Finite Element Methods. QS 4
1. Let K = [0, 1], V = P3 (K), and let L = {`0 , `1 , `00 , `01 }, where
`0 : v 7→ v(0),
`1 : v 7→ v(1),
`00 : v 7→ v 0 (0),
`01 : v 7→ v 0 (1).
This defines the first Hermite finite element in one dimension.
(a) Show that L determines V.
(b) Construct the nodal basis {φ0 , φ1 , φ00 , φ01 } for this finite element. Express your answers
in the monomial basis.
2. A quadrature rule on a cell K has degree of precision m if it computes the exact answer
for all polynomials of degree m or less.
What is the minimum degree of precision required to exactly assemble the matrix for the
following bilinear forms? Furthermore, in one dimension, how many quadrature points
n are required when using Gaussian quadrature?
(a) quadratic Lagrange elements, constant coordinate transformation Jacobian JK (x̂).
Form:
Z
a(u, v) =
∇u · ∇v dx.
Ω
(b) quintic Lagrange elements, linear coordinate transformation Jacobian JK (x̂). Form:
Z
a(u, v) =
uv dx.
Ω
1
3. Let K be a non-degenerate triangle, and let V = P2 (K). Denote the vertices of K by zi ,
i = 1, . . . , 3. Let
`1 : v 7→ v(z1 ),
`2 : v 7→ v(z2 ),
`3 : v 7→ v(z3 ),
z 1 + z2
`4 : v 7→ v(
),
2
z 1 + z3
),
`5 : v 7→ v(
2
z 2 + z3
`6 : v 7→ v(
).
2
Show that L = {`1 , . . . , `6 } determines V.
Hint: the easiest way to do this is with Lemma (3.1.10) of Brenner and Scott, reproduced
hereafter for your convenience.
2
3.2 Triangular Finite Elements
71
v ∈ Pk and 0 = Ni (v) = v(a + (b − a)i/k) ∀i = 0, 1, . . . , k then v vanishes
identically by the fundamental theorem of algebra. Thus, (K, Pk , Nk ) is a
finite element.
We will use the following terminology in subsequent sections.
(3.1.8) Definition. We say that N determines P if ψ ∈ P with N (ψ) =
0 ∀N ∈ N implies that ψ = 0.
(3.1.9) Remark. We will often refer to the hyperplane {x : L(x) = 0}, where
L is a non-degenerate linear function, simply as L.
(3.1.10) Lemma. Let P be a polynomial of degree d ≥ 1 that vanishes on
a hyperplane L. Then we can write P = LQ, where Q is a polynomial of
degree (d − 1).
Proof. Make an affine change of coordinates such that L(x̂, xn ) = xn and
the hyperplane L(x̂, xn ) = 0 is the x̂-axis. Therefore, P (x̂, 0) ≡ 0. Since
degree(P ) = d, we have
P (x̂, xn ) =
d
cîj x̂î xjn
j=0 |î|≤d−j
where x̂ = (x1 , . . . , xn−1 ) and î = (i1 , . . . , in−1 ). Letting xn = 0, we obtain
0 ≡ P (x̂, 0) = |î|≤d cî0 x̂î , which implies that cî0 = 0 for |î| ≤ d. Therefore,
P (x̂, xn ) =
d
cîj x̂î xjn
j=1 |î|≤d−j
= xn
d
cîj x̂î xj−1
n
j=1 |î|≤d−j
= xn Q
= L Q,
where degree Q = d − 1.
3.2 Triangular Finite Elements
Let K be any triangle. Let Pk denote the set of all polynomials in two
variables of degree ≤ k. The following table gives the dimension of Pk .