## 2 Transforming the Original Dynamics to Quadratic or to Bilinear Format
### Bilinear Systems
#### 2.1.1 Carleman's Linearization
,\]  
_and hence, write the differential equation as: \(\dot{x}(t)=-2x(t)-\sum_{k=3}^{\infty}c_{k}x^{k}(t)+u(t).\) We identify the following values:_  
\[\mathbf{F}_{j,1}=-2j,\ \mathbf{F}_{j,k}= \begin{cases}0,\ k\ even\\ j\frac{(-1)^{(k-1)/2}}{(k-1)!},\ \ k\ odd,\ k>1\end{cases} ,\] \[\mathbf{G}_{j,0}=j,\ \mathbf{G}_{j,k}=0,\ j,k\geqslant 1,\ \mathbf{C}_{1}=1,\ \mathbf{C}_{h}=0,\ h\geqslant 2.\]  
_First, we truncate at \(N=3\) and by using that \(x^{(k)}(t)=x^{k}(t)\), the new state variable is given by \(\mathbf{x}_{1}^{\otimes}(t)=\left[ \begin{array}{cc}x(t)&x^{2}(t)&x^{3}(t) \end{array} \right]^{T}\in\mathbb{R}^{3}\); the dynamics of the bilinear system corresponding to \(N=3\) are written as_  
\[\dot{\mathbf{x}}_{1}^{\otimes}(t)=\mathbf{A}_{1}^{\otimes}\mathbf{x}_{1}^{ \otimes}(t)+\mathbf{N}_{1}^{\otimes}\mathbf{x}_{1}^{\otimes}(t)u(t)+\mathbf{B }_{1}^{\otimes}u(t),\ \ y(t)=\mathbf{C}_{1}^{\otimes}\mathbf{x}_{1}^{\otimes}(t),\]  
_with the system matrices computed as below:_  
\[\mathbf{A}_{1}^{\otimes}=\left[ \begin{array}{ccc}-2&0&-\frac{1}{2!}\\ 0&-4&0\\ 0&0&-6\end{array} \right],\ \mathbf{N}_{1}^{\otimes}=\left[ \begin{array}{ccc}0&0&0\\ 2&0&0\\ 0&3&0\end{array} \right],\ \mathbf{B}_{1}^{\otimes}=\left[ \begin{array}{c}1\\ 0\\ 0\end{array} \right],\ \mathbf{C}_{1}^{\otimes}=\left[ \begin{array}{ccc}2&0&0\end{array} \right].\]_Next, we increase the truncation index to \(N=5\), i.e., \(\mathbf{x}_{2}^{\otimes}(t)=\left[ \begin{array}{cccc}x(t)&x^{2}(t)&\ldots&x^{ 5}(t)\end{array} \right]^{T}\) with \(\mathbf{x}_{2}^{\otimes}(t)\in\mathbb{R}^{5}\). Hence, write the state and output equation as_  
\[\hat{\mathbf{x}}_{2}^{\otimes}(t)=\mathbf{A}_{2}^{\otimes}\mathbf{x}_{2}^{ \otimes}(t)+\mathbf{N}_{2}^{\otimes}\mathbf{x}_{2}^{\otimes}(t)u(t)+\mathbf{ B}_{2}^{\otimes}u(t),\quad y(t)=\mathbf{C}_{2}^{\otimes}\mathbf{x}_{2}^{\otimes}(t),\]  
_with the following system matrices:_  
\[\mathbf{A}_{2}^{\otimes}=\left[ \begin{array}{cccc}-2&0&-\frac{1}{2}&0&\frac {1}{4!}\\ 0&-4&0&-\frac{1}{2}&0\\ 0&0&-6&0&-\frac{3}{2}\\ 0&0&0&-8&0\\ 0&0&0&0&-10\end{array} \right],\,\mathbf{N}_{2}^{\otimes}=\left[ \begin{array}[ ]{cccc}0&0&0&0\\ 2&0&0&0&0\\ 0&3&0&0&0\\ 0&0&4&0&0\\ 0&0&0&5&0\end{array} \right], \tag{11}\]  
\[\mathbf{B}_{2}^{\otimes}=\left[ \begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array} \right],\,\mathbf{C}_{2}^{\otimes}=\left[\  
**Example 2**. : _Start with the nonlinear differential equation \(\dot{x}=x+x^{3}+u\). By introducing the new variable \(z=x^{2}\), it follows that \(\dot{x}=x+xz+u\), and hence the two are equivalent_  
\[\dot{x}=x+x^{3}+u\Leftrightarrow \begin{cases}\dot{x}=x+xz+u,\\ 0=z-x^{2}.\end{cases} \]  
For the second category, we present a probative example that illustrates the lifting approach appropriately. **Example 3**. : _Consider the same nonlinear system as before, i.e.,_  
\[\mathbf{\Sigma}_{\mathrm{N1}}:\ \begin{cases}\dot{x}(t)=-3x(t)+x(t)\cos x(t)+u(t),\\ y(t)=2x(t).\end{cases}  \tag{14}\]  
_By introducing two additional variables: \(z(t)=\sin x(t)\) and \(w(t)=\cos x(t)\), compute the time derivatives and then substitute the given terms_  
\[\dot{x}(t) =-3x(t)+x(t)w(t)+u(t), \tag{15}\] \[\dot{z}(t) =\dot{x}(t)\cos x(t)=[-3x(t)+x(t)w(t)+u(t)]w(t),\] \[=-3x(t)w(t)+x(t)w^{2}(t)+w(t)u(t),\] \[w(t) =-x(t)\sin x(t)=-[-3x(t)+x(t)w(t)+u(t)]z(t)\] \[=3x(t)z(t)-x(t)z(t)w(t)-z(t)u(t).\]  
_We managed to equivalently rewrite the original nonlinear system as a polynomial system but with cubical nonlinearities as can be observed in (15). However, there is one more step until the transformation to QBTI format is completed. We need to choose another two surrogate variables, i.e., \(p(t)=x(t)z(t)\) and \(r(t)=x(t)w(t)\), to be able to equivalently rewrite the above CTI (cubic time-invariant) system as a QBTI system.