Linearization around equilibrium

Start with a small state change

Supply two dimensionless states u and v, with time t measured in seconds and a positive time scale T = 1 s. Their exact rules are du/dt = [−2(u − 1) + v + (u − 1)²]/T and dv/dt = [−v + (u − 1)²]/T. The rates have units s⁻¹. These are invented normalized dynamics, not measured hardware.

At (u, v) = (1, 0), both rates vanish, so this state can remain constant: it is an equilibrium. Move to (1.1, 0). The first rate is (−0.2 + 0.01)/(1 s) = −0.19 s⁻¹ and the second is 0.01 s⁻¹. If we neglect the square term, the estimates become −0.2 s⁻¹ and zero. We can check the neglected part instead of calling the change small without a reference.

Algebra, functions, graphs, and logarithms prepares the substitutions and graph reading; logarithms are used only in one later worked example. First follow the exact terms and the comparison figure. Derivatives, Taylor series, and local approximation, Integrals and multivariable calculus, and Matrices, linear algebra, and eigenvalue problems offer an optional return path for the Jacobian and decay-rate reasoning. The needed definitions are introduced below.

Measure deviations from the chosen equilibrium

Write ξ = u − 1 and η = v. The Greek letters xi and eta label deviations from (1, 0); they are still dimensionless. A constant coordinate shift does not change a time derivative. Our exact equations therefore become dξ/dt = f₁(ξ, η) = (−2ξ + η + ξ²)/T and dη/dt = f₂(ξ, η) = (−η + ξ²)/T.

The first-order terms are those containing one power of a deviation: −2ξ + η and −η. Retain them to define a separate linearized model, dξL/dt = (−2ξL + ηL)/T and dηL/dt = −ηL/T. The subscript L distinguishes its solution from the solution of the original model.

To compare rates at the same state, write L₁(ξ, η) = (−2ξ + η)/T and L₂(ξ, η) = −η/T. Then f₁ − L₁ = ξ²/T and f₂ − L₂ = ξ²/T exactly. The chosen equilibrium is the zero deviation pair, and there is no constant rate term in its linearization. Linearizing at an arbitrary non-equilibrium state would generally retain a nonzero constant term.

Build the Jacobian one rate at a time

A partial derivative asks how one rate changes when one state coordinate changes while the other is held fixed. The Jacobian arranges those derivatives in a matrix: row i belongs to rate fᵢ, and column j belongs to the state coordinate being varied. Evaluate the matrix at the chosen equilibrium, rather than at an unspecified nearby state.

Here ∂f₁/∂ξ = (−2 + 2ξ)/T, ∂f₁/∂η = 1/T, ∂f₂/∂ξ = 2ξ/T and ∂f₂/∂η = −1/T. At (ξ, η) = (0, 0), the matrix is A = (1/T)[[−2, 1], [0, −1]]. Multiplying it by the column of deviations gives the two linearized rates: first (−2ξ + η)/T, then −η/T.

All entries of this particular A have units s⁻¹ because both coordinates are dimensionless. In general, entry Aᵢⱼ has units of rate i divided by state coordinate j. If a model mixes different quantity kinds, keep those coordinate units; its individual Jacobian entries need not all have the same units.

For a differentiable vector rate, linearization retains its derivative at the equilibrium. The remainder becomes small relative to the deviation as the deviation tends to zero. That statement does not give every model a quadratic remainder. Our supplied polynomial has the stronger, directly checked remainder (ξ², ξ²)/T.

Read the tangent and its correction

The upper panel evaluates the first rate on the slice η = 0. Its horizontal axis is deviation ξ, not time. The solid nonlinear curve and dashed linear rate meet at ξ = 0 with the same slope. Circles and squares compare their endpoint rates; the cross marks their shared zero.

The lower panel shows the exact common correction R₁ = R₂ = ξ²/T. At ξ = 0.1 it is 0.01 s⁻¹; at ξ = 0.2 it is 0.04 s⁻¹. Doubling this deviation doubles the linear rate but quadruples the square correction. The lower vertical scale is magnified: compare the labeled values, rather than vertical distances across the two panels.

