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I want to find the value of $\vec{p}$, $p_s$, $p_t$ each of which is a function of the form $f:\mathbb{R}^2 \to \mathbb{R}$ that maximize the following function :

$$\begin{align} \int_\mathbb{R^2} \{ p_s &- \alpha(x,y)|\vec{p}| - \beta(x,y)(p_s - C_s) \\ &- \gamma(x,y)(p_t - C_t) + \lambda(x,y)(\vec{\nabla}\cdot\vec{p} - p_s + p_t)\} \end{align} $$

Each of $\alpha, \beta, \gamma$ is constrained to be non negative, $\lambda$ is unconstrained. $C_s$ and $C_t$ are given functions of the form $f:\mathbb{R}^2 \to \mathbb{R}$.

Edit :

Also we want the sup to be finite and therefore can put constraints on $\alpha, \beta, \gamma, \lambda$ accordingly.

The problem is a convex optimization problem.

I am hoping to minimize it by using a derivative approach : minimize function for a given variable by finding a derivative with respect to the variable and equating it to zero. I just dont know how to do that here.

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    @amWhy thanks for editing, I am still learning to use latex properly.2011-06-04
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    I think the idea should be to find partial derivatives with respect to the variables and then solve.2011-06-04
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    Since $p_s$, $p_t$, $p_1$, and $p_2$ are real-valued functions, this is a calculus of variations problem. I have therefore added the "calculus-of-variations" tag.2011-06-04
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    Thanks for tagging it correctly2011-06-04
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    @AnkurVj: no problem; actually, you did a fine job with your formatting, given your learning LaTeX...I just fine-tuned it, but the problem was fully comprehensible and well presented as you posted it.2011-06-04
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    Is it true that you want $p_s$ and $p_t$ to be non-negative? Are there any other constraints on these functions?2011-06-04
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    @Jim the two variables are not restricted to be non negative. I don't think there is any other constraint on these functions.2011-06-04
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    I was wondering if i can find a partial derivative of the whole equation with respect to say $p_t$ and then equate it to zero and then find some relationships. I ask so because this is the general way to go about getting the maximum value of G (i.e. the expression) in the process of finding duals to given primals2011-06-04

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There is no maximum unless $\beta+\lambda\equiv1$ and $\gamma\equiv\lambda$. The terms containing $p_s$ add up to

$$\iint_{\mathbb R_2}p_s(1-\beta-\lambda)\mathrm dx\mathrm dy\;,$$

so you can make the integral arbitrarily large by choosing $p_s$ arbitrarily large (positive or negative depending on the sign of $1-\beta-\lambda$). Likewise, the terms containing $p_t$ add up to

$$\iint_{\mathbb R_2}p_t(\lambda-\gamma)\mathrm dx\mathrm dy\;.$$

This fixes $\beta$ and $\gamma$ in terms of $\lambda$, and then $p_s$ and $p_t$ no longer occur in the integral. We can drop the constant terms containing $\beta$ and $\gamma$, leaving us with

$$I=\iint_{\mathbb R^2}\left[- \alpha|\vec{p}|+ \lambda(\vec{\nabla}\cdot\vec{p})\right]\mathrm dx\mathrm dy\;.$$

Let's restrict ourselves to solutions $\vec{p}$ that decay sufficiently rapidly at infinity that we can integrate by parts and omit the boundary term:

$$I=\iint_{\mathbb R^2}\left[- \alpha|\vec{p}|- (\vec{\nabla}\lambda)\cdot\vec{p}\right]\mathrm dx\mathrm dy\;.$$

So we want to find $\vec{p}$ such that its magnitude (weighted by $\alpha$) is small but its component along $-\vec{\nabla}\lambda$ is large. This will work best if we choose $\vec{p}$ to always point along $-\vec{\nabla}\lambda$, i.e. $\vec{p}=-\mu\vec{\nabla}\lambda$, with as yet undetermined $\mu(x,y)$. With $\vec{g}:=\vec{\nabla}\lambda$, this gives

$$ \begin{eqnarray} I &=& \iint_{\mathbb R^2}\left[- \alpha|-\mu\vec{g}|- \vec{g}\cdot(-\mu\vec{g})\right]\mathrm dx\mathrm dy \\ &=& \iint_{\mathbb R^2}\left[- \alpha|\mu||\vec{g}|+\mu|\vec{g}|^2\right]\mathrm dx\mathrm dy \\ &=& \iint_{\mathbb R^2}|\vec{g}|\left[- \alpha|\mu|+\mu|\vec{g}|\right]\mathrm dx\mathrm dy \;. \end{eqnarray}$$

Thus, we must have $\alpha\ge|\vec{\nabla}\lambda|$ everywhere; else we could make the integral arbitrarily large by choosing $\vec{p}$ to vanish wherever that inequality holds and to make the integral arbitrarily large where it doesn't.

