Improved trajectory tracing of underwater vehicles for flow field map**

Abstract

The Motion Tomography (MT) algorithm maps an ocean flow field using sporadically measured positions of underwater vehicles. A key step of the MT algorithm, called trajectory tracing, is to estimate the underwater trajectories of the vehicles based on the estimated flow field and known start and end positions. This paper extends the MT algorithm by develo** a set of analytical formulas to compute the underwater trajectories. These analytical formulas enable us to study the convergence of the MT algorithm, and we prove that the estimated trajectory and measured trajectory end positions converge as the MT algorithm proceeds. Experimental results are collected on the Georgia Tech Miniature Autonomous Blimps to demonstrate that the MT algorithm can be applied to reconstruct a wind field in an indoor environment using nothing but sparse position measurements. We further validate the MT algorithm using data collected by an underwater glider deployed in the South Atlantic Bight. We demonstrate MT is able to reconstruct the ocean flow field using recorded glider surfacing positions and improve the spatial distribution of a dead reckoning flow field map.

Appendix

We derive in the following Lemma an analytic expression for the predicted AUV trajectory, which will serve as a replacement for the numerical simulator used previously in Chang et al. (2017).

Lemma 3

Let ${\mathcal {G}}_i \cap {\mathcal {G}}_{i-1}=\left\{ C^1,\cdots ,C^m \right\}$ and $k\le m$. Then $r_{i}^{k}-r_{i-1}^{k}$ can be computed according to one of the four cases:

$$\begin{aligned} \mathrm{A1} : x_{i}^{k}-x_{i-1}^{k}&=-\frac{ \omega ({t}^{k}_{i-1})^2 }{\Vert { T_{i-1}} \Vert ^2 } \frac{V_{i-1}^{k} \times d_{i-1}}{{V_{y,{i}}^{k}}} \\&\quad +( x_{i}^{k-1}-x_{i-1}^{k-1}) \nonumber \\ y_{i}^{k}-y_{i-1}^{k}&=0. \end{aligned}$$

(19)

$$\begin{aligned} \mathrm{A2} : x_{i}^{k}-x_{i-1}^{k}&= 0 \nonumber \\ y_{i}^{k}-y_{i-1}^{k}&= \frac{ \omega ({t}^{k}_{i-1})^2 }{\Vert { T_{i-1}} \Vert ^2 } \frac{V_{i-1}^{k} \times d_{i-1}}{{V_{x,{i}}^{k}}} \\&\quad +( y_{i}^{k-1}-y_{i-1}^{k-1}) . \end{aligned}$$

(20)

$$\begin{aligned} \mathrm{A3} : x_{i}^{k}-x_{i-1}^{k}&=0 \nonumber \\ y_{i}^{k}-y_{i-1}^{k}&=\frac{ \omega ({t}^{k}_{i-1})^2 }{\Vert { T_{i-1}} \Vert ^2 } \frac{V_{i-1}^{k} \times d_{i-1}}{{V_{x,{i}}^{k}}}\nonumber \\&\quad -\frac{V_{y,i}^{k}}{V_{x,i}^{k}} ( x_{i}^{k-1}-x_{i-1}^{k-1}). \end{aligned}$$

(21)

$$\begin{aligned} \mathrm{A4}: x_{i}^{k}-x_{i-1}^{k}&= -\frac{ \omega ({t}^{k}_{i-1})^2 }{\Vert { T_{i-1}} \Vert ^2 } \frac{V_{i-1}^{k} \times d_{i-1}}{{V_{y,{i}}^{k}}} \nonumber \\&\quad -\frac{V_{x,i}^{k}}{V_{y,i}^{k}} ( y_{i}^{k-1}-y_{i-1}^{k-1}) \nonumber \\ y_{i}^{k}-y_{i-1}^{k}&=0. \end{aligned}$$

(22)

Proof

In order to simplify the calculations, we define the position change in one cell as $\beta ^{k}_{i}= r_{i}^{k}-r_{i-1}^{k}$ and let $\varepsilon _{i}^{k}=\omega \frac{ t^k_i}{\Vert { T_i} \Vert ^2} d_{i}$. Suppose the AUV crosses cell $C^k$ according to case A1. Hence, $y_{i}^k=y_{i-1}^k$ is constant and $x_i^k$ varies as follows:

$$\begin{aligned} x_{i}^{k}&= {t}^{k}_i V_{x,i}^{k}+x_{i}^{k-1} =\varDelta \frac{V_{x,i}^{k}}{V_{y,i}^{k}}+x_{i}^{k-1}. \end{aligned}$$

(23)

From Eq.(23), we obtain the following equality:

$$\varDelta \frac{V_{x,i-1}^{k}}{V_{y,i-1}^{k}}+x_{i-1}^{k-1}-x_{i-1}^{k}=0$$

and $x_{i}^{k}$ follows:

$$\begin{aligned} x_{i}^{k}&=\varDelta ( \frac{V_{x,i}^{k}}{V_{y,i}^{k}}-\frac{V_{x,{i-1}}^{k}}{V_{y,{i-1}}^{k}})+x_{i}^{k-1}-x_{i-1}^{k-1}+x_{i-1}^{k}. \end{aligned}$$

Let $\beta _{x,i}^{k-1}=x_{i}^{k-1}-x_{i-1}^{k-1}$, $x_{i}^{k}$ follows:

$$\begin{aligned} x_{i}^{k}&=x_{i-1}^{k} + {t}^{k}_{i-1} ( \frac{V_{x,i}^{k}V_{y,{i-1}}^{k}-{V_{x,{i-1}}^{k}V_{y,{i}}^{k}}}{{V_{y,{i}}^{k}}})+\beta _{x,i}^{k-1}. \end{aligned}$$

Substituting $V_{{i}}^{k}=V_{{i-1}}^{k}+\varepsilon _{i-1}^{k}$ yields:

$$\begin{aligned} x_{i}^{k}&=x_{i-1}^{k} - \frac{ {t}^{k}_{i-1} }{{V_{y,{i}}^{k}}}V_{i-1}^{k} \times \varepsilon ^{k}_{i-1} +\beta _{x,i}^{k-1}. \end{aligned}$$

Similarly to case A1, we derive Eq. (20) for case A2. Concerning case A3, the AUV position follows:

$$\begin{aligned} y_{i}^{k}&= {t}^{k}_i V_{y,i}^{k}+y_{i}^{k-1} = \frac{\varDelta - x_{i}^{k-1}}{V_{x,i}^{k}}V_{y,i}^{k}+y_{i}^{k-1} . \end{aligned}$$

