At which maximum step size is the method stable for real $\lambda$?

Question

I have this Runge-Kutta method:

\begin{array} {c|cccc} 0\\ \frac{1}{3} & \frac{1}{3}\\ \frac{2}{3} &0 &\frac{2}{3} \\ \hline & \frac{1}{4} &0 &\frac{3}{4} \end{array}

I have managed to calculate the region of stability, which is:

$$|1+(\lambda h)+\frac{1}{2}(\lambda h)^2 + \frac{1}{6}(\lambda h)^3| \lt 1$$

We know that $Re \lambda < 0$.

Now I have the question: at which maximum step size ($h$) is the method stable for real $\lambda$? I am unsure how to go about this? Thanks for any help.

Answer 1

For a real $\lambda < 0$, you just fix $z = h\lambda$ and rewrite $$ -1 < 1 + z + \frac12 z^2 + \frac16 z^3 < 1, $$ you will find that the right inequality is always true, while the left inequality is true for $z > \bar z$ where $\bar z < 0$ (for a value of $\bar z$, look here). Once you have $\bar z$, you can substitute back and get $$ \lambda h > \bar z \implies h < \frac{\bar z}{\lambda}, $$ which is your maximum stepsize. For complex $\lambda$ with $\mathrm{Re}(\lambda) < 0$ it is a bit more messy than that.

Answer 2

The case $λ>0$ is not stable (meaning bounded) in the exact solution, so it makes no sense to ask for the stability of numerical solutions.

For $λ<0$ you have found that $x=λh$ has to satisfy the inequalities $$ -1 < 1+x+\frac12x^2+\frac16x^3 < 1 $$ which gives two polynomials to solve. The left one has one real root at $x=-2.5127453266183286$, the right polynomial equation has one real root $0$. This means that the stability condition is satisfied for $$-2.5127453266183286<λh<0.$$