Lecture2 Cornell

8/13/2019 Lecture2 Cornell

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N. Zabaras 1

THE FEM FOR ELLIPTIC PROBLEMS

Abstract formulation: A unified treatment The variational (Vh) and minimization (Mh) problems

Error estimates

The energy norm Applications to various elliptic boundary value problems

Supplemental mathematical background

1. Sobolev spacesHk(), k= 1, 2, . . .

2. The Poincare inequality

Lecture 2: A unified abstract treatment of the FEM for elliptic problems Materials Process Design and Control Laboratory


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ABSTRACT FORMULATION OF THE FEM FOR ELLIPTIC

PROBLEMS

LetVbe a Hilbert space with scalar product (., .)Vand correspondingnorm V.

a(., .) is a bilinear form onV V andLa linear form onV such that:1.a(., .) is symmetric

2.a(., .) is continuous, i.e., there is constant >0 such that

|a(v, w)

|

v

V

w

V,

v, w

V

3.a(., .) isV-elliptic, i.e., there is a constant >0 such thata(v, v) v2V, v V

4.Lis continuous, i.e., there is a constant >0 such that

|L(v)

|

v

V,

v

V



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The abstract minimization (M) and variational (V) problems

The minimization problem (M):Findu V such thatF(u) = minvVF(v),where

F(v) = 12a(v, v) L(v) Variational problem (V):

Findu V such thata(u, v) =L(v) v V.

Theorem

The problems (M) and (V) are equivalent. Thereexists a unique solution u V of these problemsand the following stability estimate holds:

uV Stability estimate



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Proof of the theorem

1. The existence of a solution follows from the Lax-Milgram theorem.

2.Ifu V is a solution of (M), then u is a solution of(V). Letv V and be arbitrary. Then (u+v) V. Sinceuis a minimum,F(u) F(u+v) Letg() F(u+v), . Theng(0) g(), . Henceg(0) = 0 (ifg(0) exists). Using the symmetry ofa(., .):

g() = 1

2a(u+v, u+v) L(u+v)=

1

2a(u, u) +

2a(u, v) +

2a(v, u) +

2

2a(v, v) L(u) L(v)

=

1

2a(u, u) L(u) +a(u, v) L(v) +2

2a (v, v)

g(0) = 0 g(0) =a(u, v) L(v) = 0uis a solution of (V).



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3. Ifu V is a solution of(V), then u is a solution of (M). Letv V and be arbitrary. Then (u+v) V with

a(u, v) =L(v),v V. Using the earlier given expression for g() and theV-elliptic condition

fora(., .), we can write the following:

F(u+v) F(u) = (a(u, v) L(v)) +2

2a(v, v)

= 2

2a(v, v) 2

2vV 0

where we useda(u, v) =L(v).

It follows thatuis a solution of (M).



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5. Uniqueness of solution of(M) and (V)

Letu1 andu2be two solutions of (V), then:a(ui, v) =L(v) v V, i= 1, 2

By subtraction, we see that: a(u1 u2, v) = 0,v V Applying the stability condition for the case ofu = u1 u2,L= 0 (i.e. = 0), we obtain:

u1 u2V 0 u1=u2Important note

Even without the symmetry condition a(u, v) = a(v, u), thereexists a uniqueu V such that

a(u, v) =L(v) v V (V)and the stability estimate uV holds. However, in this case,we cannot define an associated minimization problem (M).



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Discrete problems (Mh) and (Vh)

LetVhbe a finite-dimensional subspace ofV of dimensionM. Let{1, . . . , M}be a basis forVh so thati Vh.

Anyv Vh has the unique representationv= M

i=1

ii, wherei

We can now formulate the following discrete analogues of the prob-lems (M) and (V):

Problem (Mh): Finduh Vh such thatF(uh) F(v) v Vh Discrete optimization problem

Problem (Vh): Finduh

Vh such that

a(uh, j) =L(j), j = 1, . . . , M . Discrete variational problem



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Matrix form of(Vh)

Letuh= Mi=1

ii, i . (Vh) takes the matrix form A=b:M

i=1a(i, j)i=L(j), i ,j = 1, . . . , M , withAji=a(i, j), bj =L(j)

a(v, v) =a Mi=1

ii, M

j=1jj

= Mi,j=1

ia (i, j) j = AL(v) =L

M

j=1jj

= M

j=1jL (j) =b ,v Vh,

where

is the dot product in

M.

