Mario Zanon & S ebastien Gros · 01-03-2019 · f]: Transform the OCP into the NLP min w f(w) s:t: g(w) = 0; h(w) 0: and solve it The transformation from OCP to NLP is called transcription

M. ZanonandS.Gros

Numerical Methods for Optimal Control:Introduction

Mario Zanon & Sebastien Gros

M. ZanonandS.Gros

Outline

1 Introduction

2 An Overview about OCP Solution Approaches

3 A Motivating Example

M. Zanon & S. Gros Introduction 2 / 25

M. ZanonandS.Gros

Outline

1 Introduction




M. ZanonandS.Gros

MotivationOptimization and optimal control are everywhere

We don’t just want systems to work......we want them to work in the best possible way!

Optimization is

a great tool to support decision-making

used in automation, solid mechanics, economics, and many more fields

Why direct optimal control?

very flexible and efficient

hard to use for analysis but very useful to solve practical problems

Why a full course?

optimization (especially nonconvex) does not work out of the box!

perfect recipe for disaster: hit run and hope for the best

Warning!

If there is a flaw in the problem formulation, the optimization algorithm will find it.a

J. BettsaThe proof is left to the student.


M. ZanonandS.Gros



Optimization is






Why a full course?



Warning!




M. ZanonandS.Gros



Optimization is






Why a full course?



Warning!




M. ZanonandS.Gros



Optimization is






Why a full course?



Warning!




M. ZanonandS.Gros



Optimization is






Why a full course?



Warning!




M. ZanonandS.Gros

Organization

Goals

build a sound understanding of the fundamentals

learn the different approaches and how to use them effectively

theory is provided as a support to problem solving

Setup

10 lectures, mostly twice a week, mostly 10:30 - 12:30I Discussions are highly welcome (within reasonable limits)I There are no stupid questions, only bad explanationsI Slides will be put on the website, please do not print them

AssignmentsI some theoretical work, some programmingI collaboration is highly encouraged but please don’t copy (it’s in your own interest)I coding can be challenging, ask for advice if you are unsure about how to do thingsI deadline for the assignments: end of January

All information and material will be available on the website

https://mariozanon.wordpress.com/course-doc/


https://mariozanon.wordpress.com/course-doc/

M. ZanonandS.Gros

Notation

set of integers in the interval [a, b]: Iba = { x ∈ Z | a ≤ x ≤ b }scalar / vector a ∈ Rna unless specified otherwise

all vectors are column vectors unless specified otherwise

ai indicates component i of vector a

depending on the context, ak can indicate vector a at time k

∇xa> = ∂a

∂x∈ Rna×nx

F (x , u) continuous dynamics

f (x , u) discrete dynamics


M. ZanonandS.Gros

Optimal Control Problem (OCP)

Continuous time:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Discrete-time:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .

Components of an OCP

1 cost functional

2 initial condition

3 system dynamics

4 path constraintsI actuator limitationsI physical limitationsI user-defined

Remark1 ODE can be replaced by

I DAEI PDE

2 terminal constraint can be added3 initial condition can be replaced by

I periodic conditionsI some components can be free

4 independent variable does not need tobe time

I formulations in space


M. ZanonandS.Gros


Continuous time:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Discrete-time:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .


1 cost functional

2 initial condition

3 system dynamics



I DAEI PDE






M. ZanonandS.Gros


Continuous time:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Discrete-time:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .


1 cost functional

2 initial condition

3 system dynamics



I DAEI PDE






M. ZanonandS.Gros


Continuous time:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Discrete-time:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .


1 cost functional

2 initial condition

3 system dynamics



I DAEI PDE






M. ZanonandS.Gros

Direct Optimal Control - The Basic Idea

OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Transform the OCP into the NLP

minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it

The transformation from OCP to NLP is called transcription or discretization

Then why is it not trivial?

The choice of transcription strategy has a huge impact on both efficiency andreliability of the NLP solution...

... and the best solution strategy depends on the transcription

NLPs arising from OCPs have a specific structure which must be exploited

each problem is different and needs to be formulated, discretized and solved byaccurately choosing the best method

We will try to understand all of this together!


