a numerical method for solving a system of volterra integral equations

8/10/2019 A Numerical Method for Solving a System of Volterra Integral Equations

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Finding the exact solution of the system of integral equations by classical methods is some-times too difficult, and it is usually very useful to find a numerical estimation of the exact

solution.Consider the following system of Volterra integral equations:

u1(x, t) =f1(x) +x0

k1(x1, t)1

u1(x1, t), ... ,un(x1, t)

dx1,

u2(x, t) =f2(x) +x0

k2(x1, t)2

(u1(x1, t), ... ,un(x1, t)

dx1,

::

un(x, t) =fn(x) +x0

kn(x1, t)n

(u1(x1, t), ... ,un(x1, t)

dx1,

where Ki(x, t) : R2 R(i = 1..n) are the kernels which are a known functions, fi(.) :

R R(i = 1..n) are given function. We are trying to find approximate solutions of the

above system for ui(.)(i = 1..n) and (ui(.)), are linear or nonlinear functions of ui(.)(i =1..n). Some different techniques may be used for solving our system such as Adomians

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World Applied Programming, Vol (2), No (1), January 2012. 18-33

ISSN: 2222-2510

2011 WAP journal. www.waprogramming.com

A Numerical Method for Solving a System ofVolterra Integral Equations

Mohammad Reza Yaghouti

Department of Mathematics,

Faculty of Science

The University of Guilan,

Rasht, Iran

[email protected]

Abstract:In this paper we propose a new approach for solving linear and nonlinear systems of Volterra

integral equations (LISVIE, NSVIE) of the first and the second kinds. This approach is very easy to use for

some another kinds of equations.

Keywords:Systems of Volterra integral equations, Discrimination, Nonlinear programming

I.Introduction


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decomposition method, Hes Homotopy perturbation method and etc. In this paper wepropose a new numerical approach for solving the above system, both linear and nonlinearby discretisation and using an interpolation method to find formulas for solutions of such

system of integral equations.

II.Theorems

In this section we bring some theorems and results that can be found in [1].

Theorem 2.1. Letfi(x, t)be given functions anda andb are constant, and let{t1, t2, ... , tn}

be a set of supporting points in[a, b], wherea= t1 < ... < tn= b, then ba

fi(x, t)dt = limn

j=n1j=1

fi(x, j)tj(i= 1..n) (1)

Remark 1: If we choose the same distance between support points, from (1) we obtain thefollowing formula: b

a

fi(x, t)dt= limn

h

j=n1j=0

fi(x, j) ; i= 1..n (2)

whereh= ban and j(i= 0..n 1) are arbitrary points in the interval [tj, tj+1].

Theorem 2.2: Letf(x) be a convex function on a convex set, then any local minimum off is a global one [2].

Theorem 2.3: Consider n convex functions f1, f2,...,fn, then g(x) =n

i=1ifi(x) is alsoa convex function fori(i= 1..n).

Proof. We have fi(x) are convex functions for i= 1..n. It means

fi(x+ (1 )y) fi(x) + (1 )fi(y) ;i= 1..n

Now, from

g(x) =n

i=1

ifi(x)

19

Mohammad Reza Yaghouti, World Applied Programming (WAP),Vol (2), No (1), January 2012.


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We take

g(x+ (1 y)) =n

i=1

ifi(x+ (1 y))n

i=1

ifi(x) + (1 )fi(y)=

ni=1

ifi(x) + (1 )ni=1

ifi(y) =g(x) + (1 )g(y).

III.An approach for solving SNVIE and SLVIE

In this section we propose our method to find the numerical solutions of system of nonlinearVolterra integral equations of the form

u1(x, t) =f1(x) +x0

k1(x1, t)1

u1(x1, t), ... ,un(x1, t)

dx1,

u2(x, t) =f2(x) +x0

k2(x1, t)2

(u1(x1, t), ... ,un(x1, t)

dx1,

::

un(x, t) =fn(x) +x0

kn(x1, t)n

(u1(x1, t), ... ,un(x1, t)

dx1,

where fi(x), ki(x, t) are given functions and u and fare continuous functions on [0, 1] Rand [0, 1].