η = 0 is held only for evaluating this slice. In the original dynamics, dη/dt = ξ²/T there, which is positive when ξ ≠ 0. A trajectory generally leaves the slice. Neither plotted curve traces an evolving state, and the display interval is not an established approximation radius.

State what small means for the question

Suppose the question requires each neglected component rate to be at most 0.01 s⁻¹ in absolute value. In this model ξ²/(1 s) ≤ 0.01 s⁻¹ is equivalent to |ξ| ≤ 0.1. This is an exact instantaneous component-rate condition. It imposes no condition on η, because this particular remainder does not depend on η.

For a whole-vector comparison, define the Euclidean deviation size r = √(ξ² + η²) on these normalized coordinates. The remainder size is √2 ξ²/T ≤ √2 r²/T. Thus its size divided by r tends to zero as r tends to zero. Adding unlike physical quantities inside such a norm would require a specified normalization first.

A relative comparison needs a nonzero reference. On η = 0 and ξ ≠ 0, the linear-rate size is 2|ξ|/T, so the remainder-to-linear-rate size ratio is |ξ|/√2. At ξ = 0 both rates vanish and that ratio is undefined. An acceptable absolute or relative tolerance depends on the intended use; no universal cutoff follows from the graph.

A bound on a rate difference at one state is not a bound on the difference between two trajectories over time. As the trajectories move, their states differ too. A time-dependent error claim must specify its interval and justify its own bound.

Check a time history of the linearized model

The simpler model can be solved exactly. Let ξ₀ and η₀ be its initial deviations and set τ = t/T. Then ηL(t) = η₀e^(−τ) and ξL(t) = (ξ₀ − η₀)e^(−2τ) + η₀e^(−τ). At t = 0 these give the specified initial state. Differentiation checks both linearized equations.

The two exponential rates come from A. A nonzero direction w is an eigenvector if Aw = λw. For w = (1, 0), λ = −2/T; for w = (1, 1), λ = −1/T. Along each of those linear directions the multiplying factor is e^(λt). Both decay because T is positive.

These are exact solutions and directions of the linearized model. Substituting a nonzero ξL into the original equations leaves the square correction. The label linearized therefore remains part of every time-history prediction; solving the simpler equations does not remove the approximation.

Use the stability conclusion with its conditions

Local stability means all sufficiently close initial states stay close for every future time; asymptotic stability also requires their approach to the equilibrium. Exponential stability supplies a local exponentially decreasing bound on their distance. It need not make that distance decrease at every instant.

For an autonomous finite-dimensional continuous-time system with continuously differentiable rates near an equilibrium, strictly negative real parts of all Jacobian eigenvalues are sufficient for local exponential stability. In our smooth polynomial model, −2/T and −1/T meet that condition, so (u, v) = (1, 0) is locally exponentially stable in the original nonlinear dynamics.

The theorem gives a qualitative local verdict. It does not assert that the nonlinear trajectories equal the linearized ones or that every permitted start approaches this equilibrium. Both original rates also vanish at (u, v) = (2, 1); starting exactly there remains there. Smoothness gives local existence and uniqueness, while arbitrary nonlinear models can still fail to have solutions for all future time.

Recognize an inconclusive linearization

A zero real part does not satisfy the strict negative-real-part condition. Consider two separate models near their equilibrium (ξ, η) = (0, 0): dξ/dt = −ξ³/T with dη/dt = −η/T, and dξ/dt = +ξ³/T with dη/dt = −η/T. Both Jacobians at zero are [[0, 0], [0, −1/T]]. The linearizations are identical.

In the minus-cubic model, the ξ magnitude decreases toward zero while η decays exponentially, giving an asymptotically stable equilibrium. In the plus-cubic model, arbitrarily small nonzero ξ starts grow and leave a fixed small neighborhood, giving an unstable equilibrium. The actual cubic terms settle these two cases; the shared zero eigenvalue does not.

Keep the chosen equilibrium, coordinates, rate units and smoothness with the conclusion. Use the original nonlinear terms or another justified method when the linear test is inconclusive. Linearization is a local description of a specified continuous-time model, with neither a general numerical-step stability rule nor an empirical prediction of hardware performance.

