Ivar Ekeland
email: ekeland@math.ubc.ca

Yiming Long
email: longym@nankai.edu.cn

Qinglong Zhou
email: zhou.qinglong.1985@gmail.com

A new class of problems in the calculus of variations

Introduction

In economic theory, and in optimal control, it has been customary to discount future gains at a constant rate δ > 0. If an individual with utility function u (c) has the choice between several streams of consumption c (t), 0 ≤ t, he or she will choose the one which maximises the present value, given by:

∫ ∞ 0 u (c (t)) e -δt dt (1)
That future gains should be discounted is well grounded in fact. On the one hand, humans prefer to enjoy goods sooner than later (and to suffer bads later than sooner), as every child-rearing parent 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 good reasons why people are willing to pay a little bit extra to hasten the delivery date, or will require compensation 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 the discount factor should be an exponential e -δt . The practice probably arises from the compound interest formula lim ε→0 (1 -εδ)

t/ε = e -δt , when a constant interest rate δ is assumed, but even in finance, interest rates vary with the horizon: long-term rates can be widely different from shortterm ones. As for economics, there is by now a huge amount of evidence that individuals use higher discount rates for the near future than for the long-term (see [START_REF] Frederick | Time discounting and time preference: a critical review[END_REF] for a review up to 2002). There 1 is also an aggregation problem: in a society where individuals use constant (but different) discount rates, the collective discount rate may be non-constant (see [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF]). So the present value formula [START_REF] Aghion | Endogeneous growth theory[END_REF] should be replaced by the more general one:

∫ ∞ 0 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 our knowledge 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 two streams of consumption c 1 (t) and c 2 (t), starting at time T > 0. At time t = 0, he or she finds c 1 (t) yields the highest present value:

∫ ∞ T u (c 1 (t)) h (t) dt > ∫ ∞ T u (c 2 (t)) h (t) dt. ( 3
)
He or she then chooses c 1 (t). When time T is reached, the present values are now:

∫ ∞ T u (c 1 (t)) h (t -T ) dt and ∫ ∞ T u (c 2 (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:

∫ ∞ T u (c (t)) e -δ(t-T ) dt = e δT ∫ ∞ T u (c (t)) e -δt dt
so that the two terms in (4) are proportional to the two terms in [START_REF] Asheim | Justifying, Characterizing and Indicating Sustainability[END_REF]. However, this is a peculiarity of the exponential function, and it is not to be expected with more general discount rates. The decision-maker then faces a basic rationality problem: what should he or she do ? To be more specific, assume the state k (t) is related to the control c (t) by the dynamics:

dk dt = f (k) -c (t) , k (0) = k 0 (5) c (t) ≥ 0, k (t) ≥ 0 (6)
and the decision-maker is interested in maximising [START_REF] Asheim | Ethical preferences in the presence of resource constraints[END_REF]. How should he or she behave ? In the exponential case, when h (t) = e -δt , the answer is to pick the optimal solution: if it is optimal at time t = 0, it will still be optimal at all times T > 0 (this, by the way, is the content of the dynamic programming principle). But in the non-exponential case, the notion of optimality changes with time: each observer, from time t = 0 on, has his or her own optimal solution. No one agrees on what the optimal solution is, so optimality no longer provides an answer to the decision-making process, and one must look for other concepts to describe rational behaviour.

A clear requirement for rationality is that any strategy put forward be implementable. Suppose a 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 should be a subgame-perfect Nash equilibrium of the leader-follower game played by the successive decision-makers. This idea has been introduced by Phelps ( [START_REF] Phelps | On second-best national saving and game-equilibrium growth[END_REF], [START_REF] Phelps | The indeterminacy of game-equilibrium growth[END_REF]) in models with discrete time (see [START_REF] Krusell | Consumption-savings decisions with quasi-geometric discounting[END_REF] and [START_REF] Harris | Dynamic choices of hyperbolic consumers[END_REF] for further developments), and adapted by Karp ([22], [START_REF] Karp | Time-consistent policies[END_REF]), and by Ekeland and Lazrak ([12], [START_REF] Ekeland | Being serious about non-commitment: subgame perfect equilibrium in continuous time[END_REF], [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF], [START_REF] Ekeland | Equilibrium management of fisheries with overlapping altruistic generations[END_REF]) to the case of continuous time.

In this paper, we will follow the approach by Ekeland and Lazrak. It consists of introducing a value function V (k), which is very similar to the value function in optimal control, and of showing 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 suitable boundary 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 1 t) + (1 -α) exp (-r 2 t), and it was showed that the corresponding problem had a continuum of equilibrium strategies. In the present paper, in view of applications to economics, and of the mathematical interest, we aim to extend the analysis to the more general case:

(1 -α) ∫ ∞ 0 e -r1t u (c (t) , k (t)) dt + α ∫ ∞ 0 e -r2t U (c (t) , k (t)) dt (7)
As a by-product of our analysis, we will treat the problem:

(1 -α) δ ∫ ∞ 0 e -δt u (c (t) , k (t)) dt + α lim t→∞ U (k (t) , c (t)) (8)
which was introduced by Chichilnisky (see [START_REF] Chichilnisky | An axiomatic approach to sustainable development[END_REF], [START_REF] Chichilnisky | What is sustainable development?[END_REF]) to model sustainable development. Note that, if u (c, k) ≥ 0 and:

sup {U (c, k) | c > 0, k > 0, c = f (k)} = ∞
then maximising (8) under the dynamics (5), ( 6) leads to the value +∞, so that optimisation is clearly 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 an equilibrium strategy. This support, however, is one-sided, that is, k ∞ can be reached only from the initial level of capitals k 0 lying on its left (or from its right). To our knowledge, this is the first time such strategies have been identified. The structure of the paper is as follows. In the next section, we consider the problem of maximising

∫ ∞ 0 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 the corresponding HJB equation, and we show that it has a C 2 solution. Next, we define equilibrium strategies. With each such strategy we associate a value function V (k), and we show that it satisfies an integro-differential equation which generalizes the HJB equation, and we prove a verification theorem: any solution of this equation with suitable boundary conditions gives an equilibrium strategy. We show that the trajectories satisfy an integro-differential equation which generalizes the classical Euler-Lagrange equations, and we connect the Ekeland-Lazrak approach with the Karp approach.

In section 4, we apply the theory to problem [START_REF] Caplin | The recursive approach to time inconsistency[END_REF], and show that it has a continuum of equilibrium strategies, thereby extending the results of [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF]. It should be noted that the equations for V (k) are given in implicit form, that is, they cannot be solved with respect to V ′ (k), so that finding a C 2 solution requires special techniques (first a blow-up, and then the central manifold theorem).

We then consider the criterion:

(1 -α) δ ∫ ∞ 0 e -δt u (c (t) , k (t)) dt + αr ∫ ∞ 0 e -rt U (c (t) , k (t)) dt ( 10
)
which belongs to the class [START_REF] Caplin | The recursive approach to time inconsistency[END_REF] and we let r → 0. In the limit, we get equilibrium strategies for the Chichilnisky problem ( 8), which we describe explicitly.

The Ramsey problem

This is the classical model for economic growth, originating with the seminal paper of Ramsey [START_REF] Frank | A mathematical theory of saving[END_REF] in 1928, and developed by Cass [START_REF] Cass | Optimal growth in an aggregative model of capital accumulation[END_REF], Koopmans [START_REF] Koopmans | On the concept of optimal economic growth[END_REF] and many others (see [START_REF] Barro | Economic growth[END_REF] for a modern exposition). We are given a point k 0 > 0 and a concave continuous function f on [0, ∞), which is C ∞ on ]0, ∞[ and satisfies the Inada conditions:

lim x→0 f ′ (x) = +∞, lim x→∞ f ′ (x) ≤ 0. ( 11
)
Definition 1 A capital-consumption path (c (t) , k (t)) , t ≥ 0, is admissible if: k (t) > 0 and c (t) > 0 for all t, ( 12
) dk dt = f (k) -c, k (0) = k 0 , (13)
The set of all admissible paths starting from k 0 (i.e, such that k (0) = k 0 ) will be denoted by A (k 0 ).

We are given a number δ > 0 and another concave, increasing function u on ]0, ∞), which is C ∞ on the interior, with u ′′ (c) > 0 everywhere. We introduce the following criterion on A (k 0 ):

I (c, k) = ∫ ∞ 0 u(c(t), k(t))e -δt dt, ( 14
)
and we consider the optimization problem:

sup {I (c, k) | (c, k) ∈ A (k 0 )} . ( 15
)
The Euler-Lagrange equation is given by:

u ′′ 11 dc dt = (δ -f ′ (k)) u ′ 1 -u ′ 2 -(f (k) -c) u ′′ 12 . ( 16
)
Equation ( 16), together with equation ( 13), constitute a system of two first-order ODEs for the unknown functions (c (t) , k (t)). In the particular case when u = u (c) does not depend on k, the last equation simplifies to:

u ′′ 11 dc dt = (δ -f ′ (k)) u ′ 1
and the ( 16), [START_REF] Ekeland | Being serious about non-commitment: subgame perfect equilibrium in continuous time[END_REF] 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 Ramsey problem in that case then is the solution of ( 16), ( 13) which converges to (c ∞ , k ∞ ) (see [START_REF] Barro | Economic growth[END_REF] for instance).

In the general case where u (c, k) depends on k, the situation is not as simple, and to our knowledge has not been investigated. Stationary points (c ∞ , k ∞ ) of the dynamics (if any) are given by:

c ∞ -f (k ∞ ) = 0 (17) (δ -f ′ (k ∞ )) u ′ 1 (f (k ∞ ) , k ∞ ) -u ′ 2 (f (k ∞ ) , k ∞ ) = 0. ( 18
)
To prove the existence of an optimal strategy, we do not use the Euler equation. We use the Hamilton-Jacobi-Bellman (HJB) equation instead. Introduce the optimal value as a function of the initial point:

V (k 0 ) := sup {I (c, k) | (c, k) ∈ A (k 0 )}
If there is an optimal solution (c (t) , k (t)), and it converges to (c ∞ , k ∞ ) when t → ∞, then, substituting in [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF], we must have:

V (k ∞ ) = ∫ ∞ 0 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 1
, 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 2 solution satisfying [START_REF] Hartman | Ordinary differential equations[END_REF] for some (c ∞ , k ∞ ), and define a strategy c = σ (k) by:

u ′ 1 (σ (k) , k) = V ′ (k) ( 21
)
Suppose moreover that the solution of:

dk dt = f (k) -σ (k) , k (0) = k 0 , ( 22
)
converges to k ∞ for all initial points k 0 . Then σ (k) is an optimal solution of the generalized Ramsey problem [START_REF] Ekeland | a classical problem in the calculus of variations leading to an implicit differential equation[END_REF] This is the so-called verification theorem, which is classical (see [START_REF] Bardi | Optimal control and viscosity solutions of Hamilton-Jacobi-Bellman equations[END_REF]). We need V to be C 2 , so that σ is defined. Since u ′′ 1 > 0, we can use the implicit function theorem on equation [START_REF] Heal | Intertemporal welfare economics and the environment[END_REF] to define σ. If V is C 2 , then σ is C 1 , and the initial-value problem [START_REF] Karp | Non-constant discounting in continuous time[END_REF] has a unique solution. Note also that everything is local: the functions V and σ are defined in some neighbourhood of k ∞ only, and the initial value k 0 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 HJB equation has a