(24)

From Eq.(24), we obtain at iteration $i-1$:

$$\frac{\varDelta - x_{i-1}^{k-1}}{V_{x,i-1}^{k}}V_{y,i-1}^{k}+y_{i-1}^{k-1} -y_{i-1}^{k}=0$$

and $y_{i}^{k}$ follows:

$$\begin{aligned} y_{i}^{k}&= \frac{\varDelta - x_{i-1}^{k-1}}{V_{x,i}^{k}}V_{y,i}^{k}-\frac{\varDelta - x_{i-1}^{k-1}}{V_{x,{i-1}}^{k}}V_{y,{i-1}}^{k}\\&+y_{i-1}^{k}-\frac{V_{y,i}^{k}}{V_{x,i}^{k}}\beta _{x,i}^{k-1} \\&=y_{i-1}^{k} + {t}^{k}_{i-1} ( \frac{{V_{x,{i-1}}^{k} V_{y,{i}}^{k}}-V_{x,i}^{k}V_{y,{i-1}}^{k}}{{ V_{x,{i}}^{k}}})-\frac{V_{y,i}^{k}}{V_{x,i}^{k}}\beta _{x,i}^{k-1}. \end{aligned}$$

Finally, we substitute $V_{x,{i}}^{k}=V_{x,{i-1}}^{k} +\varepsilon ^{k}_{i-1}$:

$$\begin{aligned} y_{i}^{k}&=y_{i-1}^{k} + {t}^{k}_{i-1} ( \frac{\varepsilon _{y,i-1}^{k}V_{x,{i-1}}^{k}-{V_{y,{i-1}}^{k}\varepsilon _{x,{i-1}}^{k}} }{{V_{x,{i}}^{k}}}) \\&\quad -\frac{V_{y,i}^{k}}{V_{x,i}^{k}}\beta _{x,i}^{k-1}\\&=y_{i-1}^{k}+\frac{ {t}^{k}_{i-1} }{{V_{x,{i}}^{k}}} V_{i-1}^{k} \times \varepsilon ^{k}_{i-1} -\frac{V_{y,i}^{k}}{V_{x,i}^{k}}\beta _{x,i}^{k-1}. \end{aligned}$$

Similarly to case A3, we derive Eq.(22) for case A4. $\square$

Based on Lemma 3, we expand the analysis from one cell to a set of cells. As the traced trajectory involves complicated expressions, we divide the set of crossed cells ${\mathcal {G}}_i$ into subsets ${\mathcal {S}}^j$ that combine different cases of cell crossing. We denote with R1 a segment ${\mathcal {S}}^j$ that comprises the crossed cells in one column and with $\mathrm {R2}$ a segment ${\mathcal {S}}^j$ that comprises the crossed cells in one row, see Fig11. We use the decomposition into segments to mathematically formulate the MT error dynamics and generalize the resulting analysis for trajectories with multiple segments.

Fig. 11

Decomposition of the cells ${\mathcal {G}}_i$ into 5 segments $\mathrm{R2}$. Each segment ${\mathcal {S}}^q$ is included in one row and presented with a different color (color figure online)

We use the segment definition to present the set of crossed cells ${\mathcal {G}}_i$. Let ${\mathcal {S}}^j=\lbrace C^s,\cdots , C^e \rbrace$, we define with $\lambda _i^j=r^e_i-r^e_{i-1}$ the difference between the predicted position in the last cell $C^e$ in ${\mathcal {S}}^j$ in iterations i and $i-1$. We derive a recursive formula to update $\lambda _i^{j}$ after each segment.

Theorem 3

Let ${\mathcal {S}}^j=\lbrace C^s,\cdots , C^e \rbrace$ and ${\mathcal {G}}_i = {\mathcal {G}}_{i-1}$, $\lambda ^{j}_{i}$ follows:

$$\begin{aligned} \mathrm{R1}: \lambda ^{j}_{x,i}&= \frac{V_{x,i}^{e}}{V_{y,i}^{e}}\frac{V_{y,i}^{s}}{V_{x,i}^{s}} \lambda _{x,i}^{j-1} \\&\quad -\omega \frac{V_{x,i}^{e}}{V_{y,i}^{e}} \sum ^{e}_{k=s} \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{k} \times d_{i-1} }{ {V_{x,{i}}^{k}}} \nonumber \\ \lambda ^{j}_{y,i}&=0. \end{aligned}$$

(25)

$$\begin{aligned} \mathrm{R2}: \lambda ^{j}_{x,i}&=0\\ \lambda ^{j}_{y,i}&= \frac{V_{y,i}^{e}}{V_{x,i}^{e}}\frac{V_{x,i}^{s}}{V_{y,i}^{s}} \lambda _{y,i}^{j-1} \\&\quad + \omega \frac{V_{y,i}^{e}}{V_{x,i}^{e}} \sum ^{e}_{k=s} \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{k} \times d_{i-1} }{ {V_{y,{i}}^{k}}} \end{aligned}$$

Proof

Given ${\mathcal {G}}_{i-1}={\mathcal {G}}_i$, then the predicted trajectory crosses the same side of each cell, which implies that the AUV trajectory crosses the same set of segments in iterations i and $i-1$. First, we consider that the traced trajectory crosses n columns and we prove the theorem by induction. Recall that a R1 segment ${\mathcal {S}}=\lbrace C^s,\cdots , C^e \rbrace$ requires $y^e_i - y^{s-1}_i=\varDelta$. Hence, there are two ways to reconstruct a R1 segment. Either the segment is composed of one cell if the cell is crossed according to case A1 ${\mathcal {S}}= C^e$ or the segment contains multiple cells in the same row. In this scenario, the AUV crosses the first cell $C^s$ according to case A3 then multiple cells according to case A2 and finally the last cell $C^e$ according to A4. For instance, the first segment in Fig. 11 is represented by case A1, the second segment is composed of three cells, etc. For clarity, we consider the scenario when the AUV crosses the row according to case A1, hence we can use Lemma 3 to compute $\lambda ^{1}$ in one cell. Given the initial position $r^0$ is constant, we simplify Eq.(19) with $x_{i}^{0}=x_{i-1}^{0}$:

$$\begin{aligned} \lambda ^{1}_{x,i}&= - \frac{ ( t^{1}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{1} \times d_{i-1} }{ {V_{y,{i}}^{1}}}. \end{aligned}$$

(26)

