I9 - Solving coupled models by elimination

I9 - Solving coupled models by elimination

A coupled model needs two functions of the same independent variable, with each function satisfying its own rate equation. I9 extends elimination to oscillatory and forced systems, reconstructs the second variable and interprets a predator–prey approximation. You will check the full pair and distinguish deviations from actual populations.

Eliminating one dependent variable

Can a rise in prey numbers be followed by a later rise in predators, even when both populations repeatedly return to their starting levels? Coupled rates can create that delay. First see how two first-order equations turn into one oscillation equation.

Let x(t),y(t)x(t),y(t) share the independent variable tt, and use primes for time derivatives. From x′=yx'=y, y′=−xy'=-x, differentiation gives x′′=y′=−xx''=y'=-x. The resulting equation has complex auxiliary roots, so its solutions oscillate. This is different from the real exponential modes in the earlier tank model.

Elimination with constant coefficients

For

x′=ax+by+f(t),y′=cx+dy+g(t),b≠0x'=ax+by+f(t),\qquad y'=cx+dy+g(t),\qquad b\ne0

with constant a,b,c,da,b,c,d and differentiable ff, use

y=x′−ax−f(t)by=\frac{x'-ax-f(t)}b

Differentiating the first equation and substituting gives

x′′−(a+d)x′+(ad−bc)x=f′(t)−df(t)+bg(t)x''-(a+d)x'+(ad-bc)x=f'(t)-df(t)+bg(t)

Not in the formula booklet: derive it by substitution rather than relying on memory.

Indeed, x′′=ax′+b(cx+dy+g)+f′x''=ax'+b(cx+dy+g)+f'. Replace byby by x′−ax−fx'-ax-f, collect the x′x' and xx terms, and the displayed equation follows. Notice the derivative f′f': a time-dependent input must also be differentiated.

If b=0b=0, do not divide by it. The first equation is already independent of yy; solve it first and then solve the second first-order equation, or eliminate xx instead when c≠0c\ne0. The method must respect the actual coefficients.

Worked example 1: elimination produces an oscillation

Question. For x′=yx'=y, y′=−xy'=-x, obtain a second-order equation for xx and find its general solution.

Plan. Differentiate x′=yx'=y and substitute the other rate equation.

x′′=y′=−xx′′+x=0m2+1=0,m=±ix=Acos⁡t+Bsin⁡t\begin{aligned} x''&=y'=-x\\ x''+x&=0\\ m^2+1&=0,\qquad m=\pm\mathrm{i}\\ x&=A\cos t+B\sin t \end{aligned}

Answer. x′′+x=0x''+x=0, with x=Acos⁡t+Bsin⁡tx=A\cos t+B\sin t.

Check. Differentiating the proposed xx twice gives −Acos⁡t−Bsin⁡t=−x-A\cos t-B\sin t=-x. We have checked the eliminated equation; the original pair still needs yy, which we reconstruct next.

Add only the time-dependent input tt to the first rate equation. Its derivative will supply a constant forcing term.

Worked example 2: an input becomes a forcing term

Question. For x′=y+tx'=y+t, y′=−xy'=-x, obtain a second-order equation for xx and find its general solution.

Plan. Differentiate the entire first equation, including tt, then use a constant particular integral.

x′′=y′+1=−x+1x′′+x=1xC=Acos⁡t+Bsin⁡txP=1\begin{aligned} x''&=y'+1=-x+1\\ x''+x&=1\\ x_{\mathrm C}&=A\cos t+B\sin t\\ x_{\mathrm P}&=1 \end{aligned}

Answer. x=1+Acos⁡t+Bsin⁡tx=1+A\cos t+B\sin t.

Check. Here x′′=−Acos⁡t−Bsin⁡t=1−xx''=-A\cos t-B\sin t=1-x, so the second-order equation holds. Omitting the derivative of tt would wrongly remove the constant response.

An eliminated equation is a tool for finding one component, not the finished coupled solution. The relation used to eliminate the other component must be retained.

Watch out

Wrong: replacing y′y' by yy, or differentiating x′=y+tx'=y+t as x′′=y′x''=y'. Every term is a function of the same time; differentiate the input as well.

Change the input rate and feedback coefficient in a fresh system.

Reconstructing the second solution

The first original equation tells us how to reconstruct yy from xx and x′x'. It also turns starting values for x,yx,y into a starting slope for xx. Neither operation introduces an extra arbitrary constant.

For example, if x′=y+tx'=y+t, then y=x′−ty=x'-t. At t=0t=0, x′(0)=y(0)x'(0)=y(0). At a different starting time t0t_0, the input is not zero: x′(t0)=y(t0)+t0x'(t_0)=y(t_0)+t_0.

Worked example 3: reconstructing an unforced pair

Question. Solve x′=yx'=y, y′=−xy'=-x, given x(0)=0x(0)=0, y(0)=1y(0)=1.

Plan. Use the eliminated solution, obtain x′(0)x'(0) from the first equation, then reconstruct yy.

x=Acos⁡t+Bsin⁡tA=0x(0)=0x′(0)=B=y(0)=1x=sin⁡ty=x′=cos⁡t\begin{aligned} x&=A\cos t+B\sin t\\ A&=0 && x(0)=0\\ x'(0)&=B=y(0)=1\\ x&=\sin t\\ y&=x'=\cos t \end{aligned}

Answer. x=sin⁡tx=\sin t, y=cos⁡ty=\cos t.

Check. x′=cos⁡t=yx'=\cos t=y and y′=−sin⁡t=−xy'=-\sin t=-x. At zero, x=0,y=1x=0,y=1. Both equations and both conditions hold.

Keep the initial values and add only the input tt from Worked example 2. The reconstruction now includes its subtraction.

Worked example 4: reconstructing a forced pair

Question. Solve x′=y+tx'=y+t, y′=−xy'=-x, given x(0)=0x(0)=0, y(0)=1y(0)=1.

Plan. Fix the constants in x=1+Acos⁡t+Bsin⁡tx=1+A\cos t+B\sin t, then use y=x′−ty=x'-t.

1+A=0initial xB=x′(0)=y(0)+0=1x=1−cos⁡t+sin⁡tx′=sin⁡t+cos⁡ty=sin⁡t+cos⁡t−t\begin{aligned} 1+A&=0 && \text{initial }x\\ B&=x'(0)=y(0)+0=1\\ x&=1-\cos t+\sin t\\ x'&=\sin t+\cos t\\ y&=\sin t+\cos t-t \end{aligned}

Answer. x=1−cos⁡t+sin⁡tx=1-\cos t+\sin t, y=sin⁡t+cos⁡t−ty=\sin t+\cos t-t.

Check. The first right side is y+t=sin⁡t+cos⁡t=x′y+t=\sin t+\cos t=x'. The second derivative is y′=cos⁡t−sin⁡t−1=−xy'=\cos t-\sin t-1=-x. The initial values are 0,10,1. Both original equations work, including the input.

The two constants in xx control the whole pair. Solving an independent second-order equation for yy and assigning it two more unrelated constants would usually produce pairs that fail the original system.

The elimination formula is reversible when b≠0b\ne0: reconstructing yy enforces the first equation, and the eliminated equation then enforces the second. Checking both directly is still a quick way to catch a sign or input error.

Watch out

Wrong: taking y=x′y=x' in a system with x′=y+tx'=y+t. The correct relation is y=x′−ty=x'-t. Also derive the starting slope from the original rate equation, rather than confusing y(0)y(0) with y′(0)y'(0).

Solve a new system with changed input and initial values.

Continue this lesson.

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