C 2 solution V with V (k ∞ ) = δ -1 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 results Theorem 3 Suppose there is some k ∞ > 0 satisfying [START_REF] Frederick | Time discounting and time preference: a critical review[END_REF] and:

u ′ 1 f ′′ + u ′′ 22 + f ′ ( u ′′ 12 - u ′ 2 u ′ 1 u ′′ 11 ) - u ′ 2 u ′ 1 u ′′ 12 < 0 ( 23
)
(all values to be taken at k ∞ and c ∞ = f (k ∞ )). Then there is an optimal strategy c = σ (k) [START_REF] Chichilnisky | What is sustainable development?[END_REF] on f , and condition [START_REF] Karp | Time-consistent policies[END_REF] becomes

converging to k ∞ If u = u (c) does not depend on k, then u ′ 2 = 0. Equation (18) becomes f ′ (k ∞ ) = δ, which defines k ∞ uniquely because of the Inada conditions
u ′ 1 (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 is different. Proof. We follow the method of [START_REF] Ekeland | a classical problem in the calculus of variations leading to an implicit differential equation[END_REF]. By the inverse function theorem, the equation u ′ 1 (c, k)) = p defines c as a C 1 function of p and k:

c = φ(p, k), u ′ 1 (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 2 solution V satisfying:

V (k ∞ ) = 1 δ u(f (k ∞ ), k ∞ ). (27)
Differentiating [START_REF] Kuznetsov | Elements of applied bifurcation theory[END_REF] leads to:

δdV = (f (k) -φ(p, k))dp + pf ′ (k)dk + u ′ 2 (φ(p, k), k)dk ( 28
)
Plugging ( 28) into (25), we get:

(f (k) -φ(p, k))dp = (δp -pf ′ (k) -u ′ 2 (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):

dk dt = f (k) -φ(p, k) dp dt = δp -pf ′ (k) -u ′ 2 (φ(p, k), k) (30)
with the initial condition (k(0), p(0

)) = (k ∞ , p ∞ )
By [START_REF] Krusell | Consumption-savings decisions with quasi-geometric discounting[END_REF], we must have

p ∞ = u ′ 1 (φ(p ∞ , k ∞ ), k ∞ ) = u ′ 1 (c ∞ , k ∞ ) = u ′ 1 (f (k ∞ ) , k ∞ )
Differentiating [START_REF] Krusell | Consumption-savings decisions with quasi-geometric discounting[END_REF] with respect to p and k respectively, we derive the formulas:

φ ′ 1 (p ∞ , k ∞ ) = 1 u ′′ 11 (φ(p ∞ , k ∞ ), k ∞ ) = 1 u ′′ 11 (f (k ∞ ), k ∞ ) , ( 31
)
φ ′ 2 (p ∞ , k ∞ ) = - u ′′ 11 (φ(p ∞ , k ∞ ), k ∞ ) u ′′ 11 (φ(p ∞ , k ∞ ), k ∞ ) = - u ′′ 12 (f (k ∞ ), k ∞ ) u ′′ 11 (f (k ∞ ), k ∞ ) , ( 32
)
We can now linearizd [START_REF] Phelps | The indeterminacy of game-equilibrium growth[END_REF] at the point (p ∞ , k ∞ ). We get:

d dt ( k -k ∞ p -p ∞ ) = A ∞ ( k -k ∞ p -p ∞
) .

where the constant matrix A ∞ is given by:

A ∞ :=   f ′ + u ′′ 12 u ′′ 11 , -1 u ′′ 11 -u ′ 1 f ′′ -u ′′ 22 + (u ′′ 12 ) 2 u ′′ 11 , δ -f ′ - u ′′ 12 u ′′ 11   .
and all the values are to be taken at (k ∞ , p ∞ ). The characteristic polynomial is: [START_REF] Ekeland | Convex analysis and variational problems[END_REF]. Because of assumption [START_REF] Karp | Time-consistent policies[END_REF], it has two real roots with different signs, λ + > 0 and λ -< 0. Thus (k ∞ , p ∞ ) is a hyperbolic fixed point of [START_REF] Phelps | The indeterminacy of game-equilibrium growth[END_REF], 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:

λ 2 -δλ -[ u ′ 1∞ f ′′ ∞ + u ′′ 22∞ u ′′ 11∞ + f ′ ∞ ( u ′′ 12∞ u ′′ 11∞ - u ′ 2∞ u ′ 1∞ ) - u ′ 2∞ u ′ 1∞ u ′′ 12∞ u ′′ 11∞ ] = 0 where we used f ′ ∞ + u ′ 2∞ u ′ 1∞ = δ by
dp s dk s (k ∞ ) = u ′′ 11 (f ′ ∞ -λ -) + u ′′ 12
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 [START_REF] Kuznetsov | Elements of applied bifurcation theory[END_REF], and using ( 23) again, we find:

dV s dk s (k ∞ ) = u ′ 1 (f (k ∞ ), k ∞ ) > 0, 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 [START_REF] Harris | Dynamic choices of hyperbolic consumers[END_REF]. By [START_REF] Koopmans | On the concept of optimal economic growth[END_REF], we have

d 2 V s dk s 2 (k ∞ ) = dp s dk s (k ∞ ) = u ′′ 11∞ (f ′ ∞ -λ -) + u ′′ 12∞ . ( 33
)
Thus V s (k) is C 2 in k at k ∞ ,
and the C 2 property at other points near k ∞ follows from [START_REF] Phelps | The indeterminacy of game-equilibrium growth[END_REF].

It remains to show that the strategy σ defined by

u ′ 1 (σ (k) , k) = V ′ (k) converges to (f (k ∞ ) , k ∞ ). We rewrite σ as: σ(k) = φ(V ′ (k), k).
Linearizing the equation

dk dt = 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 optimal strategy [START_REF] Heal | Intertemporal welfare economics and the environment[END_REF]. Setting c (t) := σ (k (t)), differentiating [START_REF] Heal | Valuing the future: economic theory and sustainability[END_REF] with respect to k, and applying the envelope theorem, we get:

δV ′ (k) = u ′ 2 (c, k) + f ′ (k) V ′ (k) + (f (k) -c) V ′′ (k) ,
and hence, noting that

dk dt = f (k) -c (t) : (δ -f ′ (k)) V ′ (k) -u ′ 2 (c, k) = (f (k) -c) d dk V ′ (k) = d dt V ′ (k) . Replacing V ′ (k) by u ′ 1 (σ (k) , k) = u ′ 1 (c, k), we get: (δ -f ′ (k)) u ′ 1 (c, k) -u ′ 2 (c, k) = d dt u ′ 1 (c, k),
which is precisely the Euler equation.

3 Time-inconsistency.

Equilibrium strategies

We consider the intertemporal decision problem (as it seen at time t = 0)

J (c, k) = ∫ ∞ 0 [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 ∞ nonincreasing 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, ∞) 2 , with u ′′ < 0 and U ′′ < 0 everywhere. We shall also assume that they decay exponentially, so that there is some ρ > 0 and some T > 0 such that 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 optimization problem. There is no way for the decision-maker at time 0 to achieve what is, from her point of view, the first-best solution of the problem, and she must turn to a second-best policy: the best she can 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 leaderfollower game played by successive generations.

The equilibrium policy was described in [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF] for the case when the criterion (34) did not include the 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 [START_REF] Karp | Non-constant discounting in continuous time[END_REF].

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 0 ∈ 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 public knowledge σ. The decision-maker at T has capital stock k T . If all future decision-maker apply the strategy σ, the resulting future capital stock flow k(t) obeys:

dk dt = 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 to take T = 0. Suppose the decision-maker at time 0 holds power for 0 ≤ t < ε, and expects all later decision-makers to apply the strategy σ. He or she then explores whether it is in his or her interest to apply the strategy σ, that is, to play

c 0 = σ (k 0 ) for 0 ≤ t < ε. If he or she applies the constant control 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 0 ) + U (c, k 0 )]ε + o(ε) where o(ε) is a higher order term of ε. At time ε, the resulting capital will be k 0 + (f (k 0 ) -c)ε + o(ε).
From then on, the strategy σ will be applied, which results in a capital stock k c satisfying (we omit higher-order terms):

dk c dt = f (k c ) -σ (k c ) , k c (ε) = k 0 + (f (k 0 ) -c)ε, t ≥ ε (36)
The capital stock k c can be written as

k c (t) = k 0 (t) + k 1 (t)ε, (37)
where k 0 (t) is the unperturbed solution, and k 1 (t) is given by the linearized equation:

dk 0 dt = f (k 0 ) -σ (k 0 ) , k 0 (0) = k 0 (38) dk 1 dt = (f ′ (k 0 ) -σ ′ (k 0 )) k 1 , k 1 (0) = σ (k 0 ) -c (39)
Evaluating the integral (34) we get:

J (ε) = u(c, k 0 )ε + ∫ ∞ ε h(s)u(σ(k 0 (t) + εk 1 (t)), k 0 (t) + εk 1 (t))dt +U (c, k 0 )ε + ∫ ∞ ε H(s)U (σ(k 0 (t) + εk 1 (t)), k 0 (t) + εk 1 (t)
)dt Letting ε → 0, so that commitment span of the decision-maker vanishes, we get:

lim ε→0 1 ε (J (ε) -J (0)) = P (k 0 , σ, c)
where

P (k 0 , σ, c) = u(c, k 0 ) -u(σ(k 0 ), k 0 ) + (U (c, k 0 ) -U (σ(k 0 ), k 0 )) + ∫ ∞ 0 h(t)u ′ 1 (σ(k 0 (t)), k 0 (t))σ ′ (k 0 (t))k 1 (t)dt + ∫ ∞ 0 h(t)u ′ 2 (σ(k 0 (t)), k 0 (t))k 1 (t)dt + ∫ ∞ 0 H(t)U ′ 1 (σ(k 0 (t)), k 0 (t))σ ′ (k 0 (t))k 1 (t)dt + ∫ ∞ 0 H(t)U ′ 2 (σ(k 0 (t)), k 0 (t))k 1 (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.

The HJB approach

We now characterizes the equilibrium strategy. We write k c (t) = K(t; k 0 , σ) where K is the flow associated with the differential equation (38). We also define a function φ by:

u ′ 1 (φ(x, k), k) + U ′ 1 (φ(x, k), k) = x, φ(u ′ 1 (c, k) + U ′ 1 (c, k), k) = c
Since u ′′ 1 and U ′′ 1 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 0 ) = ∫ ∞ 0 h(t)u(σ(K(t; k 0 , σ)), K(t; k 0 , σ))dt + ∫ ∞ 0 H(t)U (σ(K(t; k 0 , σ)), K(t; k 0 , σ))dt (40)
satisfies the integral equation:

V (k 0 ) = ∫ ∞ 0 h(t)u(φ • V ′ (K(t; k 0 , φ • V ′ )), K(t; k 0 , φ • V ′ ))dt + ∫ ∞ 0 H(t)U (φ • V ′ (K(t; k 0 , φ • V ′ )), K(t; k 0 , φ • V ′ ))dt (IE)
and the instantaneous optimality condition

u ′ 1 (σ(k 0 ), k 0 ) + U ′ 1 (σ(k 0 ), k 0 ) = V ′ (k 0 ), σ(k 0 ) = φ(V ′ (k 0 ), k 0 ), (41)
Conversely, suppose a function V is twice continuously differentiable, satisfies (IE), and the strategy σ(k 0 ) := φ(V ′ (k 0 ), k 0 ) is convergent. Then σ is an equilibrium strategy.

