Compared with the other task-space PID control, (3.9) has exactly the same form as the classical linear PID control of robot manipulators. In order to prove the stability, in [128] the PID is modified as

$u_x=K_p\tilde{x}+K_i{\int }_0^{t}y\left(\tau \right)d\tau +K_d\stackrel{\cdot }{\tilde{x}}$

where the integral term is changed as $y=\dot{q}+\alpha \tilde{x}$. In [27] the linear PID control is modified as

$u_x=K_{p}s\left(\tilde{x}\right)+K_i{\int }_0^{t}y\left(\tau \right)d\tau +K_d\stackrel{\cdot }{\tilde{x}}$

where $y\left(\tau \right)=\dot{q}+\alpha s\left(\tilde{x}\right)$, the position error $\tilde{x}$ is filter by a scalar potential function $s(\cdot )$. In [40], $s(\cdot )$ is a saturation function.

Consider the robot dynamic (3.2) controlled by the linear PID controller (3.13), the closed-loop system (3.15) is semiglobally asymptotically stable at the equilibrium ${[\xi -g\left(x^d\right),\tilde{x},\stackrel{\cdot }{\tilde{x}}]}^T=0$, provided that control gains satisfy

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

where $\beta =\sqrt{\frac{{\lambda }_m\left(M_x\right){\lambda }_m\left(K_p\right)}{3}}$, $k_g$ satisfies (3.8).

We construct a Lyapunov function as

$\begin{array}{c}V=\frac{1}{2}{\dot{x}}^T{M}_x\dot{x}+\frac{1}{2}{\tilde{x}}^T{K}_p\tilde{x}+U_x\left(x\right)-k_u+{\tilde{x}}^T{g}_x\left(x^d\right)\hfill \\ +\frac{3}{2}{g}_x{\left(x^d\right)}^T{K}_p^{-1}{g}_x\left(x^d\right)+\frac{\alpha }{2}{\tilde{\xi }}^T{K}_i^{-1}\tilde{\xi }+{\tilde{x}}^T\tilde{\xi }-\alpha {\tilde{x}}^{T}M\dot{x}+\frac{\alpha }{2}{\tilde{x}}^T{K}_d\tilde{x}\hfill \end{array}$

where $k_u={\min }_x\left\{U_x\left(x\right)\right\}$, $U_x\left(x\right)={\int }_0^t{g}_x\left(x\right)$, $k_u$ is added such that $V\left(0\right)=0$. α is a design positive constant.

1) We first prove that V is a Lyapunov function, $V⩾0$. The term $\frac{1}{2}{\tilde{x}}^T{K}_p\tilde{x}$ is separated into three parts, and $V={\sum }_{i=1}^4{V}_i$:

$\begin{array}{c}{V}_1=\frac{1}{6}{\tilde{x}}^T{K}_p\tilde{x}+{\tilde{x}}^T{g}_x\left(x^d\right)+\frac{3}{2}{g}_x{\left(x^d\right)}^T{K}_p^{-1}{g}_x\left(x^d\right)\hfill \\ V_2=\frac{1}{6}{\tilde{x}}^T{K}_p\tilde{x}+{\tilde{x}}^T\tilde{\xi }+\frac{\alpha }{2}{\tilde{\xi }}^T{K}_i^{-1}\tilde{\xi }\hfill \\ V_3=\frac{1}{6}{\tilde{x}}^T{K}_p\tilde{x}-\alpha {\tilde{x}}^T{M}_x\dot{x}+\frac{1}{2}{\dot{x}}^T{M}_x\dot{x}\hfill \\ V_4=U_x\left(x\right)-k_u+\frac{\alpha }{2}{\tilde{x}}^T{K}_d\tilde{x}\hfill \end{array}$

By $k_u={\min }_x\left\{U_x\left(x\right)\right\}$, $V_4⩾0$. It is easy to find

$V_1=\frac{1}{2}{\left[\begin{array}{c}\hfill \tilde{x}\hfill \\ \hfill g_x\left(x^d\right)\hfill \end{array}\right]}^T\left[\begin{array}{cc}\hfill \frac{1}{3}{K}_p\hfill & \hfill I\hfill \\ \hfill I\hfill & \hfill 3K_p^{-1}\hfill \end{array}\right]\left[\begin{array}{c}\hfill \tilde{x}\hfill \\ \hfill g_x\left(x^d\right)\hfill \end{array}\right]⩾0$

