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The turnpike property
Emmanuel Trelat1
1Sorbonne Universite (Paris 6), Labo. J.-L. Lions
Works with Gontran Lance, Can Zhang, Enrique Zuazua
FAU, Kolloquium, 2020 January 7
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Turnpike property
The solution of an optimal control problem in large time should spend most of itstime near a steady-state.
In infinite horizon the solution should converge to that steady-state.
Historically: discovered in econometry (Von Neumann points).
The first turnpike result was discovered in 1958 by Dorfman,Samuelson and Solow, in view of deriving efficient programsof capital accumulation, in the context of a Von Neumannmodel in which labor is treated as an intermediate product.
Paul Samuelson (1915–2009)
Nobel Prize in EconomicScience, 1970
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Turnpike property
The solution of an optimal control problem in large time should spend most of itstime near a steady-state.
In infinite horizon the solution should converge to that steady-state.
final
turnpike (highway)
initial
finalinitial
Excerpt from: Dorfman - Samuelson - Solow (1958)
It is exactly like a turnpike paralleled by a network of minor roads. There is a fastest routebetween any two points; and if the origin and destination are close together and far from theturnpike, the best route may not touch the turnpike. But if origin and destination are farenough apart, it will always pay to get on to the turnpike and cover distance at the best rateof travel, even if this means adding a little mileage at either end.
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Turnpike property
The solution of an optimal control problem in large time should spend most of itstime near a steady-state.
In infinite horizon the solution should converge to that steady-state.
- Turnpike theorems have been derived in the 60’s for discrete-time optimal control prob-lems arising in econometry (Mac Kenzie, 1963).
- Continous versions by Haurie for particular dynamics (economic growth models). Seealso Carlson Haurie Leizarowitz 1991, Zaslavski 2000, Faulwasser Bonvin 2015, Grune2018, Lou Wang 2018.
- More recently, in biology: Rapaport 2005, Coron Gabriel Shang 2014; human locomo-tion: Chitour Jean Mason 2012; MPC: Grune et al. 2012–2019.
- Linear heat and wave equations: Porretta Zuazua 2013.
- Rockafellar 1973, Samuelson 1972: saddle point feature of the extremal equations ofoptimal control.
- Different point of view by Anderson Kokotovic (1987), Wilde Kokotovic (1972):exponential dichotomy property→ hyperbolicity phenomenon.
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General nonlinear optimal control problem
f : X × U → X dynamics (in finite or infinite dimension)
R : X × X → Y terminal conditions (example: point-to-point, point-to-free, periodic, ...)
f 0 : X × U → IR instantaneous cost
Dynamical optimal control problem (OCP)T
For T > 0 fixed, find uT (·) ∈ L∞(0,T ; U) such that
x(t) = f (x(t), u(t))
R(x(0), x(T )) = 0
min∫ T
0f 0(x(t), u(t)) dt
x(·): state u(·): control
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Intuitive idea
min1T
∫ T
0f 0(x(t), u(t)) dt
x(t) = f (x(t), u(t)), t ∈ [0,T ]
s = t/T−→ε = 1/T
min∫ 1
0f 0(x(s), u(s)) ds
εx ′(s) = f (x(s), u(s)), s ∈ [0, 1]
We expect that, as ε→ 0, there is some convergence to the static optimization problem
min f 0(x , u) subject to f (x , u) = 0
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Dynamical optimal control problem (OCP)T
x(t) = f (x(t), u(t))
