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].

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