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In the wake of a prior question, I've solidified my understanding that for a one-dimensional function $f(x)$, when its first derivative has a jump discontinuity of height $h$ at $x_0$, we can regard the second derivative (for the purpose of eventually integrating it) as having a $\delta(x-x_0)$ factor plus a locally continuous function. What if, as in my former question, a continuous function is nonzero everywhere outside of the bounded region $\Omega \subset \mathbb{R}^2$, and its first-order partials have jump discontinuities at the boundary: how then should we derive the form of $\Delta f$ ?

My guess is that if we define $\nabla f$ on the boundary by taking the limit from inside $\Omega$, then we'll get something akin to $\Delta f = \delta(0) \| \nabla f \|$ on the boundary, and $0$ otherwise. Or is it possible that when integrating $\Delta f$ over a path hitting a single boundary point $x_0$, the resulting "$\delta$-factor" depends on what direction the path hits $x_0$?

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up vote 4 down vote accepted

Let $\Omega\subset\mathbb{R}^2$ be a bounded domain with a $C^1$ boundary. Let $f$ be zero outside of $\Omega$ and $C^2$ inside, with $f=0$ on $\partial\Omega$. Also assume that all of the partial derivatives of $f$ extend continuously to $\partial\Omega$. Let us define $$ g(x) = \begin{cases} \Delta f(x) &\text{if $x\in\Omega$},\\ 0 &\text{otherwise}. \end{cases} $$ Now, $g$ is not equal to $\Delta f$ on all of $\mathbb{R}^2$. In fact, $\Delta f$ may not even exist as a function on $\mathbb{R}^2$. But it does exist as a distribution. Let us see how $\Delta f$, the distribution, compares with $g$.

Let $\varphi$ be a smooth test function. Then $$ \langle\Delta f,\varphi\rangle = \langle f,\Delta\varphi\rangle = \int_{\mathbb{R}^2}f(x)\Delta\varphi(x)\,dx. $$ Since $f=0$ outside of $\Omega$, we need only integrate over $\Omega$. We can then use Green's second identity to obtain $$ \int_\Omega f(x)\Delta\varphi(x)\,dx = \int_\Omega g(x)\varphi(x)\,dx + \int_{\partial\Omega} (f\partial_{\bf n}\varphi - \varphi\partial_{\bf n}f)\,d\sigma, $$ where $\partial_{\bf n}$ is the derivative in the direction of the outward normal, and $\sigma$ is the surface measure on $\partial\Omega$. (Since we are in $\mathbb{R}^2$, this is just the arclength measure.) Since $f=0$ on $\partial\Omega$, we get \begin{align*} \langle\Delta f,\varphi\rangle &= \int_\Omega g\varphi\,dx -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\,d\sigma\\ &= \int_{\mathbb{R}^2} g\varphi\,dx -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\,d\sigma\\ &= \langle g,\varphi\rangle -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\,d\sigma. \end{align*} What this shows is that $\Delta f = g + \nu$, where $\nu$ is the distribution that maps $\varphi$ to $$ \langle\nu,\varphi\rangle = -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\,d\sigma. $$ For example, if $\partial_{\bf n}f(x)=-1$ for all $x\in\partial\Omega$, then $\nu$ is just the surface measure on $\partial\Omega$. This is the case in the example I gave in my answer to the other question. And in that case, the surface measure on the boundary of $[0,\pi]$ in $\mathbb{R}$ is just two point masses, one at $0$ and one at $\pi$.

Lastly, to address a comment from the other question, the Laplacian is still rotationally invariant when interpreted as a distributional derivative. To prove this, we apply $\Delta f$ to a smooth test function, then move the Laplacian over to the test function (where it acts in the classical way), and then utilize the rotational invariance of the classical Laplacian.

A fairly short and accessible reference for tempered distributions, which is free online, is Chapter 11 of Applied Analysis by Hunter and Nachtergaele. Also, there is a chapter on the Laplace operator in Folland.


To address the question in the comments, the Radon transform of a tempered distribution is defined by $\langle Rf,\varphi\rangle=\langle f,R^*\varphi\rangle$. More precisely, it is $\langle Rf,\varphi\rangle_{S^1\times\mathbb{R}}=\langle f,R^*\varphi\rangle_{\mathbb{R}^2}$. The inner product on the right is the usual $L^2$ inner product on $\mathbb{R}^2$; the inner product on the left is defined by $$ \langle f,g\rangle_{S^1\times\mathbb{R}} = \frac1{2\pi}\int_0^{2\pi}\int_{\mathbb{R}} f(\theta,s)g(\theta,s)\,ds\,d\theta. $$ It follows that \begin{align*} \langle R(\Delta f),\varphi\rangle &= \langle \Delta f,R^*\varphi\rangle = \langle f,\Delta(R^*\varphi)\rangle\\ &= \langle f,R^*(\partial_s^2\varphi)\rangle = \langle Rf,\partial_s^2\varphi\rangle = \langle \partial_s^2(Rf),\varphi\rangle, \end{align*} and so $R(\Delta f)=\partial_s^2(Rf)$, even in the distributional sense.