Given that inequality, the integral is forced to be non-positive, and thus it is maximized for $\vec{p}\equiv0$.

The first conclusion, $\alpha\ge|\vec{\nabla}\lambda|$, holds independent of the assumption about the behaviour at infinity, since we can choose $\vec{p}$ to have that behaviour if we want. The second conclusion, $\vec{p}\equiv0$, however, holds only under that assumption, since the boundary term might compensate a negative value of the integral. I'm not sure how to treat the general case, without decay at infinity.

[Edit:] Quite generally speaking, the parts of the integral that depend on $\vec{p}$ all scale linearly if you scale $\vec{p}$ by a positive scale factor, so there can't be any maximum other than $\vec{p}=0$ -- all you can do by imposing constraints on $\alpha$, $\beta$, $\gamma$ and $\lambda$ is to ensure that $\vec{p}=0$ is indeed a maximum.

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    This is what i landed up with too, making the above substitutions eliminates $p_s$ and $p_t$ from the equation. Now I just need to find $\vec{p}$ that maximizes the rest of the equation. Am I on the right track ? could you help me get a bit more further .. $ \vec{p}= p_1 i + p_2 j $2011-06-04
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    From the question, it seemed that $\beta$, $\gamma$ and $\lambda$ were given, so I don't understand what you mean by "making the above substitutions". If $\beta$, $\gamma$ and $\lambda$ are also unknown, then the problem is wildly underdetermined.2011-06-04
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    Basically, we are allowed to put restrictions on $\beta, \gamma, \lambda$ so that the equation has a finite sup. That is we can assume such conditions on these variables so that the sup is finite2011-06-04
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    @AnkurVj: That's a very important part of the problem that you should state in the question.2011-06-04
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    Okay, should I edit the question now to reflect the same ?2011-06-04
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    @AnkurVj: Also you didn't introduce $i$ and $j$.2011-06-04
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    i and j are the normal unit vectors in the x and y direction respectively. I do not know how to put a cap on i and j and dont know what to search on google for that. So basically the terms are $|\vec{p}|$ and $\vec{\nabla}\cdot\vec{p}$2011-06-04
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    @AnkurVj: Yes, it's a good idea to edit the question so that people don't have to read through all the comments to understand the question. Usually when there are already answers that are based on the unedited question, you should mark the edit visibly so it doesn't look like the answers were incorrect. In this case, since I know you're editing and will adjust my answer accordingly, there's no need to do that.2011-06-04
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    @AnkurVj: I was thinking that's what you might mean by $i$ and $j$, but that doesn't make any sense -- then each of $p_1i$ and $p_2 j$ would be a dot product, i.e. a scalar, and applying $\nabla$ to it would result in a vector. Perhaps the first sentence is in error, and $p_1$ and $p_2$ are scalars?2011-06-04
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    I have modified the statement to reflect the actual problem. Please tell me if it makes sense now.2011-06-04
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    @AnkurVj: You've got a bit of a mixup in the formatting on the first line. Also, there is again the problem that $\nabla$ applied to $|\vec{p}|$ would be a vector -- don't you mean $\nabla\cdot\vec{p}$?2011-06-04
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    Sorry yeah .. bad cut paste job from the first part of the equation2011-06-04
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    @AnkurVj: OK, I think the question makes sense now.2011-06-04
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    Joriki, Thanks for the complete precise answer. However, Having worked it out some more, I realize that the original equation should have had the term $\alpha(x,y)(|\vec{p}| - C)$ instead of just $\alpha(x,y)(|\vec{p}|)$. This means that the integral would be of the form $$I=\iint_{\mathbb R^2}\left[- \alpha(|\vec{p}|-C) - (\vec{\nabla}\lambda)\cdot\vec{p}\right]\mathrm dx\mathrm dy\;.$$ Now, can I still say that it is maximized when $\vec{p}=0$ ? I believe that it still holds. However what impact does it have on the values $\alpha$ is allowed to take?2011-06-05
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    Further, if we make the assumption that $\vec{p}$ is zero outside a rectangular region $\Omega$, does that simplify things ?2011-06-05