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From a delay system, I obtain the following as part of a characteristic equation: $$f(\lambda) = \lambda - a + be^{-c\lambda},$$ where $a, b,$ and $c$ are positive number and $a<b, ac<1$. My goal is to find the sign of the real part of the root to $f(\lambda) = 0$. Taking $\lambda = x + iy$, I obtain: \begin{align} x &= a - be^{-cx}cos(cy)\\ y &= be^{-cx}sin(cy). \end{align} I am looking for a condition for $x$ to be negative, but I fail to obtain anything meaningful based on the given constraints (for instance, if $a<0$, then $x<0$). I also narrow down $y \in [-\frac{\pi}{2c},\frac{\pi}{2c}],$ but fail to get something from that as well. Any help would be much appreciated!

All of my simulations show that under the given conditions, $x$ is always negative.

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    $\begingroup$ as stated the question is difficult to parse: why do you identify the real part of $\lambda$ with the variable $x$? are you trying to solve $f(x)=0$ ? $\endgroup$
    – Carlo Beenakker
    Feb 26, 2020 at 20:57
  • $\begingroup$ @CarloBeenakker Yes. I am trying to solve for $f(x) = 0$. $\endgroup$
    – Paichu
    Feb 26, 2020 at 21:11
  • $\begingroup$ The argument principle counts the zeros inside a closed contour. Consider applying it to a sequence of contours whose interiors exhaust the region with $x>0$. This way, you could potentially exclude the existence of zeros with $x>0$ (or count them, if they do exist). $\endgroup$
    – Igor Khavkine
    Feb 26, 2020 at 21:13
  • $\begingroup$ Stating a consistent problem might help get meaningful answers. From $\lambda-a+b\exp(c\lambda)=0$, with $\lambda=x+iy$, I get $x=a-b\exp(cx)\cos(cy)$ and $y=-b\exp(cx)\sin(cy)$, but those are not the equations you state. In a problem that is all about signs, it helps to keep them straight. $\endgroup$
    – Michael Renardy
    Feb 27, 2020 at 3:33
  • $\begingroup$ @MichaelRenardy Thank you for pointing that out. I fixed the equation. $\endgroup$
    – Paichu
    Feb 27, 2020 at 18:26

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I think your equation is $f(\lambda) = \lambda - a + be^{-c \lambda}$. Let us consider the parametrized family $f_{\varepsilon}(\lambda) = \lambda - a + b e^{-\varepsilon c \lambda}$, where $\varepsilon \in [0,1]$. First of all note that the number of roots $f_{\varepsilon}(\lambda)=0$ in every half-plane $\operatorname{Re}\lambda > \beta$, $\beta \in \mathbb{R}$, is finite and uniformly (in $\varepsilon$) bounded. Suppose we know that for all $\varepsilon$ there is no roots $f_{\varepsilon}(\lambda)=0$ lying on the imaginary axis. Then we can consider a closed contour in $\operatorname{Re}\lambda \geq 0$, which encloses all the roots with $\operatorname{Re}\lambda > 0$ for all $\varepsilon$. From the Rouché theorem it follows that all equations $f_{\varepsilon}(\lambda)=0$ has the same number of roots in this contonur for all $\varepsilon$. It is clear that for $\varepsilon=0$ there is only one root $\lambda = -b + a$, which is negative due to $a < b$. Therefore, $f_{1}(\lambda)=f(\lambda) = \lambda - a + be^{-c \lambda}$ has no roots with $\operatorname{Re}\lambda > 0$.

So now we should provide conditions for the equations $f_{\varepsilon}(\lambda)=0$ to not have purely imaginary roots. I leave this for you.

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  • $\begingroup$ Thank you very much for your help! Could you please clarify how the family of $f_\epsilon$ satisfies the conditions on Rouche's theorem on the boundary of the closed contour? I apologize if this is trivial. $\endgroup$
    – Paichu
    Feb 27, 2020 at 20:56
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    $\begingroup$ The dependence of $f_{\varepsilon}$ on $\varepsilon$ is continuous. Thus every $\varepsilon_{0}$ has a small neighbourhood in which $f_{\varepsilon}$ has the same number of roots as $f_{\varepsilon_{0}}$. Indeed, we apply the Rouché theorem to $g:=f_{\varepsilon} - f_{\varepsilon_{0}}$ and $f:=f_{\varepsilon}$; if $\varepsilon$ is sufficiently close to $\varepsilon_{0}$ then on the boundary we have $|g| < |f|$. This proves the statement. Now use the compactness of $[0,1]$ to prove that $f_{\varepsilon}$ has the same number of roots inside the contour for all $\varepsilon$. $\endgroup$
    – demolishka
    Feb 27, 2020 at 21:50

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