Define the second order linear differential operator associated with $X$ (Here $X$ is the unique strong solution to appropriate Ito SDE) by $$A = \frac{1}{2} \sigma^2(x) \frac{d^2}{dx^2} + \mu(x) \frac{d}{dx}. $$ 


For sure some conditions on $\mu$ and $\sigma$ are needed. However I’m unsure what conditions are needed but at least basic regularity conditions are not too restrictive like continuity or Lipschitz conditions.


Are there results known how the solutions to the differential equation
$$
Af = rf \Leftrightarrow (A-r)f=0
$$
behave when $r \to \infty$? I know that all the solutions to the equation are linear combination of $\varphi\_r(x)$ and $\psi\_r(x)$, where both are positive and $\varphi\_r(x)$ is decreasing (in $x$) and $\psi\_r(x)$ is increasing (in $x$).


I'm more interested in the behaviour of the decreasing solution $\varphi\_r(x)$ than the increasing in the limit $r \to \infty$ but I think if something can be said it easily follows to both of them. My quess is that some kind of Feynman-Kac type of formula might be useful?



Behaviour of solutions to $(A-r)f=0$ in the limit $r \to \infty$
First of all, by a suitable change of variables, we can rewrite the equation as $$f''(y) = r a(y) f(y) ,$$
for an appropriate coefficient $a(y)$. Indeed, if $y(x)$ is an increasing solution of $A y(x) = 0$, then
$$ \begin{aligned} A f(y(x)) & = \tfrac{1}{2} \sigma^2 (f''(y) (y')^2 + f'(y) y'') + \mu f'(y) y' \\ & = \tfrac{1}{2} \sigma^2(x) (y'(x))^2 f''(y(x)) , \end{aligned}$$
and we can set $a(y) = (\tfrac{1}{2} \sigma^2(y^{-1}(x)) (y')^2(y^{-1}(x)))^{-1}$.


The above equation has been considered for quite general (and highly irregular) coefficients $a(y)$. Let me refer to [[Kotani–Watanabe]](https://link.springer.com/content/pdf/10.1007/BFb0093046.pdf) again.




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Now suppose that $f''(y) = r a(y) f(y)$ and $f$ is decreasing. Consider $y\_1 < y\_2$. By Taylor's formula,
$$ f(y\_1) - f(y\_2) = (y\_1 - y\_2) f'(y\_2) + \int\_{y\_1}^{y\_2} (y - y\_1) f''(y) dy . $$
Since $(y\_1 - y\_2) f'(y\_2) \ge 0$ and $f''(y) = r a(y) f(y)$, we have
$$ f(y\_1) - f(y\_2) \ge r \int\_{y\_1}^{y\_2} (y - y\_1) a(y) f(y) dy . $$
Furthermore, $f(y) \ge f(y\_2)$, and so
$$ f(y\_1) - f(y\_2) \ge r f(y\_2) \int\_{y\_1}^{y\_2} a(y) dy . $$
Equivalently,
$$ \frac{f(y\_1)}{f(y\_2)} \ge 1 + r \int\_{y\_1}^{y\_2} a(y) dy . $$
This clearly goes to infinity as $r \to \infty$, provided that $a(y)$ is not constant zero.


