The Central Idea
Stability is about the response of a system to small changes in its state.
If a system begins near an equilibrium \(x^\ast\), we ask whether the resulting trajectory:
- stays near \(x^\ast\);
- returns toward \(x^\ast\);
- moves away from \(x^\ast\);
- approaches in some directions but departs in others.
These possibilities lead to several distinct notions of stability.
Equilibrium Is Not the Same as Stability
An equilibrium is a state that does not move when the system begins exactly there.
For
\[ \frac{dx}{dt}=f(x), \]an equilibrium satisfies
\[ f(x^\ast)=0. \]But this condition alone says nothing about nearby states.
Stability concerns the surrounding trajectories.
A Physical Picture
Imagine a small ball in three different landscapes.
- At the bottom of a bowl, a small displacement tends to bring the ball back.
- At the top of a hill, a small displacement sends the ball farther away.
- On a perfectly flat surface, a small displacement may leave it nearby without restoring it.
These pictures correspond broadly to attracting, unstable, and neutrally stable behavior.
Local Stability
Stability is usually first studied locally, meaning near a particular equilibrium.
We do not initially require every possible state in the entire state space to behave the same way.
Instead, we ask what happens when the initial state is sufficiently close to the equilibrium.
Lyapunov Stability
An equilibrium \(x^\ast\) is called stable in the sense of Lyapunov if trajectories that start sufficiently close to \(x^\ast\) remain close to \(x^\ast\) for all future time.
Informally,
\[ \boxed{ \text{start close} \Longrightarrow \text{stay close} }. \]Stability does not necessarily require convergence back to the equilibrium.
Formal Stability Language
For a continuous system, \(x^\ast\) is Lyapunov stable if for every
\[ \varepsilon>0 \]there exists
\[ \delta>0 \]such that
\[ |x(0)-x^\ast|<\delta \]implies
\[ |x(t)-x^\ast|<\varepsilon \]for all
\[ t\ge0. \]What the \(\varepsilon\)-\(\delta\) Definition Means
The quantity \(\varepsilon\) specifies how close we want the future trajectory to remain to the equilibrium.
The quantity \(\delta\) tells us how close the initial state must begin.
Stability means that no matter how small a permitted neighborhood \(\varepsilon\) is chosen, some sufficiently small initial neighborhood \(\delta\) can be found that keeps the trajectory inside it.
Asymptotic Stability
An equilibrium is asymptotically stable if it is stable and nearby trajectories also converge to it.
In continuous time,
\[ \boxed{ x(t)\to x^\ast \quad\text{as}\quad t\to\infty }. \]In discrete time,
\[ \boxed{ x_n\to x^\ast \quad\text{as}\quad n\to\infty }. \]Thus asymptotic stability combines staying near with eventually returning.
Attracting versus Stable
An equilibrium is attracting if nearby trajectories converge to it.
An equilibrium is stable if nearby trajectories remain nearby.
In many standard examples, attracting equilibria are also stable and hence asymptotically stable.
The logical distinction is still important because attraction and stability describe different properties.
Instability
An equilibrium is unstable if it fails the stability condition.
This means that arbitrarily small perturbations can lead to trajectories that eventually move a significant distance away.
Informally,
\[ \boxed{ \text{small disturbance} \Longrightarrow \text{possible large departure} }. \]One-Dimensional Continuous Systems
For
\[ \frac{dx}{dt}=f(x), \]one-dimensional stability can often be understood by examining the sign of \(f(x)\) on either side of an equilibrium.
If arrows point toward the equilibrium from both sides, it is attracting.
If arrows point away from both sides, it is repelling.
Example: Stable Continuous Equilibrium
Consider
\[ \frac{dx}{dt}=-(x-3). \]The equilibrium is
\[ x^\ast=3. \]If \(x<3\), then
\[ -(x-3)>0, \]so \(x\) increases toward \(3\).
If \(x>3\), then
\[ -(x-3)<0, \]so \(x\) decreases toward \(3\).
The equilibrium is asymptotically stable.
Solving the Stable Example
Write
\[ \frac{dx}{dt}=-(x-3). \]The solution can be written
\[ \boxed{ x(t)=3+(x_0-3)e^{-t} }. \]Since
\[ e^{-t}\to0, \]we have
\[ x(t)\to3. \]The exact solution confirms the phase-line analysis.
