I have read about the following example from Muller:

$(M) \begin{cases} x' = f(t,x) \\[1mm] x(0) = 0 \end{cases}$ where $f: \mathbb{R}\times\mathbb{R} \rightarrow \mathbb{R}$ is the function:

$f(t,x) = \begin{cases} 0 & t \leq 0, x \in \mathbb{R} \\ 2t & t>0,x < 0 \\ 2t - \frac{4x}{t} & t >0,0 \leq x \leq t^2 \\ -2t & t>0,x > t^2 \, . \end{cases}$

apparently it should be an example of an IVP whose Picard iterates do not convergent but that enjoys unique solution. However, I haven't been able to find studies about this function in English literature (the article is german which I do not currently read).

So, I wonder, are there any good references that:

  1. Motivate the example.
  2. Prove that it has a unique solution, compute the solution, proof that $f$ is not lipschitz (so that one cannot use Picard-Lindelof theorem, compute Picard's iterates and study its convergence (is it convergent? does it have a convergent subsequence? some subsequence converges to the solution?)
  3. Give the graph of the function (what software could I use to do it by myself?) and any other further interesting properties.

Note: I think this is not the example in its full generality, but a particular case to show the above properties.


1 Answer 1


I do not know how to fully answer your questions, but I have some remarks.

  1. Max Müller's motivation was to give an example of an equation for which no subsequence of successive approximations converges to a solution.
  2. The proof that the RHS is not Lipschitz is straightforward: compute $\lvert f(t,0) - f(t,t^2) \rvert/t^2$ and observe that it is unbounded as $t \to 0^+$. It is even more straightforward to find the Picard iterates, and show that the accumulation points of any convergent subsequence, that is, $-t^2$ or $t^2$, are not solutions of the IVP. The most tricky part is the proof of the uniqueness. As far as I can see, there is no mention of that in Müller's paper (but I checked it very cursorily). I propose to use the following theorem:

Assume that $f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous function such that it is nonincreasing in $x$ for any (fixed) $t$. Then the IVP $x' = f(t, x)$, $x(t_0) = x_0$ has a unique solution on $[t_0, \tau_{\mathrm{max}})$.

Proof. Suppose to the contrary that there are solutions $\varphi(\cdot)$ and $\psi(\cdot)$ of the IVP such that for some $\theta > t_0$ there holds $\varphi(\theta) < \psi(\theta)$. The set of those $t \in [t_0, \theta)$ at which $\varphi(t) = \psi(t)$ is nonempty (since it contains $t_0$) and closed (because both $\varphi(\cdot)$ and $\psi(\cdot)$ are continuous). Denote the supremum of that set by $\sigma$. We have thus $\varphi(\sigma) = \psi(\sigma)$ and $\varphi(t) < \psi(t)$ for $t \in (\sigma, \theta]$. By the MVT applied to the difference, there is $t_1 \in (\sigma, \theta)$ such that $\varphi'(t_1) < \psi'(t_1)$. But, as $\varphi(t_1) < \psi(t_1)$, it follows from the monotonicity assumption that $\varphi'(t_1) \ge \psi'(t_1)$, a contradition. Q.E.D.

In Müller's example the function is clearly nonincreasing in $x$ for a fixed $t$.

  1. Picard's iterative procedure does not converge to the solution, but from this it does not follow that methods used in Mathematica NDSolve, for example, do not give an approximate solution. I do not know.
  • $\begingroup$ i believe the result you quote is peano's uniqueness theorem $\endgroup$ Commented Mar 25, 2018 at 15:16
  • $\begingroup$ if i take $x_0 = 0$ and compute successive approximations then the even elements form a convergent subsequence... $\endgroup$ Commented Mar 25, 2018 at 16:48
  • $\begingroup$ and $f(t,0) = f(t,t^2)$ so i cannot use that sequence for showing unboundness $\endgroup$ Commented Mar 25, 2018 at 17:05
  • $\begingroup$ $f(t, t^2) = 2t - \tfrac{4 t^2}{t} = 2t - 4t = -2t$, and $f(t,0) = 2t$, so $\lvert f(t,t^2) - f(t,0) \rvert = 4t$, which divided by the difference in the $x$-coordinates, that is, by $t^2$, diverges to $\infty$ as $t \to 0^+$. Regarding the theorem, I just googled that it is sometimes called Peano existence theorem (I didn't know that). $\endgroup$
    – user539887
    Commented Mar 25, 2018 at 18:42
  • $\begingroup$ @Javier See another example: math.stackexchange.com/questions/2714094/…. $\endgroup$
    – user539887
    Commented Mar 30, 2018 at 8:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .