Solution 9 to problem N1t6s10f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s10f1

Expressions

The solution is given through the following expressions:


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


     1
    ---*p4*q5
     2
q8=-----------
       p2


q7=0


q6=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:

 q5, 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.

 
{g0055*p4 + g0057*p2,

 p2,

                                                         2
 g0017*p4*q5 + 2*g0020*p2*q5 + 2*g0027*p2*p4 + 2*g0029*p2 ,

 q5,

 g0044*p4 + 2*g0047*p2}

Relevance for the application:

The equation:


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

The symmetry:

            3            1      4
    Df *(Df) *f*p2*q5 + ---*(Df) *f *p4*q5
      x                  2         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))**3*f(1)*p2*q5 + 1/2*d(1,f(1))**4*df(f(1),x
)*p4*q5)/p2$