Partial differential equations and the finite element method 2--SECOND-ORDER ELLIPTIC PROBLEMS

This part is devoted to the discussion of linear second-order elliptic problems.

Consider the model equation
(1)

−∇(a1∇u)+a0u=f, − ∇ ( a 1 ∇ u ) + a 0 u = f ,

with homogeneous Dirichlet boundary conditions
(2)

u(x)=0 u ( x ) = 0

on ∂Ω. ∂ Ω .

Assuming aij(x)=a1(x)δij a i j ( x ) = a 1 ( x ) δ i j , and b(x)=c(x)=0 b ( x ) = c ( x ) = 0 .

Classical solution: The solution of (1) is a function u∈C2(Ω)∩C(Ω⎯⎯⎯⎯⎯) u ∈ C 2 ( Ω ) ∩ C ( Ω ¯ ) satisfying the equation (1) everywhere in Ω Ω and fulfilling the boundary condition (2)
at every x∈∂Ω x ∈ ∂ Ω .

C(Ω⎯⎯⎯⎯⎯) C ( Ω ¯ ) = {ϕ∈C(Ω) { ϕ ∈ C ( Ω ) , ϕ ϕ is bounded and uniformly continuous function } } : because Ω⎯⎯⎯⎯⎯ Ω ¯ is closed, the functions in closed sets are uniformly continuous.

Find the solution of (1) using the weak formulation:

1. Multiply (1) with a test function v∈C∞0(Ω) v ∈ C 0 ∞ ( Ω )
   
   −∇(a1∇u)v+a0uv=fv, − ∇ ( a 1 ∇ u ) v + a 0 u v = f v ,
2. Integrate over Ω Ω
   
   ∫Ω−∇(a1∇u)vdx+∫Ωa0uvdx=∫Ωfvdx, ∫ Ω − ∇ ( a 1 ∇ u ) v d x + ∫ Ω a 0 u v d x = ∫ Ω f v d x ,

3. reduce the maximum order of the partial derivatives

      By Green's first identity:
   

   <f,−Δx>Ω=<∇f,∇x>Ω−<f,∂nx>∂Ω < f , − Δ x > Ω =< ∇ f , ∇ x > Ω − < f , ∂ n x > ∂ Ω
   
   we have
   
   ∫Ω−∇(a1∇u)vdx=∫Ω∇a1u∇vdx−∫∂Ωv∂n(a1u)dS ∫ Ω − ∇ ( a 1 ∇ u ) v d x = ∫ Ω ∇ a 1 u ∇ v d x − ∫ ∂ Ω v ∂ n ( a 1 u ) d S
   
   The fact that v v vanishes on the boundary ∂Ω ∂ Ω removes the boundary term, then
   (3)
   ∫Ω∇a1u∇vdx+∫Ωa0uvdx=∫Ωfvdx, ∫ Ω ∇ a 1 u ∇ v d x + ∫ Ω a 0 u v d x = ∫ Ω f v d x ,

4. Find the largest possible function spaces for u,v u , v s.t.all integrals are finite
   Originally, identity (3) was derived under very strong regularity assumptions u∈C2(Ω)∩C(Ω⎯⎯⎯⎯⎯) u ∈ C 2 ( Ω ) ∩ C ( Ω ¯ ) and v∈C∞0(Ω) v ∈ C 0 ∞ ( Ω ) .
   All integrals in (3) remain finite when these assumptions are weakened to
   
   u,v∈W1,20(Ω),f∈L2(Ω) u , v ∈ W 0 1 , 2 ( Ω ) , f ∈ L 2 ( Ω )
   
   Similarly the regularity assumptions for the coefficients a1 a 1 and a0 a 0 can be reduced to
   
   a1,a0∈L∞(Ω) a 1 , a 0 ∈ L ∞ ( Ω )
   
   Note: Wm,p={u∈LP(Ω):D|a|u∈LP(Ω):∀0<|a|≤m} W m , p = { u ∈ L P ( Ω ) : D | a | u ∈ L P ( Ω ) : ∀ 0 < | a | ≤ m }
   Wm,p0 W 0 m , p contains the functions in Wm,p W m , p that have compact support in Ω Ω .

