We consider the following Butcher Tableau:
\begin{array}{c|cc} \frac{1}{3} & \frac{5}{12}& -\frac{1}{12} \\ 1 & \frac{3}{4} & \frac{1}{4}\\ \hline & \frac{3}{4} & \frac{1}{4} \\ \end{array}
What do I have to do?: First I wrote down, everything I need:
For the $A$-stability I need:
the stability function $g(z)$
an explanation why $g$ has no pole in {$ z\in \mathbb{C}: \Re(z) \leq 0 $}
$\lim_{|z| \rightarrow \infty } |g(z)| \leq 1$
$ g(it) \leq 1 $ $ \forall t \in \mathbb{R} $
equation for the stability function $g(z)$ :
$ g(z) = 1 + zb^T(I_2 -zA)^{-1}e $
In general we shouldn't invert that matrix, but in this special case we have the formula for the inverse:
$\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$
So we get : $(I_2 -zA)^{-1} = \frac{1}{(1-\frac{5}{12}z)(1-\frac{1}{4}z)-(-\frac{1}{12}z)\frac{3}{4}z}\begin{bmatrix}1-\frac{1}{4}z&-\frac{1}{12}z\\\frac{3}{4}z&1-\frac{5}{12}z\end{bmatrix}$
simplify it a bit:
$ (I_2 -zA)^{-1} = \frac{6}{z^2-4z+6}\begin{bmatrix}1-\frac{1}{4}z&-\frac{1}{12}z\\\frac{3}{4}z&1-\frac{5}{12}z\end{bmatrix} $
Then:
$(I_2 -zA)^{-1}e = \begin{bmatrix} \frac{2(3-z)}{z^2-4z+6}\\ \frac{2(3+z)}{z^2-4z+6} \end{bmatrix} $
Then:
$ 1+ z (\frac{3}{4} , \frac{1}{4} )^T (I_2 -zA)^{-1}e = 1 + \frac{z(6-z)}{z^2-4z+6}$
Theres is a chance that I miscalculated somewhere ( if I find a mistake, I will edit this post ). But to be honest, the remaining steps for $ A $ Stability are my greater concern. I tried something for 3. and 4., but this wasnt effective.
Now examine the absolute value for $z=iy$ $$ |g(iy)|^2=\left|\frac{iy+6}{-y^2-4iy+6}\right|^2=\frac{y^2+36}{(6-y^2)+16y^2}=\frac{(y^2+1)+35}{(y^2+1)^2+35}\le 1. $$ By the maximum modulus principle the maximum absolute value for the holomorphic domain $Re(z)\le 0$ has to be on the boundary (including here the infinite point), thus $|g(z)|\le 1$.
As the value at infinity is $0$, the method is, unlike the implicit trapezoidal method, A-stable and L-stable