Pizzadililehz

2022-03-27

How do I get an estimate for this nonlocal ODE?

Consider the following nonlocal ODE on $[1,\mathrm{\infty})$:

$r}^{2}f{}^{\u2033}\left(r\right)+2r{f}^{\prime}\left(r\right)-l(l+1)f\left(r\right)=-\frac{({f}^{\prime}\left(1\right)+f\left(1\right))}{{r}^{2}$

$f\left(1\right)=\alpha$

$\underset{r\to \mathrm{\infty}}{lim}f\left(r\right)=0$

where l is a positive integer and $\alpha$ is a real number.

Define the following norm $\Vert \xb7\Vert $

$\Vert f{\Vert}^{2}:={\int}_{1}^{\infty}{r}^{2}f\text{'}(r{)}^{2}dr+l(l+1\left){\int}_{1}^{\infty}f\right(r{)}^{2}dr$

I want to prove the estimate:

$\Vert f\Vert \le C\sqrt{l(l+1)}\left|\alpha \right|$

for some constant C independent of $\alpha$, l and f. But I am stuck.

Here is what I tried. Multiply both sides by f and integrate by parts to get:

$$\begin{array}{rl}\Vert f{\Vert}^{2}={\int}_{1}^{\mathrm{\infty}}{r}^{2}{f}^{2}+{\int}_{1}^{\mathrm{\infty}}l(l+1){f}^{2}& ={f}^{\prime}(1)f(1)+({f}^{\prime}(1)+f(1)){\int}_{1}^{\mathrm{\infty}}\frac{f}{{r}^{2}}\\ & \le {f}^{\prime}(1)\alpha +\frac{({f}^{\prime}(1)+\alpha )}{3}\sqrt{{\int}_{1}^{\mathrm{\infty}}{f}^{2}}\\ & \le {f}^{\prime}(1)\alpha +\frac{({f}^{\prime}(1)+\alpha )}{3}\Vert f\Vert \end{array}$$

where I used Cauchy-Schwartz in the before last line. I am not sure how to continue and how to get rid of the f'(1) term.

Any help is appreciated.

Jonathon Hanson

Beginner2022-03-28Added 9 answers

Step 1

Here is how we deal with the f'(1) term:

Step 2

Since ${\int}_{1}^{\mathrm{\infty}}\frac{1}{{r}^{4}}dr<1$, we can just solve for f'(1) and use Cauchy Schwartz to estimate it by |f(1)| and ∥f∥. Then I can easily get the estimate I wanted.

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