A New Class of Problems in the Calculus of Variations
aa r X i v : . [ q -f i n . E C ] O c t A new class of problems in the calculus of variations
Ivar Ekeland ∗ , Yiming Long † , Qinglong Zhou ‡ CEREMADE and Institut de FinanceUniversit´e de Paris-IX, Dauphine, Paris, France Chern Institute of Mathematics and LPMCNankai University, Tianjin 300071, ChinaNovember 3, 2015Dedicated to Professor Alain Chenciner on his 70th birthday
In economic theory, and in optimal control, it has been customary to discount future gains at aconstant rate δ >
0. If an individual with utility function u ( c ) has the choice between severalstreams of consumption c ( t ), 0 ≤ t , he or she will choose the one which maximises the present value,given by: Z ∞ u ( c ( t )) e − δt dt (1)That future gains should be discounted is well grounded in fact. On the one hand, humansprefer to enjoy goods sooner than later (and to suffer bads later than sooner), as every child-rearingparent knows. On the other hand, it is also a reflection of our own mortality: 10 years from now,I may simply no longer be around to enjoy whatever I have been promised. These are two goodreasons why people are willing to pay a little bit extra to hasten the delivery date, or will requirecompensation for postponement, which is the essence of discounting.On the other hand, there is no reason why the discount rate should be constant, i.e. why thediscount factor should be an exponential e − δt . The practice probably arises from the compoundinterest formula lim ε → (1 − εδ ) t/ε = e − δt , when a constant interest rate δ is assumed, but even infinance, interest rates vary with the horizon: long-term rates can be widely different from short-term ones. As for economics, there is by now a huge amount of evidence that individuals use higherdiscount rates for the near future than for the long-term (see [18] for a review up to 2002). There ∗ E-mail: [email protected] † Partially supported by NNSF, MCME, RFDP, LPMC of MOE of China, and Nankai University. E-mail:[email protected] ‡ Partially supported by the Chern Institute of Mathematics, Nankai University, CEREMADE of Universit´e ParisDauphine, and the IHES. E-mail: [email protected]
1s also an aggregation problem: in a society where individuals use constant (but different) discountrates, the collective discount rate may be non-constant (see [14]). So the present value formula (1)should be replaced by the more general one: Z ∞ u ( c ( t )) h ( t ) dt (2)where h is a decreasing function, with h (0) = 1.But then a new problem arises, which is now well recognized in economic theory, but to ourknowledge has not yet received the attention it deserves in control theory. It is the problem of time-inconsistency , which runs as follows. Suppose the decision-maker has the choice between twostreams of consumption c ( t ) and c ( t ), starting at time T >
0. At time t = 0, he or she finds c ( t )yields the highest present value: Z ∞ T u ( c ( t )) h ( t ) dt > Z ∞ T u ( c ( t )) h ( t ) dt. (3)He or she then chooses c ( t ). When time T is reached, the present values are now: Z ∞ T u ( c ( t )) h ( t − T ) dt and Z ∞ T u ( c ( t )) h ( t − T ) dt (4)If h ( t ) = e − δt , then the ordering found at time t = 0 will persist at time t = T . Indeed: Z ∞ T u ( c ( t )) e − δ ( t − T ) dt = e δT Z ∞ T u ( c ( t )) e − δt dt so that the two terms in (4) are proportional to the two terms in (3). However, this is a peculiarityof the exponential function, and it is not to be expected with more general discount rates. Thedecision-maker then faces a basic rationality problem: what should he or she do ? To be morespecific, assume the state k ( t ) is related to the control c ( t ) by the dynamics: dkdt = f ( k ) − c ( t ) , k (0) = k (5) c ( t ) ≥ , k ( t ) ≥ h ( t ) = e − δt , the answer is to pick the optimal solution: if it isoptimal at time t = 0, it will still be optimal at all times T > t = 0 on, has his or her own optimal solution. No one agreeson what the optimal solution is, so optimality no longer provides an answer to the decision-makingprocess, and one must look for other concepts to describe rational behaviour.A clear requirement for rationality is that any strategy put forward be implementable. Supposea Markov strategy c = σ ( k ) is put forward at time t = 0. If it is to be followed at all later times t >
0, then it must be the case that the decision-maker at that time finds no incentive to deviate.More precisely, if he/she assumes that at all later times the strategy (closed-loop feedback) c = σ ( k )will be applied, then he/she should find it in his/her interest to apply σ as well. In other words, σ V ( k ), which is very similar to the value function in optimal control, and ofshowing that it satisfies a functional-differential equation which is reminescent of the Hamilton-Jacobi-Bellman (HJB) of optimal control. Conversely, any solution of that equation with suitableboundary conditions will give us an equilibrium strategy.In the work by Ekeland and Lazrak, this approach was applied to (2), with h ( t ) = α exp ( − r t ) +(1 − α ) exp ( − r t ), and it was showed that the corresponding problem had a continuum of equilib-rium strategies. In the present paper, in view of applications to economics, and of the mathematicalinterest, we aim to extend the analysis to the more general case:(1 − α ) Z ∞ e − r t u ( c ( t ) , k ( t )) dt + α Z ∞ e − r t U ( c ( t ) , k ( t )) dt (7)As a by-product of our analysis, we will treat the problem:(1 − α ) δ Z ∞ e − δt u ( c ( t ) , k ( t )) dt + α lim t →∞ U ( k ( t ) , c ( t )) (8)which was introduced by Chichilnisky (see [10], [11]) to model sustainable development. Note that,if u ( c, k ) ≥ { U ( c, k ) | c > , k > , c = f ( k ) } = ∞ then maximising (8) under the dynamics (5), (6) leads to the value + ∞ , so that optimisation isclearly not an answer to the problem. Instead, we find equilibrium strategies. To our knowledge,this is an entirely new result. We show that there is a continuum of such strategies. More precisely,there is a continuum of points k ∞ which can be realized as the long-term level of capital by anequilibrium strategy. This support, however, is one-sided, that is, k ∞ can be reached only from theinitial level of capitals k lying on its left (or from its right). To our knowledge, this is the first timesuch strategies have been identified.The structure of the paper is as follows. In the next section, we consider the problem of max-imising Z ∞ e − δt u ( c ( t ) , k ( t )) dt (9)under the dynamics (5), (6), and we show that it has a solution. On the way, we introduce thecorresponding HJB equation, and we show that it has a C solution. Next, we define equilibriumstrategies. With each such strategy we associate a value function V ( k ), and we show that it satisfiesan integro-differential equation which generalizes the HJB equation, and we prove a verificationtheorem: any solution of this equation with suitable boundary conditions gives an equilibriumstrategy. We show that the trajectories satisfy an integro-differential equation which generalizesthe classical Euler-Lagrange equations, and we connect the Ekeland-Lazrak approach with the Karpapproach.In section 4, we apply the theory to problem (7), and show that it has a continuum of equilibriumstrategies, thereby extending the results of [14]. It should be noted that the equations for V ( k ) are3iven in implicit form, that is, they cannot be solved with respect to V ′ ( k ), so that finding a C solution requires special techniques (first a blow-up, and then the central manifold theorem).We then consider the criterion:(1 − α ) δ Z ∞ e − δt u ( c ( t ) , k ( t )) dt + αr Z ∞ e − rt U ( c ( t ) , k ( t )) dt (10)which belongs to the class (7) and we let r →
0. In the limit, we get equilibrium strategies for theChichilnisky problem (8), which we describe explicitly.
This is the classical model for economic growth, originating with the seminal paper of Ramsey [31] in1928, and developed by Cass [9], Koopmans [25] and many others (see [5] for a modern exposition).We are given a point k > f on [0 , ∞ ), which is C ∞ on ]0 , ∞ [and satisfies the Inada conditions:lim x → f ′ ( x ) = + ∞ , lim x →∞ f ′ ( x ) ≤ . (11) Definition 1
A capital-consumption path ( c ( t ) , k ( t )) , t ≥ , is admissible if: k ( t ) > and c ( t ) > for all t, (12) dkdt = f ( k ) − c, k (0) = k , (13)The set of all admissible paths starting from k (i.e, such that k (0) = k ) will be denoted by A ( k ).We are given a number δ > u on ]0 , ∞ ), which is C ∞ on the interior, with u ′′ ( c ) > A ( k ): I ( c, k ) = Z ∞ u ( c ( t ) , k ( t )) e − δt dt, (14)and we consider the optimization problem:sup { I ( c, k ) | ( c, k ) ∈ A ( k ) } . (15)The Euler-Lagrange equation is given by: u ′′ dcdt = ( δ − f ′ ( k )) u ′ − u ′ − ( f ( k ) − c ) u ′′ . (16)Equation (16), together with equation (13), constitute a system of two first-order ODEs for theunknown functions ( c ( t ) , k ( t )). In the particular case when u = u ( c ) does not depend on k , thelast equation simplifies to: u ′′ dcdt = ( δ − f ′ ( k )) u ′ and the (16), (13) gives rise to a well-known phase diagram, with a hyperbolic stationary point( c ∞ , k ∞ ) characterized by f ′ ( k ∞ ) = δ and f ( k ∞ ) = c ∞ . The optimal solution of the Ramsey4roblem in that case then is the solution of (16), (13) which converges to ( c ∞ , k ∞ ) (see [5] forinstance).In the general case where u ( c, k ) depends on k , the situation is not as simple, and to ourknowledge has not been investigated. Stationary points ( c ∞ , k ∞ ) of the dynamics (if any) are givenby: c ∞ − f ( k ∞ ) = 0 (17)( δ − f ′ ( k ∞ )) u ′ ( f ( k ∞ ) , k ∞ ) − u ′ ( f ( k ∞ ) , k ∞ ) = 0 . (18)To prove the existence of an optimal strategy, we do not use the Euler equation. We use theHamilton-Jacobi-Bellman (HJB) equation instead. Introduce the optimal value as a function of theinitial point: V ( k ) := sup { I ( c, k ) | ( c, k ) ∈ A ( k ) } If there is an optimal solution ( c ( t ) , k ( t )), and it converges to ( c ∞ , k ∞ ) when t → ∞ , then,substituting in (14), we must have: V ( k ∞ ) = Z ∞ e − δt u ( c ∞ , k ∞ ) dt = 1 δ u ( c ∞ , k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) (19) Theorem 2 If V ( k ) is C , it satisfies the HJB equation, namely: δV ( k ) = max c { u ( c, k ) + ( f ( k ) − c ) V ′ ( k ) } . (20) Conversely, suppose the HJB equation has a C solution satisfying (19) for some ( c ∞ , k ∞ ) , anddefine a strategy c = σ ( k ) by: u ′ ( σ ( k ) , k ) = V ′ ( k ) (21) Suppose moreover that the solution of: dkdt = f ( k ) − σ ( k ) , k (0) = k , (22) converges to k ∞ for all initial points k . Then σ ( k ) is an optimal solution of the generalized Ramseyproblem (15) This is the so-called verification theorem, which is classical (see [4]). We need V to be C , sothat σ is defined. Since u ′′ >
0, we can use the implicit function theorem on equation (21) to define σ . If V is C , then σ is C , and the initial-value problem (22) has a unique solution. Note also thateverything is local: the functions V and σ are defined in some neighbourhood of k ∞ only, and theinitial value k is assumed to belong to that neighbourhood of k ∞ .So, to prove the (local) existence of an optimal strategy σ ( k ), we have to prove that the HJBequation has a C solution V with V ( k ∞ ) = δ − u ( f ( k ∞ ) , k ∞ ), and that the corresponding path k ( t ) converges to k ∞ . Then the right-hand side of (22) converges to 0, so that c ( t ) = σ ( k ( t ))converges to σ ( k ∞ ) = f ( k ∞ ). This is the content of the following two results5 heorem 3 Suppose there is some k ∞ > satisfying (18) and: u ′ f ′′ + u ′′ + f ′ (cid:18) u ′′ − u ′ u ′ u ′′ (cid:19) − u ′ u ′ u ′′ < (all values to be taken at k ∞ and c ∞ = f ( k ∞ ) ). Then there is an optimal strategy c = σ ( k ) converging to k ∞ If u = u ( c ) does not depend on k , then u ′ = 0. Equation (18) becomes f ′ ( k ∞ ) = δ ,which defines k ∞ uniquely because of the Inada conditions (11) on f , and condition (23) becomes u ′ ( c ∞ ) f ′′ ( k ∞ ) <
0, which is satisfied automatically. So there is an optimal strategy in that case,and one can even show that it is globally defined. For u depending on k , however, the situation isdifferent. Proof.
