Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x_0$?

share|improve this question
add comment

2 Answers

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.

Edit:

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 $x$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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