Solution 1 to problem N1f0b1o35w2

Expressions | Parameters | Inequalities | Relevance | Back to problem N1f0b1o35w2

Expressions

The solution is given through the following expressions:


q15=0


      5
     ---*p5*q16
      3
q14=------------
         p6


q13=0


      10    2
     ----*p5 *q16
      9
q12=--------------
           2
         p6


q11=0


q10=0


     5
    ---*p5*q16
     3
q9=------------
        p6


q8=0


q7=0


q6=0


     10
    ----*p5*q16
     3
q5=-------------
        p6


q4=0


q3=0


     20    2
    ----*p5 *q16
     9
q2=--------------
          2
        p6


q1=0


p4=0


p3=p5


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:

 q16, p5, p6

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.

 
{p6,

 p5,

 q16,

 p3,

                                               2
 15*g0006*p6*q16 + 20*g0008*p5*q16 + 9*g0010*p6 ,

 g0027*p6 + g0028*p5 + g0030*p5,

           2                               2
 9*g0011*p6  + 15*g0013*p5*p6 + 10*g0015*p5  + 15*g0018*p5*p6 + 30*g0022*p5*p6

               2
  + 20*g0025*p5 }

Relevance for the application:

The equation:


                          2
b =Db *Db*p5 + b  *p6 + b  *p5
 t   x          3x       x

The symmetry:

     5                       20              2             2
b =(---*Db  *Db*p5*p6*q16 + ----*Db *Db*b *p5 *q16 + b  *p6 *q16
 s   3    3x                 9     x     x            5x

        10                      5     2              10    3   2        2
     + ----*b  *b *p5*p6*q16 + ---*b   *p5*p6*q16 + ----*b  *p5 *q16)/p6
        3    3x  x              3   2x               9    x

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,3)*p6 + df(b(1),x)**2*p5$

The symmetry:

df(b(1),s)=(5/3*d(1,df(b(1),x,3))*d(1,b(1))*p5*p6*q16 + 20/9*d(1,df(b(1),x))*d(1
,b(1))*df(b(1),x)*p5**2*q16 + df(b(1),x,5)*p6**2*q16 + 10/3*df(b(1),x,3)*df(b(1)
,x)*p5*p6*q16 + 5/3*df(b(1),x,2)**2*p5*p6*q16 + 10/9*df(b(1),x)**3*p5**2*q16)/p6
**2$