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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.

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  • $\begingroup$ I added the PDE tag since differential equations in one variable with boundary or growth conditions (rather than initial conditions only) behave, in some respects, much more PDE-like than ODE-like. $\endgroup$ Dec 13, 2021 at 8:39
  • $\begingroup$ Thank you Jochen. $\endgroup$
    – Laithy
    Dec 13, 2021 at 14:04

1 Answer 1

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You have a linear ODE with explicit coefficients: the solutions can actually be written down explicitly via variation of constants. 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 $5/2$.

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  • $\begingroup$ The ODE that I am actually working with is more complicated (though still linear with explicit coefficients), and it will be very difficult to apply this approach. Is there a way to prove this $O(\ell)$ result without using the simplicity of the ODE? $\endgroup$
    – Laithy
    Dec 13, 2021 at 19:23
  • $\begingroup$ (a) I don't like to deal with moving goal posts. (b) It is easy enough to write down a homogeneous ODE of the form $p(r) f''(r) + 2 q(r) f'(r) + h(\ell,r) f(r) = 0$, depending on the parameter $\ell$, whose solutions look like $\frac{1}{(r+1)(1 + r^2)^{\ell/2}}$. We can even guarantee that $p(r) \sim r^2$ and $q(r) \sim r$ as $r\nearrow \infty$, and that the leading order term of $h$ behaves like $-\ell(\ell+1)$. So something qualitatively similar to what you have. If you try to solve the equation $$p(r) f'' + 2q(r) f' + h(\ell,r) f = (f(0) + f'(0)) g(r) $$ ... $\endgroup$ Dec 13, 2021 at 19:59
  • $\begingroup$ on the interval $[0,\infty)$ with boundary conditions at $0$, and require $g(r) \sim 1/ r^2$ as $r\nearrow \infty$, you will find that the Dirichlet-Newmann map will have norm independent of $\ell$ (in fact, equal 1). So there is a possibility that the norm you are looking for can be sensitive to the precise terms that show up. $\endgroup$ Dec 13, 2021 at 20:04
  • $\begingroup$ The goal hasn't changed. I was asking for a different approach for the $\textbf{same}$ thing. Anyhow, thank you for your answer. $\endgroup$
    – Laithy
    Dec 13, 2021 at 20:17

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