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Abnormal extremals in the sub-Riemannian problem for a general model of a robot with a trailer A. A. Ardentova, E. M. Artemovab a Ailamazyan Program Systems Institute of Russian Academy of Sciences, Ves'kovo, Pereslavl' district, Yaroslavl' oblast', Russia b Ural Mathematical Center, Udmurt State University, Izhevsk, Russia
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 10, DOI: https://doi.org/10.4213/sm9829 
Bibliographic databases:
    § 1. IntroductionAbnormal trajectories are of fundamental importance for investigations of control problems (see [1]); they arise as solutions of optimal control problems independently of the functional to be minimized, in particular, in sub-Finsler problems (see [2] and [3]), including sub-Riemannian ones (see [4]–[6]). The question of the optimality of abnormal trajectories is quite acute for sub-Riemannian geometry (see [7]), and the structure properties of abnormal sub-Riemannian geodesics (see [8]), connected with the subanalyticity of the corresponding metric (see [9] and [10]), are of great interest. For example, in nilpotent sub-Riemannian problems on the Engel and Cartan groups (see [11] and [12]) — as well as in the not left-invariant three-dimensional sub-Riemannian problem in the Martinet flat case (see [13]) — the subanalyticity condition fails just in a neighbourhood of the points where abnormal trajectories arrive. A long-standing question concerns the optimality of abnormal geodesics (see [14]–[16]). In the recent paper [17] its authors advanced significantly towards the solution of this problem: they showed that optimal abnormal trajectories cannot have corners. In this paper we investigate abnormal extremals in the control problem for a wheeled robot with a trailer in the framework of a sub-Riemannian problem, in which it is shown that such extremals are nonstrictly abnormal (see [18]) and, as a consequence, their smoothness is proved. Any abnormal trajectory can be parametrized arbitrarily with respect to time (even the direction can be switched to the opposite one), so for definiteness we fix a parametrization with control on an ellipse, the same as the parametrization of controls in the normal case. We also discard the case when the motion along a trajectory changes direction, for it is certainly not optimal. Note that the sub-Riemannian problem under consideration is open and difficult. The description of abnormal trajectories is only known in the case when the trailer is coupled to the robot at the centre of the drive axis of the robot [19]. Subsequently, this result was extended in [20], where the extremal trajectories were described in a sub-Finsler problem for the same model of a robot with a trailer (coupled at the centre of the drive axis) with control on a square, as for the Reeds-Shepp car. The sub-Riemannian problem (that is, a sub-Finsler problem with control in a disc) was considered for the model of a robot without trailer [21]–[23], when the configuration space has dimension three and the control is two-dimensional. To date, only one four-dimensional sub-Riemannian problem with two controls has been solved, namely, the nilpotent sub-Riemannian problem on the Engel group [24]. Our work lays foundation for the solution of a two-parameter family (where the parameters are the distances of the coupling point to the centres of the robot and the trailer) of not left-invariant sub-Riemannian problems with two-dimensional control, which have so far been only solved approximately in the class of sub-optimal controls by means of nilpotent approximation [25], [26]. § 2. Optimal control problemConsider a controlled motion without slipping of a mobile wheeled robot on a plane. Assume that the robot consists of the main (driving) wheelset and a trailer coupled to it. The trailer is meant to be a passive wheelset attached to the robot at some point (Figure 1). We introduce a fixed coordinate system $Oxy$ in order to describe the motion of the robot with a trailer. We describe the position of the wheelset in $Oxy$ by coordinates $(x, y)$, and we describe its orientation by a rotation angle $\vartheta$. The position of the trailer relative to the main unit is described by an angle $\varphi$. Thus, the position of the robot with a trailer on the plane is described by a vector
$$
\begin{equation*}
q=(x, y, \vartheta, \varphi) \in M=\mathbb{R}_{x, y}^2 \times S_{\vartheta}^1 \times S_{\varphi}^1.
\end{equation*}
\notag
$$
The no-slipping conditions on the wheels of the robot and trailer impose two nonholonomic constraints on the system in question (see [27]):
$$
\begin{equation}
v_2=0\quad\text{and}\quad v_1 \sin \varphi+v_2 \cos \varphi+\omega l_r \cos \varphi+l_t (\omega+ \dot{\varphi})=0.
\end{equation}
\tag{2.1}
$$
Here $(v_1, v_2)$ are the components of the linear velocity of the centre of the robot in the moving coordinate system attached to the robot, $\omega$ is the angular velocity of the robot, and $l_r \geqslant 0$ and $l_t > 0$ are the distances of the centres of the robot and the trailer, respectively, to the coupling point. In what follows, for brevity we identify in view of (2.1) the first component of the linear velocity $v_1$ with the linear velocity, so that $v:=v_1$. Taking the constraints (2.1) into account, the system of equations describing the motion of the robot with a trailer assumes the following form (see [28]):
$$
\begin{equation}
\begin{cases} \dot{x}=v \cos \vartheta, \\ \dot{y}=v \sin \vartheta, \\ \dot{\vartheta}=\omega, \\ \displaystyle\dot{\varphi}=-\frac{1}{l_t}(v \sin \varphi+\omega (l_t+l_r \cos \varphi)). \end{cases}
\end{equation}
\tag{2.2}
$$
For the system under consideration we consider the sub-Riemannian problem
$$
\begin{equation}
\dot{q}=v X_1+\omega X_2, \qquad q \in M, \quad u=(v, \omega) \in \mathbb{R}^2,
\end{equation}
\tag{2.3}
$$
with the boundary conditions
$$
\begin{equation}
q(0)=q_0=(x_0, y_0, \vartheta_0, \varphi_0)\quad\text{and} \quad q(t_1)=q_1=(x_1, y_1, \vartheta_1, \varphi_1)
\end{equation}
\tag{2.4}
$$
and the cost functional
$$
\begin{equation}
\int_0^{t_1} \sqrt{v^2+\mu^2 \omega^2} \, dt \to \min, \qquad \mu= \mathrm{const},
\end{equation}
\tag{2.5}
$$
where $u=(v, \omega)$ is the vector of controls, the positive parameter $\mu$ determines a balance between the linear and angular velocities in the functional, and $X_1$ and $X_2$ are vector fields of the form
$$
\begin{equation}
X_1=\biggl(\cos \vartheta, \sin \vartheta, 0, -\frac{\sin \varphi}{l_t}\biggr)^\top\quad\text{and} \quad X_2=\biggl(0, 0, 1, -\frac{l_r \cos \varphi}{l_t}-1\biggr)^\top.
\end{equation}
\tag{2.6}
$$
Since the problem is invariant with respect to translations and rotations of the plane, we can assume that By extending the natural scaling symmetry in the plane, we obtain continuous symmetries of system (2.2) under consideration:
$$
\begin{equation}
(x, y, \vartheta, \varphi, v, \omega, l_t, l_r) \mapsto (\eta x, \eta y, \vartheta, \varphi, \eta v, \omega, \eta l_t, \eta l_r), \qquad \eta=\mathrm{const} > 0.