For the sake of convenience, we have shortened σ(k

0 ) := φ(V ′ (k 0 ), k 0 ) to σ = φ • V ′ .
Proof. Since the system is autonomous, we have:

K(s; K(t; k 0 , σ), σ) = K(s + t; k 0 , σ). ( 42
)
Next, denote the fundamental solution of the linearized equation of (35) at k 0 by R(k 0 ; t) so that:

k 1 (t) = R(k 0 ; t)(σ(k 0 ) -c) dR dt = (f ′ (K(t; k 0 , σ)) -σ ′ (K(t; k 0 , σ)))R(t), R(k 0 ; 0) = I,
R and K are related by:

∂K(t; k 0 , σ) ∂k 0 = R(k 0 ; t), ( 43
)
∂K(t; k 0 , σ) ∂t = f (K(t; k 0 , σ)) -σ(K(t; k 0 , σ)). ( 44
)
Let us now turn to the first part of the theorem. Differentiating (40)with respect to k 0 :

V ′ (k 0 ) = ∫ ∞ 0 h(t)u ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ))R(k 0 ; t)dt + ∫ ∞ 0 h(t)u ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))R(k 0 ; t)dt + ∫ ∞ 0 H(t)U ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ))R(k 0 ; t)dt + ∫ ∞ 0 H(t)U ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))R(k 0 ; t)dt. ( 45
)
Substituting k 0 (t) = K(t; k 0 , σ) and k 1 (t) in the definition of P , we get:

P (k 0 , σ, c) = [u(c, k 0 ) + U (c, k 0 )] -[u(σ(k 0 ), k 0 ) + U (σ(k 0 ), k 0 )] + ∫ ∞ 0 h(t)u ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ))R(k 0 ; t)(σ(k 0 ) -c)dt + ∫ ∞ 0 h(t)u ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))R(k 0 ; t)(σ(k 0 ) -c)dt + ∫ ∞ 0 H(t)U ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ))R(k 0 ; t)(σ(k 0 ) -c)dt + ∫ ∞ 0 H(t)U ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))R(k 0 ; t)(σ(k 0 ) -c)dt.
Since u and U are strictly concave and differentiable with respect to c, the necessary and sufficient condition to maximize P (k 0 , σ, c) with respect to c is that the derivative vanishes at c = σ(k 0 ), that is:

u ′ 1 (σ(k 0 ), k 0 ) + U ′ 1 (σ(k 0 ), k 0 ) = ∫ ∞ 0 h(t)u ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ))R(k 0 ; t)dt + ∫ ∞ 0 h(t)u ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))R(k 0 ; t)dt + ∫ ∞ 0 H(t)U ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ))R(k 0 ; t)dt + ∫ ∞ 0 H(t)U ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))R(k 0 ; t)dt,
The right-hand side is precisely V ′ (k 0 ), as we wanted. Therefore, the equilibrium strategy satisfies

u ′ 1 (σ(k 0 ), k 0 ) + U ′ 1 (σ(k 0 ), k 0 ) = V ′ (k 0 )
and we have σ(k 0 ) = φ(V ′ (k 0 ), k 0 ). Substituting back into equation (40), we get the functional equation (IE). This prove the first part of the theorem (necessity). We refer to [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF] for the second part (sufficiency).

The following theorem gives an alternative characterization, the differential equation, which resembles the usual HJB equation from the calculus of variation.

Theorem 7 Let V be a C 2 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 0 ), k 0 ) + U (φ • V ′ (k 0 ), k 0 ) + V ′ (k 0 )(f (k 0 ) -φ(V ′ (k 0 ), k 0 )) = - ∫ ∞ 0 h ′ (t)u(φ • V ′ (K(t; k 0 , φ • V ′ )), K(t; k 0 , φ • V ′ ))dt - ∫ ∞ 0 H ′ (t)U (φ • V ′ (K(t; k 0 , φ • V ′ )), K(t; k 0 , φ • V ′ ))dt (DE)
together with the boundary condition

V ( k) = u(f ( k), k) ∫ ∞ 0 h(t)dt + U (f ( k), k) ∫ ∞ 0 H(t)dt. (BC)
Proof. Introduce the function ϕ defined by

ϕ(k 0 ) = V (k 0 ) - ∫ ∞ 0 h(t)u(σ(K(t; k 0 , σ)), K(t; k 0 , σ))dt - ∫ ∞ 0 H(t)U (σ(K(t; k 0 , σ)), K(t; k 0 , σ))dt,
where σ(k 0 ) = φ(V ′ (k 0 ), k 0 ). Consider the value ψ(t, k 0 ) of ϕ along the trajectory t → K(t; k 0 , σ) originating from k 0 at time 0, that is

ψ(t, k 0 ) = ϕ(K(t; k 0 , σ)) = V (K(t; k 0 , σ)) - ∫ ∞ t h(s -t)u(σ(K(σ; s, k 0 )), K(σ; s, k 0 ))ds - ∫ ∞ t H(s -t)U (σ(K(σ; s, k 0 )), K(σ; s, k 0 ))ds,
Then the derivative of ψ with respect to t is given by:

∂ψ(t, k 0 ) ∂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 ) + ∫ ∞ 0 h ′ (s)u(φ • V ′ (K(s; K t , φ • V ′ )), K(s; K t , φ • V ′ ))ds + ∫ ∞ 0 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 0 , φ • V ′ ) by K t for convenience. If (DE) holds, then the right hand side is identically zero along the trajectory, so that ψ(t, k 0 ) does not depend on t, thus ψ(s, k 0 ) = ψ(t, k 0 ) for all s, t ≥ 0. Letting t → ∞ in the definition of ψ, we get

ψ(s, k 0 ) = lim t→∞ ψ(t, k 0 ) = V ( k) - ∫ ∞ 0 h(s)u(σ( k), k)ds - ∫ ∞ 0 H(s)U (σ( k), k)ds, (46)
and hence, if (BC) holds, then ψ = ϕ ≡ 0 and so equation (IE) holds. Conversely, if V (k) satisfies equation (IE), then the same lines of reasoning shows that equation (DE) and the boundary condition (BC) are satisfied.

The Euler equations

To obtain the Euler-Lagrange-like equation, we differentiate the both side of (DE) with respect to k 0 . We get:

-

∫ ∞ 0 h ′ (t)u ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ)) ∂K(t; k 0 , σ) ∂k 0 dt - ∫ ∞ 0 h ′ (t)u ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ)) ∂K(t; k 0 , σ) ∂k 0 dt - ∫ ∞ 0 H ′ (t)U ′ 1 (σ(K(t; k 0 , σ)), K(t; k 0 , σ))σ ′ (K(t; k 0 , σ)) ∂K(t; k 0 , σ) ∂k 0 dt - ∫ ∞ 0 H ′ (t)U ′ 2 (σ(K(t; k 0 , σ)), K(t; k 0 , σ)) ∂K(t; k 0 , σ) ∂k 0 dt = [u ′ 1 (σ(k 0 ), k 0 ) + U ′ 1 (σ(k 0 ), k 0 )]σ ′ (k 0 ) + u ′ 2 (σ(k 0 ), k 0 ) + U ′ 2 (σ(k 0 ), k 0 ) + V ′ (k 0 )f ′ (k 0 ) -V ′ (k 0 )σ ′ (k 0 ) + V ′′ (k 0 )(f (k 0 ) -σ(k 0 )).
Plugging k 0 = k(t), σ(k(t)) = c(t) and using (41) to cancel the first and the fifth terms, together with ( 13) and ( 43), we have

- ∫ ∞ 0 h ′ (s)u ′ 1 (σ(K(s; k(t), σ)), K(s; k(t), σ))σ ′ (K(s; k(t), σ))R(k (t) , s)ds - ∫ ∞ 0 h ′ (s)u ′ 2 (σ(K(s; k(t), σ)), K(s; k(t), σ))R(k (t) , s)ds - ∫ ∞ 0 H ′ (s)U ′ 1 (σ(K(s; k(t), σ)), K(s; k(t), σ))σ ′ (K(s; k(t), σ))R(k (t) , s)ds - ∫ ∞ 0 H ′ (s)U ′ 2 (σ(K(s; k(t), σ)), K(s; k(t), σ))R(k (t) , s)ds = u ′ 2 (c(t), k(t)) + U ′ 2 (c(t), k(t)) + V ′ (k(t))f ′ (k(t)) + V ′′ (k(t))(f (k(t)) -c(t)) = u ′ 2 (c(t), k(t)) + U ′ 2 (c(t), k(t)) + [u ′ 1 (c(t), k(t)) + U ′ 1 (c(t), k(t))]f ′ (k(t)) + d dt [u ′ 1 (c(t), k(t)) + U ′ 1 (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 0 ; s + t) σ ′ (k (t)) = 1 f (k(t)) -c(t) dc dt (47) R(k 0 ; s + t) = exp(f ′ (k (t)) - 1 f (k(t)) -c(t) dc dt ),
Writing this into the preceding equations, we finally get:

-

∫ ∞ t h ′ (s -t) [u ′ 1 (c(s), k(s))γ (s) + u ′ 2 (c(s), k(s))] e f ′ (k(s))-γ(s) ds (48) - ∫ ∞ t H ′ (s -t) [U ′ 1 (c(s), k(s))γ (s) + U ′ 2 (c(s), k(s))] e f ′ (k(s))-γ(s) ds (49) = u ′ 2 (c(t), k(t)) + U ′ 2 (c(t), k(t)) + [u ′ 1 (c(t), k(t)) + U ′ 1 (c(t), k(t))] f ′ (k(t)) (50)
+ d dt [u ′ 1 (c(t), k(t)) + U ′ 1 (c(t), k(t))] (51)
with:

γ (s) := 1 f (k(s)) -c(s) dc ds ( 52
)
which is the Euler-Lagrange-like equation for the time-inconsistent case.