We follow the method of [15]. By the inverse function theorem, the equation u ′ ( c, k )) = p defines c as a C function of p and k : c = ϕ ( p, k ), u ′ ( ϕ ( p, k ) , k )) = p (24)We rewrite (20) as a Pfaff system: dV = pdk, (25) p ( f ( k ) − ϕ ( p, k )) + u ( ϕ ( p, k ) , k ) = δV, (26)and we seek a C solution V satisfying: V ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) . (27)Differentiating (26) leads to: δdV = ( f ( k ) − ϕ ( p, k )) dp + pf ′ ( k ) dk + u ′ ( ϕ ( p, k ) , k ) dk (28)Plugging (28) into (25), we get:( f ( k ) − ϕ ( p, k )) dp = ( δp − pf ′ ( k ) − u ′ ( ϕ ( p, k ) , k )) dk. (29)Introducing an auxiliary variable t , we rewrite this as a system of two ODES for two functions p ( t ) and k ( t ): dkdt = f ( k ) − ϕ ( p, k ) dpdt = δp − pf ′ ( k ) − u ′ ( ϕ ( p, k ) , k ) (30)with the initial condition ( k (0) , p (0)) = ( k ∞ , p ∞ )By (24), we must have p ∞ = u ′ ( ϕ ( p ∞ , k ∞ ) , k ∞ ) = u ′ ( c ∞ , k ∞ ) = u ′ ( f ( k ∞ ) , k ∞ )6ifferentiating (24) with respect to p and k respectively, we derive the formulas: ϕ ′ ( p ∞ , k ∞ ) = 1 u ′′ ( ϕ ( p ∞ , k ∞ ) , k ∞ ) = 1 u ′′ ( f ( k ∞ ) , k ∞ ) , (31) ϕ ′ ( p ∞ , k ∞ ) = − u ′′ ( ϕ ( p ∞ , k ∞ ) , k ∞ ) u ′′ ( ϕ ( p ∞ , k ∞ ) , k ∞ ) = − u ′′ ( f ( k ∞ ) , k ∞ ) u ′′ ( f ( k ∞ ) , k ∞ ) , (32)We can now linearizd (30) at the point ( p ∞ , k ∞ ). We get: ddt (cid:18) k − k ∞ p − p ∞ (cid:19) = A ∞ (cid:18) k − k ∞ p − p ∞ (cid:19) . where the constant matrix A ∞ is given by: A ∞ := f ′ + u ′′ u ′′ , − u ′′ − u ′ f ′′ − u ′′ + ( u ′′ ) u ′′ , δ − f ′ − u ′′ u ′′ . and all the values are to be taken at ( k ∞ , p ∞ ). The characteristic polynomial is: λ − δλ − [ u ′ ∞ f ′′∞ + u ′′ ∞ u ′′ ∞ + f ′∞ ( u ′′ ∞ u ′′ ∞ − u ′ ∞ u ′ ∞ ) − u ′ ∞ u ′ ∞ u ′′ ∞ u ′′ ∞ ] = 0where we used f ′∞ + u ′ ∞ u ′ ∞ = δ by (17). Because of assumption (23), it has two real roots withdifferent signs, λ + > λ − <
0. Thus ( k ∞ , p ∞ ) is a hyperbolic fixed point of (30), with a stable C ∞ -manifold S which corresponds to λ − and an unstable C ∞ -manifold U which corresponds to λ + .Choose a smooth parametrization ( k s ( x ) , p s ( x )) for the curve S . The tangent at the fixed point is: dp s dk s ( k ∞ ) = u ′′ ( f ′∞ − λ − ) + u ′′ and plugging k = k s ( x ) , p = p s ( x ) into equation (26), we get V s = p s ( x )( f ( k s ( x )) − ϕ ( p s ( x ) , k s ( x ))) + u ( ϕ ( p s ( x ) , k s ( x )) , k s ( x )) δ . Moreover, differentiating (26), and using (23) again, we find: dV s dk s ( k ∞ ) = u ′ ( f ( k ∞ ) , k ∞ ) > , It follows that the curve in parametric form x → ( k ( x ) , V ( x )) is in fact the graph of a function V ( k ) which solves HJB and satisfies (27). By (25), we have d V s dk s ( k ∞ ) = dp s dk s ( k ∞ ) = u ′′ ∞ ( f ′∞ − λ − ) + u ′′ ∞ . (33)Thus V s ( k ) is C in k at k ∞ , and the C property at other points near k ∞ follows from (30).7t remains to show that the strategy σ defined by u ′ ( σ ( k ) , k ) = V ′ ( k ) converges to ( f ( k ∞ ) , k ∞ ).We rewrite σ as: σ ( k ) = ϕ ( V ′ ( k ) , k ) . Linearizing the equation dkdt = f ( k ) − c = f ( k ) − ϕ ( V ′ ( k ) , k )gives d ( k − k ∞ ) dt = λ − ( k − k ∞ )and this concludes the proof.Let us show how to deduce the Euler equation (16) from the HJB equation (20) and the optimalstrategy (21). Setting c ( t ) := σ ( k ( t )), differentiating (20) with respect to k , and applying theenvelope theorem, we get: δV ′ ( k ) = u ′ ( c, k ) + f ′ ( k ) V ′ ( k ) + ( f ( k ) − c ) V ′′ ( k ) , and hence, noting that dkdt = f ( k ) − c ( t ) :( δ − f ′ ( k )) V ′ ( k ) − u ′ ( c, k ) = ( f ( k ) − c ) ddk V ′ ( k ) = ddt V ′ ( k ) . Replacing V ′ ( k ) by u ′ ( σ ( k ) , k ) = u ′ ( c, k ), we get:( δ − f ′ ( k )) u ′ ( c, k ) − u ′ ( c, k ) = ddt u ′ ( c, k ) , which is precisely the Euler equation. We consider the intertemporal decision problem (as it seen at time t = 0) J ( c, k ) = Z ∞ [ h ( t ) u ( c ( t ) , k ( t )) + H ( t ) U ( c ( t ) , k ( t ))] dt (34)under the dynamics described by (5) and (6). Here h and H are discount factors, i.e. C ∞ non-increasing functions on [0 , ∞ ), such that h (0) = H (0) = 1 and h ( ∞ ) = H ( ∞ ) = 0, while u and U are utility functions. They are assumed to be C ∞ on ]0 , ∞ ) , with u ′′ < U ′′ < ρ > T > h ( t ) < e − ρt and H ( t ) < e − ρt for t ≥ T .Because of time-inconsistency, the decision problem can no longer be seen as an optimizationproblem. There is no way for the decision-maker at time 0 to achieve what is, from her point ofview, the first-best solution of the problem, and she must turn to a second-best policy: the best she8an do is to guess what her successor are planning to do, and then to plan her own consumption c (0) accordingly. In other word, we will be looking for a subgame-perfect equilibrium of the leader-follower game played by successive generations.The equilibrium policy was described in [14] for the case when the criterion (34) did not includethe second term, and u did not depend on k . We will extend this analysis to the present situation,and then compare it with the approach in [22]. Definition 4
A Markov strategy c = σ ( k ) is convergent if there is a point k ∞ and a neighbourhood N of k ∞ such that, for every k ∈ N the solution k ( t ) of (22) converges to k ∞ (and so c ( t ) = σ ( k ( t )) converges to c ∞ = f ( k ∞ ) ). If this is the case, we shall say that ( c ∞ , k ∞ ) is supported by σ . Let us suppose that a convergent Markov strategy σ has been announced and is the publicknowledge σ . The decision-maker at T has capital stock k T . If all future decision-maker apply thestrategy σ , the resulting future capital stock flow k ( t ) obeys: dkdt = f ( k ) − σ ( k ) , k ( T ) = k T , T ≤ t (35)Since every decision-maker faces the same problem (with different stock levels) it is enough totake T = 0. Suppose the decision-maker at time 0 holds power for 0 ≤ t < ε , and expects all laterdecision-makers to apply the strategy σ . He or she then explores whether it is in his or her interestto apply the strategy σ , that is, to play c = σ ( k ) for 0 ≤ t < ε . If he or she applies the constantcontrol c for 0 ≤ t ≤ ε .Suppose the constant control c is use on 0 ≤ t ≤ ε . The immediate utility flow during [0 , ε ] is[ u ( c, k ) + U ( c, k )] ε + o ( ε ) where o ( ε ) is a higher order term of ε . At time ε , the resulting capitalwill be k + ( f ( k ) − c ) ε + o ( ε ). From then on, the strategy σ will be applied, which results in acapital stock k c satisfying (we omit higher-order terms): dk c dt = f ( k c ) − σ ( k c ) , k c ( ε ) = k + ( f ( k ) − c ) ε, t ≥ ε (36)The capital stock k c can be written as k c ( t ) = k ( t ) + k ( t ) ε, (37)where k ( t ) is the unperturbed solution, and k ( t ) is given by the linearized equation: dk dt = f ( k ) − σ ( k ) , k (0) = k (38) dk dt = ( f ′ ( k ) − σ ′ ( k )) k , k (0) = σ ( k ) − c (39)Evaluating the integral (34) we get: J ( ε ) = u ( c, k ) ε + R ∞ ε h ( s ) u ( σ ( k ( t ) + εk ( t )) , k ( t ) + εk ( t )) dt + U ( c, k ) ε + R ∞ ε H ( s ) U ( σ ( k ( t ) + εk ( t )) , k ( t ) + εk ( t )) dt Letting ε →
0, so that commitment span of the decision-maker vanishes, we get:lim ε → ε ( J ( ε ) − J (0)) = P ( k , σ, c )9here P ( k , σ, c ) = u ( c, k ) − u ( σ ( k ) , k ) + ( U ( c, k ) − U ( σ ( k ) , k ))+ Z ∞ h ( t ) u ′ ( σ ( k ( t )) , k ( t )) σ ′ ( k ( t )) k ( t ) dt + Z ∞ h ( t ) u ′ ( σ ( k ( t )) , k ( t )) k ( t ) dt + Z ∞ H ( t ) U ′ ( σ ( k ( t )) , k ( t )) σ ′ ( k ( t )) k ( t ) dt + Z ∞ H ( t ) U ′ ( σ ( k ( t )) , k ( t )) k ( t ) dt, Definition 5
A convergent Markov strategy σ is an equilibrium if we have: max c P ( k, σ, c ) = P ( k, σ, σ ( k )) , ∀ k. We now characterizes the equilibrium strategy. We write k c ( t ) = K ( t ; k , σ ) where K is the flowassociated with the differential equation (38). We also define a function ϕ by: u ′ ( ϕ ( x, k ) , k ) + U ′ ( ϕ ( x, k ) , k ) = x,ϕ ( u ′ ( c, k ) + U ′ ( c, k ) , k ) = c Since u ′′ and U ′′ are both negative, the function ϕ is well-defined by the implicit function theorem. Theorem 6
Let σ be an equilibrium strategy. The function: V ( k ) = Z ∞ h ( t ) u ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt + Z ∞ H ( t ) U ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt (40) satisfies the integral equation: V ( k ) = R ∞ h ( t ) u ( ϕ ◦ V ′ ( K ( t ; k , ϕ ◦ V ′ )) , K ( t ; k , ϕ ◦ V ′ )) dt + R ∞ H ( t ) U ( ϕ ◦ V ′ ( K ( t ; k , ϕ ◦ V ′ )) , K ( t ; k , ϕ ◦ V ′ )) dt (IE) and the instantaneous optimality condition u ′ ( σ ( k ) , k ) + U ′ ( σ ( k ) , k ) = V ′ ( k ) , σ ( k ) = ϕ ( V ′ ( k ) , k ) , (41) Conversely, suppose a function V is twice continuously differentiable, satisfies (IE), and thestrategy σ ( k ) := ϕ ( V ′ ( k ) , k ) is convergent. Then σ is an equilibrium strategy. For the sake of convenience, we have shortened σ ( k ) := ϕ ( V ′ ( k ) , k ) to σ = ϕ ◦ V ′ . Proof.
