Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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
show 4 more comments

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.