Consider the robot dynamic (5.1) controlled by the neural PID control (5.19). The closed-loop system (5.24) is semiglobally asymptotically stable at the equilibrium $x={[\xi -\varphi \left(q^d\right),\tilde{q},\stackrel{\cdot }{\tilde{q}}]}^T=0$, provided that control gains satisfy

$\begin{array}{c}{\lambda }_m\left(K_p\right)⩾\frac{3}{2}{k}_{\varphi }\hfill \\ {\lambda }_M\left(K_i\right)⩽\beta \frac{{\lambda }_m\left(K_p\right)}{{\lambda }_M\left(M\right)}\hfill \\ {\lambda }_m\left(K_d\right)⩾\beta +{\lambda }_M\left(M\right)\hfill \end{array}$

where $\beta =\sqrt{\frac{{\lambda }_m\left(M\right){\lambda }_m\left(K_p\right)}{3}}$, $k_{\varphi }$ satisfies (5.10), and the weight of the neural networks (5.4) is tuned by

$\stackrel{\cdot }{\hat{W}}=-K_w\sigma (q,\stackrel{\cdot }{q}){(\dot{q}+\alpha \tilde{q})}^T$

where α is positive design constant, it satisfies

$\frac{\sqrt{\frac{1}{3}{\lambda }_m\left(M\right){\lambda }_m\left(K_p\right)}}{{\lambda }_M\left(M\right)}⩾\alpha ⩾\frac{3}{{\lambda }_m\left(K_i^{-1}\right){\lambda }_m\left(K_p\right)}$

We construct a Lyapunov function as

$\begin{array}{c}V=\frac{1}{2}{\dot{q}}^{T}M\dot{q}+\frac{1}{2}{\tilde{q}}^T{K}_p\tilde{q}+{\int }_0^t\varphi (q,\stackrel{\cdot }{q})dq-k_{\varphi }+{\tilde{q}}^T\varphi \left(q^d\right)\hfill \\ +\frac{3}{2}\varphi {\left(q^d\right)}^T{K}_p^{-1}\varphi \left(q^d\right)+\frac{\alpha }{2}{\tilde{\xi }}^T{K}_i^{-1}\tilde{\xi }\hfill \\ +{\tilde{q}}^T\tilde{\xi }-\alpha {\tilde{q}}^{T}M\dot{q}+\frac{\alpha }{2}{\tilde{q}}^T{K}_d\tilde{q}+\frac{1}{2}tr\left({\tilde{W}}^T{K}_w^{-1}\tilde{W}\right)\hfill \end{array}$

where $k_{\varphi }$ is defined in (5.13) such that $V\left(0\right)=0$. α is a design positive constant. We first prove V is a Lyapunov function, $V⩾0$. The term $\frac{1}{2}{\tilde{q}}^T{K}_p\tilde{q}$ is separated into three parts, and $V={\sum }_{i=1}^4{V}_i$:

$\begin{array}{c}{V}_1=\frac{1}{6}{\tilde{q}}^T{K}_p\tilde{q}+{\tilde{q}}^T\varphi \left(q^d\right)+\frac{3}{2}\varphi {\left(q^d\right)}^T{K}_p^{-1}\varphi \left(q^d\right)\hfill \\ V_2=\frac{1}{6}{\tilde{q}}^T{K}_p\tilde{q}+{\tilde{q}}^T\tilde{\xi }+\frac{\alpha }{2}{\tilde{\xi }}^T{K}_i^{-1}\tilde{\xi }\hfill \\ V_3=\frac{1}{6}{\tilde{q}}^T{K}_p\tilde{q}-\alpha {\tilde{q}}^{T}M\dot{q}+\frac{1}{2}{\dot{q}}^{T}M\dot{q}\hfill \\ V_4={\int }_0^t\varphi \left(q\right)d\tau -k_{\varphi }+\frac{\alpha }{2}{\tilde{q}}^T{K}_d\tilde{q}+\frac{1}{2}tr\left({\tilde{W}}^T{K}_w^{-1}\tilde{W}\right)⩾0\hfill \end{array}$

It is easy to find

$V_1=\frac{1}{2}{\left[\begin{array}{c}\hfill \tilde{q}\hfill \\ \hfill \varphi \left(q^d\right)\hfill \end{array}\right]}^T\left[\begin{array}{cc}\hfill \frac{1}{3}{K}_p\hfill & \hfill I\hfill \\ \hfill I\hfill & \hfill 3K_p^{-1}\hfill \end{array}\right]\left[\begin{array}{c}\hfill \tilde{q}\hfill \\ \hfill \varphi \left(q^d\right)\hfill \end{array}\right]$

