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 JAMP  Vol.9 No.2 , February 2021
Singular Hammerstein-Volterra Integral Equation and Its Numerical Processing
Abstract: In this paper, the existence and uniqueness of solution of singular Hammerstein-Volterra integral equation (H-VIE) are considered. Toeplitz matrix (TMM) and product Nystrom method (PNM) to solve the H-VIE with singular logarithmic kernel are used. The absolute error is calculated.

1. Introduction

The singular integral equations are considered to be of more interest than the others and a close form of solution is generally not available. Therefore, great attention must be considered for the numerical solution of these equations. Abdou in [1], studied Fredholm-Volterra integral equation with singular kernel. Al-Bugami, in [2], studied some numerical methods for solving singular and nonsingular integral equations. Abdou, El-Sayed and Deebs, in [3], obtained a solution of nonlinear integral equation. Also in [4], Abdou and Hendi used numerical solution for solving Fredholm integral equation with Hilbert kernel. In [5], Al-Bugami used TMM and Volterra-Hammerstein integral equation with a generalized singular kernel. In [6], Abdou, Borai, and El-Kojok used TMM and nonlinear integral equation of Hammerstein type. Al-Bugami, in [7], studied the error analysis for numerical solution of HIE with a generalized singular kernel. A. Shahsavaran in [8], studied Lagrange functions method for solving nonlinear F-VIE. In [9], Darwish, studied the nonlinear Fredholm-Volterra integral equations with hysteresis. In [10], Mirzaee used numerical solution of nonlinear F-VIEs via Bell polynomials. In [11], Raad studied linear F-VIE with logarithmic kernel and solved the linear system of Fredholm integral equations numerical with logarithmic form.

2. Existence and Uniqueness of the Solution of H-VIE

Consider:

μ ϕ ( x , t ) = f ( x , t ) + λ a a K ( x , y ) γ ( y , t , ϕ ( y , t ) ) d y + λ 0 t F ( t , τ ) ϕ ( x , τ ) d τ (1)

This formula is measured in L 2 [ a , a ] × C [ 0 , T ] , T < , where the FI term is measured with respect to position. While the VI term is considered in time, and f ( x , t ) is known function. λ is the parameter, while μ defines the kind of the integral Equation (1).

We assume:

1) K ( x , y ) C ( [ a , a ] × [ a , a ] ) , and satisfies:

[ a a a a | K ( x , y ) | 2 d y d x ] 1 2 = A 1 < , ( A 1 is a constant)

2) F ( t , τ ) C ( [ 0 , T ] × [ 0 , T ] ) , 0 τ t T , satisfies:

| F ( t , τ ) | A 2

3) f ( x , t ) is continuous in L 2 [ a , a ] × C [ 0 , T ] where:

f ( x , t ) = max 0 t T 0 t [ a b | f ( x , τ ) | 2 d x ] 1 2 d τ = A 3

4) γ ( x , t , ϕ ( x , t ) ) , satisfies for the constant B > B 1 , B > p , the following conditions:

a) 0 t a b ( | γ ( x , t , ϕ ( x , t ) ) | 2 d x d t ) 1 2 B 1 ϕ ( x , t ) L 2 [ a , b ] × C [ 0 , T ]

b) γ ( x , t , ϕ 1 ( x , t ) ) γ ( x , t , ϕ 2 ( x , t ) ) N ( x , t ) | ϕ 1 ( x , t ) ϕ 2 ( x , t ) |

where N ( x , t ) L 2 [ a , b ] × C [ 0 , T ] = p

In other words, we prove that the solution exists using the successive approximation method, also called the Picard method, that we pick up any real continuous function ϕ 0 ( x , t ) in L 2 [ a , a ] × C [ 0 , T ] , we assume ϕ 0 ( x , t ) = f ( x , t ) , then construct a sequence ϕ n defined by

ϕ n ( x , t ) = f ( x , t ) + λ a a K ( x , y ) γ ( y , t , ϕ n 1 ( y , t ) ) d y + λ 0 t F ( t , τ ) ϕ n 1 ( x , τ ) d τ , ( μ = 1 )

