Solution 1 to problem N1t6s8f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s8f1

Expressions

The solution is given through the following expressions: 

       3
     p5 *q6
q13=--------
         3
     4*p4


q12=0


       2
     p5 *q6
q11=--------
        2
      p4


     3*p5*q6
q10=---------
      2*p4


        2
    3*p5 *q6
q9=----------
         2
     2*p4


    3*p5*q6
q7=---------
      p4


q5=0


q4=0


q3=0


q2=0


q1=0


p6=p5


p3=0


p2=0


p1=0


      2
    p5
p7=------
    3*p4


      2
    p5 *q6
q8=--------
       2
     p4

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: 
 q6, p4, 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. 
 
{p4}

Relevance for the application:



The equation: 



                       2      2                   1        2
    Df *f *p4*p5 + (Df) *f *p4  + Df*f  *p4*p5 + ---*f  *p5
      x  x                x           2x          3   3x
f =----------------------------------------------------------
 t                             p4


The symmetry: 

                 2                    2          3               2
f =(Df  *f *p4*p5 *q6 + 3*Df *Df*f *p4 *p5*q6 + ---*Df *f  *p4*p5 *q6
 s    2x  x                 x     x              2    x  2x

           3      3       3      2       2                     2
     + (Df) *f *p4 *q6 + ---*(Df) *f  *p4 *p5*q6 + Df*f  *p4*p5 *q6
              x           2         2x                 3x

        1        3       3
     + ---*f  *p5 *q6)/p4
        4   4x


And now in machine readable form: 

The system: 

df(f(1),t)=(d(1,df(f(1),x))*df(f(1),x)*p4*p5 + d(1,f(1))**2*df(f(1),x)*p4**2 + d
(1,f(1))*df(f(1),x,2)*p4*p5 + 1/3*df(f(1),x,3)*p5**2)/p4$


The symmetry: 

df(f(1),s)=(d(1,df(f(1),x,2))*df(f(1),x)*p4*p5**2*q6 + 3*d(1,df(f(1),x))*d(1,f(1
))*df(f(1),x)*p4**2*p5*q6 + 3/2*d(1,df(f(1),x))*df(f(1),x,2)*p4*p5**2*q6 + d(1,f
(1))**3*df(f(1),x)*p4**3*q6 + 3/2*d(1,f(1))**2*df(f(1),x,2)*p4**2*p5*q6 + d(1,f(
1))*df(f(1),x,3)*p4*p5**2*q6 + 1/4*df(f(1),x,4)*p5**3*q6)/p4**3$