Solution 1 to problem N1t2s10b2


Expressions | Parameters | Inequalities | Relevance | Back to problem N1t2s10b2

Expressions

The solution is given through the following expressions:

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:
 q16, q15, q14, q13, q12, q11, q10, q9, q8, q7, q6, q5, 
q4, q3, q2, q1, p2

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.
 
{g0005*q15 + g0006*q14 + g0007*q3 + g0008*q2 + g0009*q1,p2}


Relevance for the application:



The equation: 


b =b *p2
 t  x
The symmetry:
    6                                                           2
b =b *q13 + Db  *Db*q14 + Db  *Db*b*q3 + Db  *Db *q15 + Db *Db*b *q1
 s            3x            2x             2x   x         x

                                            2                     2
 + Db *Db*b *q2 + b  *q16 + b  *b*q4 + b  *b *q6 + b  *b *q5 + b   *q9
     x     x       5x        4x         3x          3x  x       2x

        3                      3         2  2           4
 + b  *b *q8 + b  *b *b*q7 + b  *q12 + b  *b *q10 + b *b *q11
    2x          2x  x         x         x            x
And now in machine readable form:

The system:

df(b(1),t)=df(b(1),x)*p2$
The symmetry:
df(b(1),s)=b(1)**6*q13 + d(1,df(b(1),x,3))*d(1,b(1))*q14 + d(1,df(b(1),x,2))*d(1
,b(1))*b(1)*q3 + d(1,df(b(1),x,2))*d(1,df(b(1),x))*q15 + d(1,df(b(1),x))*d(1,b(1
))*b(1)**2*q1 + d(1,df(b(1),x))*d(1,b(1))*df(b(1),x)*q2 + df(b(1),x,5)*q16 + df(
b(1),x,4)*b(1)*q4 + df(b(1),x,3)*b(1)**2*q6 + df(b(1),x,3)*df(b(1),x)*q5 + df(b(
1),x,2)**2*q9 + df(b(1),x,2)*b(1)**3*q8 + df(b(1),x,2)*df(b(1),x)*b(1)*q7 + df(b
(1),x)**3*q12 + df(b(1),x)**2*b(1)**2*q10 + df(b(1),x)*b(1)**4*q11$