I am reading a paper and trying to understand a calculation and all of a sudden I bump into the following term:

$D^2_y p(t,y(t,x))(\partial_t y(t,x),y^\epsilon(t,x))$

where $p$ is a scalar field (pressure), $D_y^2p$ is the Hessian matrix of $p$ wrt the spacial variables, $y$ and $y^\epsilon$ are vector fields, and $\partial_t y$ is the partial derivative of $y$ wrt $t$.

Now, for understanding the calculations I am assuming that the brackets are notation for:

$\left(D^2_y p(t,y(t,x))\cdot\partial_t y(t,x) \right) \cdot y^\epsilon(t,x))$

This makes sense in many ways, specially since this term ought to be a scalar.

I think I get a confirmation further on in the paper where it is stated that

$D^2_y p (t,y) \left((v \cdot \nabla )y, y^\epsilon-y \right)=\sum_{ijk} \partial^2_{ij}p(t,y) \: v_k \: \partial_k y_i \: (y^\epsilon-y)_j$,

here $v$ is a (velocity) vector field.

My question is: has anybody seen something like this before? Maybe its a notation physicists use. I have never seen it before and I couldn't find it anywhere. So far I couldn't think of any other interpretation (nor I think any other is possible). However, I am having trouble following the author's calculations and maybe this could be the source of my problem.

By the way, the paper I am referring to is the following one: http://math1.unice.fr/~brenier/fichiers.ps.pageperso/cms-semigeo-xz-rigorous.pdf

Thank you very much in advance!

  • $\begingroup$ What paper? $\phantom{}$ $\endgroup$ – J. M. isn't a mathematician Apr 19 '12 at 15:09
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    $\begingroup$ Please include that link into your question. $\endgroup$ – J. M. isn't a mathematician Apr 19 '12 at 15:16
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    $\begingroup$ I think it may help to include the specific equation (as I believe you made a typo) (page 4): $$I_{3}=-\int_{D} [(\partial_{t} \nabla p^{\ast} )(t,y(t,x)) \cdot y^{\epsilon}(t,x) +(D_{y}^{2}p^{\ast})(t,y(t,x))(\partial_{t}y(t,x),y^{\epsilon}(t,x)) ]dx$$ Note that: $p$ is an implicit function defined $p=p(t,x)$ for the pressure field, $y=y(t,x)\in R^{d}$ is the "vector-valued forcing term", $D$ is a smooth bounded domain in $R^{d}$ $(d=2,3)$, and $\epsilon$ and $\alpha$ (both greater than $0$) are scaling factors. I believe $\partial$ and $\nabla$ have traditional definitions in this context. $\endgroup$ – 000 Apr 19 '12 at 16:07
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    $\begingroup$ There is also two other integral terms in $I_{3}$, which makes the full definition of $I_{3}$ read: $$I_{3}=-\int_{D} [(\partial_{t} \nabla p^{\ast} )(t,y(t,x)) \cdot y^{\epsilon} (t,x)+(D_{y}^{2}p^{\ast})(t,y(t,x))(\partial_{t}y(t,x),y^{\epsilon} (t,x)) ]dx -\int_{D}(\nabla p^{\ast})(t,y(t,x)) \cdot G(x,y^{\epsilon}(t,x)) dx+I_{5}$$ $$I_{5}=\int_{D} x \cdot (v^{\epsilon}(t,x) \cdot \nabla)y^{\epsilon}(t,x) dx$$ $G(x,y)$ is a given smooth vector-valued source term $D \times R^{d} \to R^{d}$ and $v=v(t,x)\in R^{d}$ is the velocity field, and $x \in D$. $\endgroup$ – 000 Apr 19 '12 at 16:14
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    $\begingroup$ I wasn't aware if it made any difference or not--I do not understand these equations. However, I provided the full thing in hopes of making things more easily referenced. $\endgroup$ – 000 Apr 19 '12 at 16:19

I believe the Hessian is being treated as a bilinear form; see for instance here, here and here.

If so, your interpretation in terms of products is correct (though it's slightly unusual to denote the product of the matrix $D^2_yp$ and the vector $\partial_t y$ with a dot).

  • $\begingroup$ Thank you. So it is exactly what I thought it was. I had never seen this notation used before. $\endgroup$ – chango Apr 20 '12 at 8:54
  • $\begingroup$ @mathscat: Neither had I (and I'm a physicist). You're welcome. $\endgroup$ – joriki Apr 20 '12 at 9:08

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