# Definitions and Types of Integral Equations

In this article, I will explain what Integral Equations are, how they are structured and what are certain types of Integral Equations.

## What is an Integral Equation?

An integral equation is an equation in which an unknown function appears under one or more integration signs.

Any integral calculus statement like — $y= \int_a^b \phi(x) dx$ can be considered as an integral equation.

If you noticed, I have used two integration limits (a and b) in the above integral equation – they are special in this discussion, and their significance will be discussed later in the article.

## Linear Integral Equations

A general type of integral equation, $g(x) y(x) = f(x) + \lambda \int_a^\Box K(x, t) y(t) dt$ is called linear integral equation as only linear operations are performed in the equation.

The one, which is not linear, is obviously called a “Non-linear integral equation”.

We generally mean linear integral equation when we say integral equation. When a non-linear integral equation is to be called, it is called exclusively.

In the general type of the linear equation

$g(x) y(x) = f(x) + \lambda \int_a^\Box K(x, t) y(t) dt$

we have used a ‘box $\Box$’ to indicate the higher limit of the integration.

## Types of Integral Equations

Integral Equations can be of two types according to whether the box $\Box$ (the upper limit) is a constant, $b$ or a variable, $x$.

The first type of integral equations which involve constants as both the limits — are called Fredholm Type Integral equations.

On the other hand, when one of the limits is a variable ($x$, the independent variable of which $y$, $f$ and $K$ are functions), such integral equations are called Volterra’s Integral Equations.

Thus

• $g(x) y(x) = f(x) + \lambda \int_a^b K(x, t) y(t) dt$ is a Fredholm Integral Equation and
• $g(x) y(x) = f(x) + \lambda \int_a^x K(x, t) y(t) dt$ is a Volterra Integral Equation.

In an integral equation, $y$ is to be determined with $g$, $f$ and $K$ being known and $\lambda$ being a non-zero complex parameter. The function $K (x,t)$ is called the ‘kernel’ of the integral equation.

## Types of Fredholm Integral Equations

As the general form of Fredholm Integral Equation is $g(x) y(x) = f(x) + \lambda \int_a^b K(x, t) y(t) dt$, there may be following other types of it according to the values of $g$ and $f$ :

1. Fredholm Integral Equation of First Kind  —when — $g(x) = 0$
$f(x) + \lambda \int_a^b K(x, t) y(t) dt=0$
2. Fredholm Integral Equation of Second Kind  —when — $g(x) =1$
$y(x) = f(x) + \lambda \int_a^b K(x, t) y(t) dt$
3. Fredholm Integral Equation of Homogeneous Second Kind —when $f(x)=0$
and $g(x)=1$
$y(x) = \lambda \int_a^b K(x, t) y(t) dt$
4. The general equation of Fredholm equation is also called Fredholm Equation of Third/Final kind, with $f(x) \neq 0, 1 \neq g(x)\neq 0$.

## Types of Volterra Integral Equations

As the general form of Volterra Integral Equation is $g(x) y(x) = f(x) + \lambda \int_a^x K(x, t) y(t) dt$, there may be following other types of it according to the values of $g$ and $f$ :

1. Volterra Integral Equation of First Kind  —when — $g(x) = 0$
$f(x) + \lambda \int_a^x K(x, t) y(t) dt=0$
2. Volterra Integral Equation of Second Kind  —when — $g(x) =1$
$y(x) = f(x) + \lambda \int_a^x K(x, t) y(t) dt$
3. Volterra Integral Equation of Homogeneous Second Kind —when $f(x)=0$
and $g(x)=1$
$y(x) = \lambda \int_a^x K(x, t) y(t) dt$
4. The general equation of Volterra equation is also called the Volterra Equation of Third/Final kind, with $f(x) \neq 0, 1 \neq g(x)\neq 0$.

## Singular Integral equations

In the general Fredholm/Volterra Integral equations, there arise two singular situations:

• the limit $a \to -\infty$ and $\Box \to \infty$.
• the kernel $K(x,t) = \pm \infty$ at some points in the integration limit $[a, \Box]$.

Then, such integral equations are called Singular (Linear) Integral Equations. Based on these two singular situations, here are two examples of the Singular Integral equations.

Type-1

$a \to -\infty$ and $\Box \to \infty$

General Form: $g(x) y(x) = f(x) + \lambda \int_{-\infty}^{\infty} K(x, t) y(t) dt$

Example: $y(x) = 3x^2 + \lambda \int_{-\infty}^{\infty} e^{-|x-t|} y(t) dt$

Type-2: $K(x,t) = \pm \infty$ at some points in the integration limit $[a, \Box]$

Example: $y(x) = f(x) + \int_0^x \dfrac{1}{(x-t)^n} y(t)$ is a singular integral equation as the integrand reaches to $\infty$ at $t=x$.

