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UPDATE:

I was able to solve this problem using iterative integration by parts. However, I still cannot find how Green's first identity would apply here. Suppose I had a multiple integral over $p$-dimension, could I get the result faster than by integrating by parts $p$ times?


Let $x_1, x_2 \in \mathcal{X} \subset\mathbb{R}$, and $S_1, S_2 \in \mathbb{L}^2$. Consider the following maximization problem

$$ \max_{S_1,S_2} \int\limits_{\mathcal{X}} \int\limits_{\mathcal{X}} F(x_1,x_2,S_1(x_1),S_2(x_2))dG(x_1,x_2,S_1(x_1),S_2(x_2)) $$

where $F$ and $G$ are differentiable with respect to all their arguments. In my specific problem, $G$ is a multivariate cumulative distribution function of $(x_1,x_2)$, conditioned on the functions $(S_1,S_2)$.

My objective is to write the maximization problem in $S_1,S_2$ under the form

$$ \min_{S_1,S_2} \int\limits_{\mathcal{X}} \int\limits_{\mathcal{X}} H(x_1,x_2,S_1(x_1),S_2(x_2),S_1'(x_1),S_2'(x_2))dx_1dx_2, $$

from which I will be able to derive the system of Euler-Lagrange equations I am after:

$$ H_{S_{1}} - \frac{\partial H_{S_{1}'}}{\partial x_1} = 0 \\ H_{S_{2}} - \frac{\partial H_{S_{2}'}}{\partial x_2} = 0 \\ $$ So far, I have attempted to solve the problem using the same approach as for the unidimensional case. Let $x$ and $S$ be similarly defined. The unidimensional analog writes $$ \max_{S} \int\limits_{\mathcal{X}} F(x,S(x))dG(x,S(x)). $$ In order to solve this, I first integrate by parts to obtain the equivalent problem $$ \max_{S} F(x,S(x))G(x,S(x))|_{\mathcal{X}} - \int\limits_{\mathcal{X}} dF(x,S(x))G(x,S(x))dx, $$ where the first term is constant, and the second term can be rewritten to obtain $$ \min_{S} \int\limits_{\mathcal{X}} H(x,S(x),S'(x))dx, $$ a canonical equation from which I can derive an Euler-Lagrange equation.

In a similar fashion, my strategy to solve the multidimensional problem goes this way:

  1. Integrate by parts using Green's first identity
  2. Derive the Euler-Lagrange equation of the resulting variational problem

My main difficulty here lies in the use of Green's first identity. I am not familiar with this theory and thus not sure how to apply it to my problem. It seems to me that it is a standard context, since the double integral is square. However, I could not find simple step-by-step examples for using this identity to practical problems.


UPDATE:

I couldn't find how to make use of Green's identities or the Divergence theorem to do it, so I tried to it iteratively. Here's my attempt. It seems to work. Any comment? Is there a way to use Green's results to get the solution more rapidly?

I think this answer is still relevant in a general context, even though it does not use the desired results.


We want to rewrite

$$ \int\limits_{\mathcal{c}}^{d} \int\limits_{\mathcal{a}}^{b} F(x_1,x_2,S_1(x_1),S_2(x_2))dG(x_1,x_2,S_1(x_1),S_2(x_2)) $$

in the form stated at the end of the question. Let us drop the $S_1,S_2$ arguments for notational simplicity. Define

$$ G(x_1,x_2) = \int\limits_{\mathcal{c}}^{x_2} \int\limits_{\mathcal{a}}^{x_1} g(s,t)ds dt $$

and rewrite the integral of interest as

$$ \int\limits_{\mathcal{c}}^{d} \int\limits_{\mathcal{a}}^{b} F(x_1,x_2)g(x_1,x_2)dx_1 dx_2. $$

Let us focus on the inner integral and perform standard integration by parts to obtain

