Page 121 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Assume that Lipschitz condition with respect to z , z , ... z is satisfied in D, that is, there exists
1 2 n
)
x
positive constant k such that for every pair of points ( ,x 1 , 2 ... n ), ( , 1, 2,... n in D
n
x
x
f i ( , 1 , 2 ,.... ) f i ( , 1 , 2 ,..., n ) K ( m m )
3
m 1
for every i = 1, 2, ..., n. Further let
h = min (a, b/m)
...(8)
x
,
M = SUP f i ( ,z z z 3 ,.... )
z
,
n
2
1
( ,z ,...., n z ) D
x
i 1 1,2,3,....m
x
x
x
z
Then there exists one and only one set of solution z 1 ( ), z 2 ( )... ( ) of (5) on the interval
n
x x 0 h ...(9)
satisfying the initial conditions (6).
This theorem implies the following:
Assume that ( ,f x z z ,.... ) is real valued and continuous on the domain D and satisfies the
z
,
1 2 n
x
Lipschitz condition on D, that is for every pair of points ( ,x , ,...., ),( , , ,.... ) of D,
1 2 1 2 n
n
f ( , 1 , 2 ,... ) f ( , 1 ,... n ) K n m .
x
x
n
m 1
Then there exists one and only one solution y(x) of the equation (1) satisfying the initial conditions
(2) on the interval.
x x 0 h.
b
z
a
m
x
where h min( , / ) and m SUP f ( ,z z 2 ,... )
,
n
1
x
( ,z 1 ,z 2,...zn ) D
Proof of the theorem 1
The proof of the theorem 1 is entirely the same as in the case of the first order differential
equation in unit 6. The initial value problem for (5) with (6) can be reduced to the system of
integral equations.
x
1)
(m
x
z m ( ) = y 0 f m t , ,( ), ( ),... ( )dt (m = 1, 2, ... n)
z
t
z
z
t
t
n
2
0 x
and solved by the method of successive approximations. In this case the successive approximation
functions are defined by
x
(m 1) 1 2 (n 1)
,
z (x) = y f ( , y y y ,...,y )dt
,
t
m, 1 0 m 0 0 0 0
0 x
x
t
t
z (x) = y (m 1) f m ( , z 1 ,( ), z 1 ,1( ),....z n ,1 ( )dt
t
t
0
m, 2
0 x
..........................................................................................
x
t
t
z (x) = y (m 1) f , t z ,( ) z ( ),z ( )...z ( ) dt
t
t
m, k 0 m 1, k 1 2,k 1 3,k 1 , n k 1
0 x
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