M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros


OCP:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].


minw

f (w)

s.t. g(w) = 0,

h(w) ≤ 0.

and solve it









M. ZanonandS.Gros

References

Optimal Control

L. Betts. Practical Methods for Optimal Control Using Nonlinear Programming,Advances in Design and Control 2010

L. Biegler. Nonlinear Programming, MOS-SIAM Series on Optimization 2010

NLPs

J. Nocedal and S. Wright. Numerical Optimization, Springer 2006

S. Boyd and L. Vandenberghe. Convex Optimization, University Press 2004

Integrators

J. C. Butcher Numerical Methods for Ordinary Differential Equations, Wiley 2016

E. Hairer, S. P. Nørsett, and G. Wanner. Solving Ordinary Differential EquationsI, Springer 1993

E. Hairer ,and G. Wanner. Solving Ordinary Differential Equations II, Springer1996


M. ZanonandS.Gros

Survival Map of Direct Optimal Control

OCP Single Shooting

Collocation

Multiple Shooting

Inte

gra

tor

NLP

Interior-Point

SQP

Active-setQP solver

Interior-PointQP solver

Sen

siti

viti

es

We will have to jump around this tree......but we’ll come back to this map to keep in mind the big picture!


M. ZanonandS.Gros

Outline

1 Introduction




M. ZanonandS.Gros

Overview - 1

Continuous Equations Discrete Equations

Global Hamilton-Jacobi-Bellman (HJB) Dynamic Programming (DP)

Local Pontryagin (PMP) Direct Optimal Control (DOC)


M. ZanonandS.Gros

Overview - 1





M. ZanonandS.Gros

Overview - 1




Bellman Principle of Optimality

Thm: Tails of optimal trajectories are optimal.

Proof: Optimal solution

x∗(t) =

{x1(t) t0 ≤ t < t1

x2(t) t1 ≤ t < tf.

Assume by absurd that

x3(t), t1 ≤ t < tf , x3(t1) = x2(t1),

has a lower cost than x2(t).Then x∗(t) can not be optimal.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

t

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

x1x2x3


M. ZanonandS.Gros

Overview - 2





M. ZanonandS.Gros

Overview - 2




Tails of optimal trajectories are optimal.

DP: find the cost-to-go v0(x)

vj(xj) = minu

m (xN) +N−1∑k=j

`(xk , uk)

s.t. xj = xj ,

xk+1 = f (xk , uk), k ∈ IN−1j ,

h(xk , uk) ≤ 0, k ∈ IN−1j .

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

t

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

Global optimum

Feedback solution (closed-loop)

Curse of dimensionality

Can be deployed with 2− 4 states/ controls

Mixed-integer “easily handled”


M. ZanonandS.Gros

Overview - 2






vj(xj) = minu

m (xN) +N−1∑k=j

`(xk , uk)

s.t. xj = xj ,

xk+1 = f (xk , uk), k ∈ IN−1j ,

h(xk , uk) ≤ 0, k ∈ IN−1j .

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

t

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(xj) = minu

m (xN) +N−1∑k=j

`(xk , uk)

s.t. xj = xj ,

xk+1 = f (xk , uk), k ∈ IN−1j ,

h(xk , uk) ≤ 0, k ∈ IN−1j .

vj+1(xj+1) = minu

m (xN) +N−1∑k=j+1

`(xk , uk)

s.t. xj+1 = xj+1,

xk+1 = f (xk , uk), k ∈ IN−1j+1 ,

h(xk , uk) ≤ 0, k ∈ IN−1j+1 .

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

t

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(xj) = minu

m (xN) +N−1∑k=j

`(xk , uk)

s.t. xj = xj ,

xk+1 = f (xk , uk), k ∈ IN−1j ,

h(xk , uk) ≤ 0, k ∈ IN−1j .

vj+1(xj+1) = minu

m (xN) +N−1∑k=j+1

`(xk , uk)

s.t. xj+1 = xj+1,

xk+1 = f (xk , uk), k ∈ IN−1j+1 ,

h(xk , uk) ≤ 0, k ∈ IN−1j+1 .

vj(xj) = minu

vj+1 (xj+1) + `(xj , uj)

s.t. xj = xj ,

xj+1 = f (xj , uj),

h(xj , uj) ≤ 0.