The control theory approach

Karp [START_REF] Karp | Non-constant discounting in continuous time[END_REF], [START_REF] Karp | Time-consistent policies[END_REF] has developed a different method to deal with time-inconsistency. In this section, we connect his results with ours. Defining V (k 0 ) as above, we must have:

V (k 0 ) = max c { u(c, k 0 ) + U (c, k 0 )]ε + ∫ ∞ ε [h(t)u(σ(k c (t)), k c (t)) + H(t)U (σ(k c (t)), k c (t))]dt } (53)
On the other hand, we have

V (k c (ε)) = ∫ ∞ 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 0 ) = max c    [u(c, k 0 ) + U (c, k 0 )]ε + V (k c (ε)) + ∫ ∞ ε [h(t) -h(t -ε)]u(σ(k c (t)), k c (t))dt + ∫ ∞ ε [H(t) -H(t -ε)]U (σ(k c (t)), k c (t))dt    Letting ε → 0, we have - ∫ ∞ 0 h ′ (t)u(σ(k(t)), k(t))dt - ∫ ∞ 0 H ′ (t)U (σ(k(t)), k(t))dt = max c {u(c, k 0 ) + U (c, k 0 ) + V ′ (k 0 )(f (k 0 ) -c)}. ( 54
)
This equation was first obtained by Karp [START_REF] Karp | Non-constant discounting in continuous time[END_REF], [START_REF] Karp | Time-consistent policies[END_REF]. This is the HJB equation for a certain control problem, which he terms the auxiliary control problem. We now show that this approach is equivalent to the preceding one. approach by Ekeland-Lazrak. In the right hand side of (54), the maximum is attained at:

c = arg max c {u(c, k 0 ) + U (c, k 0 ) + V ′ (k 0 )(f (k 0 ) -c)} = φ(V ′ (k 0 ), k 0 ), thus we have - ∫ ∞ 0 h ′ (t)u(σ(k(t)), k(t))dt - ∫ ∞ 0 H ′ (t)U (σ(k(t)), k(t))dt = u(φ(V ′ (k 0 ), k 0 ), k 0 ) + U (φ(V ′ (k 0 ), k 0 ), k 0 ) + V ′ (k 0 )(f (k 0 ) -φ(V ′ (k 0 ), k 0 )).
which is exactly the same as (DE)

The biexponential case 4.1 The equations

In this section, we consider the biexponential criterion

λ ∫ ∞ 0 e -δ1s u(c(s), k(s))ds + (1 -λ) ∫ ∞ 0 e -δ2s U (c(s), k(s))ds. ( 55
)
Without loss of generality, we assume that:

δ 1 > δ 2
If λ = 0 or 1, then this is just the Ramsey criterion ( 14). Thus we are interested in the case 0 < λ < 1. Dividing by λ, we find that (55) is a special case of (34), where h(t) = e -δ1t , H(t) = e -δ2t and U is replaced by 1-λ λ U . So all the results of the preceding section hold.

The HJB-type equations

Given an equilibrium strategy σ λ , introduce the two functions:

V λ (k) := ∫ ∞ 0 e -δ1t u(σ λ (k(t)) , k(t))dt + 1 -λ λ ∫ ∞ 0 e -δ2t U (σ λ (k(t)) , k(t))dt (56) W λ (k) := ∫ ∞ 0 e -δ1t u(σ λ (k(t)) , k(t))dt - 1 -λ λ ∫ ∞ 0 e -δ2t 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 ODEs for V λ and W λ :

(f -φ λ (V ′ λ )) V ′ λ + u (V ′ λ ) + 1 -λ λ U (V ′ λ ) = δ 1 + δ 2 2 V λ + δ 1 -δ 2 2 W λ , ( 58
) (f -φ λ (V ′ λ )) W ′ λ + u (V ′ λ ) - 1 -λ λ U (V ′ λ ) = δ 1 -δ 2 2 V λ + δ 1 + δ 2 2 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).
Recall that φ λ is defined by the equivalent equations:

u ′ 1 (φ λ (x, k), k) + 1 -λ λ U ′ 1 (φ λ (x, k), k) = x, ( 60
)
φ λ (u ′ 1 (c, k) + 1 -λ λ U ′ 1 (c, k), k) = c (61)
Similarly, the boundary condition (BC) becomes:

V λ (k ∞ ) = 1 δ 1 u (f (k ∞ ) , k ∞ ) + 1 -λ λ 1 δ 2 U (f (k ∞ ) , k ∞ ) , ( 62
) W λ (k ∞ ) = 1 δ 1 u (f (k ∞ ) , k ∞ ) - 1 -λ λ 1 δ 2 U (f (k ∞ ) , k ∞ ) . ( 63
)
If λ = 1, we find the usual HJB equation ( 20) for V with the boundary condition [START_REF] Hartman | Ordinary differential equations[END_REF].

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 2 function V λ and a C 1 function W λ , both defined on some open neighbourhood of k ∞ , satisfying the equations ( 58) and ( 59) with the boundary conditions ( 62) and ( 63); suppose moreover that strategy σ λ (k

) := φ λ (V ′ λ (k) , k) converges to (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 2 while W λ needs only be C 1 . Proof. Let us simplify the notation. Write σ instead of σ λ and set

a = δ 1 + δ 2 2 , b = δ 1 -δ 2 2
Arguing as in Theorem 7, we obtain (58) and (59) by differentiating ( 56) and (57). The boundary conditions (62) and (63) follow from setting k (t) = k ∞ and c (t) = σ (k ∞ ) in ( 58) and (59).

Conversely suppose v 1 and w 1 satisfy (58), ( 59) and ( 62), (63), and suppose the strategy

σ 1 = φ λ • v ′ 1 converges to k. Consider the following functions v 2 (k 0 ) = ∫ ∞ 0 e -δ1t u(σ 1 (K(t; k 0 , σ 1 )), K(t; k 0 , σ 1 ))dt + 1 -λ λ ∫ ∞ 0 e -δ2t U (σ 1 (K(t; k 0 , σ 1 )), K(t; k 0 , σ 1 ))dt, w 2 (k 0 ) = ∫ ∞ 0 e -δ1t u(σ 1 (K(t; k 0 , σ 1 )), K(t; k 0 , σ 1 ))dt - 1 -λ λ ∫ ∞ 0 e -δ2t U (σ 1 (K(t; k 0 , σ 1 )), K(t; k 0 , σ 1 ))dt.
Arguing as in Theorem 7, we find that v 2 and w 2 also satisfy (58), ( 59) and ( 62), (63). Setting

v 3 = v 1 -v 2 , w 3 = w 1 -w 2 , then we get (f (k) -σ (k)) v ′ 3 (k) + u (σ (k) , k) + 1-λ λ U (σ (k) , k) = av 3 (k) + bw 3 (k) (f (k) -σ (k)) w ′ 3 (k) + u (σ (k) , k) + 1-λ λ U (σ (k) , k) = bv 3 (k) + aw 3 (k) (64)
with the boundary conditions

v 3 (k ∞ ) = 0 w 3 (k ∞ ) = 0 (65)
Obviously, v 3 = w 3 = 0 is a solution. Lemma 9 below shows that We need to show that it is the only one, so that v 1 = v 2 and w 1 = w 2 . This is the desired result.

Lemma 9 If (v 3 , w 3 ) is a pair of continuous functions on a neighborhood k ∞ , continuously differentiable for k ̸ = k ∞ , and which solve (64) with boundary conditions (65), then v 3 = w 3 = 0.

Proof. Set f (k) -σ 1 (k) = ξ(k), then ξ(k) → 0 as k → k ∞ .
The system (64) can be rewritten as:

( ξv ′ 3 ξw ′ 3 ) = ( a b b a )( v 3 w 3
) .

Setting:

( V W ) =   √ 1 2 √ 1 2 - √ 1 2 √ 1 2   ( v 3 w 3
) ,

we have:

ξ ( V ′ W ′ ) = ( δ 1 0 0 δ 2 )( V W
) .

The first equation yields

V (k) = V (k 0 ) exp ∫ k k0 δ1 ξ(u) du. Without loss of generality, we can assume that k 0 < k ∞ . Let S = {k | ξ(k) = 0, k 0 ≤ k ≤ k ∞ },
where k 0 is the initial stock. We consider the following two cases:

First case: S = {k ∞ }. Then ξ(k) > 0 for k ∈ [k 0 , k ∞ ). So dk dt = ξ(k) > 0, and: 0 = √ 1 2 (v 3 (k ∞ ) + w 3 (k ∞ )) = V (k ∞ ) = V (k 0 ) exp ∫ k k0 δ 1 ξ(u) du It follows that V (k) = 0 for all k. Second case: S\{k ∞ } ̸ = ∅. Then S is a closed set and V (k) = ξ(k)V ′ (k) δ1
= 0 in S. The complement of S is a countable union of disjoint intervals, and ξ vanishes at the endpoints of each interval. Arguing as above it follows that V (k) vanishes for

k 0 ≤ k ≤ k ∞ .
The same argument holds for W (k). This concludes the proof.

The Euler-type equations

The Euler-type equations ( 48) to (52) can be reduced to a non-autonomous system of four first-order ODEs. It is better to work directly on (58) and (59) to get an autonomous system. Proceeding as in the end of Section 2, we differentiate (58) and ( 59) with respect to k, and notice that

(f -σ)V ′′ λ = dk dt V ′ λ dk = d dt V ′ λ ,
and then get:

δ 1 + δ 2 2 V ′ λ + δ 1 -δ 2 2 W ′ λ = d dt V ′ λ + (f ′ -σ ′ )V ′ λ + (u ′ 1 + 1 -λ λ U ′ 1 )σ ′ + u ′ 2 + 1 -λ λ U ′ 2 , δ 1 -δ 2 2 V ′ λ + δ 1 + δ 2 2 W ′ λ = d dt W ′ λ + (f ′ -σ ′ )W ′ λ + (u ′ 1 - 1 -λ λ U ′ 1 )σ ′ + u ′ 2 - 1 -λ λ U ′ 2 ,
where 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 1 + δ 2 2 ( u ′ 1 + 1 -λ λ U ′ 1 ) + δ 1 -δ 2 2 w = d dt ( u ′ 1 + 1 -λ λ U ′ 1 ) + (f ′ -σ ′ ) ( u ′ 1 + 1 -λ λ U ′ 1 ) + ( u ′ 1 + 1 -λ λ U ′ 1 ) σ ′ + u ′ 2 + 1 -λ λ U ′ 2 , δ 1 -δ 2 2 ( u ′ 1 + 1 -λ λ U ′ 1 ) + δ 1 + δ 2 2 w = dw dt + (f ′ -σ ′ )w + ( u ′ 1 - 1 -λ λ U ′ 1 ) σ ′ + u ′ 2 - 1 -λ λ U ′ 2 ,
and hence

( δ 1 + δ 2 2 -f ′ ) (λu ′ 1 + (1 -λ) U ′ 1 ) + δ 1 -δ 2 2 λw = d dt (λu ′ 1 + (1 -λ) U ′ 1 ) + λu ′ 2 + (1 -λ) U ′ 2 , (66) δ 1 -δ 2 2 (λu ′ 1 + (1 -λ) U ′ 1 ) + ( δ 1 + δ 2 2 -f ′ )λw = λ dw dt + (-λw + λu ′ 1 -(1 -λ) U ′ 1 )σ ′ + λu ′ 2 -(1 -λ) U ′ 2 , ( 67
)
to which we should add:

dk dt = 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-Lagrange equation. For λ = 1, taking w (t) = u ′ 1 (c (t) , k (t)) gives us again two copies of the same equation.

The control theory approach

Equation (54) is the HJB equation for the following:

max c(.) ∫ ∞ 0 e -rt [ u (c (t) , k (t)) + 1 -λ λ U (c (t) , k (t)) -K (k (t)) ] dt ( 69
)
dk dt = f (k) -c (t) , k (0) = k 0 (70)
where we seek a feedback σ (k) such that:

K(k 0 ) = { (δ -r) ∫ ∞ 0 e -δt u(σ (k (t)) , k (t))dt | dk dt = f (k) -σ (k) k (0) = k 0 } (71)
Solving ( 69), (70) under the constraint ( 71) is a fixed-point problem for the feedback c = σ (k).

