This paper concerns the implementation of the orthogonal polynomials using the Galerkin method for solving Volterra integro-differential and Fredholm integro-differential equations. The constructed orthogonal polynomials are used as basis functions in the assumed solution employed. Numerical examples for some selected problems are provided and the results obtained show that the Galerkin method with orthogonal polynomials as basis functions performed creditably well in terms of absolute errors obtained.
Integro-differential equations (IDEs) have attracted growing interest over the years because of the mathematical application in real life problems. Mathematical modeling of real life problems usually resulted in fractional equations. Many mathematical formulations of physical phenomena contain integro-differential equations. These equations arise in many fields like Physics, Astronomy, Potential theory, Fluid dynamics, Biological models and Chemical kinetics. Integro-differential equations contain both integral and differential operators. The derivatives of the unknown functions may appear to any order (see [
We consider the higher order linear integro-differential equation as follows:
∑ i = 0 n P i y ( i ) + λ ∫ g ( x ) h ( x ) k ( x , t ) y ( t ) d t = f ( x ) (1)
Subject to the following conditions
y ( k ) ( a ) = α k , k = 1 , 2 , ⋯ , n (2)
where α k ( k ≥ 0 ) are constant coefficients, g ( x ) and h ( x ) are lower and upper limits of integration, λ is a constant parameter and k ( x , t ) is a function of two variables x and t called the kernel, f ( x ) is a known function and y ( x ) is the unknown function to be determined.
Integro-differential equation
An integro-differential equation is an equation involving both integral and derivatives of a function. Example of such equation is stated below:
a 2 y ″ ( x ) + a 1 y ′ ( x ) + a 0 y ( x ) + λ ∫ a b H ( x , t ) y ( t ) d t = f ( x ) (3)
Galerkin method
Galerkin method is a method of determining coefficient a k in a power series solution of the form:
y ( x ) ≅ y 0 ( x ) + ∑ k = 0 n a k y k ( x ) (4)
of the ordinary differential equation L [ y ( x ) ] = 0 so that L [ y ( x ) ] , the result of applying the ordinary differential operator to y ( x ) , is orthogonal to every y k ( x ) for k = 1 , 2 , ⋯ , n
Chebyshev Polynomial
The Chebyshev polynomials of the first kind are a set of orthogonal polynomials defined as the solutions to the Chebyshev differential equation and denoted by T n ( x ) . The Chebyshev polynomial of the first kind denoted by T n ( x ) is defined by the contour integral
T n ( x ) = 1 4 π i ∮ ( 1 − t 2 ) t − n − 1 ( 1 − 2 t z + t 2 ) d t
Where the contour encloses the origin and is traversed in a counter clockwise direction.
Orthogonal over a set
A set of function { ϕ r ( x ) } is said to be orthogonal over a set of points { x i } with respect to the weight function w ( x ) , if
∑ i = 0 N w ( x i ) ϕ j ( x i ) ϕ k ( x i ) = 0 , i ≠ k
Orthogonal over an interval
A set of functions { ϕ r ( x ) } is said to orthogonal on an interval [ a , b ] with respect to the weight function w ( x ) , if
∫ a b w ( x ) ϕ i ( x ) ϕ j ( x ) d x = 0 , i ≠ j
Approximate solution
Approximate solution is used for the expression obtained after the unknown constants have been generated and substituting back into the assumed solution. It is hereby call approximate solution since it is a reasonable approximation to the exact solution.
In this section, we constructed orthogonal polynomials f i ( x ) , valid on the interval [ a , b ] with the leading term x i
Then, starting with
f 0 ( x ) = 1 , (5)
Thus, we find the linear polynomials f 1 ( x ) , with leading term x, is written as
f 1 ( x ) = x + k 1 , 0 f 0 ( x ) , (6)
where, k 1 , 0 is a constant to be determined. Since f 0 ( x ) and f 1 ( x ) are orthogonal, we have,
∫ a b w ( x ) f 0 ( x ) f 1 ( x ) d x = 0 = ∫ a b x w ( x ) f 0 ( x ) d x + k 1 , 0 ∫ a b w ( x ) ( f 0 ( x ) ) 2 d x
using (5) and (6).
