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Optimal consumption of the stochastic Ramsey problem for non-Lipschitz diffusion

Abstract

The stochastic Ramsey problem is considered in a growth model with the production function of a Cobb-Douglas form. The existence of a unique classical solution is proved for the Hamilton-Jacobi-Bellman equation associated with the optimization problem. A synthesis of the optimal consumption policy in terms of its solution is proposed.

MSC:49L20, 49L25, 91B62.

1 Introduction

We are concerned with the stochastic Ramsey problem in a growth model discussed by Merton [1]. For recent contribution in this direction, we refer to [2]. A firm produces goods according to the Cobb-Douglas production function x γ for capital x, where 0<γ<1 (cf. Barro and Sala-i-Martin [3]). The stock of capital X t at time t is modeled by the stochastic differential equation

d X t = X t γ dt+σ X t d B t ,t>0, X 0 =x>0,σ0,

on a complete probability space (Ω,F,P) carrying a standard Brownian motion { B t } endowed with the natural filtration F t generated by σ( B s ,st).

The capital stock can be consumed and the flow of consumption at time t is assumed to be written as c t X t . The rate of consumption c={ c t } per capital stock is called an admissible policy if { c t } is an { F t }-progressively measurable process such that

0 c t 1for all t0 a.s.
(1.1)

We denote by the set of admissible policies. Given a policy cA, the capital stock process { X t } obeys the equation

d X t = [ X t γ c t X t ] dt+σ X t d B t , X 0 =x>0.
(1.2)

The objective is to find an optimal policy c ={ c t } so as to maximize the expected discounted utility of consumption

J x (c)=E [ 0 e α t U ( c t X t ) d t ]
(1.3)

over cA, where α>0 is a discount rate and U(x) is a utility function in C 2 (0,)C[0,), which is assumed to have the following properties:

U ()=U(0)=0, U (0+)=U()=, U <0.
(1.4)

The Hamilton-Jacobi-Bellman (HJB for short) equation associated with this problem is given by

αu(x)= 1 2 σ 2 x 2 u (x)+ x γ u (x)+ U ˜ ( x , u ( x ) ) ,x>0,
(1.5)

where

U ˜ (x,y)= max 0 c 1 { U ( c x ) c x y } for x,y>0.
(1.6)

This kind of economic growth problem has been studied by Kamien and Schwartz [4] and Sethi and Thompson [[5], Chapter 11]. However, the optimization problem is unsolved. It is not guaranteed that (1.2) admits a unique positive solution { X t } and the value function is a viscosity solution of the HJB equation. The main difficulty stems from the fact that (1.5) is a degenerate nonlinear equation of elliptic type with the non-Lipschitz coefficient x γ in (0,). It is also analytically studied by [6], nevertheless in the finite time horizon. The resulting HJB equation is a parabolic partial differential equation (PDE, for short), which is very different from the elliptic PDE dealt with in the present paper.

In this paper, we provide the existence results on a unique positive solution { X t } to (1.2) and a classical solution u of (1.5) by the theory of viscosity solutions. For the detail of the theory of viscosity solutions, we mention the works [7, 8] and [9]. An optimal policy is shown to exist in terms of u.

This paper is organized as follows. In Section 2, we show that (1.2) admits a unique positive solution. In Section 3, we show the existence of a viscosity solution u of the HJB equation (1.5). Section 4 is devoted to the C 2 -regularity of its solution. In Section 5, we present a synthesis of the optimal consumption policy.

2 Preliminaries

In this section, we show the existence of a unique solution { X t } to (1.2).