\end{equation}
\tag{2.8}
$$
Theorem 1. Let $l_r \geqslant 0$ and $l_t > 0$. Then $\bullet$ if $l_r \neq l_t$, then system (2.2) is completely controllable and has the growth vector $(2,3,4)$; $\bullet$ if $l_r=l_t$, then system (2.2) is not completely controllable and has two invariant sets, $\{\varphi=\pi\}$ and $\{\varphi \neq \pi\}$ (with growth vectors $(2,3)$ and $(2,3,4)$, respectively). The restriction of (2.2) to each invariant set is completely controllable. Proof. We calculate the commutators of the vector fields in (2.6):
$$
\begin{equation}
\begin{gathered} \, X_3=[X_1, X_2]=\biggl(\sin \vartheta, -\cos \vartheta, 0, \frac{-l_r-l_t \cos \varphi}{l_t^2}\biggr)^\top, \\ X_4=[X_1, X_3]=\biggl(0, 0, 0, \frac{-l_t-l_r \cos \varphi}{l_t^3}\biggr)^\top, \\ X_5=[X_2, X_3]=\biggl(\cos \vartheta, \sin \vartheta, 0, \frac{(l_r^2-l_t^2) \sin \varphi}{l_t^3}\biggr)^\top. \end{gathered}
\end{equation}
\tag{2.9}
$$
The matrix formed by the vector fields (2.6) and (2.9) has rank four if $l_r \neq l_t$ or $l_r=l_t$ and $\varphi \neq \pi$. Otherwise the rank is three. For $l_r=l_t$ and $\varphi=\pi$, from (2.2) we obtain $\dot{\varphi}=0$, which means the lack of control for the angle $\varphi$.
Thus, by the Rashevskii-Chow theorem (see [29]), for $l_r \geqslant 0$ and $l_t > 0$, provided that $l_r \neq l_t$, system (2.2) is completely controllable. That (2.2) is controllable for $l_r=l_t$ after restricting to the invariant sets $\{\varphi=\pi\}$ and $\{\varphi \neq \pi\}$ is also a consequence of the Rashevskii-Chow theorem. The proof is complete. The sub-Riemannian problem on the group of motions of a plane (that is, for the robot without trailer) is closely connected with ‘bicycle mathematics’: see [30]. By analogy, we can represent the model of the robot with a trailer under consideration by two bicycles with a common ‘front wheel’ at the coupling point and two ‘rear wheels’ at the centres of the robot and the trailer. We can write an equivalent statement of problem (2.3)–(2.6) in question as a problem of finding the shortest trajectory for the coupling point. Proposition 1. Problem (2.3)–(2.6) for $\mu\!=\!l_r$ is equivalent to the sub-Riemannian problem with control system Remark 1. In problem (2.10)–(2.12) the equations for $\vartheta$ and $\phi$ coincide up to changes of the notation for the angles and the corresponding quantities $l_r$ and $l_t$. Using this symmetry between the robot and the trailer and the equivalence between problems (2.3)–(2.6) for $\mu=l_r$ and (2.10)–(2.12), in the original problem (2.3)–(2.6) for $\mu=l_r$ we have symmetry between the trajectories of the centre of the wheelset of the trailer and the trajectories of the robot $(x,y)$ (in a similar problem with $l_t$ and $l_r$ interchanged). This symmetry corresponds to the reflection in the bisector of the angle between the arms to the robot and trailer at the coupling point. § 3. Pontryagin maximum principleThat optimal trajectories exist in problem (2.3)–(2.6) is a consequence of Filippov’s theorem (see [18]) and Theorem 1 on solvability (after the restriction to the invariant sets $\{\varphi=\pi\}$ and $\{\varphi \neq \pi\}$ for $l_r=l_t$). By the Cauchy-Schwarz-Bunyakowskii inequality the problem of the minimization of sub-Riemannian length (2.5) is equivalent to the action minimization problem
$$
\begin{equation}
\int_0^{t_1} \frac{v^2+\mu^2 \omega^2}{2} \,dt \to \min,
\end{equation}
\tag{3.1}
$$
for fixed terminal time $t_1$. To the equivalent optimal control problem (2.3), (2.4), (2.6), (3.1) we can apply the Pontryagin maximum principle; see [18] and [31]. To this end we introduce a number $\nu \leqslant 0$, the vector of dual variables $\psi=(\psi_1, \psi_2, \psi_3, \psi_4)$ and the Pontryagin function
$$
\begin{equation*}
\begin{aligned} \, H^{\nu}(\psi, q, u) &= \frac{\nu (v^2+\mu^2 \omega^2)}{2}+\psi_1v \cos \vartheta+ \psi_2 v \sin \vartheta+\psi_3\omega \\ &\qquad-\frac{\psi_4}{l_t} \bigl(v \sin \varphi+\omega (l_t+l_r \cos \varphi)\bigr). \end{aligned}
\end{equation*}
\notag
$$
The equations for the dual variables can be represented in the following form:
$$
\begin{equation}
\begin{gathered} \, \dot{\psi}_1=-\dfrac{\partial H^{\nu}}{\partial x}=0, \qquad \dot{\psi}_3= -\dfrac{\partial H^{\nu}}{\partial \vartheta}=\psi_1 v \sin \vartheta-\psi_2 v \cos \vartheta, \\ \dot{\psi}_2=-\dfrac{\partial H^{\nu}}{\partial y}=0, \qquad \dot{\psi}_4=-\dfrac{\partial H^{\nu}}{\partial \varphi}=\frac{\psi_4}{l_t} (v \cos \varphi-l_r \omega \sin \varphi). \end{gathered}
\end{equation}
\tag{3.2}
$$
The Pontryagin maximum principle yields the following maximality condition:
$$
\begin{equation}
\max_{u \in \mathbb{R}^2} H^{\nu}(\psi (t), \mathbf{q}(t), u)= H^{\nu}(\psi (t), \mathbf{q}(t), \mathbf{u}(t)),
\end{equation}
\tag{3.3}
$$
where $\mathbf{u}(t)$ and $\mathbf{q}(t)$ are the optimal control and optimal trajectory, respectively. The following nontriviality condition also holds: For further examination of extremal trajectories we go over to the variables $h_i (\lambda)=\langle \lambda, X_i \rangle$, $i=1, 2, 3$, $\lambda=(\psi, q)$, expressed as follows:
$$
\begin{equation}
\begin{gathered} \, h_1=\psi_1 \cos \vartheta+\psi_2 \sin \vartheta-\frac{\psi_4}{l_t} \sin \varphi, \qquad h_2=\psi_3-\psi_4-\frac{\psi_4}{l_t} l_r \cos \varphi, \\ h_3=\psi_1 \sin \vartheta-\psi_2 \cos \vartheta-\frac{\psi_4}{l_t^2} (l_r+l_t \cos \varphi) \end{gathered}
\end{equation}
\tag{3.5}
$$
and $h_4=\psi_4$ (because the expressions $\langle \lambda, X_4 \rangle$ and $\langle \lambda, X_5 \rangle$ have singularities). In the variables $h_i$, $i=1,\dots,4,$ the vertical subsystem (3.2) reads
$$
\begin{equation}
\begin{gathered} \, \dot{h}_1=-\omega h_3, \qquad \dot{h}_3=h_1 \omega+\frac{l_r^2 h_4 \omega}{l_t^3} \sin \varphi- \frac{h_4 v}{l_t^3} (l_t+l_r \cos \varphi), \\ \dot{h}_2=v h_3, \qquad \dot{h}_4=\frac{h_4}{l_t} (v \cos \varphi-\omega l_r \sin \varphi), \end{gathered}
\end{equation}
\tag{3.6}
$$
and the Pontryagin function takes the form
$$
\begin{equation}
H^{\nu} (\lambda, u)=\frac{\nu}{2} (v^2+\mu^2 \omega^2)+v h_1(\lambda)+\omega h_2(\lambda), \qquad \lambda=(\psi,q).
