Solution 6 to problem N1t6s10f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s10f1

Expressions

The solution is given through the following expressions:


q20=0


q19=0


q18=0


      5
     ---*p4*q21
      3
q17=------------
         p7


q16=0


      10
     ----*p4*q21
      3
q15=-------------
         p7


q14=0


q13=0


q12=0


      10
     ----*p4*q21
      3
q11=-------------
         p7


      5
     ---*p4*q21
      3
q10=------------
         p7


q9=0


     5    2
    ---*p4 *q21
     6
q8=-------------
          2
        p7


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p6=0


p5=0


p3=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:

 q21, p4, p7

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.

 
{q21,

 p7,

 g0052*p7 + g0055*p4,

           2
 6*g0031*p7  + 10*g0035*p4*p7 + 20*g0037*p4*p7 + 20*g0041*p4*p7 + 10*g0042*p4*p7

              2
  + 5*g0044*p4 ,

 p4,

 10*g0008*p7*q21 + 20*g0010*p7*q21 + 20*g0014*p7*q21 + 10*g0015*p7*q21

                               2
  + 5*g0017*p4*q21 + 6*g0027*p7 }

Relevance for the application:

The equation:


       2
f =(Df) *f *p4 + f  *p7
 t        x       3x

The symmetry:

     10                          5       2
f =(----*Df  *Df*f *p4*p7*q21 + ---*(Df ) *f *p4*p7*q21
 s   3     2x     x              3     x    x

        10                          5      4      2
     + ----*Df *Df*f  *p4*p7*q21 + ---*(Df) *f *p4 *q21
        3     x     2x              6         x

        5      2                       2        2
     + ---*(Df) *f  *p4*p7*q21 + f  *p7 *q21)/p7
        3         3x              5x

And now in machine readable form:

The system:

df(f(1),t)=d(1,f(1))**2*df(f(1),x)*p4 + df(f(1),x,3)*p7$

The symmetry:

df(f(1),s)=(10/3*d(1,df(f(1),x,2))*d(1,f(1))*df(f(1),x)*p4*p7*q21 + 5/3*d(1,df(f
(1),x))**2*df(f(1),x)*p4*p7*q21 + 10/3*d(1,df(f(1),x))*d(1,f(1))*df(f(1),x,2)*p4
*p7*q21 + 5/6*d(1,f(1))**4*df(f(1),x)*p4**2*q21 + 5/3*d(1,f(1))**2*df(f(1),x,3)*
p4*p7*q21 + df(f(1),x,5)*p7**2*q21)/p7**2$