齐次线性系统稳定性类型：焦点型稳定性参数范围求解

Alright, let's break down the focus-type fixed point analysis for your given homogeneous linear system step by step:
$$ \frac{d}{du} \begin{bmatrix}x\y\end{bmatrix} = \begin{bmatrix}-5&a\2&1\end{bmatrix} \begin{bmatrix}x\y\end{bmatrix}$$

Step 1: Derive the Characteristic Equation #

For a 2x2 matrix ( A = \begin{bmatrix}a_{11}&a_{12}\a_{21}&a_{22}\end{bmatrix} ), we find the characteristic equation by calculating ( \det(A - \lambda I) = 0 ) (where ( I ) is the identity matrix).

Applying this to your matrix:

det(
  [-5 - λ,   a     ]
  [ 2    ,  1 - λ ]
) = 0

Expanding the determinant gives us:
$$(-5 - \lambda)(1 - \lambda) - (a \times 2) = 0$$
$$\lambda^2 + 4\lambda - 5 - 2a = 0$$

Step 2: Define Focus Conditions #

A fixed point is a focus when the characteristic equation has conjugate complex eigenvalues (no real eigenvalues). For a quadratic equation ( \lambda^2 + p\lambda + q = 0 ), this occurs when the discriminant is negative:
$$\Delta = p^2 - 4q < 0$$

Stability of the focus depends on the real part of these complex eigenvalues:

• Negative real part: Stable focus (solutions spiral toward (0,0))
• Positive real part: Unstable focus (solutions spiral away from (0,0))

Step 3: Calculate the Discriminant and Solve for ( a ) #

From our characteristic equation, ( p = 4 ) and ( q = -5 - 2a ). Plug these into the discriminant formula:
$$\Delta = 4^2 - 4(1)(-5 - 2a) = 16 + 20 + 8a = 36 + 8a$$

Set ( \Delta < 0 ) to get conjugate complex eigenvalues:
$$36 + 8a < 0$$
$$8a < -36$$
$$a < -\frac{9}{2} \quad (\text{or } a < -4.5)$$

Step 4: Analyze Stability of the Focus #

For a quadratic characteristic equation ( \lambda^2 + p\lambda + q = 0 ), the real part of the complex eigenvalues is ( -\frac{p}{2} ). Here, ( p = 4 ), so:
$$\text{Real part} = -\frac{4}{2} = -2$$

Since the real part is negative (and this value is constant for all ( a ) satisfying the focus condition), the fixed point (0,0) is a stable focus whenever ( a < -\frac{9}{2} ).

Quick Correction to Your Initial Understanding #

You mentioned "focus corresponds to conjugate complex eigenvalues with positive real part"—that's only one subset of foci (unstable ones). Foci can be stable or unstable depending on the sign of the real part. In this specific system, the real part of the complex eigenvalues is fixed at -2, so we only get stable foci here.

内容的提问来源于stack exchange，提问作者princetongirl818