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(xj) = minuj

vj+1 (f (xj , uj)) + `(xj , uj)

+Ih(xj , uj),

with vN(x) = m(x) and

Ih(x , u) =

{0 if h(x , u) ≤ 0

∞ otherwise

the indicator function.

vj(xj) = minu

vj+1 (xj+1) + `(xj , uj)

s.t. xj = xj ,

xj+1 = f (xj , uj),

h(xj , uj) ≤ 0.

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).

Then the optimal control is

u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

DPLQR

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

DPLQR

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 2






vj(x) = minu

vj+1 (f (x , u)) + `(x , u)

+Ih(x , u).


u∗(x) = arg minu

v0(x).

f (x, u) = 1.1x + 0.1u, `(x, u) = x2 + u2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

DPLQR

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

0

1

2

3

Global optimum






M. ZanonandS.Gros

Overview - 3





HJB: find the cost-to-go V (x , t) by solving the Partial Differential Equation

−∂V (x , t)

∂t= min

u

{L(x , u) + IH(x , u) +

∂V (x , t)

∂xF (x , u)

}, V (x , tf) = M(x).

minu(·)

M(x(tf )

)+

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Global optimum

Feedback solution(closed-loop)

Difficult to solve



M. ZanonandS.Gros

Overview - 3






−∂V (x , t)

∂t= min

u

{L(x , u) + IH(x , u) +

∂V (x , t)

∂xF (x , u)

}, V (x , tf) = M(x).

minu(·)

M(x(tf )

)+

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Global optimum


Difficult to solve



M. ZanonandS.Gros

Overview - 3






−∂V (x , t)

∂t= min

u

{L(x , u) + IH(x , u) +

∂V (x , t)

∂xF (x , u)

}, V (x , tf) = M(x).

Interpretation:

vj (x) = minu`(x, u) + Ih(x, u) + vj+1 (f (x, u)) , vN (x) = m(x),

take the limit:

limδ→0

vj+1(x(t + δ))− vj (x(t))

δ=∂V (x, t)

∂t+∂V (x, t)

∂xF (x, u)

limδ→0

`δ(x(t), u(t))

δ= limδ→0

∫ t+δt L(x(τ), u(τ)) dτ

δ= L(x(t), u(t))

minu(·)

M(x(tf )

)+

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Global optimum


Difficult to solve



M. ZanonandS.Gros

Overview - 3






−∂V (x , t)

∂t= min

u

{L(x , u) + IH(x , u) +

∂V (x , t)

∂xF (x , u)

}, V (x , tf) = M(x).

Interpretation:

vj (x) = minu`(x, u) + Ih(x, u) + vj+1 (f (x, u)) , vN (x) = m(x),

take the limit:

limδ→0

vj+1(x(t + δ))− vj (x(t))

δ=∂V (x, t)

∂t+∂V (x, t)

∂xF (x, u)

limδ→0

`δ(x(t), u(t))

δ= limδ→0

∫ t+δt L(x(τ), u(τ)) dτ

δ= L(x(t), u(t))

minu(·)

M(x(tf )

)+

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Global optimum


Difficult to solve



M. ZanonandS.Gros

Overview - 4




Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Eliminate the inputs

Single trajectory (open loop)

Local optimum

State constraints not easily handled

Two-point BVP problem

PMP: Solve 1st-order necessary conditions

Input : u∗ = u∗ (x , λ, u) = argminu∈UH (x , λ, u, µ) , U := { u |H(x , u) ≤ 0 }

States : x = F (x , u∗) , x(t0) = x0,

Costates : λ = −∂H (x , λ, u∗, µ)

∂x, λ(tf) =

∂M

∂x(x (tf)) ,

Algebraic cond. : 0 = µ>H (x , u∗) , µ ≥ 0,

where we rely on the Hamiltonian function

H(x , λ, u, µ) = L (x , u) + λ>F (x , u) + µ>H (x , u) .


M. ZanonandS.Gros

Overview - 4




Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].



