In this paper, we derive a simple and efficient matrix formulation using Laguerre polynomials to solve the singular integral equation with degenerate kernel. This method is based on replacement of the unknown function by truncated series of well known Laguerre expansion of functions. This lead s to a system of algebraic equations with Laguerre coefficients. Thus, by solving the matrix equation, the coefficients are obtained. Some numerical examples are included to demonstrate the validity and applicability of the proposed method.
Recent years, there has been a growing interest in the Fredholm and Volterra integral equations. This is an important branch of modern mathematics and arises frequently in many applied areas which include engineering, mechanics, physics, chemistry, astronomy, biology [
φ ( x ) = f ( x ) + λ ∫ 0 ∞ k ( x , t ) φ ( t ) d t (1)
where f ( x ) is a continuous function for x ≥ 0 and the kernel k ( x , t ) is a function defined on the domain D = { ( x , t ) : x ≥ 0, t < ∞ } and φ ( x ) is the unknown function that will be determined. In [
Sequences of orthogonal polynomials appear frequently used as applications in mathematics, mathematical physics, engineering and computer science, in particular during the resolution of partial differential equations (Laplace, Schrödinger) by the method of separation of variables, also these polynomials can be used to solve integral equations of first and second kind [
x y ″ ( x ) + ( 1 − x ) y ′ ( x ) + n y ( x ) = 0 , n = 0 , 1 , 2 , ⋯
The solutions of this equation are the Laguerre polynomials, expressed by the equation following differential:
L n ( x ) = e x d n d x n ( x n e − x )
Then we get an approximation of the exactly integral, let say:
I n ( Φ ) = ∫ 0 ∞ K ( x , y ) φ n ( y ) d y (2)
is an approximation of the exact integral. This type of approximation must be chosen so that the integral (2) can be evaluated (either explicitly or by an efficient numerical technique). The functions L 0 ( x ) , L 1 ( x ) , ⋯ , L n ( x ) will be called interpolating elements. In this paper, the interpolating function L n will be assumed to be the interpolating polynomial
φ n ( x ) = ∑ j = 0 n c j L j ( x ) (3)
where L j are Laguerre polynomials of degree j, n is the number of Laguerre polynomials, and c j are unknown parameters, to be determined.
Consider the following systems
{ L 0 ( x ) , L 1 ( x ) , ⋯ , L n ( x ) }
where
L 0 ( x ) = 1 , L 1 ( x ) = 1 − x , ⋯ , L n ( x ) = ∑ k = 0 n ( n k ) ( − 1 ) k k ! x k n ≥ 2.
This system forms an orthonormal basis in L 2 ( 0, + ∞ ) . In fact, we check that
〈 L n ( x ) , L m ( x ) 〉 = ∫ 0 ∞ L n ( x ) L m ( x ) e − x d x = 0 , m ≠ n
‖ L n ( x ) ‖ = 1 , n = 0 , 1 , 2 , ⋯
The previous system is called the Laguerre polynomial system. To solve the integral Equation (1) we use the projection method. Using an approximation φ n ( x ) of the solution of Equation (1) which is a finite linear combination of orthogonal polynomials ( L n ( x ) ) and also solution of the integral equation
φ n ( x ) = f ( x ) + λ ∫ 0 ∞ K ( x , t ) φ n ( t ) d t (4)
By taking the linear combination of the Laguerre polynomials
φ n ( x ) = ∑ j = 0 n c j L j ( x ) (5)
Substituting (5) in (4) we get
∑ j = 0 n c j L j ( x ) = f ( x ) + λ ∫ 0 ∞ K ( x , t ) ( ∑ j = 0 n c j L j ( t ) ) d t (6)
Let
H j ( x ) = ∫ 0 ∞ K ( x , t ) L j ( t ) d t
Then equation (6) can be written in the form
∑ j = 0 n c j ( L j ( x ) − λ H j ( x ) ) = f ( x ) (7)
By multiplying (7) by L i ( x ) , we get
∑ j = 0 n c j 〈 L j ( x ) − λ H j ( x ) , L i ( x ) 〉 = 〈 f ( x ) , L i ( x ) 〉 (8)
Using the orthogonalization condition in Equation (8) we get
c i − λ ∑ j = 0 n c j 〈 H j ( x ) , L i ( x ) 〉 = 〈 f ( x ) , L i ( x ) 〉 , i = 0 , 1 , ⋯ , n (9)
The system of Equation (9) has a unique solution if D ( λ ) ≠ 0 , where
D ( λ ) = | 1 − λ 〈 H 0 ( x ) , L 0 ( x ) 〉 − λ 〈 H 1 ( x ) , L 0 ( x ) 〉 − λ 〈 H n ( x ) , L 0 ( x ) 〉 − λ 〈 H 0 ( x ) , L 1 ( x ) 〉 1 − λ 〈 H 1 ( x ) , L 1 ( x ) 〉 − λ 〈 H n ( x ) , L 1 ( x ) 〉 ⋮ ⋮ ⋱ ⋮ − λ 〈 H 0 ( x ) , L n ( x ) 〉 − λ 〈 H 1 ( x ) , L n ( x ) 〉 1 − λ 〈 H n ( x ) , L n ( x ) 〉 |
this makes it possible to determine the coefficients ( c j ) 0 ≤ j ≤ n .