\end{equation}
\tag{3.7}
$$
Now we look at the normal ($\nu=-1$) and abnormal ($\nu=0$) cases. § 4. Normal Hamiltonian systemFrom maximality condition (3.3) we obtain the following expressions for the extremal controls in the normal case of the Pontryagin maximum principle: Taking the velocities (4.1) into account, the Hamiltonian system assumes the form
$$
\begin{equation}
\begin{gathered} \, \dot{h}_1=-\frac{h_2 h_3}{\mu^2}, \qquad \dot{h}_2=h_1 h_3, \\ \dot{h}_3=\frac{h_1 h_2}{\mu^2}+\frac{l_r^2 h_4 h_2}{l_t^3 \mu^2} \sin \varphi- \frac{h_1 h_4}{l_t^3} (l_t+l_r \cos \varphi), \\ \dot{h}_4=\frac{h_4}{l_t \mu^2} (h_1 \mu^2 \cos \varphi-h_2 l_r \sin \varphi), \\ \dot{x}=h_1 \cos \vartheta, \qquad \dot{y}=h_1 \sin \vartheta, \qquad \dot{\vartheta}= \frac{h_2}{\mu^2}, \\ \dot{\varphi}=-\frac{1}{l_t \mu^2}\bigl(h_1 \mu^2 \sin \varphi+h_2 (l_t+l_r \cos \varphi)\bigr). \end{gathered}
\end{equation}
\tag{4.2}
$$
System (4.2) has an integral coinciding with the Hamiltonian
$$
\begin{equation}
\mathcal{H}=\frac{1}{2} \biggl( h_1^2+\frac{h_2^2}{\mu^2} \biggr).
\end{equation}
\tag{4.3}
$$
For $h_4=0$ the system of equations (4.2) looks simpler:
$$
\begin{equation}
\dot{h}_1=-\frac{h_2 h_3}{\mu^2}, \qquad \dot{h}_2=h_1 h_3, \qquad \dot{h}_3=\frac{h_1 h_2}{\mu^2},
\end{equation}
\tag{4.4}
$$
$$
\begin{equation}
\dot{x}=h_1 \cos \vartheta, \qquad \dot{y}=h_1 \sin \vartheta, \qquad \dot{\vartheta}= \frac{h_2}{\mu^2},
\end{equation}
\tag{4.5}
$$
$$
\begin{equation}
\dot{\varphi}=-\frac{1}{l_t \mu^2}\bigl(h_1 \mu^2 \sin \varphi+h_2 (l_t+l_r \cos \varphi)\bigr).
\end{equation}
\tag{4.6}
$$
Remark 2. The system of equations (4.4), (4.5) defines a normal Hamiltonian system in the analogous problem for a robot without trailer (the sub-Riemannian problem on the group of motions of a plane). For $\mu=1$ system (4.4), (4.5) was fully investigated in the series of papers [21]–[23]. Equations (4.4) with respect to $h_1$, $h_2$ and $h_3$ can be split off from system (4.4)–(4.6). We fix the level $\mathcal{H}=1/2$ of the integral and perform reduction. To do this, we make the change of variables
$$
\begin{equation}
h_1=\cos \beta, \quad h_2=\mu \sin \beta, \qquad \text{where }\ \beta \in S_\beta^1, \quad \mu>0.
\end{equation}
\tag{4.7}
$$
Then the vertical subsystem (4.4) acquires the form
$$
\begin{equation}
\dot{\beta}=\frac{h_3}{\mu}, \qquad \dot{h}_3=\frac{1}{2 \mu} \sin 2 \beta.
\end{equation}
\tag{4.8}
$$
In the variables $\delta=2 \beta+\pi$, $c=2 h_3$ system (4.8) coincides with the classical system describing the motion of a mathematical pendulum:
$$
\begin{equation}
\dot{\delta}=\frac{c}{\mu}, \qquad \dot{c}=-\frac{\sin \delta}{\mu}, \quad\text{where } \delta \in 2 S_{\delta}^1=\mathbb{R}/(4 \pi \mathbb{Z}), \qquad c\in \mathbb{R}.
\end{equation}
\tag{4.9}
$$
This system has the energy integral We show level curves of this integral in Figure 2. The horizontal subsystem (4.5), (4.6) reads
$$
\begin{equation}
\dot{x} =\sin \frac{\delta}{2} \cos \vartheta, \qquad \dot{y}=\sin \frac{\delta}{2} \sin \vartheta, \qquad \dot{\vartheta}=-\frac{1}{\mu} \cos \frac{\delta}{2},
\end{equation}
\tag{4.11}
$$
$$
\begin{equation}
\dot{\varphi} =-\frac{1}{l_t \mu}\biggl(\mu \sin \frac{\delta}{2} \sin \varphi-\cos \frac{\delta}{2} (l_t+l_r \cos \varphi)\biggr).
\end{equation}
\tag{4.12}
$$
For $h_4=0$ the system of normal extremal trajectories is parametrized by points in the cylinder
$$
\begin{equation}
C_0=T^*_{q_0} M \cap \biggl\{\mathcal{H}=\frac1 2\biggr\} \cap \{h_4=0\}=\{(\delta, c) \in (2 S_{\delta}^1) \times \mathbb{R}\},
\end{equation}
\tag{4.13}
$$
which can be partitioned as follows into subsets depending on the energy $E$ (see Figure 2):
$$
\begin{equation*}
\begin{aligned} \, C_0&=\bigcup_{i=1}^5 C_{0i}, \qquad C_{0i} \cap C_{0j}=\varnothing, \qquad i \neq j, \\ C_{01}&=\{ (\delta, c) \in C_0 \mid E \in (-1, 1) \}, \\ C_{02}&=\{ (\delta, c) \in C_0 \mid E \in (1,+\infty) \}, \\ C_{03}&=\{ (\delta, c) \in C_0 \mid E=1,\, c \neq 0 \}, \\ C_{04}&=\{ (\delta, c) \in C_0 \mid E=1,\, c=0,\, \delta=(2 n+1) \pi \}, \\ C_{05}&=\{ (\delta, c) \in C_0 \mid E=-1,\, c=0,\, \delta=2 n \pi \}. \end{aligned}
\end{equation*}
\notag
$$
For $(\delta, c) \in C_{04}$ the robot rotates at one place, so that the coupling point draws a circle for $l_r \neq 0$ (for $l_r=0$ this point is fixed). For $(\delta, c) \in C_{05}$ the robot and the coupling point move along a straight line. In the general case $(\delta, c) \in \bigcup_{i=1}^3 C_{0i}$ we present the rectifying coordinates, the solutions of (4.9) and explicit expressions for trajectories in § 7. In Figures 3–5 we show trajectories of the robot calculated analytically (continuous black lines) and trajectories of the coupling point (dashed black lines) and the centre of the trailer (grey lines) calculated numerically. Explicit formulae for trajectories of the coupling point and the trailer are only known in a few special cases; in general, we do not know formulae for these trajectories in the case when $\mu \neq l_r$. Remark 3. For $\mu=l_r \,{\neq}\, 0$ the coupling point draws a non-inflectional elastica (see [30]). However, for $\mu \neq l_r$, generically for $(\delta, c) \in \bigcup_{i=1}^3 C_{0i}$ the trajectories of the coupling point are not elasticae. The more general case of $h_4\neq 0$ is still to be investigated. In what follows we show that abnormal extremal trajectories satisfy the system of equations (4.9), (4.11), (4.12). § 5. Abnormal extremalsConsider the case $\nu=0$. Then from the maximality condition (3.3), taking (3.7) and (3.6) into account it follows that
$$
\begin{equation}
\frac{\partial H^{\nu}}{\partial v}=h_1=0\quad\text{and} \quad \frac{\partial H^{\nu}}{\partial \omega}=h_2=0.
