1
$\begingroup$

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?

Improve this question
$\endgroup$
6
  • $\begingroup$ You should probably give some hypotheses about the behavior of $\sigma$ and $\mu$ as $x\to\infty$. $\endgroup$
    – Michael Renardy
    Commented Jul 10, 2018 at 14:18
  • $\begingroup$ @MichaelRenardy Well almost any basic regularity conditions can be assumed. Say continuity or Lipschitz conditions etc. I didnt state any because I really dont have idea what should be assumed to achieve some results. $\endgroup$
    – Harto Saarinen
    Commented Jul 10, 2018 at 15:20
  • $\begingroup$ What kind of description would you expect? That is, what is the interesting property of the solution in the simplest case, $A f = f''$, when we have $\varphi_r(x) = c_r e^{-\sqrt{r} \, x}$? The question seems to be somehow related to Krein's theory of second-order operators (Krein's strings). I do not really know nice references; you may want to take a look at an article by Kotani and Watanabe. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 10, 2018 at 22:29
  • $\begingroup$ @MateuszKwaśnicki Well at least in that case (and many other cases) it seems that the limit is $0$ for all $x$ as $r \to \infty$. In which cases does this happen? (In general almost all results that tell something how the solutions depend on $r$ are interesting.) $\endgroup$
    – Harto Saarinen
    Commented Jul 11, 2018 at 5:56
  • $\begingroup$ @HartoSaarinen: The limit depends on the choice of $c_r$; if $c_r = 1$, it it zero only for $x > 0$. The same will be true for any reasonable choice of $\sigma$ and $\mu$: $\varphi_r(y) / \varphi_r(x)$ will go to zero as $r \to \infty$ whenever $y > x$. If this is what you wanted, I can write the details as an answer. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 11, 2018 at 9:20

1 Answer 1

Reset to default
2
$\begingroup$

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] again.

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.

Improve this answer
$\endgroup$
7
  • $\begingroup$ How does it follow from this that $f(x) \to 0$ when $r \to \infty$ only if $x > 0$? I think when $x<0$ then $f(x) \to \infty$, but what about $x=0$? $\endgroup$
    – Harto Saarinen
    Commented Jul 11, 2018 at 11:13
  • $\begingroup$ Also is there a particular reason to specify that $y(x)$ is increasing? $\endgroup$
    – Harto Saarinen
    Commented Jul 11, 2018 at 13:39
  • $\begingroup$ @HartoSaarinen: As I understand, $\varphi_r(x)$ is only defined up to multiplication by a constant, so there is really no way to say what happens with $\varphi_r(x)$ as $x \to \infty$. However, the above argument shows that $\varphi_r(x_1) / \varphi_r(x_2) \to \infty$ as $r \to \infty$ whenever $x_1 < x_2$. If we set $\varphi_r(0) = 1$, then indeed $\varphi_r(x) \to 0$ for $x > 0$ and $\varphi_r(x) \to \infty$ if $x < 0$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 11, 2018 at 16:45
  • $\begingroup$ Regarding $y(x)$, of course you can take it decreasing, if you prefer. Avoid constants, however. (There are two linearly independent solutions of $A f = 0$: one constant, and another one strictly monotone). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 11, 2018 at 16:58
  • $\begingroup$ All the solutions to $(A-r)f=0$ are $f(x)=A_1 \varphi_r(x)+A_2 \psi_r(x)$, where $A_1$ and $A_2$ are constants. So the constant is not part of the $\varphi_r(x)$ really (I think I didnt define phi and psi properly, or is this completely off the point?) $\endgroup$
    – Harto Saarinen
    Commented Jul 11, 2018 at 17:19

You must log in to answer this question.