Since the system is autonomous, we have: K ( s ; K ( t ; k , σ ) , σ ) = K ( s + t ; k , σ ) . (42)Next, denote the fundamental solution of the linearized equation of (35) at k by R ( k ; t ) sothat: k ( t ) = R ( k ; t )( σ ( k ) − c ) d R dt = ( f ′ ( K ( t ; k , σ )) − σ ′ ( K ( t ; k , σ ))) R ( t ) , R ( k ; 0) = I, and K are related by: ∂ K ( t ; k , σ ) ∂k = R ( k ; t ) , (43) ∂ K ( t ; k , σ ) ∂t = f ( K ( t ; k , σ )) − σ ( K ( t ; k , σ )) . (44)Let us now turn to the first part of the theorem. Differentiating (40)with respect to k : V ′ ( k ) = Z ∞ h ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) R ( k ; t ) dt + Z ∞ h ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) R ( k ; t ) dt + Z ∞ H ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) R ( k ; t ) dt + Z ∞ H ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) R ( k ; t ) dt. (45)Substituting k ( t ) = K ( t ; k , σ ) and k ( t ) in the definition of P , we get: P ( k , σ, c ) = [ u ( c, k ) + U ( c, k )] − [ u ( σ ( k ) , k ) + U ( σ ( k ) , k )]+ Z ∞ h ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) R ( k ; t )( σ ( k ) − c ) dt + Z ∞ h ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) R ( k ; t )( σ ( k ) − c ) dt + Z ∞ H ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) R ( k ; t )( σ ( k ) − c ) dt + Z ∞ H ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) R ( k ; t )( σ ( k ) − c ) dt. Since u and U are strictly concave and differentiable with respect to c , the necessary and sufficientcondition to maximize P ( k , σ, c ) with respect to c is that the derivative vanishes at c = σ ( k ), thatis: u ′ ( σ ( k ) , k ) + U ′ ( σ ( k ) , k ) = Z ∞ h ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) R ( k ; t ) dt + Z ∞ h ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) R ( k ; t ) dt + Z ∞ H ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) R ( k ; t ) dt + Z ∞ H ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) R ( k ; t ) dt, The right-hand side is precisely V ′ ( k ), as we wanted. Therefore, the equilibrium strategy satisfies u ′ ( σ ( k ) , k ) + U ′ ( σ ( k ) , k ) = V ′ ( k )11nd we have σ ( k ) = ϕ ( V ′ ( k ) , k ). Substituting back into equation (40), we get the functionalequation (IE). This prove the first part of the theorem (necessity). We refer to [14] for the secondpart (sufficiency).The following theorem gives an alternative characterization, the differential equation, whichresembles the usual HJB equation from the calculus of variation. Theorem 7
Let V be a C function such that the strategy σ = ϕ ◦ V ′ converges to ¯ k . Then V satisfies the integral equation (IE) if and only if it satisfies the following integro-differential equation u ( ϕ ◦ V ′ ( k ) , k ) + U ( ϕ ◦ V ′ ( k ) , k ) + V ′ ( k )( f ( k ) − ϕ ( V ′ ( k ) , k ))= − R ∞ h ′ ( t ) u ( ϕ ◦ V ′ ( K ( t ; k , ϕ ◦ V ′ )) , K ( t ; k , ϕ ◦ V ′ )) dt − R ∞ H ′ ( t ) U ( ϕ ◦ V ′ ( K ( t ; k , ϕ ◦ V ′ )) , K ( t ; k , ϕ ◦ V ′ )) dt (DE) together with the boundary condition V (¯ k ) = u ( f (¯ k ) , ¯ k ) Z ∞ h ( t ) dt + U ( f (¯ k ) , ¯ k ) Z ∞ H ( t ) dt. (BC) Proof.
Introduce the function φ defined by φ ( k ) = V ( k ) − Z ∞ h ( t ) u ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt − Z ∞ H ( t ) U ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt, where σ ( k ) = ϕ ( V ′ ( k ) , k ). Consider the value ψ ( t, k ) of φ along the trajectory t → K ( t ; k , σ )originating from k at time 0, that is ψ ( t, k ) = φ ( K ( t ; k , σ ))= V ( K ( t ; k , σ )) − Z ∞ t h ( s − t ) u ( σ ( K ( σ ; s, k )) , K ( σ ; s, k )) ds − Z ∞ t H ( s − t ) U ( σ ( K ( σ ; s, k )) , K ( σ ; s, k )) ds, Then the derivative of ψ with respect to t is given by: ∂ψ ( t, k ) ∂t = V ′ ( K t )[ f ( K t ) − ϕ ◦ V ′ ( K t )] + u ( ϕ ◦ V ′ ( K t ) , K t ) + U ( ϕ ◦ V ′ ( K t ) , K t )+ Z ∞ h ′ ( s ) u ( ϕ ◦ V ′ ( K ( s ; K t , ϕ ◦ V ′ )) , K ( s ; K t , ϕ ◦ V ′ )) ds + Z ∞ H ′ ( s ) U ( ϕ ◦ V ′ ( K ( s ; K t , ϕ ◦ V ′ )) , K ( s ; K t , ϕ ◦ V ′ )) ds, where we have denoted K ( t ; k , ϕ ◦ V ′ ) by K t for convenience. If (DE) holds, then the right hand sideis identically zero along the trajectory, so that ψ ( t, k ) does not depend on t , thus ψ ( s, k ) = ψ ( t, k )for all s, t ≥
0. Letting t → ∞ in the definition of ψ , we get ψ ( s, k ) = lim t →∞ ψ ( t, k )= V (¯ k ) − Z ∞ h ( s ) u ( σ (¯ k ) , ¯ k ) ds − Z ∞ H ( s ) U ( σ (¯ k ) , ¯ k ) ds, (46)and hence, if (BC) holds, then ψ = φ ≡ V ( k ) satisfiesequation (IE), then the same lines of reasoning shows that equation (DE) and the boundary condition(BC) are satisfied. 12 .3 The Euler equations To obtain the Euler-Lagrange-like equation, we differentiate the both side of (DE) with respect to k . We get: − Z ∞ h ′ ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) ∂ K ( t ; k , σ ) ∂k dt − Z ∞ h ′ ( t ) u ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) ∂ K ( t ; k , σ ) ∂k dt − Z ∞ H ′ ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) σ ′ ( K ( t ; k , σ )) ∂ K ( t ; k , σ ) ∂k dt − Z ∞ H ′ ( t ) U ′ ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) ∂ K ( t ; k , σ ) ∂k dt = [ u ′ ( σ ( k ) , k ) + U ′ ( σ ( k ) , k )] σ ′ ( k ) + u ′ ( σ ( k ) , k ) + U ′ ( σ ( k ) , k )+ V ′ ( k ) f ′ ( k ) − V ′ ( k ) σ ′ ( k ) + V ′′ ( k )( f ( k ) − σ ( k )) . Plugging k = k ( t ), σ ( k ( t )) = c ( t ) and using (41) to cancel the first and the fifth terms, togetherwith (13) and (43), we have − Z ∞ h ′ ( s ) u ′ ( σ ( K ( s ; k ( t ) , σ )) , K ( s ; k ( t ) , σ )) σ ′ ( K ( s ; k ( t ) , σ )) R ( k ( t ) , s ) ds − Z ∞ h ′ ( s ) u ′ ( σ ( K ( s ; k ( t ) , σ )) , K ( s ; k ( t ) , σ )) R ( k ( t ) , s ) ds − Z ∞ H ′ ( s ) U ′ ( σ ( K ( s ; k ( t ) , σ )) , K ( s ; k ( t ) , σ )) σ ′ ( K ( s ; k ( t ) , σ )) R ( k ( t ) , s ) ds − Z ∞ H ′ ( s ) U ′ ( σ ( K ( s ; k ( t ) , σ )) , K ( s ; k ( t ) , σ )) R ( k ( t ) , s ) ds = u ′ ( c ( t ) , k ( t )) + U ′ ( c ( t ) , k ( t )) + V ′ ( k ( t )) f ′ ( k ( t )) + V ′′ ( k ( t ))( f ( k ( t )) − c ( t ))= u ′ ( c ( t ) , k ( t )) + U ′ ( c ( t ) , k ( t )) + [ u ′ ( c ( t ) , k ( t )) + U ′ ( c ( t ) , k ( t ))] f ′ ( k ( t ))+ ddt [ u ′ ( c ( t ) , k ( t )) + U ′ ( c ( t ) , k ( t ))] , where to get the last two terms in the right hand side, we have used (41) again.We have K ( s ; k ( t ) , σ ) = k ( s + t ), and R ( k ( t ) ; s ) = R ( k ; s + t ) σ ′ ( k ( t )) = 1 f ( k ( t )) − c ( t ) dcdt (47) R ( k ; s + t ) = exp( f ′ ( k ( t )) − f ( k ( t )) − c ( t ) dcdt ) , − Z ∞ t h ′ ( s − t ) [ u ′ ( c ( s ) , k ( s )) γ ( s ) + u ′ ( c ( s ) , k ( s ))] e f ′ ( k ( s )) − γ ( s ) ds (48) − Z ∞ t H ′ ( s − t ) [ U ′ ( c ( s ) , k ( s )) γ ( s ) + U ′ ( c ( s ) , k ( s ))] e f ′ ( k ( s )) − γ ( s ) ds (49)= u ′ ( c ( t ) , k ( t )) + U ′ ( c ( t ) , k ( t )) + [ u ′ ( c ( t ) , k ( t )) + U ′ ( c ( t ) , k ( t ))] f ′ ( k ( t )) (50)+ ddt [ u ′ ( c ( t ) , k ( t )) + U ′ ( c ( t ) , k ( t ))] (51)with: γ ( s ) := 1 f ( k ( s )) − c ( s ) dcds (52)which is the Euler-Lagrange-like equation for the time-inconsistent case. Karp [22], [23] has developed a different method to deal with time-inconsistency. In this section, weconnect his results with ours.Defining V ( k ) as above, we must have: V ( k ) = max c (cid:26) u ( c, k ) + U ( c, k )] ε + Z ∞ ε [ h ( t ) u ( σ ( k c ( t )) , k c ( t )) + H ( t ) U ( σ ( k c ( t )) , k c ( t ))] dt (cid:27) (53)On the other hand, we have V ( k c ( ε )) = Z ∞ ε [ h ( s − ε ) u ( σ ( k c ( s )) , k c ( s )) + H ( s − ε ) U ( σ ( k c ( s )) , k c ( s ))] ds, Substituting: V ( k ) = max c [ u ( c, k ) + U ( c, k )] ε + V ( k c ( ε ))+ R ∞ ε [ h ( t ) − h ( t − ε )] u ( σ ( k c ( t )) , k c ( t )) dt + R ∞ ε [ H ( t ) − H ( t − ε )] U ( σ ( k c ( t )) , k c ( t )) dt Letting ε →
