Discontinuous Galerkin methods Lecture 1 - Brown University

sic understanding of the schemes through a simple example.
The first DG schemes
firstx schemes ∈ D k : u k k x − xk+1
h(x) =u
xk x − xk
+ uk+1
− xk+1 xk+1 1
= u
− xk k+i ℓ k i (x) ∈ Vh,
So let us consider the scalar problem
e linear scalar wave equation
ise for the flux, f
∂u ∂f(u)
+ =0, x ∈ [L, R] =Ω,
∂t ∂x
near flux is given as f(u) =au. This is subject to the appropriate initial condit
We form the local residual
u(x, 0) = u0(x).
onditions are given when the boundary is an inflow boundary, that is
k h . The space of basis functions is defined as Vh = ⊕K
k
k=1 ℓi of piecewise polynomial functions. Note in particular that there is no restric
ss of the test functions between elements.
the finite element case, we now assume that the local solution can be well re
pproximation uh ∈ Vh and form the local residual
x ∈ D k : Rh(x, t) = ∂uk h
∂t + ∂f k h − g(x, t),
∂x
u(L, t) =g(t) if a ≥ 0,
u(R, t) =g(t) if a ≤ 0.
ate Ω by K nonoverlapping elements, x ∈ [xk l ,xkr]=D k lement. Going back to the finite element scheme, we recall that the global c
and require this to vanish locally in a Galerkin sense
ual are the source of the global nature of the operators M and S in Eq. (1.4).
equire that the residual is orthogonal to all test functions φh ∈ Vh, leading t
D , as illustrated in Fig.
e elements we express the local solution as a polynomial of order N
k
Rh(x, t)ℓ k j (x) dx =0,
x ∈ D k : u k Np
h(x, t) = û k Np
n(t)ψn(x) = u k h(x k i ,t)ℓ k and the fact that we have duplicated solutions at all interface nodes.
is locality also appears problematic as this statement does not
i (x). allow one to recov
obal solution. Furthermore, the points at the ends of the elements are shared by
n=1
i=1
i=0
1 Introduction
The problem is that all elements are now disconnected
due to the local statement on the residual!
t functions, ℓ k j (x). The strictly local statement is a direct consequence of Vh bein