The Central Idea
A fixed point or equilibrium is a state that remains unchanged under the evolution rule.
For a discrete system,
\[ x_{n+1}=f(x_n), \]an equilibrium \(x^\ast\) satisfies
\[ \boxed{f(x^\ast)=x^\ast}. \]For a continuous system,
\[ \frac{dx}{dt}=f(x), \]an equilibrium \(x^\ast\) satisfies
\[ \boxed{f(x^\ast)=0}. \]In either case, beginning exactly at the equilibrium means remaining there.
Why Equilibria Matter
Dynamical systems often have complicated trajectories, but equilibria give us natural reference states around which that behavior can be organized.
We can ask:
- how many equilibria exist;
- where they are located;
- whether nearby trajectories move toward or away from them;
- whether parameter changes create or destroy equilibria;
- whether an equilibrium represents a meaningful steady operating condition.
These questions form a major part of qualitative dynamical analysis.
Fixed Points in Discrete Systems
Consider the discrete system
\[ x_{n+1}=f(x_n). \]If \(x^\ast\) is a fixed point, then
\[ f(x^\ast)=x^\ast. \]Therefore,
\[ x_0=x^\ast \]gives
\[ x_1=x^\ast,\qquad x_2=x^\ast,\qquad x_3=x^\ast,\ldots \]The entire orbit remains at one state.
Example: A Linear Discrete Fixed Point
Consider
\[ x_{n+1}=0.8x_n+6. \]A fixed point satisfies
\[ x^\ast=0.8x^\ast+6. \]Subtract \(0.8x^\ast\):
\[ 0.2x^\ast=6. \]Therefore,
\[ \boxed{x^\ast=30}. \]Check:
\[ 0.8(30)+6=30. \]The General Affine Fixed Point
Consider the discrete affine system
\[ x_{n+1}=ax_n+b. \]A fixed point satisfies
\[ x^\ast=ax^\ast+b. \]Therefore,
\[ (1-a)x^\ast=b. \]If
\[ a\ne1, \]then
\[ \boxed{ x^\ast=\frac{b}{1-a} }. \]Special Case: \(a=1\)
If
\[ x_{n+1}=x_n+b, \]the fixed-point equation becomes
\[ x^\ast=x^\ast+b. \]If \(b\ne0\), there is no fixed point.
If \(b=0\), then
\[ x_{n+1}=x_n, \]so every state is fixed.
This illustrates that a system may have no equilibrium, one equilibrium, several equilibria, or infinitely many.
Graphical Interpretation of a Discrete Fixed Point
A fixed point of
\[ x_{n+1}=f(x_n) \]satisfies
\[ f(x)=x. \]Graphically, fixed points occur where the curve
\[ y=f(x) \]intersects the diagonal line
\[ y=x. \]Each intersection corresponds to a state that maps back to itself.
Example: Several Discrete Fixed Points
Consider
\[ x_{n+1}=x_n^2. \]Fixed points satisfy
\[ x^\ast=(x^\ast)^2. \]Rearrange:
\[ (x^\ast)^2-x^\ast=0. \]Factor:
\[ x^\ast(x^\ast-1)=0. \]Therefore,
\[ \boxed{x^\ast=0\quad\text{or}\quad x^\ast=1}. \]The Logistic Map
The logistic map is
\[ x_{n+1}=rx_n(1-x_n). \]Its fixed points satisfy
\[ x^\ast=rx^\ast(1-x^\ast). \]Move everything to one side:
\[ rx^\ast(1-x^\ast)-x^\ast=0. \]Factor \(x^\ast\):
\[ x^\ast[r(1-x^\ast)-1]=0. \]Fixed Points of the Logistic Map
The first fixed point is
\[ \boxed{x^\ast=0}. \]The second comes from
\[ r(1-x^\ast)-1=0. \]Solve:
\[ r-rx^\ast=1, \] \[ rx^\ast=r-1, \] \[ \boxed{ x^\ast=1-\frac1r }. \]The location of this second fixed point depends on the parameter \(r\).
Continuous Equilibria
For a continuous autonomous system
\[ \frac{dx}{dt}=f(x), \]an equilibrium satisfies
\[ \boxed{f(x^\ast)=0}. \]At such a state,
\[ \frac{dx}{dt}=0, \]so the system has no instantaneous motion away from the state.
