Is the first variation of a Jacobian determinant always zero?

Question

I'm trying to find the Euler-Lagrange equations for the functional $$F[\mathbf{u}]=\iint \det{(D\mathbf{u})} \, dx \, dy$$ Where $\mathbf{u}:\mathbb{R}^{2} \to \mathbb{R}^{2}$ and $\det{(D\mathbf{u})}$ is the determinant of the Jacobian of $\mathbf{u}$. In other words, if we write $\mathbf{u}=(u^{1},u^{2})$, then $$F[\mathbf{u}]=\iint (u^{1}_{x} u^{2}_{y}-u^{1}_{y} u^{2}_{x}) \, dx \, dy$$ where subscripts are partial derivatives.
I varied this by a functional $(\delta u^{1},\delta u^{2})$ and expanded the integrand. Then, as is standard, I discarded higher order terms (leaving four terms, all of the form $u^{1}_{x}\delta u^{2}_{y}$ for swaps of $1,2$ and $x,y$) and integrated by parts to get 'plain' $\delta u$s with no derivatives. But when I did this I ended up with $$\delta F=\iint\left[\delta u^{2}(u^{1}_{xy}-u^{1}_{xy})+\delta u^{1}(u_{xy}^{2}-u^{2}_{xy})\right] \, dx \, dy$$ Which is trivially $0$. Is this correct? Close to correct? I would have expected something more like $u^{1}_{xy}-u^{2}_{xy}$, is that correct?

Answer 1

This is correct: such Lagrangians are called null Lagrangians. For an unnecessarily detailed explanation, see this book.

For an easier way to think about it, what is the area of the region $A$ over which we integrate? It's obviously $$ \iint_A dx \, dy. \tag{1} $$ But we can write this, using Green's theorem, as $$ \int_{\partial A} \tfrac{1}{2}( x\, dy-y \, dx) = \frac{1}{2}\int_a^b x(t)y'(t)-x'(t)y(t)) \, dt, $$ thinking of the boundary as parametrised by $t \mapsto (x(t),y(t))$. Hence the area only depends on the boundary (shocking, I know).

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Now let's look at your integral. Let's do a change of variables, $u=u^1(x,y)$, $v=u^2(x,y)$. Then $$ du \, dv = \det{D\mathbf{u}} \, dx \, dy $$

But hang on, that's the whole integrand! If we set $f(x,y)=(u,v)$, then we have $$ \iint_B \det{D\mathbf{u}} \, dx \, dy = \iint_{f^{-1}(A)} du \, dv. $$ But we know from above that this integral calculates an area, so it only depends on surface terms. Therefore the Euler-Lagrange equations are automatically satisfied, as expected.