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This question stems from a step of a proof in this paper: http://www.cmap.polytechnique.fr/~ammari/papers/04AKS.pdf

The question itself I didn't feel to be "research level" as it is only an asymptotic estimate, so I'm posting this here.

The step occurs near the bottom of page 8:

"Since $\displaystyle-\frac{1}{4\pi}\frac{t}{(|x-x^\prime |^2+t^2)^{3/2}}$ is the Poisson kernel for the half space and $f(\cdot,h\zeta^\prime)$ has compact support in $\Omega$,

$$-\frac{1}{4\pi}\int_\Omega \frac{h^2 (\zeta-\zeta^\prime)}{(|x-x^\prime|^2 + h^2|\zeta-\zeta^\prime |^2 )^{3/2}} f(x^\prime , h\zeta^\prime) \,dx^\prime =-\frac{h}{2} f(x,0) \frac{\zeta - \zeta^\prime}{ |\zeta-\zeta^\prime |} +O(h^2)\text{"}$$

$x,x^\prime \in \Omega\subset\mathbb{R}^2$ where $\zeta,\zeta^\prime \in \mathbb{R}$. So as a big picture, we are looking at functions of $(x,h\zeta)$ and $(x^\prime , h\zeta^\prime)$ which are in $\mathbb R^3$. As $h$ goes to zero, the third component of both go to the boundary (i.e. 0). The Poisson theorem about $x$ going to the boundary that I mention below requires the third component of both to equal 0.

In Evans' PDE book chapter 2.2 theorem 14, we have that if $\zeta^\prime=0$ and (ignoring one of the $h$ in front) then we would have

$$-\frac{1}{4\pi}\int_\Omega \frac{h (\zeta-\zeta^\prime)}{(|x-x^\prime|^2 + h^2|\zeta-\zeta^\prime |^2 )^{3/2}} f(x^\prime , h\zeta^\prime) \,dx^\prime =-f(x,0)$$ as $h\to 0$. I'm confused on (a) how they get the $O$ estimate and (b) how the fraction $\frac{\zeta - \zeta^\prime}{ |\zeta-\zeta^\prime |}$ pops out. I think it has to do with $\zeta^\prime$ possibly not being on $\partial\mathbb{R}^{n}_+$ which is what is required in Evan's theorem, but I'm stuck there.

For (a): Let $\displaystyle u(x)= \frac{2 x_n}{n\alpha(n)}\int_{\partial\mathbb{R}^n_+}\frac{g(y)}{|x-y|^n}\, dy$ with $x\in\mathbb{R}^n_+$ and $y\in\partial\mathbb{R}^n_+$. For convenience we write: $u(x)=\int_{\partial\mathbb{R}^n_+}K(x,y)g(y)\, dy$ with $K(x,y)$ defined accordingly.

Following the steps in Evan's proof: Let $\epsilon>0$ be given and $\delta>0$ such that $|g(y)-g(x_0)|<\epsilon$ if $|y-x_0|<\delta$ and $y\in\partial\mathbb{R}^n_+$. we can find that for $x_0$ on the boundary, if $|x-x_0|<\delta/2$ So as $x\to x_0\in\partial\mathbb{R}^n_+$, $$|u(x)-g(x_0)|=\left |\int_{\partial\mathbb{R}^n_+}K(x,y)[g(y)-g(x_0)]\, dy \right| \leq \int_{\partial\mathbb{R}^n_+ \cap B(x_0,\delta)}K(x,y)|g(y)-g(x_0)|\, dy +\int_{\partial\mathbb{R}^n_+ \setminus B(x_0,\delta)}K(x,y)|g(y)-g(x_0)|\, dy$$ $$=I+J$$

In the book it is shown $I\leq\epsilon$ and that $J=O(x_n)$. So what I'm wondering is this: does this mean I can write:

$$u(x) = g(x_0) + O(x_n)$$ as $x_n \to 0$?

