For mixed boundary value Laplace problem with \(\Gamma=\Gamma_{\rm D}\cup\Gamma_{\rm N}\), involved boundary integral operators are \(I, V, K, K', D\). The corresponding bilinear forms are \(b_I, b_V, b_K, b_{K'}, b_D\). The matrices for their discretization are \(\mathscr{I}, \mathscr{V}, \mathscr{K}, \mathscr{K'}, \mathscr{D}\). Then the system equation to be solved is as below.

\[\begin{equation} \begin{pmatrix} -\mathscr{V} & \mathscr{K} \\ \mathscr{K}' & \mathscr{D} \end{pmatrix} \begin{pmatrix} t \big\vert_{\Gamma_{\rm D}} \\ u \big\vert_{\Gamma_{\rm N}} \end{pmatrix}= \begin{pmatrix} -\frac{1}{2}\mathscr{I} - \mathscr{K} & \mathscr{V} \\ \mathscr{-D} & \frac{1}{2}\mathscr{I} - \mathscr{K}' \end{pmatrix} \begin{pmatrix} g_D \\ g_N \end{pmatrix} \end{equation}\]

The bilinear forms related to the left hand side matrix blocks are:

\[\begin{equation} \begin{aligned} b_V: H^{-\frac{1}{2}+s}(\Gamma_{\rm D}) \times H^{-\frac{1}{2}+s}(\Gamma_{\rm D}) \rightarrow \mathbb{C}\\ b_K: H^{\frac{1}{2}+s}(\Gamma_{\rm N}) \times H^{-\frac{1}{2}+s}(\Gamma_{\rm D}) \rightarrow \mathbb{C}\\ b_{K'}: H^{-\frac{1}{2}+s}(\Gamma_{\rm D}) \times H^{\frac{1}{2}+s}(\Gamma_{\rm N}) \rightarrow \mathbb{C}\\ b_D: H^{\frac{1}{2}+s}(\Gamma_{\rm N}) \times H^{\frac{1}{2}+s}(\Gamma_{\rm N}) \rightarrow \mathbb{C} \end{aligned} \end{equation}\]

For the first row of the equation, the test function space, corresponding to the rows, is \(H^{-\frac{1}{2}+s}(\Gamma_{\rm D})\), which should be the same as the test function space for the right hand side. For the second row of the equation, the test function space is \(H^{\frac{1}{2}+s}(\Gamma_{\rm N})\). The matrices \(\mathscr{V}\) and \(\mathscr{D}\) corresponding to \(b_V\) and \(b_D\) respectively are symmetric. The matrix symmetry requires both the function space and the spatial domain to be the same for the two components in a bilinear form.

On the right hand side, the two mass matrices \(\mathscr{I}\) in the block matrix are actually different. We call them \(\mathscr{I}_1\) and \(\mathscr{I}_2\) from now on. The bilinear forms related to the matrix blocks on the right hand side are:

\[\begin{equation} \begin{aligned} b_{I_1}:& H^{\frac{1}{2}+s}(\Gamma_{\rm D})\times H^{-\frac{1}{2}+s}(\Gamma_{\rm D}) \rightarrow \mathbb{C} \\ b_K:& H^{\frac{1}{2}+s}(\Gamma_{\rm D})\times H^{-\frac{1}{2}+s}(\Gamma_{\rm D}) \rightarrow \mathbb{C} \\ b_V:& H^{-\frac{1}{2}+s}(\Gamma_{\rm N}) \times H^{-\frac{1}{2}+s}(\Gamma_{\rm D}) \rightarrow \mathbb{C}\\ b_D:& H^{\frac{1}{2}+s}(\Gamma_{\rm D}) \times H^{\frac{1}{2}+s}(\Gamma_{\rm N}) \rightarrow \mathbb{C}\\ b_{I_2}:& H^{-\frac{1}{2}+s}(\Gamma_{\rm N})\times H^{\frac{1}{2}+s}(\Gamma_{\rm N}) \rightarrow \mathbb{C} \\ b_{K'}:& H^{-\frac{1}{2}+s}(\Gamma_{\rm N}) \times H^{\frac{1}{2}+s}(\Gamma_{\rm N}) \rightarrow \mathbb{C} \end{aligned} \end{equation}\]

We can see \(\mathscr{I}_1\) and \(\mathscr{I}_2\) are not identity matrices. This further strengthens our understanding about the fact that they are mass matrices instead of identity matrices. Meanwhile, \(\mathscr{K}\) on the right hand side is different from \(\mathscr{K}\) on the left hand side. This also holds for \(\mathscr{K}'\), \(\mathscr{V}\) and \(\mathscr{D}\). Moreover, \(\mathscr{V}\) and \(\mathscr{D}\) on the right hand side are not symmetric matrices.