From the above, we have,
k 1 , 0 = − ∫ a b x w ( x ) f 0 ( x ) ∫ a b w ( x ) ( f 0 ( x ) ) 2 d x
Hence, (6) gives,
f 1 ( x ) = x − ∫ a b x w ( x ) f 0 ( x ) ∫ a b w ( x ) ( f 0 ( x ) ) 2 d x
Now, the polynomials f 2 ( x ) , of degree 2 and the leading term x 2 is written as
f 2 ( x ) = x 2 + k 2 , 0 f 0 ( x ) + k 2 , 1 f 1 ( x ) (7)
where the constants k 2 , 0 and k 2 , 1 are determined by using orthogonality conditions
∫ a b w ( x ) f p ( x ) f q ( x ) d x = { 0 , p ≠ q ∫ a b w ( x ) f p 2 ( x ) d x , p = q (8)
Since f 2 ( x ) is orthogonal to f 0 ( x ) , we have
∫ a b w ( x ) f 0 ( x ) [ x 2 + k 2 , 0 f 0 ( x ) + k 2 , 1 f 1 ( x ) ] d x = 0 (9)
Since,
∫ a b w ( x ) f 0 ( x ) f 1 ( x ) d x = 0
The above equation gives
k 2 , 0 = − ∫ a b x 2 w ( x ) f 0 ( x ) ∫ a b w ( x ) ( f 0 ( x ) ) 2 d x = − ∫ a b x 2 w ( x ) d x ∫ a b w ( x ) d x (10)
Again, since f 2 ( x ) is orthogonal to f 1 ( x ) , we have
∫ a b w ( x ) f 1 ( x ) [ x 2 + k 2 , 0 f 0 ( x ) + k 2 , 1 f 1 ( x ) ] d x = 0
Thus, using (7), we obtain
k 2 , 1 = − ∫ a b x 2 w ( x ) f 1 ( x ) ∫ a b w ( x ) ( f 1 ( x ) ) 2 d x (11)
Since k 2 , 1 and k 2 , 0 are known, (7) determines f 2 ( x ) . Proceeding in the same way, the method is generalized and we have,
f j ( x ) = x j + k j , 0 f 0 ( x ) + k j , 1 f 1 ( x ) + ⋯ + k j , j − 1 (12)
where the constants k j , i and so chosen that f j ( x ) is orthogonal to
f 0 ( x ) , f 1 ( x ) , ⋯ , f j − 1 ( x )
These conditions yield,
k j , i = − ∫ a b x j w ( x ) f i ( x ) ∫ a b w ( x ) ( f i ( x ) ) 2 d x (13)
Few terms of orthogonal polynomials valid in the interval [−1, 1] are given below.
f 0 ( x ) = 1 f 1 ( x ) = x f 2 ( x ) = x 2 − 1 3 f 3 ( x ) = x 3 − 3 5 x f 4 ( x ) = x 4 − 6 7 x 2 + 3 35 ⋮
etc.
In this section, we considered (1) and (2).
Here we assumed an approximate solution of the form
u ( x ) ≅ u N ( x ) = ∑ i = 0 N a i f i ( x ) , − 1 ≤ x ≤ 1 (14)
where f i ( x ) ( i ≥ 0 ) are the orthogonal polynomial constructed and valid in the interval [−1, 1].
Thus, differentiating (14)/with respect to x, n times, we have
u ( n ) ( x ) ≅ u N ( n ) ( x ) = ∑ i = 0 N a i f i ( n ) ( x ) (15)
Substituting (14) and (15) into (1), we obtain
∑ k = 0 n ∑ i = 0 N P k a i f i ( n ) ( x ) = f ( x ) + λ ∑ i = 0 N a i ∫ 0 x k ( x , t ) ∑ i = 0 N f i ( t ) d t (16)
We determined the unknown coefficients a i using the Galerkin idea by multiplying both sides of (16) by f j ( x ) and then integrating with respect to x from −1 to 1.
Thus, we obtain
∑ k = 0 n ∑ i = 0 N P k a i ∫ − 1 1 f i ( n ) ( x ) f j ( x ) d x = ∫ − 1 1 f j ( x ) f ( x ) d x + λ ∑ i = 0 N a i ∫ 0 x ∫ − 1 1 k ( x , t ) ∑ i = 0 N f i ( t ) f j ( x ) d t d x , j = 0 , 1 , ⋯ , N (17)
This process generates a system of linear equations for the unknown { a i } i = 0 N together with the conditions
∑ i = 0 N a i f i ( j ) ( a ) = α j , j = 1 , 2 , ⋯ , n (18)
for the same number of equations in the linear system.
