# Input-to-State Stability of Variable Impedance Control for Robotic Manipulator
## 3 ISS of Variable Impedance Model
### ISS Proof  
By using aforementioned notations and theorem, new theorem regarding the sufficient conditions for the variable impedance model (VIM) to be ISS is suggested as follow:  
**Theorem 2**. : _(Sufficient Conditions of VIM to be ISS) Assume that the variable impedance model of Equation (10) is piecewise continuous in \(t\) and locally Lipschitz in \(x\) and \(d\), and also assume that the joint torque \(d\) is piecewise continuous in \(t\) and bounded function for all \(t\geq 0\). If the following conditions hold:_  
\[\dot{K}(t)-\frac{(1-\alpha)}{\alpha}K(t)\leq 0 \tag{12}\] \[\left\|\frac{(1-\alpha)}{2}K(t)+D\right\|_{l} \geq \left\|\frac{1}{2}\dot{H}(q)\right\|, \tag{13}\]  
_where \(\left\|\bullet\right\|_{l}\) is the lower bound of matrix \(\bullet\) and \(\alpha\) is an arbitrary constant chosen within \(0<\alpha<1\), then the VIM is ISS._  
Proof. : (Step 1. Setting Lyapunov function candidate) Let us take Lyapunov function candidate using kinetic plus potential energies with scaling factor as follow:  
\[V(x,K)=\frac{1}{2}x_{2}^{T}H(q)x_{2}+\frac{\alpha}{2}x_{1}^{T}K(t)x_{1} \tag{14}\]  
where \(0<\alpha<1\). Since \(H(q)=H^{T}(q)>0\) and \(K(t)=K^{T}(t)>0\), the Lyapunov function candidate \(V(x,K)\) must be positive definite for all \(x\) and \(K\). (Step 2. Derivative of Lyapunov function candidate) Let us take time derivative of the Lyapunov function candidate as follow:  
\[\dot{V}=\frac{1}{2}x_{2}^{T}H(q)x_{2}+x_{2}^{T}H(q)\dot{x}_{2}+\frac{\alpha}{ 2}x_{1}^{T}\dot{K}(t)x_{1}+\alpha x_{1}^{T}K(t)\dot{x}_{1}. \tag{15}\]  
Now, if the VIM of Equation (10) is applied to above equation, then we have:  
\[\dot{V}=\frac{1}{2}x_{2}^{T}H(q)x_{2}+x_{2}^{T}\left[-Dx_{2}-K(t)x_{1}+d \right]+\frac{\alpha}{2}x_{1}^{T}\dot{K}(t)x_{1}+\alpha x_{1}^{T}K(t)x_{2} \tag{16}\]  
Let us rearrange above equation as follow:  
\[\dot{V}=\frac{\alpha}{2}x_{1}^{T}\dot{K}(t)x_{1}+(\alpha-1)x_{2}^{T}K(t)x_{1} +x_{2}^{T}\left[\frac{1}{2}H(q)-D\right]x_{2}+x_{2}^{T}d. \tag{17}\]  
If the inequality of \(x_{2}^{T}K(t)x_{1}\leq\frac{1}{2}x_{1}^{T}K(t)x_{1}+\frac{1}{2}x_{2}^{T}K(t)x _{2}\) is applied to above equation, then we can get  
\[\dot{V}\leq\frac{\alpha}{2}x_{1}^{T}\dot{K}(t)x_{1}+\frac{(\alpha-1)}{2}x_{1 }^{T}K(t)x_{1}+\frac{(\alpha-1)}{2}x_{2}^{T}K(t)x_{2}+x_{2}^{T}\left[\frac{1} {2}\dot{H}(q)-D\right]x_{2}+x_{2}^{T}d. \tag{18}\]Once again, let us rearrange above equation as follow:  
\[\dot{V}\leq x_{1}^{T}\left[\frac{\alpha}{2}\dot{K}(t)+\frac{(\alpha-1)}{2}K(t) \right]x_{1}-x_{2}^{T}\left[\frac{(1-\alpha)}{2}K(t)+D-\frac{1}{2}\dot{H}(q) \right]x_{2}+x_{2}^{T}d. \tag{19}\]  
Using the completion of squares for last two terms as follows:  
\[ \begin{split}-x_{2}^{T}Ax_{2}+x_{2}d&=-\frac{1}{2 }x_{2}^{T}Ax_{2}-\left(\frac{1}{2}x_{2}^{T}Ax_{2}-x_{2}^{T}d+\frac{1}{2}d^{T}A ^{-1}d\right)+\frac{1}{2}d^{T}A^{-1}d\\ &=-\frac{1}{2}x_{2}^{T}Ax_{2}-\frac{1}{2}(x_{2}-A^{-1}d)^{T}A(x_ {2}-A^{-1}d)+\frac{1}{2}d^{T}A^{-1}d\\ &\leq-\frac{1}{2}x_{2}^{T}Ax_{2}+\frac{1}{2}d^{T}A^{-1}d\end{split}  \tag{20}\]  
where \(A=\frac{(1-\alpha)}{2}K(t)+D-\frac{1}{2}\dot{H}(q)\), then above Equation (19) is rearranged as following from:  
\[ \begin{split}\dot{V}\leq&-x_{1}^{T}\left[\frac{(1- \alpha)}{2}K(t)-\frac{\alpha}{2}\dot{K}(t)\right]x_{1}\\ &-\frac{1}{2}x_{2}^{T}\left[\frac{(1-\alpha)}{2}K(t)+D-\frac{1}{2 }\dot{H}(q)\right]x_{2}\\ &+\frac{1}{2}d^{T}\left[\frac{(1-\alpha)}{2}K(t)+D-\frac{1}{2} \dot{H}(q)\right]^{-1}d.\end{split}  \tag{21}\]  
Since first two terms of the right-hand side of Equation (21) should be negative semi-definite, the sufficient conditions can be chosen as follows,  
\[ \begin{split}\frac{(1-\alpha)}{2}K(t)&\geq\frac{ \alpha}{2}\dot{K}(t)\\ \left\|\frac{(1-\alpha)}{2}K(t)+D\right\|_{l}&\geq \left\|\frac{1}{2}\dot{H}(q)\right\|,\end{split}  \tag{22}\]  
where \(\left\|\bullet\right\|_{l}\) is the lower bound of matrix \(\bullet\).