Supplementary material related to
“Characterizing the sustainability problem
in an exhaustible resource model”
http://dx.doi.org/10.1016/j.jet.2013.04.019
Tapan Mitraa
Geir B. Asheimb
Wolfgang Buchholzc
Cees Withagend
March 11, 2013
a
Department of Economics, 418 Uris Hall, Cornell University, Ithaca, New York 14853, USA
(email: tm19@cornell.edu)
b
Department of Economics, University of Oslo, P.O. Box 1095 Blindern, 0317 Oslo, Norway
(email: g.b.asheim@econ.uio.no) Corresponding author (tel: 47 45505136)
c
Department of Economics, University of Regensburg, 93040 Regensburg, Germany. (email:
wolfgang.buchholz@wiwi.uni-regensburg.de)
d
Department of Economics, VU University Amsterdam, De Boelelaan 1105, 1081 HV Amster-
dam, The Netherlands (email: cwithagen@feweb.vu.nl)
Online appendix A: Proofs of Lemmas
Proof of Lemma 1. We break up the proof of Lemma 1 into several steps.
Step 1: There exists a solution to (4). Since (c, k) ∈ D, there is some r0 > 0 such
that F (k, r0 ) > c. Since F (k, r) is continuous and increasing in r with F (k, 0) − c < 0 and
F (k, r0 ) − c > 0, there is a unique r0 ∈ (0, r0 ) such that F (k, r0 ) = c. Define:
r
r0
0
0
R ≡ r>r :
≤
.
F (k, r) − c
F (k, r0 ) − c
Then r0 ∈ R0 , so that R0 is non-empty. By the definition of r0 and the continuity of F ,
lim
r↓r 0
F (k, r0 ) − c
F (k, r) − c
=0<
,
r
r0
and there is r > r0 such that r ∈
/ R0 for r ∈ (r0 , r). It follows from F (k, 0) = 0 and the
concavity of F that F (k, r) ≤ F (k/r, 1)r if r > 1. By F (0, r) = 0 and the continuity of F ,
lim
r→∞
F (k, r)
F (k, r) − c
≤ lim
≤ lim F
r→∞
r→∞
r
r
k
r,1
=0<
F (k, r0 ) − c
,
r0
and there is r̄ > 0 such that r ∈
/ R0 for r ∈ (r̄, ∞). Hence, R0 ⊂ [r , r̄], so that R0
is bounded. Since r/(F (k, r) − c) is a continuous function of r on [r, r̄], the set R0 is
closed. As r/(F (k, r) − c) is a continuous function of r on R0 , there exists a solution to
minr∈R0 r/(F (k, r) − c). By the definition of r, this is also a solution to (4).
Step 2: There is at most one solution to (HaR). Define:
V (r) ≡ F (k, r) − c − F2 (k, r)r
for all r > 0
(A1)
Note that V is a C 1 function on R++ .
Suppose r0 and r00 were both solutions to (HaR), with 0 < r0 < r00 . Then V (r0 ) =
V (r00 ) = 0. We can find 0 < a < r0 and b > r00 , and define
Z r
U (r) ≡
V (s)ds for r ∈ [a, b] .
a
Then, we have U 0 (r) = V (r) for all r ∈ (a, b).1 By using (A1), we can infer that U is a
C 2 function on (a, b), and for all r ∈ (a, b), U 00 (r) = V 0 (r) = −F22 (k, r)r > 0. Thus, U is a
strictly convex C 2 function on (a, b), and we get the contradiction:
0 = U 0 (r0 )(r00 − r0 ) < U (r00 ) − U (r0 ) < U 0 (r00 )(r00 − r0 ) = 0 .
Step 3: There is a unique solution to (4), and this uniquely solves (HaR). Since
r/(F (k, r) − c) is a continuously differentiable function of r on {r : F (k, r) > c}, any
solution r to (4) satisfies the first-order condition
F2 (k, r)r
1
= 0,
−
F (k, r) − c (F (k, r) − c)2
1
See Rudin [4, Theorem 6.20, p. 133].
1
and therefore is also a solution to (HaR). By Steps 1 and 2, there is a unique solution to (4).
This implies that p : D → R++ and r : D → R++ , as defined in the statement of Lemma 1,
are single-valued functions.
