Numerical Methods for PDEs
MATH 610 - 600
Alan Demlow
Homework 4
Spring 2015
Homework 4
Exercise 1
70%
Consider the boundary value problem
−∆u + qu = 1 in Ω,
∂u
= g on Γ
∂n
where Ω ⊂ R2 is a polygonal domain with boundary Γ, q is a smooth function on Ω satisfying
q1 ≥ q ≥ q0 > 0 for fixed constants q0 , q1 , and g is the restriction to Γ of a smooth function defined
on all of Ω. The weak form of this problem is to find u ∈ H 1 (Ω) such that a(u, v) = F (v), all
v ∈ H 1 (Ω), where
Z
Z
Z
a(u, v) = (∇u∇v + quv)dx, F (v) =
v dx + gv dx.
Ω
Ω
Γ
Let Th be a quasi-uniform triangulation of Ω having elements of diameter h, and for p ∈ C(T ),
PN
T ∈ Th , let QT (p) = |T | i=1 wi p(xi ) be a quadrature rule with positive weights which is exact on
Pr for some r ≥ 0. Let
Z
X
ah (u, v) =
∇u∇v dx +
QT (quv).
Ω
T ∈Th
Let E∂ be the set of edges of triangles in Th which lie on ∂Ω, given e ∈ E∂ let Qe (p) = |e|
be a quadrature rule that is exact on Ps (e). Let then
Z
X
Fh (v) =
v dx +
Qe (gv).
Ω
PM
i=1
w̃i p(x̃i )
e∈E∂
Finally, let Vh be the usual degree-1 (piecewise linear) Lagrange finite element space on Th , and
let uh ∈ Vh solve
ah (uh , vh ) = Fh (vh ), all vh ∈ Vh .
a. (25 points) Show that
ku − uh kH 1 (Ω) ≤ C
+
h
(
inf
χ∈Vh
)
ku − χkH 1 (Ω) +
sup
|a(χ, vh ) − ah (χ, vh )|
vh ∈Vh ,kvh kH 1 (Ω) =1
i
|F (vh ) − Fh (vh )| .
sup
vh ∈Vh ,kvh kH 1 (Ω) =1
b. (45 points) Estimate the error in the FEM, assuming that u ∈ H k+1 (Ω) and q, g are smooth.
You may use without proof the trace inequality recently given in class, which states that if
v ∈ H 1 (T ) with T ∈ Th , then
−1/2
kvkL2 (∂T ) ≤ C(hT
1/2
kvkL2 (T ) + hT k∇vkL2 (T ) ).
Exercise 2
30%
Consider the boundary value problem
−∆u = f in Ω,
u = 0 on ∂Ω.
R
R
Let a(u, v) = Ω ∇u∇v dx and F (v) = Ω f v dx. Let uh ∈ Vh be a finite element approximation
given by ah (uh , vh ) = Fh (vh ), all vh ∈ Vh . Also, given g ∈ L2 (Ω), let zg ∈ H01 (Ω) solve −∆z = g.
Prove that
n
1
inf Cku − uh kH 1 (Ω) kzg − zh kH 1 (Ω)
ku − uh kL2 (Ω) ≤ sup
g∈L2 (Ω) kgkL2 (Ω) zh ∈Vh
o
+ |a(uh , zh ) − ah (uh , zh )| + |f (zh ) − fh (zh )| .