# Definitions and Types of Integral Equations

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

Browse by Sections

## 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 types of integration limits in above integral equations –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 ‘Non-linear integral equation’.

In this article, when you read ‘integral equation’ understand it as ‘linear integral equation’.

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. 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), the 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.

**Structure of an 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$

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$

The general equation of Volterra equation is also called **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.

**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]$

*$ 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$.*

**Example:**The nature of 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*x*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 integral sign**

Let $ F(x,t)$ and $ \dfrac{\partial F}{\partial x}$ be continuous functions of both *x *and *t *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 integral into standard simple integral

(Multiple Integral Into Simple Integral — The magical formula)

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$