Consider the following initial value problem $$ y^{\prime} (t) = \lambda y (t) \, \text{for} \, t \in [0; T] ; y (0) = y_0; $$ with $T > 0, y_0 \in \mathbb{R} $ and $ \lambda \in \mathbb{C}$. Consider the explicit Euler method. Is it possible to specify a boundry for the stepsize $h > 0$ for $$ a) \lambda = 3 + 3i; \, b) \lambda = -3 + 3i \, \text{where} \, i^2 = -1; $$ so that the method outputs a stable numerical solution?
I have tried to solve this exercise over the stability function of the Runge Kutta Method. It is known that $$ R(z) = \frac{1+\frac{1}{2}z + \frac{1}{12}z^2}{1-\frac{1}{2}z + \frac{1}{12}z^2}$$ with $z = \lambda h$. And it is also known that $\vert Re(z) \vert < 1$ sould hold, if we want a stable method.
Thus, $$ \vert Re(z) \vert = \vert \frac{1+\frac{1}{2}z + \frac{1}{12}z^2}{1-\frac{1}{2}z + \frac{1}{12}z^2} \vert = \frac{ \vert 1+\frac{1}{2}z + \frac{1}{12}z^2 \vert}{ \vert 1-\frac{1}{2}z + \frac{1}{12}z^2 \vert} \overset{!}{<} 1 \\ \Leftrightarrow \vert 1+\frac{1}{2}z + \frac{1}{12}z^2 \vert < \vert 1-\frac{1}{2}z + \frac{1}{12}z^2 \vert $$
For the left part and $\lambda = 3+3i$ I get:
$$ \vert (1 + \frac{3}{2} h) + (\frac{3}{2} h + \frac{3}{2}h^2)i \vert = \sqrt{1+3h+4.5h^2+4.5h^3 +2.25h^4} $$ for the right part and $\lambda = 3+3i$:
$$ \vert (1 - \frac{3}{2} h) + (\frac{3}{2} h^2 - \frac{3}{2}h)i \vert = \sqrt{1-3h+4.5h^2-4.5h^3 +2.25h^4} $$
Square both sides and sum uo yields to
$$ 6h+9h^3<0 \Leftrightarrow h(6+9h^2)<0 \Rightarrow h<0 \, \text{or} \, 6+9h^2<0 \Leftrightarrow h^2 < -\frac{2}{3} $$
I get the same result when I use $\lambda = 3-3i$.
This would mean that this method is unstable for $\lambda = 3+3i$ and $\lambda = 3-3i$. This is not possible, because is stability polinomial is the one of this implicit Runge Kutta Method: $$ \begin{array} {c|cccc} \frac{3-\sqrt{3}}{6} & \frac{1}{4} & \frac{3-2\sqrt{3}}{12}\\ \frac{3+\sqrt{3}}{6} &\frac{3-2\sqrt{3}}{12} &\frac{1}{4} \\ \hline & \frac{1}{2} &\frac{1}{2} \end{array} $$
This Runge Kutta Method is A-Stable. How can I calculate a upper boundary for $h$? Or can I just say, that this method have to fullfill $Re(z)<0$ of course of the A-stability and argue like follows:
$$ Re(z)<0 \Leftrightarrow Re(\lambda h) <0 \Leftrightarrow Re(\lambda <0 ) $$ because $h>0$. Thus $$ Re(3+3i) = 3 > 0 \\ Re(3-3i)=3>0 $$ so this method can not be stable for $\lambda = 3+3i$ and $\lambda = 3-3i$.
The definition of stability only applies to cases where the exact solution converges towards a stable fixed point, or at least stays bounded. That is, for $Re(λ)\le 0$. Case a) does not apply.