Answer to Question #108969 in Differential Equations for Gazal

Question

Consider the following system of differential equations representing a prey and predator
population model

dx/dt=x square -y

dy/dt= x+y
i) Identify all the real critical points of the system
ii) Obtain the type and stability of these critical points.

Answer 1

QUESTION(i)

To find critical points, we must solve the system

$$\left\{\begin{array}{l} \displaystyle\frac{dx}{dt}=x^2-y\\[0.3cm] \displaystyle\frac{dy}{dt}=x+y\end{array}\right.\longrightarrow \left\{\begin{array}{l} u(x,y)=x^2-y\\[0.3cm] v(x,y)=x+y\end{array}\right.\\[0.3cm] \left\{\begin{array}{l} x^2-y=0\\[0.3cm] x+y=0 \end{array}\right.\longrightarrow \left\{\begin{array}{l} y=x^2\\[0.3cm] x+x^2=0 \end{array}\right.\longrightarrow\left\{\begin{array}{l} y=x^2\\[0.3cm] x(x+1)=0 \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} y=x^2\\[0.3cm] x=0\quad\text{or}\quad x=-1 \end{array}\right.\longrightarrow\\[0.3cm] \boxed{A(0,0)\,\,\,\text{and}\,\,\,B(-1,1)-\text{the real critical points of the system }}$$

QUESTION(ii)

Next we find the Jacobian matrix of 

$$J(x,y)=\left(\begin{array}{cc} \displaystyle\frac{\partial u}{\partial x} &\displaystyle\frac{\partial u}{\partial y}\\[0.3cm] \displaystyle\frac{\partial v}{\partial x} &\displaystyle\frac{\partial v}{\partial y} \end{array}\right)= \left(\begin{array}{cc} 2x&-1\\[0.3cm] 1 &1 \end{array}\right)$$

Hints on remaining parts

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

In our case,

$$A(0,0) : J(0,0)= \left(\begin{array}{cc} 0&-1\\[0.3cm] 1 &1 \end{array}\right)\\[0.3cm] \left|\begin{array}{cc} 0-\lambda&-1\\[0.3cm] 1 &1-\lambda \end{array}\right|=0-\text{characteristic equation}\\[0.3cm] -\lambda(1-\lambda)-1\cdot(-1)=-\lambda+\lambda^2+1=0\\[0.3cm] D=(-1)^2-4\cdot1\cdot1=1-4=-3\to\sqrt{D}=i\sqrt{3}\\[0.3cm] \left[\begin{array}{l} \lambda_1=\displaystyle\frac{1+i\sqrt{3}}{2}\\[0.3cm] \lambda_2=\displaystyle\frac{1-i\sqrt{3}}{2}\end{array}\right.\longrightarrow Re(\lambda_1,\lambda_2)=\frac{1}{2}>0$$

Conclusion,

The eigenvalues are complex conjugates. 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.

$$B(-1,1) : J(-1,1)= \left(\begin{array}{cc} -2&-1\\[0.3cm] 1 &1 \end{array}\right)\\[0.3cm] \left|\begin{array}{cc} -2-\lambda&-1\\[0.3cm] 1 &1-\lambda \end{array}\right|=0-\text{characteristic equation}\\[0.3cm] (-2-\lambda)(1-\lambda)-1\cdot(-1)=\\[0.3cm] -2+2\lambda-\lambda+\lambda^2+1=0\\[0.3cm] \lambda^2+\lambda-1=0\\[0.3cm] D=1^2-4\cdot1\cdot(-1)=1+4=5\to\sqrt{D}=\sqrt{5}\\[0.3cm] \left[\begin{array}{l} \lambda_1=\displaystyle\frac{-1+\sqrt{5}}{2}>0,\lambda_1\in\mathbb{R}\\[0.3cm] \lambda_2=\displaystyle\frac{-1-\sqrt{5}}{2}<0,\lambda_2\in\mathbb{R}\end{array}\right.$$

Conclusion,

The eigenvalues are real. Eigenvalues opposite sign - An Unstable Saddle Node : rajectories 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.

ANSWER

(i)

$$\boxed{A(0,0)\,\,\,\text{and}\,\,\,B(-1,1)-\text{the real critical points of the system }}$$

(ii)

$$\boxed{A(0,0)-\text{An Unstable Spiral}}\\[0.3cm] \boxed{B(-1,1)-\text{An Unstable Saddle Node}}$$