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Submitted on 15 Jul 2014
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Optimization of running strategies based on anaerobic energy and variations of velocity
J. Frederic Bonnans
To cite this version:
J. Frederic Bonnans. Optimization of running strategies based on anaerobic energy and variations of velocity. NETCO 2014, Jun 2014, Tours, France. �hal-01024231�
Optimization of running strategies based on anaerobic energy and variations of velocity
J. Fr´ed´eric Bonnans
Inria-Saclay and CMAP, Ecole Polytechnique, France
Netco2014 Conf., Tours, June 23-27, 2014
Outline
1 Keller’s model
2 Variable energy recreation
3 Bounding the derivative of f
Sources
Talk based on the joint work [1] with Amandine Aftalion, Lab.
Math. Versailles, UVSQ.
Pioneering reference: J.B. Keller [2].
[1] A. Aftalion and J.F. Bonnans,Optimization of running strategies based on anaerobic energy and variations of velocity. SIAM J. Applied Math., to appear (HAL preprint).
[2] Keller, J.B.,Optimal velocity in a race, Amer. Math. Monthly 81 (1974), 474–480.
1 Keller’s model
2 Variable energy recreation
3 Bounding the derivative of f
Keller’s model: dynamics
Consider the followingstate equation:
(1.1) v(t) =˙ f(t)−φ(v(t)); e˙(t) = ¯σ−f(t)v(t),
Cost function: −RT
0 v(t)dt.
energy recreation: ¯σ >0
drag functionφ(v) =v/τ, withτ >0.
Generalized drag function: φC2 on (0,∞);φ(0) = 0;
φ′ positive; vφ′(v) increasing
Example: φ(v) =avβ,a>0, β≥1.
Keller’s model: constraints
Control constraint: 0≤f(t)≤fM. Energy constraint: e(t)≥0.
Maximal force strategy: optimal for short horizons T ≤Tc. In the sequel T >Tc.
Keller’s model: discussion
Maximal speed (never reached): 0 =fM −vM/τ, i.e.
vM =τfM.
Constant energy speed: f =v/τ,σ =fv, and so:
vE =√ τ σ.
Energy variation when constant speed:
e(t)↓ ifv >vE, and e(t)↑ otherwise.
Keller’s model: Keller’s conjecture
Optimal strategy has three arcs First arc: maximal forcef =fM. Second arc: constant speedv>vE. Last arc: zero energy e = 0.
Negative jump of force at junction times Unclear proofs in the literature for φ(v) =v/τ.
Keller’s model: numerical resultsI
Mechanical parameters: fmax = 9.6m s−2,τ = 0.67s Energy parameters: ean0 = 1400m2 s−3, ¯σ= 49m2 s−3. d = 1500m.
2000 discretization steps, i.e. time step close to 0.12s.
Free softwarewww.bocop.org.
Optimal time is 248.21s
Keller’s model: numerical results with bocop.org
0 1 2 3 4 5 6 7
0 50 100 150 200 250
velocity v(t)
Velocity
8.0 8.2 8.4 8.6 8.8 9.0 9.2 9.4 9.6 9.8
0 50 100 150 200 250
force f(t)
Force
0 500 1000 1500
0 50 100 150 200 250
AOD(t)
Reverse energy
Zoom on end of race: zero energy arc
8.4 8.5 8.6 8.7 8.8 8.9 9.0
242 243 244 245 246 247 248 249
force f(t)
Final force
1340 1350 1360 1370 1380 1390 1400 1410 1420 1430
242 243 244 245 246 247 248
AOD(t)
Final reverse energy
Keller’s model: main result
So the numerical results agree with Keller’s conjecture, and indeed:
Theorem
Keller’s conjecture holds, even with the generalized drag function.
We next present the proof of this result.
Keller’s model: Hamiltonian and costate
Hamiltonianfunction:
H :=−v+pv(f −φ(v)) +pe(¯σ−fv).
Costate equation:
−p˙v =−1−pvφ′(v)−pef
−dpe =−dµ,
Final conditionspv(T) =pe(T) = 0, and so: pe =µ.
Multiplierto the state constraint: say µ(T) = 0 and dµ≥0; supp(dµ)⊂ {t; e(t) = 0}. Thereforepe =µ≤0.
Switching function
Theswitching function is
Ψ :=Hu=pv−pev.
ByPontryagin’s principle, we have:
f(t) =
fM if Ψ(t)>0, 0 if Ψ(t)<0,
pe has negative values
Lemma
pe has negative values.
Proof.
Ifpe(τ) = 0 then over (τ,T):
pe = 0, since it is nondecreasing andpe(T) = 0.
˙
pv = 1 +pvφ′(v) (is >0 ifpv ≥0) andpv(T) = 0 Sopv <0⇒Ψ =pv <0⇒f =fM
Contradiction by lemma 3.7 of paper.
pv has negative values
Lemma
pv has negative values.
Proof.
pv ≥0⇒Ψ>0⇒f = 0⇒p˙v >0; but pv(T) = 0.
Corollary
An optimal trajectory starts with a maximal force arc.
Derivative of switching function
When the state constraint is not active,pe is constant and so:
Ψ :=Hu=pv−pev.
Ψ = (1 +˙ pvφ′(v) +pef)−pe(f −φ(v))
= 1 +pvφ′(v) +peφ(v) and so,
Ψ˙ −Ψφ′(v) = 1 +pe(φ(v) +vφ′(v)).
Lemma
On a singular arc, v is constant.
No zero force arc
∆ := ˙Ψ−Ψφ′(v) = 1 +pe(φ(v) +vφ′(v)).
Lemma
No zero force arc occurs.
Let (ta,tb) be such an arc; then ta>0.
Iftb=T then e(T)>0: impossible.
