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Application of Lissajous curves in trajectory planning of multiple agents

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Abstract

Lissajous curves have been used in various engineering applications such as optics, imaging, antenna scan, machining, as well as mobile robotics. In this article, we propose and analytically justify a Lissajous curve based trajectory planning strategy for aerial multi-agent systems to achieve the following objectives simultaneously: (i) Collision free paths for repeated coverage of a region while maintaining a closed sensor ring around a specified center for all time. (ii) Guaranteed detection of any stationary or moving object enclosed within the ring in finite time without the possibility of undetected escape. This leverages known and some novel properties of Lissajous curves that we establish as a part of this work. This has several potential applications in civil and military missions such as search and surveillance, repeated patrolling, target detection and capture, and the proposed strategy meets all these objectives simultaneously. We validate the proposed strategy through simulations and experiments using differential drive ground robots. We also demonstrate the applicability of this strategy for aerial surveillance through a Software-In-Loop-Simulation for quadrotors.

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Notes

  1. http://www.nex-robotics.com/products/fire-bird-v-robots/fire-bird-v-atmega2560-robotic-research-platform.html/.

  2. https://www.vicon.com/products/camera-systems/vantage.

  3. http://wiki.ros.org/indigo.

  4. http://wiki.ros.org/vicon_bridge.

  5. https://3dr.com/wp-content/uploads/2017/03/IRIS-Operation-Manual-v6.pdf.

  6. https://pixhawk.org/.

  7. http://gazebosim.org/.

  8. http://wiki.ros.org/kinetic/.

  9. https://dev.px4.io/en/simulation/gazebo.html.

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Correspondence to Aseem Vivek Borkar.

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Appendix

Appendix

Theorem 1

(Andreescu Titu Andrica and Cucuruzeanu 2010) Let abc be integers with \(a,b\ne 0\). Consider the Linear Diophantine equation \(ax + by = c\):

  1. 1.

    The equation has an integer solution if and only if \(d=GCD(a,b)\) divides c.

  2. 2.

    If \((x,y)=(x_o,y_o)\) is a particular solution of the equation, then infinitely many integer solutions can be constructed of the form \(x'=x_o+\frac{b}{d}t\), \(y'=y_o-\frac{a}{d}t\), where t is an integer.

1.1 Proofs of Properties in Sect. 2

Property P3:

  1. (a)

    Suppose there exists \(s' \in {\mathcal {B}}_x \cap {\mathcal {B}}_y\), then \(s'=q\frac{\pi }{a}=(2p-1)\frac{\pi }{2b}\) for some q and p as described in Property P2. This implies that \(\frac{2q}{2p-1}=\frac{a}{b}\). From Lemma 1, this is possible only for a degenerate Lissajous curve.

  2. (b)

    For a Lissajous curve described by (1) if \(s=0\), then \(s\in {\mathcal {B}}_x\) by Property P2. Also by Property P1, \(s=2\pi \) and \(s=0\) represent the same point. Thus if \(s\in {\mathcal {I}}\), then \(s \in (0, 2\pi )\). Let \(s_1,s_2 \in {\mathcal {I}}\) with \(s_1< s_2\) correspond to the same intersection point. Then \(x(s_1) = x(s_2) \) and \( y(s_1) = y(s_2)\) hold simultaneously. Now \(x(s_1) = x(s_2) \) and \(s_1< s_2\) means that either \(\cos (a s_2 ) = \cos \left( a s_1 + 2n\pi \right) \) or \(\cos (a s_2 ) =\cos \left( 2m\pi - a s_1\right) \) for \(n,m\in {{\mathbb {N}}}\), which implies that either \(s_2-s_1=\frac{2n \pi }{a} \text{ or } s_2+s_1=\frac{2 m\pi }{a}\).

Similarly \(y(s_1) = y(s_2)\) and \(s_1< s_2\) implies either \(\sin (b s_2) = \sin (b s_1 + 2p\pi )\) or \(\sin (b s_2) =\sin ((2q-1)\pi -b s_1)\) for \(p,q\in {{\mathbb {N}}}\), which implies either \(s_2-s_1=\frac{2 p\pi }{b} \text{ or } s_2+s_1=\frac{(2q-1) \pi }{b} \).

Since \(s_1,s_2\in (0,\ 2\pi )\), we have \(s_2-s_1<2\pi \). Thus \(n\in \{1,2, \ldots ,a-1\}\) and \(p\in \{1,2, \ldots ,b-1\}\). From the above discussion, two equations are obtained in each case. The four cases of linear system of equations for \(s_1\) and \(s_2\), obtained by picking one equation for each coordinate are discussed below:

Case 1:

\(s_2-s_1=\frac{2n \pi }{a}=\frac{2 p\pi }{b}\), implies that \(\frac{a}{b}=\frac{n}{p}\). This is not possible because \(n<a\) and \(p<b\) and ab are mutually co-prime by choice. Therefore this case is not considered.

