Solution 4 to problem over
Expressions |
Parameters |
Inequalities |
Relevance |
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Expressions
The solution is given through the following expressions:
r30=0
r31=0
r32=0
r33=0
r34=0
r35=0
r36=0
r37=0
r38=0
r39=0
1
---*a33*r320
2
r310=--------------
a23
r311=0
1
---*a33*r322
2
r312=--------------
a23
r313=0
r314=0
1 1 1
- ---*a33*c11*r320 + ---*a33*c11*r322 + ---*a33*c22*r320
2 2 2
r315=-----------------------------------------------------------
a23*c22
r316=0
r317=0
r318=0
1 2 1 2 1 3
---*a23 *a33*r354 + ---*a23*a33 *r333 - ---*a33 *r354
2 4 8
r319=-------------------------------------------------------
3
a23
r321=0
r323=0
r324=0
- c11*r320 + c11*r322 + c22*r320
r325=-----------------------------------
c22
r326=0
r327=0
r328=0
2 1 2
a23 *r354 + a23*a33*r333 - ---*a33 *r354
2
r329=------------------------------------------
2
a23
r330=0
r331=0
r332=0
r334=r354
r335=0
r336=0
r337=0
r338=0
r339=0
r340=0
r341=0
r342=0
r343=0
r344=0
r345=0
r346=0
r347=0
r348=0
r349=0
r350=0
r351=0
r352=0
1
---*a33*r354
2
r353=--------------
a23
r355=0
b33=0
b32=0
b31=0
Parameters
Apart from the condition that they must not vanish to give
a non-trivial solution and a non-singular solution with
non-vanishing denominators, the following parameters are free:
r320, r322, r333, r354, a33, c11, c22, a23
Inequalities
In the following not identically vanishing expressions are shown.
Any auxiliary variables g00?? are used to express that at least
one of their coefficients must not vanish, e.g. g0019*p4 + g0020*p3
means that either p4 or p3 or both are non-vanishing.
{a23,c22,c11}
Relevance for the application:
The system of equations related to the Hamiltonian HAM:
2 2 2
HAM=2*u2*u3*a23 + u3 *a33 + v1 *c11 + v2 *c22
has apart from the Hamiltonian and Casimirs the following 4 first integrals:
2 3 1 2 2 2 3 3
FI=u1 *u2*u3*a23 + ---*u1 *u3 *a23 *a33 + u2 *u3*a23
2
3 3 1 2 4 1 2 1 3
+ u2*u3 *(a23 - ---*a23*a33 ) + u3 *(---*a23 *a33 - ---*a33 )
2 2 8
1
= a product of the elements of: {---,
8
u3,
2*u2*a23 + u3*a33,
2 2 2 2 2 2 2
4*u1 *a23 + 4*u2 *a23 - 2*u2*u3*a23*a33 + u3 *(4*a23 - a33 )}
{HAM,FI} = 0
2 2 2 3 1 4 2
FI=u2 *u3 *a23 + u2*u3 *a23*a33 + ---*u3 *a33
4
1
= a product of the elements of: {---,
4
u3,
u3,
2*u2*a23 + u3*a33,
2*u2*a23 + u3*a33}
{HAM,FI} = 0
2 2 1 2 2
FI=u2*u3*v1 *a23*c11 + u2*u3*v2 *a23*c22 + ---*u3 *v1 *a33*c11
2
1 2 2
+ ---*u3 *v2 *a33*c22
2
1
= a product of the elements of: {---,
2
u3,
2 2
v1 *c11 + v2 *c22,
2*u2*a23 + u3*a33}
{HAM,FI} = 0
2 2
FI=u2*u3*v1 *( - a23*c11 + a23*c22) + u2*u3*v3 *a23*c22
2 2 1 1 1 2 2
+ u3 *v1 *( - ---*a33*c11 + ---*a33*c22) + ---*u3 *v3 *a33*c22
2 2 2
1
= a product of the elements of: { - ---,
2
u3,
2 2
v1 *(c11 - c22) - v3 *c22,
2*u2*a23 + u3*a33}
{HAM,FI} = 0
And again in machine readable form:
HAM=2*u2*u3*a23 + u3**2*a33 + v1**2*c11 + v2**2*c22$
FI=u1**2*u2*u3*a23**3 + 1/2*u1**2*u3**2*a23**2*a33 + u2**3*u3*a23**3 + u2*u3**3*
(a23**3 - 1/2*a23*a33**2) + u3**4*(1/2*a23**2*a33 - 1/8*a33**3)$
FI=u2**2*u3**2*a23**2 + u2*u3**3*a23*a33 + 1/4*u3**4*a33**2$
FI=u2*u3*v1**2*a23*c11 + u2*u3*v2**2*a23*c22 + 1/2*u3**2*v1**2*a33*c11 + 1/2*u3
**2*v2**2*a33*c22$
FI=u2*u3*v1**2*( - a23*c11 + a23*c22) + u2*u3*v3**2*a23*c22 + u3**2*v1**2*( - 1/
2*a33*c11 + 1/2*a33*c22) + 1/2*u3**2*v3**2*a33*c22$