INF-SUP CONDITIONS

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LONG CHEN

1. Inf-sup Conditions

1.1. Variational problem in the mixed form

1.2. Babu

ska theory I2

1.3. Brezzi theory I

1.4. Application to Stokes equations

References

1. INF-SUPCONDITIONS

In this section, we shall study the well posedness of the weak formulation of the steady- state Stokes equations u+rp=f;(1) divu= 0;(2) whereucan be interpreted as the velocity field of an incompressible fluid motion, and pis then the associated pressure, the constantis the viscosity coefficient of the fluid. For simplicity, we consider homogenous Dirichlet boundary condition for the velocity, i.e. uj@ = 0and= 1. The conditions for the well posedness is known as inf-sup condition or Ladyzhenskaya-Babu ska-Breezi (LBB) condition.

Multiplying test functionv2H10(

)to the momentum equation (1) andq2L2( to the mass equation ( 2 ), and applying integration by part for the momentum equation, we obtain the weak formulation of the Stokes equations: Findu2H10( )and a pressure p2L2( )such that (ru;rv)(p;divv) = (f;v);for allv2H10( (divu;q) = 0for allq2L2(

1.1.Variational problem in the mixed form.We shall consider an abstract mixed vari-

ational problem first. LetVandPbe two Hilbert spaces. For given(f;g)2V0P0, find (u;p)2VPsuch that: a(u;v) +b(v;p) =hf;vi;for allv2V; b(u;q) =hg;qi;for allq2P:

Let us introduce linear operators

A:V7!V0;ashAu;vi=a(u;v)Date: February 22, 2014.

2 LONG CHEN

and

Written in the operator form, the problem becomes

Au+B0p=f;(3)

B u=g;(4)

or (5)A B0 B0 u p =f g We shall study the well posedness of this abstract mixed problem.

1.2.Babuska theory I.Let

a(;) :UV7!R be a bilinear form on two Hilbert spacesUandV. It will introduce two linear operators

A:U7!V0;andA0:V7!U0

byhAu;vi=hu;A0vi=a(u;v): We consider the operator equation: Given af2V0, findu2Usuch that (6)Au=f;inV0; or equivalently a(u;v) =hf;vifor allv2V: To begin with, we have to assumeAis continuous. We skip the subscript of the norm for different spaces. It should be clear from the context. (C)The bilinear forma(;)is continuous in the sense that a(u;v)Ckukkvk;for allu2U;v2V: The minimal constant satisfies the above inequality will be denoted bykak. With this condition, it is easy to check thatAandA0are bounded operators andkAk=kA0k=kak. The following conditions discuss whenA1is well defined and the norm ofkA1k.

Existence of a solution to (

6 )()Ais onto()A0is into() (E) infv2Vsup u2Ua(u;v)kukkvk=E>0:

Uniqueness of the solution to (

6 )()Ais into()A0is onto() (U) infu2Usup v2Va(u;v)kukkvk=U>0: The equivalence:Ais onto()A0is into, can be easily verified using the definition of the dual operator. The difficulty is to characterize the into by the inf-sup condition.

Let us introduce the notation

N(A) = ker(A) =fu2U:Au= 0gwhich forms a linear subspace ofU.

For a subsetZU,Z:=ff2U0;hf;ui= 0;for allu2Zg.

For a subsetZU,Z?:=fv2U;(v;u) = 0;for allu2Zg.

Roughly speaking, bothZandZ?are "orthogonal" toZ. But they are in different spaces. Exercise 1.1.For a linear and continuous operatorBdefined on a Hilbert spaceU, write the projection operatorP:U!ker(B)andP?:U!ker(B)?in terms ofB.

INF-SUP CONDITIONS 3

Theorem 1.2.For a continuous bilinear forma(;), the problem (6) is well-posed if and only if (E) and (U) hold. Furthermore if (E) and (U) hold, then kA1k=k(A0)1k=1 U=1 E=1; and thus kuk 1 kfkV0: Proof.We will prove the following conditions are equivalent. (1) ( E) (2)kA0vkU0Ekvk;for allv2V: (3)A0:V7!N(A)is an isomorphism. (4)A:N(A)?7!V0is an isomorphism. (1)()(2). It can be proved by the definition of the dual norm kA0vkU0= sup u2Uhu;A0vikuk= sup u2Ua(u;v)kuk: (2)=)(3). An obvious consequence of (2) isA0is an injection. We now prove that (2) also implies that the rangeR(A0)is closed and thus form a linear subspace ofU0. Choosing a convergent sequencefA0vkg, by (2), we knowfvkgis also a Cauchy sequence and thus converges to somev2V. The continuity ofA0shows thatA0vkconverges toA0vand thus

R(A0)is closed.