Solving the boundary-value problem

Define a function g λ (k) by:

g λ (k) := λδ 1 u ′ 1 (f (k) , k) + (1 -λ)δ 2 U ′ 1 (f (k) , k) λu ′ 1 (f (k) , k) + (1 -λ)U ′ 1 (f (k) , k) - λu ′ 2 (f (k) , k) + (1 -λ)U ′ 2 (f (k) , k) λu ′ 1 (f (k) , k) + (1 -λ)U ′ 1 (f (k) , k) (72) Theorem 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 2 and W of class C 1.
Proof. We adapt the argument in [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF] to the present situation. We note that the boundary-value problem (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 -λ λ U (φ (V ′ λ , k) , k) = aV λ + bW λ . Since the function c → u(c, k) + 1-λ λ U (c, k) is concave, the function: c → u(c, k) + 1 -λ λ 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 { u(c, k) + 1 -λ λ U (c, k) -cx } = u (φ λ (x, k) , k)+ 1 -λ λ 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 [START_REF] Ekeland | Convex analysis and variational problems[END_REF]), we find that:

min y {f (k) y + u * λ (y, k)} = u(f (k) , k) + 1 -λ λ U (f (k) , k).
Note that f (k)y + u * λ (y, k) is convex in y with minimal value u(f (k), k) + 1-λ λ 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 -λ λ 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 -λ λ U (f (k) , k)
then equation (75) has precisely one solution, namely:

V ′ λ = u ′ 1 (f (k) , k) + 1 -λ λ U ′ 1 (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 -λ λ U (f (k ∞ ) , k ∞ ) ,
so that we are exactly on the boundary case. Equation (75) has precisely one solution, by (84) below, which satisfies

V ′ λ (k ∞ ) = u ′ 1 (f (k ∞ ) , k ∞ ) + 1 -λ λ U ′ 1 (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 differential equation, and special techniques are needed to solve it. The difficulty is further increased by the fact that we need only V (but not W ) to be C 2 .

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) - 1 -λ λ U (f (k), k) µ(k) := aV λ (k) + bW λ (k) -u(f (k), k) - 1 -λ λ U (f (k), k) Note that x → F (x, k) is a linear perturbation of the convex function x → u * λ (x, k).
To shorten the notations, let us set:

y * (k) := u ′ 1 (f (k) , k) + 1 -λ λ U ′ 1 (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) < 0 < x + (k, µ) for µ(k) > 0, none for µ(k) < 0 and a single solution 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) + 1-λ λ U (σ λ (k) , k) aV λ + bW λ -u (σ λ (k) , k) -1-λ λ U (σ λ (k) , k) Differentiating µ(k) with respect to k yields dµ dk = V ′ λ (a 2 + b 2 )V λ + 2abW λ -δ 1 u (σ λ (k) , k) -(1-λ)δ2 λ U (σ λ (k) , k) aV λ + bW λ -u (σ λ (k) , k) -1-λ λ U (σ λ (k) , k) -y * f ′ -u ′ 2 (f (k) , k) - 1 -λ λ U ′ 2 (f (k) , k) 20
We now take (V λ (k), µ(k)) as our new unknown functions. They satisfy the equations:

dV dk = y * (k) + x (k, µ (k)) , dµ dk = (y * + x (k, µ (k))) (a 2 + b 2 )V λ + 2abW λ -δ 1 u (σ λ (k) , k) -(1-λ)δ2 λ U (σ λ (k) , k) aV λ + bW λ -u (σ λ (k) , k) -1-λ λ U (σ λ (k) , k) -y * f ′ -u ′ 2 (f (k) , k) - 1 -λ λ U ′ 2 (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 > 0.

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 shorten our notation by setting x(k) = x(µ(k), k). We get:

dk dx = f (k) -φ λ (y * + x, k) D(x, k, V λ , W λ ) [ aV λ + bW λ -u(φ λ (y * + x, k), k) - 1 -λ λ U (φ λ (y * + x, k), k) ] (77)
where (here φ λ stands for φ λ (y

* + x, k)) D(x, k, V λ , W λ ) = (y * + x) [ (a 2 + b 2 )V λ + 2abW λ -δ 1 u (σ λ (k) , k) - (1 -λ)δ 2 λ U (σ λ (k) , k) ] -A(aV λ + bW λ -u(φ λ , k) - 1 -λ λ U (φ λ , k)) A(x, k, V λ , W λ ) = (f -φ λ ) [( u ′′ 11 (f (k) , k) + 1 -λ λ U ′′ 11 (f (k) , k) ) f ′ + u ′′ 12 (f (k) , k) + 1 -λ λ U ′′ 12 (f (k) , k) ] + (y * + x) f ′ (k) + u ′ 2 (φ λ , k) + 1 -λ λ U ′ 2 (φ λ , k)
Further more, we have

dV λ dx = dV λ dk dk dx = (y * + x) f (k) -φ λ D(x, k, V λ , W λ ) [aV λ + bW λ -u(φ λ , k) - 1 -λ λ 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

dx ds = D(x, k, V λ , W λ ) dk dx = (f (k) -φ λ ) [ aV λ + bW λ -u(φ λ (y * + x, k), k) -1-λ λ U (φ λ (y * + x, k), k) ] dV λ dx = (y * + x) (f (k) -φ λ (y * + x, k)) [aV λ + bW λ -u(φ λ , k) -1-λ λ 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 -λ λ U (f (k), k) -aV λ (k)
The dynamics of (x (s) , k (s) , V λ (s)) are given by:

dx ds = D(x, k, V λ ) dk dx = (f (k) -φ λ ) [ F (y * (k) + x, k) + u(f (k), k) + 1-λ λ U (f (k), k) -u(φ λ (y * + x, k), k) -1-λ λ U (φ λ (y * + x, k), k) ] dV λ dx = (y * + x) (f (k) -φ λ (y * + x, k)) [F (y * (k) + x, k) + u(f (k), k) + 1-λ λ U (f (k), k) -u(φ λ (y * + x, k), k) -1-λ λ 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 2 , we need f to be C 3 and u, U to be C 4 .

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 one solution 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 ′ 1 (f (k ∞ ), k ∞ ) + 1 -λ λ U ′ 1 (f (k ∞ ), k ∞ )
Computing the linearized system at (0, k ∞ , v ∞ ), we find:

d ds   x k -k ∞ V λ -v ∞   =   a ∞ b ∞ c ∞ 0 0 0 0 0 0     x k -k ∞ V λ -v ∞   .
where:

a ∞ = y * (k ∞ ) 2 ∂φ λ ∂y (y * (k ∞ ), k ∞ ) (f ′ (k ∞ ) -g λ (k ∞ )) g λ (k ∞ ) = λδ 1 u ′ 1 (f (k ∞ ) , k ∞ ) + (1 -λ)δ 2 U ′ 1 (f (k ∞ ) , k ∞ ) λu ′ 1 (f (k ∞ ) , k ∞ ) + (1 -λ)U ′ 1 (f (k ∞ ) , k ∞ ) - λu ′ 2 (f (k ∞ ) , k ∞ ) + (1 -λ)U ′ 2 (f (k ∞ ) , k ∞ ) λu ′ 1 (f (k ∞ ) , k ∞ ) + (1 -λ)U ′ 1 (f (k ∞ ) , k ∞ )
By the concavity assumptions, we find:

y ∞ = u ′ 1∞ + 1 -λ λ U ′ 1∞ = u ′ 1 (f (k ∞ ), k ∞ ) + 1 -λ λ U ′ 1 (f (k ∞ ), k ∞ ) > 0, u ′ 11 (f (k ∞ ), k ∞ ) + 1 -λ λ U ′ 11 (f (k ∞ ), k ∞ ) < 0 ∂φ λ ∂y (y * (k ∞ ), k ∞ ) = ( u ′′ 1 (f (k ∞ ), k ∞ ) + λ 1 -λ U ′′ 1 (f (k ∞ ), k ∞ ) ) -1 < 0 For any 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:

d ds   x k -k ∞ V λ -v ∞   =   -y 2 ∞ φ ′ 1∞ (g λ (k ∞ ) -f ′ (k ∞ )) 0 0 0 0 0 0 0 0     x k -k ∞ V λ -v ∞   .
Step 6: Applying the center manifold theorem By the center manifold theorem (cf. Theorem 1 of [START_REF] Carr | Applications of centre manifold theory[END_REF]), there exist an ϵ > 0, and a map h

(k, V λ ), defined in a neighborhood O = (k ∞ -ϵ, k ∞ + ϵ) × (v ∞ -ϵ, v ∞ + ϵ) of (k ∞ , v ∞ ) such that h(k ∞ , v ∞ ) = 0, ∂h ∂k (k ∞ , v ∞ ) = 0, ∂h ∂V λ (k ∞ , v ∞ , 0) = 0
and the manifold M defined by

M = {( h(k, V λ ) - b ∞ a ∞ (k -k ∞ ) - c ∞ a ∞ (V λ -v ∞ ), k, V λ ) | (k, V λ ) ∈ O } ,
is invariant under the flow associated to the system (79). The map h and the central manifold M are C 2 , 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 ∞ ) = 0
We 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 dV dk = y * (k) + x, we get

{ dV λ dk = u ′ 1 (f (k) , k) + 1-λ λ U ′ 1 (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 differentiable in O 1 = (k ∞ -ϵ, k ∞ + ϵ), therefore, is locally Lipschiz continuous. By dV λ dk | k=k∞ = y ∞ ̸ = 0 which follows from (76) and the first equation of (81), the nonconstant solution of this initial-value problem exist in O 1 which we denote by

V λ (k) = ζ(k) ,where ζ(k ∞ ) = v ∞ and ζ ∈ C 2 (O 1 ) if h ∈ C 2 (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 ′ 1 (f (k), k) + 1 -λ λ U ′ 1 (f (k), k) + x(k), k) and W λ (k) = 1 b (µ(k) + u(f (k), k) + 1 -λ λ U (f (k), k) -aζ(k)
is also C 2 , so we have found a C 2 solution of the system (58),(59) with boundary conditions (62),(63).

