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2 votes
1 answer
247 views

Leibniz rule for square root of Laplacian

Let $(M,g)$ be a compact Riemannian manifold (e.g. $M=S^3$ the 3-sphere) and let $\Delta$ be the metric Laplacian on $M$. Then $\Delta$ has an $L^2(M)$ basis of eigenfunctions $\pi_m$, $$ \Delta \pi_m ...
onamoonlessnight's user avatar
2 votes
1 answer
555 views

Resolvent of the Laplacian as a pseudodifferential operator and its single layer potential

In M.Taylor's book "Partial differential equations II" it is shown that the fundamental solution $E(x,y)$ of the Laplacian equation gives rise to an elliptic pseudodifferential operator $S$ on the ...
Appliqué's user avatar
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3 votes
2 answers
579 views

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}...
Martin's user avatar
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