The basis of our method exclusively focuses upon discretisation. We can rewrite theabove system as follows:

u1(x, t) f1(x) x0

k1(x1, t)1

u1(x1, t), ... ,un(x1, t)

dx1= 0,

u2(x, t) f2(x) x0

k2(x1, t)2

(u1(x1, t), ... ,un(x1, t)

dx1 = 0,

::

un(x, t) fn(x) x0

kn(x1, t)n

(u1(x1, t), ... ,un(x1, t)

dx1= 0, 0 x 1, t R

(1)

It is important to note that all closed open interval can change to interval [0, 1]. Let

Eu1(x, t) =u1(x, t) f1(x) x0

k1(x1, t)1

u1(x1, t), ... ,un(x1, t)

dx1 = 0,

Eu2(x, t) =u2(x, t) f2(x) x0

k2(x1, t)2

(u1(x1, t), ... ,un(x1, t)

dx1= 0,

::

Eun(x, t) =un(x, t) fn(x) x0

kn(x1, t)n

(u1(x1, t), ... ,un(x1, t)

dx1= 0, 0 x 1, t

(2)

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where Eui(x, t) are functional and depend on the unknown functions ui(x, t), so Eui :P C[0, 1]R Rwhere PC means they are piecewise continuous on the interval [0, 1]R.To solve (1), consider the following problem:

M inuiPC[0,1]R

ni=1

Eui 22, 0 x 1, t R (3)

where Eui 2=

X=[0,1]R| (Eui)|

2 d 12

.

Theorem 3.1. Let U(x, t) = (u1(x, t), u2(x, t), . . . , un(x, t))T be continuous functions on

[0, 1]Rand a solution for (1), thenU(x, t)is the optimal solution of (2) with zero objective

function and vice versaProof. Let ui(x, t);i= 1..nbe solutions for (1), which are continuous on [0, 1] R, so

ui(x, t) f(x)

x0

k(x1, t)(u1(x1, t),...,un(x1, t))dx1 = 0, 0 x 1, t R; i= 1..n

Hence,

ui(x, t) f(x) x0

k(x1, t)(u1(x1, t),...,un(x1, t))dx1

= 0 ; i= 1..n.Sinceuiand fare continuous on their domain, by integrating both sides of the last equation,

we obtain[0,1]R

ui(x, t) f(x) x0

k(x1, t)(u1(x1, t),...,un(x1, t))dx1

dxdt= 0, ; i= 1..n,Thus, ||Eui||1 = 0 and this means that ui(x, t); i = 1..n are the optimal solutions of (2)with zero objective function.

For the converse part of the proof, we let ui(x, t) be the optimal solutions of (2) withzero objective functions, then

[0,1]R

ui(x, t) f(x) x

0

k(x1, t)(u1(x1, t),...,un(x1, t))dx1dxdt= 0, ; i= 1..n, (4)Since ui(x, t) f(x)

x0

k(x1, t)(u1(x1, t),...,un(x1, t))dx1

;i= 1..n.are absolute functions, by using Lebesgue integral theorems, we see that the following equal-ity must be held

ui(x, t) f(x) x0

k(x1, t)(u1(x1, t),...,un(x1, t))dx1

= 0, 0 x 1, t R; i= 1..n.21



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which

ui(x, t) f(x) x

0

k(x1, t)(u1(x1, t),...,un(x1, t))dx1 = 0 ; i= 1..n.

so ui(x, t) are the solutions of (1).

By theorem 2.1, we have

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ui(x, t) fi(x)

x0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

2dx (5)

= limn

n1j=0

xj+1xj

ui(x, t) fi(x)

x0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

2dx

= limn

n1j=0

h

ui(j , t) fi(j) j0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx12

, i= 1..n

wherej for j = 0 .. n 1 are arbitrary points in the interval [xj , xj+1], h= 1n

and xj =j h.

We can choose j = xj, the lower bound in each interval: thus, (5) changes to thefollowing:

limn

1

n

n1j=0

ui(xj, t) fi(xj)

xj0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

2i= 1..n (6)

If we use the same method as we discussed above to approximation the inner integral in (6),then we have

xj0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1 =

jh0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

=

h0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

+

2hh

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

...

