To show that the Douglas formula does not come from applying a LMM to the method of lines, we look at the one stage LMM
\[y_{n+1} - y_{n} = h(\beta_{0}f_{n} + \beta_{1}f_{n+1})\]
If we translate this to the method of lines where we advance in time, we get
\begin{eqnarray}
U_{m}^{n+1} - U_{m}^{n} = \frac{k}{h^{2}}\left[\beta_{0}\left(
U_{m-1}^{n} - 2U_{m}^{n} + U_{m+1}^{n}
\right) + \beta_{1} \left(
U_{m-1}^{n+1} - 2U_{m}^{n+1} + U_{m+1}^{n+1}
\right) \right]
\end{eqnarray}
Keep in mind that the factor $1/h^{2}$ comes from the difference approximation.  It is clear that if we pick $\beta_{0} = \beta_{1} = 1/2$, we get the Crank-Nicolson formula.  What we have to show is that, for the Douglas formula, there are no $\beta$ values that are independent of $r = k/h^{2}$.  The Douglas formula is given as
\begin{eqnarray}
U_{m}^{n+1} - 
\frac{1}{2}\left(r - \frac{1}{6}\right) \left(U_{m-1}^{n+1} - 2U_{m}^{n+1} + U_{m+1}^{n+1}\right)
=
U_{m}^{n} + 
\frac{1}{2}\left(r + \frac{1}{6}\right) \left(U_{m-1}^{n} - 2U_{m}^{n} + U_{m+1}^{n}\right)
\end{eqnarray}
If we rewrite this so that it looks more like the LMM version, we get
\begin{eqnarray}
U_{m}^{n+1} - U_{m}^{n} =
\frac{1}{2}\left(r + \frac{1}{6}\right) \left(U_{m-1}^{n} - 2U_{m}^{n} + U_{m+1}^{n}\right)
+
\frac{1}{2}\left(r - \frac{1}{6}\right) \left(U_{m-1}^{n+1} - 2U_{m}^{n+1} + U_{m+1}^{n+1}\right)
\end{eqnarray}
We could again rewrite this to
\begin{eqnarray}
U_{m}^{n+1} - U_{m}^{n} =
\frac{1}{2}\left[
\left(r + \frac{1}{6}\right) \left(U_{m-1}^{n} - 2U_{m}^{n} + U_{m+1}^{n}\right)
+
\left(r - \frac{1}{6}\right) \left(U_{m-1}^{n+1} - 2U_{m}^{n+1} + U_{m+1}^{n+1}\right)
\right]
\end{eqnarray}
This would mean that $\beta_{0} = r + 1/6$ and $\beta_{1} = r - 1/6$, since none of the $\beta$ are independent of $r$, the Douglas formula cannot be thought of as a LMM applied to the method of lines.
