Solution 2 to problem N1f0b1o35w1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1f0b1o35w1

Expressions

The solution is given through the following expressions:


q19=0


q18=0


q17=0


q16=0


      1           1
     ---*p1*q2 + ---*p3*q2
      6           3
q15=-----------------------
              p1


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


p6=0


p5=0


p4=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, 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.

 
{g0017*p1 + 2*g0017*p3 + 6*g0030*p1,q2,g0010*q2 + g0012*p1,g0035*p3 + g0037*p1,

 p1}

Relevance for the application:

The equation:


                   2
b =Db *Db*b*p1 + b  *b*p3
 t   x            x

The symmetry:

               2          1    3  2          1    3  2
    Db *Db*b *b *p1*q2 + ---*b  *b *p1*q2 + ---*b  *b *p3*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)*p3$

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*p3*q2)/p1$