Page 419 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes with the initial conditions that at x = x , y = y . We assume that
0 0
1. The function f(x, y) is real valued and continuous on a domain D of the xy plane given by
x 0 a x x 0 , a y 0 b y y 0 b ...(2)

where a and b are positive numbers
2. f(x, y) satisfies the Lipschitz condition with respect to y in D, that is, there exists a positive
constant k such that

x
f ( ,y ) f ( ,y ) K y y ...(3)
x
1 2 1 2
)
x
for every pair of points ( ,x y 2 ), ( , y 2 of D,
with the help of Picard s method of successive approximation, it is then seen that y(x)
satisfies the integral equation
x
y
t
t
x
y ( ) y 0 f ( , ( ))dt ...(4)
0 x
The integrand f (t, y(t)) on the right hand side of (4) is a continuous function, hence y(x) is
differentiable with respect to x, and its derivatives is equal to f(x, y(x)). Here the integral equation
(4) can be solved by the method of successive approximation.
Uniqueness of the Solution: We have obtained the integral equations (4) for the solution y(x) of
(1) satisfying the initial conditions x = x , y = y . There remains an other important problem, the
0 0
problem of uniqueness. Is there any other solution satisfying the same initial condition.
Fortunately under our two assumptions, we can prove the uniqueness of the solution. To see this
let z(x) be another solution of (1) such that x = x , z(x ) = y . Then
0 0 0
x
z(x) = y 0 f ( , ( )) .
t
t
z
dt
0 x
By the assumption 2, we obtain for x x b
0
x
t
t
x
x
y ( ) z ( ) K y ( ) z ( ) dt ...(5)
0 x
Therefore, we also obtain x x h
0
y ( ) z ( ) KN x x
x
x
0
where
N = Sup y ( ) z ( )
x
x
x 0 x h
Substituting the above estimate for ( )y t z ( ) on the right side of (5), we obtain further
t
x
y ( ) z ( ) NK x x 0 2 / 2,
x

412 LOVELY PROFESSIONAL UNIVERSITY