Since $K_p⩾0$, $V_1$ is a semipositive definite matrix, $V_1⩾0$. When $\alpha ⩾\frac{3}{{\lambda }_m\left(K_i^{-1}\right){\lambda }_m\left(K_p\right)}$,

$V_2⩾\frac{1}{2}{(\sqrt{\frac{1}{3}{\lambda }_m\left(K_p\right)}\Vert \tilde{q}\Vert -\sqrt{\frac{3}{{\lambda }_m\left(K_p\right)}}\Vert \tilde{\xi }\Vert )}^2⩾0$

Because

$y^{T}Ax⩽\Vert y\Vert \Vert Ax\Vert ⩽\Vert y\Vert \Vert A\Vert \Vert x\Vert ⩽\left|{\lambda }_M\left(A\right)\right|\Vert y\Vert \Vert x\Vert$

when $\alpha ⩽\frac{\sqrt{\frac{1}{3}{\lambda }_m\left(M\right){\lambda }_m\left(K_p\right)}}{{\lambda }_M\left(M\right)}$,

$V_3⩾\frac{1}{2}{(\sqrt{{\lambda }_m\left(M\right)}\Vert \dot{q}\Vert -\sqrt{\frac{1}{3}{\lambda }_m\left(K_p\right)}\Vert \tilde{q}\Vert )}^2⩾0$

Obviously, if

$\sqrt{\frac{1}{3}}{\lambda }_m\left(K_i^{-1}\right){\lambda }_m^{\frac{3}{2}}\left(K_p\right){\lambda }_m^{\frac{1}{2}}\left(M\right)⩾{\lambda }_M\left(M\right)$

there exists

$\frac{\sqrt{\frac{1}{3}{\lambda }_m\left(M\right){\lambda }_m\left(K_p\right)}}{{\lambda }_M\left(M\right)}⩾\alpha ⩾\frac{3}{{\lambda }_m\left(K_i^{-1}\right){\lambda }_m\left(K_p\right)}$

This means if $K_p$ is sufficiently large or $K_i$ is sufficiently small, (5.34) is established, and $V(\dot{q},\tilde{q},\tilde{\xi })$ is globally positive definite. Using $\frac{d}{dt}{\int }_0^t\varphi (q,\stackrel{\cdot }{q})dq=\frac{\partial {\int }_0^t\varphi (q,\stackrel{\cdot }{q})dq}{\partial q}\frac{\partial q}{\partial t}={\dot{q}}^T\varphi (q,\stackrel{\cdot }{q})$, $\frac{d}{dt}\varphi \left(q^d\right)=0$ and $\frac{d}{dt}\left[{\tilde{q}}^T\varphi \left(q^d\right)\right]={\stackrel{\cdot }{\tilde{q}}}^T\varphi \left(q^d\right)$, the derivative of V is

$\begin{array}{c}\dot{V}={\dot{q}}^{T}M\ddot{q}+\frac{1}{2}{\dot{q}}^T\stackrel{\cdot }{M}\dot{q}+{\stackrel{\cdot }{\tilde{q}}}^T{K}_p\tilde{q}+\varphi {(q,\stackrel{\cdot }{q})}^T\dot{q}+{\stackrel{\cdot }{\tilde{q}}}^T\varphi \left(q^d\right)+tr\left({\tilde{W}}^T{K}_w^{-1}\stackrel{\cdot }{\tilde{W}}\right)\hfill \\ +\stackrel{\cdot }{\alpha {\tilde{\xi }}^T}{K}_i^{-1}\tilde{\xi }+{\stackrel{\cdot }{\tilde{q}}}^T\tilde{\xi }+{\tilde{q}}^T\stackrel{\cdot }{\tilde{\xi }}-\alpha ({\stackrel{\cdot }{\tilde{q}}}^{T}M\dot{q}+{\tilde{q}}^T\stackrel{\cdot }{M}\dot{q}+{\tilde{q}}^{T}M\ddot{q})+\alpha {\tilde{q}}^T{K}_d\stackrel{\cdot }{\tilde{q}}\hfill \end{array}$

Using (5.8) the first three terms of (5.36) become

$-{\dot{q}}^T\varphi \left(q\right)-{\dot{q}}^T{K}_d\dot{q}+{\dot{q}}^T\tilde{\xi }+{\dot{q}}^T\varphi \left(q^d\right)+{\dot{q}}^T\tilde{W}\sigma (q,\stackrel{\cdot }{q})$

