Vector Curl and Gradient, Lecture Notes - Mathematics - 1, Study notes for Mathematics. University of Oxford
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andreasphd9 September 2011

Vector Curl and Gradient, Lecture Notes - Mathematics - 1, Study notes for Mathematics. University of Oxford

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Finite Element methods, Divergence theorem, strong/classical solutions, weak solutions
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FEM – p.1/16

Finite Element Methods

FEM – p.2/16

Finite Element Methods

Boundary Value Problems Given a domain Ω ⊂ Rd, d = 1, 2 or 3 with boundary ∂Ω

−∇ · (k∇u) = f in Ω

αu + β ∂u

∂n = g on ∂Ω

(P )

k = 1 ⇒ Poisson’s equation: ∇ · (∇u) = ∂ 2u

∂x2 + ∂2u ∂y2 +

∂2u ∂z2

β = 0, (wlog α = 1) ⇒ Dirichlet boundary conditions, (P ) is the Dirichlet Problem for the Poisson equation

α = 0, (wlog β = 1) ⇒ Neumann boundary conditions, (P ) is the Neumann Problem for the Poisson equation

FEM – p.2/16

Examples: • steady heat conduction:

u = temperature, k = thermal conductivity, f = heating Neumann boundary conditions ↔ insulation, Dirichlet boundary conditions ↔ fixed boundary temperature

• k = 1, u = gravitational potential (−∇u = force), f = mass distribution

• k = 1, u = fluid potential (−∇u = velocity) •

k =

[

k1,1 k1,2 k1,2 k2,2

]

(or in 3-dimensions) permeability tensor arises in groundwater flow (flow in porous media): −k∇u = (Darcy) flow velocity

Always xTkx > 0, x 6= 0 ie. k is positive definite FEM – p.3/16

The Divergence Theorem If the vector field w and the appropriate derivatives are defined in Ω and R ⊂ Ω is sufficiently regular with ∂R being the boundary of R then

R ∇ · w dV =

∂R w · n dS

where n is the outward pointing normal to R on ∂R.

(dV is an increment of volume, dS is an increment of surface: these will usually be omitted)

FEM – p.4/16

Strong/Classical Solutions Denote

C0(Ω) = {f : Ω → R | f is continuous}

C2(Ω) = {f : Ω → R | ∂2f

∂xi∂xj ∈ C0(Ω) each i, j}

If Ω does not include cusps or other irregularities, a solution of (P) satisfying u ∈ C2(Ω) is called a strong solution.

Not even all simple problems have strong solutions:

−u′′ =

{

1, 0 ≤ x < 1 2

0, 1 2 < x ≤ 1

u(0) = 0, u(1) = 0

clearly has u′′ discontinuous! FEM – p.5/16

FEM – p.6/16

Weak solutions (Poisson’s equation) For appropriate test functions v, consider u satisfying

(∇2u + f)v = 0

Employing the derivative of a product rule:

∇ · (v∇u) = v∇2u + ∇u · ∇v

integrating and using the Divergence Theorem ∫

∇ · (v∇u) =

v∇2u +

∇u · ∇v

so ∫

∂Ω v ∂u

∂n =

v∇2u +

∇u · ∇v FEM – p.7/16

∇ · (v∇u) =

v∇2u +

∇u · ∇v

∂Ω v ∂u

∂n =

v∇2u +

∇u · ∇v

for the boundary value problem (P) with k = 1 gives ∫

∇u · ∇v =

vf +

∂Ω v ∂u

∂n

which may have a solution u — a weak solution — which is not smooth enough to be a strong solution.

For the Neumann problem ∫

∇u · ∇v =

vf +

∂Ω vg

A strong solution will certainly satisfy the weaker form. FEM – p.8/16

eg. if Ω ⊂ R2

∇u · ∇v =

∫ ∫

∂u

∂x

∂v

∂x +

∂u

∂y

∂v

∂y

Important questions: what are suitable v? Where do we find weak solutions u?

