Problem: Consider the following population model: $\frac{dN}{dt}=\alpha N(t-\tau)e^{-\beta N(t-\tau)}-\delta N(t)$
Determine equilibrium and Linearize about the equilibrium.
From Linearization, draw boundaries in $(\delta\tau,\alpha\tau)$ plane that distinguish between regions where small perturbations:
i. grow monotonically
ii. decay monotonically
iii. decay in an oscillatory manner
iv. grow in an oscillatory manner
Solution: I was able to determine that the equilibrium that we need to linearize is $N^*_2=\frac{1}{\beta}\ln(\alpha/\delta)$, we can ignore the other equilibrium $N^*_1=0$
To linearize, we set
$x(t)=N(t)-N^*_2\implies N(t)=x(t)+N^*_2$
So now, using the following method(don't know the name of it):
$\frac{dx}{dt}=f(N^*,N^*)+\frac{\partial f}{\partial N(t)}|_{(N^*,N^*)}(N(t)-N^*)+\frac{\partial f}{\partial N(t-\tau)}|_{(N^*,N^*)}(N(t-\tau)-N^*)$
which results in the linearization of the ODE near $N^*_2$,
$\frac{dx}{dt}=-\delta x(t)+\delta(1-\ln(\alpha/\delta))x(t-\tau)$.
(Let me know if you want me to code my computations deriving in each step)
From here, we are solving for solutions of the form
$x(t)=x_0e^{\lambda t}\implies\frac{dx}{dt}=x_0\lambda e^{\lambda t}$
Setting both equal to each other and solving for $\lambda$ results in
$\lambda=-\delta+\delta(1-\ln(\alpha/\delta))e^{-\lambda\tau}$
Now, we are to make a change of variables $\mu=\lambda\tau$ which results in
$\mu=-\delta\tau+\delta\tau(1-\ln(\alpha/\delta))e^{-\mu}$
From here, I don't know how to make the $(\delta\tau,\alpha\tau)$ plane to determine the behavior given the equation I have obtained for $\mu$, where do I go from here? My best guess is that I made the wrong change of variables or did a mistake in my computation. I need an equation that provides a relationship between $\delta\tau$ and $\alpha\tau$
Since no one wants to lend me a hand, I guess I'll give my take on how to plot the graphs after thinking about this problem for a few more hours and playing around with it. So after understanding what every constant was describing, I came to the conclusion that $\delta=\lambda$ in this case(I was going step by step with an example but this was the wrong thing to do). Since $\delta$ is the measure of mortality(or measure of growth), it would make sense for the solution to either decay or increase exponentially as a function of this constant. So considering this to be the truth, you'll arrive at the following pair of equations, one for the case of $\alpha\tau$ and the other for the case of $\delta\tau$. Note: I won't work out step by step derivation just state what I got.
$\begin{align*}\delta\tau=\ln(1/2(1-\ln(v)))\\ \alpha\tau=v\ln(1/2(1-\ln(v)))\end{align*}$ where we made the change of variable of the form $v=\frac{\alpha}{\delta}$.
I obtained this by setting the linearized form of the derivative and its exponential counterpart equal to each other. Initially solving for an expression that I could use, not necessarily the $\alpha\tau$ or $\delta\tau$ pair, but eventually saw an expression that would allow me to express the pair into equations that were both dependent on $v$.
Note: we also get $\delta\tau=\frac{\alpha\tau}{v}$, since $v$ is just a constant of proportionality, we can deduce more about the behavior we seek. We can then plot both expressions to see if where they intersect, from here we can deduce the behavior we need according to each region.