Page 443 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
K ( , )
x
u
x f ( )
x
u
or Z ( ) Z ( )du = K ( , ) ...(3)
x
u
x
x
K ( , )
x
x
0
which is Volterra equation of second kind with Kernel
x
u
K ( , )
u
x
K ( , )
x
x
f ( )
and the function Here it is assumed that K(x, x) 0. Applying the techniques of last
K ( , ).
x
x
unit we can write the solution of Z(x) as
x
u
x
f ( ) f ( )
Z(x) = H ( , ,1) du ...(4)
u
x
u
x
x
u
K ( , ) K ( , )
0
d
where H (x, u, 1) is the resolvent Kernel corresponding to the Kernel K ( , )/ ( , ).
K
x
x
x
u
du
Example 1: Consider the Volterra integral equation of the first kind
x
u
y
x
u
K ( , ) ( )du = f(x) ...(1)
0
with the Kernel K(x, u) given by
K(x, u) = e x u
So the equation (1) becomes
x
e x u y ( )du = f(x) ...(2)
u
0
Let us put
x
u
y ( )du = Z(x) ...(3)
0
x
dz ( )
So that Z(0) = 0 and ( )y x .
dx
Substituting this value of y in (2) we have
x x u dz
e ( )du = f(x)
u
du
0
Integrating L.H.S. by parts once we have
x
x
u
e x u Z ( ) e x u Z ( )du = f(x)
u
0
0
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