And

$\dot{V}⩽-[{\lambda }_m\left(K_d\right)-\alpha {\lambda }_M\left(M\right)-\alpha k_c\Vert \tilde{q}\Vert ]{\Vert \dot{q}\Vert }^2-[\alpha {\lambda }_m\left(K_p\right)-{\lambda }_M\left(K_i\right)-\alpha k_g]{\Vert \tilde{q}\Vert }^2$

If

$\Vert \tilde{q}\Vert ⩽\frac{{\lambda }_M\left(M\right)}{\alpha k_c}$

and

$\begin{array}{c}{\lambda }_m\left(K_d\right)⩾(1+\alpha ){\lambda }_M\left(M\right)\hfill \\ {\lambda }_m\left(K_p\right)⩾\frac{1}{\alpha }{\lambda }_M\left(K_i\right)+k_g\hfill \end{array}$

then $\dot{V}⩽0$, $\Vert \tilde{q}\Vert$ decreases. Then (5.40) is established. Using (5.34) and ${\lambda }_m\left(K_i^{-1}\right)=\frac{1}{{\lambda }_M\left(K_i\right)}$, (5.40) is (5.25).

$\dot{V}$ is negative semidefinite. Define a ball Σ of radius $\sigma >0$ centered at the origin of the state space, which satisfies these conditions:

$\mathrm{\Sigma }=\{\tilde{q}:\Vert \tilde{q}\Vert ⩽\frac{{\lambda }_M\left(M\right)}{\alpha k_c}=\sigma \}$

$\dot{V}$ is negative semidefinite on the ball Σ. There exists a ball Σ of radius $\sigma >0$ centered at the origin of the state space on which $\dot{V}⩽0$. The origin of the closed-loop equation (5.24) is a stable equilibrium. Since the closed-loop equation is autonomous, we use LaSalle's theorem. Define Ω as

$\begin{array}{c}\mathrm{\Omega }=\{x\left(t\right)=[\tilde{q},\dot{q},\tilde{\xi }]\in R^{3n}:\dot{V}=0\}\hfill \\ =\{\tilde{\xi }\in R^n:\tilde{q}=0\in R^n,\dot{q}=0\in R^n\}\hfill \end{array}$

From (5.36), $\dot{V}=0$ if and only if $\tilde{q}=\dot{q}=0$. For a solution $x\left(t\right)$ to belong to Ω for all $t⩾0$, it is necessary and sufficient that $\tilde{q}=\dot{q}=0$ for all $t⩾0$. Therefore it must also hold that $\ddot{q}=0$ for all $t⩾0$. We conclude from the closed-loop system (5.24) that if $x\left(t\right)\in \mathrm{\Omega }$ for all $t⩾0$, then

$\begin{array}{c}\varphi (q,\stackrel{\cdot }{q})=\varphi (q^d,0)=\tilde{\xi }+\varphi (q^d,0)\hfill \\ \stackrel{\cdot }{\tilde{\xi }}=0\hfill \end{array}$

implies that $\tilde{\xi }=0$ for all $t⩾0$. So $x\left(t\right)=[\tilde{q},\dot{q},\tilde{\xi }]=0\in R^{3n}$ is the only initial condition in Ω for which $x\left(t\right)\in \mathrm{\Omega }$ for all $t⩾0$.

Finally, we conclude from all of this that the origin of the closed-loop system (5.24) is locally asymptotically stable. Because $\frac{1}{\alpha }⩽{\lambda }_m\left(K_i^{-1}\right){\lambda }_m\left(K_p\right)$, the upper bound for $\Vert \tilde{q}\Vert$ can be

$\Vert \tilde{q}\Vert ⩽\frac{{\lambda }_M\left(M\right)}{k_c}{\lambda }_M\left(K_i\right){\lambda }_m\left(K_p\right)$

It establishes the semiglobal stability of our controller, in the sense that the domain of attraction can be arbitrarily enlarged with a suitable choice of the gains. Namely, increasing $K_p$ the basin of attraction will grow. □

From the above stability analysis, we see that the gain matrices of the neural PID control (5.19) can be chosen directly from the conditions (5.25). The tuning procedure of the PID parameters is more simple than [2,4,71,101,117]. No modeling information is needed. The upper or lower bounds of PID gains need the maximum eigenvalue of M in (5.25); it can be estimated without calculating M. For a robot with only revolute joints [131],

${\lambda }_M\left(M\right)⩽\beta ,\beta ⩾n(\underset{i,j}{\max }\left|m_{ij}\right|)$

where $m_{ij}$ stands the ijth element of M, $M\in R^{n\times n}$. A β can be selected such that it is much bigger than all elements.