ϕ n 1 ( x , t ) = f ( x , t ) + λ a a K ( x , y ) γ ( y , t , ϕ n 2 ( y , t ) ) d y + λ 0 t F ( t , τ ) ϕ n 2 ( x , τ ) d τ , ( μ = 1 )

ψ n ( x , t ) = ϕ n ( x , t ) ϕ n 1 ( x , t ) = λ a a K ( x , y ) [ γ ( y , t , ϕ n 1 ( y , t ) ) γ ( y , t , ϕ n 2 ( y , t ) ) ] d y + λ 0 t F ( t , τ ) [ ϕ n 1 ( x , τ ) ϕ n 2 ( x , τ ) ] d τ , n = 1 , 2 ,

Then:

ϕ n ( x , t ) = i = 0 n ψ i ( x , t ) (2)

Hence

ψ n ( x , t ) = f ( x , t ) + λ a a K ( x , y ) γ ( y , t , ψ n 1 ( y , t ) ) d y + λ 0 t F ( t , τ ) ψ n 1 ( x , τ ) d τ

Using the properties of the norm, we obtain:

ψ n ( x , t ) | λ | a a K ( x , y ) γ ( y , t , ψ n 1 ( y , t ) ) d y + | λ | 0 t F ( t , τ ) ψ n 1 ( x , τ ) d τ

For n = 1 , we get

ψ 1 ( x , t ) | λ | a a K ( x , y ) γ ( y , t , ψ 0 ( y , t ) ) d y + | λ | 0 t F ( t , τ ) ψ 0 ( x , τ ) d τ | λ | ( a a | K ( x , y ) | 2 d y ) 1 2 ( a a | γ ( y , t , ψ 0 ( y , t ) ) | 2 d y ) 1 2 + | λ | 0 t | F ( t , τ ) | | ψ 0 ( x , τ ) | d τ

Using Cauchy Schwarz inequality and from conditions (i)-(iv-a) with ψ 0 = f ( x , t ) and f = A 3 , we get

ψ 1 ( x , t ) | λ | max 0 t [ a a ( a a | K ( x , y ) | 2 d y a a | γ ( y , t , ψ 0 ( y , t ) ) | 2 d y ) d x ] 1 2 d τ 0 t T + | λ | A 2 0 t ψ 0 ( x , τ ) d τ | λ | A 1 A 3 B 1 + | λ | A 2 A 3 t

We have 0 τ t T , then m a x | t | = T = L , and then we have:

ψ 1 ( x , t ) | λ | A 3 ( A 1 B 1 + A 2 L )

In general, we get:

ψ 1 ( x , t ) | λ | n A 3 ( A 1 B 1 + A 2 L ) n = A 3 α n , α = | λ | ( A 1 B 1 + A 2 L ) (3)

This bound makes the sequence ψ n ( x , t ) converges if

α < 1 | λ | < 1 A 1 B 1 + A 2 L (4)

The result (4), leads us to say that the formula (2) has a convergent solution. So let n , we have:

ϕ ( x , t ) = i = 0 ψ i ( x , t ) = A 3 1 α , ( α < 1 ) (5)

The infinite series of (5) is convergent, and ϕ ( x , t ) represents the convergent solution of Equation (1). Also each of ψ i is continuous, therefore ϕ ( x , t ) is also continuous.

