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0Russian Journal of Nonlinear Dynamics, 2024, vol. 20, no. 1, pp. 27-42. Full-texts are available at http://nd.ics.org.ru DOI: 10.20537/nd231207
NONLINEAR PHYSICS AND MECHANICS
MSC 2010: 49K15, 58E25, 93C15
Extremals in the Markov -Dubins Problem with Control on a Triangle
A. A. Ardentov
We formulate a time-optimal problem for a differential drive robot with bounded positive velocities of the driving wheels. This problem is equivalent to a generalization of the classical Markov - Dubins problem with an extended domain of control. We classify all extremal controls via the Pontryagin maximum principle. Some optimality conditions are obtained; therefore, the optimal synthesis is reduced to the enumeration of a finite number of possible solutions.
Keywords: differential drive robot, Markov-Dubins problem, extremal trajectories, Pon-tryagin maximum principle
Introduction
We consider a kinematic model for a differential drive robot. The robot is driven by two coaxial wheels with a limited maximum and minimum rotation speed. The wheels can only rotate forward (in one direction) with an unbounded acceleration.
The configuration set of robot positions is described by the group of motions of a plane
SE(2) = {(x, y, 0) e MX,y x Si},
where x, y are coordinates of the center of the driving wheels and the angle 0 defines the direction of the wheel motion and belongs to a circle S^, see Fig. 1.
Received September 20, 2023 Accepted November 30, 2023
The work of A. Ardentov is supported by the Russian Science Foundation under grant 22-21-00877 (https://rscf.ru/en/project/22-21-00877/) and performed at Ailamazyan Program Systems Institute of Russian Academy of Sciences.
Andrei A. Ardentov aaa@pereslavl.ru
Ailamazyan Program Systems Institute, Russian Academy of Sciences Pereslavl-Zalessky, Yaroslavl Region, 152020 Russia
Fig. 1. Configuration coordinates for the robot
We formulate the time-optimal problem for the described model of the robot. Wheel speed bounds impose a bound on the track curvature similarly to the classical Markov - Dubins problem.
In [9], A.A.Markov considered the problem of connecting two settlements by a shortest railroad path with bounded curvature and a fixed direction at the beginning. Obviously, a straight line is optimal up to infinity and a circle path is not optimal after the full loop. A. A. Markov proved that the general optimal solution is a certain concatenation of a circle arc and a straight line segment or two circle arcs.
In [6], L. Dubins studied the same problem but with fixed directions both at the beginning and at the end. He proved that the optimal solution is a certain concatenation of circle arcs and a straight line segment with two switchings at most. The optimal synthesis of the problem had been reduced to comparison of length for 6 possible paths. The problem studied by L. Dubins is often called in the literature as the Markov - Dubins problem honoring the contributions of the both mathematicians.
In [4], the authors calculate the partition of the configuration space in the Markov - Dubins problem with respect to the type of the optimal paths. They specify optimality conditions for the candidate optimal paths and describe the partition for any plane $ = const (with a fixed orientation) of the configuration space SE(2).
There are several known modifications of the Markov - Dubins problem. In [3], the authors considered an asymmetric sinistral/dextral variation of the classical problem, when the absolute values of lower and upper bounds of the curvature are not necessarily equal to one another. Recently, in [8], a generalization of the asymmetric sinistral/dextral Markov - Dubins problem was proposed. It can be used to model an Unmanned Aerial Vehicle with a penalty associated with a turn due to a loss in altitude while turning or a UAV with different costs for the sinistral and dextral turns due to hardware failures or environmental conditions. Using optimal control theory, the authors showed that the optimal path belongs to a set of at most 21 candidate paths, each comprising of at most four switchings between straight line segments and circle arcs.
Another modification of the Markov - Dubins problem allows a direction-dependent speed, which accounts for the effects of waves, winds and slope of the terrain on the robot motions [2, 5, 12]. Such modifications are mostly oriented towards aerial or water robotics.
Our aim is to investigate a natural generalization of the Markov - Dubins problem appearing in ground robotics. In Section 1, we reduce the time-optimal problem for the differential drive robot to the time-optimal problem on SE(2) with linear and angular velocities in a symmetric triangle, which modifies the Markov - Dubins problem. The generalization extends the set of admissible controls from a segment (of a triangle side) to the triangle. However, it preserves
the coincidence of the absolute value of maximum and minimum curvature for the admissible trajectory of motion, which arose in the classical (symmetric) Markov - Dubins model.
