Harry G. Kwatny - Information Technologyhgk22/courses/MEM733/OptimalControl_1.pdf · I The theory...

INTRODUCTION THE OPTIMAL CONTROL PROBLEM OPTIMIZATION BASICS VARIATIONAL CALCULUS

INTRODUCTION TO OPTIMAL CONTROL

SYSTEMS

Harry G. Kwatny

Department of Mechanical Engineering & MechanicsDrexel University

OPTIMAL CONTROL


OUTLINEINTRODUCTION

PreliminariesBibliography

THE OPTIMAL CONTROL PROBLEMDefinitionExamples

OPTIMIZATION BASICSLocal ExtremaConstraints

VARIATIONAL CALCULUSClassical Variational CalculusFree Terminal TimeSystems with Constraints

OPTIMAL CONTROL


PRELIMINARIES

WHAT IS OPTIMAL CONTROL?

I Optimal control is an approach to control systems design thatseeks the best possible control with respect to a performancemetric.

I The theory of optimal control began to develop in the WW IIyears. The main result of this period was the Wiener-Kolmogorovtheory that addresses linear SISO systems with Gaussian noise.

I A more general theory began to emerge in the 1950’s and 60’sI In 1957 Bellman published his book on Dynamic ProgrammingI In 1960 Kalman published his multivariable generalization of

Wiener-KolmogorovI In 1962 Pontryagin et al published the maximal principleI In 1965 Isaacs published his book on differential games

OPTIMAL CONTROL


PRELIMINARIES

COURSE CONTENT1. Introduction

I Optimization basicsI Intro to Variational Calculus

2. Variational Calculus and the Minimum PrincipleI Unconstrained Control problemsI Control and State Constraints

3. Dynamic programmingI Principle of OptimalityI The Hamilton-Jacobi-Bellman Equation

4. Min-Max Optimal ControlI Min-Max ControlI Game Theory

5. Hybrid SystemsI Hybrid Systems BasicsI Hybrid Systems Optimal Control

OPTIMAL CONTROL


BIBLIOGRAPHY

BIBLIOGRAPHY

I E. B. Lee and L. Markus, Foundations of Optimal ControlTheory. New York: John Wiley and Sons, Inc., 1967.

I A. E. Bryson and Y.-C. Ho, Applied Optimal Control.Waltham: Blaisdell, 1969.

I S. J. Citron, Elements of Optimal Control. New York: Holt,Rinehart and Winston, Inc., 1969.

I D. E. Kirk, Optimal Control Theory: An Introduction.Englewood Cliffs, NJ: Prentice-Hall, 1970. (now availablefrom Dover)

OPTIMAL CONTROL


DEFINITION

PROBLEM DEFINITION

We will define the basic optimal control problem:I Given system dynamics

x = f (x, u) , x ∈ X ⊂ Rn, u ∈ U ⊂ Rm

I Find a control u (t) , t ∈ [0,T] that steers the system froman initial state x (0) = x0 to a target set G and minimizes thecost

J (u (·)) = gT (x (T)) +

∫ T

0g (x (t) , u (t)) dt

REMARKgT is called the terminal cost and g is the running cost. The terminaltime T can be fixed or free. The target set can be fixed or moving.

OPTIMAL CONTROL


DEFINITION

OPEN LOOP VS. CLOSED LOOP

I If we are concerned with a single specified initial state x0, thenwe might seek the optimal control control u (t), u : R→ Rm thatsteers the system from the initial state to the target. This is anopen loop control.

I On the other hand, we might seek the optimal control as afunction of the state u (x), u : Rn → Rm. This is a closed loopcontrol; sometimes called a synthesis.

I The open loop control is sometimes easier to compute, and thecomputations are sometimes performed online – a methodknown as model predictive control.

I The closed loop control has the important advantage that it isrobust with respect to model uncertainty, and that once the(sometimes difficult) computations are performed off-line, thecontrol is easily implemented online.