Local optimum





States : x = F (x , u∗) , x(t0) = x0,


∂x, λ(tf) =

∂M

∂x(x (tf)) ,





M. ZanonandS.Gros

Overview - 4




Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].



Local optimum





States : x = F (x , u∗) , x(t0) = x0,


∂x, λ(tf) =

∂M

∂x(x (tf)) ,





M. ZanonandS.Gros

HJB and Pontryagin (no path constraints)HJB:

−∂V (x , t)

∂t= min

u

{L(x , u) +

∂V (x , t)

∂xF (x , u)

}, V (x , tf) = M(x).

PMP:

H(x , λ, u, µ) = L (x , u) + λ>F (x , u)

u∗ = argminuH (x , λ, u, µ) = argmin

u

{L (x , u) + λ>F (x , u)

}x = F (x , u∗) x(t0) = x0

λ = −∇xH (x , λ, u∗, µ) λ(tf) = ∇xM (x (tf))

Observe: λ(tf) = ∇xV (x(tf), tf). Use the costate equation to obtain

λ = −∇xH (x , λ, u∗, µ) = ∇x

(minu

{L(x , u) + λ(t)>F (x , u)

})= ∇x

∂V (x , t)

∂talong the optimal trajectory

Then

λ(t) = ∇xV (x , t)


M. ZanonandS.Gros


−∂V (x , t)

∂t= min

u

{L(x , u) +

∂V (x , t)

∂xF (x , u)

}, V (x , tf) = M(x).

PMP:

H(x , λ, u, µ) = L (x , u) + λ>F (x , u)


u

{L (x , u) + λ>F (x , u)

}x = F (x , u∗) x(t0) = x0



λ = −∇xH (x , λ, u∗, µ) = ∇x

(minu

{L(x , u) + λ(t)>F (x , u)

})= ∇x

∂V (x , t)


Then

λ(t) = ∇xV (x , t)


M. ZanonandS.Gros


−∂V (x , t)

∂t= min

u

{L(x , u) +

∂V (x , t)

∂xF (x , u)

}, V (x , tf) = M(x).

PMP:

H(x , λ, u, µ) = L (x , u) + λ>F (x , u)


u

{L (x , u) + λ>F (x , u)

}x = F (x , u∗) x(t0) = x0



λ = −∇xH (x , λ, u∗, µ) = ∇x

(minu

{L(x , u) + λ(t)>F (x , u)

})= ∇x

∂V (x , t)


Then

λ(t) = ∇xV (x , t)


M. ZanonandS.Gros

Overview - 5




Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .

Discretize Continuous Problem:

1 Parametrize u(·), e.g.

u(t) = uk , t ∈ [kts, (k + 1)ts]

2 Integrate

f (xk , uk ) = x(tk+1) with

{x(t) = F (x(t), u(t))

x(tk ) = xk

`(xk , uk ) =

∫ tk+1

tk

L(x(t), u(t)) dt

3 Relax path constraints

h(xk , uk ) =

H(x(tk,0, u(tk,0)))

...

H(x(tk,n, u(tk,n)))

tk,0, . . . , tk,n ∈ [tk , tk+1]

typically: n = 0


M. ZanonandS.Gros

Overview - 5




Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .

Discretize Continuous Problem:

1 Parametrize u(·), e.g.

u(t) = uk , t ∈ [kts, (k + 1)ts]

2 Integrate

f (xk , uk ) = x(tk+1) with

{x(t) = F (x(t), u(t))

x(tk ) = xk

`(xk , uk ) =

∫ tk+1

tk

L(x(t), u(t)) dt

3 Relax path constraints

h(xk , uk ) =

H(x(tk,0, u(tk,0)))

...

H(x(tk,n, u(tk,n)))

tk,0, . . . , tk,n ∈ [tk , tk+1]

typically: n = 0


M. ZanonandS.Gros

Overview - 5




Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ],

H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .

