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3 questions
2
votes
1
answer
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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 $\pi_m$, $$ \Delta \pi_m ...
2
votes
1
answer
555
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Resolvent of the Laplacian as a pseudodifferential operator and its single layer potential
In M.Taylor's book "Partial differential equations II" it is shown that the fundamental solution $E(x,y)$ of the Laplacian equation gives rise to an elliptic pseudodifferential operator $S$ on the ...
3
votes
2
answers
579
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The square root of Laplacian with nonconstant coefficent
I am still a newbie to $\Psi$DO-Operators. As far as i understood, one can easily compute the square root of the Laplace operator $\Delta$ by
$$(-\Delta)^{1/2} \ u=\mathcal{F}^{-1}(\|\xi\| \widehat{u}...