R(x(0), x(T )) = 0
min∫ T
0f 0(x(t), u(t)) dt
Static optimal control problem
min(x,u)∈X×Uf (x,u)=0
f 0(x , u)
Optimal (assumed) solution (xT (·), uT (·)) Optimal (assumed) solution (x , u)
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Dynamical optimal control problem (OCP)T
x(t) = f (x(t), u(t))
R(x(0), x(T )) = 0
min∫ T
0f 0(x(t), u(t)) dt
Static optimal control problem
min(x,u)∈X×Uf (x,u)=0
f 0(x , u)
Optimal (assumed) solution (xT (·), uT (·)) Optimal (assumed) solution (x , u)
First-order optimality conditions:
Pontryagin maximum principle
(generic...) assumption:no abnormal⇒ λ0
T = −1
Lagrange multiplier rule
(generic...) assumption:no abnormal⇒ λ0
T = −1(Mangasarian-Fromowitz)
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Dynamical optimal control problem (OCP)T
x(t) = f (x(t), u(t))
R(x(0), x(T )) = 0
min∫ T
0f 0(x(t), u(t)) dt
Static optimal control problem
min(x,u)∈X×Uf (x,u)=0
f 0(x , u)
λT (·): adjoint
xT (t) =∂H∂λ
(xT (t), λT (t),−1, uT (t))
λT (t) = −∂H∂x
(xT (t), λT (t),−1, uT (t))
∂H∂u
(xT (t), λT (t),−1, uT (t)) = 0
+ transversality conditions(−λT (0)λT (T )
)=
k∑i=1
γi∇Ri (xT (0), xT (T ))
λ: adjoint (Lagrange multiplier)
∂H∂λ
(x , λ,−1, u) = 0
−∂H∂x
(x , λ,−1, u) = 0
∂H∂u
(x , λ,−1, u) = 0
H(x , λ, λ0, u) = 〈λ, f (x , u)〉+ λ0f 0(x , u)
(x , λ, u): equilibrium point of the extremal equations
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Exponential turnpike
Theorem
Under admissibility, exponential stabilizability and detectability assumptions:
‖xT (t)− x‖+∥∥λT (t)− λ
∥∥+ ‖uT (t)− u‖ 6 C1
(e−νt + e−ν(T−t)
)∀t ∈ [0,T ]
(Finite dim.: Trelat Zuazua, JDE 2015. Infinite dim.: Trelat Zhang Zuazua, SICON 2018)
Moreover:ν = −max
Re(µ) | µ ∈ Spec(A− BH−1
uu B∗E−)> 0
where
E−A + A∗E− − E−BH−1uu B∗E− −W = 0 minimal solution of Riccati
In some sense, this result shows that:
1) (dynamic) control 2) T → +∞ ⇐⇒ 1) T → +∞ 2) (static) control
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Exponential turnpike
Theorem
Under admissibility, exponential stabilizability and detectability assumptions:
‖xT (t)− x‖+∥∥λT (t)− λ
∥∥+ ‖uT (t)− u‖ 6 C1
(e−νt + e−ν(T−t)
)∀t ∈ [0,T ]
(Finite dim.: Trelat Zuazua, JDE 2015. Infinite dim.: Trelat Zhang Zuazua, SICON 2018)
Consequence: in large time T , the optimal extremal solution (xT (·), λT (·), uT (·)) of(OCP)T approximately consists of 3 pieces:
1 short-time: (xT (0), λT (0), uT (0))→ (x , λ, u) (transient arc, on [0, ε])
2 long-time, stationary: (x , λ, u) (on [ε,T − ε])
3 short-time: (x , λ, u)→ (xT (T ), λT (T ), uT (T )) (transient arc, on [T − ε,T ])
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Exponential turnpike
Theorem
Under admissibility, exponential stabilizability and detectability assumptions:
‖xT (t)− x‖+∥∥λT (t)− λ
∥∥+ ‖uT (t)− u‖ 6 C1
(e−νt + e−ν(T−t)
)∀t ∈ [0,T ]
(Finite dim.: Trelat Zuazua, JDE 2015. Infinite dim.: Trelat Zhang Zuazua, SICON 2018)
Extension to periodic turnpike:Equilibrium point replaced with a periodic trajectory (itself optimal for a periodic optimalcontrol problem). (needs a “periodic Riccati theory”)
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Main idea of the proof: hyperbolicity of extremals
Linearizing the first-order optimality system (Pontryagin extremals in the cotangentbundle) at the equilibrium:
z(t) =
(xT (t)− xλT (t)− λ
)⇒ z(t) =
first orderM z(t) with M =
(A BU−1B∗Q −A∗
)
Key lemma
M is Hamiltonian, hence:
µ ∈ Spec(M) ⇒ −µ, µ,−µ ∈ Spec(M)
Moreover, under controllability assumptions:
Re(Spec(M)) 6= 0 (no pure imaginary eigenvalue)
⇒ M is hyperbolic.