Example: A Continuous Linear Equilibrium
Consider
\[ \frac{dx}{dt}=6-2x. \]Set the right side equal to zero:
\[ 6-2x^\ast=0. \]Therefore,
\[ 2x^\ast=6 \]and
\[ \boxed{x^\ast=3}. \]If the system begins at \(x=3\), then its derivative remains zero.
Equilibrium as a Steady State
The term steady state is often used when an equilibrium represents a condition in which the modeled quantities remain constant.
For example, if
\[ \frac{dT}{dt}=k(T_{\mathrm{env}}-T), \]then equilibrium occurs when
\[ T=T_{\mathrm{env}}. \]At that point, the modeled temperature no longer changes.
Multiple Continuous Equilibria
Consider
\[ \frac{dx}{dt}=x(1-x). \]Equilibria satisfy
\[ x(1-x)=0. \]Therefore,
\[ \boxed{x^\ast=0,\qquad x^\ast=1}. \]The same system can therefore have more than one steady state.
Equilibria from Factored Forms
Factored evolution laws make equilibria especially easy to identify.
Suppose
\[ \frac{dx}{dt}=(x+2)(x-1)(x-4). \]Then equilibrium requires
\[ (x+2)(x-1)(x-4)=0. \]Therefore,
\[ \boxed{x^\ast=-2,\quad1,\quad4}. \]Each zero of the rate function is an equilibrium.
Direction of Motion Between Equilibria
Once equilibria are found in a one-dimensional continuous system, the sign of
\[ f(x) \]tells us the direction of motion between them.
If
\[ f(x)>0, \]then
\[ \frac{dx}{dt}>0, \]so \(x\) increases.
If
\[ f(x)<0, \]then \(x\) decreases.
Example: Sign Analysis
Consider
\[ \frac{dx}{dt}=x(1-x). \]The equilibria are \(0\) and \(1\).
If
\[ x<0, \]then \(x(1-x)<0\), so the system moves left.
If
\[ 0<x<1, \]then \(x(1-x)>0\), so the system moves right.
If
\[ x>1, \]then \(x(1-x)<0\), so the system moves left.
Phase-Line Representation
In one-dimensional continuous systems, the sign information can be summarized on a phase line.
For
\[ \frac{dx}{dt}=x(1-x), \]the directional pattern is
\[ \leftarrow\quad 0\quad \rightarrow\quad 1\quad \leftarrow. \]This compact diagram shows how trajectories move relative to the equilibria.
Equilibrium Does Not Automatically Mean Stable
A state can be an equilibrium without attracting nearby trajectories.
The equation
\[ f(x^\ast)=0 \]tells us only that the state itself does not move.
It does not tell us what happens to a nearby state.
That is the central distinction between equilibrium and stability.
A First Preview of Stable Equilibrium
Consider
\[ \frac{dx}{dt}=-(x-2). \]The equilibrium is
\[ x^\ast=2. \]If \(x>2\), then
\[ \frac{dx}{dt}<0, \]so the state moves downward toward \(2\).
If \(x<2\), then
\[ \frac{dx}{dt}>0, \]so the state moves upward toward \(2\).
A First Preview of Unstable Equilibrium
Now consider
\[ \frac{dx}{dt}=x-2. \]The equilibrium is still
\[ x^\ast=2. \]But if \(x>2\), the derivative is positive and the state moves farther right.
If \(x<2\), the derivative is negative and the state moves farther left.
Nearby trajectories move away from the equilibrium.
Fixed Points in Several Dimensions
For a discrete vector system
\[ \mathbf{x}_{n+1} = \mathbf{F}(\mathbf{x}_n), \]a fixed point \(\mathbf{x}^\ast\) satisfies
\[ \boxed{ \mathbf{F}(\mathbf{x}^\ast) = \mathbf{x}^\ast }. \]Every component of the state must reproduce itself after one update.
Example: Two-Dimensional Discrete Fixed Point
Consider
\[ \begin{aligned} x_{n+1}&=\frac12x_n+2,\\ y_{n+1}&=\frac14y_n+3. \end{aligned} \]At a fixed point,
\[ x^\ast=\frac12x^\ast+2, \] \[ y^\ast=\frac14y^\ast+3. \]Solving gives
\[ x^\ast=4, \qquad y^\ast=4. \]Therefore,
\[ \boxed{ \mathbf{x}^\ast= \begin{pmatrix} 4\\ 4 \end{pmatrix} }. \]Equilibria in Continuous Vector Systems
For
\[ \frac{d\mathbf{x}}{dt} = \mathbf{F}(\mathbf{x}), \]an equilibrium satisfies
\[ \boxed{ \mathbf{F}(\mathbf{x}^\ast)=\mathbf{0} }. \]Every component of the state-space velocity must be zero.
