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# Adomian decomposition method for solving twodimensional nonlinear Volterra fuzzy integral equations

## 1.

RESEARCH ARTICLE | DECEMBER 10 2018Adomian decomposition method for solving twodimensional nonlinear Volterra fuzzy integral equations

Artan Alidema; Atanaska Georgieva

AIP Conf. Proc. 2048, 050009 (2018)

https://doi.org/10.1063/1.5082108

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09 March 2024 19:34:42

## 2.

Adomian decomposition method for solving two-dimensionalnonlinear Volterra fuzzy integral equations

Artan Alidema1,b) and Atanaska Georgieva2,a)

2

1

FMNS, University of Prishtina ”Hasan Prishtina”, str. Nena Tereze 1000, Prishtina, Kosovo

Faculty of Mathematics and Informatics, University of Plovdiv ”Paisii Hilendarski”, 24 Tzar Asen, 4000 Plovdiv,

Bulgaria

a)

Corresponding author: [email protected]

b)

[email protected]

Abstract. A numerical method for solving two-dimensional nonlinear Volterra fuzzy integral equation (2D-NVFIE) of the second

kind will is introduced. We convert a nonlinear Volterra fuzzy integral equation to a nonlinear system of Volterra integral equation

in crisp case. We use Adomian Decomposition Method (ADM) to find the approximate solution of this system and hence obtain

an approximation for fuzzy solution of the nonlinear Volterra fuzzy integral equation. Also, the existence and uniqueness of the

solution and convergence of the proposed method are proved. Some numerical examples are included to demonstrate the validity

and applicability of the proposed technique.

The fuzzy differential and integral equations are one of the important part of the fuzzy analysis theory that play major

role in numerical analysis. The concept of integration of fuzzy functions has been introduced by Dubois and Prade

[8], Goetschel and Voxman [16], Kaleva [18] and others. The first applications of fuzzy integration was given by Wu

and Ma who investigated the fuzzy Fredholm integral equation of the second kind.

The study of fuzzy integral equations begins in Kaleva [18], Seikkala [24] and Mordeson and Newman [19],

such integral equations being applied in control mathematical models. The problems posed in the study of fuzzy

integral equations are: existence and uniqueness, boundedness of the solutions [10, 11, 14] and the construction of

numerical methods for the approximate solution.

The existing numerical methods for fuzzy integral equations are based on various techniques: successive approximations and iterative methods [7, 12, 15], analytic-numeric methods like Adomian decomposition, homotopy

analysis and homotopy perturbation [3, 6, 13], quadrature rules and Nystrom techniques [23], Bernstein polynomials

[20], Chebyshev polynomials [5], fuzzy differential transforms [22], expansion methods such as fuzzy collocation,

Galerkin techniques and Taylor series [17].

Adomian decomposition method (ADM) has been recently intensively studied by scientists and engineers and

used for solving nonlinear differential and integral problems. The ADM intruduced by Adomian [1, 2] for solving

different kind of functional equations and has been subject of extensive numerical and analytical studies. ADM consist

of decomposing un-knwon function od any equations into a sum of an infinite number of components define by the

decompositon series. The method essentially is a power series method similar to the perturbation technique. Babolian,

Goghary and Abbasbandy [4] used ADM to solve fuzzy Fredholm integral equations, respectively Allahviranloo,

Abbasbandy [3], solved fuzzy system of Fredholm integral equations. Also, Behzadi [6] solving nonlinear fuzzy

Volterra-Fredholm integral equations with ADM.

In this paper we propose ADM for solving two-dimensional nonlinear Volterra fuzzy integral equation (2DZ t

Z s

NVFIE)

u(s, t) = g(s, t) ⊕ (FR)

(FR)

k(s, t, x, y) G(u(x, y)))dxdy,

(1)

c

a

where g, u : A = [a, b]×[c, d] → E are continuous fuzzy-number valued functions and k : A× A → E 1 , G : E 1 → E 1

are continuous functions on E 1 . The set E 1 is the set of all fuzzy numbers.

