Solution 3 to problem N1f1b0o24w1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1f1b0o24w1

Expressions

The solution is given through the following expressions:


q13=0


q12=2*q2


q11=0


q10=0


q9=0


q8=0


q7= - q2


q5=0


q4= - 3*q6


q3=q2


q1=0


p4=0


p2=0


p1= - p3

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:

 q6, q2, p3

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.

 
{q2,

 p3,

 p1,

 2*g0005*q2 - g0010*q2 + g0011*q6 - 3*g0013*q6 + g0014*q2 + g0015*q2 + g0017*p3

  - g0019*p3,

 2*g0021*q2 - g0026*q2 + g0027*q6 - 3*g0029*q6 + g0030*q2 + g0031*q2}

Relevance for the application:

The equation:


f = - Df *f*p3 + Df*f *p3
 t      x            x

The symmetry:

                       2                  2                           3
f =Df  *Df*f*q2 + (Df ) *f*q2 - 3*Df *(Df) *f*q6 - Df *Df*f *q2 + (Df) *f *q6
 s   2x              x              x                x     x             x

 + 2*f  *f *f*q2
      2x  x

And now in machine readable form:

The system:

df(f(1),t)= - d(1,df(f(1),x))*f(1)*p3 + d(1,f(1))*df(f(1),x)*p3$

The symmetry:

df(f(1),s)=d(1,df(f(1),x,2))*d(1,f(1))*f(1)*q2 + d(1,df(f(1),x))**2*f(1)*q2 - 3*
d(1,df(f(1),x))*d(1,f(1))**2*f(1)*q6 - d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*q2 +
 d(1,f(1))**3*df(f(1),x)*q6 + 2*df(f(1),x,2)*df(f(1),x)*f(1)*q2$