Variational Formulation of Boundary Value Problems .Chapter 4 Variational Formulation of Boundary

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Transcript of Variational Formulation of Boundary Value Problems .Chapter 4 Variational Formulation of Boundary

  • Chapter 4

    Variational Formulation ofBoundary Value Problems

    4.1 Elements of Function Spaces

    4.1.1 Space of Continuous Functions

    N is a set of non-negative integers.

    1) An n-tuple = (1, , n) Nn is called a multi-index.

    2) The length of is

    || :=n

    j=1

    j.

    3) 0

    = (0, , 0).

    Set D := (/x1)1 (/xn)n .

    Example 4.1.1. Assume n = 3, = (1, 2, 3) N3, u(x1, x2, x3) : R3 R.

    29

    What is

    ||=3 Du ?

    || = 3 =3

    j=1

    j = 3.

    = = (3,0,0), (0,3,0), (0,0,3), (2,1,0), (2,0,1), (0,2,1),(1, 2, 0), (1, 0, 2), (0, 1, 2), (1, 1, 1).

    =

    ||=3

    Du =3u

    x31+

    3u

    x32+

    3u

    x33+

    3u

    x21x2+

    3u

    x21x3+

    3u

    x22x3

    +3u

    x1x22+

    3u

    x1x23+

    3u

    x2x23+

    3u

    x1x2x3.

    This sort of list can get very long. Hence D is useful notation.

    Definition 4.1.2. Let be an open set in Rn. Let k N. Define spaces Ck(),Ck() and C() by

    Ck() := {u : R | Du is continuous in for all || k},Ck() := {u : R | Du is continuous in for all || k},

    C() := {u : R | Du is continuous in for all Nn},

    where is the closure of . If is bounded, = , where is the boundaryof . We denote C() = C0() and C() = C0().

    Example 4.1.3. Set I := (0, 1) and u(x) := 1/x2 x I. Then clearly for allk 0 u Ck(I). However, in I = [0, 1] u is not continuous at 0. Thus, u / C(I).

    Definition 4.1.4. For a bounded open set Rn, k N and u Ck(), thenorm uCk() is defined by

    uCk() :=

    ||k

    supx

    |Du(x)|.

    Example 4.1.5. Let I = (0, 1), u(x) := x, u C(I). Then, supxI |u(x)| = 1.

    Definition 4.1.6. For an open set Rn and u C() the support of u denotedby supportu ( Rn) is defined by

    supportu := the closure of {x | u(x) *= 0}.

    Remark. The support of u is the smallest closed subset of such that u = 0 in\supportu.

    30

  • Example 4.1.7. 1) Let 0 = x0 < x1 < < xn = 1 be a patition of [0, 1]. Definej(x) : [0, 1] R by

    j(x) :=

    x xj1h

    x (xj1, xj),xj+1 x

    hx (xj , xj+1),

    0 elsewhere.

    Then we see j C(I) and supportj = [xj1, xj+1].

    2) Define w(x) : Rn R by

    w(x) :=

    e

    11 |x|2 |x| < 1,

    0 otherwise.

    Then we see w C(Rn) and supportw = {x Rn | |x| 1}.

    Definition 4.1.8. Define Ck0 () ( Ck()) by

    Ck0 () := {u Ck() | supportu is a bounded subset of }.

    4.1.2 Spaces of Integrable Functions

    Definition 4.1.9. Let denote an open subset of Rn and assume 1 p < . Wedefine a space of integrable functions Lp() by

    Lp() :={

    v : R |

    |v(x)|pdx < +

    }.

    The space Lp() is a Banach space with norm Lp() defined by

    vLp() =(

    |v(x)|pdx

    )1/p.

    Especially the space L2() is a Hilbert space with inner product , L2() definedby

    u, vL2() :=

    u(x)v(x)dx

    and norm L2() defined by uL2() :=u, vL2().

    31

    We have Minkowskis inequality as follows. For u, v Lp(), 1 p <

    u + vLp() uLp() + vLp().