Step 4: The functions p : D → R++ and r : D → R++ are continuously differentiable
on D. Note that by continuity of F , the set D is open in R2 . Define Y ≡ D × R++ . Clearly,
Y is an open set in R3 , and we can define:
H(c, k, r) = F (k, r) − c − F2 (k, r)r
for (c, k, r) ∈ Y .
Then, H is continuously differentiable on Y , and by Step 3, we have:
H(c, k, r) = 0
for r = r(c, k) .
Furthermore, for r = r(c, k),
∂H(c, k, r)
= −F22 (k, r)r > 0 .
∂r
Thus, by the implicit function theorem,2 there is an open set N ⊂ X ≡ (0, c0 ) × (k 0 , ∞)
containing (c, k) and an open set M ⊂ Y , containing (c, k, r(c, k)), and a unique function
g : N → R++ , such that
(i) for all (c̃, k̃) ∈ N , we have H(c̃, k̃, g(c̃, k̃)) = 0, and
(ii) g(c, k) = r(c, k).
Furthermore, g is continuously differentiable on N . Since, by Step 3, we certainly have
H(c̃, k̃, r(c̃, k̃)) = 0 for all (c̃, k̃) ∈ N ⊂ D, we can infer that g(c̃, k̃) = r(c̃, k̃) for all
(c̃, k̃) ∈ N . Thus, r is continuously differentiable on N . Since (c, k) ∈ D was arbitrary, r
is continuously differentiable on D. As p(c, k) = 1/F2 (k, r(c, k)) for (k, r) ∈ R2++ and F is
twice continuously differentiable on R2++ , also p is continuously differentiable on D.
Step 5: p1 (c, k) and r1 (c, k) are positive. By definition of r on D, we have:
F (k, r(c, k)) − c − F2 (k, r(c, k))r(c, k) = 0
for all (c, k) ∈ D .
Thus, differentiating this equation w.r.t. c yields
F2 (k, r(c, k))r1 (c, k) − 1 − F22 (k, r(c, k))r1 (c, k)r(c, k) − F2 (k, r(c, k))r1 (c, k) = 0 ,
from which r1 (c, k) > 0 can be inferred. Since p(c, k) = 1/F2 (k, r(c, k)) and F22 (k, r) < 0
for (k, r) ∈ R2++ , we now obtain that p1 (c, k) > 0.
Proof of Lemma 2.
c
Since (c, k0 ) ∈ D implies that F (k0 , r(c, k0 )) − c > 0, we
c
have k̇ (0) > 0. Since k̇ is continuous, there is ε ∈ (0, T ) such that k̇ c (t) > 0 for all
t ∈ [0, ε], and so by the Mean Value theorem, k c (t) > k0 for all t ∈ (0, ε]. We claim
2
See Rudin [4, Theorem 9.28, p. 224–225].
2
that k c (t) ≥ k0 for all t ∈ [0, T ). If not, there is τ 0 ∈ (ε, T ) such that k c (τ 0 ) < k0 . Let
τ ≡ inf{t ∈ (ε, T ) : k c (t) < k0 }. Then, ε ≤ τ ≤ τ 0 < T , and k c (τ ) ≤ k0 by continuity of
k c . Also, by definition of τ , k c (t) ≥ k0 for all t ∈ [0, τ ). Thus, k c (τ ) ≥ k0 by continuity of
k c , and consequently k c (τ ) = k0 . Then, using the fact that k c (t) ≥ k c (τ ) for all t ∈ [0, τ ),
we must have k̇ c (τ ) ≤ 0. On the other hand, since k c (τ ) = k0 , we must have k̇ c (τ ) > 0 by
using (6). This contradiction establishes our claim.
Since (c, k0 ) ∈ D and k c (t) ≥ k0 for all t ∈ [0, t), it follows that (c, k c (t)) ∈ D for all
t ∈ [0, T ). Thus, by definition of r, we must have k̇ c (t) > 0 for all t ∈ [0, T ).
Proof of Lemma 3. For every T 0 ∈ (0, T ), r(c, k c (t)) is continuous on [0, T 0 ] and the
R T0
Riemann integral 0 r(c, k c (t))dt is well-defined. By the change of variable formula,3
Z
kc (T 0 )
Z
T0
p(c, x)dx =
k0
p(c, k c (t))k̇ c (t)dt .
0
Thus, (10) follows from (6) and the definition of the function p.
Assume now that (11) holds. Write rc (t) = r(c, k c (t)) for t ∈ [0, T ) and
λ = max{F (1, 1/k0 )/k0 , F (k0 , 1), F (k0 , 1)/k0 } .
