91Ó°ÊÓ

To determine the critical point \(\mathbf{x}^0\), we set the rate of change \(\frac{d \mathbf{x}}{d t}\) to zero and solve for \(\mathbf{x}\): $$\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right) = \mathbf{0}$$ To find \(\mathbf{x}^0\), we can solve the linear system: $$\left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) \mathbf{x}^0 = -\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right)$$ This yields the following system of linear equations: $$\begin{cases} -\beta x_2^0 = -\alpha \\ \delta x_1^0 = \gamma \end{cases}$$ From this system, we can solve for \(x_1^0\) and \(x_2^0\) as follows: $$x_1^0 = \frac{\gamma}{\delta}, \quad x_2^0 = \frac{\alpha}{\beta}$$ Therefore, the critical point is: $$\mathbf{x}^0 = \left(\begin{array}{r}{\dfrac{\gamma}{\delta}} \\\ {\dfrac{\alpha}{\beta}}\end{array}\right)$$

Now, we will make the transformation \(\mathbf{x} = \mathbf{x}^0 + \mathbf{u}\) and find the resulting system in terms of \(\mathbf{u}\): $$\frac{d (\mathbf{x}^0 + \mathbf{u})}{d t} = \left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) (\mathbf{x}^0 + \mathbf{u})+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right)$$ Since \(\frac{d \mathbf{x}^0}{d t} = \mathbf{0}\), we have \(\frac{d \mathbf{u}}{d t} = \frac{d (\mathbf{x}^0 + \mathbf{u})}{d t}\) and rewrite the system in terms of \(\mathbf{u}\): $$\frac{d \mathbf{u}}{d t} = \left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) \mathbf{u}$$

To examine the stability and type of the critical point, we can analyze the eigenvalues of the system matrix. We first need to find the characteristic equation of the matrix: $$\begin{vmatrix}{-\lambda} & {-\beta} \\\ {\delta} & {-\lambda}\end{vmatrix} = \lambda^2 + \beta \delta = 0$$ The eigenvalues \(\lambda\) are given by the following equation: $$\lambda^2 = -\beta \delta$$ or $$\lambda = \pm \sqrt{-\beta \delta} i$$ Since the eigenvalues are purely imaginary and have zero real parts, the critical point \(\mathbf{x}^0\) is a center, meaning that trajectories near the critical point are periodic and closed. The system is neutrally stable, as perturbations neither grow nor decay in amplitude. In summary, the critical point \(\mathbf{x}^0 = \left(\begin{array}{r}{\dfrac{\gamma}{\delta}} \\\ {\dfrac{\alpha}{\beta}}\end{array}\right)\) is a center with neutral stability.