The weak form of the problem (1)-(2) is stated as follows: Given f∈L2(Ω) f ∈ L 2 ( Ω ) , find a
function u∈W1,20(Ω) u ∈ W 0 1 , 2 ( Ω ) such that

∫Ω∇a1u∇vdx+∫Ωa0uvdx=∫Ωfvdx,∀v∈W1,20(Ω) ∫ Ω ∇ a 1 u ∇ v d x + ∫ Ω a 0 u v d x = ∫ Ω f v d x , ∀ v ∈ W 0 1 , 2 ( Ω )

Let V=W1,20(Ω) V = W 0 1 , 2 ( Ω ) . We define a bilinear form a(.,.):V×V→R a ( . , . ) : V × V → R :

a(u,v)=∫Ω∇a1u∇vdx+∫Ωa0uvdx a ( u , v ) = ∫ Ω ∇ a 1 u ∇ v d x + ∫ Ω a 0 u v d x

and a linear form l∈V′ l ∈ V ′

l(v)=<l,v>=∫Ωfvdx l ( v ) =< l , v >= ∫ Ω f v d x

Then the weak formulation of the problem (1), ( 2) reads: Find a function u∈V u ∈ V such that

a(u,v)=l(v) a ( u , v ) = l ( v )

This notation is common in the study of partial differential equations and finite element methods.

Bilinear forms, energy norm, and energetic inner product

Every bilinear form a:V×V→R a : V × V → R in a Banach space V V is associated with a unique linear operator A:V→V′ A : V → V ′ defined by

(4)

(A(u))v=<Au,v>=a(u,v) ( A ( u ) ) v =< A u , v >= a ( u , v )

Note: (4) defines a one-to-one correspondence between continuous bilinear forms a:V×V→R a : V × V → R and linear continuous operators A:V→V′ A : V → V ′ .

Definition 1 (Energetic inner product, energy norm) Let V V be a Hilbert space and
a:V×V→R a : V × V → R a bounded symmetric V-elliptic bilinear form. The bilinear form defines
an inner product
(5)

(u,v)e=a(u,v) ( u , v ) e = a ( u , v )

in V V , called energetic inner product. The norm induced by the energetic inner product,
(6)

∥u∥e=a(u,u)‾‾‾‾‾‾√ ∥ u ∥ e = a ( u , u )

is called energy norm.

Theorem 1.5 (Lax-Milgram lemma) Let V V be a Hilbert space, a:V×V→R a : V × V → R a bounded
V-elliptic bilinear form and l∈V′ l ∈ V ′ . Then there exists a unique solution to the problem

a(u,v)=l(v). a ( u , v ) = l ( v ) .

Note: the unique solution is the unique represent of the linear form l∈V′ l ∈ V ′ with respect to the energetic inner product (u,v)e=a(u,v) ( u , v ) e = a ( u , v ) . In this sense the Lax-Milgram lemma is a special case of the Riesz representation theorem.

---

Riesz representation theorem.

This theorem establishes an important connection between a Hilbert space and its (continuous) dual space.

Let H H be a Hilbert space, and let H∗ H ∗ denote its dual space, consisting of all continuous linear functionals from H H into the field ℝ R or ℂ C . If x x is an element of H H , then the function φx φ x , for all y y in H H defined by:

φx(y)=⟨y,x⟩, φ x ( y ) = ⟨ y , x ⟩ ,

where ⟨⋅,⋅⟩ ⟨ ⋅ , ⋅ ⟩ denotes the inner product of the Hilbert space, is an element of H∗ H ∗ .

