# Solutions-set first order ODE's without uniqueness

In short: What can we say about the collection of all solutions of an ODE when we don't have uniqueness?

When we teach a first course in ODE's, we look at the equation

$f:D\to \mathbb{R}, \quad D\subseteq \mathbb{R}^2,$

$y'(x) = f(x,y),\quad y(x_0 ) = y_0, \quad (x_0,y_0 )\in D$

and prove two theorems

1. If $f\in C(D)$, then there exists a neighbourhood of $x_0$ for which there is a solution $y(x)$ Peano Theorem.
2. If $f$ is also Lipschitz in $y$, then there exists a neighbourhood of $x_0$ in which $y(x)$ exist and is a unique solution.Picard Lindelöf.

The natural question, which I tried to "google" and did not find an answer to, is

What can be generally said about the set of all solutions when there is no uniqueness, i.e. $f$ is continuous but not Lipschitz?

1. In most common applications, uniqueness and the Lipschitz condition are violated only at certain points which form a curve which is a "singular solution". The keyword is Clairault Equation The simplest example is $$y'=y^{1/3}$$, the singular solution is $$y=0$$. Such situation usually occurs when one considers $$F(y',y,t)=0$$ where $$F$$ is analytic. For a modern exposition, see MR1503299 Ritt, J. F. On the singular solutions of algebraic differential equations. Ann. of Math. (2) 37 (1936), no. 3, 552–617.

2. Lipschitz condition in the uniqueness theorem can be slightly relaxed. $$|F(x,t)-F(x',t)|\leq\omega(x-x'),$$ where $$\int_{0+}\frac{ds}{\omega(s)}=\infty.$$ This is due to Osgood. (See Ph. Hartman, Ordinary DE, Birkhauser 1982).

3. A remarkable theorem of Władysław Orlicz says the following. Consider the differential equation $$x'=f(x,t)$$, where $$x$$ and $$f$$ are vector functions. In the space of all continuous vector functions $$f(x,t)$$ those functions for which the differential equation has at least one non-uniqueness point is of first category. (Convergence in the space is uniform on compact subsets). This shows that in the generic situation you do not need Lipschitz condition for uniqueness. Ref. W. Orlicz, Zur Theorie der Differentialgleichung $$y'=f(x,y)$$, Bull. de Acad. Polon. des Sciences, Ser A, 1932, 221-228. (Of course the set of Lipschitz functions is also of the first category:-) Same about functions satisfying Osgood's condition above.

4. A theorem of H. Kneser says that if you have non-uniqueness at some point then the set of solutions passing through this point is a continuum (closed connected set). Über die Lösungen eines Systems gewöhnlicher Differentialgleichungen, das der Lipschitzschen Bedingung nicht genügt. (German) JFM 49.0302.03 Berl. Ber. 1923, 171-174 (1923).

5. Scalar autonomous equation $$x'=f(x)$$ is exceptional: if $$f(x_0)\neq 0$$ and $$f$$ is continuous then the solution with $$x(0)=x_0$$ is unique.

A reference for general discussion of non-uniqueness (and its relevance for science) is V. I. Yudovich, Mathematical models for natural sciences. Lecture course. Moscow, 2004. (In Russian, unfortunately not translated into Latin-alphabet languages, and not reviewed on Mathscinet). In particular, Yudovich asks for an explicit condition of uniqueness which is satisfied by a residual set of continuous functions. No such condition is known. (Residual means that the complement is of the 1-st category).

• Thank you! So, if I get it right - references 2-5 are not in English? Mar 22 '16 at 21:17
• I am not sure abut your count 2-5, but Orlicz, Kneser and Yudovich are not in English:-) But German is of the same language family as English, and uses the same alphabet, so it should not be too difficult. Osgood's result is in the book of Ph. Hartman, Ordinary differentil equations (in English). Mar 22 '16 at 21:33
• Each of these answers is worth a point! Oct 5 '20 at 15:18
• As to the result 4, it can be added that if $n=1$ (that is for scalar equations) this continuum $\Gamma$ of solutions is also path connected (while for $n>1$ there are counterexamples). The reason is that in dimension $1$ the set $\Gamma$ is a lattice (for the minimum and the maximum of two solutions are again solutions), and any closed connected lattice of bounded functions is always path connected (a fact that I was mentioned in this question: mathoverflow.net/questions/362736/… ) Oct 6 '20 at 6:56

Check the exercises in chapter existence theorem in Dieudonne Foundations on Modern Analysis you will get a nice overview for an answer to your question. best jorge