In Galerkin FEM, we are dealing with two function spaces, the trial space \(X\) and the test space \(Y\). The test function \(v\in Y\) is directly applied to the two sides of the operator equation. If we take the Laplace problem as example, we’ll then apply the Green identity to obtain the variational form or weak form. The weak form contains an integration term \(\left( \nabla u,\nabla v \right)\), which is product of the differentials of \(v\) and \(u\). If \(u\) and \(v\) belong to \(H_0^1(\Omega)\), which means both of them as well as their first order partial derivatives are square integrable, the integration term is the \(L_2\)-inner product of the gradient of \(v\) and \(u\) in the Cartesian product space \(\left[ L_2(\Omega) \right]^d\), where \(d\) is the spatial dimension.

In Galerkin BEM, a boundary integral equation will be considered, which is equivalent to the original PDE. An integral operator, such as the single layer potential (SLP) integral operator \(V\), operates on the solution function \(u\) in the PDE

\[\begin{equation} (Vu)(x) = \int_{\Gamma} U^{*}(x,y) u(y) ds_y, \end{equation}\]

where \(U^{*}(x, y)\) is the fundamental solution.

The test function \(v\) will not be directly applied to the solution function \(u\) or its differential as in FEM, but to the result function returned from the integral operator. Let the domain space and range space of \(V\) be \(X\) and \(Z\). For example, \(V: H^{-1/2}(\Gamma) \rightarrow H^{1/2}(\Gamma)\). Obviously, the trial space is the domain space of the integral operator. The test space \(Y\) is not always the same as the domain space as in FEM. Actually, the test function should belong to the dual space \(Z'\) of the range space \(Z\) and is applied to a function in the range space via duality pairing between \(Z\) and \(Z'\). Therefore, the test function space for the integral operator \(V\) is \(Y=H^{-1/2}(\Gamma)\), which happens to be the same as the domain space \(X\). This explains why the discretized matrix associated with \(V\) is symmetric.

However, for the double layer integral operator \(K: H^{1/2}(\Gamma) \rightarrow H^{1/2}(\Gamma)\), the trial function space is \(H^{1/2}(\Gamma)\) and the test space is \(H^{-1/2}(\Gamma)\), which are different. Hence, the matrix for \(K\) is generally not a square matrix.

See also Difference between interpolation and projection in FEM.

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