In Section 2, we apply the Pontryagin maximum principle in order to obtain a description of abnormal and normal extremals. The normal extremals are classified into two types: bangbang and mixed ones. We specify extremal controls as certain piecewise constant functions. In Section 3, we provide an upper bound for the cut time of each geodesic. Finally, we clarify the condition for the closed geodesics.
1. Formulation of the optimal control problem
1.1. Time-optimal problem for the differential drive robot
We assume that the robot is controlled by the velocities of the left and right wheels v1, v2 G e [vmin, vmax], where 0 ^ vmin < vmax, see Fig. 2. The distance d between the driving wheels is fixed.
v2
Vl
Fig. 2. Square domain V = [vmin, vmax]2 of velocities v1; v2
The time-optimal problem for the described model of the robot is formulated as follows:
■ cos 0,
( dx v1 + v2
dt 2
dy = V! + v2 dt ~ 2 dû v2 — v1
dt ~ d :
sin 0,
(x, y, û) G SE(2) = MX,y X Si, (Vl , vr,) G V = [v • , v l x [v • , v ],
Vv1j v2/ G " L mm max mm maxJ '
x(0)= xo, y(0) = yo, û(0) = ûo, x(T)= xT, y(T)= yT, û(T) = Ût, T —> min .
(1.1)
(1.2)
(1.3)
(1.4)
(1.5)
(1.6)
Problem (1.1)—(1.6) has 9 constant parameters:
max; xo, yo, $o, xt, yt>
1.2. Invariant formulation of the Markov —Dubins problem with control on a triangle
Problem (1.1)—(1.6) has the following continuous symmetries: parallel translations, rotations, dilations of the (x, y)-plane and dilations of the time. Using these symmetries, we reduce the number of the constant parameters and reformulate the problem considered in the following invariant form:
q = ul X1 + u2X2, (1.7)
q = (x,y,d) G SE(2) = R^ x (u1, u2) G U, (1.8)
X1 = (cos6, sin6, 0), X2 = (0, 0, 1), (1.9)
umax u
U = Ua) G M2 | u, G [1, uT], M = Jmax."1}. (L1°)
q(0) = (0, 0, 0), q(T) = (xT, yT, 6T), (1.11)
T — min, (1.12)
where a new control
«... «,) - ^p) (1.13)
is the linear and angular velocities of the robot with umax > 1, see Fig. 3 for the domain U. Moreover, we assume that the limiting case with umax = 1 is the classical Markov-Dubins problem with
U = Umd := {(ui, u2) G R2 | ui = 1, |u2| < 1}. (1.14)
-1
Fig. 3. The control domain U defined by (1.10) (two black segments with three colored vertices) Problem (1.7)-(1.12) has 4 constant parameters:
umax, Xt, yT, 6t.
Remark 1. In this article, we call U a triangle, although formally set (1.10) is not one and corresponds only to its two outer sides.
1.3. Equivalence of the formulations
Problem (1.1)—(1.6) is equivalent to problem (1.7)—(1.12) in the following sense.
Proposition 1. Suppose we have problem (1.1) —(1.6) with fixed parameters d7 vmin7 vmax7 x0> y0; xT> yT>
Set the parameters for problem (1.7)—(1.12) as follows :
x.
T
= aR
\VTJ Ot = ÛT — ûo,
yT
a
xo yo
u
= ¿vmax G (1, 2];
(1.15)
(1.16)
where a = 2j7msx , Vminj, ö =
d(vmax+vmin/
and Ri =
,xd min are coefficients of plane and time dilations, respectively, is the clockwise rotation matrix through angle 90 in the plane.
Let (x(t), v(t), O(t)), t G [0, T], be an optimal solution with optimal controls (u1(t), u2(t)) for problem (1.7) —(1.12) with the parameters Xt, Vt, Ot, umax-
Then the optimal solution to the initial problem (1.1)—(1.6) is expressed as follows :
x(t) =
y(t) = fvi(t)N
xo yo
= 5
+
/Ui(St)
+
x(5ty vv(St),
du2(St)\
àUn(St) J ■
û(t) = ûo + O(St),
t
0, T
T
J
(1.17)
(1.18)
Proof. The parallel translation (—x0, —y0) 1 and the rotation R$ of the (x, y)-plane moves the initial condition (x(0), y(0), 0(0)) = (x0, y0, 00) to q(0) = (0, 0, 0).
From formula (1.13), we obtain a quadrilateral control domain (see Fig. 4).