Example: Unstable Continuous Equilibrium
Consider
\[ \frac{dx}{dt}=x-3. \]The equilibrium is again
\[ x^\ast=3. \]If \(x<3\), then
\[ \frac{dx}{dt}<0, \]so the state moves farther left.
If \(x>3\), then
\[ \frac{dx}{dt}>0, \]so the state moves farther right.
The equilibrium is unstable.
Exact Solution of the Unstable Example
The solution is
\[ \boxed{ x(t)=3+(x_0-3)e^t }. \]Unless
\[ x_0=3, \]the difference from equilibrium grows in magnitude:
\[ |x(t)-3| = |x_0-3|e^t. \]Small initial disturbances are exponentially amplified.
Phase-Line Test
For a one-dimensional continuous system, the phase line provides a quick qualitative stability test.
Around an equilibrium \(x^\ast\):
- arrows pointing inward from both sides indicate attraction;
- arrows pointing outward from both sides indicate repulsion;
- arrows pointing the same direction on both sides indicate one-sided behavior.
Semistable Equilibrium
Consider
\[ \frac{dx}{dt}=(x-1)^2. \]The equilibrium is
\[ x^\ast=1. \]Since
\[ (x-1)^2>0 \]for every \(x\ne1\), trajectories move to the right on both sides.
States to the left move toward \(1\), while states to the right move away.
This is an example of semistable behavior.
The Derivative Test for Continuous Systems
Suppose
\[ \frac{dx}{dt}=f(x) \]and \(x^\ast\) is an equilibrium.
If
\[ f'(x^\ast)<0, \]then nearby trajectories are typically pulled toward the equilibrium.
If
\[ f'(x^\ast)>0, \]nearby trajectories are typically pushed away.
Continuous Local Stability Rule
For a one-dimensional autonomous differential equation,
\[ \boxed{ f'(x^\ast)<0 \Longrightarrow x^\ast\text{ is locally asymptotically stable} } \]and
\[ \boxed{ f'(x^\ast)>0 \Longrightarrow x^\ast\text{ is unstable} }. \]If
\[ f'(x^\ast)=0, \]this simple test is inconclusive.
Why the Derivative Test Works
Near an equilibrium \(x^\ast\), we can approximate
\[ f(x) \approx f(x^\ast)+f'(x^\ast)(x-x^\ast). \]Since
\[ f(x^\ast)=0, \]this becomes
\[ f(x) \approx f'(x^\ast)(x-x^\ast). \]The local system therefore resembles
\[ \frac{du}{dt} = f'(x^\ast)u, \]where
\[ u=x-x^\ast. \]Example: Derivative Stability Test
Consider
\[ \frac{dx}{dt}=x(1-x). \]The equilibria are
\[ x^\ast=0 \]and
\[ x^\ast=1. \]Since
\[ f'(x)=1-2x, \]we can classify both equilibria.
Classifying the Logistic Equilibria
At
\[ x^\ast=0, \]we have
\[ f'(0)=1>0. \]Therefore \(0\) is unstable.
At
\[ x^\ast=1, \]we have
\[ f'(1)=-1<0. \]Therefore \(1\) is locally asymptotically stable.
When the Derivative Test Is Inconclusive
Consider
\[ \frac{dx}{dt}=-x^3. \]The equilibrium is
\[ x^\ast=0. \]Here
\[ f'(x)=-3x^2, \]so
\[ f'(0)=0. \]The derivative test is inconclusive, but sign analysis shows that trajectories on both sides move toward zero.
Discrete Stability
Now consider a discrete dynamical system
\[ x_{n+1}=f(x_n) \]with fixed point
\[ f(x^\ast)=x^\ast. \]Stability depends on whether iteration reduces or amplifies small deviations from \(x^\ast\).
Perturbation from a Fixed Point
Write
\[ x_n=x^\ast+u_n, \]where \(u_n\) is a small deviation.
Near the fixed point,
\[ f(x^\ast+u_n) \approx f(x^\ast)+f'(x^\ast)u_n. \]Because
\[ f(x^\ast)=x^\ast, \]the perturbation approximately satisfies
\[ \boxed{ u_{n+1}\approx f'(x^\ast)u_n }. \]Discrete Derivative Test
The local multiplier is
\[ f'(x^\ast). \]If
\[ |f'(x^\ast)|<1, \]small deviations shrink under iteration.
