Solution 1 to problem N1t8s14b4

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s14b4

Expressions

The solution is given through the following expressions: 

q12=0


q11=0


q10=0


     1           1
    ---*p2*q1 + ---*p4*q1
     3           6
q9=-----------------------
             p4


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


p5=0


p3=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: 
 q1, p2, 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. 
 
{p4,g0008*q1 + g0009*p4,q1,2*g0013*p2 + g0013*p4 + 6*g0021*p4,g0023*p4

  + g0025*p2}

Relevance for the application:



The equation: 



                 2
b =Db *Db*p4 + b  *p2
 t   x          x


The symmetry: 

                       1    3          1    3
    Db *Db*b *p4*q1 + ---*b  *p2*q1 + ---*b  *p4*q1
      x     x          3   x           6   x
b =-------------------------------------------------
 s                        p4


And now in machine readable form: 

The system: 

df(b(1),t)=d(1,df(b(1),x))*d(1,b(1))*p4 + df(b(1),x)**2*p2$


The symmetry: 

df(b(1),s)=(d(1,df(b(1),x))*d(1,b(1))*df(b(1),x)*p4*q1 + 1/3*df(b(1),x)**3*p2*q1
 + 1/6*df(b(1),x)**3*p4*q1)/p4$