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