Nystrom method for solution of fredholm integral equations of the second kind under interval data

Abstract

In this paper, Nystrom method was used to solve the Fredholm integral equations of the second kind under interval data. The interval analysis is first presented along with integral equations of the second kind and then Nystrom method is used to find solution of integral equations in the intervals instead of nodes. Also the existence of solution and error analysis is presented. Numerical examples show that the method is efficient and demonstrate a good accuracy.

Keywords

intervals Interval computations Fredholm integral equation of the second kind Nystrom method under interval data.

1. Introduction

Interval analysis has been studied and applied for solving different problems (see [1 , 24]). Usually intervals are used in the problems that to have a maximum and minimum values are required. Such as forecasting of maximum and minimum of temperature, rainfall, population, cost and benefit and so on.

Also, integral equations are used in industry, economics and other sciences. In recent years, many different methods have been used to approximate the solution of Fredholm integral equations of the second kind [3 , 19]. However, in reality, despite the fact that the solution may be found in the points, one needs to find the solution in a range.

In this paper, Nystrom method under interval data contains a range of solutions around a point.

2. Interval Computations

We consider following definitions and theorems from [4]:

Definition 2.1. A real interval is a nonempty, continuous, closed and bounded subset of real numbers $ℝ$ . $X = [\underline{X}, \bar{X}] = {x \in ℝ, \underline{X}, \bar{X} \in ℝ | \underline{X} \leq x \leq \bar{X}},$ (1) where $\underline{X}$ and $\bar{X}$ denote the Lower and Upper bounds of the interval X, respectively.

Intervals $X = [\underline{X}, \bar{X}]$ and $Y = [\underline{Y}, \bar{Y}]$ are equal if and only if $\underline{X} = \underline{Y}$ and $\bar{X} = \bar{Y}$ .

Definition 2.2. The width of an interval X is defined and denoted by $w (X) = \bar{X} - \underline{X}$ .

Definition 2.3. The absolute value of X, denoted |X|, is scalar values and defiend by $| X | = max {| \underline{X} |, | \bar{X} |}$ .

If x ∈ X then we have |x| ≤ |X|.

Definition 2.4. The dual of an interval X, is denoted by $dual (X) = [\bar{X}, \underline{X}]$ .

Definition 2.5. The midpoint of an interval is scalar values and defined by $m (X) = \frac{(\underline{X} + \bar{X})}{2} .$

Definition 2.6. The radius of an interval X is defined and denoted by $rad (X) = \frac{(\bar{X} - \underline{X})}{2}$ .

From Definitions 2.5 and 2.6 we have $m (X) - rad (X) = \underline{X}$ and $m (X) + rad (X) = \bar{X}$ .

Definition 2.7. ([22]) The Hausdorff distance between two intervals $X = [\underline{X}, \bar{X}]$ and $Y = [\underline{Y}, \bar{Y}]$ is: $\begin{matrix} d (Y, X) = \\ d (X, Y) = \max {| \underline{X} - \underline{Y} |, | \bar{X} - \bar{Y} |} \\ = | m (Y) - m (X) | + | rad (Y) - rad (X) | . \end{matrix}$

Remark 2.1. Note that the interval number system represents an extension of the real number system by mapping [x, x] ↔ x.

2.1. Interval operations:

If X and Y are real intervals then specific equations for interval operations are: $\begin{matrix} X + Y = [\underline{X} + \underline{Y}, \bar{X} + \bar{Y}] \\ X - Y = [\underline{X} - \bar{Y}, \bar{X} - \underline{Y}] \\ X . Y = [\underline{X}, \bar{X}] . [\underline{Y}, \bar{Y}] = \\ [\min (\underline{X} \underline{Y}, \underline{X} \bar{Y}, \bar{X} \underline{Y}, \bar{X} \bar{Y}), \max (\underline{X} \underline{Y}, \underline{X} \bar{Y}, \bar{X} \underline{Y}, \bar{X} \bar{Y})] \\ 1 / Y = [1 / \bar{Y}, 1 / \underline{Y}] if 0 \notin Y \\ X / Y = X . [1 / \bar{Y}, 1 / \underline{Y}] if 0 \notin Y \end{matrix}$ and if 0 ∈ Y, then

$\begin{matrix} X / Y = \\ {\begin{matrix} [\bar{X} / \underline{Y}, \infty) & if \bar{X} \leq 0, \bar{Y} = 0 \\ (- \infty, \bar{X} / \bar{Y}] \cup [\bar{X} / \underline{Y}, \infty) & if \bar{X} \leq 0, \underline{Y} < 0 < \bar{Y} \\ (- \infty, \bar{X} / \bar{Y}) & if \bar{X} \leq 0, \underline{Y} = 0 \\ (- \infty, \infty) & if \underline{X} < 0 < \bar{X} \\ (- \infty, \underline{X} / \underline{Y}] & if \underline{X} \geq 0, \bar{Y} = 0 \\ (- \infty, \underline{X} / \underline{Y}] \cup [\underline{X} / \underline{Y}, \infty) & if \underline{X} \geq 0, \underline{Y} < 0 < \bar{Y} \\ [\underline{X} / \bar{Y}, \infty) & if \bar{X} \geq 0, \underline{Y} = 0 \end{matrix} \end{matrix}$

