Optimal Control of Newton-Type
Problems of Minimal Resistance
Alexander Yu. Plakhov
Delfim F. M. Torres
[email protected]
[email protected]
Department of Mathematics
University of Aveiro
3810–193 Aveiro, Portugal
Second Junior European Meeting
Historical Facts and How one tells the Story
[1686] Isaac Newton propound and give solution to the problem of
the body of minimal resistance in his Principia Mathematica.
➜ “Newton’s problem is one of the first applications of the
Calculus of Variations.”
[1697] Solutions (by Johann Bernoulli, Newton, Leibniz,
Tschirnhaus, l’Hopital, and Jakob Bernoulli) to the
brachystochrone problem are published.
➜ “The Calculus of Variations was born in 1697.”
[1956] The proof of the Pontryagin Maximum Principle is
published (by Boltyanski, Gamkrelidze, and Pontryagin).
➜ “Optimal Control Theory was born in the fifties with the
Pontryagin Maximum Principle.”
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Recent Thesis and Our Claim
In the beautiful paper
H. J. Sussmann and J. C. Willems, 300 years of
optimal control: from the brachystochrone to the
maximum principle, IEEE Control Systems Magazine,
1997, pp. 32–44.
the authors defend the (polemic) thesis that the
brachystochrone marks also the birth of Optimal Control
(1697).
Here we go a step further (Juniors can be polemic ;-))
➜ Optimal Control was born before the Calculus of Variations!
➜ Newton’s problem of Minimal Resistance marks the birth of
Optimal Control (1686)
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Newton’s Problem of Minimal Resistance
Problem. Find the shape of a solid of revolution, moving in a
fluid along its axis with constant velocity, which would offer
minimum resistance.
The mathematical formulation of the problem depends on the
resistance law of the medium.
Newton has considered:
➜ A “rare” medium, consisting of perfectly elastic particles
with constant mass.
➜ The resisting pressure at a surface point to be proportional
to the square of the normal component of its velocity.
High-speed and high-altitude flying vehicles
Newton’s problem is applicable to the design of missiles
traveling hypersonically in the thin air of our upper atmosphere.
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Newton’s problem belongs to Optimal Control
Resistance force is given by R [ẋ(·)] =
T
0
t
1
dt
2
1 + ẋ(t)
In 1788 Legendre noted that the problem R [ẋ(·)] → min,
x(0) = 0, x(T ) = H, has no solution (the infimum is zero):
➜ The Resistance integral is non-negative
➜ If one chooses a zig-zag function x(·) wildly oscillating, then
R [ẋ(·)] becomes arbitrarily small
✔ To make the problem physically consistent, one must take into
account the monotonicity of the profile Optimal Control
T
1
R [u(·)] =
t
dt −→ min
2
1
+
u(t)
0
ẋ(t) = u(t) ,
x(0) = 0 ,
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u(t) ≥ 0
x(T ) = H
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Newton’s Problem has been widely studied
Newton has found the solution in 1687, but did not explained
how he obtained it [He didn’t wrote, however, “I have a great
proof, but no space for it in the margins of this book” ;-)]
Main difficulty: standard existence proofs fail because the
Lagrangian L(t, u) = t/(1 + u2 ) is not convex with respect to u
for u ≥ 0; neither coercive
In order to prove existence, several different classes of admissible
functions have been proposed by subsequent mathematicians
✔ There has been a recent revival of interest in Newton’s problem
of the body of minimal resistance:
➜ Since 1993, many variations began to be investigated
(France, Germany, Italy, Portugal-Russia)
➜ Applications to artificial satellites (h ∼ 200 ÷ 250 Km)
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Some Recent Results on Newton-type Problems
Bodies without rotational symmetry (nonsymmetric cases):
[Buttazzo & Ferone & Kawohl, 1993] [Lachand-Robert &
Peletier, 1999]
Unbounded body (resistance per unit area) with one-impact
assumption: [Comte & Lachand-Robert, 2001]
Bodies with rotational symmetry and one-impact assumption,
but not convex: [Comte & Lachand-Robert, 2002]
Friction between particles and body (non-elastic collisions):
[Horstmann & Kawohl & Villaggio, 2002]
Bodies with prescribed volume: [Belloni & Wagner, preprint]
Multiple collisions allowed: [Plakhov, Doklady, 2003] [Plakhov,
Russian Math. Surveys, 2003]
Unbounded body, multiple collisions allowed: [Plakhov, 2004]
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Our Contribution
We consider convex d-dimensional bodies of revolution with Height
H and radius of maximal cross section T .
