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In analogy to a Dirac operator, it seems to me that formally, the equation

$\frac{\partial^n}{\partial x^n}f(x,y)=D_yf(x,y)$

is solved by

$f(x,y)=\exp{(x \sqrt[n]{D_y})}\ g(y).$

Is there a theory surronding the $\sqrt[n]{D_y}$-idea?

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    Is [this](http://en.wikipedia.org/wiki/Fractional_derivative) the kind of thing you're after?2012-08-10

2 Answers 2

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The short answer is yes, absolutely, and the theory of such operators is part of microlocal analysis. The basic ingredient is that differential operators can be written as integral operators (in an appropriate generalized sense) via the Fourier transform. E.g.

$ \frac{d}{dx} f(x) = \frac{d}{dx} \int e^{2\pi i kx} \hat{f}(k) dk = \int e^{2 \pi i k x} (2\pi ik) \hat{f}(k) dk. $ Since $\hat{f}(k) = \int e^{-2\pi i k y} f(y) dy$ (forgive me if I forgot a $2\pi$ somewhere), we have $ \frac{d}{dx} = \int \int (2\pi i k) e^{2 \pi i k(x-y)} dy dk. $ The right hand side has to be interpreted in a certain distributional sense, but if we are careful such formulae are correct and rigorous. Let's consider your example, $ \frac{\partial^n}{\partial x^n} f(x,y) = D_y f(x,y) $ and let's assume that $D_y$ is an ordinary polynomial differential operator in $y$ with constant coefficients. Since $D_y$ is a polynomial differential operator with constant coefficients, then $ \widehat{D_y g}(k) = P(k) \hat{g}(k)$ for some polynomial $P$. This suggests that whatever $\sqrt[n]{D_y}$ might be, it should satisfy $ \widehat{\sqrt[n]{D_y} g}(k) = \sqrt[n]{P(k)} \hat{g}(k). $ But using the Fourier transform, we can take this as the definition of $\sqrt[n]{D_y}$: $ \sqrt[n]{D_y} g(y) := \int e^{2\pi i ky} \sqrt[n]{P(k)} \hat{g}(k) dk = \int \int e^{2\pi i k(y-y')} \sqrt[n]{P(k)} g(y') dy' dk. $ This leads to $ \exp(x \sqrt[n]{D_y}) g(y) = \int \int e^{2\pi i k(y-y')} \exp(x\sqrt[n]{P(k)}) \hat{g}(k) dy' dk. $ As long as $P(k)$ and $g(y)$ are nice enough that this expression makes sense (and converges in an appropriate sense), this will solve the given PDE.

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    Yes, if you write $\sqrt[1/2]{D_y} \sqrt[1/2]{D_y} g(y)$ as a multiple integral, there will be a term like $\int e^{2\pi i y(k-k')} dy$ which is $\delta(k-k')$ and everything takes care of itself. This is what makes the pseudodifferential calculus work.2012-08-10
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Since you mentioned the Dirac operator, which works not by analysis but by extending the scalars in a noncommutative way; consider $\left( \begin{array}{ccc} 0 & 0 & D_y \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)^3 $ and generalize.

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    This technique depends sensitively on both the given differential operator, as well as the space of functions on which it acts. In the case of the Dirac operator, it is not the Laplacian on *functions* that admits an algebraic square root, but rather the Laplacian on spinors. This also works for vector-valued functions, since we can decompose vectors into products of spinors. These kinds of $n$th roots are completely different than what I described in my answer, but both can be useful depending on the situation.2012-08-10