    We also have Holders inequality. For u Lp() and v Lq(), 1 p, q < with1/p + 1/q = 1

    u(x)v(x)dx

    uLp()vLq().

    Now, any two integrable functions are equal if they are equal almost everywhere,that is, they are equal except on a set of zero measure. Strictly speaking, Lp()consists of equivalent classes of functions.

    Example 4.1.10. Let u, v : (1, 1) R be

    u(x) ={

    1 x (0, 1),0 x (1, 0], v(x) =

    {1 x [0, 1),0 x (1, 0),

    The functions u and v are equal almost everywhere, since the set {0} where u(0) *=v(0) has zero measure in the interval (1, 1). So u and v are equal as integrablefunctions in (1, 1).

    Suppose that u Ck(), where is an open set of Rn. Let v C0 (). Then wesee by integration by parts

    Du(x)v(x)dx = (1)||

    u(x)Dv(x)dx,

    where || k.

    Definition 4.1.11. A function u : R is locally integrable if u L1(U) forevery bounded open set U such that U .

    Definition 4.1.12. Suppose u : R is locally integrable and there is a locallyintegrable function w : R such that

    w(x)(x)dx = (1)||

    u(x)D(x)dx

    for all C0 ().

    Then the weak derivative of u of order denoted by Du is defined by Du = w.

    32

  • Note that at most only one w satisfies (4.1.12) so the weak derivative of u is well-defined. Indeed, the following DuBois-Raymond lemma shows such w is unique.

    Lemma 4.1.13. (DuBois-Raymond) Suppose is an open set in Rn and w : Ris locally integrable. If

    w(x)(x)dx = 0

    for all C0 (), then w(x) = 0 for a.e x .

    We will useD for both classical and weak derivatives.

    Example 4.1.14. Let = R. Set u(x) = (1 |x|)+, x , where

    (x)+ :={

    x x > 0,0 x 0.

    Thus,

    33

    u(x) :=

    0 x 1,1 + x 1 x 0,1 x 0 x 1,0 1 x.

    !

    "y

    x##

    ###

    $$

    $$$

    -1 10

    1

    Clearly we see that u is locally integrable, u C() and u / C1(). However, itmay have a weak derivative. Take any C0 () and = 1. Then,

    (1)||

    u(x)D(x)dx =

    u(x)(x)dx

    = 1

    1(1 |x|)(x)dx

    = 0

    1(1 + x)(x)dx

    1

    0(1 x)(x)dx

    = 0

    11 (x)dx +

    1

    0(1)(x)dx

    =

    w(x)(x)dx,

    where

    w(x) :=

    0 x < 1,1 1 < x < 0,1 0 < x < 1,0 1 < x.

    Here we do not worry about the points x = 1, 0, 1, since they have zero measure.Thus, u has its weak derivative Du = w.

    Definition 4.1.15. Let k be a non-negative integer and p [0,). The spaceW k,p() defined by

    W k,p() := {u Lp() | Du Lp() || k}

    34

  • is called a Sobolev space. It is a Banach space with the norm

    uW k,p() :=

    ||k

    DupLp()

    1/p

    .

    Especially, when p = 2, we denote Hk() as W k,2(). It is a Hilbert space with theinner product

    u, vHk() :=

    ||k

    Du, DvL2().

    Of special interest are H1() and H2(). If = (a, b), R, we see that

    u, vH1() = u, vL2() + Du,DvL2()

    = b

    au(x)v(x)dx +

    b

    aDu(x)Dv(x)dx.

    u, vH2() = u, vL2() + Du,DvL2() + D2u, D2vL2()

    = b

    au(x)v(x)dx +

    b

    aDu(x)Dv(x)dx +

    b

    aD2u(x)D2v(x)dx.

