complex valued metric space.
Next, we consider a system Equation (5) under the following conditions,
( H 1 ) h ( t ) Ω ,
( H 2 ) χ 1 , χ 2 : [ a , b ] × [ a , b ] × R n R n are continuous functions satisfying
χ 1 ( t , s , u ( s ) ) χ 1 ( t , s , v ( s ) ) 1 ( b a ) e a b Θ ( u , v ) ,
where,
Θ ( u , v ) = max ω ( u , v ) , ω ( u , A u ) ω ( v , B v ) 1 + ω ( u , v ) , ω ( u , B v ) ω ( v , A u ) 1 + ω ( u , v ) .
Next, we state and prove our main theorem of this section.
Theorem 2.
System Equation (5) has a unique common solution provided that the conditions ( H 1 ) and ( H 2 ) are satisfied.
Proof. 
For κ , μ ( C [ a , b ] , R n ) and t [ a , b ] , we define the continuous mappings A , B : Ω Ω by
A κ ( t ) = h ( t ) + a b χ 1 ( t , s , κ ( s ) ) d s , B μ ( t ) = h ( t ) + a b χ 2 ( t , s , μ ( s ) ) d s .
By this, we have
A κ ( t ) B μ ( t ) = a b χ 1 ( t , s , κ ( s ) ) χ 2 ( t , s , μ ( s ) ) d s a b 1 ( b a ) e a b Θ ( κ , μ ) d s = 1 ( b a ) e a b a b e i   cot 1 b 1 + b 3 3 Θ ( κ , μ ) 1 + b 3 3 e i   cot 1 b d s 1 ( b a ) e a b e i   cot 1 b 1 + b 3 3 Θ ( κ , μ ) a b d s = 1 e a b e i   cot 1 b 1 + b 3 3 Θ ( κ , μ ) .
This gives,
1 + b 3 3 A κ ( t ) B μ ( t ) e i   cot 1 b 1 e a b Θ ( κ , μ ) ,
or, equivalently,
A κ ( t ) B μ ( t ) 1 e a b Θ ( κ , μ ) ,
or,
ω ( A κ , B μ ) α Θ ( κ , μ ) .
So, the condition Equation (1) of Theorem 1 is satisfied with 0 < α = 1 e a b < 1 , Therefore the system Equation (5) has a unique common solution on Ω . ☐
Finally, we verify all conditions of Theorem 2 by the following example.
Example 6.
Let Ω = C ( [ a , b ] , R ) and the following nonlinear integral equation as the form
κ ( t ) = e 4 i t + a b e 1 4 4 t + i s 1 + i s + κ ( s ) d s μ ( t ) = e 4 i t + a b e 1 4 4 t + i s 1 + i s + μ ( s ) d s .
Problem Equation (6) is a special case of problem Equation (5), where h ( t ) = e 4 i t and
χ j ( t , s , u j ( s ) ) = e 1 4 4 t + i s 1 + i s + u j ( s ) , j = 1 , 2 .
It’s obvious that ( H 1 ) is satisfied, for ( H 2 ) , we get
χ 1 ( t , s , κ ( s ) ) χ 2 ( t , s , μ ( s ) ) = 1 4 e 1 4 κ ( s ) μ ( s ) t + i s 1 + i s + κ ( s ) t + i s 1 + i s + μ ( s ) 1 4 e 1 4 κ ( s ) μ ( s ) .
Therefore, ( H 2 ) is hold with α = 1 4 e 1 4 < 1 and Θ ( κ , μ ) = κ ( s ) μ ( s ) . By Theorem 2, the problem Equation (6) has a unique solution.

5. Conclusions

By changing the definition of real valued metric into complex valued metric, complex valued-metric spaces are considered as a generalization of ordinary metric spaces. This change is expected to bring wider applications of fixed point theorems. In this paper, we prove some fixed point theorem in complex valued metric spaces under contractive condition for single-valued mappings. Also, we find a unique solution of a nonlinear system of integral equations and we support our theoretical results by some explained examples.

Author Contributions

H.A.H contributed in conceptualization, investigation, methodology, validation and wrting the original draft; M.D.l.S. contributed in funding adquisition, methodology, project administration, supervision, validation, visualization, writing and editing. Both Aurhors agree and aprove the final version of this manuscript.

Funding

This research received funding from the Basque Government through project IT1207-19.

Acknowledgments

This work was supported in part by the Basque Government through project IT1207-19.

Conflicts of Interest

The authors declare no conflicts of interest.

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MDPI and ACS Style

Hammad, H.A.; De la Sen, M. Analytical Solution of Urysohn Integral Equations by Fixed Point Technique in Complex Valued Metric Spaces. Mathematics 2019, 7, 852. https://0-doi-org.brum.beds.ac.uk/10.3390/math7090852

AMA Style

Hammad HA, De la Sen M. Analytical Solution of Urysohn Integral Equations by Fixed Point Technique in Complex Valued Metric Spaces. Mathematics. 2019; 7(9):852. https://0-doi-org.brum.beds.ac.uk/10.3390/math7090852

Chicago/Turabian Style

Hammad, Hasanen A., and Manuel De la Sen. 2019. "Analytical Solution of Urysohn Integral Equations by Fixed Point Technique in Complex Valued Metric Spaces" Mathematics 7, no. 9: 852. https://0-doi-org.brum.beds.ac.uk/10.3390/math7090852

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