Proposition 2.1 There exists a unique positive solution { X t }={ X t x } to (1.2), for each cA, such that

E[ X t ] { ( 1 γ ) t + x 1 γ } 1 / ( 1 γ ) ,
(2.1)
E [ X t 2 ] e σ 2 t { 2 ( 1 λ ) t + x 2 ( 1 λ ) } 1 / ( 1 λ ) ,λ:=(1+γ)/2,
(2.2)
ε>0, C ε >0s.t. E [ | X t x X t y | ] C ε |xy|+ε ( 1 + t 1 / ( 1 γ ) + x + y ) ,x,y>0.
(2.3)

Proof We set x t = X t 1 γ . Then, by Ito’s formula and (1.2),

d x t = ( 1 γ ) X t γ d X t + σ 2 2 ( 1 γ ) ( γ ) X t 1 γ d t = ( 1 γ ) [ 1 ( c t + σ 2 2 γ ) x t ] d t + ( 1 γ ) σ x t d B t , x 0 = x 1 γ .
(2.4)

By linearity, (2.4) has a unique solution { x t }. Since

d x ˆ t =(1γ) [ ( c t + σ 2 2 γ ) x ˆ t ] dt+(1γ)σ x ˆ t d B t , x ˆ 0 = x 1 γ
(2.5)

has a positive solution { x ˆ t }, we see by the comparison theorem [[10], Chapter 6, Theorem 1.1] that x t x ˆ t >0 holds almost surely (a.s.). Therefore, (1.2) admits a unique positive solution { X t } of the form X t = x t 1 / ( 1 γ ) , which satisfies sup 0 t T E[ X t 4 ]< for each T0.

Let θ t be the right-hand side of (2.1) and ϕ t =E[ X t ]. Obviously, we see that θ t is a unique solution of

d θ t = θ t γ dt, θ 0 =x>0.

By (1.2) and Jensen’s inequality,

d ϕ t =dE[ X t ]=E [ X t γ c t X t ] dt ϕ t γ dt.

Since θ 0 = ϕ 0 =x, we deduce ϕ t θ t , which implies (2.1).

Similarly, let Θ t be the right-hand side of (2.2) and Φ t =E[ X t 2 ]. By substitution, it is easy to see that Θ ¯ t := e σ 2 t Θ t is a unique solution of

d Θ ¯ t =2 Θ ¯ t λ dt, Θ ¯ 0 = x 2 >0.

Hence

d Θ t = e σ 2 t ( 2 Θ ¯ t λ + σ 2 Θ ¯ t ) dt ( 2 Θ t λ + σ 2 Θ t ) dt.

Furthermore, by (1.2), Ito’s formula and Jensen’s inequality,

d Φ t = d E [ X t 2 ] = E [ 2 X t 2 λ 2 c t X t 2 + σ 2 X t 2 ] d t ( 2 Φ t λ + σ 2 Φ t ) d t .

Thus, we deduce Φ t Θ t and Φ 0 = Θ 0 , which implies (2.2).

Next, let { Y t } denote the solution { X t y } of (1.2) with Y 0 =y and y t = Y t 1 γ . Then, by (2.4)

d( x t y t )=(1γ) ( c t + σ 2 2 γ ) ( x t y t )dt+(1γ)σ( x t y t )d B t ,

which implies

x t y t =( x 0 y 0 )exp { ( 1 γ ) ( 0 t c s d s + σ 2 2 γ t ) + ( 1 γ ) σ B t σ 2 2 ( 1 γ ) 2 t } .

Setting β=1/(1γ)>1, we have

E [ | x t y t | β ] | x 0 y 0 | β E [ exp { σ B t σ 2 2 t } ] = | x 1 γ y 1 γ | 1 / ( 1 γ ) | x y | .
(2.6)

By Young’s inequality [11], for any ε 0 >0,

| x β y β | β ( x β 1 + y β 1 ) | x y | β [ 1 β ( 1 ε 0 ) β | x y | β + β 1 β { ε 0 ( x β 1 + y β 1 ) } β / ( β 1 ) ] ( 1 ε 0 ) β | x y | β + ( β 1 ) ( 2 ε 0 ) β / ( β 1 ) ( x β + y β ) , x , y 0 .

Hence, for any ε>0, we choose C ε >0 such that

| x β y β | C ε | x y | β +ε ( 1 + x β + y β ) ,x,y0.

