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Consider the function $u(\vec{x})=\ln|\vec{x}|$ as a distribution on $\mathbb{R}^3$ and $\mathbb{R}^2$. We want to determine $\Delta u$ in the distribution sense. First calculate $\Delta u$ as a regular function outside the singularity at the origin, using polar coordinates. Then write down the integral that defines the effect of $\Delta u$ on a test function. On this integral one can use Green's second identity. Choose a bounded set whose complement contains a small sphere around the origin. Let the radius of the small sphere approach 0. Is the distribution $\Delta u$ given by the pointwise values or not?

What I've done so far

Following the problem description pretty much verbatim to begin with, I could reach the conclusion that, in the function sense, $\Delta u = \frac{1}{r^2}$ on $\mathbb{R}^3$ and $\Delta u=0$ on $\mathbb{R}^2$. Now, following the problem description I should calculate $(\Delta u)[\phi]$ on both $\mathbb{R}^3$ and $\mathbb{R}^2$ using the given set, i.e. I have to calculate $$(\Delta u)[\phi] = u[\Delta\phi] = \lim_{r\to0}\int_r^R\int_0^{2\pi}r\ln r\Delta\phi\mathrm{d}\theta\mathrm{d}r$$ and $$(\Delta u)[\phi] = u[\Delta\phi] = \lim_{r\to0}\int_r^R\int_0^{2\pi}\int_0^\pi r^2\sin\theta\ln r\Delta\phi\mathrm{d}\theta\mathrm{d}\varphi\mathrm{d}r.$$

Now, this is where I get stuck. I think that, in the $\mathbb{R}^2$ case, the integral actually converges to $0$: $$\lim_{r\to0}\int_r^R\int_0^{2\pi}r\ln r\Delta\phi\mathrm{d}\theta\mathrm{d}r = \lim_{r\to0}\int_r^Rr\ln r\int_0^{2\pi}\Delta\phi\mathrm{d}\theta\mathrm{d}r =\\= \lim_{r\to0}\int_r^Rr\ln r\left[\int\Delta\phi\mathrm{d}\theta\right]_0^{2\pi}\mathrm{d}r = \lim_{r\to0}\int_r^Rr\ln r \cdot 0\mathrm{d}r = 0.$$ In $\mathbb{R}^3$, I am completely stumped. It seems like I could make it converge to $0$ using the same trick I used in $\mathbb{R}^2$, but that seems wrong (as it doesn't make sense that the distribution behave that differently in $\mathbb{R}^2$ and $\mathbb{R}^3)$, which makes me question doing to in the $\mathbb{R}^2$ case. I would expect that the integral computes to $\frac{1}{r^2}$, but that may be wrong.


How do I compute the two integrals (especially the $\mathbb{R}^3$ one) properly?

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You're missing a factor $\sin\theta$ from the Jacobian in the $\mathbb R^3$ integral (not that it matters much). –  joriki Aug 6 '12 at 11:46
You can't make the integral vanish like that; the integrand being periodic in $\theta$ doesn't imply that the anti-derivative is periodic in $\theta$; just think of a constant function for example. –  joriki Aug 6 '12 at 11:50
I don't think $(\Delta u)[\phi]=u[\Delta\phi]$ is right in $\mathbb R^2$ -- shouldn't you get a contribution proportional to $\phi$ from $\int r(\ln r)'\,\mathrm d\theta$ from Green's second identity? –  joriki Aug 6 '12 at 11:59
@joriki: I believe $(\Delta u)[\phi] = u[\Delta \phi]$ is correct, because if $u$ is a distribution, and $\phi$ is a smooth function with a compact support, then this equality coincides with the definition of a derivative of a distribution. –  Konrad Sakowski Aug 6 '12 at 13:09
Sorry, you're right, I was confused about that. Part of what confused me was that you wrote "I was also able to show, using Green's second identity, that $(\Delta u)[\phi]=u[\Delta\phi]$". You can't show that using Green's identity, since as you rightly point out it's true by definition. The idea is to use Green's identity to show that in $\mathbb R^2$ the limit is non-zero even though $\Delta u=0$ pointwise away from the origin. By the way, do you disagree about the $\sin\theta$ factor? It's still missing. –  joriki Aug 6 '12 at 14:57
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1 Answer

up vote 4 down vote accepted

By this time probably you have made this already. On the other hand, multidimensional analytic integration is not my middle name. Anyway, here it comes.

