Solution 1 to problem N1t13s14f2

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t13s14f2

Expressions

The solution is given through the following expressions:


q14=0


q13=0


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q4=0


q3=0


q2=0


q1= - q5


p11=0


p10=0


p9=0


p8=0


p7=0


       5
p6= - ---*p5
       4


p4=0


       1
p3= - ---*p5
       2


       1
p2= - ---*p5
       4


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:

 q5, 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.

 
{p2,q5,p5}

Relevance for the application:

The equation:


       1           2         1       2                   2
f = - ---*Df  *(Df) *f*p5 - ---*(Df ) *Df*f*p5 + Df *(Df) *f *p5
 t     4    2x               2     x               x        x

    5
 - ---*Df*f  *f *f*p5
    4      2x  x

The symmetry:

              3            4
f = - Df *(Df) *f*q5 + (Df) *f *q5
 s      x                     x

And now in machine readable form:

The system:

df(f(1),t)= - 1/4*d(1,df(f(1),x,2))*d(1,f(1))**2*f(1)*p5 - 1/2*d(1,df(f(1),x))**
2*d(1,f(1))*f(1)*p5 + d(1,df(f(1),x))*d(1,f(1))**2*df(f(1),x)*p5 - 5/4*d(1,f(1))
*df(f(1),x,2)*df(f(1),x)*f(1)*p5$

The symmetry:

df(f(1),s)= - d(1,df(f(1),x))*d(1,f(1))**3*f(1)*q5 + d(1,f(1))**4*df(f(1),x)*q5$