# Two-level control scheme for stabilisation of periodic orbits for planar monopedal running

**ABSTRACT** This study presents an online motion planning algorithm for generating reference trajectories during flight phases of a planar monopedal robot to transfer the configuration of the mechanical system from a specified initial pose to a specified final one. The algorithm developed in this research is based on the reachability and optimal control formulations of a time-varying linear system with input and state constraints. A two-level control scheme is developed for asymptotic stabilisation of a desired period-one orbit during running of the robot. Within-stride controllers, including stance and flight phase controllers, are employed at the first level. The flight phase controller is a feedback law to track the reference trajectories generated by the proposed algorithm. To reduce the dimension of the full-order model of running, the stance phase controller is chosen to be a parameterised time-invariant feedback law that produces a family of two-dimensional finite-time attractive and invariant submanifolds. At the second level, the parameters of the stance phase controller are updated by an event-based update law to achieve hybrid invariance and stabilisation. To illustrate the analytical results developed for the behaviour of the closed-loop system, a detailed numerical example is presented.

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Published in IET Control Theory and Applications

Received on 3rd September 2010

Revised on 9th February 2011

doi:10.1049/iet-cta.2010.0512

ISSN 1751-8644

Two-level control scheme for stabilisation of periodic

orbits for planar monopedal running

N. Sadati1,2

G.A. Dumont2

K.Akbari Hamed1

W.A. Gruver3

1Intelligent Systems Laboratory, Electrical Engineering Department, Sharif University of Technology, Tehran, Iran

2Department of Electrical and Computer Engineering, The University of British Columbia, Vancouver, BC, Canada V6T 1Z4

3School of Engineering Science, Simon Fraser University, Burnaby, BC, Canada V5A 1S6

E-mail: sadati@sharif.edu; sadati@ece.ubc.ca

Abstract: This study presents an online motion planning algorithm for generating reference trajectories during flight phases of a

planar monopedal robot to transfer the configuration of the mechanical system from a specified initial pose to a specified final one.

The algorithm developed in this research is based on the reachability and optimal control formulations of a time-varying linear

system with input and state constraints. A two-level control scheme is developed for asymptotic stabilisation of a desired period-

one orbit during running of the robot. Within-stride controllers, including stance and flight phase controllers, are employed at the

first level. The flight phase controller is a feedback law to track the reference trajectories generated by the proposed algorithm. To

reduce the dimension of the full-order model of running, the stance phase controller is chosen to be a parameterised time-invariant

feedback law that produces a family of two-dimensional finite-time attractive and invariant submanifolds. At the second level, the

parameters of the stance phase controller are updated by an event-based update law to achieve hybrid invariance and stabilisation.

To illustrate the analytical results developed for the behaviour of the closed-loop system, a detailed numerical example is

presented.

1Introduction

This paper presents an analytical approach for designing a

two-level control law to asymptotically stabilise a desired

period-one orbit during running by a planar monopedal

robot. The monopedal robot is a three-link, two-actuator

planar mechanism in the sagittal plane with point foot. It is

assumed that the model of monopedal running can be

expressed by a hybrid system with two continuous phases,

including stance phase (one leg on the ground) and flight

phase (no leg on the ground), and discrete transitions

between the continuous phases, including takeoff and

landing (impact) [1, 2, Chap. 9].

The configuration of the mechanical system is specified

by the absolute orientation with respect to an inertial

world frame and by the joint angles determining the

shape of the robot. During the flight phase, the angular

momentum of the mechanical system about its centre of

mass (COM) is conserved. To reduce the dimension of

the full-order hybrid model of running, which in turn

simplifies the stabilisation problem of the desired orbit, as

proposed by Chevallereau et al. [1], we desire that the

configuration of the mechanical system can be transferred

from a specified initial pose (immediately after the

takeoff) to a specified final pose (immediately before the

landing) during flight phases. This problem is referred to

as ‘landing in a fixed configuration or configuration

determinism at landing’ [1, 2, p. 252]. However, the flight

time and angular momentum about the COM may differ

during

reconfiguration problem must be solved online. A number

of control problems for reconfiguration of a planar

multilink robot with zero angular momentum have been

considered in the literature, for example [3–6]. For the

case that the angular momentum is not necessarily zero,

Kolmanovsky et al. [7] presented a method based on the

averaging theorem [8, Theorem 2.1] such that for any

value of the angular momentum, joint motions can

reorient the multilink arbitrarily over an arbitrary time

interval. However, when the angular momentum is not

zero, this method cannot be employed online for solving

the reconfiguration problem for monopedal running. For

this reason, we present an online reconfiguration algorithm

that solves this problem for given flight times and angular

momentums. The algorithm proposed in this paper is

expressed using the methodology of reachability and

optimal control for time-varying linear systems with input

and state constraints. This algorithm can also be utilised

for online generation of C2trajectories for free open

kinematic chains, conserving angular momentum about

their COM.

