ODE equivalent to a system of Difference Equations (Discrete to Continuous time)

Question

Consider the following gradient-descent ascent system of equations: $$\begin{cases} x_{k+1} = x_{k} - \eta \nabla_{x} g(x_{k}, y_{k}) \\ y_{k+1} = y_{k} + \eta \nabla_{y} g(x_{k+1}, y_{k}) \end{cases}$$ The main idea is that player $x$ plays first and then $y$ gets to respond; observe that $y_k$ depends on $x_{k+1}$. Were it not for the latter, the system could become a simple ode by letting the stepsize $\eta \rightarrow 0$: $$\begin{cases} \dot x = - \nabla_x g(x, y) \\ \dot y = + \nabla_yg(x,y) \end{cases} $$ But this is not the case. What should I do may be obvious but I cannot see it yet. Any ideas how could I convert it to a continuous problem?

Answer 1

Following the lines of https://math.stackexchange.com/a/3642860/115115, try to use a perturbation approach, set \begin{align} \dot x&=-g_x+ηf\\ \dot y&=g_y+ηh \end{align} Then \begin{align} x(t+η)&=x(t)-ηg_x+η^2f-\tfrac12η^2(g_{xx}\dot x+g_{xy}\dot y)+O(η^3) \\ y(t+η)&=y(t)+ηg_y+η^2h+\tfrac12η^2(g_{yx}\dot x+g_{yy}\dot y)+O(η^3) \end{align} To get this mostly similar to the step equation of the discrete method, one would have to chose $$ f=\tfrac12(-g_{xx}g_x+g_{xy}g_y) $$ and using that $$ y_{k+1}=y_k+ηg_y(x_k-ηg_x,y_k)=y_k+ηg_y-η^2g_{yx}g_x+O(η^3) $$ one gets to $$ h=-\tfrac12(g_{yx}g_x+g_{yy}g_y) $$ So in total, the exact solution of \begin{align} \dot x&=-g_x-\tfrac12η(g_{xx}g_x-g_{xy}g_y)\\ \dot y&=+g_y-\tfrac12η(g_{yx}g_x+g_{yy}g_y) \end{align} should follow $O(η^2)$ close the path of the discrete system. Or reverting the Taylor expansion to avoid using second order derivatives \begin{align} \dot x&=-g_x(x+\tfrac12ηg_x(x,y),\,y-\tfrac12ηg_y(x,y))\\ \dot y&=+g_y(x-\tfrac12ηg_x(x,y),\,y-\tfrac12ηg_y(x,y)) \end{align}