Solution 2 to problem N1t6s8f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s8f1

Expressions

The solution is given through the following expressions:


q13=0


q12=3*q3


q11=0


q10=0


q9=0


q8=0


q7= - 2*q3


       2
q6= - ---*q4
       3


q5=0


q2=q3


q1=0


p7=0


p6=0


p5=0


p3=0


p2= - p4


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:

 q4, q3, 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.

 
{p2,

 p4,

 q3,

 9*g0005*q3 - 6*g0010*q3 - 2*g0011*q4 + 3*g0013*q4 + 3*g0014*q3 + 3*g0015*q3

  + 3*g0019*p4 - 3*g0021*p4,

 q12,

 9*g0024*q3 - 6*g0029*q3 - 2*g0030*q4 + 3*g0032*q4 + 3*g0033*q3 + 3*g0034*q3}

Relevance for the application:

The equation:


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

The symmetry:

                       2                2
f =Df  *Df*f*q3 + (Df ) *f*q3 + Df *(Df) *f*q4 - 2*Df *Df*f *q3
 s   2x              x            x                  x     x

    2      3
 - ---*(Df) *f *q4 + 3*f  *f *f*q3
    3         x         2x  x

And now in machine readable form:

The system:

df(f(1),t)= - d(1,df(f(1),x))*d(1,f(1))*f(1)*p4 + d(1,f(1))**2*df(f(1),x)*p4$

The symmetry:

df(f(1),s)=d(1,df(f(1),x,2))*d(1,f(1))*f(1)*q3 + d(1,df(f(1),x))**2*f(1)*q3 + d(
1,df(f(1),x))*d(1,f(1))**2*f(1)*q4 - 2*d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*q3 -
 2/3*d(1,f(1))**3*df(f(1),x)*q4 + 3*df(f(1),x,2)*df(f(1),x)*f(1)*q3$