Solution 3 to problem N1f1b0o57w3

Expressions | Parameters | Inequalities | Relevance | Back to problem N1f1b0o57w3

Expressions

The solution is given through the following expressions: 

         3
     3*p5 *q5
q16=----------
           3
      28*p3


q15=0


         2
     3*p5 *q5
q14=----------
          2
      4*p3


         2
     3*p5 *q5
q13=----------
          2
      2*p3


         2
     3*p5 *q5
q12=----------
          2
      2*p3


     3*p5*q5
q11=---------
      2*p3


     3*p5*q5
q10=---------
       p3


q8=0


    3*p5*q5
q7=---------
     2*p3


    3*p5*q5
q6=---------
     4*p3


q4=0


q3=0


q2=0


q1=0


p6=p5


p4=0


p2=0


p1=0


      2
    p5
p7=------
    5*p3


        2
    3*p5 *q5
q9=----------
         2
     4*p3

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: 
 q5, 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. 
 
{p3}

Relevance for the application:



The equation: 



                        2      2                   1        2
    Df *f  *p3*p5 + (Df) *f *p3  + Df*f  *p3*p5 + ---*f  *p5
      x  2x                x           3x          5   5x
f =-----------------------------------------------------------
 t                             p3


The symmetry: 

     3                2       3               2          3                2
f =(---*Df  *f  *p3*p5 *q5 + ---*Df  *Df*f *p3 *p5*q5 + ---*Df  *f  *p3*p5 *q5
 s   4    3x  2x              2    2x     x              2    2x  3x

        3       2      2                        2          3               2
     + ---*(Df ) *f *p3 *p5*q5 + 3*Df *Df*f  *p3 *p5*q5 + ---*Df *f  *p3*p5 *q5
        4     x    x                 x     2x              2    x  4x

           3      3       3      2       2          3              2
     + (Df) *f *p3 *q5 + ---*(Df) *f  *p3 *p5*q5 + ---*Df*f  *p3*p5 *q5
              x           2         3x              4      5x

        3         3       3
     + ----*f  *p5 *q5)/p3
        28   7x


And now in machine readable form: 

The system: 

df(f(1),t)=(d(1,df(f(1),x))*df(f(1),x,2)*p3*p5 + d(1,f(1))**2*df(f(1),x)*p3**2 +
 d(1,f(1))*df(f(1),x,3)*p3*p5 + 1/5*df(f(1),x,5)*p5**2)/p3$


The symmetry: 

df(f(1),s)=(3/4*d(1,df(f(1),x,3))*df(f(1),x,2)*p3*p5**2*q5 + 3/2*d(1,df(f(1),x,2
))*d(1,f(1))*df(f(1),x)*p3**2*p5*q5 + 3/2*d(1,df(f(1),x,2))*df(f(1),x,3)*p3*p5**
2*q5 + 3/4*d(1,df(f(1),x))**2*df(f(1),x)*p3**2*p5*q5 + 3*d(1,df(f(1),x))*d(1,f(1
))*df(f(1),x,2)*p3**2*p5*q5 + 3/2*d(1,df(f(1),x))*df(f(1),x,4)*p3*p5**2*q5 + d(1
,f(1))**3*df(f(1),x)*p3**3*q5 + 3/2*d(1,f(1))**2*df(f(1),x,3)*p3**2*p5*q5 + 3/4*
d(1,f(1))*df(f(1),x,5)*p3*p5**2*q5 + 3/28*df(f(1),x,7)*p5**3*q5)/p3**3$