Case 2:

\(s_2+s_1=\frac{2 m\pi }{a}=\frac{(2q-1) \pi }{b}\), implies that \(\frac{a}{b}=\frac{2m}{2q-1}\). From Lemma 1 this case corresponds to a degenerate Lissajous curve. Hence this case is also not considered.

Case 3:

Solving \(s_2-s_1=\frac{2n \pi }{a} \) and \(s_2+s_1=\frac{(2q-1) \pi }{b}\) results in \(s_1=\frac{(2q-1)\pi }{2b}-\frac{n\pi }{a}\), and \(s_2= \frac{(2q-1)\pi }{2b}+\frac{n\pi }{a}\).

Case 4:

Solving \(s_2+s_1=\frac{2 m\pi }{a}\) and \(s_2-s_1=\frac{2 p\pi }{b}\) results in \(s_1=\frac{m \pi }{a}-\frac{p \pi }{b}\), and \(s_2=\frac{m \pi }{a}+\frac{p \pi }{b}\).

Now suppose that we also have \(s_1,\ s_2 \in {\mathcal {B}}_x\). Then from Property P2, \(s_1=q_1\frac{\pi }{a}\) and \(s_2=q_2\frac{\pi }{a}\) for some \(q_1<q_2\) and \(q_1,q_2 \in \{0, \ldots ,2a-1\}\). Since \(s_1,s_2\) correspond to the same intersection point as argued above, considering only \(s_1\), one of the following must be true:

  1. (1)

    Case 3 is satisfied. This implies \(s_1=\frac{q_1\pi }{a}=\frac{(2q-1)\pi }{2b}-\frac{n\pi }{a}\) for some \(q\in {{\mathbb {N}}}\) and \(n \in \{1,...,a-1\}\), which implies that \(\frac{2(q_1+n)}{2q-1}=\frac{a}{b}\). By Lemma 1, this is possible only for a degenerate Lissajous curve.

  2. (2)

    Case 4 is satisfied. This implies \(s_1=\frac{q_1\pi }{a}=\frac{m \pi }{a}-\frac{p \pi }{b}\) for some \(m \in {{\mathbb {N}}}\) and \(p\in \{1, \ldots ,b-1\}\). This in turn implies that \(\frac{m-q_1}{p}=\frac{a}{b}\). Since ab are mutually co-prime and \(p<b\), this equality cannot hold for any \(m-q_1\in {{\mathbb {N}}}\).

Similar arguments hold for \(s_2\). Thus for non degenerate Lissajous curves \({\mathcal {B}}_x \cap {\mathcal {I}}=\emptyset \). \(B_y\cap {\mathcal {I}}=\emptyset \) can be proved along the same lines.

(c) The parameter values of \(s_1\) and \(s_2\) for intersection points obtained above in Case 3 and Case 4 can be expressed as integer multiples of \(\frac{\pi }{2ab}\). From Property P2, the same is true for the boundary points of the Lissajous curve. Since \({\mathcal {N}}\) contains all integer multiples of \(\frac{\pi }{2ab}\) in \([0,\ 2\pi )\), it follows that

$$\begin{aligned} {\mathcal {I}}\cup {\mathcal {B}}_x \cup {\mathcal {B}}_y\subset {\mathcal {N}} \end{aligned}$$
(22)

Consider

$$\begin{aligned} {\mathcal {N}}_1= & {} \left\{ 2p\frac{\pi }{2ab} \ \vert \ p\in \{1,\dotsc ,2ab\}\right\} , \\ {\mathcal {N}}_2= & {} \left\{ (2p-1)\frac{\pi }{2ab}\vert p\in \{1,2,\dotsc ,2ab\}\right\} , \end{aligned}$$

such that \({\mathcal {N}}= {\mathcal {N}}_1 \cup {\mathcal {N}}_2\). We claim that for any \(s_1 \in {\mathcal {N}}_1\) and a corresponding \(p\in \{1,\dotsc ,2ab\}\), there exists an \(m\in \{0, \ldots ,b-1\}\) such that \(x(s_1)=x(s_1+\frac{2m\pi }{b})\) and \(y(s_1)=y(s_1+\frac{2m\pi }{b})\). This is equivalent to showing that there exists an \(m\in \{0,\ldots ,b-1\}\) satisfying

$$\begin{aligned} y(s_1)=B\sin \left( b\left( \frac{2p\pi }{2ab}\right) \right) =B\sin \left( b\left( \frac{p\pi }{ab}+\frac{2m\pi }{b}\right) \right) , \end{aligned}$$
(23)
$$\begin{aligned} x(s_1)=A\cos \left( a\left( \frac{2p\pi }{2ab}\right) \right) =A\cos \left( a\left( \frac{p\pi }{ab}+\frac{2m\pi }{b}\right) \right) . \end{aligned}$$
(24)

