Discontinuous Galerkin methods Lecture 1 - Brown University

where the piecewise linear shape function, N
Finite element methods
We begin by splitting the solution into elements as
i (xj )=δij is the basis function
and uk = u(xk ) remain as the unknowns.
To recover the scheme to solve Eq. (1.1), we define a space of test functions,
x ∈ D
Vh, and require that the residual is orthogonal to all test functions in this
space as
∂uh ∂fh
+ − gh φh(x) dx =0, ∀φh ∈ Vh.
Ω ∂t ∂x
The details of the scheme is determined by how this space of test functions is
defined. A classic choice, leading to a Galerkin scheme, is to require the that
spaces spanned by the basis functions and test functions are the same. In this
particular case we thus assume that
k : uh(x) =u(x k x − xk+1
)
xk − xk+1 + u(xk+1 )
x
where the linear Lagrange polynomial, ℓk i (x), i
ℓ k x − xk+1−
i (x) =
xk+i − xk+ With this local element-based model, each elem
other element (e.g., D k−1 and D k share xk ). W
of uh as
• The solution is defined in a nonlocal manner
φh(x) =
K
v(x k )N k (x).
uh(x) =
k=1
• The equation is satisfied globally
K
u(x k )N k (x) =
Since the residual has to vanish for all φh ∈ Vh, this amounts to
∂uh ∂fh
+ − gh N
Ω ∂t ∂x j where the piecewise linear shape function, N
(x) dx =0,
for j =1...K. Straightforward manipulations yield the scheme
i (
and uk = u(xk ) remain as the unknowns.
To recover the scheme to solve Eq. (1.1), we
Vh, and require that the residual is orthogon
k=1
k