Equilibrium Jacobian Linearization: Local Linear Models

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Guides | FAQ

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The Principle

The move: write the state as $\mathbf{x}=\mathbf{x}_*+\mathbf{u}$, then replace $\mathbf{F}(\mathbf{x}_*+\mathbf{u})$ by its first-order Jacobian term at the equilibrium.

The invariant: the equilibrium becomes the origin in $\mathbf{u}$-coordinates, and the linear model matches the vector field’s first-order Taylor term at that equilibrium. Higher-order nonlinear terms are deliberately discarded, so this is a local approximation, not an equivalent rewrite.

Pattern:

$$\mathbf{x}^{\prime}=\mathbf{F}(\mathbf{x}),\quad \mathbf{x}=\mathbf{x}_*+\mathbf{u},\quad \mathbf{F}(\mathbf{x}_*)=\mathbf{0} \Longrightarrow \mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$$

Legal route	Illegal route
$\mathbf{F}(\mathbf{0})=\mathbf{0},\ J_F(\mathbf{0})=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\Longrightarrow \mathbf{u}^{\prime}\approx \begin{bmatrix}1&0\\0&-1\end{bmatrix}\mathbf{u}$	$\mathbf{F}(\mathbf{x}_*)\neq\mathbf{0}\Longrightarrow \mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$

The illegal route skips the equilibrium check. Around a non-equilibrium point, the shifted equation has a constant forcing term and is not this equilibrium linearization.

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Conditions of Applicability

Condition: $F differentiable near \mathbf{x}_*$; $\mathbf{F}(\mathbf{x}_*)=0$

Before applying, check: verify that the candidate point makes every component of $\mathbf{F}$ zero, then compute the Jacobian at that same point.

If the condition is violated: the local system may need an affine constant term, or the Jacobian may not exist, so the equilibrium linear model is not licensed.

• The system must be autonomous in the form $\mathbf{x}^{\prime}=\mathbf{F}(\mathbf{x})$.
• $F$ must be differentiable near the equilibrium so the Jacobian is meaningful as the first-order local term.
• The same $\mathbf{x}_*$ must be used in the equilibrium check, coordinate shift, and Jacobian evaluation.
• Linearization predicts nearby local behavior. It does not prove global behavior far from the equilibrium.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: compute $J_F(\mathbf{x}_*)$ before checking $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$ -> the result may be a linearization about a moving point, not an equilibrium model.

Debug: plug $\mathbf{x}_*$ into the vector field first; if any component is nonzero, stop and do not use the equilibrium formula.

Failure mode: treat $\mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$ as an exact nonlinear solution -> the higher-order terms disappear from the model.

Debug: say “local first-order model” whenever you write the approximation sign.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• What does the perturbation vector $\mathbf{u}=\mathbf{x}-\mathbf{x}_*$ measure?
• Why does the constant term vanish only when $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$?

For the Principle

• Before linearizing, what two checks protect the move from becoming an invalid local model?
• Why is the Jacobian evaluated at the equilibrium rather than at a nearby moving state?

Between Principles

• How does this principle depend on System Equilibrium Condition before it connects to First-Order Linear System Form?

Generate an Example

• Create a two-dimensional autonomous system where $\mathbf{F}(\mathbf{0})=\mathbf{0}$, then change one constant term so the same point becomes a near miss.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Shift an autonomous system to perturbation coordinates around an equilibrium, then use the Jacobian at that equilibrium as the local linear model.

Write the canonical pattern: _____$\mathbf{x}^{\prime}=\mathbf{F}(\mathbf{x}),\ \mathbf{x}=\mathbf{x}_*+\mathbf{u},\ \mathbf{F}(\mathbf{x}_*)=\mathbf{0} \Rightarrow \mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$

State the canonical condition: _____$F differentiable near \mathbf{x}_*; \mathbf{F}(\mathbf{x}_*)=0$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to learn from worked examples.)

Procedure Walkthrough

Starting from the autonomous system

$$\mathbf{x}^{\prime}= \begin{bmatrix} x+y^2\\ -y+xy \end{bmatrix}, \qquad \mathbf{x}_*=\begin{bmatrix}0\\0\end{bmatrix},$$

reach the local linear model in perturbation coordinates.

Step	Expression	Operation
0	$\mathbf{F}(\mathbf{0})=\begin{bmatrix}0\\0\end{bmatrix}$	Check that the candidate is an equilibrium.
1	$J_F(x,y)$ has rows $(1,2y)$ and $(y,-1+x)$.	Compute the Jacobian matrix.
2	$J_F(\mathbf{0})=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$	Evaluate the Jacobian at the equilibrium.
3	$\mathbf{x}=\mathbf{0}+\mathbf{u}$	Shift the equilibrium to the origin.
4	$\mathbf{u}^{\prime}\approx \begin{bmatrix}1&0\\0&-1\end{bmatrix}\mathbf{u}$	Write the local linearized system.

