# Stability Analysis¶

Given a system of first-order differential equations

$\frac{d \mathbf{x}(t)}{dt} = \mathbf{f}(\mathbf x(t)),$

the equilibrium points are determined by $$\frac{d \mathbf{x}(t)}{dt} = 0$$, i.e.,

$\mathbf{f}(\mathbf x(t)) = 0,$

from which we solve the equilibrium point $$\mathbf x(t) = \mathbf x_0$$

## Linear Stability Analysis¶

The system of differential equations is linearized using perturbations. For simplicity, we consider the following two-dimensional system,

(1)$\begin{split}\frac{x_1(t)}{dt} &= f_1(x_1, x_2) \\ \frac{x_2(t)}{dt} &= f_2(x_1, x_2).\end{split}$

The equilibrium point $$x_{1,0}, x_{2,0}$$ is determined by

$\begin{split}f_1(x_{1,0}, x_{2,0}) &= 0 \\ f_2(x_{1,0}, x_{2,0}) &= 0.\end{split}$

We perturb the equations around its equilibrium point,

$\begin{split}x_1(t) &\to x_{1,0} + \delta x_1(t) \\ x_2(t) &\to x_{2,0} + \delta x_2(t),\end{split}$

where $$\delta x_1(t)$$ and $$\delta x_2(t)$$ are small quantities.

The equation (1) becomes

(2)$\begin{split}\frac{d\delta x_1(t)}{dt} &= f_1(x_{1,0} + \delta x_1(t), x_{2,0} + \delta x_2(t)) \\ \frac{d\delta x_2(t)}{dt} &= f_2(x_{1,0} + \delta x_1(t), x_{2,0} + \delta x_2(t)).\end{split}$

We apply Taylor series expansion to $$f_1(x_{1,0} + \delta x_1(t), x_{2,0} + \delta x_2(t))$$,

$\begin{split}&f_1(x_{1,0} + \delta x_1(t), x_{2,0} + \delta x_2(t)) \\ =& \left.\frac{d f_1(x_1, x_2)}{d x_1} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}} \delta x_1 + \left.\frac{d f_1(x_1, x_2)}{d x_2} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}} \delta x_2 + \mathscr{O},\end{split}$

where we have used $$f_1(x_{1,0}, x_{2,0}) = 0$$. The perturbed equation (2) is linearized

(3)$\begin{split}\frac{d\delta x_1(t)}{dt} &= \left.\frac{d f_1(x_1, x_2)}{d x_1} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}} \delta x_1 + \left.\frac{d f_1(x_1, x_2)}{d x_2} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}} \delta x_2 \\ \frac{d\delta x_2(t)}{dt} &= \left.\frac{d f_2(x_1, x_2)}{d x_1} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}} \delta x_1 + \left.\frac{d f_2(x_1, x_2)}{d x_2} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}} \delta x_2.\end{split}$

The equation (3) can be rewritten into a matrix form

(4)$\begin{split}\frac{d}{dt} \begin{pmatrix} \delta x_1(t) \\ \delta x_2(t) \end{pmatrix} = \left. \begin{pmatrix} \frac{d f_1(x_1, x_2)}{d x_1} & \frac{d f_1(x_1, x_2)}{d x_2} \\ \frac{d f_2(x_1, x_2)}{d x_1} & \frac{d f_2(x_1, x_2)}{d x_2} \end{pmatrix} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}}\begin{pmatrix} \delta x_1(t) \\ \delta x_2(t) \end{pmatrix}.\end{split}$

For simplicity, we define the Jacobian matrix

$\begin{split}\mathbf J = \left. \begin{pmatrix} \frac{d f_1(x_1, x_2)}{d x_1} & \frac{d f_1(x_1, x_2)}{d x_2} \\ \frac{d f_2(x_1, x_2)}{d x_1} & \frac{d f_2(x_1, x_2)}{d x_2} \end{pmatrix} \right\vert_{x_1=x_{1,0}, x_2=x_{2,0}},\end{split}$

so that the matrix form of the linearized equation (4) becomes,

$\begin{split}\frac{d}{dt} \begin{pmatrix} \delta x_1(t) \\ \delta x_2(t) \end{pmatrix} = \mathbf J \begin{pmatrix} \delta x_1(t) \\ \delta x_2(t) \end{pmatrix}.\end{split}$

To investigate the stability of the differential system, we assume that

$\begin{split}\begin{pmatrix} \delta x_1(t) \\ \delta x_2(t) \end{pmatrix} = \begin{pmatrix} \delta x_1(t_0) \\ \delta x_2(t_0) \end{pmatrix} e^{\lambda t},\end{split}$

$\begin{split}\begin{pmatrix} \delta x_1(t_0) \\ \delta x_2(t_0) \end{pmatrix} \lambda = \mathbf J \begin{pmatrix} \delta x_1(t_0) \\ \delta x_2(t_0) \end{pmatrix}.\end{split}$

For non-trivial solutions, we require

$\operatorname{Det}(\mathbf J - \lambda \mathbf I) = 0,$

which is also the eigen value problem of the Jacobian matrix. We expand the determinant

(5)$\lambda^2 - (J_{11} + J_{22}) \lambda + (J_{11}J_{22} - J_{12}J_{21}) = 0.$

For real positive solutions $$\lambda>0$$, we get an exponential growing result for the linearized equation. Any deviation from the equilibrium point leads to a run-away process and the system moves further away from the equilibrium point. For real negative solutions $$\lambda < 0$$, the system will move back to the equilibrium point given any deviations from the equilibrium. Imaginary solutions of $$\lambda$$ leads to oscillations.

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