# Two Dimensional Hodge Decomposition

In deriving the equation governing propagation of wave in a wave guide, a specific form of Hodge decomposition theorem is used. The Hodge decomposition theorem applied in two dimensions allows us to write a smooth vector field $$\vec V$$ on a bounded region $$S$$ with boundary $$C$$, such that the component of the vector field tangential to boundary $$\vec V_{\parallel}|_C=0$$ vanishes, as follows $$\vec V=\nabla A+\hat z\times\nabla B+\vec X$$ where $$\hat z$$ is orthogonal to $$S$$, and $$A|_C=0$$, $$B|_C=(\hat n\cdot\nabla B)|_C=0$$ (Dirichlet and Neumann B.C. respectively). Given $$\nabla^2A=\nabla\cdot\vec V$$ and $$\nabla^2B=-\nabla\cdot(\hat z\times\vec V)$$ and define $$\vec X=\vec V-\nabla A-\hat z\times\nabla B$$ I wonder how to prove $$\vec X$$ is harmonic, in a sense that it satisfy the Laplace's equation $$\nabla^2\vec X=0$$? I argued

Since $$\nabla^2\vec X=\nabla(\nabla\cdot\vec X)-\nabla\times(\nabla\times\vec X)$$ we can show \begin{align}\nabla(\nabla\cdot \vec X)=&\nabla(\nabla\cdot\vec V-\nabla^2A-\nabla\cdot(\hat z\times\nabla B))\\=&-(\nabla B\cdot(\nabla\times\hat z)-\hat z\cdot(\nabla\times \nabla B ))\\=&0\end{align} to get the second line, the Poisson's equation is used, and the third line follows from $$\nabla\times\hat z=0$$ and the curl of gradient is $$0$$ then we have to show the following equation equals $$0$$. \begin{align}\nabla\times(\nabla\times\vec X)=&\nabla\times(\nabla\times(\vec V-\nabla A-\hat z\times\nabla B))\\=&\nabla\times(\nabla\times \vec V-\nabla\times \nabla A-(\hat z(\nabla\cdot\nabla B)-\nabla B(\nabla\cdot\hat z)+\nabla (B\cdot\nabla)\hat z-(\hat z\cdot\nabla)\nabla B))\end{align} At this stage, I don't know how to proceed anymore, except for the identity $$\nabla\times\nabla\times \vec V=\nabla(\nabla\cdot \vec V)-\nabla^2\vec V$$. Can someone help me to fill the missing steps and show $$\nabla\times(\nabla\times\vec X)$$is indeed zero?

We have that $$\nabla \times \hat z = 0$$, $$\nabla \cdot \hat z = 0$$, $$(Y\cdot \nabla)\hat z = 0$$ for any vector field $$Y$$.
Also, we can assume that $$(\hat z\cdot \nabla) B = 0$$, because we are only interested on the value of $$B$$ on $$S$$.
Using all this, we can compute : \begin{align} \nabla\times X &= \nabla \times V - \nabla\times (\nabla A) -\nabla\times (\hat z\times \nabla B) \\ &= \nabla \times V -\hat z \nabla^2B \\ &= \nabla\times V + \hat z\nabla \cdot (\hat z\times V) \\ &= \nabla\times V - \hat z\hat z \cdot( \nabla \times V) \end{align}
Then : \begin{align} \nabla \times (\nabla \times X) &= \nabla \times (\nabla \times V)- \nabla\times (\hat z\hat z\cdot(\nabla\times V))\\ &= \nabla \times (\nabla \times V) + \hat z\times \nabla (\hat z \cdot (\nabla\times V)) \\ &=\nabla \times (\nabla \times V) +\hat z\times (\hat z\times (\nabla \times (\nabla\times V))) \end{align} Since $$\nabla\times (\nabla\times V)$$ is orthogonal to $$\hat z$$, we hat : $$\hat z\times (\hat z\times (\nabla \times (\nabla\times V)) = -\nabla \times (\nabla\times V)$$ and : $$\nabla \times (\nabla \times V)= 0$$