The unknown parameters are determined by solving the system (17) and (18). The values of the constants obtained are then substituted back into (14) to get the required approximate solution for the appropriate order.
In this section, we consider four selected problems for experimenting and compare our results with existing results.
Numerical example 1
We consider the Volterra integro-differential equations of the second kind of the form:
y ′ ( x ) = 1 − 2 x sin x + ∫ 0 x y ( t ) d t (19)
together with the condition given as
y ( 0 ) = 0 (20)
The exact solution is given as
y ( x ) = x cos x
Here we solved example 1 for case N = 4 .
Thus, Equation (14) becomes
y 4 ( x ) = ∑ i = 0 4 a i f i ( x ) (21)
Substituting the values of f i ( x ) , 0 ≤ i ≤ 4 , we obtain
y 4 ( x ) = a 0 + ( 2 x − 1 ) a 1 + ( ( 2 x − 1 ) 2 − 1 3 ) a 2 + ( ( 2 x − 1 ) 3 − 3 5 ( 2 x − 1 ) ) a 3 + ( ( 2 x − 1 ) 4 − 6 7 ( 2 x − 1 ) 2 + 3 35 ) a 4 (22)
and,
y ′ 4 ( x ) = 2 a 1 + ( 8 x − 4 ) a 2 + ( 6 ( 2 x − 1 ) 2 − 6 5 ) a 3 + ( 8 ( 2 x − 1 ) 3 − 48 7 x + 24 35 ) a 4 (23)
Substituting (23) into (19) for case N = 4, we obtain
2 a 1 + ( 8 x − 4 ) a 2 + ( 6 ( 2 x − 1 ) 2 − 6 5 ) a 3 + ( 8 ( 2 x − 1 ) 3 − 48 7 x + 24 35 ) a 4 − ∫ a x { a 0 + ( 2 t − 1 ) a 1 + ( ( 2 t − 1 ) 2 − 1 2 ) a 2 + ( ( 2 t − 1 ) 3 − 3 5 ( 2 t − 1 ) ) a 3 + ( ( 2 t − 1 ) 4 − 6 7 ( 2 t − 1 ) 2 + 3 35 ) a 4 } d t = 1 − 2 x sin x (24)
Thus, evaluating the integral in (24) and simplifying, we obtain
− x a 0 + ( 2 + x − x 2 ) a 1 + ( 15 2 x + 2 x 2 − 4 3 x 3 + 4 ) a 2 + ( 6 ( 2 x − 1 ) 2 + 2 3 x − 2 x 4 + 4 x 3 − 12 5 x 2 − 6 5 ) a 3 + ( 8 ( 2 x − 1 ) 3 − 248 35 x − 16 5 x 5 + 8 x 4 − 48 7 x 3 + 16 7 x 2 + 24 35 ) a 4 = 1 − 2 x sin x (25)
The unknown coefficients a i ( i ≤ 4 ) are determined using the Galerkin idea by multiplying both sides of (25) by f j ( 2 x − 1 ) and then integrating the resulted equation between x = −1 to x = 1.