1

Proceedings of the 44th International Conference on Applications of Mathematics in Engineering and Economics

AIP Conf. Proc. 2048, 050009-1–050009-8; https://doi.org/10.1063/1.5082108

Published by AIP Publishing. 978-0-7354-1774-8/$30.00

050009-1

09 March 2024 19:34:42

INTRODUCTION

## 3.

The remainder of this paper is organized as follows: in Section 2, we present the basic notations of fuzzy numbers,fuzzy functions and fuzzy integrals. In Section 3, the parametric from of 2D-NVFIE is introduced and then ADM is

applied for solving this equation. We aim existence and uniqueness of the solution and convergence of the proposed

method in Section 4. Finally, in Section 5, we illustrate the accuracy of method by solving numerical example.

PRELIMINARIES

In this section, we briefly state some definition and results related to fuzzy numbers and fuzzy-number-valued functions, which will be referred throughout this paper.

Definition 1

[16] A fuzzy number is a function u : R → [0, 1] satisfying the following properties:

(i) u is upper semi-continuous on R,

(ii) u(x) = 0 outside of some interval [c, d],

(iii) there are the real numbers a and b with c ≤ a ≤ b ≤ d, such that u is increasing on [c, a], decreasing on [b, d]

and u(x) = 1 for each x ∈ [a, b],

(iv) u is fuzzy convex set i. e. that is u(λx + (1 − λ)y) ≥ min{u(x), u(y)} for all x, y ∈ R and λ ∈ [0, 1].

r∈[0,1]

Lemma 1

[25] The Hausdorff metric has the following properties:

(i) (E 1 , D) is a complete metric space,

(ii) D(u ⊕ w, v ⊕ w) = D(u, v) for all u, v, w ∈ E 1 ,

(iii) D(u ⊕ v, w ⊕ e) ≤ D(u, w) + D(v, e) for all u, v, w, e ∈ E 1 ,

(iv) D(u ⊕ v, 0̃) ≤ D(u, 0̃) + D(v, 0̃) for all u, v ∈ E 1 ,

(v) D(k u, k v) = |k|D(u, v) for all u, v ∈ E 1 , for allk ∈ R.

(vi) D(k1 u, k2 u) = |k1 − k2 |D(u, 0̃) for all k1 , k2 ∈ E 1 , with k1 k2 ≥ 0 and for allu ∈ E 1 .

For any fuzzy-number-valued function f : A → E 1 we can define the functions f (., ., r), f (., ., r) : A → R, by

f (s, t, r) = ( f (s, t))r− , f (s, t, r) = ( f (s, t))r+ for each (s, t) ∈ A, for each r ∈ [0, 1]. These functions are called the left

and right r−level functions of f.

Definition 2

[26] A fuzzy-number-valued function f : A → E 1 is said to be continuous at (s0 , t0 ) ∈ A if for

1

each ε > 0 there is δ > 0 such that D( f (s, t), f (s0 , t0 )) < ε whenever ((s − s0 )2 + (t − t0 )2 ) 2 < δ. If f be continuous for

each (s, t) ∈ A then we say that f is continuous on A. A fuzzy number u ∈ E 1 is upper bound for a fuzzy-number-valued

function f : A → E 1 if f (s, t, r)r ≤ ur− and f (s, t, r) ≤ ur+ for all (s, t) ∈ A, r ∈ [0, 1]. A fuzzy number u ∈ E 1 is lower

bound for a fuzzy number-valued function f : A → E 1 if ur+ ≤ f (s, t, r) and ur− ≤ f (s, t, r) for all (s, t) ∈ A, r ∈ [0, 1].

A fuzzy-number-valued function f : A → E 1 is said to be bounded it has a lower bound and an upper bound.

Lemma 2

[26] If f : A → E 1 is continuous then it is bounded and its supremum sup f (s, t) must exist and

(s,t)∈A

is determined by u ∈ E 1 with ur− = sup f (s, t, r) and ur+ = sup f (s, t, r). A similar conclusion for the infinum is also

(s,t)∈A

(s,t)∈A

true.

On the set C(A, E 1 ) = { f : A → E 1 ; f is continuous} there is defined the metric

D ( f, g) = max D( f (s, t), g(s, t)), for all f, g ∈ C(A, E 1 ). We see that (C(A, E 1 ), D∗ ) is a complete metric space.

