Solution 4 to problem N1t8s14f1


Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s14f1

Expressions

The solution is given through the following expressions:

q55=0


q54=0


q53=0


q52=0


q51=0


q50=0


q49=0


q48=0


q47=0


q46=0


q45=0


q44=0


q43=0


q42=0


q41=0


q40=0


q39=0


q38=0


q37=0


q36=0


q35=0


q34=0


q33=0


q32=0


q31=0


q30=0


q29=0


q28=0


q27=0


q26=0


q25=0


q24=0


q23=0


q22=0


q21=0


q20=0


q19=0


q18=0


q17=0


q15=0


q14=0


q13=0


q12= - q16


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p13=0


p12=7*p3


p11=0


p10=0


p9=0


p8=0


p7= - 6*p3


p6= - 2*p4


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:
 q16, 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,

 p7,

 q16,

 p2,

 7*g0127*p3 - 6*g0132*p3 - 2*g0133*p4 + g0135*p4 + g0136*p3 + g0137*p3,

 p12,

 g0043*q16 - g0047*q16 + 7*g0059*p3 - 6*g0064*p3 - 2*g0065*p4 + g0067*p4

  + g0068*p3 + g0069*p3}


Relevance for the application:



The equation: 


                       2                2                               3
f =Df  *Df*f*p3 + (Df ) *f*p3 + Df *(Df) *f*p4 - 6*Df *Df*f *p3 - 2*(Df) *f *p4
 t   2x              x            x                  x     x               x

 + 7*f  *f *f*p3
      2x  x
The symmetry:
              5             6
f = - Df *(Df) *f*q16 + (Df) *f *q16
 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 - 6*d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*p3 -
 2*d(1,f(1))**3*df(f(1),x)*p4 + 7*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))**5*f(1)*q16 + d(1,f(1))**6*df(f(1),x)*
q16$