We are not restricted to two-dimensional or three-dimensional
bodies! (d ≥ 2)
We consider a different point of view (physically, more realistic):
the body does not move and is situated in a flux of infinitesimal
particles. The flux is invariant with respect to translations and
rotations around the symmetry axis of the body.
We address the situation when the flux of particles is not
necessarily falling vertically downwards: non-parallel flux
The pressure will be a function of ẋ+ (·) or of ẋ− (·), as the point
belongs to the front or to the rear part of the body. We denote
by p+ (ẋ+ ) and p− (ẋ− ) the corresponding pressure functions.
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Our Mathematical Model
Resistance of the flux on the front part of the body:
T
R+ [x+ ] =
td−2 p+ (x+ (t)) dt, x+ (0) = 0, x+ (T ) = H − c
0
Resistance of the flux on the rear part of the body:
T
R− [x− ] =
td−2 p− (x− (t)) dt, x− (0) = 0, x− (T ) = c
0
Resistance of the body to the flux: R[x+ , x− ] = R+ − R−
It is required to minimize R[x+ , x− ] over all pairs (x+ (·), x− (·))
of convex monotone increasing functions defined on [0, T ]
We assume the following assumptions (all physically relevant):
(i) p± ∈ C 1 [0, +∞);
(ii) ∃ limu→+∞ p± (u) ≥ 0;
(iii) p± (0) = limu→+∞ p± (u) = 0;
(iv) ∃ ū± > 0 s.t. p± is strictly monotone decreasing on [0, ū± ],
and strictly monotone increasing on [ū± , +∞)
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Our Approach
First Step: we minimize R+ [x+ (·)] and maximize R− [x− (·)]
with c fixed, over monotone increasing functions
x− : [0, T ] → [0, c], x+ : [0, T ] → [0, H − c], and verify that
among all the solutions, the convex ones are unique; we denote
them by xc− (·) and xc+ (·).
Second Step: we minimize the difference R+ [xc+ (·)] − R− [xc− (·)]
with respect to c ∈ [0, H].
✔ Difficult part is the First Step. We obtain complete solution:
➜ First we assume that there exists a solution to the problems.
➜ We apply the Pontryagin Maximum Principle to the
hypothetical solutions
➜ Finally, by means of a direct calculation, we prove that the
candidates are indeed optimum.
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Main Result
Theorem. Under our assumptions on p(·), the control ũ(·) is an
absolute minimizing (maximizing) control for the problem
T
R [u(·)] =
td−2 p(u(t)) dt −→ extr
0
x (t) = u(t) ,
u(t) ≥ 0 ,
x(0) = 0 , x(T ) = β
()
(β > 0)
if, and only if, it is a Pontryagin extremal control.
Corollary. Finding the solutions to problem () amounts to find
the minimum of the function h(u) = td−2 p(u) + λu for u ≥ 0,
t ∈ [0, T ], λ > 0.
➜ Roughly speaking, to solve a Newton-type problem of the kind
we are considering, one just need to solve three static
optimization problems!
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Proof of the Main Result (idea)
We prove that all Pontryagin extremals (x(·), u(·), ψ0 , ψ(·)) of
the problem are normal extremals (ψ0 = −1) with ψ(·) a
negative constant: ψ(t) ≡ −λ, λ > 0. (Ingredients: adjoint
system; maximality condition; and hypothesis on pressure p(·))
From previous step, one can write the maximality condition as
−td−2 p(ũ(t)) − λũ(t) ≥ −td−2 p(u(t)) − λu(t)
Having in mind that all admissible processes (x(·), u(·)) of the
T
T problem satisfy 0 u(t)dt = 0 x (t)dt = β, it is enough to
integrate in order to obtain the conclusion:
T
T
td−2 p(ũ(t))dt ≤
td−2 p(u(t))dt .