Wecan thenconclude thatA0:V7!R(A0)isan isomorphism. Next weproveR(A0) = N(A). For a subsetZU, let us recall the definitionZ:=ff2U0;hf;ui=

0;for allu2Zg. Using the definition ofA0

hu;A0vi=hAu;vi; we see thatR(A0)N(A). IfR(A0)N(A), i.e. there existsf2N(A)nR(A0). SinceR(A0)is closed, by Hahn-Banch theorem and Risez representation theorem, there existsu2Usuch thathu;A0vi= 0;for allv2Vandhu;fi= 1. Buthu;A0vi= hAu;vi= 0;for allv2Vimplies thatAu= 0;i.e.u2N(A)and thushu;fi= 0for f2N(A). Contradiction. (3)=)(2). By the assumption,(A0)1:N(A)7!Vis a well defined and bounded linear operator. Thus kvk=k(A0)1A0vk CkA0vkU0: (3)()(4). Obviously (4)()A0:V7!(N(A)?)0is an isomorphism. Thus we only need to show the isomorphism(N(A)?)0=N(A). For anyf2(N(A)?)0, we definefsuch thathf;vi:=hf;P?vifor allv2V, whereP?:U!N(A)?is the projection. Thenf2N(A). One can easily provef!fdefines an isometric isomorphism. The uniqueness is obtained by the dual argument. If both (E) and (U) hold, then kA1k=k(A0)1k=1 U=1 E=1:

4 LONG CHEN

Let us take the inf-sup condition (E) as an example to show how to verify it. To verify (E), one way is (7) for allv2V;findu2U; s:t: a(u;v)kukkvk: We shall present a slightly different characterization of (E). With this characterization, the verification is then transformed to a construction of a suitable function. Theorem 1.3.The inf-sup condition (E) is equivalent to that for anyv2V, there exists u2U, such that (8)a(u;v)C1kvk2;andkuk C2kvk: Proof.Obviously (8) will imply (7) with=C1=C2. We now prove (E) implies (8). Recall that (E) is equivalent toA:N(A)?7!V0is an isomorphism. We identifyVasV0 by the Riesz mapJ:V7!V0such thathJv;vi= (v;v) =kvk2. Then for a givenv2V, we can findu2Usuch thatAu=Jvand thusa(u;v) =hAu;vi=hJv;vi=kvk2. Sinceu2N(A)?, we also haveA1is bounded and thuskuk=kA1vk Ckvk. In ( 8 )ucould dependent onvin a subtle way. A special case isu=vwhenU=V. It is known as the corcevity a(u;u)kuk2: The corresponding result is known as Lax-Milgram Theorem. Corollary 1.4(Lax-Milgram).For a bilinear forma(;)onVV, if it satisfies (1)

Continuity: a(u;v)C1kukkvk;

(2)

Cor cevity:a(u;u)C2kuk2,

then for anyf2V0, there exists a uniqueu2Vsuch that a(u;v) =hf;vi; and kuk C1=C2kfkV0: The most simplest case is the bilinear forma(;)is symmetric and positive definite. Thena(;)defines a new inner product. Lax-Milgram theorem is simply the Riesz repre- sentation theorem.

1.3.Brezzi theory I.We consider the mixed problem

Au+B0p=f;(9)

B u=g;(10)

First we assume all bilinear forms are continuous. (C)The bilinear forma(;);andb(;)are continuous a(u;v)Ckukkvk;for allu;v2V; b(v;q)Ckvkkqk;for allv2V;q2P: We use the decompositionV=N(B)N(B)?to writeu=u0+u1; u02N(B) andu12N(B)?. Then (10) becomesBu1=g. Sinceu12N(B)?, the existence and uniqueness ofu1is equivalent toBis onto orB0is into, i.e. the following inf-sup condition

INF-SUP CONDITIONS 5

(B) infq2Psup v2Vb(v;q)kvkkqk= >0 After we get a uniqueu1, to determine a uniqueu0, we restrict the test function space of ( 9 ) toN(B). Sincehv;B0qi=hBv;qi= 0forv2N(B), we get the following variational form: findu02N(B)such that (11)a(u0;v) =hf;vi a(u1;v);for allv2N(B): The existence and uniqueness ofu0is then equivalent to the two inf-sup conditions for a(u;v)on spaceZ=N(B). (A) inf u2Zsup v2Za(u;v)kukkvk= infv2Zsup u2Za(u;v)kukkvk= >0:

