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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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  • $\begingroup$ In more generality than you're asking about, positive operators always have unique positive $n^{th}$ roots by the continuous functional calculus. But it seems that most differential operators have some negative and complex eigenvalues, so this may not be any use. $\endgroup$ – Kevin Carlson Aug 10 '12 at 12:41
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    $\begingroup$ Is this the kind of thing you're after? $\endgroup$ – Clive Newstead Aug 10 '12 at 12:46
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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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  • $\begingroup$ Thanks for the answer. With that definition of $\sqrt[n]{D_y}$, why would $\sqrt[1/2]{D_y}\sqrt[1/2]{D_y}g(y)$ be equal to $D_yg(y)$? You have this integration variable $k$ there and so it seems to me as it's different for all the applications of the operators. (I think maybe its a delta-function integral trick though, which makes all the integral variables them equal.) $\endgroup$ – Nikolaj-K Aug 10 '12 at 13:44
  • $\begingroup$ 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. $\endgroup$ – Jonathan Aug 10 '12 at 13:50
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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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  • $\begingroup$ Mhm, so the matrix is the root of the differential operator $D_y$ to the power of three (times unity), but I'm not really sure how this helps me to compute the root of a given differential operator like $D_y$. The problem complexity will hight depend on that specific situation, e.g. like having to construc a clifford algebra in case of the d'Alambert operator. $\endgroup$ – Nikolaj-K Aug 10 '12 at 14:38
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    $\begingroup$ 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. $\endgroup$ – Jonathan Aug 10 '12 at 18:11

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