0, we have − Z ∞ h ′ ( t ) u ( σ ( k ( t )) , k ( t )) dt − Z ∞ H ′ ( t ) U ( σ ( k ( t )) , k ( t )) dt = max c { u ( c, k ) + U ( c, k ) + V ′ ( k )( f ( k ) − c ) } . (54)This equation was first obtained by Karp [22], [23]. This is the HJB equation for a certaincontrol problem, which he terms the auxiliary control problem. We now show that this approach isequivalent to the preceding one. approach by Ekeland-Lazrak. In the right hand side of (54), themaximum is attained at: c = arg max c { u ( c, k ) + U ( c, k ) + V ′ ( k )( f ( k ) − c ) } = ϕ ( V ′ ( k ) , k ) , − Z ∞ h ′ ( t ) u ( σ ( k ( t )) , k ( t )) dt − Z ∞ H ′ ( t ) U ( σ ( k ( t )) , k ( t )) dt = u ( ϕ ( V ′ ( k ) , k ) , k ) + U ( ϕ ( V ′ ( k ) , k ) , k ) + V ′ ( k )( f ( k ) − ϕ ( V ′ ( k ) , k )) . which is exactly the same as (DE) In this section, we consider the biexponential criterion λ Z ∞ e − δ s u ( c ( s ) , k ( s )) ds + (1 − λ ) Z ∞ e − δ s U ( c ( s ) , k ( s )) ds. (55)Without loss of generality, we assume that: δ > δ If λ = 0 or 1, then this is just the Ramsey criterion (14). Thus we are interested in the case0 < λ <
1. Dividing by λ , we find that (55) is a special case of (34), where h ( t ) = e − δ t , H ( t ) = e − δ t and U is replaced by − λλ U . So all the results of the preceding section hold. Given an equilibrium strategy σ λ , introduce the two functions: V λ ( k ) := Z ∞ e − δ t u ( σ λ ( k ( t )) , k ( t )) dt + 1 − λλ Z ∞ e − δ t U ( σ λ ( k ( t )) , k ( t )) dt (56) W λ ( k ) := Z ∞ e − δ t u ( σ λ ( k ( t )) , k ( t )) dt − − λλ Z ∞ e − δ t U ( σ λ ( k ( t )) , k ( t )) dt (57)In Proposition 8, we prove that the HJB-type equation (DE) reduces to a system of two ODEsfor V λ and W λ : ( f − ϕ λ ( V ′ λ )) V ′ λ + u ( V ′ λ ) + 1 − λλ U ( V ′ λ ) = δ + δ V λ + δ − δ W λ , (58)( f − ϕ λ ( V ′ λ )) W ′ λ + u ( V ′ λ ) − − λλ U ( V ′ λ ) = δ − δ V λ + δ + δ W λ (59)where ϕ λ ( V ′ λ ) and u ( V ′ λ ) denote the functions k → ϕ λ ( V ′ λ ( k ) , k ) and k → u ( ϕ λ ( V ′ ) , k ). Recallthat ϕ λ is defined by the equivalent equations: u ′ ( ϕ λ ( x, k ) , k ) + 1 − λλ U ′ ( ϕ λ ( x, k ) , k ) = x, (60) ϕ λ ( u ′ ( c, k ) + 1 − λλ U ′ ( c, k ) , k ) = c (61)15imilarly, the boundary condition (BC) becomes: V λ ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) + 1 − λλ δ U ( f ( k ∞ ) , k ∞ ) , (62) W λ ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) − − λλ δ U ( f ( k ∞ ) , k ∞ ) . (63)If λ = 1, we find the usual HJB equation (20) for V with the boundary condition (19). Proposition 8
Suppose that σ λ is an equilibrium strategy such that k ( t ) converges to k ∞ . Then V λ and W λ defined by (56) and (57) satisfy the equations (58) and (59) with the boundary conditions(62) and (63). Conversely, suppose there is a point k ∞ , a C function V λ and a C function W λ , both defined on some open neighbourhood of k ∞ , satisfying the equations (58) and (59) with theboundary conditions (62) and (63); suppose moreover that strategy σ λ ( k ) := ϕ λ ( V ′ λ ( k ) , k ) convergesto ( f ( k ∞ ) , k ∞ ) . Then σ λ is an equilibrium strategy. Let us draw the reader’s attention to the fact that V λ must be C while W λ needs only be C . Proof.
Let us simplify the notation. Write σ instead of σ λ and set a = δ + δ , b = δ − δ k ( t ) = k ∞ and c ( t ) = σ ( k ∞ ) in (58) and (59).Conversely suppose v and w satisfy (58), (59) and (62), (63), and suppose the strategy σ = ϕ λ ◦ v ′ converges to k . Consider the following functions v ( k ) = Z ∞ e − δ t u ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt + 1 − λλ Z ∞ e − δ t U ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt,w ( k ) = Z ∞ e − δ t u ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt − − λλ Z ∞ e − δ t U ( σ ( K ( t ; k , σ )) , K ( t ; k , σ )) dt. Arguing as in Theorem 7, we find that v and w also satisfy (58), (59) and (62), (63). Setting v = v − v , w = w − w , then we get( f ( k ) − σ ( k )) v ′ ( k ) + u ( σ ( k ) , k ) + − λλ U ( σ ( k ) , k ) = av ( k ) + bw ( k )( f ( k ) − σ ( k )) w ′ ( k ) + u ( σ ( k ) , k ) + − λλ U ( σ ( k ) , k ) = bv ( k ) + aw ( k ) (64)with the boundary conditions v ( k ∞ ) = 0 w ( k ∞ ) = 0 (65)Obviously, v = w = 0 is a solution. Lemma 9 below shows that We need to show that it is theonly one, so that v = v and w = w . This is the desired result.16 emma 9 If ( v , w ) is a pair of continuous functions on a neighborhood k ∞ , continuously differ-entiable for k = k ∞ , and which solve (64) with boundary conditions (65), then v = w = 0 . Proof.
Set f ( k ) − σ ( k ) = ξ ( k ), then ξ ( k ) → k → k ∞ . The system (64) can be rewritten as: (cid:18) ξv ′ ξw ′ (cid:19) = (cid:18) a bb a (cid:19)(cid:18) v w (cid:19) . Setting: (cid:18) VW (cid:19) = q q − q q (cid:18) v w (cid:19) , we have: ξ (cid:18) V ′ W ′ (cid:19) = (cid:18) δ δ (cid:19)(cid:18) VW (cid:19) . The first equation yields V ( k ) = V ( k ) exp R kk δ ξ ( u ) du . Without loss of generality, we can assumethat k < k ∞ .Let S = { k | ξ ( k ) = 0 , k ≤ k ≤ k ∞ } , where k is the initial stock. We consider the followingtwo cases: First case : S = { k ∞ } . Then ξ ( k ) > k ∈ [ k , k ∞ ). So dkdt = ξ ( k ) >
0, and:0 = r
12 ( v ( k ∞ ) + w ( k ∞ )) = V ( k ∞ ) = V ( k ) exp Z kk δ ξ ( u ) du It follows that V ( k ) = 0 for all k . Second case: S \{ k ∞ } 6 = ∅ . Then S is a closed set and V ( k ) = ξ ( k ) V ′ ( k ) δ = 0 in S . Thecomplement of S is a countable union of disjoint intervals, and ξ vanishes at the endpoints of eachinterval. Arguing as above it follows that V ( k ) vanishes for k ≤ k ≤ k ∞ .The same argument holds for W ( k ). This concludes the proof. The Euler-type equations (48) to (52) can be reduced to a non-autonomous system of four first-orderODEs. It is better to work directly on (58) and (59) to get an autonomous system. Proceeding asin the end of Section 2, we differentiate (58) and (59) w.r.t. k , and notice that ( f − σ ) V ′′ λ = dkdt V ′ λ dk = ddt V ′ λ , we get: δ + δ V ′ λ + δ − δ W ′ λ = ddt V ′ λ + ( f ′ − σ ′ ) V ′ λ + ( u ′ + 1 − λλ U ′ ) σ ′ + u ′ + 1 − λλ U ′ δ − δ V ′ λ + δ + δ W ′ λ = ddt W ′ λ + ( f ′ − σ ′ ) W ′ λ + ( u ′ − − λλ U ′ ) σ ′ + u ′ − − λλ U ′ c = ϕ ( V ′ λ ( k ) , k ) and σ ′ is given by (47). Using formula (60), and setting w ( t ) := W ′ λ ( k ( t )),this becomes: δ + δ (cid:18) u ′ + 1 − λλ U ′ (cid:19) + δ − δ w = ddt (cid:18) u ′ + 1 − λλ U ′ (cid:19) + ( f ′ − σ ′ ) (cid:18) u ′ + 1 − λλ U ′ (cid:19) + (cid:18) u ′ + 1 − λλ U ′ (cid:19) σ ′ + u ′ + 1 − λλ U ′ ,δ − δ (cid:18) u ′ + 1 − λλ U ′ (cid:19) + δ + δ w = dwdt + ( f ′ − σ ′ ) w + (cid:18) u ′ − − λλ U ′ (cid:19) σ ′ + u ′ − − λλ U ′ , and hence( δ + δ − f ′ ) ( λu ′ + (1 − λ ) U ′ ) + δ − δ λw = ddt ( λu ′ + (1 − λ ) U ′ ) + λu ′ + (1 − λ ) U ′ , (66) δ − δ λu ′ + (1 − λ ) U ′ ) + ( δ + δ − f ′ ) λw = λ dwdt + ( − λw + λu ′ − (1 − λ ) U ′ ) σ ′ + λu ′ − (1 − λ ) U ′ . (67)to which we should add: dkdt = f ( k ) − c (68)The system (66) to (68) is a system of three first-order ODEs for the unknown functions k ( t ) , c ( t )and w ( t ). Note that for λ = 0, (66) and (67) reduce to two copies of the usual Euler-Lagrangeequation. For λ = 1, taking w ( t ) = u ′ ( c ( t ) , k ( t )) gives us again two copies of the same equation. Equation (54) is the HJB equation for the following:max c ( . ) Z ∞ e − rt (cid:20) u ( c ( t ) , k ( t )) + 1 − λλ U ( c ( t ) , k ( t )) − K ( k ( t )) (cid:21) dt (69) dkdt = f ( k ) − c ( t ) , k (0) = k (70)where we seek a feedback σ ( k ) such that: K ( k ) = (cid:26) ( δ − r ) Z ∞ e − δt u ( σ ( k ( t )) , k ( t )) dt | dkdt = f ( k ) − σ ( k ) k (0) = k (cid:27) (71)Solving (69), (70) under the constraint (71) is a fixed-point problem for the feedback c = σ ( k ). Define a function g λ ( k ) by: g λ ( k ) := λδ u ′ ( f ( k ) , k ) + (1 − λ ) δ U ′ ( f ( k ) , k ) λu ′ ( f ( k ) , k ) + (1 − λ ) U ′ ( f ( k ) , k ) − λu ′ ( f ( k ) , k ) + (1 − λ ) U ′ ( f ( k ) , k ) λu ′ ( f ( k ) , k ) + (1 − λ ) U ′ ( f ( k ) , k ) (72)18 heorem 10 Assume g λ ( k ) = 0 . If there is some k ∞ such that f ′ ( k ∞ ) = g λ ( k ∞ ) , (73) then the equations (58) and (59) with the boundary conditions (62) and (63) have a solution ( V, W ) near the point k ∞ with V of class C and W of class C . Proof.
