- #1
csco
- 14
- 0
Hello everyone, I have asked a similar question in the DE forum but couldn't get an answer so I'm hoping the mods will be tolerant and let me post it here even though it's not strictly analysis.
I'm considering a DE of the form x' = f(t, x) where f is a continuous function defined on an open set D of R2. More importantly f is locally Lipschitz with respect to x which implies there exists a unique maximal solution defined in the maximal interval I(t0, x0) to the initial value problem x' = f(t, x), x(t0) = x0. In short these are first order differential equations in one dimension which verify existence, uniqueness and continuous dependence with respect to initial conditions.
Now for nonlinear equations like x' = x2 the solution cannot be extended beyond a certain time (the maximal interval I(t0, x0) is not all R). The solution to x' = x^2, x(t0) = x0 is x(t) = x0/(1 +x0(t0 - t)) which blows up when t = 1/x0 + t0. In this case the blow up time is a continuous function of the initial conditions. My question is if the Lipschitz condition is enough to ensure this always happens or if it isn't and what would be a counterexample in that case.
I'm hoping someone can at least tell me if this is a too difficult problem and I'm wasting time with it or if I should look for a counterexample or a proof of the above.
I'm considering a DE of the form x' = f(t, x) where f is a continuous function defined on an open set D of R2. More importantly f is locally Lipschitz with respect to x which implies there exists a unique maximal solution defined in the maximal interval I(t0, x0) to the initial value problem x' = f(t, x), x(t0) = x0. In short these are first order differential equations in one dimension which verify existence, uniqueness and continuous dependence with respect to initial conditions.
Now for nonlinear equations like x' = x2 the solution cannot be extended beyond a certain time (the maximal interval I(t0, x0) is not all R). The solution to x' = x^2, x(t0) = x0 is x(t) = x0/(1 +x0(t0 - t)) which blows up when t = 1/x0 + t0. In this case the blow up time is a continuous function of the initial conditions. My question is if the Lipschitz condition is enough to ensure this always happens or if it isn't and what would be a counterexample in that case.
I'm hoping someone can at least tell me if this is a too difficult problem and I'm wasting time with it or if I should look for a counterexample or a proof of the above.