For case j = 1, we multiplied both sides of (25) by (2x − 1) and then integrating the resulted equation between x = −1 to x = 1, to obtain
− 4 3 a 0 − 2 a 1 − 2 5 a 2 − 460 9 a 3 + 6032 35 a 4 = − 0.7953 (26)
For case j = 2, we multiplied both sides of (25) by ( 2 x − 1 ) 2 − 1 3 and then integrating the resulted equation between x = −1 to x = 1, to obtain
8 3 a 0 + 148 45 a 1 + 20 9 a 2 + 183272 1575 a 3 − 128032 315 a 4 = 0.363 (27)
For case j = 3, we multiplied both sides of (25) by ( 2 x − 1 ) 3 − 3 5 ( 2 x − 1 ) and then integrating the resulted equation between x = −1 to x = 1, to obtain
− 32 5 a 0 − 92 15 a 1 − 1544 525 a 2 − 20776 75 a 3 + 11053408 11025 a 4 = 0.18 (28)
For case j = 4, we multiplied both sides of (25) by ( 2 x − 1 ) 4 − 6 7 ( 2 x − 1 ) 2 + 3 35 and then integrating the resulted equation between x = −1 to x = 1, to obtain
1664 105 a 0 + 432 35 a 1 + 416 105 a 2 + 2518688 3675 a 3 − 124288 49 a 4 = − 2.3 (29)
Now, using the condition given in (22), we obtain
a 0 − a 1 + 2 3 a 2 − 2 5 a 3 + 8 25 a 4 = 0 (30)
Hence, (26)-(30) are then solved to obtain the unknown constants a i ( i = 0 , 1 , 2 , 3 , 4 ) which are then substituted to the approximate Equation (22).
Again, we solved (1) and (2) for case N = 6 by re-writing (21) as:
y 6 ( x ) = ∑ i = 0 6 a i f i ( x ) (31)
Hence, (31) becomes
y 6 ( x ) = a 0 + ( 2 x − 1 ) a 1 + ( ( 2 x − 1 ) 2 − 1 3 ) a 2 + ( ( 2 x − 1 ) 3 − 3 5 ( 2 x − 1 ) ) a 3 + ( ( 2 x − 1 ) 4 − 6 7 ( 2 x − 1 ) 2 + 3 35 ) a 4 + ( ( 2 x − 1 ) 5 − 10 9 ( 2 x − 1 ) 3 + 5 21 ( 2 x − 1 ) ) a 5 + ( ( 2 x − 1 ) 6 − 15 11 ( 2 x − 1 ) 4 + 5 11 ( 2 x − 1 ) 2 − 5 231 ) a 6 (32)
And,
y ′ 6 ( x ) = 2 a 1 + ( 8 x − 4 ) a 2 + ( 6 ( 2 x − 1 ) 2 − 6 5 ) a 3 + ( 8 ( 2 x − 1 ) 3 − 48 7 x + 24 35 ) a 4 + ( 10 ( 2 x − 1 ) 4 − 20 ( 2 x − 1 ) 3 3 + 10 21 ) a 5 + ( 12 ( 2 x − 1 ) 5 − 120 ( 2 x − 1 ) 3 11 + 40 x 11 − 20 11 ) a 6 (33)
Thus substituting (32) and (33) into (19), we obtain
2 a 1 + ( 8 x − 4 ) a 2 + ( 6 ( 2 x − 1 ) 2 − 6 5 ) a 3 + ( 8 ( 2 x − 1 ) 3 − 48 7 x + 24 35 ) a 4 + ( 10 ( 2 x − 1 ) 4 − 20 ( 2 x − 1 ) 3 3 + 10 21 ) a 5 + ( 12 ( 2 x − 1 ) 5 − 120 ( 2 x − 1 ) 3 11 + 40 x 11 − 20 11 ) a 6 − ∫ 0 x { a 0 + ( 2 t − 1 ) a 1 + ( ( 2 t − 1 ) 2 − 1 2 ) a 2 + ( ( 2 t − 1 ) 3 − 3 5 ( 2 t − 1 ) ) a 3 + ( ( 2 t − 1 ) 4 − 6 7 ( 2 t − 1 ) 2 + 3 35 ) a 4 + ( 10 ( 2 t − 1 ) 4 − 20 ( 2 t − 1 ) 3 3 + 10 21 ) a 5 + ( 12 ( 2 t − 1 ) 5 − 120 ( 2 t − 1 ) 3 11 + 40 t 11 − 20 11 ) a 6 } d t = 1 − 2 x sin x (34)
Thus, evaluating the integral in (34) and simplifying, we obtain
− x a 0 + ( 2 + x − x 2 ) a 1 + ( 15 2 x + 2 x 2 − 4 3 x 3 + 4 ) a 2 + ( 6 ( 2 x − 1 ) 2 + 2 3 x − 2 x 4 + 4 x 3 − 12 5 x 2 − 6 5 ) a 3 + ( 8 ( 2 x − 1 ) 3 − 248 35 x − 16 5 x 5 + 8 x 4 − 48 7 x 3 + 16 7 x 2 + 24 35 ) a 4 + ( 10 ( 2 x − 1 ) 4 − 20 3 ( 2 x − 1 ) 3 + 1200 147 x − 32 x 5 + 280 3 x 4 − 320 3 x 3 + 60 x 2 ) a 5 + ( 12 ( 2 x − 1 ) 5 − 120 11 ( 2 x − 1 ) 3 + 72 11 x − 1280 11 x 6 + 192 x 5 − 2400 11 x 4 + 1280 11 x 3 − 320 11 x 2 ) a 6 = 1 − 2 x sin x (35)
The unknown coefficients a i ( i ≤ 4 ) are determined using the Galerkin idea by multiplying both sides of (35) by f j ( 2 x − 1 ) and then integrating the resulted equation between x = −1 to x = 1.