When $\alpha ⩾\frac{3}{{\lambda }_m\left(K_i^{-1}\right){\lambda }_m\left(K_p\right)}$,

$\begin{array}{c}{V}_2⩾\frac{1}{2}\frac{1}{6}{\lambda }_m\left(K_p\right){\Vert \tilde{x}\Vert }^2-\Vert \tilde{x}\Vert \Vert \tilde{\xi }\Vert +\frac{\alpha {\lambda }_m\left(K_i^{-1}\right)}{2}{\Vert \tilde{\xi }\Vert }^2\hfill \\ =\frac{1}{2}{(\sqrt{\frac{1}{3}{\lambda }_m\left(K_p\right)}\Vert \tilde{x}\Vert -\sqrt{\frac{3}{{\lambda }_m\left(K_p\right)}}\Vert \tilde{\xi }\Vert )}^2⩾0\hfill \end{array}$

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

$\begin{array}{c}{V}_3⩾\frac{1}{2}[{\lambda }_m\left(M_x\right){\Vert \dot{x}\Vert }^2-2\alpha {\lambda }_M\left(M_x\right)\Vert \tilde{x}\Vert \Vert \dot{x}\Vert +\frac{1}{3}{\lambda }_m\left(K_p\right){\Vert \tilde{x}\Vert }^2]\hfill \\ =\frac{1}{2}{(\sqrt{{\lambda }_m\left(M_x\right)}\Vert \dot{x}\Vert -\sqrt{\frac{1}{3}{\lambda }_m\left(K_p\right)}\Vert \tilde{x}\Vert )}^2⩾0\hfill \end{array}$

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

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

This means if $K_p$ is sufficiently large or $K_i$ is sufficiently small, (3.20) is established, and $V(\dot{x},\tilde{x},\tilde{\xi })$ is globally positive definite.

2) We now prove $\dot{V}⩽0$. Using $\frac{d}{dt}U\left(x\right)={\dot{x}}^T{g}_x\left(x\right)$, $\frac{d}{dt}{g}_x\left(x^d\right)=0$ and $\frac{d}{dt}\left[{\tilde{x}}^{T}g\left(x^d\right)\right]={\left(\frac{d}{dt}\tilde{x}\right)}^T{g}_x\left(x^d\right)$, the derivative of V is

$\begin{array}{c}\dot{V}={\dot{x}}^T{M}_x\ddot{x}+\frac{1}{2}{\dot{x}}^T{\stackrel{\cdot }{M}}_x\dot{x}+{\stackrel{\cdot }{\tilde{x}}}^T{K}_p\tilde{x}+g_x{\left(x\right)}^T\dot{x}\hfill \\ +{\stackrel{\cdot }{\tilde{x}}}^T{g}_x\left(x^d\right)+\stackrel{\cdot }{\alpha {\tilde{\xi }}^T}{K}_i^{-1}\tilde{\xi }+{\stackrel{\cdot }{\tilde{x}}}^T\tilde{\xi }+{\tilde{x}}^T\stackrel{\cdot }{\tilde{\xi }}\hfill \\ -\alpha ({\stackrel{\cdot }{\tilde{x}}}^{T}M\dot{x}+{\tilde{x}}^T\stackrel{\cdot }{M}\dot{x}+{\tilde{x}}^{T}M\ddot{x})+\alpha {\tilde{x}}^T{K}_d\stackrel{\cdot }{\tilde{x}}\hfill \end{array}$

Using (3.6), the first three terms of (3.22) become

$-x^T{g}_x\left(x\right)-{\dot{x}}^T{K}_d\dot{x}+{\dot{x}}^T\tilde{\xi }+{\dot{x}}^T{g}_x\left(x^d\right)$

Because ${\left(\frac{d}{dt}\tilde{x}\right)}^T{g}_x\left(x^d\right)=-x^T{g}_x\left(x^d\right)$, and $\stackrel{\cdot }{\tilde{\xi }}=K_i\tilde{x}$, the first eight terms of (3.22) are

