Solution 6 to problem N1t8s10f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s10f1

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


     3
    ---*p6*q5
     4
q8=-----------
       p4


q7=0


q6=0


q4=0


q3=0


q2=0


q1=0


p13=0


p12=0


p11=0


p10=0


p9=0


p8=0


p7=0


p5=0


p3=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:

 q5, p6, 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.

 
{3*g0050*p6 + 4*g0053*p4,

                                                           2
 3*g0017*p6*q5 + 4*g0020*p4*q5 + 4*g0031*p4*p6 + 4*g0033*p4 ,

 g0065*p6 + g0067*p4,

 p4,

 q5}

Relevance for the application:

The equation:


           2            3
f =Df *(Df) *f*p4 + (Df) *f *p6
 t   x                     x

The symmetry:

            3            3      4
    Df *(Df) *f*p4*q5 + ---*(Df) *f *p6*q5
      x                  4         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))**2*f(1)*p4 + d(1,f(1))**3*df(f(1),x)*p6$

The symmetry:

df(f(1),s)=(d(1,df(f(1),x))*d(1,f(1))**3*f(1)*p4*q5 + 3/4*d(1,f(1))**4*df(f(1),x
)*p6*q5)/p4$