$$ \int\limits_{\mathcal{a}}^{b} F(x_1,x_2)g(x_1,x_2)dx_1 = F(b,x_2)\int\limits_{\mathcal{a}}^{b} g(s,x_2)ds - \int\limits_{\mathcal{a}}^{b} \frac{dF(x_1,x_2)}{dx_1}\int\limits_{\mathcal{a}}^{x_1} g(s,x_2)dsdx_1, $$ and substitute this expression back into the integral. We obtain $$ \int\limits_{\mathcal{c}}^{d} ( F(b,x_2)\int\limits_{\mathcal{a}}^{b} g(s,x_2)ds )dx_2 - \int\limits_{\mathcal{c}}^{d} (\int\limits_{\mathcal{a}}^{b} \frac{dF(x_1,x_2)}{dx_1}\int\limits_{\mathcal{a}}^{x_1} g(s,x_2)dsdx_1 ) dx_2, $$ and change the order of integration on the second term to obtain $$ \int\limits_{\mathcal{c}}^{d} ( F(b,x_2)\int\limits_{\mathcal{a}}^{b} g(s,x_2)ds )dx_2 - \int\limits_{\mathcal{a}}^{b} (\int\limits_{\mathcal{c}}^{d} \frac{dF(x_1,x_2)}{dx_1}\int\limits_{\mathcal{a}}^{x_1} g(s,x_2)dsdx_2 ) dx_1. $$ Let us integrate the first term by parts $$ \int\limits_{\mathcal{c}}^{d} ( F(b,x_2)\int\limits_{\mathcal{a}}^{b} g(s,x_2)ds )dx_2 = F(b,d)G(b,d) - \int\limits_{\mathcal{c}}^{d} \frac{dF(b,x_2)}{dx_2}G(b,x_2)dx_2, $$ then the inner integral in the second term $$ \int\limits_{\mathcal{c}}^{d} \frac{dF(x_1,x_2)}{dx_1}\int\limits_{\mathcal{a}}^{x_1} g(s,x_2)dsdx_2 = \frac{dF(x_1,d)}{dx_1}G(x_1,d)- \int\limits_{\mathcal{c}}^{d} \frac{d^2F(x_1,x_2)}{dx_1dx_2} G(x_1,x_2)dx_2. $$ Therefore, the functional of interest can be rewritten as the sum of four terms $$ F(b,d)G(b,d) - \int\limits_{\mathcal{c}}^{d} \frac{dF(b,x_2)}{dx_2}G(b,x_2)dx_2 - \int\limits_{\mathcal{a}}^{b}\frac{dF(x_1,d)}{dx_1}G(x_1,d)dx_1 + (\int\limits_{\mathcal{a}}^{b}\int\limits_{\mathcal{c}}^{d} \frac{d^2F(x_1,x_2)}{dx_1dx_2} G(x_1,x_2)dx_2 dx_1), $$ or equivalently $$ F(b,d)G(b,d) + (b-a+d-c) - \int\limits_{\mathcal{a}}^{b}\int\limits_{\mathcal{c}}^{d} (\frac{dF(b,x_2)}{dx_2}G(b,x_2) + \frac{dF(x_1,d)}{dx_1}G(x_1,d) - \frac{d^2F(x_1,x_2)}{dx_1dx_2} G(x_1,x_2))dx_2 dx_1. $$ Therefore, the maximization problem of interest is equivalent to the following minimization problem $$ \min_{S_1,S_2} \int\limits_{\mathcal{a}}^{b}\int\limits_{\mathcal{c}}^{d} (\frac{dF(b,x_2,S_1(b),S_2(x_2))}{dx_2}G(b,x_2,S_1(b),S_2(x_2)) + \frac{dF(x_1,d,S_1(x_1),S_2(d))}{dx_1}G(x_1,d,S_1(x_1),S_2(d)) - \frac{d^2F(x_1,x_2,S_1(x_1),S_2(x_2))}{dx_1dx_2} G(x_1,x_2,S_1(x_1),S_2(x_2)))dx_2 dx_1. $$ This is a variational problem of the form $$ \min_{S_1,S_2} \int\limits_{\mathcal{a}}^{b}\int\limits_{\mathcal{c}}^{d} H(x_1,x_2,S_1(x_1),S_2(x_2),S_1'(x_1),S_2'(x_2)))dx_1 dx_2. $$


UPDATE:

Since we have the solution, can be infer how Green's identity would apply to this problem?

Green's first identity states $$ \oint\limits_{C} \phi \langle \nabla \psi, d\vec a\rangle = \int\limits_{V} \phi \nabla^2 \psi + \langle \nabla \phi, \nabla \psi\rangle dV $$


We used integration by parts iteratively to transform

$$ \int\limits_{\mathcal{c}}^{d} \int\limits_{\mathcal{a}}^{b} F(x_1,x_2)dG(x_1,x_2) $$

into

$$ F(b,d)G(b,d) + (b-a+d-c) - \int\limits_{\mathcal{a}}^{b}\int\limits_{\mathcal{c}}^{d} (\frac{dF(b,x_2)}{dx_2}G(b,x_2) + \frac{dF(x_1,d)}{dx_1}G(x_1,d) - \frac{d^2F(x_1,x_2)}{dx_1dx_2} G(x_1,x_2))dx_2 dx_1. $$

Is there a correspondence between those three terms and the result I derived? Although it seems totally wrong, something like...

$$ \oint\limits_{C} \phi \langle \nabla \psi, d\vec a\rangle = F(b,d)G(b,d) + (b-a+d-c) \\ \int\limits_{V} \phi \nabla^2 \psi dV = \int\limits_{\mathcal{a}}^{b}\int\limits_{\mathcal{c}}^{d} \frac{d^2F(x_1,x_2)}{dx_1dx_2} G(x_1,x_2) dx_2 dx_1 \\ \int\limits_{V} \langle \nabla \phi, \nabla \psi\rangle dV = \int\limits_{\mathcal{a}}^{b}\int\limits_{\mathcal{c}}^{d} \frac{dF(x_1,d)}{dx_1}G(x_1,d) + \frac{dF(b,x_2)}{dx_2}G(b,x_2)dx_2 dx_1 $$

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