---
tag: FZ64956KX2
kind: definition
name: Well-posed problem
msc: [35A01, 35A02, 35B30]
defines: W09B9NZ9B5
uses: [03J5RAKB30, 0X3WGX95FE]
sources:
  - url: https://en.wikipedia.org/wiki/Well-posed_problem
---

## Definition

Let $X$ and $Y$ be [[normed vector spaces|03J5RAKB30]], the space of data and the space of solutions.
Let $D \subseteq Y$ and let $A \colon D \to X$ be a map.
The problem of finding, for given $f \in X$, some $u \in D$ with $A(u) = f$ is *well posed* if:

1. for every $f \in X$ there is a solution $u \in D$ of $A(u) = f$;
2. for every $f \in X$ this solution is unique;
3. the solution map $X \to Y$, $f \mapsto u$, is [[continuous|0X3WGX95FE]].

Otherwise the problem is *ill posed*.

## Remarks

The choice of the spaces $X$ and $Y$ and of their norms is part of the problem.
A problem can be well posed for one choice and ill posed for another.
Example: the [[Dirichlet problem|8D56GQ4AG8]] for the Laplace equation on the closed unit disc $\overline{B}$, with $X = C(\partial B)$ and $Y = C(\overline{B})$ both with the supremum norm.
The Poisson integral gives a solution for each boundary function $\varphi$.
If $u$ and $v$ solve the problem with boundary values $\varphi$ and $\psi$, the [[maximum principle|D2JF19HF91]] applied to $\pm(u - v)$ gives $\max_{\overline{B}} \abs{u - v} \le \max_{\partial B} \abs{\varphi - \psi}$.
This bound gives both uniqueness and continuity, so the problem is well posed.
Non-example: the Cauchy problem for the Laplace equation fails clause 3; see [[Hadamard's example|SZ6MEVM89M]].
The backward heat equation also fails clause 3, and on bounded domains it fails clause 1 for most data.

## Explanations

### L1 human

A well-posed problem has exactly one solution for every datum, and the solution changes little when the datum changes little.