How do I derive the characteristic equation around a fixed point $x_0$, when the DDE is defined as $\tau\, dx(t)/dt=-x(t) + f(p-w\,x(t-d)) $, where p,w,d $\in R$ are constants and $f:R\rightarrow R$ is a nonlinear function with $f(x)=1+tan(x)$ ?
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You first replace $f$ with the tangent at $x_0$ namely $y=f'(x_0)(x-x_0)+f(x_0)$. That gives you a linear, non-homogeneous DDE $$ \frac{d}{dt}x(t)=-x(t)-f'(x_0)wx(t-d)+f'(x_0)p-f'(x_0)x_0+f(x_0). $$ Take the homogeneous part $$ \frac{d}{dt}x(t)=-x(t)-f'(x_0)wx(t-d) $$ and substitute $x(t):=e^{\lambda t}$ to get $$ \lambda e^{\lambda t} = -e^{\lambda t}-f'(x_0)we^{\lambda t}e^{-\lambda d}. $$ Divide by $e^{\lambda t}$ and obtain the characteristic equation $$ \lambda+1+f'(x_0)we^{-\lambda d}=0. $$
Uwe Stroinski
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Hi Uwe, many thanks! Very clear explanation. One follow-up question: the characteristic equation wouldn't change, if p was also a variable and we had to do multivariable Taylor-series expansion, would it? – user120162 Feb 09 '14 at 20:03
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I am not sure whether I understand your follow up correctly, so take my comment with a grain of salt. If $p$ depends on $t$ and is an unknown, then the whole equation changes and thus the characteristic equation too. If $p$ depends on $t$ and is known, then you get a non-autonomous equation. Here characteristic equations are not useful. If $p$ does not depend on $t$ the characteristic equation is as above. Its dependence on $p$ (through $f'(x_0)$) seems to be quite complicated. – Uwe Stroinski Feb 10 '14 at 08:28
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Thanks again for your reply. P is a constant and therefore the equation is autonomous. However, I want to study linear stability that's why I assume small perturbations in p that cause small perturbations in x around its fixed point $x_0$. And I need the characteristic equation to get the location of the roots and see how far they are from the imaginary axis. But with your help now I think I now how to proceed. – user120162 Feb 10 '14 at 19:46