Chapter 2

Finite element methods for elliptic boundary value problems

2.1 Lax-Milgram lemma 2.2 Sobolev spaces

In this section, we compile a number of important definitions and results from analysis without proofs.

Goal: Construct completions of “classical” function spaces (such as C1(Ω) ∩ C(Ω)).

Proposition 2.2.1 (Completions) If V = (V, h·, ·i) is a with inner product h·, ·i, then there is a unique (up to isometric isomorphisms) (V, hh·, ·ii) such that V is dense in V and

hhu, vii = hu, vi for all u, v ∈ V ⊆ V.

The space (V, hh·, ·ii) is called the completion of (V, h·, ·i). The completion of V is the smallest complete space which contains V .

Idea of the proof: Consider equivalence classes of Cauchy sequences in V .

(vn)n ∼ (un)n ⇐⇒ lim kvn − unk = 0 n→∞

Similar to the completion Q −→ R.

Assumption for the rest of this chapter: Ω ⊂ Rd is a bounded Lipschitz domain with boundary Γ. This means: • Ω is open, bounded, connected, and non-empty.

1 2 Finite element methods (Version: November 2, 2016)

• For every boundary point x ∈ Γ, there is a neighbourhood U such that, after an affine change of coordinates (translation and/or rotation), Γ ∩ U is described by the equation xd = χ(x1, . . . , xd−1) with a uniformly Lipschitz continuous function χ. Moreover, Ω ∩ U is on one side of Γ ∩ U. Example: Circles and rectangles are Lipschitz domains, but a circle with a cut is not.

Definition 2.2.2 (Lp(Ω) spaces) For p ∈ N:  Z  p Lp(Ω) := v :Ω → R measurable and |v(x)| dx < ∞ Ω 1 Z  p p kvkLp(Ω) := |v(x)| dx Ω For p = ∞:

L∞(Ω) := {v :Ω → R measurable and |v(x)| < ∞ a.e.}

kvkL∞(Ω) := inf{c > 0 : |v(x)| ≤ c a.e.}

The integral is the Lebesgue integral, and the abbreviation “a.e.” means “almost every- where”, i.e. for all x ∈ Ω \ N for null sets N.

For convenience, we write kvkLp instead of kvkLp(Ω) if it is clear which domain is meant.

The elements of Lp(Ω) are not functions but equivalence classes of functions: u, v ∈ Lp(Ω) are equivalent if u = v a.e.. It is common usage to speak of “Lp functions”, but some care is required: It does not make sense to speak about “the value of a Lp function in a point x” because a single point is a null set.

Important properties:

• k·kLp is a norm on Lp(Ω) for every p ∈ {1, 2, ..., ∞}, and (Lp(Ω), k·kLp ) is a (complete).

• Special case p = 2: The space L2(Ω) is a Hilbert space with inner product Z hu, vi = u(x)v(x) dx. L2(Ω) Ω

Definition 2.2.3

loc n o L1 (Ω) := u :Ω −→ R is measurable and locally integrable

“Locally integrable” means that u ∈ L1(S) for all compact subsets S ⊂ Ω. Tobias Jahnke 3

loc Definition 2.2.4 (Weak derivative) A function u ∈ L1 (Ω) is weakly differen- loc tiable with respect to xi if there is a v ∈ L1 (Ω) such that Z Z ∞ v(x)φ(x) dx = − u(x)∂xi φ(x) dx for all φ ∈ Cc (Ω) Ω Ω or equivalently hv, φi = − hu, ∂ φi for all φ ∈ C∞(Ω). L2(Ω) xi L2(Ω) c

Then, v is called the weak (partial) derivative and is denoted by v = ∂xi u. Uniqueness: If v = ∂ u ∈ Lloc(Ω) and v = ∂ u ∈ Lloc(Ω) are both weak partial xi 1 e xi 1 loc derivatives of u ∈ L1 (Ω), then v = ve a.e. Consistency of the definition: If u ∈ C1(Ω)∩C(Ω), then the weak derivative coincides with the classical derivative (exercise).

d Notation: For a multi-index α = (α1, . . . , αd) ∈ N0 we set ∂αu := ∂α1 . . . ∂αd u, |α| = α + ... + α . x1 xd 1 1 d The above definition can now be extended to higher-order weak derivatives: v(x) = ∂αu(x) is the weak partial derivative of u(x) if Z Z |α|1 α ∞ v(x)φ(x) dx = (−1) u(x)∂ φ(x) dx for all φ ∈ Cc (Ω). Ω Ω Definition 2.2.5 ( Hm(Ω)) For m ∈ N the Sobolev space Hm(Ω) is the set m n α d o H (Ω) := v ∈ L2(Ω) : ∂ v ∈ L2(Ω) exists for all α ∈ N0 with |α|1 ≤ m

with inner product X α α hu, vi m := h∂ u, ∂ vi . H (Ω) L2(Ω) |α|1≤m q and norm kvkHm(Ω) = hv, viHm(Ω).

