Optimal consumption models in economic growth

Optimal consumption models in economic growth

J. Math. Anal. Appl. 337 (2008) 480–492 www.elsevier.com/locate/jmaa Optimal consumption models in economic growth Hiroaki Morimoto Department of Mat...

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J. Math. Anal. Appl. 337 (2008) 480–492 www.elsevier.com/locate/jmaa

Optimal consumption models in economic growth Hiroaki Morimoto Department of Mathematical Sciences, Faculty of Science, Ehime University, Matsuyama 790-0826, Japan Received 5 October 2005 Available online 21 April 2007 Submitted by Goong Chen

Abstract We study the optimal consumption problem in the one-sector model of economic growth under uncertainty. We show the existence of a classical solution of the Hamilton–Jacobi–Bellman equation associated with the stochastic optimization problem, and then give an optimal consumption policy in terms of its solution. © 2007 Elsevier Inc. All rights reserved. Keywords: Hamilton–Jacobi–Bellman equation; Viscosity solutions; Economic growth

1. Introduction We are concerned with the one-sector model of optimal economic growth under uncertainty discussed by R.C. Merton [8]. Define the following quantities: y(t) = labour supply at time t  0, x(t) = capital stock at time t  0, λ = the constant rate of depreciation, λ  0, c(t) = consumption rate per person at time t  0, c(t)y(t) = the totality of consumption rate, F (x, y) = constant-returns-to-scale production function producing the commodity for the capital stock x  0 and the labour force y  0, U (c) = utility function for the consumption rate c  0. We now state the conditions of the model. The labour supply y(t) and the capital stock x(t) are governed by the stochastic differential equation dy(t) = ny(t) dt + σy(t) dW (t), y(0) = y > 0, n, σ = 0,   x(t) ˙ = F x(t), y(t) − λx(t) − c(t)y(t), x(0) = x > 0, E-mail address: [email protected]. 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.04.024

(1) (2)

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on a complete probability space (Ω, F, P ) carrying a standard Brownian motion {W (t)}. The objective function for c = {c(t)} is given by the discounted expected utilities with discount rate β > 0:  τx    −βt J (c) = E e U c(t) dt , (3) 0

where τx = inf{t  0: x(t) = 0}. The purpose of this paper is to present a synthesis of optimal consumption policy c∗ so as to maximize J (c) over the class A of non-negative consumption policies c = {c(t)} such that c(t) is progressively measurable w.r.t. the filtration Ft = σ (Ws , s  t), t c(s) ds < ∞, ∀t  0, a.s.,

(4)

0

and (2) has a non-negative solution x(t) for x  0. This optimal consumption problem has been studied by [3,8] and [6, pp. 105–110]. But, the difficulty stems from the degeneracy in the associated Hamilton–Jacobi–Bellman (HJB, for short) equation. Our method consists in finding the viscosity solution V and further the regularity V of the HJB equation     1 βV (x, y) = σ 2 y 2 Vyy (x, y) + nyVy (x, y) + F (x, y) − λx Vx (x, y) + U˜ Vx (x, y)y , 2 V (0, y) = 0, x > 0, y > 0,

(5)

under the mild conditions on n, λ, σ , where U˜ (x) is the Legendre transform of −U (−x), i.e., U˜ (x) = maxc>0 {U (c) − cx}, and U (c) is assumed to have the following properties: U ∈ C[0, ∞) ∩ C 2 (0, ∞), 

U (c):

U (c):

strictly decreasing,

strictly concave on [0, ∞),



U (∞) = U (0+) = 0,

U  (0+) = U (∞) = ∞.

(6)

According to [8], we make the assumption on the production function F (x, y), F (γ x, γ y) = γ F (x, y) Fx > 0,

Fy > 0,

for γ > 0,

Fxx < 0,

F (0, y) = F (x, 0) = 0,

Fyy < 0,

Fx (0+, y) < ∞,

Fx (∞, y) = 0,

y > 0.

