Solution 1 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


        1
q38= - ---*q28
        3


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


q17=0


q16=0


q15=0


q14=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=0


p6=0


p4=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: 
 q28, p12, p5

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. 
 
{q28,

 p5,

 g0021*q28 - 3*g0031*q28 - 3*g0059*p12 - 3*g0066*p5,

 g0127*p12 + g0134*p5}

Relevance for the application:



The equation: 



       4
f =(Df) *f*p5 + f  *f *f*p12
 t               2x  x


The symmetry: 

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


And now in machine readable form: 

The system: 

df(f(1),t)=d(1,f(1))**4*f(1)*p5 + df(f(1),x,2)*df(f(1),x)*f(1)*p12$


The symmetry: 

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