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Drills

Micro-Chains

Reach the local linear model. Assume $F$ is differentiable near $\mathbf{x}_*$ and verify $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$ before linearizing.

$$\mathbf{F}(x,y)=\begin{bmatrix}2x+y^2\\-3y+x^2\end{bmatrix},\qquad \mathbf{x}_*=\begin{bmatrix}0\\0\end{bmatrix}$$

Reveal

First,

$$\mathbf{F}(\mathbf{0})=\begin{bmatrix}0\\0\end{bmatrix}.$$

Then

$$J_F(x,y)=\begin{bmatrix}2&2y\\2x&-3\end{bmatrix}, \qquad J_F(\mathbf{0})=\begin{bmatrix}2&0\\0&-3\end{bmatrix}.$$

So the local linear model is

$$\mathbf{u}^{\prime}\approx \begin{bmatrix}2&0\\0&-3\end{bmatrix}\mathbf{u}.$$

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Reach the local linear model near the equilibrium. Assume differentiability where needed.

$$\mathbf{F}(x,y)=\begin{bmatrix}y\\-x+x^3\end{bmatrix},\qquad \mathbf{x}_*=\begin{bmatrix}0\\0\end{bmatrix}$$

Reveal

The equilibrium check passes because $\mathbf{F}(\mathbf{0})=\mathbf{0}$.

$$J_F(x,y)=\begin{bmatrix}0&1\\-1+3x^2&0\end{bmatrix}, \qquad J_F(\mathbf{0})=\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$$

Therefore

$$\mathbf{u}^{\prime}\approx \begin{bmatrix}0&1\\-1&0\end{bmatrix}\mathbf{u}.$$

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Reject or linearize. Assume $F$ is differentiable near the candidate point.

$$\mathbf{F}(x,y)=\begin{bmatrix}x+1\\-y\end{bmatrix},\qquad \mathbf{x}_*=\begin{bmatrix}0\\0\end{bmatrix}$$

Reveal

Reject this as equilibrium Jacobian linearization at $\mathbf{0}$.

$$\mathbf{F}(\mathbf{0})=\begin{bmatrix}1\\0\end{bmatrix}\neq \mathbf{0}.$$

The point is not an equilibrium, so the condition fails before the Jacobian matters.

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Reach the local linear model near the equilibrium. Assume differentiability where needed.

$$\mathbf{F}(x,y)=\begin{bmatrix}x-y\\2x+y^2-y\end{bmatrix},\qquad \mathbf{x}_*=\begin{bmatrix}0\\0\end{bmatrix}$$

Reveal

The equilibrium check passes:

$$\mathbf{F}(\mathbf{0})=\begin{bmatrix}0\\0\end{bmatrix}.$$

The Jacobian is

$$J_F(x,y)=\begin{bmatrix}1&-1\\2&2y-1\end{bmatrix}, \qquad J_F(\mathbf{0})=\begin{bmatrix}1&-1\\2&-1\end{bmatrix}.$$

So

$$\mathbf{u}^{\prime}\approx \begin{bmatrix}1&-1\\2&-1\end{bmatrix}\mathbf{u}.$$

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Action Labels

What was done between these two steps?

$$\mathbf{x}^{\prime}=\mathbf{F}(\mathbf{x}),\quad \mathbf{x}=\mathbf{x}_*+\mathbf{u} \quad \Longrightarrow \quad \mathbf{u}^{\prime}=\mathbf{F}(\mathbf{x}_*+\mathbf{u})$$

Reveal

The state was shifted to perturbation coordinates around the equilibrium. Since $\mathbf{x}_*$ is constant, $\mathbf{x}^{\prime}=\mathbf{u}^{\prime}$.

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What was done between these two steps, assuming $F$ is differentiable near $\mathbf{x}_*$ and $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$?

$$\mathbf{u}^{\prime}=\mathbf{F}(\mathbf{x}_*+\mathbf{u}) \quad \Longrightarrow \quad \mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$$

Reveal

The nonlinear vector field was replaced by its first-order Taylor term at the equilibrium. The constant term is zero because $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$.

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Which proposed steps are eligible equilibrium Jacobian linearizations?

A. $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$ and $F$ is differentiable near $\mathbf{x}_*$, so use $\mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$.
B. $\mathbf{F}(\mathbf{x}_*)\neq\mathbf{0}$, but compute $J_F(\mathbf{x}_*)$ and use $\mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$ anyway.
C. $F$ has a corner at $\mathbf{x}_*$, so estimate slopes from one side and call that $J_F(\mathbf{x}_*)$.

Reveal

A is eligible.

B fails the equilibrium condition. C fails the differentiability condition, so the Jacobian linearization is not licensed.

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What was done between these two steps?

$$J_F(x,y):\ (2x,1),(y,x-y) \quad \Longrightarrow \quad J_F(1,0):\ (2,1),(0,1)$$

Reveal

The Jacobian was evaluated at the chosen equilibrium point. This is only the right matrix for equilibrium linearization if the same point also satisfies $\mathbf{F}(1,0)=\mathbf{0}$.