OPTIMAL CONTROL


EXAMPLES

EXAMPLE – A MINIMUM TIME PROBLEM

Consider steering a unit mass, with bounded applied controlforce, from an arbitrary initial position and velocity to rest at theorigin in minimum time. Specifically,

x = vv = u, |u| ≤ 1

The cost function is

J =

∫ T

0dt ≡ T

REMARKThis is an example of a problem with a control constraint.

OPTIMAL CONTROL


EXAMPLES

EXAMPLE – A MINIMUM FUEL PROBLEMConsider the decent of a moon lander.

h = vv = −g + u

mm = −ku

The thrust u is used to steer the system to h = 0, v = 0. Inaddition we wish to minimize the fuel used during landing, i.e.

J =

∫ t

0kudt

Furthermore, u is constrained, 0 ≤ u ≤ c, and the stateconstraint h ≥ 0 must be respected.

REMARKThis problem has both control and state constraints.

OPTIMAL CONTROL


EXAMPLES

EXAMPLE – A LINEAR REGULATOR PROBLEM

Consider a system with linear dynamics

x = Ax + Bu

We seek a feedback control that steers the system from anarbitrary initial state x0 towards the origin in such a way as tominimize the cost

J = xT (T) QTx (T) +1

2T

∫ T

0

{xT (t) Qx (t) + uT (t) Ru (t)

}dt

The final time T is considered fixed.

OPTIMAL CONTROL


EXAMPLES

EXAMPLE – A ROBUST SERVO PROBLEMConsider a system with dynamics

x = Ax + B1w + B2uz = C1x + D11w + D12uy = C2x + D21w + D22u

where w is an external disturbance. The goal is to find an output (y)feedback synthesis such that the performance variables (processerrors) z remain close to zero. Note that w (t) can be characterized inseveral ways, stochastic (the H2 problem)

J = E[∫ ∞−∞

zT (t) z (t) dt]

or deterministic (the H∞ problem)

J = max‖w‖2=1

∫ ∞−∞

zT(t)z(t)dt

OPTIMAL CONTROL


LOCAL EXTREMA

DEFINITIONS

Consider the scalar function

f (x) , x ∈ Rn

which is defined and smooth on a domain D ⊂ Rn. We furtherassume that the region D is defined by a scalar inequalityψ (x) ≤ 0, i.e.,

intD = {x ∈ Rn |ψ (x) < 0}∂D = {x ∈ Rn |ψ (x) = 0}

OPTIMAL CONTROL


LOCAL EXTREMA

DEFINITIONS

DEFINITION (LOCAL MINIMA & MAXIMA)An interior point x∗ ∈ intD is a local minimum if there exists aneighborhood U of x∗ such that

f (x) ≥ f (x∗) ∀x ∈ U

It is a local maximum if

f (x) ≤ f (x∗) ∀x ∈ U

Similarly, for a point x∗ ∈ ∂D, we use a neighborhood U of x∗

within ∂D. With this modification boundary local minima andmaxima are defined as above.

OPTIMAL CONTROL


LOCAL EXTREMA

OPTIMAL INTERIOR POINTS

I Necessary conditions. A point x∗ ∈ intD is an extremalpoint (minimum or maximum) only if

∂f∂x

(x∗) = 0

I Sufficient conditions. x∗ is a minimum if

∂2f∂x2 (x∗) > 0

a maximum if∂2f∂x2 (x∗) < 0

OPTIMAL CONTROL


CONSTRAINTS

OPTIMIZATION WITH CONSTRAINTS – NECESSARY

CONDITIONSWe need to find extremal points of f (x), with x ∈ ∂D. i.e., findextremal points of f (x) subject to the constraint ψ (x) = 0.

Consider a more general problem where there are m constraints, i.e.,ψ : Rn → Rm.