The main difference between our neural PID control with the other neural PD controllers is that the stability condition is changed. We require the regulation error:

$\Vert \tilde{q}\Vert <k_1{\lambda }_M\left(K_i\right){\lambda }_m\left(K_p\right)$

The other neural PD controllers need

$\Vert \tilde{q}\Vert >\frac{k_2}{K_v}$

where $k_1$ and $k_2$ are positive constants. Obviously, if the initial condition is not worse and satisfies (5.46), (5.46) is always satisfied, and $\Vert \tilde{q}\Vert$ will decrease to zero. But (5.47) cannot be satisfied when $\Vert \tilde{q}\Vert$ becomes small, so $K_v$ has to be increased.

If the unknown $f\left(q\right)$ is estimated by the multilayer neural network (5.5) the modeling error (5.21) becomes

$\begin{array}{c}\tilde{f}=f-\hat{f}=W^{⁎}\sigma (V^{⁎}[q,\stackrel{\cdot }{q}])+\varphi (q,\stackrel{\cdot }{q})-\hat{W}\sigma (\hat{V}[q,\stackrel{\cdot }{q}])\hfill \\ =\tilde{W}\sigma (\hat{V}[q,\stackrel{\cdot }{q}])-W^{⁎}\sigma (\hat{V}[q,\stackrel{\cdot }{q}])+W^{⁎}\sigma (V^{⁎}[q,\stackrel{\cdot }{q}])+\varphi (q,\stackrel{\cdot }{q})\hfill \\ =\tilde{W}\sigma (\hat{V}[q,\stackrel{\cdot }{q}])+W^{⁎}{\sigma }^{\prime }\tilde{V}[q,\stackrel{\cdot }{q}]+{ϵ}_1+\varphi (q,\stackrel{\cdot }{q})\hfill \\ =\tilde{W}[\sigma (\hat{V}[q,\stackrel{\cdot }{q}])+{\sigma }^{\prime }\tilde{V}[q,\stackrel{\cdot }{q}]]+\hat{W}{\sigma }^{\prime }\tilde{V}[q,\stackrel{\cdot }{q}]+{\varphi }_1(q,\stackrel{\cdot }{q})\hfill \end{array}$

where ${\varphi }_1\left(q\right)={ϵ}_1+\varphi (q,\stackrel{\cdot }{q})$, ${ϵ}_1$ is the Taylor approximation error. The closed-loop equation (5.24) becomes

$\begin{array}{cc}\hfill & M\left(q\right)\ddot{q}+C(q,\dot{q})\dot{q}+\tilde{W}\{\sigma (\hat{V}[q,\stackrel{\cdot }{q}])+{\sigma }^{\prime }\tilde{V}[q,\stackrel{\cdot }{q}]\}+\hat{W}{\sigma }^{\prime }\tilde{V}[q,\stackrel{\cdot }{q}]+{\varphi }_1(q,\stackrel{\cdot }{q})\hfill \\ \hfill & =K_p\tilde{q}-K_d\dot{q}+\tilde{\xi }+\varphi \left(q^d\right)\hfill \\ \hfill & \stackrel{\cdot }{\tilde{\xi }}=K_i\tilde{q}\hfill \end{array}$

If the Lyapunov function in (5.28) is changed as

$V_m=V+\frac{1}{2}tr\left({\tilde{V}}^T{K}_v^{-1}\tilde{V}\right)$

then the derivative of (5.50) is

${\stackrel{\cdot }{V}}_m=\dot{V}-{\dot{q}}^T\tilde{W}\sigma (q[q,\stackrel{\cdot }{q}])+{\dot{q}}^T\tilde{W}[\sigma (\hat{V}[q,\stackrel{\cdot }{q}])+{\sigma }^{\prime }\tilde{V}[q,\stackrel{\cdot }{q}]]+tr\left({\tilde{V}}^T{K}_v^{-1}\stackrel{\cdot }{\tilde{V}}\right)$

If the training rule (5.26) is changed as

$\begin{array}{c}\stackrel{\cdot }{\hat{W}}=-K_w\{\sigma (\hat{V}[q,\stackrel{\cdot }{q}])+{\sigma }^{\prime }\tilde{V}[q,\stackrel{\cdot }{q}]\}{(\dot{q}+\alpha \tilde{q})}^T\hfill \\ \stackrel{\cdot }{\hat{V}}=-K_v\hat{W}{\sigma }^{\prime }q{(\dot{q}+\alpha \tilde{q})}^T\hfill \end{array}$

Theorem 5.1 is also established.