Solution 4 to problem N1t6s12f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s12f1

Expressions

The solution is given through the following expressions: 

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


q16=0


q15=0


q14=0


q13=0


      2
     ---*p4*q9
      5
q12=-----------
        p2


q11=0


q10=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: 
 q9, p4, p2

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*g0068*p4 + 5*g0071*p2,

 q9,

                                                           2
 2*g0027*p4*q9 + 5*g0030*p2*q9 + 5*g0041*p2*p4 + 5*g0043*p2 ,

 p2 + p4,

 p2,

 g0083*p4 + g0085*p2}

Relevance for the application:



The equation: 



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


The symmetry: 

            4            2      5
    Df *(Df) *f*p2*q9 + ---*(Df) *f *p4*q9
      x                  5         x
f =----------------------------------------
 s                    p2


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)=(d(1,df(f(1),x))*d(1,f(1))**4*f(1)*p2*q9 + 2/5*d(1,f(1))**5*df(f(1),x
)*p4*q9)/p2$