Show that two solutions $x_n$ and $y_n$ generated by implicit Euler satisfy the inequality $|x_n - y_n|$ $\leq$ $|x_0 - y_0|$.
Can someone push me in the right direction by explaining to me how exactly implicit Euler's method works? I understand that it is some iteration method where $$u_{k+1} = u_k + hf(t_{k+1}, u_{k+1}).$$ And the $u_{k+1}$ is often found by some other iteration method (kind of shaky on this part here). I was wondering if there were some way to manipulate the equation to rewrite $x_n = x_{n+1} - hf(t_{k+1}, x_{k+1})$ and $y_n$ similarly but I'm not sure if that would get anything done. We are also given that $f$ satisfies the condition where $(x-y)(f(t,x)-f(t,y)) \leq 0$, which means $|x(t)-y(t)| \leq |x(0)-y(0)|$. Any help would be greatly appreciated. Thank you!
If you just write down the difference of the two iterations and multiply with the right difference of arguments, you find that \begin{align} (x_+-y_+)^2&=(x_+-y_+)(x-y)+h(x_+-y_+)(f(t_+,x_+)-f(t_+,y_+)) \\ &\le (x_+-y_+)(x-y) \end{align} Now dividing by $|x_+-y_+|$ gives the desired result in one step and thus also in the general case.
Using Cauchy-Schwarz, this can also be extended to vector-valued ODE.