I'm trying to sketch the phase portrait for a version of Lotka-Volterra given by $$\begin{cases} \dot{x} = x(3-x-2y)\\ \dot{y} = y(2-x-y) \end{cases}.$$
I can sketch this just fine except for the brown curve (shown below).
- The fixed points are: $(0,0), (0,2), (3,0), (1,1)$.
- The linearised system is given by $Df(x,y) = \begin{pmatrix} 3-2x-2x & -2x \\ -y & 2-x-2y \end{pmatrix}$.
So we have:
- $Df(0,0) = \begin{pmatrix}3 & 0 \\ 0 & 2 \end{pmatrix}$, $\lambda_1 = 3, \lambda_2 = 2, e_1 = \begin{pmatrix}1 \\ 0 \end{pmatrix}, e_2 = \begin{pmatrix}0 \\ 1 \end{pmatrix}$
- $Df(0,2) = \begin{pmatrix}-1 & 0 \\ 2 & -2 \end{pmatrix}$, $\lambda_1 = -1, \lambda_2 = -2, e_1 = \begin{pmatrix}1 \\ -2 \end{pmatrix}, e_2 = \begin{pmatrix}0 \\ 1 \end{pmatrix}$
- $Df(3,0) = \begin{pmatrix}3 & -6 \\ 0 & -1 \end{pmatrix}$, $\lambda_1 = -3, \lambda_2 = -1, e_1 = \begin{pmatrix}1 \\ 0 \end{pmatrix}, e_2 = \begin{pmatrix}3 \\ -1 \end{pmatrix}$
- $Df(1,1) = \begin{pmatrix}-1 & -2 \\ -1 & -1 \end{pmatrix}$, $\lambda_1 = -1 + \sqrt{2}, \lambda_2 = -1-\sqrt{2}, e_1 = \begin{pmatrix}1 \\ -\frac{1}{\sqrt{2}} \end{pmatrix}, e_2 = \begin{pmatrix}1 \\ \frac{1}{\sqrt{2}} \end{pmatrix}$
So respectively we have an unstable node, a stable node, a stable node and a saddle point.
I'm having difficulty sketching the stable manifold for the saddle point. I know that the manifold must contain $(1,1)$ and is tangent to $(1,1) + E^s$, where $E^s$ is the space spanned by the eigenvector $\begin{pmatrix}1 \\ \frac{1}{\sqrt{2}} \end{pmatrix}$, i.e. the manifold must be tangent to the line $y = \frac{1}{\sqrt{2}}(x-1)+1$.
Here is a sketch of the phase portrait by the lecturer:
Since the manifold needs to be tangent to $y = \frac{1}{\sqrt{2}}(x-1)+1$, why is the brown curve not concave downward then concave upward (rather than the way it's drawn, i.e. concave upward then concave downward)?