Solution 1 to problem N1t10s14f2

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t10s14f2

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


q5= - q1


q4=0


q3=0


q2=0


p6=0


       5
p5= - ---*p3
       4


p4=0


       1
p2= - ---*p3
       4


       1
p1= - ---*p3
       4

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:

 q1, 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.

 
{p5,

 p3,

 q1,

 p1,

 4*g0013*q1 - 4*g0017*q1 + 5*g0018*p3 - 4*g0020*p3 + g0021*p3 + g0022*p3,

 p2}

Relevance for the application:

The equation:


       1                  1       2                        5
f = - ---*Df  *Df*f*p3 - ---*(Df ) *f*p3 + Df *Df*f *p3 - ---*f  *f *f*p3
 t     4    2x            4     x            x     x       4   2x  x

The symmetry:

           3            4
f =Df *(Df) *f*q1 - (Df) *f *q1
 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))*f(1)*p3 - 1/4*d(1,df(f(1),x))**2*f
(1)*p3 + d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*p3 - 5/4*df(f(1),x,2)*df(f(1),x)*f
(1)*p3$

The symmetry:

df(f(1),s)=d(1,df(f(1),x))*d(1,f(1))**3*f(1)*q1 - d(1,f(1))**4*df(f(1),x)*q1$