Problem (Mh): 12 A b = minM

12 A b

Properties of matrixA: A=a(v, v) v2V >0 (forv= 0) A >0, for= 0

Ais positive definite.

a(i, j) =a(j, i) Ais symmetric.There is a unique solutionof the system A=b.



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Stability condition for the solution of problems (Vh) and (Mh)

There exists a unique solutionuh Vhto the equivalent problems(Vh) and (Mh). Further, the following stability estimate holds:

uhV

The stability estimate follows by choosing v = uh ina(uh, vh) = L(vh) and using the V -ellipticity condition fora(., .) and the L-continuity condition, i.e.

uh2V a(uh, uh) =L(uh) uhV From the above equation for uhV= 0, the stability condition

is derived.

The above stability estimate can be viewed as the theoreticalbasis for the success of the FE method



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Error estimate

TheoremLetu Vbe the solution of (V) anduh Vhthat of (Vh) whereVh V.Then:

u

uh

V

u

v

V

v

Vh Error estimate

Recall that:

appears in thea(., .)-continuity condition:

|a(v, w)

|

v

V

w

V,

v, w

V

appears in the ellipticity condition for a(., .):a(v, v) v2V, v V



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Proof of the error estimate

We can easily show that: a(u uh, w) = 0 w Vh. For an arbitraryv Vh, definew=uh v. Using theVellipticity and

continuity conditions ofa(., .) and thata(u uh, w) = 0, we obtain:

u uh2Vellipticity a(u uh, u uh) +

= 0 a(u uh, w)

= a(u uh, u uh+w) =a(u uh, u v)continuity

u

uh

V

u

v

V

Division byu uhV= 0 provides the required error estimate. Can choosev =huwhere huVh is a suitable interpolant ofu. We

would then like to calculate the interpolation error

u

hu

V.

The FEM error estimate can now take the following form:u uhV u huV Error estimate



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The energy norm:va=

a(v, v), v V

Define the energy norm.a as follows:v2a=a(v, v), v V. .a is equivalent to.V:

c

v

V

v

a

C

v

V,

v

V, withc=

andC=

The corresponding inner product is defined as: (v, w)a=a(v, w). We showed thatuh is the projection ofuontoVh, i.e.:

(u

uh, w)a= 0

w

Vh

Using (uuh, w)a= 0, w Vhand the Cauchy inequality, we obtain:u uh2a = (u uh, u uh)a= (u uh, u uh)a+ (u uh, w)a

= (u

uh, u

uh+w)a= (u

uh, u

v)a

u uhau va u uha u va, v Vhi.e. uh is a best approximation ofuin the energy norm.



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Example 1:u+u=f in , un = 0 on andf L2()

V =H1() and the following variational problem is obtained:a(v, w) =

[v w+vw] dx

L(v) = fvdx

a(., .) is a symmetric bilinear form on V V. Lis a linear form.

a(v, v) =

v

2

H

1

() a(., .) is V

elliptic and using Cauchys in-

equality we obtain:

a(v, w) a(v, v)12 a(w, w)12 = vH1()wH1() ais continuous

The continuity ofLis shown using the Cauchy inequality in L2:

|L(v)| fvdx fL2()vL2() fL2()vH1()



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Example 2:u=fonI= (0, 1), u(0) =u(1) = 0 and f L2(I)

V =H10 (I) and the following variational problem is obtained:a(v, w) =

Iv

wdx, L(v) =If vdx

a(., .) is obviously symmetric and bilinear andLis linear. The continuity

ofLis derived as in Example 1.

The continuity ofa(., .) is shown as follows:|a(v, w)| vL2(I)wL2(I) vH10 (I)wH10 (I)

TheVelliptic condition fora(., .) can be shown using the fact thatIv

2dx I(v)2dx v H10 (I) a(v, v) =I(v)2dx 12

Iv

2dx+

I(v)2dx

= 12vH10 (I),v H10 (I).

