^{1}

^{1}

^{2}

^{3}

This paper proposes the Laplace Discrete Adomian Decomposition Method and its application for solving nonlinear integro-differential equations. This method is based upon the Laplace Adomian decomposition method coupled with some quadrature rules of numerical integration. Four numerical examples of integro-differential equations in both Volterra and Fredholm integrals are used to be solved by the proposed method. The performance of the proposed method is verified through absolute error measures between the approximated solutions and exact solutions. The series of experimental numerical results show that our proposed method performs in high accuracy and efficiency. The study clearly highlights that the proposed method could be used to overcome the analytical approaches in solving nonlinear integro-differential equations.

One of the issues in finding the solutions of integro-differential equations is the choices of approaches used that are either analytical or numerical. Previous literature suggested that analytical solutions of integro-differential equations are not usually obtainable especially when the equations are entangled with nonlinear terms. Therefore, it is required to find an efficient approximate solution numerically. Much attention has been devoted for the search of better and more efficient methods in recent years with the introduction of several numerical methods such as the Galerkin methods [

The LADM is known for its rapid convergence in solution and also uses only little iteration as successfully applied in Kiymaz [

In this section, we present the Laplace discrete Adomian decomposition method over the integro-differential equation of the form

u ″ ( x ) = f ( x ) + ∫ a b k ( x , t ) ⋅ ( R u ( t ) + N ( u ( t ) ) ) d t , (2.1)

Subject to the following initial conditions

u ( 0 ) = α , u ′ ( 0 ) = β .

To solve the nonlinear integro-differential Equation (2.1) using the Laplace transform method, we recall the Laplace transform of the second derivative of u ( x ) , that is L { u ″ ( x ) } = s 2 L { u ( x ) } − s u ( 0 ) − u ′ ( 0 ) . Thus on applying the Laplace transform to both sides of Equation (2.1) we obtain

s 2 L { u ( x ) } − s u ( 0 ) − u ′ ( 0 ) = L { f ( x ) } + L { ∫ a b k ( x , t ) ⋅ ( R u ( t ) + N ( u ( t ) ) ) d t } (2.2)

We thus obtain the following equation after using the above prescribed initial conditions

L { u ( x ) } = α s + β s 2 + 1 s 2 L { f ( x ) } + 1 s 2 L ∫ a b k ( x , t ) ( R u ( t ) + N ( u ( t ) ) ) d t . (2.3)

The decomposition method represents the solution u ( x ) as a series of the form:

u ( x ) = ∑ n = 0 ∞ u n ( x ) , (2.4)

and the nonlinear term N u ( t ) is decomposed into an infinite series of the form:

N ( u ( t ) ) = ∑ n = 0 ∞ A n , (2.5)

where A n ’s are the Adomian polynomials of u 0 , u 1 , ⋯ , u n given by the formula

A n = 1 n ! d n d λ n [ N ( ∑ i = 0 n λ i u i ) ] λ = 0 , i = 0 , 1 , 2 , ⋯ (2.6)

On substituting Equation’s (2.4) and (2.5) in Equation (2.3) and making comparison between the right and left hand sides, we thus obtain:

{ L { u 0 ( x ) } = α s + β s 2 + 1 s 2 L ( f ( x ) ) L { u k + 1 ( x ) } = 1 s 2 L ∫ a b k ( x , t ) ( R u k ( x ) + A k ( x ) ) , k ≥ 0. (2.7)

Furthermore, if the evaluation of the integrals in Equation (2.7) is analytically possible, the ADM can be applied directly. However, in the case where the evaluation of integrals is analytically impossible we consider the numerical integration scheme given by the following formula:

∫ a b f ( t ) d t ≈ ∑ j = 0 0 w n , i f ( t n , i ) , (2.8)

where f ( t ) is continuous function on [ a , b ] , t n , i = a + i h are the nodes of the numerical integration, h = b − a n is the fixed step length and w n , i , i = 0 , 1 , 2 , ⋯ , n are the weights functions. Now, applying the formula given in Equation (2.8) on Equation (2.7) to obtain

