Coupled Differential Equations

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4 illustrated lessons, each teaching the why before the how.

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Coupled Differential Equations

Two rates that each need the other quantity.

Two rates that each depend on both quantities form one matrix equation, dX/dt = M X

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neither rate can be worked out alone

Two quantities whose rates each depend on the other are coupled, so they must be solved together.

dx/dtdy/dtM1141xy=×

Stack them and the pair is a single matrix equation: dX/dt = M X.

xy

Divide one rate by the other to get dy/dx, and the pair prescribes a direction at every point.

xyO

Both rates vanish at the origin, so a pair starting there never moves at all.

for example, migration between two towns

Population models take the same form: each quantity appears in the rate of the other.

both use the values at step n

Euler’s method still works: step both together, each from the current pair of values.

Now you

dx/dtdy/dtM1141xy=×

What makes a pair of differential equations coupled?

dx/dtdy/dtM1141xy=×

dx/dt = x + y and dy/dt = 4x + y. Where does the pair stay still?

Solving a Coupled System with Eigenvalues

One exponential per eigenvalue, added together.

With two real eigenvalues the solution is A e^(λ₁t) v₁ plus B e^(λ₂t) v₂

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try
a fixed direction, growing at one rate

Guess a solution that keeps one direction and only changes length: X = e^(λt) v.

the eigenvector equation

Cancel e^(λt) and the guess works exactly when v is an eigenvector of M.

M1141

For this M the characteristic polynomial is λ² − 2λ − 3, so λ = 3 or λ = −1.

gives v = (1, 2)
the two directions the flow keeps
gives v = (1, −2)

Solving (M − λI)v = 0 twice gives the directions (1, 2) and (1, −2).

the equation is linear
two solutions added is a solution

Add the two, each with its own constant, and that is the general solution.

X(0) = (2, 0) gives A + B = 22A − 2B = 0, so A = B = 1

The initial values fix the constants: from X(0) = (2, 0), both A and B come out as 1.

grows, decays
the larger eigenvalue dominates
X lines up along (1, 2)

In the long run the larger eigenvalue dominates, and the pair lines up along (1, 2).

Now you

M1141xy×

M has eigenvalues 3 and −1 with eigenvectors (1, 2) and (1, −2). The general solution is

M1141xy×

Feeding X = e^(λt) v into dX/dt = M X requires

Phase Portraits and Equilibrium Types

Saddle, stable node, unstable node, spiral.

The signs of the two eigenvalues decide whether paths run into the origin, away from it, or past it

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xy

Plot y against x rather than against t, and each solution becomes one path.

xy

The two eigenvector directions are straight paths: a path that starts on one never leaves it.

xyoutin

With one eigenvalue positive and one negative, paths come in, turn, and leave again. This is a saddle point.

xy

With both eigenvalues negative every path runs into the origin. This is a stable node.

xy

With both eigenvalues positive every path runs away from the origin instead. This is an unstable node.

xy

Complex eigenvalues turn every path; with a negative real part, each path spirals in to the origin.

opposite signs saddle point
both negative stable node
both positive unstable node

So the eigenvalues predict the shape of the portrait before a single path is drawn.

Now you

xy

The eigenvalues are −1 + 2i and −1 − 2i. The paths

xy

The eigenvalues are −2 and −4. The origin is

A Second-Order Equation as a Coupled System

Name the derivative and the order drops.

Naming the first derivative turns one second-order equation into two first-order ones

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give the derivative a name of its own
let

Name the derivative v and one second-order equation becomes two first-order ones.

the original equation, rearranged

The first line is the definition v = dx/dt, and the second line is the original equation solved for dv/dt.

dx/dtdv/dtM01−2−3xv=×

So dX/dt = M X with X = (x, v), and M reads straight off the coefficients.

the same quadratic as before
this is the auxiliary equation

Its characteristic polynomial is λ² + 3λ + 2, which is the auxiliary equation.

or

The roots −1 and −2 give the same solution the auxiliary equation gives.

xy

Drawn as v against x, every path tends to the origin: the motion dies away.

one step of the pair

Euler’s method applies too: one step of the pair is one step of the original equation.

Now you

dx/dtdv/dtM01−2−3xv=×

To turn x'' + 3x' + 2x = 0 into a pair, what is named v?

dx/dtdv/dtM01−2−3xv=×

The characteristic polynomial of that M is

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