$\begin{matrix} X \subseteq Y \Leftrightarrow {\begin{matrix} \underline{Y} \leq \underline{X}; \\ \bar{X} \leq \bar{Y} . \end{matrix} \Leftrightarrow \\ {\begin{matrix} m (Y) - rad (Y) \leq m (X) - rad (X); \\ m (X) + rad (X) \leq m (Y) + rad (Y) . \end{matrix} \Leftrightarrow \\ {\begin{matrix} m (Y) - m (X) \leq rad (Y) - rad (X); \\ m (X) - m (Y) \leq rad (Y) - rad (X) . \end{matrix} \end{matrix}$

Then

$X \subseteq Y \Leftrightarrow | m (Y) - m (X) | \leq rad (Y) - rad (X)$ (2)

2.2. Interval vectors and matrices

Definition 2.8. An n-dimensional interval vector is a vector whose elements are interval numbers and is denoted by X = (X₁, …, X_n).

We have the following properties for interval vectors: If $x = (x_{1}, . . ., x_{n}) \in ℝ^{n}$ and X = (X₁, …, X_n) is an interval vector, then we have x ∈ X if x_i ∈ X_i for i = 1, …, n. The width of an interval vector X is defined by $w (X) = max_{i} w (X_{i})$ . The midpoint of an interval vector X is defined by m (X) = (m (X₁), …, m (X_n)). The norm of an interval vector X is defined by $∥ X ∥ = max_{i} | X_{i} |$ .The radius of an interval vector X is rad (X) = (rad (X₁), …, rad (X_n)).

Definition 2.9. An interval matrix is a matrix whose elements are interval numbers.

If A is an interval matrix with elements A_ij, and B is a matrix with real elements B_ij such that B_ij ∈ A_ij for all i and j, then B ∈ A.

Definition 2.10. The width of an interval matrix A define as: $w (A) = max_{i, j} w (A_{ij}) .$

Definition 2.11. The midpoint of A is the real matrix m (A) whose elements are the midpoints of the corresponding elements of A: $(m (A))_{ij} = m (A_{ij}),$ that, m (A) ∈ A.

Definition 2.12. The radius of A is the real matrix rad (A) whose elements are the radius of the the corresponding elements of A: $(rad (A))_{ij} = rad (A_{ij}) .$

Noted that if A is an invertible matrix, then $A = [\underline{A}, \bar{A}] = [m (A) - rad (A), m (A) + rad (A)] .$

Definition 2.13. The ijth element C_ij of the product C = AB of a m by p interval matrix A and a p by n interval matrix B gives sharp bounds on the range of $C_{ij} = {M_{ij} = \sum_{k = 1}^{p} P_{ik} Q_{kj} : P_{ik} \in A_{ik}$ and Q_kj ∈ B_kj for 1 ≤ k ≤ p}

for each i, 1 ≤ i ≤ m, and each j, 1 ≤ j ≤ n.

Definition 2.14. The determinant of a square interval matrix defines as in the case of real square matrix except that the determinant of an interval matrix is an interval number. That is det(A) = ∑a_ijA_ij, where A_ij is the cofactor of a_ij with usual meaning.

Most of the properties of determinants of classical matrices are held for the determinants of interval matrices under the modified interval arithmetic.

Definition 2.15. A square interval matrix A is said to be nonsingular or regular if det(A) is invertible (i.e. 0 ∉ det(A)). Alternatively, a square interval matrix A is said to be invertible if det(A) is invertible (i.e. 0 ∉ det(A)).

Definition 2.16. An interval matrix A is regular if every point matrix A ∈ A is nonsingular.

Definition 2.17. Let A be a square interval matrix. The adjoint matrix A^* of A is the transpose of the matrix of cofactors of the elements of A. That is A^* = adj (A) = (b_ij), where b_ij = det(A_ji), for all i, j = 1, 2, 3, …, n.

Definition 2.18. For any $A \in ℝ^{n \times n}$ , if det(A) is invertible, then the common solution of equations AX = I and XA = I is called the inverse of A and is denoted by $A^{- 1} = \frac{adj (A)}{det (A)} = \frac{A^{*}}{det (A)} .$

Noted that if A is invertible, then m (A^-1) = [m (A)] ^-1.

Theorem 2.1. For any interval matrices $A, B \in ℝ^{n \times n}$ , we have:i) m (A + B) = m (A) + m (B) and rad (A + B) = rad (A) + rad (B), ii) m (A - B) = m (A) - m (B) and rad (A - B) = rad (A) + rad (B), iii) m (AB) = m (A) m (B).

