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Mar 31, 2019 at 0:11 comment added EagleToLearn Dear DCM, Thank you so much for your explanation!
Mar 30, 2019 at 10:44 comment added DCM I think the word `Hessian' might be enough to help you out here ;)
Mar 30, 2019 at 10:39 comment added DCM I believe that the operator spoken of in the second part is the `Hessian' (the integral disappears because it's part of the inner product). I think this is usually defined by requiring that $f''(u)\phi \psi = (\phi,(\mathrm{Hess}f)(u)\psi)$ for all $\phi$ and $\psi$, where $(.,.)$ is the inner product for the space in which you're working.
Mar 29, 2019 at 20:00 review First posts
Mar 29, 2019 at 20:32
Mar 29, 2019 at 19:59 history asked EagleToLearn CC BY-SA 4.0