Let λ be an m-dimensional constant vector (called Lagrangemultipliers) and define the function

H (x, λ)∆= f (x) + λTψ (x)

Then x∗ is an extremal point only if

∂H (x∗, λ)

∂x= 0,

∂H (x∗, λ)

∂λ≡ ψ (x) = 0

Note there are n + m equations in n + m unknowns x, λ

OPTIMAL CONTROL


CONSTRAINTS

SIGNIFICANCE OF THE LAGRANGE MULTIPLIERConsider extremal points of f (x1, x2) subject to the single constraintψ (x1, x2) = 0. At an extremal point (x∗1 , x

∗2 ) we must have

df (x∗1 , x∗2 ) =

∂f (x∗1 , x∗2 )

∂x1dx1 +

∂f (x∗1 , x∗2 )

∂x2dx2 = 0 (1)

but dx1 and dx2 are not independent. They satisfy

dψ (x∗1 , x∗2 ) =

∂ψ (x∗1 , x∗2 )

∂x1dx1 +

∂ψ (x∗1 , x∗2 )

∂x2dx2 = 0 (2)

From (1) and (2) it must be that∂f (x∗1 , x

∗2 )/∂x1

∂ψ (x∗1 , x∗2 )/∂x1

=∂f (x∗1 , x

∗2 )/∂x2

∂ψ (x∗1 , x∗2 )/∂x2

∆= −λ

Accordingly, (1) and (2) yield∂f (x∗1 , x

∗2 )

∂x1+ λ

∂ψ (x∗1 , x∗2 )

∂x1= 0,

∂f (x∗1 , x∗2 )

∂x2+ λ

∂ψ (x∗1 , x∗2 )

∂x2= 0

REMARKNote that these are the form of the previous slide.

OPTIMAL CONTROL


CONSTRAINTS

OPTIMIZATION WITH CONSTRAINTS – SUFFICIENT

CONDITIONS

df =∂H∂x

dx +12

dxT ∂2H∂x2 dx− λTdψ + h.o.t.

butdψ =

∂ψ

∂xdx = 0⇒ dx ∈ ker

∂ψ

∂x⇒ dx = Ψdα

Ψ = span ker∂ψ (x)

∂xThus, for x∗ extremal

df (x∗) =12

dαTΨT ∂2H (x∗)∂x2 Ψdα+ h.o.t.

ΨT ∂2H (x∗)∂x2 Ψ > 0⇒ min

ΨT ∂2H (x∗)∂x2 Ψ < 0⇒ max

OPTIMAL CONTROL


CONSTRAINTS

EXAMPLE

f (x1, x2) = x1

(x2

1 + 2x22 − 1

), ψ (x1, x2) = x2

1 + x22 − 1

Interior:(x1, x2, f ) = (0,−0.707, 0) ∨ (0, 0.707, 0) ∨ (−0.577, 0, 0.385) ∨ (0.577, 0.− 0.385)

Boundary:

(x1, x2, λ, f ) = (−0.577,−0.8165, 1.155,−0.385) ∨ (−0.577, 0.8165, 1.155,−0.385)∨ (0.577,−0.8165,−1.155, 0.385) ∨ (0.577, 0.8165,−1.155, 0.385)∨ (−1, 0, 1, 0) ∨ (1, 0,−1, 0)

OPTIMAL CONTROL


CLASSICAL VARIATIONAL CALCULUS

OPTIMIZING A TIME TRAJECTORY

I We are interested in steering a controllable system along atrajectory that is optimal in some sense.

I Three methods are commonly used to address suchproblems:

I The ‘calculus of variations’I The Pontryagin ‘maximal Principle’I The ‘principle of optimality’ and dynamic programming

I The calculus of variations was first invented to characterizethe dynamical behavior of physical systems governed by aconservation law.

OPTIMAL CONTROL



CALCULUS OF VARIATIONS: LAGRANGIAN SYSTEMSA Lagrangian System is characterized as follows:

I The system is define in terms of a vector of configurationcoordinates, q, associated with velocities q.