{ L { u 0 ( x ) } = α s + β s 2 + 1 s 2 L ( f ( x ) ) L { u k + 1 ( x ) } = 1 s 2 L ( ∑ i = 0 n w n , i k ( x , t n , i ) ⋅ ( R ( u k ( t n , i ) ) + A k ( t n , i ) ) ) , k ≥ 0. (2.9)

Finally, on applying the inverse Laplace transform to the first part of Equation (2.9) gives u 0 ( x ) , and, consequently will define A 0 . Also, using A 0 enables us to evaluate u 1 ( x ) . The determination of u 0 ( x ) and u 1 ( x ) leads to the determination of A 1 that will allow us to determine u 2 ( x ) , and so on. This in turn will lead to the complete determination of the components of u k , k ≥ 0 upon using the second part of Equation (2.9).

{ u 0 ( x ) = L − 1 { α s + β s 2 + 1 s 2 L ( f ( x ) ) } u k + 1 ( x ) = L − 1 { 1 s 2 L ( ∑ i = 0 n w n , i k ( x , t n , i ) ⋅ ( R ( u k ( t n , i ) ) + A k ( t n , i ) ) ) } , k ≥ 0 (2.10)

From this recursive relation in Equation (2.10), u 0 , u 1 , u 2 , ⋯ can be calculated. The solution of Equation (2.1) is now determined in Equation (2.10). However, in practice the series ∑ n = 0 ∞ u n must be truncated to the series:

φ n = ∑ i = 0 n u i with lim n → ∞ φ n = u (2.11)

We couple the trapezoidal method (TR) to Equation (2.10) to obtain:

{ u 0 ( x ) = α + β x + L − 1 f ( x ) u k + 1 ( x ) = L − 1 ( h 2 ( k ( x , a ) ( R ( u k ( a ) ) + A k ( a ) ) + 2 ∑ i = 1 n − 1 k ( x , x i ) ( R ( u k ( x i ) ) + A k ( x i ) ) + k ( x , b ) ( R ( u k ( b ) ) + A k ( b ) ) ) , k ≥ 0. (2.12)

Also, on coupling the Simpson’s method (SM) to Equation (2.10), we get:

{ u 0 ( x ) = α + β x + L − 1 f ( x ) u k + 1 ( x ) = L − 1 ( h 2 ( k ( x , a ) ( R ( u k ( a ) ) + A k ( a ) ) + 4 ∑ i = 1 n 2 k ( x , x 2 i − 1 ) ( R ( u ( x 2 i − 1 ) ) + A k ( x 2 i − 1 ) ) + 2 ∑ i = 1 n 2 − 1 k ( x , x 2 i ) ( R ( u ( x 2 i ) ) + A k ( x 2 i ) ) + k ( x , b ) ( R ( u k ( b ) ) + A k ( b ) ) ) , k ≥ 0. (2.13)

In this section, we consider several nonlinear integro-differential equations as examples in order to show the efficiency and the simplicity of the proposed method. We start with the nonlinear Volterra and Fredholm integro-differential equations down to their systems, respectively.

Example 3.1

Consider the nonlinear Volterra integro-differential equation

u ′ ( x ) = 1 5 x 5 − ∫ 0 x ( u 2 ( t ) − 2 ) d t , u ( 0 ) = 0 , (3.1)

with the exact solution given by u ( x ) = x 2 .

In order to use the quadrature rule for Equation (3.1), let t = x ⋅ v , we get

u ′ ( x ) = 1 5 x 5 − x ∫ 0 1 ( u 2 ( x ⋅ v ) − 2 ) d v , u ( 0 ) = 0.

Taking the Laplace transform of both sides of the above equation gives

L { u ′ ( x ) } = L ( 1 5 x 5 ) − L ( x ∫ 0 1 ( u 2 ( x ⋅ v ) − 2 ) d v ) .