∗

(s,t)∈A

050009-2

09 March 2024 19:34:42

The set of all fuzzy numbers is denoted by E 1 . Any real number a ∈ R can be interpreted as a fuzzy number

ã = χ[a] and therefore R ⊂ E 1 .

According to [25] for any 0 < r ≤ 1 we denote the r−level set [u]r = {x ∈ R : u(x) ≥ r} that is a closed

interval and [u]r = [ur− , ur+ ] for all r ∈ [0, 1]. These lead to the usual parametric representation of a fuzzy number,

by an ordered pair of functions (ur− , ur+ ), which satisfies the following properties: ur− is bounded left continuous nondecreasing function over [0, 1], ur+ is bounded left continuous non-increasing function over [0, 1] and ur− < ur+ .

For u, v ∈ E 1 , k ∈ R, the addition and the scalar multiplication

by

( are defined

[kur− , kur+ ], if k ≥ 0

r

r

r

r

r

r

r

r

r

[u ⊕ v] = [u] + [v] = [u− + v− , u+ + v+ ] and [k u] = k.[u] =

for all r ∈ [0, 1].

[kur+ , kur− ], if k < 0

The neutral element respect to ⊕ in E 1 , denoted by 0̃ = χ[0] . The algebraic properties of addition and scalar

multiplication of fuzzy numbers are given in [25].

As a distance between fuzzy numbers we use the Hausdorff metric [25] defined by D(u, v) =

sup max{|ur− − vr− |, |ur+ − vr+ |} for any u, v ∈ E 1 .

## 4.

Definition 3[25] Let f : A → E 1 , for ∆nx : a = x0 < x1 < ... < xn = b and ∆ny : c = y0 < y1 < ... <

yn = d, be two partitions of the intervals [a, b] and [c, d], respectively. Let one consider the intermediates points

ξi ∈ [xi−1 , xi ] and η j ∈ [y j−1 , y j ], i = 1, ..., n; j = 1, ..., n, and δ : [a, b] → R+ and σ : [c, d] → R+ . The divisions

P x = ([xi−1 , xi ]; ξi ), i = 1, ..., n, and Py = ([y j−1 , yi ]; η j ), j = 1, ..., n are said to be δ-fine and σ-fine, respectively, if

[xi−1 , xi ] ⊆ (ξi − δ(ξi ), ξi + δ(ξi )) and [y j−1 , y j ] ⊆ (η j − σ(η j ), η j + σ(η j )).

The function f is said to be two-dimensional Henstock integrable to I ∈ E 1 if for every ε > 0 there are functions

n P

n

P

δ : [a, b] → R+ and σ : [c, d] → R+ such that for any δ-fine and σ-fine divisions we have D(

(xi − xi−1 )(y j −y j−1 )

j=1 i=1

P

f (ξi , η j ), I) < ε, where denotes the fuzzy summation. Then, I is called the two-dimensional Henstock integral of f

Rd

Rb

and is denoted by I( f ) = (FH) (FH) f (s, t)dsdt. If the above δ and σ are constant functions, then one recaptures

c

a

the concept of Riemann integral. In this case, I ∈ E 1 will be called two-dimensional integral of f on A and will be

Rd

Rb

denoted by (FR) (FR) f (s, t)dsdt.

c

a

Lemma 3

[21] Let f : A → E 1 , then f is (FH)-integrable if and only if f and f are Henstock integrable for

any r ∈ [0, 1]. Moreover

r

Zb

Zd

Zb

Zd

Zb

Zd

f

(H)

f

(s,

t)dsdt

=

(H)

(FH)

(FH)

(H)

(s,

t,

r)dsdt,

(H)

f

(s,

t,

r)dsdt

.

a

c

a

c

a

c

Also, if f is continuous then f (., ., r) and f (., ., r) are continuous for any r ∈ [0, 1] and consequently, they are Henstock

integrable. Using Lemma 2. we deduce that f is (FH)-integrable.

c

a

c

a

c

a

Lemma 5

[21] If f : A → E 1 is an integrable bounded function then for any fixed (u, v) ∈ A, the function

ϕu,v : A → R+ defined by ϕu,v (s, t) = D( f (u, v), f (s, t)) is Lebesgue integrable on A.

