APPLICATION OF HYBRID METHODS TO THE
SOLUTION
OF VOLTERRA INTEGRAL EQUATIONS
Ramil Mirzayev1, Galina Mehdiyeva2, Vagif Ibrahimov3
Baku State University, Baku, Azerbaijan
r_mirzoyev@mail.ru, 2, 3 ibvag@yahoo.com
1
As is known, many problems of scientific-technical problems
are reduced to the solution of integral equations with variable
boundaries. The scientists are engaged in approximate solutions of
such equations for a long time. One of the papers in this field belongs
to V.Volterra. He is the initiator in investigation and application of
integral equations with variable boundary. Therefore, in honour of
Volterra, these equations are called Volterra equations. Mainly,
different variations of the quadrature method are used in numerical
solution of these equations. Here the hybrid type concrete onestep
method is applied to numerical solution of Volterra integral equations.
Introduction. Consider the following Volterra integral
equation of second kind:
x
y ( x)  f ( x)   K ( x, s, y ( s))ds, x [ x0 , X ],
(1)
x0
We assume that integral equation (1) has a unique continuous
solution determined on the segment [ x0 , X ] . Denote this solution by
the function y (x) and find its approximate values. For that we partition the segment [ x0 , X ] into N equal parts by means of the constant step 0  h and determine the partitioning points in the form:
xm  x0  mh (m  0,1,2,..., N ) . Approximate values of the function
y (x) at the points xm (m  0,1,2,..., N ) are denoted by y m , the exact
ones by y ( xm ) .
The classic method of solution to integral equations (1) is the
quadrature method that in one variant may be written in the form
(see, for example, [1], [2]):
n
y n  f n  h ai K ( xn , xi , yi ) (n  0,1,2,..., N ),
(2)
i 0
where f n  f ( xn ), ai (i  0,1,...,n) are the coefficients of the
quadrature formula. As it follows from (2), the number of inversions
to calculations of the kernel K ( x, s, y(s)) increases due to increase
of the values of the quantity n . In order to preserve the amount of
computational works in solving integral equations (1) at each step,
we suggest a multistep method with constant coefficients, of the
form:
k
 y
i
i 0
k
n i
k
k
  i f ni  h  i( j ) K ( xn j , xni yni ),
i 0
(3)
j 0 i 0
where the coefficients  i , i( j ) (i, j  0,1,2,..., k ) are some real
numbers, moreover  k  0 , that are determined from the
homogeneous system of linear algebraic equations.
Usually, in the theory of numerical methods, onestep and
multistep methods are investigated. Each of them has its advantage
and lack. Taking into account what has been noted, the scientists
suggested to construct the methods on the joint of these directions
that could preserve the best properties of onestep and multistep
methods and called them hybrid methods. Therefore, here we attempt
to apply hybrid methods to the numerical solution of equation (1).
1. Construction of the hybrid method for solving Volterra
integral equations.
Consider a special case when the function K ( x, s, y(s)) is
independent of the argument x and denote it by F (s, y)  K ( x, s, y) .
Then from equations (1) we can write:
y ( x)  f ( x)  F ( x, y ( s )), y ( x0 )  f ( x0 ) .
(1.1)
This is an onestep method and has degree of accuracy p  3 .
Formally, method (1.2) may be considered as a twostep one taking
into account that three points x n , x n  h / 3, x n  h participate in it.
Since xn  h / 3 is not contained in the set of partitioning points,
method (1.2) may be considered as an onestep method. However,
method (1.2) may be replaced by the following one:
This method is an onestep method and has degree of accuracy
p  3 . Here, on the base of method (1.2) we construct a method of
type (3). For that, we show that it is possible to apply method (1.2) to
finding the numerical solution of equation (1). To that end, we use
some values of the solution of the function y (x) determined as:
xn 1
y ( xn 1 )  f n 1 
 K (x
n 1
, s, y ( s ))ds .
x0
Using the Lagrange theorem, we can write:
K ( xn 1 , s, y ( s ))  K ( xn , s, y ( s))  hK x ( n , s, y ( s )),
where x n   n  x n 1 .
Consequently, we can write:
xn
xn
x0
x0
 ( K ( xn1 , s, y(s))  K ( xn , s, y(s)))ds  h  K x ( n , s, y(s))ds.
Then we can write
xn
h  K x ( n , s, y ( s ))ds  O(h).
(1.5)
x0
If we take into account the ones obtained in (1.4), we can write
x n 1
y n 1  y n  f n 1  f n 
 K (x
n 1
, s, y ( s ))ds .
(1.6)
xn
The integral replaced by a small quantity of first order with
respect to h may be replaced by a small quantity of higher order
with respect to h by adding an intermediate point while investigating
the difference of the function y (x) , for example in the following
way:
y ( xn1 )  2 y ( xn1/ 2 )  y ( xn ) .
Indeed, in this case we have:
y( xn1 )  2 y( xn1/ 2 )  y ( xn )  f n1  2 f n1/ 2  f n 
xn
 (K( x
n 1
, s, y( s ))  2 K ( xn1/ 2 , s, y ( s ))  K ( xn , s, y( s )))ds 
x0
xn 1
xn 1 / 2
xn
xn
 K ( xn1 , s, y( s))ds  2
 K(x
n 1 / 2
, s, y ( s ))ds.
Thus, we get that in estimating the first integral participating
in (1.4), we apply (1.8) and repeat what has been written above,
change the integral by second order quantity of the following form:
xn
 (K (x
n 1
, s, y ( s))  2 K ( xn1 / 2 , s, y ( s ))  K ( xn , s, y ( s )))  O(h 2 ).
x0
Notice that while approximating the differential equation
y ( x)  f ( x, y ) , at the point x n , its left hand side, i.e. the derivatives
y ( xn ) may be replaced as: yn1  yn  hyn  O(h 2 ) . A method for
its solution may be constructed as follows:
k
y n 1  y n  h  i f ( xni , y n i ) .
i 0
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