Solution 112 to problem e3null
Expressions |
Parameters |
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Expressions
The solution is given through the following expressions:
1
a22= - ---*a33
3
b22=0
b31=0
b32=0
b33=0
c12=0
c13=0
c22=0
c23=0
n1=0
n2=0
m1=0
m2=0
m3=0
r6=0
r5=0
r4=0
r2=0
r1=0
q19=0
q18=0
q16=0
q15=0
3*c33*q1
q14=----------
a33
q13=0
q12=0
q11=0
q10=0
3*c33*q1
q9=----------
a33
q8=0
q7=0
q5=0
q4=0
q3=q1
q2=0
p50=0
p49=0
p48=0
a33*p15 - 3*c33*p5
p47=--------------------
a33
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=p15
p27=0
p26=0
p25=0
p24=0
p23=0
p22=0
p21=0
p20=0
p19=0
p18=0
p17=0
p16=0
p14=0
p13=0
p12=0
p11=0
p9=0
p8=0
p7=p5
p6=0
p4=0
p3=0
p2=0
p1=0
k104=0
k103=0
k101=0
k100=0
2 2
- 18*c33 *k1 + 9*c33 *k10
k99=----------------------------
2
a33
k98=0
k97=0
k96=0
k95=0
2 2
- 18*c33 *k1 + 9*c33 *k10
k94=----------------------------
2
a33
k93=0
k92=0
k90=0
k89=0
- 6*c33*k1 + 3*c33*k10
k88=-------------------------
a33
k87=0
- 6*c33*k1 + 3*c33*k10
k86=-------------------------
a33
k85=0
k84=0
k83=0
k82=0
k81=0
k80=0
k79=0
k78=0
k77=0
6*c33*k10
k76=-----------
a33
k75=0
k74=0
k73=0
k72=0
k71=0
k70=0
k69=0
k68=0
k67=0
k66=0
k65=0
k64=0
k63=0
k62=0
k61=0
k60=0
k59=0
k58=0
k57=0
k56=0
2 2
- 9*c33 *k1 + 9*c33 *k10
k55=---------------------------
2
a33
k54=0
k53=0
k52=0
k51=0
2 2
- 18*c33 *k1 + 18*c33 *k10
k50=-----------------------------
2
a33
k49=0
k48=0
k47=0
k46=0
k45=0
6*c33*k10
k44=-----------
a33
k43=0
3*c33*k10
k42=-----------
a33
k41=0
k40=0
k39=0
k38=0
k37=0
k36=0
k35=0
k34=0
k33=0
k32=0
k31=0
k30=0
k28=0
k27=0
k26=0
2 2
- 9*c33 *k1 + 9*c33 *k10
k25=---------------------------
2
a33
k24=0
k23=0
k21=0
3*c33*k10
k20=-----------
a33
k19=0
k18=0
k17=0
k16=0
k14=0
k13=0
k12=k10
k11=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:
c33,n3,r3,q1,p15,p10,k1,p5,k10,a33
Relevance for the application:
The following expression INT is a first
integral for the Hamiltonian HAM:
2 2 2 2
- a33*u1 - a33*u2 + 3*a33*u3 + 3*c33*v3 + 3*n3*u3
HAM=--------------------------------------------------------
3
2 4 2 2 2 2 4 2 2 2
INT=(a33 *k1*u1 + 2*a33 *k1*u1 *u2 + a33 *k1*u2 + a33 *k10*u1 *u3
2 2 2 2 3 2 2 2 2
+ a33 *k10*u2 *u3 + a33 *p10*u3 + a33 *p15*u3*v1 + a33 *p15*u3*v2
2 2 2 2 2 2 2 2
+ a33 *p15*u3*v3 + a33 *p5*u1 *u3 + a33 *p5*u2 *u3 + a33 *q1*u1
2 2 2 2 2 2 2
+ a33 *q1*u2 + a33 *r3*u3 - 6*a33*c33*k1*u1 *v3 - 6*a33*c33*k1*u2 *v3
2 2 2 2 2 2
+ 3*a33*c33*k10*u1 *v2 + 3*a33*c33*k10*u1 *v3 + 3*a33*c33*k10*u2 *v1
2 2 2 2
+ 6*a33*c33*k10*u2 *v2 + 3*a33*c33*k10*u2 *v3
2 2
+ 6*a33*c33*k10*u2*u3*v2*v3 - 3*a33*c33*p5*u3*v3 + 3*a33*c33*q1*v1
2 2 4 2 2 2
+ 3*a33*c33*q1*v2 - 9*c33 *k1*v1 - 18*c33 *k1*v1 *v2
2 2 2 2 4 2 2 2 2 4
- 18*c33 *k1*v1 *v3 - 9*c33 *k1*v2 - 18*c33 *k1*v2 *v3 + 9*c33 *k10*v1
2 2 2 2 2 2 2 4
+ 18*c33 *k10*v1 *v2 + 9*c33 *k10*v1 *v3 + 9*c33 *k10*v2
2 2 2 2
+ 9*c33 *k10*v2 *v3 )/a33
And again in machine readable form:
HAM=( - a33*u1**2 - a33*u2**2 + 3*a33*u3**2 + 3*c33*v3**2 + 3*n3*u3)/3$
INT=(a33**2*k1*u1**4 + 2*a33**2*k1*u1**2*u2**2 + a33**2*k1*u2**4 + a33**2*k10*u1
**2*u3**2 + a33**2*k10*u2**2*u3**2 + a33**2*p10*u3**3 + a33**2*p15*u3*v1**2 +
a33**2*p15*u3*v2**2 + a33**2*p15*u3*v3**2 + a33**2*p5*u1**2*u3 + a33**2*p5*u2**2
*u3 + a33**2*q1*u1**2 + a33**2*q1*u2**2 + a33**2*r3*u3 - 6*a33*c33*k1*u1**2*v3**
2 - 6*a33*c33*k1*u2**2*v3**2 + 3*a33*c33*k10*u1**2*v2**2 + 3*a33*c33*k10*u1**2*
v3**2 + 3*a33*c33*k10*u2**2*v1**2 + 6*a33*c33*k10*u2**2*v2**2 + 3*a33*c33*k10*u2
**2*v3**2 + 6*a33*c33*k10*u2*u3*v2*v3 - 3*a33*c33*p5*u3*v3**2 + 3*a33*c33*q1*v1
**2 + 3*a33*c33*q1*v2**2 - 9*c33**2*k1*v1**4 - 18*c33**2*k1*v1**2*v2**2 - 18*c33
**2*k1*v1**2*v3**2 - 9*c33**2*k1*v2**4 - 18*c33**2*k1*v2**2*v3**2 + 9*c33**2*k10
*v1**4 + 18*c33**2*k10*v1**2*v2**2 + 9*c33**2*k10*v1**2*v3**2 + 9*c33**2*k10*v2
**4 + 9*c33**2*k10*v2**2*v3**2)/a33**2$