To show that ϕ ( x , t ) is unique, we assume that ϕ ¯ ( x , t ) is also a continuous solution of (1) then, we write

ϕ ( x , t ) ϕ ¯ ( x , t ) = λ a a K ( x , y ) [ γ ( y , t , ϕ ( y , t ) ) γ ( y , t , ϕ ¯ ( y , t ) ) ] d y + λ 0 t F ( t , τ ) [ ϕ ( x , τ ) ϕ ¯ ( x , τ ) ] d τ , ( μ = 1 )

which leads us to the following:

ϕ ( x , t ) ϕ ¯ ( x , t ) | λ | a a | K ( x , y ) | | γ ( y , t , ϕ ( y , t ) ) γ ( y , t , ϕ ¯ ( y , t ) ) | d y + | λ | 0 t | F ( t , τ ) | | ϕ ( x , τ ) ϕ ¯ ( x , τ ) | d τ

Using conditions (iv-b), then we have:

ϕ ( x , t ) ϕ ¯ ( x , t ) | λ | max 0 t T 0 t [ a a a a ( | K ( x , y ) | d x d y ) 1 2 ( a a N 2 ( x , t ) | ϕ ( x , t ) ϕ ¯ ( x , t ) | 2 d y ) 1 2 ] d τ + | λ | 0 t | F ( t , τ ) | | ϕ ( x , t ) ϕ ¯ ( x , t ) | d τ

Finally, with the aid of conditions (i) and (ii):

ϕ ( x , t ) ϕ ¯ ( x , t ) α ϕ ( x , t ) ϕ ¯ ( x , t )

Then:

( 1 α ) ϕ ( x , t ) ϕ ¯ ( x , t ) 0

Since ϕ ( x , t ) ϕ ¯ ( x , t ) is necessarily non-negative, and α < 1 :

ϕ ( x , t ) ϕ ¯ ( x , t ) = 0 ϕ ( x , t ) = ϕ ¯ ( x , t )

It follows that if (1) has a solution it must be unique.

3. SHIEs

Consider:

ϕ ( x , t ) = f ( x , t ) + λ a a K ( x , y ) γ ( y , t , ϕ ( y , t ) ) d y + λ 0 t F ( t , τ ) ϕ ( x , τ ) d τ (6)

when t = 0 Equation (13) becomes:

ϕ 0 ( x ) = f 0 ( x ) + λ a a K ( x , y ) γ ( y , ϕ 0 ( y ) ) d y (7)

where ϕ 0 ( x ) = ϕ ( x , 0 ) , f 0 ( x ) = f ( x , 0 ) .

The formula (7) represents HIE of the second kind at t = 0 . Divide the interval [ 0 , T ] , 0 t T < as 0 = t 0 t 1 < < t k < < t N = T , then using the quadrature formula, the Volterra integral term in (6) becomes:

0 t k F ( t , τ ) ϕ ( x , τ ) d τ = j = 0 k u j F ( t k , t j ) ϕ ( x , t j ) + o ( i p ˜ + 1 ) , ( k 0 , p ˜ > 0 ) (8)

where k = max 0 j k h j , h j = t j + 1 t j

Using (8) in (6), we have:

ϕ k ( x ) = f k ( x ) + λ a a K ( x , y ) γ ( y , t k , ϕ k ( y ) ) d y + λ j = 0 k u j F k j ϕ j ( x ) (9)

where ϕ k ( x ) = ϕ ( x , t k ) , f k ( x ) = f ( x , t k ) , F k j = F ( t k , t j ) .

μ n ϕ n ( x ) = G n ( x ) + λ a a K ( x , y ) ϕ n ( y ) d y (10)

where μ n = 1 λ F n n u n , G n ( x ) = f n ( x ) + λ j = 0 n 1 u j F n j γ ( x , t j , ϕ j ( x ) ) , n = 0 , 1 , , N .

The formula (10) represents SHIEs of the second kind, and we have N unknown ϕ n ( x ) .

