Solution 4 to problem N1t6s14f3


Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s14f3

Expressions

The solution is given through the following expressions:

q16=0


q15=0


q14=0


q13=0


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


     1
    ---*p2*q4
     3
q5=-----------
       p1


q3=0


q2=0


q1=0


p3=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:
 q4, p2, p1

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.
 
{g0033*p2 + 3*g0034*p1,

 q4,

                                                         2
 g0015*p2*q4 + 3*g0016*p1*q4 + 3*g0020*p1*p2 + 3*g0021*p1 ,

 g0039*p2 + g0040*p1,

 p1}


Relevance for the application:



The equation: 


f =Df *f*p1 + Df*f *p2
 t   x            x
The symmetry:
            2            1      3
    Df *(Df) *f*p1*q4 + ---*(Df) *f *p2*q4
      x                  3         x
f =----------------------------------------
 s                    p1
And now in machine readable form:

The system:

df(f(1),t)=d(1,df(f(1),x))*f(1)*p1 + d(1,f(1))*df(f(1),x)*p2$
The symmetry:
df(f(1),s)=(d(1,df(f(1),x))*d(1,f(1))**2*f(1)*p1*q4 + 1/3*d(1,f(1))**3*df(f(1),x
)*p2*q4)/p1$