Proof. See [7]. □

From [13] we have for any two intervals X and Y: $\begin{matrix} rad (XY) = \\ | m (X) | rad (Y) + rad (X) | m (Y) | + rad (X) rad (Y) . \end{matrix}$ Also from [7] we have for any two intervals X and Y and $a, b \in ℝ$ :

$rad (a X + b Y) = | a | rad (X) + | b | rad (Y) .$ (3)

2.3. Systems of interval linear equations

System of interval linear equations is considered as:

$A x = b,$ (4) where A is square n × n interval matrix, b and x are n-dimensional interval vectors. Generally, these systems have no exact solutions. However, there are methods for approximate solution of Eq.4. The dominant approaches to the solution of interval linear system are NP hard problems (see [15]). In interval analysis, the following four solution sets have been the subject of more or less vigorous enquiry, so far (see [21]):

Definition 2.19. The united solution set formed by solutions of all point systems Ax = b with A ∈ A and b ∈ b.

Definition 2.20. The tolerance solution set, formed by all point vectors x such that the product Ax ∈ b for any A ∈ A (tolerance solutions are defined based on Ax ⊆ b).

Definition 2.21. The control solution set formed by all point vectors $x \in ℝ^{n}$ , such that for any desired, b ∈ b one can find a corresponding A ∈ A satisfying Ax = b (control solutions are defined based on b ⊆ Ax).

Definition 2.22. The algebraic solution is an interval vector x such that, substituting into Eq.4 and executing of all interval arithmetic operations results in the valid equality Ax = b.

There are two types of methods for the numerical solution of systems of interval linear equations including direct and indirect/iterative methods (see [2 , 24]). By the following theorem an algebraic solution of Eq.4 can be found.

Theorem 2.2. Let Ax = b be a system of linear equations involving interval numbers. If the (n × n) interval matrix A is invertible, then it is possible to find a smallest box x = (x₁, x₂, x₃, …, x_n), where each $x_{i} = \frac{det (A_{i})}{det (A)}$ , A_i is the interval matrix obtained when the i-th column of A is replaced by the vector b = (b₁, b₂, b₃, …, b_n).

Proof. See [24]. □

2.4. Interval Functions

Let f be a real-valued function of a single real variable x. The image of the set X under the mapping f is: $f (X) = {f (x) : x \in X} .$ (5) For some functions, Eq.19 is rather easy to compute. For example, consider f (x) = x², x ∈ R. If $X = [\underline{X}, \bar{X}]$ , then the image of the set X under the mapping f is: $f (X) = {x^{2} : x \in X},$ where: $\begin{matrix} {\begin{matrix} [\underline{X^{2}}, \bar{X^{2}}] 0 \leq \underline{X} \leq \bar{X}, \\ - [\bar{X^{2}}, \underline{X^{2}}] \underline{X} \leq \bar{X} \leq 0, \\ [0, \max (\underline{X^{2}}, \bar{X^{2}})] \underline{X} < 0 < \bar{X}, \end{matrix} \end{matrix}$ For the exponential function of $f (x) = \exp (x) = e^{x} (x \in ℝ);$ we have: $\exp (X) = [\exp (\underline{X}), \exp (\bar{X})] .$ For the logarithmic function of f (x) = logx (x> 0); we have: $\log X = [\log \underline{X}, \log \bar{X}],$ for $\underline{X} > 0$ . The square root of an interval is given by: $\sqrt{X} = [\sqrt{\underline{X}}, \sqrt{\bar{X}}],$ for $\underline{X} > 0$ .

For the exponential function of f (x) = exp (- x),we have: $\exp (- X) = [\exp (- \bar{X}), \exp (- \underline{X})] .$ In general, If f is increasing, then $f (X) = [f (\underline{X}), f (\bar{X})]$ and if f is decreasing, then $f (X) = [f (\bar{X}), f (\underline{X})]$ .

2.5. Interval integrals

The integral $\int_{a}^{b} f (t) dt$ can be represented by either ∫_[a,b]f (t) dt or ∫_Xf (t) dt whereX = [a, b].

Lemma 2.1. If f is continuous in X = [a, b], then $\int_{X} f (t) dt \in f (X) w (X) .$ (6)

Proof. From mean value theorem for integrals, we have ∫_[a,b]f (x) dx = (b - a) f (τ) where τ ∈ [a, b]. But f (τ) ∈ f (X) and w (x) = b - a.□

From [9] we have: $\int_{[a, b]} f (t) dt \in \sum_{i = 1}^{n} f (X_{i}) w (X_{i}),$ (7) where X₁, …, X_n are sub-intervals of interval X and $a \leq {\underline{X}}_{1} < {\bar{X}}_{1} = {\underline{X}}_{2} < {\bar{X}}_{2} < \dots < {\bar{X}}_{n} = b$ and $w (X_{i}) = \frac{(b - a)}{n} (i = 1, \dots, n)$ .

Lemma 2.2. Given a real interval [a, b], a < b, and f ∈ C¹ [a, b], there exists τ ∈ [a, b] such that $\int_{a}^{b} f (x) dx - (b - a) f (a) = \frac{(b - a)^{2}}{2} f^{'} (τ) .$

Proof. See [12].□

The quadrature method $\int_{a}^{b} f (x) dx ≃ (b - a) f (a)$ is called right rectangles rule. If we consider $h = \frac{b - a}{n}$ , using Lemma Lemma.1, we have $| E | = | \int_{a}^{b} f (x) dx - (b - a) f (a) | \leq | \frac{(b - a)}{2} h max f^{'} (x) |$ for all x ∈ [a, b] and method convergent. Based on section 2.4 we have

If f (X_i) is increasing then $f (X_{i}) = [f ({\underline{X}}_{i}), f ({\bar{X}}_{i})]$ .

