Solution 6 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


q27=0


q26=0


q25=0


q24=0


q23=0


q22=0


q21=0


q20=0


q19=0


q18=0


        1
q17= - ---*q28
        2


q16=0


q15=0


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


p13=0


p11=0


p10=0


p9=0


p8=0


p7= - p12


p6=0


p5=0


p4=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:

 q14, q28, p12, 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.

 
                                                                 2
{6*g0031*p12*q28 - 3*g0042*p12*q28 + 4*g0045*p3*q28 + 6*g0059*p12

               2
  - 6*g0064*p12  + 6*g0068*p12*p3,

 6*g0098*p12 - 3*g0109*p12 + 4*g0112*p3,

 q14,

 p12,

 p3,

 g0127*p12 - g0132*p12 + g0136*p3,

 p7,

 q28}

Relevance for the application:

The equation:


        2
f =(Df ) *f*p3 - Df *Df*f *p12 + f  *f *f*p12
 t    x            x     x        2x  x

The symmetry:

        3             1       2     2
f =(Df ) *Df*f*q14 - ---*(Df ) *(Df) *f *q28 + Df *Df*f  *f *f*q28
 s    x               2     x          x         x     2x  x

And now in machine readable form:

The system:

df(f(1),t)=d(1,df(f(1),x))**2*f(1)*p3 - d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*p12
 + df(f(1),x,2)*df(f(1),x)*f(1)*p12$

The symmetry:

df(f(1),s)=d(1,df(f(1),x))**3*d(1,f(1))*f(1)*q14 - 1/2*d(1,df(f(1),x))**2*d(1,f(
1))**2*df(f(1),x)*q28 + d(1,df(f(1),x))*d(1,f(1))*df(f(1),x,2)*df(f(1),x)*f(1)*
q28$