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For example, consider a Verhulst model with delay

$$ \boxed{\dot N(t) = r N(t) \left( 1 - \frac{N(t-T)}{K} \right)} $$

where $r$ gives the reproduction rate and $K$ means the carrying capacity of the environment.

The non-trival equilibrium is $N^*=K$. To study the stability of this equilibrium, we introduce $h$, where $N(t) = K + h(t)$, and use the following linearisation

$$ \dot N(t)= \dot h (t) \approx -r h(t-T) $$

The solution will be in the form $ae^{\lambda t}$ and thus

$$a\lambda e^{\lambda t}=-are^{\lambda t}e^{-\lambda T}\\ \lambda=-re^{-\lambda T}$$

Thus, we have introduced an 'eigenvalue' $\lambda$ which is a complex number because there is no reason to put any restrictions. So generally $$\lambda=-re^{-\lambda T}=\nu+i\omega$$ However, how should I understand the biological meaning of this value? As well as its real and imaginary components?

I have seen another definition for 'eigenvalue' in linear algebra, but I can't see the relation with the one I encountered here.

m1ssing
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  • imaginary means oscillations – user619894 Nov 17 '22 at 18:36
  • Aren't DDEs infinite-dimensional? – Rodrigo de Azevedo Nov 19 '22 at 01:35
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    What makes you think there is a nonzero imaginary component? – Rodrigo de Azevedo Nov 19 '22 at 11:32
  • You don't have an ODE. You have a DDE. Why are you looking for solutions of the exponential form? – Rodrigo de Azevedo Nov 19 '22 at 11:37
  • @RodrigodeAzevedo So is the reason why this characteristic equation differs from ones from gereral linear algebra is that, there are infinitely many initial conditions, and as a result, the asymptomatic behavior near equilibria is influenced? – m1ssing Nov 20 '22 at 15:51
  • @user619894 I'm still a little confused. Does this mean the real part results in a monotone behavior, and the imaginary part results in a oscillation, the overall behavior is a combination of the two? – m1ssing Nov 20 '22 at 15:55
  • I agree with @RodrigodeAzevedo, that $\lambda=-re^{-\lambda T}$ does not imply complex $\lambda$ – user619894 Nov 20 '22 at 16:13
  • @m1ssing I hadn't thought of it that way but it does make sense. I have very little experience with DDEs, but in control systems one encounters DDEs and even linear DDEs are somewhat nightmarish. Hopefully, someone who actually knows something about the topic will provide some input. – Rodrigo de Azevedo Nov 20 '22 at 16:54
  • @m1ssing The Laplace transform of the solution of the linearised DDE is $$\hat h (s) = \frac{h_0}{s + r e^{-sT}}$$ Do you want to try to compute the inverse Laplace transform? Note that the exponential can be written in power series form, which means that the denominator has "degree $\infty$" rather than degree $1$. Infinitely many "modes of oscillation". Sounds like fun? – Rodrigo de Azevedo Nov 20 '22 at 17:04
  • @m1ssing In other words, you may have infinitely many "eigenvalues". Or, maybe, I am spewing nonsense. Take a look at this – Rodrigo de Azevedo Nov 20 '22 at 17:55
  • have you looked at floquet theory? – GeauxMath Jan 03 '23 at 08:52

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