# curl identity on $\mathcal{D}'\left(\mathbb{R}^2\right)$ for $v\in H^1\left(\mathbb{R}^2\right)^2$

I need to show that if $$u\in H^1\left(\mathbb{R}^2\right)^2$$ such that $$\text{div }u=0$$, then if $$w=\text{curl }u=\partial_1 u_2-\partial_2 u_1$$ in $$\mathcal{D}'\left(\mathbb{R}^2\right)$$ (space of distributions), we have $$\text{curl }\left(\left(u\cdot\nabla\right)u\right)=u\cdot \nabla w$$ where $$\left(u\cdot\nabla\right)u=\left(\sum_j u_j\partial_j u_i\right)_i$$ I have shown this holds if $$u\in\mathcal{C}^2\left(\mathbb{R}^2\right)^2$$, how can I pass it to the Sobolev space ? I thought about using density of smooth functions with compact support, but I do not know how this works in details.

I cannot pass all the derivatives to the test functions because I have a dot product.

• What does $u\cdot w$ mean? Just multiplication? Dec 1, 2020 at 9:11
• @JackyChong It is a scalar product. Dec 1, 2020 at 13:55
• It might be a dumb question, but isn't curl is a scalar in the plane? Dec 1, 2020 at 20:43
• @JackyChong Sorry ! There is obviously a $\nabla$ operator missing in front of $w$... Fixed on the original post. Dec 2, 2020 at 9:45
• i can't make sense of $u\cdot \nabla w$, since $\nabla w\in H^{-1}$...? Dec 10, 2020 at 4:22

If $$v\in H^{1}\left(\mathbb{R}^{2}\right)^{2}$$, let $$\varphi\in\mathcal{D}\left(\mathbb{R}^{2}\right)$$.
On one side, we have $$\begin{equation*} \int_{\mathbb{R}^{2}}\left(v\cdot \omega\right)\varphi=\int_{\mathbb{R}^{2}}\text{div}\left(v w\right)\varphi=-\int_{\mathbb{R}^{2}}wv\cdot\nabla\varphi =-\int_{\mathbb{R}^{2}}\text{curl}\left(v\right)v\cdot\nabla\varphi \end{equation*}$$ as $$\text{div}(v)=0$$. and on the other side $$\begin{equation*} \int_{\mathbb{R}^{2}}\text{curl}\left(\left(v\cdot\nabla\right)v\right)\varphi=\int_{\mathbb{R}^{2}}\left(v\cdot\nabla\right)v\wedge\nabla\varphi=\int_{\mathbb{R}^{2}}\left(v\cdot\nabla\right)v\cdot\left(\nabla\varphi\right)^{\perp} \end{equation*}$$ where we noted $$\begin{equation*} \left(\nabla\varphi\right)^{\perp}= \begin{pmatrix} \partial_{2}\varphi\\ -\partial_{1}\varphi \end{pmatrix} \end{equation*}$$ As such
\begin{align*} \int_{\mathbb{R}^{2}}\left(v\cdot\nabla\right)v\cdot\left(\nabla\varphi\right)^{\perp}+\text{curl}\left(v\right)v\cdot\nabla\varphi &=\int_{\mathbb{R}^{2}}\left[-v_{1}\partial_{1}v_{2}-v_{2}\partial_{2}v_{2}+v_{1}\partial_{1}v_{2}-v_{1}\partial_{2}v_{1}\right]\partial_{1}\varphi\\ &~~~~~~~~+\left[v_{1}\partial_{1}v_{1}+v_{2}\partial_{2}v_{1}+v_{2}\partial_{1}v_{2}-v_{2}\partial_{2}v_{1}\right]\partial_{2}\varphi\\ &=\int_{\mathbb{R}^{2}}-\left[v_{2}\partial_{2}v_{2}+v_{1}\partial_{2}v_{1}\right]\partial_{1}\varphi\\ &~~~~~~~~+\left[v_{1}\partial_{1}v_{1}+v_{2}\partial_{1}v_{2}\right]\partial_{2}\varphi\\ &=\int_{\mathbb{R}^{2}}\left[\frac{v_{2}^{2}}{2}+\frac{v_{1}^{2}}{2}\right]\partial_{2}\partial_{1}\varphi-\left[\frac{v_{1}^{2}}{2}+\frac{v_{2}^{2}}{2}\right]\partial_{1}\partial_{2}\varphi\\ &=0 \end{align*} Everything is well defined as $$v_{i}\in H^{1}\left(\mathbb{R}^{2}\right)$$ for $$i=1,2$$ (use Cauchy-Schwarz inequality and the fact that $$\partial_{i}\varphi$$ is bounded for the first two lines, or the fact that $$\partial_{1}\partial_{2}\varphi$$ is bounded for the third).