4. Some Numerical Techniques for Solving SHIEs

4.1. The TMM

In this section, we present the TMM to obtain numerical solution for HIE of the second kind with singular kernel. Consider:

ϕ ( x ) = f ( x ) + λ a a K ( | x y | ) γ ( y , ϕ ( y ) ) d y (11)

Write the integral term in the form:

a a K ( | x y | ) γ ( y , ϕ ( y ) ) d y = n = N N 1 n h n h + h K ( | x y | ) γ ( y , ϕ ( y ) ) d y , ( h = 2 a N ) (12)

Approximate the integral in the right hand side of Equation (12) by:

n h n h + h K ( | x y | ) γ ( y , ϕ ( y ) ) d y = A n ( x ) γ ( n h , ϕ ( n h ) ) + B n ( x ) γ ( n h + h , ϕ ( n h + h ) ) + R (13)

where A n ( x ) and B n ( x ) are two arbitrary functions. Putting ϕ ( x ) = 1 , x in Equation (13), where in this case we choose R = 0 . By solving the result, then we take:

A n ( x ) = 1 h [ γ ( n h + h , n h + h ) I ( x ) γ ( n h + h , 1 ) J ( x ) ] (14)

And

B n ( x ) = 1 h [ γ ( n h + h , 1 ) J ( x ) γ ( n h , n h ) I ( x ) ] (15)

where:

I ( x ) = n h n h + h K ( | x y | ) γ ( y , 1 ) d y (16)

J ( x ) = n h n h + h K ( | x y | ) γ ( y , y ) d y (17)

The relation (12), becomes:

a a K ( | x y | ) γ ( y , ϕ ( y ) ) d y = n = N N D n ( x ) γ ( n h , ϕ ( n h ) )

where

D n ( x ) = { A N ( x ) , n = N A n ( x ) + B n ( x ) , N < n < N B N 1 ( x ) , n = N (18)

The IE (11) becomes:

ϕ ( x ) λ n = N N D n ( x ) γ ( n h , ϕ ( n h ) ) = f ( x ) (19)

Putting x = m h , we have:

ϕ m λ n = N N D n , m γ n ( ϕ n ) = f m , N m N (20)

where ϕ m = ϕ ( m h ) , D n , m = D n ( m h ) , f m = f ( m h ) .

The matrix D n , m may be written as D n , m = G n , m + E n , m , where:

G n , m = A n ( m h ) + B n 1 ( m h ) , N n , m N (21)

Is a Toeplitz matrix of order 2 N + 1 and:

E n , m ( x ) = { B N 1 ( x ) , n = N , m = N + i 0 , N < n < N A N ( x ) , n = N , m = N + i (22)

where 0 i 2 n . The solution of the formula (20):

ϕ m = [ I λ ( G n , m + E n , m ) ] 1 f m , | I λ ( G n , m + E n , m ) | 0 (23)

Also

R = | a a K ( | x y | ) γ ( y , ϕ ( y ) ) d y n = N N D n m γ ( n h , ϕ ( n h ) ) | (24)

4.2. The PNM

Consider:

ϕ ( x ) λ a a p ( x , y ) K ¯ ( x , y ) γ ( y , ϕ ( y ) ) d y = f ( x ) (25)

where p and K ¯ are badly behaved and well-behaved functions of their arguments, respectively. Then, we get:

ϕ ( x i ) λ j = 0 N w i j K ¯ ( x i , y j ) γ ( y j , ϕ ( y j ) ) = f ( x i ) (26)

where x i = y i = a + i h , i = 0 , 1 , , N with h = 2 a N , N even and w i j are the weights. When x = x i , we write:

a a p ( x i , y ) K ¯ ( x i , y ) γ ( y , ϕ ( y ) ) d y = j = 0 N 2 2 y 2 j y 2 j + 2 p ( x i , y ) K ¯ ( x i , y ) γ ( y , ϕ ( y ) ) d y (27)

Form relation (25) through (27) we find:

j = 0 N w i j K ¯ ( x i , y j ) γ ( y j , ϕ ( y j ) ) = j = 0 N 2 2 y 2 j y 2 j + 2 p ( x i , y ) K ¯ ( x i , y ) γ ( y , ϕ ( y ) ) d y (28)

