Solution 1 to problem N1t6s13b4+5

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s13b4+5

Expressions

The solution is given through the following expressions:


q46=0


q45=0


q44=0


q43=0


q42=0


q41=0


q40=0


q39=0


q38=0


q37=0


q36=0


q35=0


q34=0


q33=0


q32=0


q31=0


q30=0


q29=0


q28=0


q27=q1


q26=0


q25=0


q24=0


q23=0


q22=0


q21=0


q20=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


q6=0


q5=0


q4=0


q3=0


q2=0


p7=0


p4=0


p3=p5


p2=0


p1=p5

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:

 q1, p5, p6

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.

 
{g0010*q1 + g0020*q1 + g0021*p5,p3,p6,p5,q1,p1}

Relevance for the application:

The system:


          2
b(1) =b(2) *p6 + b(1) *b(1)*p5
    t                x

b(2) =Db(2)*Db(1)*p5 + b(2) *b(1)*p5
    t                      x

The symmetry:

b(1) =Db(2)*Db(1)*b(1) *q1
    s                 x

b(2) =Db(2)*Db(1)*b(2) *q1
    s                 x

And now in machine readable form:

The system:

df(b(1),t)=b(2)**2*p6 + df(b(1),x)*b(1)*p5$

df(b(2),t)=d(1,b(2))*d(1,b(1))*p5 + df(b(2),x)*b(1)*p5$

The symmetry:

df(b(1),s)=d(1,b(2))*d(1,b(1))*df(b(1),x)*q1$

df(b(2),s)=d(1,b(2))*d(1,b(1))*df(b(2),x)*q1$