So that,

s L { u ( x ) } − u ( 0 ) = 24 s 6 − L ( x ∫ 0 1 ( u 2 ( x ⋅ v ) − 2 ) d v ) ,

or equivalently

L { u ( x ) } = 24 s 7 − 1 s L ( x ∫ 0 1 ( u 2 ( x ⋅ v ) − 2 ) d v ) .

1) Trapezoidal Method (TM)

We divide the interval (0, 1) into subinterval of equal lengths h = 0.2 , n = 5 and denote v i = a + i h , 0 ≤ i ≤ 5 .

The recursive relation is given by

{ L { u 0 ( x ) } = 24 s 7 , L { u k + 1 ( x ) } = − 0.1 s L ( x ⋅ ( x A k ( v 0 ) + 2 ∑ i = 1 4 x A k ( v i ) + x A k ( v 5 ) ) ) , k ≥ 0

or

{ L { u 1 ( x ) } = 2 s 3 − 7.900194170 s 15 , L { u 2 ( x ) } = − 3315.398639 s 11 − 3.15801038 s 23 , ⋮

Taking the inverse Laplace transform of both sides of the first part of recursive relation, and using this recursive relation gives

{ u 0 ( x ) = 1 30 x 6 u 1 ( x ) = x 2 − 0.00000906 x 14 ⋮

The series solution is obtained by summing the following iterates

u ( x ) = u 0 + u 1 + u 2 + ⋯ .

The results produced by the present method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

2) Simpson’s Method (SM)

We divide the interval (0, 1) into subinterval of equal lengths h = 0.1 , n = 10 and denote x i = a + i h , 0 ≤ i ≤ 10 . The recursive relation is given by

{ L { u 0 ( x ) } = 24 s 7 , L { u k + 1 ( x ) } = − 0.1 3 s L ( x A k ( v 0 ) + 4 ∑ i = 1 5 x A k ( v 2 i − 1 ) + 2 ∑ i = 1 4 x A k ( v 2 i ) + x A k ( v 10 ) ) , k ≥ 0 ,

x | Exact | LDADM | Absolute Error |
---|---|---|---|

0 | 0 | 0 | 0 |

0.20 | 0.04000000 | 0.03999986 | 1.41640000e−07 |

0.40 | 0.16000000 | 0.15999094 | 9.05930000e−06 |

0.60 | 0.36000000 | 0.35989712 | 1.02878600e−04 |

0.80 | 0.64000000 | 0.63942742 | 5.72582800e−04 |

1 | 1.00000000 | 0.99787295 | 2.12705460e−03 |

Or

{ L { u 1 ( x ) } = 2 s 3 − 5.36881755 s 15 , L { u 2 ( x ) } = − 2692.40899737 s 11 + 1.05763414 s 23 , ⋮

Taking the inverse Laplace transform of both sides of the first part of recursive relation, and using this recursive relation gives

{ u 0 ( x ) = 1 30 x 6 , u 1 ( x ) = x 2 − 0.00000616 x 14 , ⋮

The series solution is obtained by summing the following iterates

u ( x ) = u 0 + u 1 + u 2 + ⋯ .

The results produced by the present method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

Example 3.2

Consider the nonlinear Fredholm integro-differential equation

x | Exact | LDADM | Absolute Error |
---|---|---|---|

0 | 0 | 0 | 0 |

0.10 | 0.01000000 | 0.01000000 | 2.22221713e−12 |

0.20 | 0.04000000 | 0.04000000 | 1.42214800e−10 |

0.30 | 0.09000000 | 0.09000000 | 1.61901986e−09 |

0.40 | 0.16000000 | 0.15999999 | 9.05840967e−09 |

0.50 | 0.25000000 | 0.24999997 | 3.37952007e−08 |

u ″ ( x ) = 1 2 e x + 1 2 ∫ 0 1 e x − 2 t ⋅ u 2 ( t ) d t , u ′ ( 0 ) = u ( 0 ) = 1 ,

with the exact solution is u ( x ) = e x .

