MA 2327
Assignment 7
Due 7 December 2015
Id:
2327-f2015-7.m4,v 1.1 2015/12/01 23:32:47 john Exp john
1. The second order scalar equation x′′ (t) + tx(t) = 0 is equivalent to the
first order system
d
dt
x(t)
x′ (t)
=
0 1
−t 0
x(t)
.
x′ (t)
(a) Calculate the first few approximations Wk (t, t0 ), but only for t0 =
0.
Solution:
Wk =
k
X
Dk
j=0
where Dj satisfies the recursive relations D0 = I,
Dj+1 (t, t0 ) =
Z t
t0
0 1
D( s, t0 ) ds.
−s 0
The first few are
D0 (t, 0) =
D2 (t, 0) =
D4 (t, 0) =
1 0
,
0 1
−t3 /6
0
,
0
−t3 /3
t6 /180
0
,
6
0
t /72
1
D1 (t, 0) =
D3 (t, 0) =
D5 (t, 0) =
0
t
,
2
−t /2 0
0
−t4 /12
,
5
t /30
0
0
t7 /504
.
−t8 /1440
0
Id:
2327-f2015-7.m4,v 1.1 2015/12/01 23:32:47 john Exp john 2
(b) Find the fundamental matrix W (t, t0 ), but again only for t0 = 0.
Note: You will get a power series in t. Its sum is not any function
you are likely to be familiar with. It’s fine to leave it as an infinite
sum, but the expression for the general term in the sum should be
explicit.
Solution:
The general summand above is

D2m (t, 0) = 

m t3m
(−1)
Qm−1
l=0
0
(3l+2)(3l+3)
(−1)m t3m
Qm−1
(3l+1)(3l+3)
l=0
0

D2m+1 (t, 0) = 




m t3m+1
0
(−1)
Qm−1
(3l+3)(3l+4)
l=0
(−1)m+1 t3m+2
Q
m−1
2 l=0 (3l)(3l+1)
0
so summing over all values of m gives
P
∞
m=0
(3l+2)(3l+3)
l=0
(−1)m+1 t3m+2
m=0 2 Qm−1 (3l)(3l+1)
l=0
W (t, 0) =  P∞

m t3m
(−1)
Qm−1
(c) Compute det W (t, t0 ).
Solution:
det W (t, t0 ) = exp
Z
t
t0
tr
P∞
m=0



m t3m+1
(−1)
Qm−1
(3l+3)(3l+4)
(−1)m t3m
m=0 Qm−1 (3l+1)(3l+3)
l=0
P∞
l=0


.
0 1
ds = 1.
−s 0
(d) Compute W (t, t0 ).
Note: Once again, you won’t be able to do this completely explicitly. The best you can hope for a power series in t and t0 .
Solution:
W (t, t0 ) = W (t, 0)W (t0, 0)−1
W (t, 0) was computed previously. W (t0 , 0) is obtained from it by
substitution:
P
∞
n=0
W (t0 , 0) =  P
∞

(−1)n t3n
0
Qn−1
(3k+2)(3k+3)
(−1)n+1 t3n+2
0
k=0
n=0 2 Qn−1 (3k)(3k+1)
k=0
P∞
n=0
P∞
n=0
(−1)n t3n+1
0
Qn−1
k=0
(3k+3)(3k+4)
(−1)n t3n
0
Qn−1
k=0
(3k+1)(3k+3)


.
The indices have been changed to make subsequent calculations
easier. The determinant is, as we just showed, equal to 1, so
P
∞
n=0
W (t0 , 0)−1 =  P
∞

(−1)n t3n
0
Qn−1
(3k+1)(3k+3)
(−1)n t3n+2
0
k=0
n=0 2 Qn−1 (3k)(3k+1)
k=0
P∞
n=0
P∞
n=0
(−1)n+1 t3n+1
0
Qn−1
k=0
(3k+3)(3k+4)
(−1)n t3n
0
Qn−1
k=0
(3k+2)(3k+3)


.
Id:
2327-f2015-7.m4,v 1.1 2015/12/01 23:32:47 john Exp john 3
Multiplying,
W (t, t0 ) =
w11 (t, t0 ) w12 (t, t0 )
,
w21 (t, t0 ) w22 (t, t0 )
where
m t3m
(−1)n t3n
0
Qn−1
(3l+2)(3l+3)
(3k+1)(3k+3)
l=0
k=0
P
P∞
(−1)n t3n+2
(−1)m t3m+1
0
+ ∞
m=0
n=0 Qm−1 (3l+3)(3l+4) 2 Qn−1 (3k)(3k+1)
k=0
l=0
w11 (t, t0 ) =
P∞
P∞
w12 (t, t0 ) =
P∞
P∞
m=0
m=0
+
n=0
m t3m
n=0
P∞
m=0
(−1)
Qm−1
(−1)
Qm−1
(−1)n+1 t3n+1
0
Qn−1
(3l+2)(3l+3)
(3k+3)(3k+4)
k=0
(−1)n t3n
(−1)m t3m+1
0
n=0 Qm−1 (3l+3)(3l+4) Qn−1 (3k+2)(3k+3)
l=0
k=0
P∞l=0
(−1)n t3n
(−1)m+1 t3m+2
0
n=0 2 Qm−1 (3l)(3l+1) Qn−1 (3k+1)(3k+3)
l=0
k=0
P
P∞
(−1)n t3n+2
(−1)m t3m
0
+ ∞
m=0
n=0 Qm−1 (3l+1)(3l+3) 2 Qn−1 (3k)(3k+1)
k=0
l=0
w21 (t, t0 ) =
P∞
w22 (t, t0 ) =
P∞
m=0
P∞
(−1)n+1 t3n+1
(−1)m+1 t3m+2
0
n=0 2 Qm−1 (3l)(3l+1) Qn−1 (3k+3)(3k+4)
P
P∞ l=0 (−1)m t3m k=0
(−1)n t3n
0
+ ∞
m=0
n=0 Qm−1 (3l+1)(3l+3) Qn−1 (3k+2)(3k+3)
l=0
k=0
m=0
P∞