20 -
2 Lagrange portrait, H amilton portrait, Lie series, C lairaut constant 43 1
Box 20.5 (Coefficients of the solution of the initial value problem with respect to the Hamilton portrait
η
2
0 := E
2 /(1 − E
2 ) cos
2 B 0 , V
2
0 := 1 + η
2
0 , t 0 := tan B 0 ).
[
10 ]=
1
N 0 cos B 0
, [
0
1]=
V
2
0
N 0
, [ 0
2]= −
3V
2
0 η
2
0 t 0
2N
2
0
, [
20 ]= −
V
2
0 t 0
2N
2
0
, [
11]=
t 0
N
2
0 cos B 0
, (20 .8 3
)
[ 12]=
1 + 3 t
2
0 + η
2
0
3N
3
0 cos B 0
, [
21]= −
V
2
0 (1 + 3 t
2
0 + η
2
0 − 9η
2
0 t
2
0 )
6N
3
0
,
(20 .8 4)
[
0
3
]=
−
V
2
0 η
2
0 (1 − t
2
0 + η
2
0 − 5η
2
0 t
2
0 )
2N
3
0
, [
3
0
]=
−
t
2
0
3N
3
0 cos B 0
,
(20 .8 5)
[
13 ]=
t 0 (2 + 3 t
2
0 + η
2
0 − η
4
0 )
3N
4
0 cos B 0
, [ 3
1]= −
t 0 (1 + 3 t
2
0 + η
2
0 )
3N
4
0 cos B 0
,
(20 .8 6
)
[
40 ]=
V
2
0 t 0 (1 + 3 t
2
0 + η
2
0 − 9η
2
0 t
2
0 )
24N
4
0
,
(20 .8 7
)
[
22]=
−V
2
0 t 0 (4 + 6 t
2
0 − 13η
2
0 − 9η
2
0 t
2
0 − 17η
4
0 + 45 η
4
0 t
2
0 )
12N
4
0
,
(20 .8 8
)
[
0
4]=
V
2
0 η
2
0 t 0 (12 + 6
9 η
2
0 − 45η
2
0 t
2
0 + 57 η
4
0 − 10 5 η
4
0 t
2
0 )
24N
4
0
,
(20 .8 9)
[
50 ]=
t
2
0 (1 + 3 t
2
0 + η
2
0 )
15N
5
0 cos B 0
,
(20 .90 )
[
41]=
V
2
0
120N
5
0
[
1 + 3
0
t
2
0 + 45 t
4
0 + η
2
0 (2 − 7
2 t
2
0 − 90t
4
0 ) + η
4
0 (1 − 10 2 t
2
0 + 225 t
4
0 )] ,
(20 .91)
[
3
2]=
1 + 20 t
2
0 + 3
0 t
4
0 + η
2
0 (2 + 13 t
2
0 ) + η
4
0 (1 − 7t
2
0 )
15N
5
0 cos B 0
,
(20 .92)
[
23 ]=
V
2
0
6
0 N
5
0
[−4 − 3
0 t
2
0 (1 + t
2
0 ) + 9 η
2
0 (1 + 2 t
2
0 + 5 t
4
0 )+
+
2 η
4
0 (15 − 17 7 t
2
0 ) + η
6
0 (17 − 40 2 t
2
0 + 525 t
4
0 )] ,
(20 .93 )
[
14]=
1
15N
5
0 cos B 0
[
2 + 15 t
2
0 + 15 t
4
0 + 3 η
2
0 (1 + 2 t
2
0 ) − 3η
4
0 t
2
0 − η
6
0 (1 − 6t
2
0 )] ,
(20 .94)
[
0
5]=
V
2
0 η
2
0
40N
5
0
[
4 − 4t
2
0 + η
2
0 (27 − 142t
2
0 + 15 t
4
0 )+
+
2 η
4
0 (21 − 226t
2
0 + 10 5 t
4
0 ) + η
6
0 (19 − 3
14 t
2
0 + 3
15 t
4
0 )] ,
(20 .95)
[
10 ] α = −
t 0
N 0
, [
11] α = −
1 + 2 t
2
0 + η
2
0
2N
2
0
, [
3
0
] α =
t 0 (1 + 2 t
2
0 )
6N
3
0
,
[
12] α = −
t 0 (5 + 6 t
2
0 )
6N
3
0
, [
3
1] α =
1 + 20 t
2
0 + 24 t
4
0
24N
4
0
, [
13 ] α = −
5 + 28 t
2
0 + 24 t
4
0
24N
4
0
.