Probably the most basic tool for analysing the stability of

periodicorbitsoftime-invariant

described by ordinary differential equations is the Poincare ´

return map. Grizzle et al. [9] showed that the Poincare ´

return map can be applied to systems with impulse effects

for analysing the stability of periodic orbits. To reduce the

dimension of the Poincare ´ return map during bipedal

consecutive steps.Consequently, the

dynamicalsystems

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walking with one degree of underactuation, the strategy of

using virtual constraints has been developed by Grizzle

et al. in [1, 2, 9–17]. For coordination of robot links, a set

of holonomic output functions, referred to as virtual

constraints, are defined and imposed to be zero by a

feedback law [18]. For the case that the corresponding zero

dynamics manifold is impact invariant, Westervelt et al.

[10] introduced the concept of hybrid zero dynamics (HZD)

which in turn results in a one-dimensional restricted

Poincare ´ returnmapwith

By using the virtual constraints approach, Chevallereau

et al. [1] proposed the configuration determinism at landing

to obtain a closed-form expression for the one-dimensional

restricted Poincare ´ return map of running by a five-link,

four-actuator planar bipedal robot. Moreover, to ensure that

the stance phase zero dynamics manifold is hybrid invariant

[2, p. 96] under the closed-loop hybrid model of running,

an additional constraint was imposed on the vector of

generalised velocities at the end of flight phases. To satisfy

the configuration determinism at landing and hybrid

invariance, [1] utilised the approach of parameterised HZD.

Specifically, on the basis of the implicit function theorem

and a numerical non-linear optimisation problem with an

equality constraint, the parameters of the virtual constraints

of the flight phase were updated in a step-by-step fashion

during the discrete transition from stance to flight (i.e.

takeoff). However, the stance phase controller was assumed

to be fixed.

The main contribution of this paper is to present an

analytical approach for online generation of modified

reference trajectories during flight phases of running to

satisfytheconfiguration

Moreover, by relaxing the constraint of [1] on the vector

of generalised velocities at the end of the flight phases,

we present a two-level control scheme based on the

reconfiguration algorithm to asymptotically stabilise a

desiredperiodic orbit.In

controllers, including stance and flight phase controllers,

areemployedatthefirst

controllerisanalogousto

differences. In particular, it is chosen as a time-invariant

and parameterised feedback law to generate a family of

finite-time attractive, zero dynamics manifolds. Unlike the

approach of [1], a continuous feedback law is employed

to track the modified reference trajectories generated by

the reconfiguration algorithm during the flight phase. To

generate a family of hybrid invariant manifolds, an event-

based controller updates the parameters of the stance

phase controller during the transition from flight to stance

(i.e. impact). The terminology of an event-based controller

is taken from [2, p. 199, 11–13].

This paper is organised as follows. Section 2 summarises

the equations of motion during the stance and flight phases

and assembles them into a hybrid model of running. In

Section 3, we treat the reconfiguration problem and an

online reconfiguration algorithm is presented. Constructive

proofs are given based on the formulation of reachability

for a time-varying linear system with input and state

constraints. The control laws for the stance and flight

phases are developed in Section 4. The time-invariant

feedback law during the stance phase is similar to those

developed in [1, 10]. In Section 5, the stabilisation

problem is studied. Finally, simulation results of the

closed-loop system in Section 6 confirm the validity of the

analyticalresults and Section 7 contains concluding

remarks.

a closed-formexpression.

determinismatlanding.

this scheme,within-stride

level.

that

The

of

stance

with

phase

some[1]

2 Mechanical model of a monopedal runner

2.1Monopedal runner

A planar three-link monopedal robot with two ideal

revolute joints and point foot (see Fig. 1) is considered

throughout this paper. The joints are controlled by internal

actuators. Also, it is assumed that torques cannot be applied

at the leg end.

2.2Dynamics of the flight and stance phases

A convenient choice of the configuration variables consists of

thebody angles,theabsolute

absolute position of the monopedal with respect to the

world frame. The body angles represented by w :¼ (w1, w2)′

describe the shape of the robot, where prime denotes

matrix transpose. The absolute orientation of the robot is

represented by u, and the absolute position is represented by

the Cartesian coordinates of its COM, pcm:¼ (xcm, ycm)′.

Consequently,thegeneralised

flight phase are defined as qf:¼ (w′, u, pcm′)′¼ (q′, pcm′)′,

where q :¼ (w′, u)′. Following the notations used in

[1, 2, Chap. 3], the dynamical model during the flight

phasecanbeexpressedasthefollowingsecond-orderequation

orientationandthe

coordinates duringthe

A(w)¨ q +?C(w, ˙ q)˙ q = Bu

(1)

¨ xcm= 0(2)