Example: Two-Dimensional Continuous Equilibrium
Consider
\[ \frac{dx}{dt}=4-x, \] \[ \frac{dy}{dt}=6-2y. \]At equilibrium,
\[ 4-x^\ast=0, \] \[ 6-2y^\ast=0. \]Therefore,
\[ x^\ast=4, \qquad y^\ast=3. \]The equilibrium state is
\[ \boxed{(4,3)}. \]Coupled Equilibrium Equations
In interacting systems, equilibrium equations may be coupled.
Consider
\[ \frac{dx}{dt}=x-y, \] \[ \frac{dy}{dt}=x+y-2. \]Equilibrium requires
\[ x-y=0 \]and
\[ x+y-2=0. \]The equilibrium problem becomes a system of algebraic equations.
Solving the Coupled Equilibrium
From
\[ x-y=0, \]we obtain
\[ x=y. \]Substitute into
\[ x+y-2=0: \] \[ 2x-2=0. \]Therefore,
\[ x=1,\qquad y=1. \]Hence
\[ \boxed{(x^\ast,y^\ast)=(1,1)}. \]Nullclines
In a two-dimensional continuous system
\[ \frac{dx}{dt}=f(x,y), \qquad \frac{dy}{dt}=g(x,y), \]an \(x\)-nullcline is a curve where
\[ f(x,y)=0. \]A \(y\)-nullcline is a curve where
\[ g(x,y)=0. \]Equilibria occur where the relevant nullclines intersect.
Example: Finding Nullclines
Consider
\[ \frac{dx}{dt}=x(2-y), \] \[ \frac{dy}{dt}=y(x-1). \]The \(x\)-nullclines satisfy
\[ x(2-y)=0, \]so
\[ x=0 \qquad\text{or}\qquad y=2. \]The \(y\)-nullclines satisfy
\[ y(x-1)=0, \]so
\[ y=0 \qquad\text{or}\qquad x=1. \]Equilibria from Nullcline Intersections
To be an equilibrium, both derivatives must vanish simultaneously.
From the nullclines
\[ x=0,\qquad y=2, \]and
\[ y=0,\qquad x=1, \]the simultaneous intersections are
\[ \boxed{(0,0)} \]and
\[ \boxed{(1,2)}. \]Equilibrium Sets
Not every system has isolated equilibrium points.
Consider
\[ \frac{dx}{dt}=0, \] \[ \frac{dy}{dt}=-y. \]Equilibrium requires
\[ y=0. \]But \(x\) can be any real number.
Therefore the entire line
\[ \boxed{y=0} \]consists of equilibria.
Isolated versus Nonisolated Equilibria
An equilibrium is isolated if there is a neighborhood around it containing no other equilibria.
An equilibrium lying on a continuous curve or surface of equilibrium states is nonisolated.
This distinction matters because some standard stability and linearization methods are simplest when the equilibrium is isolated.
Equilibrium and Conservation
A conservation law can constrain where equilibria occur.
Suppose a system preserves
\[ x+y=M. \]Even if the algebraic equilibrium equations admit several solutions, only those lying on the conserved set corresponding to the initial condition may be dynamically accessible.
Equilibrium analysis should therefore be interpreted within the actual state space and model constraints.
Physical Meaning of an Equilibrium
The mathematical condition for equilibrium is precise, but its interpretation depends on the application.
An equilibrium may represent:
- a constant population;
- a steady temperature;
- a stationary mechanical configuration;
- a steady chemical concentration;
- a constant market or inventory level;
- a balanced flow in a network;
- a controlled operating point.
Equilibrium in a Population Model
Consider the continuous logistic equation
\[ \frac{dP}{dt} = rP\left(1-\frac{P}{K}\right). \]Equilibrium requires
\[ rP\left(1-\frac{P}{K}\right)=0. \]Assuming \(r\ne0\), the equilibria are
\[ \boxed{P^\ast=0} \]and
\[ \boxed{P^\ast=K}. \]Interpreting Logistic Equilibria
The equilibrium
\[ P^\ast=0 \]represents absence of the modeled population.