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Main idea of the proof: hyperbolicity of extremals
Linearizing the first-order optimality system (Pontryagin extremals in the cotangentbundle) at the equilibrium:
z(t) =
(xT (t)− xλT (t)− λ
)⇒ z(t) =
first orderM z(t) with M =
(A BU−1B∗Q −A∗
)
In appropriate coordinates:
v(t) = P−v(t) Re(Spec(P−)) < 0
w(t) = P+w(t) Re(Spec(P+)) > 0
hence
‖v(t)‖ 6 ‖v(0)‖e−νt
‖w(t)‖ 6 ‖w(T )‖e−ν(T−t)
(click on the figure to see time evolution)
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Consequences for the numerical computations
Direct methods (full discretization): initialization with the solution of the static problem
⇒ successful convergence
Indirect method (shooting):
solve z(t) = F (z(t)), G(z(0), z(T )) = 0
Usual implementation: z(0) unknown, tuned such that G(z(0), z(T )) = 0.
Here we propose the following variant:
z(0) ←− z(T/2) unknown −→ z(T )
backward forwardintegration integration
↓tuned s.t.
G(z(0), z(T )) = 0
(initialization with the solution of the static problem)
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Example
x1(t) = x2(t), x1(0) = 1
x2(t) = 1− x1(t) + x2(t)3 + u(t), x2(0) = 1
min12
∫ T
0
((x1(t)− 1)2 + (x2(t)− 1)2 + (u(t)− 2)2
)dt
Optimal solution of the static problem:
x2 = 0, 1− x1 + x32 + u = 0
minx2=0
1−x1+x32 +u=0
((x1 − 1)2 + (x2 − 1)2 + (u − 2)2
)
whencex = (2, 0) , u = 1, λ = (−1,−1)
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Example
x1(t) = x2(t), x1(0) = 1
x2(t) = 1− x1(t) + x2(t)3 + u(t), x2(0) = 1
min12
∫ T
0
((x1(t)− 1)2 + (x2(t)− 1)2 + (u(t)− 2)2
)dt
Oscillation of (x1(·), x2(·)) around
the steady-state (2, 0)
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Example
x1(t) = x2(t), x1(0) = 1
x2(t) = 1− x1(t) + x2(t)3 + u(t), x2(0) = 1
min12
∫ T
0
((x1(t)− 1)2 + (x2(t)− 1)2 + (u(t)− 2)2
)dt
Impossible to make converge the usualshooting method if T > 3(explosive term + too high sensitivity)
Easy convergence with the variant, ∀T
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Further comments, extensions and open issues
More applications:aerospace (bi-impulse orbit transfers), motion planning and robotics, optimization ofchemotherapies (Pouchol Clairambault Lorz Trelat, JMPA 2018), model of a runner (Aftalion Trelat, ongoing)
Competition between turnpikes (see also Rapaport Cartigny, JOTA 2005)
Periodic turnpike (with a “periodic Riccati theory”) (Trelat Zhang, MCSS 2018)
Extensions to infinite dimension(Gugat Trelat Zuazua, SCL 2016; Trelat Zhang Zuazua, SICON 2018; Grune Schaller Schiela 2019)
“Measure” turnpike results for dissipative systems involving state and/or control constraints(Bonvin Faulwasser 2016; Grune 2018; Trelat Zhang MCSS 2018)
Interaction with discretizations (also discrete-time turnpike) (Grune; Grune Guglielmi 2017)
Shape turnpike: turnpike in optimal design, adiabatic theory (Trelat Zhang Zuazua, PAFA 2018;
ongoing PhD thesis of G. Lance)
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Shape turnpike (PhD thesis of Gontran Lance)
Dynamical shape optimization problem:
Ω ⊂ IRn bounded domain
y0, yd ∈ L2(Ω)
0 < L < 1
T > 0
Dirichlet heat equation on Ω with source term:
∂t y = 4y + χω(t) y|∂Ω = 0
y(0) = y0
minω(·)
∫ T
0‖y(t)− yd‖2
L2(Ω)dt
over all ω(t) ⊂ Ω s.t. |ω(t)| = L|Ω|
χω(x) =
1 if x ∈ ω0 otherwise
More generally: parabolic PDE satisfying the maximum principle.