Equation (23) holds for any integer value of m. Using the trigonometric relation \(\cos (\frac{p\pi }{b})=\cos (2\pi k-\frac{p\pi }{b})\) for \(k\in {{\mathbb {Z}}}\), (24) can be reduced to the equation \(2\pi k- \frac{p\pi }{b}=\frac{p\pi }{b}+\frac{2ma\pi }{b}\), which further simplifies to:

$$\begin{aligned} bk-ma=p. \end{aligned}$$
(25)

Thus we must show that given p, (25) has an integer solution for (mk) with \(m\in \{0,\ldots ,b-1\}\).

Equation (25) is a linear Diophantine equation (Andreescu Titu Andrica and Cucuruzeanu 2010). Using the fact that a and b are co-prime and from the result on Diophantine equations given by Theorem 1 in Appendix, there exist infinitely many integer solutions of (mk) satisfying (25) for each \(p\in \{1,\ldots ,2ab\}\). Also if \((m_o,\ k_o)\) is one such solution of (25) for a fixed value of p, say \(p=p_o\in \{1,\ldots ,2ab\}\), then \((m_o+bt,\ k_o+at)\) are also solutions of (25) at \(p=p_o\), where \(t\in {{\mathbb {Z}}}\). We can write \(m_o=m_qb+m_r\) for some \(m_q\in {{\mathbb {Z}}}\) and \(m_r\in \{0,\ldots ,b-1\}\). Now by taking \(t=-m_q\), a new solution \((m_o',k_o')\) satisfying (25) at \(p=p_o\) can be generated such that \(m_o'=m_r\in \{0,\ldots ,b-1\}\).

Thus in general for all \(s_1 \in {\mathcal {N}}_1\) with \(p \in \{1,\ldots ,2ab\}\), there exists \(m_o'\in \{0,\ldots ,b-1\}\) such that \(x(s_1)=x(s_2)\) and \(y(s_1)=y(s_2)\) where \(s_2=s_1+\frac{2m_o'\pi }{b}\). If for some \(s_1\in {\mathcal {N}}_1 \), \((m_1,k_1)\) satisfy (25) such that \(m_1\in \{1,\ldots ,b-1\}\) then \(\vert s_2-s_1\vert \in \left\{ \left. \frac{2m_1\pi }{b}\right| m_1 \in \{1,\ldots ,b-1\} \right\} \) and \(s_1\ne s_2\) such that \(x(s_1)=x(s_2)\) and \(y(s_1)=y(s_2)\) are satisfied, thus from Property P6 \(s_1,s_2\in {\mathcal {I}} \).

There exists \(s_1\in {\mathcal {N}}_1\) such that for \(p=p_o\), \((m_o',k_o')\) satisfy (25) and \(m_o'=0\), then from (25) \(p=bk_o'\). Thus in (24) \(x(s_1)=A\cos (k_o'\pi )=(-1)^{k_o'}A\). Thus in this case \(s_1\in {\mathcal {B}}_x\). This proves that \({\mathcal {N}}_1\subset {\mathcal {B}}_x \cup {\mathcal {I}}\). By analogous arguments it can be shown that, for all \(s_1 \in {\mathcal {N}}_2\) with \(p_o \in \{1,\ldots ,2ab\}\), there exists \(m_o'\in \{0,\ldots ,a-1\}\) such that \(x(s_1)=x(s_2)\) and \(y(s_1)=y(s_2)\) where \(s_2=s_1+\frac{2m_o'\pi }{a}\), and as a consequence \({\mathcal {N}}_2\subset {\mathcal {B}}_y \cup {\mathcal {I}}\). Thus \({\mathcal {N}}_1 \cup {\mathcal {N}}_2={\mathcal {N}}\subset {\mathcal {B}}_x\cup {\mathcal {B}}_y \cup {\mathcal {I}}\) and thus with (22) it implies that \({\mathcal {I}} \cup {\mathcal {B}}_x \cup {\mathcal {B}}_y = {\mathcal {N}}\). \(\square \)

Property P5: Consider the following two alternative parametric representations of the node points (set \({\mathcal {N}}\)) of the Lissajous curve:

