I am plotting the Lyapunov exponent as a function of a parameter $r$ with an initial condition $x_0$. The equation looks like this:
$$x_{n+1} =4rx_n (1-x_n)$$
When I try different initial conditions I get the same plot. I have read that the Lyapunov exponent is sensitive to changes in the initial condition. If so should't I get a different plot when I change $x_0$ ?
On
What you describe happens in many examples.
Even simpler: say the equation $x'=x$ has all Lyapunov exponents equal (to $1$), but it does have sensitive dependence with respect to the initial conditions.
If you like geodesic flows, take for example the upper half place with constant negative curvature. The Lyapunov exponents are constant (independent of the point in the unit tangent bundle), one positive and one negative, and again the flow has sensitive dependence with respect to the initial conditions.
On the other hand, in "real life" examples it is very common for the Lyapunov exponent to be discontinuous, but again, this is not directly related to sensitive dependence. Essentially the latter has to do with the existence of positive (or sometimes nonzero, depends on the notions) Lyapunov exponents.
No, you should not get different plots. You are just mixing two notions. The Lyapunov exponent is not sensitive to the initial conditions. Quite the opposite, very often, it can be shown that the set of Lyapunov exponents for any orbits belonging to the same invariant set is the same.
The sensitive dependence for a given parameter $r$, and hence for a given calculated Lyapunov exponent bigger than zero, means that no matter how close you start two orbits, given enough time they will "significantly" separate.