¨ ycm+ g0= 0 (3)

in which A is a (3 × 3) mass-inertia matrix,?C is a (3 × 3)

matrix containing the Coriolis and centrifugal terms,

u :¼ (u1, u2)′is a vector of actuator torques, g0 is the

gravitational constant and B :¼ [I2×202×1]′. By introducing

xf:= (q′

mechanical system during the flight phase can be expressed

in the state-space form: ˙ xf= ff(xf) + gf(xf)u. Moreover, the

state manifold for the flight phase is chosen as Xf:= TQf,

where Qf denotes the configuration space of the flight

phase. Using the principle of virtual work, a reduced-order

model for describing the evolution of the mechanical system

during the stance phase can also be obtained as follows

[2, p. 74]

f, ˙ q′

f)′as the state vector, the evolution of the

D(w)¨ q + C(w, ˙ q)˙ q + G(q) = Bu

(4)

whereD is a (3 × 3) mass-inertia matrix, C is a (3 × 3) matrix

containing the Coriolis and centrifugal terms and G is a

(3 × 1) gravity vector. By defining the state vector of the

stance phase as xs:= (q′

(4)canberepresented

˙ xs= fs(xs) + gs(xs)u. The state manifold is chosen as

Xs:= TQs, in which Qsis the configuration space of the

stance phase.

s, ˙ q′

s)′, where qs:¼ q and ˙ qs:= ˙ q,

instate-spaceformby

2.3Open-loop hybrid model of running

Following the modelling method presented by Chevallereau

et al. [1, 2, p. 249], the open-loop model of monopedal

running can be expressed by the following non-linear

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hybrid system

Ss:

˙ xs= fs(xs) + gs(xs)u,

x+

x−

x−

s? Sf

s[ Sf

s

f= Df

s(x−

s),

s

?

Sf:

˙ xf= ff(xf) + gf(xf)u,

x+

f),

x−

x−

f? Ss

f[ Ss

f

s= Ds

f(x−

f

?

(5)

In (5), the superscripts ‘–’ and ‘ + ’ denote the state of the

hybrid system immediately before and after the switching

between the state manifolds, respectively. As in [1, 2,

p. 77], we assume that the takeoff switching hypersurface

can be defined as Sf

gs(qs) is the angle of the virtual leg with respect to the

world frame (see Fig. 1) and g−

Moreover, the impact switching hypersurface is defined as

Ss

f:= {xf[ Xf|y1(qf) = 0}, where y1denotes the height of

the leg end with respect to the world frame. Df

and Ds

f? Xsalso represent the takeoff and impact

maps, respectively.

s:= {xs[ Xs|gs(qs) − g−

s= 0}, where

s

is a threshold value.

s: Sf

s? Xf

f: Ss

3

flight phase

Reconfiguration algorithm for the

The conservation of angular momentum about the COM of the

monopedalrobotduringtheflight phaseisexpressedinthethird

line of matrix equation (1) which can be rewritten as follows

˙u =

scm

A3,3(w)− J(w)˙ w

(6)

where scmis a constant representing the angular momentum of

themechanicalsystemabout itsCOM,

J(w) :=

(1/A3,3(w))[A3,1(w) A3,2(w)] [ R1×2and Qb is the body

configuration space. Because matrix A(w) is positive definite,

A3,3(w) . 0 for any w [ Qb. As in [1], we will assume that

the takeoff and landing occur in fixed configurations.

In particular, assume that a C2

w∗: [t∗

2] ? Qb(the evolution of body angles on the

desired orbit) can transfer the configuration of the monoped

robot during the flight phase from the initial condition

q∗

when the angular momentum about its COM is identically

equal to s∗

cm. Now, let the angular momentum about the

COM be scm, where scm= s∗

section is to present an online algorithm for generating

the modified reference trajectory w: [t1, t2] ? Qbbased on

the nominal trajectory w∗such that the configuration of the

mechanical system can be transferred from the initial

condition q∗

t2= t∗

Section 5 to simplify the stabilisation problem.

Integrating of (6) over the time interval [t1, t2] results in

nominal trajectory

1, t∗

1:= [w′∗(t∗

1), u1]′to the final condition q∗

2:= [w′∗(t∗

2), u2]′

cm. The objective of this

1to the final condition q∗

2. The results of this section will be utilised in

2, where t1= t∗

1and

u(t2) = u1+

?t2

t1

scm

A3,3(w(t))dt −

?

C

J(w)dw

(7)

where C := {c [ Qb|c = w(t), t1≤ t ≤ t2} , Qb is the

path of w (t) in the body configuration space. By assuming

w(t) :¼ w∗(t(t)), where t: [t1, t2] ? [t∗

time fulfilling the following constraints:

1, t∗

2] is the virtual

1. t(t1) = t∗

2. t(t2) = t∗

3.inf

t1≤t≤t2˙ t(t) . 0

1

2

Fig. 1

Block diagram of the online reconfiguration algorithm over the time interval [t1, t2] during flight phases of monopedal running

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Equation (7) can be rewritten as follows

u(t2) = u1+

?t∗

2

t∗

1

scm

A3,3(w∗(s))

ds

˙ tWt−1(s)−

?

