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Consider the finite element spaces Vh =nvh ∈ C0(Ω)2: vh ˜ T ∈ P12 for all ˜T ∈ Th 2 and vh=0 on ∂Ωo , Wh = ph ∈ C0(Ω): ph T ∈ P1 for all T ∈ Th and Z Ωphdx=0

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Exercises to Wissenschaftliches Rechnen I/Scientific Computing I (V3E1/F4E1)

Winter 2016/17

Prof. Dr. Martin Rumpf

Alexander Effland — Stefanie Heyden — Stefan Simon — Sascha Tölkes

Problem sheet 12

Please hand in the solutions on Tuesday January31!

Exercise39 6Points

Letbe a polygonal domain with regular triangulationTh. The triangulationTh 2 is obtained by adding new vertices at the midpoints of the edges of each element and joining the vertices correspondingly (see Figure1). Consider the finite element spaces

Vh =nvh C0()2: vh ˜

T ∈ P12 for all ˜T ∈ Th

2 and vh=0 on ∂Ωo

, Wh =

ph C0(): ph

T ∈ P1 for all T ∈ Th and Z

phdx=0

.

Figure1: TriangulationsTh (thick) andTh 2.

(i.) Show that the discrete inf-sup conditon for Stokes problem is satisfied.

Hint: Follow the lines of Lemma4.14.

(ii.) Let (v,p) H2,2()×H1,2() be the solution to Stokes problem and (vh,ph) the associated discrete solution. Show the estimate

kvvhk1,2,Ω+kpphk0,2,Ω Ch(kvk2,2,Ω+kpk1,2,Ω).

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Exercise40 4 Points Let ρ > 0, n 1, p,q [1,] and 0 m l be fixed. Consider a bounded domain Rn with polygonal boundary and Th a triangulation of such that ρh diam(T) hfor everyT ∈ Th. Show that for anyvh Vh Hl,p()Hm,q(), whereVh is a finite element space, and any T ∈ Th the inequality

kvhkl,p,T Chml+npnqkvhkm,q,T

holds true, whereCdoes neither depend on vhnor on h.

Exercise41 6Points

Let = [0, 1]2, Mh be a regular mesh on composed of quadratic elements and Vh :=nvh L2(Ω,R2) : vh

K ∈ Q21 for all K∈ Mh, andvh=0 on ∂Ωo , Wh :=

wh L2(Ω,R) : wh

K ∈ Q1for all K ∈ Mh, Z

wh =0

. Recall thatQ1was introduced in exercise 10. Consider the bilinear forms

a : Vh×Vh R, a(uh,vh) = Z

2 i=1

∇(uh)i· ∇(vh)idx, b: Vh×Wh R, b(vh,qh) =

Z

(divvh)·qhdx

associated with Stokes problem. Construct a counterexample showing that the inf-sup-condition is not satisfied, i.e.

qhinfWh sup

vhVh

|b(vh,qh)|

kvhk1,2,Ωkqhk0,2,Ω =0 .

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