# System of ODEs tricky equilibrium expression!

I have the following system of ODEs that I am trying to find the equilibrium points to. I firstly begin by taking (2) and rearranging for $x$. I then insert this into (3) to find a relation for $z$. Then my strategy is to insert the newly found expressions for $x$ and $z$ into $(1)$ to generate an expression for $y$. However when I do this, everything cancels and I cannot generate my expression for $y$.

Where am I going wrong?

\begin{align} (1) \qquad \frac{dx}{dt} &= -\beta x(y + z) + \gamma y + \alpha z + \epsilon\eta '(1-\phi)y \\ (2) \qquad \frac{dy}{dt} &= \beta yx - \gamma y - \eta \epsilon y \\ (3) \qquad \frac{dz}{dt} &= \beta zx - \alpha z + \epsilon\eta \phi y \end{align}

The expression for $x$ is:

$$x^* = \frac{1}{\beta}(\gamma + \eta\epsilon)$$

The expression for $z^*$ is:

$$z = - \frac{\epsilon \eta\phi}{\gamma + \eta\epsilon - \alpha } y$$

Inserting these into (1) cancels everything out and I cannot obtain an independent expression for $y^*$! I am struggling to complete further analysis without this crucial step!

• Everything cancels in all three equations regardless of the value of $y^*$? In that case you have a whole line of equilibria, which is not that unusual of a situation.
– Ian
Commented Oct 2, 2017 at 0:35
• Yes there are many non-isolated equilibria in this system. I see another with $(x^*,0,0)$. Commented Oct 2, 2017 at 0:37
• Noticed that $y = 0$ was a solution of the (2) equation and that $y$ and $z$ always appear in terms. Commented Oct 2, 2017 at 1:02
• You are missing two critical points and I agree with the third you found. The two missing CPs are $$x = \dfrac{\alpha}{\beta}, y = 0, ~\mbox{and}~ y = 0, z = 0$$
– Moo
Commented Oct 2, 2017 at 1:08
• When I show two values for the CP, it means the third can be anything, so yes, for example, the second CP I show is $$(x, y, z) = (x, 0, 0)$$ Substitute those two values into each of the 3-equations. Clear?
– Moo
Commented Oct 2, 2017 at 1:22

Equation $(2)$ gives $$y^* = 0\, , \qquad\text{or}\qquad x^* = \frac{\gamma+\eta\epsilon}{\beta}$$ at equilibrium. Injecting these two possibilities in equation $(3)$ yields either $$z^* = 0 \quad\text{or}\quad x^* = \frac{\alpha}{\beta}\, , \qquad\text{or}\qquad z^* = \frac{\epsilon\eta\phi\, y^*}{\alpha-\gamma-\eta\epsilon} .$$ Now, we use $(1)$ at equilibrium:
• the case $y^* = 0 = z^*$ gives $x^* \in \mathbb{R}$. There is an equilibrium line;
• the case $y^* = 0$, $x^* = {\alpha}/{\beta}$ gives $z^* \in \mathbb{R}$. There is a second equilibrium line;
• the case $x^* = ({\gamma+\eta\epsilon})/{\beta}$, $z^* = {\epsilon\eta\phi\, y^*}/({\alpha-\gamma-\eta\epsilon})$ gives $y^*=0=z^*$.
The last equilibrium point is conditioned by $\alpha \neq \gamma + \eta\epsilon$, which was already assumed when writing $z^* = {\epsilon\eta\phi\, y^*}/({\alpha-\gamma-\eta\epsilon})$. This equilibrium point belongs to the first equilibrium line, and is therefore not isolated.
• Thank you!, in the third case doesn't $y^*=0$ then mean that $z^*=0$ too? Commented Oct 2, 2017 at 11:12