b = -1 We begin with a stability analysis of

$$\frac{\dd y}{\dd t}=-y-y^2$$ This dynamical system can also be written as $$\frac{\dd y}{\dd t}=-y\,(y+1)$$ Now you can see immediately that there are two equilibria: $$y=-1\quad\text{and}\quad y=0$$ The stability of these equilibria can be determined by local linearization. If $$\varphi(y) = -y-y^2$$ then $$\frac{\dd \varphi}{\dd y}=-1-2y$$ and thus $$\frac{\dd \varphi}{\dd y}(-1)=1\quad\text{and}\quad \frac{\dd \varphi}{\dd y}(0)=-1$$ It follows that $y=-1$ is a repeller and that $y=0$ is an attractor. The phase line is as follows:

b= 0 If

$$\frac{\dd y}{\dd t}=-y^2$$ then there exists only one equilibrium, namely $y=0$. Because $\varphi(y)\le 0$ is equal to $0$ if and only if $y=0$, we have a so-called semi-stable equilibrium. This means here that for an initial value $y_0>0$ the solution of the differential equation approaches the equilibrium, but that for an initial value $y_0<0$ the solution curve of the differential equation moves away from the equilibrium. The phase line is as follows:

b = 1 We end with

$$\frac{\dd y}{\dd t}=y-y^2$$ This dynamical system can also be written as $$\frac{\dd y}{\dd t}=-y\,(1-y)$$ Now you can see immediately that there are two equilibria: $$y=0\quad\text{and}\quad y=1$$ The stability of these equilibria can be determined by local linearization. If $$\varphi(y) = y-y^2$$ then $$\frac{\dd \varphi}{\dd y}=1-2y$$ and thus $$\frac{\dd \varphi}{\dd y}(0)=1\quad\text{and}\quad \frac{\dd \varphi}{\dd y}(1)=-1$$ It follows that $y=0$ is a repeller and that $y=1$ is an attractor. The phase line is as follows:

Transcritical bifurcation In the previous three examples we have seen that the number of equilibria and their nature for the differential equation

$$\frac{\dd y}{\dd t}=b\cdot y-y^2$$ depends on the values of $b$. In general, the following is true:

• if $b<0$ then there are two equilibria of which $y=0$ is attracting and $y=b$ is repelling;
• if $b=0$ then a single semi-stable equilibrium exist;
• if $b>0$ then there are two equilibria of which $y=0$ is repelling and $y=b$ is attracting.

Three types of phase lines can be distinguished:

The types of the equilibria changes from negative to positive parameter value. The following animation in a looping illustrates the change in the phase line when the bifurcation parameters moves from negative values to positive values.

The bifurcation diagram looks as follows:

This kind of bifurcation, in which two equilibria get close together when the parameter approaches the bifurcation value and exchange stability at his point, is called a transcritical bifurcation.

In the interactive version below you can vary the parameter $b$ by dragging the corresponding green triangle along the horizontal axis. In this way you can explore how the slope field and the behaviour of solution curves depend on the bifurcation parameter $b$. By displaying the derivative $y'$ as a function of $y$ you can read off the sign of an equilibrium value and thus determine the location of an equilibrium with respect to the origin.