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I'm somewhat troubled by how the notion of a well-posed problem extends to ordinary differential equations. It is commonly said that a problem is well-posed if following three criteria are met:-

  1. The problem has a solution.
  2. The solution is unique.
  3. The solution depends continuously on the initial conditions.

My confusion is specifically with the final criterion. Consider as an example the simplest non-trivial ODE:

$$ x' = \alpha x, \quad x(0) = c, \tag{1} $$

where $\alpha > 0$. Now, consider (1) with two initial conditions, $x_1(0) = c$ and $x_2(0) = c + \epsilon$. The corresponding solutions for the two initial-value problems satisfy the following equation

$$ x_2(t) - x_1(t) = \epsilon e^{\alpha t} \tag{2} $$

Since $x_1$ and $x_2$ are unbounded, it is clear that regardless of how small $\epsilon$ is chosen, the two solutions will diverge. Does this not mean that (1) is ill-posed? Am I conflating the notions of stability and well-posedness? If so, what is the fundamental difference?

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    $\begingroup$ Ill-posedness = diverging solutions in arbitrarily small time. Well-posed but separating trajectories = diverging solutions as $t\to\infty$ but can be made arbitrarily close for some small time $t$ $\endgroup$
    – whpowell96
    Dec 12, 2023 at 21:57

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Continuous dependence requires a metric of a norm. In the case of ODEs, usually the supremum norm on some interval $[0,T]$ is used for this. Depending on your literature, you will recall that existence proofs (e.g. using a fixed point argument) are often done in the space $C([0,T]; \mathbb R)$ of continuous functions.

In the case of your example, you do have continuous dependence on the initial condition in $C([0,T]; \mathbb R)$: $$\|x_2 - x_1\|_\infty \le \epsilon e^{\alpha T},$$ which will tend to zero as $\epsilon\to 0$.

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  • $\begingroup$ Excellent, thank ya $\endgroup$
    – zaccandels
    Dec 13, 2023 at 10:46

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