Nonparametric dynamics modeling for underwater vehicles using local adaptive moment estimation Gaussian processes learning

Abstract

This paper investigates a nonparametric modeling scheme for underwater vehicles to achieve continuous-time dynamics modeling, which is essential for various marine missions, control, and navigation of these vehicles. The proposed scheme addresses the challenges posed by the nonlinearity, strong coupling, and complex structure of underwater vehicles through local adaptive moment estimation Gaussian processes learning. This approach constructs mappings between hydrodynamics and motion states while providing uncertainty estimates of the dynamics model. A local weighted strategy is used to construct local models to localize Gaussian processes learning, and an adaptive moment estimation method is designed using gradients of innovation to tune hyperparameters of Gaussian processes automatically. Moreover, a subspace index is created and updated based on feature distance measures to improve the computational efficiency of Gaussian processes learning in each local model. The developed scheme can perform real-time simulation considering environmental disturbances and is applied to a 6 degree-of-freedom autonomous underwater vehicle. The results demonstrate that this scheme is an effective mathematical modeling tool for underwater vehicles dynamics.

Appendices

Appendix A

Proof of Theorem 1

In order to make the following statement clearer, as mentioned in Theorem 1, here we will take one-dimensional space as an example to illustrate, i.e., $\varvec{\vartheta }_{k,i} \rightarrow \vartheta _k, \varvec{m}_{k,i} \rightarrow m_k, \varvec{n}_{k,i} \rightarrow n_k$, etc. Based on the above formulation of ‘adaptive learning problem’ and Eq. (33), ${\vartheta }_{k+1}$ is constructed as follows

$$\begin{aligned} \begin{aligned} {\vartheta }_{k+1}&={\prod }_{\Omega , \sqrt{{n}_{k}}}{\vartheta }_{k} - \mathcal {Q} n_k^{-1 / 2} m_k \\&= \min _{{\vartheta } \in \Omega }\{n_k^{1 / 4}[{\vartheta }-({\vartheta }_k-\mathcal {Q} n_k^{-1 / 2} m_k)]\} \\ \end{aligned} \end{aligned}$$

(A1)

From Lemma 2, using $u_1 = {\vartheta }_{k+1}$, $u_2 = {\vartheta }_{*}$, $x={\vartheta }$ and $\mathcal {E} = n_k^{1/2}$, we have:

$$\begin{aligned} \begin{aligned}&[n_k^{1 / 4}({\vartheta }_{k+1}-{\vartheta }_*)]^2 \\&\quad \le [n_k^{1 / 4}({\vartheta }_k-\mathcal {Q} n_k^{-1 / 2} m_k-{\vartheta }_*)]^2 \\&\quad =[n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)]^2+\mathcal {Q}^2 [n_k^{-1 / 4} m_k]^2 \\&\quad -2 \mathcal {Q}[m_k({\vartheta }_k-{\vartheta }_*)] \\&\quad =[n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)]^2+\mathcal {Q}^2 [n_k^{-1 / 4} m_k]^2 \\&\quad -2 \mathcal {Q}[(\beta _{1} m_{k-1}+(1-\beta _{1}){\mathcal {A}}_k)({\vartheta }_k-{\vartheta }_*)] \end{aligned} \end{aligned}$$

(A2)

Since ${\mathcal {A}}_k$ is the k-th iteration element about ${\mathcal {A}}$, the term $[(\beta _{1} m_{k-1}+(1-\beta _{1}){\mathcal {A}}_k)({\vartheta }_k-{\vartheta }_*)]$ approximates to $[k{{g}}_k({\vartheta }_k-{\vartheta }_*)]$. Note that $\beta _{1} \in [0,1)$ and $\beta _{2} \in [0,1)$, and using Cauchy–Schwartz and Young’s inequality rearranging inequality Eq. (A2), we have:

