Solution 7 to problem N1t6s10f1

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

Expressions

The solution is given through the following expressions:


q21=0


q20=0


q19=0


q18=0


q17=0


q16=0


q15=0


q14=0


q13=0


q12=0


q11=0


    2*p4*q10
q9=----------
       p5


      2
    p4 *q10
q8=---------
        2
      p5


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p7=0


p6=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:

 q10, 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.

 
{q10,

 p5,

 g0054*p5 + g0055*p4,

         2                                   2               3              2
 g0015*p5 *q10 + 2*g0016*p4*p5*q10 + g0017*p4 *q10 + g0026*p5  + g0027*p4*p5 ,

         2                           2
 g0042*p5  + 2*g0043*p4*p5 + g0044*p4 }

Relevance for the application:

The equation:


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

The symmetry:

         2      2                 2                    4      2
    (Df ) *f *p5 *q10 + 2*Df *(Df) *f *p4*p5*q10 + (Df) *f *p4 *q10
       x    x               x        x                    x
f =-----------------------------------------------------------------
 s                                  2
                                  p5

And now in machine readable form:

The system:

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

The symmetry:

df(f(1),s)=(d(1,df(f(1),x))**2*df(f(1),x)*p5**2*q10 + 2*d(1,df(f(1),x))*d(1,f(1)
)**2*df(f(1),x)*p4*p5*q10 + d(1,f(1))**4*df(f(1),x)*p4**2*q10)/p5**2$