What is relevant for your other question is how to compute $R\nu$. My guess is that you are asking if $$ R\nu = -\sum_{x\in\partial\Omega\cap L}\partial_{\bf n}f(x). $$ The intuition behind this formula does not work, because it does not account for the angle between $L$ and $\partial\Omega$ at the point of intersection. I will leave it as an exercise to show that if $\sigma$ is the arclength measure on $S^1$, that is, if $$ \langle\sigma,\varphi\rangle = \int_0^{2\pi}\varphi(e^{i\theta})\,d\theta, $$ then $$ R\sigma(\theta,s) = 2\frac1{\sqrt{1 - s^2}}\chi_{[-1,1]}(s). $$ The $2$ comes from the two points where the line intersects the circle, and the factor of $|1-s^2|^{-1/2}$ comes from the angle of intersection. If $L$ is the line corresponding to $(\theta,s)$ and $\alpha(L)\in[0,\pi/2]$ is the angle with which $L$ intersects $S^1$, then $|s|=\cos\alpha(L)$. We therefore have $$ R\sigma(\theta,s) = 2\csc\alpha(L)\chi_{[-1,1]}(s). $$ The natural generalization to $\nu$ would be $$ R\nu(L) = -\sum_{x\in\partial\Omega\cap L} \partial_{\bf n}f(x)\csc\alpha(x), $$ where, for $x\in\partial\Omega\cap L$, we define $\alpha(x)\in[0,\pi/2]$ to be the angle of intersection between $L$ and the tangent line to $\partial\Omega$ at $x$.

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I apologize for so many questions, but would I be correct in deducing from this formula the following modification of the Radon-Laplacian formula I mentioned? $$(R\Delta - \partial_s^2 R) u=-\sum_{x\in\partial\Omega\cap L} \partial_{\mathbf{n}} u(x),$$ where $L$ is the line defined by the parameters $(\theta,s)$? – anon Jul 17 '11 at 10:27
See the edit above. – Jason Swanson Jul 17 '11 at 18:35
I don't think $\langle Rf,\phi\rangle=\langle f,R^*\phi\rangle$ is valid. The "duality" between the transforms is, I believe, in their geometric constructions and their inversion formulas, but I may also be wrong about that. Note that you figured out $v$ is a distribution, so that means $Rv$ is an integral over a distribution $v$, which should be well-defined right? If we let $\phi$ be the distribution such that $$\langle \phi,f\rangle=\int_Lf(x)dx,$$ then although it isn't a test function, it seems formally that $Rv=\langle v,\phi\rangle=-\sum_{\partial\Omega\cap L} \partial_nf$ – anon Jul 17 '11 at 23:01
I have confirmed with a colleague who works in this area that $\langle Rf,\varphi\rangle=\langle f,R^*\varphi\rangle$ is indeed the definition of the Radon transform on tempered distributions. I also confirmed that there should, in fact, be correction factors to account for the angle of intersection between $L$ and $\partial\Omega$. I will edit the answer to reflect this. – Jason Swanson Jul 18 '11 at 2:03
AWESOMESAUCE. $\text{ }$ – anon Jul 19 '11 at 12:01

Let the boundary $S=\partial \Omega$ be smooth enough. Applying the second Green's formula for test function $\varphi$ we get $$ \int_{\mathbb R^n}(f\Delta \varphi -\Delta f\varphi)\mathrm dx= \int_{\Omega}(f\Delta \varphi -\Delta f\varphi)\mathrm dx= $$ $$ \int_{S}\left(f\frac{\partial \varphi}{\partial\bar n}- \varphi\frac{\partial f}{\partial \bar n}\right)\mathrm dx= -\int_{S}\frac{\partial f}{\partial\bar n}\mathrm dx, $$ where $\bar n$ is the unit outword normal to $S$. So in the sense of distributions $\Delta f=\{\Delta f\}-\frac{\partial\Delta f^+}{\partial\bar n}\delta_S$, where $\{\Delta f\}$ is $\Delta f$ outside $S$. The derivative in the last term is the limite values of $ \frac{\partial\Delta f}{\partial\bar n}$ inside the domain.

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I don't quite understand how you went from the equations to your expression for $\Delta f$. Anyway, I'm now pretty certain that $\Delta$, defined as $\partial_1^2+\cdots+\partial_n^2$ doesn't satisfy rotational invariance in the distributional sense, so I'm thinking my original question is ill-defined. Given that, I'll have to address my former query by modifying the Radon transform formula $\partial_s^2 R = R \Delta$ to handle impulse functions accordingly, and probably delete this question... – anon Jul 17 '11 at 9:37
By definition $(\Delta f,\varphi)=(f,\Delta \varphi)$ and the rhs is the integral $\int_{\mathbb R^n}f\Delta \varphi\; dx\ $ since $f$ is continuous and therefore defines a regular distribution. – Andrew Jul 17 '11 at 9:51
Oh, okay, I didn't realize that was the definition of the Laplacian for distributions. Wikipedia doesn't have that and if I've seen the definition somewhere I must have forgot it. Anyways, you might want to check Green's identities again because you misused them - you have some extra $\nabla$'s in there. Other than that, your answer agrees with user11867's and is more succinct. – anon Jul 17 '11 at 10:20

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