$$\begin{aligned} \begin{aligned}&[{{g}}_k({\vartheta }_k-{\vartheta }_*)]\\&\quad \le \frac{1}{2 \mathcal {Q}(1-\beta _{1})}\left( [n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)]^2-[n_k^{1 / 4}({\vartheta }_{k+1}-{\vartheta }_*)]^2\right) \\&\quad +\frac{\mathcal {Q}}{2(1-\beta _{1})}(n_k^{-1 / 4} m_k)^2-\frac{\beta _{1}}{1-\beta _{1}}[m_{k-1}({\vartheta }_k-{\vartheta }_*)] \\&\quad \le \frac{1}{2 \mathcal {Q}(1-\beta _{1})}\left( [n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)]^2-[n_k^{1 / 4}({\vartheta }_{k+1}-{\vartheta }_*)]^2\right) \\&\quad +\frac{\mathcal {Q}}{2(1-\beta _{1})}(n_k^{-1 / 4} m_k)^2 \\&\quad +\frac{\beta _{1}}{2(1-\beta _{1})} \mathcal {Q}(n_k^{-1 / 4} m_{k-1})^2+\frac{\beta _{1}}{2 \mathcal {Q}(1-\beta _{1})}[n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)]^2 \end{aligned} \end{aligned}$$

(A3)

where Cauchy–Schwartz and Young’s inequality is $ab \le a^2\epsilon /2 + b^2 / 2\epsilon$, $\forall \epsilon >0$.

According to the purpose in ‘Adaptive learning problem,’ by convexity of $f(\bullet )$ and Eq. (A3), we have:

$$\begin{aligned}{} & {} {\sum }_{k=1}^\mathcal {K} f_k({\vartheta }_k)-f_k({\vartheta }_*) \nonumber \\{} & {} \quad \le {\sum }_{k=1}^\mathcal {K} [{g}_k({\vartheta }_k-{\vartheta }_*)]\nonumber \\{} & {} \quad \le {\sum }_{k=1}^\mathcal {K} \left\{ \frac{1}{2 \mathcal {Q}(1-\beta _{1})}\left( [n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)]^2-[n_k^{1 / 4}({\vartheta }_{k+1}-{\vartheta }_*)]^2\right) \right. \nonumber \\{} & {} \quad +\frac{\mathcal {Q}}{2(1-\beta _{1})}(n_k^{-1 / 4} m_k)^2 \nonumber \\{} & {} \quad \left. +\frac{\beta _{1} \mathcal {Q}}{2(1-\beta _{1})}(n_k^{-1 / 4} m_{k-1})^2+\frac{\beta _{1}}{2 \mathcal {Q}(1-\beta _{1})}[n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)]^2 \right\} \nonumber \\ \end{aligned}$$

(A4)

Since $\beta _{1} \in [0,1)$ and $\beta _{2} \in [0,1)$, $0 \le n_{k-1} \le n_{k}$,

$$\begin{aligned} \begin{aligned}&{\sum }_{k=1}^\mathcal {K} f_k({\vartheta }_k)-f_k({\vartheta }_*) \le {\sum }_{k=1}^\mathcal {K} \left[ {g}_k({\vartheta }_k-{\vartheta }_*)\right] \\&\le \frac{1}{2\mathcal {Q}\left( 1-\beta _1\right) }\left[ n_1^{1 / 4}\left( \vartheta _1-\vartheta _*\right) \right] ^2 \\&+ \frac{1}{2\left( 1-\beta _1\right) } {\sum }_{k=2}^\mathcal {K}\left( \vartheta _k-\vartheta _*\right) ^2 \left[ \frac{n_k^{1 / 2}-n_{k-1}^{1 / 2}}{\mathcal {Q}}\right] \\&+\frac{1+\beta _1}{2\left( 1-\beta _1\right) } {\sum }_{k=1}^\mathcal {K} \mathcal {Q}\left( n_k^{-1 / 4} m_k\right) ^2 \\&+\frac{\beta _{1}}{2\mathcal {Q}\left( 1-\beta _1\right) } {\sum }_{k=1}^\mathcal {K} \left[ n_k^{1 / 4}\left( \vartheta _k-\vartheta _*\right) \right] ^2 \end{aligned} \end{aligned}$$

(A5)

where for ${\sum }_{t=1}^T \mathcal {Q}(n_k^{-1 / 4} m_k)^2$ in Eq. (A5), assuming $0< c < n_k$, $\forall k \in [\mathcal {K}]$, according to the definition of $m_k$ in Eq. (33), then