The existence of equilibrium strategies

Introduce another function g λ (k) defined by:

g λ (k) = λu ′ 1 (f (k) , k) + (1 -λ)U ′ 1 (f (k) , k) λ δ1 u ′ 1 (f (k) , k) + 1-λ δ2 U ′ 1 (f (k) , k) - λδ 2 u ′ 2 (f (k) , k) + (1 -λ) δ 1 U ′ 2 (f (k) , k) λδ 2 u ′ 1 (f (k) , k) + (1 -λ) δ 1 U ′ 1 (f (k) , k) (82)
Theorem 11 Suppose f ′ (k ∞ ) lies between g λ (k ∞ ) and ḡλ (k ∞ ). Then there exists an equilibrium strategy converging to k ∞ Comparing (72) and (82), we find that:

g λ (k) -g λ (k) = λ(1 -λ)(δ 1 -δ 2 ) (δ 1 -δ 2 ) u ′ 1 U ′ 1 -u ′ 2 U ′ 1 + u ′ 1 U ′ 2 (λu ′ 1 + (1 -λ)U ′ 1 )(λδ 2 u ′ 1 + (1 -λ) δ 1 U ′ 1 ) So the sign of g λ (k) -g λ (k) is the sign of (δ 1 -δ 2 ) u ′ 1 U ′ 1 -u ′ 2 U ′ 1 + u ′ 1 U ′ 2 : δ 1 -δ 2 > u ′ 2 u ′ 1 - U ′ 2 U ′ 1 =⇒ g λ (k) > g λ (k) δ 1 -δ 2 < u ′ 2 u ′ 1 - U ′ 2 U ′ 1 =⇒ 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 + (1 -λ)δ 2 g λ (k) = 1 λ δ1 + (1-λ) δ2
These formulas do not depend on the the utility function u (c). They were first derived in [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF] for the special case u (c) = ln c. Note that λδ 1 + (1 -λ)δ 2 is the arithmetic mean, and

( λ δ1 + (1-λ) δ2 ) -1 is the geometric mean. Example 2: u (c, k) = U (c, k)
In that case, the criterion (55) becomes:

∫ ∞ 0 ( λe -δ1t + (1 -λ) e -δ2t ) u (c (t) , k (t)) dt ( 83
)
This problem was studied in [START_REF] Ekeland | The golden rule when preferences are time-inconsistent[END_REF] for u (c, k) = U (c, k) = ln c. In the case at hand, (83), we find:

g λ (k) = λδ 1 + (1 -λ)δ 2 - u ′ 2 (f (k) , k) u ′ 1 (f (k) , k) g λ (k) = 1 λ δ1 + (1-λ) δ2 - u ′ 2 (f (k) , k) u ′ 1 (f (k) , k)
These are the same as the preceding ones, with the corrective term -u ′ 2 /u ′ 1 . Example 3: u = u (c) and U = U (c) In that case, the criterion (55) becomes:

λ ∫ ∞ 0 e -δ1t u (c (t)) dt + (1 -λ) ∫ ∞ 0 e -δ2t U (c (t)) dt
We find:

g λ (k) = λδ 1 u ′ 1 (f (k)) + (1 -λ)δ 2 U ′ 1 (f (k)) λu ′ 1 (f (k)) + (1 -λ)U ′ 1 (f (k)) g λ (k) = λu ′ 1 (f (k)) + (1 -λ)U ′ 1 (f (k)) λ δ1 u ′ 1 (f (k)) + (1-λ) δ2 U ′ 1 (f (k))
Here 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 corresponding trajectory k (t), defined by (35), converges to k ∞ . By Proposition 8, this will prove that σ λ is an 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 -λ λ U (φ λ (V ′ λ (k ∞ ), k ∞ ), k ∞ ) = u(f (k ∞ ), k ∞ ) + 1 -λ λ U (f (k ∞ ), k ∞ ), thus F (V ′ λ (k ∞ ), k ∞ ) = 0 and hence: V ′ λ (k ∞ ) = y * (k ∞ ) = u ′ 1 (f (k ∞ ), k ∞ ) + 1 -λ λ U ′ 1 (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 -δ2t , and replace U with (1 -λ) λ -1 . Evaluating (45) at k = k ∞ , we get:

V ′ λ (k ∞ ) = u ′ 1 (f (k ∞ ), k ∞ )σ ′ λ (k ∞ ) + u ′ 2 (f (k ∞ ), k ∞ ) δ 1 -f ′ (k ∞ ) + σ ′ λ (k ∞ ) + 1 -λ λ U ′ 1 (f (k ∞ ), k ∞ )σ ′ λ (k ∞ ) + U ′ 2 (f (k ∞ ), k ∞ ) δ 2 -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 1 δ 2 ( 1 δ1 u ′ 1 (f (k ∞ ), k ∞ ) + 1-λ λδ2 δU ′ 1 (f (k ∞ ), k ∞ ))(g λ (k ∞ ) -f ′ (k ∞ )) (u ′ 1 (f (k ∞ ), k ∞ ) + 1-λ λ U ′ 1 (f (k ∞ ), k ∞ ))(g λ (k ∞ ) -f ′ (k ∞ )) , ( 86
)
Linearizing the equation of motion

dk dt = f (k) -σ λ (k) at k = k ∞ yields dk dt = (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 the expressions of g λ (k ∞ ) and g λ (k ∞ ) yields:

δ 1 δ 2 ( 1 δ1 u ′ 1 (f (k ∞ ), k ∞ ) + 1-λ λδ2 U ′ 1 (f (k ∞ ), k ∞ ))(g λ (k ∞ ) -f ′ (k ∞ )) (u ′ 1 (f (k ∞ ), k ∞ ) + 1-λ λ U ′ 1 (f (k ∞ ), k ∞ ))(g λ (k ∞ ) -f ′ (k ∞ )) < 0.
Because u and U are increasing with respect to their first variable, the first factor in the numerator and denominator are positive. We are left with:

g λ (k ∞ ) -f ′ (k ∞ ) g λ (k ∞ ) -f ′ (k ∞ ) < 0
and the proof is complete.

The Chichilnisky criterion

In two influential papers [START_REF] Chichilnisky | What is sustainable development?[END_REF], [START_REF] Chichilnisky | An axiomatic approach to sustainable development[END_REF], Chichilnisky has proposed an axiomatic approach to sustainable development, based on the twin ideas that there should be no dictatorship of the present and no dictatorship of the future. She suggests to use the following criterion

I α (c (•) , k (•)) := δ ∫ ∞ 0 u(c(t), k(t))e -δt dt + α lim t→∞ U (c(t), k(t)), (87)
The coefficient δ > 0 in front of the integral will make later formulas simpler.

Lemma 12 Suppose u (c, k) ≥ 0 and

sup {U (c, k) | c > 0, k > 0, c = f (k)} = ∞.
Then, for any α > 0,

sup {I α (c (•) , k (•)) | (c (•) , k (•)) ∈ A (k 0 )} = ∞,
where A(k 0 ) is defined below Definition 1.

Proof. For fixed k 0 > 0, pick any A > 0, and choose some constants (c 1 , k 1 ) such that c 1 = f (k 1 ) and U (c 1 , k 1 ) > Aα -1 . Then choose some c 0 ∈ (0, c 1 ) such that f (k 0 ) -c 0 > 0. With every T > 0, we associate the path (c T (t) , k T (t)) defined by:

c T (t) = { c 0 for 0 ≤ t ≤ T, c 1 for T ≤ t,
and denote the consumption-capital pair by (c T , k T ). Since f is increasing, we have dk dt ≥ f (k 0 ) -c 0 for all t, so eventually k (t) will reach the value k 1 . Choose for T the first time when k (t) = k 1 . Writing this into the criterion, and remembering c 1 = f (k 1 ), we find:

I α (c T (•) , k T (•)) ≥ α lim t→∞ U (c T (t), k T (t)) = αU (c 1 , k 1 ) ≥ A.
This result shows that very often it is not possible to optimize I α . Even when it is, there is the time-inconsistency problem: successive decision-makers will not agree on what the optimal solution is. This is seen most easily by considering the following criterion:

(1 -α) δ ∫ ∞ 0 u(c(t), k(t))e -δt dt + αr ∫ ∞ 0 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 the other hand, Criterion (88) is a special case of the biexponential criterion (55) with δ 1 = δ, δ 2 = r , and:

αr (1 -α) δ = 1 -λ λ , λ = (1 -α) δ αr + (1 -α) δ , 1 -λ = αr αr + (1 -α) δ (89)
We have to investigate this system near the point k = k ∞ and V = V ∞ . Writing (105) at this point, 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 to have 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 1 ̸ = p 2 . Note that neither p 1 nor p 2 minimize the left hand side so f (k ∞ ) -φ(p i , k ∞ ) ̸ = 0 for i = 1, 2. We may therefore consider the initial value problem:

dp dk = δp -pf ′ (k) -u ′ 2 (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 a function 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 1 (k) and V 2 (k) of the same initial value problem, with

V ′ 1 (k ∞ ) ̸ = V ′ 2 (k ∞ ). Case 2. δV ∞ = u(f (k ∞ ), k ∞ ) and f ′ (k ∞ ) ̸ = δ -u ′
2 /u ′ 1 . Equation (107) then has a single solution p 0 , and we have f (k ∞ ) = φ(p 0 , k ∞ ), so that:

p 0 = u ′ 1 (φ(p 0 , k ∞ ), k ∞ ) = u ′ 1 (f (k ∞ ), k ∞ ) > 0.
In this case, we shall use the same system (104)-( 105), but we will take p instead of k as the independent variable. We consider the initial value problem

dk dp = f (k) -φ(p, k) p(δ -f ′ (k) - u ′ 2 (φ(p,k),k) p ) , k(p 0 ) = k ∞ .
It has a C 2 solution k(p), defined in a neighborhood of p = p 0 . We associate with it a curve in the phase space (k, V ), defined in parametric form by the equations Since we are looking for a strategy which converges to k ∞ , we pick the strategy σ i , and the solution V i , associated with the branch oriented towards k ∞ . We will denote them by σ and V . We have proved the following: Theorem 15 Suppose f ′ (k ∞ ) lies between g 0 (k ∞ ) and g 0 (k ∞ ). If g 0 (k ∞ ) > g 0 (k ∞ ), then there exists an equilibrium strategy σ (k), defined on some interval ]k ∞ -κ, k ∞ ] , which converges to k ∞ . It is continuous on the interval, C 1 on its interior, with σ (k ∞ ) = f (k ∞ ). If If g 0 (k ∞ ) < g 0 (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 0 (k ∞ ), corresponds to the solution of the Ramsey problem (α = 0). Indeed, the equation f ′ (k ∞ ) = g 0 (k ∞ ) coincides with equation [START_REF] Frederick | Time discounting and time preference: a critical review[END_REF].

k = k(p), V = 1 δ [(f (k) -φ(p, k))p + u(φ(p, k), k)]. k k ∞ k f (k) c c c ∞ σ(k) σ 1 (k)
We have thus identified a class of equilibrium strategies for the Chichilnisky problem. They are one-sided, except when f ′ (k ∞ ) = g 0 (k ∞ ), where we can apply Theorem 15 to get a strategy defined on 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 ′ 1 (σ (k) , k) = V ′ (k) are defined only on one of the two half-intervals limited by k ∞ . Suppose for instance it is the right one, [k ∞ , k ∞ + κ[. Then, if k ∞ ≤ k 0 < k ∞ + κ, the equilibrium strategy will bring k 0 to k ∞ in finite time and stay there.

To our knowledge, this is the first time equilibrium strategies have been found for the Chichilinisky criterion. Their economic interpretation, and their detailed study, will be the subject of forthcoming work.

Let us give some examples.

Example 1: u (c) = U (c)

• if U ′ 2 (f (k),k) u ′
1 (f (k),k) < 0. The equilibrium strategy then is defined on the left hand side of k ∞ .

Remember that u ′ 1 > 0. If U ′ 2 > 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 accumulating capital accumulates pollution. In that case, the situation is inverted with respect to the preceding examples: 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 ′ 2 (f (k),k) u ′
1 (f (k),k) . But there is no turning back: one can never achieve a stationary level k ∞ < k 0 . The situation is described in Figure 3

k k ∞ k f (k) c c c ∞ k k ∞ k 0 k f (k) c c c ∞ k k ∞ k 0 k f (k)

Figure 2 .