Define the Lagrangian:

L = λ(x0 − x0) + m(xN ) +

N−1∑k=0

Lk ,

Lk = `(xk , uk ) + λ>k (f (xk , uk )− xk+1)

+ µ>k (h(xk , uk ))

Write the KKT conditions:

0 = ∇ukL, k ∈ IN−1

0

xk+1 = f (xk , uk ), k ∈ IN−10

0 = ∇xkL, k ∈ IN−1

0

0 ≥ h(xk , uk ), k ∈ IN−10

0 ≤ µk , k ∈ IN−10

0 = hk,iµk,i , k ∈ IN−10 , i ∈ Inh0 .


M. ZanonandS.Gros

Overview - 5




Easy to formulate

Very large and difficult problems

Many solvers available

Local optimum

Mixed-Integer not as easy as in DP

Discrete problem:

minu

m (xN) +N−1∑k=0

`(xk , uk)

s.t. x0 = x0,

xk+1 = f (xk , uk), k ∈ IN−10 ,

h(xk , uk) ≤ 0, k ∈ IN−10 .

Define the Lagrangian:

L = λ(x0 − x0) + m(xN ) +

N−1∑k=0

Lk ,


+ µ>k (h(xk , uk ))

Write the KKT conditions:

0 = ∇ukL, k ∈ IN−1

0

xk+1 = f (xk , uk ), k ∈ IN−10

0 = ∇xkL, k ∈ IN−1

0

0 ≥ h(xk , uk ), k ∈ IN−10

0 ≤ µk , k ∈ IN−10

0 = hk,iµk,i , k ∈ IN−10 , i ∈ Inh0 .


M. ZanonandS.Gros

PMP and DOC (no path constraints)

Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)

Euler: δ = tk+1 − tk

(1b)⇒ xk+1 = xk + δF (xk , uk ) := f (xk , uk )

(1c)⇒(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0

0 = ∇xkL λN = ∇xNm (xN)


L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP

Specific choice of integration

Considering discrete OC ≈ continuous OC: Multiple Shooting


M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ λk−1 = λk + δ∇xkH (xk , λk , uk )

(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ λk−1 = λk + δ (∇xk L(xk , uk ) +∇xkF (xk , uk )λk )

(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = δ∇xk L(xk , uk ) + (δ∇xkF (xk , uk ) + I )λk − λk−1

(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = δ∇xk L(xk , uk ) +∇xk (δF (xk , uk ) + xk )λk − λk−1

(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xk `(xk , uk ) +∇xk f (xk , uk )λk −∇xk xkλk−1

(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xkL

(1a)⇒

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xkL(1a)⇒ 0 = ∇ukH (xk , λk , uk ) = δ∇ukH (xk , λk , uk )

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xkL(1a)⇒ 0 = δ∇uk L(xk , uk ) + δ∇ukF (xk , uk )λk

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xkL(1a)⇒ 0 = ∇uk `(xk , uk ) +∇uk f (xk , uk )λk

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xkL(1a)⇒ 0 = ∇ukL

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xkL(1a)⇒ 0 = ∇ukL

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros


Continuous problem:

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0,

x(t) = F (x(t), u(t)), t ∈ [t0, tf ].

Discrete problem:

minu

m (xN) +

N−1∑k=0

`(xk , uk )

s.t. x0 = x0,

xk+1 = f (xk , uk ), k ∈ IN−10 .

PMP:

0 = ∇uH (1a)

x = F (x , u) x(t0) = x0 (1b)

−λ = ∇xH λ(tf) = ∇xM (x (tf)) (1c)

H = L (x , u) + λ>F (x , u)



(1c)⇒ 0 = ∇xkL(1a)⇒ 0 = ∇ukL

KKT:

0 = ∇ukLxk+1 = f (xk , uk ) x0 = x0



L = λ(x0 − x0) + m(xN) +

N−1∑k=0

Lk

DOC ≈ PMP




M. ZanonandS.Gros

Overview - 6




Global methods:


Limited applicability

Good for analysis

Local methods:

Very powerful

Applicable to wider range of problems

Used for practical problems

DOC

straightforward to apply (compared to the other methods)

solvers available

the most flexible


M. ZanonandS.Gros

Outline

1 Introduction




M. ZanonandS.Gros

Pontryagin Maximum Principle (PMP) - an example

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0

x(t) = F (x(t), u(t)), t ∈ [t0, tf ]

u∗ = argminuH (x , λ, u)

x = F (x , u∗)

x(t0) = x0

λ = −∇xH (x , λ, u∗)

λ(tf) = ∇xM (x (tf))

Is Newton flawed?

