Why does it seem I can't apply the Radon transform to the Helmholtz equation?

Say we a function $u$ and a bounded region $\Omega \subset \mathbb{R}^2$, such that $(\Delta+\lambda)u = 0$ everywhere, and $u=0$ on the boundary. We extend it to the entire plane by defining $u=0$ everywhere outside of $\Omega$. Wikipedia states that, for the Radon transform, $R\Delta = \partial^2 / \partial s^2 R$, hence we can derive the ODE

$$\left( \frac{\partial^2}{\partial s^2} + \lambda \right) Ru(\theta,s) = 0,$$

therefore we can reason that

$$Ru(\theta,s) = a(\theta) e^{i \sqrt{\lambda} s} + b(\theta) e^{-i \sqrt{\lambda} s}.$$

But this is periodic in $s$ with period $2\pi/\sqrt{\lambda}$, which would imply either that the above vanishes everywhere, or that it's possible to integrate $u$ over lines arbitrarily far away from $\Omega$ and still get a positive value, despite $u$ vanishing identically on said line. Contradiction.

So does this mean:

1. I can't use $R\Delta = \partial^2 / \partial s^2 R$ for functions that aren't sufficiently smooth, or
2. The Radon transform is discontinuous at any line tangent to $\partial \Omega$?
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Maybe after extending $u$ to the whole plane, you have $(\Delta + \lambda)u = \nu$, where $\nu$ is a distribution supported on $\partial\Omega$. Then the right-hand side of the ODE should be $R\nu$, provided that is well-defined. – Jason Swanson Jul 17 '11 at 5:58
That makes sense to me. Technically, taking $(\Delta+\lambda)u$ as $x \to x_0\in\partial\Omega$ from a path strictly inside $\Omega$ (or strictly outside) would make it 0, but precisely at the point $x=x_0$ the quantity is actually equal to $Hu(\vec{N},\vec{N})+Hu(\vec{T},\vec{T})$, where $H$ is the Hessian matrix of second-order partial derivatives interpreted as a bilinear function of vectors $\in\mathbb{R}^2$, and $\vec{N}$ and $\vec{T}$ are $\partial\Omega$'s normal and tangent vectors at $x_0$. I'll accept this answer if it's submitted sometime soon, and type it myself if need be. – anon Jul 17 '11 at 6:39
Okay, I'll type something up. – Jason Swanson Jul 17 '11 at 7:28
I've never edited someone else's question before, so I'm not sure of the protocol, but I just added the "analysis" tag to your question. Hope that's okay. – Jason Swanson Jul 17 '11 at 7:59

I suggested in a comment that after extending $u$ to the whole plane, you have $(\Delta+\lambda)u=\nu$, where $\nu$ is a distribution supported on $\partial\Omega$. Then the right-hand side of the ODE should be $R\nu$, provided that is well-defined. Since this seems a little skimpy for an answer, let me describe a simpler situation (or at least a situation I am more familiar with) where something analogous happens.

Consider $u(x)=\sin(x)$ on $[0,\pi]$, and $0$ everywhere else on $\mathbb{R}$. Suppose we reason that $(\partial^2 + I)u=0$ everywhere on $\mathbb{R}$, and then take the Fourier transform of both sides. If we use the normalization that defines $$\widehat{\varphi}(k) = \frac1{\sqrt{2\pi}}\int_{\mathbb{R}} \varphi(x)e^{-ikx}\,dx$$ for every Schwartz function, then we get $(-k^2+1)\widehat{u}(k)=0$. Since $u$ is integrable, $\widehat{u}$ is a continuous function, and so this identity implies that $\widehat{u}=0$ everywhere, which gives $u=0$ everywhere.

The mistake is that we should have used $(\partial^2 + I)u=\delta_0+\delta_\pi$, where $\delta_a$ is the delta function centered at $a$. The derivative of $u$ has a jump discontinuity at $x=0$ of size $1$, so the second derivative gives a unit point mass there, and similarly at $x=\pi$. Now the Fourier transform gives $$(-k^2 + 1)\widehat{u}(k) = \frac1{\sqrt{2\pi}}(1 + e^{-i\pi k}),$$ so that $$\widehat{u}(k) = \frac{1 + e^{-i\pi k}}{\sqrt{2\pi}(1 - k^2)}.$$ Taking limits as $k\to\pm1$, we can readily verify that this function is continuous, as it should be.

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I'm wondering, how might one generalize "has a jump discontinuity ...of size $1$, so the second derivative gives a unit point mass" to a two-dimensional function? Should I heuristically regard the Laplacian as $\delta(0) \| \nabla u \|$ on $\partial\Omega$ and $0$ everywhere else? How might that be justified? EDIT: I'm just going to ask that as a follow-up question, because it deserves separate treatment in my opinion. – anon Jul 17 '11 at 8:08
The part you quoted is formalized by saying that $\delta_0+\delta_\pi$ is the second distributional derivative of $u$. In $\mathbb{R}^n$, $\Delta=\partial_1^2+\cdots+\partial_n^2$, and these can all be interpreted as distributional derivatives. – Jason Swanson Jul 17 '11 at 8:14
That would seem to imply $(\Delta+\lambda) u(x) = \sum (\partial_iu) \delta(x_i-y_i)$ near the point $y\in\partial\Omega$, for $x\not\in\partial\Omega$. That's not rotationally invariant, as the Laplacian usually is. If that breaks down, I'm no longer sure that $\Delta R = \partial_s^2 R$ holds in a distributional sense... – anon Jul 17 '11 at 8:34

As another example, addressing the comments, in 2D (or higher) consider the distributional derivative $\Delta u$ in $\mathbb R^2$, where $u$ is the characteristic function of the disk. The exercise is to see that this is the distribution $f\rightarrow$ integrate-$\nu f$-along-the-unit-circle, where $\nu$ is the normal derivative.

The comments of the earlier answer about distributions supported at the boundary are right on the money as accounting for seeming discrepancies between naive forms of these computations and the slightly wised-up versions taking into account correct (a.k.a. "distributional") computation of derivatives.

Indeed, rotational symmetry (or equivariance) is preserved by $\Delta$, etc., which provides a "check" on heuristic computations, and also provides a very efficient way to do complete computations, in some cases! For example, one can prove that every distribution in $\mathbb R^2$ supported on the unit circle is of the form $f\rightarrow \nu f \rightarrow u(\nu f)$ for $\nu$ a differential operator in the normal direction, and $u$ a distribution on the circle. This might be viewed as (and is, essentially) a higher-dimensional version of the manifestation of Taylor-Maclaurin series in proving that the distributions supported at a point are linear combinations of Dirac delta and its derivatives. Then the rotation-invariant among these distributions are easy to characterize. (Note, distributions on the circle have Fourier expansions with coefficients of moderate growth.)

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