# The square root of Laplacian with nonconstant coefficent

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

## 2 Answers

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.

• If $c$ and $1/c$ are bounded, then $-c\Delta$ is conjugate to $-c^{1/2}\Delta(\cdot c^{1/2})$, which is self-adjoint and positive. So the question reduces to the square root of a positive operator. – Denis Serre May 15 '16 at 17:29
• My question arrises from Thermoacoustic tomography arising in brain imaging (with G. Uhlmann). Inverse Problems, 27(4):045004, 26, 2011. (<math.purdue.edu/~stefanov/publications/…) (page 9, 4.2). Here $c$ is smooth, strictly greater than zero and bounded They use a weighted $L^2$ with $c^{-2}$. By which it is formally positive and self-adjoint. – Martin May 17 '16 at 13:54
• Well, the definition of "square root" of an operator $A$ is: the unique positive definite operator $B$ such that $A=B^2$. This is certainly not satisfied fir $B=c\nabla$. – Delio Mugnolo May 17 '16 at 15:42
• Is this a comment to the second part of my answer? I did not speak of square root when talking of B – Piero D'Ancona May 17 '16 at 15:59

The central question in this area was Kato's conjecture.

From Wikipedia: Tosio Kato asked whether the square root of certain elliptic operators, defined via functional calculus, are analytic. The problem remained unresolved for nearly a half-century, until it was jointly solved in 2001 by Pascal Auscher, Steve Hofmann, Michael Lacey, Alan McIntosh, and Philippe Tchamitchian.

See their paper "The solution of the Kato square root problem for second order elliptic operators on ${\mathbb R}^n$". Annals of Mathematics 156 (2002), pp 633–654

• Hey Denis, first, thanks for you comment. My question is, is there any explicit computation possible? In order to get the squareroot, i need to know $(c(x)\Delta (\lambda +c(x) \Delta)^{-1})$, right ? – Martin May 13 '16 at 15:50