Ngô Quốc Anh

September 7, 2010

Asympotic behavior of integrals

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 10:52

Long time ago, we studied [here] the following fact

Suppose f \in L^1(\mathbb R^n) \cap L_{loc}^\infty (\mathbb R^n) with f \geq 0. Define

\displaystyle Sf\left( x \right) = \int_{\mathbb{R}^n } {\log \frac{{\left| y \right|}}{{\left| {x - y} \right|}}f\left( y \right)dy}.

Show that Sf(x) is finite for all x \in \mathbb R^n and Sf \in L_{loc}^1(\mathbb R^n).

In this entry, from now on we continue to prove several useful results appearing in PDE. We shall prove the following

Theorem. Assume u is a solution to

\displaystyle (-\Delta)^\frac{3}{2} u(x)=-2e^{3u(x)}, \quad x \in \mathbb R^3

with finite energy

\displaystyle \int_{{\mathbb{R}^3}} {{e^{3u(x)}}dx} < \infty.


\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \frac{{u(x)}}{{\log |x|}} = - \frac{1}{{{\pi ^2}}}\int_{{\mathbb{R}^3}} {{e^{3u(y)}}dy} .

The proof is elementary. Let us presume some knowledge of potential analysis. If we denote

\displaystyle v(x) = \frac{1}{{{\pi ^2}}}\int_{{\mathbb{R}^3}} {\log \frac{{|x - y|}}{{|y|}}{e^{3u(y)}}dy}

then v(x) satisfies

\displaystyle (-\Delta)^\frac{3}{2} v(x)=-2e^{3u(x)}, \quad x \in \mathbb R^3.

It then follows that

\displaystyle u(x)+v(x)={\rm const.}.

Thus, it suffices to prove

\displaystyle\mathop {\lim }\limits_{|x| \to \infty }  \frac{{v(x)}}{{\log |x|}} = \frac{1}{{{\pi ^2}}}\int_{{\mathbb{R}^3}}  {{e^{3u(y)}}dy} .

For simplicity, we denote

\displaystyle\alpha = \frac{1}{{{\pi ^2}}}\int_{{\mathbb{R}^3}} {{e^{3u(x)}}dx} .


\displaystyle\begin{gathered} \left| { - \frac{{v(x)}}{{\log |x|}} + \alpha } \right| = \frac{1}{{{\pi ^2}}}\frac{1}{{\left| {\log |x|} \right|}}\left| {\int_{{\mathbb{R}^3}} {\log \frac{{|x - y|}}{{|y|}}{e^{3u(y)}}dy} - \log |x|\int_{{\mathbb{R}^3}} {{e^{3u(y)}}dy} } \right| \hfill \\ \qquad= \frac{1}{{{\pi ^2}}}\frac{1}{{\left| {\log |x|} \right|}}\left| {\int_{{\mathbb{R}^3}} {\left[ {\log |x - y| - \log |y| - \log |x|} \right]{e^{3u(y)}}dy} } \right| \hfill \\ \qquad= \frac{1}{{{\pi ^2}}}\frac{1}{{\left| {\log |x|} \right|}}\left| {\int_{{\mathbb{R}^3}} {\log \frac{{|x - y|}}{{|x||y|}}{e^{3u(y)}}dy} } \right|. \hfill \\ \end{gathered}

We now split

\displaystyle\int_{{\mathbb{R}^3}} {} = \int_{|y| \leqslant 1} {} + \int_{|y| \geqslant 1} {} .

We have

\displaystyle\left| {\int_{{\mathbb{R}^3}} {\log \frac{{|x - y|}}{{|x||y|}}{e^{3u(y)}}dy} } \right| \leqslant \int_{|y| \leqslant 1} {\left| {\log \frac{{|x - y|}}{{|x||y|}}} \right|{e^{3u(y)}}dy} + \int_{|y| \geqslant 1} {\left| {\log \frac{{|x - y|}}{{|x||y|}}} \right|{e^{3u(y)}}dy} .

Observe that

\displaystyle\int_{|y| \leqslant 1} {\left| {\log \frac{{|x - y|}}{{|x||y|}}} \right|{e^{3u(y)}}dy} \leqslant {e^C}\int_{|y| \leqslant 1} {\log \left( {\frac{1}{{|x|}} + \frac{1}{{|y|}}} \right)dy}


\displaystyle\int_{|y| \geqslant 1} {\left| {\log \frac{{|x - y|}}{{|x||y|}}} \right|{e^{3u(y)}}dy} \leqslant \log \left( {1 + \frac{1}{{|x|}}} \right)\int_{|y| \geqslant 1} {{e^{3u(y)}}dy} .


\displaystyle\left| { - \frac{{v(x)}}{{\log |x|}} + \alpha } \right| \leqslant \frac{1}{{{\pi ^2}}}\frac{1}{{\left| {\log |x|} \right|}}\left[ {{e^C}\int_0^1 {\log \left( {\frac{1}{{|x|}} + \frac{1}{r}} \right){r^2}dr} + \log \left( {1 + \frac{1}{{|x|}}} \right)\alpha } \right].

The desired result follows easily since thing sitting within the square bracket is uniformly bounded. For a good reference, we refer the reader to a paper due to N. Zhu [here] published in Comm. Partial Differential Equations (2004).

It turns out that whenever the following

\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \left( {u(x) + \alpha \log |x|} \right) = \frac{1}{{{\pi ^2}}}\int_{{\mathbb{R}^3}} {\log |y|{e^{3u(y)}}dy}


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