Solution 1 to problem N1f1b0o37w1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1f1b0o37w1

Expressions

The solution is given through the following expressions:


q55=0


q54=0


q53=0


q52=0


q51=0


q50=0


q49=0


q48=0


q47=0


q46=0


q45=0


q44=0


q43=0


q42=0


q41=0


q40=0


        2
q39= - ---*q27
        3


q38=0


     1
q37=---*q27
     3


q36=0


q35=0


q34=0


q33=0


q32=0


q31=0


q30=0


q29=0


q28=0


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


q1=0


p7=0


p6=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:

 q27, p3

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,q27,2*g0018*q27 - g0020*q27 - 3*g0030*q27 - 3*g0060*p3}

Relevance for the application:

The equation:


       3
f =(Df) *f*p3
 t

The symmetry:

                        2                      1
f =Df  *f  *f *f*q27 - ---*Df *f  *f *f*q27 + ---*Df*f  *f  *f*q27
 s   2x  2x  x          3    x  3x  x          3      3x  2x

And now in machine readable form:

The system:

df(f(1),t)=d(1,f(1))**3*f(1)*p3$

The symmetry:

df(f(1),s)=d(1,df(f(1),x,2))*df(f(1),x,2)*df(f(1),x)*f(1)*q27 - 2/3*d(1,df(f(1),
x))*df(f(1),x,3)*df(f(1),x)*f(1)*q27 + 1/3*d(1,f(1))*df(f(1),x,3)*df(f(1),x,2)*f
(1)*q27$