I The system has kinetic energy T (q, q) = qTM (q) q/2, andpotential energy V (q) from which we define the Lagrangian

L (q, q) = T (q, q)− V (q)

I The system moves along a trajectory q (t), between initialand final times t1, t2 in such a way as to minimize theintegral

J (q (t)) =

∫ t2

t1L (q (t) , q (t)) dt

OPTIMAL CONTROL



EXAMPLES OF LAGRANGIAN SYSTEMS

T = 12 qTM (q) q

M (q) =(`2

1 (m1 + m2) + `22m2 + 2`1`2m2 cos θ2 `2m2 (`2 + `1 cos θ2)

`2m2 (`2 + `1 cos θ2) `22m2

)V (q) = m1g (g`1 (m1 + m2) sin θ1 + g`2m2 sin (θ1 + θ2))

m1

m2

�1

� 2

1θ

2θ

OPTIMAL CONTROL



NECESSARY CONDITIONS: FIXED TERMINAL TIME –1I A real-valued, continuously differentiable function q (t) on the

interval [t1, t2] will be called admissible.I Let q∗ (t) be an optimal admissible trajectory and q (t, ε) a not

necessarily optimal trajectory, with

q (t, ε) = q∗ (t) + εη (t)

where ε > 0 is a small parameter and η (t) is arbitrary.I Then

J (q (t, ε)) =

∫ t2

t1L (q∗ (t) + εη (t) , q∗ (t) + εη (t)) dt

I An extremal of J is obtained from

δJ (q (t)) =∂J (q (t, ε))

∂ε

∣∣∣∣ε=0

= 0

OPTIMAL CONTROL



FIXED TERMINAL TIME – 2

δJ (q (t)) =

∫ t2

t1

(∂L (q∗ (t) , q∗ (t))

∂q∗η (t) +

∂L (q∗ (t) , q∗ (t))q∗

η (t))

dt

We can apply the ‘integration by parts’ formula∫udv = uv−

∫vdu

to the first term to obtain

δJ (q (t)) =∫ t2

t1

(− d

dt∂L(q∗(t),q∗(t))

∂q∗ + ∂L(q∗(t),q∗(t))q∗

)η (t) dt

+ ∂L(q∗(t),q∗(t))∂q∗ η (t)

∣∣∣t2t1

OPTIMAL CONTROL



FIXED TERMINAL TIME – 3Now, set δJ (q (t)) = 0 to obtain:

I the Euler-Lagrange equations

ddt∂L (q∗ (t) , q∗ (t))

∂q∗− ∂L (q∗ (t) , q∗ (t))

q∗= 0

I the transversality conditions

∂L (q∗ (t1) , q∗ (t1))

∂q∗δq (t1) = 0,

∂L (q∗ (t2) , q∗ (t2))

∂q∗δq (t2) = 0

REMARKThese results allow us to treat problems in which the initial and terminal times are fixedand individual components of q(t1) and q(t2) are fixed or free. Other cases of interestinclude: 1) the terminal time is free, and 2) the terminal time is related to the terminalconfiguration, e.g., by a relation ϕ (q (t2) , t2) = 0.

OPTIMAL CONTROL


FREE TERMINAL TIME

NECESSARY CONDITIONS: FREE TERMINAL TIME – 1Consider the case of fixed initial state and free terminal time. Let q∗(t) be theoptimal trajectory with optimal terminal time t∗2 . The perturbed trajectoryterminates at time t∗2 + δt2. Its end state is q∗ (t∗2 ) + εη (t∗2 + ετ). Theperturbed cost is

J (q∗, δq, δt) =∫ t∗2 +δt

t1

L (q∗ (t) + δq (t) , q∗ (t) + δq (t)) dt

From this we obtain:

δJ =∫ t∗2

t1

(− d


∂q∗ +∂L(q∗(t),q∗(t))

q∗

)δq (t) dt − ∂L(q∗(t∗1 ),q

∗(t∗1 ))∂q∗ δq (t∗1 )