ADM FOR SOLVING 2D-NVFIE

In this section we introduce the parametric form of integral equation (1) and then apply ADM for solving this

equation. For solving in parametric form of equation (1), consider u(s, t, r) = (u(s, t, r), u(s, t, r)) and u(s, t, r) =

(g(s, t, r), g(s, t, r)), 0 ≤ r ≤ 1 and (s, t) ∈ A are parametric form of u(s, t) and g(s, t), respectively. So the parametric

form of equation (1) is as follows:

Z t Zs

u(s, t, r) = g(s, t, r) +

k(s, t, x, y)G(u(x, y, r))dxdy,

u(s, t, r) = g(s, t, r) +

c

a

c

a

Z t Zs

k(s, t, x, y)G(u(x, y, r))dxdy.

Let for (x, y) ∈ A, we have

H(u(x, y, r), u(x, y, r)) = min{G(β) : u(x, y, r) ≤ β ≤ u(x, y, r)},

Then,

F(u(x, y, r), u(x, y, r)) = max{G(β) : u(x, y, r) ≤ β ≤ u(x, y, r)}.

(

k(s, t, x, y)F(u(x, y, r), u(x, y, r)), if k(s, t, x, y) ≥ 0

k(s, t, x, y)G(u(x, y, r)) =

k(s, t, x, y)H(u(x, y, r), u(x, y, r)), if k(s, t, x, y) < 0

(

k(s, t, x, y)H(u(x, y, r), u(x, y, r)), if k(s, t, x, y) ≥ 0

k(s, t, x, y)G(u(x, y, r)) =

k(s, t, x, y)F(u(x, y, r), u(x, y, r)), if k(s, t, x, y) < 0

050009-3

09 March 2024 19:34:42

Lemma 4

[21] If f and g are fuzzy Henstock integrable functions on A and if the function given by

D( f (s, t), g(s, t)) is Lebesgue integrable, then

Zd

Zb

Zd

Zb

Zd

Zb

D (FH) (FH)

f (s, t)dsdt, (FH) (FH) g(s, t)dsdt ≤ (L) (L) D( f (s, t), g(s, t))dsdt.

## 5.

for a ≤ x ≤ s ≤ b, c ≤ y ≤ t ≤ d and 0 ≤ r ≤ 1. The ADM method is applied to solve 2D-NVFIE. Let for alla ≤ x ≤ s ≤ b, c ≤ y ≤ t ≤ d and 0 ≤ r ≤ 1 the function G(β) is increasing for β ∈ [u(x, y, r), u(x, y, r)] and

k(s, t, x, y) ≥ 0 . Then the parametric form of the equation (1) is,

Rt Rs

u(s, t, r) = g(s, t, r) +

k(s, t, x, y)G(u(x, y, r))dxdy

c a

(2)

t

s

R R

u(s, t, r) = g(s, t, r) +

k(s, t, x, y)G(u(x, y, r))dxdy.

c a

Now, we explain ADM as a numerical algorithm for approximating solution of this system of nonlinear integral

equations in crisp case. Then, we find approximate solution for u(s, t, r)

The ADM assume an infinite series solution for the unknowns functions (u(s, t, r), u(s, t, r)), given by

∞

∞

X

X

ui (s, t, r), u(s, t, r) =

u(s, t, r) =

ui (s, t, r)

(3)

i=0

i=0

The nonlinear operator G(u(x, y, r)) and G(u(x, y, r)) into an infinite series of polynomials given by

G(u(x, y, r)) =

∞

X

An (u0 (x, y, r), u1 (x, y, r), ..., un (x, y, r)), G(u(x, y, r)) =

n=0

∞

X

An (u0 (x, y, r), u1 (x, y, r), ..., un (x, y, r)),

n=0

where the An = (An , An ), n ≥ 0 are the so-called Adomian polynomial defined by

∞

∞

1 dn X i

1 dn X i

An (u0 , u1 , ..., un ) =

G λ ui , An (u0 , u1 , ..., un ) =

G λ ui

n! dλn i=0

n! dλn i=0

(4)

,

(5)

λ=0

λ=0

09 March 2024 19:34:42

where λ is formal parameter.

