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 October 4th, 2014, 04:21 PM #1 Member   Joined: May 2012 Posts: 86 Thanks: 0 Diff. Equations problem I got to prove the following: Suppose f:R->R is continuously differentiable and consider the differential equation x'=f(x) (DE) If f(x)=0 for at least two values of x, then (DE) has at least one unstable equilibrium. (By an unstable equilibrium c it is meant that f(c)=0 and for any epsilon>0 there is some delta>0 such that any solutions that are at a distance less than delta of c will from there onwards remain at a distance less than epsilon from c). However I fail to see how this is always true, since if we let f=0, then DE has only stable equilibriums. Is there something I am not seeing? Answers are appreciated. October 4th, 2014, 04:29 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 Math Focus: Mainly analysis and algebra I assume that $f = 0$ is excluded as being the trivial solution. You have written $x^\prime = f(x)$. That suggest that $x$ is itself a function of some other variable. Is that correct? October 4th, 2014, 04:37 PM #3 Member   Joined: May 2012 Posts: 86 Thanks: 0 Yes, that is correct. However the function defined piecewise by f(x)=0 if x<0, f(x)=-x^2 if x>=0 Also appears to have only stable equilibriums, doesn't it? Last edited by Jakarta; October 4th, 2014 at 04:49 PM. Tags diff, equations, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Niko Bellic Calculus 2 July 8th, 2013 10:01 AM mathkid Calculus 5 September 10th, 2012 03:10 PM BonTrust Calculus 10 January 21st, 2012 05:24 PM thummel1 Calculus 2 September 1st, 2009 05:58 PM Polle Applied Math 0 December 5th, 2007 09:38 AM

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