Linearisation
Linearisation replaces a nonlinear system by a simpler linear approximation near a particular point, usually an equilibrium. The approximation is local: it is designed to describe small departures from that state.
First see the idea with an ordinary function
For a differentiable function \(f(x)\), near \(x=x^*\),
\[f(x)\approx f(x^*)+f'(x^*)(x-x^*).\]The right-hand side is the tangent-line approximation. It uses the value and slope at \(x^*\) to approximate the nearby curve.
Curved function and local tangent
Apply this idea to a differential equation
Suppose
\[\frac{dx}{dt}=f(x)\]and \(x^*\) is an equilibrium. Therefore
\[f(x^*)=0.\]Write a nearby state as
\[x=x^*+u,\]where \(u\) is a small displacement from equilibrium.
What does the perturbation u mean?
The quantity
\[u=x-x^*\]is simply the distance from the equilibrium in state space.
If \(u=0\), the system is exactly at equilibrium. If \(u\) is small, the system is nearby.
Derive the one-dimensional linearisation
Using Taylor expansion,
\[f(x^*+u)=f(x^*)+f'(x^*)u+\frac12f''(x^*)u^2+\cdots.\]Because \(x^*\) is an equilibrium, \(f(x^*)=0\). For sufficiently small \(u\), the terms involving \(u^2,u^3,\ldots\) are smaller than the first-order term.
Therefore
\[f(x^*+u)\approx f'(x^*)u.\]Since \(x=x^*+u\) and \(x^*\) is constant,
\[\frac{du}{dt}=\frac{dx}{dt}.\]Hence
\[\boxed{\frac{du}{dt}\approx f'(x^*)u}.\]Why this immediately tells us stability
The linear equation
\[\frac{du}{dt}=au\]has solution
\[u(t)=u(0)e^{at}.\]Here \(a=f'(x^*)\). Therefore, if \(f'(x^*)\lt0\), the displacement decays. If \(f'(x^*)\gt0\), it grows.
Worked example: logistic growth
Consider
\[\frac{dN}{dt}=rN\left(1-\frac{N}{K}\right),\qquad r\gt0.\]Define
\[f(N)=rN\left(1-\frac{N}{K}\right).\]The positive equilibrium is
\[N^*=K.\]The derivative is
\[f'(N)=r-\frac{2rN}{K}.\]At \(N=K\),
\[f'(K)=-r.\]So near carrying capacity, if
\[u=N-K,\]then
\[\boxed{\frac{du}{dt}\approx-r u}.\]Thus
\[u(t)\approx u(0)e^{-rt}.\]A small deviation from carrying capacity decays approximately exponentially.
What happened to the nonlinear term?
Substitute \(N=K+u\) directly into logistic growth:
\[\frac{du}{dt}=r(K+u)\left(1-\frac{K+u}{K}\right).\]Simplifying gives
\[\frac{du}{dt}=-ru-\frac{r}{K}u^2.\]The exact perturbation equation contains both a linear term and a quadratic term. Near equilibrium, \(|u|\) is small, so \(u^2\) is much smaller than \(|u|\). Dropping the quadratic term gives
\[\frac{du}{dt}\approx-ru.\]Two-dimensional systems
Now consider
\[\frac{dx}{dt}=f(x,y),\qquad\frac{dy}{dt}=g(x,y).\]Let \((x^*,y^*)\) be an equilibrium and define small displacements
\[u=x-x^*,\qquad v=y-y^*.\]Then
\[x=x^*+u,\qquad y=y^*+v.\]Multivariable Taylor approximation
Near the equilibrium,
\[f(x^*+u,y^*+v)\approx f(x^*,y^*)+\frac{\partial f}{\partial x}u+\frac{\partial f}{\partial y}v,\]and similarly
\[g(x^*+u,y^*+v)\approx g(x^*,y^*)+\frac{\partial g}{\partial x}u+\frac{\partial g}{\partial y}v.\]All partial derivatives here are evaluated at the equilibrium.
