Solution 3 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           1
     ---*p3*q3 + ---*p5*q3
      6           6
q34=-----------------------
              p5


q33=0


q32=0


q31=0


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


q2=0


q1=0


p6=0


p4=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: 
 q3, p3, 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. 
 
{q3,p3,p5,g0063*p5 + g0065*p3,g0019*q3 + g0022*p5,g0028*p3 + g0028*p5

  + 6*g0059*p5}

Relevance for the application:



The equation: 



                 2
b =Db *Db*p5 + b  *p3
 t   x          x


The symmetry: 

             2          1    4          1    4
    Db *Db*b  *p5*q3 + ---*b  *p3*q3 + ---*b  *p5*q3
      x     x           6   x           6   x
b =--------------------------------------------------
 s                         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*p3$


The symmetry: 

df(b(1),s)=(d(1,df(b(1),x))*d(1,b(1))*df(b(1),x)**2*p5*q3 + 1/6*df(b(1),x)**4*p3
*q3 + 1/6*df(b(1),x)**4*p5*q3)/p5$