Page 442 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 26: Volterra Integral Equation of the First Kind

Also suppose that the Kernel K(x, u) and the function f(x) have continuous derivatives on the Notes
interval a x b and a u b i.e.

u
d K ( , ) K ( , )
u
x
x
x
f ( ), ,
dx x u
exist and continuous, the equation (1) can be reduced to that of second kind provided k(x, x) 0.
To see that differentiate (1) with respect to x,
x
u
K ( , ) df
x
x
x
y
y
x
u
K ( , ) ( ) . ( )du =
x dx
0
dt
a K ( , )
u
x
x dx
u
or y ( ) y ( )du = ...(2)
x
x
x
x
K ( , ) K ( , )
x
0
which is the Volterra equation of the second kind with Kernel
u
x
K
[ ( , )]
x
K ( , )
x
x
and the function
df
dx .
K ( , )
x
x
If K(x, x) = 0 then we have to differentiate twice to reduce the equation to that of second kind.
There is a second method of reducing the Volterra equation of the first kind to Volterra equation
of the second kind. For this consider the equation (1)
x
x
K ( , ) ( )du = f(x) ...(1)
y
u
u
0
x
If we set y ( )du = Z(x) ...(2)
u
0
Clearly Z(0) = 0
Now integrate by parts of L.H.S. of the integral i.e.
x
dz
u
K ( , ) ( )du = f(x)
x
u
du
0
x
u
r
u x K ( , )
u
Z
u
x
u
or K ( , ) ( ) Z ( )du = f(x)
u 0 u
0
x
K ( , )
x
u
x
Z
or K ( , ) ( ) Z ( )du = f(x)
x
x
u
u
0
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