For t ∈ [0, T ), either (i) rc (t) ≤ k c (t)/k0 , or (ii) rc (t) > k c (t)/k0 . Case (i) can be divided
into two subcases: (i)(a) k c (t) ≤ 1 and (i)(b) k c (t) > 1. In case (i)(a),
k̇ c (t)/k c (t) ≤ F (k c (t), rc (t))/k c (t) ≤ F (1, 1/k0 )/k0 ≤ λ .
(A2)
k̇ c (t)/k c (t) ≤ F (k c (t), rc (t))/k c (t) ≤ F (1, rc (t)/k c (t)) ≤ F (1, 1/k0 ) ≤ λ .
(A3)
In case (i)(b),
In case (ii), we have that rc (t) > k c (t)/k0 ≥ 1. So
k̇ c (t) ≤ F (k c (t), rc (t)) ≤ rc (t)F (k c (t)/rc (t), 1) ≤ rc (t)F (k0 , 1)
and
k̇ c (t)/k c (t) ≤ rc (t)F (k0 , 1)/k0 ≤ λrc (t) .
Let Λ = {t ∈ [0, T ) : rc (t) > k c (t)/k0 }. Then, by (A2)–(A4), we have:
Z
Z
Z
k̇ c (t)/k c (t) dt ≤
λrc (t)dt +
λdt ≤ λ(S + T ) .
[0,T )
Λ
[0,T )\Λ
Thus, for every t ∈ (0, T ), ln k c (t) ≤ ln k0 + λ(S + T ), showing that k c (T ) < ∞.
3
See Apostol [1] Theorem 7.36, p. 164.
3
(A4)
Proof of Lemma 4. By the definition of C(k0 , m0 ), there is a feasible path (c0 (t), k 0 (t),
0
r (t)) from (k0 , m0 ) with c0 (t) ≥ c0 for t ≥ 0. Construct (c(t), k(t), r(t)) as follows:
c(t) = F (k(t), r(t)) − k̇(t)
for all t ≥ 0
0
k(t) = (1 − λ)k0 + λk (t)
for all t ≥ 0
r(t) = λ[r0 (t) + (/et )]
for all t ≥ 0 ,
(A5)
where λ = c/c0 < 1 and = (1 − λ)m0 /λ > 0. We must show that (c(t), k(t), r(t)) is a
feasible interior path from (k0 , m0 ) with c(t) > c for t ≥ 0.
Clearly, k(t) is a differentiable function of t, with k̇(t) = λk̇ 0 (t) for t ≥ 0, and (c(t), r(t))
are continuous functions of t. Using (A5), for t ≥ 0,
c(t) = F (k(t), r(t)) − k̇(t) > F (λk 0 (t), λr0 (t)) − λk̇ 0 (t)
≥ λ[F (k 0 (t), r0 (t)) − k̇ 0 (t)] = λc0 (t) ≥ λc0 = c .
Also, k(t) ≥ (1 − λ)k0 > 0 for t ≥ 0, and (1) is satisfied since k(0) = (1 − λ)k0 + λk0 = k0 .
Again using (A5), r(t) > 0 for t ≥ 0. So the path (c(t), k(t), r(t)) is interior. Moreover,
Z t
Z t
r(τ )dτ ≤ λ
r0 (τ )dτ + λ < λm0 + (1 − λ)m0 = m0 .
0
0
Thus, the path (c(t), k(t), r(t)) is feasible.
Proof of Lemma 5.
Suppose on the contrary that there is k̄ ∈ (0, ∞) such that
k(t) ≤ k̄ for all t ≥ 0. We have that F (k̄, 0) = 0 and F (k̄, ·) is continuous, concave and
increasing on R+ . Using Jensen’s inequality, we have for all T > 0:
!
Z
Z
m 1 T
1 T
0
.
F k̄, r(t) dt ≤ F k̄,
r(t)dt ≤ F k̄,
T 0
T 0
T
Then we get:
Z
T
k(T ) − k0 =
Z
k̇(t)dt ≤
0
0
T
i
h m 0
−c .
F (k̄, r(t))dt − T c ≤ T F k̄,
T
Since limT →∞ F k̄, m0 /T = 0, this implies k(T ) < 0 for large T , contradicting that k(t) ≥
0 for t ≥ 0.