However if the segment includes more than one cell, the expression of $\lambda ^{1}$ becomes more complicated. We reformulate ${\mathcal {S}}^1= \lbrace C^{s},\cdots , C^k,\cdots , C^{e} \rbrace$, where the start cell $C^{s}$ represents case A3, $C^k$ case A2 and the ending cell $C^{e}$ case A4. Consider $C^{s}$ and insert $\lambda ^{0}_{x,i}=0$ in Eq.(21):

$$\begin{aligned} y^{1}_i-y^{1}_{i-1}&=- \frac{ ( t^{1}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{1} \times d_{i-1} }{ {V_{y,{i}}^{1}}} . \end{aligned}$$

Concerning cells $C^k$ that are between $C^{s}$ and $C^{e}$, the predicted trajectory follows case A2, (20):

$$\begin{aligned} y^{k}_i-y^{k}_{i-1}&= \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{k} \times d_{i-1} }{ {V_{x,{i}}^{k}}} +(y^{k-1}_i-y^{k-1}_{i-1}) . \end{aligned}$$

(27)

Finally plugging (21) and (27) in Eq.(22) provides $\lambda ^{1}_i$ in the last cell in segment ${\mathcal {S}}^{1}$ :

$$\begin{aligned} \lambda ^{1}_{x,i}&=- \frac{ ( t^{e}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{e} \times d_{i-1} }{ {V_{y,{i}}^{e}}} -\frac{V_{x,i}^{e}}{V_{y,i}^{e}} (y^{e-1}_i-y^{e-1}_{i-1})\\&=- \frac{V_{x,i}^{e}}{V_{y,i}^{e}} \sum ^{e}_{k=1} \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{k} \times d_{i-1} }{ {V_{x,{i}}^{k}}} \end{aligned}$$

Suppose $\lambda ^{j-1}_{i}$ is given and consider $\lambda ^{j}_i$ corresponding to ${\mathcal {S}}^j= \lbrace C^{s},\cdots ,C^{e} \rbrace$. Combining Eq.(27) and (21) leads to $y^{e-1}_i-y^{e-1}_{i-1}$:

$$\begin{aligned} (y^{e-1}_i-y^{e-1}_{i-1}) = \sum ^{e-1}_{k=s} \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{k} \times d_{i-1} }{ {V_{x,{i}}^{k}}} -\frac{V_{y,i}^{s}}{V_{x,i}^{s}}\lambda _{x,i}^{j-1}. \end{aligned}$$

(28)

Inserting (28) in (22) results in $\lambda ^{j}_{x,i}$:

$$\begin{aligned} \lambda ^{j}_{x,i}&=- \frac{ ( t^{e}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{e} \times d_{i-1} }{ {V_{y,{i}}^{e}}} -\frac{V_{x,i}^{e}}{V_{y,i}^{e}} (y^{e-1}_i-y^{e-1}_{i-1}) \nonumber \\&= \frac{V_{x,i}^{e}}{V_{y,i}^{e}}\frac{V_{y,i}^{s}}{V_{x,i}^{s}} \lambda _{x,i}^{j-1} -\omega \frac{V_{x,i}^{e}}{V_{y,i}^{e}} \sum ^{e}_{k=s} \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ V_{i-1}^{k} \times d_{i-1} }{ {V_{x,{i}}^{k}}} \end{aligned}$$

(29)

If a row is crossed according to case A1, then $C^{s}=C^{e}$ and Eq.(29) can be reduced to (26). Hence, the induction statement holds for any scenarios. Consider the scenario when the AUV trajectory crosses n segments type of $\mathrm {R2}$, then each segment can be composed of one cell if the cell is crossed according to case A2 ${\mathcal {S}}= C^e$ or the segment contains multiple cells in the same column. In this scenario, the AUV crosses the first cell $C^s$ according to case A4 then multiple cells according to case A1 and finally the last cell $C^e$ according to A3. Hence the same reasoning leads to Eq.(25). $\square$

We show in the following that the MT error $d_i$ has a constant direction $\forall i$ when the flow field is initially constant.

Lemma 4

Suppose Assumption 2 is true, then $V_{i}^{k} \times d_{i}=V_{0}^{k} \times d_{i}$ and $d_{i}=\kappa _i d_0$ hold where: $\kappa _{i+1}= (1- \omega \sum ^{k=f}_{k=1} \sum ^{i}_{j=0} \frac{ {t}^k_{i+1} {t}^k_j}{\Vert {T_j} \Vert ^2} \kappa _j )$

Proof

We prove the Lemma by induction. Given $d_1={\tilde{r}}^f-r^f_1$:

$$\begin{aligned} d_1&={\tilde{r}}^f-r^0- \sum ^{k=f}_{k=1} {t}^k_1 ( V^k_{0}+ \omega \frac{ {t}^k_0}{\Vert {T_0} \Vert ^2} d_0) \nonumber \\&=(1- \omega \sum ^{k=f}_{k=1} \frac{ {t}^k_1 {t}^k_0}{\Vert {T_0} \Vert ^2} ) d_0 = \kappa _1 d_0 \end{aligned}$$

(30)

Assuming that $\kappa _{i}= (1- \omega \sum ^{k=f}_{k=1} \sum ^{i-1}_{j=0} \frac{ {t}^k_{i} {t}^k_j}{\Vert {T_j} \Vert ^2} \kappa _j )$ and $d_{i}=\kappa _{i} d_0$, let us consider the MT error for $i+1$:

$$\begin{aligned} d_{i+1}&={\tilde{r}}^f-r^0- \sum ^{k=f}_{k=1} {t}^k_{i+1} ( V^k_{0}+ \omega \sum ^{i}_{j=0} \frac{ {t}^k_j}{\Vert {T_j} \Vert ^2} d_j) \nonumber \\&=(1- \omega \sum ^{k=f}_{k=1} \sum ^{i}_{j=0} \frac{ {t}^k_{i+1} {t}^k_j}{\Vert {T_j} \Vert ^2} \kappa _j ) d_0 =\kappa _{i+1} d_0 \end{aligned}$$

(31)

Finally, insert Eq.(31) in the flow update (8) simplifies $V_{i}^{k} \times d_{i}$ to:

$$\begin{aligned} V_{i+1}^{k} \times d_{i+1}&=(V_0+ \sum ^{i}_{j=0} \frac{ {t}^k_j}{\Vert {T_j} \Vert ^2} \kappa _j ) d_{0} )\times \kappa _{i+1} d_{0}\\&=V_{0}^{k} \times d_{i+1} \end{aligned}$$

