Critical points (will be a function of beta and delta)

[Rubik]

Homework Statement

Find the critical points where [itex]\alpha[/itex] = 1.

dv/dt = v² + [itex]\alpha[/itex]v - u + [itex]\delta[/itex] [I will call this (1)]

du/dt = [itex]\beta[/itex]v - u [call this (2)]

For what values of [itex]\beta[/itex] and [itex]\delta[/itex] are there no critical points?

Homework Equations

The Attempt at a Solution

So firstly I set (1) and (2) to = 0 and from (2) I get:
u = [itex]\beta[/itex]v and now I subst. this back into (1)

0 = v² + v(1 - [itex]\beta[/itex]) + [itex]\delta[/itex]

Now from the quadratic formula I have this

v = {-(1 - [itex]\beta[/itex]) ± √[(1 - [itex]\beta[/itex])² - 4[itex]\delta[/itex]]}/2

Now here I am stuck and not sure how to simplify this or interpret this?

 

[mighty2000]

I would like to start by clarifying that the problem statement is not complete. It is missing important information such as the initial conditions and the domain of the variables. Without this information, it is difficult to provide a complete and accurate solution.

That being said, I will try to provide some general guidance on finding critical points for this system of equations.

To find the critical points, we need to solve the system of equations (1) and (2) simultaneously. This means that we need to find values of v and u that satisfy both equations at the same time. One way to approach this is to substitute the expression for u from equation (2) into equation (1), as you have done. This will give you a quadratic equation in terms of v.

Once you have this quadratic equation, you can use the quadratic formula to solve for v. This will give you two possible values for v, which we can call v1 and v2. These values represent the potential critical points.

To determine which of these values are actually critical points, we need to substitute them back into both equations (1) and (2) and see if they satisfy both equations. If they do, then they are critical points. If they do not, then they are not critical points.

Now, to answer the question about the values of β and δ for which there are no critical points, we need to consider the discriminant of the quadratic equation we obtained earlier. The discriminant is the part inside the square root in the quadratic formula, and it determines the nature of the solutions to the equation. If the discriminant is negative, then there are no real solutions, which means there are no critical points in this case.

So, to summarize, to find the critical points for this system of equations, we need to solve the equations simultaneously and check if the solutions satisfy both equations. And to determine the values of β and δ for which there are no critical points, we need to consider the discriminant of the quadratic equation obtained from the first step.

I hope this helps you in finding the critical points and understanding the role of β and δ in determining the existence of critical points. Remember to always provide all necessary information in a problem statement to ensure a complete and accurate solution.