Proof of Lemma 6. Assume that (c(t), k(t), r(t)) is a feasible and interior path from
(k0 , m0 ).
Part 1: If (c(t), k(t), r(t)) satisfies (HoR), then it is competitive. Since F is twice continuously differentiable and k(t) and r(t) are differentiable, we may define p(t) by (P) and set
q1 (t) = p(t) and q2 (t) = 1 for all t ≥ 0. Note that, for each t ≥ 0, (c(t), k(t), m(t), k̇(t), ṁ(t)) ∈
Y . Furthermore, for all (c0 , k 0 , m0 , k̇ 0 , ṁ0 ) ∈ Y , we have by A2 and A3:
(c0 + k̇ 0 ) − (c(t) + k̇(t)) ≤ F (k 0 , −ṁ0 ) − F (k(t), −ṁ(t))
≤ F1 (k(t), r(t)) (k 0 − k(t)) + F2 (k(t), r(t)) (−ṁ0 + ṁ(t)) .
4
(A6)
Multiplying through (A6) by p(t) > 0, and using (HoR), (P) and the definitions of q1 (·) and
q2 (·), yields:
p(t) (c0 − c(t)) + q1 (t) k̇ 0 − k̇ c (t) ≤ −q̇1 (t) (k 0 − k(t)) + q2 (t) (−ṁ0 + ṁ(t)) .
(A7)
Transposing terms in (A7) and noting that q̇2 (t) = 0 for t ≥ 0, we obtain
p(t)c(t) + q1 (t)k̇(t) + q2 (t)ṁ(t) + q̇1 (t)k(t) + q̇2 (t)m(t)
≥ p(t)c0 + q1 (t)k̇ 0 + q2 (t)ṁ0 + q̇1 (t)k 0 + q̇2 (t)m0
for all (c0 , k 0 , m0 , k̇ 0 , ṁ0 ) ∈ Y and all t ≥ 0.
Part 2: If (c(t), k(t), r(t)) is competitive, then it satisfies (HoR). By the premise, there
are functions p(·) : [0, ∞) → R++ and (q1 (·), q2 (·)) : [0, ∞) → R2 , where p(t) is continuous
and (q1 (t), q2 (t)) are differentiable, such that, for all t ≥ 0, (c(t), k(t), m(t), k̇(t), ṁ(t)) maximizes instantaneous profits p(t)c0 + q1 (t)k̇ 0 + q2 (t)ṁ0 + q̇1 (t)k 0 + q̇2 (t)m0 over all quintuples
(c0 , k 0 , m0 , k̇ 0 , ṁ0 ) ∈ Y .
Since (c(t), k(t), r(t)) is interior, m(t) > 0 for all t ≥ 0, implying that q̇2 (t) = 0 for all
t ≥ 0, since otherwise there is some τ ≥ 0 such that instantaneous profit could be increased
at time τ by m0 6= m(τ ). Furthermore, q1 (t) ≥ p(t) > 0 for all t ≥ 0, since otherwise there
is some τ ≥ 0 such that instantaneous profit could be increased by k̇ 0 < k(τ ) and c0 > c(τ )
with c0 + k̇ 0 = c(τ )+k(τ ). Finally, q1 (t)F1 (k(t), r(t)) = −q̇1 (t) and q1 (t)F2 (k(t), r(t)) = q2 (t)
for all t ≥ 0, since otherwise there is some τ ≥ 0 such that instantaneous profit could be
increased at time τ by k 0 6= k(τ ) or ṁ0 6= ṁ(τ ).
Differentiating q1 (t)F2 (k(t), r(t)) = q2 (t) w.r.t. t and applying q̇2 (t) = 0 yields
q̇1 (t)F2 (k(t), r(t)) + q1 (t)Ḟ2 (k(t), r(t)) = 0 ,
which combined with q1 (t)F1 (k(t), r(t)) = −q̇1 (t) implies
q̇1 (t)
Ḟ2 (k(t), r(t))
=−
= F1 (k(t), r(t)) ,
F2 (k(t), r(t))
q1 (t)
thereby establishing that (c(t), k(t), r(t)) satisfies (HoR).
Proof of Lemma 7. Assume that (c(t), k(t), r(t)) is an interior and competitive path
from (k0 , m0 ) that satisfies (CVT) and is resource exhausting.