$\square$

In order to relate the change of the AUV predicted position when ${{\mathcal {G}}}_{i}= {{\mathcal {G}}}_{i-1}$ and ${{\mathcal {G}}}_{i}\ne {{\mathcal {G}}}_{i-1}$, we define $t^m$ as the time when $\gamma _i$ crosses new cell $C^m$, ${\hat{r}}^f_i = \int _{0 }^{t^\mathrm{tot}} V_i(t) \mathrm{d}t$, where $V(t)=V^k_0=V_0$ for $t \in [t^m,t^\mathrm{tot}]$ and $r^f_{i}$ the predicted final position if ${{\mathcal {G}}}_{i}= {{\mathcal {G}}}_{i-1}$. Apply Lemma 4 and Eq.(8) results in: $\int _{0 }^{t^m} V^k_{i}(t) - V^k_{i-1}(t) \mathrm{d}t =\alpha _2 d_0$. Suppose ${{\mathcal {G}}}_{i}$ includes new cells for $t \in [t^m,t^\mathrm{tot}]$, then $V^k_{i}(t) =V_{0}$ for $t \in [t^m,t^\mathrm{tot}]$:

$$\begin{aligned} {\hat{r}}^f_i- r^f_{i-1}&= \int _{0 }^{t^{m}} \left( V^k_{i} - V^k_{i-1} \right) (t) \mathrm{d}t \\&+ \int _{t^{m}}^{t^\mathrm{tot}} V_{0} - V^k_{i-1}(t) \mathrm{d}t \nonumber \\&=(\alpha _2- \int _{t^{m}}^{t^\mathrm{tot}} \omega \sum ^{i-1}_{j=0} \sum ^{f}_{k=m} \frac{ {t}^k_j}{\Vert { {T}_j} \Vert ^2} \kappa _j \mathrm{d}t ) d_0. \end{aligned}$$

(32)

We define

$$\begin{aligned} \alpha _1= \alpha _2- \int _{t^{m}}^{t^\mathrm{tot}} \omega \sum ^{i-1}_{j=0} \sum ^{f}_{k=m} \frac{ {t}^k_j}{\Vert { {T}_j} \Vert ^2} \kappa _j \mathrm{d}t \end{aligned}$$

so that

$$\begin{aligned} {\hat{r}}^f_i- r^f_{i-1} =\alpha _1 d_0. \end{aligned}$$

(33)

Lemma 5

Suppose that Assumption 2 is true, $\alpha _1 \ge 0$ and $\kappa _j \ge 0$ hold for all $j < i$. Then

$$\begin{aligned} \Vert {{\hat{r}}^{f}_{i}} -r^{f_{i-1}} \Vert \le \Vert {r^{f_i}}- r^{f_{i-1}} \Vert \end{aligned}$$

(34)

Proof

Consider the scenario when ${{\mathcal {G}}}_{i}= {{\mathcal {G}}}_{i-1}$ and let $\alpha _3= \int _{t^{m}}^{t^\mathrm{tot}} \omega \frac{ {t}^k_{i-1}}{\Vert { {T}_{i-1}} \Vert ^2} \kappa _{i-1} \mathrm{d}t$. Then $r^f_i -r^f_{i-1}$ follows:

$$\begin{aligned} r^f_i -r^f_{i-1}&= \int _{0 }^{t^{m}} \left( V^k_{i} - V^k_{i-1}\right) (t) \mathrm{d}t \nonumber \\&\quad + \int _{t^{m}}^{t^\mathrm{tot}} \left( V^k_{i} - V^k_{i-1} \right) (t) \mathrm{d}t \nonumber \\&=(\alpha _2+\alpha _3 )d_0. \end{aligned}$$

(35)

Since we have assumed that $\alpha _1 \ge 0$, then $\Vert {\hat{r}}^f_i- r^f_{i-1} \Vert = \alpha _1 \Vert d_0 \Vert$ holds. Further,

$$V^k_{i} = V_0+\omega \sum ^{i-1}_{j=0} \sum ^{f}_{k=m} \frac{{t}^k_j}{\Vert { {T}_j} \Vert ^2} \kappa _j d_0$$

implies $\alpha _3 \ge - \int _{t^{m}}^{t^\mathrm{tot}} \omega \sum ^{i-1}_{j=0} \sum ^{f}_{k=m} \frac{ {t}^k_j}{\Vert { {T}_j} \Vert ^2} \kappa _j \mathrm{d}t$. Thus $0 \le \alpha _1 = \alpha _2 - \int _{t^{m}}^{t^\mathrm{tot}} \omega \sum ^{i-1}_{j=0} \sum ^{f}_{k=m} \frac{ {t}^k_j}{\Vert { {T}_j} \Vert ^2} \kappa _j \mathrm{d}t \le \alpha _2+\alpha _3$. Combining (35) and (32) results in:

$$\begin{aligned} \Vert {\hat{r}}^f_i- r^f_{i-1} \Vert&\le (\alpha _2+\alpha _3)\Vert d_0\Vert =\Vert r^f_i -r^f_{i-1} \Vert \end{aligned}$$

$\square$

We derive bounds on the trajectory evolution after the estimated flow is updated in the following Lemma.

Lemma 6

Suppose ${\mathcal {G}}_i = {\mathcal {G}}_{i-1}$ and Assumption 2 is true. Let $\mu = \max \left( \mid \frac{d_{x,0}}{d_{y,0}} \frac{V_{y,0}}{V_{x,0}} \mid ,\mid \frac{d_{y,0}}{d_{x,0}} \frac{V_{x,0}}{V_{y,0}} \mid \right)$ and ${\omega }= \frac{1}{\mu ^{n}} \min (\mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } \mid )$, then an upper bound for $\Vert \lambda ^{n-1}_{i} \Vert$ follows:

R1:

$$\begin{aligned} \Vert \lambda ^{n-1}_{i}\Vert \le&\mid V_{0} \times d_{i-1} \mid \cdot \\&\sum ^{f-1}_{k=1} { \frac{ ({t_{i-1}^{k}})^2 }{\Vert T_{i-1}\Vert ^2} } \frac{ \min \left( \mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } \mid \right) }{ \mid V_{y,{i}}^{k} \mid } \end{aligned}$$

(36)

R2:

$$\begin{aligned} \Vert \lambda ^{n-1}_{i}\Vert \le&\mid V_{0} \times d_{i-1} \mid \cdot \\&\sum ^{f-1}_{k=1} \frac{ ( {t_{i-1}^{k}})^2 }{\Vert T_{i-1}\Vert ^2} \frac{ \min \left( \mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} }\mid \right) }{ \mid V_{x,{i}}^{k} \mid } \end{aligned}$$

