Linear Algebra and Its Applications · Problem 6.5.5

Computations with Matrices — Gilbert Strang

Mathematics · galerkin-method

Problem

Galerkin's method starts with the differential equation (say

-u'' = f(x)

) instead of the energy

P

. The trial solution is still

u = y_1 V_1 + y_2 V_2 + \cdots + y_n V_n

, and the

y

's are chosen to make the difference between

-u''

and

f

orthogonal to every

V_j

\textbf{Galerkin} \qquad \int (-y_1 V_1'' - y_2 V_2'' - \cdots - y_n V_n'') V_j dx = \int f(x) V_j(x) dx.

Integrate the left side by parts to reach

Ay = f

, proving that Galerkin gives the same

A

and

f

as Rayleigh-Ritz for symmetric problems.