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I am learning about mass transportation theory and the Monge-Ampere equation, to transport a function $f$ toward $g$ by a change of variable $T$. In particular, in order to solve for : $$ \min \int c(x, T(x)) f(x) dx $$ under the constraint $ f = g\circ T~(\det(DT))$ where $\det(DT)$ is the jacobian of T (here, the determinant is supposed to be positive to remove absolute values), and $c(x,y)$ is a cost function, I found in a lecture that one introduces Lagrange multipliers $\lambda = \lambda(x)$ and solve for the extended functional:

$$ \min \int \left[ c(., T) f + \lambda~ g\circ T (\det DT) \right] $$ By computing the Euler Lagrange equation of the above equation, one get : $$ f~c_{yi} = D_i[\lambda g~(cof~DT)^{i,j}] -\lambda~(\det DT) g_{yi} $$ where $cof DT$ stands for the matrix of cofactors for the jacobian matrix $DT$.

I have a few basic questions, that I'd like to be answered assuming very little knowledge on my side :
- Why would the lagrange multipliers depend on $x$ ? Usually, when I have a set of equations to minimize with a set of equations as constraints, I have one $\lambda$ per constraint, and it doesn't depend on $x$
- I don't understand at all how one arrives to this Euler Lagrange equation. I mean, I know that in general, deriving a determinant with respect to a matrix gives a cofactor matrix... but nothing more that I can use here. Could someone add 5-6 steps in between ? what are those $y_i$ and $i,j$ indices ?? Please, treat me as a newbie :)

Thank you very much in advance !

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Could you please check your equations to make sure indices weren't dropped? I think there is a missing index $y$ in the first term on the right. I assume you meant $g$ to be $g_{yj}$.

But, for now, recognize that the Euler-Lagrange equation comes from varying with respect to $T$ the action you've written. So, do you see how to the the left hand side by varying the first term in the action? If so, are you familiar with varying the determinant of a matrix with respect to that matrix? Check out this answer from Physics.SE and see if that clarifies things.


See if that clarifies. If not, let me know where you're stuck.

Rewrite your equation as $$ (f~c_{yi} + \lambda~(\det DT) g_{yi}) - D_i[\lambda g_{yj}~(cof~DT)^{i,j}] = 0 $$ and recognize that $$ \frac{\partial L}{\partial T} = (f~c_{yi} + \lambda~(\det DT) g_{yi})\,, $$ and $$ \frac{\partial L}{\partial D_iT} = \lambda g_{yj}~(cof~DT)^{i,j}\,, $$ where $$ L = c(., T) f + \lambda~ g\circ T (\det DT)\,. $$ I assume the floating indices refer to the location and direction (?) of the variation.

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Thanks for your answer, but I guess you assume too much skills from me ;) The lecture really says $g$ and not $g_{yj}$ : it can be watched online at : video.ias.edu/gpde/otnepde (first part, at 29min 19sec). In general I know that $\frac{\partial \det T}{T_{i,j}} = (cof T)_{i,j}$ but that's actually the only part I understood in the formula. I don't even understand what is the index $y_i$, or why the lagrangian multiplier depends on $x$.... –  WhitAngl Feb 26 '12 at 7:20
In particular, while I know the above formula for the derivative of the determinant of a matrix with respect to the matrix, here we differentiate the determinant of the Jacobian matrix with respect to the diffeomorphism $T$ (if I understood well) and not with respect to the Jacobian matrix of $T$... (in addition to all my other misunderstandings) –  WhitAngl Feb 26 '12 at 18:23
Thanks for the clarifications. But could you also tell me why I would differentiate $L$ with respect to $D_i T$ rather than $T$ and what is meant by "direction" of variation ? What we are trying to find is $lim_{\epsilon\rightarrow 0} \frac{L(T+\epsilon U)-L(T)}{\epsilon}$ , isn't it ? I am not sure I understood how you arrived to these results... (I haven't time to check carefully though - and I'll likely award the bounty before I have time to fully understand since I'm rather busy now, but I'll re-read the answer more carefully later). Thanks for your efforts ! –  WhitAngl Feb 28 '12 at 19:42
When computing the variational derivative of a functional which depends both on a function's profile, $\phi(x)$ say, and the profile of its derivative, $\phi^\prime(x)$, the Euler-Lagrange equation involves both $\partial L/\partial\phi$ and $\partial L/\partial\phi^\prime(x)$. Read more about the derivation of the Euler-Lagrange equations to see how this comes about. I can recommend a good source or two if the book you're pulling this material from doesn't do it. –  josh Feb 29 '12 at 1:44
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