C∗J(w∗)dw∗

where

consequently

C∗:= {c [ Qb|c = w∗(t), t∗

1≤ t ≤ t∗

2} = C,and

u(t2) − u2=

?t∗

2

t∗

1

1

A3,3(w∗(s))

scm

˙ tWt−1(s)− s∗

cm

??

ds

(8)

By defining m(s) := 1/˙ tWt−1(s) . 0 and w(s): ¼ 1/A3,

3(w∗(s)) . 0 for s [ [t∗

condition u(t2) ¼ u2 can be expressed as the following

equality constraint

1, t∗

2], and assuming scm= 0, the

?t∗

2

t∗

1

w(s)m(s)ds =s∗

cm

scm

?t∗

2

t∗

1

w(s)ds

(9)

Furthermore, from the definition of m(s), ˙ t(t) = 1/m(t(t)),

t1≤ t ≤ t2, and hence

?t∗

2

t∗

1

m(s)ds = t2− t1

(10)

Problem restatement: Determination of m(t) . 0, t∗

t∗

2such that the equality constraints in (9) and (10) are met

isequivalenttodetermining

m: [t∗

2] ? R.0which transfers the state of the following

linear time-varying system in the virtual time domain

1≤ t ≤

thecontrolinput

1, t∗

S:˙ x1= w(t)m

˙ x2= m

(11)

from (x1(t∗

where ˙ xi:= (d/dt)xifor i ¼ 1, 2 and

1), x2(t∗

1))′= (0, 0)′to (x1(t∗

2), x2(t∗

2))′= (xf

1, xf

2)′,

xf

1:=s∗

cm

scm

?t∗

2

t∗

1

w(s)ds

xf

2:= t2− t1

(12)

3.1Determination of the reachable set

The purpose of this subsection is to determine the reachable

set from the origin (at t∗

w(t) ¼ w∗(t(t)), the following relations can be obtained for

the first and second time derivatives of w(t)

1) at time t∗

2for the system S. Since

˙ w(t) =∂w∗

∂t(t(t))˙ t(t)

∂t(t(t))¨ t(t) +∂2w∗

¨ w(t) =∂w∗

∂t2(t(t))˙ t2(t)

Hence, a discontinuity of m may result in an impulsive nature

of ¨ w(t). In view of the actuator limitations, this latter fact

implies that w(t) cannot be used as a reference trajectory for

the joint angles. Thus, we present the following definition.

Definition 1: The set of admissible control inputs for the

system S is denoted by Um,Mand defined to be the set of

all continuously differentiable functions t ? (t) [ [m, M]

defined on the interval [t∗

2], where 0 , m , M.

We present a design method for obtaining an admissible

control m [ C1([t∗

2], [m, M]). For this purpose, we

consider m to be the output of a double integrator and study

the following augmented system

1, t∗

1, t∗

Sa:

˙ x1= w(t)x3

˙ x2= x3

˙ x3= x4

˙ x4= v

which can be viewed as a cascade connection of two

components. [During the continuous phases (i.e. stance and

flight), we desire that the control inputs of the mechanical

system are continuous, whereas they can be discontinuous

during the discrete transitions (i.e. takeoff and impact). For

this reason, we assume that m is the output of a double

integrator.] The first component is the system S in (11)

with x3as input and the second component is the double

integrator with a piecewise continuous function v as input.

The admissibility of m can be expressed as m ≤ x3≤ M,

which is a constraint on the state of Sa.

Definition 2: The set of admissible control inputs for the

system Sais denoted by VL1,L2and defined to be the set of

all piecewise continuous functions t ? v(t) [ [L1, L2]

defined on the interval [t∗

2], where L1, 0 , L2.

1, t∗

Definition

(x0

(xf

that the state of the system Sais transferred from the initial

point (0, 0, x0

with the constraint m ≤ x3(t) ≤ M, t∗

xf

4are free.

Itisclearthatforeveryx0

To determine Am,M,L1,L2, we study two optimal control

problems. From these problems, the optimal admissible

control inputs, vmax(t), vmin(t) [ VL1,L2, t∗

determined such that the state of the system Sa is to be

transferred from the initial point x0:= (0, 0, x0

the final point (x1(t∗

property m ≤ x3(t) ≤ M, t∗

measure Ia(v) := x1(t∗

and minimised (see point C in Fig. 2). Note that in these two

3:

For any 0 , m , M, L1, 0 , L2 and

4)′[ R2, define Am,M,L1,L2(x0

1, xf

3, x0

3, x0

4) as the set of all points

2)′[ R2for which there exists a control v [ VL1,L2such

3, x0

4)′at t∗

1to the final point (xf

1, xf

2, xf

2, where xf

3, xf

4)′at t∗

3and

2

1≤ t ≤ t∗

3? [m, M], Am,M,L1,L2(x0

3, x0

4) = f.

1≤ t ≤ t∗

2, are

3, x0

4)′at t∗

2with the

1to

2), x2(t∗

2), x3(t∗

1≤ t ≤ t∗

2) is maximised (see point D in Fig. 2)

2), x4(t∗

2, while the performance

2))′at t∗

Fig. 2

minimisation and maximisation problems for a given x2

denoted by C and D, respectively

Reachable set Am,M,L1,L2(x3

0, x4

0). The solutions of the

f

are

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optimal control problems, x2(t∗

x3(t∗

The constraint m ≤ x3(t) ≤ M can be rewritten as the

following inequality constraints

2) = xf

2is specified, whereas

2) and x4(t∗

2) are free.

