Unemployment cycles

Plug the expressions for ˙J, (42), and for J from free entry (41), into (40),
c
+ λ(Ω)γ
˙θ(Ω) (1 − η(θ(Ω)))u
m(θ(Ω)) u
(
c u + λ(Ω)γ
= −(py − w(Ω)) +
q(θ(Ω)) u
+ ˙u λ(Ω)γ
u 2 (
− λ(Ω)γ J
u
−
θc )
m(θ(Ω)) + J − ˙γ λ(Ω)
u
)
(r + δ + λ(Ω)m(θ(Ω)))
(
− θ(Ω)c
m(θ(Ω)) + J )
and solve for ˙θ, to obtain:
˙θ(Ω) =
m(θ(Ω))u
c(1 − η(θ(Ω)))(u + λ(Ω)γ)
So our dynamic system is given by:
×
[ ( λ
−
θc ) (
u m(θ(Ω)) + J − ˙u γ )
u + ˙γ − (py − w(Ω))
(
c u + λ(Ω)γ
+
− λ(Ω)γ )
]
J (r + δ + λ(Ω)m(θ(Ω)))
q(θ(Ω)) u u
˙u = δ(1 − u) − um(θ(Ω)) (43)
˙γ = um(θ(Ω)) − (δ + λ(Ω)m(θ(Ω)))γ (44)
m(θ(Ω))u
˙θ(Ω) =
× [ λ (
− θ(Ω)c ) (
c(1 − η(θ(Ω)))(u + λ(Ω)γ) u m(θ(Ω)) + J − ˙u γ )
u + ˙γ − (py − w(Ω))
(
c u + λ(Ω)γ
+
− λ(Ω)γ )
J (r + δ + λ(Ω)m(θ(Ω)))] (45)
q(θ(Ω)) u u
where w(0)and w(1) are given by (11).
To analyze the stability of system (43)-(45), we further have to specify the Jacobian matrix,
J ∗ (Ω) =
⎡
⎢
⎣
∂ ˙u ∗
∂u
∂ ˙γ ∗
∂u
∂ ˙θ ∗
∂u
where all partial derivatives are functions of Ω and are evaluated at the steady state under consideration
∂ ˙u ∗
∂γ
∂ ˙γ ∗
∂γ
∂ ˙θ ∗
∂γ
∂ ˙u ∗
∂θ
∂ ˙γ ∗
∂θ
∂ ˙θ ∗
∂θ
⎤
⎥
⎦