Currently, I am working on a system of non-linear partial integro-differential equations and trying to find stationary solutions numerically. The system can be written as \begin{equation} \begin{aligned} \frac{\partial}{\partial t} f(x;t) = F(f(x;t), l((x,y;t)) \\ \frac{\partial}{\partial t} l(x,y;t) = L(f(x;t), l((x,y;t)), \end{aligned} \end{equation} where $t$ is the time and $x,y$ are variables contained in a certain set $S$. $f$ and $l$ are unknown, real positive, functions and $F$ and $L$ are expressions containing these functions in multiple terms, amongst others double integral terms. I am struggeling to find out if the Newton-Raphson scheme is a suitable approach for finding stationary solutions.
So far I tried the following:
We write $$ \mathbf{x} = \begin{bmatrix} f(x;t) \\ l(x,y;t) \end{bmatrix} $$
and $$ \mathbf{G} = \begin{bmatrix} F(f(x;t), l((x,y;t)) \\ L(f(x;t), l((x,y;t)) \end{bmatrix} $$
such that after discretisation $$ \mathbf{x}^{\tau+1} = \mathbf{x}^\tau - J_{\mathbf{G}}^{-1}(\mathbf{x}^\tau) \cdot \mathbf{G(x^\tau)} $$ in which, $$ J_{\mathbf{G}}(\mathbf{x}^\tau) = \begin{bmatrix} \frac{\partial F(f(x;t), l((x,y;t))}{\partial f(x;t)} & \frac{\partial F(f(x;t), l((x,y;t))}{\partial f(x;t)} \\ \frac{\partial L(f(x;t), l((x,y;t))}{\partial l((x,y;t)}& \frac{\partial L(f(x;t), l((x,y;t))}{\partial l((x,y;t)} \end{bmatrix}\\ $$
Here I am getting stuck. How are these derivatives of an expression to a function defined (if they exist)? And maybe more general, does it make sense to find stationairy solutions this way?
Addition:
Would it be an idea to use $$ \frac{\partial F(f(x;t),l(x,y;t))}{\partial f(x;t)} = \frac{\partial F(f(x;t),l(x,y;t))}{\partial x} \frac{\partial x}{\partial f(x;t)} + \frac{\partial F(f(x;t),l(x,y;t))}{\partial t} \frac{\partial t}{\partial f(x;t)} = \frac{\partial F(f(x;t),l(x,y;t))}{\partial x} \frac{\partial f(x;t)}{\partial x}^{-1} + \frac{\partial F(f(x;t),l(x,y;t))}{\partial t} \frac{\partial f(x;t)}{\partial t}^{-1}$$ Under what assumptions is this valid?
Thanks in advance.