    Remark. 1) By using Holders inequality, we can prove Cauchy-Schwarz inequalityfor the inner product of Hk() as follows.

    |u, vHk()|

    ||k

    |Du, DvL2()|

    ||k

    DuL2()DvL2()

    ||k

    Du2L2()

    ||k

    Dv2L2()

    = uHk()vHk().

    2) Let = (a, b) R and u H1(). Then u C(). In higher space dimensionsthis statement is no longer true.

    4.2 One Dimensional Problem: Dirichlet condition

    Let = (0, 1), p(), q() C() and f() L2(). Note that = {x = 0} { x =1}. We consider the following problem.

    35

    Find u : R such that{ d

    dx(p

    du

    dx) + qu = f, x ,

    u(x) = 0, x .(BVP)

    Specifying the value of u at boundary points is said to be a Dirichlet boundarycondition. Now the methodology is

    1) multiply the equation by a test function, integrate by parts and use boundaryconditions appropriately,

    2) identify V , a(, ) and l(),

    3) verify, if possible, the assumptions of Lax-Milgram.

    = Unique existence to the variational formulation of the BVP.

    Let : R be sufficiently smooth. We will call our test function. Let us followthe methodology.

    1)

    f(x)(x)dx =

    ( d

    dx(p(x)

    du(x)dx

    )(x) + q(x)u(x)(x))

    dx

    =[p(x)du(x)

    dx(x)

    ]x=1

    x=0

    +

    (p(x)

    du(x)dx

    d(x)dx

    + q(x)u(x)(x))

    dx.

    We want to eliminate the term [p(x)du(x)/dx(x)]x=1x=0, so we suppose that thetest function satisfies the same Dirichlet conditions as u, i.e, (0) = (1) = 0.Then we have that

    (p(x)

    du(x)dx

    d(x)dx

    + q(x)u(x)(x))

    dx =

    f(x)(x)dx

    for any test function . We want u, v to be from the same space. For the term udx to make sense, we need u, L

    2(). For the derivatives du/dx, d/dxto make sense, we take this further, so u, H1().

    2) Let us chooseV :=

    { H1() | v(0) = v(1) = 0

    },

    36

  • whereH1() =

    { L2() | D L2()

    }.

    We equip the inner product , V := , H1(). Let us define

    a(u, v) :=

    (pDuDv + quv)dx,

    l(v) :=

    fv dx.

    Moreover, assume that p(x) p0 > 0, q(x) q0 > 0 for all x .

    3) We will verify the assumptions of Lax-Milgrams theorem.

    i) For V and f L2(), we see by Cauchy-Schwarz inequality that

    |l()| = |

    fdx|

    fL2()L2() fL2() (2L2() + D

    2L2())

    = clV ,

    where we have set cl := fL2(). Thus, l : V R is bounded. Clearly l islinear, i.e, l( + ) = l() + l() for any , V and , R.

    ii) Obviously a(, ) : V V R is bilinear. Moreover a(, ) is bounded. Indeed,

    |a(, )| |

    pDDdx| + |

    qdx|

    maxx

    |p(x)|

    |DD|dx + max

    x|q(x)|

    ||dx

    maxx

    |p(x)|DL2()DL2() + maxx

    |q(x)|L2()L2()

    C(DL2()DL2() + L2()L2())

    CD2L2() +

    2L2()

    D2L2() +

    2L2()

    = CH1()H1(),

    where we have set

    C := max{maxx

    |p(x)|,maxx

    |q(x)|}.

    37

    The bilinear form a(, ) is coercive, since for all V

    a(, ) =

    p|D|2dx +

    q||2dx

    p0

    |D|2dx + q0

    ||2dx

    = C2V ,

    where we have set C := min{p0, q0}.

    We can now apply Lax-Milgrams theorem to see that there uniquely exists asolution to the following problem

    (P) Find u V such that

    (pDuD + qu)dx =

    fdx,

    for any V .

    Remark. For V = H10 () we can use the norm V defined by

    2V =

    |D|2dx,

    since the following Poincare inequality holds:- there exist