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I am considering the following ODE \begin{equation} \begin{split} &\frac{d^2}{dy^2}u + \frac{\alpha}{(1+y^2)^{\frac{r}{2}}}u = \delta(y)\\ &\lim_{|y|\to \infty}u(y) = 0. \end{split} \end{equation} Here $\alpha$ is a constant and $r > 0$ and $\delta(y)$ is the Dirac delta function. Is it possible to solve this ODE? I expect the solution to be singular in $\alpha$ as $\alpha$ approaches zero. If we can't get the explicit expression of the solution, can we analyze its singularity in $\alpha$ asymptotically as $\alpha \to 0$? For example, when $r = 0$, the solution behaves like $\frac{e^{-|y|\sqrt{-\alpha}}}{\sqrt{-\alpha}}$ which is singular in $\alpha$.

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  • $\begingroup$ What is $\delta$? $\endgroup$ Jul 9, 2021 at 12:43
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    $\begingroup$ Dirac delta function? $\endgroup$ Jul 9, 2021 at 13:14
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    $\begingroup$ Yes, $\delta$ is the Dirac delta function. $\endgroup$
    – Jacob Lu
    Jul 9, 2021 at 14:22
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    $\begingroup$ (i) Does $\alpha$ have a definite sign? (ii) Why do you call $u$ the fundamental solution and why is it of interest? -- The ODE is not with constant coefficients, and so, the general solution of the ODE is probably not the convolution of the right-hand side with your $u$. $\endgroup$ Jul 9, 2021 at 17:03
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    $\begingroup$ "For example, when $r = 0$, the solution behaves like $\frac{e^{-|y|\sqrt{-\alpha}}}{\sqrt{-\alpha}}$ which is singular in $\alpha$." -- (i) What do you mean by the solution? (The general solution depends on two constants.) (ii) How do you get such a statement? If $\alpha=0$, then a solution is given by $u(y)=\max(0,y)$. How do you get the singularity at all? $\endgroup$ Jul 12, 2021 at 0:32

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This is essentially an elliptic second-order ODE in $(0,\infty)$: $$ (1 + y^2)^{r/2} u''(y) = (-\alpha) u(y) , $$ with boundary conditions $u'(0) = -1$ and $u(\infty) = 0$. This kind of problems are well-studied, the corresponding theory is known as Krein's spectral theory of strings.

The value $h(\alpha)$ of $u$ at $0$ for the given parameter $\alpha$ is the corresponding spectral function, and $h(-\alpha)$ is a Stieltjes function of $\alpha$ (other names: Nevanlinna–Pick function, Herglotz function). The relation between the coefficient (here: $(1+y^2)^{-r/2}$, roughly known as the string) and the spectral function $h$ is not explicit, but some comparison results are known.

I do not have time now to search for the relevant results, but I would start by looking at the following paper:

  • S. Kotani and S. Watanabe, Krein's spectral theory of strings and generalized diffusion processes, in: Functional Analysis in Markov processes (Katata/Kyoto, 1981), Lecture Notes in Math. 923, Springer, Berlin, 1982, 235–259, DOI:10.1007/BFb0093046.

Another excellent reference with a chapter on Krein's theory is

  • R. Schilling, R. Song, Z. Vondraček, Bernstein Functions: Theory and Applications. De Gruyter, Studies in Math. 37, Berlin, 2012, DOI:10.1515/9783110269338.

Edit: some additional details.

Strictly speaking, our case corresponds to the string $m(dy) = (1 + y^2)^{-r/2} dy$, which is commonly identified with its distribution function $$m(y) = m([0, y)) = \int_0^y (1 + s^2)^{-r/2} ds.$$

The spectral function can be defined in a number of equivalent ways: one of them inolves the integral of $(\phi_N)^{-2}$, where $\phi_N$ is the solution with "Neumann" initial condition $\phi_N(0) = 1$, $\phi_N'(0) = 0$. The one I am most familiar with defines it to be the reciprocal of $-\phi'(0)$, where $\phi$ is the solution with boundary conditions $\phi(0) = 1$ and $\phi(\infty) = 0$. This is clearly equivalent to the definition as $-u'(0)$ with $u$ as in the statement of the problem.

I failed once to find a simple reference for the above calculation in analytical terms, so together with Jacek Mucha we included a very brief discussion in the appendix to our paper (see Section A.3 therein):

  • J. Mucha, M. Kwaśnicki, Extension technique for complete Bernstein functions of the Laplace operator, J. Evol. Equ. 18(3) (2018): 1341–1379 DOI:10.1007/s00028-018-0444-4.
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  • $\begingroup$ Thanks a lot for your answer! It's very helpful. I read the first reference you mentioned and got two questions. It seems for the ODE in my problem, the string should be defined as $\int_0^y\frac{1}{(1+z^2)^{r/2}}dz$? I found the Krein spectral function is defined as an integral in that reference. Why is the value $u(0)$ called the spectral function in our situation? $\endgroup$
    – Jacob Lu
    Jul 13, 2021 at 15:50
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    $\begingroup$ I added some details that hopefully address your questions. $\endgroup$ Jul 13, 2021 at 17:50
  • $\begingroup$ Thanks! That answers my questions. Do you think it is possible to get the asymptotic behaviour of $u(y, \alpha)$ as $|y| \to \infty$? I guess the solution would behave like $h(\alpha)e^{-\frac{1}{h(\alpha)}|y|}$ if we pretend the behaviour is similar to the case $r = 0$. $\endgroup$
    – Jacob Lu
    Jul 14, 2021 at 0:52
  • $\begingroup$ When $y$ is large, the behaviour of $u(\alpha, y)$ should be quite different! I expect something like $c_1(\alpha) \exp(-c_2(\alpha) |y|^{1-r/2})$ — at least this is what one gets for the explicitly solvable coefficient $(1+|y|)^{-r}$ instead of $(1+y^2)^{-r/2}$. (I may have made a mistake, though.) $\endgroup$ Jul 14, 2021 at 8:51
  • $\begingroup$ This is interesting! The index $1-r/2$ in the exponential looks very reasonable. I guess the two coefficients $(1+y^2)^{-r/2}$ and $(1+|y|)^{-r}$ would give solutions with similar asymptotic behavior. But I don't know how to get the explicit solution for the $(1+|y|)^{-r}$ case. I also failed to check the function you give is a solution. Do you know if there is any reference on this case? Thanks! $\endgroup$
    – Jacob Lu
    Jul 14, 2021 at 12:21

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