The Central Idea
Consider the autonomous system
\[ \boxed{ \begin{aligned} x'&=f(x,y),\\ y'&=g(x,y). \end{aligned} } \]At every state \((x,y)\), the system determines a velocity vector
\[ \boxed{ \langle f(x,y),g(x,y)\rangle }. \]The phase plane displays these vectors and the trajectories that follow them.
What Is the Phase Plane?
The phase plane is the coordinate plane whose axes represent two state variables.
A point
\[ (x,y) \]represents one possible state of the system.
As time changes, the state moves through the plane.
Time Is Usually Not an Axis
In an ordinary graph of \(x(t)\), time appears on the horizontal axis.
In a phase-plane graph, the axes are state variables:
\[ x \]and
\[ y. \]Time determines how the point moves along the trajectory, but time is not usually shown as an axis.
Trajectories
A solution
\[ x=x(t), \qquad y=y(t) \]determines a parametric curve
\[ \boxed{ (x(t),y(t)) }. \]This curve is called a trajectory, orbit, or phase curve.
Direction of Motion Matters
A phase trajectory is not merely a geometric curve.
The system also determines the direction in which the state moves along the curve.
Arrows are therefore essential when interpreting a phase portrait.
Vector Fields
The system
\[ \begin{aligned} x'&=f(x,y),\\ y'&=g(x,y) \end{aligned} \]generates the vector field
\[ \boxed{ \mathbf{F}(x,y) = \langle f(x,y),g(x,y)\rangle }. \]The arrows of this vector field show the instantaneous direction of motion.
Reading the Direction Vector
Suppose at some point
\[ x'>0, \qquad y'>0. \]The state moves right and up.
If
\[ x'<0, \qquad y'>0, \]it moves left and up.
The signs of the two derivatives therefore determine a local directional quadrant.
Example: Local Direction
Consider
\[ \begin{aligned} x'&=x+y,\\ y'&=x-y. \end{aligned} \]At the point
\[ (1,2), \]we have
\[ x'=3 \]and
\[ y'=-1. \]The direction vector is
\[ \boxed{\langle3,-1\rangle}. \]The state moves primarily to the right and slightly downward.
Equilibrium Points
An equilibrium occurs where the state does not move.
Therefore,
\[ \boxed{ f(x,y)=0, \qquad g(x,y)=0 }. \]At an equilibrium,
\[ x'=0, \qquad y'=0. \]Example: Find the Equilibrium
For
\[ \begin{aligned} x'&=2x+y,\\ y'&=x-3y, \end{aligned} \]set
\[ 2x+y=0 \]and
\[ x-3y=0. \]The only simultaneous solution is
\[ \boxed{(0,0)}. \]Nullclines
Nullclines identify where one component of motion vanishes.
The \(x\)-nullcline satisfies
\[ \boxed{f(x,y)=0}. \]Along it,
\[ x'=0. \]The \(y\)-nullcline satisfies
\[ \boxed{g(x,y)=0}. \]Along it,
\[ y'=0. \]Geometric Meaning of Nullclines
Along an \(x\)-nullcline, horizontal motion stops instantaneously.
Therefore, the local vector is vertical.
Along a \(y\)-nullcline, vertical motion stops instantaneously.
Therefore, the local vector is horizontal.
Example: Find the Nullclines
For
\[ \begin{aligned} x'&=x+y,\\ y'&=x-y, \end{aligned} \]the \(x\)-nullcline is
\[ x+y=0, \]or
\[ \boxed{y=-x}. \]The \(y\)-nullcline is
\[ x-y=0, \]or
\[ \boxed{y=x}. \]Using Nullclines to Divide the Plane
Nullclines often divide the phase plane into regions where the signs of \(x'\) and \(y'\) are constant.
In each region, we can determine whether trajectories move:
- right or left;
- up or down.
This creates a qualitative sketch of the vector field.
Phase Portraits
A phase portrait combines:
- equilibrium points;
- direction arrows;
- nullclines;
- representative trajectories.
It summarizes the qualitative behavior of an entire family of solutions.
Linear Systems
For a linear system
\[ \boxed{ \mathbf{x}'=A\mathbf{x} }, \]the eigenvalues and eigenvectors of \(A\) largely determine the phase portrait.
