Consider the following nonlocal ODEs on $[1,\infty)$.


#1)
$$\begin{align}
r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\
f(1) &= \alpha \\
\lim\_{r\to \infty} f(r) &= 0
\end{align}$$
#2
$$\begin{align}
r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\
f'(1) &= \beta \\
\lim\_{r\to \infty} f(r) &= 0
\end{align}$$


where $l$ is a positive integer, $\alpha, \beta \in \mathbb{R}$.


Both ODEs are uniquely solvable, and so we can define the Dirichlet to Neumann operator $T: \alpha \mapsto \beta$. I am trying to prove some properties of $T$.


Define the following norm $\lVert \cdot \lVert$:
$$\lVert f \lVert^2 := \int\_1^{\infty} r^2 f'(r)^2 dr + l(l+1) \int\_1^{\infty} f(r)^2 dr$$


It can be shown that $\lVert f \lVert \leq C |f'(1)|$ for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. It then follows that $|f(1)| \leq |\int\_1^{\infty} f'(r) dr| \leq C' \sqrt{\int\_1^{\infty}r^2 f'^2} \leq C' \lVert f \lVert \leq C'C|f'(1)|$
And so we have for any $f$ solving the above ode,
$$|f(1)| \leq \bar C |f'(1)|$$
for some $\bar C$ that is independent of $f$ and $l$.


However, the other way around is not true. In fact, it holds that
$$|f'(1)| \leq C \sqrt{l(l+1)}|f(1)|$$
for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. I don't know how to prove this inequality. A weaker inequality is the following,
$$\lVert f \lVert \leq C \sqrt{l(l+1)} |f(1)|$$
which I also was not able to prove.


Any help is appreciated.



Dirichlet to Neumann operator for a nonlocal ODE
You have a linear ODE with explicit coefficients: the solutions can actually be written down explicitly via [variation of constants](https://en.wikipedia.org/wiki/Variation_of_parameters). To summarize the result1 when $\ell > 1$, let $g\_{a,b}(r)$ be given by


$$ g\_{a,b}(r) = \frac{a}{r^{\ell+1}} + \frac{b}{r^2} $$


you find that


$$ r^2 g\_{a,b}'' + 2r g\_{a,b}' - \ell(\ell+1) g\_{a,b} = -\frac{(\ell+2)(\ell-1) b}{r^2} $$


As


$$ g\_{a,b}(1) = a + b , \qquad g'\_{a,b}(1) = -(\ell+1) a - 2b $$


for $g\_{a,b}$ to solve the equation you indicated requires $(\ell+2)\ell-1)b = g\_{a,b}(1) + g'\_{a,b}(1)$. So we need (when $\ell > 1$)


$$ - \ell a - b = (\ell+2)(\ell - 1)b \implies a = \frac{1 - \ell - \ell^2}{\ell} b $$


which yields


$$ \alpha = \frac{1-\ell^2}{\ell} b, \qquad \beta = - \frac{b}{\ell}(1 + 2\ell - 2\ell^2 - \ell^3) $$


so the Dirichlet to Neumann map has norm exactly


$$ \frac{\ell^3 - 2\ell^2 + 2\ell + 1}{\ell^2 - 1} = O(\ell) = O(\sqrt{\ell(\ell+1)})$$




---


1 The cases where $\ell = 0$ and $1$ requires more care.


When $\ell = 0$, the equation you wrote is only solvable when $b = 0$, and the Dirichlet-to-Neumann map has norm 1.


When $\ell = 1$, variation of constants gives
$$ g\_{a,b} = \frac{1}{r^2}(a + b\ln(r)) $$
for which
$$ r^2 g\_{a,b}'' + 2r g'\_{a,b} - 2 g\_{a,b} = -\frac{3b}{r^2} $$
which then requires $a = -2 b$, and hence $\alpha = -2b$ and $\beta = 5b$, and the D2N map has norm $5/2$.


