Question

Use the information provided to determine the unspecified constants. The system $$ \begin{aligned} &x^{\prime}=x+\alpha x y+\beta \\ &y^{\prime}=\gamma y-3 x y+\delta \end{aligned} $$ has equilibrium points at \((x, y)=(0,0)\) and \((2,1)\). Is \((-2,-2)\) also an equilibrium point?

Short answer

Question: Determine the unspecified constants α, β, γ, and δ if the equilibrium points of the following system are (0, 0) and (2, 1): $$ \begin{aligned} &x^{\prime}=x+2 \alpha x^{2}+\beta y^{2} \\ &y^{\prime}=\gamma y+x y(3 \alpha+\delta). \end{aligned} $$ Check if the point (-2, -2) is also an equilibrium point. Answer: The unspecified constants are: α = -1, β = 0, γ = 6, and δ = 0. The point (-2, -2) is not an equilibrium point for the given system.

Step-by-step solution

Find the system of equations using the equilibrium points

At the equilibrium points, the derivatives are equal to zero, thus: For \((x, y) = (0, 0)\), the system becomes: $$ \begin{aligned} &0 = 0 + 0 + \beta \\ &0 = 0 - 0 + \delta \end{aligned} $$ Which gives us \(\beta = 0\) and \(\delta = 0\). For \((x, y) = (2, 1)\), the system becomes: $$ \begin{aligned} &0 = 2 + 2 \alpha + 0 \\ &0 = \gamma - 3(2) + 0 \end{aligned} $$ Which results in the following system of equations: $$ \begin{aligned} &2 + 2 \alpha = 0 \\ &\gamma - 6 = 0 \end{aligned} $$

Solve the system of equations

Solving the system of equations, we have: $$ \begin{aligned} &2 \alpha = -2 \Rightarrow \alpha = -1 \\ &\gamma = 6 \end{aligned} $$ So, the constants are: \(\alpha = -1\), \(\beta = 0\), \(\gamma = 6\), and \(\delta = 0\).

Check if (-2, -2) is an equilibrium point

Now, we'll plug the constants back into the original system and check if (-2, -2) is an equilibrium point. The system becomes: $$ \begin{aligned} &x^{\prime}=x-x y \\ &y^{\prime}=6y -3xy \end{aligned} $$ Checking for the equilibrium point (-2, -2), we get: $$ \begin{aligned} &x^{\prime}=(-2)-(-2)(-2) = -2 - 4 = -6 \\ &y^{\prime}=6(-2) -3(-2)(-2) = -12 -12 = -24 \end{aligned} $$ Since both derivatives are non-zero, we conclude that \((-2, -2)\) is not an equilibrium point for the given system.

Key concepts

Equilibrium Points

In differential equations, equilibrium points play a crucial role. These are points where the system remains constant, meaning there is no change over time. At an equilibrium point, the rate of change (derivative) is zero. - To find equilibrium points in the system, we set the derivatives of both equations to zero. This helps establish conditions that tell us where the system does not change. For the given problem, the equations are: 1. \( x' = x + \alpha xy + \beta \) 2. \( y' = \gamma y - 3xy + \delta \) When we tested \((x, y) = (0, 0)\) and \((2, 1)\), both equations gave zero when derivatives were calculated with the constants found during solving. Thus, these are equilibrium points.However, further testing of \((-2, -2)\) resulted in non-zero derivatives, meaning it is not an equilibrium point.

System of Equations

A system of equations consists of multiple equations that are solved together. In this context, they often describe the interactions of variables over time. - This system has two equations: one for \( x \) and one for \( y \), linked by constants \( \alpha, \beta, \gamma, \delta \). To find these constants, the equilibrium points\((0, 0)\) and \((2, 1)\) emerged as crucial since they allowed us to derive a set of equations:- From the point \((0, 0)\), we found \( \beta = 0 \) and \( \delta = 0 \).- From the point \((2, 1)\), we obtained two additional equations, \( 2 + 2\alpha = 0 \) and \( \gamma - 6 = 0 \), which we solved to find \( \alpha = -1 \) and \( \gamma = 6 \). Understanding and setting up a system of equations is key to solving complex problems like these and provides a framework to work with the dynamics of differential systems.

Stability Analysis

Stability analysis helps us understand the behavior of equilibrium points. It determines whether small deviations from these points will dampen out or escalate. - An equilibrium point is stable if, when slightly disturbed, the system returns to that equilibrium point. Through calculation, we determined the equilibrium points for this system. By examining if \((-2, -2)\) was an equilibrium point, we determined it was not, which suggests such a point could lead to instability since it does not fulfill the zero derivative condition. - By analyzing other determined equilibrium points, we can conclude whether they are stable based on further tests such as evaluating eigenvalues or examining the system's response to perturbations. Stability analysis not only tells us about the behavior at equilibrium points but also informs us how changes in initial conditions affect the system's future state.