Fig. 4. Transformation of the square domain V via mapping (1.13) (gray quadrilateral)
[J
cos lo sin lo
— sin lo cos lo
a
The a-dilation of the plane and the ¿-dilation of the time gives the transformation (u1, u2) ^ (^p, ^f), therefore the quadrilateral domain shown in Fig. 4 transforms to the invariant one illustrated in Fig. 3 with t/,fax = Zfy* e (1, 2].
Note that we can assume that optimal control belongs to the U since otherwise there exists the dilated control from U which generates the same trajectory of system (1.7) for a smaller time interval.
Therefore, formulas (1.15) and (1.16) express the described transformations and formulas (1.17) and (1.18) correspond to the inverse transformations providing the equivalence of problems (1.1)-(1.6) and (1.7)-(1.12). □
Complete controllability of system (1.7) is guaranteed because there are at least two non-collinear velocities [10]. The existence of optimal controls in problem (1.7)-(1.10) follows from Filippov theorem [1].
2. The Pontryagin maximum principle
In this section, we apply the Pontryagin maximum principle to invariant problem formulation (1.7)-(1.12). Introduce a vector of adjoint variables p = (p1, p2, p3) and write the following Pontryagin function:
H(p, q, u) = u1 (p1 cos 9 + p2 sin9) + u2p3, u = (u1, u2). (2.1)
The Hamiltonian system is expressed as follows:
' p1 = -HX = ° p2 = —Hy = 0,
2 y (2.2) p3 = —Hg = u1 (p1 sin 9 — p2 cos 9) ,
q = u^1 + ^Xj.
The Pontryagin maximum principle (PMP) states that, if a curve q(t) and a control u(t) = = (u1(t), u2(t)), t e [0, T], are optimal, then there exists a Lipschitzian curve (p(t), q(t)) G e Tfa SE(2), t G [0, +rc>) (called an extremal) with p(t) = 0, s.t. it satisfies the Hamiltonian system (2.2) and the maximality condition
H(p(t), q(t)) := maxH(p(t), q(t), u) = H(p(t), q(t), u(t)) ^ 0. (2.3)
uEU
2.1. Reduction of the Hamiltonian system
Before specifying in detail maximality condition (2.3), we define convenient coordinates for the maximized Hamiltonian H.
The Pontryagin function H depends on the control u, the vector p and the component 9 of the vector q. Also, the subsystem of equations for p and 9 decouples from whole Hamiltonian system (2.2).
Introduce the polar coordinates p1 = r cos f3, p2 = r sin /3 and denote 0 = 9 — f3, then
H(r, 0, p3, u1, u2) = u1r cos 0 + u2p3, (2.4)
r = 0,
<0 = u2, (2.5)
p3 = u1 r sin 0.
In order to exclude the first integral r from (2.4)—(2.5), we specify the application of PMP in the degenerate case r = 0 as follows:
H(p3, u2) = u2p3, H(рз) = |рз| = const = 0, Ф = sign Рз,
and conclude that this case provides only constant extremal controls u = (1, sign p3) corresponding to turns (such trajectories form circles on the (x, y)-plane and obviously are not optimal for T ^ 2n).
For the general case г ф 0, we introduce с = '-y- and rewrite (2.4), (2.5) in the coordinates (ф, c) as follows:
Ш/А, \ 'H(r, Ф, Рз, uv U2) / | /n ГЛ
пуф, С, Щ, U2) '■=-2-i-— = Щ cos ф + u2c, (2.6)
( * = и2'^ ф (2.7)
[ c = U1 Sin ф.
Thus, in order to describe extremal controls of the problem, it is enough to maximize new Pontryagin function (2.6) taking into account system (2.7).
Remark 2. If cos ф < 0, then
H
Н(ф, с) = — = max (u1 cosф + unc) = max (w1cos^>) + max (unc) =cos</>+ |c|. (2.8) r (uj ,u2)eu («i,«2 )eU (u1,u2)eu
Therefore,
H< 0 ^^ cosф + |c| < 0, (2.9)
H = 0 ^^ cos ф + |c| =0, (2.10)
since, for cos $ > 0, we have H > 0.
Further, we consider the abnormal case H = 0.
2.2. Abnormal case
Theorem 1. An abnormal extremal control (u1, u2) is a piecewise constant function with the values (1, ±1), (1, ^1), (1, ±1), ... and the time intervals between the switchings Tl7 n, n, ..., where T1 ^ n.
Proof. If H = 0, then H = cos $ + |c| =0 (see Remark 2).
If c = 0, then (u1, u2) = (1, signc) from maximization (2.8).
