= Solution
Apply the method to the <linear stability domain>[linear test equation] $y'=\lambda y$ and put $z=h\lambda$. Solving the implicit stage gives the <stability function>
$$
\boxed{
R(z)=\frac{1+(1-a)z+\tfrac12(1-2a)z^2}{1-az}}.
$$
If $a\ne1/2$, the quadratic numerator makes $|R(z)|$ unbounded as $z\to-\infty$, so the method cannot be <A-stable>. For $a=1/2$,
$$
R(z)=\frac{1+z/2}{1-z/2},
$$
which is the <trapezoidal rule> stability function and satisfies $|R(z)|\leq1$ whenever $\operatorname{Re}z\leq0$. Therefore
$$
\boxed{\text{the method is A-stable exactly when }a=\frac12}.
$$
Back to article page
