Introduction: Equilibrium and Stability

Let \(\mathbf{A}\) be a \(2 \times 2\) matrix and \(\det \mathbf{A} \neq 0\). Consider the homogeneous linear system:

\[ \vec{x}' = \mathbf{A}\vec{x}, \quad \vec{x} = \begin{bmatrix} x_{1}(t) \\ x_{2}(t) \end{bmatrix} \]

\(\vec{x} = \vec{0}\) is the equilibrium solution. Proof: \(\vec{x}' = \vec{0} \implies \mathbf{A}\vec{x} = \vec{0} \implies \vec{x} = \mathbf{A}^{-1}\vec{0} = \vec{0}\). We use the phase plane (\(x_1\)-\(x_2\) plane) to determine the stability of the equilibrium solution (the origin). Let \(\lambda_1\) and \(\lambda_2\) be the eigenvalues of \(\mathbf{A}\).

(I) \(\lambda_{1} \neq \lambda_{2}\): Distinct Real Eigenvalues

The general solution is \(\vec{x}(t) = C_1\vec{v}_1e^{\lambda_1 t} + C_2\vec{v}_2e^{\lambda_2 t}\).

Case 1: \(\lambda_1 < 0 < \lambda_2\) (Saddle Point)

Example

From Section 5.2 Example 1, where \(\lambda_1 = -2\) and \(\lambda_2 = 5\):

\[ \vec{x}(t) = C_{1} \begin{bmatrix} 1 \\ -3 \end{bmatrix} e^{-2t} + C_{2} \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{5t} \]

Analysis:

  • If \(C_2 = 0 \implies x_2 = -3x_1\). This is line \(\ell_1\) passing through \(\vec{v}_1\) and the origin.
  • If \(C_1 = 0 \implies x_1 = 2x_2\). This is line \(\ell_2\) passing through \(\vec{v}_2\) and the origin.
  • If \(C_1 \neq 0, C_2 \neq 0\): As \(t \to \infty\), solution curves extend asymptotically toward \(\ell_2\). As \(t \to -\infty\), solution curves extend asymptotically toward \(\ell_1\).
A phase portrait of a saddle point at the origin. Two intersecting lines represent eigenvectors: line L1 (y = -3x) has arrows pointing toward the origin, and line L2 (x = 2y) has arrows pointing away from the origin. Hyperbolic solution curves in the four regions between the lines approach L1 in the distant past and L2 in the distant future. The origin is unstable because most trajectories eventually move away from it.
Figure: Phase Portrait of an Unstable Saddle Point

The origin is called a saddle point. It is unstable. Note that \(\vec{v}_1\) (with \(\lambda_1 < 0\)) points to the origin, while \(\vec{v}_2\) (with \(\lambda_2 > 0\)) points outward.

Case 2: \(\lambda_1 < \lambda_2 < 0\) (Improper Nodal Sink)

Example

Consider \(\vec{x}' = \begin{bmatrix} -3 & 1 \\ 2 & -2 \end{bmatrix} \vec{x}\).

Step 1: Eigenvalues. \(\det(\mathbf{A} - \lambda \mathbf{I}) = \lambda^2 + 5\lambda + 4 = (\lambda + 1)(\lambda + 4) = 0 \implies \lambda_1 = -4, \lambda_2 = -1\).

Step 2: Eigenvectors. For \(\lambda_1 = -4, \vec{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\). For \(\lambda_2 = -1, \vec{v}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\).

Step 3: Solution Analysis. \(\vec{x}(t) = C_1 \begin{bmatrix} 1 \\ -1 \end{bmatrix} e^{-4t} + C_2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} e^{-t}\).

As \(t \to \infty\), all solutions approach the origin. Because \(e^{-4t}\) decays faster than \(e^{-t}\), trajectories become tangent to line \(\ell_2\) passing through \(\vec{v}_2\) as \(t \to \infty\).

A high-clarity phase portrait of an improper nodal sink. Two bold, solid black lines representing the eigenvectors L1 (y = -x) and L2 (y = 2x) provide the structural frame of the plot. Long, continuous solution curves in alternating red and blue colors flow from the edges of the grid into the origin. The curves exhibit a distinct bending motion: they originate parallel to the fast decay line L1 and eventually sweep around to become tangent to the slow decay line L2 as they reach the equilibrium point. Multiple arrows along the paths and eigenvector lines confirm a globally stable inward flow.
Figure: Dynamics of a Globally Stable Improper Nodal Sink
The origin is called an improper nodal sink and is stable.

