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Let $\Lambda\subseteq\mathbb R^2$ be open, $u\in C^1(\Lambda,\mathbb R^2)$ with $\nabla\cdot u=0$ and $$w:=\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}.$$

How can we show that $$\int_\Lambda w=0?\tag1$$

Since $\nabla\cdot u=0$, $$\int_\Lambda(u\cdot\nabla)\varphi=-\int_\Lambda(\nabla\cdot u)\varphi=0\;\;\;\text{for all }\varphi\in C_c^\infty(\Lambda)\tag2.$$ On the other hand, $$\int_\Lambda w\varphi=\int u_1\frac{\partial\varphi}{\partial x_2}-u_2\frac{\partial\varphi}{\partial x_1}\;\;\;\text{for all }\varphi\in C_c^\infty(\Lambda)\tag3.$$

I guess I've made a mistake at any point above, since the desired conclusion seems to require that $\int w\varphi=\int u_1\frac{\partial\varphi}{\partial x_1}-u_2\frac{\partial\varphi}{\partial x_2}$ instead (since this is equal to $\int_\Lambda(u\cdot\nabla)\varphi$).

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This is false. Take $u=(-x_2,x_1)$. Then $w=2$ everywhere.

EDIT: With periodic boundary conditions on the square $S$, by Green's Theorem $\int_S w = \int_{\partial S} u\cdot dr = 0$.

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  • $\begingroup$ I was on the wrong track then. This is what I actually want to know: math.stackexchange.com/q/3728477/47771 (I thought $Ł^2$ would denote $L^2_0:=\{f\in L^2:\int f=0\}$.) $\endgroup$
    – 0xbadf00d
    Jun 21, 2020 at 4:20
  • $\begingroup$ It does seem to denote that. Note that the authors specifically list $\int v(x)\,dx = 0$ as one of their conditions. Also, note that they're working on $T^2$, which is the $2$-torus, I presume. $\endgroup$ Jun 21, 2020 at 5:10
  • $\begingroup$ But does $\int v=0$ imply that $\int w=0$? They explicitly remark in that paragraph that. $v\in H^1$ with $\nabla\cdot v=0$ implies $v\in Ł^2$ ... And regarding the torus: If I got it right, they identify the $T^2$ with $[-\pi,\pi]^2$; I don't think that anything changes when we consider $\Lambda=(a_1,b_1)\times(a_2,b_2)$ instead or am I missing something? $\endgroup$
    – 0xbadf00d
    Jun 21, 2020 at 6:20
  • $\begingroup$ Periodicity of the functions. $\endgroup$ Jun 21, 2020 at 15:05
  • $\begingroup$ Periodicity in the sense that $f(x+2\pi)=f(x)$? $\endgroup$
    – 0xbadf00d
    Jun 21, 2020 at 15:13

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