## 3 F-M Synchronization
### Case 1: \(d=m\)  
Let us start by defining the error system between the master and slave systems in Label (6) for the \(\mathbf{F}\)-\(\mathbf{M}\) synchronization of dimension \(m\) as  
\[e\left(t\right)=\mathbf{F}\left(Y\left(t\right)\right)-\mathbf{M}\left(t \right)X(t). \tag{8}\]  
By defining the matrix  
\[\mathbf{DF}\left(Y\left(t\right)\right)=\left(\begin{array}{cccc}\frac{ \partial\mathbf{F}_{1}}{\partial y_{1}}&\frac{\partial\mathbf{F}_{1}}{ \partial y_{2}}&\cdots&\frac{\partial\mathbf{F}_{1}}{\partial y_{m}}\\ \frac{\partial\mathbf{F}_{2}}{\partial y_{1}}&\frac{\partial\mathbf{F}_{2}}{ \partial y_{2}}&\cdots&\frac{\partial\mathbf{F}_{2}}{\partial y_{m}}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial\mathbf{F}_{m}}{\partial y_{1}}&\frac{\partial\mathbf{F}_{m}}{ \partial y_{2}}&\cdots&\frac{\partial\mathbf{F}_{m}}{\partial y_{m}}\end{array} \right), \tag{9}\]  
we may reformulate (8) as  
\[e\left(t\right)=\mathbf{DF}\left(Y\left(t\right)\right)\dot{Y}\left(t\right)- \dot{\mathbf{M}}\left(t\right)X(t)-\mathbf{M}\left(t\right)\dot{X}(t). \tag{10}\]  
It was shown in [24] that the fractional derivative of the product \(\mathbf{M}\left(t\right)\dot{X}(t)\), for instance, results in an infinite sum containing integer and fractional order derivatives. Hence, we may not use Label (10) directly. Instead, let us rewrite it in the more convenient form  
\[e\left(t\right)=\left(B-C\right)e\left(t\right)+\mathbf{DF}\left(Y\left(t \right)\right)\dot{Y}\left(t\right)+R, \tag{11}\]  
with \(C\in\mathbb{R}^{m\times m}\) being our new constant control matrix to be selected later and  
\[R=\left(C-B\right)e\left(t\right)-\dot{\mathbf{M}}\left(t\right)X(t)-\mathbf{ M}\left(t\right)\dot{X}(t). \tag{12}\]  
To achieve synchronization between the systems in Label (6), we assume that \(\mathbf{DF}\left(Y\left(t\right)\right)\) is an invertible matrix with its inverse denoted by \(\mathbf{D}^{-1}\). This leads us to the following theorem.  
**Theorem 1**.: _There exists a suitable feedback gain matrix \(C\in\mathbb{R}^{m\times m}\) such that_  
\[U=-BY(t)-g(Y(t))+J^{1-q}\left(-\mathbf{D}^{-1}\times R\right), \tag{13}\]_whereby \(m\)-dimensional \(\mathbf{F}\)-\(\mathbf{M}\) synchronization is realized for the master-slave pair (6)._  
Proof.: Substituting the control law (13) into the slave system yields  
\[D_{t}^{q}Y(t)=J^{1-q}\left(-\mathbf{D}^{-1}\times R\right). \tag{14}\]  
If we use the notation \(F(s)=\mathbf{L}\left(Y(t)\right)\), the Laplace transform of (14) results in  
\[s^{q}F(s)-s^{q-1}Y(0)=s^{q-1}\mathbf{L}\left(-\mathbf{D}^{-1}\times R\right). \tag{15}\]  
Then, by multiplying Label (15) by \(s^{1-q}\) and taking the inverse Laplace transform, we obtain  
\[\dot{Y}\left(t\right)=-\mathbf{D}^{-1}\times R, \tag{16}\]  
which when substituted in the error Equation (11) gives us  
\[\dot{e}\left(t\right) = \left(B-C\right)e\left(t\right)+\mathbf{DF}\left(Y\left(t\right) \right)\left[-\mathbf{D}^{-1}\times R\right]+R \tag{17}\] \[= \left(B-C\right)e\left(t\right).\]  
The rest is simple. We can choose an arbitrary control matrix \(C\) such that all eigenvalues of \(B-C\) have negative real parts, thereby guaranteeing the convergence of the error to zero, i.e., \(\lim\limits_{t\rightarrow+\infty}\left\|e\left(t\right)\right\|=0\). Consequently, we can say that the zero solution of the error system (17) is globally asymptotically stable, and thus the master-slave pair (6) is globally \(\mathbf{F}\)-\(\mathbf{M}\) synchronized with dimension \(m\).