# If $\nabla\cdot u=0$ and $w=\operatorname{curl}u$, then $\int w=0$

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 $$\int_\Lambda(u\cdot\nabla)\varphi$$).

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 $$\int_S w = \int_{\partial S} u\cdot dr = 0$$.
• 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 $L^2_0:=\{f\in L^2:\int f=0\}$.) Jun 21, 2020 at 4:20
• 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. Jun 21, 2020 at 5:10
• 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? Jun 21, 2020 at 6:20
• Periodicity in the sense that $f(x+2\pi)=f(x)$? Jun 21, 2020 at 15:13