Table of Content
Conversion of I.E. into Differentiation under the sign of Integration. This is the most important formula to differentiate integral equation.
Integral Equation
An Integral Equation is an equation in which an unknown function appears under one or more integral sign.
If the derivative of the function is involved, it is called an Integro-differential equation.
$$ e.g., for \ a \le x \le b, \ a \le t \le b $$
The equation,
$$ f(x) = \int_a^b k(x,t) y(t) dt $$
$$ y(x) = f(x) + \int_a^b k(x,t) y(t) dt $$
General Formula
The most general form of an Integral Equation:
$$ v(x) u(x) = f(x) + \lambda \int k(x,t)u(t)dt\ ...eq^n (1) $$
is called a Linear Integral Equation where
- upper limit may be either variable \( x \) or fixed and \( u(x) \) is the unknown function,
- \( v(x) \), \( f(x) \) and kernel of the I.E. \( k(x, t) \) is known;
- \( \lambda \) is a non-zero real or complex parameter.
- If \( v(x) = 0 \), then equation (1) is called first kind integral equation.
- If \( v(x) = 1 \), then equation (1) is called second kind integral equation.
An Integral Equation is called linear if only linear operations are perform. An Integral Equation is not linear, it is called non-linear Integral Equation.
Types of Integral Equation
Fredholm I.E. :
$$ v(x)u(x) = f(x) + \lambda \int_a^b k(x,t)u(t)dt $$
If both limits are constant.
- when \( v(x) = 0 \) then \( f(x) + \lambda \int_a^b k(x,t)u(t)dt = 0 \) \( \implies \) Fredholm I.E. of first kind.
- when \( v(x) = 1 \) then \( u(x) = f(x) + \lambda \int_a^b k(x,t)u(t)dt \) \( \implies \) Non-Homogeneous Fredholm I.E. of second kind.
- when \( v(x) = 1\ and\ f(x) = 0 \) then \( u(x) = \lambda \int_a^b k(x,t)u(t)dt \) \( \implies \) Homogeneous I.E. of second kind.
Volterra I.E. :
$$ v(x)u(x) = f(x) + \lambda \int_a^x k(x,t)u(t)dt $$
If the upper limit is variable and the lower limit is constant.
- when \( v(x) = 0 \) then \( f(x) + \lambda \int_a^x k(x,t)u(t)dt = 0 \) \( \implies \) Volterra I.E. of first kind.
- when \( v(x) = 1 \) then \( u(x) = f(x) + \lambda \int_a^x k(x,t)u(t)dt \) \( \implies \) Non-Homogeous Volterra I.E. of second kind.
- when \( v(x) = 1\ and\ f(x) = 0 \) then \( u(x) = \lambda \int_a^x k(x,t)u(t)dt \) \( \implies \) Homogeneous Volterra I.E. of second kind.
Singular I.E.:
- When one or both limits of integration become infinite.
- When kernel becomes infinite at one or more point within the range of integration.
$$ u(x) = f(x) + \lambda \int_{ }^{ \infty } k(x,t)u(t)dt $$
Leibnitz's rule of differential under the integral sign
$$ { \frac {d}{dx}} \left (\int_{a(x)}^{b(x)} F(x,t)\ dt \right) = F(x,b(x)) \frac { \partial b}{ \partial x} - F(x,a(x)) \frac { \partial a} { \partial x} $$
$$ + \int_{a(x)}^{b(x)} \frac{ \partial F}{ \partial x}dt $$
If a and b are contant then $$ { \frac {d}{dx}} \left ( \int _a^b F(x,t)\ dt \right) = \int _a^b \frac { \partial F}{ \partial x}dt $$
If a is only constant then $$ { \frac {d}{dx}} \left ( \int_a^x F(x,t)\ dt \right) = \int_a^x \frac { \partial F}{ \partial x} dt + F(x,x) \frac { \partial x}{ \partial x} - F(x,a) \frac { \partial a}{ \partial x} $$
NOTE: $$ \int_a^x \int_a^x...\int_a^x y(t)\ dt\ dt...dt = \int_a^x { \frac {(x-t)^{(n-1)}}{(n-1)!}} y(t)dt $$
Now, I'm giving an example to understand the above formulas.
Example: Solve $$ f(x) = \int_{sin\ x}^{cos\ x} e^{(-t^2)} dt. $$ Find \( f'( \frac { \pi}{4}) \)
By using Leibnitz's Rule we have,
Here, \( F(x,t) = e^{(-t^2)} \) In the right hand side of the formula, we put an upper limit and lower limit in place of \( t \) in function F(x,t) and differentiate an upper limit and lower limit w.r.t.\( x \) given below:
{ NOTE: In case, either an upper or lower or both limit were constant, then we differentiate w.r.t \( x \) it becomes zero.}
$$ f'(x) = \int_{sin\ x}^{cos\ x} e^{(-t^2)} dt = F(x,{cos\ x}) \frac { \partial (cos\ x)}{ \partial x} - F(x,{sin\ x}) \frac { \partial (sin\ x)}{ \partial x} $$
$$ + \int_{sin\ x}^{cos\ x} \frac { \partial F(x,t)}{ \partial x}dt $$
$$ f'(x) = \int_{sin\ x}^{cos\ x} e^{(-t^2)} dt = e^{-cos^2(x)} \frac { \partial (cos\ x)}{ \partial x} - e^{-sin^2(x)} \frac { \partial (sin\ x)}{ \partial x} $$
$$ + \int_{sin\ x}^{cos\ x} \frac { \partial e^{(-t^2)}}{ \partial x} dt $$
Here, in the last term of RHS, \( F(x,t) = e^{(-t^2)} \) is function of \( t \) only so, when we differentiate it w.r.t. \( x \) it becomes zero. We get,
$$ f'(x) = e^{-cos^2(x)} (-sin\ x) - e^{-sin^2(x)} (cos\ x) $$
$$ f'( \frac { \pi}{4}) = e^{-cos^2( \frac { \pi}{4})} (- sin( \frac { \pi}{4})) - e^{-sin^2( \frac { \pi}{4})} (cos( \frac { \pi}{4}) $$
$$ f'( \frac { \pi}{4}) = e^{ \frac {-1}{2}} ( \frac {-1}{ \sqrt(2)}) - e^{ \frac {-1}{2}} ( \frac {1}{ \sqrt(2)}) $$
$$ f'( \frac { \pi}{4}) = - { \sqrt \frac {2}{e}}\ Answer. $$