(7)

This paper is organized as follows. In Sections 2 and 3, we reduce (5) to the 1-dimensional HJB equation associated with the stochastic Ramsey problem [8], and we show the existence of viscosity solutions of the HJB equation. Section 4 is devoted to the C 2 -regularity of the viscosity solution. In Section 5, we give a synthesis of the optimal consumption policy. 2. Hamilton–Jacobi–Bellman equations We consider the HJB equation (5) and seek the solution V (x, y) of (5) of the form V (x, y) = v(z),

z=

x . y

(8)

Then, by (7), v(z) solves the 1-dimensional HJB equation     1 βv(z) = σ 2 z2 v  (z) + f (z) − μz v  (z) + U˜ v  (z) , 2 v(0) = 0, z > 0, where

μ = n + λ − σ2

f (z):

(9)

and f (z) = F (z, 1). By (7), it is clear that

Lipschitz continuous, concave,

f (0) = 0.

(10)

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We notice that (9) is the HJB equation associated with the stochastic Ramsey problem so as to maximize  τz    −βt ¯ J (c) = E e U c(t) dt ,

(11)

0

over the class A0 , subject to     dz(t) = f z(t) − μz(t) − c(t) dt − σ z(t) dW (t),

z(0) = z  0,

(12)

where A0 denotes the class A with {z(t)} replacing {x(t)}. We rewrite (9) as

    1 1 1 v(z) = σ 2 z2 v  (z) + f (z) − μz v  (z) + U˜ v  (z) + v(z), β+ ε 2 ε v(0) = 0, z > 0,

(13)

for ε > 0 chosen later. We shall show that v is approximated by the solution u = uL for each L > 0 of

    1 1 1 u(z) = σ 2 z2 u (z) + f (z) − μz u (z) + U˜ L u (z) + u(z), β+ ε 2 ε u(0) = 0, z > 0,

(14)

where U˜ L (x) = max0 0, 2   1 βv(z)  σ 2 z2 X + f (z) − μz p + U˜ (p), ∀(p, X) ∈ J 2,− v(z), ∀z > 0, 2 where J 2,+ and J 2,− are the second-order superjets and subjets defined by v(y) − v(z) − p(y − z) − 12 X|y − z|2 2,+ 2 J v(z) = (p, X) ∈ R ; lim sup 0 , |y − z|2 y→z v(y) − v(z) − p(y − z) − 12 X|y − z|2 2,− 2 J v(z) = (p, X) ∈ R ; lim inf 0 . y→z |y − z|2 3.1. Existence We note that (14) is the HJB equation associated with the optimization problem  τz      1 1 uL (z) = sup E e−(β+ ε )t U c(t) + uL z(t) dt , L > 0, ε c∈AL

(15)

0

where AL denotes the class of all non-negative, integrable, Ft -progressively measurable processes c ∈ A0 such that 0  c(t)  L for all t  0, and the supremum is taken over all admissible control systems [2]. By (12), we have z(t) = z(t ∧ τz )  0 for each c ∈ AL , because c(t) is identified with c(t)1{tτz } in (15). We assume β + μ > 0. Taking

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H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

0 < α < β,

0 < κ < α + μ,

483

(17)

we can choose A > 0 by concavity such that f (z) − κz < A.

(18)

Furthermore, we observe by (6) and (18) that ϕ(z) := z + A¯ satisfies     1 −αϕ(z) + σ 2 z2 ϕ  (z) + f (z) − μz ϕ  (z) + U˜ ϕ  (z)  −α A¯ + A + U˜ (1) < 0, z  0, (19) 2 for some constant A¯ > 0. Let B denote the Banach space of all functions h on [0, ∞), which fulfills the relation There exists Cρ > 0, for any ρ > 0, such that

