Solution 2 to problem N1t4s14f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t4s14f1

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


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p4=0


p2=0


     6*p1*q16
q12=----------
        p3

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, p3, p1

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. 
 
{g0101*p3 + 6*g0105*p1,p3,p1 + p3,g0118*p3 + g0120*p1}

Relevance for the application:



The equation: 



f =Df *f*p1 + Df*f *p3
 t   x            x


The symmetry: 

              5                6
    6*Df *(Df) *f*p1*q16 + (Df) *f *p3*q16
        x                         x
f =----------------------------------------
 s                    p3


And now in machine readable form: 

The system: 

df(f(1),t)=d(1,df(f(1),x))*f(1)*p1 + d(1,f(1))*df(f(1),x)*p3$


The symmetry: 

df(f(1),s)=(6*d(1,df(f(1),x))*d(1,f(1))**5*f(1)*p1*q16 + d(1,f(1))**6*df(f(1),x)
*p3*q16)/p3$