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Let $f$ be a continuous function in $L^2([1,\infty)$ satisfying $\sup_{r\geq 1} r|f(r)| <\infty$. Let $\ell$ be a positive integer and $a$ be a real number.

Let $u(r)$ be a function on $[1,\infty)$ solving the following BVP: $$r^2u''(r)+2ru'(r) - \ell(\ell+1) u(r) = f(r), \quad r\in [1,\infty)$$ $$u(1)= a, \quad \lim_{r\to \infty} u(r) = 0$$

We should be able to estimate the solution $u$ by $f$ and $a$. I want to understand precisely how $\ell$ will appear in this estimate. (In a certain context, $\ell$ corresponds to an angular derivative of some function in $\mathbb{R}^3$)

I am speculating the following: there exists a constant $C$ independent of $\ell$, $a$, $u$ and $f$ satisfying $$\ell \sup_{r\geq 1}r|u(r)| + \sup_{r\geq 1}r^2|u'(r)| \leq C(\ell^{-1}\sup_{r\geq 1} r|f(r)| + \ell^{-1}\lVert f\rVert_{L^2} + \ell|a|)$$

Can anyone verify if this is true? How would I prove this estimate (or the correct one in case the above is incorrect)?

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  • $\begingroup$ This boundary value problem for the Euler equation can be solved explicitly, why do you need an estimate? What is the range of your $\ell$? Is it real, positive? $\endgroup$
    – Alexandre Eremenko
    Commented Apr 27 at 12:53
  • $\begingroup$ $\ell$ is a positive integer. I want this estimate because it will eventually imply regularity results on solutions to the Poisson equation in $\mathbb{R}^3\setminus B_1$ equipped with an asymptotically flat metric $g$. In particular, if the forcing term has k angular derivatives in $L^2$ and $C^0$, then the solution and its derivative have $k+2$ and $k+1$ angular derivatives in $L^2$ and $C^0$. I haven't found such regularity results (where only angular derivatives are addressed) in the literature. (note the solution will have only 2 radial derivatives, but possibly much more angular ones. $\endgroup$
    – Laithy
    Commented Apr 27 at 20:11

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Noticing the factorization $r^{-\ell} \frac{d}{dr} r^{2(\ell+1)} \frac{d}{dr} r^{-\ell} u(r) = f(r)$ (as a trick to avoid recalling the precise variation of constants formula), an integral representation of the solution with the desired asymptotic behavior is $$\begin{aligned} u(r) &= \frac{A}{r^{\ell+1}} - r^\ell \int_r^\infty ds \, s^{-2(\ell+1)} \int_1^s dt \, t^\ell f(t) \\ &= \frac{A}{r^{\ell+1}} - \frac{r^\ell}{2\ell+1} \int_r^\infty ds\, s^{-\ell-1} f(s) - \frac{r^{-\ell-1}}{2\ell+1} \int_1^r ds\, s^\ell f(s) \\ r|u(r)| &\le \frac{|A|}{r^{\ell}} + \left( r^{\ell+1} \int_r^\infty ds\, s^{-\ell-2} + r^{-\ell} \int_1^r ds\, s^{\ell-1} \right) \frac{\sup_{s\ge 1} s|f(s)|}{2\ell+1} \\ &\le \frac{|A|}{r^{\ell}} + \frac{\sup_{s\ge 1} s|f(s)|}{\ell(\ell+1)} \end{aligned}$$ The constant $A$ is fixed by the condition $u(1)=a$, $$\begin{aligned} u(1) &= A - \frac{1}{2\ell+1} \int_1^\infty ds\, s^{-\ell-1} f(s) = a \\ A &= a + \frac{1}{2\ell+1} \langle s^{-\ell-1}, f(s) \rangle_{L^2} \\ |A| &\le a + \frac{\|s^{-\ell-1}\|_{L^2} \|f\|_{L^2}}{2\ell+1} = a + \frac{\|f\|_{L^2}}{(2\ell+1)^{3/2}} \end{aligned}$$

Completing the bound for $r^2|u'(r)|$ can be done analogously and checking the original guessed estimate is left as an exercise.

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  • $\begingroup$ Thank you! It's interesting that both the $C^0$ and $L^2$ norm of the forcing term appear. $\endgroup$
    – Laithy
    Commented Apr 27 at 20:16
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    $\begingroup$ @Laithy It is a matter of preference. In retrospect, I switched between usual Cauchy-Schwarz $L_2$-$L_2$ estimates and $L_1$-$L_\infty$ estimates somewhat randomly. But I think you can choose either one or the other, when the choice comes up. Just be careful to keep track of the $r$-dependent finite integration domain in $\int_1^r ds\, s^\ell f(s)$. In the naive $r\to\infty$ limit, this integral diverges. Also of course, as a running assumption, the value of $\ell$ has to be sufficiently large (at least in real part) for all the steps to work. $\endgroup$
    – Igor Khavkine
    Commented Apr 27 at 22:10
  • $\begingroup$ Yes, I understand. I believe what you did above implies the estimate in the original post for any $\ell$. The hard part is to show that $C$ can be chosen independently of $\ell$ for large $\ell$. Thank you for helping me (you have also helped me several times in the past :) ). $\endgroup$
    – Laithy
    Commented Apr 27 at 22:36
  • $\begingroup$ Are there more robust techniques to prove this kind of estimate? I posted a similar question here with a similar ODE, but couldn't show it the same way. $\endgroup$
    – Laithy
    Commented May 2 at 2:42

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