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Let $p(s)\in\mathbb{R}[s]$ be s.t.

  • $p(0)=0$;

  • $p(s)$ has at least one root in the right half complex plane $\{s\in\mathbb{C}\,:\,\Re\mathrm{e}(s)>0 \}$.

Then for every $\varepsilon\in\mathbb{R}$, $$ p_\varepsilon(s):=p(s)+\varepsilon $$ has at least one root in the right half complex plane, that is, $p_\varepsilon(s)$ is not Hurwitz stable.

Do you have some ideas about how to prove the latter claim? Or can you provide a counterexample?

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    $\begingroup$ If the known root is simple, this is just the continuity. Otherwise, $x(x-1)^2$ is a counterexample (shift the graph up by $\epsilon$). $\endgroup$
    – Alex Degtyarev
    Commented Dec 6, 2014 at 19:16
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    $\begingroup$ I think I missed your point. If $p(s)=s(s-1)^2$ then, by applying the Routh-Hurwitz stability criterion, $p_\varepsilon(s)=s(s-1)^2+\varepsilon$ has always a root with positive real part for all $\varepsilon\in\mathbb{R}$. $\endgroup$
    – Ludwig
    Commented Dec 6, 2014 at 20:05
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    $\begingroup$ Sorry, I guess I misread your question. Please edit something so that I can remove the downvote :) $\endgroup$
    – Alex Degtyarev
    Commented Dec 6, 2014 at 21:18

1 Answer 1

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This is a question a while ago, however a counterexample is as follows:

(In SAGE) E=1; P=x*(x+1)(x+2)(x-0.00001+10*i)*(x-0.00001-10*i)+E; P.roots(ring=CDF)

(Output) [(-2.0047725863348935, 1), (-0.9900961078881397, 1), (-0.0050380070624979085, 1), (-3.6649357233242696e-05 - 9.99998572023422*I, 1), (-3.664935723257656e-05 + 9.999985720234221*I, 1)]

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