## Kernels

The nature of the solution of integral equations solely depends on the nature of the Kernel of the integral equation K(x,t).

Kernels are of the following special types:

### Symmetric Kernel

When the kernel $K(x,t)$ is symmetric or complex symmetric or Hermitian, if $K(x,t)= \bar{K}(t,x)$ .

Here bar  $\bar{K}(t,x)$ denotes the complex conjugate of $K(t,x)$.

That’s if there is no imaginary part of the kernel, then $K(x, t) = K(t, x)$ implies that $K$ is a symmetric kernel.

For example $K(x,t)= \sin (x+t)$ is symmetric kernel.

### Separable or Degenerate Kernel

A kernel $K(x,t)$ is called separable if it can be expressed as the sum of a finite number of terms, each of which is the product of ‘a function’ of only and ‘a function’ of t only, i.e., $K(x,t)= \displaystyle{\sum_{n=1}^{\infty}} \phi_i (x) \psi_i (t)$

### Difference Kernel

When $K(x,t) = K(x-t)$, the kernel is called difference kernel.

### Resolvent or Reciprocal Kernel

The solution of the integral equation $y(x) = f(x) + \lambda \int_a^\Box K(x, t) y(t) dt$ is of the form $y(x) = f(x) + \lambda \int_a^\Box \mathfrak{R}(x, t;\lambda) f(t) dt$.

The kernel $\mathfrak{R}(x, t;\lambda)$ of the solution is called resolvent or reciprocal kernel.

## Integral Equations of Convolution Type

The integral equation $g(x) y(x) = f(x) + \lambda \int_a^\Box K(x, t) y(t) dt$ is called of integral equation of convolution type when the kernel $K(x,t)$ is difference kernel, i.e., $K(x,t) = K(x-t)$.

Let $y_1(x)$ and $y_2(x)$ be two continuous functions defined for $x \in E \subseteq\mathbb{R}$ then the convolution of $y_1$ and $y_2$ is given by

$$y_1 * y_2 = \int_E y_1 (x-t) y_2(t) dt$$

$$=\int_E y_2 (x-t) y_1(t) dt$$

For standard convolution, the limits are $-\infty$ and $\infty$.

## Eigenvalues and Eigenfunctions of Integral Equations

The homogeneous integral equation $y(x) = \lambda \int_a^\Box K(x, t) y(t) dt$ has the obvious solution $y(x)=0$ which is called the zero solution or the trivial solution of the integral equation.

Except this, the values of $\lambda$ for which the integral equation has non-zero solution $y(x) \neq 0$, are called the eigenvalues of integral equation or eigenvalues of the kernel.

Every non-zero solution $y(x)\neq 0$ is called an eigenfunction corresponding to the obtained eigenvalue $\lambda$.

• Note that $\lambda \neq 0$
• If $y(x)$ an eigenfunction corresponding to eigenvalue $\lambda$ then $c \cdot y(x)$ is also an eigenfunction corresponding to $\lambda$

## Leibnitz Rule of Differentiation under the integral sign

Let $F(x,t)$ and $\dfrac{\partial F}{\partial x}$ be continuous functions of both and and let the first derivatives of $G(x)$ and $H(x)$ are also continuous, then

$\dfrac{d}{dx} \displaystyle {\int_{G(x)}^{H(x)}} F(x,t) dt$

$= \displaystyle {\int_{G(x)}^{H(x)}}\dfrac{\partial F}{\partial x} dt + F(x, H(x)) \dfrac{dH}{dx} – \\ F(x, G(x)) \dfrac{dG}{dx}$.

This formula is called Leibnitz’s Rule of differentiation under integration sign. In a special case, when G(x) and H(x) both are absolute (constants) –let $G(x) =a$, $H(x)=b \iff dG/dx =0=dH/dx$ ; then

$\dfrac{d}{dx} \displaystyle {\int_a^b} F(x,t) dt = \displaystyle {\int_a^b}\dfrac{\partial F}{\partial x} dt$

## Changing Integral Equation with Multiple Integrals into Standard Simple  Integral

(Multiple Integral Into Simple Integral — The magical formula)

Say, the integral of order n is given by $\displaystyle{\int_{\Delta}^{\Box}} f(x) dx^n$

We can prove that $\displaystyle{\int_{a}^{t}} f(x) dx^n = \displaystyle{\int_{a}^{t}} \dfrac{(t-x)^{n-1}}{(n-1)!} f(x) dx$

Example: Solve $\int_0^1 x^2 dx^2$

Solution: $\int_0^1 x^2 dx^2$

$= \int_0^1 \dfrac{(1-x)^{2-1}}{(2-1)!} x^2 dx$

(since t=1)

$=\int_0^1 (1-x) x^2 dx$

$=\int_0^1 (1-x) x^2 dx$

$=\int_0^1 (x^2-x^3) dx =1/12$