Solution 1 to problem N1t2s11b1


Expressions | Parameters | Inequalities | Relevance | Back to problem N1t2s11b1

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:
 q25, q24, q23, q22, q21, q20, q19, q18, q17, 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*q25 + g0006*q24 + g0007*q8 + g0008*q7 + g0009*q6 + g0010*q5 + g0011*q4

  + g0012*q3 + g0013*q2 + g0014*q1,

 p2}


Relevance for the application:



The equation: 


b =b *p2
 t  x
The symmetry:
    12                              2                              4
b =b  *q23 + Db  *Db*q24 + Db  *Db*b *q8 + Db  *Db *q25 + Db  *Db*b *q6
 s             4x            3x              3x   x         2x

                               2              6
 + Db  *Db*b *b*q7 + Db  *Db *b *q5 + Db *Db*b *q1 + Db *Db*b  *b*q4
     2x     x          2x   x           x              x     2x

            2                 3                      3
 + Db *Db*b  *q3 + Db *Db*b *b *q2 + b  *b*q9 + b  *b *q11 + b  *b *q10
     x     x         x     x          5x         4x           4x  x

        5                             2          2  2            7
 + b  *b *q14 + b  *b  *q12 + b  *b *b *q13 + b   *b *q17 + b  *b *q18
    3x           3x  2x        3x  x           2x            2x

         2                 4         4         3  3         2  6           9
 + b  *b  *b*q16 + b  *b *b *q15 + b  *q22 + b  *b *q19 + b  *b *q20 + b *b *q21
    2x  x           2x  x           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)**12*q23 + d(1,df(b(1),x,4))*d(1,b(1))*q24 + d(1,df(b(1),x,3))*d(
1,b(1))*b(1)**2*q8 + d(1,df(b(1),x,3))*d(1,df(b(1),x))*q25 + d(1,df(b(1),x,2))*d
(1,b(1))*b(1)**4*q6 + d(1,df(b(1),x,2))*d(1,b(1))*df(b(1),x)*b(1)*q7 + d(1,df(b(
1),x,2))*d(1,df(b(1),x))*b(1)**2*q5 + d(1,df(b(1),x))*d(1,b(1))*b(1)**6*q1 + d(1
,df(b(1),x))*d(1,b(1))*df(b(1),x,2)*b(1)*q4 + d(1,df(b(1),x))*d(1,b(1))*df(b(1),
x)**2*q3 + d(1,df(b(1),x))*d(1,b(1))*df(b(1),x)*b(1)**3*q2 + df(b(1),x,5)*b(1)*
q9 + df(b(1),x,4)*b(1)**3*q11 + df(b(1),x,4)*df(b(1),x)*q10 + df(b(1),x,3)*b(1)
**5*q14 + df(b(1),x,3)*df(b(1),x,2)*q12 + df(b(1),x,3)*df(b(1),x)*b(1)**2*q13 + 
df(b(1),x,2)**2*b(1)**2*q17 + df(b(1),x,2)*b(1)**7*q18 + df(b(1),x,2)*df(b(1),x)
**2*b(1)*q16 + df(b(1),x,2)*df(b(1),x)*b(1)**4*q15 + df(b(1),x)**4*q22 + df(b(1)
,x)**3*b(1)**3*q19 + df(b(1),x)**2*b(1)**6*q20 + df(b(1),x)*b(1)**9*q21$