+

jh

(j1)h

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1 (7)

=

j1=0

(+1)hh

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

i= 1..n

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Since his small, we can approximate each integral by

(+1)h

h

ki(x1, t)iu1(x1, t), ... ,un(x1, t)dt = 1n

ki(j , t)u(j , t)

= 1

nki(h,t)u(h,t), i= 1..n (8)

where= h. By use of (5), (3) and (8), finally we have

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ui(x, t) fi(x)

x0

ki(x1, t)i

u1(x1, t), ... ,un(x1, t)

dx1

2dx (9)

= limn

1

n

n1j=0

ui(xj, t) fi(xj) h

j1=0

ki(xj , t)i

u1(xj, t), ... ,un(xj, t)2

i= 1..n

IV.Some numerical examples

In this section, we solve some examples by our method, and we can compare the numericalresults with the exact solutions and Hes homotopy perturbation method[3].

Example 4.1. Consider the following nonlinear system of Volterra integral equations:

f1(x) = sin(x) x+ x0

f21 (t) +f

22 (t)

dt,

f2(x) = cos(x) 1

2sin2(x) +

x0

f1(t)f2(t)dt

with the exact solutions f1(x) =sin(x) and f2(x) =cos(x).

We explain this method for n= 200 as follows,

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f1(x) sin(x) +x

x0

f21 (t) +f

22 (t)

dt2

dx =

= 0.0050

f1(x) sin(x) +x

x0

f21 (t) +f

22 (t)

dt2dx

+

0.0100.005

f1(x) sin(x) +x

x0

f21 (t) +f

22 (t)

dt2

dx

+ ...

+

10.995

f1(x) sin(x) +x

x0

f21 (t) +f

22 (t)

dt2

dx

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and

1

0f2(x) cos(x) +

1

2

sin2(x) x

0

f1(t)f2(t)dt2

dx =

=

0.0050

f2(x) cos(x) +

1

2sin2(x)

x0

f1(t)f2(t)dt2

dx

+

0.0100.005

f2(x) cos(x) +

1

2sin2(x)

x0

f1(t)f2(t)dt2

dx

+ ...

+

10.995

f2(x) cos(x) +

1

2sin2(x)

x0

f1(t)f2(t)dt2

dx

If we discretize the inner integral as above and replace the lower bound and t in eachintegral together, then we have

10

f1(x) sin(x) +x

x0

f21 (t) +f

22 (t)

dt2

dx=

= 0.005

f1(0) sin(0) + 0 02

+ 0.005

f1(0.005) sin(0.005) + 0.005 0.005

f21 (0) +f22 (0)

2

+ 0.005

f1(0.010) sin(0.010) + 0.010 0.005

(f21 (0) +f22 (0))

+(f21 (0.005) +f22 (0.005))2

+...

+ 0.005

f1(0.995) sin(0.995) + 0.995 0.005

(f21 (0) +f22 (0)) + (f

21 (0.005) +f

22 (0.005

. . .+ (f21 (0.990) +f22 (0.990))

2

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and

1

0f2(x) cos(x) +

1

2

sin2(x) x

0f1(t)f2(t)dt

2

dx=

= 0.005

f2(0) cos(0) +1

2sin2(0) 0

2

+ 0.005

f2(0.005) cos(0.005) +1

2sin2(0.005) 0.005

f1(0)f2(0)

2

+ 0.005

f2(0.010) cos(0.010) +1

2sin2(0.010) 0.005

(f1(0)f2(0))

+(f1(0.005)f2(0.005))2

+

...

+ 0.005

f2(0.995) cos(0.995) +1

2sin2(0.995) 0.005

(f1(0)f2(0)) + (f1(0.005)f2(0.005

. . .+ (f1(0.990)f2(0.990))2

For n= 200, the approximated graphs off1(x) and f2(x) in example 1, also the graphsof error are presented in Fig.1 till Fig.4.

Example 4.2.Consider the following nonlinear system of Volterra integral equations:

f1(x) = x x4 + 4

x0

f1(t)f2(t)

dt,

f2(x) = x2 x8 + 8

x0

f4(t)f3(t)dt

f3(x) = 2

3x3

1

2x2

1

5x5

1

4x4 +

x0

f1(t) +f2(t) +f3(t)

dt,

f4(x) = x4 x6 + 6

x0

f2(t)f3(t)dt

with the exact solutions f1(x) =x, f2(x) =x2, f3(x) =x3, f4(x) =x4.

