0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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 : $$\nabla \times (\nabla \times V)= 0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.