2

Consider $$u(x_0,y_0,z_0)=\frac{z_0}{2\pi}\int_{\mathbb R^2}\frac{f(x,y)}{\left[(x-x_0)^2+(y-y_0)^2+z_0^2\right]^{3/2}}\,dx\,dy.$$ Show that (i) $u\to f(x_0,y_0)$ in the limit as $z_0\to0$ and (ii) $u\to0$ in the limit as $x_0^2+y_0^2\to\infty$.

Some motivation: this form of $u$ is the solution to the Laplace equation $\nabla^2u=0$ on the upper half-space $z>0$, subject to the boundary conditions $u(x,y,0)=f(x,y)$ and $u\to0$ as $x^2+y^2\to\infty$. This can be obtained e.g. by the reflection method. I want to verify directly that this form of $u$ does indeed satisfy those boundary conditions, but am struggling with the computation.

I have been given as a hint the substitution $z_0^2t^2=(x-x_0)^2+(y-y_0)^2$, but I'm not really sure how to use this?

Verge
  • 303

1 Answers1

1

The statement (i) can be proven following Evan's book approach (Partial Differential Equations - page 38). I will assume that the function $f$ is continuous and bounded in $\mathbb{R}^2$.
In this post, it is shown that the Poisson kernel in the Half-space given by $$ K(x,y) = \frac{2x_n}{n \alpha(n)} \frac{1}{|x-y|^n} \hspace{1 cm} (x \in \mathbb{R}^{n}_{+}, \hspace{0.3 cm} y \in \partial \mathbb{R}^{n}_{+})$$ (where $n\alpha(n)$ is the surface area of the unit sphere in $\mathbb{R}^n$) satisfies the relation $$ \int_{\partial \mathbb{R}^{n}_{+}} K(x,y) dy = 1.$$ This can be used to prove the desired result. Indeed, if you fix $(x_0,y_0)$ in $\mathbb{R}^2$ and $\epsilon >0$, we may choose $\delta >0$ so that \begin{equation} \label{2}|f(x,y)-f(x_0,y_0)|< \epsilon \hspace{0.5 cm} \text{if} \hspace{0.5 cm} |(x,y)-(x_0,y_0)|< \delta. \hspace{1 cm } (1) \end{equation}

If we take $|(x,y) - (x_0,y_0)|< \frac{\delta}{2}$, then \begin{eqnarray*} |u(x,y,z) - f(x_0,y_0)| &=& |\int_{\mathbb{R}^2} K(x,y,z,w)f(w) dw -f(x_0,y_0) | \\ &\leq& \int_{\mathbb{R}^2} K(x,y,z,w) |f(w)-f(x_0,y_0)| dw \\ &=& \int_{\mathbb{R}^2 \cap B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw \\ &+& \int_{\mathbb{R}^2 - B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw. \\ \end{eqnarray*} Now, by (1) we have $$ \int_{\mathbb{R}^2 \cap B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw \leq \epsilon.$$ On the other hand, if $|(x,y)-(x_0,y_0)| < \frac{\delta}{2}$ and $|w-(x_0,y_0)|> \delta$, we have \begin{eqnarray} |w-(x_0,y_0)| &\leq& |w-(x,y)|+|(x,y)-(x_0,y_0)|\\ &\leq & |w-(x,y)| + \frac{\delta}{2}\\ &<& |w-(x,y)|+ \frac{1}{2}|w-(x_0,y_0)| \end{eqnarray} Thus, we obtain $|w-(x,y)|> \frac{1}{2}|w-(x_0,y_0)|$ and therefore

\begin{eqnarray} \int_{\mathbb{R}^2 - B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw &\leq& 2||f||_{\infty} \int_{\mathbb{R}^2 - B((x_0,y_0),\delta))}K(x,y,z,w) dw \\ &\leq & C\frac{ z}{2 \alpha(2)} ||f||_{\infty} \int_{\mathbb{R}^2 - B((x_0,y_0),\delta))} \frac{dw}{|w-(x_0,y_0)|^{2}} \end{eqnarray} and this last integral tends to zero when $z \longrightarrow 0$, which shows that $u \to f(x_0,y_0)$ for any $(x_0,y_0) \in \mathbb{R}^2$.

I think that statement (ii) is not true. You may take, for instance, $f(x,y)=1$ and then, $$u(x,y,z) = \int_{\mathbb{R}^2} K(x,y,z,w) f(w) = \int_{\mathbb{R}^2} K(x,y,z,w) = 1$$ and this function doesn't satisfy the required condition at the limit.

J. David Beltran
  • 278