So I'm again working on doing something similar to this paper and could use some help. In the paper they worked with the equation $N(t)[(a(t)-b(t)N(t))dt + \alpha(t)dB(t)]$. It's a normal logistic equation but here the usual parameters a,b are continuous T-periodic functions (whereas usually they're considered constants). Here is the excerpt from the paper:
Now I want to do something similar, but to work with a modified equation: $N(t)[(a(t)-b(t)N(t))dt + \alpha(t)N(t)dB(t)]$. The difference is that in the original paper, the a(t) was stochastically perturbed, and not I'm perturbing the b(t). I would like to derive an $x(t)$ like they have.
I worked on it a bit and was able to derive the following:
$dx(t) = \left[ \frac{\alpha(t)^2}{x(t)} -a(t)x(t) + b(t)\right]dt - \alpha(t)dB(t)$.
What I'm stuck on being able to do is derive the $x(t)$ from this like they have. In this paper, they just gave an $x(t)$ and said it satisfies $dx(t)$. I can see the motivation of $x(t)$ if you're a bit informal and solve $dx(t)$ like it's a usual first-order differential equation. In the case I'm working on, I'm not so lucky. Can anyone please help me derive an $x(t)$ if it can be done? Thank you!
You would be better served by going the other way and not having to guess. Thus set, like in quadratic Bernoulli equations, $x=1/N$, then as in the citation, $$ dx=\frac1{N+dN}-\frac1N=-\frac{dN}{N^2}+\frac{dN^2}{N^3}+h.o.t $$ where $dN^2=d⟨N⟩=d⟨N,N⟩$ and the higher order terms do not contribute in the integration of the SDE. With $$ dN^2=(αN^2dB+h.o.t)^2=α^2N^4\,dt+h.o.t. $$ one gets thus $$ dx=-[(a·x−b)·dt+α·dB] + α^2/x·dt $$ which does not seem as helpful as the same transformation in the original problem.