I am still a newbie to $\Psi$DO-Operators. As far as i understood, one can easily compute the square root of the Laplace operator $\Delta$ by


$$(-\Delta)^{1/2} \ u=\mathcal{F}^{-1}(\|\xi\| \widehat{u}).$$


However if i want to compute the spare root of $(-c(x) \ \Delta)$ things get complicated (Lets assume $c(x)$ is sufficiently nice as positive, bounded, smooth etc.). I cannot use simple Fourier transform, since the non constant coefficient will lead to a convolution:


$$\mathcal{F}(-c(x) \Delta u(x)) = \widehat{c}(\xi) \* \|\xi\|^2 \widehat u (\xi).$$


Of course i can use symbol calculus for getting the symbol of $\sqrt{-c(x)\Delta}$, but in that case i will only get the operator modulo smooth error. 


Is there any method to get the operator exatly?


I appreciate your help
best
Martin



The square root of Laplacian with nonconstant coefficent
Note that your operator is not positive for the $L^2$ product. A better starting point might be $-Au=\nabla\cdot(c^2(x)\nabla u)$ which satisfies at least $(Au,u)\ge0$. That said, it depends how explicit you want your square root. For a selfadjoint operator $A$ with spectral measure $dE\_\lambda$ you have simply $A^{1/2}=\int\_0^\infty \lambda^{1/2}dE\_\lambda$, an abstract formula which sometimes is surprisingly effective in explicit computations since the spectral measure has some nice representations. 


Of course, you also have the cheap factorization $A=B^\*B$ where the operator $B=c(x)\nabla$ goes from $S$ to $S^n$, $S$ being the Schwartz space, but I guess this is not what you need.


