Dynamic Programming for Mean-Field Type Control

Q: How is the distribution of the oil centered at x?

At time 0 it is a Gaussian distribution centered at x = 5; at time T the distribution is concentrated around x = 0.5, so most producers have pumped 90% of the oil available to them.

Q: What is the macro-economic assumption that pt is a decreasing function of the mean?

This contains the macro-economic assumption that pt is a decreasing function of the mean production because scarcity of oil increases its price and conversely.

Q: What is the deterministic function of a(Xt, t)?

The authors suppose that at := a(Xt , t) is a deterministic function of t and Xt , meaning by this that the producers apply a feedback law to control their production.

Q: what is the correct tool for this differentiability?

The correct mathematical tool for this differentiability is the Wasserstein distance and the differentiability with respect to the probability measure rather than to its density (see e.g. [13,22]).

Q: what is the simplest way to get the time derivative?

A first order approximation of the time derivative in the Fokker-Planck equation givesδτ ρ := ρτ+δτ −ρτ = δτ [ ∇2 : (µτ ρτ)−∇ · (uτ ρ) ] +o(δτ). (14)As V is assumed to be smooth, the authors have :V (τ +δτ;ρτ+δτ) =V (τ;ρτ)+∂τV (τ;ρτ)δτ +V ′ρ(τ;ρτ) ·δτ ρ +o(δτ).

Question

1. What is the problem in the paper?

2. how long does u have to be a conjugate-gradient algorithm?

3. What is the simplest way to solve the non-linear PDE?

4. What is the way to reduce the cost functions?

Accepted Answer

The first difficulty which this paper had to deal with isDynamic Programming for Mean-Field Type Control 19the extension of the HJB dynamic programming approach.

Accepted Answer

As long as 4βert µ−Qt > 0 it leads toQt Qt −4βert µ = γe(T−t)µ γ−4β µ ⇒ Pt =4β µγe(T−t)µγe(T−t)µ − γ +4β µ . (43)Then, u is found by (36).

Accepted Answer

To solve this non-linear PDE the authors use the fact that it is embedded into an iterative loop in Algorithm 1, and semi-linearise it by evaluating the square term in the last integral as a product of the same, where one factor is evaluated at the previous iteration.

Accepted Answer

If u0 = ue, then the algorithm decreases the cost functions by introducing small oscillations, a strategy which is clearly mesh dependent.

Accepted Answer

Because the second order term vanishes at x = 0, a weak formulation ought to be in the weighted Sobolev space and would be to find ρ , withρ ∈H = {ν ∈ L2(R+) : x∂xν ∈ L2(R+)} such that ∀ν ∈H∫ R+ [ ν∂tρdx+(xσ2−u)ρ∂xν + x2σ2∂xρ∂xν ] dx = 0, for almost all t. (39)Theorem 2.2 in [29] asserts existence and uniqueness of ρ , provided that there exists uM such that u(x, t) < xuM, ∀t.

Accepted Answer

At time 0 it is a Gaussian distribution centered at x = 5; at time T the distribution is concentrated around x = 0.5, so most producers have pumped 90% of the oil available to them.

Accepted Answer

This contains the macro-economic assumption that pt is a decreasing function of the mean production because scarcity of oil increases its price and conversely.

Accepted Answer

The authors suppose that at := a(Xt , t) is a deterministic function of t and Xt , meaning by this that the producers apply a feedback law to control their production.

Accepted Answer

The correct mathematical tool for this differentiability is the Wasserstein distance and the differentiability with respect to the probability measure rather than to its density (see e.g. [13,22]).

Accepted Answer

A first order approximation of the time derivative in the Fokker-Planck equation givesδτ ρ := ρτ+δτ −ρτ = δτ [ ∇2 : (µτ ρτ)−∇ · (uτ ρ) ] +o(δτ). (14)As V is assumed to be smooth, the authors have :V (τ +δτ;ρτ+δτ) =V (τ;ρτ)+∂τV (τ;ρτ)δτ +V ′ρ(τ;ρτ) ·δτ ρ +o(δτ).

Accepted Answer

When it converges the fixed point method is preferred but there is no known way to guarantee convergence even to a local minimum; as for the steepest descent the authors found it somewhat hard to use, mainly because it generates irregular oscillating solutions; some bounds on the derivatives of u need to be added in penalised form in the criteria.

Accepted Answer

ε,Dynamic Programming for Mean-Field Type Control 7where ρ̃t is the solution of (7) at t with control ũ starting with ρτ at time τ .