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Shape turnpikeTheorem (Lance Trelat Zuazua, 2019):If either yd is convex or yd 6 0 or yd > y1 (with4y1 + 1 = 0, y1
|∂Ω= 0) then:
Dynamical shape optimization problem
∂t y = 4y + χω(t), y|∂Ω = 0
y(0) = y0
minω(·)
∫ T
0‖y(t)− yd‖2
L2(Ω)dt
over all ω(t) ⊂ Ω s.t. |ω(t)| = L|Ω|
has a unique solution ω(t)
Static shape optimization problem
4y + χω = 0, y|∂Ω = 0
minω‖y − yd‖2
L2(Ω)
over all ω ⊂ Ω s.t. |ω| = L|Ω|
has a unique solution ω
and
∃M > 0 | ∀T > 0∫ T
0
(‖y(t)− y‖2
L2(Ω)+ ‖λ(t)− λ‖2
L2(Ω)+ ‖χω(t) − χω‖L1(Ω)
)dt 6 M
⇒1T
∫ T
0y(t) dt L2
−→T→+∞
y ,1T
∫ T
0λ(t) dt L2
−→T→+∞
λ,1T
∫ T
0χω(t) dt L1
−→T→+∞
χω
(integral turnpike) Open question: exponential turnpike
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Shape turnpike
Proof:
Relaxation / convexification:
replace χω(t)(x) with a(t , x) s.t.
0 6 a(t , x) 6 1∫Ω a(t , x) dx = L|Ω|
Dynamical shape optimization problem
∂t y = 4y + a, y|∂Ω = 0
y(0) = y0
mina(·)
∫ T
0‖y(t)− yd‖2
L2(Ω)dt
a(t , x) ∈ [0, 1],
∫Ω
a(t , x) dx = L|Ω|
→ ∃! optimal a = a(t , x)
replace χω(x) with a(x) s.t.
0 6 a(x) 6 1∫Ω a(x) dx = L|Ω|
Static shape optimization problem
4y + a = 0, y|∂Ω = 0
mina‖y − yd‖2
L2(Ω)
a(x) ∈ [0, 1],
∫Ω
a(x) dx = L|Ω|
→ ∃! optimal a = a(x)
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Shape turnpikeFirst-order optimality conditions:
Pontryagin maximum principle
∂t y = 4y + a, y|∂Ω = 0, y(0) = y0
∂tλ = −4λ+ y − yd , λ|∂Ω = 0, λ(T ) = 0∫Ωλ(t , x)a(t , x) dx = max
∫Ωλ(t , x)u(x) dx
Lagrange multiplier rule
4y + a = 0, y|∂Ω = 0
4λ = y − yd , λ|∂Ω = 0∫Ωλ(x)a(x) dx = max
∫Ωλ(x)u(x) dx
over the set of u s.t. 0 6 u(x) 6 1 and∫
Ω u(x) dx = L|Ω|
(bathtub principle)
⇒ a(x) =
1 if λ(x) > µ0 if λ(x) < µ
with µ s.t.∫
Ω a(x) dx = L|Ω|
and if λ(x) = µ on a subset of positivemeasure then a(x) = −4yd (x) there.
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Shape turnpikeFirst-order optimality conditions:
Pontryagin maximum principle
∂t y = 4y + a, y|∂Ω = 0, y(0) = y0
∂tλ = −4λ+ y − yd , λ|∂Ω = 0, λ(T ) = 0∫Ωλ(t , x)a(t , x) dx = max
∫Ωλ(t , x)u(x) dx
Lagrange multiplier rule
4y + a = 0, y|∂Ω = 0
4λ = y − yd , λ|∂Ω = 0∫Ωλ(x)a(x) dx = max
∫Ωλ(x)u(x) dx
over the set of u s.t. 0 6 u(x) 6 1 and∫
Ω u(x) dx = L|Ω|
(bathtub principle)
⇒ a(x) =
1 if λ(x) > µ0 if λ(x) < µ
with µ s.t.∫
Ω a(x) dx = L|Ω|
and if λ(x) = µ on a subset of positivemeasure then a(x) = −4yd (x) there.
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Shape turnpikeFirst-order optimality conditions:
Pontryagin maximum principle
∂t y = 4y + a, y|∂Ω = 0, y(0) = y0
∂tλ = −4λ+ y − yd , λ|∂Ω = 0, λ(T ) = 0∫Ωλ(t , x)a(t , x) dx = max
∫Ωλ(t , x)u(x) dx
Lagrange multiplier rule
4y + a = 0, y|∂Ω = 0
4λ = y − yd , λ|∂Ω = 0∫Ωλ(x)a(x) dx = max
∫Ωλ(x)u(x) dx
over the set of u s.t. 0 6 u(x) 6 1 and∫
Ω u(x) dx = L|Ω|
⇒ a(t , x) =
1 if λ(t , x) > µ(t)0 if λ(t , x) < µ(t)
with µ(t) s.t.∫
Ω a(t , x) dx = L|Ω| for a.e. t
and if λ(t , x) = µ(t) on a subset of positivemeasure then a(t , x) = µ(t)−4yd (x) there.