Substituting equations (3) and (4) into equation (2) to get

Z t Zs

∞

∞

X

X

un (s, t, r) = g(s, t, r) +

An (u0 (x, y, r), u1 (x, y, r), ..., un (x, y, r))dxdy,

k(s, t, x, y)

n=0

c

∞

X

Z t Zs

un (s, t, r) = g(s, t, r) +

n=0

n=0

a

k(s, t, x, y)

c

a

∞

X

An (u0 (x, y, r), u1 (x, y, r), ..., un (x, y, r))dxdy

n=0

The components un (s, t, r) and un (s, t, r), n ≥ 0 are computed using the following recursive relations

u0 (s, t, r) = g(s, t, r)

Rt Rs

u1 (s, t, r) =

k(s, t, x, y)A0 (u0 (x, y, r))dxdy

c a

...

un+1 (s, t, r) =

Rt Rs

(6)

k(s, t, x, y)An (u0 (x, y, r), u1 (x, y, r), ..., un (x, y, r))dxdy

c a

and

u0 (s, t, r) = g(s, t, r)

Rt Rs

k(s, t, x, y)A0 (u0 (x, y, r))dxdy

u1 (s, t, r) =

un+1 (s, t, r) =

Rt Rs

k(s, t, x, y)An (u0 (x, y, r), u1 (x, y, r), ..., un (x, y, r))dxdy

c a

We approximate u(s, t, r) = (u(s, t, r), u(s, t, r)) by φ (s, t, r) =

n

lim φ (s, t, r) = u(s, t, r) and lim φn (s, t, r) = u(s, t, r)

n→∞

n

(7)

c a

...

n→∞

050009-4

n−1

P

i=0

ui (s, t, r),

φn (s, t, r) =

n−1

P

i=0

ui (s, t, r), where

## 6.

EXISTENCE AND CONVERGENCE ANALYSISIn this section we prove the existence, uniqueness of the solution of equation (1) and convergence of ADM.

Lemma 6

[21] Let the functions g ∈ C(A, E 1 ) and h ∈ C(A, R+ ). Then the function h.g : A → E 1 given by

(h.g)(s, t) = h(s, t) g(s, t), is continuous on A.

Lemma 7

[13] Let the functions k ∈ C(A × A, R+ ), u ∈ C(A, E 1 ) and G ∈ C(E 1 , E 1 ). Then the function

Rt

Rs

Fu : A → E 1 defined by Fu (s, t) = (FR) (FR) k(s, t, x, y) G(u(x, y))dxdy is continuous on A.

c

a

We introduce the following conditions:

(i) g ∈ C(A, E 1 ), k ∈ C(A × A, R+ ) and G ∈ C(E 1 , E 1 );

(ii) there exists L ≥ 0 such that D(G(u), G(v)) ≤ LD(u, v) for all u, v ∈ E 1 ;

(iii) α = Mk L∆ < 1, where |k(s, t, x, y)| ≤ Mk for all (s, t), (x, y) ∈ A, according to the continuity of k and ∆ =

(b − a)(d − c).

Theorem 1

Let the conditions (i) − (iii) are fulfilled. Then the integral equation (1) has a unique solution.

Proof: Let F(A, E 1 ) = { f : A → E 1 } and X = C(A, E 1 ). We define the operator A : X → F(A, E 1 ) by

Z t

Z s

A(u)(s, t) = g(s, t) ⊕ (FR)

(FR)

k(s, t, x, y) G(u(x, y))dxdy,

c

a

c

a

F ∈ X. Since g ∈ X, we conclude that the operator A(F) is continuous on A for any u ∈ X.

