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Consider a real sequence $(x_k)$ for $k=0,1,2,\dots,N$ as $x_0=1$ and for $k>0$ $$ x_k=x_{k-1}+\frac{\gamma}{N}x_{k-1}^2,\qquad (\gamma>0).$$ I wonder to show that the sequence is bounded as $N\to\infty$. I appreciate any idea for proving that.

Hint 1: Numerical experiments suggest me that if and only if $\gamma<1$ the sequence is bounded.

Hint 2: It is known that for $x_k=x_{k-1}+\frac{\gamma}{N}x_{k-1}$ which is linear, a bound exists as $e^{\gamma}$.

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You can view this difference equation as the Euler method for the IVP $y'=\gamma y^2$, $y(0)=1$, on the interval $0\le t\le 1$, using a grid of width $1/N$ and setting $x_k=y(k/N)$.

By solving the ODE, we find that $y$ blows up at $t=1/\gamma$. We want to know if $x_N=y(1)$ stays bounded, and it now follows that this will be the case if $\gamma<1$ (as you suspected), while $x_N$ becomes unbounded if $\gamma>1$. The borderline case $\gamma=1$ would seem to require a more careful analysis.

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    $\begingroup$ For the borderline case $\gamma = 1$: take positive integer $M > 1$, and note that the ODE solution takes the value $M$ at $t = 1 - 1/M$. For $N$ a multiple of $M$, $x_{N - N/M}$ is the Euler approximation to the solution of the ODE at $t = 1 - 1/M$, and thus approaches $M$ as $N \to \infty$, and $x_{N} > x_{N - N/M}$. Thus in this case $x_N$ is unbounded. $\endgroup$
    – Robert Israel
    Commented Aug 8, 2016 at 1:59
  • $\begingroup$ @ChristianRemling Thank you so much! I had not thought about it this way :) It remains to validate that the difference between the numerical solution and the continuous one is bounded which I think is doable since I know the regularity of the exact solution (for $\gamma<1$). $\endgroup$
    – Hamed
    Commented Aug 8, 2016 at 12:53
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    $\begingroup$ @Hamed: That's taken care of by the general theory of these numerical methods: $x_N\to y(1)$ as $N\to\infty$, so not only is the difference bounded, it actually goes to zero. $\endgroup$
    – Christian Remling
    Commented Aug 8, 2016 at 16:26
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    $\begingroup$ @Hamed: Yes, you can indeed be sure that this works; the convergence results for numerical methods for ODEs are of this type. More precisely, for small step size (for us, this means for large $N$), the discrete approximation stays close to the exact solution. On this region, we have a global Lipschitz constant, so everything is peachy. $\endgroup$
    – Christian Remling
    Commented Aug 8, 2016 at 18:49
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    $\begingroup$ ... If $\gamma>1$, you need to be slightly more careful and argue as in Robert's comment: take an interval $[0,t_0]$ on which $y(t)$ stays still bounded, but ends with a very large value, and then apply the convergence results to this situation. $\endgroup$
    – Christian Remling
    Commented Aug 8, 2016 at 18:53

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