⇒ a(x) =
1 if λ(x) > µ0 if λ(x) < µ
with µ s.t.∫
Ω a(x) dx = L|Ω|
and if λ(x) = µ on a subset of positivemeasure then a(x) = −4yd (x) there.
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Shape turnpikeSketch of proof:
Recall that
4y + a = 0 y|∂Ω = 0
4λ = y − yd λ|∂Ω = 0
Let y1 be the solution of4y1 + 1 = 0, y1|∂Ω
= 0.
Since −1 6 4y = −a 6 0, by the maximum principle: 0 6 y 6 y1.
If λ(x) = Cst on a subset E ⊂ Ω of positive measure then4λ = 0 on E and thusy = yd on E .
Contradiction if yd 6 0 or if yd > y1.
Similar arguments for the dynamical problem.
Then, use equivalence with the problem of minimizing
∫ T
0
(‖y(t)− yd‖2
L2(Ω)+ ‖a(t)‖2
L2(Ω)
)dt
which is strictly dissipative (in the sense of Willems 1972), and finally use thatstrict dissipativity implies integral turnpike (Trelat Zhang MCSS 2018).
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Shape turnpike
Example in 1D: Ω = [0, 2], y0 ≡ 0, yd ≡ 0.1, L = 0.25, T = 5
(computation with IpOpt + FreeFEM++, G. Lance)
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Shape turnpikeExample in 2D: square Ω = [−1, 1]2, y0 ≡ 0, yd ≡ 0.1, L = 0.25, T = 5
t 7→ ω(t)
t = 0 t = 0.5 1 6 t 6 4 t = 4.5 t = T = 5
(computation with IpOpt + FreeFEM++,
G. Lance)
Optimal static shape ω t 7→ ‖yT (t)−y‖L2 +‖pT (t)−p‖L2
+‖χωT (t) − χω‖L2
![Page 30: The turnpike property - FAU...The turnpike property Emmanuel Trelat´ 1 1Sorbonne Universite (Paris 6), Labo. J.-L. Lions´ Works with Gontran Lance, Can Zhang, Enrique Zuazua FAU,](https://reader036.vdocument.in/reader036/viewer/2022071418/6115a33d41084906e113b68d/html5/thumbnails/30.jpg)
Shape turnpike
Square Ω = [−1, 1]2, y0 ≡ 0, yd ≡ 0.1, L = 0.25
T = 1
T = 2
![Page 31: The turnpike property - FAU...The turnpike property Emmanuel Trelat´ 1 1Sorbonne Universite (Paris 6), Labo. J.-L. Lions´ Works with Gontran Lance, Can Zhang, Enrique Zuazua FAU,](https://reader036.vdocument.in/reader036/viewer/2022071418/6115a33d41084906e113b68d/html5/thumbnails/31.jpg)
Shape turnpike
Square Ω = [−1, 1]2, y0 ≡ 0, yd (x , y) = 120 (xy + 1), L = 3/16, T = 2
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Shape turnpikeΩ stadium-shaped, y0 ≡ 0, L = 3/16, T = 2
yd ≡ 0.1
yd = 120 (xy +1)
![Page 33: The turnpike property - FAU...The turnpike property Emmanuel Trelat´ 1 1Sorbonne Universite (Paris 6), Labo. J.-L. Lions´ Works with Gontran Lance, Can Zhang, Enrique Zuazua FAU,](https://reader036.vdocument.in/reader036/viewer/2022071418/6115a33d41084906e113b68d/html5/thumbnails/33.jpg)
Shape turnpike
Cube Ω = [0, 1]3, y0 ≡ 0, yd ≡ 0.1, L = 1/40, T = 1
optimal time-evolving shape optimal static shape
![Page 34: The turnpike property - FAU...The turnpike property Emmanuel Trelat´ 1 1Sorbonne Universite (Paris 6), Labo. J.-L. Lions´ Works with Gontran Lance, Can Zhang, Enrique Zuazua FAU,](https://reader036.vdocument.in/reader036/viewer/2022071418/6115a33d41084906e113b68d/html5/thumbnails/34.jpg)
Shape turnpike
Relaxation phenomenon:Ω = [−1, 1]2, y0 = 0, yd = − 1
20 (x2 + y2 − 2), L = 0.25, T = 5
