From: lasmith@athena.mit.edu (Lones A Smith) Newsgroups: sci.math Subject: A brief differential equations problem to chew on... Date: 15 Dec 1994 03:53:04 GMT Let x be a smooth function of t, satisfying x'=f(x). Under what iff conditions on f is \int_0^m x(t) dt<\infty for m>0 small enough? Lones -- Lones A. Smith, Economics, MIT E52-252C, Cambridge MA 02139 v-mail: (617) 253-0914 e-mail: lones@lones.mit.edu (NeXT mail or the garden variety stuff) lasmith@athena.mit.edu (if all else fails) ============================================================================== From: lasmith@athena.mit.edu (Lones A Smith) Newsgroups: sci.math Subject: Re: A brief differential equations problem to chew on... Date: 15 Dec 1994 15:33:27 GMT In article <3coff1$8hc@mark.ucdavis.edu> chrisman@ucdmath.ucdavis.edu (Mark Chrisman) writes: >In article <3coej0$9ln@senator-bedfellow.MIT.EDU> >lasmith@athena.mit.edu (Lones A Smith) writes: > >> >> Let x be a smooth function of t, satisfying x'=f(x). >> >> Under what iff conditions on f is \int_0^m x(t) dt<\infty >> for m>0 small enough? >> Lones >> > >Huh? You don't need any conditions on f. Given that x is differentiable, it >is continuous, hence is both bounded and integrable on any interval [0,m]. > >Do you mean that x is smooth for t>0 (and may have an asymptote at 0)? > > >-------------------------------------------------------- >Mark Chrisman (if your reply is of general interest > please post it, don't email it. Thanks!) > >Do you mean that x is smooth for t>0 (and may have an asymptote at 0)? Yes, sorry. What do you expect for posts made at 1am?? To sharpen the (or one particular application) for those who like functional forms, the following is an incredible kniefedge knifeedge knife-edge case: f(x)=-x^2 (\log x)^2 I would love to see how to approach that one, i.e. does x'=-x^2 (\log x)^2 => \int_0^1 x(t)<\infty Note as previously observed that x blows up near the origin. Lones -- Lones A. Smith, Economics, MIT || The `v' in Massachvsetts Institvte E52-252C, Cambridge MA 02139 || of Technology is a Latin affectation 617-253-0914 (voice), -6915 (fax) || over which I have no control. e-mail: lones@lones.mit.edu || Feel free to substitute `u' as desired. ============================================================================== From: edgar@math.ohio-state.edu (Gerald Edgar) Newsgroups: sci.math Subject: Re: A brief differential equations problem to chew on... Date: Thu, 15 Dec 1994 11:13:55 -0500 In article <3cpnk7$mjq@senator-bedfellow.MIT.EDU>, lasmith@athena.mit.edu (Lones A Smith) wrote: > > f(x)=-x^2 (\log x)^2 > > I would love to see how to approach that one, i.e. does > > x'=-x^2 (\log x)^2 => \int_0^1 x(t)<\infty > > Note as previously observed that x blows up near the origin. > A bit of fiddling with Maple seems to show that the solution behaves asymptotically like t = 1/(x log(x)^2), so the integral converges. . . . . Gerald A. Edgar edgar@math.ohio-state.edu Department of Mathematics The Ohio State University telephone: 614-292-0395(Office) Columbus, OH 43210 614-292-4975 (Math. Dept.) ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: A brief differential equations problem to chew on... Date: 15 Dec 1994 16:42:30 GMT In article <3cpnk7$mjq@senator-bedfellow.MIT.EDU>, lasmith@athena.mit.edu (Lones A Smith) wrote: > f(x)=-x^2 (\log x)^2 > > I would love to see how to approach that one, i.e. does > > x'=-x^2 (\log x)^2 => \int_0^1 x(t)<\infty > > Note as previously observed that x blows up near the origin. In article , Gerald Edgar wrote: >A bit of fiddling with Maple seems to show that the solution >behaves asymptotically like t = 1/(x log(x)^2), so the integral converges. As in a related post I just made, you have dt=dx/[-x^2 log(x)^2] so the integral is equal to \int_x(1) ^\infty dx/[xlog(x)^2] = 1/log x(1). This is finite, assuming that x(1) <> 1. This brings up a technical issue I didn't put in the previous post. I hope it's OK to assume that in your differential equation x'=f(x) the function f itself is also differentiable. That way the uniqueness theorem for solutions of first-order ODE's applies. As a consequence, if there is a point on the curve x=x(t) at which f(x)=0, then x will be constant. (In the example above, for instance, you can't have x(1)=1 and expect to continue the solution x=x(t) past t=1 unless x is the constant function x(t)=1.) Furthermore, the assumption that f is differentiable implies f will be continuous, so the fact that it's never zero will mean it keeps the same sign. I had all this in mind when I suggested in the previous post that you need x/f(x) integrable over [M, oo), but I didn't make it explicit. While we're getting technical, let me point out that the solution to the ODE above need _not_ blow up at the origin. It does if x(t)>1 for some t>0, but not if x(t)=1 for some t (which makes x constant) or if x(t)<1 for some t (in which case 0< x(t) < 1 for all t). dave (still grading an ODE final). ============================================================================== From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math Subject: Re: A brief differential equations problem to chew on... Date: 15 Dec 1994 20:45:58 GMT In article <3cpnk7$mjq@senator-bedfellow.MIT.EDU>, lasmith@athena.mit.edu (Lones A Smith) writes: |> To sharpen the (or one particular application) for those who like functional |> forms, the following is an incredible kniefedge knifeedge knife-edge case: |> |> f(x)=-x^2 (\log x)^2 |> |> I would love to see how to approach that one, i.e. does |> |> x'=-x^2 (\log x)^2 => \int_0^1 x(t)<\infty |> |> Note as previously observed that x blows up near the origin. The equation x' = f(x) is separable, and has solutions of the form G(x) = t + c, where G'(x) = 1/f(x). We can assume wlog that c=0. Now you want x -> infinity as t -> 0, i.e. G(x) -> 0 as x -> infinity. So we must have G(x) = - int_x^infinity 1/f(s) ds. Now int_0^1 (x(t) - x(1)) dt = int_x(1)^infinity G(x) dx (reflect the graph about x-x(1)=t) = - int_x(1)^infinity int_x^infinity 1/f(s) ds dx = - int_x(1)^infinity (s - x(1))/f(s) ds So the criterion you want is that this should be finite. In the case above, it is true that int_x(1)^infinity (s-x(1))/(s^2 (log s)^2 ds < infinity. -- Robert Israel israel@math.ubc.ca Department of Mathematics University of British Columbia Vancouver, BC, Canada V6T 1Y4