$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\Aex}{{A_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\X}{\boldsymbol{X}}
\renewcommand{\u}{\boldsymbol{u}}
\renewcommand{\v}{\boldsymbol{v}}
\newcommand{\e}{\boldsymbol{e}}
\newcommand{\f}{\boldsymbol{f}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\Iz}{\mathcal{I}_z}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\Ifb}{{I_b}} % for FEM
\newcommand{\sequencei}[1]{\left\{ {#1}_i \right\}_{i\in\If}}
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\psib}{\boldsymbol{\psi}}
\newcommand{\sinL}[1]{\sin\left((#1+1)\pi\frac{x}{L}\right)}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\Xno}[1]{X_{(#1)}}
\newcommand{\xdno}[1]{\boldsymbol{x}_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
The projection (or Galerkin) method
- Last slide: \( \min E \) is equivalent with \( (\e, \psib_0)=0 \)
- \( (\e, \psib_0)=0 \) implies \( (\e, \v)=0 \) for any \( \v\in V \)
- That is: instead of using the least-squares principle, we can
require that \( \e \) is orthogonal to any \( \v\in V \)
(visually clear, but can easily be computed too)
- Precise math: find \( c_0 \) such that \( (\e, \v) = 0,\quad\forall\v\in V \)
- Equivalent (see notes): find \( c_0 \) such that \( (\e, \psib_0)=0 \)
- Insert \( \e = \f - c_0\psib_0 \) and solve for \( c_0 \)
- Same equation for \( c_0 \) and hence same solution as in the least squares
method