S1(x) := m − x3≤ 0

S2(x) := x3− M ≤ 0

Next, we take successive virtual time derivatives of S1(x) and

S2(x) until obtaining an expression that is explicitly dependent

on v [19, p. 118]. This process will result in¨S1= −v and

¨S2= v, where¨Si:= (d2/dt2)Si, i = 1, 2. Now, define the

following Hamiltonian function

H(x,p,l,v,t):= p1w(t)x3+p2x3+p3x4+p4v+l1¨S1+l2¨S2

= p1w(t)x3+p2x3+p3x4+(p4−l1+l2)v

(13)

where x :¼ (x1, x2, x3, x4)′, p :¼ (p1, p2, p3, p4)′and

l :¼ (l1,l2)′are the state, costate and multiplier vectors,

respectively. Furthermore, in (13)

¨Si= 0,

li= 0,

ontheconstraintboundary (i.e. Si= 0)

off theconstraintboundary (i.e. Si, 0)

for i ¼ 1, 2, which can also be expressed as

v = 0, ontheconstraintboundary (i.e. Si= 0)

off theconstraintboundary (i.e. Si, 0)

li= 0,

(14)

Necessary condition for multipliers li(t), i ¼ 1, 2, to

minimise the performance measures is

li(t) ≥ 0,ontheconstraintboundary (i.e.Si= 0)(15)

Note that the maximisation of the performance measure

Ia(v) := x1(t∗

−Ia(v). The costates satisfy the following differential

equations

2) can be expressed as the minimisation of

˙ p1= 0

˙ p2= 0

˙ p3= −p1w(t) − p2

˙ p4= −p3

in which ˙ pi:= (d/dt)pi, i = 1, ..., 4. From here on, the

superscripts ‘max’ and ‘min’ will denote the solutions of the

maximisation and minimisation problems, respectively. We

first study the maximisation problem. Since the final values

xmax

1

(t∗

3

(t∗

4

(t∗

p. 200], pmax

1

(t∗

boundary conditions in combination with the costate

equations yield

2), xmax

2) and xmax

2) = −1 and pmax

2) are free, from Table 5-1 of [20,

(t∗

432) = pmax

(t∗

2) = 0. These

pmax

3

(t; pmax

2

) = −

?t∗

?t∗

2

t

w(s)ds − pmax

2

(t − t∗

2)

pmax

4

(t; pmax

2

) = −

2

t

?t∗

2

s

w(h)dhds +pmax

2

2

(t − t∗

2)2

From Pontryagin’s minimum principle [21], vmax(t) is

given by

vmax(t) =

L1,

L2,

undetermined,

pmax

4

pmax

4

pmax

4

− lmax

− lmax

− lmax

1

+ lmax

+ lmax

+ lmax

2

. 0

, 0

= 0

12

12

(16)

Notice

through the zero, a switching of the optimal control input

vmax(t) occurs. Assume that w(t) satisfies the following

hypothesis:

H: ˙ w(t) := (d/dt)w(t) is not zero on the open set (t∗

It will be shown that by hypothesis H the optimal control

inputs vmax(t) and vmin(t) can switch at most once, and the

singular condition does not occur. For this purpose, we

present the following result.

thatif

pmax

4

(t; pmax

2

) − lmax

1

(t) + lmax

2

(t)passes

1, t∗

2).

Lemma 1: Let m , x0

hypothesis H holds and x0

trajectories xmax(t) and xmin(t) of the system Saexist. Then,

the following statements are true:

3, M and L1, 0 , L2. Assume that

4 is such that the optimal

(a) The optimal trajectories do not enter onto the boundaries

S1¼ 0 and S2¼ 0.

(b) The optimal control inputs vmax(t) and vmin(t) can switch

at most once.

(c) The singular condition does not occur. In other words, the

sets Tmax

0

:= {t [ [t∗

4

and

Tmin

0

:= {t [ [t∗

are Lebesgue negligible.

1, t∗

1, t∗

2]|pmax

2]|pmin

(t) − lmax

4 (t) − lmin

1

(t) + lmax

1 (t) + lmin

2

2 (t) = 0}

(t) = 0}

Proof: To prove the statements (a), (b) and (c) of Lemma 1,

themaximisationproblem

reasonings can also be presented for the minimisation

problem.

If the optimal trajectory enters onto the constraint boundary

S1¼ 0, S2will be negative and consequently from condition

(14), vmax¼ 0 and lmax

2

= 0. Since L1, 0 , L2, from the

control input vmax(t) in (16), vmax¼ 0 results in lmax

which in combination with the necessary conditions given in

(15) yields pmax

4

≥ 0. Similarly, if the optimal trajectory

enters onto the constraint boundary S2¼ 0, then vmax¼ 0

and lmax

1

= 0. Moreover, lmax

2

pmax

4

≤ 0.

