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(This is a crossed post where physical considerations can be helpful.)

Let $\Omega\subset \mathbb{R}^2$ be some simply connected domain and consider the functional

\begin{align} V\left[u(x,y),v(x,y),w(x,y)\right]=\int_\Omega dxdy\:\left(u_x v_y-u_y v_x\right)w, \end{align}

We wish to find the functions that maximize $V$ constrained to

\begin{align} (u_x)^2+(v_x)^2+(w_x)^2&=1\\ (u_y)^2+(v_y)^2+(w_y)^2&=1. \end{align}

If we introduce the multipliers $\lambda_1$ and $\lambda_2$ for the constraints, we have the Lagrangian

\begin{align} \mathcal{L}=\left(u_x v_y-u_y v_x\right)w-\frac{\lambda_1}{2}\Big[(u_x)^2+(v_x)^2+(w_x)^2-1 \Big]-\frac{\lambda_2}{2}\Big[(u_y)^2+(v_y)^2+(w_y)^2-1 \Big] \end{align}

that leads to the equations

\begin{align} 0&=u_x v_y-u_y v_x+\left(\lambda_1 w_x \right)_x+\left(\lambda_2 w_y \right)_y\\ 0&=-\left(wv_y-\lambda_1 u_x\right)_x+\left(wv_x+\lambda_2 u_y\right)_y\\ 0&=\left(wu_y+\lambda_1 v_x\right)_x-\left(w u_x-\lambda_2 v_y\right)_y \end{align}

The question is, how to solve for the multipliers $\lambda_1$ and $\lambda_2$ ?

In the typical classroom examples the constraints do not depend on the derivatives of the dependent variables, thus the derivatives of the multipliers don't appear in the equations and one can solve for them algebraically. Here, however, that's not the case.

(We impose just the boundary condition $w\left(\partial \Omega\right)=0$, leaving $u$ and $v$ free. This should not be relevant for the question, which is only about how to consistently eliminate the multipliers.)

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The $\lambda_i$ in your system are constants, so $(\lambda_1 v_x)_x= \lambda_1 \partial_x v_x$.

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    $\begingroup$ The multipliers are constants only when the constraint is an integral constraint. Since here the constraint holds at every point they (the multipliers) depend on the coordinates as well. $\endgroup$ Jan 10, 2023 at 13:58
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    $\begingroup$ Ah. I did not appreciate that. You might look at a hw question where this occurs See q3 in hw3 courses.physics.illinois.edu/phys508/fa2022 A solution is avaiable there also, $\endgroup$
    – mike stone
    Jan 10, 2023 at 14:01
  • $\begingroup$ Thank you. It's a nice problem. The multiplier there was easily eliminated because it appeared inside a total derivative. My question is which boundary conditions should one impose on the multipliers (and why) when such a complete elimination is not possible. $\endgroup$ Jan 12, 2023 at 12:01

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