For case j = 1, we multiplied both sides of (35) by (2x − 1) and then integrating the resulted equation between x = −1 to x = 1, to obtain
− 4 3 a 0 − 2 a 1 − 2 5 a 2 − 460 9 a 3 + 6032 35 a 4 − 97264 189 a 5 + 360160 231 a 6 = − 0.7953 (36)
For case j = 2, we multiplied both sides of (35) by ( 2 x − 1 ) 2 − 1 3 and then integrating the resulted equation between x = −1 to x = 1, to obtain
8 3 a 0 + 148 45 a 1 + 20 9 a 2 + 183272 1575 a 3 − 128032 315 a 4 + 238816 189 a 5 − 1507904 385 a 6 = 0.363 (37)
For case j = 3, we multiplied both sides of (35) by ( 2 x − 1 ) 3 − 3 5 ( 2 x − 1 ) and then integrating the resulted equation between x = −1 to x = 1, to obtain
− 32 5 a 0 − 92 15 a 1 − 1544 525 a 2 − 20776 75 a 3 + 11053408 11025 a 4 − 1007648 315 a 5 + 2330560 231 a 6 = 0.18 (38)
For case j = 4, we multiplied both sides of (35) by ( 2 x − 1 ) 4 − 6 7 ( 2 x − 1 ) 2 + 3 35 and then integrating the resulted equation between x = −1 to x = 1, to obtain
1664 105 a 0 + 432 35 a 1 + 416 105 a 2 + 2518688 3675 a 3 − 124288 49 a 4 + 360321152 47659 a 5 − 63937952 24255 a 6 = − 2.3 (39)
For case j = 5, we multiplied both sides of (25) by ( 2 x − 1 ) 5 − 10 9 ( 2 x − 1 ) 3 + 5 21 ( 2 x − 1 ) and then integrating the resulted equation between x = −1 to x = 1, to obtain
− 2528 63 a 0 − 4976 180 a 1 + 2720 7 a 2 − 550112 315 a 3 + 1396705664 218295 a 4 − 85672064 3969 a 5 + 48377661184 693693 a 6 = − 3152 63 − 2002592 63 cos ( 1 ) + 144 7 (40)
For case j = 6, we multiplied both sides of (35) by ( 2 x − 1 ) 6 − 15 11 ( 2 x − 1 ) 4 + 5 11 ( 2 x − 1 ) 2 − 5 231 and then integrating the resulted equation between x = −1 to x = 1, to obtain
− 2528 63 a 0 − 4976 180 a 1 + 2720 7 a 2 − 550112 315 a 3 + 1396705664 218295 a 4 − 85672064 3969 a 5 + 48377661184 693693 a 6 = − 1376 11 − 230568512 231 cos ( 1 ) + 1290272 63 sin ( 1 ) (41)
Now, using the condition given in (22), we obtain
a 0 − a 1 + 2 3 a 2 − 2 5 a 3 + 8 25 a 4 − 8 63 a 5 + 16 231 a 6 = 0 (42)
Hence, (36)-(42) are then solved to obtain the unknown constants a i ( i = 0 , 1 , 2 , 3 , 4 , 5 , 5 , 6 ) which are then substituted to the approximate equation (32). More values of N are computed follow the same procedure and the results obtained are tabulated below.
Example 2:
y ″ ( x ) + x y ′ ( x ) − x y ( x ) = e x − 2 sin x + ∫ − 1 1 y ( t ) d t
With the conditions
y ( 0 ) = 1 and y ′ ( 0 ) = 1 , The exact solution is y ( x ) = e x .