If f (X_i) is decreasing then $f (X_{i}) = [f ({\bar{X}}_{i}), f ({\underline{X}}_{i})]$ .

If n→ ∞ then

f ({\underline{X}}_{i}) ≃ f ({\bar{X}}_{i})

and using Remark 2.1 we have

f (X_{i}) = [f ({\bar{X}}_{i}), f ({\bar{X}}_{i})] \leftrightarrow f ({\bar{X}}_{i})

. So we can approximate the exact area under a curve in the interval of X = [a, b], with a sum of right rectangles given by the following formula:

\begin{matrix} \int_{X} f (t) dt ≃ Q_{n} \\ = \frac{b - a}{n} (f ({\bar{X}}_{1}) + f ({\bar{X}}_{2}) + . . . + f ({\bar{X}}_{n})), \end{matrix}

where, n is the number of rectangles. This quadrature scheme is called Right Rectangle Rule.

3. Fredholm integral equations of the second kind

Consider the integral equation: $λ u (x) - \int_{X} K (x, t) u (t) dt = f (x), x \in X$ (8) where X is a closed and bounded interval in $ℝ^{2}$ .

Note that if K (x, t) and f (x) be continuous and λ is regular value of K (x, t) in (Eq.5), then unknown function of u (x) is continuous (see [6]). Also in this paper, it is assumed that u (x) ≥0.

3.1. Nystrom method

The Nystrom method approximates the integral operator in the Equation Eq.5 by a quadrature method. The resulting solution is found first at the set of quadrature node points, and then it is extended to all points in X by interpolation formula. Let K (x, t) be continuous for all x, t ∈ X. By using the quadrature scheme, we approximate the integral in Eq.5 and we have: $λ u_{n} (x) - \sum_{j = 1}^{n} w_{j} K (x, x_{j}) u_{n} (x_{j}) = f (x), x \in X .$ (9) This is written as an exact equation with a new unknown function u_n (x). To find the solution at the node points, let x run through the quadrature node points x_i. We have $\begin{matrix} λ u_{n} (x_{i}) - \sum_{j = 1}^{n} w_{j} K (x_{i}, x_{j}) u_{n} (x_{j}) = f (x_{i}), \\ i = 1, 2, \dots, n, \end{matrix}$ which is a linear system of order n such as: $(λ I - K) U = f,$ where $U = [\begin{matrix} u_{n} (x_{1}) \\ u_{n} (x_{2}) \\ ⋮ \\ u_{n} (x_{n}) \end{matrix}], f = [\begin{matrix} f (x_{1}) \\ f_{(} x_{2}) \\ ⋮ \\ f (x_{n}) \end{matrix}],$

$\begin{matrix} λ I - K = \\ [\begin{matrix} λ - w_{1} k (x_{1}, x_{1}) & - w_{2} k (x_{1}, x_{2}) & \dots & - w_{n} k (x_{1}, x_{n}) \\ - w_{1} k (x_{2}, x_{1}) & λ - w_{2} k (x_{2}, x_{2}) & \dots & - w_{n} k (x_{2}, x_{n}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ - w_{1} k (x_{n}, x_{1}) & - w_{2} k (x_{n}, x_{2}) & \dots & λ - w_{n} k (x_{n}, x_{n}) \end{matrix}] . \end{matrix}$ Suppose A = λI - K and then find U from the linear system of AU = f.

3.2. Nystrom method under interval data

Now Nystrom method can be considered for the following integral equation under interval data: $\begin{matrix} λ u (x) = f (x) + \int_{X} k (x, t) u (t) dt . \end{matrix}$ Let n be a positive integer and divide X = [a, b] into n sub-intervals X₁, …, X_n then, by using Right Rectangle Rule (Q_n) we have: $\begin{matrix} λ u (x) = f (x) + \sum_{j = 1}^{n} \frac{1}{n} k (x, X_{j}) u_{n} (X_{j}), \end{matrix}$ where u_n (X_j) are unknown and need to be calculated and also by substituting X_i with x, we have $\begin{matrix} λ u_{n} (X_{i}) = f (X_{i}) + \frac{1}{n} \sum_{j = 1}^{n} k (X_{i}, X_{j}) u_{n} (X_{j}), \\ i = 1, 2, \dots, n \end{matrix}$

and we have a system of interval linear equations: $\begin{matrix} (λ I - K) U = f, \end{matrix}$

where

$\begin{matrix} U = [\begin{matrix} u_{n} (X_{1}) \\ u_{n} (X_{2}) \\ ⋮ \\ u_{n} (X_{n}) \end{matrix}], f = [\begin{matrix} f (X_{1}) \\ f (X_{2}) \\ ⋮ \\ f (X_{n}) \end{matrix}], \end{matrix}$