Otherwise, c = cos $ = 0 and the control (u1, u2) is not defined from the maximality condition of PMP. However, C = u1 sin $ = 0 in the neighborhood of such points, therefore the control is undefined only at isolated points of an abnormal trajectory and does not change its behavior.
Switching of the sign of u2 occurs at c = cos $ = 0 in the time interval n, since $ = sign c for cos $ = 0. □
2.3. Normal case
Now we study the case of PMP with H > 0.
We reduce the maximization of the Pont.ryagin function 7i in the following way:
H(0, c) = max (u1 cos0 + u2c) = umaxcos0 + max (|u2| (|c|-(umax - 1)cos0)), (2.11) (u1,u2)eu |u2|e[0,1]
assuming u1 = u™ax — (u™ax — 1) |u2| and u2c = |u2c|. We decompose the domain of all normal geodesies
D = {(0, c) G S1 x R \\c\ + cos 0}
(2.12)
into subsets
i=0
D = LJ Di, (2.13)
(2.14)
(2.15)
;os (2.16)
(2.17)
(2.18)
(2.19)
(2.20)
See Fig. 5 with an illustration of decomposition (2.13)-(2.20) and the corresponding phase trajectories in each subdomain in accordance with (2.7).
Do = (0, c) G D 0 = 0, \c\ < <urx - 1},
Di = (0, c) G D 0 = 0, i^i < Curx - 1) cos 0},
D2 = (0, c) G D c = 0, «ax - 1) cos 0<\ c \< urx
D3 = (0, c) G D 0c = 0, \ c \ = (u5"ax - 1) cos 0},
D4 = (0, c) G D \c\ >u5"ax - cos 0},
D5 = (0, c) G D \c\ = urx - cos 0},
Do = (0, c) G D 0 = 0, \ c \ = umax -1}.
Fig. 5. The decomposition of the domain D for the value w™ax = f: D0 — lighter green dots, D1 — two green subdomains, D2 — two red and two blue subdomains, D3 — four dashed black curves, D4 — lighter red and lighter blue domains, D5 — solid black lines, D6 — two lighter green dots in black circles
Lemma 1. According to decomposition (2.13)-(2.20), the maximized Hamiltonian H = and the corresponding extremal controls are expressed in the following way:
(</>, c) £ D0 U Dl => H = cos </>,K, u2) = 0),
((f), c) G D2 U U D5 => H = coscf) + |c|,, u2) = (1, signc),
r, i max _ ~.
max i „, _ ^ „ "1 "1
((/), c) £ fl3 U L>6 ==> ii = t/,™ax cos 2 = sign c
umax -1
Proof. If |c| — (t/,™ax — 1) cos (f) < 0, then we have \u2\ = 0, H = t/,™ax cos <fi from maximization (2.11), also u1 = ufax.
If |c| — (t/.™ax — l)cos</> > 0, then we obtain \u2\ = 1, H = cos(f) + |c| from maximization (2.11), also u1 = 1 and signu2 = sign c, since c = 0.
If |c| — (t/.™ax — l)cos (f) = 0, then H = t/,™axcos(/>. Note that, c / 0, therefore signt/,2 = = sign c. □
Using notations of sub-Finsler geometry [2], we introduce the following sets in order to distinguish different reasons for Cauchy non-uniqueness of extremals (i. e., when an extremal (p(t), q(t)) is not defined uniquely by an initial point (p(0), q(0))):
6 = {(</>, c)eD\H(<t>, c) ^ C2((j), c)},
where C2(0, c) is a class of functions with the continuous first and second derivatives. We call an extremal arc with (0(t), c(t)), t G [tb, te], tb < te:
• a bang arc if (0(t), c(t)) G S for all t G (tb, te),
• a singular arc if (0(t), c(t)) G S for all t G [tb, te]. We call an extremal with (0(t), c(t)), t G [0, +rc>):
• a bang-bang extremal if (0(t), c(t)) G S for a.e. t G [0, +rc>),
• mixed extremal if it has a singular arc. Suppose tb <te, we specify
• bang-bang extremals:
^[tb, te] C [0, {(0(t), c(t)) | t G [tb, te]} C D3 U D6,
• mixed extremals:
l[tb, te] C [0, {(0(t), c(t)) | t G [tb, te]} C D3 U D6.
Remark 3. An arc with t g [tb, te] of a normal extremal is called singular when (^(t), c(t)) g D3 U D6, Vt G [tb,te].
