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Is there an algorithm which I can use to find an approximate solution of the following second-order difference equation in order to run simulations of the solution :

$\nu(1 - \beta) f(u_{n+1})-g(u_{n+1}-u_{n}) + \beta g (u_{n+2} - u_{n+1}) = 0 $

with $\nu > 0$ and $0 < \beta < 1$ and $u_0 = 0$ and $u_N = C > 0$ and $N \in \mathbb{N}^{*}$ and $f$ and $g$ two real, positive and increasing functions.

Thank you.

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Rewrite as $$ u_n = u_{n+1} - g^{-1}\bigl[(1-\beta)\nu f(u_{n+1})+\beta g(u_{n+2}-u_{n+1})\bigr]. $$

If you fix $u_N=C$ (as given in the problem) and also $u_{N-1}$, then you can use the previous recurrence formula to solve backward for $x_{N-2},x_{N-3},\dots,x_1,x_0$. Then the problem becomes to find an appropriate value for $x_{N-1}$ so that you end up with $x_0=0$.

This is what is called shooting method for the boundary value problem $x_0=0$, $x_N=C$.

In precise terms, through the above backward solution, you get a map $\phi:\mathbb R\to\mathbb R$ which gives you $x_0=\phi(x_{N-1})$. Your goal is now to solve $\phi(x)=0$. This can be done in several ways, noting in particular that if $f,g,g^{-1}$ are differentiable, then $\phi$ is also differentiable, so you may be able to use Newton's method, beside the bisection method.

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  • $\begingroup$ Thank you very much, seems to work perfectly. Great method. $\endgroup$ – AlexC75 Dec 13 '18 at 17:42
  • $\begingroup$ @AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you. $\endgroup$ – Federico Dec 13 '18 at 17:48

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