Taking the Laplace transform of both sides of the given equation gives

L { u ″ ( x ) } = L ( 1 2 e x ) + 1 2 L ( ∫ 0 1 e x − 2 t ⋅ u 2 ( t ) d t ) .

So that

s 2 L { u ( x ) } − s u ( 0 ) − u ′ ( 0 ) = 1 2 ( s − 1 ) + 1 2 L ( ∫ 0 1 e x − 2 t ⋅ u 2 ( t ) d t ) ,

or equivalently

L { u ( x ) } = 1 s + 1 s 2 + 1 2 s 2 ( s − 1 ) + 1 2 s 2 L ( ∫ 0 1 e x − 2 t ⋅ u 2 ( t ) d t ) .

1) Trapezoidal Method (TM)

Let h = 0.2 , n = 5 , the recursive relation is given by

{ L { u 0 ( x ) } = 1 s + 1 s 2 + 1 2 s 2 ( s − 1 ) , L { u k + 1 ( x ) } = 0.2 4 s 2 L ( e x − 2 t 0 ⋅ A k ( t 0 ) + 2 ∑ i = 1 4 e x − 2 t i ⋅ A k ( t i ) + e x − 2 t 5 ⋅ A k ( t 5 ) ) , k ≥ 0.

Taking the inverse Laplace transform of both sides of the first part of recursive relation, and using this recursive relation gives

{ u 0 ( x ) = 1 2 + 1 2 x + 1 2 e x , u 1 ( x ) = − 0.44983021 − 0.44983021 x + 0.44983021 e x , ⋮

The series solution is obtained by summing the following iterates

u ( x ) = u 0 + u 1 + u 2 + ⋯ .

The results produced by the present method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

2) Simpson’s Method (SM)

Let h = 0.1 , n = 10 , the recursive relation is given by

{ L { u 0 ( x ) } = 1 s + 1 s 2 + 1 2 s 2 ( s − 1 ) , L { u k + 1 ( x ) } = 0.1 6 s 2 L ( e x − 2 t 0 ⋅ A k ( t 0 ) + 4 ∑ i = 1 5 e x − 2 t 2 i − 1 ⋅ A k ( t 2 i − 1 ) + 2 ∑ i = 1 2 e x − 2 t 2 i ⋅ A k ( t 2 i ) + e x − 2 t 5 ⋅ A k ( t 5 ) ) , k ≥ 0.

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.20 | 1.22140276 | 1.22140206 | 7.03000000e−07 |

0.40 | 1.49182470 | 1.49182168 | 3.01500000e−06 |

0.60 | 1.82211880 | 1.82211150 | 7.29500000e−06 |

0.80 | 2.22554093 | 2.22552695 | 1.39750000e−05 |

1 | 2.71828183 | 2.71825824 | 2.35900000e−05 |

{ u 0 ( x ) = 1 2 + 1 2 x + 1 2 e x , u 1 ( x ) = − 0.45035999 − 0.45035999 x + 0.450359991 e x , ⋮

The series solution is obtained by summing the following iterates

u ( x ) = u 0 + u 1 + u 2 + ⋯ .

The results produced by the present method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

Example 3.3

Consider the system of nonlinear Volterra integro differential equation

{ u ′ ( x ) = 3 x 2 − 2 3 x 3 − 1 126 x 9 + ∫ 0 x ( x − t ) 2 ( u 2 ( t ) + v 2 ( t ) ) d t , u ( 0 ) = 1 , v ′ ( x ) = − 3 x 2 − 1 35 x 7 + ∫ 0 x ( x − t ) 2 ( u 2 ( t ) − v 2 ( t ) ) d t , v ( 0 ) = 1 , (3.2)

with the exact solution ( u ( x ) , v ( x ) ) = ( 1 + x 3 , 1 − x 3 ) .

