# stability of fixed/equilibrium points in system of differential equations

Given is the following system $$(a>0)$$: $$\dot{x}=x(1-x)-xy$$ $$\dot{y}=y(ax-1)$$ In order to find the fixed points I have set $$\dot{x}$$ and $$\dot{y}$$ equal to zero and found $$(0,0)$$, $$(1,0)$$ and $$(\frac{1}{a},1-\frac{1}{a})$$. To inspect the stability of these I have set up the Jacobian matrix as follows: $$\begin{bmatrix} \frac{\partial \dot{x}}{\partial x} & \frac{\partial \dot{x}}{\partial y} \\[1ex] % <-- 1ex more space between rows of matrix \frac{\partial \dot{y}}{\partial x} & \frac{\partial \dot{y}}{\partial y} \\ \end{bmatrix}$$ Which is $$\begin{bmatrix} 1-2x-y & -x \\[1ex] % <-- 1ex more space between rows of matrix ay & ax-1 \\ \end{bmatrix}$$ For the point $$(1,0)$$ this would yield: $$\begin{bmatrix} -1 & -1 \\[1ex] % <-- 1ex more space between rows of matrix 0 & a-1 \\ \end{bmatrix}$$ Which has the eigenvalues $$\lambda_1=-1$$ and $$\lambda_2=a-1$$.

Going through the same process for the point $$(\frac{1}{a},1-\frac{1}{a})$$ gives the eigenvalues: Which is massive and I have no idea what to do with this.

For the point $$(0,0)$$ I get the eigenvalues $$\lambda_1=1$$ and $$\lambda_2=-1$$ which at least is clear to me that is a unstable saddle point.

• Sketch the linearised system for each fixed point where $$a<1$$ and make clear what kind of fixed point it is (saddle, centre, focus).

As far as I know I have linearised the system already to find the stability of the found fixed points so this question confuses me. For $$(0,0)$$ I believe it's a saddle and for $$(1,0)$$ where $$a<1$$ we end up with two negative eigenvalues causing a stable node. But $$(\frac{1}{a},1-\frac{1}{a})$$ is still a puzzle to me.

• Take $$a<1$$. Sketch the nullclines. Find the signs of $$\dot{x}$$ and $$\dot{y}$$.

For the points $$(0,0)$$ and $$(1,0)$$ this is clear for me but the other one not so much.

• Look at the solutions with initial values $$(x_0,y_0)$$ for $$0 < a \leq 1$$ and where $$x_0>0$$ and $$y_0\geq 0$$. Find the limit for all these solutions for when $$t$$ goes to infinity.

Any explanation or hint would be much appreciated.

• I think you have miscalculated the points of equilibrium, since I get these: $$P_1\left(0,0\right)\qquad P_2\left(1,0\right)\qquad P_3\left(\frac{1}{a},1-\frac{1}{a}\right)$$
– user588775
Commented Jan 1, 2019 at 13:15
• yup you're right. I've edited the question with the right points now. Commented Jan 1, 2019 at 13:51
• Sorry but... I think something is wrong with your eigenvalues for $P_3$
– user588775
Commented Jan 1, 2019 at 15:13

For $$\,P_3\left(\frac{1}{a},1-\frac{1}{a}\right)\,$$ your Jacobian matrix is $$\begin{pmatrix}-\frac{1}{a}&-\frac{1}{a}\\\;\,a-1&\;\;0\;\end{pmatrix}$$
so, when you look for the eigenvalues you should get $$\lambda=\frac{-1\pm\sqrt{1-4a\left(a-1\right)}}{2a}$$ which is a real solution only if $$\;0
• putting $\dot{y}=0$ gives $y(ax-1)=0$ leading to $y=0$ and $x=\frac{1}{a}$. I can't see how you have found the last point unless you use a different method to find the fixed points? In that case could you elaborate on that a bit? Commented Jan 1, 2019 at 13:19