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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.

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