Next, we study the roots of the non-linear equation

pmax

4

(t; pmax

2

) = 0 on the interval [t∗

also be expressed as follows

willbestudied.Similar

1

= pmax

4

= −pmax

4

which implies that

1, t∗

2]. This equation can

W(t) =pmax

2

2

(t − t∗

2)2

(17)

where W(t) :=?t∗

has at most one root in the interval [t∗

? t [ [t∗

unique and can be given by

2

t

?t∗

2

sw(h)dhds. For any pmax

2

[ R, t∗

(t; pmax

2is the

) = 0

solution of (17). We claim that the equation pmax

4

2

1, t∗

2). To show this, let

) = 0. Then, pmax

1, t∗

2) exist such that pmax

4

(? t; pmax

22

is

pmax

2

=

2

(? t − t∗

2)2W(? t)(18)

We remark that pmax

2

is positive. If there exists ˜ t [ [t∗

1, t∗

2)

1532

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Page 6

such that ˜ t = ? t and pmax

4

(˜ t; pmax

2

) = 0, then (18) implies that

W(? t)

(? t − t∗

2)2=

W(˜ t)

(˜ t − t∗

2)2

Hence, it is sufficient to show that the function k:[t∗

by k(t) := W(t)/(t − t∗

derivative of k(t) can be obtained as follows

1, t∗

2) ? R

2)2is strictly monotonic. The first

˙ k(t) =

˙W(t)(t − t∗

2) − 2W(t)

(t − t∗

2)3

=:

F(t)

(t − t∗

2)3

(19)

Assume that there exists h1[ (t∗

Since F(h1) = F(t∗

thatthere is

¨W(h2)(h2− t∗

zero.Hence,from

h3[ (h2, t∗

follows that

˙ w(h3) = 0, and this contradicts hypothesis

H. Therefore

k(t)isstrictly

consequence, the equation pmax

4

root in the interval [t∗

2).

Let

? t [ (t∗

2)bethe

pmax

4

(t; pmax

2

) = 0. Substituting (18) into pmax

1, t∗

2) such that F(h1) ¼ 0.

2) = 0, the Rolle’s theorem implies

h2[ (h1, t∗

such

2) −˙W(h2) = 0. Furthermore, ˙F(t∗

the Rolle’s theorem,

2) such that¨F(h3) = ˙ w(h3)(h3− t∗

2)that

˙F(h2) =

2) is also

there

2) = 0 which

exists

monotonic

(t; pmax

2

) = 0 has at most one

and,asa

1, t∗

1, t∗

rootofthe

(t; pmax

equation

) yields

4

2

˙ pmax

4

(? t; pmax

2

) =−˙W(? t)(? t − t∗

2) + 2W(? t)

? t − t∗

2

=F(? t)

t∗

2−? t= 0

Therefore the condition ˙ pmax

expressed as

4

(? t; pmax

2

) , 0 which can be

2

t∗

2−? t

?t∗

(t; pmax

2

? t

?t∗

2

s

w(h)dhds .

?t∗

2

? t

w(s)ds

(20)

implies that pmax

pmax

4

(t; pmax

2

not satisfied, then pmax

pmax

4

(t; pmax

2

Since m , x0

S1(x0) and

lmax

1

(t∗

that pmax

4

(t∗

Since the non-linear equation pmax

onerootinthe

(t∗

4

(t; pmax

2

as a consequence, without loss of generality, we can assume

that pmax

4

(t∗

2

) = 0. Then there are four possible cases.

4

2

) . 0 for any t [ [t∗

2]. If condition (20) is

(t; pmax

2

) , 0 for any t [ [t∗

) . 0 for any t [ (? t, t∗

2].

3, M and xmax(t) is the optimal trajectory,

S2(x0)arenegative

1) = lmax

2

(t∗

1; pmax

2

) = 0. If pmax

4

(t∗

1, ? t) and

) , 0 for any t [ (? t, t∗

4

1, ? t) and

whichresultin

1) = 0. Without loss of generality, assume

1; pmax

2

(t; pmax

interval[t∗

2),

) = 0. In addition, vmax(t∗

) = 0, then ? t = t∗

) = 0 has at most

onthe

1) is finite and

1.

4

2

1, t∗

open set

1, t∗

2), pmax

1; pmax

Case 1: Assume that there exists ? t [ (t∗

pmax

4

(? t; pmax

2

) = 0 and inequality (20) holds. In this case, on

the interval [t∗

1, t∗

2) such that

1, ? t), vmax(t) = L1and consequently

xmax

4

(t) = x0

4+ L1(t − t∗

1)

xmax

3

(t) = x0

3+ x0

4(t − t∗

1) +L1

2(t − t∗

1)2

xmax

2

(t) = x0

3(t − t∗

1) +x0

4

2(t − t∗

1)2+L1

6(t − t∗

1)3.

Because pmax

trajectory xmax(t), t∗

S2¼ 0. From [19, p. 118], since the control of S1is obtained

only by changing¨S1, no finite control can keep the optimal

4

(t; pmax

2

) . 0 on the interval [t∗

1≤ t , ? t cannot enter onto the boundary

1, ? t), the optimal

trajectory of the system Sa on the constraint boundary

S1¼ 0, unless the following tangency constraints hold.

