Solution 1 to problem N1t4s6f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t4s6f1

Expressions

The solution is given through the following expressions:


     2
    ---*p4*q6
     3
q7=-----------
       p3


q5=q6


     1
    ---*p3*q6
     2
q4=-----------
       p4


q3=0


q2=0


q1=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:

 q6, p4, p3

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.

 
{q6,

 p4,

 p3,

 g0021*p4 + g0022*p3,

 2*g0005*p4*q6 + 2*g0006*p4*q6 + g0007*p3*q6 + 2*g0011*p3*p4,

           2                                             2
 4*g0014*p4  + 6*g0015*p3*p4 + 6*g0016*p3*p4 + 3*g0017*p3 }

Relevance for the application:

The equation:


f =Df*f *p3 + f  *p4
 t     x       2x

The symmetry:

                       1      2      2                         2        2
    Df *f *p3*p4*q6 + ---*(Df) *f *p3 *q6 + Df*f  *p3*p4*q6 + ---*f  *p4 *q6
      x  x             2         x              2x             3   3x
f =--------------------------------------------------------------------------
 s                                   p3*p4

And now in machine readable form:

The system:

df(f(1),t)=d(1,f(1))*df(f(1),x)*p3 + df(f(1),x,2)*p4$

The symmetry:

df(f(1),s)=(d(1,df(f(1),x))*df(f(1),x)*p3*p4*q6 + 1/2*d(1,f(1))**2*df(f(1),x)*p3
**2*q6 + d(1,f(1))*df(f(1),x,2)*p3*p4*q6 + 2/3*df(f(1),x,3)*p4**2*q6)/(p3*p4)$