# Determine all the critical points of the given system and discuss their type and stability.

Determine all the critical points of the given system and discuss their type and stability.

$$\frac{\mathrm{d} x}{\mathrm{d} t}=x(1-x-y)$$$$\frac{\mathrm{d} y}{\mathrm{d} t}=y({\frac{3}{4}-y-\frac{1}{2}x})$$$$(1,0),(\frac{1}{2},\frac{1}{2})$$$$A=\begin{pmatrix} 1-2x-y, &-x \\-\frac{y}{2}, &\frac{3}{4}-2y-\frac{x}{2} \end{pmatrix}$$$$\begin{pmatrix} \frac{1}{2} &-\frac{1}{2} \\-\frac{1}{4} &-\frac{1}{2} \end{pmatrix}$$

$$(\lambda-\frac{1}{2})(\lambda+\frac{1}{2})=\frac{1}{8}$$ $$V_y=\frac{3}{4}-2y-\frac{x}{2}$$$$V_{xx}=1-2x-y$$$$V_{yy}=-\frac{5}{2}$$$$V_{xy}=-x$$$$R(x,y)=V_{xx}V_{yy}-2V_{xy}^2$$$$=-\frac{5}{2}+5x+\frac{5}{2}y-2x^2$$$$R(1,0)=\frac{1}{2}$$

But where to from here?

There is an issue with your critical points. We want to simultaneously solve

$$\tag 1 x(1-x-y) = 0 \\ y\left(\dfrac{3}{4}-y-\dfrac{1}{2}x\right) = 0$$

We see that for $x = 0$ in the first equation, we solve the second and find $y = 0, y = \dfrac{3}{4}$.

We see that for $y = 0$ in the second equation, we solve the first and find $x = 0, x = 1$.

Next, we assume that $x \ne 0, y \ne 0$, and have to solve the system

$$1-x-y = 0 \\ \dfrac{3}{4}-y-\dfrac{1}{2}x = 0$$

Negating the first equation and adding gives $x = \dfrac{1}{2}, y = \dfrac{1}{2}$.

Summing up, we have four critical points as

$$(x, y) = (0, 0), \left(0, \dfrac{3}{4}\right), (1, 0), \left(\dfrac{1}{2}, \dfrac{1}{2} \right)$$

Next we find the Jacobian matrix of $(1)$

$$J(x, y) = \begin{pmatrix} \dfrac{\partial u}{\partial x}& \dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}& \dfrac{\partial v}{\partial y} \end{pmatrix} =\begin{pmatrix} 1-2x-y &-x \\-\dfrac{y}{2} &\dfrac{3}{4}-2y-\dfrac{x}{2} \end{pmatrix}$$

Hints on remaining parts

1. Find the eigenvalues of the Jacobian at each critical point.
2. Classify the stability using the following

If the eigenvalues are real

• Eigenvalues both positive = An Unstable Node: All trajectories in the neighborhood of the fixed point will be directed outwards and away from the fixed point.

• Eigenvalues both negative = A Stable Node: All trajectories in the neighborhood of the fixed point will be directed towards the fixed point.

• Eigenvalues opposite sign = An Unstable Saddle Node: Trajectories in the general direction of the negative eigenvalue's eigenvector will initially approach the fixed point but will diverge as they approach a region dominated by the positive (unstable) eigenvalue.

If the eigenvalues are complex conjugates - their real parts are equal and their imaginary parts have equal magnitudes but opposite sign.

• Real parts positive = An Unstable Spiral: All trajectories in the neighborhood of the fixed point spiral away from the fixed point with ever increasing radius.

• Real parts negative = An Stable Spiral: All trajectories in the neighborhood of the fixed point spiral into the fixed point with ever decreasing radius.

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Look at a phase portrait to verify the results 