I have been struggling lately with solving numerically an equation of the form:

$$ g(x\pm x_{0}) = F[ g(x) ] $$

where $g(x)$ is a matrix satisfying the condition $g(x\to\pm\infty)=0$. My question is on how does one solve this recursively. I am running over some predefined sets $S=\{x_{n}\}_{n=1}^{N}$ of arguments where the function is supposed to be evaluated, so ideally the algorithm should run over a loop that yields: $$g(x_{1}\pm x_{0})$$ $$g(x_{2}\pm x_{0})$$ $$ \vdots$$ $$g(x_{N}\pm x_{0})$$

in this order with $x_{1}<x_{2}< ... <x_{N}$ for $x<0$ $\forall x\in S$ and $x_{1}>x_{2}> ... >x_{N}$ for $x>0$ $\forall x\in S$. Any help would be greatly appreciated thanks!

Edit: For clarity, consider the example:

$$A(x-x_{0})=B(x) + C(x)A(x)D(x)$$ and we know $A(x\to\infty)=0$ and B,C,D are matrices that also depend on $x$.

More of an example:

Let's say I start with a value $ x = -a<0$. My desired goal is to calculate a matrix product for that specific value of x that has the form $$ M(x) = inv( D(x) + T(x)g_{1}(x - x_{0})C(x) + S(x)g_{2}(x+x_{0})P(x) ) $$ where all quantities are matrices but the only unknown ones are $g_{1},g_{2}$. Note that $g_{1},g_{2}$ represent two different matrices. Now, what we know about these matrices is that they satisfy $g_{1,2}(x \to \pm \infty) = 0$. And, additionally, the recursion relation of each of them ( $g_{1} $ and $g_{2}$) has the form written in the edit. More specifically:

$$g_{1}(x-x_{0}) = H( g_{1}(x) )$$ $$ g_{2}(x+x_{0}) = H(g_{2}(x))$$

and both need to be evaluated for that specific value of $x=-a$. My question is on how to do that, since eventually, I want to do it not only for $x=-a<0$, but for an entire grid of $\{x\}$ that includes both positive and negative values.

  • $\begingroup$ What is the meaning of $\pm$ in your $x\pm x_0$? Do the signs generate two different equations that have to be considered simultaneously? If not, choose $x+x_0$ for example and rewrite your equation as the recurrence $g(x) = F[g(x-x_0)]$. Then, assigning an arbitrary value to $g(x)$ on the interval $[a,a+x_0)$, the recurrence uniquely defines $g(x)$ for any $x>a+x_0$. $\endgroup$ May 6, 2020 at 16:41
  • $\begingroup$ Thanks for the quick reply. The $\pm$ sign represent independent cases. For a fixed value of $x$, I want to solve for the case $x-x_{0}$ and $x+x_{0}$. $\endgroup$ May 6, 2020 at 16:47
  • $\begingroup$ So I guess my suggestion was not what you were looking for? But then your "for clarity" edit writes exactly the same kind of equation as I wrote. There's still something not very clear about the question. What is known (the input to the problem) and what is unknown (the desired output) about $g(x)$? $\endgroup$ May 6, 2020 at 19:24
  • $\begingroup$ Aha, the problem is that the recurrence relation leaves $g_1$ and $g_2$ highly underdetermined, unless their values are known on some interval of width $x_0$. It seems that you want to fix that underdeterminacy by the boundary conditions $g_{1,2}(x\to\pm \infty) = 0$. That might work, but it means that you need to solve the recurrence relation on the entire real line (or at least an infinite sub-grid) before being able to determine the value of $g_{1,2}(x)$ at any given $x$. So there is no algorithm that will give you the solution that involves only finitely many points $\{x_n\}_{n=1}^N$. $\endgroup$ May 7, 2020 at 10:43
  • $\begingroup$ So a proposed solution was to start with the boundary condition as initial value, and then iterate the solution, but I am having serious difficulties trying to understand this and how this might work numerically. Because what I want to do eventually is to evaluate $g_{1,2}$ on my fixed set of points $\{\x_{n}}$ to later carry out the matrix product, which has to be one to one, that is, for a fixed value of $x$ of my original grid $\{x_{n}\}$ I want to know $g_{1}(x-x_{0})$ in order to carry out the products. $\endgroup$ May 7, 2020 at 16:27


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