Page 116 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

dy
Unit 6: Existence Theorem for the Solution of the Equation = f(x, y)
dx

,
In other words, f(x, y) is regular function in the domain D . From this assumption it follows that Notes
y
f  ( , ) , , ,
x
is also regular in D . Therefore, for any positive numbers a, b such that a < a and b < b ,
dy
y
x
f  ( , )
both |f(x, y)| and are continuous on the closed domain D given by
dy
|x x | a, |y y | b
0 0
Thus there exist positive numbers M and K such that
f
y
SUP | ( , )| M
x
y

( , ) D
x
x
f  ( , )
y
SUP ( , ) D | | K ...(2)
x
y

y 
x
y
f  ( , )
Integrating along the segment connecting y and y , we obtain
y  1 2
2 y f  ( , )
x
y
f(x, y ) f(x, y ) = dy .
1 2
1 y y 
Hence the Lipschitz condition
|f(x, y ) f(x, y )| K|y y | ...(3)
2 1 2 1
holds on D. Therefore, under the above assumption, we can apply to the equation (1), the
method of successive approximations and the domain
|x x | h = min|a, b/M| ...(4)
0
as follows, we write
x
x
x
y 1 ( ) y 0 f ( , y 0 )dt
0 x
x
y 2 ( ) y 0 f ( , y 1 )dt
x
x
0 x
.............................................
.............................................
x
y n ( ) y 0 f { , y n 1 ( )}dt
x
t
x
0 x
where the integration means complex integration along a smooth curve connecting x and x in
0
the domain (4). Since f(x, y ) is regular in the domain |x x | < h, the first integral is well-
0 0
defined, independent of the curves, and hence so is y . Taking the first integral along the segment
1
connecting x and x, we obtain,
0
|y, (x) y | hM b
0
Hence f{x, y (x)} is well defined for |x x | < h as a function of x.
1 0
Since y (x) is given by the integral of the regular function f(x, y ), y (x) is regular in the domain
1 0 1
|x x | < h. Hence f{(f, y (x)} is also regular. Therefore the second integral is well defined and
0 1
hence y (x) is well defined and regular. Taking the integral along the segment connecting x and
2 0
x, we obtain further
|y (x) y (x)| hM b.
2 0
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