[The terminology of a ‘tangency constraint’ is taken from

[20, p. 118].]

N1(xmax(t)) :=

S1(xmax(t))

˙S1(xmax(t))

??

=

m − xmax

−xmax

3

(t)

4

(t)

??

=

0

0

? ?

(21)

Because L1, 0, xmax

concave function with respect to t. Thus, the condition

m , x0

3, M implies that the tangency constraints (21)

cannot be satisfied on the interval [t∗

3

(t), t∗

1≤ t , ? t is a quadratic and

1, ? t). Now, define

? x4:= xmax

4

(? t) = x0

4+ L1(? t − t∗

1)

? x3:= xmax

3

(? t) = x0

3+ x0

4(? t − t∗

1) +L1

2(? t − t∗

1)2

? x2:= xmax

2

(? t) = x0

3(? t − t∗

1) +x0

4

2(? t − t∗

1)2+L1

6(? t − t∗

1)3

Also, on the interval (? t, t∗

2], vmax(t) ¼ L2and thus

xmax

4

(t) = ? x4+ L2(t −? t)

xmax

3

(t) = ? x3+? x4(t −? t) +L2

2(t −? t)2

2(t −? t)2+L2

xmax

2

(t) = ? x2+? x3(t −? t) +? x4

6(t −? t)3

Since pmax

xmax(t), ? t , t ≤ t∗

Moreover, the tangency constraints to remain on the

boundary S2¼ 0 can be expressed as

4

(t; pmax

2

) , 0 on (? t, t∗

2cannot enter onto the boundary S1¼ 0.

2], the optimal trajectory

N2(xmax(t)) :=

S2(xmax(t))

˙S2(xmax(t))

??

=

xmax

3

(t) − M

xmax

4

(t)

??

=

0

0

? ?

(22)

The fact that xmax

function with respect to t in combination with the condition

m , ? x3, M implies that the tangency constraints in (22)

cannot be satisfied on the interval (? t, t∗

S1(xmax(t)), S2(xmax(t)) , 0 for any t [ [t∗

optimal trajectory is feasible (i.e. m ≤ xmax

L1, 0, ? x3= m will result in ? x4, 0, which in turn implies

the existence of ˆ t [ (? t, t∗

t [ (? t, ˆ t). This contradicts the feasibility of the optimal

trajectory xmax(t). In a similar manner, it can be shown that

? x3= M.

The final constraint xmax

2

(t∗

the following third-degree equation

3

(t), ? t , t ≤ t∗

2is a quadratic and convex

2]. Consequently,

1, t∗

(t) ≤ M) and

2]. Since the

3

2] such that xmax

3

(t) , m for any

2) = xf

2can also be expressed as

L1− L2

6

(? t − t∗

2)3+ x0

3lmax+x0

4

2l2

max+L1

6l3

max= xf

2

(23)

where lmax:= t∗

calculated as follows

2− t∗

1. From (23), ? t [ R is unique and can be

? t = t∗

2+

???????????????????????????????????????????? ?

L1− L2

6

xf

2− x0

3lmax−x0

4

2l2

max−L1

6l3

max

??

3

?

IET Control Theory Appl., 2011, Vol. 5, Iss. 13, pp. 1528–1543

doi:10.1049/iet-cta.2010.0512

1533

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Page 7

Ift∗

solution and, as a consequence, the validity of Case 1 is

confirmed.

1, ? t , t∗

2andinequality(20)issatisfied,then? tisafeasible

Case 2: There exists ? t [ (t∗

and

1, t∗

2) such that pmax

4

(? t; pmax

2

) = 0

2

t∗

2−? t

?t∗

2

? t

?t∗

2

s

w(h)dhds ,

?t∗

2

? t

w(s)ds

(24)

An analysis similar to that presented for Case 1 can be

performed. However, the third-degree equation in (23) is

given by

L2− L1

6

(? t − t∗

2)3+ x0

3lmax+x0

4

2l2

max+L2

6l3

max= xf

2

(25)

Equation (25) has the following real and unique root

? t = t∗

2+

???????????????????????????????????????????? ?

L2− L1

6

xf

2− x0

3lmax−x0

4

2l2

max−L2

6l3

max

??

3

?

which is feasible if t∗

satisfied.

1, ? t , t∗

2and inequality (24) is

Case 3: The function pmax

[t∗

vmax(t) ; L1and similar to the analysis performed for the

interval [t∗

S2(xmax(t)) , 0. Also, the final condition xmax

be satisfied only for the following specific value of xf

4

(t; pmax

2

) is positive on the interval

is not unique. In addition,

1, t∗

2]. This implies that pmax

2

1, ? t) in Case 1, it can be shown that S1(xmax(t)),

2

(t∗

2) = xf

2can

2

xf

2= xf

2:= x0

3lmax+x0

4

2l2

max+L1

6l3

max

Case 4: If the function pmax

interval[t∗

pmax

4

(t; pmax

2

) , 0

S2(xmax(t)) , 0, t∗

xf

4

(t; pmax

not

L2. 0

2. The final condition xmax

2

) is negative on the

uniqueand

implythat

1, t∗

2], pmax

2

ismoreover,

S1(xmax(t)),

(t∗

and

1≤ t ≤ t∗

2

2) =

2can be satisfied only for the following specific value of xf

2

xf

2= ? xf

2:= x0

3lmax+x0

4

2l2

max+L2

6l3

max

The proof of Lemma 1 follows from the results obtained in

Cases 1–4.