(37)

Proof

Since Eq.(29) depends on $\frac{V_{x,i}^{e}}{V_{y,i}^{e}}\frac{V_{y,i}^{s}}{V_{x,i}^{s}}$, we show first that $\frac{V_{x,i}^{e}}{V_{y,i}^{e}}\frac{V_{y,i}^{s}}{V_{x,i}^{s}}$ is bounded. Notice that $\frac{V_{x,i}^{k}}{V_{y,i}^{k}}-\frac{d_{x,0}}{d_{y,0}} =\frac{V_{x,i}^{k} d_{y,0}- V_{y,i}^{k}d_{x,0} }{ V_{y,i}^{k} d_{y,0} }$ and $\frac{V_{x,i}^{k}}{V_{y,i}^{k}}-\frac{V_{x,0}}{V_{y,0}} =\frac{V_{x,i}^{k} V_{y,0}- V_{y,i}^{k}V_{x,0} }{ V_{y,i}^{k} V_{y,0} }$. Hence if $\frac{d_{x,0}}{d_{y,0}} \le \frac{V_{x,0}}{V_{y,0}}$, then $\frac{d_{x,0}}{d_{y,0}} \le \frac{V_{x,i}^{k}}{V_{y,i}^{k}} \le \frac{V_{x,0}}{V_{y,0}}$ and vice versa. Therefore, $\min (\frac{d_{x,0}}{d_{y,0}} ,\frac{V_{x,0}}{V_{y,0}}) \le \frac{V_{x,i}^{k}}{V_{y,i}^{k}} \le \max (\frac{d_{x,0}}{d_{y,0}} ,\frac{V_{x,0}}{V_{y,0}})$ and $\mid \frac{V_{x,i}^{m}}{V_{y,i}^{m}}\frac{V_{y,i}^{n}}{V_{x,i}^{n}} \mid \le \mid \frac{d_{x,0}}{d_{y,0}} \frac{V_{y,0}}{V_{x,0}} \mid$ is true $\forall 0<n,m \le f$. Combining the latter with $\mu = \mid \max ( \frac{d_{x,0}}{d_{y,0}} \frac{V_{y,0}}{V_{x,0}} , \frac{d_{y,0}}{d_{x,0}} \frac{V_{x,0}}{V_{y,0}} ) \mid$, then $\mu > \mid \frac{V_{x,i}^{m}}{V_{y,i}^{m}}\frac{V_{y,i}^{n}}{V_{x,i}^{n}} \mid$ is true. Thus, inserting $\mu > \mid \frac{V_{x,i}^{e}}{V_{y,i}^{e}}\frac{V_{y,i}^{s}}{V_{x,i}^{s}} \mid$ and $V_{i-1}^{k} \times d_{i-1}=V_{0}^{k} \times d_{i-1}$ in Eq.(29) results in the following upper bound:

$$\begin{aligned} \mid \lambda ^{j}_{x,i} \mid&\le \frac{V_{x,i}^{e}}{V_{y,i}^{e}}\frac{V_{y,i}^{s}}{V_{x,i}^{s}} \mid \lambda _{x,i}^{j-1} \mid \\&\quad +\omega \frac{V_{x,i}^{e}}{V_{y,i}^{e}} \sum ^{e}_{k=s} \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ \mid V_{0}^{k} \times d_{i-1} \mid }{ {\mid V_{x,{i}}^{k} \mid } } \nonumber \\&\le \mu \mid \lambda _{x,i}^{j-1} \mid \\&\quad +\omega \frac{V_{x,i}^{e}}{V_{y,i}^{e}} \sum ^{e}_{k=s} \frac{ ( t^{k}_{i-1})^2 }{\Vert { T_i} \Vert ^2} \frac{ \mid V_{0}^{k} \times d_{i-1} \mid }{ \mid {V_{x,{i}}^{k}}\mid }. \end{aligned}$$

(38)

Consider the second tern in Eq.(38) and use $\mu > \mid \frac{V_{x,i}^{e}}{V_{y,i}^{e}}\frac{V_{y,i}^{k}}{V_{x,i}^{k}} \mid$, an upper bound for Eq.(38) follows:

$$\begin{aligned} \mid \lambda ^{j}_{x,i} \mid&\le \mu \mid \lambda _{x,i}^{j-1} \mid + \omega \mu \sum ^{e}_{k=s} \mid \frac{ (t^{k}_{i-1})^2 }{\Vert T_{i-1}\Vert ^2 } \mid \frac{V_{0} \times d_{i-1}}{{ V_{y,{i}}^{k} }} \mid . \end{aligned}$$

(39)

Applying ${\omega }= \frac{1}{\mu ^{n}} \min (\mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } \mid )$, $\lambda ^{0}_{x,i}=0$ and $j=n-1$ results in the following bounds:

$$\begin{aligned} \mid \lambda ^{n-1}_{x,i} \mid \le \mid V_{0} \times d_{i-1} \mid \sum ^{f-1}_{k=1} { \frac{ ({t_{i-1}^{k}})^2 }{\Vert T_{i-1}\Vert ^2} } \frac{ \min \left( \mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } \mid \right) }{ \mid V_{y,{i}}^{k} \mid } \end{aligned}$$

Using $\lambda ^{n-1}_{y,i} =0$ results in the inequality (36). Since $\mu = \max ( \mid \frac{d_{x,0}}{d_{y,0}} \frac{V_{y,0}}{V_{x,0}}\mid , \mid \frac{d_{y,0}}{d_{x,0}} \frac{V_{x,0}}{V_{y,0}} \mid )$, the same reasoning results in the following upper bound for

$$\begin{aligned} \mid \lambda ^{j}_{y,i} \mid&\le \mu \mid \lambda _{y,i}^{j-1}\mid + \omega \mu \sum ^{e }_{k=s } { \mid \frac{ ({t_{i-1}^{k}})^2 }{\Vert T_{i-1}\Vert ^2} } \frac{V_{0} \times d_{i-1} }{{V_{x,{i}}^{k}}} \mid . \end{aligned}$$

$\square$

Further, an analytic expression for $y^{f}_{i}$ is computed using $\lambda ^{n-1}_{x,i}$ in the following Lemma.