Now, we prove that A : X → X is a contraction. Let arbitrary u, v ∈ X. From conditions (ii) and (iii) we have

Rt

Rs

Rt

Rs

D(A(u)(s, t), A(v)(s, t)) ≤ D((FR) (FR) k(s, t, x, y) G(u(x, y))dxdy, (FR) (FR) k(s, t, x, y) G(v(x, y))dxdy) ≤

c

≤

Rt Rs

a

c

|k(s, t, x, y)|D(G(u(x, y)), G(v(x, y)))dxdy ≤ Mk L

c a

Rd Rb

a

D(u(x, y), v(x, y))dxdy ≤ Mk L∆D∗ (u, v) for all (s, t) ∈ A.

c a

Under the condition α < 1 the operator A is contraction therefore, by the Banach fixed-point theorem for contraction, there exist a unique solution to problem (1) and this completes the proof.

∞

P

Theorem 2

The series solution u(s, t, r) = ui (s, t, r) of the integral equation (1) using ADM converges if

i=0

α < 1 and u1 (s, t, r) bounded for all (s, t) ∈ A, 0 ≤ r ≤ 1.

Proof: Denoting E = (C(A × [0, 1], R), k.k) the Banach space of all continuous functions on A with the norm

n

P

kuk = max sup |u(s, t, r)|. Define the sequence of partial sums sn (s, t, r) =

ui (s, t, r) for all (s, t) ∈ A, 0 ≤ r ≤ 1.

(s,t)∈A r∈[0,1]

i=0

Let sn and sm be arbitrary partial sums with n ≥ m. We are going to prove that {sn } is a Cauchy sequence in E.

n

m

n

P

P

P

ksn − sm k = max sup |sn (s, t, r) − sm (s, t, r)| = max sup | ui (s, t, r) − ui (s, t, r)| = max sup |

ui (s, t, r)| =

(s,t)∈A r∈[0,1]

= max sup |

(s,t)∈A r∈[0,1] i=0

n Rt Rs

P

(s,t)∈A r∈[0,1] i=m+1 c a

Rt Rs

= max sup |

(s,t)∈A r∈[0,1] c a

i=0

(s,t)∈A r∈[0,1] i=m+1

k(s, t, x, y)Ai−1 (u0 (x, y, r), u1 (x, y, r), ..., ui−1 (x, y, r))dxdy| =

k(s, t, x, y)

n−1

P

Ai (u0 (x, y, r), u1 (x, y, r), ..., ui (x, y, r))dxdy|.

i=m

From [9], we have

n−1

X

Ai (u0 (x, y, r), u1 (x, y, r), ..., ui (x, y, r)) = G(sn−1 (x, y, r)) − G(sm−1 (x, y, r)).

i=m

050009-5

09 March 2024 19:34:42

(s, t) ∈ A, u ∈ X. Firstly, we prove that A(X) ⊂ X. To this purpose, we show that the operator A is uniformly continuous.

Let arbitrary u ∈ X, (t0 , s0 ) ∈ A and ε > 0. Since u is continuous it follows that for Lε > 0 there exists δ(ε) > 0

such that D(u(s, t), u(s0 , t0 )) ≤ Lε for any (s, t) ∈ A with |s − s0 | + |t − t0 | < δ(ε). From condition (ii) we have

D(G(u(s, t)), G(u(s0 , t0 ))) ≤ LD(u(s, t), u(s0 , t0 )) ≤ ε and the function Gu : A → E 1 , defined by Gu (s, t) = G(u(s, t))

is continuous in (s0 , t0 ). We conclude that Gu is continuous on A for any u ∈ X. From Lemma 6. it follows that the

function k(s.t, ., .) Gu (., .) : A → E 1 is continuous on A for any u ∈ X, (s, t) ∈ A. Applying Lemma 7. it follows that

Rt

Rs

the function Fu : A → E 1 , defined by Fu (s, t) = (FR) (FR) k(s, t, x, y) Gu (x, y)dxdy is continuous on A for any

## 7.

So,Rt Rs

ksn − sm k = max sup |

k(s, t, x, y)[G(sn−1 (x, y, r)) − G(sm−1 (x, y, r))]dxdy| ≤

(s,t)∈A r∈[0,1] c a

Rd Rb

≤ Mk

max sup |G(sn−1 (x, y, r)) − G(sm−1 (x, y, r))|dxdy ≤

c a (x,y)∈A r∈[0,1]

Rd Rb

≤ Mk L

Let, n = m + 1 then

max sup |sn−1 (x, y, r) − sm−1 (x, y, r)|dxdy ≤ Mk L∆ksn−1 − sm−1 k = αksn−1 − sm−1 k

c a (x,y)∈A r∈[0,1]

ksm+1 − sm k ≤ αksm − sm−1 k ≤ α2 ksm−1 − sm−2 k ≤ ... ≤ αm ks1 − s0 k.