Therefore, by (2.1) and (2.6), we have

E [ | X t Y t | ] = E [ | x t β y t β | ] C ε E [ | x t y t | β ] + ε E [ 1 + x t β + y t β ] C ε | x y | + ε E [ 1 + X t + Y t ] C ε | x y | + ε { 1 + 2 β ( t β + x ) + 2 β ( t β + y ) } ,

which implies (2.3). □

Remark 2.1 The uniqueness for (1.2) is violated if x=0 and c t is deterministic since 0 and the limit process χ t := lim x 0 + X t x satisfy (1.2) with X 0 =0, and

E [ χ t 1 γ ] =E [ 0 t ( 1 γ ) { 1 ( c s + σ 2 2 γ ) χ s 1 γ } d s ] 0.
(2.7)

3 Viscosity solutions of the HJB equation

Definition 3.1 Let uC(0,). Then u is called a viscosity solution of (1.5) if the following relations are satisfied:

α u ( x ) 1 2 σ 2 x 2 q + x γ p + U ˜ ( x , p ) , ( p , q ) J 2 , + u ( x ) , x > 0 , α u ( x ) 1 2 σ 2 x 2 q + x γ p + U ˜ ( x , p ) , ( p , q ) J 2 , u ( x ) , x > 0 ,

where J 2 , + u(x) and J 2 , u(x) are the second-order superjets and subjets [7].

Define the value function u by

u(x)= sup c A E [ 0 e α t U ( c t X t ) d t ] ,
(3.1)

where the supremum is taken over all systems (Ω,F,P,{ F t };{ B t },{ c t }).

In this section, we study the viscosity solution u of the HJB equation (1.5). Due to Proposition 2.1, we can show the value function u with the following properties.

Lemma 3.1 We assume (1.4). Then we have

0u(x)ζ(x):=x+ ζ 0 ,x>0
(3.2)

for some constant ζ 0 >0, and there exists C ρ >0 for any ρ>0 such that

|u(x)u(y)| C ρ |xy|+ρ(1+x+y),x,y>0.
(3.3)

Proof Clearly, u is nonnegative. By concavity, there is C ¯ >0 such that

U(x)α 2 1 / ( 1 γ ) x+ C ¯ ,x0.

By (1.1) and (2.1), we have

E [ 0 e α t U ( c t X t ) d t ] E [ 0 e α t ( α 2 1 / ( 1 γ ) X t + C ¯ ) d t ] 0 e α t { α ( t 1 / ( 1 γ ) + x ) + C ¯ } d t = x + α 0 e α t t 1 / ( 1 γ ) d t + C ¯ / α ,

which implies (3.2).

Now, let ρ>0 be arbitrary. By (1.4), there is δ>0 such that U(x)ρ for all x[0,δ]. Furthermore,

|U(x)U(y)| U (δ)|xy|,x,yδ.

Thus, we obtain a constant C ρ >0, depending on ρ>0, such that

|U(x)U(y)| C ρ |xy|+ρ,x,y0.
(3.4)

By (1.1), (2.3) and (3.4), we get

| u ( x ) u ( y ) | sup c A E [ 0 e α t | U ( c t X t ) U ( c t Y t ) | d t ] sup c A E [ 0 e α t { C ρ | X t Y t | + ρ } d t ] C ρ 0 e α t { C ε | x y | + ε ( 1 + t 1 / ( 1 γ ) + x + y ) } d t + ρ / α C { C ρ C ε | x y | + ( ε + ρ ) ( 1 + x + y ) } , x , y > 0 ,
(3.5)

where the constant C>0 is independent of ε, ρ>0. Replacing ρ by ρ/2C and choosing sufficiently small ε>0, we deduce (3.3). □

Remark 3.1 The continuity of u is immediate from (3.3).

Theorem 3.1 We assume (1.4). Then the value function u is a viscosity solution of (1.5).