 

h(z) − h(˜z)  Cρ |z − z˜ | + ρ ϕ(z) + ϕ(˜z) ,

z, z˜ ∈ [0, ∞),

(20)

with norm h = supz0 |h(z)|/ϕ(z) < ∞. Lemma 3.2. Under (6), (7) and (16), there exists a unique solution u = uL ∈ B of (15) for some ε > 0. Proof. We first show that  τz    1   −(α+ 1ε )t U c(t) + ϕ z(t) dt  ϕ(z), e sup E ε c∈AL

(21)

0

  1 sup E e−(α+ ε )τ z(τ ) − z˜ (τ )  |z − z˜ |,

(22)

c∈AL

for any stopping time τ , where {˜z(t)} is the solution of (12) to c ∈ AL with z˜ (0) = z˜ . It is easily seen that  t E

  t    2  2 z(s) ds < ∞. z(s)ϕ z(s) ds = E

0

0

This yields that Mε (t) := gives

t 0

e

−(α+ 1ε )s

ϕ  (z(s))σ z(s) dW (s) is a martingale. Hence, by (19) and (12), Ito’s formula

   1 0  E e−(α+ ε )(t∧τ ) ϕ z(t ∧ τ )  t∧τ

       1 1  ϕ z(s) + f z(s) − μz(s) − c(s) ϕ  z(s) e−(α+ ε )s − α + = ϕ(z) + E ε 0    1 2 2  + σ z(s) ϕ z(s) ds + Mε (t ∧ τ ) 2  t∧τ    1   −(α+ 1ε )s U c(s) + ϕ z(s) ds .  ϕ(z) − E e ε 0 1

Thus, we deduce (21). We set Z(t) = z(t) − z˜ (t) and hξ (z) = (z2 + ξ ) 2 for ξ > 0. It is clear that       dZ(t) = f z(t) − f z˜ (t) − μZ(t) dt − σ Z(t) dW (t)

   f  (0) + |μ| Z(t) dt − σ Z(t) dW (t), Z(0) = z − z˜ . Then, by Ito’s formula

(23)

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H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

   1 E e−(α+ ε )τ hξ Z(τ ) = hξ (z − z˜ ) + E

 τ



  1 1 hξ Z(s) e−(α+ ε )s − α + ε

0

+ hξ

        1 2   2  Z(s) f z(s) − f z˜ (s) − μZ(s) + σ Z(s) hξ Z(s) ds 2

 hξ (z − z˜ ), where we take ε > 0 such that

1   1 − α+ hξ (z) + f  (0) + |μ| zhξ (z) + σ 2 z2 hξ (z) ε 2

 1   2 1 1 + f  (0) + |μ| + σ 2 < 0,  z +ξ 2 − α+ ε 2

∀z ∈ R.

Letting ξ → 0, we get

  1 E e−(α+ ε )τ Z(τ )  |z − z˜ |, which implies (22). Next, we define  τz TL h(z) = sup E c∈AL

e

−(β+ 1ε )t

   1   U c(t) + h z(t) dt for h ∈ B, ε

(24)

0

and show that TL : B ϕ → B ϕ ,

(25)

where Bϕ is the closed subset of B defined by Bϕ = {h ∈ B: 0  h  ϕ, h(0) = 0}. By (21), it is easy to see that TL h(0) = 0  TL h  ϕ and TL h < ∞ for h ∈ Bϕ . Since z(t)  0, we note by (12) that z(t) = 0 if t > τz . Hence, by (20), we have  τz   1   1 TL h(z) − TL h(˜z)  sup E e−(β+ ε )t U c(t) + h z(t) dt ε c 0

τz˜ −

e 0

−(β+ 1ε )t

 τz

 sup E c

e

 sup E

+

−(β+ 1ε )t

τz ∧τz˜

 τz c

   1   U c(t) + h z˜ (t) dt ε



e

−(β+ 1ε )t

τz ∧τz˜

ρ sup E ε c

  U c(t) dt +

 ∞

∞ e 0

  Cρ sup E U c(t) dt + ε c

0

 ∞

   1   e−(β+ ε )t ϕ z(t) + ϕ z˜ (t) dt say.