Eigenvectors provide special directions, while eigenvalues determine growth, decay, or oscillatory behavior along those modes.
Eigenvector Trajectories
If
\[ A\mathbf{v}=\lambda\mathbf{v}, \]then
\[ \boxed{ \mathbf{x}(t)=C\mathbf{v}e^{\lambda t} } \]is a solution.
Its trajectory lies along the line spanned by the eigenvector \(\mathbf{v}\).
Stable Node
Suppose a linear system has two distinct negative real eigenvalues.
Then both exponential modes decay.
Trajectories move toward the equilibrium.
The origin is called a
\[ \boxed{\text{stable node}}. \]Example: Stable Node
Consider
\[ \begin{aligned} x'&=-2x,\\ y'&=-5y. \end{aligned} \]The solutions are
\[ x=C_1e^{-2t} \]and
\[ y=C_2e^{-5t}. \]Both approach zero, so every trajectory approaches
\[ \boxed{(0,0)}. \]Unstable Node
If both eigenvalues are positive real numbers, both modes grow.
Trajectories move away from the equilibrium.
The equilibrium is an
\[ \boxed{\text{unstable node}}. \]Saddle Point
If one eigenvalue is positive and the other is negative, then one mode grows while the other decays.
The equilibrium is a
\[ \boxed{\text{saddle point}}. \]A saddle is unstable overall even though some trajectories may approach it along a special direction.
Stable and Unstable Eigendirections
For a saddle point, the eigenvector associated with the negative eigenvalue defines the stable direction.
The eigenvector associated with the positive eigenvalue defines the unstable direction.
These directions organize the local phase portrait.
Example: Saddle Point
Consider
\[ \begin{aligned} x'&=2x,\\ y'&=-y. \end{aligned} \]Then
\[ x=C_1e^{2t}, \qquad y=C_2e^{-t}. \]The \(y\)-axis is a stable eigendirection, while the \(x\)-axis is an unstable eigendirection.
Complex Eigenvalues
Suppose the eigenvalues are
\[ \lambda=\alpha\pm\beta i. \]The imaginary part \(\beta\) produces rotational behavior.
The real part \(\alpha\) determines whether the rotation grows, decays, or remains neutral.
Stable Spiral
If
\[ \alpha<0, \]then the oscillatory modes decay.
Trajectories spiral inward toward the equilibrium.
The equilibrium is a
\[ \boxed{\text{stable spiral}}. \]Unstable Spiral
If
\[ \alpha>0, \]the oscillatory modes grow.
Trajectories spiral outward.
The equilibrium is an
\[ \boxed{\text{unstable spiral}}. \]Centers
If the eigenvalues are purely imaginary,
\[ \lambda=\pm\beta i, \]an ideal linear system may have closed trajectories around the equilibrium.
The equilibrium is called a
\[ \boxed{\text{center}}. \]A center is stable in a neutral sense but is not asymptotically stable.
Example: Center
Consider
\[ \begin{aligned} x'&=y,\\ y'&=-x. \end{aligned} \]Differentiate the first equation:
\[ x''=y'. \]Therefore,
\[ x''=-x. \]The motion is harmonic.
The trajectories are closed curves around the origin.
A Conserved Quantity for the Center
For
\[ x'=y, \qquad y'=-x, \]compute
\[ \frac{d}{dt}(x^2+y^2). \]We get
\[ 2xx'+2yy' = 2xy+2y(-x) = 0. \]Therefore,
\[ \boxed{x^2+y^2=C}. \]The trajectories are circles.
Trace and Determinant Classification
For
\[ A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}, \]define
\[ \tau=\operatorname{tr}(A) \]and
\[ \Delta=\det(A). \]The characteristic equation is
\[ \lambda^2-\tau\lambda+\Delta=0. \]The Discriminant for a Linear System
The eigenvalue discriminant is
\[ \boxed{ D=\tau^2-4\Delta }. \]Then:
- \(D>0\): real eigenvalues;
- \(D=0\): repeated real eigenvalue;
- \(D<0\): complex-conjugate eigenvalues.
Quick Stability Clues
For a two-dimensional linear system:
- \(\Delta<0\) implies a saddle;
- \(\Delta>0\) and \(\tau<0\) suggest stable behavior;
- \(\Delta>0\) and \(\tau>0\) suggest unstable behavior;
- \(\tau=0\) with \(D<0\) gives purely imaginary eigenvalues in the linear case.