Example 3: Consider the Fredholm integro-differential equation (See [
y ' ' ' ' ( x ) = 1 + ∫ 0 1 e − x y 2 ( t ) d t , 0 < x < 1
Together with the conditions y ( 0 ) = y ′ ( 0 ) = 1 ; y ( 1 ) = e ; y ′ ( 1 ) = e . The exact solution is y ( x ) = e x .
· Denotes the results are not available for comparison
· Denotes Results are not available for comparison
Example 4: Consider the Fredholm integro-differential equation (See [
y ' ' ' ' ( x ) = x + ( x + 3 ) e x + y ( x ) − ∫ 0 x y ( t ) d t , 0 < x < 1
With the following conditions
y ( 0 ) = 1 ; y ( 1 ) = 1 + e ; y ″ ( 0 ) = 2 ; y ″ ( 1 ) = 3 e . The exact solution is y ( x ) = 1 + x e x .
· Denotes Results are not available for comparison
· Denotes Results are not available for comparison
The approximate solution obtained by means of Galerkin method is a finite power series which can be in turn expressed in closed form of exact solution as the degree of the approximant increases. The finite series solution is obtained for each problem considered by increasing the value of N, which in turn converges to closed form of exact solution, the absolute errors obtained tend to zero and ensures stability of our method (See Tables 1-8). Also, from the results obtained by [
| X | Exact solution | Approximate solution | Approximate solution |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0.1 | 0.09999984769 | 0.1007787777 | 7.7893 × 10−4 |
| 0.2 | 0.19999871500 | 0.20007989915 | 8.0021 × 10−4 |
| 0.3 | 0.29999588772 | 0.30082048742 | 8.2460 × 10−4 |
| 0.4 | 0.39990252364 | 0.40085698231 | 9.5446 × 10−4 |
| 0.5 | 0.49998096153 | 0.50093900554 | 9.5813 × 10−4 |
| 0.6 | 0.59996710167 | 0.60128870164 | 1.2920 × 10−3 |
| 0.7 | 0.69994775882 | 0.70214095881 | 2.1932 × 10−3 |
| 0.8 | 0.79992201922 | 0.80415158192 | 4.2296 × 10−3 |
| 0.9 | 0.89988896922 | 0.90630266921 | 6.4113 × 10−3 |
| 1.0 | 0.99984769523 | 1.00008011995 | 2.3242 × 10−4 |
| X | Exact solution | Approximate solution | Approximate solution |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0.1 | 0.09999984769 | 0.1000123167 | 1.2409 × 10−5 |
| 0.2 | 0.19999871500 | 0.2000261375 | 2.7350 × 10−5 |
| 0.3 | 0.29999588772 | 0.3000306197 | 3.4732 × 10−5 |
| 0.4 | 0.39990252364 | 0.4000469833 | 5.6731 × 10−5 |
| 0.5 | 0.49998096153 | 0.5000608685 | 7.9907 × 10−5 |
| 0.6 | 0.59996710167 | 0.6000487216 | 8.1620 × 10−5 |
| 0.7 | 0.69994775882 | 0.7000377598 | 9.0001 × 10−5 |
| 0.8 | 0.79992201922 | 0.8003215592 | 3.9951 × 10−5 |
| 0.9 | 0.89988896922 | 0.9002501792 | 3.6121 × 10−5 |
| 1.0 | 0.99984769523 | 1.0003425534 | 5.5778 × 10−5 |
| X | Exact solution | Approximate solution | Approximate solution |
|---|---|---|---|
| −1 | 0.36787944 | 0.37418684 | 6.3074 × 10−3 |
| −0.8 | 0.44932896 | 0.45641056 | 7.0816 × 10−3 |
| −0.6 | 0.54881164 | 0.55712374 | 8.3121 × 10−3 |
