# Integral Equations

• I. N. Bronshtein
• K. A. Semendyayev

## Abstract

By an integral equation is meant an equation to determine an unknown function φ(x) (φ(x) is defined in a < x < b) such that in the equation appears an integral in which the integrand is dependent on the desired function φ(x). Naturally in such an equation there can occur other terms—not necessarily in the form of an integral—which depend directly on φ(x).

## Keywords

Integral Equation Characteristic Function Product Kernel Fredholm Integral Equation Volterra Integral Equation
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

## Notes

1. (1).
This integral can be evaluated through the substitution x = y + (η − y)u. It follows that $$\int\limits_{0}^{1}\frac{du}{\sqrt{u-u^2}}=[\textup{arc}\;\textup{sin}\;(2u-1)]_{0}^{1}=\pi$$ (cf. p. 432, formula (264)).Google Scholar
2. (1).
In order to avoid confusion we replace y by η.Google Scholar
3. (1).
In this example the exceptionally small number n = 3 suffices.Google Scholar
4. (1).
From the process on p. 693 it follows that for $$\lambda =\frac{5}{2}\;[25\pm \sqrt{601}]$$ the integral equation has the following two linearly independent characteristic solutions: $$\varphi _{1}=C_{1}[6x-119\sqrt{x}-5\sqrt{601}\sqrt{x}]$$, $$\varphi _{2}=C_{2}[6x-119\sqrt{x} +5\sqrt{601}\sqrt{x}]$$. Since an inhomogeneous integral equation in the case of its characteristic values is solvable if and only if the perturbation function f(x) is orthogonal to all the characteristic solutions of the transposed integral equation, the conditions of solvability are, when one notices, that the integral equation coincides with its transpose: $$\int\limits_{0}^{1}f(x)\varphi _{1}(x)\;dx=0,\;\;\;\int\limits_{0}^{1}f(x)\varphi _{2}(x)\;dx=0$$. One sees easily that in the case f(x) = x this condition will not be fulfilled.Google Scholar
5. (1).
A zero x0 of a function is said to be of order n if it is also a zero of the first n − 1 derivatives but not of the n-th. Accordingly, a simple zero is a zero for which the value of the first derivative is different from zero.Google Scholar
6. (1).
The reader should compare this with the section on degenerate kernels.Google Scholar
7. (1).
If the function φ(x) were only integrable (cf. the footnote on p. 454) then from the continuity of the function K(x, y) considered as a function of x it would follow that the integral $$\int\limits_{a}^{b}K(x,y)\varphi (y)dy$$ is also a continuous function of x. On the left side of (1) would therefore stand the sum of two continuous functions of x, and therefore the function φ(x) would have to be continuous also.Google Scholar
8. (1).
δij is the Kronecker symbol. That is, it equals 1 for i = j and zero for ij.Google Scholar
9. (1).
Where (x, g) denotes the scalar product of the function x with the function g(x); $$(x,g)=\int\limits_{0}^{1}xg(x)dx=\int\limits_{0}^{1}yg(y)dy$$.Google Scholar
10. (1).
Where (g, sin x) again is the scalar product $$(g,\textup{sin}\;x)=\int\limits_{-\pi}^{+\pi}g(x)\;\textup{sin}\;x\;dx=\int\limits_{-\pi}^{+\pi}g(y)\;\textup{sin}\;y\;dy$$ and analogously $$(g,\textup{cos}\;x)=\int\limits_{-\pi}^{+\pi}g(x)\;\textup{cos}\;x\;dx=\int\limits_{-\pi}^{+\pi}g(y)\;\textup{cos}\;y\;dy$$.Google Scholar
11. (1).
Since $$\varphi _{1}=\tilde{\varphi _{1}}/||\tilde{\varphi} ||$$ the conditions $$(g,\tilde{\varphi _{1}})=0$$ and (g, φ1) = 0 are equivalent.Google Scholar
12. (1).
There can be only finitely many K such that λK = λ. Thus for each characteristic value there can exist only finitely many linearly independent characteristic functions.Google Scholar