We adapt the argument in [14] to the present situation. We note that the boundary-valueproblem (58), (59), (62), (63) cannot be reduced to a standard initial-value problem for the pair( V λ , W λ ). To see that, rewrite equation (58) as follows:( f ( k ) − ϕ ( V ′ λ , k )) V ′ λ + u ( ϕ ( V ′ λ , k ) , k ) + 1 − λλ U ( ϕ ( V ′ λ , k ) , k ) = aV λ + bW λ . Since the function c → u ( c, k ) + − λλ U ( c, k ) is concave, the function: c → u ( c, k ) + 1 − λλ U ( c, k ) − cx attains its maximum at the point c = ϕ λ ( x, k ) defined by (61). We set: u ∗ λ ( x, k ) = max c (cid:26) u ( c, k ) + 1 − λλ U ( c, k ) − cx (cid:27) = u ( ϕ λ ( x, k ) , k )+ 1 − λλ U ( ϕ λ ( x, k ) , k ) − ϕ λ ( x, k ) x (74)The function x → u ∗ λ ( x, k ) is convex, and the equation (58) becomes: f ( k ) V ′ λ + u ∗ λ ( V ′ λ , k ) = aV λ + bW λ . (75)This is an equation for V ′ λ . From the basic duality results in convex analysis (see for instance[17]), we find that: min y { f ( k ) y + u ∗ λ ( y, k ) } = u ( f ( k ) , k ) + 1 − λλ U ( f ( k ) , k ) . Note that f ( k ) y + u ∗ λ ( y, k ) is convex in y with minimal value u ( f ( k ) , k ) + − λλ U ( f ( k ) , k ). Then(75), considered as an equation for V ′ λ , has two solutions if: aV λ ( k ) + bW λ ( k ) > u ( f ( k ) , k ) + 1 − λλ U ( f ( k ) , k ) , and no solutions if the inverse inequality holds. If we have equality: aV λ ( k ) + bW λ ( k ) = u ( f ( k ) , k ) + 1 − λλ U ( f ( k ) , k )then equation (75) has precisely one solution, namely: V ′ λ = u ′ ( f ( k ) , k ) + 1 − λλ U ′ ( f ( k ) , k )At the point k ∞ , with the values (62), (63), we find that: aV λ ( k ∞ ) + bW λ ( k ) = u ( f ( k ∞ ) , k ∞ ) + 1 − λλ U ( f ( k ∞ ) , k ∞ ) ,
19o that we are exactly on the boundary case. Equation (75) has precisely one solution, by (84)below, which satisfies V ′ λ ( k ∞ ) = u ′ ( f ( k ∞ ) , k ∞ ) + 1 − λλ U ′ ( f ( k ∞ ) , k ∞ ) , but it is degenerate: ∂∂y [ f ( k ∞ ) y + u ∗ λ ( y, k ∞ )] | y = V ′ λ ( k ∞ ) = 0 , so that equation (75) cannot be written in the form V ′ λ = ψ ( k, V λ ). It is an implicit differentialequation, and special techniques are needed to solve it. The difficulty is further increased by the factthat we need only V (but not W ) to be C .The proof of Theorem 10 proceeds in several steps. Step 1: Changing the unknown functions from ( V λ ( k ) , W λ ( k )) to ( V λ ( k ) , µ ( k ))Using (74), we find that the equation (58) is equivalent to F ( V ′ ( k ) , k ) = µ ( k )with: F ( x, k ) := u ∗ λ ( x, k ) + xf ( k ) − u ( f ( k ) , k ) − − λλ U ( f ( k ) , k ) µ ( k ) := aV λ ( k ) + bW λ ( k ) − u ( f ( k ) , k ) − − λλ U ( f ( k ) , k )Note that x → F ( x, k ) is a linear perturbation of the convex function x → u ∗ λ ( x, k ). To shortenthe notations, let us set: y ∗ ( k ) := u ′ ( f ( k ) , k ) + 1 − λλ U ′ ( f ( k ) , k ) (76)As we pointed out, the equation F ( x + y ∗ ( k ) , k ) = µ in the variable x has two solutions x − ( k, µ ) < < x + ( k, µ ) for µ ( k ) >
0, none for µ ( k ) < x = 0 for µ ( k ) = 0.From the equation (59), we have W ′ λ = V ′ λ bV λ + aW λ − u ( σ λ ( k ) , k ) + − λλ U ( σ λ ( k ) , k ) aV λ + bW λ − u ( σ λ ( k ) , k ) − − λλ U ( σ λ ( k ) , k )Differentiating µ ( k ) with respect to k yields dµdk = V ′ λ ( a + b ) V λ + 2 abW λ − δ u ( σ λ ( k ) , k ) − (1 − λ ) δ λ U ( σ λ ( k ) , k ) aV λ + bW λ − u ( σ λ ( k ) , k ) − − λλ U ( σ λ ( k ) , k ) − y ∗ f ′ − u ′ ( f ( k ) , k ) − − λλ U ′ ( f ( k ) , k )20e now take ( V λ ( k ) , µ ( k )) as our new unknown functions. They satisfy the equations: dVdk = y ∗ ( k ) + x ( k, µ ( k )) ,dµdk = ( y ∗ + x ( k, µ ( k ))) ( a + b ) V λ + 2 abW λ − δ u ( σ λ ( k ) , k ) − (1 − λ ) δ λ U ( σ λ ( k ) , k ) aV λ + bW λ − u ( σ λ ( k ) , k ) − − λλ U ( σ λ ( k ) , k ) − y ∗ f ′ − u ′ ( f ( k ) , k ) − − λλ U ′ ( f ( k ) , k ) . In fact, according to which determination ( x ( k, µ ( k ))) is chosen, x + ( k, µ ( k )) or x − ( k, µ ( k )),these equations define two distinct dynamical systems on the region µ > Step 3: Taking x instead of k as the independent variable .To get rid of the indetermination, we pick x instead of k as the independent variable. We shortenour notation by setting x ( k ) = x ( µ ( k ) , k ). We get: dkdx = f ( k ) − ϕ λ ( y ∗ + x, k ) D ( x, k, V λ , W λ ) (cid:20) aV λ + bW λ − u ( ϕ λ ( y ∗ + x, k ) , k ) − − λλ U ( ϕ λ ( y ∗ + x, k ) , k ) (cid:21) (77)where (here ϕ λ stands for ϕ λ ( y ∗ + x, k )) D ( x, k, V λ , W λ ) = ( y ∗ + x ) (cid:20) ( a + b ) V λ + 2 abW λ − δ u ( σ λ ( k ) , k ) − (1 − λ ) δ λ U ( σ λ ( k ) , k ) (cid:21) − A ( aV λ + bW λ − u ( ϕ λ , k ) − − λλ U ( ϕ λ , k )) A ( x, k, V λ , W λ ) = ( f − ϕ λ ) (cid:20)(cid:18) u ′′ ( f ( k ) , k ) + 1 − λλ U ′′ ( f ( k ) , k ) (cid:19) f ′ + u ′′ ( f ( k ) , k )+ 1 − λλ U ′′ ( f ( k ) , k ) (cid:21) + ( y ∗ + x ) f ′ ( k ) + u ′ ( ϕ λ , k ) + 1 − λλ U ′ ( ϕ λ , k )Further more, we have dV λ dx = dV λ dk dkdx = ( y ∗ + x ) f ( k ) − ϕ λ D ( x, k, V λ , W λ ) [ aV λ + bW λ − u ( ϕ λ , k ) − − λλ U ( ϕ λ , k )] . (78) Step 4: Rescaling the time
We introduce a new variable s such that D ( x, k, V λ , W λ ) ds = dx , then the system becomes dxds = D ( x, k, V λ , W λ ) dkdx = ( f ( k ) − ϕ λ ) (cid:2) aV λ + bW λ − u ( ϕ λ ( y ∗ + x, k ) , k ) − − λλ U ( ϕ λ ( y ∗ + x, k ) , k ) (cid:3) dV λ dx = ( y ∗ + x ) ( f ( k ) − ϕ λ ( y ∗ + x, k )) [ aV λ + bW λ − u ( ϕ λ , k ) − − λλ U ( ϕ λ , k )]We now eliminate W λ , to get an equation in ( x, k, V ) only. From the equation of µ ( k ) = F ( x, k ),we have bW λ ( k ) = F ( y ∗ ( k ) + x, k ) + u ( f ( k ) , k ) + 1 − λλ U ( f ( k ) , k ) − aV λ ( k )21he dynamics of ( x ( s ) , k ( s ) , V λ ( s )) are given by: dxds = ˜ D ( x, k, V λ ) dkdx = ( f ( k ) − ϕ λ ) (cid:2) F ( y ∗ ( k ) + x, k ) + u ( f ( k ) , k ) + − λλ U ( f ( k ) , k ) − u ( ϕ λ ( y ∗ + x, k ) , k ) − − λλ U ( ϕ λ ( y ∗ + x, k ) , k ) (cid:3) dV λ dx = ( y ∗ + x ) ( f ( k ) − ϕ λ ( y ∗ + x, k )) [ F ( y ∗ ( k ) + x, k ) + u ( f ( k ) , k ) + − λλ U ( f ( k ) , k ) − u ( ϕ λ ( y ∗ + x, k ) , k ) − − λλ U ( ϕ λ ( y ∗ + x, k ) , k )] (79)where ˜ D ( x, k, V λ ) = D ( x, k, V λ , W λ ). For ˜ D to be C , we need f to be C and u, U to be C . Step 5: Linearizing the system
As we already noted, if k = k ∞ , then µ ( k ∞ ) = 0 and F ( y ∗ ( k ∞ )+ x, k ∞ ) = µ ( k ∞ ) = 0 has only onesolution x = 0. Set v ∞ = V λ ( k ∞ ). We consider the system near the point ( x, k, V λ ) = (0 , k ∞ , v ∞ ).For simplicity, we write y ∞ = y ∗ ( k ∞ ) = u ′ ( f ( k ∞ ) , k ∞ ) + 1 − λλ U ′ ( f ( k ∞ ) , k ∞ )Computing the linearized system at (0 , k ∞ , v ∞ ), we find: dds xk − k ∞ V λ − v ∞ = a ∞ b ∞ c ∞ xk − k ∞ V λ − v ∞ . where: a ∞ = y ∗ ( k ∞ ) ∂ϕ λ ∂y ( y ∗ ( k ∞ ) , k ∞ ) ( f ′ ( k ∞ ) − g λ ( k ∞ )) g λ ( k ∞ ) = λδ u ′ ( f ( k ∞ ) , k ∞ ) + (1 − λ ) δ U ′ ( f ( k ∞ ) , k ∞ ) λu ′ ( f ( k ∞ ) , k ∞ ) + (1 − λ ) U ′ ( f ( k ∞ ) , k ∞ ) − λu ′ ( f ( k ∞ ) , k ∞ ) + (1 − λ ) U ′ ( f ( k ∞ ) , k ∞ ) λu ′ ( f ( k ∞ ) , k ∞ ) + (1 − λ ) U ′ ( f ( k ∞ ) , k ∞ )By the concavity assumptions, we find: y ∞ = u ′ ∞ + 1 − λλ U ′ ∞ = u ′ ( f ( k ∞ ) , k ∞ ) + 1 − λλ U ′ ( f ( k ∞ ) , k ∞ ) > ,u ′ ( f ( k ∞ ) , k ∞ ) + 1 − λλ U ′ ( f ( k ∞ ) , k ∞ ) < ∂ϕ λ ∂y ( y ∗ ( k ∞ ) , k ∞ ) = (cid:18) u ′′ ( f ( k ∞ ) , k ∞ ) + λ − λ U ′′ ( f ( k ∞ ) , k ∞ ) (cid:19) − < k ∞ satisfying (73) with g λ ( k ∞ ) − f ′ ( k ∞ ) = 0, we then have a ∞ = 0. Introducing:˜ x = x + b ∞ a ∞ ( k − k ∞ ) + c ∞ a ∞ ( V λ − v ∞ ) , (80)we transform the system (79) for (˜ x ( s ) , k ( s ) , V λ ( s )). The linearization at the origin is given by: dds ˜ xk − k ∞ V λ − v ∞ = − y ∞ ϕ ′ ∞ ( g λ ( k ∞ ) − f ′ ( k ∞ )) 0 00 0 00 0 0 ˜ xk − k ∞ V λ − v ∞ . tep 6: Applying the center manifold theorem By the center manifold theorem (cf. Theorem 1 of [8]), there exist an ǫ >
