13
$\begingroup$

As is well known, the space of solutions of a linear ODE with, say, $\mathcal{C}^\infty$ coefficients on $\mathbb{R}$ is a finite dimensional affine space (a vector space, in the homogeneus case).

What is the structure of the space of solutions of a non linear ODE? In which cases does it have a "natural" structure of finite dimensional smooth manifold?

For the question to make sense, some conditions on the form of the equation must be put. Let's assume the ODE is of the form

$$F(t,u(t),u'(t),u''(t),\dots,u^{(n)}(t))=0$$

where $F:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}$ is a smooth function.

$\endgroup$
5
  • 7
    $\begingroup$ You need to put more assumptions on $F$ in order for this to have a meaningful answer. For example, you haven't ruled out that $F$ is identically $0$ (all functions $u$ satisfy the equation) or $1$ (no functions $u$ satisfy the equation). Usually, in the case of a higher order ODE for one unknown function, one assumes that the equation can be solved smoothly for the highest derivative. Then one locally has the structure of a smooth manifold of dimension $n$, with the 'initial values' $\bigl(u^{(j)}(t_0)\bigr)$ for $0\le j< n$ giving local coordinates. $\endgroup$ Dec 8, 2012 at 20:25
  • 3
    $\begingroup$ In addition to what Robert already said, there's also the issue of whether you care about the size of the interval on which a solution exists. For a linear ODE, the solution exists for all time (both positive and negative), but for a nonlinear ODE, solutions may exist for only a finite time and the size of the time interval might depend on the initial data. $\endgroup$
    – Deane Yang
    Dec 8, 2012 at 20:30
  • $\begingroup$ @Robert Bryant: your comment comes very close to what I would have expected as an answer! Are there cases in which the manifold is compact? And in which it has nontrivial topology? $\endgroup$
    – Qfwfq
    Dec 8, 2012 at 20:55
  • 2
    $\begingroup$ @Dean Yang: yes, that is an issue too; that's why I didn't specify in which functional space the solutions have to live ($\mathcal{C}^{\infty}(\mathbb{R})$ or $\mathcal{C}^{\infty}(t_0-\epsilon,t_0+\epsilon)$); should I have considered solutions as living in the space of germs of smooth functions at $t_0$ (which is still infinite dimensional)? ( @both commenters: I realize my question is perhaps a bit elementary for people who know differential equations). $\endgroup$
    – Qfwfq
    Dec 8, 2012 at 20:55
  • $\begingroup$ I didn't know whether to accept R.Bryant's or F.Ziegler's very well written answer. The latter just seemed a bit more general, so I've clicked on it. $\endgroup$
    – Qfwfq
    Dec 19, 2012 at 17:34

3 Answers 3

12
$\begingroup$

As Robert Bryant observed, something like the solubility of $F$ for $u^{(n)}$ needs to be assumed. Then by the trick (due I believe to D'Alembert) of setting $x=(t,u,\dots,u^{(n-1)})$ and dt/ds=1, the equation can always be rewritten $$ \frac{dx}{ds}=f(x). $$ This is what most geometers would call the "standard ODE", wherein $f$ is a smooth vector field on the manifold where $x$ evolves.

In this setting, the space of (maximal connected) solution curves is indeed always a (not necessarily Hausdorff) manifold. I don't know who first wrote this, but according to this recent obituary a contender would be J.-M. Souriau in his 1970 book "Structure des systèmes dynamiques". Translation of the relevant passage:

The manifold of motions of a system

Jean-Marie Souriau has observed that the set of maximal solutions of a differential system on a differentiable manifold possesses itself a natural structure of differentiable manifold, not always separated: thus one can speak of the system's manifold of motions. This very simple property is rarely mentioned in the usual courses on differential calculus.

$\endgroup$
9
$\begingroup$

Deep and interesting theory exists when $F$ is analytic, not just smooth. It was developed by J. F. Ritt in his books on Differential algebra. (In particular, Ritt rigorously defined the things like "singular solutions").

Some parts of the theory also apply to PDE.

Most interesting and useful theory is obtained when $F$ is a polynomial, in which case the set of solution is a "differential-algebraic manifold", and a theory of DA manifolds is parallel to the theory of the usual algebraic manifolds.