Proof According to [12], the viscosity property of u follows from the dynamic programming principle for u, that is,

u(x)= sup c A E [ 0 τ e α t U ( c t X t ) d t + e α τ u ( X τ ) ] ,x>0
(3.6)

for any bounded stopping time τ0, where the supremum is taken over all systems (Ω,F,P,{ F t };{ B t },{ c t }). Let u ¯ (x) be the right-hand side of (3.6). Let X ˜ t = X t + r and B ˜ t = B t + r B r , when τ=r is non-random. Then we have

d X ˜ t = [ X ˜ t γ c ˜ t X ˜ t ] dt+σ X ˜ t d B ˜ t , X ˜ 0 = X r

for the shifted process c ˜ ={ c ˜ t } of c by r, i.e., c ˜ t = c t + r . It is easy to see that

e α r E [ r e α t U ( c t X t ) d t | F r ] =E [ 0 e α t U ( c ˜ t X ˜ t ) d t | F r ] = J X r ( c ˜ )a.s.

with respect to the conditional probability P(| F r ). We take ζ 1 >0 such that x γ αx+ ζ 1 and sufficiently large ζ 0 >0 to obtain

αζ+ 1 2 σ 2 x 2 ζ + x γ ζ α ζ 0 + ζ 1 0.

By (3.2) in Lemma 3.1, Ito’s formula and Doob’s inequality, we have

E [ sup 0 t T e α t J X t ( c ˜ ) ] E [ sup 0 t T e α t ζ ( X t ) ] ζ(x)+C,T>0

for some constant C>0. Hence, approximating τ by a sequence of countably valued stopping times, we see that

E [ e α τ J X τ ( c ˜ ) ] =E [ τ e α t U ( c t X t ) d t ] .

Thus

J x ( c ) = E [ 0 τ e α t U ( c t X t ) d t + τ e α t U ( c t X t ) d t ] E [ 0 τ e α t U ( c t X t ) d t + e α τ u ( X τ ) ] .

Taking the supremum, we deduce u u ¯ .

We shall show the reverse inequality in the case that τ=r is constant. For any ε>0, we consider a sequence { S j :j=1,,n+1} of disjoint subsets of (0,) such that

diam( S j )<δ, j = 1 n S j =(0,R)and S n + 1 =[R,)

for δ,R>0 chosen later. We take x j S j and c ( j ) A such that

u( x j )ε J x j ( c ( j ) ) ,j=1,,n+1.
(3.7)

By the same argument as (3.5), we note that

| J x ( c ( j ) ) J y ( c ( j ) ) |+|u(x)u(y)| C ε |xy|+ ε 4 (1+x+y),x,y>0

for some constant C ε >0. We choose 0<δ<1 such that C ε δ<ε/2. Then we have

| J x ( c ( j ) ) J y ( c ( j ) ) |+|u(x)u(y)|ε(1+x),x,y S j ,j=1,2,,n.
(3.8)

Hence, by (3.7) and (3.8),

J X r ( c ( j ) ) = J X r ( c ( j ) ) J x j ( c ( j ) ) + J x j ( c ( j ) ) ε ( 1 + X r ) + u ( x j ) ε 2 ε ( 1 + X r ) + u ( X r ) ε 3 ε ( 1 + X r ) + u ( X r ) if  X r S j , j = 1 , , n .
(3.9)

By definition, we find cA such that

u ¯ (x)εE [ 0 r e α t U ( c t X t ) d t + e α r u ( X r ) ] .

In view of [[10], Chapter 6, Theorem 1.1], we can take c, c ( j ) on the same probability space. Define

c t r = c t 1 { t < r } + c t r ( j ) 1 { r t } if  X r S j ,j=1,,n+1,

where 1 { } denotes the indicator function. Then { c t r } belongs to . Let { X t r } be the solution of

d X t r = [ ( X t r ) γ c t r X t r ] dt+σ X t r d B t , X 0 r =x>0.

Clearly, X t r = X t a.s. if t<r. Further, for each j=1,,n+1, we have on { X r S j }

X t + r r = X r + r t + r [ ( X s r ) γ c s r X s r ] d s + r t + r σ X s r d B s = X r + 0 t [ ( X s + r r ) γ c s ( j ) X s + r r ] d s + 0 t σ X s + r r d B ˜ s a.s.