Then, we obtain:

a a p ( x i , y ) K ¯ ( x i , y ) γ ( y , ϕ ( y ) ) d y = j = 0 N 2 2 y 2 j y 2 j + 2 p ( x i , y ) { ( y 2 j + 1 y ) ( y 2 j + 2 y ) 2 h 2 γ ( y 2 j , ϕ ( y 2 j ) ) + ( y y 2 j ) ( y 2 j + 2 y ) h 2 γ ( y 2 j + 1 , ϕ ( y 2 j + 1 ) ) + ( y 2 j + 1 y ) ( y 2 j y ) 2 h 2 γ ( y 2 j + 2 , ϕ ( y 2 j + 2 ) ) } d y

Therefore:

w i , 0 = β 1 ( y i ) w i , 2 j + 1 = 2 γ j + 1 ( y i ) w i , 2 j = α i ( y i ) + β j + 1 ( y i ) w i , N ( y i ) = α N 2 ( y i ) (29)

where:

α j ( y i ) = 1 2 h 2 y 2 j 2 y 2 j p ( y i , y ) ( y y 2 j 2 ) ( y y 2 j 1 ) d y β j ( y i ) = 1 2 h 2 y 2 j 2 y 2 j p ( y i , y ) ( y y 2 j 1 ) ( y y 2 j ) d y γ j ( y i ) = 1 2 h 2 y 2 j 2 y 2 j p ( y i , y ) ( y y 2 j 2 ) ( y 2 j y ) d y (30)

We now introduce the change of variable y = y 2 j 2 + ζ h , 0 ζ 2 thus the system (30) becomes:

α j ( y i ) = h 2 0 2 ζ ( ζ 1 ) p ( y 2 j 2 + ζ h , y i ) d ζ

β j ( y i ) = h 2 0 2 ( ζ 1 ) ( ζ 2 ) p ( y 2 j 2 + ζ h , y i ) d ζ

γ j ( y i ) = h 2 0 2 ζ ( 2 ζ ) p ( y 2 j 2 + ζ h , y i ) d ζ

If we define:

ψ i = 0 2 ζ i p ( y 2 j 2 + ζ h , y i ) d ζ

For p ( x , y ) = p ( x y ) , we have:

ψ i = 0 2 ζ i p ( y i , y 2 j 2 + ζ h ) d ζ , i = 0 , 1 , 2 (31)

When y i y 2 j 2 = ( i 2 j + 2 ) h . If we assume z = i 2 j + 2 , then:

α j ( y i ) = h 2 0 2 ζ ( ζ 1 ) p ( z ζ ) d ζ β j ( y i ) = h 2 0 2 ( ζ 1 ) ( ζ 2 ) p ( z ζ ) d ζ γ j ( y i ) = h 2 0 2 ζ ( 2 ζ ) p ( z ζ ) d ζ (32)

Hence, the system (29) becomes:

w i , 0 = h 2 [ 2 ψ 0 ( z ) 3 ψ 1 ( z ) + ψ 2 ( z ) ] , z = i w i , 2 j + 1 = h [ 2 ψ 1 ( z ) ψ 2 ( z ) ] , z = i 2 j

w i , 2 j = h 2 [ ψ 2 ( z ) ψ 1 ( z ) + 2 ψ 0 ( z 2 ) 3 ψ 1 ( z 2 ) + ψ 2 ( z 2 ) ] , z = i 2 j + 2 w i , N = h 2 [ ψ 2 ( z ) ψ 1 ( z ) ] , z = i N + 2 (33)

Therefore, the integral Equation (25) is reduced to SLAEs as in (26) or: ( I λ W ) ϕ = F

Which has the solution:

ϕ = ( I λ W ) 1 F , | I λ W | 0 (34)

The PNM is said to be convergent of order r in [ a , a ] . If for N sufficiently large, there exists a constant C > 0 independent of N such that:

ϕ ( x ) ϕ N ( x ) C N r

5. Numerical Applications

We using TMM and PNM at N = 20 , 40 , T = 0.03 , 0.7 , λ = 1 , and μ = 1 . In Tables 1-4:

ϕ E x a c t ® Exact solution, ϕ T ® appro. sol. of TMM, E T ® the absolute error of TMM, ϕ N ® appro. sol. of PNM, E N ® the absolute error of PNM.