Lemma 7

Suppose that Assumption 2 is true, ${\mathcal {G}}_i={\mathcal {G}}_{i-1}$ and $y^{f-1}_{i}=y^{f-1}_{i-1}$, then

$$\begin{aligned} {y^{f}_{i}}&= y^f_{i-1} + \left( \omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2} - \frac{ \lambda ^{n-1}_{x,i} V^f_{y,i}}{ V_0\times d_{i-1} } \right) d_{y,i-1} \end{aligned}$$

(40)

Proof

Let $\xi _i=t^f_{i}V^f_{y,i}$, then $y^{f-1}_{i}=y^{f-1}_{i-1}$ implies $y^{f}_{i}=y^{f}_{i-1}+\xi _i - \xi _{i-1}$ and we compute $\xi _i$ as a function of $\xi _{i-1}$ and $\lambda ^{n-1}_{x,i}$.

Let $L_1$ be the line that joins $r^{f}_{i}$ and ${\tilde{r}}^{f}$, $M=L_1 \cap [r^{f-1}_{i} r^{f-1}_{i-1}]$ and $r^f_{i}r^{f-1}_{i} M$ the triangle with the angles $\theta$, $\alpha$ and $\beta$ respectively, Fig 12. We define with $\psi$ the angle between $L_1$ and $[r^{f-1}_{i} r^{f-1}_{i-1}]$, then $\xi _i=\sin (\psi ) {\Vert r^f_i M\Vert }$. Applying $\theta =180-\psi$ and the sinus Law in triangle $r^f_{i}r^{f-1}_{i} M$ results in:

$$\begin{aligned} {\Vert r^{f_{i}} M\Vert }&=\frac{\sin (\alpha )}{\sin (\beta )} {\Vert r^{f-1}_i M\Vert } =\frac{\sin (\alpha )}{\sin (\psi -\alpha )} {\Vert r^{f-1}_i M\Vert }\nonumber \\&= \frac{1}{{\cos (\psi )} -{\sin (\psi )} \frac{\cos (\alpha )}{\sin (\alpha )} } {\Vert r^{f-1}_i M\Vert } \end{aligned}$$

(41)

Since $r^f_{i}-r^{f-1}_{i}=\xi _i$ and $M,r^f_{i} \in L_1$, then $\tan (\psi )=\frac{d_{y}}{d_x}$ and $\tan (\alpha )=\frac{V^f_{y,i}}{V^f_{x,i}}$ hold. Apply $\xi _i=\sin (\psi )\Vert r^f_i M\Vert$ results in:

$$\begin{aligned} \xi _i&= \frac{1}{\frac{\cos (\psi )}{\sin (\psi )} - \frac{\cos (\alpha )}{\sin (\alpha )} } {\Vert r^{f-1}_i M\Vert } = \frac{1}{\frac{d_{x,i-1}}{d_{y,i-1}} - \frac{V^f_{x,i}}{V^f_{y,i}} } {\Vert r^{f-1}_i M\Vert }, \end{aligned}$$

(42)

where $\Vert r^{f-1}_i M\Vert = \lambda ^{n-1}_{x,i}+\Vert r^{f-1}_{i-1} M\Vert$. Consider $\Vert r^{f-1}_{i-1} M\Vert$ and applying Eq. (41) and (42) for triangle $r^f_{i-1}r^{f-1}_{i-1} M$ results in $\xi _{i-1}$:

$$\begin{aligned} \xi _{i-1}&= \frac{1}{\frac{d_{x,i-1}}{d_{y,i-1}} - \frac{V^f_{x,i-1}}{V^f_{y,i-1}} } {\Vert r^{f-1}_{i-1} M\Vert } \end{aligned}$$

(43)

Combining (42) and (43) results in:

$$\begin{aligned} \xi _i&= \frac{1}{\frac{d_{x,i-1}}{d_{y,i-1}} - \frac{V^f_{x,i-1}}{V^f_{y,i-1}} } \lambda ^{n-1}_{x,i} +\frac{ \frac{d_{x,i-1}}{d_{y,i-1}} - \frac{V^f_{x,i-1}}{V^f_{y,i-1}} }{\frac{d_{x,i-1}}{d_{y,i-1}} - \frac{V^f_{x,i}}{V^f_{y,i}} } \xi _{i-1} \end{aligned}$$

Inserting $V_{i} \times d_{i-1}= V_{i-1} \times d_{i-1}$ simplifies $\xi _i$:

$$\begin{aligned} \xi _i&= \xi _{i-1} - \frac{ \lambda ^{n-1}_{x,i} V^f_{y,i}}{ V_0\times d_{i-1} } +\omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2}d_{y,i-1} \end{aligned}$$

Recall $y^f_i=y^{f}_{i-1}+ \xi _i- \xi _{i-1}$, then $y^f_i$ is reduced to:

$$\begin{aligned} y^f_i&= y^f_{i-1} + ( \omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2} - \frac{ \lambda ^{n-1}_{x,i} V^f_{y,i}}{ V_0\times d_{i-1} } ) d_{y,i-1} \end{aligned}$$

$\square$

Theorem 4

Suppose Assumption 2 and 1 are true. Let $\mu = \max \left( \mid \frac{d_{x,0}}{d_{y,0}} \frac{V_{y,0}}{V_{x,0}} \mid ,\mid \frac{d_{y,0}}{d_{x,0}} \frac{V_{x,0}}{V_{y,0}} \mid \right)$ and ${\omega }= \frac{1}{\mu ^{n}} \min (\mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } \mid )$, then the MT error $d_i$ converges to 0, as $i \rightarrow \infty$.

Fig. 12

Illustration of the traced trajectories $\gamma _i$ (purple line), $\gamma _{i-1}$ (blue line) and the change of $r^f_i$ as a function of $r^f_{i-1}$ and $\lambda ^{n-1}_{x,i}$ (color figure online)

Proof

The proof consists of two parts. Given that $d_{j}=\kappa _j d_0$ for all $j\le i$ according to Lemma 4, we use the induction method to show that $\kappa _i \ge 0$ in part 1 of the proof. Then we show $\Vert d_i\Vert$ is monotonically decreasing and accordingly $\Vert d_{i} \Vert \longrightarrow 0$ when $i \rightarrow \infty$ in part 2.

As the following analysis holds for $\forall i \ge 0$.