Hence, X t + r r coincides with the solution X t ( j ) of (1.2) for ( Ω ˜ , F ˜ , P ˜ ,{ F ˜ t };{ B ˜ t },{ c t ( j ) }) a.s. on { X r S j } with X 0 ( j ) = X r . Thus, we get

J X r ( c ˜ r ) = E P ˜ [ 0 e α t U ( c t + r r X t + r r ) d t | F ˜ r ] = E P ˜ [ 0 e α t U ( c t ( j ) X t ( j ) ) d t | F ˜ r ] = J X r ( c ( j ) ) a.s. on  { X r S j } , j = 1 , , n + 1 ,
(3.10)

where E P ˜ denotes the expectation with respect to P ˜ .

Now, we fix x>0 and choose R>0, by (2.1), (2.2) and (3.1), such that

sup c A E [ u ( X r ) 1 { X r R } ] sup c A E [ ζ ( X r ) 1 { X r R } ] sup c A 1 R E [ X r 2 + ζ 0 X r ] C 0 R ( 1 + x + x 2 ) < ε ,
(3.11)

where the constant C 0 >0 depends only on r, ζ 0 . By (3.9), (3.10) and (3.11), we have

E [ r e α t U ( c t r X t r ) d t ] = E [ E [ r e α t U ( c t r X t r ) d t | F r ] ] = E [ e α r J X r ( c ˜ r ) ] = E [ j = 1 n + 1 e α r J X r ( c ( j ) ) 1 { X r S j } ] E [ j = 1 n e α r { u ( X r ) 3 ε ( 1 + X r ) } 1 { X r S j } ] E [ e α r { u ( X r ) u ( X r ) 1 { X r R } } ] 3 ε E [ 1 + X r ] E [ e α r u ( X r ) ] ε 3 ε C ( 1 + x )

for some constant C>0 independent of ε. Thus

u ( x ) E [ 0 r e α t U ( c t r X t r ) d t + r e α t U ( c t r X t r ) d t ] E [ 0 r e α t U ( c t X t ) d t + e α r u ( X r ) ] ε 3 ε C ( 1 + x ) u ¯ ( x ) 2 ε 3 ε C ( 1 + x ) .

Letting ε0, we get u ¯ u.

In the general case, by the above argument, we note that

u ( X r ) = u ( X ˜ 0 ) E [ 0 s e α t U ( c ˜ t X ˜ t ) d t + e α s u ( X ˜ s ) | F r ] = E [ 0 s e α t U ( c t + r X t + r ) d t + e α s u ( X s + r ) | F r ] a.s.  s , r 0 .

Hence { e α s u( X s )+ 0 s e α t U( c t X t )dt} is a supermartingale. By the optional sampling theorem,

u( X 0 )E [ 0 τ e α t U ( c t X t ) d t + e α τ u ( X τ ) | F 0 ] a.s.

Taking the expectation and then the supremum over , we conclude that u ¯ u. Noting the continuity of u, we obtain (3.6). □

4 Classical solutions

In this section, using the viscosity solutions technique, we show the C 2 -regularity of the viscosity solution u of (1.5). For any fixed 0<a<b, we consider the boundary value problem

αw= 1 2 σ 2 x 2 w + x γ w + U ˜ ( x , w ) in (a,b),
(4.1)

with boundary condition

w(a)=u(a),w(b)=u(b),
(4.2)

given by u.

Proposition 4.1 Let w i C[a,b], i=1,2, be two viscosity solutions of (3.1), (4.2). Then, under (1.4), we have

w 1 = w 2 .

Proof It is sufficient to show that w 1 w 2 . Suppose that there exists x 0 [a,b] such that w 1 ( x 0 ) w 2 ( x 0 )>0. Clearly, by (4.2), x 0 a,b, and we find x ¯ (a,b) such that

ϱ:= sup x [ a , b ] { w 1 ( x ) w 2 ( x ) } = w 1 ( x ¯ ) w 2 ( x ¯ )>0.