If
\[ |f'(x^\ast)|>1, \]small deviations grow.
Discrete Local Stability Rule
For a one-dimensional discrete map,
\[ \boxed{ |f'(x^\ast)|<1 \Longrightarrow x^\ast\text{ is locally asymptotically stable} } \]and
\[ \boxed{ |f'(x^\ast)|>1 \Longrightarrow x^\ast\text{ is unstable} }. \]If
\[ |f'(x^\ast)|=1, \]the basic derivative test is inconclusive.
Example: Stable Discrete Fixed Point
Consider
\[ x_{n+1}=0.5x_n+3. \]The fixed point satisfies
\[ x^\ast=0.5x^\ast+3, \]so
\[ x^\ast=6. \]Since
\[ f'(x)=0.5, \]we have
\[ |f'(6)|=0.5<1. \]Therefore the fixed point is locally asymptotically stable.
Example: Unstable Discrete Fixed Point
Consider
\[ x_{n+1}=2x_n-3. \]The fixed point satisfies
\[ x^\ast=2x^\ast-3, \]so
\[ x^\ast=3. \]Since
\[ f'(x)=2, \]we have
\[ |f'(3)|=2>1. \]Therefore the fixed point is unstable.
Positive versus Negative Multipliers
In a discrete system, the sign of \(f'(x^\ast)\) affects how trajectories approach or leave a fixed point.
If
\[ 0<f'(x^\ast)<1, \]nearby states typically approach without alternating sides.
If
\[ -1<f'(x^\ast)<0, \]nearby states typically alternate from one side of the fixed point to the other while converging.
Example: Alternating Convergence
Consider
\[ x_{n+1}=-0.5x_n. \]The fixed point is
\[ x^\ast=0. \]Starting from
\[ x_0=8, \]we obtain
\[ 8,\ -4,\ 2,\ -1,\ 0.5,\ldots \]The sign alternates, but the magnitude decreases toward zero.
Example: Alternating Instability
Consider
\[ x_{n+1}=-2x_n. \]The fixed point is \(0\).
Starting from
\[ x_0=0.1, \]we obtain
\[ 0.1,\ -0.2,\ 0.4,\ -0.8,\ 1.6,\ldots \]The orbit alternates sides while moving farther from the fixed point.
The Logistic Map Stability Test
Consider
\[ f(x)=rx(1-x). \]The derivative is
\[ f'(x)=r(1-2x). \]The fixed point
\[ x^\ast=0 \]has multiplier
\[ f'(0)=r. \]Stability of \(x^\ast=0\) in the Logistic Map
The condition for local asymptotic stability is
\[ |r|<1. \]Thus the zero fixed point is locally asymptotically stable when
\[ \boxed{|r|<1}. \]In the standard population interpretation, where \(r\) is often taken nonnegative, this becomes
\[ 0\le r<1. \]The Second Logistic Fixed Point
The second fixed point is
\[ x^\ast=1-\frac1r, \]when \(r\ne0\).
Evaluate the derivative:
\[ f'(x^\ast) = r\left(1-2\left(1-\frac1r\right)\right). \]Simplifying,
\[ f'(x^\ast)=2-r. \]Stability of the Second Logistic Fixed Point
Local asymptotic stability requires
\[ |2-r|<1. \]This gives
\[ -1<2-r<1. \]Solving,
\[ \boxed{1<r<3}. \]At \(r=3\), the fixed point reaches the boundary of this local stability condition, foreshadowing a bifurcation.
Global versus Local Stability
Local stability concerns initial states sufficiently close to an equilibrium.
Global stability concerns a much larger region, sometimes the entire admissible state space.
A locally asymptotically stable equilibrium need not attract every possible trajectory.
Basin of Attraction
The basin of attraction of an attracting equilibrium is the set of initial states whose trajectories converge to that equilibrium.
If \(x^\ast\) is attracting, its basin may be written conceptually as
\[ \boxed{ \mathcal{B}(x^\ast) = \{x_0:\ x(t;x_0)\to x^\ast\} }. \]Basins can be small, large, disconnected, or geometrically complicated.
Multiple Stable Equilibria
A system can contain more than one attracting equilibrium.
In such cases, the long-term state depends on the initial condition.
