Considering the numerical scheme
$$u^{n+1}=u^n+ku^n(1-u^n)$$
where $u'(t)=u(t)(1-u(t))$ and $u(0)=a$. Also, assume that $0<k<1$, and $0<a<1$.
How can I prove that if $0 \le u^n \le 1$, then $0 \le u^{n+1} \le 1$ using mathematical induction.
Thank you.
First observe Property 1, $\min_{x\in[0,1]} x+kx(1-x) = 0$ and $\max_{x\in[0,1]} x+kx(1-x) = 1$ if $k<1$.
So for $n=0$, since $u^0=a$ and $0<a<1$, $0\leq u^1\leq 1$ by Property 1.
For $n=k$, assume it is true, e.g. $u^{n} = u^{n-1} + ku^{n-1}(1-u^{n-1})$ with $0\leq u^{n-1}\leq 1$ and $0\leq u^{n}\leq 1$.
For $n=k+1$, $$\begin{align} u^{n+1} &= u^{n} + ku^{n}(1-u^{n}) \end{align}.$$ Since $0\leq u^n \leq 1$, by Property 1, we conclude $0\leq u^{n+1} \leq 1$.
Only missing part is to show that Property 1 is actually correct. But that is easy and I am leaving it to you