Suppose that (c0 (t), k 0 (t), r0 (t)) is another feasible path from (k0 , m0 ). By the proof
of Lemma 6, (c(t), k(t), r(t)) is competitive at prices defined by (P) and q1 (t) = p(t) and
q2 (t) = 1 for all t ≥ 0. We therefore obtain:
Z T
Z T
Z T
d
p(t) (c0 (t) − c(t)) dt ≤
(p(t) (k(t) − k 0 (t))) dt +
(ṁ(t) − ṁ0 (t)) dt
dt
0
0
0
Z T
0
0
= p(T ) (k(T ) − k (T )) +
(r (t) − r(t)) dt
0
Z ∞
≤ p(T )k(T ) + m0 −
r(t)dt ,
0
5
since p(T )k 0 (T ) ≥ 0 and
RT
0
r0 (t)dt ≤ m0 for all T ≥ 0. From the premise that (c(t), k(t), r(t))
satisfies (CVT) and is resource exhausting, it now follows that
Z T
lim sup
p(t) (c0 (t) − c(t)) dt ≤ 0 .
T →∞
0
Since p(t) > 0 for all t ≥ 0, this establishes that (c(t), k(t), r(t)) is efficient.
Online appendix B: An example without maximin existence
Consider the case where F is given by F (k, r) = k + r and (k0 , m0 ) = (1, 1).
Sustainability. We first claim that C(k0 , m0 ) is non-empty. To establish this, simply
define:
k(t) = 1, r(t) = 0, c(t) = 1
for all t ≥ 0 ,
and note that k̇(t) = 0 for t ≥ 0. Thus, (c(t), k(t), r(t)) is a path from (k0 , m0 ) = (1, 1), and
c(t) = 1 > 0 for all t ≥ 0. This establishes our claim.
An upper bound on sustainable consumption. We now claim that that there is
no path (c(t), k(t), r(t)) satisfying:
c(t) ≥ 2
for all t ≥ 0 .
(B1)
Suppose, there were such a path. We then establish the following steps.
Step 1: We must have k(t) < 2 for all t ≥ 0. For if k(t) ≥ 2 for some t ≥ 0, then we
can define T = inf{t ≥ 0 : k(t) ≥ 2}. By continuity of k(t), we must have k(T ) = 2. Since
k(0) = 1, we know that T > 0. Furthermore,
for all t ∈ [0, T ) .
(B2)
R T /2
Denote (2 − k(t)) by α(t) for all t ∈ [0, T ], and let σ ≡ 0 α(t)dt. Then, by (B2), σ > 0,
k(t) < 2
and using the fact that for all t ∈ [0, T ),
k̇(t) = k(t) + r(t) − c(t) ≤ k(t) + r(t) − 2 = r(t) − α(t) ,
we obtain for all τ ∈ [T /2, T ),
Z τ
Z τ
Z τ
Z τ
Z
k(τ ) − k(0) = k̇(t)dt = r(t)dt − α(t)dt ≤ r(t)dt −
0
0
0
0
T /2
α(t)dt ≤ m(0) − σ = 1 − σ.
0
Thus, for all τ ∈ [T /2, T ), we get k(τ ) ≤ k(0) + (1 − σ) = (2 − σ). So, by continuity of k(t),
we obtain k(T ) ≤ 2 − σ, and this contradicts the fact that k(T ) = 2. This completes Step 1.
Define α(t) = 2 − k(t) for all t ≥ 0. Then, α(t) > 0 for all t ≥ 0 by Step 1, and therefore:
Z 1
β≡
α(t)dt > 0 .
0
Step 2: We must have k(t) ≤ 2 − β for all t ≥ 1. To see this, note that for all t ≥ 0,
k̇(t) = k(t) + r(t) − c(t) ≤ k(t) + r(t) − 2 = r(t) − α(t) .
6
so that for all T ≥ 1,
Z
k(T ) − k(0) =
T
Z
T
Z
Z
Z
1
α(t)dt ≤ m(0) − β = 1 − β,
0
0
0
0
T
r(t)dt −
α(t)dt ≤
r(t)dt −
k̇(t)dt =
0
T
and consequently, k(T ) ≤ k(0) + (1 − β) = 2 − β for all T ≥ 1. This completes Step 2.
