# How to convert Integral Equations into Differential Equations?

## Introduction

In earlier parts we discussed the basics of integral equations and how they can be derived from ordinary differential equations. We also solved a linear integral equation using the trial method in the second part. Now we are in a situation from where the main job of solving Integral Equations can be started. But before we go ahead with that mission, it will be better to learn how integral equations can be converted into differential equations.

**Table of Contents**

See

- Part- I

Basics of integral equations - Part- II

Square Integrable Functions, Norms, Trial Method - Part- III

Changing Differential Equations into Integral Equations

## Integral Equation to Differential Equation

Converting an integral equation into a differential equation is exactly opposite to what we did in the last part, where we converted **boundary value differential equations into respective integral equations**. In last workout, initial value problems always ended up as **Volterra Integrals **and boundary value problems resulted as **Fredholm Integrals.**

In the converse process, we will get **initial value problems from Volterra Integrals** and **boundary value problems from Fredholm Integral Equations**. Also, as in earlier conversion *we continuously integrated the differentials* within given boundary values, we will *continuously differentiate* provided integral equations and refine the results by putting all constant integration limits.

The above instructions can be practically understood by following two examples. First problem involves the conversion of *Volterra Integral Equation into differential equation *and the second problem displays the conversion of *Fredholm Integral Equation into differential equation*.

### Problem 1: Converting Volterra Integral Equation into Ordinary Differential Equation with initial values

*Convert *$$y(x) = – \int_{0}^x (x-t) y(t) dt$$ *into initial value problem.*

Please note that this was the same integral equation we obtained after converting the initial value problem: $$y”+y=0$$ when $$y(0)=y'(0)=0$$ *( See Problem 1 of Part 3 )*

#### Solution:

We have, $$y(x) = – \int_{0}^x (x-t) y(t) dt \ldots (1)$$

**Differentiating (1) **with respect to $x$ will give

$$y'(x) = -\frac{d}{dx} \int_{0}^x (x-t) y(t) dt$$

$$ \Rightarrow y'(x)=-\int_{0}^x y(t) dt \ldots (2)$$

Again **differentiating (2)** w.r.t. $x$ will give

$$ y”(x)=-\frac{d}{dx}\int_{0}^x y(t) dt$$

$$ \Rightarrow y”(x)=-y(x) \ldots (3′)$$

$$ \iff y”(x)+y(x)=0 \ldots (3) $$

Putting *the lower limit* $x=0$ (i.e., the initial value) in equation (1) and (2) will give, respectively the following:

$$y(0) = – \int_{0}^0 (0-t) y(t) dt$$

$$y(0)=0 \ldots (4)$$

And, $$ y'(0)=-\int_{0}^0 y(t) dt$$

$$y'(0)=0 \ldots (5)$$

These equations (3), (4) and (5) form the ordinary differential form of given integral equation. $\Box$

### Problem 2: Converting Fredholm Integral Equation into Ordinary Differential Equation with boundary values

*Convert *$$ y(x) =\lambda \int_{0}^{l} K(x,t) y(t) dt$$ *into boundary value problem where* $$ K(x,t)=\frac{t(l-x)}{l} \qquad \mathbf{0<t<x} $$ *and* $$ K(x,t)=\frac{x(l-t)}{l} \qquad \mathbf{x<t<l}$$

#### Solution:

Please see Example 2 of Part 3.

The given integral equation is $$ y(x) =\lambda \int_{0}^{l} K(x,t) y(t) dt \ldots (1)$$ or $$y(x) =\lambda (\int_{0}^{x} \frac{(l-x)t}{l} y(t) dt + \int_{x}^{l} \frac{x(l-t)}{l} y(t) dt) \ldots (2)$$

**Differentiating (2) **with respect to $x$ will give $$ y'(x) = -\frac{\lambda}{l} \int_{0}^x t y(t) dt + \frac{\lambda}{l} \int_{x}^l (l-t) y(t) dt \ldots (3)$$

Continued differentiation of (3) will give $$ y”(x) = -\lambda y(x)$$ That’s $$ y”(x) +\lambda y(x) =0 \ldots (4)$$

To get the boundary values, we place $x$ equal to both integration limits in (1) or (2).

$x =0 \Rightarrow$ $$y(0)=0 \ldots (5)$$

$x=l \Rightarrow$ $$y(l)=0 \ldots (6)$$

The ODE (4) with boundary values (5) & (6) is the exact conversion of given integral equation. $\Box$