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I'm trying to solve an optimal stopping problem which led me to an obstacle problem involving the following family of ODE's

$$(x^2+d)y'(x)-2xy(x) = 1.$$

For simplicity I first considered the case $d = 0$ before moving to the much more relevant case $d > 0$. This revealed a somewhat unexpected phenomenon to me: The solutions are of the form $C(x^2+d) + f(x)$, where $C$ is some constant. For $d = 0$ we have $f(x) = -\frac{1}{3x}$ which is negative for $x > 0$. On the other hand, for $d > 0$ we have $$f(x) = \frac{x}{2 d}+ \frac{(x^2+d)\tan^{-1}\left(\frac x {\sqrt d}\right)}{2 d^{3/2}},$$ which is positive and $f(x)$ diverges as $d\to 0$; instead of converging to, say $x\mapsto -\frac 1 {3x}$.

Is there some theoretic result that shows that this must be the case? If not is there some result that would show convergence, but we can verify that its assumptions are violated?

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    $\begingroup$ There is no reason why your particular choice of $f_d$ should converge towards $f_0$, you must take into account the coefficient $C_d$ to built a converging family of solutions starting from a given (fixed) initial value. $\endgroup$
    – Loïc Teyssier
    Commented Nov 16, 2020 at 11:07
  • $\begingroup$ Also as $d\to 0$ you have a merging of two regular (=simple) singular points into an irregular (=multiple) singular point (in the complex line). This kind of bifurcation can be wild sometimes. Try looking into the complex domain to have convergence. $\endgroup$
    – Loïc Teyssier
    Commented Nov 16, 2020 at 11:12
  • $\begingroup$ @LoïcTeyssier Yeah, I know I ignored the coefficient $C_d$ - mostly because in the original problem I was actually looking for $\int y$ and then there was an argument for $d = 0$ that let you conclude that $C_0 = 0$. I wanted to extend this somehow to the $d > 0$ case (which doesn't seem possible anymore), so I thought it could be a little bit sloppy in my explanation without serious harm. $\endgroup$
    – Stefan Perko
    Commented Nov 16, 2020 at 11:43
  • $\begingroup$ @LoïcTeyssier Do you have any readings you can point me to? I'm not familiar with singular points or the passage to the complex domain in this context (I haven't studied ODE's all that much). $\endgroup$
    – Stefan Perko
    Commented Nov 16, 2020 at 11:44
  • $\begingroup$ @LoïcTeyssier I'm still very interested in your answer - if you like to write one, no pressure. $\endgroup$
    – Stefan Perko
    Commented Nov 19, 2020 at 8:28

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Using the expansion for $\arctan(t)$ as $t\to +\infty$ in the form $$\arctan(t) = \frac{\pi}{2}-\frac{1}{t}+\frac{1}{3t^3}+\cdots$$ we get, for fixed $x>0$ and positive $d\to 0$: $$ f_d(x) \sim \frac{\pi(x^2+d)}{4d^{3/2}} =: C_d(x^2+d) $$ and further (that's where the magic cancellation happens) $$g_d(x):=f_d(x)-C_d(x^2+d)\sim \frac{x}{2d}-\frac{(x^2+d)}{2d^{3/2}}\left( \frac{\sqrt{d}}{x}-\frac{d^{3/2}}{3x^3} \right) \\ = \frac{d^2-2dx^2}{6dx^3}\to -\frac{1}{3x}$$

There's nothing deep happening here, except the necessity to keep track of the constant $C$ as the parameter varies (and also of the intervals definition for maximal solutions, which hasn't been checked above).

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  • $\begingroup$ Thank you, I understand now. It teaches me not to handle the constant $C$ sloppily. I guess I just never expected this strange switch as $d\to 0$. $\endgroup$
    – Stefan Perko
    Commented Nov 28, 2020 at 9:45

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