$\begin{matrix} λ I - K = \\ [\begin{matrix} λ - \frac{1}{n} k (X_{1}, X_{1}) & - \frac{1}{n} k (X_{1}, X_{2}) & \dots & - \frac{1}{n} k (X_{1}, X_{n}) \\ - \frac{1}{n} k (X_{2}, X_{1}) & λ - \frac{1}{n} k (X_{2}, X_{2}) & \dots & - \frac{1}{n} k (X_{2}, X_{n}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ - \frac{1}{n} k (X_{n}, X_{1}) & - \frac{1}{n} k (X_{n}, X_{2}) & \dots & λ - \frac{1}{n} k (X_{n}, X_{n}) \end{matrix}] . \end{matrix}$ Here I is used instead of I, because it is acceptable to write [1, 1] ↔1 and [0, 0] ↔0. Suppose A = λI - K and then solve system of interval linear equations as: $\begin{matrix} A U = f, \end{matrix}$ and find interval vector U(algebraic solution).

4. Existence of solution and error analysis

We write the integral Equation Eq.5 in operator form as: $λ u - Ku = f,$ which is approximated by the solution of: $λ u_{n} - K_{n} u_{n} = f .$ We consider the following theorems:

Theorem 4.1. Let u_n be a solution of

u_{n} (x) - \sum_{j = 1}^{n} w_{j} K (x, x_{j}) u_{n} (x_{j}) = f (x), x \in X .

(10)

Then the values $u_{i}^{(n)} = u_{n} (x_{i}), i = 1, . . ., n$ , at the quadrature points satisfy the linear system $u_{i}^{(n)} - \sum_{j = 1}^{n} w_{j} K (x_{i}, x_{j}) u_{j} (n) = f (x_{i}), i = 1, . . ., n .$ (11) Conversely, let $u_{i}^{(n)}, i = 1, . . ., n$ be a solution of the system Eq.7. Then the function u_n is defined by: $u_{n} (x) = f (x) + \sum_{j = 1}^{n} w_{j} K (x, x_{j}) u_{j}^{(n)} x \in X,$ (12) solves Equation Eq.8.

Proof. See [8]. □

Theorem 4.2. Assume the quadrature formulas (Q_n) are convergent. Then the sequence {K_n} is collectively compact and pointwise convergent (i. e., K_nu→ Ku, n → ∞, for all u ∈ C (X)), but not norm convergent.

Proof. See [8]□

Corollary 1. For a uniquely solvable integral equation of the second kind with a continuous kernel and a continuous f (x), the Nystrom method with a convergent sequence of quadrature formulas is uniformly convergent.

Proof. See [8]. □

Theorem 4.3. Let X be a closed, bounded interval in $ℝ^{2}$ , and let K (x, t) be continuous for x, t ∈ X. Assume the quadrature scheme (Q_n) is convergent for all continuous functions on X. Further, assume that the integral Equation Eq.5 is uniquely solvable for givenf ∈ C (X), with λ ≠ 0. Then for all sufficiently large n, say n ≥ N, the approximate inverses (λI - K_n) ^-1 exist.

Proof. See [3]. □

Theorem 4.4. The system of interval linear equations (Eq.5) has an algebraic solution, where X be a closed, bounded interval in $ℝ^{2}$ ,K (x, t) be continuous for x, t ∈ X and f ∈ C (X).

Proof. From Corollary 1 and Theorem 4.3 the inverse of all point matrices A = λI - K_n ∈ A exist, also point matrix m (A) ∈ A is invertible and A is regular by Definition 2.16. Then system of interval linear Equations (Eq.5) has an algebraic solution by Theorems 2.2. □

We approximate solution of Eq.5 and consider m (A) U ≃ f then we have

$U_{exc} ≃ U_{appr} = m (A)^{- 1} f .$ (13) Exact solution is denoted by U = U_exc and approximate solution is denoted by U_appr.

4.1. Error analysis

For error analysis, let d (u_appr (X_i), u_exc (X_i)) be the Hausdorff distance, where u_appr (X_i) is the estimation of the true solution u_exc (X_i). We have from Definition 2.7:

$\begin{matrix} d (u_{appr} (X_{i}), u_{exc} (X_{i})) = \\ | m (u_{exc} (X_{i})) - m (u_{appr} (X_{i})) | \\ + | rad (u_{exc} (X_{i})) - rad (u_{appr} (X_{i})) | . \end{matrix}$ (14)

