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 ============================================================================== Date: Thu, 15 Dec 94 11:23:59 CST From: rusin (Dave Rusin) To: lones@lones.mit.edu, rusin@math.niu.edu Subject: chewed-up differential equation Saw your subsequent post, responded, saw Edgar's post, followed up, then noticed my first post was blank. Don't know what happened. Anyway, given x'(t) = f(x) with f differentiable and x not constant, we can compute integral_0^m x(t) dt by using a change of variables. Since dx/dt = f(x) we have dt = dx / f(x) and in terms of x the integral \int_r^m x dt becomes \int_x(r) ^x(m) [x/f(x)] dx. Since f is only of one sign, x is either increasing or decreasing. Hence lim_{t->0^+} x(t) either exists or goes to infinity or minus infinity respectively. If the limit exists, the original integral is bounded near 0 and so certainly exists. If not, we now have \int_M^\infty x/|f(x)| dx (where M=x(m) ) if f<0 and \int_{-\infty}^M x/f(x) dx if f>0. Either way, the original integral exists on some interval [0,m] iff x/f(x) is integrable on some interval [M, oo) or (-oo, M] respectively. In other words, the functions you want are precisely the (differentiable, one-signed) functions f(x) = x/g(x) where g is in L^1([M,oo)) (or L^1(-oo,M]). Typically such functions g are on the order of x^{-b} for b>1 (making f(x) ~ x^{1+b} ), but your previously posted question and the example in your present post illustrate that there is a delicate area here around g(x) = x^-1. Really there isn't any other way to characterize functions in L^1 except to say they're functions in L^1 ! My students are always amazed that these topics can be of real use. Can you indicate how this question arose in your work? dave ============================================================================== Date: Thu, 15 Dec 94 13:18:07 -0500 From: [Permission pending] To: rusin@math.niu.edu Subject: Re: chewed-up differential equation Hey, thanks a bundle! My coauthor and I were trying to construct a pathological example of sorts, and My first guess is that your proof shows that one does not exist. This is helpful (although the pathological example would have been more striking! One point: in your solution, you write: the functions you want are precisely the (differentiable, one-signed) functions f(x) = x/g(x) where g is in L^1([M,oo)) Now I presume the f being diff'ble is not needed, am I right? As for where this is needed: The paper is called Pathological Outcomes of Observational Learning and I could send you a copy if you wish. It is a really nice tour of how to apply neat math (we develop a theory of stochastic difference equations in ana appendix) in an easily understood "model". Here is the basic idea: Suppose that a countable number of individuals each must make a once-in-a-lifetime binary decision\footnote{% A historical instance might have been the decision to enjoy the maiden voyage of the Titanic.} --- encumbered solely by uncertainty about the state of the world. Assume that preferences are identical, and that there are no congestion effects or network externalities from acting alike. Then in a world of complete and symmetric information, all would ideally wish to make the same decision. But life is more complicated than that. Assume instead that the individuals must decide sequentially, all in some preordained order. Suppose that each individual may condition his decision both on his (endowed) private signal about the state of the world and on all his predecessors' decisions, but not their private signals. ... then with actions and not signals publicly observable, there is a positive chance that a `herd' eventually arises: Everyone after some period would make the identical less profitable decision. ====== The DE help I solicited was for later in the paper. Suppose that 10% of individuals are `crazy' and they make decisions at random. Then we can still show that "public" beliefs converge a.s. to the true state of the world. The question is: Can beliefs converge to confidence that individuals should take action 1 and still have an infinite subsequence of sufficiently pigheaded (sane) individuals who do not. I believe the DE result implies the answer is unifromly NO. (That is why I wish to relax the diff'bility assumption on f.) [sig deleted--djr] ============================================================================== Date: Thu, 15 Dec 94 12:59:04 CST From: rusin (Dave Rusin) To: lones@lones.mit.edu, rusin@math.niu.edu Subject: Re: chewed-up differential equation >As for where this is needed: The paper is called >Pathological Outcomes of Observational Learning >and I could send you a copy if you wish. It is a really nice tour of how to Yes, I'd appreciate that. It appears you're using TeX; a .tex file would be fine > My coauthor and I were trying to construct a pathological >example of sorts, and My first guess is that your proof shows that one does not >exist. This is helpful (although the pathological example would have been more How pathological do you want? You can find functions in L^1 which are just barely in there, for example g(x) = 1/[x log(x) loglog(x)^r ] for any r>1; the antiderivative is loglog(x)^{1-r} /(1-r), so as x->oo you get a bounded integral (when r=1 the antiderivative is logloglog(x) which does not stay bounded but takes a heck of a long time even passing the value 2 !) And of course more log terms makes the situation more dramatic. >One point: in your solution, you write: > >the functions you want are precisely the (differentiable, >one-signed) functions f(x) = x/g(x) where g is in L^1([M,oo)) > >Now I presume the f being diff'ble is not needed, am I right? Gee, is my back really against the wall here? The main problem for my solution is that I want to avoid having f(x)=0 for any x which occurs in the solution of the ODE. Otherwise I can't make the substitution dt=dx/f(x) quite so glibly, nor is the integrability of x/f(x) likely. My way out was to note that x'=f(x) so if x(t) happens to be a value x_0 of x where f(x_0)=0, then we'd have a horizontal tangent in the graph there, suggesting a (locally) constant function x. Indeed, the constant function x(t) = x_0 IS a solution to the ODE passing through this point. But must it be the solution you have at hand? The uniqueness of solutions to ODEs ought to say so, but the usual statement of this theorem is: "If F(t,x) and dF/dx are continuous in a neighborhood of (t_0, x_0) then there is a unique solution to the DE dx/dt = F(t, x) passing through the point (t_0, x_0)". That's why I want f differentiable. Maybe if you want something "pathological" this would be the place to look. Examples include ODEs like x'=sqrt(x) (both x(t)=0 and x(t)=t^2/4 pass through the origin) and x'=2x/t, admittedly not of the form you want, but having infinitely many solutions x=Ct^2 all passing through (0,0). I guess if you can assume that f is always of the same sign you'd be OK; that would prevent f from ever being zero and so division by zero would be avoided. I'm making this up as I go along, but try this: Let F(x) = \int_M^x u/f(u) du. Then by Fundamental Theorem of Calc., F is differentiable [oh, let's assume u/f(u) is at least continuous,OK?] and F'(x) = x/f(x). Then if x is a solution of the DE x'=f(x), we use the Chain Rule to see (d/dt) (F(x(t)) = (x/f(x)).(x'(t)) = x(t). Now, G(t) = \int_t ^m x(u) du likewise has G'(t) = -x(t), so F(x(t)) and -G(t) must differ by a constant. Taking t=m makes both equal to zero, so F(x(t)) = -G(t). Thus your original integral is bounded iff F(x(t)) is bounded. I think it's easy to show this is equivalent to F itself being bounded as x--> oo. I'll have to double-check that, but it looks like this makes a valid proof with the weaker assumption that u/f(u) is continuous -- which in particular still requires f(x) not be zero. If that's still too strong an assumption let me know. Off to class. Another final today. dave