Solution 1 to problem N1t8s14b4
Expressions |
Parameters |
Inequalities |
Relevance |
Back to problem N1t8s14b4
Expressions
The solution is given through the following expressions:
q12=0
q11=0
q10=0
1 1
---*p2*q1 + ---*p4*q1
3 6
q9=-----------------------
p4
q8=0
q7=0
q6=0
q5=0
q4=0
q3=0
q2=0
p5=0
p3=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:
q1, p2, p4
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.
{p4,g0008*q1 + g0009*p4,q1,2*g0013*p2 + g0013*p4 + 6*g0021*p4,g0023*p4
+ g0025*p2}
Relevance for the application:
The equation:
2
b =Db *Db*p4 + b *p2
t x x
The symmetry:
1 3 1 3
Db *Db*b *p4*q1 + ---*b *p2*q1 + ---*b *p4*q1
x x 3 x 6 x
b =-------------------------------------------------
s p4
And now in machine readable form:
The system:
df(b(1),t)=d(1,df(b(1),x))*d(1,b(1))*p4 + df(b(1),x)**2*p2$
The symmetry:
df(b(1),s)=(d(1,df(b(1),x))*d(1,b(1))*df(b(1),x)*p4*q1 + 1/3*df(b(1),x)**3*p2*q1
+ 1/6*df(b(1),x)**3*p4*q1)/p4$