Why is u=0 always an unstable fixed point?

Question

In the following analysis, I cannot see why $u$ is always an unstable fixed point. Calculating $f'(u)= 2R_{0}u-R_{0}-\ln(u)-1 $ at $ u=0 : -R_{0}-1 $ and this in fact suggests that u is a stable fixed point.

Any help would be much appreciated!

Answer 1

While the component $\ln u$ of the formula is not defined at $u=0$ we know that $$\lim_{u>0,~ u\to 0}u\ln u=0$$ so that the ODE is defined at $u=0$ with $f(0)=0$ in this continuous continuation. We also know that $\ln u$ diverges towards $-\infty$ for $u\to 0$ so that this term dominates in the factor $R_0(1-u)+\ln u$. Which means that on some interval $(0,\omega)$ this factor will be negative and the full right side $f(u)=-u(R_0(1-u)+\ln u)$ thus positive. Any solution that starts in that interval is monotonically increasing, moving away from zero, which is the characterization of an unstable point.

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To get a qualitative idea of the way solutions behave close to $0$, reduce the ODE to the terms that are dominant for $u\approx 0$, $$u'=-u\ln u,~~u(0)=u_0\approx 0$$ of the dominant terms. By separation one finds $$ -dt=\frac{du}{u\ln u}= \frac{d(\ln u)}{\ln u}=d(\ln|\ln u|)\\~\\ \ln|\ln u(t)|-\ln|\ln u_0|=-t\implies \ln u(t)=\ln(u_0)e^{-t},~~u(t)={u_0}^{e^{-t}} $$ which moves $\ln u(t)$ from the very large negative value $\ln u_0$ towards zero, that is, from the very small values of $u_0$ towards $1$, away from zero.

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You can find a quasi-explicit form of the second root besides $u=1$ using the Lambert-W function, as $$ 0=R_0(1-u)+\ln u\iff -R_0ue^{-R_0u}=-R_0e^{-R_0} $$ which gives the second solution as $u=-\frac1{R_0}W_k(-R_0e^{-R_0})$ with branch index $k=-1$, $u>\frac1{R_0}$ for $R_0< 1$ and $k=0$, $u<\frac1{R_0}$ for $R_0>1$.

Answer 2

$\ln 0$ isn't zero, it's undefined.

(But $\ln u$ tends to $-\infty$ as $u \to 0^+$, while $u \ln u \to 0$).