Let $\Omega \subseteq \mathbb{R}^3$ be a bounded, connected domain, and set $$\mathcal{V} = \{ \vec\phi = (\phi_1, \phi_2, \phi_3) \in C_c^\infty(\Omega): \nabla\cdot \vec\phi=0\}.$$ Denote $V$ to be the closure of $\mathcal{V}$ in $H^1(\Omega)$, and let $\mathbb{L}^2(\Omega)$ be the space of vector-valued functions from $\Omega$ to $\mathbb{R}^3$ with each component in $L^2(\Omega)$.
Since the curl operator $\nabla \times : V \to\mathbb{L}^2(\Omega)$ is continuous, there exists some adjoint operator $(\nabla \times)^*$ for which, given $\vec u, \vec v\in V$,
$$(\nabla \times \vec u, \vec v )_{\mathbb{L}^2} = (\vec u, (\nabla \times)^* \vec v)_{\mathbb{L}^2} .$$
What is an expression for the adjoint for the curl operator?
Best Answer
This is an exercise in integration by parts. Since the Levi-Civita symbol satisfies $\epsilon_{ijk} = -\epsilon_{kji}$,
$$\sum_i\int v_i(\nabla\times u)_i dx= \sum_{i,j,k}\int v_i \epsilon_{ijk}\partial_j u_k dx = \sum_{i,j,k}\int \epsilon_{kji} \partial_j v_i u_k dx = \sum_k \int u_k (\nabla\times v)_k dx$$
Which means the curl operator is self-adjoint.
You can also think of curl as left matrix multiplication by a skew-symmetric “matrix” of partial derivatives, which picks up one minus sign when transferring onto $v$, and differentiation is skew-adjoint (by integration by parts) which picks up another minus sign, hence making curl self-adjoint.