(a) From (2), the parameter values corresponding to node points are of the form \(\frac{m\pi }{2ab}\) with \(m\in \{0, \ldots ,4ab\}\). For any node point m can be expressed as \(2bq_m+r_m\) where \(q_m\) and \(r_m\) are the quotient and remainder obtained by dividing m by 2b. Thus the parameter value of any node point can be expressed as

$$\begin{aligned} \frac{m\pi }{2ab}=\frac{q_m\pi }{a}+\frac{r_m\pi }{2ab}=\frac{(q_m+1)\pi }{a}-\frac{(2b-r_m)\pi }{2ab}. \end{aligned}$$

Since \(r_m,2b-r_m\in \{0,\ldots ,2b\}\) and either \(q_m\) or \(q_m+1\) is a positive even integer, the parameter value of any node point can be represented as follows:

$$\begin{aligned} s=\frac{q_e\pi }{a}\pm \frac{v\pi }{2ab}, \end{aligned}$$
(26)

where \(q_e\) is an even integer and \(v\in \{0,\ldots ,2b\}\).

From (26) and (1), we observe that since \(q_e\) is an even integer and for all node points corresponding to a fixed value of v, \(x(s_v)=A\cos (q_e\pi \pm \frac{v\pi }{2b})=A\cos (\frac{v\pi }{2b})\). In particular \(v=0\) and \(v=2b\) correspond to the vertical boundary points \({\mathcal {B}}_x\) at \(x(s)=\pm A\). Thus we can group the node points into vertical level sets

$$\begin{aligned} V_v=\left\{ s\in {\mathcal {N}} \ \vert \ x(s)=A\cos \left( \frac{v\pi }{2b}\right) \right\} \end{aligned}$$
(27)

where \(v\in \{0,\ldots ,2b\}\).

(b) For some \(m\in {{\mathbb {N}}}\) and fixed values of a, consider the linear Diophantine equation (Andreescu Titu Andrica and Cucuruzeanu 2010) in the variables n and r: \(2an+r=m+a\), where factors of n and r (that is, 2a and 1) are mutually co-prime. Thus by Theorem 1 in the Appendix, this equation has infinite integer solutions. Also if \((n',r')\) is one such solution then the other solutions are of the form \(({\tilde{n}},{\tilde{r}})=(n'+t,r'-2at)\) for \(t\in {{\mathbb {Z}}}\). By appropriately selecting \(t=t_o\), we can ensure that \({\tilde{r}}\in \{0,\ldots ,2a-1\}\). Substituting \(({\tilde{n}},\ {\tilde{r}})\) as the solution at \(t=t_o\) and multiplying by \(\frac{\pi }{2ab}\), we get \( \frac{(2{\tilde{n}}-1)\pi }{2b}+\frac{{\tilde{r}}\pi }{2ab}=\frac{m\pi }{2ab}\). Similarly, substituting the solution \(({\tilde{n}}+1,{\tilde{r}}-2a)\) at \(t=t_o+1\) and multiplying by \(\frac{\pi }{2ab}\), we get \(\frac{(2({\tilde{n}}+1)-1)\pi }{2b}-\frac{(2a-{\tilde{r}})\pi }{2ab}=\frac{m\pi }{2ab}\). Since \({\tilde{r}}, 2a-{\tilde{r}} \in \{0,\ldots ,2a\}\) and either \({\tilde{n}}\) or \({\tilde{n}}+1\) is an odd integer for all \(m\in {{\mathbb {N}}}\). From (2), the parameter value of any node point can be represented as follows:

$$\begin{aligned} s=\frac{(2n_o-1)\pi }{2b}\pm \frac{h\pi }{2ab}, \end{aligned}$$
(28)

where \(n_o\) is an odd integer and \(h\in \{0,\ldots ,2a\}\). From (28) and (1), we observe that \(n_o\) is an odd integer and for all node points corresponding to a particular value of h, \(y(s_h)=B\sin (\frac{(2n_o-1)\pi }{2} \pm \frac{h\pi }{2a})=B\cos (\frac{h\pi }{2a})\). In particular, \(h=0\) and \(h=2a\) correspond to the horizontal boundary points \({\mathcal {B}}_y\) at \(y(s)=\pm B\). Thus we can group the node points into horizontal level sets

$$\begin{aligned} H_h=\left\{ s\in {\mathcal {N}} \ \vert \ y(s)=B\cos \left( \frac{h\pi }{2a}\right) \right\} \end{aligned}$$
(29)

where \(h\in \{0,\ldots ,2a\}\).

For both representations given by (26) and (28), the parameter of every node point belongs to the sets \(H_h\) and \(V_v\) for some \(h\in \{0,\ldots ,2a\}\) and \(v\in \{0,\ldots ,2b\}\). Next, we find the pairs of (hv) that correspond to node points coordinates.

