Is the Blow Up Time of an ODE with Lipschitz Condition Continuous?

[csco]

I would like to know if the blow up time of a ordinary differential equation with the lipschitz condition is a continuous function (in its domain whatever it might be) of the initial conditions and parameters. With blow up time I mean the length of the time interval to the future of the inital time of the maximal solution for given initial conditions and parameters.

I have examples in two dimensions were the blow up time in the sense defined above is not continuous but they are not good examples because I'm not choosing the natural domain for the function (the solution reaches the boundary of the domain of the function which defines the equation but it has a well defined limit as the blow up time is approached).

I don't have examples like the above for the one dimensional case and I would like to know what happens in the nth dimensional case when the function is defined in the whole space (so that the maximal solution is forced to go to infinity in the case of a blow up). I tried to prove that the blow up time is a continuous function in this case but for the proof to work I should bound some terms which I think can't be bounded at all so I'm stuck.

Can someone show me a counterexample or a proof of any of the above? Any help would be really appreciated.

 

[AlephZero]

I'm not quite sure how you are defining your "blowup time", but I don't see why it should be a continuous function of the initial conditions.

Suppose the solutions of a DE are linear combinations of a set of basis functions. You can make an arbitary choice of the basis functions and then construct the DE which has them as its solutions. It should be possible to make a counter example by working backwards like that.

 

[csco]

I'm not quite sure how you are defining your "blowup time", but I don't see why it should be a continuous function of the initial conditions.

Suppose the solutions of a DE are linear combinations of a set of basis functions. You can make an arbitary choice of the basis functions and then construct the DE which has them as its solutions. It should be possible to make a counter example by working backwards like that.

Hi AlephZero and thanks for your answer. I don't understand very well where are you trying to get at with what you are saying but I wasn't very clear in my first post so I'll define the terms I'm using more clearly.

I'm considering an ODE of the form x' = f(t,x) (I'm ignoring dependence on parameters to simplify), f:D->Rⁿ with D an open set included in RxRⁿ. The function f is continuous and locally Lipschitz with respect to x. Given (t₀, x₀) in D there exists an unique maximal solution of the problem that verifies the initial condition x(t₀) = x₀. Let I(t₀, x₀) be the maximal interval in which this solution is defined. Consider the (t₀, x₀) in D such that I(t₀, x₀) is bounded above. For these (t₀, x₀) in D I define the blow-up time t⁺(t₀, x₀) as the right end of I(t₀, x₀). What I'm saying is that I(t₀, x₀) = (t^-(t₀, x₀), t⁺(t₀, x₀)) when they exist.

It is well known that the solution must escape every compact subset of D in both the future and the past when f is locally Lipschitz. For the solution to end in finite time this implies that x(t) must approach the boundary of D or go to infinity as t->t⁺. It is easy to construct an example where t⁺ is discontinuous (for example a constant vector field in a L shaped open set) but of all these examples I have come up with the discontinuity is actually caused by technical details, the solution approaches the boundary of D but the function f admits an obvious extension which is locally Lipschitz. What I would like to know is if there is an example where the discontinuity in t⁺ is not caused by technical reasons like above, a case where f behaves badly near the boundary of D and does not admit an extension.

Because the question above seemed too difficult I considered some simpler cases: 1. the unidimensional case: D included in RxR (I don't have a "techincal" example like the above for this case), 2. the case where f is globally defined: D = RxRⁿ.

The simplest examples of unidimensional equations that blow up like x' = x^2 and x' = 1/x have a continuous t⁺ or t^-. Maybe someone knows a counterexample or a proof of this fact for the unidimensional case.

In the case when f is globally defined the solution must go to infinity in finite time. It seems to me that the Lipschitz condition should be enough to make t⁺ continuous in this case. I tried to prove it but I cannot bound some terms which I'm supposed to bound for the proof to work. Can someone think of a counterexample in this case?

The thing I have a proof of is: t⁺ cannot have jump discontinuities that jump down. With this I mean in (t₀, x₀) there can be a discontinuity only if the nearby initial conditions have a larger t⁺.

AlephZero could you please explain with more detail how to construct the counterexample you suggest? Does anybody else have a thought on all of this? Any other comments are more than welcome.

 

[AlephZero]

Ahhh. You didn't say you were talking about a first order DE.

I was thinking about higher order DEs, and counter-examples where the general solution of was $Af_1(t) + Bf_2(t)$, and the maximum of the solution could "jump" between local maxima of $f_1$ and $f_2$.

 

[csco]

Ahhh. You didn't say you were talking about a first order DE.

But that's not a problem because you can turn a higher order equation into a first order one by increasing the dimensions of the space. So I'm considering higher order ODEs too because I'm not fixing the n in Rⁿ.

I was thinking about higher order DEs, and counter-examples where the general solution of was $Af_1(t) + Bf_2(t)$, and the maximum of the solution could "jump" between local maxima of $f_1$ and $f_2$.

I don't understand how are you relating the maximum of the solution with the blow-up time but I can see how the maximum can be discontinuous with respect to the initial conditions. For example the maximum of the autonomous equation x' = x near the initial condition x₀ = 0 (the maximum is not really defined except in 0 but it does make an infinite jump, and I think if you the add the quadratic term -x² it becomes finite). But when you are considering solutions where the interval I(t₀, x₀) is finite (blow-up) doesn't the continuity of the general solution imply that the supremum (if defined) is continuous with respect to the initial conditions anyways?

Maybe if I reformulate the question it becomes more clear. What I want to know is simply if the length of interval I(t₀, x₀) can be discontinuous with respect to the initial conditions. I know it can't have a sudden decrease but a sudden increase might be possible.

 

[csco]

Any suggestions or comments on all of this anyone? I expected I would get a definitive answer at least for the unidimensional case but maybe the problem is too difficult and I'm just wasting my time trying to solve it? I'm hoping the question is being understood and if not please let me know and I'll try to make it more clear!