Different regions of the state space may belong to different basins of attraction.
This phenomenon is called multistability.
Example: Multiple Equilibria
Consider
\[ \frac{dx}{dt}=x-x^3. \]Factor:
\[ \frac{dx}{dt}=x(1-x^2). \]The equilibria are
\[ \boxed{x^\ast=-1,\quad0,\quad1}. \]Classifying the Three Equilibria
Let
\[ f(x)=x-x^3. \]Then
\[ f'(x)=1-3x^2. \]At \(x^\ast=0\),
\[ f'(0)=1>0, \]so \(0\) is unstable.
At \(x^\ast=\pm1\),
\[ f'(\pm1)=-2<0, \]so both are locally asymptotically stable.
Basins in the Three-Equilibrium Example
In
\[ \frac{dx}{dt}=x-x^3, \]initial states with
\[ x_0>0 \]tend toward \(1\), while initial states with
\[ x_0<0 \]tend toward \(-1\).
The unstable equilibrium at \(0\) separates the two basins.
Linear Systems in Several Dimensions
Consider the continuous linear system
\[ \frac{d\mathbf{x}}{dt} = A\mathbf{x}. \]The origin
\[ \mathbf{x}^\ast=\mathbf{0} \]is always an equilibrium.
Its stability is governed by the eigenvalues of the matrix \(A\).
Eigenvalues and Continuous Stability
For
\[ \frac{d\mathbf{x}}{dt}=A\mathbf{x}, \]the broad rule is:
- if every eigenvalue has negative real part, the origin is asymptotically stable;
- if at least one eigenvalue has positive real part, the origin is unstable;
- if eigenvalues lie on the imaginary axis, further analysis may be required.
Example: Stable Linear System
Consider
\[ \frac{d}{dt} \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} -2 & 0\\ 0 & -5 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}. \]The eigenvalues are
\[ \lambda_1=-2, \qquad \lambda_2=-5. \]Both are negative, so the origin is asymptotically stable.
Example: Unstable Linear System
Consider
\[ \frac{d}{dt} \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 3 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}. \]The eigenvalues are
\[ \lambda_1=1, \qquad \lambda_2=3. \]Both are positive, so trajectories move away from the origin.
The equilibrium is unstable.
Saddle Behavior
Consider
\[ \frac{d}{dt} \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} -1 & 0\\ 0 & 2 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}. \]The eigenvalues are
\[ -1 \qquad\text{and}\qquad 2. \]One direction contracts toward the origin, while another expands away from it.
This equilibrium is a saddle and is unstable.
Stable and Unstable Directions
At a saddle equilibrium, some directions may be attracted while others are repelled.
In the diagonal system
\[ x'=-x, \qquad y'=2y, \]the \(x\)-direction is stable because
\[ x(t)=x_0e^{-t}, \]while the \(y\)-direction is unstable because
\[ y(t)=y_0e^{2t}. \]Discrete Linear Systems
Consider
\[ \mathbf{x}_{n+1}=A\mathbf{x}_n. \]The origin is again a fixed point.
Stability is now governed by the magnitudes of the eigenvalues of \(A\).
Eigenvalues and Discrete Stability
For a discrete linear system:
- if every eigenvalue satisfies \(|\lambda|<1\), the origin is asymptotically stable;
- if at least one eigenvalue satisfies \(|\lambda|>1\), the origin is unstable;
- if eigenvalues lie on the unit circle, further analysis may be required.
The unit circle plays for discrete systems a role similar to the imaginary axis in continuous systems.
Example: Stable Discrete Matrix System
Consider
\[ \mathbf{x}_{n+1} = \begin{pmatrix} 0.5 & 0\\ 0 & -0.25 \end{pmatrix} \mathbf{x}_n. \]The eigenvalues are
\[ \lambda_1=0.5, \qquad \lambda_2=-0.25. \]Since both magnitudes are less than \(1\), the origin is asymptotically stable.
Example: Discrete Saddle-Type Behavior
Consider
\[ \mathbf{x}_{n+1} = \begin{pmatrix} 0.5 & 0\\ 0 & 2 \end{pmatrix} \mathbf{x}_n. \]One eigenvalue has magnitude less than \(1\), while the other has magnitude greater than \(1\).
The origin attracts in one direction and repels in another.
