From duality pairing to operator preconditioning
Contents
1 Duality pairing
In Galerkin BEM, the mass matrix \(\mathcal {M}\) for the identity operator \(I: H^{1/2}(\Gamma ) \rightarrow H^{1/2}(\Gamma )\) 1 is the duality pairing between trial basis functions \(\{ \varphi _j \}_{j=1}^n\) in \(V_h \subset H^{1/2}(\Gamma )\) and test basis functions \(\{ \psi _i \}_{i=1}^m\) in \(W_h \subset H^{-1/2}(\Gamma )\). To compute the matrix element \(\mathcal {M}_{ij}\), we directly compute the \(L_2(\Gamma )\) inner product of \(\varphi _j\) and \(\psi _i\).
The reason why we can do this is: \(H^{1/2}(\Gamma )\) is a subspace of \(L_2(\Gamma )\), so the trial basis function \(\varphi _j\) in the finite dimensional subspace \(V_h\) is also an \(L_2\) function. Meanwhile, we are using a finite dimensional polynomial space \(W_h\) to approximate \(H^{-1/2}(\Gamma )\), such as the subspace spanned by piecewise constant functions. Therefore, the test basis function \(\psi _i\) is also an \(L_2\) function and computing \(\langle \varphi _j,\psi _i \rangle = \int _{\Gamma } \varphi _j\psi _i \intd s_x\) is meaningful. From this we can see that even though the original Sobolev spaces adopted in the weak form are relatively large, in practice, we are building matrices and searching the solution in smaller spaces.
2 Gelfand triple
Because the duality pairing between \(H^{1/2}\) and \(H^{-1/2}\) inherits the duality pairing defined in \(L_2\), and \(H^{1/2}\) is dense in \(L_2\), according to (Brezis, page 136), \(L_2\) is self-dual or identified with itself and \(H^{-1/2}\) is a space larger than \(H^{1/2}\). There is the chain of embedded function spaces on the boundary \(\Gamma \): \begin{equation} H^{1/2}(\Gamma ) \subset L_2(\Gamma ) \simeq L_2^{*}(\Gamma ) \subset H^{-1/2}(\Gamma ). \end{equation} Similarly, in Galerkin FEM, function spaces are defined in the domain \(\Omega \) and we also have \begin{equation} H_0^1(\Omega ) \subset H^1(\Omega ) \subset L_2(\Omega ) \simeq L_2^{*}(\Omega ) \subset H^{-1}(\Omega ). \end{equation} The central space \(L_2\) in the above two chains is called the pivot space, which is wrapped by a dense subspace of \(L_2\) and its dual space. Such a “sandwich” structure like \(( H^{1/2}(\Gamma ), L_2(\Gamma ), H^{-1/2}(\Gamma ) )\) or \(( H_0^1(\Omega ), L_2(\Omega ), H^{-1}(\Omega ) )\) is called the Gelfand triple. In general, this triple is written as \begin{equation} \label {eq:gelfand-triple} V\subset H \simeq H^{*} \subset V^{*}. \end{equation} Even though the Riesz-Fréchet representation theorem tells us that for a Hilbert space we can identify it with its dual space, we cannot simultaneously identify \(V\) with \(V^{\ast }\) and \(H\) with \(H^{\ast }\). This is because when we define bounded linear functionals on the dense subspace \(V\) by restricting the domain of the linear functionals on \(H\), the duality pairing \(\langle \cdot ,\cdot \rangle _{V^{\ast },V}\) inherits the original duality pairing \(\langle \cdot ,\cdot \rangle _{H^{\ast },H}\), the latter of which further depends on the inner product in \(H\). So the inner product in \(V\) is not used, and \(V\) cannot be identified with \(V^{\ast }\) when \(H\) has been identified with \(H^{\ast }\). On the other hand, if we want to identify \(V\) with \(V^{\ast }\), the inner product in \(V\) should be used. Then the inner product in \(H\) as well as the identification between \(H\) and \(H^{\ast }\) should be abandoned.
Here is the important point: if we want to identify a Hilbert space \(H\) with its dual space \(H^{\ast }\), according to the Riesz-Fréchet representation theorem, duality pairing describes the way of applying a bounded linear functional in the dual space onto an element in the Hilbert space. This duality pairing is actually determined by the inner product assigned to \(H\).
