Solution 5 to problem N2f0b1o23w2

Expressions | Parameters | Inequalities | Relevance | Back to problem N2f0b1o23w2

Expressions

The solution is given through the following expressions: 

q13=0


q12=0


q11=0


q10=0


q9=0


       1
q8= - ---*i*q4
       2


q7=0


    1
q6=---*i*q4
    2


q5=0


q3=0


q2=0


q1=0


p5=i*p6


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: 
 q15, q14, q4, p6

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. 
 
{q4,

 p6,

 2*g0003*q14 - g0007*i*q4 + 2*g0009*q4 + 2*g0012*i*p6,

 p5,

 2*g0015*q14 - g0019*i*q4 + 2*g0021*q4 + 2*g0024*i*p6,

 2*g0044*q15 + 2*g0045*q14 - g0051*i*q4 + g0053*i*q4 + 2*g0055*q4}

Relevance for the application:



The equation: 



b =D D b *i*p6 + b  *p6
 t  1 2 x         2x


The symmetry: 

       1         2                                               1    2
b = - ---*(D D b) *i*q4 + D D b*b *q4 + D D b  *q14 + b  *q15 + ---*b  *i*q4
 s     2    1 2            1 2   x       1 2 2x        3x        2   x


And now in machine readable form: 

The system: 

df(b(1),t)=d(1,d(2,df(b(1),x)))*i*p6 + df(b(1),x,2)*p6$


The symmetry: 

df(b(1),s)= - 1/2*d(1,d(2,b(1)))**2*i*q4 + d(1,d(2,b(1)))*df(b(1),x)*q4 + d(1,d(
2,df(b(1),x,2)))*q14 + df(b(1),x,3)*q15 + 1/2*df(b(1),x)**2*i*q4$