In order to use the quadrature rule for Equation (3.2), let t = x ⋅ v , we get

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.10 | 1.10517092 | 1.10517076 | 1.58713332e−07 |

0.20 | 1.22140276 | 1.22140210 | 6.56924555e−07 |

0.30 | 1.34985881 | 1.34985728 | 1.53033898e−06 |

0.40 | 1.49182470 | 1.49182188 | 2.81841706e−06 |

0.50 | 1.64872127 | 1.64871671 | 4.56476936e−06 |

{ u ′ ( x ) = 3 x 2 − 2 3 x 3 − 1 126 x 9 + x ∫ 0 1 ( x − x ⋅ v ) 2 ( u 2 ( x ⋅ v ) + v 2 ( x ⋅ v ) ) d v , u ( 0 ) = 1 , v ′ ( x ) = − 3 x 2 − 1 35 x 7 + x ∫ 0 1 ( x − x ⋅ v ) 2 ( u 2 ( x ⋅ v ) − v 2 ( x ⋅ v ) ) d v , v ( 0 ) = 1.

Taking the Laplace transform of both sides of the above equation gives

{ s L { u ( x ) } − 1 = L ( 3 x 2 − 2 3 x 3 − 1 126 x 9 ) + L ( x ∫ 0 1 ( x − x ⋅ v ) 2 ( u 2 ( x ⋅ v ) + v 2 ( x ⋅ v ) ) d v ) , s L { v ( x ) } − 1 = L ( − 3 x 2 − 1 35 x 7 ) + L ( x ∫ 0 1 ( x − x ⋅ v ) 2 ( u 2 ( x ⋅ v ) − v 2 ( x ⋅ v ) ) d v ) .

So that

{ L { u ( x ) } = 1 s + 6 s 4 − 4 s 5 − 2880 s 11 + 1 s L ( x ∫ 0 1 ( x − x ⋅ v ) 2 ( u 2 ( x ⋅ v ) + v 2 ( x ⋅ v ) ) d v ) , L { v ( x ) } = 1 s − 6 s 4 − 144 s 9 + 1 s L ( x ∫ 0 1 ( x − x ⋅ v ) 2 ( u 2 ( x ⋅ v ) − v 2 ( x ⋅ v ) ) d v ) .

1) Trapezoidal Method (TM)

We divide the interval (0, 1) into subinterval of equal lengths h = 0.2 , n = 5 and denote v i = a + i h , 0 ≤ i ≤ 5 . The recursive relations as

{ L { u 0 ( x ) } = 1 s + 6 s 4 − 4 s 5 − 2880 s 11 , L { v 0 ( x ) } = 1 s − 6 s 4 − 144 s 9 ,

{ L ( u k + 1 ( x ) ) = 1 s L ( x ( ( x − x ⋅ v 0 ) 2 ( A k ( v 0 ) + B k ( v 0 ) ) ) + 2 ∑ i = 1 4 ( x − x ⋅ v i ) 2 ( A k ( v i ) + B k ( v i ) ) + ( x − x ⋅ v 5 ) 2 ( A k ( v 5 ) + B k ( v 5 ) ) ) , k ≥ 0 L ( v k + 1 ( x ) ) = 1 s L ( x ( ( x − x ⋅ v 0 ) 2 ( A k ( v 0 ) − B k ( v 0 ) ) ) + 2 ∑ i = 1 4 ( x − x ⋅ v i ) 2 ( A k ( v i ) − B k ( v i ) ) + ( x − x ⋅ v 5 ) 2 ( A k ( v 5 ) − B k ( v 5 ) ) ) , k ≥ 0

{ u 0 = 1 + x 3 − 1 6 x 4 − 1 1260 x 10 v 0 = 1 − x 3 − 1 280 x 8 u 1 = 0.17 x 4 − 0.00039467 x 8 v 1 = 0.00950857 x 7 − 0.00039467 x 8 ⋮