I believe your approach is wrong, because I agree with joriki, that $\int_0^{2 \pi} \Delta u \;d\theta$ is generally not zero.

Let $x \in \mathbb{R}^2$, $u(x) := \log(|x|)$. We want to compute $\Delta u$ in a distributional sense. First we compute $\Delta u(x)$ for $x \neq 0$ to and we obtain $\Delta u(x)=0$ (direct calculus, I omit this). The only "suspicious" point is then $x = 0$, the origin. Let then $\phi$ be a smooth function with a compact support. We want to compute \begin{equation} (\Delta u)[\phi] := u[\Delta \phi] = \int_{\mathbb{R}^2} \log|x| \; \Delta \phi(x) dx = \lim_{r \rightarrow 0} \int_{B(0;R)-B(0;r)}\log|x| \; \Delta \phi(x) dx\;. \end{equation} where $R$ is big enough so that $\mathrm{supp}(\phi) \subset B(0;R)$. Then we use the Green's identity, but we do not yet switch to the polar coordinates \begin{eqnarray} \int_{B(0;R)-B(0;r)}\log|x| \; \Delta \phi(x) dx &=& \int_{B(0;R)-B(0;r)}\Delta \log|x| \; \phi(x) dx \\&&+ \int_{S(0;R)} \left( \log|x| \; \frac{\partial \phi(x)}{\partial n} - \frac{\partial \log|x|}{\partial n} \; \phi(x)\right)dS(x) \\ & &- \int_{S(0;r)} \left( \log|x| \; \frac{\partial \phi(x)}{\partial n} - \frac{\partial \log|x|}{\partial n} \; \phi(x)\right)dS(x)\;. \end{eqnarray} Then the first integral is zero, as we know that $\Delta\log|x|=0$ for $x\neq 0$. Also the second integral is zero, as we choosen $R$ big enough for $\phi$ and its derivatives to vanish. Thus there is only third integral left. Then we switch to the polar coordinates, when calculating the spherical integrals \begin{equation} - \int_{S(0;r)} \log|x| \; \frac{\partial \phi(x)}{\partial n} dS(x) = -\int_0^{2\pi} r \log(r) \; \frac{\partial \phi(r,\theta)}{\partial r} d\theta = -r \log(r) \int_0^{2\pi} \frac{\partial \phi(r,\theta)}{\partial r} d\theta \rightarrow 0, \end{equation} as $r \rightarrow 0$, since $\phi$ and its derivatives are bounded. Still we have \begin{equation} \int_{S(0;r)} \frac{\partial \log|x|}{\partial n} \; \phi(x)dS(x) = \int_{0}^{2\pi} r \frac{\partial \log(r)}{\partial r} \; \phi(r,\theta)d\theta = \int_{0}^{2\pi} r \frac{1}{r} \; \phi(r,\theta)d\theta. \end{equation} Then $r$ and $1/r$ cancel and by Lebesgue's dominated convergence theorem \begin{equation} \int_{0}^{2\pi} \phi(r,\theta)d\theta \rightarrow \int_{0}^{2\pi} \phi(0)d\theta = 2 \pi \phi(0). \end{equation} In conclusion, we obtain $\Delta u [\phi] = 2 \pi \delta [\phi]$. This is, I believe, some standard fact, as $\frac{1}{2 \pi}\log |x|$ is a fundamental solution of Laplacean in $\mathbb{R}^2$.

In $\mathbb{R}^3$ the case is different. I would only denote main differences. First, as you have noted, $\Delta u = |x|^{-2}$, so the first integral does not vanish. Then in the boundary integrals we would have $r^2 \sin\theta$ instead of $r$. So now the third integral vanishes, if my calculus is correct. The second integral vanishes as in $\mathbb{R}^2$. We therefore obtain \begin{equation} \int_{B(0;R)-B(0;r)}\log|x| \; \Delta \phi(x) dx = \int_{B(0;R)-B(0;r)} |x|^{-2} \; \phi(x) dx. \end{equation} Then we pass $r$ to the limit and we conclude that $\Delta u = |x|^{-2}$.

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Great; I was meaning to spell this out when I find the time, but I see you've already done it in detail :-) –  joriki Aug 12 '12 at 10:24
What is your middle name? –  NiftyKitty95 Aug 12 '12 at 11:23
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