Please let me know if this is not the appropriate site for this question. I found questions regarding EFIE/MFIE/CFIE on this site, so I thought my question might fit.

I am studying the paper by [Putnam & Medgyesi on the CFIE method for BoR targets with junctions][1]. In Section V (second paragraph) of the paper, the authors claim that by equating coefficients associated with half-triangle functions at the junction leads to a cancellation of boundary terms that one would get from integration-by-parts in the computation of the EFIE $L$ operator (see eqn. (5)) when the testing surface is open. 

My issue is that I can't seem to reproduce this claim. I outline my approach below. Let us take eqn. (25) from the paper with $\beta = 0$. This formula is for the geometry in Fig (2). If I understand the discussion in Section V correctly, the boundary terms that arise in obtaining formulas for $\mathcal{L} (S_1^c , S_1^c ; R_1)$ and $\mathcal{L} (S_1^c , S_1^d ; R_1)$ should cancel if the expansion coefficients in the currents associated with the half-triangle functions are equated. After discretizing the generating curve of the BoR into $N_1 > 1$ straight line segments $[t_{1k}^- , t_{1 , k+1}^-]$ (the subscript $1$ here indicates we are on the "outer" surface of the BoR) and parametrizing the surface by arclength along the generating curve $t$ and the polar angle $\varphi$, we expand the currents $\mathbf{J}^c_1$ according to 

$$ \mathbf{J}_1^c \simeq \sum_{n \in \mathbb{Z}} \sum_{k = 1}^{N_1^c} \left( c_{1nk}^t \mathbf{J}_{1nk}^t + c_{1nk}^{\varphi} \mathbf{J}_{1nk}^{\varphi} \right) , $$

and likewise for $\mathbf{J}^d_1$. Above, $\mathbf{J}_{1nk}^t (t , \varphi)$ and $\mathbf{J}_{1nk}^{\varphi} (t , \varphi)$ are $\mathbf{u}_t$-directed and $\mathbf{u}_{\varphi}$-directed expansion functions which are both harmonic in $\varphi$ and use triangle functions in the $t$ variable, and $N_1^c$ is the number of segments making up the conducting part of the BoR outer surface (likewise, $N_1^d$). Above, $\mathbf{u}_t$ is the unit tangent vector along the BoR generating curve and $\mathbf{u}_{\varphi}$ is the usual polar unit vector. Equating the coefficients associated with half-triangle functions at the junctions means to me that $c_{1 n , N_1^c}^t = d_{1 n , 1}^t$ and $c_{1 n , N_1^c}^{\varphi} = d_{1 n , 1}^{\varphi}$. (This assumes the generating curve for the conducting part of the surface ends at the junction, and the dielectric part begins at the junction.)

The general formula for a matrix element is

$$ \mathcal{L}^{\alpha \beta}_{\ell k} (S_1^c, S_1^d ; R) = i k \eta \int_{S_1^c} ( \mathbf{J}_{1 n \ell}^{\alpha} )^* (\mathbf{r}) \cdot \int_{S_1^d} \{ \mathbf{J}_{1nk}^{\beta} (\mathbf{r}') G (\mathbf{r} - \mathbf{r}') + k^{-2} (\nabla G) (\mathbf{r} - \mathbf{r}') \mathrm{div}_S~{\mathbf{J}_{1nk}^{\beta}} (\mathbf{r}') \} d S' d S , $$

where $G (\mathbf{r}) = e^{- i k |\mathbf{r}|} / (4 \pi |\mathbf{r}|)$, $\mathrm{div}_S$ is the surface divergence, and $\alpha , \beta \in \{ t , \varphi \}$. In the second term in the expression above, we integrate-by-parts to put the gradient on $G$ onto the testing function $(\mathbf{J}_{i n \ell}^{\alpha} )^* (\mathbf{r})$. Since $S_1^c$ is open, this will produce the boundary term

$$ \oint_{\partial S_1^c} [ ( \mathbf{J}_{1 n \ell}^{\alpha} )^* \cdot \mathbf{u}_t ] (\mathbf{r}) \left( \int_{S_1^d} G (\mathbf{r} - \mathbf{r}') \mathrm{div}_S~{\mathbf{J}_{1nk}^{\beta}} (\mathbf{r}') d S' \right) d s .$$

This boundary term is nonzero only when $\alpha = t$ and $\ell = N_1^c$. Based on the above formula, my understanding is that the authors are claiming the term 

$$ \sum_{\beta \in \{t, \varphi\}} \sum_{k = 1}^{N_1^c} \oint_{\partial S_1^c} ( J_{1 n , N_1^c}^t )^* (\mathbf{r}) \left\lbrace c^{\beta}_{1nk} \left( \int_{S_1^c} G (\mathbf{r} - \mathbf{r}') \mathrm{div}_S~{\mathbf{J}_{1nk}^{\beta} (\mathbf{r}')} d S' \right) + d^{\beta}_{1nk} \left( \int_{S_1^d} G (\mathbf{r} - \mathbf{r}') \mathrm{div}_S~{\mathbf{J}_{1nk}^{\beta} (\mathbf{r}')} d S' \right) \right\rbrace d s $$

that would appear in equation (25) of the paper vanishes because $c_{1 n , N_1^c}^t = d_{1 n , 1}^t$ and $c_{1 n , N_1^c}^{\varphi} = d_{1 n , 1}^{\varphi}$. I am somewhat struggling to see this. The authors claim this is true because effectively $S_1^c \cup S_1^d$ form a closed surface. This suggests to me to try integrating-by-parts on the term 

$$ \int_{S_1^c} G (\mathbf{r} - \mathbf{r}') \mathrm{div}_S~{\mathbf{J}_{1nk}^{\beta} (\mathbf{r}')} d S' = \oint_{\partial S_1^c} G (\mathbf{r} - \mathbf{r}') [ \mathbf{J}_{1 n k}^{\beta} \cdot \mathbf{u}_t ] (\mathbf{r}') d S' - \int_{S_1^c}  \nabla' G (\mathbf{r} - \mathbf{r}') \cdot \mathbf{J}_{1nk}^{\beta} (\mathbf{r}') d S' . $$

I can see how the boundary term for $\beta = t$ above would cancel out with a similar term in the similar formula for $S_1^d$ because $c_{1 n , N_1^c}^t = d_{1 n , 1}^t$. However, it is the lagging term 

$$ \int_{S_1^c}  \nabla' G (\mathbf{r} - \mathbf{r}') \cdot \mathbf{J}_{1nk}^{\beta} (\mathbf{r}') d S' $$

that doesn't seem to disappear. This is making me quite confused about the authors' claim. 

My question: Why do these boundary terms cancel? What am I missing? Is there anyone on this site who has worked on Method of Moments CEM codes or the EFIE and can shed light on the junction problem in general?


  [1]: https://ieeexplore.ieee.org/document/81498