0, and a map h ( k, V λ ),defined in a neighborhood O = ( k ∞ − ǫ, k ∞ + ǫ ) × ( v ∞ − ǫ, v ∞ + ǫ ) of ( k ∞ , v ∞ ) such that h ( k ∞ , v ∞ ) = 0 , ∂h∂k ( k ∞ , v ∞ ) = 0 , ∂h∂V λ ( k ∞ , v ∞ ,
0) = 0and the manifold M defined by M = (cid:26)(cid:18) h ( k, V λ ) − b ∞ a ∞ ( k − k ∞ ) − c ∞ a ∞ ( V λ − v ∞ ) , k, V λ (cid:19) | ( k, V λ ) ∈ O (cid:27) , is invariant under the flow associated to the system (79). The map h and the central manifold M are C , and M is two-dimensional and tangent to the critical plane defined by ˜ x = 0.If k = k ∞ and V λ = v ∞ , then x = 0, and h ( k ∞ , v ∞ ) = ˜ x ( k ∞ , v ∞ ) = x ( k ∞ , v ∞ ) = 0We are interested in the solutions which lie on the central manifold M . Writing: x = h ( k, V λ ) − b ∞ a ∞ ( k − k ∞ ) − c ∞ a ∞ ( V λ − v ∞ )in the equation dVdk = y ∗ ( k ) + x , we get (cid:26) dV λ dk = u ′ ( f ( k ) , k ) + − λλ U ′ ( f ( k ) , k ) + h ( k, V λ ) − b ∞ a ∞ ( k − k ∞ ) − c ∞ a ∞ ( V λ − v ∞ ) ,V λ ( k ∞ ) = v ∞ (81)which can be viewed as eliminating the variable s from the second and third equations of the system(79). Since a ∞ = 0 and the right hand side of the first equation of (81) is continuously differentiablein O = ( k ∞ − ǫ, k ∞ + ǫ ), therefore, is locally Lipschiz continuous. By dV λ dk | k = k ∞ = y ∞ = 0 whichfollows from (76) and the first equation of (81), the nonconstant solution of this initial-value problemexist in O which we denote by V λ ( k ) = ζ ( k ) ,where ζ ( k ∞ ) = v ∞ and ζ ∈ C ( O ) if h ∈ C ( O ).Substituting V λ ( k ) = ζ ( k ) into x = h ( k, V λ ) − b ∞ a ∞ ( k − k ∞ ) − c ∞ a ∞ ( V λ − v ∞ ) yields x ( k ) = h ( k, ζ ( k )) − b ∞ a ∞ ( k − k ∞ ) − c ∞ a ∞ ( ζ ( k ) − v ∞ ) . Finally, µ ( k ) = F ( u ′ ( f ( k ) , k ) + 1 − λλ U ′ ( f ( k ) , k ) + x ( k ) , k )and W λ ( k ) = 1 b ( µ ( k ) + u ( f ( k ) , k ) + 1 − λλ U ( f ( k ) , k ) − aζ ( k )is also C , so we have found a C solution of the system (58),(59) with boundary conditions (62),(63). Introduce another function g λ ( k ) defined by: g λ ( k ) = λu ′ ( f ( k ) , k ) + (1 − λ ) U ′ ( f ( k ) , k ) λδ u ′ ( f ( k ) , k ) + − λδ U ′ ( f ( k ) , k ) − λδ u ′ ( f ( k ) , k ) + (1 − λ ) δ U ′ ( f ( k ) , k ) λδ u ′ ( f ( k ) , k ) + (1 − λ ) δ U ′ ( f ( k ) , k ) (82)23 heorem 11 Suppose f ′ ( k ∞ ) lies between g λ ( k ∞ ) and ¯ g λ ( k ∞ ) . Then there exists an equilibriumstrategy converging to k ∞ Comparing (72) and (82), we find that: g λ ( k ) − g λ ( k ) = λ (1 − λ )( δ − δ ) ( δ − δ ) u ′ U ′ − u ′ U ′ + u ′ U ′ ( λu ′ + (1 − λ ) U ′ )( λδ u ′ + (1 − λ ) δ U ′ )So the sign of g λ ( k ) − g λ ( k ) is the sign of ( δ − δ ) u ′ U ′ − u ′ U ′ + u ′ U ′ : δ − δ > u ′ u ′ − U ′ U ′ = ⇒ g λ ( k ) > g λ ( k ) δ − δ < u ′ u ′ − U ′ U ′ = ⇒ g λ ( k ) < g λ ( k )Let us give some examples: Example 1: u ( c ) = U ( c )In this case, u = U and they do not depend on k . We get: g λ ( k ) = λδ + (1 − λ ) δ g λ ( k ) = 1 λδ + (1 − λ ) δ These formulas do not depend on the the utility function u ( c ). They were first derived in [14] forthe special case u ( c ) = ln c . Note that λδ + (1 − λ ) δ is the arithmetic mean, and (cid:16) λδ + (1 − λ ) δ (cid:17) − is the geometric mean. Example 2: u ( c, k ) = U ( c, k )In that case, the criterion (55) becomes: Z ∞ (cid:0) λe − δ t + (1 − λ ) e − δ t (cid:1) u ( c ( t ) , k ( t )) dt (83)This problem was studied in [14] for u ( c, k ) = U ( c, k ) = ln c . In the case at hand, (83), we find: g λ ( k ) = λδ + (1 − λ ) δ − u ′ ( f ( k ) , k ) u ′ ( f ( k ) , k ) g λ ( k ) = 1 λδ + (1 − λ ) δ − u ′ ( f ( k ) , k ) u ′ ( f ( k ) , k )These are the same as the preceding ones, with the corrective term − u ′ /u ′ . Example 3: u = u ( c ) and U = U ( c )In that case, the criterion (55) becomes: λ Z ∞ e − δ t u ( c ( t )) dt + (1 − λ ) Z ∞ e − δ t U ( c ( t )) dt We find: g λ ( k ) = λδ u ′ ( f ( k )) + (1 − λ ) δ U ′ ( f ( k )) λu ′ ( f ( k )) + (1 − λ ) U ′ ( f ( k )) g λ ( k ) = λu ′ ( f ( k )) + (1 − λ ) U ′ ( f ( k )) λδ u ′ ( f ( k )) + (1 − λ ) δ U ′ ( f ( k ))24ere again we find g λ ( k ) < g λ ( k ). We now proceed to the proof of Theorem 11 . We apply Theorem 10, and we denote by V λ and W λ the solution of equations (58) and (59) with the boundary conditions (62) and (63). Set σ λ ( k ) := ϕ λ ( V ′ λ ( k ) , k ), where ϕ λ is defined by (60) or (61). We will now show that the correspond-ing trajectory k ( t ), defined by (35), converges to k ∞ . By Proposition 8, this will prove that σ λ isan equilibrium strategy.We need the value of V ′ λ ( k ∞ ) and σ ′ λ ( k ∞ ). Differentiating (58), we find: V ′ λ ( k ∞ ) f ( k ∞ ) − V ′ λ ( k ∞ ) ϕ λ ( V ′ λ ( k ∞ ) , k ∞ ) + u ( ϕ λ ( V ′ λ ( k ∞ ) , k ∞ ) , k ∞ )+ 1 − λλ U ( ϕ λ ( V ′ λ ( k ∞ ) , k ∞ ) , k ∞ ) = u ( f ( k ∞ ) , k ∞ ) + 1 − λλ U ( f ( k ∞ ) , k ∞ ) , thus F ( V ′ λ ( k ∞ ) , k ∞ ) = 0 and hence: V ′ λ ( k ∞ ) = y ∗ ( k ∞ ) = u ′ ( f ( k ∞ ) , k ∞ ) + 1 − λλ U ′ ( f ( k ∞ ) , k ∞ ) . (84)To compute σ ′ λ ( k ∞ ), we consider (45), the integrated form of V ′ λ ( k ) where we substitute h ( t ) = e − δt and H ( t ) = e − δ t , and replace U with (1 − λ ) λ − . Evaluating (45) at k = k ∞ , we get: V ′ λ ( k ∞ ) = u ′ ( f ( k ∞ ) , k ∞ ) σ ′ λ ( k ∞ ) + u ′ ( f ( k ∞ ) , k ∞ ) δ − f ′ ( k ∞ ) + σ ′ λ ( k ∞ )+ 1 − λλ U ′ ( f ( k ∞ ) , k ∞ ) σ ′ λ ( k ∞ ) + U ′ ( f ( k ∞ ) , k ∞ ) δ − f ′ ( k ∞ ) + σ ′ λ ( k ∞ ) . (85)Comparing (84) with (85), we can solve for σ ′ λ ( k ∞ ). We get: σ ′ λ ( k ∞ ) = f ′ ( k ∞ ) − δ δ ( δ u ′ ( f ( k ∞ ) , k ∞ ) + − λλδ δU ′ ( f ( k ∞ ) , k ∞ ))( g λ ( k ∞ ) − f ′ ( k ∞ ))( u ′ ( f ( k ∞ ) , k ∞ ) + − λλ U ′ ( f ( k ∞ ) , k ∞ ))( g λ ( k ∞ ) − f ′ ( k ∞ )) , (86)Linearizing the equation of motion dkdt = f ( k ) − σ λ ( k ) at k = k ∞ yields dkdt = ( f ′ ( k ∞ ) − σ ′ λ ( k ∞ ))( k − k ∞ ) . Then k converges to k ∞ if f ′ ( k ∞ ) − σ ′ λ ( k ∞ ) <
0. Comparing with (86), and writing in theexpressions of g λ ( k ∞ ) and g λ ( k ∞ ) yields: δ δ ( δ u ′ ( f ( k ∞ ) , k ∞ ) + − λλδ U ′ ( f ( k ∞ ) , k ∞ ))( g λ ( k ∞ ) − f ′ ( k ∞ ))( u ′ ( f ( k ∞ ) , k ∞ ) + − λλ U ′ ( f ( k ∞ ) , k ∞ ))( g λ ( k ∞ ) − f ′ ( k ∞ )) < . Because u and U are increasing with respect to their first variable, the first factor in the numeratorand denominator are positive. We are left with: g λ ( k ∞ ) − f ′ ( k ∞ ) g λ ( k ∞ ) − f ′ ( k ∞ ) < The Chichilnisky criterion
In two influential papers [11], [10], Chichilnisky has proposed an axiomatic approach to sustainabledevelopment, based on the twin ideas that there should be no dictatorship of the present and nodictatorship of the future. She suggests to use the following criterion I α ( c ( · ) , k ( · )) := δ Z ∞ u ( c ( t ) , k ( t )) e − δt dt + α lim t →∞ U ( c ( t ) , k ( t )) , (87)The coefficient δ > Lemma 12
Suppose u ( c, k ) ≥ and sup { U ( c, k ) | c > , k > , c = f ( k ) } = ∞ . Then, for any α > , sup { I α ( c ( · ) , k ( · )) | ( c ( · ) , k ( · )) ∈ A ( k ) } = ∞ , where A ( k ) is defined below Definition 1. Proof.