 2 Figure 2. The phase diagram for the Euler equation.

Figure 3 .

 3 Figure 3. The case when capital pollutes.

(f (k),k) > 0, then g 0 (k) < g 0 (k). The equilibrium strategy then is defined on the right hand side of k ∞ , as in the preceding cases.

Acknowledgements. The authors thank sincerely Professors Larry Karp and Gerd Asheim for generously sharing their knowledge and ideas with them, and pointing out several useful references. Qinglong Zhou and Yiming Long would like to express their thanks to Professors Jean-Pierre Bourguignon, Jean Dolbeault, Olivier Glass and Ivar Ekeland as well as the IHES and the CERE-MADE (Université Paris Dauphine) for supporting Qinglong's visits to IHES and CEREMADE from February to April of 2012, and their warm hospitalities during this time.

So, for each r > 0, the criterion (88) gives rise to a time-inconsistent problem. Using the results in the preceding section, we find a continuum of equilibrium strategies. Substituting (89) into (72) and (82), we find the corresponding values:

The equations ( 58), (59) become:

where φ r (x, k) is defined by:

and the boundary conditions (62), (63):

The equilibrium strategy is given by σ r (k) = φ r (V ′ r (k) , k), and the corresponding trajectory converges 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 boundary condition (92), (93), so that k - r < k ∞ < k + r . We shall now solve the Chichilnisky problem by setting r = 0 in the preceding equations. We have:

The pair (V 0 , W 0 ) has to solve the following boundary-value problem:

with:

Substracting ( 97) from (96), we get:

Similarly to Lemma 9, it follows that V ′ 0 -W ′ 0 = 0, and so V 0 -W 0 is a constant, namely:

Writing this in ( 96) and (97), we find that V 0 (k) is a solution of the boundary-value problem:

This problem has been studied in [START_REF] Ekeland | a classical problem in the calculus of variations leading to an implicit differential equation[END_REF] (Section 2.4, case 2) when u (c, k) does not depend on k. In the general case, we have:

the problem ( 101), (102) has two solutions V 1 and V 2 , defined on some non-empty half-interval

Both are C 1 on the half-interval, C 2 on the interior, and have the same derivative at k ∞ , given by:

Proof. From now on, write φ instead of φ 0 . Rewrite (101) as a Pfaff system:

Differentiating (105) leads to:

where we used (100). Together with (104), this yields

Then V is also C 2 with respect to p near p = p 0 . Moreover, we have

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 respect to p. We find:

This shows that the cusp has two branches, with a common tangent between them. The branches extend on the left of k ∞ if the left-hand side of (109) is negative, that is, if:

and on the right if it is positive.The slope m of the common tangent is given by

and it is also the one-sided derivative of V at k ∞ . Note for future use that because

so that m = p 0 does not coincide with the tangent of the curve

This is the case which was investigated in Theorem 3. We have shown that there exists a

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 plane in two regions.

• 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 two smooth solutions intersecting transversally.

1 (f (k∞),k∞) ̸ = δ, we are in Case 2. There are two C 1 solutions defined only on one side of k ∞ . They are tangent at (k ∞ , V ∞ ), and transversal to Γ.

1 (f (k∞),k∞) = δ, we are in Case 3: there is a C 2 solution defined on a neighbourhood of k ∞ .the Figure 1 gives the phase diagram in the (k, V ) plane:

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 0 (V ′ i (k) , k) with φ 0 defined by (100). The strategy σ i is C 0 on the half-interval, C 1 on its interior, with σ i (k ∞ ) = f (k ∞ ).

Proposition 14

One, and only one, of the strategies σ 1 and σ 2 , converges to k ∞ .

Proof. Consider the Euler-Lagrange equation [START_REF] Ekeland | Equilibrium management of fisheries with overlapping altruistic generations[END_REF] for the Ramsey problem. The phase diagram is given in Figure 2, where k is defined by

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 two strategies σ 1 and σ 2 , so one of them converges and the other diverges.

Neither depends on k, and we have:

We have g 0 (k) < g 0 (k), so the equilibrium strategy exists only on the right hand side of g 0 (k). The existence condition is:

(

by k and k respectively. There are three cases, depending on the position of the initial point k 0 :

there exists an equilibrium strategy starting from k 0 and converging to k ∞ .

• if k < k 0 < k, then, for any k ∞ ∈]k, k 0 [, there exists an equilibrium strategy starting from k 0 and converging to k ∞ .

• if k 0 < k, the only equilibrium strategy starting from k 0 is the optimal strategy for the Ramsey problem (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:

The existence condition is:

and the equilibrium strategy σ is defined on [k ∞ , k ∞ + κ[ for some κ > 0. The situation is similar to 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:

There are two subcases:


Infeasibility-free Inverse Kinematics Method

Wael Suleiman 1 , Fumio Kanehiro 2 and Eiichi Yoshida 3 Abstract-The problem of inverse kinematics is revisited in the present paper. The paper is focusing on the problem of solving the inverse kinematics problem while respecting velocity limits on both the robot's joints and the end-effector. Eventhough the conventional inverse kinematics algorithms have been proven to be efficient in many applications, defining an admissible trajectory for the end-effector is still a burdensome task for the user, and the problem can easily become unsolvable. The main idea behind the proposed algorithms is to consider the sampling time as a free variable, hence adding more flexibility to the optimization problem associated with the inverse kinematics. We prove that the reformulated problem has always a solution if the end-effector path is in the reachable space of the robot, thus solving the problem of infeasibility of conventional inverse kinematics methods.

To validate the proposed approach, we have conducted three simulations scenarios. The simulation results point that while the conventional inverse kinematics methods fail to track precisely a desired end-effector trajectory, the proposed algorithms always succeed.

I. INTRODUCTION

The inverse kinematics problem is one of the most studied problems in robotics, and many methods to solve it have been proposed in the literature. Those methods are extensively used in robotics, they can be also combined with other techniques, e.g. force control, to cope with the imperfect modelling of the robot and its dynamic parameters.

Generally speaking, the objective is to find a vector in the configuration space that satisfies constraints in the operational space, in other words the values of the robot's joints to make the end-effector reaches a desired goal (position and orientation) in the Cartesian space. Although in some special cases the problem can be solved analytically [START_REF] Raghaven | Kinematic analysis of the 6r manipulator of general geometry[END_REF], [START_REF] Raghavan | Solving polynomial systems for the kinematic analysis and synthesis of mechanisms and robot manipulators[END_REF], it is generally solved numerically [START_REF] Nakamura | Integrating geometric constraints into reactive leg motion generation[END_REF], [4]. Many efficient and robust numerical algorithms have been proposed for solving the inverse kinematics problem [5], [6], [7], [START_REF] Nakamura | Integrating geometric constraints into reactive leg motion generation[END_REF], [4], only to cite few.

The inverse kinematics problem was originally formulated as follows:

min qt qt T Q qt subject to J qt = ṙt (1)
1 Wael Suleiman is with Electrical and Computer Engineering Department, Faculty of Engineering, Université de Sherbrooke, 2500, boul. Universite, Sherbrooke, Quebec (Canada) J1K 2R1.

Wael.Suleiman@USherbrooke.ca 2 F. Kanehiro is with Intelligent Systems Research Institute, National Institute of Advanced Industrial Science and Technology (AIST), 1-1-1 Umezono, Tsukuba, Ibaraki 305-8568 Japan. f-kanehiro@aist.go.jp 3 E. Yoshida is with CNRS-AIST, JRL (Joint Robotics Laboratory), UMI 3218/CRT, AIST, 1-1-1 Umezono Tsukuba Ibaraki, Japan.

e.yoshida@aist.go.jp where Q is a diagonal and positive semi-definite matrix, J is the Jacobian matrix, q ∈ R n is the joint velocity and ṙt is the linear and angular velocity of the end-effector.

The optimization problem (1) can be efficiently solved using the pseudo-inverse technique [START_REF] Nakamura | Integrating geometric constraints into reactive leg motion generation[END_REF], [4] as follows:

qt = Ĵ † ṙt + (I -Ĵ † J)z (2)
where

Ĵ † = Q -1 J T J Q -1 J T -1
and z is an arbitrary vector. If Q is the identity matrix, Ĵ † becomes the wellknown Moore-Penrose pseudo-inverse. It is worth mentioning that the optimization problem (1) is time-invariant as the time derivative can be simply replaced by the difference. Thus, we obtain the following equivalent optimization problem:

min ∆q ∆q T Q∆q subject to J∆q = ∆r (3)
The solution of the above problem is given by:

∆q t = Ĵ † ∆r t + (I -Ĵ † J)ẑ (4)
However, in order to consider the velocity and joints limits as well as avoid obstacles, inequality constraints should be added. The general inverse kinematics problem can be therefore formulated as follows:

min qt qt T Q qt subject to J qt = ṙt b -≤ A qt ≤ b + q-≤ qt ≤ q+ (5)
qand q+ are respectively the lower and upper limits of the velocity. A, b -and b + represent the additional constraints that result from considering geometric constraints, such as collision avoidance, or physical constraints, such as joint limits, for more details the reader is referred to [8], [9], [START_REF] Kanoun | Real-time prioritized kinematic control under inequality constraints for redundant manipulators[END_REF], [START_REF] Escande | Fast resolution of hierarchized inverse kinematics with inequality constraints[END_REF], [START_REF]Hierarchical quadratic programming: Fast online humanoidrobot motion generation[END_REF]. The main differences between (1) and (5) are:

• There is no closed-form expression for the solution of (5). However, it is a convex optimization problem if Q is a semi-definite positive matrix, and in this case, the problem can be efficiently solved by a Quadratic Programming (QP) solver [9], [START_REF] Kanoun | Real-time prioritized kinematic control under inequality constraints for redundant manipulators[END_REF], [START_REF] Escande | Fast resolution of hierarchized inverse kinematics with inequality constraints[END_REF], [START_REF]Hierarchical quadratic programming: Fast online humanoidrobot motion generation[END_REF].

• Problem (5) is time-variant.

A major difficulty that is faced, in practice, by the user is how to define the trajectory of end-effector (r t ). It is clear that if the end-effector moves fast, the joints velocities will be saturated and as a result the end-effector trajectory diverges from the desired one. A solution to the previous problem is to ensure that the end-effector velocity is slow enough to avoid the joints velocity saturation issue, however this solution is not the optimal one in practice, as it yields to a very slow motion and it does not take full advantage of the robot's physical limits.

A well known approach in the literature to deal simultaneously with the velocity limits of the joints and the endeffector is time parameterization [START_REF] Suleiman | Time parameterization of humanoid-robot paths[END_REF], however this approach is mainly adapted for offline computation and it needs the whole path to be known in advance.

It is worth to point the difference between a path and a trajectory. A path denotes the locus of points in the joint space, or in the operational space, the robot has to follow in the execution of the desired motion, while a trajectory is a path on which a time law is specified [START_REF] Sciavicco | Modelling and Control of Robot Manipulators[END_REF].

In this paper, we are interested in a robust online implementation of inverse kinematics, which is infeasibility-free and can efficiently deal with the velocity limits of both the joints and the end-effector.