t = 0 t = 0.5 1 6 t 6 4 t = 4.5 t = T = 5
(computation with IpOpt + FreeFEM++, G. Lance)
Optimal static solution a
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Perspectives
Shape turnpike for general semilinear PDE models
∂t y = Ay + f (y) + χω(t)
Shape-moving bottom pool wave generator (wavemaker)
Nersisyan Dutykh Zuazua 2014
Dalphin Barros 2019
KdV or Saint-Venant equations
Kinetic lift formulation (Perthame)
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Competition between two global turnpikes
min∫ T
0
((x(t)− 1)2 + (u(t)− 3.47197)2
)dt
x(t) = −3x(t) + 3x(t)2 + u(t), x(0) = x0, x(T ) = xf or free
The choice of ud = 3.47197 is done so thatthe static optimal control problem
min(x,u) | −3x+3x3+u=0
((x − 1)2 + (u − 3.47197)2
)has two global minima:
x1 = −1.3473, x2 = 0.5939
Plot of x 7→ (x − 1)2 + (3x − 3x3 − 3.47197)2:
![Page 38: The turnpike property - FAU...The turnpike property Emmanuel Trelat´ 1 1Sorbonne Universite (Paris 6), Labo. J.-L. Lions´ Works with Gontran Lance, Can Zhang, Enrique Zuazua FAU,](https://reader036.vdocument.in/reader036/viewer/2022071418/6115a33d41084906e113b68d/html5/thumbnails/38.jpg)
Competition between two global turnpikes
min∫ T
0
((x(t)− 1)2 + (u(t)− 3.47197)2
)dt
x(t) = −3x(t) + 3x(t)2 + u(t), x(0) = x0, x(T ) = xf or free
Global solutions of the static problem: x1 = −1.3473 and x2 = 0.5939.
x0 = −5, xf = −1, T = 10
→ Turnpike around x1 = −1.3473
x0 = 2, xf = 1, T = 10
→ Turnpike around x2 = 0.5939
![Page 39: The turnpike property - FAU...The turnpike property Emmanuel Trelat´ 1 1Sorbonne Universite (Paris 6), Labo. J.-L. Lions´ Works with Gontran Lance, Can Zhang, Enrique Zuazua FAU,](https://reader036.vdocument.in/reader036/viewer/2022071418/6115a33d41084906e113b68d/html5/thumbnails/39.jpg)
Local versus global turnpike
min∫ T
0
((x(t)− 1)2 + (u(t)− 1)2
)dt
x(t) = −3x(t) + 3x(t)2 + u(t), x(0) = x0, x(T ) = xf or free
The static optimal control problem has
a unique globally optimal solution x = 0.7815
a locally optimal solution xloc = −1.1055
Plot of x 7→ (x − 1)2 + (3x − 3x3 − 1)2:
![Page 40: The turnpike property - FAU...The turnpike property Emmanuel Trelat´ 1 1Sorbonne Universite (Paris 6), Labo. J.-L. Lions´ Works with Gontran Lance, Can Zhang, Enrique Zuazua FAU,](https://reader036.vdocument.in/reader036/viewer/2022071418/6115a33d41084906e113b68d/html5/thumbnails/40.jpg)
Local versus global turnpike
min∫ T
0
((x(t)− 1)2 + (u(t)− 1)2
)dt
x(t) = −3x(t) + 3x(t)2 + u(t), x(0) = x0, x(T ) = xf or free
Global solution x = 0.7815. Local solution xloc = −1.1055.
x0 = −2, xf = −1, T = 10Initialization with the constant trajectory xloc .
→ Local turnpike around xloc = −1.1055.Cost C = 4.19.
x0 = −2, xf = −1, T = 10Initialization with x .
→ Global turnpike around x = 0.7815.Cost C = 2.61.
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Local versus global turnpike
min∫ T
0
((x(t)− 1)2 + (u(t)− 1)2
)dt
x(t) = −3x(t) + 3x(t)2 + u(t), x(0) = x0, x(T ) = xf or free
x0 = −2, xf = −1, T = 2 (quite small)Initialization with the constant trajectory xloc .