Using the triangle inequality we have,

ksn − sm k ≤ hksm+1 − sm k + ksm+2 − sim+1 k + ... + ksn −h sn−1 k ≤

i

n−m

ku1 k.

≤ αm + αm+1 + ... + αn−1 ks1 − s0 k ≤ αm 1 + α + α2 + ... + αn−m−1 ks1 − s0 k ≤ αm 1−α

1−α

n−m

Since 0 < α < 1 so, 1 − α

≤ 1 then

αm

max sup |u1 (s, t, r)|

ksn − sm k ≤

1 − α (s,t)∈A r∈[0,1]

(8)

But |u1 (s, t, r)| ≤ ∞ (since g(s, t, r) is bounded) then ksn − sm k → 0 as m → ∞, from which we conclude that {sn } is a

Cauchy sequence in E. Therefore

u(s, t, r) = lim sn (s, t, r).

(9)

n→∞

Similarly, we have sn is a Cauchy sequence in E, then we can write

u(s, t, r) = lim sn (s, t, r).

(10)

n→∞

u(s, t, r) = (u(s, t, r), u(s, t, r)) = ( lim sn (s, t, r), lim sn (s, t, r)).

n→∞

Theorem 3

(11)

n→∞

The maximum absolute truncation error of the series solution (3) to problem (1) is estimated to

be

max sup |u(s, t, r) −

(s,t)∈A r∈[0,1]

max sup |u(s, t, r) −

(s,t)∈A r∈[0,1]

m

X

i=0

m

X

ui (s, t, r)| ≤

1 αm+1

max sup |G(g(s, t, r)|,

L 1 − α (s,t)∈A r∈[0,1]

(12)

ui (s, t, r)| ≤

1 αm+1

max sup |G(g(s, t, r)|.

L 1 − α (s,t)∈A r∈[0,1]

(13)

i=0

Proof: From Theorem 2. inequality (8) we have

max sup |sn (s, t, r) − sm (s, t, r)| ≤

(s,t)∈A r∈[0,1]

αm

max sup |u1 (s, t, r)|.

1 − α (s,t)∈A r∈[0,1]

As n → ∞ then sn (s, t, r) → u(s, t, r). Then

αm

max sup |u(s, t, r) − sm (s, t, r)| ≤

max sup |u1 (s, t, r)|.

(s,t)∈A r∈[0,1]

1 − α (s,t)∈A r∈[0,1]

From (5), (7) and condition (iii) for u1 (s, t, r) we obtain

max sup |u1 (s, t, r)| = max sup

(s,t)∈A r∈[0,1]

Rt Rs

(s,t)∈A r∈[0,1] c a

k(s, t, x, y)A0 (u0 (x, y, r))dxdy ≤

Rd Rb

c a

|k(s, t, x, y)| max sup |G(u0 (x, y, r))|dxdy ≤

(x,y)∈A r∈[0,1]

≤ Mk ∆ max sup |G(u0 (x, y, r))| = αL max sup |G(g(s, t, r))|.[2pt]

(x,y)∈A r∈[0,1]

(s,t)∈A r∈[0,1]

(14)

Finally the maximum absolute truncation error is

m

X

1 αm+1

max sup |u(s, t, r) −

ui (s, t, r)| ≤

max sup |G(g(s, t, r))|.

(s,t)∈A r∈[0,1]

L 1 − α (s,t)∈A r∈[0,1]

i=0

Analogously, we obtain the inequality (13).