\end{equation}
\tag{5.1}
$$
Next, by (3.6) we have $\mu^2 \dot{h}_1^2+\dot{h}_2^2=(v^2+\mu^2 \omega^2) h_3^2$. Since we can assume that $v^2+\mu^2 \omega^2 \neq 0$ without loss of generality, by (5.1) we obtain Bearing in mind (5.1) and (5.2), the nontriviality condition of the Pontryagin maximum problem (3.4) can be written as Proposition 2. In problem (2.3)–(2.6) under consideration, the abnormal extremals $\lambda=(\psi, q)$ are ‘nice’ (see [32], Definition 12.29), that is, they satisfy the condition $\lambda \in (\mathcal{D}^2)^{\bot}\setminus(\mathcal{D}^3)^{\bot}$, where $\mathcal{D}=\operatorname{span}\{X_1,X_2\}$. Proof. The condition $\lambda \in (\mathcal{D}^2)^{\bot}\setminus(\mathcal{D}^3)^{\bot}$ in the proposition is equivalent to
$$
\begin{equation*}
h_1(\lambda)=h_2 (\lambda)=h_3 (\lambda)=0, \qquad h_4 (\lambda) \neq 0;
\end{equation*}
\notag
$$
see [32], Theorem 12.31. Hence it holds by (5.1)–(5.3). Corollary 1 (see [32], Theorem 12.34). Small arcs of abnormal trajectories are optimal in problem (2.3)–(2.6). Consider the following equality, which holds because $h_1=h_3=0$: Using (3.5) in identity (5.4) we return to the original coordinates $\psi_i$, $i=1,\dots,4$, and then go over to the polar variables $\psi_1=\psi_{12} \sin \alpha$ and $\psi_2=\psi_{12} \cos \alpha$, where $\alpha \in S_{\alpha}^1$ and $\psi_{12} > 0$ (it is impossible that $\psi_{12}=0$ under (5.1)–(5.3)). Simplifying the expression (5.4) in the new variables, we obtain where $\gamma=\varphi+\vartheta+\alpha \in S_{\gamma}^1$ is the new angle, describing the position of the trailer with respect to the $Ox$-axis rotated through $\alpha \equiv \mathrm{const}$.Differentiating (5.5) with respect to time we obtain the following equality for the abnormal controls: Now we consider the cases $l_r=0$ and $l_r > 0$ separately. 5.1. Coupling point on the drive axisProposition 3. Abnormal trajectories of the control problem (2.3), (2.4) for $l_r=0$ are rotations of the robot at one place with a static trailer: This follows directly from (5.5) and (5.6). 5.2. Coupling point away from the drive axisConsider the case of $l_r > 0$. Abnormal trajectories can be parametrized arbitrarily with respect to time: in other words, if $(v (t), \omega(t) )$, $t\in[0,t_1]$, is an abnormal control for a trajectory $q(t)$, then after a smooth change of the variable $t=f(s)$ satisfying
$$
\begin{equation*}
f(0)=0, \qquad f(s_1)=t_1\quad\text{and} \quad f'(s)>0,
\end{equation*}
\notag
$$
we obtain a control $(f'(s) v (s), f'(s) \omega(s) )$, $s\in[0,s_1]$, corresponding to the same abnormal trajectory $ q (f(s))$ but with reparametrized time. Because of this invariance, without loss of generality we can introduce the constraint $v^2+l_r^2 \omega^2 \equiv 1$ on the velocities, which is similar to the constraint $v^2+\mu^2 \omega^2 \equiv 1$, $\mu>0$, arising in the normal case of the Pontryagin maximum principle. From (5.6) we find expressions for abnormal controls:
$$
\begin{equation}
v=\sin \gamma\quad\text{and} \quad \omega=\frac{\cos \gamma}{l_r}.
\end{equation}
\tag{5.8}
$$
Remark 5. There exists a symmetry that takes the solution with control $(v,\omega)$ (5.8) to the solution corresponding to the control $(-v,-\omega)$. It transforms the values $(\gamma, \alpha)$ of the parameters into $(\pi+\gamma, \pi+\alpha)$. Under its action an extremal trajectory $q(t)$, $t\in[0,+\infty)$, is inverted to $q(-t)$, $t\in[0,+\infty)$. So without loss of generality, taking this symmetry into account we can consider controls of the form (5.8) alone. Proposition 4. Abnormal trajectories of the control problem (2.3), (2.4) for $l_r > 0$ satisfy the system of equations This is a direct consequence of formulae (5.8) for the controls, the original system (2.3) and the integral (5.5). Lemma 1. For $l_r > 0$ equations (5.10) and (5.11) reduce to the system of differential equations Proof. Differentiating (5.10) while taking (5.11) into account, we obtain Then, after the substitution $\gamma=\pi-\delta/2$, $\dot{\gamma}=-{c}/(2l_r)=-\dot{\delta}/2$, we derive system (5.12).
Taking the initial conditions $\vartheta(0)=0$ and $\varphi(0)=\varphi_0$ into account we obtain
$$
\begin{equation*}
\delta(0)=2(\pi-\varphi_0-\alpha), \qquad c(0)=l_r \dot{\delta}(0)=- 2 l_r \dot{\gamma}(0)= \frac{2 l_r \cos \alpha}{l_t}.
