Math Insight

1. Stability is a local property. We only care about what is happening right around the equilibrium. To determine stability, we can zoom into the region right around the equilibrium.
2. If you zoom in on a differentiable function, it begins to look indistinguishable from its tangent line slope derivative
   . The function looks constant linear quadratic cubic
   in this zoomed in view.
3. The slope of the tangent line at the equilibrium $x=x_e$ is $m =$
   . (Online, enter $x_e$ as x_e.)
4. In the zoomed in view, if we pretend the equilibrium is at zero, the differential equation looks like $\diff{x}{t}=mx$.
5. We get exponential decay if m = 0 m < 0 m > 0
   ; hence, in terms of $f'(x_e)$, the equilibrium is stable if
   .
6. We get exponential growth if m = 0 m < 0 m > 0
   ; hence, in terms of $f'(x_e)$, the equilibrium is unstable if
   .

An equivalent approach works for a two-dimensional system of equations, say \begin{align*} \diff{x}{t} &= f(x,y)\\ \diff{y}{t} &= g(x,y), \end{align*} where $f$ and $g$ are some functions of $x$ and $y$. If $(x_e,y_e)$ is an equilibrium (where $f(x_e,y_e)=0 $ and $g(x_e,y_e)=0$), then we can zoom in around the equilibrium to determine the stability. As we zoom in, the same thing happens as did for one-dimensional systems: the system begins to look quadratic linear cubic constant
. This time, rather than looking like the linear equation $\diff{x}{t}=mx$ when zooming in, the zoomed-in system of differential equations will look like a linear system of the form \begin{align*} \diff{x}{t} &= \alpha x + \beta y\\ \diff{y}{t} &= \gamma x + \delta y \end{align*} or, in matrix form, as the equation $\diff{\vc{x} }{t} = A\vc{x}$, where $A=\begin{bmatrix}\alpha & \beta\\ \gamma & \delta \end{bmatrix}$. Here, $\alpha$, $\beta$, $\gamma$, and $\delta$ are just four numbers that play the role that the slope $m$ did for the one-dimensional case. We already studied this two-dimensional linear system and determined the condition for the equilibrium to be stable. If $\lambda_1$ and $\lambda_2$ are the eigenvalues of $A$, then the equilibrium $(0,0)$ of $\diff{\vc{x} }{t}= A\vc{x}$ is stable if
. It turns out that this condition will carry over to the nonlinear system. The only question is: which matrix's eigenvalues are the relevant ones?

In the one-dimensional case, the slope $m=f'(x_e)$ played the role of the matrix $A$ and its eigenvalue. The linear approximation (i.e., tangent line) was determined by the derivative of $f$ at the equilibrium. The same idea is true for the two-dimensional system. We need to take the derivative of vector-valued function $\vc{f}(x,y)=(f(x,y),g(x,y))$ at the equilibrium. In this case, the derivative is the matrix of partial derivatives, or Jacobian matrix, of $\vc{f}$, which is $$D_{\vc{f}}(x,y) = \begin{bmatrix}\pdiff{f}{x}(x,y) & \pdiff{f}{y}(x,y)\\ \pdiff{g}{x}(x,y) & \pdiff{g}{y}(x,y) \end{bmatrix}.$$ If we let $A = D_{\vc{f}}(x_e,y_e)$ then, when we zoom in on the equilibrium $(x_e,y_e)$, the original system becomes indistinguishable from the linear system $\diff{\vc{x} }{t} =A\vc{x}$, with the understanding that the origin of the linear system corresponds to the equilibrium $(x_e,y_e)$ (as we shift our view to center the equilibrium as we zoom in). The stability of the equilibrium $(x_e,y_e)$ depends on the eigenvalues $\lambda_1$ and $\lambda_2$ of the Jacobian matrix derivative matrix of partial derivatives
of $\vc{f}$ evaluated at $(x,y)=$
. The equilibrium $(x_e,y_e)$ is stable if
. Moreover, the classification of the equilibrium (as a stable node, unstable, saddle, stable spiral or unstable spiral) is determined from these eigenvalues just like for the linear system (with the exception of the center, but we won't worry about that here).