Example 1

Consider:

ϕ ( x , t ) = f ( x , t ) + λ 1 1 ln | x y | ( y t ) 2 d y + λ 0 t τ 2 ϕ ( x , τ ) d τ

Table 1. The values of exact, approximate solutions, and errors by using TMM, PNM at N = 20.

Table 2. The values of exact, approximate solutions, and errors by using TMM, PNM at N = 40.

Table 3. The values of exact, approximate solutions, and errors by using TMM, PNM at N = 20.

Table 4. The values of exact, approximate solutions, and errors by using TMM, PNM at N = 40.

Exact solution: ϕ ( x , t ) = x t

Example 2

Consider:

ϕ ( x , t ) = f ( x , t ) + λ 1 1 ln | x y | ( y t 2 ) 2 d y + λ 0 t t τ ϕ ( x , τ ) d τ

Exact solution: ϕ ( x , t ) = x t 2

6. Conclusion

The goal of this work is to study the H-VIE with singular kernel of the second kind. TMM and PNM are successive to solve this equation numerically. As N is increasing, the errors are decreasing. As t is increasing, the errors are increasing.

Cite this paper: Al-Bugami, A. (2021) Singular Hammerstein-Volterra Integral Equation and Its Numerical Processing. Journal of Applied Mathematics and Physics, 9, 379-390. doi: 10.4236/jamp.2021.92026.
References

[1]   Abdou, M.A. (2003) Fredholm-Volterra Integral Equation with Singular Kernel. Applied Mathematics and Computation, 137, 231-243.
https://doi.org/10.1016/S0096-3003(02)00046-2

[2]   Al-Bugami, A.A. (2008) Some Numerical Method for Solving Singular and Nonlinear Integral Equation. Umm Al-Qura University, Makkah.

[3]   Abdou, M.A., Elsayed, W.G. and Deebs, E.I. (2005) A Solution of a Nonlinear Integral Equation. Applied Mathematics and Computation, 160, 1-14.
https://doi.org/10.1016/S0096-3003(03)00613-1

[4]   Abdou, M.A. and Hendi, F.A. (2005) Numerical Solution for Fredholm Integral Equation with Hilbert Kernel. The Journal of the Korean Society for Industrial and Applied Mathematics, 9, 111-123.

[5]   Al-Bugami, A.A. (2013) Toeplitz Matrix Method and Volterra-Hammerstein Integral Equation with a Generalized Singular Kernel. Progress in Applied Mathematics, 6, 16-42.
https://doi.org/10.14419/ijbas.v2i1.601

[6]   Abdou, M.A., El-Boria, M.M. and El-Kojok, M.M. (2009) Toeplitz Matriex Method and Nonlinear Integral Equation of Hammerstein Type. Journal of Computational and Applied Mathematics, 223, 765-776.
https://doi.org/10.1016/j.cam.2008.02.012

[7]   Al-Bugami, A.M. (2013) Error Analysis for Numerical Solution of Hammerstein Integral Equation with a Generalized Singular Kernel. Progress in Applied Mathematics, 6, 1-15.
https://doi.org/10.14419/ijbas.v2i1.601

[8]   Shahsavaran, A. (2011) Lagrange Functions Method for Solving Nonlinear Fredholm-Volterra Integral Equation. Applied Mathematical Sciences, 5, 2443-2450.

[9]   Darwish, M.A. (2004) On Nonlinear Fredholm-Volterra Integral Equations with Hysteresis. Journal Applied Mathematics and Computation, 156, 479-484.
https://doi.org/10.1016/j.amc.2003.08.006

[10]   Mirzaee, F. (2017) Numerical Solution of Nonlinear Fredholm-Volterra Integral Equations via Bell Polynomials. Computational Methods for Differential Equations, 5, 88-102.

[11]   Raad, S.A. (2005) Some Numerical Methods for Solving Singular Integral Equation. Umm Al-Qura University, Makkah.

 
 
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