By (12), we can take sufficiently small ε > 0 such that  ∞ 1 2 



−(α+ 1ε )s 2

E sup Mε (t)  2|σ |E e z(s) ds < ∞. t

   1

 h z(t) − h z˜ (t) dt ε

e 0

0

≡ J1 + J2 + J3 ,

−(β+ 1ε )t



−(β+ 1ε )t



z(t) − z˜ (t) dt



H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

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By the same line as (23), we have      1 1 J1  E e−(α+ ε )(τz ∧τz˜ ) ϕ z(τz ∧ τz˜ ) − e−(α+ ε )τz ϕ z(τz )       1 1 = E e−(α+ ε )(τz ∧τz˜ ) ϕ z(τz ∧ τz˜ ) − e−(α+ ε )τz ϕ z(τz ) 1{τz˜ <τz }       1 1  E e−(α+ ε )τz˜ ϕ z(τz˜ ) − e−(α+ ε )τz˜ ϕ z˜ (τz˜ ) 1{τz˜ <τz }

  1  sup E e−(α+ ε )τz˜ z(τz˜ ) − z˜ (τz˜ ) 1{τz˜ <τz } c

 |z − z˜ |. Moreover, by (22) J2  Cρ |z − z˜ |/(β − α)ε. Also, we recall that z˜ (t) = 0 if t > τz˜ . Hence, by (21)   J3  ρ ϕ(z) + ϕ(˜z) . Therefore, we get TL h ∈ B, which implies (25). Now, by (23), we have    E e−α(t∧τz ) ϕ z(t ∧ τz )  ϕ(z). Hence

TL h1 (z) − TL h2 (z) = sup E c∈AL

 τz

   1

 h1 z(t) − h2 z(t) dt ε

e

−(β+ 1ε )t

  1 h1 − h2 ϕ z(t) dt ε

0  τz

 sup E c∈AL

e

−(β+ 1ε )t





0

1  h1 − h2 ϕ(z). (β − α)ε + 1 Therefore, by the contraction mapping theorem, TL has a fixed point uL ∈ Bϕ . This completes the proof.

2

Theorem 3.3. We assume (6), (7), (16). Then u = uL ∈ B of (15) is a concave viscosity solution of (14). Proof. We recall (20) and also (23) to obtain    1 E e−(β+ ε )τ ϕ z(τ )  ϕ(z). Then, by a slight modification of the proof of [7, Theorem 2.3], we can show that the dynamic programming principle holds for u, i.e.,  τz ∧τ      1   −(β+ 1ε )t −(β+ 1ε )τz ∧τ u(z) = sup E e u z(τz ∧ τ ) U c(t) + u z(t) dt + e ε c∈AL 0

for any bounded stopping time τ . Thus, by the standard theory of viscosity solutions [2], we deduce that u is a viscosity solution of (14). To see the concavity of u, we also recall TL h(x) of (24). By the same line as [2, p. 204, Lemma 10.6], we observe that TL 0 is concave. Moreover, by induction, TLn 0 is concave for any n  1. By Lemma 3.2, we have TLn 0 → u

as n → ∞.

Therefore, this yields that u is concave.

2

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H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

Theorem 3.4. We assume (6), (7), (16). Then there exists a concave viscosity solution v of (9) such that 0  v  ϕ. Proof. Let L < L . By (24), we have TL h  TL h for h ∈ B. Hence uL has a limit, denoted by v uL ↑ v

as L → ∞.