(20 .96 )
2 Lagrange portrait, H amilton portrait, Lie series, C lairaut constant 43 1
Box 20.5 (Coefficients of the solution of the initial value problem with respect to the Hamilton portrait
η
2
0 := E
2 /(1 − E
2 ) cos
2 B 0 , V
2
0 := 1 + η
2
0 , t 0 := tan B 0 ).
[
10 ]=
1
N 0 cos B 0
, [
0
1]=
V
2
0
N 0
, [ 0
2]= −
3V
2
0 η
2
0 t 0
2N
2
0
, [
20 ]= −
V
2
0 t 0
2N
2
0
, [
11]=
t 0
N
2
0 cos B 0
, (20 .8 3
)
[ 12]=
1 + 3 t
2
0 + η
2
0
3N
3
0 cos B 0
, [
21]= −
V
2
0 (1 + 3 t
2
0 + η
2
0 − 9η
2
0 t
2
0 )
6N
3
0
,
(20 .8 4)
[
0
3
]=
−
V
2
0 η
2
0 (1 − t
2
0 + η
2
0 − 5η
2
0 t
2
0 )
2N
3
0
, [
3
0
]=
−
t
2
0
3N
3
0 cos B 0
,
(20 .8 5)
[
13 ]=
t 0 (2 + 3 t
2
0 + η
2
0 − η
4
0 )
3N
4
0 cos B 0
, [ 3
1]= −
t 0 (1 + 3 t
2
0 + η
2
0 )
3N
4
0 cos B 0
,
(20 .8 6
)
[
40 ]=
V
2
0 t 0 (1 + 3 t
2
0 + η
2
0 − 9η
2
0 t
2
0 )
24N
4
0
,
(20 .8 7
)
[
22]=
−V
2
0 t 0 (4 + 6 t
2
0 − 13η
2
0 − 9η
2
0 t
2
0 − 17η
4
0 + 45 η
4
0 t
2
0 )
12N
4
0
,
(20 .8 8
)
[
0
4]=
V
2
0 η
2
0 t 0 (12 + 6
9 η
2
0 − 45η
2
0 t
2
0 + 57 η
4
0 − 10 5 η
4
0 t
2
0 )
24N
4
0
,
(20 .8 9)
[
50 ]=
t
2
0 (1 + 3 t
2
0 + η
2
0 )
15N
5
0 cos B 0
,
(20 .90 )
[
41]=
V
2
0
120N
5
0
[
1 + 3
0
t
2
0 + 45 t
4
0 + η
2
0 (2 − 7
2 t
2
0 − 90t
4
0 ) + η
4
0 (1 − 10 2 t
2
0 + 225 t
4
0 )] ,
(20 .91)
[
3
2]=
1 + 20 t
2
0 + 3
0 t
4
0 + η
2
0 (2 + 13 t
2
0 ) + η
4
0 (1 − 7t
2
0 )
15N
5
0 cos B 0
,
(20 .92)
[
23 ]=
V
2
0
6
0 N
5
0
[−4 − 3
0 t
2
0 (1 + t
2
0 ) + 9 η
2
0 (1 + 2 t
2
0 + 5 t
4
0 )+
+
2 η
4
0 (15 − 17 7 t
2
0 ) + η
6
0 (17 − 40 2 t
2
0 + 525 t
4
0 )] ,
(20 .93 )
[
14]=
1
15N
5
0 cos B 0
[
2 + 15 t
2
0 + 15 t
4
0 + 3 η
2
0 (1 + 2 t
2
0 ) − 3η
4
0 t
2
0 − η
6
0 (1 − 6t
2
0 )] ,
(20 .94)
[
0
5]=
V
2
0 η
2
0
40N
5
0
[
4 − 4t
2
0 + η
2
0 (27 − 142t
2
0 + 15 t
4
0 )+
+
2 η
4
0 (21 − 226t
2
0 + 10 5 t
4
0 ) + η
6
0 (19 − 3
14 t
2
0 + 3
15 t
4
0 )] ,
(20 .95)
[
10 ] α = −
t 0
N 0
, [
11] α = −
1 + 2 t
2
0 + η
2
0
2N
2
0
, [
3
0
] α =
t 0 (1 + 2 t
2
0 )
6N
3
0
,
[
12] α = −
t 0 (5 + 6 t
2
0 )
6N
3
0
, [
3
1] α =
1 + 20 t
2
0 + 24 t
4
0
24N
4
0
, [
13 ] α = −
5 + 28 t
2
0 + 24 t
4
0
24N
4
0
.
(20 .96 )