The reason of my question about the fact that solution $u_{n}>0$ are requested is now illustrated. Let's write again the recursive relation...
$\displaystyle \Delta_{n}=u_{n+1}-u_{n}= (r-1)\ u_{n} -r\ u^{2}_{n}= f(u_{n})\ ,\ 0<r<1$ (1)
In the previous post it has been demonstrated that there is only one 'attractive fixed point' in $x_{0}=0$, so that the only 'steady state' can be in $x_{0}=0$. Now we have to find the set of 'initial values' $u_{0}$ for which the sequence converges to $x_{0}$. Analysing we find the following...
a) if $\displaystyle u_{0}< x_{-}= 1-\frac{1}{r}$ the sequence diverges to $- \infty$...
b) if $\displaystyle u_{0}= x_{-}= 1-\frac{1}{r}$ the sequence converges to $x_{-}$...
c) if $\displaystyle x_{-}<u_{0}<0$ the sequence converges to the 'steady state' $x_{0}=0$ and $\forall n$ is $u_{n}<0$...
d) if $\displaystyle u_{0}=0$ the sequence converges [trivially...] to the 'steady state' $x_{0}=0$ and $\forall n$ is $u_{n}=0$...
e) if $\displaystyle 0<u_{0}< x_{1}=1$ the sequence converges to the 'steady state' $x_{0}=0$ and $\forall n$ is $u_{n}>0$...
f) if $\displaystyle u_{0}=x_{1}=1$ the sequence converges [trivially...] to the 'steady state' $x_{0}=0$ and $\forall n>0 $ is $u_{n}=0$...
g) if $\displaystyle 1<u_{0}< x_{2}=\frac{1}{2}+ \frac{\sqrt{r^{2}-4 r +4}}{2 r}$ the sequence converges to the 'steady state' $x_{0}=0$ and $\forall n>0 $ is $u_{n}<0$...
h) if $\displaystyle u_{0}= x_{2}=\frac{1}{2}+ \frac{\sqrt{r^{2}-4 r +4}}{2 r}$ the sequence converges to ' $x_{-}= 1-\frac{1}{r}$ and $\forall n>0 $ is $u_{n}= x_{-}$...
i) if $\displaystyle u_{0}> x_{2}=\frac{1}{2}+ \frac{\sqrt{r^{2}-4 r +4}}{2 r}$ the sequence to $- \infty$ and $\forall n>0 $ is $u_{n}<0$...
These result are more easily understandable looking at the figure, where f(x) for $r=\frac{1}{2}$ is represented...
View attachment 41
Kind regards
$\chi$ $\sigma$