Part 1: In order to show that $\kappa _i \ge 0$, we suppose w.l.o.g that $d_{y,0} \ne 0$ and we consider $\mid {\hat{y}}^f_{i}-y^f_{i-1} \mid$, where ${\hat{r}}^f_i$ is the AUV position when ${\mathcal {G}}_{i-1} \ne {\mathcal {G}}_i$ and $r^f_i$ when ${\mathcal {G}}_{i-1} = {\mathcal {G}}_i$. Since $r^f_{i}$, $r^f_{i-1}$ and ${\hat{r}}^f_{i}$ are aligned, then ${\hat{r}}^f_i -r^f_{i-1} = \frac{ \Vert {\hat{r}}^f_i -r^f_{i-1} \Vert }{\Vert r^f_i- r^f_{i-1} \Vert } ( r^f_i- r^f_{i-1} )$ is true and Eq.(34) implies that $\frac{ \Vert {\hat{r}}^f_i -r^f_{i-1} \Vert }{\Vert r^f_i- r^f_{i-1} \Vert } \le 1$. Hence, $\mid {\hat{y}}^f_i -y^f_{i-1} \mid \le \mid y^f_i- y^f_{i-1} \mid$ holds. Apply $d_{y,i}= {\tilde{y}}^f -{\hat{y}}^f_{i}$, then an upper bound for $\mid {\hat{y}}^f_{i} - y^f_{i-1} \mid$ follows:

$$\begin{aligned} \mid {\hat{y}}^f_{i}-y^f_{i-1} \mid&=\mid d_{y,i-1}-d_{y,i} \mid \\&= \mid \kappa _{i-1}-\kappa _i \mid \mid d_{y,0}\mid \\&\le \mid {y}^f_{i}-y^f_{i-1} \mid \end{aligned}$$

Substitute $\mid {y}^f_{i}-y^f_{i-1} \mid$ using Eq.(40) which holds when ${\mathcal {G}}_i={\mathcal {G}}_{i-1}$ results in the following upper bound:

$$\begin{aligned} \mid \kappa _{i-1}-\kappa _i \mid \mid d_{y,0} \mid \le \mid \omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2} - \frac{ \lambda ^{n-1}_{x,i} V^f_{y,i}}{ V_0\times d_{i-1} } \mid \mid d_{y,i-1} \mid \end{aligned}$$

(44)

Insert $\mid d_{y,i-1} \mid = \kappa _{i-1} \mid d_{y,0} \mid$ in (44) and divide with $\kappa _{i-1}$:

$$\begin{aligned} \mid 1- \frac{\kappa _i}{\kappa _{i-1}} \mid&\le \omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2} + \mid \frac{ \mid \lambda ^{n-1}_{x,i} \mid V^f_{y,i}}{ \mid V_0\times d_{i-1} \mid } \mid \end{aligned}$$

Now substitute $\lambda ^{n-1}_{x,i}$ with Eq.(36) leads to:

$$\begin{aligned} \mid 1- \frac{\kappa _i}{\kappa _{i-1}} \mid&\le \omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2} + \mid \frac{ \mid \lambda ^{n-1}_{x,i} \mid V^f_{y,i}}{ \mid V_0\times d_{i-1} \mid } \mid \nonumber \\&\le \mid \sum ^{f-1}_{k=1} { \frac{ ({t_{i-1}^{k}})^2 }{\Vert T_{i-1}\Vert ^2} } \min ( \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } , \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } ) \frac{ V^f_{y,i} }{ V_{y,{i}}^{k} } \mid \nonumber \\&\quad +\omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2} \end{aligned}$$

(45)

Since $\mu = \max \left( \mid \frac{d_{x,0}}{d_{y,0}} \frac{V_{y,0}}{V_{x,0}} \mid ,\mid \frac{d_{y,0}}{d_{x,0}} \frac{V_{x,0}}{V_{y,0}} \mid \right) \ge 1$

and $\min (\mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } \mid ) <1$, then ${\omega }= \frac{1}{\mu ^{n}} \min (\mid \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } \mid , \mid \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } \mid ) < 1$.

Further $\sum ^{f}_{k=1} ({t_{i-1}^{k}})^2 ={\Vert T_{i-1}\Vert ^2}$ and $\frac{{V_{y,{i}}^{f}}}{{V_{y,{i}}^{k}}} \min ( \frac{ \underline{ {V}}_{y} }{ \overline{ {V}}_{y} } , \frac{ \underline{ {V}}_{x} }{ \overline{ {V}}_{x} } ) \le 1$ leads to $\omega \frac{(t^f_{i-1})^2}{\Vert T_{i-1}\Vert ^2} + \sum ^{f-1}_{k=1} \frac{ ({t_{i-1}^{k}})^2 }{\Vert T_{i-1}\Vert ^2} \le 1$.

Hence, $1- \frac{\kappa _i}{\kappa _{i-1}} \le 1$ and $\kappa _{i-1}\ge 0$ implies $\kappa _i \ge 0$. We show in the second part that $d_{y,i}$ is monotonically decreasing and accordingly $d_{y,i} \longrightarrow 0$ as $i \longrightarrow \infty$.

Part 2: We show first that $r^f_{i} \in [r^f_{i-1},{\tilde{r}}^f]$. As Lemma 4 implies that $d_{i}=\kappa _i d_0, \forall i$ then $V^k_{i}= V_{0}+ (\omega \sum ^{i-1}_{j=0} \frac{t^k_j}{\Vert {T_j} \Vert ^2} \kappa _j )d_0$ and there exists $\nu : [0,t^{ \mathrm tot }] \rightarrow {\mathcal {R}}$ such that $r_i(t) - r_{i-1}(t )=\nu (t) d_{0}$. We show in the following that $\nu (t ) > 0$.

Let $t^1_i$ and $t^1_{i-1}$ be the travel time in cell $C^1$ at iteration i and $i-1$ and we define $t^m= \min (t^1_i,t^1_{i-1})$, then $\int _{0 }^{t } \left( V_{i} - V_{i-1}\right) (s) \mathrm{d}s= \frac{ t t^1_{i-1}}{\Vert T_{i-1}\Vert ^2} \kappa _{i-1}d_0=\nu (t ) d_{0}$ with

$\nu (t) > 0$ for $0< t \le t^m$. Suppose there exists $t^m < \tau \le t^{ \mathrm tot }$ such that $\nu (\tau )=0$ and $\nu (t)>0$ for $0< t < \tau$. Hence, $r_i(\tau ) - r_{i-1}(\tau )=0$ and $\gamma _i$ and $\gamma _{i-1}$ intersect in the same cell.

