Stability of Explicit midpoint method

Question

I am trying to determine the stability region of the well known explicit midpoint method $$y_{i+1} = y_i + h f\left( t_i + \frac h 2, \ y_i + \frac h 2 f(t_i, y_i)\right)$$ and after following the links Determine a stability region? and Calculating stability and order of implicit midpoint scheme, I managed to apply the numerical method on the test equation and got $$\begin{align*}y_{i+1} & = y_i + h f\left( t_i + \frac h 2, \ y_i + \frac h 2 f(t_i, y_i)\right) \\ & = y_i + \left( \lambda h +\frac{\lambda^2h^2}{2} \right)y_i \\ & = \left( 1 + \lambda h +\frac{\lambda^2h^2}{2} \right)y_i\end{align*}$$ Stability: $\big| 1 + \lambda h +\frac{\lambda^2h^2}{2}\big|<1$ and simplifying, $-2 < (1+ \lambda h)^2<0$ is the stability. I don't know if the stability region is correct. How do I find it?

Answer 1

In some way, you already found the region of absolute stability, which is in this case $$\left \vert 1 + z + 0.5z^2 \right \vert \leq 1, \quad z = \Delta t \lambda.$$

If you want to know e.g. the boundary of the absolute region of stability, you need to get your hands dirty and split $z$ in real and imaginary part $z = a + bi$ and perform many operations or ask Wolfram Alpha for help which computes for real $a, b$

$$\sqrt{\Big(0.5 \big(a^2 - b^2\big) + a + 1\Big)^2 + \big(a b + b\big)^2} \leq 1 $$

and then solve for $b$ as a function of $a$ the boundary $$\sqrt{\Big(0.5 \big(a^2 - b^2\big) + a + 1\Big)^2 + \big(a b + b\big)^2} = 1 $$

which gives two candidates

$$ b = \pm \sqrt{-a^2 - 2 a \color{red}- 2 \sqrt{-a (a + 2)}} $$ $$ b = \pm \sqrt{-a^2 - 2 a \color{red}+ 2 \sqrt{-a (a + 2)}} $$

Based on your knowledge that for real $\lambda$ you have that $-2 \leq z^2 \leq 0$ you could then figure out that only the second possibility gives real solutions for $b$. Or, you just plot both and see what gives a real solution. Since there are only two solutions for the second and third quadrant, respectively, it is guaranteed that there are no holes within that region. Indeed,