Solution 1 to problem N1t12s14f3


Expressions | Parameters | Inequalities | Relevance | Back to problem N1t12s14f3

Expressions

The solution is given through the following expressions:

q16=0


q15=0


q14=0


q13=0


q12=0


q11=0


q10=0


q9=0


q8=0


q7=0


q6=0


q5= - q4


q3=0


q2=0


q1=0


p12=0


p11=4*p2


p10=0


p9=0


p8=0


p7=0


p6= - 3*p2


p5=0


p4=0


p3=p2


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:
 q4, 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.
 
{p6,

 p2,

 q4,

 p3,

 g0015*q4 - g0016*q4 - 4*g0020*p2 + 3*g0025*p2 - g0028*p2 - g0029*p2,

 p11}


Relevance for the application:



The equation: 


                       2
f =Df  *Df*f*p2 + (Df ) *f*p2 - 3*Df *Df*f *p2 + 4*f  *f *f*p2
 t   2x              x              x     x         2x  x
The symmetry:
           2            3
f =Df *(Df) *f*q4 - (Df) *f *q4
 s   x                     x
And now in machine readable form:

The system:

df(f(1),t)=d(1,df(f(1),x,2))*d(1,f(1))*f(1)*p2 + d(1,df(f(1),x))**2*f(1)*p2 - 3*
d(1,df(f(1),x))*d(1,f(1))*df(f(1),x)*p2 + 4*df(f(1),x,2)*df(f(1),x)*f(1)*p2$
The symmetry:
df(f(1),s)=d(1,df(f(1),x))*d(1,f(1))**2*f(1)*q4 - d(1,f(1))**3*df(f(1),x)*q4$