From Equations Eq.1 and Eq.9 we have: $\begin{matrix} d (u_{appr} (X_{i}), u_{exc} (X_{i})) \leq \\ 2 | rad (u_{exc} (X_{i})) - rad (u_{appr} (X_{i})) | . \end{matrix}$ (15) To find rad (u_exc (X_i)), let $A = [\begin{matrix} a_{11} & . . . & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{n 1} & . . . & a_{nn} \end{matrix}], U_{exc} = [\begin{matrix} u_{exc} (X_{1}) \\ ⋮ \\ u_{exc} (X_{n}) \end{matrix}], f = [\begin{matrix} f_{1} \\ ⋮ \\ f_{n} \end{matrix}],$ So that $A U_{exc} = [\begin{matrix} \sum_{j = 1}^{n} a_{1 j} u_{exc} (X_{j}) \\ ⋮ \\ \sum_{j = 1}^{n} a_{nj} u_{exc} (X_{j}) \end{matrix}] = [\begin{matrix} f_{1} \\ ⋮ \\ f_{n} \end{matrix}]$ . Then $\sum_{j = 1}^{n} a_{ij} u_{exc} (X_{j}) = f_{i}, i = 1, . . ., n$ . From Theorem 2.1 we have:

$\begin{matrix} rad (\sum_{j = 1}^{n} a_{ij} u_{exc} (X_{j})) = \\ \sum_{j = 1}^{n} rad (a_{ij} u_{exc} (X_{j})) = \\ rad (f_{i}), i = 1, . . ., n . \end{matrix}$ (16)

From Eq.11 we have:

$\begin{matrix} rad (a_{ij} u_{exc} (X_{j})) = | m (a_{ij}) | rad (u_{exc} (X_{j})) + \\ rad (a_{ij}) | m (u_{exc} (X_{j})) | + rad (a_{ij}) rad (u_{exc} (X_{j})) \\ i = 1, . . ., n . \end{matrix}$ (17) By substituting Equation (Eq.3) into Equation (Eq.2) we have: $\begin{matrix} \sum_{j = 1}^{n} {| m (a_{ij}) | rad (u_{exc} (X_{j})) + rad (a_{ij}) | m (u_{exc} (X_{j})) | \\ + rad (a_{ij}) rad (u_{exc} (X_{j}))} \\ = rad (f_{i}), i = 1, . . ., n . \end{matrix}$ (18) We suppose m = min|m (a_ij) |, r = min (rad (a_ij)). From Equation Eq.10 we have:

$\begin{matrix} \sum_{j = 1}^{n} {| m | rad (u_{exc} (X_{j})) + r | m (u_{exc} (X_{j})) | \\ + r (rad (u_{exc} (X_{j})))} \leq rad (f_{i}), i = 1, . . ., n . \end{matrix}$ (19) From Equation Eq.13 we have:

$\begin{matrix} | m | \sum_{j = 1}^{n} rad (u_{exc} (X_{j})) + | r | \sum_{j = 1}^{n} {m (u_{exc} (X_{j})) + \\ rad (u_{exc} (X_{j}))} \leq rad (f_{i}), i = 1, . . ., n . \end{matrix}$ (20) From $\bar{u_{exc} (X_{j})} = m (u_{exc} (X_{j})) + rad (u_{exc} (X_{j}))$ and Equation Eq.14 we have:

$\begin{matrix} | m | \sum_{j = 1}^{n} rad (u_{exc} (X_{j})) + | r | \sum_{j = 1}^{n} \bar{u_{exc} (X_{j})} \leq rad (f_{i}), \\ i = 1, . . ., n . \end{matrix}$ (21) Since u_exc (X_j) ≥0 then $\bar{u_{exc} (X_{j})} \geq rad (u_{exc} (X_{j})) = \frac{\bar{u_{exc} (X_{j})} - \underline{u_{exc} (X_{j})}}{2}$ And from Equation Eq.15 we have: $\begin{matrix} | m | \sum_{j = 1}^{n} rad (u_{exc} (X_{j})) + | r | \sum_{j = 1}^{n} rad (u_{exc} (X_{j})) \\ \leq rad (f_{i}), i = 1, . . ., n . \end{matrix}$ Then: $\begin{matrix} \sum_{j = 1}^{n} rad (u_{exc} (X_{j})) \leq \frac{rad (f_{i})}{| m | + | r |}, i = 1, . . ., n . \end{matrix}$ Since u_exc (X_j) ≥0 then rad (u_exc (X_j)) ≥0, j = 1, 2, …, n and from Equation Eq.16 we have: $\begin{matrix} rad (u_{exc} (X_{j})) \leq \frac{rad (f_{i})}{| m | + | r |}, i, j = 1, . . ., n . \end{matrix}$

5. Numerical examples

In the following examples, maximum distance (estimation error) of our method is denoted by $d_{est} (u_{appr} (X_{i}), u_{exc} (X_{i}))$ and exact distance(exact error) in the sub-interval is denoted by d (u_appr (X_i), u_exc (X_i)). The real value of u (x) in the interval X_i is denoted by u_exc (X_i) and the approximate value of u (x) in the interval X_i is denoted by u_appr (X_i).

Example 5.1. Exam.1 Consider the integral equation

$u (x) = \frac{1}{2} x - \frac{1}{3} + \int_{0}^{1} (x + t) u (t) dt$ with exact solution u (x) = x. Table 1 shows the numerical results for Example 5.1 using n = 20.