It is clear that v is concave and then continuous on (0, ∞). It follows from (15) and (21) that  τz    1   −(β+ 1ε )t U c(t) + v z(t) dt  ϕ(z), z  0. e v(z)  sup E ε c∈A0 0

For any ρ > 0, there exists c ∈ A0 such that    τz  τz     −(β+ 1ε )t −(β+ 1ε )t 1 v z(t) dt . e U c(t) dt + E e v(z) − ρ < E ε 0

0

By the comparison theorem [5], we have τz ↓ θ,

z(t)  q(t) ↓ 0,

a.s.

as z ↓ 0,

where q(t) is the solution of   dq(t) = f  (0) + |μ| q(t) dt − σ q(t) dW (t),

q(0) = z > 0.

Since E[sup0ts q(t)2 ] < ∞ for each s > 0, it is clear by (12) that   E z(τz ∧ s) = z + E

  τz ∧s     f z(t) − μz(t) − c(t) dt . 0

Letting z ↓ 0 and then s → 0, we get  θ    θ   E f (0) − c(t) dt  0, −c(t) dt = E 0

0

so that  θ e

E

−(β+ 1ε )t

   U c(t) dt = 0.

0

Passing to the limit, we obtain  ∞ v(0+) − ρ  E

e 0

−(β+ 1ε )t

 1 1 v(0+) dt = v(0+), ε βε + 1

which implies v(0+) = 0. Thus v ∈ C[0, ∞). By Dini’s theorem, uL converges to v locally uniformly on [0, ∞). Therefore, by the standard stability results [2], we deduce that v is a viscosity solution of (13) and then (9). 2 3.2. Comparison We give a comparison theorem for the viscosity solution v of (9). Theorem 3.5. Let fi , i = 1, 2, satisfy (10) and let vi ∈ C[0, ∞) be the concave viscosity solution of (9) for fi replacing f such that 0  vi  ϕ. Suppose f1  f2 .

(26)

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Then, under (6), (7) and (16), we have v1  v2 . Proof. By (17), we first note that there exists 1 < ν < 2 such that 1 −β + σ 2 ν(ν − 1) + κν − μν < 0. 2 Hence, by (18), ϕν (z) := zν + B satisfies   1 −βϕν (z) + σ 2 z2 ϕν (z) + f2 (z) − μz ϕν (z) 2

1 2  −β + σ ν(ν − 1) + κν − μν zν + Aνzν−1 − βB < 0, 2

z  0,

(27)

for a suitable choice of B > 0. Suppose that v1 (z0 ) − v2 (z0 ) > 0 for some z0 ∈ (0, ∞). Then there exists η > 0 such that   sup v1 (z) − v2 (z) − 2ηϕν (z) > 0. z0

Since v1 (z) − v2 (z) − 2ηϕν (z)  ϕ(z) − 2ηϕν (z) → −∞ as z → ∞, we find z¯ ∈ (0, ∞) such that   sup v1 (z) − v2 (z) − 2ηϕν (z) = v1 (¯z) − v2 (¯z) − 2ηϕν (¯z) > 0. z0

Define   n Ψn (z, y) = v1 (z) − v2 (y) − |z − y|2 − η ϕν (z) + ϕν (y) 2 for any n > 0. It is clear that   Ψn (z, y)  ϕ(z) + ϕ(y) − η ϕν (z) + ϕν (y) → −∞ as z + y → ∞. Hence we find (zn , yn ) ∈ [0, ∞)2 such that   Ψn (zn , yn ) = sup Φ(z, y): (z, y) ∈ [0, ∞)2   n = v1 (zn ) − v2 (yn ) − |zn − yn |2 − η ϕν (zn ) + ϕν (yn ) 2  v1 (¯z) − v2 (¯z) − 2ηϕν (¯z) > 0, from which   n |zn − yn |2  v1 (zn ) − v2 (yn ) − η ϕν (zn ) + ϕν (yn ) 2    ϕ(zn ) + ϕ(yn ) − η ϕν (zn ) + ϕν (yn ) . Thus we deduce that the sequences {zn + yn } and {n|zn − yn |2 } are bounded by some constant C > 0, and √ |zn − yn |  C/ n → 0 as n → ∞. Moreover, zn , yn → zˆ ∈ [0, ∞)

as n → ∞,

(28)