$-x^T{K}_{d}x+\alpha {\tilde{x}}^T\tilde{\xi }+{\tilde{x}}^T{K}_i\tilde{x}$

Now we discuss the last two terms of (3.22). From (3.7), we have

${\tilde{x}}^T\stackrel{\cdot }{M_x}\dot{x}={\tilde{x}}^T{C}_x\dot{x}+{\tilde{x}}^T{C}_x^T\dot{x}$

From (3.15),

${\tilde{x}}^T{M}_x\ddot{x}=-{\tilde{x}}^T{C}_x\dot{x}-{\tilde{x}}^T{g}_x\left(x\right)+{\tilde{x}}^T{K}_p\tilde{x}-{\tilde{x}}^T{K}_d\dot{x}+{\tilde{x}}^T\tilde{\xi }+{\tilde{x}}^T{g}_x\left(x^d\right)$

Since ${\left(\frac{d}{dt}\tilde{x}\right)}^T{M}_x\dot{q}=-{\dot{x}}^T{M}_x\dot{x}$, using (3.5) and (3.8) the last two terms of (3.22) are

$\begin{array}{cc}\hfill & -\alpha \{{\tilde{x}}^T{K}_p\tilde{x}-{\dot{x}}^T{M}_x\dot{x}+{\tilde{x}}^T{C}_x^T\dot{x}+{\tilde{x}}^T[g\left(x^d\right)-g\left(x\right)]+{\tilde{x}}^T\tilde{\xi }-{\tilde{x}}^T{K}_d\dot{x}\}+\alpha {\tilde{x}}^T{K}_d\stackrel{\cdot }{\tilde{x}}\hfill \\ \hfill & ⩽\alpha {\dot{x}}^T{M}_x\dot{x}-\alpha {\tilde{x}}^T{K}_p\tilde{x}+\alpha k_c\Vert \tilde{x}\Vert {\Vert \dot{x}\Vert }^2+\alpha k_g{\Vert \tilde{x}\Vert }^2-\alpha {\tilde{x}}^T{K}_e\tilde{x}-\alpha {\tilde{x}}^T\tilde{\xi }\hfill \end{array}$

From (3.23) and (3.24),

$\begin{array}{c}\dot{V}⩽-{\dot{x}}^T(K_d-\alpha M_x-\alpha k_c\Vert \tilde{x}\Vert )\dot{x}-{\tilde{x}}^T(\alpha (K_p+K_e)-K_i-\alpha k_g)\tilde{x}\hfill \\ ⩽-[{\lambda }_m\left(K_d\right)-\alpha {\lambda }_M\left(M_x\right)-\alpha k_c\Vert \tilde{x}\Vert ]{\Vert \dot{x}\Vert }^2-[\alpha {\lambda }_m\left(K_p\right)-{\lambda }_M\left(K_i\right)-\alpha k_g]{\Vert \tilde{x}\Vert }^2\hfill \end{array}$

If

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

and

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

then $\dot{V}⩽0$, $\Vert \tilde{x}\Vert$ decreases. From (3.21), if

$\begin{array}{c}{\lambda }_m\left(K_d\right)⩾{\lambda }_M\left(M_x\right)+\sqrt{\frac{1}{3}{\lambda }_m\left(M_x\right)}\sqrt{{\lambda }_m\left(K_p\right)}\hfill \\ {\lambda }_m\left(K_p\right)⩾\frac{1}{3}{\lambda }_m\left(K_i^{-1}\right){\lambda }_m\left(K_p\right){\lambda }_M\left(K_i\right)+k_g\hfill \end{array}$

then (3.26) is established. Using (3.20) and ${\lambda }_m\left(K_i^{-1}\right)=\frac{1}{{\lambda }_M\left(K_i\right)}$, (3.27) is (3.16). $\dot{V}$ is negative semidefinite.