\end{equation*}
\notag
$$
Finally, it follows from (5.11) that the identity
$$
\begin{equation*}
(l_t \cos \varphi_0+l_r) \sin \alpha+l_t \sin \varphi_0 \cos \alpha=0
\end{equation*}
\notag
$$
holds. From it, taking Remark 5 into account, we derive (5.14). The lemma is proved. Remark 6. System (5.12) coincides with the equation of a mathematical pendulum (4.9) for $\mu=l_r > 0$. Proposition 5. Initial condition (5.13) for the pendulum equation (5.12) fixes the energy level of the pendulum equal to $E=2 l_r^2/l_t^2-1$. Proof. This result can be verified by substituting the initial condition (5.13) directly into the expression (4.10) for the energy and taking (5.14) into account: The proof is complete. The set of pairs of abnormal extremal trajectories for problem (2.3), (2.4), (2.7), (3.1) is parametrized for $l_r>0$ by points in the set of initial data
$$
\begin{equation*}
\mathcal{A}=\{(l_r, l_t, \varphi_0) \mid l_r > 0, \, l_t > 0, \, \varphi_0 \in S^1\}.
\end{equation*}
\notag
$$
In view of Proposition 5 this set is partitioned into the subsets corresponding to types of trajectories (similarly to the partition of $C_0$ in the normal case for $h_4=0$); namely,
$$
\begin{equation*}
\begin{aligned} \, \mathcal{A} &=\bigcup_{i=1}^4 \mathcal{A}_i, \qquad \mathcal{A}_i \cap \mathcal{A}_j= \varnothing, \quad i \neq j, \\ \mathcal{A}_1 &=\{ (l_r, l_t, \varphi_0) \mid 0 < l_r < l_t \}, \\ \mathcal{A}_2 &=\{ (l_r, l_t, \varphi_0) \mid l_r > l_t \}, \\ \mathcal{A}_3 &=\{ (l_r, l_t, \varphi_0) \mid l_r=l_t > 0,\, \varphi_0 \neq \pi \} \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathcal{A}_4 =\{ (l_r, l_t, \varphi_0) \mid l_r=l_t > 0,\, \varphi_0=\pi \}.
\end{equation*}
\notag
$$
Remark 7. Although in the domain $\mathcal{A}_3 \cup \mathcal{A}_4$ the system is globally uncontrollable (because $l_r=l_t$), we also consider this case for completeness, bearing in mind that the restriction of the original system (2.2) to both $\mathcal{A}_3$ and $\mathcal{A}_4$ is controllable by Theorem 1. For $(l_r, l_t, \varphi_0)\in \mathcal{A}_4$ the robot, coupling point and trailer move along a straight line. Theorem 2. For any fixed $l_r \geqslant 0$, $l_t>0$ and $\varphi_0$, in problem (2.3), (2.4), (3.1), up to a reparametrization in time there exist two symmetric abnormal trajectories $q(t)$ and $\widetilde{q}(t)$, $t\in[0,+\infty)$, such that one trajectory is the analytic continuation of the other in the reverse direction of time: $q(t)=\widetilde{q}(-t)$ $\forall\, t \in \mathbb{R}$. Proof. For $l_r=0$ the assumptions of the theorem hold because of Proposition 3. It follows from Lemma 1 that for $l_r>0$ there exist a unique tuple of initial data $(\delta(0), c(0))$ for the trajectory of the pendulum (5.12), and therefore there exists a unique trajectory of the robot with a trailer corresponding to it, which is smooth in $M$. Thus, by Remark 5 we obtain two trajectories symmetric one to the other. The proof is complete. Proposition 6. In the sub-Riemannian problem (2.3), (2.4), (2.6), (3.1) for $\mu=l_r > 0$ abnormal trajectories are nonstrictly abnormal. The centre of the robot traverses a trajectory coinciding with a sub-Riemannian geodesic for the robot without trailer, and the coupling point of the robot with the trailer always draws a non-inflectional elastica. This is a consequence of Remarks 2 and 3, Proposition 4 and Lemma 1. Proposition 7. In the cases of the domains $\mathcal{A}_1$ and $\mathcal{A}_2$ the trajectories in the $(x,y)$-plane of the centre of the robot coincide with the trajectories of the centre of the trailer in the cases of the domains $\mathcal{A}_2$ and $\mathcal{A}_1$, respectively, up to a reflection in the plane. This is a consequence of Remark 1. We present details of the integration of system (5.9), (5.12)–(5.14) with initial condition $(x_0, y_0, \vartheta_0)=(0,0,0)$ in the general case for the domains $\mathcal{A}_1$, $\mathcal{A}_2$ and $\mathcal{A}_3$ in § 8. In Figure 6 we show trajectories of the robot (continuous black lines), the coupling point (dashed black lines) and the centre of the wheelset of the trailer for the different domains $\mathcal{A}_i$, $i= 1,2,3$. Theorem 3. Let $(l_r,l_t,\varphi_0)\in \bigcup_{i=1}^3 \mathcal{A}_i$. If in problem (2.3), (2.4), (2.6), (3.1), in place of the initial condition (2.7) one specifies $(x_0,y_0,\vartheta_0)=(l_t \cos(\varphi_0+\alpha),0,\alpha)$ (that is, one draws the $x$-axis parallel to the straight line through the centres of the robot and trailer), then the abnormal trajectory is monotone with respect to the variables $y$, $\widehat{y}$ and $\overline{y}$ and bounded (and also periodic for $\mathcal{A}_1$ and $\mathcal{A}_2$) with respect to $x$, $\widehat{x}$ and $\overline{x}$ in the following sense: Trajectories of the robot $(x_t,y_t)$ and the trailer $(\overline{x}_t,\overline{y}_t)$ coincide with the relevant sub-Riemannian geodesics on the planar motion group on which the energies of the pendulum (with respect to the dual variables) are mutually inverse. The trajectory of the coupling point $(\widehat{x}_t,\widehat{y}_t)$ is a non-inflectional elastica or a straight line. Moreover, $y_t=\overline{y}_t$ for all $ t\in\mathbb{R}$. Proof. Consider the case when $(l_r, l_t, \varphi_0) \in \mathcal{A}_1$; then from (8.8), (8.11) and (8.15) (for $(x_0, y_0, \vartheta_0)=(0, 0, 0)$) we can deduce explicit formulae for the trajectories $(x_t,y_t)$ of the robot, $(\widehat{x}_t,\widehat{y}_t)$ of the coupling point and $(\overline{x}_t,\overline{y}_t)$ of the trailer that satisfy the condition $(x_0,y_0,\vartheta_0)=(l_t \cos(\varphi_0+\alpha),0,\alpha)$:
$$
\begin{equation}
\begin{aligned} \, \begin{pmatrix}x_t \\ y_t \end{pmatrix} &=l_t \begin{pmatrix} \operatorname{dn} \sigma_t \\ \mathrm{E}(\sigma_t)-\mathrm{E}(\sigma_0)-\dfrac{t}{l_r} \end{pmatrix}, \\ \begin{pmatrix} \widehat{x}_t\\ \widehat{y}_t \end{pmatrix} &=l_t \begin{pmatrix} \operatorname{dn} \sigma_t \\ \mathrm{E}(\sigma_t)-\mathrm{E}(\sigma_0)-\dfrac{t}{l_r} \end{pmatrix}-l_r \begin{pmatrix} \operatorname{cn} \sigma_t \\ \operatorname{sn} \sigma_t \end{pmatrix}, \\ \begin{pmatrix} \overline{x}_t\\ \overline{y}_t \end{pmatrix} &=\begin{pmatrix} - l_r \operatorname{cn} \sigma_t \\ l_t \biggl( \mathrm{E}(\sigma_t)-\mathrm{E}(\sigma_0)-\dfrac{t}{l_r} \biggr) \end{pmatrix}. \end{aligned}