Newton can be improved, we willsee how

Is there something else tounderstand?

There is also a more fundamentalproblem


M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0

x(t) = F (x(t), u(t)), t ∈ [t0, tf ]


x = F (x , u∗)

x(t0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗

2 Guess λ(t0) = λ0

3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0

4 Adjust λ0 to enforce


5 If not converged, go to 2

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0

x(t) = F (x(t), u(t)), t ∈ [t0, tf ]


x = F (x , u∗)

x(t0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]


x = F (x , u∗)

x(t0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))


x = F (x , u∗)

x(t0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

∇uH = u + λ


x = F (x , u∗)

x(t0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = F (x , u∗)

x(t0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

∇xH = x − λ cos(x)

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = −∇xH (x , λ, u∗)


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x


1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




How to adjust λ0?

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




How to adjust λ0?

Newton:

λf = λ (tf , λ0 + ∆λ0)

≈ λ(tf , λ0) +∂λ(tf , λ0 + ∆λ0)

∂λ0∆λ0

then

∆λ0 =−(∂λ(tf , λ0 + ∆λ0)

∂λ0

)−1

(λf−λ(tf , λ0))

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0 0.5 1 1.5 2 2.5 3 3.5 4-20

-15

-10

-5

0

5

0 0.5 1 1.5 2 2.5 3 3.5 40

5

10

15

20

25

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0 0.5 1 1.5 2 2.5 3 3.5 4-5

-4

-3

-2

-1

0

1

0 0.5 1 1.5 2 2.5 3 3.5 40

1

2

3

4

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4-2

-1.5

-1

-0.5

0

0.5

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0 0.5 1 1.5 2 2.5 3 3.5 4-0.2

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4-0.1

0

0.1

0.2

0.3

0.4

0.5

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4-0.1

0

0.1

0.2

0.3

0.4

0.5

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 40

1

2

3

4

0 0.5 1 1.5 2 2.5 3 3.5 4-4

-3

-2

-1

0

1

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 4-10

-8

-6

-4

-2

0

2

0 0.5 1 1.5 2 2.5 3 3.5 40

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 40

100

200

300

400

500

0 0.5 1 1.5 2 2.5 3 3.5 4-500

-400

-300

-200

-100

0

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 40

50

100

150

200

250

0 0.5 1 1.5 2 2.5 3 3.5 4-250

-200

-150

-100

-50

0

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 4-25

-20

-15

-10

-5

0

5

0 0.5 1 1.5 2 2.5 3 3.5 40

5

10

15

20

25

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 4-15

-10

-5

0

5

0 0.5 1 1.5 2 2.5 3 3.5 40

2

4

6

8

10

12

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 4-4

-3

-2

-1

0

1

0 0.5 1 1.5 2 2.5 3 3.5 40

1

2

3

4

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 40

2

4

6

8

0 0.5 1 1.5 2 2.5 3 3.5 4-8

-6

-4

-2

0

2

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3 3.5 4-4

-3

-2

-1

0

1

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0 0.5 1 1.5 2 2.5 3 3.5 4-60

-40

-20

0

20

0 0.5 1 1.5 2 2.5 3 3.5 40

10

20

30

40

50

60

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0



5 If not converged, go to 2 0.2 0.3 0.4 0.5 0.6 0.7 0.8-10

-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




-8

-6

-4

-2

0

2

4

6

8

10

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Guess: λ0 = 0.4

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2-30

-20

-10

0

10

20

30

Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?





M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?Newton can be improved, we willsee how




M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1

x = u − sin(x), t ∈ [0, 4]

H (x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

u∗ = − λx = − λ− sin(x)

x(0) = x0

λ = λ cos(x)− x

λ(4) = 0

1 Solve for u∗


3 Solve IVP

x = F (x , u∗)

λ = −∇xH (x , λ, u∗, µ)

with λ(t0) = λ0, x(t0) = x0




Is Newton flawed?Newton can be improved, we willsee how

Is there something else tounderstand?There is also a more fundamentalproblem


M. ZanonandS.Gros

Pontryagin Maximum Principle (PMP) - numerical difficulties

Dynamics of states x and costates λ cannot be both stable!