IfVh consists of piecewise linear functions onIanduis smooth enough:u uhH1() Ch



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Proof of

Iv2dx I(v)2dx, v H10 (I)

Usingv(0) = 0, we can write: v(x) =v(0) + x0

v(y)dy= x0

v(y)dy

Using Cauchys inequality:

|v(x)| =|x

0v(y)dy|

10|v(y)|dy

(1

012dy)

12 (

10

(v(y))2dy)12 = (

10

(v(y))2dy)12

Taking the square and then integrating the last equation from 0 to 1:10

v2(x)dx 10(v(y))2dy

The above proof needs the boundary condition v(0) = 0 (e.g. takev= 1to see that the inequality does not stand!).

In order to control the norm of the function vby the norm of the derivativev we need a fixed point to start from.



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Example 3:u=f in 2, u= 0 on , withf L2()

Here,V =H10 (), witha(v, w) = v wdx, L(v) = fvdx a(., .) is symmetric & bilinear andLis linear. For the continuity ofa:

|a(v, w)

| |v

|L2()

|w

|L2()

v

H1

0

()

w

H1

0

()

TheVellipticity can be shown using Poincares inequality: v

2dx c |v|2dxv H10 ()

wherec is a constant. Indeed, we can write:

a(v, v) |v|2dx 1

c+ 1

v2 + |v|2 dx 1

c+ 1v2H1()

For piecewise continuous functions in and usufficiently smooth:

u uhH1() Ch



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Example 4: d4u

dx4 =f, x

I= (0, 1),

u(0) =u(0) =u(1) =u(1) = 0 with f L2(I) Introduce: H2(I) ={v L2(I) :v, v L2(I)}. Then V = H20 (I),

where:

H20 (I) =v H2(I) :v(0) =v(0) =v(1) =v(1) = 0

Findu V such thata(u, v) =L(v)v V, wherea(v, w) =

Iv

wdx, L(v) =If vdx

The symmetry ofa(., .) and continuity ofa(., .) andLare easy to show. Using twice the Poincare inequality andv(0) =v(0) = 0, we can write:

Iv

2dx

I(v

)2dx

I(v

)2dx,

v

H20 (I)

v2H2(I)=I{v2 + (v)2 + (v)2}dx 3 I(v)2dx 3a(v, v)(ellipticity condition forawith= 1/3)



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Example 5: The biharmonic problem

u=f in , u= u

n = 0 on

Physical models: Clamped plate under transverse load,Stokes equations in fluid mechanics, etc.

HereV =H20 () = v H2 () :v= vn = 0 on . Use:(u)vdx= un vds uvnds + uvdx = uvdx. (V): Find u V such that: a(u, v) =L(v), with:

a(u, v) =

uvdx andL(v) =

fvdx Easy to show the symmetry ofa(., .) and continuity ofa(., .) andL. To show the V-ellipticity ofa(., .) need to first show the Poincare inequality:

v

2H2()

c

(v)2dx,

v

H20 (), for some constant c

(see derivation at the end of the lecture).



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Example 6: A convection-diffusion problem

u+1u

x1 +2u

x2 +u=f in , u= 0 on ,, constants with >0 Convection in the direction= (1, 2). Assume||/is small. We next

simplify by taking= 1.

v

V = H1

0

() and (V): Find u

V: a(u, v) =L(v)

v

V, where:

a(v, w) ={v w+ 1 vx1 +2 vx2 +v

w}dx, L(v) =

fvdx

Note that: {1 vx1 +2

vx2

} v dx ={v2(1n1+2n2}ds

{1 vx1 +

2v

x2

}v dx

{1

vx1

+2v

x2

}v dx= 0.

Thus: a(v, v) =

|v|2 +v2

dx= v2H1() a(., .) isV-elliptic.

The stiffness matrixA is not symmetric. As noted earlier (p. 7), there stillexists a unique solutionuh

Vh to (Vh): a(uh, v) =L(v)

v

Vh.

