Petrov–Galerkin method

Let us consider an abstract problem posed as a weak formulation on a pair of Hilbert spaces ${\displaystyle V}$ and ${\displaystyle W}$, namely,

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

Here, ${\displaystyle a(\cdot ,\cdot )}$ is a bilinear form and ${\displaystyle f}$ is a bounded linear functional on ${\displaystyle W}$.

Choose subspaces ${\displaystyle V_{n}\subset V}$ of dimension n and ${\displaystyle W_{m}\subset W}$ of dimension m and solve the projected problem:

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

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

The key property of the Petrov-Galerkin approach is that the error is in some sense "orthogonal" to the chosen subspaces. Since ${\displaystyle W_{m}\subset W}$, we can use ${\displaystyle w_{m}}$ as a test vector in the original equation. Subtracting the two, we get the relation for the error, ${\displaystyle \epsilon _{n}=v-v_{n}}$ which is the error between the solution of the original problem, ${\displaystyle v}$, and the solution of the Galerkin equation, ${\displaystyle v_{n}}$, as follows

${\displaystyle a(\epsilon _{n},w_{m})=a(v,w_{m})-a(v_{n},w_{m})=f(w_{m})-f(w_{m})=0}$ for all ${\displaystyle w_{m}\in W_{m}}$.

Since the aim of the approximation is producing a linear system of equations, we build its matrix form, which can be used to compute the solution algorithmically.

Let ${\displaystyle v^{1},v^{2},\ldots ,v^{n}}$ be a basis for ${\displaystyle V_{n}}$ and ${\displaystyle w^{1},w^{2},\ldots ,w^{m}}$ be a basis for ${\displaystyle W_{m}}$. 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

${\displaystyle a(v_{n},w^{j})=f(w^{j})\quad j=1,\ldots ,m.}$

We expand ${\displaystyle v_{n}}$ with respect to the solution basis, ${\displaystyle v_{n}=\sum _{i=1}^{n}x^{i}v^{i}}$ and insert it into the equation above, to obtain

${\displaystyle a\left(\sum _{i=1}^{n}x^{i}v^{i},w^{j}\right)=\sum _{i=1}^{n}x^{i}a(v^{i},w^{j})=f(w^{j})\quad j=1,\ldots ,m.}$

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

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

Due to the definition of the matrix entries, the matrix ${\displaystyle A}$ is symmetric if ${\displaystyle V=W}$, the bilinear form ${\displaystyle a(\cdot ,\cdot )}$ is symmetric, ${\displaystyle n=m}$, ${\displaystyle V_{n}=W_{m}}$, and ${\displaystyle v^{i}=w^{j}}$ for all ${\displaystyle i=j=1,\ldots ,n=m.}$ In contrast to the case of Bubnov-Galerkin method, the system matrix ${\displaystyle A}$ is not even square, if ${\displaystyle n\neq m.}$