Consider a domain D(t) in the x(t), λ(t) space

Define the volume:

V (D(t)) =

∫D(t)

dD and its time derivatived

dtV (D(t)) =

∫D(t)

div

([x

λ

])dD

Observe:

div

([x

λ

])= Tr

(∂x

∂x+∂λ

∂λ

)= Tr

(∂F

∂x− ∂

∂λ

∂H∂x

)

= 0

and

∂

∂λ

∂H∂x

=∂

∂λ

(∂L

∂x+∂λ>F

∂x

)=∂F

∂x

Then,

d

dtV (D(t)) = 0 ⇒ V (D(t)) = const.


M. ZanonandS.Gros



Consider a domain D(t) in the x(t), λ(t) spaceDefine the volume:

V (D(t)) =

∫D(t)


dtV (D(t)) =

∫D(t)

div

([x

λ

])dD

Observe:

div

([x

λ

])= Tr

(∂x

∂x+∂λ

∂λ

)= Tr

(∂F

∂x− ∂

∂λ

∂H∂x

)

= 0

and

∂

∂λ

∂H∂x

=∂

∂λ

(∂L

∂x+∂λ>F

∂x

)=∂F

∂x

Then,

d

dtV (D(t)) = 0 ⇒ V (D(t)) = const.


M. ZanonandS.Gros




V (D(t)) =

∫D(t)


dtV (D(t)) =

∫D(t)

div

([x

λ

])dD

Observe:

div

([x

λ

])= Tr

(∂x

∂x+∂λ

∂λ

)= Tr

(∂F

∂x− ∂

∂λ

∂H∂x

)

= 0

and

∂

∂λ

∂H∂x

=∂

∂λ

(∂L

∂x+∂λ>F

∂x

)=∂F

∂x

Then,

d

dtV (D(t)) = 0 ⇒ V (D(t)) = const.


M. ZanonandS.Gros




V (D(t)) =

∫D(t)


dtV (D(t)) =

∫D(t)

div

([x

λ

])dD

Observe:

div

([x

λ

])= Tr

(∂x

∂x+∂λ

∂λ

)= Tr

(∂F

∂x− ∂

∂λ

∂H∂x

)

= 0

and

∂

∂λ

∂H∂x

=∂

∂λ

(∂L

∂x+∂λ>F

∂x

)=∂F

∂x

Then,

d

dtV (D(t)) = 0 ⇒ V (D(t)) = const.


M. ZanonandS.Gros




V (D(t)) =

∫D(t)


dtV (D(t)) =

∫D(t)

div

([x

λ

])dD

Observe:

div

([x

λ

])= Tr

(∂x

∂x+∂λ

∂λ

)= Tr

(∂F

∂x− ∂

∂λ

∂H∂x

)= 0

and

∂

∂λ

∂H∂x

=∂

∂λ

(∂L

∂x+∂λ>F

∂x

)=∂F

∂x

Then,

d

dtV (D(t)) = 0 ⇒ V (D(t)) = const.


M. ZanonandS.Gros




V (D(t)) =

∫D(t)


dtV (D(t)) =

∫D(t)

div

([x

λ

])dD

Observe:

div

([x

λ

])= Tr

(∂x

∂x+∂λ

∂λ

)= Tr

(∂F

∂x− ∂

∂λ

∂H∂x

)= 0

and

∂

∂λ

∂H∂x

=∂

∂λ

(∂L

∂x+∂λ>F

∂x

)=∂F

∂x

Then,

d

dtV (D(t)) = 0 ⇒ V (D(t)) = const.


