Assuming we have an ODE $y'_n(x) = f_n(x) y_n(x)$ with $f_n$ be Gauß-densities with mean value 0 and variance $\frac{1}{n}$, then we have that they converge in distribution to a delta peak $δ(x)$. Now, assume that we solved the ODE $y′_n(x)=f_n(x)y_n(x)$ for every n with $y(−∞)=y_0$ specified. In what sense does the solution yn(x) convergence to y(x), where y(x) solves $y′(x)=δ(x)y(x)$?
I want to understand this convergence without solving the ODE $y′_n(x)=f_n(x)y_n(x)$ explicitely(!)
You have
$$ d\log y_n= f_n dx $$
which tells you that the measures $d\log y_n$ converge weakly to $\delta$. For more precise statements I refer you to Section 1 of this paper where I and my collaborator study what happens to operators such as $\frac{d}{dx}-f_n(x)$ as $n\to\infty$. There we are mostly interested on $x$ restricted to a compact interval and $y_n$ constrained to satisfy some boundary conditions. The upshot is that the operators converge in the sense that their graphs have a limit as $n\to\infty$ and in many case we can give a precise description of this limit.
Let me give you a taste. From the fact that $d\log y_n\to \delta$ we deduce that for any $r>0$ we have
$$\log y_n(r)-\log y_n(-r)=\int_{[-r,r]} \delta (x) dx=1. $$
This implies that
$$ y_n(r)/y_n(-r)\to e. $$
Suppose that $y_n(-r)\to c$ Using the results in the above paper you can conclude that $y_n$ converges in $L^2(-r,r)$ to
$$y_\infty(x)=\begin{cases} c, & x<0 \\ ec, & x>0. \end{cases} $$