Repeated Eigenvalues
Repeated eigenvalues can produce more than one phase portrait.
If there are two independent eigenvectors, the system may form a proper or star node.
If there is only one independent eigenvector, the system is defective and trajectories may approach or leave the equilibrium tangent to the single eigendirection.
Nonlinear Systems
A nonlinear system has the form
\[ \begin{aligned} x'&=f(x,y),\\ y'&=g(x,y), \end{aligned} \]where \(f\) or \(g\) is nonlinear.
Exact solution formulas may be difficult or unavailable.
Phase-plane analysis becomes especially valuable.
Nonlinear Equilibria
Equilibria still satisfy
\[ f(x,y)=0, \qquad g(x,y)=0. \]Nonlinear systems can have several equilibria with different local behaviors.
Linearization
Near an equilibrium
\[ (x_*,y_*), \]a nonlinear system can often be approximated by a linear system.
The relevant matrix is the Jacobian
\[ \boxed{ J(x,y) = \begin{bmatrix} f_x&f_y\\ g_x&g_y \end{bmatrix} }. \]Jacobian at an Equilibrium
Evaluate the Jacobian at
\[ (x_*,y_*): \] \[ \boxed{ J_* = J(x_*,y_*) }. \]The local behavior is then approximated by
\[ \mathbf{u}'=J_*\mathbf{u}, \]where \(\mathbf{u}\) measures displacement from the equilibrium.
Why Linearization Helps
The nonlinear equations may be complicated, but near an equilibrium the first-order terms often dominate.
Eigenvalues of the Jacobian can therefore reveal whether nearby trajectories tend to:
- approach the equilibrium;
- move away;
- spiral;
- follow saddle-like directions.
Hyperbolic Equilibria
An equilibrium is called hyperbolic when no eigenvalue of the Jacobian has zero real part.
For hyperbolic equilibria, the linearized classification usually gives the correct local qualitative behavior.
When Linearization Is Inconclusive
If one or more eigenvalues have zero real part, linearization may not settle the stability question.
Higher-order nonlinear terms can become decisive.
More advanced tools may then be required.
Example: Nonlinear Logistic Interaction
Consider
\[ \begin{aligned} x'&=x(1-x-y),\\ y'&=y(1-x-y). \end{aligned} \]The nullclines include
\[ x=0, \qquad y=0, \qquad x+y=1. \]The line
\[ x+y=1 \]consists of equilibrium states.
Predator-Prey Phase Behavior
Predator-prey systems often produce trajectories that circulate around a coexistence equilibrium.
The phase plane can reveal:
- alternating rises and falls;
- cyclic behavior;
- equilibrium coexistence;
- direction of population interaction.
This behavior may be difficult to see from the equations alone.
Phase Plane versus Time-Series Plots
A time-series plot answers:
How does each variable change with time?
A phase-plane plot answers:
How do the state variables relate to one another as the system evolves?
Both views are useful and often complementary.
Phase Plane and Initial Conditions
An initial condition
\[ (x(0),y(0))=(x_0,y_0) \]selects one trajectory from the phase portrait.
Different initial conditions may produce different paths while still following the same vector field.
Invariant Sets
A set is invariant if a trajectory beginning inside it remains inside it.
Eigenvector lines in linear systems can be invariant.
Coordinate axes may also be invariant in population models when factors such as \(x\) or \(y\) appear in the equations.
Example: Invariant Axis
Suppose
\[ x'=x(1-y). \]If
\[ x=0, \]then
\[ x'=0. \]Therefore a trajectory beginning on the \(y\)-axis remains on the \(y\)-axis.
Closed Orbits
A closed phase trajectory represents periodic behavior.
The state returns to its initial point after some period.
In the phase plane, the same closed path repeats indefinitely in an ideal periodic model.
Spirals versus Closed Orbits
A spiral changes its distance from the equilibrium over time.
A closed orbit repeats the same path.
Therefore:
- inward spiral: amplitude shrinks;
- outward spiral: amplitude grows;
- closed orbit: amplitude remains bounded on the same path.