Comparing parameter values in Property P2 with (28), we see that (28) describes the relative parametric displacement of node points from some point in \({\mathcal {B}}_y\). From Property P2 the parametric displacement between a horizontal boundary point in \({\mathcal {B}}_y\) and a vertical boundary point in \({\mathcal {B}}_x\) is given by \(\vert ((2n-1)a-2bm)\frac{\pi }{2ab} \vert \). This is an odd multiple of \(\frac{\pi }{2ab}\) as a is odd for a non-degenerate Lissajous curve described by (1). Since points on \(x=A\) are contained in \({\mathcal {B}}_y\), if the parameter values of these points are represented using (28), then \(h\in \{1,3,\ldots ,2a-1\}\) in order to ensure that relative parametric displacement from the corresponding point in \({\mathcal {B}}_y\) is an odd multiple of \(\frac{\pi }{2ab}\). Furthermore the \(y(s_h)=B\cos (\frac{h\pi }{2a})\) implies that distinct boundary points on \(x=A\) have distinct h values. Now for boundary points on \(x=A\) (that is, the set \(V_0\) for \(v=0\) in (27)), there are a possible values of h. Computing the x coordinate using (1) for parameter values in \({\mathcal {B}}_x\) from Property P2, \(m\in \{0,2,\ldots ,2a-2\}\) result in boundary points on \(X=A\). Thus number of boundary points on the tangent \(x=A\) are a, having distinct y coordinates. Thus for every value of \(h\in \{1,3,\ldots ,2a-1\}\) there is a vertical boundary point on the tangent \(x=A\). Thus points in set \(V_0\) have \(h\in \{1,3,\ldots ,2a-1\}\).

A parametric shift of \(\pm \frac{\pi }{2ab}\) in the (26) for points corresponding to set \(V_0\) gives \(s=\frac{q_e\pi }{a}\pm \frac{\pi }{2ab}\) and \(x(s)=A\cos (\frac{\pi }{2ab})\). Thus from (27) and Property P7, for every node point in set \(V_0\) there are two adjacent node points in the set \(V_{1}\). From (28) representation of parameter values in \(V_0\), a parametric shift of \(\frac{\pi }{2ab}\) gives \(s=\frac{(2n_o-1)\pi }{2b}\pm \frac{(h \pm 1)\pi }{2ab}\) and \(y(s)=B\cos (\frac{(h\pm 1)\pi }{2a})\) which implies shift in horizontal level sets \(H_h\) by \(\pm 1\). This implies that the node points corresponding to the set \(V_{1}\) have values of \(h\in \{0,2,\ldots ,2a\}\) where \(h=0,2a\) correspond to points in \({\mathcal {B}}_y\) and the rest \(a-1\) points are intersection points with \(h\in \{2,\ldots ,2a-2\}\). Since for all \(h\in \{1,3,\ldots ,2a-1\}\) there is a node point in \(V_0\) (for \(v=0\)), the same is true for all \(h\in \{0,2,\ldots ,2a\}\) and node points in \(V_1\) (for \(v=1\)) as well. Thus by analogous arguements we have,

$$\begin{aligned} h\in {\left\{ \begin{array}{ll} \{1,3,\ldots ,2a-1\} \text{ for } v\in \{0,2,\ldots ,2b\},\\ \{0,2,\ldots ,2a\} \text{ for } v\in \{1,3,\ldots ,2b-1\}. \end{array}\right. } \square \end{aligned}$$
(30)

Property P8: From (1) for points on \(x=0\), \(x(s)=A\cos (as)=0\) implies \(as=(2p-1)\frac{\pi }{2}\) for some \(p\in {{\mathbb {N}}}\). Since \(0<s<2\pi \), \(\frac{1}{2}<p<2a+\frac{1}{2}\) which implies \(p\in \{1,\ldots ,2a\}\). Similarly it can be shown for points on \(y=0\).\(\square \)

Property P9: Follows from (1)\(\square \)