3 Operator preconditioning
In BEM, we also have the above “sandwich” with \(L_2(\Gamma )\) as the pivot. Taking the Laplace problem as example. There are two possible choices of key integral operators to be discretized as stiff matrices:
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The single layer potential integral operator \(V: H^{-1/2}(\Gamma ) \rightarrow H^{1/2}(\Gamma )\).
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The hyper singular integral operator \(D: H^{1/2}(\Gamma ) \rightarrow H^{-1/2}(\Gamma )\).
The “sandwich” is \(H^{1/2}(\Gamma ) \rightarrow L_2(\Gamma ) \rightarrow H^{-1/2}(\Gamma )\).
If FEM is used, the key operator is the Laplace operator \(-\Delta : H_0^1(\Omega ) \rightarrow H^{-1}(\Omega )\) and the “sandwich” is \(H_0^1(\Omega ) \rightarrow L_2(\Omega ) \rightarrow H^{-1}(\Omega )\).
According to (Mardal and Winther), the condition for Krylov space methods to be applicable, such as CG, GMRES, etc., is that the operator should be symmetric and belong to the space of bounded linear operators from a Hilbert space \(X\) to itself, i.e. \(\mathcal {L}(X,X)\). In other words, the domain space and the range space of the operator should be the same. Obviously, neither \(V\) nor \(D\) satisfies this condition. They belong to a space \(\mathcal {L}(X,Y)\) where \(X \neq Y\) and usually \(X\) and \(Y\) are a pair of dual spaces.
Generally, let a partial differential equation (PDE) or boundary integral equation (BIE) be \(Au = f\), where the symmetric operator \(A: X \rightarrow Y\) is an isomorphism from \(X\) to \(Y\) and \(X \neq Y\). If \(A=V\), \(Y\) is smaller space than \(X\). If \(A=D\) or \(A=-\Delta \), \(Y\) is a larger space. To solve this equation using Krylov space methods, we can apply a left preconditioner \(B: Y \rightarrow X\) to both sides of the equation, which is another isomorphism from \(Y\) to \(X\). Then the composition \(BA\) is an isomorphism from \(X\) to itself. The preconditioned operator equation (strong form) is \begin{equation} BA u = Bf. \end{equation} Its weak form is \begin{equation} \left \langle BA u, v \right \rangle _X = \left \langle Bf, v \right \rangle _X \quad \forall v\in X. \end{equation} Because both \(BAu\) and the test function \(v\) are in \(X\), \(\left \langle \cdot ,\cdot \right \rangle _X\) is an inner product on \(X\) instead of duality pairing as in the weak form without the preconditioner \(\left \langle Au,v \right \rangle = \left \langle f, v \right \rangle \). To ensure that the new operator \(BA: X \rightarrow X\) is symmetric so that Krylov methods can be applied, the inner product \(\left \langle \cdot ,\cdot \right \rangle _X\) should not be arbitrarily given but be induced by the inverse operator \(B^{-1}\) as \(\langle B^{-1}\cdot ,\cdot \rangle _X\). This can be proved by simply checking the equality between \begin{equation} \left \langle BAu, v \right \rangle _X = \left \langle B^{-1}BAu, v \right \rangle = \left \langle Au, v \right \rangle \end{equation} and \begin{equation} \left \langle u, BAv \right \rangle _X = \left \langle u, B^{-1}BAv \right \rangle = \left \langle u, Av \right \rangle . \end{equation} Because \(A\) is symmetric, we have \(\left \langle Au,v \right \rangle = \left \langle u,Av \right \rangle \), hence \(\left \langle BAu,v \right \rangle _X = \left \langle u, BAv \right \rangle _{X}\). Therefore, \(BA\) is symmetric.
Now, a Krylov space method can be selected to solve the equation \(BAu = Bf\). The whole method is called the preconditioned Krylov space method. It can be understood that applying a preconditioner to the equation is equivalent to defining a norm \(\langle B^{-1}\cdot ,\cdot \rangle \) on \(X\). The error estimate of the preconditioner Krylov space method is also given in this norm.
References
Haim Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer. ISBN 978-0-387-70913-0 978-0-387-70914-7. doi: 10.1007/978-0-387-70914-7.
Kent-Andre Mardal and Ragnar Winther. Preconditioning discretizations of systems of partial differential equations. 18(1):1–40. ISSN 1099-1506. doi: 10.1002/nla.716.
1N.B. The bilinear form associated with this identity operator is \(b_I: H^{1/2}(\Gamma )\times H^{-1/2}(\Gamma ) \rightarrow \mathbb {R}\).