For fixed k >
0, pick any
A >
0, and choose some constants ( c , k ) such that c = f ( k )and U ( c , k ) > Aα − . Then choose some c ∈ (0 , c ) such that f ( k ) − c >
0. With every
T > c T ( t ) , k T ( t )) defined by: c T ( t ) = (cid:26) c for 0 ≤ t ≤ T,c for T ≤ t, and denote the consumption-capital pair by ( c T , k T ).Since f is increasing, we have dkdt ≥ f ( k ) − c for all t , so eventually k ( t ) will reach the value k . Choose for T the first time when k ( t ) = k . Writing this into the criterion, and remembering c = f ( k ), we find: I α ( c T ( · ) , k T ( · )) ≥ α lim t →∞ U ( c T ( t ) , k T ( t )) = αU ( c , k ) ≥ A. This result shows that very often it is not possible to optimize I α . Even when it is, there is thetime-inconsistency problem: successive decision-makers will not agree on what the optimal solutionis. This is seen most easily by considering the following criterion:(1 − α ) δ Z ∞ u ( c ( t ) , k ( t )) e − δt dt + αr Z ∞ U ( c ( t ) , k ( t )) e − rt dt. (88)When r →
0, the last term converges to αU ( c ∞ , k ∞ ), so that (88) converges to (87). On theother hand, Criterion (88) is a special case of the biexponential criterion (55) with δ = δ , δ = r ,and: αr (1 − α ) δ = 1 − λλ , λ = (1 − α ) δαr + (1 − α ) δ , − λ = αrαr + (1 − α ) δ (89)26o, for each r >
0, the criterion (88) gives rise to a time-inconsistent problem. Using the resultsin the preceding section, we find a continuum of equilibrium strategies. Substituting (89) into (72)and (82), we find the corresponding values: g r ( k ) = (1 − α ) δ u ′ ( f ( k ) , k ) + αr U ′ ( f ( k ) , k )(1 − α ) δu ′ ( f ( k ) , k ) + αrU ′ ( f ( k ) , k ) − (1 − α ) δu ′ ( f ( k ) , k ) + αrU ′ ( f ( k ) , k )(1 − α ) δu ′ ( f ( k ) , k ) + αrU ′ ( f ( k ) , k ) g r ( k ) = (1 − α ) δu ′ ( f ( k ) , k ) + αrU ′ ( f ( k ) , k )(1 − α ) u ′ ( f ( k ) , k ) + αU ′ ( f ( k ) , k ) − (1 − α ) u ′ ( f ( k ) , k ) + αU ′ ( f ( k ) , k )(1 − α ) u ′ ( f ( k ) , k ) + αU ′ ( f ( k ) , k )The equations (58), (59) become:( f − ϕ r ( V ′ r )) V ′ r + u ( ϕ r ) + αr (1 − α ) δ U ( V ′ r ) = δ + r V r + δ − r W r , (90)( f − ϕ λ ( V ′ r )) W ′ r + u ( ϕ λ ) + αr (1 − α ) δ U ( V ′ λ ) = δ − r V r + δ + r W r (91)where ϕ r ( x, k ) is defined by: u ′ ( ϕ r ( x, k ) , k ) + αr (1 − α ) δ U ′ ( ϕ r ( x, k ) , k ) = x and the boundary conditions (62), (63): V r ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) + α (1 − α ) δ U ( f ( k ∞ ) , k ∞ ) , (92) W r ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) − α (1 − α ) δ U ( f ( k ∞ ) , k ∞ ) . (93)The equilibrium strategy is given by σ r ( k ) = ϕ r ( V ′ r ( k ) , k ), and the corresponding trajectoryconverges to k ∞ . Note that these strategies are defined locally. More precisely, denote by ] k − r , k + r [the maximal interval of existence of the solution ( V r , W r ) of the ODE (90), (91) with the boundarycondition (92), (93), so that k − r < k ∞ < k + r .We shall now solve the Chichilnisky problem by setting r = 0 in the preceding equations. Wehave: g ( k ) = δ − u ′ ( f ( k ) , k ) u ′ ( f ( k ) , k ) , (94) g ( k ) = (1 − α ) δu ′ ( f ( k ) , k )(1 − α ) u ′ ( f ( k ) , k ) + αU ′ ( f ( k ) , k ) − (1 − α ) u ′ ( f ( k ) , k ) + αU ′ ( f ( k ) , k )(1 − α ) u ′ ( f ( k ) , k ) + αU ′ ( f ( k ) , k ) (95)The pair ( V , W ) has to solve the following boundary-value problem:( f ( k ) − ϕ ( V ′ , k )) V ′ + u ( ϕ ( V ′ , k ) , k ) = δ V + W ) , (96)( f ( k ) − ϕ ( V ′ , k )) W ′ + u ( ϕ ( V ′ , k ) , k ) = δ V + W ) , (97) V ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) + α (1 − α ) δ U ( f ( k ∞ ) , k ∞ ) , (98) W ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) − α (1 − α ) δ U ( f ( k ∞ ) , k ∞ ) (99)27ith: u ′ ( ϕ ( x, k ) , k ) = x, ϕ ( u ′ ( c, k ) , k ) = c. (100)Substracting (97) from (96), we get:( f ( k ) − ϕ ( V ′ , k )) ( V ′ − W ′ ) = 0 . Similarly to Lemma 9, it follows that V ′ − W ′ = 0, and so V − W is a constant, namely: W ( k ) = V ( k ) + W ( k ∞ ) − V ( k ∞ ) = V ( k ) − α (1 − α ) δ U ( f ( k ∞ ) , k ∞ ) . Writing this in (96) and (97), we find that V ( k ) is a solution of the boundary-value problem:( f ( k ) − ϕ ( V ′ , k )) V ′ + u ( ϕ ( V ′ , k ) , k ) = δV − α − α U ( f ( k ∞ ) , k ∞ ) ,V ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) + α (1 − α ) δ U ( f ( k ∞ ) , k ∞ ) . Setting V := V − α (1 − α ) δ U ( f ( k ∞ ) , k ∞ ), we see that V is a solution of the boundary-valueproblem: ( f ( k ) − ϕ ( V ′ , k )) V ′ + u ( ϕ ( V ′ , k ) , k ) = δV, (101) V ( k ∞ ) = 1 δ u ( f ( k ∞ ) , k ∞ ) . (102)This problem has been studied in [15] (Section 2.4, case 2) when u ( c, k ) does not depend on k .In the general case, we have: Proposition 13 If f ′ ( k ∞ ) = δ − u ′ ( f ( k ∞ ) , k ∞ ) u ′ ( f ( k ∞ ) , k ∞ ) (103) the problem (101), (102) has two solutions V and V , defined on some non-empty half-interval [ k ∞ , k ∞ + a ) or ( k ∞ − a, k ∞ ] . Both are C on the half-interval, C on the interior, and have thesame derivative at k ∞ , given by: ϕ ( V ′ i ( k ∞ ) , k ∞ ) = c ∞ = f ( k ∞ ) , i = 1 , . Proof.