As for the Lp spaces, we often write hu, viHm and k · kHm instead of hu, viHm(Ω) and k · kHm(Ω).

Example (m = 1):

1 n o H (Ω) := v ∈ L2(Ω) : ∂xi v ∈ L2(Ω) exists for all i = 1, ..., d d X hu, vi 1 := hu, vi + h∂ u, ∂ vi H (Ω) L2(Ω) xi xi L2(Ω) i=1 d !1/2 2 X 2 1 kvkH (Ω) = kvkL2(Ω) + k∂xi vkL2(Ω) . i=1 4 Finite element methods (Version: November 2, 2016)

Important properties: m  0 H (Ω), h·, ·iHm is a Hilbert space for every m ∈ N0. (We define H (Ω) = L2(Ω).) In fact, the space Z ∞,m n ∞ α 2 d o C (Ω) := v ∈ C (Ω) : |∂ v(x)| dx < ∞ for all α ∈ N0 with |α|1 ≤ m . Ω

m m is dense in H (Ω) with respect to k · kHm : For every u ∈ H (Ω) and every ε > 0 there ∞,m is a vε ∈ C (Ω) such that kvε − ukHm < ε . m ∞,m It can be shown that H (Ω) is the completion of C (Ω) with respect to k · kHm , i.e.

m ∞,m u ∈ H (Ω) ⇐⇒ There are vn ∈ C (Ω) such that lim ku − vnkHm(Ω) = 0. n→∞

Example: Let Ω = (−1, 1) and u(x) = |x|. Then u ∈ H1(Ω), but u 6∈ C∞,1(Ω). However, q 2 1 ∞,1 we can define vn := x + n2 ∈ C (Ω) and show that kvn − ukH1(Ω) −→ 0 for n −→ ∞ (exercise).

m ∞ Definition 2.2.6 The Sobolev space H0 (Ω) is the completion of Cc (Ω) with respect to k · kHm(Ω), i.e.

m ∞ u ∈ H0 (Ω) ⇐⇒ There are vn ∈ Cc (Ω) such that lim ku − vnkHm(Ω) = 0. n→∞

m m 1 m H0 (Ω) is a closed subspace of H (Ω). If the boundary Γ is C , then v ∈ C(Ω) ∩ H0 (Ω) implies that v(x) = 0 for all x ∈ Γ.

Proposition 2.2.7 (Poincar´e-Friedrichs inequality) Let

1 d ! 2 X 2 1 |v|H (Ω) := k∂xi vkL2(Ω) i=1

denote the Sobolev seminorm. There is a constant cΩ > 0 such that

1 kvkH1(Ω) ≤ cΩ|v|H1(Ω) for all v ∈ H0 (Ω).

Proof. See 1.5, page 29 in [Bra07].

1 Remark: On H0 (Ω), the seminorm | · |H1(Ω) is even a norm and | · |H1(Ω) ∼ k · kH1(Ω) (exercise).

If we want to solve a PDE with inhomogeneous boundary conditions, we need a condition such as, e.g., “u(x) = g(x) for all x ∈ Γ”. Since Γ is a null set, however, we have to specify how such a condition has to be understood if u ∈ L2(Ω). Tobias Jahnke 5

Theorem 2.2.8 (Trace theorem) There is a bounded linear map

1 1 γ : H (Ω) −→ L2(Γ), kγ(v)kL2(Γ) ≤ ckvkH (Ω) (c > 0)

1 ¯ such that γ(v) = v|Γ for all v ∈ C (Ω)

Proof: page 45-47 in [Bra07]

Remark: It can be shown that

1  1 H0 (Ω) := v ∈ H (Ω) : γ(v) = 0

d Theorem 2.2.9 (Sobolev’s embedding theorem) Let Ω ⊂ R and let r, s ∈ N0 with r > s + d/2. Then

Hr(Ω) ⊂ Cs(Ω) and there is a constant c > 0 such that

X α kukCs(Ω) ≤ ckukHr(Ω), where kukCs(Ω) := sup |∂ u(x)|. x∈Ω |α|1≤s

Examples:

d = 1 =⇒ H1(Ω) ⊂ C(Ω),H2(Ω) ⊂ C1(Ω) d ∈ {2, 3} =⇒ H1(Ω) 6⊂ C(Ω),H2(Ω) ⊂ C(Ω)

