Solution 3 to problem N1t8s10f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s10f1

Expressions

The solution is given through the following expressions:


q21=0


q20=0


q19=0


q18=0


q17=0


q16=0


q15=0


q14=0


q13=0


q12=0


q11=0


q10=0


q9=0


q8= - q5


q7=0


q6=0


q4=0


q3=0


q2=0


q1=0


p13=0


p12=5*p3


p11=0


p10=0


p9=0


p8=0


p7= - 4*p3


       4
p6= - ---*p4
       3


p5=0


p2=p3


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:

 q5, 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,

 p2,

 q5,

 p7,

 15*g0059*p3 - 12*g0064*p3 - 4*g0065*p4 + 3*g0067*p4 + 3*g0068*p3 + 3*g0069*p3,

 p12,

 3*g0017*q5 - 3*g0020*q5 - 15*g0025*p3 + 12*g0030*p3 + 4*g0031*p4 - 3*g0033*p4

  - 3*g0034*p3 - 3*g0035*p3}

Relevance for the application:

The equation:


                       2                2
f =Df  *Df*f*p3 + (Df ) *f*p3 + Df *(Df) *f*p4 - 4*Df *Df*f *p3
 t   2x              x            x                  x     x

    4      3
 - ---*(Df) *f *p4 + 5*f  *f *f*p3
    3         x         2x  x

The symmetry:

           3            4
f =Df *(Df) *f*q5 - (Df) *f *q5
 s   x                     x

And now in machine readable form:

The system:

df(f(1),t)=d(1,df(f(1),x,2))*d(1,f(1))*f(1)*p3 + d(1,df(f(1),x))**2*f(1)*p3 + d(
1,df(f(1),x))*d(1,f(1))**2*f(1)*p4 - 4*d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*p3 -
 4/3*d(1,f(1))**3*df(f(1),x)*p4 + 5*df(f(1),x,2)*df(f(1),x)*f(1)*p3$

The symmetry:

df(f(1),s)=d(1,df(f(1),x))*d(1,f(1))**3*f(1)*q5 - d(1,f(1))**4*df(f(1),x)*q5$