Let $C^r= {\mathcal {G}}_{i-1} \cap {\mathcal {G}}_{i}$ and $\varepsilon >0$ such that $r_i(\tau ) \in C^r$ and $r_i(\tau -\varepsilon ) \in C^r$. As $C^r \in {\mathcal {G}}_{i-1}$, then

$V_{i}^r=V_{i-1}^r+\frac{t^r_{i-1}}{\Vert T_{i-1}\Vert ^2} \kappa _{i-1}d_0$ and accordingly

$\int _{ \tau -\varepsilon }^{ \tau } \frac{t^r_{i-1}}{\Vert T_{i-1}\Vert ^2} \mathrm{d}t = \varepsilon \frac{t^r_{i-1}}{\Vert T_{i-1}\Vert ^2} >0$. Given $\nu (t)>0$ for $0<t < \tau$, then $\nu (\tau -\varepsilon ) >0$ and $\nu (\tau ) > \nu (\tau -\varepsilon )$, Thus, $\nu (\tau ) >0$ which contradicts $\nu (\tau )=0$. Hence, $\nu (t)>0$ for $0< t \le t^{ \mathrm tot }$.

Let $\alpha _1=\nu ( t^\mathrm{{tot}} )$ such that $r^f_i - r^f_{i-1}= \alpha _1 d_{0}$, then applying $d_{i-1}=\kappa _{i-1} d_{0}$ and ${\tilde{r}}^f - r^f_{i}= d_{i}=\kappa _{i} d_{0}$ leads to $d_{i} = (1- \frac{\alpha _1}{\kappa _{i-1}}) d_{i-1}$. Recall that $d_{i}=\frac{\kappa _{i}}{\kappa _{i-1}}d_{i-1}$, then $(1- \frac{\alpha _1}{\kappa _{i-1}})= \frac{\kappa _{i}}{\kappa _{i-1}}$. Hence, $\kappa _{i-1} \ge 0$ and $\kappa _{i} \ge 0$ implies that $(1- \frac{\alpha _1}{\kappa _{i-1}}) > 0$.

Finally, $\alpha _1=\nu ( t^\mathrm{{tot}} ) > 0$ implies $(1- \frac{\alpha _1 }{\kappa _{i-1}} ) <1$. Therefore, $d_{i} = (1- \frac{\alpha _1}{\kappa _{i-1}}) d_{i-1}$ with $0< (1- \frac{\alpha _1 }{\kappa _{i-1}} ) <1$ implies $r^f_{i} \in [r^f_{i-1},{\tilde{r}}^f]$. However, we still can not draw the conclusion that $\lim _{i\rightarrow \infty } d_i=0$ because it is still possible that $\lim _{i\rightarrow \infty } (1- \frac{\alpha _1 }{\kappa _{i-1}} )=1$.

The following arguments will show that $\Vert d_i\Vert <\Vert d_{i-1}\Vert$ holds for all i, $i\rightarrow \infty$.

Given the AUV trajectory ${\gamma }_{i}$ evolves monotonically and the number of cells is bounded, then there exists p such that ${{\mathcal {G}}}_i= {{\mathcal {G}}}_{i+1}, \forall i \ge p$ and $y_i^{f-1}=y_{i+1}^{f-1}$ or $x_i^{f-1}=x_{i+1}^{f-1}$.

Suppose $y^{f-1}_{i+1} =y^{f-1}_p=y^{f-1}$ and $\delta = \min ( \mid y^{f}_p-y^{f-1} \mid , \mid {\tilde{y}}^{f}-y^{f-1} \mid )$. As $y^f_{i} \in [y^f_{i-1},{\tilde{y}}^f]$, then $\delta \le \mid y^{f}_i-y^{f-1} \mid$ is constant $\forall i \ge p$.

Let $\tau = \frac{ \delta }{ {\overline{V}}_{y} } < t^f_i,t^f_{i+1}$ and recall $\nu (t) >0$, then $r_i(t^\mathrm{{tot}}-\tau ) - r_{i-1}(t^\mathrm{{tot}}-\tau )= \frac{\nu (t^\mathrm{{tot}}-\tau )}{\kappa _{i}} d_{i}$, with $\frac{\nu (t^\mathrm{{tot}}-\tau )}{\kappa _{i}} > 0$ and a bound on $\mid d_{y,i+1} \mid$ follows:

$$\begin{aligned} \mid d_{y,i+1} \mid&= \mid d_{y,i} - \int _{0}^{t^\mathrm{tot}} \left( V^k_{y,i+1} - V^k_{y,i} \right) (t) \mathrm{d}t \mid \\&\le \mid 1 - \frac{\nu ( t^\mathrm{{tot}}-\tau ) }{\kappa _{i}} - \int _{t^\mathrm{{tot}}-\tau }^{ t^\mathrm{{tot}} } \frac{ t^f_{i} }{\Vert T_i\Vert ^2} \mathrm{d}t \mid \mid d_{y,i}\mid \\&\le (1 - \frac{\tau ^2}{\Vert T_i\Vert ^2})\mid d_{y,i} \mid \end{aligned}$$

Given ${\Vert T_{i}\Vert ^2} = \sum ^{f}_{k=1} ({t_{i}^{k}})^2 \le ( \sum ^{f}_{k=1} {t_{i}^{k}})^2=(t^\mathrm{{tot}})^2$, then $\frac{ ( \frac{ \delta }{{\overline{V}}_{y}} ) ^2}{(t^{tot})^2} \le \frac{\tau ^2}{\Vert T_{i}\Vert ^2}$ and $\mid d_{y,i+1} \mid$ is bounded as follows:

$$\begin{aligned} 0 \le \mid d_{y,i+1} \mid \le (1- \omega \frac{ ( \frac{ \delta }{{\overline{V}}_{y}} ) ^2}{(t^{tot})^2} ) \mid d_{y,i} \mid . \end{aligned}$$

Since $0< 1- \omega \frac{ ( \frac{ \delta }{{\overline{V}}_{y}} ) ^2}{(t^{tot})^2} <1$, which does not depend on i, we conclude that $\lim _{i \longrightarrow \infty } (1- \omega \frac{ ( \frac{ {y}^1-y^{0} }{{\overline{V}}_{y}} ) ^2}{(t^{tot})^2} )^i =0$ and $\lim _{i \longrightarrow \infty } d_{y,i} =0$. Recall that $d_{i}=\kappa _i d_0$ and $d_{y,0}\ne 0$ then $\lim _{i \longrightarrow \infty } \kappa _{i} =0$ and $d_{i} \longrightarrow \varvec{0}$ when $i \longrightarrow \infty$. $\square$