The following error estimate is obtained (= 1):u uhH1() u vH1() v Vh



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Example 7: A heat conduction problem:

qi= ki(x)u

xi in (no-sum oni)with divq=f in , u= 0 on 1, q n= g on 2

With the ki non-constant, this is an example of a partial dif-ferential equation with variable coefficients.

HereV = v H1() :v= 0 on 1

with

a(u, v) =L(v), v V, where:a(v, w) =

3

i=1

ki(x)vx

i

wx

i

dx,

L(v) =

fvdx+

2gvds.

The general conditions are satisfied under the following hy-pothesis: There are positive constants candCsuch that

c ki(x) C, x , i= 1, 2, 3,f L2(), g L2(2) and the area of 1is positive.



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The Poincare inequality:

v2dx c()

|v|2dx, v H10 ()

Let be an open bounded region. The region = + can be enclosedin a square regionK: K= {(x1, x2) :x1 [0, a], x2 [0, a]}.

v(x) is extended onto K be setting v = 0 on K . Hence, v iscontinuously differentiable inKand vanishes on the boundary ofK.

Using: v(x1, x2) = x10

v(, x2)dand the Cauchy inequality, we obtain:

|v(x1, x2)

|2 =

|

x1

0

v

(, x2)d

|2

((

x1

0

12d)1/2(x1

0 |

v

|2d)1/2)2

x1x10

|v

|2d aa

0|v

|2d



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The Poincare inequality

Integrating with respect tox1andx2 overK, we obtain:a

0

a0

|v(x1, x2)|2dx1dx2 aa

0

a0

(a

0|v(, x2)

|2d)dx1dx2

= a2 a0

a0|v

|2ddx2

= a2K

| vx1

|2dx1dx2

Similarly we obtain: a0 a0 |v(x1, x2)|2dx1dx2 a2

K

| vx2|2dx1dx2. From the above two inequalities and using v= 0 onK , we obtain:

|v(x1, x2)|2dx1dx2 a22

(| vx1|2 + |v

x2|2)dx1dx2

We conclude that: v2dx c() |v|2dx, v H10 (), wherec()>0 is a constant that depends on (cdictates the value ofaonly).



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The Poincare inequality:v2H2() C()(v)2dx, v H20 ()

Fromv= 0 on vs on . With

vn = 0 on

vx1 =

vx2 = 0 on .

Using these boundary conditions and integrating by parts first inx1 andthen inx2, one can easily show that:

2vx21

2vx22

dx1dx2=

( 2v

x1x2)2 dx1 dx2,

v

H20 () (a)

Using the earlier Poincare inequality for functions in H10 (), we obtain:(

vx1

)2 dx1 dx2 c(){(2vx21)2 + (

2vx1x2

)2}dx1dx2 (b)(

vx2

)2 dx1 dx2 c(){(2vx22)2 + (

2vx1x2

)2}dx1dx2 (c) v2 dx1dx2 c(){( vx1 )2 + ( vx2 )2}dx1dx2 (d)

From (b)+(c) and (a): |v|2 dx1 dx2 c()(v)2dx1dx2 (e) From (d) and (e): v2 dx1 dx2 c2()(v)2dx1dx2 (f) v2H2() = {v2 + (v)2 + (

2vx21 )

2 + (2

vx22 )2 + 2(

2vx1x2)2}dx1 dx2

v2H2()={v2+(v)2+(v)2}dx1dx2 (c2 +c+ 1) C

(v)

2dx1dx2


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The semi-norm|v|1:|v|1=

[| vx1|2 + |v

x1|2]dx1dx2

c2v1 |v|1 c1v1, v H10 () Based on the first Poincare inequality, we can write that:

v2dx1dx2 C2()

(| vx1|2 + |

vx2

|2)dx1dx2 Definec22= 12 min{1, 1C2}. We thus can conclude:

c22

[v2dx+ | vx1|2 + |v

x1|2]dx1dx2

(| v

x1|2 + | v

x2|2)dx1dx2

It is also clear that (e.g. forc1= 1):|v|1 c1v1, v H1() We can now conclude that:

c2

v

1

|v

|1

c1

v

1,

v

H10 ()

Forv H10 (), the seminorm |v|1is thus equivalent to the norm v1.


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