Therefore it is unstable.
Neutral Stability
A system can keep nearby trajectories near an equilibrium without causing them to converge to it.
This is often called neutral stability in introductory settings.
Ideal undamped oscillations provide a familiar example: nearby trajectories can remain bounded around an equilibrium without spiraling inward.
Example: Ideal Harmonic Oscillator
Consider
\[ x'=v, \qquad v'=-x. \]The origin is an equilibrium.
The quantity
\[ x^2+v^2 \]remains constant along trajectories.
Nearby trajectories remain on nearby circles rather than converging to the origin.
Damping Changes Stability
Add damping:
\[ x'=v, \] \[ v'=-x-cv, \]with
\[ c>0. \]Mechanical energy decreases over time, and trajectories can spiral or move inward toward the origin.
Damping can therefore convert persistent oscillation into asymptotic stability.
Lyapunov Functions
A Lyapunov function is a scalar function used to study stability without explicitly solving the dynamical system.
It often behaves like an energy measure.
A typical candidate
\[ V(\mathbf{x}) \]is positive away from the equilibrium and decreases along trajectories.
Basic Lyapunov Idea
Suppose the equilibrium is at the origin.
We seek a function satisfying
\[ V(\mathbf{0})=0 \]and
\[ V(\mathbf{x})>0 \]for nearby
\[ \mathbf{x}\ne\mathbf{0}. \]If \(V\) decreases along trajectories, the state is constrained from moving toward larger values of this generalized energy.
Derivative of a Lyapunov Function
For
\[ \dot{\mathbf{x}}=\mathbf{F}(\mathbf{x}), \]the rate of change of \(V\) along a trajectory is
\[ \boxed{ \dot V = \nabla V(\mathbf{x}) \cdot \mathbf{F}(\mathbf{x}) }. \]If
\[ \dot V<0 \]away from the equilibrium, that often provides strong evidence of asymptotic stability.
Example: Simple Lyapunov Function
Consider
\[ x'=-x. \]Choose
\[ V(x)=\frac12x^2. \]Then
\[ \dot V = x x'. \]Substitute \(x'=-x\):
\[ \dot V=-x^2. \]Thus
\[ \dot V<0 \]whenever
\[ x\ne0. \]Why Lyapunov Methods Matter
Exact solutions of nonlinear systems may be difficult or unavailable.
A suitable Lyapunov function can still establish stability using inequalities and monotonic decrease.
This makes Lyapunov theory central in:
- nonlinear dynamics;
- control engineering;
- robotics;
- aerospace guidance;
- power systems;
- optimization algorithms;
- economic adjustment models.
Stability Is Relative to a Model
A mathematically stable equilibrium is stable under the assumptions and variables included in the model.
Real systems may experience disturbances, unmodeled forces, parameter changes, noise, delays, or structural changes.
Stability analysis is therefore powerful, but its conclusions inherit the assumptions of the underlying model.
Structural Sensitivity
Stability can change when a parameter changes.
Consider
\[ \frac{dx}{dt}=\mu x-x^3. \]The equilibrium structure depends on \(\mu\), and so does the stability of those equilibria.
When stability changes as parameters vary, bifurcation theory becomes the natural next framework.
Stability and Resilience
In applications, stability is often interpreted as a form of resilience.
A stable operating point tolerates small disturbances without allowing the system to depart dramatically.
An asymptotically stable operating point goes further by returning toward its original state after the disturbance.
Rate of Return Matters Too
Two equilibria can both be asymptotically stable while returning at very different speeds.
Compare
\[ x'=-x \]with
\[ x'=-10x. \]Both equilibria at zero are asymptotically stable, but
\[ e^{-10t} \]decays much faster than
\[ e^{-t}. \]Stability Margin
In engineering and numerical work, one often cares not only whether a system is stable but how far it is from a stability boundary.
For continuous linear systems, eigenvalues far into the left half-plane typically indicate stronger exponential decay.
For discrete systems, eigenvalues with magnitudes well below \(1\) indicate stronger contraction per iteration.
Common Error: Calling Every Equilibrium Stable
Equilibrium means
\[ \text{no motion when exactly at the state}. \]Stability means
\[ \text{small nearby disturbances do not produce large departures}. \]These are different conditions and must not be conflated.
