Solution 11 to problem N1t8s12f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s12f1

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


q19=0


q18=0


q17=0


q16=0


q15=0


q14=0


q13=0


      3
     ---*p6*q20
      5
q12=------------
        p12


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p13=0


p11=0


p10=0


p9=0


p8=0


p7=0


p5=0


p4=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:

 q20, p6, p12

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.

 
{g0087*p12 + g0093*p6,

 p12,

 p6,

                                               2
 5*g0019*p12*q20 + 3*g0027*p6*q20 + 5*g0039*p12  + 5*g0045*p12*p6,

 q20,

 5*g0066*p12 + 3*g0074*p6}

Relevance for the application:

The equation:


       3
f =(Df) *f *p6 + f  *f *f*p12
 t        x       2x  x

The symmetry:

     3      5                 2
    ---*(Df) *f *p6*q20 + (Df) *f  *f *f*p12*q20
     5         x                 2x  x
f =----------------------------------------------
 s                      p12

And now in machine readable form:

The system:

df(f(1),t)=d(1,f(1))**3*df(f(1),x)*p6 + df(f(1),x,2)*df(f(1),x)*f(1)*p12$

The symmetry:

df(f(1),s)=(3/5*d(1,f(1))**5*df(f(1),x)*p6*q20 + d(1,f(1))**2*df(f(1),x,2)*df(f(
1),x)*f(1)*p12*q20)/p12$