FEM – p.9/16

eg. if Ω ⊂ R2

∇u · ∇v =

∫ ∫

∂u

∂x

∂v

∂x +

∂u

∂y

∂v

∂y

Important questions: what are suitable v? Where do we find weak solutions u?

Need appropriate spaces of functions: based on integration:

L2(Ω) =

{

u : Ω → R |

u2 < ∞

}

with the corresponding norm (ie. size/magnitude of a function)

‖u‖ =

( ∫

u2 )1/2

FEM – p.9/16

In fact

〈u, v〉 =

uv

defines an inner product on L2(Ω):

(i) 〈αu + βv,w〉 =

(αu + βv)w

= α

uw + β

vw

= α〈u,w〉+ β〈v, w〉

(ii) 〈u, v〉 = ∫

uv = 〈v, u〉

(iii) 〈u, u〉 = ∫

u2 > 0

unless u = 0 ∈ L2(Ω) FEM – p.10/16

So we have the Cauchy-Schwarz inequality:

〈u, v〉 =

uv ≤ ‖u‖‖v‖

FEM – p.11/16

Returning to the weak form: ∫

vf ≤ ‖v‖‖f‖

and so is defined if v, f ∈ L2(Ω) ∫

∇u · ∇v =

∂u

∂x

∂v

∂x +

∂u

∂y

∂v

∂y

≤ ‖ ∂u

∂x ‖‖

∂v

∂x ‖ + ‖

∂u

∂y ‖‖

∂v

∂y ‖

is defined if ux, uy, vx, vy ∈ L2(Ω) and ∫

∂Ω v ∂u

∂n ≤ ‖v‖∂Ω‖

∂u

∂n ‖∂Ω

is defined if v, ∂u ∂n

∈ L2(∂Ω) FEM – p.12/16

This gives us the natural space in which weak solutions live:

If f ∈ L2(Ω) and g ∈ L2(∂Ω) then weak solutions u of the above lie in the Sobolev space H1(Ω) defined by

H1(Ω) =

{

u : Ω → R |u, ∂u

∂x , ∂u

∂y ∈ L2(Ω)

}

for Ω ⊂ R2.

FEM – p.13/16

This gives us the natural space in which weak solutions live:

If f ∈ L2(Ω) and g ∈ L2(∂Ω) then weak solutions u of the above lie in the Sobolev space H1(Ω) defined by

H1(Ω) =

{

u : Ω → R |u, ∂u

∂x , ∂u

∂y ∈ L2(Ω)

}

for Ω ⊂ R2.

H1(Ω) = {

u : Ω → R |u, u′ ∈ L2(Ω) }

for Ω ⊂ R1

H1(Ω) =

{

u : Ω → R |u, ∂u

∂x , ∂u

∂y , ∂u

∂z ∈ L2(Ω)

}

for Ω ⊂ R3. FEM – p.13/16

Note Ω is an open set just as (0, 1) is an open interval

We will sometimes need to refer to the closure of Ω, and denote this by Ω ie. Ω = Ω ∪ ∂Ω

FEM – p.14/16

For definiteness consider the following problem:

Find u such that

−∇2u = f in Ω

u = gD on ∂ΩD and ∂u

∂n = gN on ∂ΩN

where ∂ΩD ∪ ∂ΩN = ∂Ω and ∂ΩD and ∂ΩN are distinct with ∂ΩD non-empty (so that we do not have a pure Neumann problem).

We need a solution space

H1E = {u ∈ H 1(Ω) |u = gD on ∂ΩD},

and a test space

H1E0 = {v ∈ H 1(Ω) | v = 0 on ∂ΩD},

FEM – p.15/16

The weak form is:

Find u ∈ H1E such that ∫

∇u · ∇v =

vf +

∂ΩN

vgN for all v ∈ H 1

E0 .

[ ∫

∇u · ∇v =

vf +

∂Ω v ∂u

∂n

]

FEM – p.16/16

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