Separatrices
A separatrix is a trajectory that separates regions with different qualitative behavior.
Stable and unstable manifolds of saddle points often act as separatrices.
They can divide the phase plane into distinct basins or flow regions.
Basins of Attraction
If a system has multiple stable equilibria, different initial conditions may approach different attractors.
The set of initial conditions that approach a particular attractor is called its basin of attraction.
Phase-plane geometry helps visualize these regions.
Common Error: Forgetting Direction Arrows
Two systems can have the same geometric trajectories but opposite direction of motion.
Therefore, a phase portrait without arrows can hide essential dynamic information.
Common Error: Treating Nullclines as Trajectories
A nullcline is a set where one derivative vanishes.
It is not automatically a trajectory.
A trajectory must remain tangent to the full vector field.
Common Error: Calling Every Equilibrium Stable
Equilibria may be:
- stable;
- unstable;
- saddles;
- centers;
- nonhyperbolic and more difficult to classify.
Stability must be analyzed.
Common Error: Ignoring Eigenvectors
Eigenvalues determine rates of growth or decay.
Eigenvectors determine geometric directions in the phase plane.
A correct phase portrait requires both.
Common Error: Confusing a Center with a Stable Spiral
A center has closed orbits in the ideal linear case.
A stable spiral moves inward.
The distinction depends on the real part of the complex eigenvalues.
Purely imaginary eigenvalues give zero real part, while a stable spiral has negative real part.
Common Error: Overtrusting Linearization
Linearization is highly effective near hyperbolic equilibria.
It may be inconclusive when eigenvalues have zero real part.
Nonlinear terms can then change the local behavior substantially.
A Reliable Phase-Plane Workflow
- Write the system as \(x'=f(x,y)\), \(y'=g(x,y)\).
- Find all equilibrium points.
- Find the \(x\)- and \(y\)-nullclines.
- Use sign analysis to determine local arrow directions.
- For a linear system, compute eigenvalues and eigenvectors.
- Classify each equilibrium.
- Sketch invariant eigendirections when present.
- Draw representative trajectories consistent with the vector field.
- Add arrows to show direction of motion.
- Use initial conditions to identify a specific trajectory.
- For nonlinear systems, evaluate the Jacobian at equilibria.
- Use numerical trajectories when analytical geometry is insufficient.
Example: Full Linear Phase-Plane Analysis
Consider
\[ \mathbf{x}' = \begin{bmatrix} 2&0\\ 0&-1 \end{bmatrix} \mathbf{x}. \]The eigenvalues are
\[ \lambda_1=2 \]and
\[ \lambda_2=-1. \]Their eigenvectors are the coordinate-axis directions.
Complete the Full Linear Analysis
The solution is
\[ x=C_1e^{2t}, \qquad y=C_2e^{-t}. \]Along the \(y\)-axis, trajectories approach the origin.
Along the \(x\)-axis, trajectories move away from the origin.
Therefore,
\[ \boxed{(0,0)\text{ is a saddle point}}. \]Example: Full Spiral Analysis
Consider
\[ \mathbf{x}' = \begin{bmatrix} -1&-2\\ 2&-1 \end{bmatrix} \mathbf{x}. \]The characteristic equation is
\[ (-1-\lambda)^2+4=0. \]Therefore,
\[ \lambda=-1\pm2i. \]The negative real part implies decay.
The imaginary part implies rotation.
Therefore,
\[ \boxed{(0,0)\text{ is a stable spiral}}. \]Practice 1: Find an Equilibrium
Find the equilibrium of
\[ \begin{aligned} x'&=x+y,\\ y'&=2x-y. \end{aligned} \]Practice 2: Find the Nullclines
Find the nullclines of
\[ \begin{aligned} x'&=2x-y,\\ y'&=x+y. \end{aligned} \]Practice 3: Local Direction
For
\[ \begin{aligned} x'&=x-y,\\ y'&=x+y, \end{aligned} \]find the direction vector at
\[ (2,1). \]Practice 4: Classify a Node
A linear system has eigenvalues
\[ -2 \]and
\[ -7. \]Classify the origin.
Practice 5: Classify a Saddle
A linear system has eigenvalues
\[ 4 \]and
\[ -3. \]Classify the origin.