Define

Ψ k (x,y)= w 1 (x) w 2 (y) k 2 | x y | 2

for k>0. Then there exists ( x k , y k ) [ a , b ] 2 such that

Ψ k ( x k , y k )= sup ( x , y ) [ a , b ] 2 Ψ k (x,y) Ψ k ( x ¯ , x ¯ )=ϱ,
(4.3)

from which

k 2 | x k y k | 2 < w 1 ( x k ) w 2 ( y k ).

Thus

| x k y k |0as k.
(4.4)

Furthermore, by the definition of ( x k , y k ),

Ψ k ( x k , y k ) Ψ k ( x k , x k ).

Hence, by uniform continuity

k 2 | x k y k | 2 w 2 ( x k ) w 2 ( y k ) sup | x y | ρ | w 2 ( x ) w 2 ( y ) | 0 as  k  and then  ρ 0 .
(4.5)

By (4.3), (4.4) and (4.5), extracting a subsequence, we have

( x k , y k )( x ˜ , x ˜ ) ( a , b ) 2 as k.
(4.6)

Now, we may consider that ( x k , y k ) ( a , b ) 2 for sufficiently large k. Applying Ishii’s lemma [7] to Ψ k (x,y), we obtain X,YR such that

( k ( x k y k ) , X ) J ¯ 2 , + w 1 ( x k ) , ( k ( x k y k ) , Y ) J ¯ 2 , w 2 ( y k ) , ( X 0 0 Y ) 3 k ( 1 1 1 1 ) .
(4.7)

By Definition 3.1,

α w 1 ( x k ) 1 2 σ 2 x k 2 X + x k γ μ + U ˜ ( x k , μ ) , α w 2 ( y k ) 1 2 σ 2 y k 2 Y + y k γ μ + U ˜ ( y k , μ ) ,

where μ=k( x k y k ). Putting these inequalities together, we get

α { w 1 ( x k ) w 2 ( y k ) } 1 2 σ 2 ( x k 2 X y k 2 Y ) + ( x k γ y k γ ) μ + { U ˜ ( x k , μ ) U ˜ ( y k , μ ) } I 1 + I 2 + I 3 , say .

By (4.5) and (4.7), it is clear that

I 1 = σ 2 2 ( x k 2 X y k 2 Y ) σ 2 2 3k ( x k y k ) 2 0as k.

Also, by (4.5)

I 2 =k ( x k γ y k γ ) ( x k y k )kγ a γ 1 | x k y k | 2 0as k.

By (1.6), (3.4), (4.5) and (4.6), we have

I 3 max 0 c 1 | U ( c x k ) U ( c y k ) | + | x k y k | | μ | C ρ | x k y k | + ρ + k | x k y k | 2 0 as  k  and then  ρ 0 .

Consequently, by (4.6), we deduce that

αϱα { w 1 ( x ˜ ) w 2 ( x ˜ ) } 0,

which is a contradiction. □

Theorem 4.1 We assume (1.4). Then there exists a solution u C 2 (0,) of (1.5).

Proof For any 0<a<b, we recall the boundary value problem (4.1), (4.2). Since

U(0) U (x)(0x)+U(x),x>0,

we have

K 0 := sup 0 < x a x U (x)<.

Hence, by (1.4)

| U ( c x 1 ) U ( c x 2 ) | c U ( c a ) | x 1 x 2 | K 0 a | x 1 x 2 | , x 1 , x 2 [ a , b ] , 0 c 1 .

Also, by (1.6)

| U ˜ ( x 1 , y 1 ) U ˜ ( x 2 , y 2 ) | max 0 c 1 | U ( c x 1 ) U ( c x 2 ) | + | x 1 y 1 x 2 y 2 | K 0 a | x 1 x 2 | + | x 1 x 2 | | y 1 | + b | y 1 y 2 | , y 1 , y 2 > 0 .