+∂L(q∗(t∗2 ),q

∗(t∗2 ))∂q∗ δq (t∗2 ) + L (q∗ (t∗2 + δt) , q∗ (t∗2 )) δt

Now, we want to allow both the final time and the end point to vary. Theactual end state is:

δq2∆= δq (t∗2 + δt) = δq (t∗2 ) + q∗ (t∗2 ) δt

OPTIMAL CONTROL


FREE TERMINAL TIME

FREE TERMINAL TIME – 2

δJ =∫ t∗2

t1

(− d


∂q∗ +∂L(q∗(t),q∗(t))

q∗

)δq (t) dt − ∂L(q∗(t∗1 ),q

∗(t∗1 ))∂q∗ δq (t∗1 )

+∂L(q∗(t∗2 ),q

∗(t∗2 ))∂q∗ δq (t∗2 ) +

(L (q∗ (t∗2 ) , q

∗ (t∗2 ))−∂L(q∗(t∗2 ),q

∗(t∗2 ))∂q∗ q∗ (t∗2 )

)δt

Thus, we have have the Euler-Lagrange Equations, as before, but thetransversality conditions become:

I the previous conditions:

∂L (q∗ (t1) , q∗ (t1))

∂q∗δq (t1) = 0,

∂L (q∗ (t∗2 ) , q∗ (t∗2 ))

∂q∗δq (t∗2 ) = 0

I plus, additional conditions:(L (q∗ (t∗2 ) , q

∗ (t∗2 ))−∂L (q∗ (t∗2 ) , q

∗ (t∗2 ))∂q∗

q∗ (t∗2 ))δt = 0

OPTIMAL CONTROL


FREE TERMINAL TIME

FREE TERMINAL TIME – 3

Ordinarily, the initial state and time are fixed so that the transversalityconditions require

δq (t1) = 0

For free terminal time δt is arbitrary. Thus from the transversality conditions:I If the terminal state is fixed, δq (t2) = 0 the terminal condition is:

L (q∗ (t∗2 ) , q∗ (t∗2 ))−

∂L (q∗ (t∗2 ) , q∗ (t∗2 ))

∂q∗q∗ (t∗2 ) = 0

I If the terminal state is free, δq2 arbitrary the terminal condition is:

∂L (q∗ (t∗2 ) , q∗ (t∗2 ))

∂q∗= 0, L (q∗ (t∗2 ) , q

∗ (t∗2 )) = 0

OPTIMAL CONTROL


FREE TERMINAL TIME

CONTROL EXAMPLE

First note that in the elementary variational calculus, we couldsimply replace q by x, and add the definition x = u, so that thecost function becomes

J (x (t)) =

∫ t2

t1L (u (t) , x (t)) dt

Hence, we have a simple control problem. We consider 3variants:

1. Free endpoint, fixed terminal time2. Fixed endpoint, fixed terminal time3. Fixed endpoint, free terminal time

OPTIMAL CONTROL


FREE TERMINAL TIME

EXAMPLE 1: FREE ENDPOINT, FIXED TERMINAL TIME

Suppose x ∈ R, t1 = 0, t2 = 1, and L =(x2 + u2

)/2:

J (x (t)) =

∫ 1

0

12(x2 + u2) dt

The Euler-Lagrange equations become:

ddt

(∂L∂u

)− ∂L∂x

= 0⇒ u− x = 0

The transversality conditions are:

x (0) = x0,∂L (x (1) , u (1))

∂u= 0

OPTIMAL CONTROL


FREE TERMINAL TIME

EXAMPLE 1, CONT’DThus, we have the equations[

xu

]=

[0 11 0

] [xu

]

x (0) = x0,∂L (x (1) , u (1))

∂u= 0⇒ u (1) = 0

Consequently,

[x (t)u (t)

]= e

0 11 0

t [ab

]⇒[

x (t)u (t)