0
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0
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Solution to the problem of dimension two (d = 2)
We present only the solution to R+ [x+ (·)] −→ min (R− is similar)
+ (u)
→ max
There exists a unique solution u+ of p+ (0)−p
u

 0
as t ∈ [0, t0 ]
c
(H − c)/T < u+ ⇒ x+ (t) =
 u+ (t − t0 ) as t ∈ [t0 , T ]
t0 = T − (H − c)/u+
(H − c)/T ≥ u+ ⇒ xc+ (t) = (H − c) t/T
R+ [xc+ (·)] = T p+ (0) − (H − c) b+ , with b+ =
p+ (0)−p+ (u+ )
.
u+
✔ There are four possible solutions (we give full characterization)
✔ All the computations can be done with our Maple procedure
> solution(p+,p-,T,H)
which plots the body of minimum resistance.
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Example – solution is a trapezium
>
>
>
>
pp := u -> 1/(1+u^2)+0.5:
pm := u -> 0.5/(1+u^2)-0.5:
T := 2: H := 1:
solution(pp,pm,T,H);
1
0.8
0.6
0.4
0.2
–2
–1
0
1
2
t
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Example – solution is a triangle
>
>
>
>
pp := u -> 1/(1+u^2)+0.5:
pm := u -> 0.5/(1+u^2)-0.5:
T := 2: H := 2:
solution(pp,pm,T,H);
2
1.5
1
0.5
–2
–1
0
1
2
t
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Example – union of a triangle and a trapezium
>
>
>
>
pp := u -> 1/(1+u^2)+0.5:
pm := u -> 0.5/(1+u^2)-0.5:
T := 2: H := 4:
solution(pp,pm,T,H);
4
3
2
1
–2
–1
1
2
t
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Example – union of two triangles with same base
>
>
>
>
pp := u -> 1/(1+u^2)+0.5:
pm := u -> 0.5/(1+u^2)-0.5:
T := 2: H := 6:
solution(pp,pm,T,H);
6
5
4
3
2
1
–2
–1
1
2
t
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Solution to the problem with d ≥ 3
There exist a unique U such that T d−2 p+ (U ) + λ = 0
U
d−1
u
p
(U
)
T d−1 +
+
du
R+ = d−1 |p+ (U )| d−2
+
1
d−1 + u
+
1
|p+ (u)| d−2
|p+ (U )| d−2
d−2
b+
0 ≤ t ≤ t0 ⇒ x(t) = 0; t0 ≤ t ≤ T ⇒ x(t) given parametrically:
t=
x=λ
1
d−2
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λ
|p+ (u)|
1
d−2
u
1
|p+ (u)| d−2
,
−
u+
u+ ≤ u ≤ U
1
b+ d−2
−
u
dν
1
u+
|p+ (ν)| d−2
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Example – Newton’s Classical Problem
d = 3; p+ (u) = 1/(1 + u2 ); p− (u) ≡ 0.
Our general expressions give the well-known classical formulas:
the optimal solution x(t) is given in parametric form by
λ
7
1
λ 3u4
, t=
u3 + 2u +
+ u2 − ln u −
x=
2
4
4
2
u
Newton’s classical problem with T = 1, H = 1, 2, 3, 4
4
3
x(t) 2
1
–1
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–0.5
0
0.5
t
1
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Newton’s Problem in higher-dimensions
We have solved the problem for an arbitrary dimension d ≥ 2,
and for a class of functions p+ and p−
For example, for d = 4; p+ (u) = 1/(1 + u2 ); p− (u) ≡ 0
(Newton’s problem in dimension four) one gets:
λ=
2 T 2U
2
U 2)
,
t0 =
√
T −5 U + 3 U 3 + 2 U
4T 2 U
2
U 2)
,
H=
5 (1 + U 2 )
(1 +
√ √
5/2
2
T U −5 u + 3 u + 2
U 1+u
, x=
t=T
u 1 + U2
5 (1 + U 2 )
2
2
T 1 + 3U
R+ =
2
2 (1 + U 2 )
(1 +
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Conclusions and Open Questions
Newton-type problems are best treated using an optimal control
approach
We have completed solved a class of Newton-type problems for
which the Pontryagin extremals give the absolute minimizer
(PMP holds as a necessary and sufficient condition)
We consider bodies of arbitrary dimension
✔ We cover problems with a non-parallel flux of particles
? There always exist a flux of particles which realizes our functions
p+ and p− ?
? If yes, how to obtain the flux from p+ and p− ? (inverse problem)
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