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$