CP 9.9 - Coupled first-order equations

CP 9.9 - Coupled first-order equations

Two changing quantities can affect each other's rates of change. You will build and interpret linear coupled models, eliminate one variable to obtain a second-order equation, and recover the other variable with the same two linked constants.

Two quantities sharing one time variable

Can you solve two first-order equations by solving each independently? If each rate depends on both quantities, the two solutions must fit together at every instant. Choosing four unrelated constants would describe motions that the original equations do not permit.

Consider two stores containing amounts xx and yy. Material transfers from the first to the second at rate 2x2x, and back at rate 3y3y. An external input of 44 enters the first store, while material leaves the second at rate yy. Each internal transfer has one minus sign and one plus sign:

x˙=4−2x+3y,y˙=2x−4y\dot x=4-2x+3y,\qquad \dot y=2x-4y

[DIAGRAM: asset_name: Transfers in a two-variable linear model; asset_slug: 9fm0-cp99-coupled-stores; caption: Every internal transfer leaves one store and enters the other; description: Two labelled boxes contain x and y. An upper arrow from x to y is labelled rate 2x, a lower arrow from y to x is labelled rate 3y, an external arrow of rate 4 enters x from the left, and an external arrow of rate y leaves the second store to the right. Below are the equations x dot=4-2x+3y and y dot=2x-4y.]
Every internal transfer leaves one store and enters the other

You already know

The independent variable is time tt; x(t)x(t) and y(t)y(t) are dependent variables. You can solve a second-order constant-coefficient equation and use initial data to determine two constants. A derivative of a time-dependent forcing term must be included when differentiating an equation.

A linear coupled model has the form

x˙=ax+by+f(t),y˙=cx+dy+g(t)\dot x=ax+by+f(t),\qquad \dot y=cx+dy+g(t)

with constant coefficients. Terms such as xyxy or x2x^2 would make it non-linear and require different methods.

In a simplified predator-prey model, a positive prey contribution to predator growth and a negative predator contribution to prey growth give opposite signs in the cross terms. A linear model may be useful over a limited range; do not assume it predicts meaningful non-negative population counts for arbitrarily long times.

For the two stores, adding the equations gives a useful surprise:

ddt(x+y)=4−y\frac{\mathrm d}{\mathrm dt}(x+y)=4-y

The internal transfers cancel completely. The total's instantaneous rate depends only on external input and loss, although each store's individual rate depends on both stores.

Key idea

Coupling links rates and constants. Check the signs against the physical interpretation, then check both original equations against any proposed pair of solutions.

Match the arrows to the equations in this diagnostic check.

Eliminating one variable

Differentiate one equation, replace the other variable's derivative using the second equation, then remove the other variable itself using the first equation. The result is a familiar second-order equation.

Keep a clear distinction between xx, x˙\dot x and x¨\ddot x. Substituting a rate equation for the variable itself is a common source of sign errors.

Worked example 1: two real auxiliary roots

Question. Given x˙=x+y\dot x=x+y, y˙=x+y\dot y=x+y, find a general expression for x(t)x(t).

Plan. Differentiate the first equation and use both original equations to eliminate yy.

x¨=x˙+y˙differentiate the first equation=x˙+x+yuse the second equation=2x˙x+y=x˙\begin{aligned} \ddot x&=\dot x+\dot y && \text{differentiate the first equation}\\ &=\dot x+x+y && \text{use the second equation}\\ &=2\dot x && x+y=\dot x \end{aligned}

Thus x¨−2x˙=0\ddot x-2\dot x=0, with auxiliary equation r(r−2)=0r(r-2)=0.

Why differentiate only one equation? It raises that variable to second order, creating a single equation with the two constants the pair originally needs.

Answer. x=A+Be2tx=A+B\mathrm{e}^{2t}.

Check. x˙=2Be2t\dot x=2B\mathrm{e}^{2t}, x¨=4Be2t=2x˙\ddot x=4B\mathrm{e}^{2t}=2\dot x.

Worked example 2 changes only the coefficient of xx in the second equation, from 11 to −1-1. The elimination steps are the same, but the auxiliary roots become complex.

Worked example 2: a complex auxiliary pair

Question. Given x˙=x+y\dot x=x+y, y˙=−x+y\dot y=-x+y, find a general expression for x(t)x(t).

Plan. Differentiate the first equation, replace y˙\dot y, then use y=x˙−xy=\dot x-x.

x¨=x˙+y˙=x˙−x+ysubstitute the second equation=2x˙−2xy=x˙−x\begin{aligned} \ddot x&=\dot x+\dot y\\ &=\dot x-x+y && \text{substitute the second equation}\\ &=2\dot x-2x && y=\dot x-x \end{aligned}

The auxiliary equation is r2−2r+2=0r^2-2r+2=0, with roots 1±i1\pm\mathrm{i}.

Why substitute y=x˙−xy=\dot x-x? That removes the final occurrence of the second dependent variable, rather than leaving a disguised coupled equation.

Answer. x=et(Acos⁡t+Bsin⁡t)x=\mathrm{e}^t(A\cos t+B\sin t).

Check. The exponential rate 11 and angular frequency 11 correspond to the real and imaginary parts of the roots.

The solution for xx alone is only an intermediate result when a question asks for both dependent variables. The original first equation will recover yy next.

Watch out

Wrong: after differentiating, replace yy by the expression for y˙\dot y. Right: first replace y˙\dot y with its rate equation, then rearrange an original equation to replace yy.

Change the cross coefficient once more.

Recovering the second variable with linked constants

Once x(t)x(t) is known, recover y(t)y(t) directly from an original equation. This automatically keeps the two constants consistent across both solutions.

In both previous examples, the first equation gives y=x˙−xy=\dot x-x. Differentiate the expression for xx, subtract xx, then simplify. Starting a new second-order solution for yy introduces extra constants that would have to be reconciled afterwards.

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