Lemma 2. If an arc with t G [tb, te], tb < te, of a normal extremal is singular, then №),c(t)) G D6, Vt G [tb,te].
Proof. Suppose we have an arc with (<fi(t), c(t)) G D3, t G [tb, te], for <fi(tb) G (0, From c(tb) = ± (umax — 1) cos 0(tb), we obtain
3e Vt G [tb, tb + e] c(t) = u1 sin 0(t) > 0, ±<j)(t) ^ 0. Therefore, te = tb. □
Remark 4. We have the following:
• D0, D6 are unstable equilibria.
• D0, D1, D2, D4, D5 contain bang arcs.
• D3 are switching points for bang-bang extremals.
• D6 contains singular arcs.
• D5, D6 contain mixed extremals.
Bang arcs contained in D0 and D4 are trivial straightforward movements and turns with the constant control. Bang-bang trajectories are contained in D1, D2, D3 with undefined control at isolated switching points in D3.
Theorem 2. Mixed extremals arise only with singular arcs of the domain C5 on the level set of the maximized Hamiltonian H = umax• The extremal controls in this case are piecewise constant functions with the values
(1, ±1), (urx, o), (1, ±1), (urx, 0), (1, ±1), (urx, 0), ...
and the corresponding time intervals
T\, TSi, 2n, Ts2 , 2n, TS3 , ...,
where T1 g [0, 2n] and TS G [0, i = 1, 2, ... (assuming that there exists i with Ts. > 0) •
Proofs Follows from Remark 4 and Lemma 1. □
Theorem 3. Bang-bang extremals have a piecewise constant control (u1, u2) with values
(1, ±1), (urx, 0), (1, ^1), (urx, 0), (1, ±1), (urx, 0), ...
(or (urx, 0), (1, ^1), (u5"ax, 0), (1, ±1), (urx, 0), (1, ±1), ...) and corresponding time intervals
tS , tC , tS , tC , tS , ... with T1 < TC (or respectively T1, Tc, Ts, Tc, Ts, Tc, ... with T1 ^ Ts),
2(«max —l) T
where Ts = — ^max—^cot. -f- and Tc G (it, 2ir).
Proof\ Follows from the phase portrait illustrated in Fig. 5, Remark 4 and Lemma 1. Every bang-bang arc is passing through points ((f), c±) G D3 with 0 G (0, and c± = = ± (ufax — 1) cos The value TS is the time required to travel from the point ($, c—) to the
2 (Umax_1) cos 0
point, ((f), c+) with the velocities (f> = 0, c = t/,™axsin0. Therefore, Ts = —. Also, it
is easy to check that TC = 2(n — 0). □
Remark 5. A bang-bang extremal (p(t), q(t)) is quasi-periodic:
p(t + TCs)= p(t), q(t + Tcs) = (x(t) + xCs,y(t) + yes, 0(t)), (2.21)
where Tcs = 2TC + 2TS is the period and constants xcs, ycs do not depend on t G [0,
We summarize that there are two types of normal extremals (mixed and bang-bang) described in detail in Theorems 2 and 3.
3. Upper bound for the cut time
In this section, we provide an upper bound for the cut time, where an extremal loses its global optimality.
Theorem 4. An abnormal extremal with 2 switchings is not optimal. An optimal mixed control is a piecewise constant function with the values (1, ±1), (-umax, 0), (1, ±1) and the corresponding time intervals T1 ^ 0, Ts > 0, T2 ^ 0, where T1 + T2 ^ Tc = 2n.
Proof. Note that projections on the (x, y)-plane of abnormal and mixed extremals for problem (1.7)—(1.12) are also the projections of abnormal and mixed extremals for the classical Markov - Dubins problem. An abnormal or mixed extremal is not optimal for (1.7)—(1.12) if it is not optimal for the Markov —Dubins problem, since case (1.14) is contained in (1.10).
Therefore, the proof of the theorem follows from results for the classical Markov —Dubins
problem, e.g., see [7, Lemma 7] for abnormal extremals and [4, Lemma 3.5] for mixed extremals.
□
Theorem 5. Suppose arcs t £ [0, te] with positive length te > 0 of two extremals (p(t), q(t)), (p(t), q(t)) differently (i. e., q(t) ^ q(t) for t £ [0, te]) connect points q(0) = q(0), q(te) = q(te) (in the configuration space SE(2)).
If for some e > 0, the composition arc t £ [0, te + e] with
q i q(t), fort £ [MeL . 2 (31)
q = < _ i = 1, 2, (3.1)
[q(t), for t £ [te,te + e],
does not coincide with any arc of extremals, then the extremal arcs q(t), t £ [0, te + e], and q(t), t £ [0, te + e], are not optimal.
