Solution 1 to problem N1t2s12b1


Expressions | Parameters | Inequalities | Relevance | Back to problem N1t2s12b1

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:
 q32, q31, q30, q29, q28, q27, q26, 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*q14 + g0006*q13 + g0007*q12 + g0008*q11 + g0009*q10 + g0010*q9 + g0011*q8

  + g0012*q7 + g0013*q6 + g0014*q5 + g0015*q4 + g0016*q3 + g0017*q2 + g0018*q1,

 p2}


Relevance for the application:



The equation: 


b =b *p2
 t  x
The symmetry:
    13                                3
b =b  *q31 + Db  *Db*b*q14 + Db  *Db*b *q13 + Db  *Db*b *q12 + Db  *Db *b*q11
 s             4x              3x               3x     x         3x   x

            5                                    2                3
 + Db  *Db*b *q10 + Db  *Db*b  *q8 + Db  *Db*b *b *q9 + Db  *Db *b *q7
     2x               2x     2x        2x     x           2x   x

                            7                                  2
 + Db  *Db *b *q6 + Db *Db*b *q5 + Db *Db*b  *q1 + Db *Db*b  *b *q2
     2x   x  x        x              x     3x        x     2x

            2                   4                     2            4
 + Db *Db*b  *b*q4 + Db *Db*b *b *q3 + b  *q32 + b  *b *q15 + b  *b *q17
     x     x           x     x          6x        5x           4x

                       6                             2               3
 + b  *b *b*q16 + b  *b *q21 + b  *b  *b*q18 + b  *b  *q19 + b  *b *b *q20
    4x  x          3x           3x  2x          3x  x         3x  x

      2  3          2               8             2  2               5
 + b   *b *q25 + b   *b *q22 + b  *b *q26 + b  *b  *b *q24 + b  *b *b *q23
    2x            2x   x        2x           2x  x            2x  x

     4           3  4         2  7           10
 + b  *b*q27 + b  *b *q28 + b  *b *q29 + b *b  *q30
    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)**13*q31 + d(1,df(b(1),x,4))*d(1,b(1))*b(1)*q14 + d(1,df(b(1),x,3
))*d(1,b(1))*b(1)**3*q13 + d(1,df(b(1),x,3))*d(1,b(1))*df(b(1),x)*q12 + d(1,df(b
(1),x,3))*d(1,df(b(1),x))*b(1)*q11 + d(1,df(b(1),x,2))*d(1,b(1))*b(1)**5*q10 + d
(1,df(b(1),x,2))*d(1,b(1))*df(b(1),x,2)*q8 + d(1,df(b(1),x,2))*d(1,b(1))*df(b(1)
,x)*b(1)**2*q9 + d(1,df(b(1),x,2))*d(1,df(b(1),x))*b(1)**3*q7 + d(1,df(b(1),x,2)
)*d(1,df(b(1),x))*df(b(1),x)*q6 + d(1,df(b(1),x))*d(1,b(1))*b(1)**7*q5 + d(1,df(
b(1),x))*d(1,b(1))*df(b(1),x,3)*q1 + d(1,df(b(1),x))*d(1,b(1))*df(b(1),x,2)*b(1)
**2*q2 + 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)*b(1)**4*q3 + df(b(1),x,6)*q32 + df(b(1),x,5)*b(1)**2*q15 + df(b(
1),x,4)*b(1)**4*q17 + df(b(1),x,4)*df(b(1),x)*b(1)*q16 + df(b(1),x,3)*b(1)**6*
q21 + df(b(1),x,3)*df(b(1),x,2)*b(1)*q18 + df(b(1),x,3)*df(b(1),x)**2*q19 + df(b
(1),x,3)*df(b(1),x)*b(1)**3*q20 + df(b(1),x,2)**2*b(1)**3*q25 + df(b(1),x,2)**2*
df(b(1),x)*q22 + df(b(1),x,2)*b(1)**8*q26 + df(b(1),x,2)*df(b(1),x)**2*b(1)**2*
q24 + df(b(1),x,2)*df(b(1),x)*b(1)**5*q23 + df(b(1),x)**4*b(1)*q27 + df(b(1),x)
**3*b(1)**4*q28 + df(b(1),x)**2*b(1)**7*q29 + df(b(1),x)*b(1)**10*q30$