Page 111 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes h | |x x | l 1
K
x
y
–
| ( ) y l 1 ( )| 0 for|x x 0 |  ...(13)
x
l
(l 1)!
We obtain for l = n + 1,
K
k | |x x | n
–
x
|y ( ) y ( )| 0 for|x x |  ...(14)
x
n 1 n 0
! n
Since (13) holds for n = 1 as mentioned above, we see, by mathematical induction, that (14) holds
for every n. Thus for m > n, we obtain
m 1 m 1 l
(k )
x
–
|y ( ) y ( )| y ( ) y ( ) k
x
x
x
m n l 1 l ! l ...(15)
l n l n
Since the right hand side of (15) tends to zero as n , {y (x)} converges uniformly to a function
n
y(x) on the interval |x x | . As the convergence is uniform, y(x) is continuous and more over,
0
evidently, y(x ) to y . To prove that y(x) is the solution, we know that as the sequence of functions
0 0
{y (x)} converges uniformly and y (x) is continuous on the interval |x x | , then the lim and
n n 0
integral can be interchanged. Thus
x x
lim y n ( )dx lim y n ( )dx
x
x
n 0 x x 0 n
Hence we obtain
x
y ( ) lim y n 1 ( )
x
n
x
t
t
y 0 lim f ( , y n ( ))dt
n 0 x
x
f
t
t
y [ lim ( , y ( ))]dt
0 n
x 0 n
x
t
y 0 f ( , ( ))dt
y
t
0 x
that is,
x
y
t
y ( ) y 0 f ( , ( ))dt ...(16)
t
x
0 x
The integrand f(t, y(t)) on the right side of (16) is a continuous function, hence y(x) is differentiable
with respect to x, and its derivative is equal to f(x, y(x)).
Hence the proof.
Integrating from x to x, we see that a solution y(x) of (6) satisfying the initial conditions (3), must
0
satisfy the integral equation (16). The above proof also shows that the integral equation can be
solved by the method of successive approximation.
Uniqueness of Solution
In the above treatment we have obtained by the method of successive approximation, a solution
y(x) of (6) satisfying the initial condition (3). We have yet to show the uniqueness of the above
solution.
Proof:
If the solution y(x) is not unique, let z(x) be another solution of (6), such that z(x ) = y . Then
0 0
x
z
x
t
z ( ) y 0 f t , ( ) dt .
0 x
104 LOVELY PROFESSIONAL UNIVERSITY