How to find the bifurcation values for rx(1-x). Please help me understand this?
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You appear to be talking about differential equations, but that isn't a differential equation. It isn't an equation. There's no equals sign. There's no derivative. There's no other side. So what was the actual equation? Was it
x' = rx(1 - x)?
Let's assume it was. For any given value of r, there is some behavior of x. At a bifurcation, the nature of the solutions changes.
Start by examining equilibrium solutions. When does x' = 0? When rx(1 - x) = 0, which means r = 0, x = 0 or x = 1. If x is 0 or 1, then x' = 0 means x doesn't change, x will just sit there at that value.
If r = 0, x is stable (unchanging) no matter what its starting point.
Now look at the signs of the factors x and 1 - x in the regions x < 0, 0 < x < 1, and x > 1.
x < 0: (1 - x) > 0 and x(1 - x) < 0. If r is positive then x' < 0 which means solutions starting < 0 will get more negative, moving toward -infinity. If r < 0 then in this region x' > 0, which means any x starting at a negative value moves TOWARD the origin.
0 < x < 1: (1 - x) > 0 and x > 0 so x(1 - x) > 0. If r > 0, then x' > 0 so any x in this region (positive) will increase, moving away from 0. If r < 0, x' < 0 and x will move toward the origin.
We just found the origin is a source for r > 0 (displace a little away from 0 and it gets farther) but a sink for r < 0 (displace a little away from 0 and it tends to decay back to 0).
That establishes r = 0 as the place where the nature of the solutions changes, i.e. a bifurcation.
To complete the analysis you should see the nature of the point x = 1 for r > 0 and r < 0. Source? Sink? Neither?
x' = rx(1 - x)?
Let's assume it was. For any given value of r, there is some behavior of x. At a bifurcation, the nature of the solutions changes.
Start by examining equilibrium solutions. When does x' = 0? When rx(1 - x) = 0, which means r = 0, x = 0 or x = 1. If x is 0 or 1, then x' = 0 means x doesn't change, x will just sit there at that value.
If r = 0, x is stable (unchanging) no matter what its starting point.
Now look at the signs of the factors x and 1 - x in the regions x < 0, 0 < x < 1, and x > 1.
x < 0: (1 - x) > 0 and x(1 - x) < 0. If r is positive then x' < 0 which means solutions starting < 0 will get more negative, moving toward -infinity. If r < 0 then in this region x' > 0, which means any x starting at a negative value moves TOWARD the origin.
0 < x < 1: (1 - x) > 0 and x > 0 so x(1 - x) > 0. If r > 0, then x' > 0 so any x in this region (positive) will increase, moving away from 0. If r < 0, x' < 0 and x will move toward the origin.
We just found the origin is a source for r > 0 (displace a little away from 0 and it gets farther) but a sink for r < 0 (displace a little away from 0 and it tends to decay back to 0).
That establishes r = 0 as the place where the nature of the solutions changes, i.e. a bifurcation.
To complete the analysis you should see the nature of the point x = 1 for r > 0 and r < 0. Source? Sink? Neither?