The equilibrium terms disappear
Because
\[f(x^*,y^*)=0,\qquad g(x^*,y^*)=0,\]we obtain
\[\frac{du}{dt}\approx\frac{\partial f}{\partial x}u+\frac{\partial f}{\partial y}v,\] \[\frac{dv}{dt}\approx\frac{\partial g}{\partial x}u+\frac{\partial g}{\partial y}v.\]Matrix form
These two equations can be written compactly as
\[\boxed{\frac{d}{dt}\begin{pmatrix}u\\v\end{pmatrix}\approx J(x^*,y^*)\begin{pmatrix}u\\v\end{pmatrix}},\]where
\[J(x^*,y^*)=\begin{pmatrix}\dfrac{\partial f}{\partial x}&\dfrac{\partial f}{\partial y}\\[6pt]\dfrac{\partial g}{\partial x}&\dfrac{\partial g}{\partial y}\end{pmatrix}_{(x^*,y^*)}.\]What the matrix equation means
The vector \((u,v)^T\) tells us how the state has been displaced from equilibrium. Multiplication by the Jacobian gives the approximate instantaneous rate at which that displacement changes.
The Jacobian therefore describes the local shape of the vector field around the equilibrium.
Nonlinear flow and its linear approximation
Why eigenvalues enter
Once the system has been linearised, we study
\[\frac{d\mathbf u}{dt}=J\mathbf u.\]If \(\mathbf v\) is an eigenvector of \(J\),
\[J\mathbf v=\lambda\mathbf v.\]A perturbation in that direction behaves like
\[e^{\lambda t}\mathbf v.\]So eigenvalues tell us whether the important local disturbance modes grow, decay or oscillate.
Local stability criterion
If every eigenvalue satisfies
\[\operatorname{Re}(\lambda_i)\lt0,\]the equilibrium is locally asymptotically stable.
If at least one eigenvalue has positive real part, the equilibrium is unstable.
What complex eigenvalues mean
If
\[\lambda=a+bi,\]the real part \(a\) controls growth or decay while the imaginary part \(b\) produces rotation or oscillation.
Thus a complex pair with negative real part produces a local spiral toward equilibrium.
When does linearisation reliably classify an equilibrium?
If no eigenvalue of the Jacobian has zero real part, the equilibrium is called hyperbolic. In this important case, the nonlinear system has the same local qualitative stability type as its linearisation.
This is why eigenvalue analysis is so powerful for nonlinear models.
When can the test fail?
If one or more eigenvalues have zero real part, the first-order linearisation may be inconclusive. Terms that were discarded as higher order can then determine the behaviour.
Local does not mean global
A linearisation describes a neighbourhood of the chosen equilibrium. Farther away, nonlinear terms may become large and produce very different behaviour.
The full system may contain other equilibria, cycles, thresholds or basins of attraction that cannot be inferred from one local linearisation.
Why linearisation is useful in biology
Biological models are often nonlinear because infection, predation, competition and density dependence involve products or nonlinear response functions.
Yet many questions concern small departures from a biologically important state: can infection invade a disease-free state? Will populations return after a small disturbance? Will a coexistence state resist perturbation?
Linearisation makes these questions mathematically manageable.
Example: disease-free equilibrium
In an epidemic model, we can linearise around a disease-free equilibrium where infected compartments are zero. The resulting local system describes what happens when a small amount of infection is introduced.
If the infected perturbation grows, the disease-free equilibrium is unstable to invasion. If it decays, the infection cannot establish under the model assumptions.
Example: ecological coexistence
For interacting species, linearising around a coexistence equilibrium tells us whether small changes in species abundances decay, grow, or produce damped oscillations.
The Jacobian entries encode local self-effects and cross-species interactions, while the eigenvalues combine these effects into local dynamical modes.
Linearisation is an approximation, not a new biological model
The original nonlinear model remains the model of the biological process. Linearisation is a mathematical tool used to understand its local behaviour.
We should therefore not interpret the linear approximation far outside the neighbourhood for which it was constructed.
A practical workflow
Find an equilibrium \(\mathbf x^*\). Define the perturbation \(\mathbf u=\mathbf x-\mathbf x^*\). Calculate the Jacobian of the nonlinear system. Evaluate it at the equilibrium. Write the local system \(d\mathbf u/dt\approx J(\mathbf x^*)\mathbf u\). Calculate its eigenvalues and interpret their real parts. If zero real parts occur, use additional nonlinear analysis.