I9 - Coupled differential equations

I9 - Coupled differential equations

Two quantities can change together, with each rate depending on both quantities. I9 represents this interaction with coupled first-order equations. You will solve constant-coefficient systems using real eigenmodes, see how elimination produces one second-order equation, and interpret initial conditions and population-model signs.

Two quantities changing together

Two tanks start with amounts 33 and 11. If material moves between them and also leaks away, why can their imbalance disappear faster than their total? One equation for the total will not tell us how much is in each tank.

Suppose the amount leaving tank 1 for tank 2 per unit time is xx, and the reverse transfer rate is yy. Each tank also loses material to the outside at a rate equal to its own amount. These are rate assumptions, with coefficients measured per unit time. Then

dxdt=y−x−x=−2x+ydydt=x−y−y=x−2y\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t}&=y-x-x=-2x+y\\ \frac{\mathrm{d}y}{\mathrm{d}t}&=x-y-y=x-2y \end{aligned}

Each transfer is a loss in one equation and a gain in the other. It should not count as a loss from both tanks.

Coupled first-order equations

In a pair of coupled first-order differential equations, two dependent variables, such as x(t)x(t) and y(t)y(t), share one independent variable tt, and their rates are linked. A solution is a pair of functions satisfying both equations at the same times.

At x=3,y=1x=3,y=1, the model gives x′=−5x'=-5, y′=1y'=1. The smaller tank initially gains material even though the total is falling: (x+y)′=−x−y=−4(x+y)'=-x-y=-4. These are derivatives with respect to time; y′y' here is not dy/dx\mathrm{d}y/\mathrm{d}x.

Using column vectors, the same system is

u′=Au,u=(xy),A=(−211−2)\mathbf{u}'=\mathbf{A}\mathbf{u}, \qquad \mathbf{u}=\begin{pmatrix}x\\y\end{pmatrix}, \qquad \mathbf{A}=\begin{pmatrix}-2&1\\1&-2\end{pmatrix}

The first row specifies x′x', and the second specifies y′y'. This is the same matrix representation used in diagonalisation, now applied to rates.

Key idea

The two variables are functions of the same time. A proposed pair must satisfy both rate equations; satisfying just one leaves half the model unchecked.

Check the signs in a fresh transfer model.

Real eigenmodes and elimination

For the opening system, adding and subtracting the two equations gives independent scalar equations. Let s=x+ys=x+y and d=x−yd=x-y. Then s′=−ss'=-s and d′=−3dd'=-3d: the total decays at one rate, and the imbalance at another.

This is diagonalisation in a useful disguise. The eigenvectors (1,1)T(1,1)^{\mathsf T} and (1,−1)T(1,-1)^{\mathsf T}, with eigenvalues −1-1 and −3-3, describe the common and opposing modes. A scalar equation z′=λzz'=\lambda z has solutions z=Ceλtz=Ce^{\lambda t}, including the zero solution.

Solution by real eigenmodes

If a constant real 2×22\times2 matrix A\mathbf A has two independent real eigenvectors v1,v2\mathbf v_1,\mathbf v_2, with real eigenvalues λ1,λ2\lambda_1,\lambda_2, then the general solution of u′=Au\mathbf u'=\mathbf A\mathbf u is

u=C1eλ1tv1+C2eλ2tv2\mathbf u=C_1e^{\lambda_1t}\mathbf v_1+C_2e^{\lambda_2t}\mathbf v_2

Not in the formula booklet: learn it.

Why does this work? Differentiating one mode gives λeλtv=eλtAv\lambda e^{\lambda t}\mathbf v=e^{\lambda t}\mathbf A\mathbf v, as required. In the eigenvector basis the two coefficients satisfy separate scalar equations, so the two constants describe every initial vector. Distinct real eigenvalues provide independent eigenvectors; repeated real eigenvalues alone do not guarantee an eigenbasis.

Worked example 1: separating total and imbalance

Question. Find the general solution of x′=−2x+yx'=-2x+y, y′=x−2yy'=x-2y.

