Solution 7 to problem N1t6s14f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s14f1

Expressions

The solution is given through the following expressions: 

q55=0


q54=0


q53=0


q52=0


q51=0


q50=0


q49=0


q48=0


q47=0


q46=0


q45=0


q44=0


q43=0


q42=0


q41=0


q40=0


q39=0


q38=0


q37=0


q36=0


q35=0


q34=0


q33=0


q32=0


q31=0


q30=0


q29=0


q28=0


q27=0


q26=0


q25=0


q24=0


q23=0


q22=0


q21=0


q20=0


         2
     3*p4 *q19
q18=-----------
          2
        p5


     3*p4*q19
q17=----------
        p5


       3
     p4 *q19
q16=---------
         3
       p5


q15=0


q14=0


q13=0


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1=0


p7=0


p6=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: 
 q19, p4, 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. 
 
{q19,

 p5,

 g0122*p5 + g0123*p4,

         3                 2                       2               3
 g0040*p5 *q19 + 3*g0041*p4 *p5*q19 + 3*g0042*p4*p5 *q19 + g0043*p4 *q19

            4              3
  + g0060*p5  + g0061*p4*p5 ,

         3             2                   2           3
 g0101*p5  + 3*g0102*p4 *p5 + 3*g0103*p4*p5  + g0104*p4 }

Relevance for the application:



The equation: 



                   2
f =Df *f *p5 + (Df) *f *p4
 t   x  x             x


The symmetry: 

         3      3              2     2         2                 4      2
f =((Df ) *f *p5 *q19 + 3*(Df ) *(Df) *f *p4*p5 *q19 + 3*Df *(Df) *f *p4 *p5*q19
 s     x    x                x          x                  x        x

           6      3        3
     + (Df) *f *p4 *q19)/p5
              x


And now in machine readable form: 

The system: 

df(f(1),t)=d(1,df(f(1),x))*df(f(1),x)*p5 + d(1,f(1))**2*df(f(1),x)*p4$


The symmetry: 

df(f(1),s)=(d(1,df(f(1),x))**3*df(f(1),x)*p5**3*q19 + 3*d(1,df(f(1),x))**2*d(1,f
(1))**2*df(f(1),x)*p4*p5**2*q19 + 3*d(1,df(f(1),x))*d(1,f(1))**4*df(f(1),x)*p4**
2*p5*q19 + d(1,f(1))**6*df(f(1),x)*p4**3*q19)/p5**3$