Table 1
The numerical results for Example 5.1

X _i u_appr (X_i) u_exc (X_i) d (u_appr (X_i), u_exc (X_i)) d_est (u_appr (X_i), u_exc (X_i))

X ₁ [-0.0722, 0.1249] [0.0000, 0.0500] 0.0749 0.1471

X ₂ [-0.0288, 0.1817] [0.0500, 0.1000] 0.0817 0.1605

X ₃ [0.0147, 0.2384] [0.1000, 0.1500] 0.0884 0.1737

X ₄ [0.0583, 0.2951] [0.1500, 0.2000] 0.0951 0.1868

X ₅ [0.1019, 0.3517] [0.2000, 0.2500] 0.1017 0.1998

X ₆ [0.1457, 0.4082] [0.2500, 0.3000] 0.1082 0.2125

X ₇ [0.1895, 0.4646] [0.3000, 0.3500] 0.1146 0.2251

X ₈ [0.2334, 0.5210] [0.3500, 0.4000] 0.1210 0.2376

X ₉ [0.2773, 0.5773] [0.4000, 0.4500] 0.1273 0.2500

X ₁₀ [0.3214, 0.6335] [0.4500, 0.5000] 0.1335 0.2621

X ₁₁ [0.3655, 0.6897] [0.5000, 0.5500] 0.1397 0.2742

X ₁₂ [0.4097, 0.7457] [0.5500, 0.6000] 0.1457 0.2860

X ₁₃ [0.4539, 0.8017] [0.6000, 0.6500] 0.1517 0.2978

X ₁₄ [0.4983, 0.8576] [0.6500, 0.7000] 0.1576 0.3093

X ₁₅ [0.5427, 0.9135] [0.7000, 0.7500] 0.1635 0.3208

X ₁₆ [0.5865, 0.9699] [0.7500, 0.8000] 0.1699 0.3334

X ₁₇ [0.6291, 1.0276] [0.8000, 0.8500] 0.1776 0.3485

X ₁₈ [0.6717, 1.0852] [0.8500, 0.9000] 0.1852 0.3635

X ₁₉ [0.7143, 1.1429] [0.9000, 0.9500] 0.1929 0.3786

X ₂₀ [0.7569, 1.2005] [0.9500, 1.0000] 0.2005 0.3936

X _i	u_appr (X_i)	u_exc (X_i)	d (u_appr (X_i), u_exc (X_i))	d_est (u_appr (X_i), u_exc (X_i))
X ₁	[-0.0722, 0.1249]	[0.0000, 0.0500]	0.0749	0.1471
X ₂	[-0.0288, 0.1817]	[0.0500, 0.1000]	0.0817	0.1605
X ₃	[0.0147, 0.2384]	[0.1000, 0.1500]	0.0884	0.1737
X ₄	[0.0583, 0.2951]	[0.1500, 0.2000]	0.0951	0.1868
X ₅	[0.1019, 0.3517]	[0.2000, 0.2500]	0.1017	0.1998
X ₆	[0.1457, 0.4082]	[0.2500, 0.3000]	0.1082	0.2125
X ₇	[0.1895, 0.4646]	[0.3000, 0.3500]	0.1146	0.2251
X ₈	[0.2334, 0.5210]	[0.3500, 0.4000]	0.1210	0.2376
X ₉	[0.2773, 0.5773]	[0.4000, 0.4500]	0.1273	0.2500
X ₁₀	[0.3214, 0.6335]	[0.4500, 0.5000]	0.1335	0.2621
X ₁₁	[0.3655, 0.6897]	[0.5000, 0.5500]	0.1397	0.2742
X ₁₂	[0.4097, 0.7457]	[0.5500, 0.6000]	0.1457	0.2860
X ₁₃	[0.4539, 0.8017]	[0.6000, 0.6500]	0.1517	0.2978
X ₁₄	[0.4983, 0.8576]	[0.6500, 0.7000]	0.1576	0.3093
X ₁₅	[0.5427, 0.9135]	[0.7000, 0.7500]	0.1635	0.3208
X ₁₆	[0.5865, 0.9699]	[0.7500, 0.8000]	0.1699	0.3334
X ₁₇	[0.6291, 1.0276]	[0.8000, 0.8500]	0.1776	0.3485
X ₁₈	[0.6717, 1.0852]	[0.8500, 0.9000]	0.1852	0.3635
X ₁₉	[0.7143, 1.1429]	[0.9000, 0.9500]	0.1929	0.3786
X ₂₀	[0.7569, 1.2005]	[0.9500, 1.0000]	0.2005	0.3936

In case of n = 40 maximum d (u_appr (X_i), u_exc (X_i)) is 0.1021.

Example 5.2. Consider the integral equation $u (x) = e^{3 x} - \frac{1}{9} (2 e^{3} + 3) x + \int_{0}^{1} xtu (t) dt,$ with exact solution x (t) = e^3x.

Table 2 shows the numerical results for Example 5.2 using n = 20.

