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I am trying to prove that when an IVP in ODEs does not enjoy uniqueness, then it has infinitely many different solutions. I know that when the Lipschitz condition is satisfied, then there is a unique solution.

I guess, I am trying to show that if the IVP possesses two solutions, then we can create infinitely many solutions using these two. However, I am slightly lost as where to start the proof.

Yiorgos S. Smyrlis
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user75514
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1 Answers1

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This is not something totally trivial.

It is Helmut Kneser's Theorem.

Let the following IVP $$ x'=f(t,x), \quad x(\tau)=\xi,\tag{1} $$ where $x,f,\xi\in\mathbb R^n$. Then the set for any $s\in\mathbb R$, for which there exists a solution of the above in the interval $[\tau,s]$ the set $$ S=\{x(s): x\,\, \text{is a solution of $(1)$}\}, $$ is connected.

In the one-dimensional case, the proof is rather simple. Assume that $\varphi,\psi:[\tau,\sigma]\to\mathbb R$ are solutions of $(1)$ and $\varphi(\sigma)<\eta<\psi(\sigma)$. Then solve "backwards" the IVP $$ x'=f(t,x), \quad x(\sigma)=\eta,\tag{2} $$ which means that you obtain a solution $\zeta$ for $t\le \sigma$. As soon as the graph of $\zeta$ hits the graph of $\varphi$ or $\psi$, say at $t_0\in(\tau,\sigma)$ you have $\zeta(t_0)=\psi(t_0)$, then you can define a new solution $\tilde\zeta$ as $$ \tilde\zeta(t)=\left\{ \begin{array}{lll} \psi(t) & \text{if} & t\in[\tau,t_0], \\ \zeta(t) & \text{if} & t\in[t_0,\sigma]. \end{array} \right. $$ Clearly $\tilde\zeta$ satisfies both $(1)$ and $(2)$, and that's how you obtain a continuum of solutions.

For a more general theorem, see Philip Hartman, Ordinary Differential Equations, page 15.

Yiorgos S. Smyrlis
  • 86,581
  • Thanks for the reference. I knew the theorem had a name, but I couldn't recall it. So my question is the other answer I received shows that you can construct additional solutions from the non-uniqueness conditions, but does this imply there is a continuum of solutions? Do I need to go through the proof of connectedness to show that it is indeed a continuum? – user75514 Jan 30 '14 at 15:21
  • @user75514: See updated answer. – Yiorgos S. Smyrlis Jan 30 '14 at 15:34
  • Thanks. That was very helpful. – user75514 Jan 31 '14 at 15:33
  • There is a late reply. Why the function $\zeta$ is defined in $[\tau ,\sigma]$? Or what's more important, why does its graph reach the line of function $\psi$ or $\phi$? – Limsup Oct 01 '20 at 18:35
  • How do we know the new constructed soln is differentiable at $t_0$? Do you need some regularity conditions on $f$? – Golden_Ratio Jan 06 '22 at 09:20
  • @YiorgosS.Smyrlis sir Kneser's Theorem holds even if function $f(t,x)$ is non continuous? – neelkanth Oct 10 '22 at 16:03
  • @neelkanth My guess is YES, since Caratheodory solutions are more general than the classical ones. – Yiorgos S. Smyrlis Oct 11 '22 at 08:45
  • @YiorgosS.Smyrlis thank you sir – neelkanth Oct 11 '22 at 12:04
  • Same question as Limsup: how can you be sure that "the graph of $\zeta$ hits the graph of $\varphi$ or $\psi$" ? – perturbation Jun 13 '25 at 02:31