Plan. Solve for the total and difference, then add and subtract to reconstruct both variables.

s′=x′+y′=−(x+y)=−sd′=x′−y′=−3(x−y)=−3ds=Ke−t,d=Le−3tx=s+d2,y=s−d2\begin{aligned} s'&=x'+y'=-(x+y)=-s\\ d'&=x'-y'=-3(x-y)=-3d\\ s&=Ke^{-t},\qquad d=Le^{-3t}\\ x&=\frac{s+d}{2},\qquad y=\frac{s-d}{2} \end{aligned}

Answer. Writing A=K/2A=K/2, B=L/2B=L/2,

x=Ae−t+Be−3t,y=Ae−t−Be−3tx=Ae^{-t}+Be^{-3t},\qquad y=Ae^{-t}-Be^{-3t}

Check. The first right side is −2x+y=−Ae−t−3Be−3t=x′-2x+y=-Ae^{-t}-3Be^{-3t}=x'. The second is x−2y=−Ae−t+3Be−3t=y′x-2y=-Ae^{-t}+3Be^{-3t}=y'.

Increase the external loss rate in each tank from one times its amount to two times its amount. Only this common loss parameter changes: each diagonal coefficient becomes −3-3, while the transfer coefficients remain 11.

Worked example 2: the same modes with extra loss

Question. Find the general solution of x′=−3x+yx'=-3x+y, y′=x−3yy'=x-3y.

Plan. Use the same total-and-difference coordinates; recompute their rates.

s′=−2(x+y)=−2sd′=−4(x−y)=−4ds=Ke−2t,d=Le−4tx=12Ke−2t+12Le−4ty=12Ke−2t−12Le−4t\begin{aligned} s'&=-2(x+y)=-2s\\ d'&=-4(x-y)=-4d\\ s&=Ke^{-2t},\qquad d=Le^{-4t}\\ x&=\frac12Ke^{-2t}+\frac12Le^{-4t}\\ y&=\frac12Ke^{-2t}-\frac12Le^{-4t} \end{aligned}

Answer. x=Ae−2t+Be−4tx=Ae^{-2t}+Be^{-4t}, y=Ae−2t−Be−4ty=Ae^{-2t}-Be^{-4t}.

Check. Substitution gives −3x+y=−2Ae−2t−4Be−4t=x′-3x+y=-2Ae^{-2t}-4Be^{-4t}=x' and x−3y=−2Ae−2t+4Be−4t=y′x-3y=-2Ae^{-2t}+4Be^{-4t}=y'. Both modes decay faster than before, consistent with the extra loss.

There is also an elimination route. For the first system, differentiate x′=−2x+yx'=-2x+y, substitute for y′y', then use y=x′+2xy=x'+2x:

x′′=−2x′+y′=−2x′+x−2y=−2x′+x−2(x′+2x)x′′+4x′+3x=0\begin{aligned} x''&=-2x'+y'=-2x'+x-2y\\ &=-2x'+x-2(x'+2x)\\ x''+4x'+3x&=0 \end{aligned}

The modes already found give x=Ae−t+Be−3tx=Ae^{-t}+Be^{-3t}. Reconstructing y=x′+2xy=x'+2x gives Ae−t−Be−3tAe^{-t}-Be^{-3t}. Thus elimination agrees with the eigenmode result. The systematic method for solving such a second-order equation is developed next.

For x′=ax+byx'=ax+by, y′=cx+dyy'=cx+dy, with constant coefficients and b≠0b\ne0, exactly the same substitutions give

y=x′−axb,x′′−(a+d)x′+(ad−bc)x=0y=\frac{x'-ax}{b},\qquad x''-(a+d)x'+(ad-bc)x=0

The coefficients involve the trace a+da+d and determinant ad−bcad-bc, linking elimination to the matrix's characteristic equation. If b=0b=0, do not divide by it: the first equation is already independent, or eliminate the other variable when possible.

Watch out

Wrong: attaching independent constants to every exponential in both variables. The equations link their coefficients. Each eigenmode has one shared constant multiplying its whole eigenvector.

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