Ψ≥0 over the arc; Ψ(ta) = Ψ(tb) = 0.
We have therefore ∆(ta)≥0≥∆(tb).
On this arc pe is constant and v decreases, and so ∆(t) increases: contradiction.
No second maximal force arc
∆ := ˙Ψ−Ψφ′(v) = 1 +pe(φ(v) +vφ′(v)).
Lemma
No second maximal force arc occurs.
Let (ta,tb) be such an arc; then ta>0.
Iftb<T, ”symmetric argument”: v increases, ∆ decreases, but ∆(ta)≤0≤∆(tb).
Iftb=T,µshould have a jump at time T, but then limt↑TΨ(t) =−pe(T−)v(T)>0 implying that f = 0 at the end of the trajectory.
Keller’s model: proof of main result
Let ta∈(0,T) be the exit point of the maximal force arc.
We know that Ψ = 0 over (ta,T).
Let tb∈(0,T) be the first time at which the energy vanishes (µhas no final jump).
(ta,tb) is a singular arc.
If the energy is not zero on (tb,T): there would exist tc,td
with tb≤tc <td ≤T such thate(tc) =e(td) = 0, and e(t)>0, for allt ∈(tc,td).
Then (tc,td) is a singular arc, over which ˙e =σ−fv is constant: contradiction.
1 Keller’s model
2 Variable energy recreation
3 Bounding the derivative of f
Variable energy recreation
Less recreation when energy close to its extreme values, and additional recreation when deceleration:
˙
e =σ(e) +η(a)−fv
where a is the acceleration: a= ˙v =f −φ(v),and η convex:
η(a)≥0, η(a) = 0 iff a≥0.
Function σ(e) regularization of a piecewise affine and continuous function, value 0 at extreme values, otherwise constant.
Numerical results with variable σ(e) and η(a) = 0
0 1 2 3 4 5 6 7
0 50 100 150 200 250
velocity v(t)
Velocity
8.8 8.9 9.0 9.1 9.2 9.3 9.4 9.5 9.6 9.7
0 50 100 150 200 250
force f(t)
Force
5 10 15 20 25 30 35 40 45 50 55
0 50 100 150 200 250
sigma(t)
σ(e)
Hamiltonian and costate with variable σ(e)
Hamiltonian function:
H:=−v+pv(f −φ(v)) +pe(σ(e) +η(f −φ(v))−fv).
Costate equation:
−p˙v =−1−pvφ′(v)−pef −peη′(a)φ′(v)
−dpe =peσ′(e)−dµ, Final conditionspv(T) =pe(T) = 0.
µ(T) = 0,dµ≥0,supp(dµ)⊂ {t; e(t) = 0}.
Minimization of Hamiltonian with variable σ(e)
Hamiltonian function:
H:= (pv −pev)f +peη(f −φ(v)) + indep. off Concave functionof the control f
Minimum attained at extreme values 0 or fM. Need of relaxed formulation; expectation of force:
f = 0×(1−θ) +θfM
We may as well takef as parameter and thenθ=f/fM.
Relaxed formulation
Recreation due to deceleration at zero speed:
R(v) :=η(−φ(v) State equation
˙
v =f −φ(v)
˙
e =σ(e)−fv+ (1−f/fM)R(v) Hamiltonian
H=pv(f −φ(v)) +pe(σ(e)−fv+ (1−f/fM)R(v)).
Relaxed formulation: theoretical analysis
Assume σ(e) = ¯σ andη(a) =εη′(a) For ε >0 small enough
Same optimal structure as for Keller’model:
maximal force, constant speed, zero energy.
1 Keller’s model
2 Variable energy recreation
3 Bounding the derivative of f
Numerical results when bounding ˙f
0 1 2 3 4 5 6 7
0 50 100 150 200 250
velocity v(t)
Velocity
7.0 7.5 8.0 8.5 9.0 9.5 10.0
0 50 100 150 200 250
force f(t)
Force
Numerical results: zoom with variable σ(e) and η(a) 6= 0
5.80 5.85 5.90 5.95 6.00 6.05 6.10 6.15 6.20
113 114 115 116 117 118 119
velocity v(t)
Zoom on velocity
7.8 8.0 8.2 8.4 8.6 8.8 9.0 9.2 9.4 9.6 9.8
113 114 115 116 117 118 119
force f(t)
Zoom on force
Periodic problem
Maximize average speed: (1/T)RT
0 v(t)dt.
Periodic speed and force, loss of energy e(0) =e(T) +Ted. State equation and constraints
v(0) =v(T); f(0) =f(T).
e(0) =e(T) +Ted;
˙
v =f −vτ;
˙
e =σ+η(a)−fv;
0≤f ≤fM; |f˙| ≤1, for a.e. t∈(0,T).
Zoom vs periodic: velocity
5.80 5.85 5.90 5.95 6.00 6.05 6.10 6.15 6.20
113 114 115 116 117 118 119
velocity v(t)
Zoom on velocity
5.7 5.8 5.9 6.0 6.1 6.2 6.3 6.4
2 3 4 5 6 7 8 9 10
velocity v(t)
Periodic velocity
Zoom vs periodic: force
7.8 8.0 8.2 8.4 8.6 8.8 9.0 9.2 9.4 9.6 9.8
113 114 115 116 117 118 119
force f(t)
Zoom on force
8.0 8.2 8.4 8.6 8.8 9.0 9.2 9.4 9.6 9.8 10.0
2 3 4 5 6 7 8 9 10
force f(t)
Periodic force
Related work
[1] S. Aronna, J.F. Bonnans and B.S. Goh,Second order necessary conditions for control-affine problems with state constraints. Research report, to appear (HAL preprint).
Open questions
Asymptotic analysis η(a) =εη′(a).
Shape of oscillations ? Expansion of cost function
THE END !
