Galerkin method

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

find ${\displaystyle u\in V}$ such that for all ${\displaystyle v\in V,a(u,v)=f(v)}$.

Here, ${\displaystyle a(\cdot ,\cdot )}$ is a bilinear form (the exact requirements on ${\displaystyle a(\cdot ,\cdot )}$ will be specified later) and ${\displaystyle f}$ is a bounded linear functional on ${\displaystyle V}$.

Choose a subspace ${\displaystyle V_{n}\subset V}$ of dimension n and solve the projected problem:

Find ${\displaystyle u_{n}\in V_{n}}$ such that for all ${\displaystyle v_{n}\in V_{n},a(u_{n},v_{n})=f(v_{n})}$.

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 ${\displaystyle u_{n}}$ as a finite linear combination of the basis vectors in ${\displaystyle V_{n}}$.

The key property of the Galerkin approach is that the error is orthogonal to the chosen subspaces. Since ${\displaystyle V_{n}\subset V}$, we can use ${\displaystyle v_{n}}$ as a test vector in the original equation. Subtracting the two, we get the Galerkin orthogonality relation for the error, ${\displaystyle \epsilon _{n}=u-u_{n}}$ which is the error between the solution of the original problem, ${\displaystyle u}$, and the solution of the Galerkin equation, ${\displaystyle u_{n}}$

${\displaystyle a(\epsilon _{n},v_{n})=a(u,v_{n})-a(u_{n},v_{n})=f(v_{n})-f(v_{n})=0.}$

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 ${\displaystyle e_{1},e_{2},\ldots ,e_{n}}$ be a basis for ${\displaystyle V_{n}}$. Then, it is sufficient to use these in turn for testing the Galerkin equation, i.e.: find ${\displaystyle u_{n}\in V_{n}}$ such that

${\displaystyle a(u_{n},e_{i})=f(e_{i})\quad i=1,\ldots ,n.}$

We expand ${\displaystyle u_{n}}$ with respect to this basis, ${\displaystyle u_{n}=\sum _{j=1}^{n}u_{j}e_{j}}$ and insert it into the equation above, to obtain

${\displaystyle a\left(\sum _{j=1}^{n}u_{j}e_{j},e_{i}\right)=\sum _{j=1}^{n}u_{j}a(e_{j},e_{i})=f(e_{i})\quad i=1,\ldots ,n.}$

This previous equation is actually a linear system of equations ${\displaystyle Au=f}$, where

${\displaystyle A_{ij}=a(e_{j},e_{i}),\quad f_{i}=f(e_{i}).}$

Due to the definition of the matrix entries, the matrix of the Galerkin equation is symmetric if and only if the bilinear form ${\displaystyle a(\cdot ,\cdot )}$ is symmetric.

Since ${\displaystyle V_{n}\subset V}$, boundedness and ellipticity of the bilinear form apply to ${\displaystyle V_{n}}$. Therefore, the well-posedness of the Galerkin problem is actually inherited from the well-posedness of the original problem.

The error ${\displaystyle u-u_{n}}$ between the original and the Galerkin solution admits the estimate

${\displaystyle \|u-u_{n}\|\leq {\frac {C}{c}}\inf _{v_{n}\in V_{n}}\|u-v_{n}\|.}$

This means, that up to the constant ${\displaystyle C/c}$, the Galerkin solution ${\displaystyle u_{n}}$ is as close to the original solution ${\displaystyle u}$ as any other vector in ${\displaystyle V_{n}}$. In particular, it will be sufficient to study approximation by spaces ${\displaystyle V_{n}}$, completely forgetting about the equation being solved.

Since the proof is very simple and the basic principle behind all Galerkin methods, we include it here: by ellipticity and boundedness of the bilinear form (inequalities) and Galerkin orthogonality (equals sign in the middle), we have for arbitrary ${\displaystyle v_{n}\in V_{n}}$:

${\displaystyle c\|u-u_{n}\|^{2}\leq a(u-u_{n},u-u_{n})=a(u-u_{n},u-v_{n})\leq C\|u-u_{n}\|\,\|u-v_{n}\|.}$

Dividing by ${\displaystyle c\|u-u_{n}\|}$ and taking the infimum over all possible ${\displaystyle v_{n}}$ yields the lemma.