Page 444 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 26: Volterra Integral Equation of the First Kind

Notes
x
u
x
or z ( ) e x u Z ( )du = f(x) ...(4)
0
Which is the Volterra integral equation of the second kind. The equation (4) can be solved by the
method developed in the last unit. Here

x
K ( , ) = e x .e u ...(5)
u
u
= K 1 ( , )
x
Example 2: Consider the Volterra equation of the first kind

x
K ( , ) ( ) = f(x) ...(1)
y
u
x
u
0
Where K(x, u) is a degenerate Kernel of the form
u
x
K ( , ) = g ( )g ( ) h ( )h ( ) ...(2)
x
u
u
x
1 2 1 2
Substituting in (1) we get
x x
u
x
g 1 ( )g 2 ( ) ( )du h 1 ( )h 2 ( ) ( )du = f(x)
x
x
u
u
y
y
0 0
or
x x
u
y
u
g 1 ( ) g 2 ( ) ( )du h 1 ( ) h 2 ( ) ( )du = f(x) ...(3)
u
x
x
u
y
0 0
Let us introduce an other variable Z(x) by the relation
x
y
u
Z(x) = g ( ) ( )du ...(4)
u
2
0
where Z(0) = 0
dz ( )
x
and = g 2 ( ) ( ) ...(5)
y
x
x
dx
So equation (3) becomes
dz
x h
2
Z
g ( ) ( ) h ( ) dx du = f(x)
x
x
x
1 1
g ( )
u
0 z
Now integrating by parts the integral on L.H.S. we have
x x
h ( ) d h ( )
u
u
x
x
Z
x
x
u
u
g 1 ( ) ( ) h 1 ( ) 2 Z ( ) h 1 ( ) 2 Z ( )du = f(x)
g ( ) du g ( )
u
u
2 0 0 2
LOVELY PROFESSIONAL UNIVERSITY 437