The series solutions are

{ u ( x ) = u 0 + u 1 + u 2 + ⋯ v ( x ) = v 0 + v 1 + v 2 + ⋯

The results produced by the method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

2) Simpson’s Method (SM)

We divide the interval (0, 1) into subinterval with equal length h = 0.1 , n = 10 and denote x i = a + i h , 0 ≤ i ≤ 10 . The recursive relation is given by

{ L { u 0 ( x ) } = 1 s + 6 s 4 − 4 s 5 − 2880 s 11 , L { v 0 ( x ) } = 1 s − 6 s 4 − 144 s 9 ,

{ L ( u k + 1 ( x ) ) = 0.1 3 s L ( x ( ( x − x ⋅ v 0 ) 2 ( A k ( v 0 ) + B k ( v 0 ) ) ) + 4 ∑ i = 1 4 ( x − x ⋅ v 2 i − 1 ) 2 ( A k ( v 2 i − 1 ) + B k ( v 2 i − 1 ) ) + 2 ∑ i = 1 4 ( x − x ⋅ v 2 i ) 2 ( A k ( v 2 i ) + B k ( v 2 i ) ) + ( x − x ⋅ v 10 ) 2 ( A k ( v 10 ) + B k ( v 10 ) ) ) , k ≥ 0 L ( v k + 1 ( x ) ) = 0.1 3 s L ( x ( ( x − x ⋅ v 0 ) 2 ( A k ( v 0 ) − B k ( v 0 ) ) ) + 4 ∑ i = 1 4 ( x − x ⋅ v 2 i − 1 ) 2 ( A k ( v 2 i − 1 ) − B k ( v 2 i − 1 ) ) + 2 ∑ i = 1 4 ( x − x ⋅ v 2 i ) 2 ( A k ( v 2 i ) − B k ( v 2 i ) ) + ( x − x ⋅ v 10 ) 2 ( A k ( v 10 ) − B k ( v 10 ) ) ) , k ≥ 0

{ u 0 = 1 + x 3 − 1 6 x 4 − 1 1260 x 10 , v 0 = 1 − x 3 − 1 280 x 8 , u 1 = 0.16666667 x 4 − 0.00039736 x 8 , v 1 = 0.00952761 x 7 − 0.00039736 x 8 , ⋮

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.20 | 1.00800000 | 1.00800533 | 5.33335213e−06 |

0.40 | 1.06400000 | 1.06408534 | 8.53371745e−05 |

0.60 | 1.21600000 | 1.21643206 | 4.32061253e−04 |

0.80 | 1.51200000 | 1.51336546 | 1.36545876e−03 |

1 | 2.00000000 | 2.00333032 | 3.33031586e−03 |

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.20 | 0.99200000 | 0.99200011 | 1.12587007e−07 |

0.40 | 0.93600000 | 0.93601324 | 1.32433409e−05 |

0.60 | 0.78400000 | 0.78420632 | 2.06317915e−04 |

0.80 | 0.48800000 | 0.48939610 | 1.39609894e−03 |

1 | 0.00000000 | 0.00594422 | 5.94421577e−03 |

The series solutions are

{ u ( x ) = u 0 + u 1 + u 2 + ⋯ v ( x ) = v 0 + v 1 + v 2 + ⋯

The results produced by the method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.10 | 1.00100000 | 1.00100000 | 4.16058963e−16 |

0.20 | 1.00800000 | 1.00800000 | 4.71714828e−13 |

0.30 | 1.02700000 | 1.02700000 | 2.96484081e−11 |

0.40 | 1.06400000 | 1.06400000 | 5.65791067e−10 |

0.50 | 1.12500000 | 1.12500001 | 5.58673293e−09 |

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.10 | 0.99900000 | 0.99900000 | 9.17047572e−10 |