After we determine a uniqueuin this way, we solve

(12)B0p=fAu to getp. Sinceu0is the solution to (11), the right hand sidefAu2N(B). Thus we requireB0:V7!N(B)is an isomorphism which is also equivalent to the condition (B). Theorem 1.5.The continuous variational problem (5) is well-posed if and only if (A) and (B) hold. When (A) and (B) hold, we have the stability result kukV+kpkP.kfkV0+kgkP0: The following characterization of the inf-sup condition for the operator B is useful. The verification is again transfered to a construction of a suitable function. The proof is similar to that in Theorem 1.3 and thus skipped here. Theorem 1.6.The inf-sup condition (B) is equivalent to that: for anyq2P, there exists v2V, such that (13)b(v;q)C1kqk2;andkvk C2kqk: Note thatv=v(q)and the construction ofvmay not be straightforward for some problems.

1.4.Application to Stokes equations.Let us return to the Stokes equations. The setting

for the Stokes equations:

Spaces:

V=H10(

);P=L20( ) =fq2L2( );Z q= 0:g:

Bilinear form:

a(u;v) =Z ru:rv; b(v;q) =Z (divv)q:

Operator:

A= :H10(

)7!H1( );hAu;vi=a(u;v) =(ru;rv);

B=div :H10(

)7!L20( );hBv;qi=b(v;q) =(divv;q); B

0=r:L20(

)7!H1( );hv;rqi=(divv;q):

6 LONG CHEN

Remark 1.7.A natural choice of the pressure space isL2( ). Note thatZ divvdx=Z vndS= 0 due to the boundary condition. Thus div operator will mapH10( )into the subspace L 20( ). InL20( )the pressure of the Stokes equations is unique. But inL2( ), it is unique up to a constant. Remark 1.8.By the same reason, for Stokes equations with non-homogenous Dirichlet boundary conditionuj@ =g, the datagshould satisfy the compatible conditionZ gndS=Z divudx= 0: The continuity ofa(;)is trivial. The continuity ofb(;)can be proved using the identity in the following exercise.

Exercise 1.9.Prove

=grad div+curlcurl holds as an operator fromH10!H1. Namely for allu;v2H10 (ru;rv) = (divu;divv) + (curlu;curlv): We need to verify two inf-sup conditions. (A) is easy by the Poinc

´are inequality.

Lemma 1.10.Inf-sup conditions (A) is satisfied since the following inequalityZ ru:ruCkuk1;for allu2H10( The key is the inf-sup condition (B) which is equivalent to eitherquotesdbs_dbs19.pdfusesText_25

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What is DLST and how does it work?

What does the 2501st DLD do?

What is a DL1 instance?

INF-SUP CONDITIONS

LONG CHEN

CONTENTS

1. Inf-sup Conditions

1.1. Variational problem in the mixed form

1.2. Babu

1.3. Brezzi theory I

1.4. Application to Stokes equations

References

1. INF-SUPCONDITIONS

Multiplying test functionv2H10(

1.1.Variational problem in the mixed form.We shall consider an abstract mixed vari-

Let us introduce linear operators

A:V7!V0;ashAu;vi=a(u;v)Date: February 22, 2014.

2 LONG CHEN

Written in the operator form, the problem becomes

Au+B0p=f;(3)

B u=g;(4)

1.2.Babuska theory I.Let

A:U7!V0;andA0:V7!U0

Existence of a solution to (

Uniqueness of the solution to (

Let us introduce the notation

For a subsetZU,Z:=ff2U0;hf;ui= 0;for allu2Zg.

For a subsetZU,Z?:=fv2U;(v;u) = 0;for allu2Zg.

INF-SUP CONDITIONS 3

R(A0)is closed.

0;for allu2Zg. Using the definition ofA0

4 LONG CHEN

Continuity: a(u;v)C1kukkvk;

Cor cevity:a(u;u)C2kuk2,

1.3.Brezzi theory I.We consider the mixed problem

Au+B0p=f;(9)

B u=g;(10)

INF-SUP CONDITIONS 5

After we determine a uniqueuin this way, we solve

1.4.Application to Stokes equations.Let us return to the Stokes equations. The setting

Spaces:

V=H10(

Bilinear form:

Operator:

A= :H10(

B=div :H10(

0=r:L20(

6 LONG CHEN

Exercise 1.9.Prove

´are inequality.