# Problem in a system of ODEs

I'm running into a problem. This is mainly for reading over the summer and I'm working on getting through a dynamical systems book on my own. I've come across a system that I'm not too sure on the procedure.

Consider the following system of differential equations:
$\frac{dX}{dt} = 1 - X - XY - XZ, \\ \frac{dY}{dt} = aXY - Y,\\ \frac{dZ}{dt} = aXZ - Z.$

Here, 'a' is a real positive constant and $X,Y,Z \geq 0 .$

I can find all of the three steady states except for the one where all three exist, i.e. when $\bar{X},\bar{Y},\bar{Z} \neq 0 .$

When solving for this steady state, I arrive at $\bar{X} = \frac{1}{a} .$

Then, plugging this back into the first equation, I get the following

$0 = a -1 - \bar{Y} - \bar{Z},$

or,

$a-1 = \bar{Y}+\bar{Z}.$

Is there a way to solve for $\bar{Y}$ and $\bar{Z}$?

-as a side note, I've explored this numerically with matlab and it looks like it depends on the initial conditions. Could it be a saddle?

Continuing to the Jacobian anyways, we arrive at

$J = \left[ \begin{array}{ccc} -1 - (\bar{Y} + \bar{Z}) & -\bar{X} & -\bar{X} \\ a\bar{Y} & a\bar{X}-1 & 0 \\ a\bar{Z} & 0 & a\bar{X}-1 \end{array} \right].$

Plugging in what we know,

$J = \left[ \begin{array}{ccc} -a & -\frac{1}{a} &-\frac{1}{a} \\ a\bar{Y} & 0 & 0 \\ a\bar{Z} & 0 & 0 \end{array} \right].$

This means that the trace of the Jacobian is negative, i.e. $tr(J) = -a < 0$. Also that the determinant is equal to zero. My book does not talk about what happens when the determinant of the Jacobian is zero. Any insights?

AlephZero
Homework Helper
If you let U = Y+Z and W - Y-Z, the equations become
dX/dt = 1 - X - XU
dU/dt = aXU - U
dW/dt = aXW - W

Then, you have to consider more than one possibiltiy. From the third equation, if dW/dt = 0 then either aX = 1 or W = 0 .....

Thanks, that helps me understand it a bit better.

So, continuing that thought, if $\bar{X} = \frac{1}{a}$, then that implies $\bar{U} = a -1,\textrm{ and } \bar{W} = 0.$ This means that $\bar{Y} = \bar{Z}.$

So $a-1 = 2 \bar{Y}.$ Hence $\bar{Y} = \bar{Z} = \frac{a-1}{2}$.

But I'm still confused about the stability. The Jacobian of the new system is technically only slightly different but still results in a determinant of zero and a trace of $-a$.

Looking at the characteristic equation of such a matrix, we get

$det[J-I\lambda] = -\lambda^{3} + \lambda^{2}\textrm{trace}[J] + \frac{1}{2} \lambda \left[ \textrm{trace}[J^{2}] - \textrm{trace}[J]^{2} \right] + \textrm{det}[J].$

The trace and determinant are the same for both systems. I just hand wrote out the square of the Jacobian and the trace of it comes out to be just $a^{2}$.

So the characteristic equation is

$det[J-I\lambda] = -\lambda^{3} - a \lambda^{2} = 0$,

and

$\lambda_{1,2}= 0, \textrm{ and } \lambda_{3} = -a$.

And all the rules I know only apply when the eigenvalues are non-zero. I've read what two books I have on this and they don't go into this situation.

Any thoughts on the stability?

pasmith
Homework Helper
I think you've fallen into error here.

For a fixed point, all of $\dot X$, $\dot Y$ and $\dot Z$ must vanish. Looking at the Y equation, we see that $Y = 0$ or $aX-1 = 0$. Looking at the Z equation, we see that $Z = 0$ or $aX-1 = 0$.

Taking the case $Y = Z = 0$, the X equation gives $X = 1$, so there is a fixed point at $(1,0,0)$.

In the case $aX-1=0$, there will be a fixed point if $Y + Z = a - 1$. Thus every point on the line $(X,Y,Z) = (\frac{1}{a}, s, a-1-s)$ is a fixed point.

(Making the substitutions proposed by AlephZero gives $U = Y + Z= a -1$ and $X = \frac{1}{a}$, but $W = Y - Z$ is arbitrary, not 0 as you have assumed.)

The characteristic polynomial of the Jacobian is then
$$0=\mathrm{det}(J - \lambda I) = \left|\begin{array}{ccc}-a-\lambda & -a^{-1} & -a^{-1} \\ aY & -\lambda & 0 \\ aZ & 0 & -\lambda \end{array}\right| = (-a-\lambda)\left|\begin{array}{cc} -\lambda & 0 \\ 0 & -\lambda \end{array} \right| +a^{-1} \left|\begin{array}{cc} aY & 0 \\ aZ & -\lambda \end{array} \right| -a^{-1} \left|\begin{array}{cc} aY & -\lambda \\ aZ & 0 \end{array}\right|$$
which reduces to
$$\lambda^2(\lambda+a)+\lambda Y + \lambda Z = \lambda(\lambda^2 + a\lambda + (a-1)) = \lambda(\lambda + 1)(\lambda+ a - 1) = 0$$
There is therefore one zero eigenvalue (corresponding to the line of fixed points).

In general, when the jacobian at a fixed point has a zero eigenvalue, there are two possibilities. The first is that the system is degenerate and has non-isolated fixed points. The second is that the fixed point is not hyperbolic, and you must look at higher-order terms to determine the stability of the fixed point (the key term is "center manifold").

Here we have the first case: the system is degenerate. Stability is therefore determined by the other two eigenvalues: -1 (stable) and $1 - a$ (unstable if $a < 1$, indeterminate if $a = 1$, and stable if $a > 1$).

pasmith
Homework Helper
Now I'm up to 10 posts, I can give you an actual link: Center manifold