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October 4th, 2014, 04:21 PM   #1
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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.
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October 4th, 2014, 04:29 PM   #2
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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?
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October 4th, 2014, 04:37 PM   #3
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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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