My math for engineering teacher has made a claim using the Cauchy-Lipschitz theorem that I fail to justify using other sources. Please keep in mind that my question relates the Cauchy-Lipschitz theorem in the context of differential equations defined on intervals of $\mathbb{R}$ as I seem to understand this theorem may be applicable to broader subjects.

For what I understand the Cauchy-Lipschitz theorem claims that for $(E)$ such that : $$ (E) : \left\{ \begin{array}{ccc} \frac{dy}{dt} &=& f(t, y)\\ y(x_0) &=& y_0\\ \end{array} \right. $$

with $y \in C_1$ and $y$ defined on some interval of $A \subset\mathbb{R}$. Let, for readability purposes, define $L$ to be the set of all local Lipschitz functions with respect to the second variable. Then

$$ ( \forall t \in A, f \in L) \Rightarrow \exists!\ \ g,\ \ g\texttt{ satisfy } (E). $$

My teacher then uses this theorem in a way I fail to grasp : in one of their exercises they claim that the following differential equation defined for $t \in [0, +\infty]$ :

$$ (E') : \left\{ \begin{array}{ccc} \frac{dy}{dt} &=& 2\sqrt{|y|}\\ y(0) &=& 0\\ \end{array} \right. $$

has infinitely many solutions because $2\sqrt{|y|}$ is not a Lipschitz function with respect to $y$ for $t\rightarrow 0$. This is where I'm getting confused because it seems to me that they mean

$$ ( \forall t \in A, f \not\in L) \Rightarrow \exists^{\infty}\ \ g,\ \ g\texttt{ satisfy } (E), $$

which doesn't fit my understanding of the Cauchy Lipschitz theorem. Would you mind help me clear the confusion ?

  • $\begingroup$ I assume you mean $y(0) = 0$ (otherwise the solution is actually unique)? Anyway, it's hard to judge what your teacher intended, but you don't need any theorem to show that this has infinitely many solutions. You can just write them down. $\endgroup$
    – Klaus
    Oct 7, 2022 at 22:23
  • $\begingroup$ You are right, i misremembered the initial value. It is now fixed, thank you. My question is not about how to solve but rather wether I misunderstood my teacher or the definition of the Cauchy-Lipschitz theorem. $\endgroup$
    – NRagot
    Oct 8, 2022 at 0:48
  • $\begingroup$ The Cauchy-Lipschitz theorem does not imply the claim that is made. The given example rather illustrates that the claim of the theorem no longer holds (in general) if one removes the assumption of local Lipschitzness. $\endgroup$
    – PhoemueX
    Oct 8, 2022 at 4:49


You must log in to answer this question.

Browse other questions tagged .