Case 3: \(0 < \lambda_2 < \lambda_1\) (Improper Nodal Source)

Similar to Case 2, but trajectories move away from the origin as \(t\) increases (the direction of motion along each solution curve is reversed). The origin is an improper nodal source and is unstable.

(II) \(\lambda_1 = \lambda_2 = \lambda\): Repeated Real Eigenvalues

Case 1: Matrix \(\mathbf{A}\) has 2 Linearly Independent Eigenvectors (Complete)

Example

Solve \(\vec{x}' = \begin{bmatrix} -3 & 0 \\ 0 & -3 \end{bmatrix} \vec{x}\).

\(\lambda = -3\) (multiplicity 2). Any vector is an eigenvector. Choose \(\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\) and \(\vec{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\). General solution: \[\vec{x}(t) = \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} e^{-3t}\]

A phase portrait of a proper nodal sink, also known as a star node. The origin acts as the equilibrium point. Numerous straight-line trajectories, colored in alternating red and blue, originate from the edges of the plot and point directly toward the origin from every angle. Bold black axes represent the standard coordinate system. Every trajectory is a perfectly straight line, indicating that every direction is an eigenvector direction for the repeated eigenvalue lambda equals negative three. Arrows on every line point inward, showing global asymptotic stability.
Figure: Phase Portrait of a Proper Nodal Sink (Star Node)
The origin is called a proper nodal sink (stable if \(\lambda < 0\)) or proper nodal source (unstable if \(\lambda > 0\)). Solutions are a family of straight lines through the origin.

Case 2: Matrix \(\mathbf{A}\) has only 1 Linearly Independent Eigenvector (Defective)

Example

Recall Section 5.5 Example 2: \(\vec{x}' = \begin{bmatrix} 1 & -3 \\ 3 & 7 \end{bmatrix} \vec{x}\) has \(\lambda = 4\) (multiplicity 2). General solution: \[\vec{x}(t) = C_1 \begin{bmatrix} -3 \\ 3 \end{bmatrix} e^{4t} + C_2 \begin{bmatrix} -3t + 1 \\ 3t \end{bmatrix} e^{4t}\]

Analysis: If \(C_2 = 0\), the solution is the line \(x_1 = -x_2\). If \(C_2 \neq 0\), as \(t \to \infty\), the derivative \(\vec{x}'(t) \to \text{scalar multiple of } \vec{v}_1\). The line \(x_1 = -x_2\) is the tangent line to the solution curves.

A clear phase portrait of a defective improper source at the origin. A single bold, solid black line represents the only eigenvector direction along the line x1 equals negative x2. Numerous continuous solution curves in alternating red and blue colors emerge from the center, all initially tangent to the black eigenvector line. These curves then sweep outward and away from the origin in a characteristically twisted 'C' shape. Arrows along every curve and the eigenvector line point outward, indicating global instability and exponential growth of all solutions.
Figure: Phase Portrait of an Unstable Defective Improper Source

The origin is called an improper nodal source (unstable if \(\lambda > 0\)) or improper nodal sink (stable if \(\lambda < 0\)).

(III) Complex Conjugate Eigenvalues \(\lambda_{1,2} = p \pm iq\)

Case 1: \(p = 0\) (Purely Imaginary)

Example

Solve \(\vec{x}' = \begin{bmatrix} 6 & -17 \\ 8 & -6 \end{bmatrix} \vec{x}\) with \(\vec{x}(0) = \begin{bmatrix} 4 \\ 2 \end{bmatrix}\).

Step 1: Eigenvalues. \(\det(\mathbf{A} - \lambda \mathbf{I}) = \lambda^2 + 100 = 0 \implies \lambda = \pm 10i\).

Step 2: Eigenvector for \(\lambda = 10i\). \(4a = (3 + 5i)b\). Let \(b = 4 \implies a = 3 + 5i\). Thus \(\vec{v} = \begin{bmatrix} 3+5i \\ 4 \end{bmatrix}\).

