Solution 4 to problem N1t8s12f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s12f1

Expressions

The solution is given through the following expressions:


q35=0


q34=0


q33=0


q32=0


q31=0


q30=0


q29=0


q28=0


q27= - q3


q26= - 6*q3


q25=0


q24=0


q23=0


q22=0


q21=0


q20=0


q19=0


q18=0


q17=0


q16=2*q3


q15=0


q14=0


q13= - q3


q12=0


q11=0


q10=0


q9=0


q8=q3


q7= - 2*q3


q6=0


q5=0


q4=0


q2=0


q1=0


p13=0


p12=2*p3


p11=0


p10=0


p9=0


p8=0


p7= - p3


       1
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:

 q3, 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.

 
{p4,

 p2,

 q3,

 p3,

 6*g0087*p3 - 3*g0092*p3 - g0093*p4 + 3*g0095*p4 + 3*g0096*p3 + 3*g0097*p3,

 3*g0012*q3 + 18*g0013*q3 - 6*g0023*q3 + 3*g0026*q3 - 3*g0031*q3 + 6*g0032*q3

  - 3*g0036*q3 - 6*g0039*p3 + 3*g0044*p3 + g0045*p4 - 3*g0047*p4 - 3*g0048*p3

  - 3*g0049*p3,

 q7}

Relevance for the application:

The equation:


                       2                2                        1      3
f =Df  *Df*f*p3 + (Df ) *f*p3 + Df *(Df) *f*p4 - Df *Df*f *p3 - ---*(Df) *f *p4
 t   2x              x            x                x     x       3         x

 + 2*f  *f *f*p3
      2x  x

The symmetry:

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

        2                3
 - (Df ) *Df*f *q3 - (Df) *f  *q3 - 6*Df*f  *f *f*q3
      x       x             3x            3x  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 - d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*p3 - 1
/3*d(1,f(1))**3*df(f(1),x)*p4 + 2*df(f(1),x,2)*df(f(1),x)*f(1)*p3$

The symmetry:

df(f(1),s)=d(1,df(f(1),x,3))*d(1,f(1))**2*f(1)*q3 - 2*d(1,df(f(1),x,2))*d(1,df(f
(1),x))*d(1,f(1))*f(1)*q3 + 2*d(1,df(f(1),x,2))*d(1,f(1))**2*df(f(1),x)*q3 + d(1
,df(f(1),x))**3*f(1)*q3 - d(1,df(f(1),x))**2*d(1,f(1))*df(f(1),x)*q3 - d(1,f(1))
**3*df(f(1),x,3)*q3 - 6*d(1,f(1))*df(f(1),x,3)*df(f(1),x)*f(1)*q3$