=========================preview======================
(math303)[2009](s)midterm~ma_yxf^_10496.pdf
Back to MATH303 Login to download
======================================================
Solution of MATH 303 Midterm Exam, Spring 2009
dx
1. (10 points) Consider the ODE = X(x, t), where X(x, t) is a continuous function and satis.es
dt
1
|X(x, t) . X(y, t)| |x . y|.
.|t|
Suppose that x(t) and y(t) are two solutions of this ODE in .1 t 0. Show that


2
|x(t) . y(t)| e |t||x(0) . y(0)|, .1 t 0.
Proof: Denote (t)= | x(t) . y(t) |2 0. Then
.(t) = 2[x(t) . y(t)][x .(t) . y .(t)] = 2[x(t) . y(t)][X(x, t) . X(y, t)]
Thus
..(t) | .(t) |=2 | (X(x, t) . X(y, t))(x(t) . y(t)) |22
| x . y |2 = (t),
.|t|.|t|
or
2
.(t)+ (t) 0.
.|t|
. . 0 2 ds
|s|
For t 0, multiplying both sides of the above inequality by e t , we have
. . 0 2 ds 2 . . 0 2 ds
tt
|s||s|
.(t)e + (t)e 0,
.|t|
which implies
d . . 0 2 ds
|s|
((t)e t ) 0.
dt
Since
. . 0 2 ds . . 0 2
t |s| t .s ds .4 .t .4 |t|
e = e = e = e,

.4
d ((t)e |t|) 0.
dt
Integrating it from t to 0 (Note that t 0),


.4
(0)e 0 . (t)e |t| 0.
Therefore for .1 t 0,
4
(t) e |t|(0),
or

2
|x(t) . y(t)| e |t||x(0) . y(0)|.
2
2. (10 points) Consider the ODE system
.dx1
.
=2x1
. dt .dx2 = x2.
dt
with the initial condition x1(0) = x2(0) = 1.

(1)
Find the equivalent integral equation of this initial value problem.
. x1(t) .


(2)
Let x(t) = . Find the .rst four approximations x0(t), x1(t), x2(t), and x3(t) of the Picard


x2(t)
iteration.
(3)
Guess the formula of the n-th approximation of the Picard iteration and use mathematical induction
to justify your formula.


(4)
Find the solution of the initial value problem from the formula of the Picard iteration obtained in


(3).
Solution: (1) The equivalent integral equation is

. x1(t) .. x1(0) .. t . 2x1(s) .. 1 .. t . 2x1(s) .
=+ ds =+ ds.
x2(t) x2(0) x2(s)1 x2(s)
00
(2) In the Picard iteration,
0
. x(t) .. 1 .
1
x0(t)= = ,
0
x(t)1
2
10

. x(t) .. 1 .. t . 2x(s) .. 1 .. t . 2 .. 1+2t .
11
x1(t)= =+ ds =+ ds =
10
x(t)1 x(s) 1 11+ t
2020
21

. x(t) .. 1 .. t . 2x(s) .. 1 .. t . 2+4s .. 1+2t +2t2 .
11
x2(t)= 2 =+ 1 ds =+ ds = t2
x(t)1 x(s) 11+ s 1+ t + 1
2020 2 32 2
. x(t) .. 1 .. t . 2x(s) .. 1 .. t . 2+4s +4s.. 1+2t +2t2 + 4 t3 .
113
x3(t)= 3 =+2 ds =+2 ds = t2 + 1 t3
x(t)1 x(s) 11+ s + 1 s1+ t + 1
20202 26
(3) The formula is
+ (2t)n
. 1+2t + (2t)2 + .
2! n!
xn(t)=
+ tn
1+ t + t2 +
2! n!
For n = 0, the formula gives
. 1 .

x0(t)= ,
1
which is correct from the result obtained in question (2).
Assume that n = k, the formula is correct, i.e.,
+ (2t)k .
. 1+2t + (2t)2 +
2! k!
xk(t)= .
+ tk
1+ t + t2 +
2! k!
For n = k + 1,
k
xk+1(t) . 1 .. t . 2x1(s) .
=+ ds
k
1 x(s)
02. t + (2s)k .
. 1 .. 2[1 + 2s + (2s)2 + ]
2! k!
=+ k ds
1
2
1+ s + s
+ + sk!
0
2!
+ (2t)k+1
. 1+2t + (2t)2 + .
2! (k+1)!
= 1+