Table 2

The numerical results for Example 5.2

X _i	u_appr (X_i)	u_exc (X_i)	d (u_appr (X_i), u_exc (X_i))	d_est (u_appr (X_i), u_exc (X_i))
X ₁	[1.0628, 1.1075]	[1.0000, 1.1618]	0.1075	0.1171
X ₂	[1.2484, 1.2727]	[1.1618, 1.3498]	0.0866	0.1637
X ₃	[1.4155, 1.5131]	[1.3498, 1.5683]	0.0657	0.1209
X ₄	[1.6130, 1.7888]	[1.5683, 1.8221]	0.0447	0.078
X ₅	[1.8459, 2.1056]	[1.8221, 2.1170]	0.0235	0.0352
X ₆	[2.1169, 2.4702]	[2.1170, 2.4596]	0.0106	0.0107
X ₇	[2.4415, 2.8901]	[2.4596, 2.8576]	0.0325	0.0506
X ₈	[2.8186, 3.3745]	[2.8576, 3.3201]	0.0544	0.0934
X ₉	[3.2602, 3.9337]	[3.3201, 3.8574]	0.0763	0.1362
X ₁₀	[3.7765, 4.5799]	[3.8574, 4.4816]	0.0983	0.1792
X ₁₁	[4.3799, 5.3271]	[4.4816, 5.2069]	0.1202	0.2219
X ₁₂	[5.0842, 6.1917]	[5.2069, 6.0496]	0.1421	0.2648
X ₁₃	[5.9060, 7.1927]	[6.0496, 7.0286]	0.1641	0.3077
X ₁₄	[6.8641, 8.3521]	[7.0286, 8.1661]	0.1860	0.3505
X ₁₅	[7.9806, 9.6956]	[8.1661, 9.4877]	0.2079	0.3934
X ₁₆	[9.2813, 11.2529]	[9.4877, 11.0231]	0.2298	0.4362
X ₁₇	[10.7958, 13.0588]	[11.0231, 12.8071]	0.2517	0.479
X ₁₈	[12.5588, 15.1533]	[12.8071, 14.8797]	0.2736	0.5219
X ₁₉	[14.6104, 17.5833]	[14.8797, 17.2877]	0.2956	0.5649
X ₂₀	[16.9976, 20.4030]	[17.2877, 20.0855]	0.3175	0.6076

In case of n = 40 maximum d (u_appr (X_i), u_exc (X_i)) is 0.1580.

Example 5.3. Consider the following integral equation $u (x) = \sin^{- 1} x + (\frac{π}{2} - 1) x - \int_{0}^{1} xu (t) dt,$ with exact solution u (x) = sin^-1x. Table 3 shows the numerical results for Example 5.3 using n = 20.

Table 3

The numerical results for Example 5.3

X _i	u_appr (X_i)	u_exc (X_i)	d (u_appr (X_i), u_exc (X_i))	d_est (u_appr (X_i), u_exc (X_i))
X ₁	[-0.0151, 0.0651]	[0.0000, 0.0200]	0.0451	0.0602
X ₂	[0.0131, 0.1168]	[0.0200, 0.1001]	0.0167	0.0236
X ₃	[0.0815, 0.1687]	[0.1001, 0.1505]	0.0186	0.0368
X ₄	[0.1302, 0.2211]	[0.1505, 0.2013]	0.0203	0.0401
X ₅	[0.1792, 0.2739]	[0.2013, 0.2526]	0.0221	0.0434
X ₆	[0.2288, 0.3275]	[0.2526, 0.3046]	0.0238	0.0467
X ₇	[0.2790, 0.3819]	[0.3046, 0.3575]	0.0256	0.0500
X ₈	[0.3302, 0.4374]	[0.3575, 0.4115]	0.0273	0.0532
X ₉	[0.3824, 0.4941]	[0.4115, 0.4667]	0.0291	0.0565
X ₁₀	[0.4359, 0.5525]	[0.4667, 0.5235]	0.0308	0.0598
X ₁₁	[0.4910, 0.6128]	[0.5235, 0.5823]	0.0325	0.0630
X ₁₂	[0.5481, 0.6754]	[0.5823, 0.6435]	0.0342	0.0661
X ₁₃	[0.6076, 0.7409]	[0.6435, 0.7075]	0.0377	0.0693
X ₁₄	[0.6700, 0.8102]	[0.7075, 0.7753]	0.0375	0.0724
X ₁₅	[0.7362, 0.8843]	[0.7753, 0.8480]	0.0391	0.0754
X ₁₆	[0.8073, 0.9649]	[0.8480, 0.9272]	0.0407	0.0784
X ₁₇	[0.8851, 1.0548]	[0.9272, 1.0160]	0.0421	0.0809
X ₁₈	[0.9725, 1.1596]	[1.0160, 1.1198]	0.0435	0.0833
X ₁₉	[1.0756, 1.2936]	[1.1198, 1.2532]	0.0442	0.0846
X ₂₀	[1.2134, 1.6066]	[1.2532, 1.5708]	0.0398	0.0756

In case of n = 40 maximum d (u_appr (X_i), u_exc (X_i)) is 0.0074.

6. Conclusion

Using the Nystrom method, the present study aimed to solve integral equations under interval data. All illustrated numerical examples and their results vigorously strengthen both efficiency and accuracy of the proposed method.

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