2.3 Weak solutions of elliptic boundary value prob- lems

Let Ω ⊆ Rd be a bounded Lipschitz domain with d ∈ {2, 3} and assume that Γ is piecewise C1. Consider the elliptic PDE

Au = f for f ∈ L2(Ω) and with the second-order differential operator  Au(x) = −div κ(x)∇u(x) + κ0(x)u(x) d X   = − ∂xi κij(x)∂xj u(x) + κ0(x)u(x) i,j=1

 d×d with κ(x) = κij(x) i,j ∈ R for all x ∈ Ω. The coefficient functions κij :Ω −→ R and κ0 :Ω −→ R are supposed to have the following properties: 1. |κ0(x)|, |κij(x)| ≤ M for all x ∈ Ω 6 Finite element methods (Version: November 2, 2016)

2. κij = κji for all i, j = 1, ..., d

3. There are constants c0 ≥ 0 and c1 > 0 such that κ0(x) ≥ c0 and

d d d X X X 2 T d κij(x)ξiξj ≥ c1 ξi for all x ∈ Ω and ξ = (ξ1, . . . , ξd) ∈ R . (2.1) i=1 j=1 i=1 The second assertion is equivalent to

T 2 T ξ κξ ≥ c1kξk2, ξ = (ξ1, ..., ξd) . A differential operator with the properties2 and3 is called elliptic.

Example: For κ0(x) = 0, κii = 1 and κij = 0 for i 6= j, we obtain A = −∆ .

1. Homogeneous Dirichlet boundary conditions. Consider the boundary value problem

Au = f in Ω u = 0 on Γ.

Assume that u ∈ C2(Ω) ∩ C(Ω) is a classical solution. Multiply both sides of the first line ∞ with a test function v ∈ Cc (Ω) and integrate over Ω:

d X Z   Z Z − ∂xi κij(x)∂xj u(x) v(x) dx + κ0(x)u(x)v(x) dx = f(x)v(x) dx i,j=1 Ω Ω Ω Apply Green’s formula:

d X Z Z Z κij(x)∂xj u(x)∂xi v(x) dx + κ0(x)u(x)v(x) dx = f(x)v(x) dx i,j=1 Ω Ω Ω | {z } | {z } =: a(u, v) =: `(v) Equivalent notation: Z Z T a(u, v) = (∇v(x)) κ(x)∇u(x) dx + κ0(x)u(x)v(x) dx Ω Ω

1 Variational (weak) formulation: Find u ∈ H0 (Ω) such that

1 a(u, v) = `(v) for all v ∈ H0 (Ω). (2.2) Such an u is called a weak solution of the boundary value problem. 1 Since we seek a solution in H0 (Ω), the boundary condition is fulfilled. Show that the assumptions of the Lax-Milgram theorem (Theorem ??) are true: Tobias Jahnke 7

1 • Show that ` : H0 (Ω) → R is a continuous linear form:

Cauchy- 1 |`(v)| = | hf, viL | ≤ kfkL2 kvkL2 ≤ kfkL2 kvkH . 2 Schwarz | {z } =:c

1 • Show that a(·, ·) is a H0 (Ω)-elliptic . Boundedness:

d X Z |a(u, v)| ≤ |κij(x)| ·|∂xj u(x)| · |∂xi v(x)| + |κ0(x)| ·|u(x)| · |v(x)| dx i,j=1 Ω | {z } | {z } ≤M ≤M d X ≤ M |∂x u|, |∂x v| + M h|u|, |v|i j i L2 L2 i,j=1 d Cauchy- X ≤ M k∂xj ukL2 · k∂xi vkL2 + MkukL2 · kvkL2 Schwarz i,j=1 2 ≤ M(d + 1) · kukH1 kvkH1

Coercivity:

Z d X 2 a(v, v) = κij(x)∂xi v(x)∂xj v(x) + κ0(x)v (x) dx Ω i,j=1 Z d X 2 2 ≥ c1 (∂xi v(x)) + c0v (x) dx (A is elliptic) Ω i=1 2 2 2 1 1 = c1|v|H + c0kvkL2 ≥ c1|v|H c1 2 ≥ 2 kvkH1 (Poincar´e-Friedrichs) cΩ

1 Applying Lax-Milgram (Theorem ??) with V = H0 (Ω) yields that the problem (2.2) has 1 a unique (weak) solution u ∈ H0 (Ω).