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09 March 2024 19:34:42

Therefore,

## 8.

NUMERICAL EXAMPLESThe Adomian‘s polynomials can be generated using the traditional formula (5). For example, if G(u) = u2 the

polynomials using the formula (5) are

A0 = u20 ,

A1 = 2u0 u1 ,

A2 = u21 + 2u0 u2 ,

A3 = 2u1 u2 + 2u0 u3

A4 = u22 + 2u1 u3 + 2u0 u4 ,

A5 = 2u2 u3 + 2u1 u4 + 2u0 u5 .

By induction we find for m = 0 A0 = u20 for m = 1, 2, 3, ..., 2ν + 1, where ν ≥ 0 A2ν+1 = 2

m = 2, 4, 6, ..., 2ν, where ν ≥ 1 A2ν = u2ν + 2

Example 1

ν−1

P

ν

P

un u2ν+1−n , and for

n=0

un u2ν−n .

n=0

Let A = [0, 1] × [0, 1]. Consider the following 2D-NVFIE

Zt

Zs

u(s, t) = g(s, t) ⊕ (FR) (FR) k(s, t, x, y) u2 (x, y)dxdy,

0

0

1 5 5

1 5 5

s t (2 + r)2 , st(4 − r) − 16

s t (4 − r)2 ), k(s, t, x, y) = stxy.

where g(s, t, r) = (st(2 + r) − 16

The exact solution is uexact (s, t, r) = (st(2 + r), st(4 − r)).

0 0

0 0

Using ADM, we have

1 5 5

u0 (s, t, r) = st(2 + r) − 16

s t (2 + r)2 ,

Rt Rs

1 9 9

1

1 5 5

s t (2 + r)2 + 512

s t (2 + r)3 + 36864

s13 t13 (2 + r)4

u1 (s, t, r) =

stxyu20 (x, y, r)dxdy = 16

0 0

u2 (s, t, r) =

Rt Rs

stxy2u0 (x, y, r)u1 (x, y, r)dxdy =

0 0

1 8 8

7

7

1

= 392

s t (2 + r)3 − 123904

s12 t12 (2 + r)4 − 33177600

s16 t16 (2 + r)5 − 425852928

s20 t20 (2 + r)6 ,

u(s, t, r)

1 9 9

1 8 8

1

7

= (st + ( 512

s t + 392

s t )(2 + r)2 + ( 36864

s13 t13 − 123904

s12 t12 )(2 + r)3 −

7

1

− 33177600

s16 t16 (2 + r)4 − 425852928

s20 t20 (2 + r)5 + ...)(2 + r)

With the same procedure we can obtain,

1 9 9

1 8 8

1

7

u(s, t, r) = (st + ( 512

s t + 392

s t )(4 − r)2 + ( 36864

s13 t13 − 123904

s12 t12 )(4 − r)3 −

7

1

− 33177600

s16 t16 (4 − r)4 − 425852928

s20 t20 (4 − r)5 + ...)(4 − r).

So the approximate solution u(s, t, r) = (u(s, t, r), u(s, t, r)). The numerical results obtained with the ADM with

threee iterations are presented in Table 1.

TABLE 1. Comparison of approximation solutions with exact solution for (s0 , t0 ) = (0.2, 0.7) and r = 0.5

m

um (0.2, 0.7, 0.5)

um (0.2, 0.7, 0.5)

u(0.2, 0.7, 0.5)

u(0.2, 0.7, 0.5)

uexact (0.2, 0.7, 0.5)

uexact (0.2, 0.7, 0.5)

0

1

2

0,349978991

7,56378E-06

1,6141E-08

0,489958823

4,11789E-059

5,88232E-097

0,349978991

0,3499865555

0,349986571

0,489958823

0,490000002

0,490000008

0,35

0,35

0,35

0,49

0,49

0,49

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09 March 2024 19:34:42

The general form of the equation is

Rt Rs

Rt Rs

1 5 5

1 5 5

u(s, t, r) = st(2+r)− 16

s t (2+r)2 +

stxyu2 (x, y, r)dxdy, u(s, t, r) = st(4−r)− 16

s t (4−r)2 )+

stxyu2 (x, y, r)dxdy.

## 9.

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050009-8

09 March 2024 19:34:42

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