For a more modern exposition (including fields of non-zero characteristic) see Kolchin, Differential algebra and algebraic groups.

$\endgroup$
8
$\begingroup$

There is yet another direction in which one could take the question, one that makes more contact with the linear and affine structures that one observes in the standard treatment of linear equations (both homogeneous and inhomogeneous). For example, consider the Riccati equation $$ u'(t) = a_0(t) + a_1(t)u(t) + a_2(t)u(t)^2,\tag{1} $$ which is nonlinear (when $a_2$ is not zero). A standard fact that is proved in ODE courses is that there is a 'trick' for 'linearizing' this equation, yielding the result that the general solution can be written in the form $$ u(t) = \frac{c_1b_1(t)+c_2b_2(t)}{c_1b_3(t)+c_2b_4(t)} $$ for some functions $b_i(t)$ (defined on the entire interval on which the $a_i$ are defined) and constants $c_1$ and $c_2$ (not both zero). Obviously, only the ratio of $c_1$ to $c_2$ matters for specifying $u$, so the ($1$-dimensional) space of solutions to $(1)$ is 'naturally' diffeomorphic to the real projective line $\mathbb{RP}^1$. Note that this representation works even when the solution $u$ has a 'pole'. (For example, consider the Riccati equation $u' = 1 + u^2$, which has $u(t) = \tan t$ as a solution.)

Sophus Lie developed a theory that associated a type of ODE to each homogeneous space of what we now call Lie groups, one that generalized the theory of linear ODE (both homogeneous and inhomogeneous) and Riccati equations in their various forms (which correspond to vector spaces, affine spaces, and projective spaces, thought of as homogeneous spaces of appropriate Lie groups).

It's possible that the OP is interested in learning more about the theory of ODE (and PDE) of Lie type, which is what this theory developed into. For this theory, putting a (possibly non-Hausdorff) manifold structure on the space of solutions is a necessary first step, but it is only the beginning.

$\endgroup$
7
  • 1
    $\begingroup$ The formula given in this answer is related to the fact any four solutions of the Ricatti equation are in constant Cross-ratio. $\endgroup$ Dec 9, 2012 at 13:47
  • 1
    $\begingroup$ That's true. The point is that the 'evaluation map' from the space of solutions with its natural $\mathbb{RP}^1$-structure to $\mathbb{R}$ given by $u\mapsto u(t_0)$ is an affine chart on the $\mathbb{RP}^1$ for each $t_0$. The constancy of the cross-ratio is then a reflection of the fact that projective transformations of $\mathbb{RP}^1$ preserve the cross-ratio of $4$ points. Of course, this generalizes to all equations of Lie type, replacing $\mathbb{RP}^1$ with the appropriate homogeneous space $M$ and the cross-ratio with appropriate invariant function(s) on products of $M$ with itself. $\endgroup$ Dec 9, 2012 at 14:36
  • $\begingroup$ For more on Lie and Ricatti: 1) "An introduction to Lie groups and symplectic geometry" - Bryant, math.duke.edu/~bryant/ParkCityLectures.pdf, 2) "Elie Cartan and geometric duality" - Bryant, math.duke.edu/~bryant/Cartan.pdf, 3) numerous papers by Carinena and associates, e.g., "Integrability of Lie systems through Ricatti equations" arxiv.org/abs/1002.0530. $\endgroup$ Oct 10, 2015 at 18:16
  • $\begingroup$ Two others with interesting links: 4) "Bernoulli numbers and solitons - revisited" - Rzadkowski, and 5) "Lie algebras, representations, and analytic semigoups through dual vector fields" - Feinsilver (pp. 44-45), chanoir.math.siu.edu/MATH/Merida/PDF/Merida.pdf. $\endgroup$ Oct 10, 2015 at 18:44
  • $\begingroup$ In my quest for links: Bernoulli numbers and solitons—revisited. It's not on the arXiv, and I can't find a home page for Rzadkowski, so I'm not sure if there's a free version out there. $\endgroup$
    – LSpice
    Oct 15, 2015 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.