Solution 7 to problem N2t6s8f3

Equations

The following unsolved equations remain:

    2     2
0=p3  + p5

Expressions

The solution is given through the following expressions:


      - p3*q15
q14=-----------
        p5


q13=0


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p6= - p3


p4=p5


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:

 q15, p8, p7, p5, 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.

 
{p4,

 q15,

 p3,

 p5,

 g0067*p8 + g0068*p7 - g0069*p3 + g0070*p5 + g0071*p5 + g0072*p3,

 p6,

                                                    2
 g0018*p3*q15 - g0028*p5*p7 + g0029*p3*p5 - g0030*p5  - g0031*p3*p5,

 g0046*p3 - g0047*p5 - g0048*p5 - g0049*p3,

                                      2           2
 g0003*p3*q15 - g0013*p5*p7 - g0014*p5  - g0015*p5  - g0016*p3*p5,

 g0052*p5 - g0053*p3}

Relevance for the application:

The equation:


f =D f*D D f*p3 + D f*f *p5 + D D f  *p7 + D D f*D f*p5 - D f*f *p3 + f  *p8
 t  2   1 2        2   x       1 2 2x       1 2   1        1   x       3x

The symmetry:

     - D D f  *p3*q15 + f  *p5*q15
        1 2 3x           4x
f =--------------------------------
 s                p5

And now in machine readable form:

The system:

df(f(1),t)=d(2,f(1))*d(1,d(2,f(1)))*p3 + d(2,f(1))*df(f(1),x)*p5 + d(1,d(2,df(f(
1),x,2)))*p7 + d(1,d(2,f(1)))*d(1,f(1))*p5 - d(1,f(1))*df(f(1),x)*p3 + df(f(1),x
,3)*p8$

The symmetry:

df(f(1),s)=( - d(1,d(2,df(f(1),x,3)))*p3*q15 + df(f(1),x,4)*p5*q15)/p5$