| −0.4 | 0.67032005 | 0.68009815 | 9.7781 × 10−3 |
| −0.2 | 0.81873075 | 0.82014445 | 1.4137 × 10−2 |
| 0 | 1.00000000 | 1.00180376 | 1.8937 × 10−2 |
| 0.2 | 1.22140283 | 1.24357182 | 2.2169 × 10−3 |
| 0.4 | 1.47182472 | 1.49774274 | 2.5918 × 10−2 |
| 0.6 | 1.82211881 | 1.85630581 | 3.4187 × 10−2 |
| 0.8 | 2.22551000 | 2.26616893 | 4.0928 × 10−2 |
| 1.0 | 2.71828182 | 2.78212785 | 6.3846 × 10−3 |
| X | Exact solution | Approximate solution | Approximate solution |
|---|---|---|---|
| −1 | 0.36787944 | 0.367966169 | 8.6729 × 10−5 |
| −0.8 | 0.44932896 | 0.449409094 | 8.0134 × 10−5 |
| −0.6 | 0.54881164 | 0.548889417 | 7.7837 × 10−5 |
| −0.4 | 0.67032005 | 0.676389371 | 6.9321 × 10−5 |
| −0.2 | 0.81873075 | 0.818758949 | 7.8199 × 10−5 |
| 0 | 1.00000000 | 1.000966532 | 7.6653 × 10−4 |
| 0.2 | 1.22140283 | 1.222229894 | 8.9614 × 10−4 |
| 0.4 | 1.47182472 | 1.472514031 | 6.8933 × 10−4 |
| 0.6 | 1.82211881 | 1.822781972 | 5.9397 × 10−4 |
| 0.8 | 2.22551000 | 2.226029824 | 4.8892 × 10−4 |
| 1.0 | 2.71828182 | 2.718738011 | 4.5621 × 10−4 |
| X | Exact | Approximate of [ | Approx. of Our Method | Absolute errors of [ | Absolute errors of Our Method |
|---|---|---|---|---|---|
| 0.0 | 1.0000000 | 1.0000000 | 1.00000000 | 0 | 0 |
| 0.1 | 1.105171 | 1.105173451 | * | 2.451 × 10−6 | |
| 0.2 | 1.2214027 | 1.2214 | 1.221409351 | 1.0270 × 10−4 | 6.651 × 10−6 |
| 0.3 | 1.349859 | * | 1.349868872 | * | 9.872 × 10−6 |
| 0.4 | 1.4918246 | 1.4918 | 1.491856800 | 1.1246 × 10−3 | 3.220 × 10−5 |
| 0.5 | 1.648721 | * | 1.648800850 | * | 7.985 × 10−5 |
| 0.6 | 1.8221188 | 1.8221 | 1.822700800 | 6.1188 × 10−3 | 5.820 × 10−4 |
| 0.7 | 2.013753 | * | 2.014370200 | * | 6.172 × 10−4 |
| 0.8 | 2.2255409 | 2.2255 | 2.228210900 | 2.0241 × 10−2 | 2.670 × 10−3 |
| 0.9 | 2.459603 | * | 2.465275000 | * | 5.672 × 10−3 |
| 1.0 | 2.71828183 | 2.7183 | 2.725281830 | 5.1282 × 10−2 | 7.000 × 10−3 |
*Denotes the results are not available for comparison.
| X | Exact | Approximate of [ | Approx. of Our Method | Absolute errors of [ | Absolute errors of Our Method |
|---|---|---|---|---|---|
| 0.0 | 1.0000000 | 1.0000 | 1.00000000000 | 0 | 0 |
| 0.1 | 1.105171 | * | 1.10517109874 | * | 9.874 × 10−8 |
| 0.2 | 1.2214027 | 1.2214 | 1.22140278125 | 2.700 × 10−6 | 8.125 × 10−8 |
| 0.3 | 1.349859 | * | 1.34985906846 | * | 6.845 × 10−8 |
| 0.4 | 1.4918246 | 1.4918 | 1.49182466533 | 2.460 × 10−5 | 5.329 × 10−8 |
| 0.5 | 1.648721 | * | 1.64872104101 | * | 4.101 × 10−8 |
| 0.6 | 1.8221188 | 1.8221 | 1.82211884674 | 1.880 × 10−5 | 4.674 × 10−8 |
| 0.7 | 2.013753 | * | 2.01375304115 | * | 4.115 × 10−8 |
| 0.8 | 2.2255409 | 2.2255 | 2.22554093985 | 4.090 × 10−5 | 3.985 × 10−8 |
| 0.9 | 2.459603 | * | 2.45960302679 | * | 2.679 × 10−8 |
| 1.0 | 2.71828183 | 2.7183 | 2.71828184068 | 1.820 × 10−5 | 1.068 × 10−8 |
*Denotes the results are not available for comparison.
