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?

  • 1
    $\begingroup$ You're missing a factor $\sin\theta$ from the Jacobian in the $\mathbb R^3$ integral (not that it matters much). $\endgroup$ – joriki Aug 6 '12 at 11:46
  • 1
    $\begingroup$ 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. $\endgroup$ – joriki Aug 6 '12 at 11:50
  • $\begingroup$ 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? $\endgroup$ – joriki Aug 6 '12 at 11:59
  • $\begingroup$ @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. $\endgroup$ – Konrad Sakowski Aug 6 '12 at 13:09
  • $\begingroup$ 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. $\endgroup$ – joriki Aug 6 '12 at 14:57

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 $\Delta u = |x|^{-2}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.