Solution 1 to problem N2t4s6f3

Equations

The following unsolved equations remain:

    2     2
0=p3  + p4

Expressions

The solution is given through the following expressions:


     - p4*q4
q6=----------
       p3


q5=q4


    p4*q4
q3=-------
     p3


q2=0


q1=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:

 q8, q7, q4, p4, 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.

 
{p3,

 g0033*p4 + g0034*p3,

                                                                 2
 g0003*p3*q7 + g0004*p3*q4 + g0005*p3*q4 + g0006*p4*q4 + g0008*p3 ,

                                                                 2
 g0010*p3*q7 - g0011*p4*q4 + g0012*p3*q4 + g0013*p4*q4 + g0015*p3 ,

 g0025*p3*q8 + g0026*p3*q7 - g0027*p4*q4 + g0028*p3*q4 + g0029*p3*q4

  + g0030*p4*q4,

 q4,

 g0017*p4 - g0018*p3 - g0019*p3 - g0020*p4,

 q3,

 p4}

Relevance for the application:

The equation:


f =D D f *p3 + f  *p4
 t  1 2 x       2x

The symmetry:

f =(D f*D D f*p4*q4 + D f*f *p3*q4 + D D f  *p3*q7 + D D f*D f*p3*q4
 s   2   1 2           2   x          1 2 2x          1 2   1

     - D f*f *p4*q4 + f  *p3*q8)/p3
        1   x          3x

And now in machine readable form:

The system:

df(f(1),t)=d(1,d(2,df(f(1),x)))*p3 + df(f(1),x,2)*p4$

The symmetry:

df(f(1),s)=(d(2,f(1))*d(1,d(2,f(1)))*p4*q4 + d(2,f(1))*df(f(1),x)*p3*q4 + d(1,d(
2,df(f(1),x,2)))*p3*q7 + d(1,d(2,f(1)))*d(1,f(1))*p3*q4 - d(1,f(1))*df(f(1),x)*
p4*q4 + df(f(1),x,3)*p3*q8)/p3$