Solution 2 to problem N1t8s14b2

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s14b2

Expressions

The solution is given through the following expressions: 

q39=0


q38=0


q37=0


q36=0


q35=0


q34=0


q33=0


q32=0


      1           1
     ---*p1*q2 + ---*p5*q2
      6           3
q31=-----------------------
              p1


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


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


q1=0


p9=0


p8=0


p7=0


p6=0


p4=0


p3=0


p2=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: 
 q2, p1, 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. 
 
{g0032*p1 + 2*g0032*p5 + 6*g0061*p1,q2,g0067*p5 + g0071*p1,g0020*q2 + g0023*p1,

 p1}

Relevance for the application:



The equation: 



                   2
b =Db *Db*b*p1 + b  *b*p5
 t   x            x


The symmetry: 

               2          1    3  2          1    3  2
    Db *Db*b *b *p1*q2 + ---*b  *b *p1*q2 + ---*b  *b *p5*q2
      x     x             6   x              3   x
b =----------------------------------------------------------
 s                             p1


And now in machine readable form: 

The system: 

df(b(1),t)=d(1,df(b(1),x))*d(1,b(1))*b(1)*p1 + df(b(1),x)**2*b(1)*p5$


The symmetry: 

df(b(1),s)=(d(1,df(b(1),x))*d(1,b(1))*df(b(1),x)*b(1)**2*p1*q2 + 1/6*df(b(1),x)
**3*b(1)**2*p1*q2 + 1/3*df(b(1),x)**3*b(1)**2*p5*q2)/p1$