Timeline for Adjoint operator of OU generator
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2022 at 8:50 | comment | added | Giorgio Metafune | My previous comment applies also to the degenerate case, but in this situation $A$ cannot be self-adjoint, this I meant in the final part. There is a nice paper by A. Lunardi and M. Geissert on the domain of OU operators in $L^2$ of the invariant measure in the degenerate case. I think there you find more basic information. | |
| Jan 31, 2022 at 3:14 | comment | added | Kung Yao | thank you. It seems that the degenerate case is quite interesting, too. I read that if $\ker(Q)$ does not contain any non-trivial subspace invariant under $B^T$ , see Corollary 12 here arxiv.org/pdf/1510.05936.pdf, that there exists still an invariant measure. This situation allow even for $Q$ to be degenerate. Do you know if one can understand the situation of adjoints in that case, too? Or is this condition precisely such that $Q_{\infty}$ is still invertible and the invariant measure is as in my question?-The author of that paper does not seem to specify the invariant measure. | |
| Jan 30, 2022 at 9:22 | comment | added | Giorgio Metafune | When $Q$ is non degenerate, through a linear change of variables one reduces to the case above. However one can use the same argument in the answer, modifying the form to $a(u,v)=\int Q\nabla u\nabla v$ which gives $A_1=tr(QD^2)-\frac 12 QQ_\infty^{-1}x \cdot \nabla $ so that $C=(Bx+\frac 12 Q Q_\infty^{-1} x)\cdot \nabla$. The $A$ is self adjoint iff $B=-\frac 12 QQ_\infty^{-1}$. Note that if $Q$ is degenerate this cannot happen. In fact the determinant of $B$ would be zero and $Q_\infty$ would not exist. | |
| Jan 30, 2022 at 2:32 | vote | accept | Kung Yao | ||
| Jan 30, 2022 at 2:32 | comment | added | Kung Yao | thanks a lot for this proper explanation, but just to be sure. What does this assumption $Q=I$ you make in the beginning mean? Is the adjoint otherwise still given by $A_1-C$ where $\Delta$ is replaced by $tr(QD^2)$ in $A_1$? | |
| Jan 30, 2022 at 0:58 | history | answered | Giorgio Metafune | CC BY-SA 4.0 |