Maximizing a function by finding derivative

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. – AnkurVijay Jun 4 '11 at 1:27
I think the idea should be to find partial derivatives with respect to the variables and then solve. – AnkurVijay Jun 4 '11 at 1:32
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. – Jim Belk Jun 4 '11 at 1:42
Thanks for tagging it correctly – AnkurVijay Jun 4 '11 at 1:43
@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. – amWhy Jun 4 '11 at 2:34

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 .. $\vec{p}= p_1 i + p_2 j$ – AnkurVijay Jun 4 '11 at 5:39
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. – joriki Jun 4 '11 at 5:43
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 finite – AnkurVijay Jun 4 '11 at 5:46
@AnkurVj: That's a very important part of the problem that you should state in the question. – joriki Jun 4 '11 at 5:49
Okay, should I edit the question now to reflect the same ? – AnkurVijay Jun 4 '11 at 5:50