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H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

taking a subsequence if necessary. By (28) we recall that   n Ψn (zn , yn ) = v1 (zn ) − v2 (yn ) − |zn − yn |2 − η ϕν (zn ) + ϕν (yn ) 2  v1 (zn ) − v2 (zn ) − 2ηϕν (zn ), and hence   n |zn − yn |2  v2 (zn ) − v2 (yn ) + η ϕν (zn ) − ϕν (yn ) → 0 as n → ∞. 2 Passing to the limit in (28), we get v1 (ˆz) − v2 (ˆz) − 2ηϕν (ˆz) > 0 and zˆ = 0.

(30)

Next, let V1 (z) = v1 (z) − ηϕν (z) and V2 (y) = v2 (y) + ηϕν (y). Applying Ishii’s lemma [1,2] to n Ψn (z, y) = V1 (z) − V2 (y) − |z − y|2 , 2 we obtain X1 , X2 ∈ R such that   n(zn − yn ), X1 ∈ J¯2,+ V1 (zn ),   n(zn − yn ), X2 ∈ J¯2,− V2 (yn ),





0 1 −1 1 0 X1  3n , −3n  0 −X2 −1 1 0 1 where



¯2,±

J

∃zr → z, ∃(pr , Xr ) ∈ J 2,± Vi (zr ),    Vi (z) = (p, X):  Vi (zr ), pr , Xr → Vi (z), p, X

(29)

(31)

 ,

i = 1, 2.

Recall that    p + ηϕν (z), X + ηϕν (z) : (p, X) ∈ J 2,+ V1 (z) ,    J 2,− v2 (y) = p − ηϕν (y), X − ηϕν (y) : (p, X) ∈ J 2,− V2 (y) . J 2,+ v1 (z) =

Hence   (p1 , X¯ 1 ) := n(zn − yn ) + ηϕν (zn ), X1 + ηϕν (zn ) ∈ J¯2,+ v1 (zn ),   (p2 , X¯ 2 ) := n(zn − yn ) − ηϕν (yn ), X2 − ηϕν (yn ) ∈ J¯2,− v2 (yn ). By the definition of viscosity solutions, we have   1 −βv1 (zn ) + σ 2 zn2 X¯ 1 + f1 (zn ) − μzn p1 + U˜ (p1 )  0, 2   1 −βv2 (yn ) + σ 2 yn2 X¯ 2 + f2 (yn ) − μyn p2 + U˜ (p2 )  0. 2 Putting these inequalities together, we get        1     β v1 (zn ) − v2 (yn )  σ 2 zn2 X¯ 1 − yn2 X¯ 2 + f1 (zn ) − μzn p1 − f2 (yn ) − μyn p2 + U˜ (p1 ) − U˜ (p2 ) 2 ≡ I1 + I2 + I3 , say. We consider the case when there are infinitely many zn  yn  0. By (29) and (31), it is easy to see that   1  I1  σ 2 3n|zn − yn |2 + η zn2 ϕν (zn ) + yn2 ϕν (yn ) → σ 2 ηˆz2 ϕν (ˆz) 2 By monotonicity and p1  p2 , I3  0.

as n → ∞.

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By (26), we have f1 (zn )(zn − yn )  f2 (zn )(zn − yn ). Hence        I2  f2 (zn ) − f2 (yn ) n(zn − yn ) − μn(zn − yn )2 + η f1 (zn ) − μzn ϕν (zn ) + f2 (yn ) − μyn ϕν (yn )   → η f1 (ˆz) + f2 (ˆz) − 2μˆz ϕν (ˆz) as n → ∞. Thus, by (26) and (27)

     1 2 2  β v1 (ˆz) − v2 (ˆz)  2η σ zˆ ϕν (ˆz) + f2 (ˆz) − μˆz ϕν (ˆz) 2  2ηβϕν (ˆz).

