Coupled Differential Equations

Two rates, each depending on both

Take $\frac{dx}{dt}=x+y$dx/dt = x + y and $\frac{dy}{dt}=4x+y$dy/dt = 4x + y. The rate of x depends on y as well as on x, and the rate of y depends on x as well as on y. So neither equation can be separated and solved on its own: to know how x changes you must already know y, and the other way round. A pair like this is coupled, and the two equations are solved together.

At the point (1, 0) the rates are 1 + 0 = 1 and 4 + 0 = 4, so x grows slowly while y grows four times as fast. At (0, 1) they are both 1. The pair gives the velocity of the point (x, y) wherever it is.

One matrix equation

Stack the two quantities into one column, X = (x; y), and the two rates into its derivative $\frac{dX}{dt}$dX/dt, the column whose entries are $\frac{dx}{dt}$dx/dt and $\frac{dy}{dt}$dy/dt. The right-hand sides are then a matrix times X: (1 1; 4 1) times (x; y) is (x + y; 4x + y). So the pair is the single equation $\frac{dX}{dt}=MX$dX/dt = M X with M = (1 1; 4 1).

Each equation fills one row of M, with the coefficient of x first and the coefficient of y second. A rate that leaves out a letter has a 0 in that place: $\frac{dx}{dt}=3x$dx/dt = 3x and $\frac{dy}{dt}=x-2y$dy/dt = x − 2y give M = (3 0; 1 −2).

A direction at every point

A solution is a pair of functions x(t) and y(t), and as t runs the point (x, y) traces a path. Dividing one rate by the other removes t: $\frac{dy}{dx}=\left(\frac{dy}{dt}\right)/\left(\frac{dx}{dt}\right)=\frac{4x+y}{x+y}$dy/dx = (dy/dt) / (dx/dt) = (4x + y)/(x + y). That is a gradient at every point, so the pair draws a slope field, and every path runs along it.

The slope field has lost one thing: which way along the segment the point moves. The rates themselves still say. At (1, 0) the velocity is (1, 4), up and to the right. At (−1, 0) it is (−1, −4), down and to the left, along a segment of the same gradient. On the line y = −x, $\frac{dx}{dt}=0$dx/dt = 0, so the path there moves straight up or down.

The slope field $\frac{dy}{dx}=\frac{4x+y}{x+y}$dy/dx = (4x + y)/(x + y), with the velocity (x + y, 4x + y) drawn at a quarter of its length at six points: (1, 4) at (1, 0), (1, 1) at (0, 1), and (−1, 2) at (1, −2), each with its opposite at the opposite point. On y = −x the field is vertical and no segment is drawn.

Where nothing moves

A point where both rates are 0 is an equilibrium: a pair that starts there stays there. Here x + y = 0 and 4x + y = 0. Subtracting the first from the second gives 3x = 0, so x = 0 and then y = 0. The origin is the only equilibrium.

That is the case whenever det M is not 0: here det M = 1 × 1 − 1 × 4 = −3, so M X = 0 has only the solution X = 0. A pair with det M = 0 has a whole line of points at rest, and a pair with constant terms, such as $\frac{dx}{dt}=x+y-2$dx/dt = x + y − 2, has its equilibrium somewhere else.

Two towns

Models of two connected quantities take the same form. Take $\frac{dx}{dt}=-0.2x+0.1y$dx/dt = −0.2x + 0.1y and $\frac{dy}{dt}=0.3x-0.4y$dy/dt = 0.3x − 0.4y, for two populations that each lose people at a rate set by their own size and gain them at a rate set by the other’s. In each equation the coefficient of its own quantity is negative and the coefficient of the other is positive, and M = (−0.2 0.1; 0.3 −0.4).

Its determinant is 0.08 − 0.03 = 0.05, which is not 0, so again the only equilibrium is the origin. The next lesson solves this pair: with x = 100 and y = 50 at the start, it gives $x=87.5e^{-0.1t}+12.5e^{-0.5t}$x = 87.5e^(−0.1t) + 12.5e^(−0.5t) and $y=87.5e^{-0.1t}-37.5e^{-0.5t}$y = 87.5e^(−0.1t) − 37.5e^(−0.5t), and by t = 10 the two are almost equal, at about 32.3 and 31.9.

Euler’s method for a pair

Euler’s method still works, one step for both quantities at once. From the values $x_n$xₙ and $y_n$yₙ, work out both rates there, then move each quantity by h times its own rate: $x_{n+1}=x_n+h\left(x_n+y_n\right)$xₙ₊₁ = xₙ + h(xₙ + yₙ) and $y_{n+1}=y_n+h\left(4x_n+y_n\right)$yₙ₊₁ = yₙ + h(4xₙ + yₙ).

From (1, 0) with h = 0.1: the rates are 1 and 4, so the first step reaches (1.1, 0.4). There the rates are 1.5 and 4.8, giving (1.25, 0.88). Two more steps give (1.463, 1.468) and (1.756, 2.2).

The exact solution from (1, 0), found in the next lesson, is $x=0.5e^{3t}+0.5e^{-t}$x = 0.5e^(3t) + 0.5e^(−t), $y=e^{3t}-e^{-t}$y = e^(3t) − e^(−t). At t = 0.2 it is (1.320, 1.003), so the two Euler steps are low. As for a single equation, smaller steps miss by less.

Four Euler steps of h = 0.1 from (1, 0), drawn as arrows: to (1.1, 0.4), (1.25, 0.88), (1.463, 1.468) and (1.756, 2.2). The gold curve is the exact path, through (1.320, 1.003) at t = 0.2 and (1.995, 2.650) at t = 0.4. The arrows run above and to the left of it, and each Euler point falls short of where the exact path is at the same time.

Predators and prey

Coupled pairs need not be linear. In the instrument below, x is a number of prey and y a number of predators: $\frac{dx}{dt}=x-xy$dx/dt = x − xy and $\frac{dy}{dt}=xy-y$dy/dt = xy − y. The prey grow on their own and are eaten when they meet predators; the predators die out on their own and grow when they meet prey. The point (1, 1) is an equilibrium, and every other path circles it.

The path from 2.2 prey and 0.6 predators closes up on itself, because x − ln x + y − ln y keeps the value it starts with. Beside it, the two numbers against time: each peak in the prey comes before a peak in the predators. Drag the start onto the line y = 1, right of (1, 1): there $\frac{dx}{dt}=x-x=0$dx/dt = x − x = 0, so the path starts at a peak in the prey.

The usual mistakes

Filling M by columns. The first row of M is the first equation: $\frac{dx}{dt}=3x-y$dx/dt = 3x − y and $\frac{dy}{dt}=x+2y$dy/dt = x + 2y give (3 −1; 1 2), not (3 1; −1 2).

Dropping a sign or a zero. A minus sign travels with its coefficient into M, and a missing letter is a 0 entry, not a blank.

Using a new value too soon. In an Euler step both rates are worked out at step n. Updating x first and then using the new x in the rate of y is a different method, and gives a different answer.

Expecting the slope field to show the direction of travel. $\frac{dy}{dx}=\frac{4x+y}{x+y}$dy/dx = (4x + y)/(x + y) is the same at (1, 0) and at (−1, 0), but the point moves up the segment at one and down it at the other.

Brine in two tanks, and a technetium generator

In the applications below, salt passes between two tanks in both directions, so each tank’s salt changes at a rate set by both; eliminating one tank turns the pair into one second-order equation. In the generator, one isotope decays into another, so the second rate depends on the first quantity but the first does not depend on the second, and the pair can be solved one equation at a time.