I am somewhat familiar with linearization of non-linear ODEs for local stability analysis.

It's been explained to me like so. Let $\eta(t) = x(t) - x_0$ represent a small change of the solution away from $x_0$.

Then we have:

$$x(t) = \eta(t) + x_0$$


$$\dot{x} = \dot{\eta} = f(\eta(t) + x_0)$$

Then if we expand $f(\eta(t) + x_0)$ about $x_0$:

$$f(\eta(t) + x_0) = f(x_0) + f'(x_0)\eta(t) + O(\eta^2)$$

Now here, in my dynamics class we were referring to $x_0$ as not just any point, but specifically one that is a fixed point or equilibrium. In that case the first term disappears and neglecting higher order terms we get:

$$f(x) = \dot{\eta} = \eta f'(x_0)$$

Which is an ODE in $\eta$. We see by solving it that we can determine stability by the sign of $f'(x_0)$. If it is negative we have a decaying exponential, and if it is positive, the solution tends to infinity.

Ok, so with that preamble out of the way, I'm reading some notes I found online from a numerical methods class, that seem to be applying this exact line of thinking to the stability of a numerical method. I'm trying to find the connection:

If a numerical method is stable in the above sense for a certain range of values of λ, then it is possible to show that it will be stable for the ODE $y' = f(t, y)$ as long as $\frac{\partial f}{\partial y}$ is in that range of λ (and f is smooth enough). We won’t prove this theorem here.

Let us consider a few examples.

Example 14. For the forward Euler method applied to $y_0 = λy$, we get $y_n+1 = y_n + hλy_n = (1 + hλ)y_n$. The iterates $y_n$ tend to zero provided $|1 + hλ| < 1$, where the | · | denote the complex modulus. Write $hλ = x + iy$, so that the inequality becomes $(1 + x)^2 + y^2 < 1$. This is the equation of a disc in the complex (hλ)-plane, with center at −1, and radius 1. If hλ sits inside this disk, the method is stable, and otherwise it isn’t. We say that the forward Euler method is conditionally stable: typically, we require both Re(λ) < 0 and a small step size h in order to guarantee stability.

Question 1: It appears to me that the logic behind $\lambda$ here is arrived at with the same thought process as in my example above. Is this true?

Question 2: When applying Euler's method to the ODE $y' = \lambda y$, how is it this statement true: "The iterates $y_n$ tend to zero provided $|1 + hλ| < 1$"

I'm afraid the answer to question 2 is something so easy and obvious, and I hate to ask, but I just can't see it.

  • $\begingroup$ Ah, perfect. I know the answer is simple, but I'd like to give you credit for it, if you would like to post an answer. $\endgroup$ Commented Nov 21, 2019 at 15:11
  • $\begingroup$ Thank you, I'll remove the comment. $\endgroup$ Commented Nov 21, 2019 at 17:57

1 Answer 1


Regarding question 1, the answer is yes, it is the same reasoning that motivates this stability definition. Regarding question 2, you just need to write down what would be Euler's method applied to the ode $y'=\lambda y, y(0)=1$:

$$ y_{n+1} = y_n + h f(t_n,y_n) = y_n + h \lambda y_n = (1+\lambda h) y_n. $$

In this case you can get the explicit formula $y_n= (1+\lambda h)^n$, and you conclude that $y_n$ converges to zero iif $|1+\lambda h| < 1$.


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