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.

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