Note: The Riesz representation theorem states that every element of H∗ H ∗ can be written uniquely in this form. Given any continuous linear functional g∈H∗ g ∈ H ∗ , the corresponding element xg∈H x g ∈ H can be constructed uniquely by xg=g(e1)e1+g(e2)e2+... x g = g ( e 1 ) e 1 + g ( e 2 ) e 2 + . . . , where {ei} { e i } is an orthonormal basis of H H , and the value of xg x g does not vary by choice of basis. Thus, if y∈H,y=a1e1+a2e2+... y∈H,y=a1e1+a2e2+... y ∈ H , y = a 1 e 1 + a 2 e 2 + . . . , then g(y)=a1g(e1)+a2g(e2)+...=⟨xg,y⟩ g ( y ) = a 1 g ( e 1 ) + a 2 g ( e 2 ) + . . . = ⟨ x g , y ⟩ .

---

Lemma 1. Assume that a1(x)≥Cmin>0 a 1 ( x ) ≥ C m i n > 0 and a0(x)≥0 a 0 ( x ) ≥ 0 a.e. in Ω Ω . Then the weak
problem has a unique solution u∈V u ∈ V .

Nonhomogeneous Dirichlet boundary conditions

More general Dirichlet boundary conditions
u(x)=g(x),x∈∂Ω

u ( x ) = g ( x ) , x ∈ ∂ Ω

and g∈C(∂Ω) g ∈ C ( ∂ Ω ) . Consider a new function G∈C2(Ω)∩C(¯Ω) G ∈ C 2 ( Ω ) ∩ C ( Ω ¯ ) , so that g=G g = G on ∂Ω ∂ Ω .

Note: G G is not unique, but we will show later that the solution is invariant under its choice.

Denote u=G+U, then the problem (1) can be rewrited as finding U∈C20(Ω) U ∈ C 0 2 ( Ω )
−∇(a1∇(G+U))+a0(G+U)=f,x∈Ω

− ∇ ( a 1 ∇ ( G + U ) ) + a 0 ( G + U ) = f , x ∈ Ω

s.t.
G+U=g,x∈∂Ω

G + U = g , x ∈ ∂ Ω

or, equivalently,
−∇(a1∇U)+a0U=f+∇(a1∇G)−a0G,x∈Ω

− ∇ ( a 1 ∇ U ) + a 0 U = f + ∇ ( a 1 ∇ G ) − a 0 G , x ∈ Ω

s.t.
U=0,x∈∂Ω

U = 0 , x ∈ ∂ Ω

Independence of the solution u=U+G u = U + G on the Dirichlet lift G G :
Assume that
U1+GI=u1∈W1,20(Ω) U 1 + G I = u 1 ∈ W 0 1 , 2 ( Ω ) and U2+G2=u2∈W1,20(Ω) U 2 + G 2 = u 2 ∈ W 0 1 , 2 ( Ω ) are two weak solutions.
So,
a(u1,v)−a(u2,v)=0

a ( u 1 , v ) − a ( u 2 , v ) = 0

set v=u1−u2 v = u 1 − u 2 , then
0=a(u1−u2,u1−u2)≥C∥u1−u2∥

0 = a ( u 1 − u 2 , u 1 − u 2 ) ≥ C ∥ u 1 − u 2 ∥

that is u1=u2 u 1 = u 2 .

Neumann boundary conditions

Neumann boundary conditions of the form

∂u∂ν=g,x∈∂Ω

∂ u ∂ ν = g , x ∈ ∂ Ω

1.Assume u∈ u ∈ Multiply (1) with a test function v∈C∞0(Ω) v ∈ C 0 ∞ ( Ω )
−∇(a1∇u)v+a0uv=fv,

− ∇ ( a 1 ∇ u ) v + a 0 u v = f v ,

2. Integrate over Ω Ω
∫Ω−∇(a1∇u)vdx+∫Ωa0uvdx=∫Ωfvdx,

∫ Ω − ∇ ( a 1 ∇ u ) v d x + ∫ Ω a 0 u v d x = ∫ Ω f v d x ,

3. reduce the maximum order of the partial derivatives

       By Green's first identity:

