Thinking further about this I thought it would be beneficial to the student if what might be coming across as an oddball exception is more integrated in its context of physicochemical applications and that of homogeneous linear d.e.'s with constant coefficients.
The equation, far from anything unusual, could stand e.g. for the simplest mechanism and kinetics of a chemical reaction of X converting to Y reversibly.
X \stackrel{k_{xy}}{\rightarrow} Y
...\stackrel{\leftarrow}{k_{yx}}
The main interest would be in the time domain where, solving the equation , you find a (negative) exponential decrease with eigenvalue -(k
xy + k
yx)* of X or Y towards the equilibrium level X
eq/Y
eq = k
yx/k
xy = K
eq. The equilibrium is approached from one side or the other depending on the initial X/Y and never overshot, according to what we said before.
I just failed to associate with that before, because this reaction is never represented in the 'phase plane' (the XY plane) no doubt because that would be almost totally uninformative! The X that disappears in reaction equals the Y that appears or vice versa - that is just conservation of mass - so we always would have the slope of the phase paths is -1 as we said , or -45° angle, we noticed previously. But we would have that with any physically possible mechanism, can have any number of mechanisms and of rate laws (the differential equations) but the picture would be the same for all of them as long as no intermediate or by-product of the reaction is present in concentrations significant compared to X and Y. The forms of time dependence could then differ from that mentioned above; however X
eq/Y
eq = K
eq would always be true when equilibrium is reached, the value of K
eq is independent of reaction mechanism.
In your example your final equilibrium point (X
eq, Y
eq) on the Y=2X line, though not the ratio, depends on your initial conditions, in other words on the total amount (X
0 + Y
0) of reacting stuff that is there. You'd get a single stationary point, like you had been expecting, at (0, 0) if you introduced an additional irreversible reaction Y \stackrel{k_{yz}}{\rightarrow} Z .
Then you've got
X' = -k
xyX + k
yxY
Y' = k
xyX - (k
yx+k
yz)Y ...(eq.1)
(An equation for Z' would be redundant because of conservation of mass). No longer is the determinant of your 2X2 matrix 0. Your previous equilibrium line is replaced by another strong line - the characteristic or eigen- vector to which the phase paths tend - see fig 1 (phase paths are now a bit more informative) which is the typical pattern when eigenvalues are real negative. (For the given chemical example everything happens in the positive quadrant of course).
This pattern morphs towards one that seems to resemble your original problem as you make k
yz small, understandably. See figs 2, 3. What is happening there with small k
yz is that X and Y interconvert to close to the equilibrium ratio then your point slowly creeps down in a path near the eigenvector to (0, 0) as Y slowly converts to Z. As k
yz gets small the eigenvector gets close to the equilibrium line and the isoclines get close to the eigenvector too.
It would be instructive if this is what you are on to now, to solve eq. 1.
Instructive too to work out the eigenvalues when k
yz is small and see that one becomes close to -(k
xy+kyx) i.e. the same as above for the mechanism without the Y -> Z step, which makes sense. Slightly more difficult, you can work out that the other eigenvalue is approximately – kxykyz/ (kxy+kyx) . That also makes sense because it equals -kyz which would be the eigenvalue (constant in the exponent in the exponential) for the simple process of disappearance of Y in reaction Y -> Z multiplied by the fraction Yeq/(Xeq + Yeq)of Y present in (X + Y) .
I mention these because this is often done in practice. For complicated mechanisms not rarely the differential equation is not solved in full, but parts of it are assumed much faster than others so that a quasi-equilibrium approximation can be made and the problem split up into two or more simpler ones. Not rarely either the experimental conditions, concentrations etc. are arranged to make the equations simpler!
It is also instructive and useful to solve the full equation for the mechanism.
X \stackrel{k_{xy}}{\rightarrow} Y \stackrel{k_{yz}}{\rightarrow} Z
...\stackrel{\leftarrow}{k_{yx}} ...\stackrel{\leftarrow}{k_{zy}}
X \stackrel{k_{xy}}{\rightarrow} Y \stackrel{k_{yz}}{\rightarrow} Z
...\stackrel{\leftarrow}{k_{yx}}\stackrel{\leftarrow}{k_{zy}}
I give up on Tex
here. Just trying to say X -> Y -> Z and the reverse with obvious notation for rate constants.
In order however to have homogeneous d.e.’s (without constant terms so that your stationary (equilibrium) point is at (0, 0)) you will have to make the variables ‘distance from equilibrium' ones (X – Xeq), (Y – Yeq). You can eliminate Z from equations by mass conservation. You can also check their simplifications as above in cases of gross inequalities of rate constants.
I don’t say you should do these exercises. No should. I say that if you do it now you are doing these simultaneous linear d.e.‘s and keep your notes you will be glad later because you will surely need it again in chemistry, biochemistry or physics and by the time you need them they will have gotten worryingly hazy and confused, but that way you can get back on top of it fast. Of same nature is the Bateman equations for successive radioactive decays, subject of ongoing thread here, which are like the chemical ones with all reactions irreversible. The math of coupled vibrations is much the same math by the way.
* (X - Xeq) = (X - X0) exp[-(kxy + kyx)]
Fig. 1
Fig 2
Fig 3Phaseplane plotter: http://www.math.rutgers.edu/curses/ODE/sherod/phaselocal.html
. I'm understanding by x', y'. dx/dt and dy/dt.
