If you have finished your course in Calculus and Differential Equations, you should head to your next milestone: the Integral Equations. This marathon series (planned to be of 6 or 8 parts) is dedicated to interactive learning of integral equations for the beginners —starting with just definitions and demos —and the pros— taking it to the heights of problem solving. Comments and feedback are invited.

Also read:

### 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.

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)*.

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. Kernels are of 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 *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 the 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$

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**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$

gold price

We will derive and then solve a renewal equation for \( u_y \) by conditioning on the time of the first arrival. We can then find integral equations that describe the distribution of the current age and the joint distribution of the current and remaining ages. We need some additional notation. Let \( \bar{F}(t) = 1 – F(t) = \P(X \gt t) \) for \( t \ge 0 \) (the right-tail distribution function of an interarrival time), and for \( y \ge 0 \), let \( \bar{F}_y(t) = \bar{F}(t y) \).

Gaurav Tiwari

Interesting spam comment. 🙂

Siddharth Sharma

Hey Gaurav,

This post is really mind blowing. Actually I’m in class 12th and these days i am solving equations of Differentials and Integrals . So, after read your post i learnt many things about , means Deeply knowledge .

Thanks for Share 🙂

Samuel Gichimu

Hello thanks very much. Have a question does the following equation lie under itergral equation . How do we solve it

Evaluate ∫_1^3▒〖3x^2+ x〗^(5 ) dx

Gaurav Tiwari

This appears to be a simple integration problem of type $$ \int (ax^2+bx+c)^n \ dx$$. There is no integral equations-method is required to solve it. The solution can be obtained by using simple integration.

PS:Try using WolframAlpha for a demonstrative solution for the same.FontsDownloadFree

It is worth noting that Integral Equations often do not have an analytical solution, and must be solved numerically.

Aania

any one can explain what is the physical use of integral equation now in my last semester of BS i’m studying this subject

Jesse Francis

Just what I was looking for. Integral Equation in a nutshell! Thanks a lot! 🙂

Rohit Khanna

The best guide on integral equations on the internet. Thank you.

RAJESH DAS

Satisfactory explanation.

Abdul hafeez

i got ; what i want. thanks

pooja

How do i solve this equation

∫(xy + x^2y^2) f(y) dy

limits are 1 and 0