Solution 4 to problem N1t6s14b2

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s14b2

Expressions

The solution is given through the following expressions: 

q39=0


q38=0


q37=0


q36=0


q35=0


     1
q34=---*q3
     6


q33=0


q32=0


q31=0


q30=0


q29=0


q28=0


q27=0


        1
q26= - ---*q5
        8


q25=0


q24=0


q23=0


     1
q22=---*q5
     4


q21=0


q20=0


q19=0


q18=0


q17=0


q16=0


q15=0


q14=0


     1
q13=---*q5
     4


q12=0


q11=0


q10=0


q9=0


       1
q8= - ---*q5
       2


    1
q7=---*q5
    4


q6=0


q4=0


q2=0


q1=0


p6=0


p4=0


p3=0


p2=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: 
 q5, q3, p5

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. 
 
{p5,

 g0009*q5 - 2*g0014*q5 + g0015*q5 + 4*g0017*q5 + 4*g0019*q3 + 4*g0022*p5,

 4*g0028*q3 - 3*g0036*q5 + 6*g0040*q5 + 6*g0049*q5 - 12*g0054*q5 + 6*g0055*q5

  + 24*g0057*q5 + 24*g0059*q3}

Relevance for the application:



The equation: 



b =Db *Db*p5
 t   x


The symmetry: 

    1                   1                    1
b =---*Db  *Db*b *q5 - ---*Db  *Db*b  *q5 + ---*Db  *Db *b *q5 + Db *Db*b  *q5
 s  4    3x     x       2    2x     2x       4    2x   x  x        x     3x

            2       1        2       1     2          1    4
 + Db *Db*b  *q3 + ---*b  *b  *q5 - ---*b   *b *q5 + ---*b  *q3
     x     x        4   3x  x        8   2x   x       6   x


And now in machine readable form: 

The system: 

df(b(1),t)=d(1,df(b(1),x))*d(1,b(1))*p5$


The symmetry: 

df(b(1),s)=1/4*d(1,df(b(1),x,3))*d(1,b(1))*df(b(1),x)*q5 - 1/2*d(1,df(b(1),x,2))
*d(1,b(1))*df(b(1),x,2)*q5 + 1/4*d(1,df(b(1),x,2))*d(1,df(b(1),x))*df(b(1),x)*q5
 + d(1,df(b(1),x))*d(1,b(1))*df(b(1),x,3)*q5 + d(1,df(b(1),x))*d(1,b(1))*df(b(1)
,x)**2*q3 + 1/4*df(b(1),x,3)*df(b(1),x)**2*q5 - 1/8*df(b(1),x,2)**2*df(b(1),x)*
q5 + 1/6*df(b(1),x)**4*q3$