| X | Exact | Approx. of [ | Approx. of Our Method | Absolute errors of [ | Absolute errors of Our Method |
|---|---|---|---|---|---|
| 0.0 | 1.0000000 | 1.0000 | 1.0000000000 | 0 | 0 |
| 0.1 | 1.110517 | * | 1.1105179874 | * | 9.874 × 10−7 |
| 0.2 | 1.2442805 | 1.244 | 1.2442922210 | 2.8055 × 10−4 | 1.172 × 10−6 |
| 0.3 | 1.404958 | * | 1.4049590990 | * | 1.099 × 10−6 |
| 0.4 | 1.5967298 | 1.592 | 1.4967570200 | 2.7299 × 10−4 | 9.722 × 10−5 |
| 0.5 | 1.824361 | * | 1.8244327200 | * | 7.172 × 10−5 |
| 0.6 | 2.0932712 | 2.068 | 2.0933164710 | 2.5270 × 10−2 | 4.527 × 10−5 |
| 0.7 | 2.409627 | * | 2.4096387200 | * | 1.172 × 10−5 |
| 0.8 | 2.7804327 | 2.696 | 2.7805028800 | 8.4430 × 10−2 | 9.018 × 10−4 |
| 0.9 | 3.213943 | * | 3.2140147700 | * | 7.177 × 10−4 |
| 1.0 | 3.71828183 | 3.5 | 2.7183814900 | 2.1820 × 10−1 | 6.966 × 10−4 |
*Denotes the results are not available for comparison.
| X | Exact | Approx. of [ | Approx. of Our Method | Absolute errors of [ | Absolute errors of Our Method |
|---|---|---|---|---|---|
| 0.0 | 1.0000000 | 1.0000 | 1.000000000000 | 0 | 0 |
| 0.1 | 1.110517 | * | 1.1105170009231 | * | 9.231 × 10−10 |
| 0.2 | 1.2442805 | 1.2443 | 1.2442805007638 | 1.950 × 10−5 | 7.638 × 10−10 |
| 0.3 | 1.404958 | * | 1.4049580006618 | * | 6.618 × 10−10 |
| 0.4 | 1.5967298 | 1.5967 | 1.5967298002963 | 3.000 × 10−10 | 2.963 × 10−10 |
| 0.5 | 1.824361 | * | 1.8243610001316 | * | 1.316 × 10−10 |
| 0.6 | 2.0932712 | 2.0933 | 2.0932712009316 | 1.772 × 10−8 | 9.316 × 10−9 |
| 0.7 | 2.409627 | * | 2.4096270492700 | * | 4.927 × 10−8 |
| 0.8 | 2.7804327 | 2.7804 | 2.7804327297800 | 3.214 × 10−7 | 2.978 × 10−8 |
| 0.9 | 3.213643 | * | 3.2136430198200 | * | 1.982 × 10−8 |
| 1.0 | 3.71828183 | 3.7184 | 2.7182827690000 | 1.820 × 10−5 | 9.390 × 10−7 |
*Denotes the results are not available for comparison.
In this work, we have proposed the Galerkin method for solving both the boundary and initial value problems for a class of higher order linear and nonlinear Volterra and Fredholm integro-differential based on the constructed orthogonal polynomials as basis function. Illustrative examples are included to demonstrate the validity and applicability of the technique and the tables of results presented reveal that the absolute error decreases when the degree of approximation increases. Furthermore, since basis functions constructed are polynomials, the values of the integrals for the nonlinear integro differential equations are calculated as approximately close to the exact solutions.
The authors declare no conflicts of interest regarding the publication of this paper.
Taiwo, O.A., Alhassan, L.K., Odetunde, O.S. and Alabi, O.O. (2023) Galerkin Method for Numerical Solution of Volterra Integro-Differential Equations with Certain Orthogonal Basis Function. International Journal of Modern Nonlinear Theory and Application, 12, 68-80. https://doi.org/10.4236/ijmnta.2023.122005