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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 1

Please hand in the solutions in the exercise classes on Wednesday/Thursday November 2 / 3 !

Exercise 1 4 Points

Let Ω ⊂ R 2 a polygonal domain and let T = ( T i ) i = 1,...,I be a triangulation of Ω. Prove that for a continuous function u h : Ω → R, which is piecewise affine w.r.t. T (i.e. u h is affine on each T i ), the surface area of the graph of u h is given by

E[ u h ] =

∑ I i = 1

| T i | q 1 + |∇ u h | T

i

| 2 .

Exercise 2 4 Points

Let Ω ⊂ R n and p > 1. Consider the energy E[ u ] =

Z

Ω |∇ u | p dx .

(i) Compute the derivate D E[ u ]( v ) : = ds d E [ u + sv ] s = 0 for a test function v. What are suitable function spaces for u and v?

(ii) Derive from D E [ u ]( v ) = 0 a partial differential equation for u.

(2)

Exercise 3 4 Points For any N ≥ 2 and h = N 1 let

x i = i · h , i = 0, . . . , N ,

x i + 1 3

= x i + 1 3 h , i = 0, . . . , N − 1 ,

x i + 2 3 = x i + 2 3 h , i = 0, . . . , N − 1 . Consider the finite element space

V h 3 ([ 0, 1 ]) = n v ∈ C 0 ([ 0, 1 ] , R ) : v

[ x

i

,x

i+1

] ∈ P 3 ([ x i , x i + 1 ]) ∀ i = 0, . . . , N − 1 o . Here, P 3 ([ x i , x i + 1 ]) is the set of all cubic polynomials on [ x i , x i + 1 ] . Compute the set of base functions ( φ i h ) i = 0,...,N , ( φ i +

1 3

h ) i = 0,...,N 1 , ( φ i +

2 3

h ) i = 0,...,N 1 ⊂ V h 3 ([ 0, 1 ]) , which satisfy φ i +

k 3

h ( x j +

l

3

) = δ i,j δ k,l for k, l ∈ { 0, 1, 2 } .

Hint: Split the construction of the base functions on the different intervals [ x j , x j + 1 ] .

Exercise 4 4 Points

Let a : ( 0, 1 ) → R be the following function:

a ( x ) =

c 1 if x ∈ ( 0, b 1 ) , c 2 if x ∈ [ b 1 , b 2 ) , c 3 if x ∈ [ b 2 , 1 ) ,

for c 1 , c 2 , c 3R and 0 < b 1 < b 2 < 1. Determine a weak solution u ∈ H 1 ( 0, 1 ) = W 1,2 ( 0, 1 ) of the boundary value problem

−( a ( x ) u 0 ( x )) 0 = 0 ∀ x ∈ ( 0, 1 ) , u ( 0 ) = 0 ,

u ( 1 ) = 1 .

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