The map T associated to System (1) is given by
The Jacobian matrix of the map T has the form:
(28)
The value of the Jacobian matrix of T at the equilibrium point is
(29)
The determinant of (29) is given by
The trace of (29) is
The characteristic equation has the form
Theorem 9 Assume that A
1 > β
1. Then there exists the equilibrium point E
1
and:
-
(i)
E 1 is locally asymptotically stable if .
-
(ii)
E 1 is a saddle point if . The eigenvalues are
The corresponding eigenvectors, respectively, are
-
(iii)
E 1 is non-hyperbolic if The eigenvalues are , λ 2 = 1. The corresponding eigenvectors are and (1, 0).
Proof. Evaluating Jacobian (29) at the equilibrium point E
1(α
1/(A
1 - β
1), 0),
(30)
The determinant of (30) is given by
The trace of (30) is
The characteristic equation associated to System (1) at E
1 has the form
(31)
From Equation 31 we have
-
(i)
If A 1 > β 1 and then λ 1 < 1 and λ 2 < 1. Hence, E 1 is a sink.
-
(ii)
If A 1 > β 1 and . Then λ 1 < 1, and λ 2 < 1. Hence, E 1 is a saddle.
-
(iii)
If A 1 > β 1 and . Then, using Equation 31, we have that λ 1 < 1 and λ 2 < 1.
From (30) we obtain the eigenvectors that correspond to these eigenvalues. □
We now perform a similar analysis for the other cases in table.
Theorem 10
Assume
Then E
1, E
2, E
3
exist and:
-
(i)
Equilibrium E 1 is locally asymptotically stable.
-
(ii)
Equilibrium E
3
is a saddle point. The eigenvalues are
and
and |λ
1| < 1, |λ
2| > 1, where
The corresponding eigenvectors, respectively, are
-
(iii)
Equilibrium E 2 is locally asymptotically stable.
Proof. By Theorem 9 (i) holds.
Equilibrium E
3 is a saddle if and only if the following conditions are satisfied
The first condition is equivalent to
which is equivalent to
This is equivalent to
We have to prove that . Notice that
Now,
is equivalent to . This implies which is true. Condition
is equivalent to
which is clearly satisfied. Hence, E
3 is a saddle.
Now, we prove that E
2 is locally asymptotically stable. Notice that
implies which is true.
The second condition is equivalent to
This implies the following
Now, using Equation 5, we obtain
which is true, since the left side is always negative, while the right side is always positive.
Theorem 11
Assume
Then E
1(α
1/(A
1 - β
1), 0) and
exist and
-
(i)
Equilibrium E 1 is locally asymptotically stable.
-
(ii)
Equilibrium E
2
is non-hyperbolic. The eigenvalues are
The corresponding eigenvectors are
Proof. By Theorem 9, E
1 is locally asymptotically stable.
Now, we prove that E
2 is non-hyperbolic.
Evaluating Jacobian (29) at the equilibrium point ,
(32)
The eigenvalues of (32) are
Notice that |λ
2| < 1. Hence, E
2 is non-hyperbolic.
Theorem 12
Assume
Then there exists a unique equilibrium E
1 (α
1/(A
1 - β),0) which is locally asymptotically stable.
Proof. Observe that the assumption of Theorem 12 implies that the y coordinates of the equilibrium E
2 and E
3 are less then zero. By Theorem 9 E
1 is locally asymptotically stable.
Theorem 13
Assume
Then then there exist two equilibrium points E
1
and E
2. E
1
is a saddle point. The eigenvalues are
The corresponding eigenvectors, respectively, are
The equilibrium E
2
is locally asymptotically stable.
Proof. By Theorem 9 (ii), E
1 is a saddle point.
Now, we check the sign of coordinates of the equilibrium point E
2. We have that , since all parameters are positive. Consider Since
we have that (γ
2 - A
2 + A
1 - β
1)2 - 4α
1
B
2 > 0.
This implies
(33)
From Equation 33, we see that inequality is always true if A
1 - β
1 < γ
2 - A
2. If A
1 - β
1 > γ
2 - A
2, then
which is true, since . So, in both cases and
Notice, that Now, we check the sign of Assume that Then, we have
This is a contradiction with the assumption of theorem and so E
3 is not in considered domain.
By Theorem 10, E
2 is a locally asymptotically stable.
Theorem 14
Assume
Then there exist two equilibrium points and and E
1
≡ E
3
is non-hyperbolic. The eigenvalues are , λ
2 = 1. The corresponding eigenvectors are and (1. 0) The equilibrium point E
2
is locally asymptotically stable.
Proof. By Theorem 10, E
2 is locally asymptotically stable. By Theorem 9 (iii), E
1 is non-hyperbolic.
Now, we consider the special case of System (1) when A
1 = β
1.
In this case, System (1) becomes
(34)
The map T associated to System (34) is given by
The Jacobian matrix of the map T has the form: