How Do You Solve for Non-Zero Steady States in a Tri-Variable ODE System?

[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, \\<br /> \frac{dY}{dt} = aXY - Y,\\<br /> \frac{dZ}{dt} = aXZ - Z.<br />

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?

 

[toastermm]

Continuing to the Jacobian anyways, we arrive at

<br /> J = \left[<br /> \begin{array}{ccc}<br /> -1 - (\bar{Y} + \bar{Z}) &amp; -\bar{X} &amp; -\bar{X} \\<br /> a\bar{Y} &amp; a\bar{X}-1 &amp; 0 \\<br /> a\bar{Z} &amp; 0 &amp; a\bar{X}-1 <br /> \end{array}<br /> \right].<br />

Plugging in what we know,

<br /> J = \left[<br /> \begin{array}{ccc}<br /> -a &amp; -\frac{1}{a} &amp;-\frac{1}{a} \\<br /> a\bar{Y} &amp; 0 &amp; 0 \\<br /> a\bar{Z} &amp; 0 &amp; 0 <br /> \end{array}<br /> \right].<br />

This means that the trace of the Jacobian is negative, i.e. tr(J) = -a &lt; 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]

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 ...

 

[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?

 

[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) =<br /> \left|\begin{array}{ccc}-a-\lambda &amp; -a^{-1} &amp; -a^{-1} \\<br /> aY &amp; -\lambda &amp; 0 \\<br /> aZ &amp; 0 &amp; -\lambda \end{array}\right| = <br /> (-a-\lambda)\left|\begin{array}{cc} -\lambda &amp; 0 \\ 0 &amp; -\lambda \end{array} \right|<br /> +a^{-1} \left|\begin{array}{cc} aY &amp; 0 \\ aZ &amp; -\lambda \end{array} \right|<br /> -a^{-1} \left|\begin{array}{cc} aY &amp; -\lambda \\ aZ &amp; 0 \end{array}\right|<br />
which reduces to
<br /> \lambda^2(\lambda+a)+\lambda Y + \lambda Z = <br /> \lambda(\lambda^2 + a\lambda + (a-1)) = <br /> \lambda(\lambda + 1)(\lambda+ a - 1) = 0<br />
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 &lt; 1, indeterminate if a = 1, and stable if a &gt; 1).

 

[pasmith]

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