The equilibrium
\[ P^\ast=K \]represents the carrying-capacity level.
Which equilibrium is approached depends on the dynamics and the initial condition.
Equilibrium in Newton's Cooling Model
Consider
\[ \frac{dT}{dt} = -k(T-T_a), \]where \(T_a\) is the ambient temperature.
Set the derivative equal to zero:
\[ -k(T^\ast-T_a)=0. \]Assuming \(k\ne0\),
\[ \boxed{T^\ast=T_a}. \]The equilibrium occurs when the object and environment have the same temperature.
Equilibrium in a Simple Inventory Model
Suppose inventory obeys
\[ I_{n+1}=0.9I_n+20. \]A fixed inventory level satisfies
\[ I^\ast=0.9I^\ast+20. \]Thus
\[ 0.1I^\ast=20 \]and
\[ \boxed{I^\ast=200}. \]At this level, the modeled losses and additions balance from one period to the next.
Equilibria Depend on Parameters
In parameterized systems, the number and location of equilibria may change as a parameter changes.
Consider
\[ \frac{dx}{dt}=\mu-x^2. \]Equilibrium requires
\[ x^2=\mu. \]Therefore the number of real equilibria depends on the sign of \(\mu\).
Parameter Cases
For
\[ \frac{dx}{dt}=\mu-x^2, \]we have:
- if \(\mu>0\), there are two equilibria \(x^\ast=\pm\sqrt{\mu}\);
- if \(\mu=0\), there is one equilibrium \(x^\ast=0\);
- if \(\mu<0\), there are no real equilibria.
This kind of parameter-dependent change is a first preview of a bifurcation.
Equilibrium Branches
When an equilibrium depends on a parameter, we can think of its location as a function of that parameter.
For example,
\[ x^\ast=\sqrt{\mu} \]and
\[ x^\ast=-\sqrt{\mu} \]define two equilibrium branches for
\[ \mu\ge0. \]Later, bifurcation diagrams will plot such branches and indicate their stability.
Equilibrium versus Periodic Motion
An equilibrium remains at one state.
A periodic orbit returns to a previous state after some positive amount of time or number of iterations.
Thus an equilibrium is stationary, while a nontrivial periodic orbit moves through multiple states.
In discrete time, a fixed point can be viewed as a period-one orbit.
Equilibrium versus Turning Point
A turning point in one plotted variable is not necessarily an equilibrium of the full system.
Suppose a mechanical system has state
\[ (x,v). \]At a turning point, the velocity may satisfy
\[ v=0, \]while the acceleration is nonzero.
The object momentarily stops moving in position, but the full state is not necessarily an equilibrium.
Example: Turning Point Is Not Equilibrium
For the harmonic oscillator
\[ \frac{dx}{dt}=v, \qquad \frac{dv}{dt}=-x, \]consider the state
\[ (x,v)=(1,0). \]Then
\[ \frac{dx}{dt}=0, \]but
\[ \frac{dv}{dt}=-1. \]The state-space velocity is not zero, so \((1,0)\) is not an equilibrium.
Mechanical Equilibrium
For the same harmonic oscillator,
\[ \frac{dx}{dt}=v, \qquad \frac{dv}{dt}=-x, \]equilibrium requires
\[ v=0 \]and
\[ x=0. \]Therefore,
\[ \boxed{(x^\ast,v^\ast)=(0,0)}. \]Equilibria and State-Space Geometry
In state space, an equilibrium is represented by a stationary point.
Other trajectories may:
- approach that point;
- move away from it;
- pass nearby and continue elsewhere;
- circle around it;
- approach it only from certain directions.
These patterns provide the geometric foundation of stability theory.
Equilibrium and Long-Term Behavior
One of the most important questions is whether
\[ \mathbf{x}(t)\to\mathbf{x}^\ast \]as
\[ t\to\infty, \]or, in discrete time,
\[ \mathbf{x}_n\to\mathbf{x}^\ast \]as
\[ n\to\infty. \]Convergence toward an equilibrium is one possible form of long-term behavior, but it is not the only one.
Checking a Candidate Equilibrium
A proposed equilibrium should always be substituted back into the original evolution rule.