Worked examples

Try it yourself

1. At (u, v) = (0.9, 0) in the main model, find the original and linearized rate pairs. Is the common correction negative because ξ is negative?

   Show the answer and reasoning

   The original rates are (+0.21, +0.01) s⁻¹ and the linearized rates are (+0.2, 0) s⁻¹. Both corrections are +0.01 s⁻¹.

   ξ = −0.1 and η = 0. The square ξ² = 0.01 is nonnegative on either side of the equilibrium; f₁ = (0.2 + 0.01)/(1 s) and f₂ = 0.01/(1 s). The correction is original minus linearized, with the same chosen order in both components.

2. Find ∂f₂/∂ξ at the chosen equilibrium and at ξ = 0.1. Which value enters A for linearization around (u, v) = (1, 0)?

   Show the answer and reasoning

   At the equilibrium it is 0 s⁻¹; at ξ = 0.1 it is 0.2 s⁻¹. The zero equilibrium value enters A.

   Differentiate f₂ = (−η + ξ²)/T while holding η fixed: ∂f₂/∂ξ = 2ξ/T. The linearization matrix is evaluated at (ξ, η) = (0, 0), not at the state whose later rate you want to estimate.

3. A correction of 0.01 s⁻¹ occupies 1 px on the upper rate scale and 8 px on the lower correction scale. Has the correction changed? Is the lower curve a time history?

   Show the answer and reasoning

   The correction has not changed; it is still 0.01 s⁻¹. The lower curve is a rate correction versus ξ, not time.

   100 px per s⁻¹ times 0.01 s⁻¹ gives 1 px, while 800 px per s⁻¹ gives 8 px. These are different drawing scales for a rate quantity. The horizontal axis is a dimensionless state deviation and η is fixed only for the comparison slice.

4. Require both component rate corrections in the main model to be no greater than 0.01 s⁻¹ in absolute value at a state. What restriction on ξ is sufficient and necessary here? Does it prove a trajectory error is no greater than 0.01?

   Show the answer and reasoning

   |ξ| ≤ 0.1 is sufficient and necessary for this component-rate condition, with any η. It does not prove a trajectory-error bound.

   Each correction is ξ²/T, with T = 1 s, so ξ² ≤ 0.01 gives −0.1 ≤ ξ ≤ 0.1. This statement compares two vector fields at the same instantaneous state. A trajectory error concerns different evolving states over a specified time interval and requires a separate argument.

5. At (ξ, η) = (0, 0.1), the original and linearized rate pairs are equal. Does that establish equal future solutions? Also, is the whole slice η = 0 invariant in the original model?

   Show the answer and reasoning

   No to both claims. Equal rates at one state do not establish equal trajectories; η = 0 is not an invariant slice away from ξ = 0.

   At the selected state dξ/dt = 0.1 s⁻¹, so ξ starts changing and the square term need not remain zero. On η = 0 with ξ ≠ 0, dη/dt = ξ²/T > 0, so a solution leaves that slice. The fixed slice used to draw the figure is not a constraint imposed on the dynamics.

6. The two separate cubic models in the last section have the same Jacobian with eigenvalues 0 and −1/T. Must they have the same nonlinear stability?

   Show the answer and reasoning

   No. The minus-cubic model is asymptotically stable at zero, while the plus-cubic model is unstable.

   For the minus model, ξ(t) = ξ₀/√(1 + 2ξ₀²t/T) decreases in magnitude toward zero, and η(t) = η₀e^(−t/T) also tends to zero. For the plus model, a nonzero ξ start instead has denominator √(1 − 2ξ₀²t/T) while it is positive; its growing magnitude eventually leaves a fixed small neighborhood. The identical zero eigenvalue leaves the linear criterion inconclusive.

What to carry forward

• Shift to deviations from a specified equilibrium, then retain the first-order rate terms.
• The Jacobian is evaluated at that equilibrium, with coordinate and rate units kept consistent.
• Our exact square remainder quantifies a rate approximation; it does not itself bound accumulated trajectory error.
• Strictly negative real Jacobian eigenvalues give a local nonlinear stability conclusion under the smooth autonomous continuous-time conditions.
• A zero real part can leave the linear test inconclusive, and a local verdict does not describe every allowed start.