# Ngô Quốc Anh

## September 14, 2010

### Asympotic behavior of integrals, 2

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 15:07

We now prove the following result

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

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

Proof.

Let

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

It is easy to see that

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

We claim that

$\displaystyle\frac{{w(x)}}{{\log |x|}} \to \overline f : = \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {f(y)dy} ,\quad |x| \to \infty$.

To see this, we need only to verify that

$\displaystyle I = \int_{{\mathbb{R}^2}} {\frac{{\log |x - y| - \log (|y| + 1) - \log |x|}}{{\log |x|}}f(y)dy} \to 0$

as $|x| \to \infty$.

Decompose $\mathbb R^2$ into three domains

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

and write $I=I_1+I_2+I_3$ where $I_i$ are integrals over these domains, respectively. We consider $|x| \geqslant 2$, then we estimate as follows.

• For the case of $I_1$, note that

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

which implies

$\displaystyle\begin{gathered} |{I_1}| = \left| {\int_{|x - y| \leqslant 1} {\frac{{\log |x - y| - \log (|y| + 1) - \log |x|}}{{\log |x|}}f(y)dy} } \right| \hfill \\ \qquad\leqslant \int_{|x - y| \leqslant 1} {\left| {\frac{{\log |x - y|}}{{\log |x|}}} \right||f(y)|dy} + \int_{|x - y| \leqslant 1} {\left| {\frac{{\log (|y| + 1) + \log |x|}}{{\log |x|}}} \right||f(y)|dy} \hfill \\ \qquad\leqslant \frac{M}{{\log |x|}}\int_{|x - y| \leqslant 1} {\left| {\log |x - y|} \right|dy} + 3\int_{|x - y| \leqslant 1} {|f(y)|dy} \to 0 \hfill \\ \end{gathered}$

as $|x| \to \infty$.

• For the case of $I_2$, we fix $R$ and observe that

$\displaystyle |x - y| \leqslant |x| + |y| \leqslant |x| + R$

which implies

$\displaystyle\begin{gathered} \left| {\frac{{\log |x - y| - \log (|y| + 1) - \log |x|}}{{\log |x|}}} \right| \leqslant \frac{{\left| {\log \frac{{|x - y|}}{{|x|}}} \right| + \log (R + 1)}}{{\log |x|}} \hfill \\ \qquad\leqslant \frac{{\left| {\log \frac{{|x| + R}}{{|x|}}} \right| + \log (R + 1)}}{{\log |x|}} \to 0 \hfill \\ \end{gathered}$

as $|x| \to \infty$. Therefore

$\displaystyle |{I_2}| \leqslant \left| {\frac{{\log |x - y| - \log (|y| + 1) - \log |x|}}{{\log |x|}}} \right|\int_{{D_2}} {|f(y)|dy} \to 0$

as $|x| \to \infty$.

• For the case of $I_3$, we firstly fix $x$ and consider $R$ sufficiently large such that

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

This is possible since

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

and

$\displaystyle |y| + 1 \leqslant |x - y| + |x| + 1 \leqslant |x - y|\left( {|x| + 1} \right)$

provided

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

since

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

Thus

$\displaystyle\begin{gathered} |{I_3}| \leqslant \left| {\frac{{\log |x - y| - \log (|y| + 1) - \log |x|}}{{\log |x|}}} \right|\int_{{D_3}} {|f(y)|dy} \hfill \\ \qquad\leqslant \frac{{\left| {\log \frac{{|x - y|}}{{|y| + 1}}} \right| + \log |x|}}{{\log |x|}}\int_{{D_3}} {|f(y)|dy} \hfill \\ \qquad\leqslant \frac{{\left| {\log (|x| + 1)} \right| + \log |x|}}{{\log |x|}}\int_{{D_3}} {|f(y)|dy} . \hfill \\ \end{gathered}$

We now let $R \to \infty$ and then $|x| \to \infty$ to get the desired result.

From above, we deduce that

$\displaystyle |I| \leqslant |{I_1}| + |{I_2}| + |{I_3}| \to 0$

as $|x| \to \infty$.

Now by using the potential analysis together with asymtotic behavior of $u$ we deduce that

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

which completes the proof.

For other estimates, we refer the reader to a paper due to Wang and Zhu published in Duke Math. J. (2000) [here].