Solution 10 to problem N1t6s14f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s14f1

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


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=0


q26=0


q25=0


q24=0


q23=0


q22=0


q21=0


q20=0


q19=0


q18=0


q17=0


q15=0


q14=0


q13=0


     3*p2*q16
q12=----------
        p4


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


p3=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:

 q16, p2, p4

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.

 
{g0104*p4 + 3*g0108*p2,

 q16,

                                         2
 g0043*p4*q16 + 3*g0047*p2*q16 + g0061*p4  + g0063*p2*p4,

 p2 + p4,

 p4,

 g0123*p4 + g0125*p2,

 p2}

Relevance for the application:

The equation:


                     2
f =Df *Df*f*p2 + (Df) *f *p4
 t   x                  x

The symmetry:

              5                6
    3*Df *(Df) *f*p2*q16 + (Df) *f *p4*q16
        x                         x
f =----------------------------------------
 s                    p4

And now in machine readable form:

The system:

df(f(1),t)=d(1,df(f(1),x))*d(1,f(1))*f(1)*p2 + d(1,f(1))**2*df(f(1),x)*p4$

The symmetry:

df(f(1),s)=(3*d(1,df(f(1),x))*d(1,f(1))**5*f(1)*p2*q16 + d(1,f(1))**6*df(f(1),x)
*p4*q16)/p4$