Thus the nonlinear term of (4.1) is Lipschitz. By uniform ellipticity, a standard theory of nonlinear elliptic equations yields that there exists a unique solution w C 2 (a,b)C[a,b] of (4.1), (4.2). For details, we refer to [[13], Theorem 17.18] and [[14], Chapter 5, Theorem 3.7]. Clearly, by Theorem 3.1, u is a viscosity solution of (4.1), (4.2). Therefore, by Proposition 4.1, we have w=u and u is smooth. Since a, b are arbitrary, we obtain the assertion. □

5 Optimal consumption

In this section, we give a synthesis of the optimal policy c ={ c t } for the optimization problem (1.4) subject to (1.2). We consider the stochastic differential equation

d X t = [ ( X t ) γ η ( X t ) X t ] dt+σ X t d B t , X 0 =x>0,
(5.1)

where η(x)=I(x, u (x)) and I(x,y) denotes the maximizer of (1.6) for x,y>0, i.e.,

I(x,y)= { ( U ) 1 ( y ) / x if  U ( x ) y , 1 otherwise .
(5.2)

Our objective is to prove the following.

Theorem 5.1 We assume (1.4). Then the optimal consumption policy { c t } is given by

c t =η ( X t ) .
(5.3)

To obtain the optimal consumption policy { c t }, we should study the properties of the value function u and the existence of strong solution { X t } of (5.1). We need the following lemmas.

Lemma 5.1 Under (1.4), the value function u is concave. In addition, we have

u (x)>0for x>0,
(5.4)
u (0+)=.
(5.5)

Proof Let x i >0, i=1,2. For any ε>0, there exists c ( i ) A such that

u( x i )ε<E [ 0 e α t U ( c t ( i ) X t ( i ) ) d t ] ,

where { X t ( i ) } is the solution of (1.2) corresponding to c ( i ) with X 0 ( i ) = x i . Let 0ξ1, and we set

c ¯ t = ξ c t ( 1 ) X t ( 1 ) + ( 1 ξ ) c t ( 2 ) X t ( 2 ) ξ X t ( 1 ) + ( 1 ξ ) X t ( 2 ) ,

which belongs to . Define { X ¯ t } and { X ˜ t } by

d X ¯ t = [ ( X ¯ t ) γ c ¯ t X ¯ t ] d t + σ X ¯ t d B t , X ¯ 0 = ξ x 1 + ( 1 ξ ) x 2 , X ˜ t = ξ X t ( 1 ) + ( 1 ξ ) X t ( 2 ) .

By concavity,

X ˜ t ξ x 1 +(1ξ) x 2 + 0 t [ ( X ˜ s ) γ c ¯ s X ˜ s ] ds+ 0 t σ X ˜ s d B s a.s.

By the comparison theorem, we have

X ˜ t X ¯ t for all t0 a.s.

Thus, by (1.4)

u ( ξ x 1 + ( 1 ξ ) x 2 ) E [ 0 e α t U ( c ¯ t X ¯ t ) d t ] E [ 0 e α t U ( c ¯ t X ˜ t ) d t ] = E [ 0 e α t U ( ξ c t ( 1 ) X t ( 1 ) + ( 1 ξ ) c t ( 2 ) X t ( 2 ) ) d t ] ξ E [ 0 e α t U ( c t ( 1 ) X t ( 1 ) ) d t ] + ( 1 ξ ) E [ 0 e α t U ( c t ( 2 ) X t ( 2 ) ) d t ] ξ u ( x 1 ) + ( 1 ξ ) u ( x 2 ) ε .

Therefore, letting ε0, we obtain the concavity of u.

To prove (5.4), by Theorem 4.1, we recall that u is smooth. Furthermore, we get u (x)0 for x>0. If not, then u ( a 0 )<0 for some a 0 >0. By concavity,

0u(x) u ( a 0 )(x a 0 )+u( a 0 )as x,

which is a contradiction. Suppose that u (z)=0 for some z>0. Then, by concavity, we have u (x)=0 for all xz. Hence, by (1.5) and (1.6),

αu(z)=αu(x)= U ˜ (x,0)=U(x),xz.

This is contrary to (1.4). Thus, we obtain (5.4).