Practice 6: Classify a Spiral
A system has eigenvalues
\[ -2\pm5i. \]Classify the equilibrium.
Practice 7: Classify a Center
A linear system has eigenvalues
\[ \pm4i. \]What phase-plane behavior is expected?
Practice 8: Trace and Determinant
For
\[ A= \begin{bmatrix} -2&1\\ 3&-4 \end{bmatrix}, \]find the trace and determinant.
Practice 9: Conserved Quantity
For
\[ x'=y, \qquad y'=-x, \]show that
\[ x^2+y^2 \]remains constant.
Practice 10: Linearization Matrix
For
\[ \begin{aligned} x'&=x-x^2-xy,\\ y'&=2y-xy-y^2, \end{aligned} \]write the Jacobian matrix.
Practice Answers
Practice 1
Set
\[ x+y=0 \]and
\[ 2x-y=0. \]From the first,
\[ y=-x. \]Substitute:
\[ 2x+x=0. \]Therefore,
\[ x=0, \qquad y=0. \]The equilibrium is
\[ \boxed{(0,0)}. \]Practice 2
The \(x\)-nullcline satisfies
\[ 2x-y=0. \]Thus,
\[ \boxed{y=2x}. \]The \(y\)-nullcline satisfies
\[ x+y=0. \]Thus,
\[ \boxed{y=-x}. \]Practice 3
At
\[ (2,1), \]we have
\[ x'=2-1=1 \]and
\[ y'=2+1=3. \]Therefore,
\[ \boxed{\langle1,3\rangle}. \]Practice 4
Both eigenvalues are real and negative.
Therefore, the origin is a
\[ \boxed{\text{stable node}}. \]Practice 5
One eigenvalue is positive and one is negative.
Therefore, the origin is a
\[ \boxed{\text{saddle point}}. \]Practice 6
The eigenvalues have negative real part and nonzero imaginary part.
Therefore, the equilibrium is a
\[ \boxed{\text{stable spiral}}. \]Practice 7
The eigenvalues are purely imaginary.
In the ideal linear case, the expected behavior is a
\[ \boxed{\text{center with closed trajectories}}. \]Practice 8
The trace is
\[ -2+(-4)=-6. \]The determinant is
\[ (-2)(-4)-(1)(3)=8-3=5. \]Therefore,
\[ \boxed{ \operatorname{tr}(A)=-6, \qquad \det(A)=5 }. \]Practice 9
Differentiate:
\[ \frac{d}{dt}(x^2+y^2) = 2xx'+2yy'. \]Substitute
\[ x'=y, \qquad y'=-x. \]Then
\[ 2xy+2y(-x)=0. \]Therefore,
\[ \boxed{x^2+y^2=C}. \]Practice 10
Let
\[ f(x,y)=x-x^2-xy \]and
\[ g(x,y)=2y-xy-y^2. \]Then
\[ f_x=1-2x-y, \qquad f_y=-x, \] \[ g_x=-y, \qquad g_y=2-x-2y. \]Therefore,
\[ \boxed{ J(x,y) = \begin{bmatrix} 1-2x-y&-x\\ -y&2-x-2y \end{bmatrix} }. \]What to Remember
The phase plane represents the joint state of a two-variable differential system.
For
\[ \begin{aligned} x'&=f(x,y),\\ y'&=g(x,y), \end{aligned} \]the vector
\[ \boxed{ \langle f(x,y),g(x,y)\rangle } \]gives the local direction of motion.
- equilibria occur where both derivatives vanish;
- nullclines show where one derivative vanishes;
- trajectories trace the evolution of the system state;
- eigenvalues and eigenvectors classify linear phase portraits;
- nodes, saddles, spirals, and centers describe major equilibrium types;
- the Jacobian linearizes nonlinear systems near equilibria;
- phase portraits reveal behavior even when explicit solutions are unavailable.
The central structural idea is
\[ \boxed{ \text{local vectors} \longrightarrow \text{trajectories} \longrightarrow \text{global phase portrait} }. \]Where We Go Next
Phase-plane representations reveal the geometry of differential-equation solutions, but many models still cannot be solved exactly.
The next lesson, Numerical Solution Methods, develops Euler's method, improved Euler methods, Runge-Kutta ideas, step-size selection, local and global error, stability, and numerical solution of systems.