Step 3: Expansion. \(\vec{v}e^{\lambda t} = \begin{bmatrix} 3+5i \\ 4 \end{bmatrix} (\cos 10t + i \sin 10t) = \begin{bmatrix} 3\cos 10t - 5\sin 10t \\ 4\cos 10t \end{bmatrix} + i \begin{bmatrix} 5\cos 10t + 3\sin 10t \\ 4\sin 10t \end{bmatrix}\).

Step 4: General Solution:

\[ \vec{x}(t) = C_1 \begin{bmatrix} 3\cos 10t - 5\sin 10t \\ 4\cos 10t \end{bmatrix} + C_2 \begin{bmatrix} 5\cos 10t + 3\sin 10t \\ 4\sin 10t \end{bmatrix} \]

Step 5: Particular Solution: Using \(\vec{x}(0) = \begin{bmatrix} 4 \\ 2 \end{bmatrix}\) yields \(C_1 = 1/2, C_2 = 1/2\). Resulting in:

\[ x_1(t) = 4\cos 10t - \sin 10t, \quad x_2(t) = 2\cos 10t + 2\sin 10t \]
A high-clarity phase portrait of a center equilibrium point at the origin. The plot features several concentric, closed elliptical trajectories. A single bold, solid red ellipse highlights the particular solution that passes exactly through the initial point (4, 2). Other trajectories are shown as dashed blue ellipses of various sizes. Light silver streamlines in the background indicate the overall clockwise flow of the vector field. The axes are bold black lines, and the plot confirms that all solutions are periodic and remain bounded as time increases.
Figure: Phase Portrait of a Stable Center (Elliptical Orbits)

The solution curve is an ellipse rotated by \(\alpha = \arctan(3/4)\). The origin is called a center. It is stable, but not asymptotically stable.

Case 2: \(p \neq 0\) (Spiral Points)

Example

From Section 5.2 Example 4: \(\lambda = 4 \pm 3i\). The general solution is \(x_1(t) = e^{4t}(C_1\cos 3t + C_2\sin 3t)\) and \(x_2(t) = e^{4t}(C_1\sin 3t - C_2\cos 3t)\). As \(t \to \infty, e^{4t} \to \infty\).

A detailed phase portrait of an unstable spiral source at the origin. Numerous continuous, smooth trajectories in alternating red and blue spiral outward in a counter-clockwise direction. The background features a light gray vector field of streamlines indicating the spiraling outward flow. Bold black horizontal and vertical axes represent the coordinate system. Arrows along each spiral path confirm that as time increases, the solutions rotate and grow in magnitude, moving further from the origin toward the edges of the plot. This visualization clearly demonstrates the exponential growth and rotational behavior caused by the complex eigenvalues with a positive real part.
Figure: Phase Portrait of an Unstable Spiral Source

If \(p > 0\), the origin is called a spiral source (unstable) . If \(p < 0\), the origin is called a spiral sink (stable).

Summary of Classification

Eigenvalues (\(\lambda_1, \lambda_2\)) Condition Critical Point Type Stability
Real, Distinct \(\lambda_1 < 0 < \lambda_2\) Saddle Point Unstable
Real, Distinct \(\lambda_1 < \lambda_2 < 0\) Improper Nodal Sink Stable
Real, Distinct \(0 < \lambda_1 < \lambda_2\) Improper Nodal Source Unstable
Repeated Real Complete Proper Nodal Sink (\(\lambda < 0\)) / Source (\(\lambda > 0\)) Stable / Unstable
Repeated Real Defective Improper Nodal Sink (\(\lambda < 0\)) / Source (\(\lambda > 0\)) Stable / Unstable
Complex Conjugate \(p \pm iq\) \(p = 0\) Center Stable (not asym.)
Complex Conjugate \(p \pm iq\) \(p < 0\) Spiral Sink Stable
Complex Conjugate \(p \pm iq\) \(p > 0\) Spiral Source Unstable

Stability Classifications

  • Stable: Trajectories that start near the equilibrium point remain near it for all \(t > 0\).
  • Asymptotically Stable: Trajectories that start near the equilibrium point approach the origin as \(t \to \infty\).
  • Unstable: Trajectories that start near the equilibrium point move away from it.

Classification of Nodes

  • Proper Node (Star Node): Occurs when there is a repeated eigenvalue with two linearly independent eigenvectors. Every trajectory is a straight line through the origin.
  • Improper Node: Occurs when eigenvalues have the same sign (either distinct or repeated defective). Most trajectories are tangent to a specific straight line through the origin.