Linearisation and the Jacobian · Differential Equations

Intuition

Near an equilibrium a smooth system looks almost linear, just as a smooth curve looks almost straight close to a point. Replace the rates by their best linear approximations and what is left is a linear system in the small displacements from the equilibrium, whose matrix is the Jacobian: the partial derivatives of the rates. Everything the linear chapter learned about that matrix then describes the nonlinear system near that point — with one exception, the subject of the next lesson.

The Earth is round, but a street plan of one town is flat and perfectly good for getting about in it. The linearisation is the street plan of an equilibrium.

The pendulum's swings, solid, and those of its linearisation $\theta'=\omega$ , $\omega'=-\theta$ , dashed circles, from the same starts. A small swing is almost exactly its circle; a large one is not, because $\sin\theta$ is close to $\theta$ only near $0$ .

The Jacobian at an equilibrium

At an equilibrium $(x^{*},y^{*})$ put $u=x-x^{*}$ and . Since and vanish there, their first-order Taylor approximations give the linear system below, whose matrix is the Jacobian of evaluated at the equilibrium. The neglected terms are of second order in and , small compared with the linear ones close to the equilibrium.

\begin{aligned}&\begin{pmatrix}u\\v\end{pmatrix}'=J\begin{pmatrix}u\\v\end{pmatrix}\\&J=\begin{pmatrix}f_{x}&f_{y}\\g_{x}&g_{y}\end{pmatrix}\ \text{at}\ (x^{*},y^{*})\end{aligned}

Computing it

Pendulum at $(0,0)$ : $f=\omega$ and $g=-\sin\theta$ give , the small swings .

Near an equilibrium the rates are linear plus smaller terms

\begin{aligned}&f(x^{*}+u,\,y^{*}+v)\\&=f_{x}u+f_{y}v+R,\qquad|R|\le C\,(u^{2}+v^{2})\end{aligned}

Expand each rate about the equilibrium by Taylor’s theorem, to first order. The constant term vanishes because the point is an equilibrium, and the remainder is of second order in the displacement when the second derivatives are continuous. What is left is a linear system in the displacements, whose matrix is the Jacobian; near the equilibrium the remainders are small compared with it.

Proof steps

f(x^{*}+u,y^{*}+v)=f(x^{*},y^{*})+f_{x}u+f_{y}v+R

The first-order Taylor expansion, with a remainder R.

f(x^{*},y^{*})=0

At an equilibrium the rate vanishes.

|R|\le C\,(u^{2}+v^{2})\ \text{near}\ (0,0)

With continuous second derivatives the remainder is of second order.

u'=f_{x}u+f_{y}v+R,\qquad v'=g_{x}u+g_{y}v+S

The same for g; dropping R and S leaves the linear system with matrix J.

Applications

Practice

Partial Derivatives in a Square

Row one of the Jacobian differentiates f, row two differentiates g. Column one is the derivative in x, column two in y.

J=\begin{pmatrix}f_{x}&f_{y}\\g_{x}&g_{y}\end{pmatrix}

Try it

For $x'=x-xy$ and $y'=-y+xy$ , the Jacobian at $(1,1)$ is below. Select the entry that is $\partial g/\partial x$ , the derivative of $-y+xy$ in $x$ .

$J=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$