$$\begin{aligned} \begin{aligned}&{\sum }_{k=1}^\mathcal {K} \mathcal {Q}\left( n_k^{-1 / 4} m_k\right) ^2\\&\quad ={\sum }_{k=1}^{\mathcal {K}-1} \mathcal {Q}\left( n_k^{-1 / 4} m_k\right) ^2+\mathcal {Q}\left( n_\mathcal {K}^{-1 / 4} m_\mathcal {K}\right) ^2 \\&\quad \le {\sum }_{k=1}^{\mathcal {K}-1} \mathcal {Q}\left( s_k^{-1 / 4} m_k\right) ^2+\frac{\mathcal {Q}}{\sqrt{c\mathcal {K}}}m_\mathcal {K}^2\\&\quad \le {\sum }_{k=1}^{\mathcal {K}-1} \mathcal {Q}\left( n_k^{-1 / 4} m_k\right) ^2+\frac{\mathcal {Q}\beta _1}{\sqrt{c\mathcal {K}}\left( 1-\beta _1\right) }{\sum }_{k=1}^\mathcal {K} g_{k}^2 \\&\quad \le \frac{\mathcal {Q}\beta _1}{\sqrt{c}\left( 1-\beta _1\right) }{\sum }_{k=1}^\mathcal {K} g_{k}^2 \frac{1}{\sqrt{k}}\\ \end{aligned} \end{aligned}$$

(A6)

According to Cauchy–Schwartz inequality $\langle x, y \rangle \le \Vert x\Vert \Vert y\Vert$, and ${\sum }_{k=1}^\mathcal {K}1/\sqrt{k} \le 1+\log \mathcal {K}$, then Eq. (A6) becomes

$$\begin{aligned} \begin{aligned}&{\sum }_{k=1}^\mathcal {K} \mathcal {Q}\left( n_k^{-1 / 4} m_k\right) ^2 \le \frac{\mathcal {Q} \sqrt{1+\log \mathcal {K}}}{\sqrt{c}\left( 1-\beta _1\right) ^2} \Vert \varvec{\mathcal {A}}^2\Vert _2 \\ \end{aligned} \end{aligned}$$

(A7)

Apply Eqs. (A7) to (A5), we have:

$$\begin{aligned} \begin{aligned}&{\sum }_{k=1}^\mathcal {K} f_k({\vartheta }_k)-f_k({\vartheta }_*) \le {\sum }_{k=1}^\mathcal {K} [ {g}_k({\vartheta }_k-{\vartheta }_*) ]\\&\le \frac{1}{2(1-\beta _1)} \frac{\left[ n_1^{1 / 4}({\vartheta }_1-{\vartheta }_*)\right] ^2}{\mathcal {Q}} \\&+\frac{1}{2(1-\beta _1)} {\sum }_{k=2}^\mathcal {K} ({\vartheta }_k-{\vartheta }_* )^2\left[ \frac{n_k^{1 / 2} - n_{k-1}^{1 / 2}}{\mathcal {Q}}\right] \\&+\frac{(1+\beta _1) \mathcal {Q} \sqrt{1+\log \mathcal {K}}}{2 \sqrt{c}(1-\beta _1)^3} \Vert \varvec{\mathcal {A}}^2\Vert _2 \\&+ \frac{\beta _1}{2\mathcal {Q}(1-\beta _1)} {\sum }_{k=1}^\mathcal {K} \left[ n_k^{1 / 4}({\vartheta }_k-{\vartheta }_*)\right] ^2 \\&\le \frac{\mathcal {G}_{\infty }^2 \sqrt{\mathcal {K}}}{2 \mathcal {Q} (1-\beta _1)} n_{\mathcal {K}}^{1 / 2}+\frac{(1+\beta _1) \mathcal {Q} \sqrt{1+\log \mathcal {K}}}{2 \sqrt{c}(1-\beta _1)^3} \Vert \varvec{\mathcal {A}}^2\Vert _2 \\&+\frac{\beta _1 \mathcal {G}_{\infty }^2}{2 \mathcal {Q}(1-\beta _1)} {\sum }_{k=1}^\mathcal {K}n_{k}^{1 / 2}\\ \end{aligned} \end{aligned}$$

(A8)

It can be seen from Eq. (A8) that $f_k({\vartheta }_k)-f_k({\vartheta }_*)$ is converged, which means $\lim _{k \rightarrow \infty } \left[ f_k({\vartheta }_k)-f_k({\vartheta }_*)\right] ^2=0$ can be obtained. For the assumed convex case, Theorem 1 means that the convergence time complexity of AME is $O(\sqrt{\mathcal {K}})$. $\square$

Appendix B

To supplement the details of the process in Fig. 2, the pseudocode of our proposed LAGPL algorithm is presented here as shown in Algorithm 1, which includes the training and prediction stages.

Algorithm 1

Pseudocode for local adaptive moment estimation Gaussian processes learning