Solution 1 to problem N1t12s14f2

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t12s14f2

Expressions

The solution is given through the following expressions:


q14=0


q13=0


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5= - q1


q4=0


q3=0


q2=0


p9=0


p8=3*p2


p7=0


p6=0


p5= - p2


p4= - 2*p2


    1
p3=---*p2
    2


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:

 q1, 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.

 
{p8,

 p2,

 p3,

 q5,

 q1,

 p4,

 p5,

 2*g0013*q1 - 2*g0017*q1 - 6*g0018*p2 + 2*g0021*p2 + 4*g0022*p2 - g0023*p2

  - 2*g0024*p2}

Relevance for the application:

The equation:


    1                                                        2
f =---*Df  *Df*f*p2 + Df  *Df *f*p2 - 2*Df  *Df*f *p2 - (Df ) *f *p2
 t  2    3x             2x   x            2x     x         x    x

 + 3*f  *f *f*p2
      3x  x

The symmetry:

           3            4
f =Df *(Df) *f*q1 - (Df) *f *q1
 s   x                     x

And now in machine readable form:

The system:

df(f(1),t)=1/2*d(1,df(f(1),x,3))*d(1,f(1))*f(1)*p2 + d(1,df(f(1),x,2))*d(1,df(f(
1),x))*f(1)*p2 - 2*d(1,df(f(1),x,2))*d(1,f(1))*df(f(1),x)*p2 - d(1,df(f(1),x))**
2*df(f(1),x)*p2 + 3*df(f(1),x,3)*df(f(1),x)*f(1)*p2$

The symmetry:

df(f(1),s)=d(1,df(f(1),x))*d(1,f(1))**3*f(1)*q1 - d(1,f(1))**4*df(f(1),x)*q1$