Property P10 and P11:\(X_{d}(s)=\vert x(s+\frac{\pi }{4ab})- x(s-\frac{\pi }{4ab}) \vert \) and from (1), \(X_{d}(s)=\left| 2A\sin \left( \frac{\pi }{4b}\right) \right| \left| \sin (as) \right| \). Since \(b \ge 1\), \(2A \sin (\frac{\pi }{4b})>0\), so \( X_d(s)\in \left[ 0,\ 2A \sin (\frac{\pi }{4b})\right] \). As a consequence we have \(\min \nolimits _{s} X_{d}(s)=0\) which only holds for \(s=\frac{m\pi }{a}\) for some \(m \in {{\mathbb {N}}}\). Now \(0\le s<2\pi \) implies \(0\le m< 2a\). Thus from Property P2, \(\arg \min \nolimits _{s} X_d(s)= {\mathcal {B}}_x\). Similarly \(\max \nolimits _{s} X_{d}(s)=2A \sin (\frac{\pi }{4b})\) holds only for \(s=\frac{(2p-1)\pi }{2a}\) for some \(p \in {{\mathbb {N}}}\). Now \(0<s<2\pi \), means \(\frac{1}{2}<p<2a+\frac{1}{2}\) which implies \(p\in \{1, \ldots ,2a\}\). Thus by the definition of \({\mathcal {X}}_o\) in Property P8, \(\arg \max \nolimits _{s} X_d(s)={\mathcal {X}}_o\). Property P11 can be proved along the same lines.\(\square \)

The proofs of Properties S1 to S6 are as follows:

Property S1 and S2: Since parametric rate \({\dot{s}}(t)=K\), by differentiating (1) \(\vert {\dot{x}}(s)\vert =AaK\left| \sin (as)\right| \) and \(\vert {\dot{y}}(s)\vert =BbK\left| \cos (bs) \right| \). Thus it clearly follows that

$$\begin{aligned} \max \limits _s \vert {\dot{x}}(s)\vert =AaK. \end{aligned}$$

Suppose \(s'=\arg \max \nolimits _s \vert {\dot{x}}(s)\vert \), then \(s'= (2p- 1)\frac{\pi }{2a}\) and since \(0\le s'<2\pi \), it means \(\frac{1}{2}<p<2a+\frac{1}{2}\), which implies that \(p\in \{1, \ldots 2a\}\). Thus from Property P8, \(\arg \max \limits _s \vert {\dot{x}}(s)\vert \in {\mathcal {X}}_o\). Also, \(\max \limits _s \vert {\dot{y}}(s)\vert =BbK\). Suppose \(s'=\arg \max \limits _s \vert {\dot{y}}(s)\vert \), then \(s'= q\frac{\pi }{b}\) and since \(0\le s'<2\pi \), it implies that \(q\in \{0,\ldots ,2b-1\}\). Thus from Property P8, \(\arg \max \limits _s \vert {\dot{y}}(s)\vert \in {\mathcal {Y}}_o\). At \(s_o\in \{\frac{\pi }{2}, \frac{3\pi }{2}\},\) since a is odd and b is even, \(\vert {\dot{x}}(s_o)\vert =AaK=\max \limits _s \vert {\dot{x}}(s)\vert \) and \(\vert {\dot{y}}(s_o)\vert =BbK=\max \limits _s \vert {\dot{y}}(s)\vert \). Thus as a consequence from (5),

$$\begin{aligned} V(s_o)= \max \limits _s V(s)=\sqrt{(AaK)^2+(BbK)^2} \square \end{aligned}$$

Property S3:Sufficiency: If \(s' \in {\mathcal {X}}_o \cap {\mathcal {Y}}_o\) then from Property P8, \(s'=(2p-1)\frac{\pi }{2a}\) for \(p\in \{1,\ldots ,2a\}\) and \(s'=\frac{q\pi }{b}\) for \(q\in \{0,\ldots 2b-1\}\). This implies that \(\frac{a}{b}=\frac{2p-1}{2q}\). Since a and b are mutually co-prime, it implies \(2p-1=ma\) and \(2q=mb\) for \(m \in { {\mathbb {N}}}\). Note that for the Lissajous curves having \((a,b)\in {\mathcal {C}}_1\cup {\mathcal {C}}_2\), a is odd and b is even. Thus for each value of m:

  1. (1)

    for \(m=1\), \(p=\frac{a+1}{2}\in \{1,\ldots ,2a\}\), \(q=\frac{b}{2}\in \{0,\ldots 2b-1\}\). Thus \(s'=\frac{q\pi }{b}=\frac{(2p-1)\pi }{2a}=\frac{\pi }{2}\).

  2. (2)

    for \(m=2\), \(p=\frac{2a+1}{2}\) is not an integer and is not considered.

  3. (3)

    for \(m=3\), \(p=\frac{3a+1}{2}\in \{1,\ldots ,2a\}\), \(q=\frac{3b}{2}\in \{0,\ldots 2b-1\}\). Thus \(s'=\frac{q\pi }{b}=\frac{(2p-1)\pi }{2a}=\frac{3\pi }{2}\).