This is contrary with (30). Suppose that there are infinitely many yn  zn  0. By concavity, we have vi (yn )  vi (zn ), i = 1, 2, so that v1 (yn ) − v2 (zn )  v1 (zn ) − v2 (yn ). Hence the maximum of Ψn (z, y) is attained at (yn , zn ). Interchanging yn and zn in the above argument, we get a contradiction. Thus the proof is complete. 2 4. Classical solutions In this section, we shall show the smoothness of the viscosity solution v of (9). Theorem 4.1. Under (6), (7) and (16), we have v ∈ C 2 (0, ∞) and v  (0+) = ∞. Proof. Let [a, b] ⊂ (0, ∞) be arbitrary and we consider the boundary value problem     1  βw(z) = σ 2 z2 w  (z) + f (z) − μz w  (z) + U˜ w  (z) ∨ v+ (b) 2 w(a) = v(a), w(b) = v(b),

in (a, b), (32)

 (x) v+

where denotes the right-hand derivative of the concave viscosity solution v. By concavity, we see that p   (b) > 0 for any (p, X) ∈ J 2,+ v(z) with a < z < b. Hence v is a viscosity solution of (32). v+  (b)), it is known in [4] that (32) has a Now, by the uniform ellipticity and the Lipschitz continuity of U˜ (· ∨ v+ smooth solution w. By the same line as the proof of Theorem 3.5, we can show the uniqueness of the viscosity solution of (32). Therefore we deduce that v = w ∈ C 2 (a, b), and hence v ∈ C 2 (0, ∞). Next, suppose v  (0+) < ∞. By (9), we have v  (z)  0 for all z > 0, since U˜ (v  (z)) = ∞ if v  (z) < 0. Moreover, by L’Hospital’s rule z2 v  (z) = 0. z→0 z

lim z2 v  (z) = lim z2 v  (z) + 2zv  (z) = lim

z→0

z→0

Passing to the limit in (9) as z → 0, we have that U˜ (v  (0+)) = 0. This is contrary with (6). Therefore v  (0+) = ∞.

2

Corollary 4.2. Under the assumptions of Theorem 4.1, we have the solution V (x, y) ∈ C 2 ((0, ∞)2 ) of (5). Proof. We define V (x, y) by (8). Then the proof is immediate from Theorem 4.1.

2

5. Optimal consumption In this section we give a synthesis of optimal consumption policy c∗ for the optimization problem (3). Lemma 5.1. Under (6), (7) and (16), we have    lim inf E e−βt V x(t), y(t) = 0 t→∞

for every c ∈ A.

(33)

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H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

Proof. We set Φ(x, y) = ϕ( xy ). By (19), we have     1 −αΦ(x, y) + σ 2 y 2 Φyy (x, y) + nyΦy (x, y) + F (x, y) − λx Φx + U˜ Φx (x, y)y < 0, 2

x  0, y > 0. (34)

Hence, Ito’s formula gives    E e−βt Φ x(t), y(t)  Φ(x, y) + E

 t e

−βs

   (−β + α)Φ x(s), y(s) ds ,

0

from which

 ∞

(β − α)E

e

−βs

   Φ x(s), y(s) ds < ∞.

0

Thus    lim inf E e−βt Φ x(t), y(t) = 0. t→∞

By (8) V (x, y)  Φ(x, y), which implies (33).

2

Now we consider the equation of the form:   x˙ ∗ (t) = F x ∗ (t), y(t) − λx ∗ (t) − c∗ (t)y(t),

x ∗ (0) = x > 0,

(35)

where     c∗ (t) = (U  )−1 Vx x ∗ (t), y(t) y(t) 1{tτx ∗ } .

