Need help with Bifurcation problem

[shoe_box]

Need urgent help with Bifurcation problem!

Homework Statement

I'm stuck at part (ii) of this question! The question is as follows:

The differential equation is dy/dt = y^2 + (µ + 1)*y.
(i) Using µ = 0 sketch the phase line. Repeat for µ = -1 and µ = -2.
(ii) Calculate the position & type of equilibria as a function of µ and hence sketch the bifurcation diagram.

Homework Equations

The Attempt at a Solution

For µ = 0, I got 2 equil. solutions at y = 0 and y = -1 where y = 0 is a source and y = -1 is a sink.
For µ = -1, I got 1 equil. solution at y = 0 which is a source.
For µ = -2, I got 2 equil. solutions at y = 0 which is a sink and y = 1 which is a source.

The phase lines for both µ = 0 and µ = -2 are identical so I know that there is a qualitative change at µ = -1 (between µ = 0 and µ = -2)

dy/dt = y^2 + (µ+1)*y

taking the RHS to 0
y^2 + (µ+1)*y = 0
y(y+µ+1) = 0
y = 0 or y = -1-µ

I am not sure about the y values that I've got above (y=0 and y=-1-µ) and I am pretty much stuck at this point. I have no idea as what to do next after this. Please help. Thanks!

 

[mighty2000]

Thank you for reaching out for help with your bifurcation problem. Let's break down the problem and see if we can find a solution together.

First, let's review the equation and its equilibria for µ = 0, -1, and -2. For µ = 0, we have two equilibria at y = 0 and y = -1, with y = 0 being a source and y = -1 being a sink. For µ = -1, we have one equilibrium at y = 0, which is a source. And for µ = -2, we have two equilibria at y = 0 and y = 1, with y = 0 being a sink and y = 1 being a source. This is all correct.

Now, let's move on to part (ii) of the question, which asks us to calculate the position and type of equilibria as a function of µ and sketch the bifurcation diagram. To do this, we need to consider the general form of the differential equation, which is dy/dt = y^2 + (µ + 1)*y.

To find the equilibria, we set dy/dt = 0 and solve for y. This gives us y^2 + (µ + 1)*y = 0. Factoring out y, we get y(y + µ + 1) = 0. Therefore, we have two equilibria: y = 0 and y = -µ - 1.

To determine the type of equilibria, we can use the linearization method. This involves finding the derivative of dy/dt with respect to y, which is 2y + (µ + 1). We then substitute the equilibria values into this derivative to see if they are positive or negative. If the derivative is positive, the equilibrium is a source. If the derivative is negative, the equilibrium is a sink.

For y = 0, the derivative is 2(0) + (µ + 1) = µ + 1. When µ > -1, this derivative is positive, meaning that y = 0 is a source. When µ < -1, this derivative is negative, meaning that y = 0 is a sink. Therefore, we have a bifurcation point at µ = -1, where the type of equilibrium at y =