samples/texts/2487562/page_3.md

But, if the integral in the above Eq. (1.1) is replaced by a quadrature formula (see Fox and Goodwin [8]), we get

where $w_k$ are the weights and $t_k$'s are appropriately chosen interpolation points.

The Eq. (1.12) represents an over-determined system of linear algebraic equations for the determination of N + 1 un- knowns φ(t_k) (k = 0, ..., N).

So, if from theoretical considerations it is already known that the given integral Eq. (1.1) possesses a unique solution, then varieties of methods can be used to cast the over-determined system of Eq. (1.12) into a system of (N + 1) equations and the method of least-squares provides the most appropriate procedure to handle the situation completely.

Note that one can obtain exactly (N + 1) equations for the (N + 1) unknowns φ₀, ..., φ_N from the over-determined system of Eq. (1.12) by selecting (N + 1) interpolating points x = t_k, k = 0, 1, 2, ..., N, (0 < x < 1).

Substituting the approximate solution (1.11) into the integral Eq. (1.1) we obtain the relation

giving rise to an over-determined system of linear algebraic equations for the determination of the unknown constants $a_{i-1}$ ($i = 1, 2, \dots, N+1$) where

$${\psi }_{i-1}(x)=x^{i-1}+{\int }_{\alpha }^{\beta }k(x,t)t^{i-1}dt,i=1,2,\dots ,N+1.(1.14)\backslash psi\_\{i-1\}(x)\; =\; x^\{i-1\}\; +\; \backslash int\_\{\backslash alpha\}^\{\backslash beta\}\; k(x,t)t^\{i-1\}\; dt,\; \backslash quad\; i\; =\; 1,\; 2,\; \backslash dots,\; N+1.\; \backslash qquad\; (1.14)$$

On using the least-squares method, we obtain the normal equations

$$\begin{array}{cccc}& \sum _{i=1}^{N+1}{a}_{i-1}{c}_{ij}=b_j,j=1,2,\dots ,N+1.& & \text{(1.15)}\end{array}\backslash sum\_\{i=1\}^\{N+1\}\; a\_\{i-1\}\; c\_\{ij\}\; =\; b\_j,\; \backslash quad\; j\; =\; 1,\; 2,\; \backslash dots,\; N+1.\; \backslash tag\{1.15\}$$

where

$$\begin{array}{cccc}& c_{ij}={\int }_{\alpha }^{\beta }{\psi }_{i-1}(x){\psi }_{j-1}(x)dx,i=1,2,\dots ,N+1,j=1,2,\dots ,N+1,& & \text{(1.16)}\end{array}c\_\{ij\}\; =\; \backslash int\_\{\backslash alpha\}^\{\backslash beta\}\; \backslash psi\_\{i-1\}(x)\; \backslash psi\_\{j-1\}(x)\; dx,\; \backslash quad\; i\; =\; 1,\; 2,\; \backslash dots,\; N+1,\; \backslash quad\; j\; =\; 1,\; 2,\; \backslash dots,\; N+1,\; \backslash tag\{1.16\}$$

and

$$\begin{array}{cccc}& b_j={\int }_{\alpha }^{\beta }f(x){\psi }_{j-1}(x)dx,j=1,2,\dots ,N+1.& & \text{(1.17)}\end{array}b\_j\; =\; \backslash int\_\{\backslash alpha\}^\{\backslash beta\}\; f(x)\; \backslash psi\_\{j-1\}(x)\; dx,\; \backslash quad\; j\; =\; 1,\; 2,\; \backslash dots,\; N+1.\; \backslash tag\{1.17\}$$

The solution of the system of Eq. (1.15) along with the relation (1.11), finally determines an approximate solution $\phi(x)$.

2. Illustrative examples

We illustrate the above procedure through the following examples.

Examples

(i) $k(x, t) = -(xt + x^2t^2)$, $f(x) = 1$, $\alpha = -1$, $\beta = 1$.

(ii) $k(x,t) = -(x^2 + t^2)$, $f(x) = x^2$, $\alpha = 0$, $\beta = 1$.

(iii) $k(x,t) = -(\sqrt{x} + \sqrt{t})$, $f(x) = 1+x$, $\alpha = 0$, $\beta = 1$.

(iv) $k(x,t) = -(\cos x + \cos t)$, $f(x) = \sin x$, $\alpha = 0$, $\beta = \pi$.

It can be verified that all the above integral equations possess a unique solution, by examining the eigenvalues of the associated operators.

Solution:

Using the method described in Section 1, if $\phi(x)$ is approximated by the relation (1.11), then we find that the constants $a_{i-1}$ ($i = 1, 2, \dots, N+1$) satisfy the system of Eq. (1.15) where

in example 2(i),

$$c_{ij}=\frac{1-(-1{)}^{i+j-1}}{i+j-1}-\frac{4}{3}\frac{\{1-(-1{)}^{i+1}\}\{1-(-1{)}^{j+1}\}}{(i+1)(j+1)}-\frac{8}{5}\frac{\{1-(-1{)}^{i+2}\}\{1-(-1{)}^{j+2}\}}{(i+2)(j+2)},(2.18)c\_\{ij\}\; =\; \backslash frac\{1\; -\; (-1)^\{i+j-1\}\}\{i+j-1\}\; -\; \backslash frac\{4\}\{3\}\; \backslash frac\{\backslash \{1\; -\; (-1)^\{i+1\}\backslash \}\; \backslash \{1\; -\; (-1)^\{j+1\}\backslash \}\}\{(i+1)(j+1)\}\; -\; \backslash frac\{8\}\{5\}\; \backslash frac\{\backslash \{1\; -\; (-1)^\{i+2\}\backslash \}\; \backslash \{1\; -\; (-1)^\{j+2\}\backslash \}\}\{(i+2)(j+2)\},\; \backslash quad\; (2.18)$$

$$\begin{array}{cccc}& b_j=\frac{1-(-1{)}^j}{j}-\frac{2}{3}\frac{\{1-(-1{)}^{j+2}\}}{j+2}\mathrm{.}& & \text{(2.19)}\end{array}b\_j\; =\; \backslash frac\{1\; -\; (-1)^j\}\{j\}\; -\; \backslash frac\{2\}\{3\}\; \backslash frac\{\backslash \{1\; -\; (-1)^\{j+2\}\backslash \}\}\{j\; +\; 2\}.\; \backslash tag\{2.19\}$$