  4. (4)

    for \(m\ge 4\), \(p>2a \) so \(p\not \in \{1,\ldots ,2a\}\) and \(q>2b\) so \(q\not \in \{0,\ldots ,2b-1\}\). Thus these are not considered. Thus \(s'\in {\mathcal {X}}_o \cap {\mathcal {Y}}_o\) implies that \(s'\in \{\frac{\pi }{2},\frac{3\pi }{2}\}.\)

Necessity: This can be easily verified by computing \(x(s')\) and \(y(s')\) at \(s'\in \{\frac{\pi }{2},\frac{3\pi }{2}\}\) using (1). Since \(s'\in {\mathcal {X}}_o\) and \(s'\in {\mathcal {Y}}_o\), from Properties P10 and P11, \(X_d(s')=X_{dm}\) and \(Y_d(s')=Y_{dm}\) respectively.\(\square \)

Property S4: For Lissajous curve having \((a,b)\in {\mathcal {C}}_1\), (3) implies \(a-1=b\). Thus for \(s_p\in {\mathcal {X}}_o\) with \(p \in \{1,\ldots ,2a\}\), from Property P8, \(s_p=\frac{(2p-1)\pi }{2a}\) and from (1),

$$\begin{aligned} y(s_p) = B \sin \left( (a-1)\frac{(2p-1)\pi }{2a} \right) = (-1)^{p+1}B \cos \left( s_p \right) . \end{aligned}$$

Thus \(y(s_p) y(s_{p+1})=(-1)^{2p+3} B^2 \cos ( s_p) \cos (s_{p+1})=-B^2\cos (s_p)\cos (s_{p+1}).\) Now consider the following cases:

Case 1:\(p\in \{\frac{a+1}{2},\ldots , \frac{3a-1}{2} \}\). This implies \(p+1 \in \{\frac{a+3}{2},\ldots , \frac{3a+1}{2} \}\). This means \(s_p=\frac{(2p-1)\pi }{2a} \in \{\frac{\pi }{2},\ldots ,\frac{3\pi }{2}-\frac{\pi }{a}\}\) and similarly \(s_{p+1}\in \{\frac{\pi }{2}+\frac{\pi }{a},\ldots ,\frac{3\pi }{2}\}\). For all \(s_p\) and \(s_{p+1} \in (\frac{\pi }{2}, \frac{3\pi }{2})\) from trigonometry it is known that \(\cos (s_p)\cos (s_{p+1})>0\). Also at \(s_p=\frac{\pi }{2}\) or \(s_{p+1}=\frac{3\pi }{2}\) then \(\cos (s_p)\cos (s_{p+1})=0\). Thus at these values of p and \(p+1\), \(y(s_p)y(s_{p+1})\le 0\).

Case 2:\(p \in \{ 1,\ldots ,\frac{a-1}{2}\} \cup \{\frac{3a+1}{2} ,\ldots ,2a-1,2a\}\). Here by analysis similar to Case 1 it can be shown that \(s_{p}\in (0,\frac{\pi }{2})\cup [\frac{3\pi }{2}, 2\pi )\) and \(s_{p+1}\in (0,\frac{\pi }{2}]\cup (\frac{3\pi }{2}, 2\pi )\), for which by basic trigonometry \(\cos (s_p)\cos (s_{p+1})\ge 0\). Thus for these values of p and \(p+1\), \(y(s_p)y(s_{p+1})\le 0\).\(\square \)

Property S5: For a Lissajous curve having \((a,b)\in {\mathcal {C}}_2\), (4) implies \(a=b-1\). Thus for \(s_q\in {\mathcal {Y}}_o\) with \(q \in \{0,\ldots ,2b-1\}\), \(s_q=\frac{q\pi }{b}\) and \(x(s_q) = A\cos \left( (b-1)\frac{q\pi }{b} \right) = (-1)^{q}A \cos \left( s_q \right) \). Thus

$$\begin{aligned} x(s_p) x(s_{p+1})&= (-1)^{2q+1} A^2 \cos ( s_q) \cos (s_{q+1})\\&= -A^2\cos (s_q)\cos (s_{q+1}). \end{aligned}$$

Now consider the following cases:

Case 1:\(q\in \{\frac{b}{2},\ldots ,\frac{3b}{2}-1\}\), it implies \(s_q\in [\frac{\pi }{2}, \frac{3\pi }{2})\)\(s_{q+1}\in (\frac{\pi }{2}, \frac{3\pi }{2}]\), thus from trigonometry \(\cos (s_q)\cos (s_{q+1})\ge 0\) and hence \(x(s_q)x(s_{q+1})\le 0\).