From now on, write ϕ instead of ϕ . Rewrite (101) as a Pfaff system: dV = pdk, (104) p ( f ( k ) − ϕ ( p, k )) + u ( ϕ ( p, k ) , k ) = δV (105)Differentiating (105) leads to: δdV = ( f ( k ) − ϕ ( p, k )) dp + pf ′ ( k ) dk + u ′ ( ϕ ( p, k ) , k ) dk, where we used (100). Together with (104), this yields( f ( k ) − ϕ ( p, k )) dp = [ δp − pf ′ ( k ) − u ′ ( ϕ ( p, k ) , k )] dk. (106)28e have to investigate this system near the point k = k ∞ and V = V ∞ . Writing (105) at thispoint, we get: p ( f ( k ∞ ) − ϕ ( p, k ∞ )) + u ( ϕ ( p, k ∞ ) , k ∞ ) = δV ∞ (107)which has to be solved for p . Note that, because of (100), we have:min p { ( f ( k ∞ ) − ϕ ( p, k ∞ )) p + u ( ϕ ( p, k ∞ ) , k ∞ ) } = u ( f ( k ∞ ) , k ∞ ) (108)In the case at hand, we have δV ∞ = u ( f ( k ∞ ) , k ∞ ), but for the sake of completeness, and tohave a full description of the phase space in the ( k, V ) plane, we will first investigate the cases δV ∞ < u ( f ( k ∞ ) , k ∞ ) and δV ∞ > u ( f ( k ∞ ) , k ∞ ). Case 0 : δV ∞ < u ( f ( k ∞ ) , k ∞ )Because of (108), equation (107) has no solution. So there are no solutions going through( k ∞ , V ∞ ) . Case 1. δV ∞ > u ( f ( k ∞ ) , k ∞ ).Equation (107) has two distinct solutions p = p . Note that neither p nor p minimize theleft hand side so f ( k ∞ ) − ϕ ( p i , k ∞ ) = 0 for i = 1 ,
2. We may therefore consider the initial valueproblem: dpdk = δp − pf ′ ( k ) − u ′ ( ϕ ( p, k ) , k ) f ( k ) − ϕ ( p, k ) , p ( k ∞ ) = p i . It has a well-defined smooth solution p i ( k ), defined in a neighborhood of k ∞ . We then define afunction V i by: V i ( k ) := 1 δ [( f ( k ) − ϕ ( p, k )) p + u ( ϕ ( p, k ) , k )] . Notice that V i ( k ∞ ) = V ∞ , so V i ( k ) solves the initial value problem (101), (102), with V ′ i ( k ∞ ) = p i .Taking i = 1 ,
2, we find two solutions V ( k ) and V ( k ) of the same initial value problem, with V ′ ( k ∞ ) = V ′ ( k ∞ ). Case 2. δV ∞ = u ( f ( k ∞ ) , k ∞ ) and f ′ ( k ∞ ) = δ − u ′ /u ′ .Equation (107) then has a single solution p , and we have f ( k ∞ ) = ϕ ( p , k ∞ ), so that: p = u ′ ( ϕ ( p , k ∞ ) , k ∞ ) = u ′ ( f ( k ∞ ) , k ∞ ) > . In this case, we shall use the same system (104)-(105), but we will take p instead of k as theindependent variable. We consider the initial value problem dkdp = f ( k ) − ϕ ( p, k ) p ( δ − f ′ ( k ) − u ′ ( ϕ ( p,k ) ,k ) p ) , k ( p ) = k ∞ . It has a C solution k ( p ), defined in a neighborhood of p = p . We associate with it a curve inthe phase space ( k, V ), defined in parametric form by the equations k = k ( p ) ,V = 1 δ [( f ( k ) − ϕ ( p, k )) p + u ( ϕ ( p, k ) , k )] . V is also C with respect to p near p = p . Moreover, we have dVdp = 1 δ [( f ( k ) − ϕ ( p, k ) − ϕ ′ ( p, k ) p + u ′ ( ϕ ( p, k ) , k ) ϕ ′ ( p, k )]+ 1 δ [( f ′ ( k ) − ϕ ′ ( p, k )) p + u ′ ( ϕ ( p, k ) , k ) ϕ ′ ( p, k ) + u ′ ( ϕ ( p, k ) , k )] dkdp = f ( k ) − ϕ ( p, k ) δ − f ′ ( k ) − u ′ ( ϕ ( p,k ) ,k ) p Since f ( k ∞ ) = ϕ ( p , k ∞ ), we obtain dkdp ( p ) = 0 ,dVdp ( p ) = 0 . This shows that the parametric curve p → ( k ( p ) , V ( p )) in the ( k, V )-plane has a cusp at( k ∞ , V ∞ ). To find the type of the cusp, we compute the second order derivatives with respectto p . We find: d kdp ( p ) = − ϕ ′ ( p , k ∞ ) p [ δ − f ′ ( k ∞ ) − u ′ ( f ( k ∞ ) ,k ∞ ) u ′ ( f ( k ∞ ) ,k ∞ ) ] = − u ′′ ( f ( k ∞ ) , k ∞ ) u ′ ( f ( k ∞ ) , k ∞ )[ δ − f ′ ( k ∞ ) − u ′ ( f ( k ∞ ) ,k ∞ ) u ′ ( f ( k ∞ ) ,k ∞ ) ] = 0 , (109) d Vdp ( p ) = − ϕ ′ ( p , k ∞ ) δ − f ′ ( k ∞ ) − u ′ ( f ( k ∞ ) ,k ∞ ) u ′ ( f ( k ∞ ) ,k ∞ ) = − u ′′ ( f ( k ∞ ) , k ∞ ) δ − f ′ ( k ∞ ) − u ′ ( f ( k ∞ ) ,k ∞ ) u ′ ( f ( k ∞ ) ,k ∞ ) = 0 , (110)This shows that the cusp has two branches, with a common tangent between them. The branchesextend on the left of k ∞ if the left-hand side of (109) is negative, that is, if: δ − f ′ ( k ∞ ) − u ′ ( f ( k ∞ ) , k ∞ ) u ′ ( f ( k ∞ ) , k ∞ ) < m of the common tangent is given by m = d Vdp ( k ) / d kdp ( p ) = p , and it is also the one-sided derivative of V at k ∞ . Note for future use that because p = u ′ ( f ( k ∞ ) , k ∞ )we have: ddk δ u ( f ( k ) , k ) | k = k ∞ = 1 δ ( u ′ ( f ( k ∞ ) , k ∞ ) f ′ ( k ∞ ) + u ′ ( f ( k ∞ ) , k ∞ ) , so that m = p does not coincide with the tangent of the curve δV = u ( f ( k )) at k = k ∞ . Case 3 : δV ∞ = u ( f ( k ∞ ) , k ∞ ) and f ′ ( k ∞ ) = δ − u ′ /u ′ This is the case which was investigated in Theorem 3. We have shown that there exists a C solution V ( k ) on a neighbourhood of k ∞ . This concludes the proof of Proposition 13 (in fact, we only need Case 2).Let us summarize these results. The curve Γ = { ( k, V ) | δV = u ( f ( k ) , k ) } separates the planein two regions. 30 The region below the curve corresponds to Case 0: there are no solution there. • The region above the curve corresponds to Case 1: through each point ( k ∞ , V ∞ ) there are twosmooth solutions intersecting transversally. • If ( k ∞ , V ∞ ) is on the curve, but f ′ ( k ∞ ) + u ′ ( f ( k ∞ ) ,k ∞ ) u ′ ( f ( k ∞ ) ,k ∞ ) = δ , we are in Case 2. There are two C solutions defined only on one side of k ∞ . They are tangent at ( k ∞ , V ∞ ), and transversalto Γ. • If ( k ∞ , V ∞ ) is on the curve, and f ′ ( k ∞ ) + u ′ ( f ( k ∞ ) ,k ∞ ) u ′ ( f ( k ∞ ) ,k ∞ ) = δ , we are in Case 3: there is a C solution defined on a neighbourhood of k ∞ .theFigure 1 gives the phase diagram in the ( k, V ) plane: Case 1 Case 2Case 3 kV δ u ( f ( k ) , k ) V ( k ) V ( k ) V ( k ) V ( k ) V ( k ) Figure 1. The illustration of the solutions.Proposition 13 gives us two solutions, V i , i = 1 ,
2. Each of them gives rise to an strategy σ i through the formula σ i ( k ) = ϕ ( V ′ i ( k ) , k ) with ϕ defined by (100). The strategy σ i is C on thehalf-interval, C on its interior, with σ i ( k ∞ ) = f ( k ∞ ). Proposition 14
One, and only one, of the strategies σ and σ , converges to k ∞ . Proof.
Consider the Euler-Lagrange equation (16) for the Ramsey problem. The phase diagram isgiven in Figure 2, where k is defined by f ′ ( k ) = δ − u ′ /u ′ .If k ∞ = k , there is one trajectory T going through ( k ∞ , u ( f ( k ∞ ) , k ∞ )). The point ( k ∞ , u ( f ( k ∞ ) , k ∞ ))separates it into two branches, the upper one and the lower one. One of them goes to ( k ∞ , u ( f ( k ∞ ) , k ∞ )),and the other one leaves it. These two branches are also the trajectories associated with the twostrategies σ and σ , so one of them converges and the other diverges.31 k ∞ kf ( k ) ccc ∞ σ ( k ) σ ( k ) Figure 2. The phase diagram for the Euler equation.Since we are looking for a strategy which converges to k ∞ , we pick the strategy σ i , and thesolution V i , associated with the branch oriented towards k ∞ . We will denote them by σ and V . Wehave proved the following: Theorem 15
Suppose f ′ ( k ∞ ) lies between g ( k ∞ ) and g ( k ∞ ) . If g ( k ∞ ) > g ( k ∞ ) , then thereexists an equilibrium strategy σ ( k ) , defined on some interval ] k ∞ − κ, k ∞ ] , which converges to k ∞ .It is continuous on the interval, C on its interior, with σ ( k ∞ ) = f ( k ∞ ) . If If g ( k ∞ ) < g ( k ∞ ) ,there exists an equilibrium strategy with the same properties, defined on some interval [ k ∞ , k ∞ + κ [ . Note that one of the boundary values for f ′ ( k ∞ ), namely g ( k ∞ ), corresponds to the solution ofthe Ramsey problem ( α = 0). Indeed, the equation f ′ ( k ∞ ) = g ( k ∞ ) coincides with equation (18).We have thus identified a class of equilibrium strategies for the Chichilnisky problem. They areone-sided, except when f ′ ( k ∞ ) = g ( k ∞ ), where we can apply Theorem 15 to get a strategy definedon a neighbourhood of k ∞ . For every other value of k ∞ satisfying (94) and (95), the function V ( k ) and the strategy σ ( k ) defined by u ′ ( σ ( k ) , k ) = V ′ ( k ) are defined only on one of the twohalf-intervals limited by k ∞ . Suppose for instance it is the right one, [ k ∞ , k ∞ + κ [. Then, if k ∞ ≤ k < k ∞ + κ , the equilibrium strategy will bring k to k ∞ in finite time and stay there.To our knowledge, this is the first time equilibrium strategies have been found for the Chichiliniskycriterion. Their economic interpretation, and their detailed study, will be the subject of forthcomingwork.Let us give some examples. Example 1 : u ( c ) = U ( c ) 32either depends on k , and we have: g ( k ) = δ,g ( k ) = (1 − α ) δ We have g ( k ) < g ( k ), so the equilibrium strategy exists only on the right hand side of g ( k ).The existence condition is: (1 − α ) δ < f ′ ( k ∞ ) < δ and the equilibrium strategy σ is defined on [ k ∞ , ∞ [. We denote f ′− ( δ ) and f ′− ((1 − α ) δ ) by k and k respectively. There are three cases, depending on the position of the initial point k : • If k > k , then, for any k ∞ ∈ ] k, k [, there exists an equilibrium strategy starting from k andconverging to k ∞ . • if k < k < k , then, for any k ∞ ∈ ] k, k [, there exists an equilibrium strategy starting from k and converging to k ∞ . • if k < k , the only equilibrium strategy starting from k is the optimal strategy for the Ramseyproblem (that is, for the case α = 0 ) which converges to the level k where f ′ ( k ) = δ . Example 2 : u ( c, k ) = U ( c, k )We find: g ( k ) = δ − u ′ ( f ( k ) , k ) u ′ ( f ( k ) , k ) g ( k ) = (1 − α ) δ − u ′ ( f ( k ) , k ) u ′ ( f ( k ) , k )The existence condition is:(1 − α ) δ − u ′ ( f ( k ) , k ) u ′ ( f ( k ) , k ) < f ′ ( k ∞ ) < δ − u ′ ( f ( k ) , k ) u ′ ( f ( k ) , k )and the equilibrium strategy σ is defined on [ k ∞ , k ∞ + κ [ for some κ >
0. The situation is similarto the preceding one, bearing in mind that now the strategy σ may be defined locally only Example 3: u = u ( c ) and U = U ( k )In that case, we find: g ( k ) = δ,g ( k ) = δ − α (1 − α ) U ′ ( f ( k ) , k ) u ′ ( f ( k ) , k )There are two subcases: • if U ′ ( f ( k ) ,k ) u ′ ( f ( k ) ,k ) >
0, then g ( k ) < g ( k ). The equilibrium strategy then is defined on the righthand side of k ∞ , as in the preceding cases.33 if U ′ ( f ( k ) ,k ) u ′ ( f ( k ) ,k ) <
0. The equilibrium strategy then is defined on the left hand side of k ∞ .Remember that u ′ >
0. If U ′ >
0, that is, if capital is beneficial, we are in the first case.The second case corresponds to the case when capital is detrimental, for instance if accumulatingcapital accumulates pollution. In that case, the situation is inverted with respect to the precedingexamples: if the initial stock of capital is small enough, one may stop capital accumulation at a level k ∞ smaller than the level k where f ′ ( k ) = δ − α (1 − α ) U ′ ( f ( k ) ,k ) u ′ ( f ( k ) ,k ) . But there is no turning back: onecan never achieve a stationary level k ∞ < k . The situation is described in Figure 3 kk ∞ kf ( k ) ccc ∞ kk ∞ k kf ( k ) ccc ∞ kk ∞ k kf ( k ) ccc ∞ Figure 3. The case when capital pollutes.
Acknowledgements.
The authors thank sincerely Professors Larry Karp and Gerd Asheimfor generously sharing their knowledge and ideas with them, and pointing out several useful refer-ences. Qinglong Zhou and Yiming Long would like to express their thanks to Professors Jean-PierreBourguignon, Jean Dolbeault, Olivier Glass and Ivar Ekeland as well as the IHES and the CERE-MADE (Universit´e Paris Dauphine) for supporting Qinglong’s visits to IHES and CEREMADE fromFebruary to April of 2012, and their warm hospitalities during this time.
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