1 Remark: The definition of the differential operator A only makes sense if κij ∈ C (Ω), whereas only κij ∈ L∞(Ω) is required in the weak formulation.

2. Inhomogeneous Dirichlet boundary conditions. For a given function g :Γ → R we consider the problem

Au = f in Ω u = g on Γ

1 Assumption: There is a u∗ ∈ H (Ω) such that g = u∗ Γ = γ(u∗) in the sense of the trace theorem 2.2.8. 8 Finite element methods (Version: November 2, 2016)

1 1 Variational (weak) formulation: Find u ∈ H (Ω) (instead of H0 (Ω)) such that w := 1 u − u∗ ∈ H0 (Ω) and

1 a(u, v) = `(v) for all v ∈ H0 (Ω) (2.3)

1 (same derivation as before). Equivalent: Find w ∈ H0 (Ω) such that

1 a(w, v) = `(v) − a(u∗, v) =: `e(v) for all v ∈ H0 (Ω). (2.4)

1 Verify the conditions of the Lax-Milgram theorem: `e: H0 (Ω) → R is continuous, because

|`e(v)| ≤ |`(v)| +|a(u∗, v)| | {z } = hf,vi | L2 | 2 1 1 1 ≤ kfkL2 kvkH + M(d + 1)ku∗kH kvkH 2  1 1 = kfkL2 + M(d + 1)ku∗kH kvkH

1 The Lax-Milgram theorem yields that (2.4) has a unique solution w ∈ H0 (Ω). Hence, 1 (2.3) has a unique solution u = u∗ + w ∈ H (Ω).

3. Neumann boundary conditions. For every x ∈ Γ, we define the conormal derivative

d ∂u X (x) := κ (x)∂ u(x)η (x) = η(x)T κ(x)∇u(x), ∂η ij xj i κ i,j=1

d where η :Γ →∈ R is again the outer unit normal. Let f ∈ L2(Ω) and g ∈ L2(Γ) and consider the boundary value problem

Au = f in Ω (2.5a) ∂u = g on Γ. (2.5b) ∂ηκ

Variational (weak) formulation? Let u ∈ C2(Ω) ∩ C(Ω) is a classical solution and proceed as in 1.: Multiply both sides of the first line with a test function v ∈ C∞(Ω), integrate Tobias Jahnke 9 over Ω and apply Green’s formula. This yields Z Z `(v) = f(x)v(x) dx = Au(x)v(x) dx Ω Ω d X Z   Z = − ∂xi κij(x)∂xj u(x) v(x) dx + κ0(x)u(x)v(x) dx i,j=1 Ω Ω d ! Z X = − κij(x)∂xj u(x)ηi(x) v(x)dσ(x) Γ i,j=1 | {z } = ∂u (x)=g(x) ∂ηκ d Z X Z + κij(x)∂xj u(x)∂xi v(x) + κ0(x)u(x)v(x) dx Ω i,j=1 Ω Z = − g(x)v(x)dσ(x) + a(u, v) Γ

1 1 Weak formulation: Find u ∈ H (Ω) (instead of H0 (Ω)) such that Z a(u, v) = `(v) + g(x)v(x) dσ(x) =: `e(v) for all v ∈ H1(Ω) (2.6) Γ with a(·, ·) as before. Lax-Milgram conditions:

• Show that ` : H1(Ω) → R is continuous: The trace theorem (Theorem 2.2.8) yields

1 |`e(v)| ≤ kfkL2(Ω) · kvkH (Ω) + kgkL2(Γ) · kγ(v)kL2(Γ) | {z } ≤ckvkH1(Ω)

≤ CkvkH1(Ω).

• Show that a(·, ·) is H1(Ω)-elliptic. Boundedness can be shown as before. To prove coercivity, it can be shown as in 1. that

2 2 1 a(v, v) ≥ c1|v|H + c0kvkL2 . This time, the Poincar´e-Friedrichs inequality cannot be applied. In order to obtain coercivity, we have to assume that c0 > 0 (instead of c0 ≥ 0) and use that

2 2 2 1 1 c1|v|H + c0kvkL2 ≥ min{c1, c0} · kvkH .

Lax-Milgram −→ unique solution.

Remark. Let κ0(x) = 0 for all x ∈ Ω, i.e. c0 = 0. In this case, the solution of the boundary value problem (2.5) is not unique (exercise). This indicates that the case c0 = 0 may also cause problems in the original formulation of the problem. Bibliography

[Bra07] Dietrich Braess. Finite Elemente. Theorie, schnelle L¨oserund Anwendungen in der Elastizit¨atstheorie. Springer, Berlin, 4nd ed. edition, 2007.

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