]=

[a cosh (t) + b sinh (t)b cosh (t) + a sinh (t)

]

a = x0, b =x0 − e2x0

1 + e2

OPTIMAL CONTROL


FREE TERMINAL TIME

EXAMPLE 1: FIXED ENDPOINT, FIXED TERMINAL TIME

Suppose we consider a fixed end point, say x (t2) = 0 with theterminal time still fixed at t2 = 1. Then the Euler-Lagrangeequations remain the same, but the boundary conditionschange to:

x (0) = x0, x (1) = 0

Thus, we compute

a = x0, b = −x0 + e2x0

−1 + e2

OPTIMAL CONTROL


FREE TERMINAL TIME

EXAMPLE 1: FIXED ENDPOINT, FREE TERMINAL TIME

Suppose we consider a fixed end point, say x (t2) = xf with theterminal time t2 free. Then the Euler-Lagrange equationsremain the same, but the boundary conditions change to:

x (0) = x0, x (t2) = xf , (L− Luu)|t2 = 0

From this we compute

a = x0, a2 = b2, a cosh (t2) + b sinh (t2) = xf

OPTIMAL CONTROL


FREE TERMINAL TIME

EXAMPLE 1, CONT’D

The figures below show the optimal control and statetrajectories from the initial state x0 = 1,xf = 0, with t2 = 1 for thefixed time case. For the free time case xf = 0.01.

Free terminal state, Fixedtime:

0.2 0.4 0.6 0.8 1.0t

-0.5

0.5

1.0

x,u

Fixed terminal state,Fixed Time:

0.2 0.4 0.6 0.8 1.0t

-1.0

-0.5

0.5

1.0

x,u

Fixed terminal state,Free time:

1 2 3 4 5t

-1.0

-0.5

0.5

1.0

x,u

REMARKIn the free time case, with xf = 0.01, the final time is t2 = 4.60517. With xf = 0 the finaltime is t2 =∞. Note that the case of free terminal state and free terminal time (notshown) is trivial with t2 = 0.

OPTIMAL CONTROL


SYSTEMS WITH CONSTRAINTS

VARIATIONAL CALCULUS WITH DIFFERENTIAL

CONSTRAINTS

Systems with ‘nonintegrable’ differential , i.e., nonholonomic,constraints have been treated by variational methods. Asbefore, we seek extremals of the functional:

J (q (t)) =

∫ t2

t1L (q (t) , q (t) , t) dt

But now subject to the constraints:

ϕ (q (t) , q (t) , t) = 0

where ϕ : Rn × Rn × R→ Rm.

OPTIMAL CONTROL



DIFFERENTIAL CONSTRAINTS, CONT’D

The constraints are enforce for all time, so introduce the mLagrange multipliers λ (t) and consider the modified functional

J (q (t)) =

∫ t2

t1L (q (t) , q (t) , t) + λT (t)ϕ (q (t) , q (t) , t) dt

We allow variations in both q and λ and t2, Taking the variationand integrating by parts yields

δJ =∫ t2

t1

({− d

dt

[Lq + λTϕq

]+[Lq + λTϕq

]}δq (t) + ϕTδλ

)dt

+[Lq + λTϕq

]t2δq2 +

([L + λTϕq

]−[Lq + λTϕq

]q)

t2δt2

OPTIMAL CONTROL



NECESSARY CONDITIONS WITH CONSTRAINTSAgain we have the Euler-Lagrange equations, now in the form:

ddt

[Lq + λTϕq

]−[Lq + λTϕq

]= 0

The differential constraints:

