Solution 55 to problem e3null
Expressions |
Parameters |
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Back to problem e3null
Expressions
The solution is given through the following expressions:
1
a22=---*a33
2
b22=0
b31=0
b32=0
b33=0
c13=0
c23=0
1
c33=---*c22
2
n1=0
n2=0
m1=0
m2=0
m3=0
r6=0
r5=0
r4=0
r3=0
r2=0
r1=0
q19=0
q18=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
2
- 4*k1*n3
q3=-------------
2
a33
q2=0
2
- 4*k1*n3
q1=-------------
2
a33
p50=0
p49=0
p48=0
4*c22*k1*n3
p47=-------------
2
a33
p46=0
p45=0
p44=0
p43=0
p42=0
8*c22*k1*n3
p41=-------------
2
a33
8*c12*k1*n3
p40=-------------
2
a33
p39=0
p38=0
8*c12*k1*n3
p37=-------------
2
a33
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
- 4*k1*n3
p7=------------
a33
p6=0
- 4*k1*n3
p5=------------
a33
p4=0
p3=0
p2=0
p1=0
k104=0
k103=0
k102=0
k101=0
k100=0
2 2
- 8*c12 *k1 - 2*c22 *k1
k99=--------------------------
2
a33
k98=0
k97=0
k96=0
k95=0
2 2
- 8*c12 *k1 - 2*c22 *k1
k94=--------------------------
2
a33
k93=0
k92=0
k91=0
k90=0
k89=0
2*c22*k1
k88=----------
a33
8*c12*k1
k87=----------
a33
- 2*c22*k1
k86=-------------
a33
k85=0
k84=0
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
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
- 4*c12 *k1 - c22 *k1
k55=------------------------
2
a33
k54=0
k53=0
k52=0
k51=0
2 2
- 8*c12 *k1 - 2*c22 *k1
k50=--------------------------
2
a33
k49=0
k48=0
k47=0
k46=0
k45=0
k44=0
k43=0
k42=0
k41=0
k40=0
k39=0
k38=0
k37=0
k36=0
k35=0
k34=0
k33=0
k32=0
k31=0
k30=0
k29=0
k28=0
k27=0
k26=0
2 2
- 4*c12 *k1 - c22 *k1
k25=------------------------
2
a33
k24=0
k23=0
k21=0
k20=0
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,c22,c12,n3,a33
Relevance for the application:
The following expression INT is a first
integral for the Hamiltonian HAM:
2 2 2 2 2
HAM=(a33*u1 + a33*u2 + 2*a33*u3 + 4*c12*v1*v2 + 2*c22*v2 + c22*v3 + 2*n3*u3
)/2
2 4 2 2 2 2 4 2
INT=(k1*(a33 *u1 + 2*a33 *u1 *u2 + a33 *u2 + 8*a33*c12*u1*u2*v3
2 2 2 2 2
- 2*a33*c22*u1 *v3 + 2*a33*c22*u2 *v3 - 4*a33*n3*u1 *u3
2 2 4 2 2 2 2 2 2
- 4*a33*n3*u2 *u3 - 4*c12 *v1 - 8*c12 *v1 *v2 - 8*c12 *v1 *v3
2 4 2 2 2
- 4*c12 *v2 - 8*c12 *v2 *v3 + 8*c12*n3*u1*v2*v3 + 8*c12*n3*u2*v1*v3
2 4 2 2 2 2 2 2 2 4
- c22 *v1 - 2*c22 *v1 *v2 - 2*c22 *v1 *v3 - c22 *v2
2 2 2 2 2 2
- 2*c22 *v2 *v3 + 8*c22*n3*u2*v2*v3 + 4*c22*n3*u3*v3 - 4*n3 *u1
2 2 2
- 4*n3 *u2 ))/a33
And again in machine readable form:
HAM=(a33*u1**2 + a33*u2**2 + 2*a33*u3**2 + 4*c12*v1*v2 + 2*c22*v2**2 + c22*v3**2
+ 2*n3*u3)/2$
INT=(k1*(a33**2*u1**4 + 2*a33**2*u1**2*u2**2 + a33**2*u2**4 + 8*a33*c12*u1*u2*v3
**2 - 2*a33*c22*u1**2*v3**2 + 2*a33*c22*u2**2*v3**2 - 4*a33*n3*u1**2*u3 - 4*a33*
n3*u2**2*u3 - 4*c12**2*v1**4 - 8*c12**2*v1**2*v2**2 - 8*c12**2*v1**2*v3**2 - 4*
c12**2*v2**4 - 8*c12**2*v2**2*v3**2 + 8*c12*n3*u1*v2*v3 + 8*c12*n3*u2*v1*v3 -
c22**2*v1**4 - 2*c22**2*v1**2*v2**2 - 2*c22**2*v1**2*v3**2 - c22**2*v2**4 - 2*
c22**2*v2**2*v3**2 + 8*c22*n3*u2*v2*v3 + 4*c22*n3*u3*v3**2 - 4*n3**2*u1**2 - 4*
n3**2*u2**2))/a33**2$