M. ZanonandS.Gros

Simulation with the set

Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

0

2

4 -4-2

02

4

-4

-2

0

2

4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-40

-2

4

0

2 2

2

0

4

-24 -4


M. ZanonandS.Gros


Problem:

minu(·)

1

2

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

Hamiltonian

H(x , λ, u) =1

2

(x2 + u2

)+ λ(u − sin(x))

minimized by

u = −λ

Dynamics become

x = −λ− sin(x)

λ = λ cos(x)− x

-20

0

0

2020

20

-204


M. ZanonandS.Gros

Direct Optimal Control

minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0

x(t) = F (x(t), u(t)), t ∈ [t0, tf ]H(x(t), u(t)) ≤ 0, t ∈ [t0, tf ]

Choose a time grid t0, . . . , tN

Discretize the input w = (u0, . . . , uN−1)

u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10

Get corresponding state trajectory

x(t) = x(t,w , x0)

and the corresponding cost

φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)

for α sufficiently small

Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros


minu(·)

M (x(tf)) +

∫ tf

t0

L(x(t), u(t)) dt

s.t. x(t0) = x0




u(t) = uk , t ∈ [tk , tk+1], k ∈ IN−10


x(t) = x(t,w , x0)


φ(w)

Solve

minwφ(w)

How?

Gradient steps

w ← w − α∇φ(w)


Newton steps

w ← w − α∇2φ(w)−1∇φ(w)



M. ZanonandS.Gros

Direct Optimal Control - an example

minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

What is happening??

This approach is called Single Shooting.We will unpack what is going on and fix it!


M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 0

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 1

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 2

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 3

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 4

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 5

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 6

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 7

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 8

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 9

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 10

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 0

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 1

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 2

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 3

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 4

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 5

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 6

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 7

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 8

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 9

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4

0

0.5

1SQP iter: 10

0 1 2 3 4

-0.4

-0.2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u − sin(x), t ∈ [0, 4]

N = 20

α = 1

Great! Can we stop here? One more example...

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

Great! Can we stop here? One more example...

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

2

3

4SQP iter: 0

0 1 2 3 4-1

0

1

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-4

-2

0

2SQP iter: 1

0 1 2 3 4-2

-1

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 40

2

4

6SQP iter: 2

0 1 2 3 40

2

4

6

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-4

-2

0

2SQP iter: 3

0 1 2 3 4-2

-1

0

1

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 40

2

4

6SQP iter: 4

0 1 2 3 40

2

4

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-4

-2

0

2SQP iter: 5

0 1 2 3 4-3

-2

-1

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

2

3

4SQP iter: 6

0 1 2 3 40

2

4

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

1.5

2

2.5SQP iter: 7

0 1 2 3 4-2

-1

0

1

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-5

0

5SQP iter: 8

0 1 2 3 4-3

-2

-1

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

2

3

4SQP iter: 9

0 1 2 3 40

2

4

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

1.5

2

2.5SQP iter: 10

0 1 2 3 4-2

-1

0

1

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-5

0

5SQP iter: 11

0 1 2 3 4-3

-2

-1

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

2

3

4SQP iter: 12

0 1 2 3 40

1

2

3

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

1.5

2

2.5SQP iter: 13

0 1 2 3 4-2

-1

0

1

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-5

0

5SQP iter: 14

0 1 2 3 4-3

-2

-1

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 40

2

4

6SQP iter: 15

0 1 2 3 40

1

2

3

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 41

1.5

2

2.5SQP iter: 16

0 1 2 3 4-2

-1

0

1

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-5

0

5SQP iter: 17

0 1 2 3 4-3

-2

-1

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 40

2

4

6SQP iter: 18

0 1 2 3 40

1

2

3

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 4-10

-5

0

5SQP iter: 19

0 1 2 3 4-4

-2

0

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

0 1 2 3 40

10

20

30SQP iter: 20

0 1 2 3 40

5

10

15

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

What is happening??



M. ZanonandS.Gros


minu(·)

∫ 4

0

x2 + u2 dt

s.t. x(0) = 1x = u + sin(x), t ∈ [0, 4]

N = 20

α = 1

What is happening??



Documents

Mario Zanon & S ebastien Gros · 01-03-2019 · f]: Transform the OCP into the NLP min w f(w) s:t: g(w) = 0; h(w) 0: and solve it The transformation from OCP to NLP is called transcription