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I am trying to give meaning to the notion of second differential of total variation.

For sufficiently regular $u:\Omega \subset \mathbb{R}^2 \to \mathbb{R}$ let the total variation be given by $$TV(u)=\int_{\Omega} |Du| dx.$$ Assuming that $u$ is sufficiently smooth and $|Du|\neq 0$ everywhere on $\Omega$ we have $$ \begin{split} \partial TV(u)(v)& :=\frac{d}{d \epsilon} TV(u+\epsilon v)|_{\epsilon=0} \\ & =\int_{\Omega} Dv \cdot \frac{Du}{|Du|} dx =-\int_\Omega \text{div}\left(\frac{Du}{|Du|}\right) v~ dx, ~~ \forall v \in C^\infty_c(\Omega). \end{split}$$ Analogously $$ \begin{split} \partial^2 TV(u)(v,w) &:=\frac{d}{d \epsilon} \partial TV(u+\epsilon w)(v) \\ & =\int_\Omega \frac{((Du)^\perp \cdot Dv) ((Du)^\perp \cdot Dw) }{|Du|^3} dx,\qquad \forall v,w \in C^\infty_c(\Omega), \end{split} $$ where $(Du)^\perp=(\partial u/\partial y, - \partial u/\partial x)$.

I would like to write the above integral in terms of $v,w$, without using their derivatives, something like it was done for the first differential. Is this possible, at least for $\Omega=[0,1]^2$? I am in particular interested in the case $v=w$.

Furthermore, does anyone know any results in direction of understanding the second differential of total variation?

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    $\begingroup$ There is something in "Measure theory and fine properties of functions" by Evans and Gariepy. if I remember correctly. $\endgroup$
    – Dirk
    Commented Jun 9, 2022 at 17:59
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    $\begingroup$ And what is wrong with you expression? It should be bilinear in (v,w) and it is. $\endgroup$
    – Dirk
    Commented Jun 9, 2022 at 18:00
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    $\begingroup$ That's hopeless. Even if you are in 1D and your functional is $\int (f')^2$, then the first variation is $2\int f'u'=-2\int f''u$ but the second one is $\int u'v'$ and you cannot get rid of derivatives on both $u$ and $v$ no matter how hard you try: you can only kill one at the expense of increasing the order of the other one. $\endgroup$
    – fedja
    Commented Jun 9, 2022 at 19:40
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    $\begingroup$ @MarkoRajkovic You understand correctly, but note that I changed the functional to something non-linear to get an interesting 1D example (the absolute value of the gradient in 2D and up is also non-linear and that's where all the trouble comes from). The point was that you will have no more luck getting rid of the derivatives in your formula than in the much simpler $\int u'v'$ 1D formula and there is no way to do it with the latter. $\endgroup$
    – fedja
    Commented Jun 9, 2022 at 21:16
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    $\begingroup$ I was wrong - Evans, Gariepy do not treat the differential of the TV seminorm (but kind of second derivatives of BV functions). Also the subdifferential is a bit more complicated - you can find a thorough discussion in the book "Mathematical Image Processing" (Secion 6.3.3) by Kristian Bredies and myself. We do not treat a "second differential" though. $\endgroup$
    – Dirk
    Commented Jun 10, 2022 at 7:12

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