II. PROBLEM FORMULATION

We consider two cases, the first one focuses on transforming the original inverse kinematics problem into an equivalent optimization problem which is always feasible, while the second focuses on incorporating the end-effector velocity limits into the inverse kinematics problem. The main idea is instead of supposing that the sampling time (T ) fixed, making it a free variable, and solving the inverse kinematics problem as a function of ∆q and T . Thus:

qt ∼ = ∆q T (6)
The end-effector velocity vector is defined as follows:

ṙt = v t ω t (7)
where v t is the linear velocity of the end-effector:

v t ∼ = ∆X e T (8)
X e ∈ R 3 is the Cartesian position of the end-effector. ω t is the angular velocity of the end-effector and can be obtained using the following formula:

[ω t ] ∧ ∼ = ∆R e T R e T ( 9
)
where R e is the rotation matrix of the end-effector frame, and [.] ∧ designs the skew operator defined as follows:

[.] ∧ : ω ∈ R 3 → so(3) [ω] ∧ =   0 -ω z ω y ω z 0 -ω x -ω y ω x 0   (10)
where so(3) denotes the Lie algebra of SO(3) which is the group of rotation matrices in the Euclidean space. The inverse operator of skew operator can be defined as follows:

[.] ∨ : Ω ∈ so(3) → R 3 [Ω] ∨   0 -ω z ω y ω z 0 -ω x -ω y ω x 0   ∨ =   ω x ω y ω z   (11)
and the angular velocity can be calculated using the inverse operator:

ω t ∼ = 1 T ∆R e R e T ∨ (12)
Note that the skew operator and its inverse are linear operators.

Let us define ∆r as follows:

ṙt ∆r T (13)
As a result:

∆r ∼ = ∆X e ∆R e R e T ∨ (14)
A. Case Study 1: Sampling Time as a Free Variable

By considering T as a free variable, the optimization problem (5) can be transformed into the following equivalent problem:

min ∆q,T ∆q T Q∆q + αT 2 subject to J∆q = ∆r b -T ≤ A∆q ≤ b + T q-T ≤ ∆q ≤ q+ T ≤ T (15)
where 0 < 1, and α > 0 are user-defined constants. Let us introduce the following parameters:

X = ∆q T , Q X = Q 0 n×1 0 1×n α , J = J 0 m×1 A =       A -b + -A b - I n -q+ -I n q- 0 1×n -1       , B =       0 n×1 0 n×1 0 n×1 0 n×1 -       (16)
where I n and 0 k×l denotes, respectively, the identity matrix of dimension n and a zeros matrix of dimension k ×l. Recall that n and m are the dimension of ∆q and ∆r respectively.

Thus the optimization problem is transformed into the following classical QP problem:

min X X T Q X X subject to J X = ∆r A X ≤ B (17)
The above QP problem can be efficiently solved, in real-time, using an appropriate QP solver such as uQuadProg solver [9] or qpOASES solver [START_REF] Ferreau | qpOASES: A parametric active-set algorithm for quadratic programming[END_REF]. By adding constraints on the end-effector velocity, the optimization problem ( 15) is transformed into the following optimization problem:

min ∆q,T ∆q T Q∆q + αT 2 subject to J∆q = ∆r b -T ≤ A∆q ≤ b + T q-T ≤ ∆q ≤ q+ T ≤ T ṙ-T ≤ ∆r ≤ ṙ+ T (18)
where ṙ+ and ṙare respectively the upper and lower limits of the end-effector velocity.

Note that the new constraint can be rewritten in an equivalent and more compact form as follows:

ṙ-T ≤ ∆r ≤ ṙ+ T ⇔ |∆r| ≤ ṙ+ T (19)
By introducing the following parameters:

X = ∆q T , Q X = Q 0 n×1 0 1×n α , J = J 0 m×1 Â =         A -b + -A b - I n -q+ -I n q- 0 1×n -1 0 m×n -ṙ+         , B =         0 n×1 0 n×1 0 n×1 0 n×1 - -|∆r|         (20)
Similarly to Case Study 1, the optimization problem is transformed into the following classical QP problem:

min X X T Q X X subject to J X = ∆r  X ≤ B (21)
Algorithm 2 1: Compute the jacobian matrix (J) 2: Compute ∆r Eq. ( 14) 3: procedure INVERSE KINEMATICS(J, ∆r, A, b -,b + , q-, q+ , ṙ+ ) 4:

Solve the optimization problem (21) 5:

return ∆q and T 6: go to 1

III. COMPLEXITY ANALYSIS

A complexity comparison with conventional inverse kinematics algorithms, Eq. ( 5), can give a clear idea about the possibility of an online implementation of the proposed algorithms.

1) Algorithm 1:

• By comparing the optimization problems ( 5) and ( 17), the dimension of the optimization variable in the proposed algorithm (X ) is n + 1 where n is the dimension of the conventional joints velocity variable ( qt ). • One additional inequality constraint has been added. As a result, the computational complexity of Algorithm 1 and of a conventional inverse kinematic algorithm are almost the same. However, Algorithm 1 have a significant advantage of always providing a feasible solution.

2) Algorithm 2:

• Similarly to Algorithm 1, the dimension of the optimization variable (X ) is n + 1. • The number of inequality constraints has however been increased by m+1, where m is the dimension of the end-effector configuration vector. Consequently, the complexity of Algorithm 2 and a conventional inverse kinematics problem are comparable. Recall that Algorithm 2 can simultaneously handle constraints on the joint and the end-effector velocity limits, while conventional inverse kinematics algorithms simply cannot. The numerical simulations in Section V also confirmed the above conclusions.

IV. PRACTICAL IMPLEMENTATION

Many of robotic systems have a control loop with a fixed time step. To take this constraint into account, the sampling time T should satisfies the following condition:

T = n T s (22)
where n ∈ N is a strictly positive integer, which becomes the new optimization variable, and T s is the robot's control loop fixed time step. Theoretically speaking, the optimization problems ( 17) and ( 21) become Mixed-Integer Quadratic Programming (MIQP) problems. However, in practice, solving a MIQP is generally more complex and computationally expensive than solving a standard QP.

In our special case, however, one can figure out that we do not need to solve MIQP problems, instead once T is obtained by solving ( 17) and ( 21), the parameter n can be easily obtained such as: (n -1) T s ≤ T ≤ n T s .

V. SIMULATION RESULTS

We have conducted three simulations scenarios:

A. Scenario 1: Joint Velocity Limits

In order to validate the proposed algorithm for the case of joint velocity limits (Algorithm 1), we consider a planar redundant manipulator, which has 4 degrees of freedom (Fig. 1). The end-effector path is a Bezier curve defined by an initial, control and a goal positions. The joint velocity limits have been chosen as follows:

• q+ = -q-= 0.5 × [1 1 1 1] T 0 0 .5 1 1 .5 2 2 .5 3 3 .5 4 -2 -1.5 -1 -0.5 0 0.5 1 1.5 X (m) Y (m)

Initial position goal position control point

Fig. 1. Planar robot configurations Fig. 2(a) points out that the constraints on the joint velocity are fully respected, and Fig. 2(c) shows that the error between the executed trajectory and the desired one is less than 10 -5 m. To compare those results with a conventional inverse kinematics algorithm, the end-effector path has been first transformed into a trajectory by considering a uniform time distribution with a sampling time T = 5 × 10 -3 s over the interval [0, t f ], after several trials we found that the minimum t f that makes the problem feasible is t f = 4.40 s. It is worth to mention that, at every trial, the conventional inverse kinematics algorithm stopped after few iterations because of an infeasibility issue. From Fig. 3 and Fig. 2, one can notice that:

• Our method automatically found that t f = 3.30 s versus t f = 4.40 s for the conventional method. This is because our method adapts the sampling time according to the joint velocity constraints; on the other hand, the conventional method requires a fixed sampling rate for the whole trajectory. • Fig. 3(a) shows that the joint velocity constraints have been violated over the time interval [0.02s, 0.83s], this is because the constraints of the optimization problem (5) cannot be simultaneously satisfied. The solver, in this case, tries first to satisfy the equality constraints and the inequality constraints are considered in a second priority. It is important to point out that the high oscillations in the joint velocity trajectories are not fully transmitted to the end-effector, this is because the robot is redundant with respect to the desired task. • The end-effector tracking error in Fig. 3(c) is much bigger than that one of our method Fig. 2(c).

B. Scenario 2: Joint and End-effector Velocity Limits

To validate the proposed algorithm for the joint and endeffector velocity limits (Algorithm 2), we consider the same planar redundant manipulator in Scenario 1. The following parameters has been chosen: 4(a) and Fig. 4(b) show that the constraints on the joint and the end-effector velocity are fully respected, and Fig. 4(c) shows that the error between the executed trajectory and the desired one is less than 2.5 × 10 -5 m. Note that the total time of the trajectory became 4.30 s, as it can be seen in Fig. 4(b), this is mainly to respect the end-effector velocity limits.

• q+ = -q-= 0.5 × [1 1 1 1] T • ṙ+ = -ṙ-= [ ẋ+ t ẏ+ t ] = 0.7 × [1 1] T Fig.

C. Scenario 3: Simulated Baxter Research Robot

A simulated scenario of a Baxter research robot and an obstacle that consists of a sphere attached to a thin cylindric rod is given in Fig. 5(a). The objective is to reach a goal pose (position and orientation) from an initial pose while avoiding the collision with the obstacles.

The following constraints have been considered: 1) Collision avoidance: the Baxter robot is approximated by its collision geometry model from the Unified Robot Description Format (URDF) file, where the arms are approximated by cylinders and boxes. The collision avoidance is formulated as inequality constraints [8] having the following form: A qt ≤ b. 2) Velocity limits on the robot's joints, these values are given in the URDF file. 3) Desired velocity limits (linear and angular velocities) of the right arm end-effector. The formulation of the inverse kinematics problem in this scenario is however different, as the end-effector tries to reach the goal pose at each iteration via a direct straight line, at the same time the collision avoidance inequality constraints repulse the arm to keep a minimum clearance

VI. CONCLUSION AND FUTURE WORK

In this paper, a new method for inverse kinematics while simultaneously considering joint and end-effector velocity limits is proposed. The method is simple to implement, yet efficient to handle velocity constraints. The proposed algorithms solve the problem of infeasibility that conventional inverse kinematics algorithms suffer from. These algorithms can be implemented online, and they have as input the geometric path of the end-effector that is converted, onthe-fly, to a feasible trajectory. Simulation results revealed the efficiency of the method to solve inverse kinematics problems in several scenarios. Future work will focus on the

Algorithm 1 1 :

 1 Compute the jacobian matrix (J) 2: Compute ∆r Eq. (14) 3: procedure INVERSE KINEMATICS(J, ∆r, A, b -,b + , q-, q+ ) 4: Solve the optimization problem (17) 5: return ∆q and T 6: go to 1 B. Case Study 2: Incorporating End-effector Velocity Limits

Fig. 2 .

 2 Fig. 2. Algorithm 1: joint velocity limits

Fig. 3 .

 3 Fig. 3. Conventional inverse kinematics algorithm

ACKNOWLEDGMENT

This research is partially supported by a discovery grant (Prof. Wael Suleiman) from the Natural Sciences and Engineering Research Council of Canada (NSERC).