For the discrete map
\[ x_{n+1}=0.75x_n+5, \]suppose we claim
\[ x^\ast=20. \]Check:
\[ 0.75(20)+5=15+5=20. \]The state reproduces itself, so the fixed point is correct.
Checking a Continuous Equilibrium
Consider
\[ \frac{dx}{dt}=x^2-9. \]A candidate equilibrium is
\[ x^\ast=3. \]Substitute:
\[ 3^2-9=0. \]Therefore \(x=3\) is an equilibrium.
The second equilibrium
\[ x^\ast=-3 \]should be checked the same way.
Common Error: Using the Wrong Equilibrium Condition
For a discrete system
\[ x_{n+1}=f(x_n), \]the correct fixed-point equation is
\[ \boxed{f(x^\ast)=x^\ast}, \]not necessarily
\[ f(x^\ast)=0. \]The zero condition belongs to continuous systems of the form
\[ \frac{dx}{dt}=f(x). \]Common Error: Solving Only One Equation in a System
In a multidimensional system, every state derivative or update equation must satisfy the equilibrium condition simultaneously.
For
\[ \frac{dx}{dt}=f(x,y), \qquad \frac{dy}{dt}=g(x,y), \]equilibrium requires
\[ f(x,y)=0 \]and
\[ g(x,y)=0. \]Satisfying only one equation places the state on a nullcline, not necessarily at an equilibrium.
Common Error: Ignoring Model Constraints
Algebra may produce equilibria that lie outside the meaningful state space.
A population model may mathematically yield
\[ P^\ast=-50, \]but if the state space requires
\[ P\ge0, \]then that equilibrium is not physically admissible for the intended model.
Common Error: Assuming Every Equilibrium Is Attracting
The fact that a state satisfies the equilibrium equation tells us only that it remains stationary when the system begins exactly there.
Nearby trajectories may approach it, move away from it, or behave differently in different directions.
Stability must be analyzed separately.
Common Error: Confusing Zero Velocity in One Coordinate with Equilibrium
In a multidimensional system, one derivative may vanish while others do not.
If
\[ \frac{dx}{dt}=0 \]but
\[ \frac{dy}{dt}\ne0, \]the full state is still moving through state space.
Equilibrium requires the entire state-space velocity to be zero.
A Reliable Procedure for Discrete Fixed Points
- Write the update equation \(x_{n+1}=f(x_n)\).
- Replace both \(x_n\) and \(x_{n+1}\) by \(x^\ast\).
- Solve \(x^\ast=f(x^\ast)\).
- Find every solution permitted by the state space.
- Substitute each candidate back into the update rule.
- Interpret the result in the context of the model.
- Analyze nearby trajectories separately to determine stability.
A Reliable Procedure for Continuous Equilibria
- Write the differential equation or system.
- Set every state derivative equal to zero.
- Solve the resulting algebraic equation or system.
- Apply state-space and physical constraints.
- Check each candidate in the original equations.
- Use sign analysis or phase-space methods to examine surrounding motion.
- Proceed to stability analysis when needed.
Practice 1: Discrete Fixed Point
Find the fixed point of
\[ x_{n+1}=0.6x_n+8. \]Practice 2: Two Discrete Fixed Points
Find all fixed points of
\[ x_{n+1}=x_n^2-2x_n+2. \]Practice 3: Continuous Equilibrium
Find the equilibrium of
\[ \frac{dx}{dt}=12-3x. \]Practice 4: Multiple Continuous Equilibria
Find all equilibria of
\[ \frac{dx}{dt}=x(x-2)(x+3). \]Practice 5: Logistic Equilibria
Find the equilibria of
\[ \frac{dP}{dt} = rP\left(1-\frac{P}{K}\right), \]assuming
\[ r\ne0,\qquad K>0. \]Practice 6: Two-Dimensional Equilibrium
Find the equilibrium of
\[ \frac{dx}{dt}=5-x, \] \[ \frac{dy}{dt}=4-2y. \]Practice 7: Coupled System
Find the equilibrium of
\[ \frac{dx}{dt}=x+y-4, \] \[ \frac{dy}{dt}=x-y. \]Practice 8: Nullclines
For
\[ \frac{dx}{dt}=x(3-y), \] \[ \frac{dy}{dt}=y(x-2), \]find the \(x\)-nullclines and \(y\)-nullclines.