Step 3: k(t) < 0 for all t > (2 + β)/β. For all t ≥ 0, we have k̇(t) = k(t) + r(t) − c(t),
and we can write for all T > 1,
Z
T
Z
k(T ) − k(0) =
T
0
Z
≤
1
Z
Z
0
T
Z
r(t)dt −
Z
T
c(t)dt
0
0
T
r(t)dt − 2T
k(t)dt +
k(t)dt +
(B3)
0
1
0
T
k(t)dt +
k̇(t)dt =
< 2 + (2 − β)(T − 1) + 1 − 2T = 1 − β(T − 1) ,
the third line of (B3) following from Steps 1 and 2. Thus, k(T ) ≤ k(0) + 1 − β(T − 1) =
2 − β(T − 1) for all T > 1. For T > (2 + β)/β, we have (T − 1) > 2/β, and so β(T − 1) > 2.
Thus, for T > (2 + β)/β, k(T ) ≤ 2 − β(T − 1) < 0 and this establishes Step 3.
By Step 3, the hypothesis that there is a path (c(t), k(t), r(t)) satisfying (B1) must be
false, and this establishes our claim.
We have now demonstrated that an upper bound of C(k0 , m0 ) is 2. We will show in the
next section that this is also its least upper bound.
The supremum of C(k0 , m0 ). We now show that, given any ε ∈ (0, 1), there is a path
(c(t), k(t), r(t)) satisfying:
c(t) ≥ 2 − ε for all t ≥ 0 .
Given the ε, define:
n = (1/ε)
and T = 2/(n + 1)3 ,
(B4)
and determine the path of resource depletion by:

 2 − 2t for t ∈ [0, T ] ,
r(t) =
T
T2
 0
for t > T .
Then, r(t) ≥ 0 for t ∈ [0, T ], with r(T ) = 0, and r(t) → 0 as t → T. Thus r(t) is continuous
for t ≥ 0. Furthermore,
Z
∞
Z
r(t)dt =
0
T
Z
r(t)dt =
0
0
T
2t
2
− 2
T
T
T
2 t2
dt = 2 − 2
= 2 − 1 = 1 = m0 .
T
2 0
Denote (1 − ε2 ) by λ, and determine the capital path by:

2
 1 + λ 2t − t
for t ∈ [0, T ] ,
T
T2
k(t) =

1+λ
for t > T .
7
(B5)
Note that k(0) = 1 = k0 and k(t) ≥ 1 for all t ≥ 0. Furthermore,
2
2t
k̇(t) = λ
− 2 = λr(t) ≥ 0 for t ∈ [0, T )
T
T
with k̇(T −) = 0 = k̇(T +). Thus, k(t) is a C 1 function on R+ , and k̇(t) = λr(t) for all t ≥ 0.
If we now determine the consumption path by:
c(t) = (1 − λ)r(t) + k(t)
for all t ≥ 0 ,
(B6)
then clearly c(t) ≥ 0 for t ≥ 0, and c(t) = F (k(t), r(t)) − λr(t) = F (k(t), r(t)) − k̇(t) for
t ≥ 0. So {c(t), k(t), r(t)} is a path from (k0 , m0 ).
It remains to show that c(t) ≥ 2 − ε for all t ≥ 0. For t > T , we have r(t) = 0 and
k(t) = 1 + λ, so by (B5) and (B6),
c(t) = k(t) = 1 + λ = 2 − ε2 > 2 − ε for t > T .
(B7)
So, we now concentrate on t ∈ [0, T ].
Define N = nT /(n + 1) Then,
2N
N2
n2
2n
k(N ) = 1 + λ
− 2 =1+λ
−
T
T
(n + 1) (n + 1)2
λn
n
λn (n + 2)
1
=1+
2−
=1+
=1+λ 1−
.
(n + 1)
(n + 1)
(n + 1) (n + 1)
(n + 1)2
(B8)
Now, by choice of n in (B4), we have:
(n + 1)2 ≥ (n + 1) =
and so:
1+ε
1
+1=
ε
ε
1
ε
.
≤
(n + 1)2
1+ε
Thus,
1−
1
ε
1
≥ 1−
=
.
2
(n + 1)
1+ε
1+ε
(B9)
Using (B9) in (B8), we get:
k(N ) ≥ 1 +
λ
1 − ε2
=1+
= 2 − ε.
1+ε
1+ε
Since k̇(t) ≥ 0 for all t ≥ 0, we have k(t) ≥ 2 − ε for all t ∈ [N, T ]. Consequently, using (B6),
c(t) = (1 − λ)r(t) + k(t) ≥ k(t) ≥ 2 − ε
for all t ∈ [N, T ] .
