I posted the same question to Math Stackexchange earlier without much luck, so I am posting here.
I am dealing with a time-dependent model, which can be expressed as a function. $f$ is dependent on two variables: $ x \in \mathbb{R}^n$ (an input to the model),and $t\in \mathbb{R}$ (time). Although the function is dependent on both variables, I will use the following notation $f(x):=f(x,t)$ where the dependency to and $t$ is implied. $f : \mathbb{R}^d \rightarrow \mathbb{R}^n$.
Consider a function $\Pi: \mathcal{F} \rightarrow \mathcal{F}$:
$$ \Pi (f) (x) = \sum_{i=1}^{P} K \left(x,x_i\right) f(x_i) $$
where $x\in\mathbb{R}^d$, and $K:\mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}^{n\times n}$. To clarify, $\Pi (f) (x)$ means "evaluate $\Pi (f) \in \mathcal{F}$ at $x$". The set $\{x_1,\ldots,x_P\}$ is fixed (not time-dependent), but the input $x$ to the function $K \left(x,x_i\right)$ can be any value in $\mathbb{R}^d$. $K$ is a symmetric tensor in $\mathcal{F}\otimes \mathcal{F}$. (side note: In fact, $K_{kk'}(x,x')=\mathbb{E}_g[g_k(x)g_{k'}(x')]$, where $g_k(x)$ indicates $k^{th}$ coordinate of $g(x)$. $g$ is a function that is randomly sampled from $\mathcal{F}$ according to some distribution. The sampling/averaging has nothing to do with the time $t$. Also, $K$ does not change over time $t$. Equivalently, $g$ is independent of $t$.)
Our dynamical equation for a function $f$ is:
$$ \frac{\partial f_t}{\partial t}=\Pi(f_t) $$
Apparently, assuming $f_0$ as the initial condition, the solution is
$$ f_t = \exp(t\Pi) f_0$$
but I am not sure how to prove this. The solution for a matrix differential equation has the same form and I can prove it for that, but I am not sure how to do it for this functional case.
To provide the context, this is from the neural tangent kernel paper by Jacot et al.