Forn= 200, the approximated graphs off1(x), f2(x), f3(x) andf4(x) also the graphs oferror are presented in Fig.5 till Fig.12.

It is important to note that our method can extended to any finite interval. For this matter,we have just to change our wanted interval in the program.

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In this paper, we extended a new method for a system of Volterra integral equations, thatwas introduced in [1] for a Volterra integral equation. Also two examples are presented toshow the ability of this method.

Table 1: The results of our method for n= 200 in example 1 by exact solution.

Our numerical solutions Error

Points xi f1(xi) f2(xi) f1(xi) exactf1(xi) f2(xi) exactf2(xi)

0 0.3187380978e-21 1.0 0.3187380978e-21 0.00.1 0.9980962836e-1 0.9947543056 -0.2378829e-4 -0.2498597e-3

0.2 0.1985710278 0.9795671240 -0.983030e-4 -0.4994538e-30.3 0.2952945123 0.9545875554 -0.2256944e-3 -0.7489337e-30.4 0.3890085388 0.9200622849 -0.4098035e-3 -0.9987091e-30.5 0.4787691437 0.8763330070 -0.6563949e-3 -0.12495549e-20.6 0.5636689987 0.8238329424 -0.9734747e-3 -0.15026725e-20.7 0.6428460150 0.7630825152 -0.13716722e-2 -0.17596721e-20.8 0.7154914334 0.6946842732 -0.18646575e-2 -0.20224361e-20.9 0.7808573608 0.6193171472 -0.24695488e-2 -0.22928211e-21 0.8382637337 0.5377301892 -0.32072511e-2 -0.25721167e-2

For example 2, because we do not have enough space in table, we bring just error valuesin table 2 .

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V.Conclusion


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Table 2: The difference of our method for n= 200 in example 2 by exact solution.

Error

Points xi f1(xi) exactf1(xi) f2(xi) exactf2(xi) f3(xi) exactf3(xi) f4(xi) exactf4(xi)

0 0.1567010952e-19 0.9613823951e-19 0.9952637130e-19 0.5295337169e-260.1 -0.975555e-5 -0.5138e-8 -0.290305625e-3 -0.5235482e-60.2 -0.793991e-4 -0.51957e-6 -0.68242003e-3 -0.11979564e-40.3 -0.2726887e-3 -0.827148e-5 -0.121236953e-2 -0.7885978e-40.4 -0.6670297e-3 -0.618086e-4 -0.19290648e-2 -0.31008957e-30.5 -0.13827626e-2 -0.3057475e-3 -0.29046236e-2 -0.9214861e-30.6 -0.26674134e-2 -0.11700524e-2 -0.42632859e-2 -0.23206338e-20.7 -0.51370041e-2 -0.37834746e-2 -0.62570420e-2 -0.53439430e-20.8 -0.104681288e-1 -0.110009642e-1 -0.9460342e-2 -0.12010337e-1

0.9 -0.234588528e-1 -0.304139877e-1 -0.15295304e-1 -0.28018865e-11 -0.586183397e-1 -0.848858099e-1 -0.275988574e-1 -0.716213703e-1

References

[1] A. Vahidian Kamyad, M. Mehrabinezhad, J. Saberi-Najafi. A Numerical Approach forSolving Linear and Nonlinear Volterra Integral Equations with Controlled Error.IAENG International Journal of Applied Mathematics, IJAM, in press (2010).

[2] M. S. Bazara, C.M. Shetty, Nonlinear programming: Theory and Algoritms,JonWiley and Sons, Inc., New York, (1979).

[3] J. Biazar, H. Ghazvini,Hes homotopy perturbation method for solving systemsof Volterra integral equations of the seond kind , Chaos Solutions and Fractals,39, 770-777, (2009).

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Figure 1: Approximated(point) and exact solution(line) for f1(x) in Ex.1.

Figure 2: Error graph for f1(x) in Ex.1.

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Figure 11: Approximated(point) and exact solution(line) forf4(x) in Ex.2.


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a numerical method for solving a system of volterra integral equations

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