# Problem in a system of ODEs

1. Jun 15, 2012

### toastermm

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?

2. Jun 15, 2012

### toastermm

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?

3. Jun 15, 2012

### AlephZero

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 .....

4. Jun 15, 2012

### toastermm

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?

5. Jun 23, 2012

### pasmith

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 $a > 1$).

6. Jun 23, 2012

### pasmith

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