Proof. The theorem follows from PMP, since an optimal arc has to be extremal. □
Here te is the analogue of the Maxwell time and q(te) = q(te) is the analogue of the Maxwell point (see [11]).
Theorem 6. A bang-bang extremal is not optimal after the full period, i. e., tcut ^ 2TS + + 2Tc.
As a consequence, extremals with 5 switchings are not optimal. For the maximum 4 switchings, the optimal control is a piecewise constant function with the values (1, ±1), (u™ax, 0), (1, T1), (u5"ax, 0), (1, ±1) (or (u5"ax, 0), (1, ^1), (u5"ax, 0), (1, ±1), (u5"ax, 0) with the nonnegative time intervals T1, Ts, Tc, Ts, T2 (T1, Tc, Ts, Tc, T2), satisfying T1 + T2 ^ Tc
2 (7/max_l) T
(2\ + T2 < Ts), where Ts = - [ ¿fax J cot if for Tc e (vr, 2vr).
Proof. Suppose we have a bang-bang extremal (0(t), c(t), q(t)) with (0(0), c(0)) = (0O, c0).
If c0 = 0, then there exists another symmetric bang-bang extremal (<fi(t), c(t), q(t)) with (<q(0), c(0)) = (0O, —c0). One can easily check that q(Tcs) = q(Tcs) (for an illustration, see Fig. 6). Therefore, from Theorem 5, we prove the theorem for c0 = 0.
Otherwise c0 = 0. If such a geodesic is optimal for t £ [0, Tcs + e] for some e > 0, then there exists an optimal geodesic (<fi(t), c(t), q(t)) with (0(0), c(0)) = (<fi0, 5), 5^0 for t e [0, Tcs + |], which contradicts the above case c0 = 0. □
Fig. 6. Examples of the reflection for the full period providing the upper bound for the cut time
y
0.51.0 1.5 -1.0
Fig. 7. Projections of one period of bang-bang extremals on the (x, y)-plane for the value umax =
l+cos( J71-)
1.1716 with T1 = ±f-
We distinguish two types of bang-bang arcs with maximum 4 switchings: ► CSCSC (starting from a circle arc) parameterized by
Tc G (n, 2n), T G (0,Tc], T G (0, 2TC + 2TS];
2
-20 -15 -10 -5
Fig. 8. Projections of one period of bang-bang extremals on the (x, y)-plane for the value u5"ax = 2 with
rp _ TC
J-i- —
• SCSCS (starting from a straight segment) parameterized by
Tc G (n, 2n), Ti G (0, Ts], T G (0, 2TC + 2TS].
Figures 7 and 8 demonstrate bang-bang extremals starting from a circle arc with TC =
= 7T + "if, i = 1
Tr,
., 7, and Tl = We note that for each value t/,™ax there is a unique value of TC, s.t. the corresponding extremal is not only quasi-periodic (see Remark 5), but also periodic, providing the closed extremal trajectory in the configuration space (figure of eight on the (x, y)-plane).
Remark 6. Periodic extremals satisfy the condition
TC = 2n — arccos
2 - u
.max X1
(3.2)
Figures 10 and 11 demonstrate bang-bang extremals starting from a line segment with TC =
= 7T + -if, i = 1, ..., 7, and 2\ = Ts.
4. Conclusion
We obtain a complete description of the extremals for the generalized Markov-Dubins problem with control on a triangle. We distinguish the following types of extremals: abnormal,
1
-0.5 /0.5 ~
\ -2
Fig. 10. The projections of one period of bang-bang extremals on the (x, y)-plane for the value = --« 1.1716 with T-, = To
mixed and bang-bang. For each type, we specify the upper bound for the cut time, where geodesies lose their optimality. The abnormal and mixed extremals inherit the estimate for the cut time from the classical Markov-Dubins problem. We prove that the bang-bang extremals are quasi-periodic and are not optimal after one period. As in Euler's elastic problem, there are two shapes of closed extremals: a circle and a figure of eight. The shape of the figure of
eight depends on u5"ax (in the limit for the classical Markov-Dubins problem, it consists of two tangent circles).
This study lays the foundation for further research into the problem. In particular, the full period of bang-bang extremals illustrated in Figs. 7 and 8 is obviously not optimal, since there is the optimal straight line solution.
Conflict of interest
The author declares that he has no conflict of interest.
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