Q:

Order of $\theta$-methods for IVP

Given the initial value problem
$$
y' = f(y)
$$
consider the family of $\theta$-methods
$$
Y^{n+1} = Y^n + \Delta t ( (1-\theta) f(Y^n) + \theta f(Y^{n+1}) )
$$
so that $\theta=0$ corresponds to the Euler method, $\theta=1$ to the implicit Euler method and $\theta=\frac12$ to the implicit trapezium (Crank-Nicolson) method. 
We know that these methods all have global order 1, except from the case
$\theta=\frac12$ for which the order is 2.  On the other hand, the solutions $Y$ that they produce are manifestly continuous as a function of $\theta$.
Indeed, if I try $\theta=.5$ on a simple problem I empirically observe an order of convergence of 2, whereas if I try $\theta=.51$, I really see that the order of convergence is destroyed and is really 1. 
My question is: how is this behavior (continuity of solutions and discontinuity of order) possible?  What is the intuition behind this fact? 
Thanks a lot in advance! 

A:

The truncation error of the method is
$$
\epsilon = \Delta t\left(\frac{1}{2} - \theta\right) y''(t_{n+1/2}) 
-\frac{\Delta t^2}{12} y'''(t_{n+1/2}) + O(\Delta t^3)
$$
When $\theta$ approaches $\frac{1}{2}$ the linear term of the error tends to zero and finally is zero when $\theta = \frac{1}{2}$.
Let $\Delta t_0 = \left|(6-12\theta) \frac{y''(t_{n+1/2})}{y'''(t_{n+1/2})}\right|$.
For $\Delta t \gg \Delta t_0$ the linear term is small and the method effectively behaves like a second order one, but when $\Delta t \ll \Delta t_0$ the linear term dominates and the method behaves like a first order one.  Note the closer $\theta$ is to $1/2$ the smaller is $\Delta t_0$. Finally for $\theta = 1/2$, $\Delta t_0 = 0$.