Cost C = 8.88.→ Globally optimal!
x0 = −2, xf = −1, T = 2Initialization with x .
Cost C = 9.64.→ Locally but not globally optimal!
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Local versus global turnpike
min∫ T
0
((x(t)− 1)2 + (u(t)− 1)2
)dt
x(t) = −3x(t) + 3x(t)2 + u(t), x(0) = x0, x(T ) = xf or free
x0 = −2, xf = −1
T ∈ 2.5, 2.7, 2.9, 3.1, 3.3
Global optimal trajectory:Bifurcation at T = T0 ' 2.9.
T < T0 ⇒ turnpike around xloc
T > T0 ⇒ turnpike around x
→ in accordance with the globalturnpike result for T large enough
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Periodic turnpike
X , U, V Hilbert spaces
A : D(A)→ X operator generating a C0 semigroup on X
B ∈ L(U,X) linear bounded control operator
C ∈ L(X ,V ) linear bounded observation operator
Q ∈ L(U,U) positive definite
Tracking trajectory: yd (·) ∈ C([0,+∞); X), ud (·) ∈ L2loc(0,+∞; U), Π-periodic:
yd (t + Π) = yd (t), ud (t + Π) = ud (t) ∀t
Optimal control problem (OCP)T
For T > 0 fixed, find uT (·) ∈ L2(0,T ; U) such that
y(t) = Ay(t) + Bu(t)
y(0) = y0
min12
∫ T
0
(‖C(y(t)− yd (t))‖2
V + 〈Q(u(t)− ud (t)), u(t)− ud (t)〉U)
dt
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Periodic turnpike
We replace the steady-state optimal control problem with:
Periodic optimal control problem
min12
∫ Π
0
(‖C(y(t)− yd (t))‖2
V + 〈Q(u(t)− ud (t)), u(t)− ud (t)〉U)
dt
y(t) = Ay(t) + Bu(t)
y(0) = y(Π)
Theorem (Trelat Zhang Zuazua, SICON 2018)
If (A,B) is exponentially stabilizable and (A,C) is exponentially detectable, then:
The periodic optimal control problem has a unique solution (yΠ(·), uΠ(·)), which has aunique extremal lift (yΠ(·), uΠ(·), λΠ(·)) (of which we have explicit expressions).
There exist c > 0, ν > 0 such that ∀T > 0
∥∥∥yT (t)− yΠ(t)∥∥∥
X+∥∥∥uT (t)− uΠ(t)
∥∥∥U
+∥∥∥λT (t)− λΠ(t)
∥∥∥X6 c
(e−νt + e−ν(T−t)
)∀t ∈ [0,T ]
ν = exponential stability rate for a C0 semigroup resulting from the (operator) algebraic Riccati equation.
⇒ The optimal extremal is almost Π-periodic (except at the beginning and at the end).
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Periodic turnpike
The proof uses in particular a kind of “periodic Riccati theory”:
Lemma
If (A,B) is exponentially stabilizable and (A,C) is exponentially detectable, then the uniquesolution of the periodic optimal control problem is
yΠ(t) = z(t)− Eq(t), λΠ(t) = −Pz(t) + (I + PE)q(t), uΠ(t) = ud (t) + Q−1B∗λΠ(t)
with:
P > 0 unique solution of the algebraic Riccati equation A∗P + PA−PBQ−1B∗P + C∗C = 0
E 6 0 unique solution of the Lyapunov equation 2(A− BQ−1B∗P)E − BQ−1B∗ = 0
z(t) = S(t)(I − S(Π))−1 ∫ Π
0 S(Π− τ)((I + EP)Bud (τ)− EC∗Cyd (τ)
)dτ
+∫ t
0 S(t − τ)((I + EP)Bud (τ)− EC∗Cyd (τ)
)dτ
q(t) = S(Π− t)∗(I − S(Π)∗)−1 ∫ Π
0 S(Π− τ)∗(− PBud (Π− τ) + C∗Cyd (Π− τ))
)dτ
+∫ Π−t
0 S(Π− t − τ)∗(− PBud (Π− τ) + C∗Cyd (Π− τ)
)dτ
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Example of periodic turnpike
min12
∫ T
0
((x(t)− cos(2πt))2 + (y(t)− sin(2πt))2 + u(t)2
)dt T = 20, Π = 1
x(t) = y(t), y(t) = u(t), x(0) = 0.1, y(0) = 0