Page 112 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

By assumption 2, we obtain Notes
x
dt
y
x
t
y
t
x
| ( ) z ( )| | ( ) z ( ) | ...(17)
0 x
Therefore we also obtain for |x x | .
0
y
x
x
| ( ) z ( )| KN |x x 0 |
where
N = SUP |y(x) z(x)|.
|x x 0 |
Substituting the above estimate for |y(t) z(t)| on the right side of (17), we obtain
2
y
x
x
| ( ) z ( )| N ( |x x 0 ) |2!
K
for |x x | . Substituting this estimate for |y(t) z(t)| once more on the right side of (17), we
0
have
3
|y(x) z(x)| N(K|x x |) /3! for |x x | .
0 0
Repeating this substitution, we obtain
m
|y(x) z(x)| N(K|x x |) /m!, m = 1, 2, ...... ...(18)
0
for |x x | . The right side of the above inequality tends to zero as m . This means that
0
N = SUP |y(x) z(x)|
|x x |
0
is equal to zero.
Hence y(x) given by (16) is a unique solution.
Example 1: Solve
dy
xy ...(1)
dx
with the initial conditions x = 0.0, y(0) = 0.1
Now y (x) = 0.1
0
x
x
y (x) = 0.1 x y 0 ( )dx
1 0
x
= 0.1 x (0.1)dx
0
x 2 x 2
= 0.1 0.1 0.1 1
2 2
x
y (x) = 0.1 x y ( )dx
x
2 2
0
x x 2
= 0.1 0.1 x 1 dx
0 2
x 2 x 4
= 0.1 0.1
2 2.4

x 2 x 4
= 0.1 1
2 2.4

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