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(!)



Convergence in distribution and ODE
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](http://www3.nd.edu/~lnicolae/cobord-degen.pdf) 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 


$$y\_\infty(x)=\begin{cases}
c, & x<0 \\
ec, & x>0.
\end{cases} $$