A

Remark 1: In the minimisation problem

pmin

3 (t; pmin

2 ) =

?t∗

?t∗

2

t

w(s)ds − pmin

2 (t − t∗

2)

pmin

4 (t; pmin

2 ) =

2

t

?t∗

2

s

w(h)dhds +pmin

2

2

(t − t∗

2)2

Moreover, the non-linear equation pmin

most one root in the interval [t∗

such that pmin

2 ) = 0. The validity of Cases 1 and 2 in

the minimisation problem is confirmed by

4 (t; pmin

1, t∗

2 ) = 0 has at

2). Let t [ [t∗

1, t∗

2) be

4 (t, pmin

2

t∗

2− t

?t∗

2

t

?t∗

2

s

w(h)dhds ,

?t∗

2

t

w(s)ds

and

2

t∗

2− t

?t∗

2

t

?t∗

2

s

w(h)dhds .

?t∗

2

t

w(s)ds

respectively. Cases 3 and 4 of the minimisation problem are

similar to the ones presented in the maximisation problem.

Remark 2: As discussed previously, by hypothesis H, the

function F(t) defined in (19) is non-zero on the interval

(t∗

2). Thus, without loss of generality, we will assume

that F(t) , 0 on the interval (t∗

imposes that Case 2 of the maximisation problem and Case

1 of the minimisation problem are not feasible.

Now let m , x0

(x0

for which the optimal solutions of the maximisation and

minimisation problems starting from the initial point

(0, 0, x0

exist. Also, denote the solutions of the

maximisation and minimisation problems corresponding to

xf

2), respectively. The functions

pmax: Vmax

R are introduced by

1, t∗

1, t∗

2). This assumption

3, M and x0

m,M,L1,L2(x0

4[ R. Define Vmax

4) to be the sets of all xf

m,M,L1,L2

2[ R

3, x0

4) and Vmin

3, x0

3, x0

4)′

2by xmax(t; xf

m,M,L1,L2(x0

2) and xmin(t; xf

3, x0

4) ? R and pmin: Vmin

m,M,L1,L2(x0

3, x0

4) ?

pmax(xf

2):=

?t∗

?t∗

2

t∗

1

w(s)xmax

3

(s; xf

2)ds = xmax

1

(t∗

2; xf

2)

pmin(xf

2):=

2

t∗

1

w(s)xmin

3 (s; xf

2)ds = xmin

1 (t∗

2, xf

2)

(26)

(see Fig. 2). We claim that if the sets Vmax

Vmin

4) are non-empty, they are connected sets.

For this purpose, we present the following lemma for which

it is shown that Vmax

A similar result can also be obtained for the set Vmin

(x0

4).

Lemma 2: Let m , x0

are two scalars such that a, b [ Vmax

any g [ (a, b), g [ Vmax

The proof is given in Appendix 1. Now we are in a position

to present the main result of this section. This result is

expressed as the following theorem which determines the

C1input m transferring the state of S from the origin at t∗

to the final point (xf

m,M,L1,L2(x0

3, x0

4) and

m,M,L1,L2(x0

3, x0

m,M,L1,L2(x0

3, x0

4) is a connected set.

m,M,L1,L2

3, x0

3, M and x0

4[ R. Assume that a , b

m,M,L1,L2(x0

m,M,L1,L2(x0

3, x0

4). Then, for

3, x0

4).

1

1, xf

2)′[ Am,M,L1,L2(x0

3, x0

4) at t∗

2.

Theorem 1: Let m , x0

L1, 0 and L2. 0 are such that

3, M and x0

4[ R. Assume that

min x0

3, x0

3+ x0

4lmax+L1

2l2

max

?

?

?

?

. m

max x0

3, x0

3+ x0

4lmax+L2

2l2

max

, M

(27)

Then, the set Am,M,L1,L2(x0

3, x0

4) is given by

Am,M,L1,L2(x0

3, x0

4)

= {(xf

1, xf

2)′[ R2|xf

2≤ xf

2≤ ? xf

2, pmin(xf

2) ≤ xf

1≤ pmax(xf

2)}

where xf

(x0

2:= x0

max+ (L2/6)l3

3lmax+ (x0

4/2)l2

max+ (L1/6)l3

max, ? xf

1.

2:= x0

3lmax+

4/2)l2

maxand lmax:= t∗

2− t∗

1534

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IET Control Theory Appl., 2011, Vol. 5, Iss. 13, pp. 1528–1543

doi:10.1049/iet-cta.2010.0512

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