# Determining Bifurcation of a Function

Hello I am trying to find analyze the bifurcation behavior of

$$\dot{N} = N(N - e^{\alpha N}) , N \geq 0, \alpha > 0$$

as $$\alpha$$ is varied and find their stability. Playing around with different values of $$\alpha$$, I see that when $$\alpha$$ is not small(e.g. $$\alpha \geq 1$$ would be considered "not small" in regards to this particular system) I find that for positive values of $$N$$ the only fixed point will be at $$N = 0$$ when considering only the non negative values of $$N$$. This fixed point is stable because all positive values of $$N$$ are decreasing towards it and more rapidly so the larger $$N$$ is relative to $$\alpha$$.

On the other hand an interesting thing seems to occur for a small enough value, say suppose $$\alpha = .1$$, assigning random nonnegative but small such as $$N = 1$$ $$\dot{N}$$ will be negative since $$\dot{N}(1) = 1 - e^.1 < 0$$, positive when $$N$$ is "just right in terms of being small or large" e.g. $$N = 10$$ causes $$\dot{N}$$ to be positive since $$\dot{N}(10) = 10(10 - e^1) > 0$$ but then negative again after $$N$$ is large enough relative to $$\alpha = .1$$ since $$\dot{N}(100) = 100(100 - e^{10}) < 0$$. Based on this quick qualitative analysis of the simple example of the value I assigned of $$\alpha = .1$$, it seems there will now be 3 fixed points, 0 will still be stable but the second fixed point occurring immediately after $$N$$ starts to be positive will be unstable since values greater than this fixed point will increase away from it. The third fixed point occurring when $$\dot{N} = 0$$for when the function then starts to decrease again will also be stable since values before it are increasing but decreasing afterwards.

The main difficulty I am having in further studying the bifurcation behavior of this system is determining what is the exact bifurcation point and what type of bifurcation it would be classified as. Currently I hypothesize that the particular bifurcation point would be considered a saddle node bifurcation since the two fixed points appear and vanish before and after alpha is small enough relative to the system and that a logarithmic scale analysis would be helpful when trying to represent the bifurcation. I want to know how to solve for the exact bifurcation value in a quantitative manner, any advice would be much appreciated.

• I added the "differential-equations" tag to your post. Cheers! – Robert Lewis Oct 12 '18 at 20:09

First, there is a constant fixed point of $$N=0$$ for all $$\alpha$$. The other fixed points are the real roots of $$N - \exp(\alpha N)$$, which are potentially $$r_0(\alpha) = \frac{\text W_0(-\alpha)}{-\alpha}\text{ and }r_{-1} (\alpha) = \frac{\text W_{-1}(-\alpha)}{-\alpha}$$ where $$\text W(x)\exp(\text W(x)) = x$$ for all $$x\in\mathbb{C}$$, and $$\text W_n(x)$$ is the $$n$$-th branch of $$\text W(x)$$.

I say potentially because, as you know, these roots are only real for a subset of possible $$\alpha$$. In particular, we can see that for $$\alpha > 0$$ close to $$0$$, both roots are real, and we have $$\lim\limits_{\alpha\to 0} r_0 = 1\text{ and }\lim\limits_{\alpha\to 0} r_{-1} = \infty$$ The $$\alpha$$ for which $$\max\limits_N [N - \exp(\alpha N)] = 0$$ is $$\exp(-1) = \frac{1}{e}$$, and $$r_0(\frac{1}{e}) = r_{-1}(\frac{1}{e}) = e$$.

Using this information, you can see there is a saddle-node bifurcation at $$\alpha = \frac{1}{e}$$. The $$N=0$$ branch and the upper $$N = \exp(\alpha N)$$ branch, $$r_{-1}(\alpha)$$, are both stable; the lower $$N = \exp(\alpha N)$$ branch, $$r_0(\alpha)$$, is unstable. Below is a graph of the bifurcation diagram on the $$(\alpha, N)$$-plane (the vertical green line is $$\alpha = \frac{1}{e}$$):

Here is a link to the graphing utility

For $$\alpha < 0$$ there are indeed two equilibria, one at $$x_1 = 0$$ (stable) and one at $$x_2 = - \frac{W(-\alpha)}{\alpha}$$ (unstable), where $$W$$ is the Lambert W function. For $$\alpha = 0$$ the unstable solution is at $$x_2 = 1$$. For $$\alpha > 0$$ a third equilibrium $$x_3= - \frac{W_{-1}(-\alpha)}{\alpha}$$ appears, which is stable. Here $$W_{-1}$$ is the lower branch of the Lambert W function.

Therefore, as $$\alpha \to 0$$, $$x_3 \to \infty$$. As $$\alpha$$ increases past $$e^{-1}$$, $$x_2$$ and $$x_3$$ disappear in a fold bifurcation and only the stable solution $$x_1 = 0$$ remains.

So there is a stable solution branch which appears from $$\infty$$ as $$\alpha$$ becomes positive, a phenomenon called "bifurcation from infinity".