Solution 1 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


     1
q30=---*q21
     2


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


     1
q13=---*q21
     2


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p7=0


p5=p1


p4=0


p3= - p1


p2=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, p1, 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.

 
{p3,q21,p6,p1,g0007*q21 - g0016*p1}

Relevance for the application:

The system:


          2
b(1) =b(2) *p6 + b(1) *b(1)*p1
    t                x

b(2) = - Db(2)*Db(1)*p1 + b(2) *b(1)*p1
    t                         x

The symmetry:

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

       1            2
b(2) =---*b(2) *b(2) *q21
    s  2      x

And now in machine readable form:

The system:

df(b(1),t)=b(2)**2*p6 + df(b(1),x)*b(1)*p1$

df(b(2),t)= - d(1,b(2))*d(1,b(1))*p1 + df(b(2),x)*b(1)*p1$

The symmetry:

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

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