Page 428 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 24: Integral Equations and Algebraic System of Linear Equations

Notes
r 1 r
n
(x) = x , (r 1,2,... )
r
n n
then equation (1) can be written in the form
n
n
K = f (r 1,2,3,.... ) ...(6)
r rs s r
s 1
Define the determinant A with elements

rs K rs for r, s = 1, 2, ... n
n

0 , r s
and =
rs 1 , r s

If the determinant A does not vanish, then (6), and therefore (5) has a unique solution for any
given step function f(x).
In the same way if we take up equation (2) and use equations (3) and (4) then equation (2) takes
up the form

n s /m r 1 r
f r = K rs ( )dy x , ...(7)
y
s 1 (s 1)/n n n
This case of (7) is different than that of (5) as here one cannot conclude that (x) is necessarily a
step function. All that can be said is that if we set

s /n
n ( )dy = x ,
y
s
(s 1)/n
then (7) becomes

n
F = K x (r 1,2,... ) ...(8)
n
r rs s
n
s 1
1 1 2 n 1 n
Here x , x , ... x give the mean values of (x) is the successive intervals 0, , , ... , ,
1 2 n
n n n n n
So there are infinitely many solutions of (x).
24.5 Summary

 In this unit we have seen how to convert a differential equation with conditions into an
integral equation.

 The existence and uniqueness of the solution of the integral equation is based on Picard s
method which puts some conditions on the Kernel as well on the function.
 It is seen that the integral equation is reduced to a algebraic system of equations if we
divide the interval into segment.

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