0.20 | 0.99200000 | 0.99200011 | 1.12810573e−07 |

0.30 | 0.97300000 | 0.97300185 | 1.84936099e−06 |

0.40 | 0.93600000 | 0.93601327 | 1.32693006e−05 |

0.50 | 0.87500000 | 0.87506048 | 6.04816460e−05 |

Example 3.4

Consider the system of nonlinear Fredholm integro differential equation

{ u ″ ( x ) = 2 + 12 5 x − ∫ 0 1 x ( u 2 + v 2 ) d t , u ( 0 ) = 1 , u ′ ( 0 ) = 0 , v ″ ( x ) = − 2 + 4 3 x − ∫ 0 1 x ( u 2 − v 2 ) d t , v ( 0 ) = 1 , v ′ ( 0 ) = 0 ,

with the exact solution ( u ( x ) , v ( x ) ) = ( 1 + x 2 , 1 − x 2 ) .

As usual, on taking the Laplace transform of both sides of the above equation gives

{ s 2 L { u ( x ) } − s = L ( 2 + 12 5 x ) − L ( ∫ 0 1 x ( u 2 + v 2 ) d t ) , s 2 L { v ( x ) } − s = L ( − 2 + 4 3 x ) − L ( ∫ 0 1 x ( u 2 − v 2 ) d t ) .

So that

{ L { u ( x ) } = 1 s + 2 s 3 + 12 5 s 4 − 1 s 2 L ( ∫ 0 1 x ( u 2 + v 2 ) d t ) , L { u ( x ) } = 1 s − 2 s 3 + 4 3 s 4 − 1 s 2 L ( ∫ 0 1 x ( u 2 − v 2 ) d t ) .

1) Trapezoidal Method (TM)

We divide the interval (0, 1) into subinterval of equal lengths h = 0.2 , n = 5 and denote v i = a + i h , 0 ≤ i ≤ 5 . The recursive relations are

{ L { u 0 ( x ) } = 1 s + 2 s 3 + 12 5 s 4 , L { v 0 ( x ) } = 1 s − 2 s 3 + 4 3 s 4 .

{ L ( u k + 1 ( x ) ) = 0.2 2 s 2 L ( x ( A k ( t 0 ) + B k ( t 0 ) ) + 2 ∑ i = 1 4 ( A k ( v i ) + B k ( v i ) ) + ( A k ( v 5 ) + B k ( v 5 ) ) ) , k ≥ 0 L ( v k + 1 ( x ) ) = 0.2 2 s 2 L ( x ( A k ( t 0 ) − B k ( t 0 ) ) + 2 ∑ i = 1 4 ( A k ( v i ) − B k ( v i ) ) + ( A k ( v 5 ) − B k ( v 5 ) ) ) , k ≥ 0

{ u 0 ( x ) = 1 + x 2 + 2 5 x 3 , v 0 ( x ) = 1 − x 2 + 2 9 x 3 , u 1 ( x ) = − 0.47488288 x 3 , v 1 ( x ) = − 0.28306869 x 3 , ⋮

The series solutions are

{ u ( x ) = u 0 + u 1 + u 2 + ⋯ v ( x ) = v 0 + v 1 + v 2 + ⋯

The results produced by the method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

2) Simpson’s Method (SM)

We divide the interval (0, 1) into subinterval of equal lengths h = 0.1 , n = 10 and denote x i = a + i h , 0 ≤ i ≤ 10 . The recursive relation is given by

{ L { u 0 ( x ) } = 1 s + 2 s 3 + 12 5 s 4 , L { v 0 ( x ) } = 1 s − 2 s 3 + 4 3 s 4 ,

