# Well-Posed ODEs

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?

• 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 $t$ Dec 12, 2023 at 21:57

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 $$\epsilon\to 0$$.