# Ngô Quốc Anh

## September 18, 2010

### Asympotic behavior of integrals, 3

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 11:56

In the previous entry, we showed the following

Theorem. Let $u$ and $f$ be two smooth functions on $\mathbb R^2$ satisfying

$\Delta u(x)=f(x), \quad x \in \mathbb R^2$.

Suppose that $f$ is bounded and also $f \in L^1(\mathbb R^2)$ and

$|u(x)| \leqslant o(|x|), \quad |x| \to \infty$.

Then

$\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \frac{{u(x)}}{{\log |x|}} = \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {f(y)dy}$.

As suggested in an earlier entry, in this topic, we show that the following limit

$\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \left[ {u(x) - \alpha \log |x|} \right]$

exists where

$\displaystyle\alpha = \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {f(y)dy}$

for some good $f$.

It is worth recalling that if we denote by $w$ the following quantity

$\displaystyle w(x) = \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\left[ {\log |x - y| - \log |y|} \right]f(y)dy}$

then

$u-w={\rm const.}$.

This is about to say that it suffice to prove

$\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \left[ {w(x) - \alpha \log |x|} \right]$

exists. Amazingly, the following

$\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \left[ {w(x) - \alpha \log |x|} \right] = -\frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\log |y|f(y)dy}$

holds.

Theorem. If $f$ satisfies $f=o(\frac{1}{|x|^3})$ as $|x| \to \infty$ then

$\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \left[ {u(x) - \alpha \log |x|} \right]$

exists and is finite.

Proof.

It suffices to show that

$\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \left[ {w(x) - \alpha \log |x|} \right]$

exists and is finite. Observe that

$\displaystyle\begin{gathered} w(x) - \alpha \log |x| + \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\log |y|f(y)dy} \hfill \\ \qquad= \frac{1}{{2\pi }}\left[ {\int_{{\mathbb{R}^2}} {\left[ {\log |x - y| - \log |y|} \right]f(y)dy} - \int_{{\mathbb{R}^2}} {\log |x|f(y)dy} + \int_{{\mathbb{R}^2}} {\log |y|f(y)dy} } \right] \hfill \\ \qquad= \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\left[ {\log |x - y| - \log |x|} \right]f(y)dy} \hfill \\ \qquad= \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\left[ {\log \frac{{|x - y|}}{{|x|}}} \right]f(y)dy}. \hfill \\ \end{gathered}$

We show that

$\displaystyle\frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\left[ {\log \frac{{|x - y|}}{{|x|}}} \right]f(y)dy} \to 0$

as $|x| \to \infty$.

For each $R$ we consider

$\displaystyle\begin{gathered} {D_1} = \left\{ {y:|y| \leqslant R} \right\}, \hfill \\ {D_2} = \left\{ {y:|x - y| \leqslant 1,|y| > R} \right\}, \hfill \\ {D_3} = \left\{ {y:|x - y| > 1,y > R} \right\}. \hfill \\ \end{gathered}$

Then we decompose $I$ into $I_1+I_2+I_3$ where $I_i$ are integrals on the region $D_i$, respectively.

• The case of $I_1$.

We start with $|x|>R$, then

$\displaystyle\frac{{|x| - R}}{{|x| + 1}} \leqslant \frac{{|x| - |y|}}{{|x|}} \leqslant \frac{{|x - y|}}{{|x|}} \leqslant \frac{{|x| + |y|}}{{|x|}} \leqslant \frac{{|x|}}{{|x| - |y|}} \leqslant \frac{{|x| + 1}}{{|x| - R}}$

which implies

$\displaystyle\left| {\log \frac{{|x - y|}}{{|x|}}} \right| \leqslant \log \frac{{|x| + 1}}{{|x| - R}}$.

Therefore

$\displaystyle |{I_1}| \leqslant \frac{1}{{2\pi }}\log \frac{{|x| + 1}}{{|x| - R}}\int_{D_1} {|f(y)|dy} \to 0$

as $|x| \to \infty$.

• The case of $I_2$. For large $R$, we observe that

$\displaystyle\log \frac{{|x - y|}}{{|x|}} < 0$

then

$\displaystyle\begin{gathered} |{I_2}| \leqslant \frac{1}{{2\pi }}\int_{{D_2}} {\log \frac{{|x|}}{{|x - y|}}|f(y)|dy} \hfill \\ \quad\leqslant \frac{1}{{2\pi }}\int_{{D_2}} {\left[ {\log \frac{{|x|}}{{|x - y|}} + \log \frac{1}{{|x|}} + \log |x|} \right]|f(y)|dy} \hfill \\ \quad= \frac{1}{{2\pi }}\int_{{D_2}} {\log \frac{1}{{|x - y|}}|f(y)|dy} + \frac{1}{{2\pi }}\int_{{D_2}} {\log |x||f(y)|dy} \hfill \\ \quad= {I_{21}} + {I_{22}}. \hfill \\ \end{gathered}$

We can further estimate. We get

$\displaystyle\begin{gathered} {I_{21}} = \frac{1}{{2\pi }}\int_{{D_2}} {\frac{1}{{|x - y{|^2}}}\log \frac{1}{{|x - y|}}|x - y{|^2}|f(y)|dy} \hfill \\ \qquad\leqslant \frac{C}{{2\pi }}\int_{{D_2}} {\frac{1}{{|x - y{|^2}}}\log \frac{1}{{|x - y|}}dy} \hfill \\ \end{gathered}$

and

$\displaystyle {I_{22}} \leqslant \frac{1}{{2\pi }}\int_{{D_2}} {\log (|y| + 1)|f(y)|dy} \to 0$

as $R \to \infty$. Therefore, $I_2 \to 0$ as $|x| \to \infty$.

• The case of $I_3$. In this situation, we first fix $x$. Then

$\displaystyle\frac{1}{{|y| + 2}} \leqslant \frac{{|x - y|}}{{|x|}} \leqslant \frac{{|x| + |y|}}{{|x|}} \leqslant |y| + 2$.

Thus

$\displaystyle |{I_3}| \leqslant \frac{1}{{2\pi }}\int_{{D_3}} {\log (|y| + 2)|f(y)|dy} \to 0$

as $R \to \infty$. Then we take $|x| \to \infty$.

We are going to prove another result

$\displaystyle\left| {w(x) - \alpha \log |x| + \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {\log |y|f(y)dy} } \right| \leqslant \frac{{C\log |x|}}{{|x|}},\quad \forall |x| \geqslant 1$

for some positive constant $C$.