Solution 8 to problem N1t8s14f1


Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s14f1

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


     p7*q29
q28=--------
       p6


q27=0


q26=0


q25=0


q24=0


q23=0


q22=0


q21=0


q20=0


q19=0


     p7*q29
q18=--------
      p12


     p6*q29
q16=--------
     2*p12


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


p13=0


p11=0


p10=0


p9=0


p8=0


p5=0


p4=0


p3=0


p2=0


p1=0


       2
     p7 *q29
q17=----------
     2*p12*p6


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:
 q29, p12, p6, p7

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.
 
                                                           2           2
{2*g0097*p12*p6 + 2*g0098*p12*p7 + 2*g0108*p6*p7 + g0109*p7  + g0110*p6 ,

 g0127*p12 + g0132*p7 + g0133*p6}


Relevance for the application:



The equation: 


                      3
f =Df *Df*f *p7 + (Df) *f *p6 + f  *f *f*p12
 t   x     x             x       2x  x
The symmetry:
     1       2     2      2               4
f =(---*(Df ) *(Df) *f *p7 *q29 + Df *(Df) *f *p6*p7*q29
 s   2     x          x             x        x

                                     1      6      2
     + Df *Df*f  *f *f*p12*p7*q29 + ---*(Df) *f *p6 *q29
         x     2x  x                 2         x

           3
     + (Df) *f  *f *f*p12*p6*q29)/(p12*p6)
              2x  x
And now in machine readable form:

The system:

df(f(1),t)=d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*p7 + d(1,f(1))**3*df(f(1),x)*p6 
+ df(f(1),x,2)*df(f(1),x)*f(1)*p12$
The symmetry:
df(f(1),s)=(1/2*d(1,df(f(1),x))**2*d(1,f(1))**2*df(f(1),x)*p7**2*q29 + d(1,df(f(
1),x))*d(1,f(1))**4*df(f(1),x)*p6*p7*q29 + d(1,df(f(1),x))*d(1,f(1))*df(f(1),x,2
)*df(f(1),x)*f(1)*p12*p7*q29 + 1/2*d(1,f(1))**6*df(f(1),x)*p6**2*q29 + d(1,f(1))
**3*df(f(1),x,2)*df(f(1),x)*f(1)*p12*p6*q29)/(p12*p6)$