Accepted Answer

Assumptions for which a strong solution is known to exist once the distribution of X0 is known to be in L2∩L∞ are given in [25] (see also [26], Proposition 4Dynamic Programming for Mean-Field Type Control 3which applies when σσT is uniformly positive definite:u ∈ L2 ( ]0,T [,L2loc(Rd) )d ,u 1+ |x|∈ L1 ( ]0,T [,L1 +L∞(Rd) )d ,∇ ·u ∈ L1 ( ]0,T [,L∞(Rd) ) ,(2)and with σ̃(x, t) := σ(x, t,u(x, t)):σ̃ ∈ L∞ ( ]0,T [,L∞loc(Rd) )d×k ,σ̃ 1+ |x|∈ L2 ( ]0,T [,L2 +L∞(Rd) )d×k (3) Then the PDF of Xt satisfiesρ ∈ L∞ ( ]0,T [,L2(Rd)∩L∞(Rd) ) ∩L2 ( ]0,T [,H1(Rd) ) , σT ∇ρ ∈ L2 ( ]0,T [,L2(Rd) ) and is the unique solution of the Fokker-Planck equation,∂tρ +∇ · (uρ)−∇2 : ( 1 2 σσ T ρ ) = 0, ρ|t=0 = ρ0(x), (4)Conversely, with (2)(3), there is a unique solution to (4) which is the PDF of a Markov process which satisfies (1).

Accepted Answer

(18)Proof Note that the Fokker-Planck equation implies the existence of a semi-group operator G such that, for all τ ≤ t, ρt = G(t− τ) ∗ρτ .

Accepted Answer

With sufficient regularity (see [28] for weaker assumptions), with the Fokker-Planck partial differential equation for ρ(x, t) := ρt(x); the problem is equivalent to the deterministic distributed control problem,min u∈UdJ = ∫ T0 ∫ Rd H̃(x, t,u(x, t),ρ(x, t),χ(t))ρ(x, t)dxdt + ∫ Rd G(x,ξ )ρ(x,T )dx,where χ(t) := ∫ Rd h̃(x, t,u(x, t),ρ(x, t))ρ(x, t)dx, ξ := ∫ Rd g(x)ρ(x,T )dxand ∂tρ +∇ · (uρ)−∇2 : (µρ) = 0, ρ|t=0 = ρ0(x), (7)where µi j = 12 ∑k σikσ jk and ∇ 2 is the d × d matrix operator of element ∂i j.

Accepted Answer

(15)Using (15) and the mean value theorem for the time integral, Bellman’s principle yields , up to o(δτ),V (τ;ρτ) = min u∈Ud { δτ ∫ Rd Hρτ dx+V (τ;ρτ)+∂τV (τ;ρτ)δτ +V ′ρ(τ;ρτ) ·δτ ρ } . (16)8 Mathieu Laurière, Olivier Pironneau∗ Communicated by Nizar TouziThe terms V (τ;ρτ) cancel; divided by δτ and combined with (14) and letting δτ→ 0, (16) gives0 = min u∈Ud {∫ Rd Hρτ dx+∂τV (τ;ρτ)+V ′ρ(τ;ρτ) · [ ∇2 : (µτ ρτ)−∇ · (uτ ρ) ]} . (17)To finalize the proof the authors need the following proposition to relate V to V ′ρ .

Accepted Answer

When the Hamiltonian depends on the distribution ρt only through the local value ρt(x) and the average of a fixed function, the authors can make explicit the link with the calculus of variations (see Section 3).

Dynamic Programming for Mean-Field Type Control

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Most frequently asked questions

1. What is the problem in the paper?

2. how long does u have to be a conjugate-gradient algorithm?

3. What is the simplest way to solve the non-linear PDE?

4. What is the way to reduce the cost functions?

5. What is the problem with the second order term?

6. How is the distribution of the oil centered at x?

7. What is the macro-economic assumption that pt is a decreasing function of the mean?

8. What is the deterministic function of a(Xt, t)?

9. what is the correct tool for this differentiability?

10. what is the simplest way to get the time derivative?

11. What is the way to guarantee convergence of the two algorithms?

12. what is the solution of t with u ud?

13. What is the PDF of a Markov process?

14. what is the simplest way to prove the existence of a semi-group operator?

15. what is the deterministic distributed control problem?

16. what is the simplest way to prove the existence of the time integral?

17. What is the inverse of the Hamiltonian?

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References

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