\end{equation}
\tag{5.18}
$$
Inequalities (5.15) are consequences of the formulae for $x_t$, $\widehat{x}_t$ and $\overline{x}_t$ in (5.18) and the definitions of the elliptic functions $\operatorname{sn}$, $\operatorname{cn}$ and $\operatorname{dn}$ (see [33]). It is also a straightforward consequence of (5.18) that $y_t=\overline{y}_t$. For $(l_r, l_t, \varphi_0) \in \mathcal{A}_2$ or $(l_r, l_t, \varphi_0) \in \mathcal{A}_3$ the proof has a similar structure. The theorem is proved. § 6. ConclusionWe have considered optimal control problem for the robot with a trailer (2.3)–(2.6). We have proved that for fixed values of $l_r$, $l_t$ and $\varphi_0$ there exist precisely two mutually symmetric (with respect to time reversal) extremal trajectories (similarly to the sub-Riemannian problem on the Engel group — see [24] — providing a nilpotent approximation to this problem; see [26]). The centre of the robot and the centre of the trailer draw trajectories coinciding with normal trajectories of the Pontryagin maximum principle in the sub-Riemannian problem for a robot without trailer, first described in [21]. All abnormal extremals in the problem under consideration are ‘nice’ (in the sense of [32]) and, as a consequence, small arcs of these trajectories are optimal. Furthermore, the cut time (after which global optimality is lost) for trajectories of a robot without trailer, which was calculated in [22], provides a lower bound for the cut time in our problem. In particular, trajectories that are straight lines or tractrices are metric lines: all segments of them are optimal. We have shown that the coupling point of the robot and the trailer draws noninflectional Euler elasticae, which agrees with the results in [30]. The shapes of trajectories of the robot, the coupling point and the trailer depend on the relation between $l_r$ and $l_t$, and the initial points of these trajectories depends monotonically on the initial angle $\varphi_0$ of the trailer. The directrices of the trajectories of the robot, the coupling point and the trailer are codirected and orthogonal to the line connecting the centres of the robot and the trailer. It is still an open question whether or not a normal Hamiltonian system (4.2) is Liouville integrable in the generic case of $h_4 \neq 0$. The authors are grateful to Yu. L. Sachkov and A. A. Kilin for useful discussions. § 7. Appendix: formulae for normal trajectories for $h_4=0$Below we use Jacobi elliptic functions (see [33]): $\operatorname{sn}$, $\operatorname{cn}$, $\operatorname{dn}$, the Jacobi amplitude $\operatorname{am}$, the incomplete elliptic integral of the second kind $\mathbf{E}$ and the complete elliptic integral of the first kind $K$; we also set $\mathrm{E} (u)=\mathbf{E}(\operatorname{am} u)$. For each domain $C_{0i}$, $i=1,2,3$, we use the following rectifying coordinates (see [21]):
$$
\begin{equation}
(\delta, c) \in C_{01} \quad\Longrightarrow\quad k=\sqrt{\dfrac{E+1}{2}}, \qquad \dfrac{c}{2}=k \operatorname{cn} (\sigma, k), \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
\sin \dfrac{\delta}{2}=s_1 k \operatorname{sn} (\sigma, k), \qquad \cos \dfrac{\delta}{2}=s_1 \operatorname{dn} (\sigma, k);
\end{equation}
\tag{7.1}
$$
$$
\begin{equation}
(\delta, c) \in C_{02} \quad\Longrightarrow\quad k=\sqrt{\dfrac{2}{E+1}}, \qquad \dfrac{c}{2}=\dfrac{s_2}{k}\operatorname{dn}\biggl(\dfrac\sigma k, k\biggr), \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
\sin \dfrac{\delta}{2}=s_2 \operatorname{sn} \biggl(\dfrac\sigma k, k\biggr), \qquad \dfrac{\delta}{2}=\operatorname{cn} \biggl(\dfrac\sigma k, k\biggr);
\end{equation}
\tag{7.2}
$$
and
$$
\begin{equation}
\begin{aligned} \, (\delta, c) \in C_{03} \quad\Longrightarrow\quad &k=1, \qquad \dfrac{c}{2}=\dfrac{s_2}{\operatorname{ch} \sigma}, \nonumber \\ &\sin \dfrac{\delta}{2}=s_1 s_2 \operatorname{th} \sigma, \qquad \cos \dfrac{\delta}{2}=\dfrac{s_1}{\operatorname{ch} \sigma}; \end{aligned}
\end{equation}
\tag{7.3}
$$
where $s_1=\operatorname{sign} \cos(\delta/2)$ and $ s_2=\operatorname{sign} c$. In these coordinates the pendulum equation (4.9) assumes the following form: and its solution can be expressed as $\sigma_t=\sigma_0+t/{\mu}$, where $\sigma_0$ specifies the initial position of the pendulum.Below we present explicit formulae for $x_t$, $y_t$ and $\vartheta_t$ which are obtained by the integration of equations in (4.11) in accordance with the partition (4.13).
Lemma 2. If $(\delta, c) \in C_{01} \cup C_{02} \cup C_{03}$, then the trajectory $(x_t, y_t)$ lies in a strip containing the origin. In each case explicit expressions for this strip look as follows: Proof. Let $(\delta, c) \in C_{01}$. We multiply the corresponding equality in (7.4) for $(x_t,y_t)^\top$ by $({k}/{\mu}) \mathbf{J}_1^{-1}$ from the left. Then the equality for the first coordinate becomes as follows:
$$
\begin{equation*}
\frac{k}{\mu}(\operatorname{cn} \sigma_0 x_t+s_1 \operatorname{sn} \sigma_0 y_t)=s_1 (\operatorname{dn} \sigma_0-\operatorname{dn} \sigma_t).
\end{equation*}
\notag
$$
Combining this with the condition for the elliptic delta amplitude
$$
\begin{equation*}
\sqrt{1-k^2} \leqslant \operatorname{dn} \sigma_t \leqslant 1
\end{equation*}
\notag
$$
we deduce the double inequality in (7.7), which describes the subset of the plane containing the whole trajectory of the robot.
For $(\delta, c) \in C_{02}$ and $(\delta, c) \in C_{03}$ we multiply by $\mathbf{J}_2^{-1}/(\mu k)$ and $\mathbf{J}_3^{-1}/\mu$. Then we combine the equation for the first coordinate in a similar way with the inequalities $-1 \leqslant \operatorname{cn} \varsigma_t \leqslant 1$ and $0<1/\cosh \sigma_t \leqslant 1$. The proof is complete. Remark 8. Equalities in (7.7) and (7.8) are attained for $t=\mu(n K(k)- \sigma_0)$ and $t=\mu (2 n k K(k)-\varsigma_0)$, respectively, for each $n \in \mathbb{Z}$. In either case the points $(x_t,y_t)$ alternate at equal distances (in a staggered manner) on the two sides of the strip. In (7.9) equality is attained at a unique point, for $t=- \mu \sigma_0$. Recall that in the fixed coordinate system $Oxy$ the coupling point traverses the trajectory
$$
\begin{equation}
\widehat{x}_t=x_t-l_r \cos \vartheta_t, \qquad \widehat{y}_t=y_t-l_r \sin \vartheta_t.