ϕ (q, q, t) = 0

and the boundary conditions[Lq + λTϕq

]t2δq2 = 0

With free terminal time, we also have([L + λTϕq

]−[Lq + λTϕq

]q)

t2= 0

OPTIMAL CONTROL



EXAMPLEOnce again consider the problem

x = u, J (x (t)) =∫ t2

0

12

(x2 + u2

)dt

But now, add the constraint

ϕ (x, x) = x2 − 1 ≡ u2 − 1 = 0→ u = ±1

The Euler-Lagrange equations now become

ddt

[Lu + λTϕu

]−[Lx + λTϕx

]= 0⇒ λ =

x2u

REMARKI Note that u = ±1→ u = 0 almost everywhere,I Also,

∫ t0 u2dt = t

OPTIMAL CONTROL



EXAMPLE CONTINUED

We now consider three cases:I free end point, fixed terminal timeI fixed end point, fixed terminal timeI fixed end point, free terminal time

OPTIMAL CONTROL



EXAMPLE: FREE END POINT, FIXED TERMINAL TIME

Steer from x (0) = x0.

I system x = u

I Euler equation λ = x2u

I constraint u = ±1

I initial condition x (0) = x0

I terminal condition[Lu + λTϕu

]t2

= 0⇒ λ (1) = − 12

OPTIMAL CONTROL



EXAMPLE: FREE END POINT, FIXED TERMINAL TIME–2The control is piecewise constant, switching between u = +1 andu = −1. Assume:

I there is only one switch that takes place at t = T with 0 < T < 1 withu(0) = u0 = ±1 and u(t) = −u0, t > T

I λ(0) = λ0

x(t) =

{x0 + u0t 0 < t < T

x0 + u0T − u0 (t − T) T < t ≤ 1

λ (t) =

{λ0 + 1

2u0

(x0 + u0

2 t2)

0 < t < T

λ0 + 12u0

(x0 + u0

2 T2)−(

x02u0

+ 12 T)

(t − T) + 14 (t − T)

2 T < t ≤ 1

λ (1) = −12⇒ λ0 +

34

+

(x0

2u0− 1)

T + T2 = 0

OPTIMAL CONTROL



EXAMPLE: FREE END POINT, FIXED TERMINAL TIME–3Consider the case x(0) = 0. The necessary conditions are satisfied with

(u0 = ±1) ∧(−

34≤ λ0 ≤ −

12

)∧(

T =12±

12

√−2− 4λ0

)

Two examples are given below. Only the case λ0 = − 12 , T = 1

2 is optimal.

λ0 = −12

:

0.0 0.2 0.4 0.6 0.8 1.0-1.0

-0.5

0.0

0.5

t

x,Λ,J

λ0 = −3

4

:

0.0 0.2 0.4 0.6 0.8 1.0-1.0

-0.5

0.0

0.5

t

x,Λ,J

OPTIMAL CONTROL



EXAMPLE: FIXED END POINT, FIXED TERMINAL TIME

Steer from x (0) = x0 to x (1) = x1.

I system x = u

I Euler equation λ = x2u

I constraint u = ±1

I boundary conditions x (0) = x0, x (1) = x1

Assume single switch at time T

x0 + kT − k(1− T) = x1 ∧ (k = ±1) ∧ 0 ≤ T ≤ 1⇒ −1 + x1 ≤ x0 ≤ 1 + x1 ∧ (k = ±1) ∧ T = k−x0+x1

2k

OPTIMAL CONTROL



EXAMPLE: FREE TERMINAL TIME, FIXED END POINT

Steer from x (0) = x0 to x (t2) = 0I system x = uI Euler equation λ = x

2uI constraint u = ±1I initial condition x (0) = x0, x (t2) = 0I terminal time condition

(L + λTϕx

)−(Lu − λTϕu

)= 0⇒ λ (t2) = 1

4k + 12

(x0 ≥ 0, t2 ≥ x0)∧ (k = ±1)∧(

T =t2 − kx0

2

)∧

(λ0 =

(t2 − 2T)2 − 1

4

)∨

(x0 ≤ 0, t2 ≥ −x0)∧(k = ±1)∧(

T =t2 − kx0

2

)∧

(λ0 =

(t2 − 2T)2 − 1

4

)

OPTIMAL CONTROL

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