We …nd the equilbrium of the system, by solving :
 
ppln = 0 (3.3)q
 2
bp  + dp3 q = 0: (3.4)
From the fact that  
ppln = 0, p 6= 0 and  6= 0; (3.5)q
introducing
 p
ln = 0 , p = q (3.6)q
and
 p
f1(p;q) = pln (3.7)q
 2
f2(p;q) = bp  + dp3 q (3.8)
2 bpClearly, if u = 0 , and q_ = 0 ) (bp ( + d(p
3))q) = 0, p = ( q
 3
2d ) : From the fact that
p = q, we get the following values:
b 3p
eq = qeq = ( )2: (3.9)d
We see that the jacobian matrix J of the system is the following:
2 3
@f1(p;q) @f1(p;q)
J = 6 @p @q 7 (3.10)4 5
@f2(p;q) @f2(p;q)
@p @q2 3
  ln p p= 6 q q 7 (3.11)
4 52
b 2 pd3 3 pq dp3
10
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