=========================preview======================
(math303)[2011](s)midterm~ma_ywmab^_20400.pdf
Back to MATH303 Login to download
======================================================
Math 303
Theory of Ordinary Di.erential Equations

Midterm Test, Spring 2011
18:30 -20:00, March 23, 2011
1. (35 Marks) Consider the initial value problem
.. .

dx
= tx +2t,
dt x(0) = 0.
(a) Calculate the Picard iterations {xn} for the induced integral op-erator U, where
t
U(x)= [sx(s)+2s]ds;
0
(b) Verify that the limit of the sequence {xn} is a solution of the initial value problem.
2. (35 Marks) Consider the initial value problem
.. .

dx = x + sin(x + t),
dt x(0) = 1,
where is a constant.
(a)
Show that the problem has a unique solution x(t) on |t| T for any T> 0;

(b)
Prove that


t
lim x(t)= e,
0
uniformly on |t| T .
(continue to the next page......)
1 3. (30 Marks) Consider the initial value problem
.
dx
.
= Pn(x) . Pn(0),
dt
.
x(0) = 0,
where Pn is a polynomial of degree n.

(a)
Prove that there is > 0 such that x(t) 0 is the unique solution on |t| ;

(b)
Prove that x(t) 0 is the only solution for all t (., +).


SOLUTIONS
1. (35 Marks) Consider the initial value problem
.
dx
.
= tx +2t,
dt
.
x(0) = 0.
(a) Calculate the Picard iterations {xn} for the induced integral op-erator U, where
. t
U(x)= [sx(s)+2s]ds;
0
(b) Verify that the limit of the sequence {xn} is a solution of the initial value problem.
Solution For any T> 0, the function X(x, t) is Lipschitz in |t| T for all x, y, since |X(x, t) . X(y, t)| = |t||x . y| T |x . y|.
By the Existence Theorem, the initial value problem has a unique solution in |t| T .
(a) The Picard iterations are given by
x0(t)=0,xn+1 = U(xn),n =1, 2, 3,....
We postulate that for n =1, 2, 3, ,
(t2/2)n (t2/2)n.1
xn(t)=2 + + +(t2/2) .
n!(n . 1)!
In fact, we know that x0(t) = 0. For n = 1,
. t . t
x1(t)= (sx0(s)+2s)ds = s ds =2 t2/2.
00
Assume that the formula holds for n = k:
(t2/2)k (t2/2)k.1
xk(t)=2 + + +(t2/2) .
k!(k . 1)!
Then
. t
xk+1(t)= (sxk(s)+2s)ds
0
. t ..
2k+1 2k.1
ss
= ++ + s 3 +2s ds
2k.1k!2k.2(k . 1)!
0 t2k+2 t2k t4 2
= ++ ++ t
2k(k + 1)! 2k.1k!4
(t2/2)k+1 (t2/2)k (t2/2)2
=2 ++ + +(t2/2)
(k + 1)! k! 2!
Thus, the formula holds for n =1, 2, 3,... , by mathematical induction.
(b) By taking n , we obtain the solution of the initial value problem:
t
x(t)=2 e 2/2 . 1 .
It is easy to see that
2
dx
t/2
=2 e t = tx +2t,
dt
and
x(0) = 0.
Hence it is a solution of the initial value problem.

2. (35 Marks) Consider the initial value problem
.
dx
.
= x + sin(x + t),
dt
.
x(0) = 1,
where is a constant.
(a)
Show that the problem has a unique solution x(t) on |t| T for any T> 0;

(b)
Prove that


t
lim x(t)= e,
0
uniformly on |t| T .
Solution For any T> 0, the function X(x, t)= x + sin(x + t) is Lipschitz in |t| T for all x, y, since
|X(x, t) . X(y, t)| = |[x + sin(x + t)] .