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.
Best Answer
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 $\;0<a<\frac{1+\sqrt{2}}{2}$