Geometric interpretation of the half-derivative?

For $f(x)=x$, the half-derivative of $f$ is $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x = 2 \sqrt{\frac{x}{\pi}} \;.$$ Is there some geometric interpretation of (Q1) this specific derivative, and, (Q2) of the half-derivative more generally? I have read that fractional derivatives are nonlocal, but it seems strange to me that integral derivatives can be described in terms of local geometry only, while fractional derivatives cannot be so described. This would suggest an odd discontinuity between, say, $d^{1}$ and $d^{1.01}$. This seems especially at odds with the many applications of fractional derivatives, which (superficially) suggests continuity should reign.

I'd appreciate someone clearing up my elementary confusions—Thanks in advance!

Addendum (5Jan14). @AlexR. found this geometric interpretation of the fractional integral in Richard Herrmann's book, Fractional Calculus: An Introduction for Physicists,World Scientific, 2011: • Well, my view of fractional derivatives is from it's kernel (by using Fourier transform), though this has nothing to do with geometric interpretation. For integral derivatives,like $\frac{d^k}{dx^k}$, the Schwartz kernel is $\delta_0^{(k)}$, which measures only on a point, however, when considering the fractional derivative,like $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$ on $\mathbb{R}$, the respective kernel is $|x-y|^{-1-\frac{1}{2}}$, which is nonlocal anymore. – Tomas Jan 4 '14 at 2:43
• I haven't found fractional powers of the first derivative $d/dx$, which is indefinite, to be of much use. Fractional powers $\Delta^\alpha$ of the (analyst's) Laplacian $\Delta = -\sum_{i=1}^n \frac{\partial}{\partial x_i^2}$, on the other hand, are much more useful (note that the Laplacian is positive definite, in contrast to the first order operator.) The square root of the Laplacian has a natural geometric interpretation as the Dirichlet-to-Neumann operator for the upper half-plane. – Terry Tao Jan 4 '14 at 2:46
• Also, fractional derivatives depend continuously on the exponent $\alpha$ (at least in the distributional topology), as can be seen on the Fourier side. The non-locality disappears as $\alpha$ approaches a natural number due to denominators such as $\Gamma(-\alpha)$ that appear in the formulae for the kernel away from the origin (which is something like $\frac{1}{\Gamma(-\alpha)} |x-y|^{-1-\alpha}$ in one dimension). – Terry Tao Jan 4 '14 at 2:49
• One obstruction to an obvious geometric interpretation is that the ordinary derivative can be regarded as generalizing to, for example, the exterior derivative, but I'm not aware of a generalization of the half-derivative this broad without extra structure (fractional powers of the Laplacian require a Riemannian metric, for example). – Qiaochu Yuan Jan 4 '14 at 6:00
• A correction to my previous comment: the first order operator $d/dx$ is indefinite on the real line (the spectral variable $\xi$ can be either positive real or negative real), and so it is not natural to consider fractional powers of this operator (one has to arbitrarily choose a branch cut for $\xi^\alpha$). But if one is working on the half-line instead, then the spectral variable $\xi$ now naturally lives on the upper half-plane (Fourier-Laplace transform) and one now has a canonical interpretation of $\xi^\alpha$. So fractional powers of $d/dx$ are reasonable in half-line settings. – Terry Tao Jan 4 '14 at 16:54

A possible mechanical interpretation of the half-derivative can be given in terms of Abel's solution to a classical problem from the calculus of variations (the tautochrone problem).

Let there be a heavy particle which is constrained to slide without friction along the curve $y=y(t)$ in uniform gravity to its lowest point. Then, given a function $T(y)$ that specifies the total time of descent for a given starting height what is an equation of the curve that yields this result?

The principle of conservation of energy implies that the distance $S=S(t)$ travelled by the particle along the curve from the initial height $y_0$ satisfies the equation $$\left(\frac{dS}{dt}\right)^2=2g(y_0-y).$$ This is equivalent to the integral equation $$T(y_0)=\frac{1}{\sqrt{2g}}\int_0^{y_0}\frac{1}{(y_0-y)^{1/2}}\frac{dS}{dy}dy.$$

The r.h.s. of the latter equation is nothing else but the Riemann–Liouville fractional integral of $f=\pi^{1/2}(2g)^{-1/2}dS/dy$, i.e. $$D^{-\alpha}f(x)=\frac{1}{\Gamma(\alpha)}\int_0^x (x-y)^{\alpha-1}f(y)dy$$ of the order $\alpha=1/2$.

The solution to Abel's integral equation $dS/dy$ can be now interpreted (up to a constant factor) as the half-derivative of $T=T(y_0)$. • And what is the heuristic interpretation of this... interpretation? – Qfwfq Jan 4 '14 at 8:35
• Beautiful!! Thanks for the lucid lesson! – Joseph O'Rourke Jan 4 '14 at 13:31