\end{equation}
\tag{7.10}
$$
Substituting the corresponding expressions into (7.10) we obtain
$$
\begin{equation*}
\begin{aligned} \, (\delta, c) \in C_{01} &\quad\Longrightarrow\quad \begin{pmatrix} \widehat{x}_t\\ \widehat{y}_t \end{pmatrix} =\mathbf{J}_1 \biggl(\frac{\mu}{k} \begin{pmatrix} s_1(\operatorname{dn} \sigma_0-\operatorname{dn} \sigma_t) \\ \mathrm{E}(\sigma_t)-\mathrm{E}(\sigma_0)-\dfrac t\mu\end{pmatrix} -l_r \begin{pmatrix}\operatorname{cn} \sigma_t\\ - s_1 \operatorname{sn} \sigma_t\end{pmatrix} \biggr), \\ (\delta, c) \in C_{02} &\quad\Longrightarrow\quad \begin{pmatrix} \widehat{x}_t\\ \widehat{y}_t \end{pmatrix} =\mathbf{J}_2 \biggl( s_2 \mu \begin{pmatrix} k (\operatorname{cn} \varsigma_0-\operatorname{cn} \varsigma_t) \\ \mathrm{E}(\varsigma_t)-\mathrm{E}(\varsigma_0)-\dfrac{t}{k \mu} \end{pmatrix} -l_r \begin{pmatrix}\operatorname{dn} \varsigma_t\\ - k \operatorname{sn} \varsigma_t\end{pmatrix}\biggr), \\ (\delta, c) \in C_{03} &\quad\Longrightarrow\quad \begin{pmatrix} \widehat{x}_t\\ \widehat{y}_t \end{pmatrix} =s_2 \mu \mathbf{J}_3 \begin{pmatrix} \dfrac{s_1}{\operatorname{cosh} \sigma_0}-\dfrac{s_1+s_2 l_r/\mu}{\operatorname{cosh} \sigma_t} \\ \biggl(1 + \dfrac{s_1 s_2 l_r}{\mu}\biggr) \operatorname{tanh} \sigma_t -\operatorname{tanh} \sigma_0-\dfrac{t}{\mu} \end{pmatrix}. \end{aligned}
\end{equation*}
\notag
$$
Since the distance between the centre of the robot and the coupling point is $l_r$, the trajectory $(\widehat{x}_t, \widehat{y}_t)$ fits in the strip constructed for $(x_t, y_t)$ (see Lemma 2) and widened by $l_r$ on the sides. One can describe this strip more accurately in a similar way to the proof of Lemma 2, since for $(\delta, c) \in C_{0i}$, $i=1,2,3$, the first coordinate of the vector $\mathbf J_i^{-1} (\widehat{x}_t, \widehat{y}_t)^\top$ is bounded. It is obvious that $(\overline{x}_t, \overline{y}_t)$ also fits in the same strip, provided that it is widened by $l_r+l_t$ on the sides. We use this observation in the next section, where we describe abnormal trajectories. § 8. Appendix: formulae for abnormal trajectoriesEquations (5.9) and (5.12) coincide with (4.9) and (4.11) for $\mu=l_r$. So in the abnormal case, as explicit expressions for the trajectory $(x_t, y_t, \vartheta_t)$ of the robot and $(\widehat{x}_t,\widehat{y}_t)$ of the coupling point, we can use the formulae presented in § 7 for $\mu=l_r$. In each of the domains $\mathcal{A}_1, \mathcal{A}_2$ and $ \mathcal{A}_3$ we introduce the coordinates (7.1)–(7.3); here $k$ is connected as follows (by Proposition 5) with the parameters $l_r$ and $l_t$ of the system:
$$
\begin{equation}
(l_r, l_t, \varphi_0) \in \mathcal{A}_1 \quad\Longrightarrow\quad k=\sqrt{\frac{E+1}2}=\frac{l_r}{l_t},
\end{equation}
\tag{8.1}
$$
$$
\begin{equation}
(l_r, l_t, \varphi_0) \in \mathcal{A}_2 \quad\Longrightarrow\quad k=\sqrt{\frac{2}{E+1}}=\frac{l_t}{l_r},
\end{equation}
\tag{8.2}
$$
$$
\begin{equation}
(l_r, l_t, \varphi_0) \in \mathcal{A}_3 \quad\Longrightarrow\quad k=1.
\end{equation}
\tag{8.3}
$$
From (5.13) and (5.14) we deduce the additional relations
$$
\begin{equation}
\begin{aligned} \, \sin \frac{\delta(0)}{2} &=\frac{l_r \sin \varphi_0}{\sqrt{l_t^2+l_r^2+2 l_t l_r \cos \varphi_0}}, \\ \cos \frac{\delta(0)}{2} &=- \frac{l_t+l_r \cos \varphi_0}{\sqrt{l_t^2+l_r^2+ 2 l_t l_r \cos \varphi_0}}. \end{aligned}
\end{equation}
\tag{8.4}
$$
Comparing (7.1)–(7.3) with (8.4) and (5.13) we arrive at the following expressions:
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_1 \quad\Longrightarrow\quad \operatorname{sn} \sigma_0=\sin \alpha, \qquad \operatorname{cn} \sigma_0=\cos \alpha, \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
\operatorname{dn} \sigma_0=\cos (\varphi_0+\alpha) > 0,
\end{equation}
\tag{8.5}
$$
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_2 \quad\Longrightarrow\quad \operatorname{sn} \varsigma_0=- \frac{l_r}{l_t} \sin \alpha, \qquad \operatorname{cn} \varsigma_0=- \cos (\varphi_0+\alpha), \nonumber
\end{equation}
\notag
$$
and
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_3 \quad\Longrightarrow\quad \operatorname{tanh} \sigma_0=\sin \alpha, \qquad \frac{1}{\operatorname{cosh} \sigma_0}=\cos \alpha=\cos (\varphi_0+\alpha),
\end{equation}
\tag{8.7}
$$
where by Lemma 1 the angle $\alpha$ is expressed by formula (5.14) in terms of the initial angle $\varphi_0$ of the trailer, and the signs are $s_1=-\operatorname{sign} \cos (\varphi_0+\alpha)=-\operatorname{sign}(l_t+l_r \cos \varphi_0)=-1$ and $s_2= \operatorname{sign} \cos \alpha=\operatorname{sign} (l_r+l_t \cos \varphi_0)=1$. The corresponding matrices $\mathbf{J}_1$, $\mathbf{J}_2$ and $\mathbf{J}_3$ coincide after the substitutions (8.5)–(8.7), so we introduce a single matrix of clockwise rotation through an angle of $\alpha$:
$$
\begin{equation*}
\mathbf{J} :=\mathbf{J}_1=\mathbf{J}_2=\mathbf{J}_3= \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix}.
