求线性化系统不稳定结点(0,0)的特征向量以绘制相图

Got it, let's walk through this problem step by step to sort out the eigenvector confusion and get you on track to sketching the phase portrait.

First, we need the Jacobian matrix of the nonlinear system, since linearization at an equilibrium point relies on evaluating this matrix at that point. The system is:
$$
\begin{align} x'(t)&=x(1-10y)\\ y'(t)&=y(1-10x) \\end{align}
$$
The Jacobian matrix $J(x,y)$ is structured as:

J(x,y) = [ ∂x'/∂x  ∂x'/∂y ]
         [ ∂y'/∂x  ∂y'/∂y ]

Calculating the partial derivatives and evaluating at $(0,0)$:

• $\partial x'/∂x = 1 - 10y$ → equals 1 at $(0,0)$
• $\partial x'/∂y = -10x$ → equals 0 at $(0,0)$
• $\partial y'/∂x = -10y$ → equals 0 at $(0,0)$
• $\partial y'/∂y = 1 - 10x$ → equals 1 at $(0,0)$

So the linearized matrix at $(0,0)$ simplifies to:

J = [[1, 0],
     [0, 1]]

As you noted, we solve the characteristic equation $\det(J - \lambda I) = 0$:
$$
\det\left(\begin{bmatrix}1-\lambda & 0 \ 0 & 1-\lambda\end{bmatrix}\right) = (1-\lambda)^2 = 0
$$
This gives a repeated positive eigenvalue $\lambda = 1$, confirming $(0,0)$ is an unstable node.

Now, to find eigenvectors, we solve $(J - \lambda I)\mathbf{v} = \mathbf{0}$. Substituting $\lambda=1$, we get:

[[0, 0],
 [0, 0]] * [v₁; v₂] = [0; 0]

This is a special case: the matrix is the zero matrix, which means every non-zero 2D vector is an eigenvector! There's no restriction on $v₁$ and $v₂$ (as long as they aren't both zero). You don't need to hunt for specific vectors—every direction through the origin is a valid eigendirection here.

The linearized system simplifies to $x' = x$, $y' = y$. The general solution is:
$$
x(t) = x_0 e^t, \quad y(t) = y_0 e^t
$$
For any initial point $(x_0, y_0) \neq (0,0)$, this means the trajectory is a straight ray starting at the origin and extending outward, since $e^t$ grows exponentially with time. Every line passing through $(0,0)$ is a trajectory for the linearized system.

Near the origin, the nonlinear terms $-10xy$ are negligible (since $x$ and $y$ are small), so the nonlinear system's trajectories will closely match the linearized system's outward rays. As you move away from $(0,0)$, the nonlinear terms will cause trajectories to deviate toward the system's other equilibrium points:

• $(1, 0)$ and $(0, 1)$ are saddle points (you can confirm this by evaluating the Jacobian at these points and checking eigenvalues)
• $(1/10, 1/10)$ is also a saddle point (its Jacobian has eigenvalues $\pm 1$)

• Start by drawing all outward rays from $(0,0)$ (these are the linearized trajectories, which approximate the nonlinear behavior near the origin)
• Mark the other equilibrium points and sketch their local behavior (saddle points have incoming and outgoing trajectories)
• Connect the trajectories smoothly, keeping in mind that near $(0,0)$, all paths diverge outward.

内容的提问来源于stack exchange，提问作者Bernhard Listing