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I am just beginning to learn ordinary differential equation. My question:

Let : $t \in \mathbb{R}$, $x_0 \in \mathbb{R}$,

Let $f:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}: (t,x) \mapsto f(t,x)$.

Let $x:I \rightarrow \mathbb{R}:t \mapsto x(t)$, where $I \subseteq \mathbb{R}$ is the maximal interval of existence such that the solution of the following ordinary differential equation initial value problem:

$\frac{d}{dt}x = f(t,x), x(0)=x_0$

exists and is unique on $I$ (well-posed in the sense of Hadamard).

Is it possible that the solution $x$ has infinitely jump discontinuities?

Any comments, feedbacks, or inputs are very welcome. Thanks in advance.

Note: I have removed $d$ as to make things clearer.

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On $I$, $x$ is differentiable, hence continuous, hence it has ZERO jump discontinuity. – Did Feb 13 '12 at 9:18
May this is what you want $y'=y^2+1$, $y(0)=0$ . Its, solution is $\tan x$. – Norbert Feb 13 '12 at 9:44
@Didier, thanks for your answer... since my example is general enough, does it mean that mostly every solutions of ode has no jump discontinuty on its interval of existence? – netsurfer Feb 13 '12 at 9:44
@Nobert... is the maximal interval of existence includes $\frac{\pi}{2}$? But even so, it's more like asymptotic discontinuity for me – netsurfer Feb 13 '12 at 9:48
Given the final question you asked, the function $d$ is just a red-herring. You are just asking about (dis)continuity of the ordinary differential equation $x' = f(t,x)$. So as long as on $I$, the function $f$ is everywhere finite (not necessarily bounded uniformly), the assumption that $x' = f$ implies that $x$ is differentiable, and so like Didier said, must be continuous. – Willie Wong Feb 13 '12 at 9:53

Any solution $t\mapsto x(t)$ of the ODE $\dot x=f(t,x)$ carries with it a solution interval $I$ (you say it yourself) such that for all $t\in I$ we have $\dot x(t)=f(t,x(t))$. In particular, the function $x(\cdot)$ is continuous on $I$, so there is no room for a jump discontinuity.

Now what about the differential equation $\dot x= 1+x^2\ $? When an initial point $(t_0,x_0)$ is given there is a unique $\alpha_0\in\bigl]-{\pi\over2},{\pi\over2}\bigr[\ $ such that $\tan\alpha_0=x_0$, and it is easy to check that $$x(t)\ :=\ \tan(t-t_0+\alpha_0)$$ satisfies the differential equation in some $t$-interval $I$ containing $t_0$ as well as the given initial condition. In order to determine $I$ we have to make sure that $$-{\pi\over 2}<t- t_0+\alpha_0 <{\pi\over2}\ ,$$ from which we deduce $I=\ \bigl]t_0-(\alpha_0+{\pi\over2}),\ t_0+({\pi\over2}-\alpha_0)\bigr[\ $.

That the full graph of the $\tan$-function consists of infinitely many (disjoint) such curves is another matter and should not bother us here.

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Thank you for your answer. Yes, I understand now.. On the interval I, the solution x(.) should be differentiable, which means it is continuous. However, I am still curious, does there exist a well-posed ODE which has a discontinuous solution? As per your analysis (and some others too), it seems it is not possible that a well-posed ODE has a discontinuous solution on its interval of existence I. :) – netsurfer Feb 14 '12 at 1:37

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