Solution 2 to problem N1t8s10f1

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

Expressions

The solution is given through the following expressions:


q21=0


q19=0


q18=0


q17=0


q16=0


q15=0


q14=0


q13=4*q3


q12=0


        3
q11= - ---*q20
        5


        3
q10= - ----*q20
        10


q9= - 3*q3


       3
q8= - ---*q5
       4


q7=0


q6=2*q3


    2
q4=---*q20
    5


    1
q2=---*q20
    5


q1=0


p13=0


p12=0


p11=0


p10=0


p9=0


p8=0


p7=0


p6= - p4


p5=0


p3=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:

 q5, q3, q20, p4

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.

 
{20*g0038*q20 + 80*g0045*q3 - 12*g0047*q20 - 6*g0048*q20 - 60*g0049*q3

  - 15*g0050*q5 + 40*g0052*q3 + 20*g0053*q5 + 8*g0054*q20 + 20*g0055*q3

  + 4*g0056*q20,

 20*g0005*q20 + 80*g0012*q3 - 12*g0014*q20 - 6*g0015*q20 - 60*g0016*q3

  - 15*g0017*q5 + 40*g0019*q3 + 20*g0020*q5 + 8*g0021*q20 + 20*g0022*q3

  + 4*g0023*q20 - 20*g0031*p4 + 20*g0033*p4,

 p4}

Relevance for the application:

The equation:


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

The symmetry:

    1                   2                            2         3
f =---*Df  *Df*f*q20 + ---*Df  *Df *f*q20 + Df  *(Df) *f*q3 - ---*Df  *Df*f *q20
 s  5    3x             5    2x   x           2x               5    2x     x

          2            3        2                  3                  2
 + 2*(Df ) *Df*f*q3 - ----*(Df ) *f *q20 + Df *(Df) *f*q5 - 3*Df *(Df) *f *q3
        x              10     x    x         x                  x        x

    3      4
 - ---*(Df) *f *q5 + 4*Df*f  *f *f*q3 + f  *f *f*q20
    4         x            2x  x         3x  x

And now in machine readable form:

The system:

df(f(1),t)=d(1,df(f(1),x))*d(1,f(1))**2*f(1)*p4 - d(1,f(1))**3*df(f(1),x)*p4$

The symmetry:

df(f(1),s)=1/5*d(1,df(f(1),x,3))*d(1,f(1))*f(1)*q20 + 2/5*d(1,df(f(1),x,2))*d(1,
df(f(1),x))*f(1)*q20 + d(1,df(f(1),x,2))*d(1,f(1))**2*f(1)*q3 - 3/5*d(1,df(f(1),
x,2))*d(1,f(1))*df(f(1),x)*q20 + 2*d(1,df(f(1),x))**2*d(1,f(1))*f(1)*q3 - 3/10*d
(1,df(f(1),x))**2*df(f(1),x)*q20 + d(1,df(f(1),x))*d(1,f(1))**3*f(1)*q5 - 3*d(1,
df(f(1),x))*d(1,f(1))**2*df(f(1),x)*q3 - 3/4*d(1,f(1))**4*df(f(1),x)*q5 + 4*d(1,
f(1))*df(f(1),x,2)*df(f(1),x)*f(1)*q3 + df(f(1),x,3)*df(f(1),x)*f(1)*q20$