3) Finally, we prove semiglobally asymptotic stability. Define a ball Σ of radius $\sigma >0$ centered at the origin of the state space, which satisfies these conditions:

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

$\dot{V}$ 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 (3.15) is a stable equilibrium. Since the closed-loop equation is autonomous, we use LaSalle's theorem. Define Ω as

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

From (3.22), $\dot{V}=0$ if and only if $\tilde{x}=\dot{x}=0$. For a solution $x\left(t\right)$ to belong to Ω for all $t⩾0$, it is necessary and sufficient that $\tilde{x}=\dot{x}=0$ for all $t⩾0$. Therefore it must also hold that $\ddot{x}=0$ for all $t⩾0$. We conclude that from the closed-loop system (3.15), if $X\left(t\right)\in \mathrm{\Omega }$ for all $t⩾0$, then

$\begin{array}{c}{g}_x\left(x\right)=g_x\left(x^d\right)=\tilde{\xi }+g_x\left(x^d\right)\hfill \\ \stackrel{\cdot }{\tilde{\xi }}=0\hfill \end{array}$

implies that $\tilde{\xi }=0$ for all $t⩾0$. So $x\left(t\right)=[\tilde{x},\dot{x},\tilde{\xi }]=0\in R^{3n}$ 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 this that the origin of the closed-loop system (3.15) 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 $\Vert \tilde{x}\Vert$ can be

$\tilde{x}⩽\frac{{\lambda }_M\left(M_x\right)}{k_c}{\lambda }_M\left(K_i\right){\lambda }_m\left(K_p\right)$

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. □

The tuning procedure of the parameters can be calculated from the conditions (3.16). It is more simple than the tuning procedures in [27,40,72,128]. PID controllers are not linear, and the conditions for the PID gains are not explicit. The upper or lower bounds of PID gains need the maximum eigenvalue of $M_x$ in (3.16), it can be estimated without calculating $M_x$. For a robot with only revolute joints,

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

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

It is well known that without the gravity force $g_x$ in (3.2), PD control with any positive gains can drive the closed-loop system asymptotic stability. The main objective of the integral action can be regarded to cancel the gravity torque. In order to decrease integral gain, the estimated gravity is applied to the PID control (3.13). The PID control with an approximate gravity compensation ${\hat{g}}_x$ is

$\begin{array}{c}{u}_x=K_p\tilde{x}-K_d\dot{x}+{\hat{g}}_x+\xi \hfill \\ \dot{\xi }=K_i\tilde{x},\xi \left(0\right)={\xi }_0\hfill \end{array}$

If we define

$\begin{array}{c}{\tilde{g}}_x=g_x-{\hat{g}}_x\hfill \\ {\tilde{U}}_x={\int }_0^T{\tilde{g}}_x,\tilde{U}\left(0\right)=0\hfill \end{array}$

${\tilde{g}}_x$ also satisfies Lipschitz condition (3.8)

$\Vert {\tilde{g}}_x\left(a\right)-{\tilde{g}}_x\left(b\right)\Vert ⩽{\tilde{k}}_{gx}\Vert a-b\Vert$

The new Lyapunov function in the above proof is

$\begin{array}{c}V=\frac{1}{2}{\dot{x}}^T{M}_x\dot{x}+\frac{1}{2}{\tilde{x}}^T{K}_p\tilde{x}+{\tilde{U}}_x-{\tilde{k}}_u+{\tilde{x}}^T{g}_x\left(x^d\right)+\frac{3}{2}{g}_x{\left(x^d\right)}^T{K}_p^{-1}{g}_x\left(x^d\right)\hfill \\ +\frac{\alpha }{2}{\tilde{\xi }}^T{K}_i^{-1}\tilde{\xi }+{\tilde{x}}^T\tilde{\xi }-\alpha {\tilde{x}}^{T}M\dot{x}+\frac{\alpha }{2}{\tilde{x}}^T{K}_d\tilde{x}\hfill \end{array}$

where ${\tilde{k}}_u={\min }_q\left\{{\tilde{U}}_x\right\}$. The above theorem is also correct for the PID control with an approximate gravity compensation (3.28). The condition for PID gains (3.16) becomes

${\lambda }_m\left(K_p\right)⩾\frac{3}{2}{\tilde{k}}_g,{\lambda }_M\left(K_i\right)⩽\frac{3\beta }{2}\frac{{\tilde{k}}_g}{{\lambda }_M\left(M\right)}$

where ${\tilde{k}}_g\ll k_g$, $\beta =\sqrt{\frac{{\lambda }_m\left(M\right){\lambda }_m\left(K_p\right)}{3}}$.