(B10)
Finally, we turn to t ∈ [0, N ). Here, we have by (B6),
2
2t
c(t) = (1 − λ)r(t) + k(t) ≥ (1 − λ)r(t) + 1 = ε2 r(t) + 1 = ε2
− 2 +1
T
T
2
2
2N
2nT
− 2 + 1 = ε2
−
+1
≥ ε2
T
T
T
(n + 1)T 2
2ε2
n
2ε2
=
1−
+1=
+ 1 = ε2 (n + 1)2 + 1 > ε2 n2 + 1 = 2 ,
T
(n + 1)
(n + 1)T
8
(B11)
the second line of (B11) following from the definition of N , and the last line of (B11) following
from the definitions of T and n in (B4).
Combining (B7), (B10) and (B11), we have c(t) ≥ 2 − ε for all t ≥ 0, and so:
inf c(t) ≥ 2 − ε .
t≥0
Combining the result of this part with the previous one, we conclude that the supremum
of C(k0 , m0 ) is equal to 2. However, as shown in the previous part, there is no path in
C(k0 , m0 ) which attains this supremum. Thus, there is no maximin path in this model.
It is of interest to note that in discussing basically the same example of the production
function, Dasgupta and Heal [3, p. 18] claim that there exists an optimal path in this case.
In fact they make the claim that the Dirac delta function is the optimal strategy for optimal
depletion of the resource. However, given that the integral of resource depletion needs to
be well-defined, at least as a Lebesgue integral, it follows that we cannot admit the Dirac
delta-function as a feasible path of resource depletion.
Online appendix C: Steepness and speed in (k, m) space
Introduction. Consider a path (i) satisfying Hotelling’s rule (HoR) and (ii) with growing consumption over time, starting from initial stocks (k0 , m0 ) ∈ R2++ . A path which is
optimal according to the undiscounted utilitarian criterion will satisfy both these properties. Let us denote this path as (cu (t), ku (t), ru (t)) with a subscript u, even though it is not
necessarily the utilitarian optimal path; its initial consumption is cu (0).
Consider, next, the Hartwick path (ch (t), kh (t), rh (t)) where consumption is kept constant at cu (0); that is:
ch (t) = cu (0) for all t ≥ 0
Such a path is well-defined, since we know that ċu (t) > 0 for all t ≥ 0. The path (ch (t), kh (t),
rh (t)) also satisfies Hotelling’s rule [2, Proposition 1]. It does not exhaust the resource stock
m0 (and therefore it is “long-run inefficient”).
We are interested in establishing two properties regarding the resource use at time t = 0.
First, we want to show that the initial resource use on the u path is not the same as that
on the h path:
ru (0) 6= rh (0)
(C1)
Second, we want to show that, in fact, the initial resource use on the u path exceeds the
resource use on the h path:
ru (0) > rh (0)
(C2)
The reason for separating the results (C1) and (C2) is that (C1) essentially reveals a
property about the steepness of the trajectories followed by the two paths (evaluated at
the same initial stocks) in (k, m) space, while (C2) reveals a property about the speed of
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accumulating the capital stock and depleting the resource stock, which cannot directly be
described in (k, m) space.
For the analysis of this appendix, we introduce the following notation:
p(t) =
1
F2 (ku (t), ru (t))
for all t ≥ 0 ,
(C3)
and V (r) ≡ F (k0 , r) − cu (0) − F2 (k0 , r)r.
The value of net investments p(t)k̇u (t) − ru (t) is positive. By the identity of [2,
Proposition 1] and the fact that (cu (t), ku (t), ru (t)) obeys Hotelling’s rule (HoR), we obtain
p(t)ċu (t) +
i
d h
p(t)k̇u (t) − ru (t) = 0 for all t ≥ 0
dt
(C4)
by applying (C3). Since p(t) > 0 and ċu (t) > 0 for all t ≥ 0, (C4) implies that:
i
d h
p(t)k̇u (t) − ru (t) < 0 for all t ≥ 0 .
dt
(C5)
p(t)k̇u (t) − ru (t) > 0 for all t ≥ 0 .
(C6)
We now claim that:
Suppose contrary to (C6), there is some τ ≥ 0 such that p(τ )k̇u (τ ) − ru (τ ) ≤ 0. Then, by
(C5), for T ≡ τ + 1, we have that θ ≡ p(T )k̇u (T ) − ru (T ) < 0 and:
p(t)k̇u (t) − ru (t) ≤ θ for all t ≥ T .