\end{equation*}
\notag
$$
We can simplify the formulae for trajectories of the robot:
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_1 \quad\Longrightarrow\quad \mathbf{J}^{-1}\begin{pmatrix} x_t\\ y_t\end{pmatrix} =l_t \begin{pmatrix} \operatorname{dn} \sigma_t-\cos (\varphi_0+\alpha)\\ \mathrm{E}(\sigma_t)-\mathrm{E}(\sigma_0)-\dfrac{t}{l_r} \end{pmatrix},
\end{equation}
\tag{8.8}
$$
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_2 \quad\Longrightarrow\quad \mathbf{J}^{-1} \begin{pmatrix} x_t\\ y_t \end{pmatrix} = \begin{pmatrix} l_t (-\operatorname{cn} \varsigma_t-\cos (\varphi_0+\alpha) )\\ l_r \biggl(\mathrm{E}(\varsigma_t)-\mathrm{E}(\varsigma_0)-\dfrac{t}{l_t} \biggr) \end{pmatrix}
\end{equation}
\tag{8.9}
$$
and
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_3 \quad\Longrightarrow\quad \mathbf{J}^{-1} \begin{pmatrix} x_t\\ y_t \end{pmatrix} =l_t \begin{pmatrix} \dfrac{1}{\operatorname{cosh} \sigma_t}-\cos (\varphi_0+\alpha) \\ \operatorname{tanh} \sigma_t-\sin \alpha-\dfrac{t}{l_t} \end{pmatrix}.
\end{equation}
\tag{8.10}
$$
The trajectory of the coupling point (7.10) has the following form in the abnormal case:
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_{1} \quad\Longrightarrow\quad \mathbf{J}^{-1}\begin{pmatrix} \widehat{x}_t\\ \widehat{y}_t\end{pmatrix} =\mathbf{J}^{-1}\begin{pmatrix}x_t\\y_t\end{pmatrix} -l_r \begin{pmatrix}\operatorname{cn} \sigma_t \\ \operatorname{sn} \sigma_t \end{pmatrix},
\end{equation}
\tag{8.11}
$$
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_{2} \quad\Longrightarrow\quad \mathbf{J}^{-1} \begin{pmatrix} \widehat{x}_t\\ \widehat{y}_t\end{pmatrix} =\mathbf{J}^{-1} \begin{pmatrix}x_t\\y_t\end{pmatrix} +\begin{pmatrix}- l_r \operatorname{dn} \varsigma_t\\ l_t \operatorname{sn} \varsigma_t\end{pmatrix}
\end{equation}
\tag{8.12}
$$
and
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_{3} \quad\Longrightarrow\quad \mathbf{J}^{-1} \begin{pmatrix}\widehat{x}_t\\\widehat{y}_t\end{pmatrix} =\begin{pmatrix}- l_t \cos (\varphi_0+\alpha)\\ -l_t \sin \alpha-t\end{pmatrix}.
\end{equation}
\tag{8.13}
$$
To determine the trajectory of the wheelset of the trailer
$$
\begin{equation}
\overline{x}_t=\widehat{x}_t-l_t \cos (\vartheta_t+\varphi_t), \qquad \overline{y}_t= \widehat{y}_t-l_t \sin (\vartheta_t+\varphi_t),
\end{equation}
\tag{8.14}
$$
we must know $\cos (\vartheta_t+\varphi_t)$ and $\sin (\vartheta_t+\varphi_t)$, which can be calculated as follows:
$$
\begin{equation*}
\begin{aligned} \, (l_r,l_t,\varphi_0) \in \mathcal{A}_1 &\quad\Longrightarrow\quad \begin{pmatrix}\cos (\vartheta_t+\varphi_t)\\ \sin (\vartheta_t+\varphi_t)\end{pmatrix} =\mathbf{J}\begin{pmatrix}\operatorname{dn} \sigma_t\\ - \dfrac{l_r}{l_t} \operatorname{sn} \sigma_t\end{pmatrix}, \\ (l_r,l_t,\varphi_0) \in \mathcal{A}_2 &\quad\Longrightarrow\quad \begin{pmatrix}\cos (\vartheta_t+\varphi_t)\\ \sin (\vartheta_t+\varphi_t)\end{pmatrix} =\mathbf{J}\begin{pmatrix}-\operatorname{cn} \varsigma_t\\ \operatorname{sn} \varsigma_t\end{pmatrix}, \\ (l_r,l_t,\varphi_0) \in \mathcal{A}_3 &\quad\Longrightarrow\quad \begin{pmatrix}\cos (\vartheta_t+\varphi_t)\\ \sin (\vartheta_t+\varphi_t)\end{pmatrix} =\mathbf{J}\begin{pmatrix}\dfrac{1}{\operatorname{cosh} \sigma_t}\\ -\operatorname{tanh} \sigma_t\end{pmatrix}. \end{aligned}
\end{equation*}
\notag
$$
Substituting these expressions into (8.14) we obtain the following equations for the motion of the trailer:
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_1 \quad\Longrightarrow\quad \mathbf{J}^{-1}\begin{pmatrix} \overline{x}_t\\ \overline{y}_t\end{pmatrix} =\begin{pmatrix} - l_r \operatorname{cn} \sigma_t-l_t \cos (\varphi_0+\alpha)\\ l_t \biggl( \mathrm{E}(\sigma_t)-\mathrm{E}(\sigma_0)-\dfrac{t}{l_r} \biggr)\end{pmatrix},
\end{equation}
\tag{8.15}
$$
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_2 \quad\Longrightarrow\quad \mathbf{J}^{-1} \begin{pmatrix}\overline{x}_t\\ \overline{y}_t\end{pmatrix} =\begin{pmatrix} -l_r \operatorname{dn} \varsigma_t-l_t \cos (\varphi_0+\alpha) \\ l_r\biggl(\mathrm{E}(\varsigma_t)-\mathrm{E}(\varsigma_0)-\dfrac{t}{l_t} \biggr) \end{pmatrix},
\end{equation}
\tag{8.16}
$$
$$
\begin{equation}
(l_r,l_t,\varphi_0) \in \mathcal{A}_3 \quad\Longrightarrow\quad \mathbf{J}^{-1} \begin{pmatrix} \overline{x}_t\\ \overline{y}_t \end{pmatrix}=l_t \begin{pmatrix} -\dfrac{1}{\operatorname{cosh} \sigma_t}-\cos (\varphi_0+\alpha) \\ \operatorname{tanh} \sigma_t-\sin \alpha-\dfrac{t}{l_t} \end{pmatrix}.
\end{equation}
\tag{8.17}
$$
Remark 9. The trajectories of the centre of the robot, the coupling point and the centre of the trailer for $(l_r,l_t,\varphi_0) \in \mathcal{A}_2$ can be expressed directly in terms of the ones for $(l_r,l_t,\varphi_0) \in \mathcal{A}_1$ by use of the following properties of elliptic functions (see [33]): 
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