Given a degenerate kernel K ( x , t ) = P 1 ( x ) P 2 ( t ) then Equation (1) becomes
φ ( x ) = f ( x ) + λ P 1 ( x ) ∫ 0 ∞ P 2 ( t ) φ ( t ) d t (10)
Let c be the number defined by
c = ∫ 0 ∞ P 2 ( t ) φ ( t ) d t (11)
Therefore
φ ( x ) = f ( x ) + λ c P 1 ( x ) (12)
substituting (12) into (11) gives
c = ∫ 0 ∞ P 2 ( t ) f ( t ) d t 1 − λ ∫ 0 ∞ P 1 ( t ) P 2 ( t ) d t (13)
from (13) and (12) we get:
φ ( x ) = f ( x ) + λ P 1 ( x ) ∫ 0 ∞ P 2 ( t ) f ( t ) d t 1 − λ ∫ 0 ∞ P 1 ( t ) P 2 ( t ) d t (14)
provided that
∫ 0 ∞ P 1 ( t ) P 2 ( t ) d t ≠ 1 λ
Note that, using some degenerate kernel one can compute exactly the integral to obtain an exact solution. Sometimes ∫ 0 ∞ K ( x , y ) L n ( y ) d y can not be evaluated exactly, for that one use quadrature rule to approximate the integral [
To confirm the validity, the accuracy and support our theoretical presentation of the proposed method, we give some computational examples. The computations, associated with the examples are performed by MATLAB. The system of algebraic system will be solved using Gauss elimination and iterative schemes will also be applied for large system.
In this example, we consider λ = 1 and K ( x , t ) = e − t − x , f ( x ) = x 3 − e − x ; here we have P 1 ( x ) = e − x and P 2 ( t ) = e − t according to equation (14) we have
φ ( x ) = x 3 − e − x + e − x ∫ 0 ∞ e − t ( t 3 − e − t ) d t 1 − ∫ 0 ∞ e − 2 t d t
Considering
I = ∫ 0 ∞ e − t ( t 3 − e − t ) d t 1 − ∫ 0 ∞ e − 2 t d t = 11
so the exact solution of Equation (1) is φ ( x ) = x 3 + 10 e − x .
We used both Gauss elimination method, and SOR method iterations to solve the linear system. To plot the solution, we truncate the spacial domain to [0; 5]. The behavior of exact and numerical solution is presented in
Let λ = 1 and K ( x , t ) = e − t 2 − x 2 , f ( x ) = x 4 − 3 π 8 e − x 2 here we have P 1 ( x ) = e − x 2 and P 2 ( t ) = e − t 2 and by (14), the exaction solution is
φ ( x ) = x 4 − 3 π 8 e − x 2 + e − x 2 ∫ 0 ∞ e − t 2 ( t 4 − 3 π 8 e − t 2 ) d t 1 − ∫ 0 ∞ e − 2 t 2 d t
φ ( x ) = x 4 − 3 π 8 e − x 2 + e − x 2 ⋅ ∫ 0 ∞ t 4 e − t 2 d t − 3 π 8 ∫ 0 ∞ e − 2 t 2 d t 1 − ∫ 0 ∞ e − 2 t 2 d t
the Gaussian integral gives
∫ 0 ∞ e − t 2 d t = π 2
by integration by parts, we have
∫ 0 ∞ t 4 e − t 2 d t = 3 π 8
so the exact solution is φ ( x ) = x 4 . For n = 6 the approximate solution
φ 6 ( x ) = ∑ j = 0 6 c j L j ( x )
with
L 0 ( x ) = 1 ,
L 1 ( x ) = 1 − x ,
L 2 ( x ) = 1 2 x 2 − 2 x + 1 ,
L 3 ( x ) = − 1 6 x 3 + 3 2 x 2 − 3 x + 1 ,
L 4 ( x ) = 1 24 x 4 − 2 3 x 3 + 3 x 2 − 4 x + 1 ,
L 5 ( x ) = − 1 120 x 5 + 5 24 x 4 − 5 3 x 3 + 5 x 2 − 5 x + 1 ,
L 6 ( x ) = 1 720 x 6 − 1 20 x 5 + 5 8 x 4 − 10 3 x 3 + 15 2 x 2 − 6 x + 1 ,
the coefficients c i are the solution of the system
c i − ∑ j = 0 6 c j 〈 h j ( x ) , L i ( x ) 〉 = 〈 f ( x ) , L i ( x ) 〉 , i = 0 , 1 , ⋯ , 6 h j ( x ) = ∫ 0 ∞ e − t 2 − x 2 L j ( t ) d t
To plot the solution, we truncate the spacial domain to [0; 1]. The behavior of exact and numerical solution is depicted in
A very simple and efficient method based on the Laguerre polynomial basis has been developed to solve singular integral equations. The result obtained confirms the strategy proposed. The method presented is tested and confirmed by two examples. A few numbers of Laguerre polynomials are needed to get convergence to the exact solution. Further, one can apply the proposed method to more general kernels or systems of integral equation.
The authors declare no conflicts of interest regarding the publication of this paper.
Alroogy, K.S. and Aigo, M.A. (2023) On the Numerical Solution of Singular Integral Equation with Degenerate Kernel Using Laguerre Polynomials. American Journal of Computational Mathematics, 13, 153-160. https://doi.org/10.4236/ajcm.2023.131007