How to integrate this differential form on the boundary of the cube

Question

The setup. Assume $u = u_1+iu_2: \mathbb{R}^3 \to \mathbb{C}$ and we have the differential 1-forms $$ \star\xi=-x_2 dx_3 + x_3 dx_2 $$ and $$ u \times du = \sum_{i=1}^3 (u \times \partial_i u) dx_i = \sum_{i=1}^3 u_1 \partial_i u_2 - u_2 \partial_i u_1 dx_i. $$ Assume further that $\Omega_n^3=[-\pi n, \pi n]^3$ which is a 3D cube.

The problem. I want to integrate $(u \times du) \wedge \star \xi$ on the boundary of $\Omega_n^3$ to obtain $$ -\frac{1}{2} \int_{\partial \Omega_n^3} (u \times du)_\top \wedge (\star \xi)_\top = n\pi \sum_{i=2}^N \int_{C_i} u_2\partial_1 u_1 - u_1\partial_1 u_2 $$ where $$ C_2=[-\pi n, \pi n] \times \lbrace -\pi n, \pi n \rbrace \times [-\pi n, \pi n]$$ $$ C_3=[-\pi n, \pi n] \times [-\pi n, \pi n] \times \lbrace -\pi n, \pi n \rbrace. $$ This result is stated in this paper on page 55. For the definition of $\top$ see below.

What I tried. As I understand it from this paper on page 70, we have $$ (\star\xi)_\top = x_3 dx_2 $$ and $$ (u \times du)_\top = \sum_{i=1}^2 u_1 \partial_i u_2 - u_2 \partial_i u_1 dx_i.$$

So I compute $$ (u \times du)_\top \wedge (\star \xi)_\top = (u_1 \partial_1 u_2 - u_2 \partial_1 u_1) x_3 dx_1 \wedge dx_2$$ and so $$ -\frac{1}{2} \int_{\partial \Omega_n^3} (u \times du)_\top \wedge (\star \xi)_\top = \frac{1}{2} \int_{\partial \Omega_n^3} (u_2 \partial_1 u_1 - u_1 \partial_1 u_2) x_3 dx_1 \wedge dx_2 $$

The question. Can anyone help me to proceed from here? I don't have any clue on how to integrate a wedge-product.

EDIT: I'm terribly sorry, but I forgot to mention that $u$ is $\pi n$ periodic in every direction, i.e. it is the same on opposing faces of the torus.

Answer 1

Your expression for $(\star\xi)_\top$ is incorrect - this form is going to have different expressions on different faces of the cube. In particular I think you used the original coordinate system with the expressions given on page 70 on that paper, while you needed to adapt your coordinate system separately for each face. Intuitively, the $\top$ operation is just taking the component of the form tangent to the surface; so the correct expression should be

$$(\star\xi)_\top = \cases{ -x_2 dx^3 + x_3 dx^2 & on $C_1$ \\ -x_2 dx^3 & on $C_2$ \\ x_3 dx^2 & on $C_3$. \\ }$$

So just split the integral over $\partial\Omega$ into the six squares and use the correct expression on each.

As for how to integrate a differential form over the surface - you can find a general description at Wikipedia or in a textbook; but in this case it's easy since the surfaces are all oriented along the axes. If $C$ is a subset of the $x^i,x^j$ plane, then $$ \int_C f \ dx^i \wedge dx^j = - \int_C f dx^j \wedge dx^i$$ is just the double integral $\iint_C f dx^i dx^j$, while the integral of $f\ dx^k \wedge dx^l$ for $\{k,l\} \ne \{i,j\}$ vanishes. Along with linearity of the integral this is enough for what you want.

Try working it through - I calculated your final integral in my head and (almost) got one of the two terms in the desired answer, so hopefully with the correct expression for $\star \xi$ it'll work out. The one discrepancy seems to be the factor of $n\pi$ - I would expect this to come out as $(n \pi)^2$, the area of each square.

Answer 2

What you need is

1. pag. 43: understand notation in Lemma 4.1.
2. Remark 4.2.; in particular formula 4.15.

Formula 4.15 is interesting because for $u\in Y^N_n$ the r.h.s. vanishes (by definition of $Y^N_n$ $u=1$ on $\partial \Omega^N_n$).

However, in Prop. 4.1. you need to consider non trivial finite energy elements in $X^3_n:=H(\mathbb T^3_n)$, though: so the boundary integrals

$$-\frac{1}{2}\int_{\partial \Omega^N_n}(u\times du)\wedge d(\star\xi). $$

do not vanish, in general.

These integrals are computed as follows. In the 2D case (I focus on this for simplicity) you integrate along $\partial \Omega^2_n$ (can you visulalize it?). Using the mean value theorem on the integrand (along the $y$-direction in our case), you reduce the integral to

$$-\frac{1}{2}2n\pi\int_{[0,\pi)\times\{\alpha\} }(u\times du)\wedge d(\star\xi),~~(*)$$

with $\partial \Omega^2_n\ni\alpha:=(-n\pi,\alpha_2)$. Please note that this is the notation used in Lemma 4.1..

The above integral is equal to (4.30) using the formula at pag.43 under (4.3) for the integrand and the explicit choice $\alpha=(-n\pi,n\pi)$.

The integral for the case $N=3$ is completely similar.