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Based on this link, in particular Figure 1, what is the exact meaning of the plot?

To my understanding, it implies that for a given differential equation:

$$ \frac {dy}{dt} = \lambda y $$

that the value $\lambda \Delta t$ has to be within the complex region shown in Figure 1 corresponding to which order of Runge-Kutta you used. To ensure I am understanding this plot correctly, must $\lambda \Delta t < 0$ if both $\lambda$ and $\Delta t$ are purely real? If $\Delta t$ is purely positive and real, wouldn't this restrict you to only considering equations where $\lambda < 0$?

If we now consider a case where $\lambda$ is complex-valued and $\Delta t$ purely positive and real, what exactly is the intuition behind the value $\lambda \Delta t = 0.01 + 2i$ being stable (based on Figure 1) yet $\lambda \Delta t = 0.01$ not being stable? Why does removing/decreasing the magnitude of the imaginary part in $\lambda$ induce instability when using a Runge-Kutta method of order 4?

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    $\begingroup$ While the imaginary part is surprising to me, I think you should not be searching for intuition in how a numerical method behaves. Intuition should only stick to the real equation, for which case it would indeed be true that stability would require a negative eigenvalue. The effects of the parameters are completely contained in the graph and that is all you should be looking for. $\endgroup$ – Vikram Nov 17 '16 at 14:17
  • $\begingroup$ @Vikram Perhaps I am simply overthinking this, but that answer is not quite so satisfying. Surely there is a reason for why the method is unstable for particular complex eigenvalues and not for others, especially if the problem one is considering requires evaluating complex eigenvalues. $\endgroup$ – Mathews24 Nov 18 '16 at 0:13
  • $\begingroup$ Think about the stability of the exact solution when $\lambda$ has positive real part. Also, I suggest reading any basic text on numerical methods. $\endgroup$ – David Ketcheson Nov 18 '16 at 4:41
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The plot basically say that it's the region where the linear differential equation

$$ \frac{dy}{dt} = \lambda y$$

with

$$\lambda < 0$$

will actually go to zero. If you play around with the Euler method for a bit, you will see that if you make the stepsize too large it will explode. What happens is it will make a big step with low accuracy, end up negative, then flip back to positive, and these oscillations will continue to explode and the solution will go to infinity. So stability is essentially a maximum stepsize for which a method will accurately solve the linear differential equation without exploding ($\lambda < 0 $ because of course if $ \lambda > 0$ the solution and the thus numerical method will always explode if you take enough steps).

This is very much related to oscillations (since it seems to come from over-stepping a boundary which it shouldn't) and so complex $\lambda$ is important since a pure complex $\lambda$ is only oscillations. So the real part measures the speed to zero, the imaginary part measures the oscillations, and the stability plot shows you how well a method can handle the combinations.

This maps to "real" problems because every smooth enough problem is locally linear. Thus you can instead think of $\lambda$ as an eigenvalue of the localized Jacobian. For a nonlinear problem, you will have stability if you always have that the "instantaneous speed and oscillations", $\lambda $, are within the stability region. This may change throughout the problem, which is why adaptive timestepping methods are crucial parts of efficient solvers.

When your problem has very large $\lambda$, either complex or real, that's when people begin to say your problem is stiff (though there isn't truly one definition around. Another way is to say if your highest eigenvalue is "large" and your smallest eigenvalue is "small", since this separation causes many numerical problems). In that case, you can see from the stability plots that most explicit methods will require a very small timesteps in order to be stable. This is where "stiffly stable", A-stable, etc. methods come in since they can be used in these cases with "normal" timesteps, but with added costs somewhere else (no free lunch!).

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