Common Error: Confusing Stability with Attraction
Stability requires nearby trajectories to remain nearby.
Attraction requires them to converge.
Asymptotic stability combines the two ideas.
A system may display bounded nearby motion without convergence.
Common Error: Using the Continuous Test for a Discrete System
In continuous time, a one-dimensional equilibrium is typically stable when
\[ f'(x^\ast)<0. \]In discrete time, the corresponding criterion is
\[ |f'(x^\ast)|<1. \]The two tests are related through linearization, but they are not the same.
Common Error: Ignoring the Absolute Value in Discrete Stability
For discrete systems, a multiplier of
\[ -2 \]is unstable even though it is negative.
What matters is
\[ |-2|=2>1. \]The negative sign causes alternation, while the magnitude causes growth.
Common Error: Treating an Inconclusive Test as Proof of Stability
If
\[ f'(x^\ast)=0 \]in a continuous system, or
\[ |f'(x^\ast)|=1 \]in a discrete system, the basic derivative test does not classify the equilibrium.
Additional sign analysis, higher-order terms, Lyapunov methods, or other tools may be required.
Common Error: Looking at Only One Direction in Several Dimensions
A multidimensional equilibrium may attract along one direction and repel along another.
A saddle is therefore unstable even though some trajectories approach it.
Stability must account for all sufficiently small perturbation directions.
A Reliable Procedure for One-Dimensional Continuous Stability
- Find the equilibria by solving \(f(x^\ast)=0\).
- Determine the sign of \(f(x)\) on intervals between equilibria.
- Draw the phase-line arrows.
- Identify inward, outward, or one-sided behavior.
- Compute \(f'(x^\ast)\) when useful.
- Use \(f'(x^\ast)<0\) for local asymptotic stability.
- Use \(f'(x^\ast)>0\) for instability.
- If \(f'(x^\ast)=0\), use additional analysis.
A Reliable Procedure for One-Dimensional Discrete Stability
- Find fixed points from \(f(x^\ast)=x^\ast\).
- Differentiate the update map \(f\).
- Evaluate \(f'(x^\ast)\).
- If \(|f'(x^\ast)|<1\), classify the fixed point as locally asymptotically stable.
- If \(|f'(x^\ast)|>1\), classify it as unstable.
- Use the sign of \(f'(x^\ast)\) to anticipate alternating or nonalternating behavior.
- If \(|f'(x^\ast)|=1\), perform further analysis.
A Reliable Procedure for Linear Systems
- Write the system in matrix form.
- Locate the equilibrium.
- Compute the eigenvalues of the system matrix.
- For continuous systems, inspect the real parts of the eigenvalues.
- For discrete systems, inspect the eigenvalue magnitudes.
- Identify stable, unstable, and mixed directions.
- Relate the algebraic classification to the geometry of trajectories.
Practice 1: Continuous Stable Equilibrium
Classify the equilibrium of
\[ \frac{dx}{dt}=-4(x-2). \]Practice 2: Continuous Unstable Equilibrium
Classify the equilibrium of
\[ \frac{dx}{dt}=3(x+1). \]Practice 3: Derivative Test
Consider
\[ \frac{dx}{dt}=x-x^3. \]Classify the equilibrium
\[ x^\ast=1. \]Practice 4: Discrete Stable Fixed Point
Consider
\[ x_{n+1}=0.25x_n+6. \]Find and classify the fixed point.
Practice 5: Discrete Unstable Fixed Point
Consider
\[ x_{n+1}=1.5x_n-2. \]Find and classify the fixed point.
Practice 6: Alternating Convergence
Consider
\[ x_{n+1}=-0.4x_n. \]Classify the fixed point \(x^\ast=0\) and describe the qualitative form of convergence.
Practice 7: Linear Continuous System
Classify the origin for
\[ \mathbf{x}' = \begin{pmatrix} -2 & 0\\ 0 & -1 \end{pmatrix} \mathbf{x}. \]Practice 8: Saddle
Classify the origin for
\[ \mathbf{x}' = \begin{pmatrix} -3 & 0\\ 0 & 2 \end{pmatrix} \mathbf{x}. \]Practice 9: Discrete Matrix Stability
Classify the origin for
\[ \mathbf{x}_{n+1} = \begin{pmatrix} 0.8 & 0\\ 0 & 0.3 \end{pmatrix} \mathbf{x}_n. \]Practice 10: Inconclusive Test
Consider
\[ \frac{dx}{dt}=-x^3. \]Explain why the derivative test is inconclusive at \(x^\ast=0\), and use sign analysis to classify the equilibrium.