(C7)
Furthermore, since ṗ(t) = −p(t)F1 (ku (t), ru (t)) < 0 (by (HoR)) and ku (t) > 0 for all t ≥ 0,
d
[p(t)ku (t)] − ru (t) = ṗ(t)ku (t) + p(t)k̇u (t) − ru (t) < p(t)k̇u (t) − ru (t) ≤ θ
dt
for all t ≥ T by (C7). Therefore, for all S > T ,
Z
S
ru (t)dt + θ(S − T ) ≤ [p(T )ku (T ) + m0 − θT ] + θS .
p(S)ku (S) < p(T )ku (T ) +
(C8)
T
However, the right hand side of (C8) goes to −∞ as S → ∞ (since θ < 0), while the right
hand side of (C8) is non-negative for all S > T . This contradiction establishes (C6).
Steepness is lower and speed is higher. The Hartwick path (ch (t), kh (t), rh (t))
which keeps consumption constant at cu (0) satisfies Hartwick’s rule (HaR), also at t = 0:
V (rh (0)) = F (k0 , rh (0)) − cu (0) − F2 (k0 , rh (0))rh (0) = 0 .
By (C6), the path (cu (t), ku (t), ru (t)) with strictly increasing consumption has positive value
of net investments, also at t = 0: p(0)k̇u (0) − ru (0) > 0, or equivalently, using (C3):
V (ru (0)) = F (k0 , ru (0)) − cu (0) − F2 (k0 , ru (0))ru (0) > 0 .
Since V 0 (r) = −F22 (k0 , r) > 0, we have established (C2) and thus (C1).
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By Lemma 1, rh (0) = r(cu (0), k0 ) is the unique solution to the problem of minimizing resource input per unit of capital accumulation, keeping consumption fixed at cu (0).
Hence, the steepness of the trajectory in (k, m) space is maximized if and only if resource
input equals rh (0). By (C1), it follows that the path (cu (t), ku (t), ru (t)) with strictly increasing consumption corresponds to a flatter trajectory in (k, m) space than the trajectory
corresponding to the Hartwick path (ch (t), kh (t), rh (t)) path, at the initial point (k0 , m0 ):
k̇u (0)
k̇u (0)
k̇h (0)
k̇h (0)
=
<
=
.
−ṁu (0)
ru (0)
rh (0)
−ṁh (0)
(C9)
We have, by (C2),
k̇u (0) = F (k0 , ru (0)) − cu (0) > F (k0 , rh (0)) − cu (0) = F (k0 , rh (0)) − ch (0) = k̇h (0) .
Thus, capital accumulation occurs at a faster rate on the u path compared to the h path,
and resource stock depletion occurs at a faster rate on the u path compared to the h path
(at least initially). Since (C9) holds, the higher resource stock depletion effect outweighs
the higher capital accumulation effect on the u path relative to the h path.
Remark. Solow [5, p. 39] compares the undiscounted utilitarian optimal path with
the maximin path from the same initial stocks (in a framework in which the production
function is of the Cobb-Douglas form, and the utility function is iso-elastic) and says that
the undiscounted utilitarian optimal path “will use up the pool of natural resources more
slowly” than the maximin path. It should be noted carefully that our exercise above is not
concerned with this comparison. In Solow’s comparison, the maximin path is a Hartwick
path which is egalitarian (and so satisfies Hotelling’s rule), and it is also efficient. The
h path in our exercise is inefficient, as it is constrained to consume forever the low initial
consumption on the u path.
References
[1] T. Apostol, Mathematical Analysis, Addison-Wesley, Reading, MA, 1974.
[2] W. Buchholz, S. Dasgupta, T. Mitra, Intertemporal equity and Hartwick’s rule in an
exhaustible resource model, Scand. J. Econ. 107 (2005) 547–561.
[3] P.S. Dasgupta, G.M. Heal, The optimal depletion of exhaustible resources, in: Symposium on the Economics of Exhaustible Resources, Rev. Econ. Stud. 41 (1974) 3–28.
[4] W. Rudin, Principles of Mathematical Analysis, 3 ed., McGraw-Hill, New York, NY,
1976.
[5] R.M. Solow, Intergenerational equity and exhaustible resources, in: Symposium on the
Economics of Exhaustible Resources, Rev. Econ. Stud. 41 (1974) 29–45.
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