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up vote 1 down vote accepted

I admire the book by Evans, but his notation for half-space Poisson kernel does not align with the way I think of it. Loosely following Wikipedia, I write the Poisson kernel for half-space $\mathbb R^3_+$ as $$ P_t(x,x')=\frac{t}{2\pi}\frac{1}{(|x-x'|^2+t^2)^{3/2}},\qquad x,x'\in\mathbb R, \ t>0 \tag{I} $$ This kernel has the property that $\int_{\mathbb R^2} P_t(x-x')g(x')\,dx'$ is the value at $(x,t)\in\mathbb R^3_+$ of the harmonic extension of $g$ to the half-space. The harmonic extension of a continuous functions is continuous up to the boundary, which means we get the original function $g$ back as $t\to 0$.

Comparing (I) with the integral (with one less $h$, as in your post): $$-\frac{1}{4\pi}\int_\Omega \frac{h (\zeta-\zeta^\prime)}{(|x-x^\prime|^2 + h^2|\zeta-\zeta^\prime |^2 )^{3/2}} f(x^\prime , h\zeta^\prime) \,dx^\prime \tag{II}$$ we decide to let $t=h|\zeta-\zeta^\prime |$. We become aware at this point that the numerator in (II) is $\pm t$, depending on the sign of $\zeta-\zeta'$. So, rewrite the integral as $$-\frac{1}{4\pi}\frac{\zeta-\zeta^\prime}{|\zeta-\zeta^\prime|}\int_\Omega \frac{h |\zeta-\zeta^\prime|}{(|x-x^\prime|^2 + h^2|\zeta-\zeta^\prime |^2 )^{3/2}} f(x^\prime , h\zeta^\prime) \,dx^\prime \\ = -\frac{1}{2}\frac{\zeta-\zeta^\prime}{|\zeta-\zeta^\prime|}\int_\Omega P_t(x,x') f(x^\prime , h\zeta^\prime) \,dx^\prime $$ Some people, including myself, would rather write $\mathrm{sign}\,(\zeta-\zeta')$ instead of the fraction.

As $h\to 0$, two things happen. First, the function whose harmonic extension we consider changes: $f(\cdot, h\zeta')\to f(\cdot, 0)$ in some way (I hope uniformly, but did not check). Second, we approach the boundary where the Poisson integral recovers the original function. This, $$-\frac12\frac{\zeta-\zeta^\prime}{|\zeta-\zeta^\prime|} f(x,0)\tag{III}$$ is exactly what we should expect in the limit.

The authors claim that the difference between (II) and (III) is bounded by $h\|f\|_{C^1}$. I don't quite believe them, because harmonic extension of a $C^1$ function is not necessarily Lipschitz. Anything a little better than $C^1$, such as $C^{1,\alpha}$, would be enough.

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Thanks for your response! I have a better understanding now. I'm going to fully work through it tomorrow to make sure I get everything. Do you have any comments about the O-estimate? –  toypajme Aug 5 '12 at 19:24
    
@toypajme The last paragraph of my answer is about the O-estimate. As I said, I don't think the authors got it right. If $f$ has two continuous derivatives, the O-estimate holds, but one derivative may not be enough. –  user31373 Aug 5 '12 at 19:58
    
Oh, right. I figured out where my confusion was. –  toypajme Aug 6 '12 at 16:01
    
I appreciate your help. I'm missing how, even with $C^{1,\alpha}$ we get the bound $h\|f\|_{C^{1}}$. I see when subtracting II from III in absolute value and substituting in the poisson kernel $f(x,0)$ as a "1" in III, we can bound to get $|f(x^\prime,h\zeta)-f(x,0)|$ inside the integral, but I don't know how to continue from here to get $h\|f\|_{C^{1}}$. –  toypajme Aug 8 '12 at 14:11
    
I'm guessing it has to do with a metric inequality and using the norm for the Lipschitz "C," but I can't write it down satisfactory. Is it just that for a fixed $h$, we may find a $C$ such that $|(x^\prime,h\zeta^\prime)-(x,0)|\leq C |h\zeta^\prime| = Ch$ ? –  toypajme Aug 8 '12 at 14:17
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