<f,−Δx>Ω=<∇f,∇x>Ω−<f,∂nx>∂Ω

< f , − Δ x > Ω =< ∇ f , ∇ x > Ω − < f , ∂ n x > ∂ Ω

we have
∫Ω−∇(a1∇u)vdx=∫Ω∇a1u∇vdx−∫∂Ωv∂n(a1u)dS

∫ Ω − ∇ ( a 1 ∇ u ) v d x = ∫ Ω ∇ a 1 u ∇ v d x − ∫ ∂ Ω v ∂ n ( a 1 u ) d S

As v v does not vanish on the boundary ∂Ω ∂ Ω , we have
∫Ω∇a1u∇vdx+∫Ωa0uvdx−∫∂Ωa1v∂u∂νdS=∫Ωfvdx,

∫ Ω ∇ a 1 u ∇ v d x + ∫ Ω a 0 u v d x − ∫ ∂ Ω a 1 v ∂ u ∂ ν d S = ∫ Ω f v d x ,

Using the boundary condition, rewrite the problem with Neumann condition: given f∈L2(Ω) f ∈ L 2 ( Ω ) , g∈L2(∂Ω) g ∈ L 2 ( ∂ Ω ) , find u u s.t.
∫Ω∇a1u∇vdx+∫Ωa0uvdx=∫Ωfvdx+∫∂Ωa1vgdS,

∫ Ω ∇ a 1 u ∇ v d x + ∫ Ω a 0 u v d x = ∫ Ω f v d x + ∫ ∂ Ω a 1 v g d S ,

4. Find the largest possible function spaces for u,v u , v s.t.all integrals are finite

All integrals remain finite when these assumptions are weakened to, as u,v u , v need not to vanish on the boundary
u,v∈W1,2(Ω),f∈L2(Ω)

u , v ∈ W 1 , 2 ( Ω ) , f ∈ L 2 ( Ω )

Remark (Essential and natural boundary conditions)
1. Dirichler boundary conditions are sometimes called essential since they essentially influence the weak formulation: They determine the function space in which the solution is sought.
2. Neumann boundary conditions do not influence the function space and can be naturally incorporated into the boundary integrals. Therefore they are called natural.

Newton (Robin) boundary conditions

boundary conditions involves a combination of function values and normal derivatives.

---

Energy of elliptic problems

In this part we introduce the explicit form of the abstract energy, at least for symmetric problems. The most important numerical scheme based on the minimization of the abstract energy.

Theorem. Let V be a linear space, a:V×V→R a symmetric V-elliptic bilinear
form and l∈V′. Then the functional of abstract energy,
(T.1)E(v)=12a(v,v)−l(v)
attains its minimum in V at an element u∈V if and only if
(T.2)a(u,v)=l(v),∀v∈V
Moreover; the minimizer u∈V is unique.

Proof: Let (T.2) holds. Then

E(u+tv)=a(u+tv,u+tv)/2−l(u+tv)=(u+tv,u+tv)e/2−l(u+tv)=(u,u)e/2+t2(v,v)e/2+t(u,v)e−l(u)−tl(v)

E ( u + t v ) = a ( u + t v , u + t v ) / 2 − l ( u + t v ) = ( u + t v , u + t v ) e / 2 − l ( u + t v ) = ( u , u ) e / 2 + t 2 ( v , v ) e / 2 + t ( u , v ) e − l ( u ) − t l ( v )

E(u+tv)=E(u)+t2(v,v)e/2>E(u)

E ( u + t v ) = E ( u ) + t 2 ( v , v ) e / 2 > E ( u )

.

if E E has a minimum at u∈V u ∈ V , then for every v∈V v ∈ V the derivative of the ϕ(t)=E(u+tv) ϕ ( t ) = E ( u + t v ) must vanish at 0
0=ϕ′(0)=E(u+tv)∂t|t=0=(u,v)e−l(v)

0 = ϕ ′ ( 0 ) = E ( u + t v ) ∂ t | t = 0 = ( u , v ) e − l ( v )

…………