Case 2: \(q\in \{0,\ldots \frac{b}{2}-1\}\cup \{\frac{3b}{2},\ldots ,2b-1\}\), it implies \(s_q\in [0, \frac{\pi }{2})\cup [\frac{3\pi }{2}, 2\pi )\) and \(s_{q+1}\in (0, \frac{\pi }{2}]\cup (\frac{3\pi }{2}, 2\pi ]\), thus from trigonometry \(\cos (s_q)\cos (s_{q+1})\ge 0\) and hence \(x(s_q)x(s_{q+1})\le 0\). \(\square \)

Property S6: From (3) and (4), for Lissajous curve having \((a,b)\in {\mathcal {C}}_1 \cup {\mathcal {C}}_2\), \(b=a\pm 1\). From Property P8, if \(s'\in {\mathcal {X}}_o\) then \(s'= (2p-1)\frac{\pi }{2a}\) for \(p\in \{1,\ldots ,2a\}\) and from (1),

$$\begin{aligned} \vert y(s')\vert= & {} \vert B\sin ((a\pm 1)\frac{(2p-1)\pi }{2a})\vert \\= & {} \vert (-1)^{p+1}B\cos \left( \frac{(2p-1)\pi }{2a}\right) \vert . \end{aligned}$$

Since \(2p-1\) is an odd integer, it is not divisible by 2a. Hence the values of \(p\in \{1,\ldots ,2a\}\) for which \(2p-1\) closest integer to 2a and its multiples, is the maximizer for \(\vert y(s_p)\vert \). This implies that \(2p-1=2a\pm 1\) or \(2p-1=4a-1\) for \(p\in \{a,a+1,2a\}\). Thus it follows that:

  1. (1)

    At \(p=a\), \(y(s')=B\sin (b\frac{(2a-1)\pi }{2a})=B\sin (b\pi -\frac{b\pi }{2a})=-B\sin (\frac{b\pi }{2a})=\min \limits _{s'\in {\mathcal {X}}_o}y(s_p)\)

  2. (2)

    At \(p=a+1\), \(y(s')=B\sin (b\frac{(2a+1)\pi }{2a})=B\sin (b\pi +\frac{b\pi }{2a})=B\sin (\frac{b\pi }{2a})=\max \limits _{s'\in {\mathcal {X}}_o}y(s_p)\)

  3. (3)

    At \(p=2a\), \(y(s')=B\sin (b\frac{(4a-1)\pi }{2a})=B\sin (2b\pi -\frac{b\pi }{2a})=-B\sin (\frac{b\pi }{2a})=\min \limits _{s'\in {\mathcal {X}}_o}y(s_p)\)

\(\square \)

Lemma 3

The function \(f(x)=\frac{\sin \left( \frac{\pi }{4(x+1)}\right) }{\sin \left( \frac{\pi }{4(x)}\right) }\) is monotone increasing for all \( x \ge 1\).

Proof

Consider the function \(g(x)=x^2 \tan \left( \frac{\pi }{4x}\right) \). Then

$$\begin{aligned} \frac{d}{dx}g(x)=\frac{x\left( \sin \left( \frac{\pi }{2x}\right) -\frac{\pi }{4x}\right) }{\cos ^2\left( \frac{\pi }{4x}\right) }. \end{aligned}$$

Since \(\frac{\pi }{2x} \in (0,\ \frac{\pi }{2}]\) for all \(x\ge 1\), \(\sin (\frac{\pi }{2x})>\frac{\pi }{4x}\). Thus \(\frac{d}{dx}g(x)>0\) for all \(x\ge 1\). This implies that g(x) is monotone increasing and thus \((x+1)^2 \tan \left( \frac{\pi }{4(x+1)}\right) - x^2 \tan \left( \frac{\pi }{4x}\right) >0\), which on rearranging gives

$$\begin{aligned} \frac{\sin \left( \frac{\pi }{4(x+1)}\right) \cos \left( \frac{\pi }{4x}\right) }{x^2} - \frac{\cos \left( \frac{\pi }{4(x+1)}\right) \sin \left( \frac{\pi }{4x}\right) }{(x+1)^2} > 0. \end{aligned}$$

Now for \(P=\frac{\pi }{4 \sin ^2\left( \frac{\pi }{4x}\right) }\) which is positive for all \(x\ge 1\), \(\frac{d}{dx}f(x)\) is given by

$$\begin{aligned} \frac{-P\cos \left( \frac{\pi }{4(x+1)}\right) \sin \left( \frac{\pi }{4x}\right) }{(x+1)^2} + \frac{P\sin \left( \frac{\pi }{4(x+1)}\right) \cos \left( \frac{\pi }{4x}\right) }{x^2} >0 . \end{aligned}$$

Thus f(x) is monotone increasing for all \(x\ge 1\). \(\square \)

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Borkar, A.V., Sinha, A., Vachhani, L. et al. Application of Lissajous curves in trajectory planning of multiple agents. Auton Robot 44, 233–250 (2020). https://doi.org/10.1007/s10514-019-09888-7

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