Practice Answers
Practice 1
The equilibrium is
\[ x^\ast=2. \]Here
\[ f'(x)=-4<0. \]Therefore,
\[ \boxed{x^\ast=2\text{ is locally asymptotically stable}}. \]Practice 2
The equilibrium is
\[ x^\ast=-1. \]Since
\[ f'(x)=3>0, \]the equilibrium is
\[ \boxed{\text{unstable}}. \]Practice 3
For
\[ f(x)=x-x^3, \]we have
\[ f'(x)=1-3x^2. \]At \(x^\ast=1\),
\[ f'(1)=-2<0. \]Therefore,
\[ \boxed{x^\ast=1\text{ is locally asymptotically stable}}. \]Practice 4
The fixed point satisfies
\[ x^\ast=0.25x^\ast+6. \]Therefore,
\[ 0.75x^\ast=6 \]and
\[ x^\ast=8. \]Since
\[ |f'(x^\ast)|=0.25<1, \]the fixed point is
\[ \boxed{\text{locally asymptotically stable}}. \]Practice 5
The fixed point satisfies
\[ x^\ast=1.5x^\ast-2. \]Thus
\[ 0.5x^\ast=2 \]and
\[ x^\ast=4. \]Since
\[ |f'(4)|=1.5>1, \]the fixed point is
\[ \boxed{\text{unstable}}. \]Practice 6
The multiplier is
\[ -0.4. \]Since
\[ |-0.4|<1, \]the origin is locally asymptotically stable.
Because the multiplier is negative, successive iterates alternate sides of the origin while their magnitudes shrink.
Practice 7
The eigenvalues are
\[ -2 \qquad\text{and}\qquad -1. \]Both have negative real part.
Therefore,
\[ \boxed{\text{the origin is asymptotically stable}}. \]Practice 8
The eigenvalues are
\[ -3 \qquad\text{and}\qquad 2. \]One direction contracts and one expands.
Therefore the origin is a
\[ \boxed{\text{saddle and is unstable}}. \]Practice 9
The eigenvalues are
\[ 0.8 \qquad\text{and}\qquad 0.3. \]Both have magnitude less than \(1\).
Therefore,
\[ \boxed{\text{the origin is asymptotically stable}}. \]Practice 10
Here
\[ f(x)=-x^3, \]so
\[ f'(x)=-3x^2 \]and
\[ f'(0)=0. \]The derivative test is therefore inconclusive.
If \(x>0\), then
\[ -x^3<0, \]so the state moves left toward \(0\).
If \(x<0\), then
\[ -x^3>0, \]so the state moves right toward \(0\).
Thus
\[ \boxed{x^\ast=0\text{ is asymptotically stable}}. \]What to Remember
Stability describes what happens to trajectories that begin near an equilibrium.
The key distinctions are:
- stable: nearby trajectories remain nearby;
- attracting: nearby trajectories converge to the equilibrium;
- asymptotically stable: both stable and attracting;
- unstable: sufficiently small disturbances can lead away;
- saddle: attracting in some directions and repelling in others.
For one-dimensional continuous systems,
\[ \boxed{ f'(x^\ast)<0 \Rightarrow \text{local asymptotic stability} } \]and
\[ \boxed{ f'(x^\ast)>0 \Rightarrow \text{instability}. } \]For one-dimensional discrete systems,
\[ \boxed{ |f'(x^\ast)|<1 \Rightarrow \text{local asymptotic stability} } \]and
\[ \boxed{ |f'(x^\ast)|>1 \Rightarrow \text{instability}. } \]In higher dimensions, eigenvalues, invariant directions, and Lyapunov methods extend these ideas.
Where We Go Next
Stability tells us how trajectories behave near equilibria, but a complete dynamical picture requires seeing many trajectories at once.
Instead of studying one solution in isolation, we can draw the directional structure of an entire state space and show how different initial conditions move through it.
The next lesson, Phase Portraits, develops this geometric representation and uses it to visualize equilibria, stable and unstable directions, cycles, separatrices, and the overall organization of a dynamical system.