{ L ( u k + 1 ( x ) ) = 0.1 3 s 2 L ( x ( A k ( t 0 ) + B k ( t 0 ) ) + 4 ∑ i = 1 5 x ⋅ ( A k ( t 2 i − 1 ) + B k ( t 2 i − 1 ) ) + 2 ∑ i = 1 4 x ⋅ ( A k ( t 2 i ) + B k ( t 2 i ) ) + x ( A k ( t 10 ) + B k ( t 10 ) ) ) , k ≥ 0 L ( v k + 1 ( x ) ) = 0.1 3 s 2 L ( x ( A k ( t 0 ) − B k ( t 0 ) ) + 4 ∑ i = 1 5 x ⋅ ( A k ( t 2 i − 1 ) − B k ( t 2 i − 1 ) ) + 2 ∑ i = 1 4 x ⋅ ( A k ( t 2 i ) − B k ( t 2 i ) ) + x ( A k ( t 10 ) − B k ( t 10 ) ) ) , k ≥ 0

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.20 | 1.04000000 | 1.03996370 | 3.62969667e−05 |

0.40 | 1.16000000 | 1.15970962 | 2.90375734e−04 |

0.60 | 1.36000000 | 1.35901998 | 9.80018101e−04 |

0.80 | 1.64000000 | 1.63767699 | 2.32300587e−03 |

1 | 2.00000000 | 1.99546288 | 4.53712084e−03 |

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.20 | 0.96000000 | 0.95996436 | 3.56392238e−05 |

0.40 | 0.84000000 | 0.83971489 | 2.85113790e−04 |

0.60 | 0.64000000 | 0.63903774 | 9.62259042e−04 |

0.80 | 0.36000000 | 0.35771909 | 2.28091032e−03 |

1 | 0 | −0.00445490 | 4.45490297e−03 |

{ u 0 ( x ) = 1 + x 2 + 2 5 x 3 , v 0 ( x ) = 1 − x 2 + 2 9 x 3 , u 1 ( x ) = − 0.46672242 x 3 , v 1 ( x ) = − 0.27424681 x 3 , ⋮

The series solutions are

{ u ( x ) = u 0 + u 1 + u 2 + ⋯ v ( x ) = v 0 + v 1 + v 2 + ⋯

The results produced by the method with only few components (m = 5) are in a very good agreement with the exact solution results as shown in

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.10 | 1.01000000 | 1.00999945 | 5.51111213e−07 |

0.20 | 1.04000000 | 1.03999559 | 4.40888971e−06 |

0.30 | 1.09000000 | 1.08998512 | 1.48800028e−05 |

0.40 | 1.16000000 | 1.15996473 | 3.52711177e−05 |

0.50 | 1.25000000 | 1.24993111 | 6.88889017e−05 |

x | Exact | DADM | Absolute Error |
---|---|---|---|

0 | 1.00000000 | 1.00000000 | 0 |

0.10 | 0.99000000 | 0.98999967 | 3.28616130e−07 |

0.20 | 0.96000000 | 0.95999737 | 2.62892904e−06 |

0.30 | 0.91000000 | 0.90999113 | 8.87263551e−06 |

0.40 | 0.84000000 | 0.83997897 | 2.10314323e−05 |

0.50 | 0.75000000 | 0.74995892 | 4.10770163e−05 |

In this paper, a modification of the Laplace Adomian decomposition method inspired by property of discretization is proposed. We developed a new Laplace Discrete Adomian decomposition method (LDADM) in which has been successfully applied to finding efficient numerical solutions of integro-differential equations featuring both nonlinear Volterra and Fredholm integrals. The method was based on the well-known Adomian decomposition method coupled with some numerical integration schemes (quadrature rules) alongside utilizing the famous and most used Laplace transform. The method gives approximate solutions iteratively with less number of computational steps. The results reveal that the proposed method is simple to execute and effective. Thus, many highly nonlinear integro-differential equations can be solved using the proposed method.

The authors declare no conflicts of interest regarding the publication of this paper.

Bakodah, H.O., Al-Mazmumy, M., Almuhalbedi, S.O. and Abdullah, L. (2019) Laplace Discrete Adomian Decomposition Method for Solving Nonlinear Integro Differential Equations. Journal of Applied Mathematics and Physics, 7, 1388-1407. https://doi.org/10.4236/jamp.2019.76093