Solution 2 to problem N1t6s14b2

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s14b2

Expressions

The solution is given through the following expressions: 

q39=0


q38=0


q37=0


q36=0


q35=0


     1
q34=---*q3
     3


      1    3
     ---*p2 *q3
      3
q33=------------
          3
        p5


       2
     p2 *q3
q32=--------
        2
      p5


     p2*q3
q31=-------
      p5


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


    2*p2*q3
q2=---------
      p5


      2
    p2 *q3
q1=--------
       2
     p5


p6=0


p4=0


p3=p5


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: 
 q3, p2, 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. 
 
{p2,

 q3,

 p5,

 g0063*p5 + g0065*p5 + g0066*p2,

 p3,

         2                                 2              3
 g0019*p5 *q3 + 2*g0020*p2*p5*q3 + g0021*p2 *q3 + g0022*p5 ,

         3           3             2                   2             3
 g0028*p5  + g0029*p2  + 3*g0030*p2 *p5 + 3*g0031*p2*p5  + 3*g0059*p5

                 2             2
  + 6*g0060*p2*p5  + 3*g0061*p2 *p5}

Relevance for the application:



The equation: 



                 2          2
b =Db *Db*p5 + b  *p5 + b *b *p2
 t   x          x        x


The symmetry: 

            4   2                  2   3                   2      2
b =(Db *Db*b *p2 *p5*q3 + Db *Db*b  *p5 *q3 + 2*Db *Db*b *b *p2*p5 *q3
 s    x                     x     x               x     x

        1    4   3        3  2      2        2  4   2          1      6   3
     + ---*b  *p5 *q3 + b  *b *p2*p5 *q3 + b  *b *p2 *p5*q3 + ---*b *b *p2 *q3)/
        3   x            x                  x                  3   x

  3
p5


And now in machine readable form: 

The system: 

df(b(1),t)=d(1,df(b(1),x))*d(1,b(1))*p5 + df(b(1),x)**2*p5 + df(b(1),x)*b(1)**2*
p2$


The symmetry: 

df(b(1),s)=(d(1,df(b(1),x))*d(1,b(1))*b(1)**4*p2**2*p5*q3 + d(1,df(b(1),x))*d(1,
b(1))*df(b(1),x)**2*p5**3*q3 + 2*d(1,df(b(1),x))*d(1,b(1))*df(b(1),x)*b(1)**2*p2
*p5**2*q3 + 1/3*df(b(1),x)**4*p5**3*q3 + df(b(1),x)**3*b(1)**2*p2*p5**2*q3 + df(
b(1),x)**2*b(1)**4*p2**2*p5*q3 + 1/3*df(b(1),x)*b(1)**6*p2**3*q3)/p5**3$