(36)

Lemma 5.2. Under (6), (7) and (16), there exists a unique solution x ∗ (t)  0 of (35). Proof. Let G(x, y) = F (x + , y) − λx − (U  )−1 (Vx (x + , y)y). Since G(x, y) is continuous and G(0, y) = 0, there exists an Ft -progressively measurable solution χ(t) of   dχ(t) = G χ(t), y(t) dt, χ(0) = x > 0. (37) Define x ∗ (t) = χ(t ∧ τχ )  0. We note by the concavity of f that G(x, y)  C1 x + + C2 y for some C1 , C2 > 0. Then we apply the comparison theorem to (37) and   d x(t) ˆ = C1 x(t) ˆ + + C2 y(t) dt, x(0) ˆ = x > 0, (38) ˆ for all t  0. Further, it is easy to see that τx ∗ = τχ , and hence to obtain 0  x ∗ (t)  x(t)   x˙ ∗ (t) = 1{tτχ } G χ(t), y(t)     = 1{tτx ∗ } F x ∗ (t), y(t) − λx ∗ (t) − c∗ (t)y(t) . ˜ be another solution of (35). We notice that the function Therefore x ∗ (t) solves (35). To prove uniqueness, let x(t) x → G(x, y) is locally Lipschitz continuous on (0, ∞). By using a standard technique, we can get x ∗ (t ∧ τx ∗ ∧ τx˜ ) = x(t ˜ ∧ τx ∗ ∧ τx˜ ).

H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

491

Hence τx ∗ = τx˜ , This implies

x ∗ (t) = x(t) ˜

˜ x ∗ (t) = x(t)

for t  τx ∗ .

for all t  0.

2

Theorem 5.3. We make the assumptions of Theorem 4.1. Then the optimal consumption c∗ = {c∗ (t)} is given by (36). Proof. We set ζ = s ∧ τx ∗ for any s > 0. By (5), (35) and Ito’s formula, we have  ζn  −βζ  ∗    −βt n −βV (x, y) + Vx (x, y)1{t<τx ∗ } F (x, y) − λx − c∗ (t)y E e V x (ζn ), y(ζn ) = V (x, y) + E e 0



1 2 2

+ nyVy (x, y) + σ y Vyy (x, y) ∗ dt + M(ζn ) (x (t),y(t)) 2  ζn   ∗  −βt = V (x, y) − E e U c (t) dt , 0

where ζn = ζ ∧ τn for a localizing sequence of stopping times τn ↑ ∞ of the local martingale M(t) with M(0) = 0. From (34), (38) and Doob’s inequalities for martingales it follows that       E sup e−βζn V x ∗ (ζn ), y(ζn )  E sup e−αr Φ x ∗ (r), y(r) n

0rs

r





−αt ∗  Φ(x, y) + E sup e σ x (t) dW (t)

0rs 

 s  Φ(x, y) + 2|σ |E

0



2 e−αt x ∗ (t) dt

0

 s  Φ(x, y) + 2|σ |E

1 2

1 2

x(t) ˆ dt 2

.

0

Therefore, letting n → ∞, we get by the dominated convergence theorem  ζ   −βζ  ∗   ∗  −βt E e V x (ζ ), y(ζ ) = V (x, y) − E e U c (t) dt . 0

By (33), we have        lim inf E e−β(s∧τx ∗ ) V x ∗ (s ∧ τx ∗ ), y(s ∧ τx ∗ ) = lim inf E e−βs V x ∗ (s), y(s) : τx ∗  s = 0. s→∞

s→∞

Passing to the limit, we deduce  τx ∗   ∗   J c =E e−βt U c∗ (t) dt = V (x, y). 0

By the same calculation as above, we can obtain  τx    −βt J (c) = E e U c(t) dt  V (x, y),

c ∈ A.

0

Since V (x, y) = supc∈A J (c), we remark that the solution V (x, y) of (5) is unique. The proof is complete.

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H. Morimoto / J. Math. Anal. Appl. 337 (2008) 480–492

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