Practice 9: Parameter-Dependent Equilibria
Consider
\[ \frac{dx}{dt}=\mu-x^2. \]How many real equilibria exist when
\[ \mu=9,\qquad \mu=0,\qquad \mu=-1? \]Practice 10: Identify the Error
A student studies
\[ x_{n+1}=x_n+4 \]and sets
\[ x_n+4=0 \]to find a fixed point.
Explain the error and determine whether the system has a fixed point.
Practice Answers
Practice 1
Set
\[ x^\ast=0.6x^\ast+8. \]Then
\[ 0.4x^\ast=8, \]so
\[ \boxed{x^\ast=20}. \]Practice 2
Set
\[ x^\ast=(x^\ast)^2-2x^\ast+2. \]Rearranging gives
\[ (x^\ast)^2-3x^\ast+2=0. \]Factor:
\[ (x^\ast-1)(x^\ast-2)=0. \]Therefore,
\[ \boxed{x^\ast=1,\qquad x^\ast=2}. \]Practice 3
Set
\[ 12-3x^\ast=0. \]Therefore,
\[ \boxed{x^\ast=4}. \]Practice 4
Set
\[ x(x-2)(x+3)=0. \]Therefore,
\[ \boxed{x^\ast=-3,\quad0,\quad2}. \]Practice 5
Set
\[ rP\left(1-\frac{P}{K}\right)=0. \]Since \(r\ne0\),
\[ P=0 \]or
\[ 1-\frac{P}{K}=0. \]Hence
\[ \boxed{P^\ast=0,\qquad P^\ast=K}. \]Practice 6
Set both derivatives equal to zero:
\[ 5-x^\ast=0, \] \[ 4-2y^\ast=0. \]Thus
\[ \boxed{(x^\ast,y^\ast)=(5,2)}. \]Practice 7
Solve
\[ x+y=4 \]and
\[ x-y=0. \]Adding the equations gives
\[ 2x=4, \]so
\[ x=2. \]Then \(y=2\).
Therefore,
\[ \boxed{(2,2)}. \]Practice 8
The \(x\)-nullclines satisfy
\[ x(3-y)=0, \]so
\[ \boxed{x=0\quad\text{or}\quad y=3}. \]The \(y\)-nullclines satisfy
\[ y(x-2)=0, \]so
\[ \boxed{y=0\quad\text{or}\quad x=2}. \]Practice 9
Equilibrium requires
\[ x^2=\mu. \]If
\[ \mu=9, \]there are two equilibria:
\[ x^\ast=\pm3. \]If
\[ \mu=0, \]there is one equilibrium:
\[ x^\ast=0. \]If
\[ \mu=-1, \]there are no real equilibria.
Practice 10
For a discrete system, a fixed point satisfies
\[ x^\ast=f(x^\ast), \]not
\[ f(x^\ast)=0. \]Here,
\[ x^\ast=x^\ast+4 \]would be required.
This simplifies to
\[ 0=4, \]which is impossible.
Therefore,
\[ \boxed{\text{the system has no fixed point}}. \]What to Remember
A fixed point or equilibrium is a state that remains unchanged under the dynamical law.
In discrete time,
\[ \boxed{f(x^\ast)=x^\ast}. \]In continuous time,
\[ \boxed{f(x^\ast)=0}. \]For multidimensional systems,
\[ \boxed{ \mathbf{F}(\mathbf{x}^\ast) = \mathbf{x}^\ast } \]in discrete time, and
\[ \boxed{ \mathbf{F}(\mathbf{x}^\ast)=\mathbf{0} } \]in continuous time.
- equilibria may be absent, unique, multiple, or continuous sets;
- nullcline intersections help locate equilibria in two-dimensional systems;
- parameters may move, create, or destroy equilibria;
- model constraints determine which algebraic equilibria are admissible;
- an equilibrium is not automatically stable;
- the behavior of nearby trajectories must be studied separately.
Equilibria are stationary states, but their true dynamical significance comes from how the surrounding system behaves.
Where We Go Next
Finding an equilibrium answers one question: where can the system remain unchanged?
The next question is more important for prediction and control: what happens if the system is disturbed slightly away from that state?
Some equilibria pull nearby trajectories back toward them. Others push nearby trajectories away. Still others behave differently depending on the direction of the disturbance.
The next lesson, Stability and Instability, develops the mathematical tools used to distinguish these possibilities.