# Critically Leveraging Theory for Optimal Control of Quadrotor Unmanned Aircraft Systems
## 2 The Structure and Mathematical Model of the Quadrotor
### Nonlinear Control \(H_{\infty}\) for Rotational Subsystem
The inertial matrix \(M(q)\) is symmetric and positively defined, \(C(q,\dot{q})\dot{q}\) is the radial vector and Coriolis, \(F(\dot{q})\) represents friction, and \(G(q)\) is gravity. As we know, the matrix \(C(q,\dot{q})\) is not unique, so for convenience, it can be written as:  
\[C(q,\dot{q})=\frac{1}{2}\dot{M}(q,\dot{q})+N(q,\dot{q}) \tag{59}\]  
where \(\dot{M}(q,\dot{q})\) and \(N(q,\dot{q})\) are calculated:  
\[\dot{M}_{\bar{q}}=\frac{d}{dt}M_{\bar{q}}=\frac{\partial M_{\bar{q}}}{\partial q }\dot{q}=\sum_{k=1}^{n}\frac{\partial M_{\bar{q}}}{\partial q_{k}}\dot{q}_{k} \tag{60}\]  
\[N_{\bar{q}}=\frac{1}{2}\sum_{k=1}^{n}\left(\frac{\partial M_{\bar{k}}}{ \partial q_{\bar{q}}}-\frac{\partial M_{\bar{q}}}{\partial q_{\bar{q}}}\right) \dot{q}_{k} \tag{61}\]  
Consider the function:  
\[V(x,t)=\frac{1}{2}x^{T}T_{0}^{T} \begin{pmatrix}M&0&0\\ 0&Y&X-Y\\ 0&X-Y&X+Y\end{pmatrix} T_{0}x \tag{62}\]  
The function \(V(x,t)\) is positive if and only if:  
\[ \begin{pmatrix}M&0&0\\ 0&Y&X-Y\\ 0&X-Y&X+Y\end{pmatrix} >0 \tag{63}\]  
When the inertial matrix \(M\) is symmetric, positive definite, and assuming the matrices \(X,Y,Z\in R^{n\times n}\) are constant, symmetric, positive definite matrices, then the above inequality is satisfied if and only if:  
\[Z+Y-(X-Y)Y^{-1}(X-Y)>0 \tag{64}\]  
Or:  
\[Z-XY^{-1}X+2X>0 \tag{65}\]  
(65) is the hypothesis. Next, notice that \(V(x,t)\) is a solution to the Hamilton-Jacobi equation:  
\[ \begin{array}{c}\frac{\partial Y}{\partial t}+\frac{\partial^{T}V}{ \partial x}f(x,t)+\frac{1}{2}\frac{\partial^{T}V}{\partial x}[\frac{1}{ \gamma^{2}}k(x,t)k^{T}(x,t)\\ -g(x,t)R^{-1}g^{T}(x,t)]\frac{\partial V}{\partial x}g(x,t)R^{-1}S^{T}h(x)\\ +\frac{1}{2}h^{T}(x)(Q-SR^{-1}S^{T})h(x)=0\end{array}  \tag{66}\]  
Vector gradient of \(V(x,t)\):  
\[\frac{\partial^{T}V(x,t)}{\partial x}=x^{T}T^{T} \begin{pmatrix}M&0&0\\ 0&Y&X-Y\\ 0&X-Y&Z+Y\end{pmatrix} T_{0}+\frac{1}{2}[0_{1n}\Omega 0_{1n}] \tag{67}\]  
where \(\Omega\in R^{1 sn}\)  
\[ \begin{pmatrix}x^{T}T_{0}^{T}\begin{pmatrix}\frac{\partial^{T}M}{\partial q_{1 }}&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}T_{0}x,\ldots,x^{T}T_{0}^{T}\begin{pmatrix}\frac{\partial^{T }M}{\partial q_{n}}&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}T_{0}x\end{pmatrix}  \tag{68}\]Deduce:  
\[[0_{1\text{+n}}\Omega 0_{1\text{+n}}][\text{g}(\text{x},\text{t})\text{u}+ \text{k}(\text{x},\text{t})\omega]=0 \tag{69}\]  
The result is:  
\[ \begin{split}\frac{\partial\text{V}(\text{x},\text{t})}{\partial \text{x}}\text{f}=&\left(\text{x}^{\text{T}}\text{T}^{\text{T}} \begin{pmatrix}\text{M}&0&0\\ 0&\text{Y}&\text{X}-\text{Y}\\ 0&\text{X}-\text{Y}&\text{Z}+\text{Y}\end{pmatrix}\text{T}_{0}+\frac{1}{2}[0_ {1\text{+n}}\Omega 0_{1\text{+n}}]\end{pmatrix}\text{f}\\ &=\text{x}^{\text{T}}\text{T}^{\text{T}}\begin{pmatrix}\text{M}&0&0\\ 0&\text{Y}&\text{X}-\text{Y}\\ 0&\text{X}-\text{Y}&\text{Z}+\text{Y}\end{pmatrix}\text{T}_{0}\text{f}+\frac{1 }{2}[0_{1\text{+n}}\Omega 0_{1\text{+n}}]\text{f}\\ &+\frac{1}{2}[0_{1\text{+n}}\Omega 0_{1\text{+n}}](\text{g}\text{u}+ \text{k}\omega)\\ &=\text{x}^{\text{T}}\text{T}_{0}^{\text{T}}\begin{pmatrix}\text{M}&0&0 \\ 0&\text{Y}&\text{X}-\text{Y}\\ 0&\text{X}-\text{Y}&\text{Z}+\text{Y}\end{pmatrix}\text{T}_{0}\text{f}+\frac{1 }{2}[0_{1\text{+n}}\Omega 0_{1\text{+n}}]\text{x}\end{split}  \tag{70}\]  
Instead, have:  
\[ \begin{split}\text{x}^{\text{T}}\text{T}_{0}^{\text{T}}\begin{pmatrix} \text{M}&0&0\\ 0&\text{Y}&\text{X}-\text{Y}\\ 0&\text{X}-\text{Y}&\text{X}+\text{Y}\end{pmatrix}\text{T}_{0}\text{f}\\ &=\text{x}^{\text{T}}\text{T}_{0}^{\text{T}}\begin{pmatrix}\text{M}&0&0 \\ 0&\text{Y}&\text{X}-\text{Y}\\ 0&\text{X}-\text{Y}&\text{X}+\text{Y}\end{pmatrix}\end{split}  \tag{71}\]  
where A= X + Z. And:  
\[