Solution 2 to problem N1t6s12b4+5

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s12b4+5

Expressions

The solution is given through the following expressions:


q36=0


q35=0


q34=0


q33=0


q32=0


q31=0


q30=q21


q29=0


q28=0


q27=0


q26=0


q25=0


q24=0


q23=0


q22=0


q20=0


q19=0


q18=0


q17=0


q16=0


q15=0


q14=0


q13=q21


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


     1
    ---*p3*q21
     2
q1=------------
        p6


p7=0


p5=0


p4=0


p2=0


p1=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:

 q21, p3, 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.

 
{2*g0027*p6 + g0039*p3,q21,p3,p6,2*g0007*p6*q21 + g0015*p3*q21 + 2*g0016*p3*p6}

Relevance for the application:

The system:


          2
b(1) =b(2) *p6
    t

b(2) =Db(2)*Db(1)*p3
    t

The symmetry:

                                       2
b(1) =Db(2)*Db(1)*b(2)*q21 + b(1) *b(2) *q21
    s                            x

        1                                       2
       ---*Db(2)*Db(1)*b(1) *p3*q21 + b(2) *b(2) *p6*q21
        2                  x              x
b(2) =---------------------------------------------------
    s                         p6

And now in machine readable form:

The system:

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

df(b(2),t)=d(1,b(2))*d(1,b(1))*p3$

The symmetry:

df(b(1),s)=d(1,b(2))*d(1,b(1))*b(2)*q21 + df(b(1),x)*b(2)**2*q21$

df(b(2),s)=(1/2*d(1,b(2))*d(1,b(1))*df(b(1),x)*p3*q21 + df(b(2),x)*b(2)**2*p6*
q21)/p6$