Solution 35 to problem e3c2new
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Parameters |
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Expressions
The solution is given through the following expressions:
1
a22=---*a33
2
b22=0
b32=0
b33=0
c12=0
c13=0
c22=0
c23=0
1 2
- ---*b31
2
c33=-------------
a33
n1=0
n2=0
n3=0
m1=0
m2=0
m3=0
r6=0
r5=0
r4=0
r3=0
r2=0
r1=0
q20=0
q19=0
q17=0
q16=0
q15=0
q14=0
q13=0
q12=0
q11=0
q10=0
q9=0
q8=0
q7=0
q5=0
q4=0
q3=0
q2=0
q1=0
p56=0
p55=0
p54=0
p53=0
p52=0
p51=0
p50=0
p49=0
p48=0
p47=0
p46=0
p45=0
p44=0
p43=0
p42=0
p41=0
p40=0
p39=0
p38=0
p37=0
p36=0
p35=0
p34=0
p33=0
p32=0
p31=0
p30=0
p29=0
p28=0
p27=0
p26=0
p25=0
p24=0
p23=0
p22=0
p21=0
p20=0
p19=0
p18=0
p17=0
p16=0
p15=0
p14=0
p13=0
p12=0
p11=0
p10=0
p9=0
p8=0
p7=0
p6=0
p5=0
p4=0
p3=0
p2=0
p1=0
k125=0
k124=0
k122=0
k121=0
4
- 2*b31 *k1
k120=--------------
4
a33
k119=0
k118=0
k117=0
k116=0
4
- 2*b31 *k1
k115=--------------
4
a33
3
- 4*b31 *k1
k114=--------------
3
a33
k113=0
k112=0
k110=0
k109=0
2
- 2*b31 *k1
k108=--------------
2
a33
k107=0
2
- 2*b31 *k1
k106=--------------
2
a33
k105=0
k104=0
k103=0
k102=0
k101=0
k100=0
k99=0
k98=0
k97=0
k96=0
2
8*b31 *k1
k95=-----------
2
a33
k94=0
k93=0
k92=0
k91=0
k90=0
k89=0
k88=0
k87=0
k86=0
k85=0
2
8*b31 *k1
k84=-----------
2
a33
k83=0
k82=0
k81=0
k80=0
k79=0
k78=0
k77=0
k76=0
k75=0
k74=0
k73=0
k72=0
k71=0
4
- b31 *k1
k70=------------
4
a33
k69=0
k68=0
k67=0
k66=0
4
- 2*b31 *k1
k65=--------------
4
a33
k64=0
k63=0
k62=0
k61=0
k60=0
k59=0
k58=0
k57=0
k56=0
k55=0
k54=0
k53=0
k52=0
k51=0
k50=0
k49=0
k48=0
k47=0
k46=0
k45=0
k44=0
k43=0
k42=0
- 8*b31*k1
k41=-------------
a33
k40=0
k38=0
k37=0
k36=0
4
- b31 *k1
k35=------------
4
a33
k34=0
k33=0
k32=0
k30=0
k29=0
k28=0
k27=0
k26=0
k25=0
k24=0
k23=0
4*b31*k1
k22=----------
a33
k21=0
- 4*b31*k1
k20=-------------
a33
k19=0
k18=0
k17=0
k16=0
k14=0
k13=0
k12=0
k11=0
k10=0
k9=0
k8=0
k7=0
k6=0
k5=k1
k4=0
k3=2*k1
k2=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:
k1,b31,a33
Relevance for the application:
The following expression INT is a first
integral for the Hamiltonian HAM:
2 2 2 2 2 2 2 2
a33 *u1 + a33 *u2 + 2*a33 *u3 + 2*a33*b31*u3*v1 - b31 *v3
HAM=---------------------------------------------------------------
2*a33
4 4 4 2 2 4 4 3 2
INT=(k1*(a33 *u1 + 2*a33 *u1 *u2 + a33 *u2 - 4*a33 *b31*u1 *u3*v1
3 3 2 2 2 2 2
- 8*a33 *b31*u1*u2*u3*v2 + 4*a33 *b31*u2 *u3*v1 - 2*a33 *b31 *u1 *v3
2 2 2 2 2 2
+ 8*a33 *b31 *u1*u3*v1*v3 - 2*a33 *b31 *u2 *v3
2 2 3 2 4 4
+ 8*a33 *b31 *u2*u3*v2*v3 - 4*a33*b31 *u3*v1*v3 - b31 *v1
4 2 2 4 2 2 4 4 4 2 2 4
- 2*b31 *v1 *v2 - 2*b31 *v1 *v3 - b31 *v2 - 2*b31 *v2 *v3 ))/a33
And again in machine readable form:
HAM=(a33**2*u1**2 + a33**2*u2**2 + 2*a33**2*u3**2 + 2*a33*b31*u3*v1 - b31**2*v3
**2)/(2*a33)$
INT=(k1*(a33**4*u1**4 + 2*a33**4*u1**2*u2**2 + a33**4*u2**4 - 4*a33**3*b31*u1**2
*u3*v1 - 8*a33**3*b31*u1*u2*u3*v2 + 4*a33**3*b31*u2**2*u3*v1 - 2*a33**2*b31**2*
u1**2*v3**2 + 8*a33**2*b31**2*u1*u3*v1*v3 - 2*a33**2*b31**2*u2**2*v3**2 + 8*a33
**2*b31**2*u2*u3*v2*v3 - 4*a33*b31**3*u3*v1*v3**2 - b31**4*v1**4 - 2*b31**4*v1**
2*v2**2 - 2*b31**4*v1**2*v3**2 - b31**4*v2**4 - 2*b31**4*v2**2*v3**2))/a33**4$