Next, by definition, we have

0<E [ 0 e α t U ( X ˇ t ) d t ] u(x),x>0,

where { X ˇ t } is the solution of (1.2) corresponding to c t =1. As in (2.7), the limit process χ ˇ t := lim x 0 + X ˇ t is different from 0. Hence

0<E [ 0 e α t U ( χ ˇ t ) d t ] u(0+).

Suppose that u (0+)<. By (1.5) and concavity, we get u(0+)=0, which is a contradiction. This implies (5.5). □

Lemma 5.2 Under (1.4), there exists a unique positive strong solution { X t } of (5.1).

Proof Let { N t } be the solution of (1.2) corresponding to c t =0. We can take the Brownian motion { B t } on the canonical probability space [[4], p.71]. Since 0η1, the probability measure P ˆ is defined by

d P ˆ /dP=exp { 0 t η ( N s ) / σ d B s 1 2 0 t ( η ( N s ) / σ ) 2 d s }

for every t0. Girsanov’s theorem yields that

B ˆ t := B t + 0 t η( N s )/σdsis a Brownian motion under  P ˆ .

Hence

d N t = [ ( N t ) γ η ( N t ) N t ] dt+σ N t d B ˆ t under  P ˆ .

Thus, (5.1) admits a weak solution.

Now, by (5.2), we have

η(x)x=min { ( U ) 1 u ( x ) , x } .

Hence, by (1.4) and concavity,

d d x ( U ) 1 u (x)= u ( x ) U ( U ) 1 u ( x ) 0.

Thus, η(x)x is nondecreasing on (0,). We rewrite (5.1) as the form of (2.4) to obtain X t >0 a.s. Then we see that the pathwise uniqueness holds for (5.1). Therefore, by the Yamada-Watanabe theorem [10], we deduce that (5.1) admits a unique strong solution { X t }. □

Proof of Theorem 5.1 Since { c t } satisfies (1.1), it belongs to . By Lemma 5.2, we note that

0< u (x)xu(x)u(0+)<u(x),x>0.

Hence, by (2.2) and (3.2),

E [ 0 t { e α s u ( X s ) X s } 2 d s ] E [ 0 t { e α s u ( X s ) } 2 d s ] E [ 0 t e α s ζ ( X s ) 2 d s ] < .

This yields that { 0 t e α s u ( X s ) X s d B s } is a martingale. By (1.6), (5.3) and Ito’s formula,

E [ e α t u ( X t ) ] = u ( x ) + E [ 0 t e α s { α u ( X s ) + ( X s ) γ u ( X s ) c s X s u ( X s ) + 1 2 σ 2 ( X s ) 2 u ( X s ) } d s ] = u ( x ) E [ 0 t e α s U ( c s X s ) d s ] .

By (2.1) and (3.2), it is clear that

E [ e α t u ( X t ) ] E [ e α t ζ ( X t ) ] e α t { ( 1 γ ) t + x ( 1 γ ) } 1 / ( 1 γ ) + e α t ζ 0 0 as  t .

Letting t, we deduce

E [ 0 e α t U ( c t X t ) d t ] =u(x).

By the same calculation as above, we obtain

E [ 0 e α t U ( c t X t ) d t ] u(x)

for any cA. The proof is complete. □

Remark 5.1 From the proof of Theorem 5.1, it follows that the solution u of the HJB equation (1.5) coincides with the value function. This implies that the uniqueness holds for (1.5).

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Acknowledgements

I would like to thank Professor H Morimoto for his useful help. The research was supported by the National Natural Science Foundation of China (11171275) and the Fundamental Research Funds for the Central Universities (XDJK2012C045).

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Liu, C. Optimal consumption of the stochastic Ramsey problem for non-Lipschitz diffusion. J Inequal Appl 2014, 391 (2014). https://doi.org/10.1186/1029-242X-2014-391

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Keywords

  • Hamilton-Jacobi-Bellman equation
  • viscosity solutions
  • Ramsey problem
  • Cobb-Douglas production function