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Galerkin method

Method for solving continuous operator problems (such as differential equations)

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Through Mathematics, Numerical analysis and Differential equation

Often when referring to a Galerkin method, one also gives the name along with typical assumptions and approximation methods used:

Examples of Galerkin methods are:

Linear equation in a Hilbert space

Weak formulation of a linear equation

Let us introduce Galerkin's method with an abstract problem posed as a weak formulation on a Hilbert space V, namely,

find u∈V such that for all v∈V:a(u,v)=f(v).

Here, a(⋅,⋅) is a bilinear form (the exact requirements on a(⋅,⋅) will be specified later) and f is a bounded linear functional on V.

Galerkin dimension reduction

Choose a subspace Vn⊂V of dimension n and solve the projected problem:

Find un∈Vn such that for all vn∈Vn,a(un,vn)=f(vn).

We call this the Galerkin equation. Notice that the equation has remained unchanged and only the spaces have changed. Reducing the problem to a finite-dimensional vector subspace allows us to numerically compute un as a finite linear combination of the basis vectors in Vn.

Galerkin orthogonality

The key property of the Galerkin approach is that the error is orthogonal to the chosen subspaces. Since Vn⊂V, we can use vn as a test vector in the original equation. Subtracting the two, we get the Galerkin orthogonality relation for the error, ϵn=u−un which is the error between the solution of the original problem, u, and the solution of the Galerkin equation, un

a(ϵn,vn)=a(u,vn)−a(un,vn)=f(vn)−f(vn)=0.

Matrix form of Galerkin's equation

Since the aim of Galerkin's method is the production of a linear system of equations, we build its matrix form, which can be used to compute the solution algorithmically.

Let e1,e2,…,en be a basis for Vn. Then, it is sufficient to use these in turn for testing the Galerkin equation, i.e.: find un∈Vn such that

a(un,ei)=f(ei)i=1,…,n.

We expand un with respect to this basis, un=∑j=1nujej and insert it into the equation above, to obtain

a(∑j=1nujej,ei)=∑j=1nuja(ej,ei)=f(ei)i=1,…,n.

This previous equation is actually a linear system of equations Au=f, where

Aij=a(ej,ei),fi=f(ei).

Symmetry of the matrix

Due to the definition of the matrix entries, the matrix of the Galerkin equation is symmetric if and only if the bilinear form a(⋅,⋅) is symmetric.

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