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Forward Steps

Apply the principle once after the equilibrium check has passed.

$$J_F(\mathbf{x}_*)=\begin{bmatrix}-2&4\\1&0\end{bmatrix}$$

Reveal

With $\mathbf{x}=\mathbf{x}_*+\mathbf{u}$ and the equilibrium condition already checked,

$$\mathbf{u}^{\prime}\approx \begin{bmatrix}-2&4\\1&0\end{bmatrix}\mathbf{u}.$$

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Apply the principle once after the equilibrium check has passed.

$$J_F(\mathbf{x}_*)=\begin{bmatrix}0&1\\-5&-2\end{bmatrix}$$

Reveal

The local linearized perturbation system is

$$\mathbf{u}^{\prime}\approx \begin{bmatrix}0&1\\-5&-2\end{bmatrix}\mathbf{u}.$$

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Reject or complete the step. A student writes $\mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$ before checking $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$.

Reveal

Reject the step as incomplete. For this principle, the equilibrium check is part of the move, not a later note.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from the autonomous system

$$\mathbf{x}^{\prime}= \begin{bmatrix} x+y\\ -2x+y-y^2 \end{bmatrix}, \qquad \mathbf{x}_*=\begin{bmatrix}0\\0\end{bmatrix},$$

reach the local linearized system using Equilibrium Jacobian Linearization.

Full solution

Step	Expression	Move
0	$\mathbf{F}(\mathbf{0})=\begin{bmatrix}0\\0\end{bmatrix}$	Check the equilibrium condition.
1	$\mathbf{x}=\mathbf{0}+\mathbf{u}$	Shift the equilibrium to perturbation coordinates.
2	$J_F(x,y)=\begin{bmatrix}1&1\\-2&1-2y\end{bmatrix}$	Compute the Jacobian.
3	$J_F(\mathbf{0})=\begin{bmatrix}1&1\\-2&1\end{bmatrix}$	Evaluate at the same equilibrium.
4	$\mathbf{u}^{\prime}\approx \begin{bmatrix}1&1\\-2&1\end{bmatrix}\mathbf{u}$	Write the local linearized system.

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Related Guides

• Differential Equations Subdomain - Return to the full systems and qualitative-analysis lane.
• System Equilibrium Condition - Check whether the candidate state is an equilibrium before linearizing.
• First-Order Linear System Form - Read the linearized perturbation equation as a matrix system.
• Matrix Exponential Solution Form - Connect the local linear system to solution behavior when the matrix is constant.
• Principle Structures - Keep the principle, condition, and equation distinct while studying.

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FAQ

What is Equilibrium Jacobian Linearization?

Equilibrium Jacobian Linearization is the move that approximates an autonomous nonlinear system near an equilibrium by a linear perturbation system. After writing $\mathbf{x}=\mathbf{x}_*+\mathbf{u}$, the local model is $\mathbf{u}^{\prime}\approx J_F(\mathbf{x}_*)\mathbf{u}$.

When is Equilibrium Jacobian Linearization valid?

It is valid when $F$ is differentiable near $\mathbf{x}_*$ and $\mathbf{F}(\mathbf{x}_*)=\mathbf{0}$. The differentiability condition gives the Jacobian, and the equilibrium condition removes the constant term from the shifted local model.

Why do we shift to perturbation coordinates?

The shift $\mathbf{x}=\mathbf{x}_*+\mathbf{u}$ makes $\mathbf{u}$ measure displacement from the equilibrium. That turns the equilibrium itself into $\mathbf{u}=\mathbf{0}$, so the nearby motion can be compared with a linear system around the origin.

Is the linearized system an exact replacement for the nonlinear system?

No. It is a first-order local approximation. The nonlinear terms beyond the Jacobian are ignored, so conclusions from the linearized system need the usual local-stability interpretation rather than a global exact-solution interpretation.

What happens if the candidate point is not an equilibrium?

Then $\mathbf{F}(\mathbf{x}_*)$ is not zero, and the shifted equation has a constant term. You may still approximate the vector field locally, but it is not the equilibrium Jacobian linearization described by this principle.

How is this related to eigenvalues?

Once the perturbation system is $\mathbf{u}^{\prime}\approx A\mathbf{u}$ with $A=J_F(\mathbf{x}_*)$, eigenvalues of $A$ describe the local linear modes. If the linearized matrix has eigenvalues with zero real part, the linear model may be inconclusive for the nonlinear system’s stability. That builds on Eigenvalue-Eigenvector Solution Mode rather than replacing the equilibrium check.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium treats equilibrium linearization as a condition-first move: find the equilibrium, shift coordinates, compute the Jacobian, then read the local system. The Unisium Study System pairs that sequence with retrieval practice, self-explanation, and compact problem-solving chains so the equilibrium condition stays visible while the matrix work becomes fluent.

Ready to practice differential equations with structure? Check access and join the Unisium waitlist or explore the complete framework in Masterful Learning.

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