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I'm stuck with a little exercise and cannot find out where I'm wrong. Maybe you can help me.

So we have a differential equation (modified logistic growth):

$$\frac{dN}{dt}=k \ N \left(1-\frac{N}{B}\right) -aN$$

and the questions "What are the steady states $x^*$ and their linear stabilities?"

I can find the steady states; we set the RHS zero and solve: $N=0$ or $$\left( 1-\frac{N}{B}-a\right)=0 \Leftrightarrow N=B(1-a).$$ But what is linear stability? I presume that it does not mean continuity, i.e. a small change in $t$ leads to a small change in $N$.

I think it means that a small change in $N$ means that the change in $N$ as $t\rightarrow \infty$ will be zero? But how do I check this rigorously? And what does linearity have to do with this?

-Marie!

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You missed a $k$ from your equation. So either $N=0$ or $k(1-\frac{N}{B})=\alpha$ etc. "linear stability" - in this context - is a term I hear for the first time. You should ask for clarifications. It sounds like the stability conclusion derived by local linearisation about these equilibrium points. –  Pantelis Sopasakis May 7 '12 at 21:35

1 Answer 1

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Your system can be written as: $$\frac{dN}{dt}=f(N) \ , \ with \ f(N)=kN\left(1-\frac{N}{B}\right)-aN$$In order to define the stability of the fixed point, take a fixed point $N^{*}$ and let $$\eta(t)=N(t)-N^{*}$$ be a small perturbation away from $N^{*}$. If the fixed point is stable, then this small perturbation will decay with time, otherwise it will grow (unstable fixed point). Differentiating the above equation we get: $$\frac{d \eta}{dt}=\frac{dN}{dt}$$ Note that $$\frac{d\eta}{dt}=\frac{dN}{dt}=f(N)=f(N^{*}+\eta)$$ Using Taylor's expansion we obtain: $$f(N^{*} + \eta)=\frac{d\eta}{dt}=f(N^{*})+\eta f'(N^{*})+\mathcal{O}(\eta^2)$$ We know that $f(N^{*})=0$ since $N^{*}$ is a fixed point.

  • Case A

If $f'(N^{*}) \neq 0$, the $\mathcal{O}(\eta^2)$ terms are negligible and we may write the approximation:

$$\frac{d\eta}{dt}\approx \eta \cdot f'(N^{*})$$

This linear equation is the linearization about the fixed point $N^{*}$ and shows the behaviour of the perturbation $\eta (t)$ with time. As you can see, $\eta (t)$ grows exponentially if $f'(N^{*})>0$ and decays if $f'(N^{*})<0$. Therefore, your fixed point is stable if $f'(N^{*})<0$.

  • Case B (not your case)

If $f'(N^{*})=0$, then the linearization fails and the $\mathcal{O}(\eta ^2)$ terms are non-negligible. You have to make a non-linear analysis in this case, solving the non-linear equation (prefferably with a numerical method like $RK4$) and determing the stability on the vector field around the fixed point for example.

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