# What is the most general mathematical object that defines a solution to an ordinary differential equation?

What is the most general object that defines a solution to an ordinary differential equation? (I don't know enough to know if this question is specific enough. I am hoping the answer will be something like "a function", "a continuous function", "a piecewise continuous function" ... or something like this.)

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I don't understand how you're using "defines". Do you mean how can one generally characterize solutions to ordinary differential equations? Differential equations by their nature require the solution to have the derivatives that occur in the equation, i.e. at least one, so the solution of an ODE is necessarily differentiable, and hence continuous. (Though you can also consider ODEs for distributions, in which case you might have a delta distribution in the equation and the solution might have a jump discontinuity at that point, for instance.) –  joriki Mar 11 '11 at 1:49
A function. ... –  The Chaz 2.0 Mar 11 '11 at 1:58
Yes, what is the most general characterization? –  Henry B. Mar 11 '11 at 1:59
(I was being somewhat facetious) If you require some general/common characteristics of solutions to ODE's, then joriki's comment should suffice. –  The Chaz 2.0 Mar 11 '11 at 2:38
@jorki: Thanks for the response. If you want to post the comment as an answer, I will vote for it. –  Henry B. Mar 11 '11 at 2:48

(All links go to Wikipedia unless stated otherwise.) Elaborating on joriki's answer: The most general spaces where it does make sense to talk about differential equations are certain classes of topological vector spaces, it is for example rather straight forward to formulate the concept of a differential equation in Banach spaces. (Here the esolution to an equation is a mapping of topological vector spaces)

For differential equations in $\mathbb{R}^n$, the solutions themselves are elements of certain topological vector spaces.The most general topological vector spaces that are considered are AFAIK Sobolev spaces, these are function spaces such that each individual function is normable with respect to a prescribed $L^p$ norm, has generalized derivatives in the sense of distributions to the order that is necessary to formulate the weak formulation (see also the Azimuth wiki) of the equation one would like to solve, such that the generalized derivatives are again normable with respect to the $L^p$ norm. Some Sobolev spaces have similar characterizations like "all piecwise continuous function that..." which are a little bit more complicated than that, and differ from space to space, so I'd rather like to refer to the extensive literature instead of repeating that here :-(

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For ODEs you don't really want to consider Sobolev spaces. You have a "Sobolev embedding" (also known as the fundamental theorem of calculus, or the Radon-Nikodym theorem) which tells you that $W^{1,1}(\mathbb{R})$ (1 derivative in $L^1$) functions are absolutely continuous. By localisation and Holder's inequality, you immediately have that for any number of derivatives $k \geq 1$ and any $p$-norm with $p \geq 1$ you have $W^{k,p}((0,1)) \subset C^0((0,1))$. So if a solution to an ODE (which is a function with one real independent variable) is in any Sobolev space, it must be continuous. –  Willie Wong Mar 11 '11 at 10:30
Along these lines, would distributions be more general than Sobolev spaces? –  timur Feb 27 '12 at 3:58

It's rather meaningless to ask for a "most general mathematical object", since it carries an implicit prediction that no progress will be made in the future. Also, it is usually pretty easy to make useless generalizations of existing notions. Anyway, I'll offer a construction from Beilinson and Drinfeld's book Chiral algebras that may be sufficiently encompassing for your current needs.

Take an ODE $F(x, y, y', \dots, y^{(n)}) = 0$, where $F$ is some function (did you want it to be smooth?) in $n+2$ real variables. Ordinarily, a solution would be a function $y(x)$ that has $n$ derivatives (defined on some open interval $I$ of reals), such that at each point $x \in I$, the values of $y$ and its $n$ derivatives satisfy the equation.

If your equation lacks solutions in the above sense, you can work with a very formal notion of solution, that arises from manipulating algebras of differential operators. First, choose your favorite category of functions closed under multiplication and differentiation (e.g., polynomial, analytic, smooth...), and define three basic sheaves:

1. $\mathscr{O}$ is the sheaf of functions on $\mathbb{R}$ in your category. To each open set, it assigns the ring of (polynomial, analytic, smooth...) functions on that set, and to each inclusion of open sets, it assigns the restriction homomorphism.
2. $\mathscr{D}$ is the sheaf of linear differential operators. It is the unital $\mathscr{O}$-algebra generated by $\frac{d}{dx}$, with the usual relation satisfied by derivatives. Commutative $\mathscr{D}$-algebras are commutative $\mathscr{O}$-algebras with integrable connection, and $\mathscr{O}$ has a natural commutative $\mathscr{D}$-algebra structure.
3. $\mathscr{P}$ is the sheaf of all finite order ordinary differential operators. It assigns to each open set $I \subset \mathbb{R}$ the commutative algebra of functions $G(x, y, y',\dots)$ for $x \in I$ that depend on only finitely many derivatives of $y$,. Global sections of $\mathscr{P}$ form a space of functions on an infinite dimensional vector space that has coordinates labeled by $x$ and the derivatives of $y$ (which is viewed as a formal variable).

The operator $F$ defines an ideal in the commutative $\mathscr{D}$-algebra $\mathscr{P}$, and the quotient $A = \mathscr{P}/(F)$ is a commutative $\mathscr{D}$-algebra. Given a commutative $\mathscr{D}$-algebra $R$, an $R$-solution to your ODE is a $\mathscr{D}$-algebra map from $A$ to $R$. If $R = \mathscr{O}$ and $\mathscr{O}$ is the sheaf of smooth functions, then $R$-solutions are ordinary solutions. Even if the set of ordinary solutions is empty, you'll get a nonempty set of $A$-solutions.

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The object you are looking for is called a phase curve. For ODE $\frac{dy}{dx}=f(x)$ with $x$, we are looking for a solution of the form:

$$y=y(t), x=x(t)$$

The result is a graph on the $xy$ plane that pass the initial condition $y=y_{0}$, $x=x_{0}$. All of this are quite standard. The best literature on this is:

Differential Geometry: Manifolds, Curves, and Surfaces

And Ordinary Differential Equations, by V.I.Arnold.