Solutions to Problem Set 5
ARE 261
November 4, 2001
Question 1
The Euler equation, using Calculus of Variations, is
¨
x
=0
(1)
If we integrate this equation twice and use the boundary conditions we get the
following solution
x
(
t
)=
t
+2
(2)
The necessary conditions using the Maximum Principle are
∂H
∂u
=
x
−
2
u
+
λ
=0
∂
2
H
∂u
2
=
−
2
≤
0
−
∂H
∂x
=
˙
λ
=
−
u
+2
x
−
λ
∂H
∂λ
=
˙
x
=
x
+
u
Now guess that the costate variable is linear in the state, or that,
λ
=
s
(
t
)
x
.
Di
f
erentiate this equation with respect to time. This yields
˙
λ
=
x
˙
s
+
s
˙
x
(3)
From the
f
rst necessary condition we know that
u
=
λ
+
x
2
. Use this to get rid
o
f
u
in the third and fourth necessary condition and then substitute these into
equation (3). This gives the following ODE in
s
2
˙
s
+6
s
+
s
2
−
3=0
(4)
Question 2
The
f
rst thing we are asked to
f
nd is the stock level that maximizes steady
state harvest. The equation for steady state level harvest is given by setting the
equation of motion for the stock of
f
sh to zero. Thus
h
ss
=
αβx
ss
−
α
(
x
ss
)
2
(5)
1
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View Full Documentwhere the superscript
ss
denotes steady state. To determine the maximum
steady state level of harvest di
f
erentiate equation 5 with respect to
x
ss
and set
the expression to zero. This implies that
h
m
=
αβ
2
4
x
m
=
β
2
Next we want to
f
nd the stock level that maximizes the steady state
F
ow of net
utility. Net utility at steady state is given by
U
(
h
ss
)
−
c
(
x
ss
)
h
ss
(6)
where
h
ss
is de
f
n
edbyequa
t
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 Fall '06
 KARP
 Thermodynamics, Equations, Steady State, Boundary value problem, Necessary and sufficient condition

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