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    16.21 Techniques of Structural Analysis andDesign

    Spring 2005Unit #9 - Calculus of Variations

    Raul Radovitzky

    March 28, 2005

    Let u be the actual configuration of a structure or mechanical system. usatisfies the displacement boundary conditions: u = u on Su. Define:

    u = u + v, where: : scalar

    v : arbitrary function such that v = 0 o n SuWe are going to define v as u, thefirst variationofu:

    u = v (1)

    Schematically:

    u1

    u2

    u

    a b

    u(a)

    u(b)

    v

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    As a first property of thefirst variation:du du dv= + dx dx dx

    so we can identify dvdx

    with the first variation of the derivative ofu:du dv =

    dx dx

    But:

    dv dv d

    dx=

    dx=

    dx(u)

    We conclude that:

    du d = (u)

    dx dx

    Consider a function of the following form:

    F= F(x, u(x), u(x))

    It depends on an independent variable x, another function ofx (u(x)) and itsderivative (u(x)). Consider the change in F, when u (therefore u) changes:

    F= F(x, u + u, u+ u) F(x, u, u)= F(x, u + v, u+ v) F(x, u, u)

    expanding in Taylor series:

    F F 1 2F 1 2FF= F+ v + v+

    u2(v)2 + (v)(v) + F

    u u 2! 2! uuF F

    = v + v+ h.o.t.u u

    First total variation of F:

    FF= lim

    0 F(x, u + v, u+ v) F(x, u, u)= lim

    0 Fv + Fv F F= lim u u = v + v 0 u u

    F FF= u + u

    u u

    2

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    Note that:

    dF(x, u + v, u+ v)F= d =0

    since:

    dF(x, u + v, u+ v) F(x, u + v, u+ v)=

    d uevaluated at = 0

    F(x, u + v, u+ v)v + v

    u

    dF(x, u + v, u+ v)=F(x, u, u)v + F(x, u, u)v

    d =0 u uNote analogy with differential calculus.

    (aF1 + bF2) = aF1 + bF2 linearity

    (F1F2) = F1F2 + F1F2

    etc

    The conclusions for F(x, u, u) can be generalized to functions of severaluiindependent variables xi and functions ui, xj :

    F xi, ui,uixj

    We will be making intensive use of these properties of the variational operator:

    d d dv du(u) = (v) = =

    dx dx dx dx

    udx = vdx = vdx = udx

    Concept of a functional

    b

    F(x, u(x), u

    (x))dxI(u) =a

    3

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    First variation of a functional:

    I= F(x, u(x), u(x))dx = F(x, u(x), u(x)) dx

    F FI= u + u dx

    u u

    Extremum of a functional

    u0 is the minumumof a functional if:

    I(u) I(u

    0

    )uA necessary condition for a functional to attain an extremum at u0 is:

    dII(u0) = 0, or (u0 + v,u0

    + v) = 0d =0

    Note analogy with differential calculus. Also difference since here we requiredF = 0 at = 0.d

    b

    F FI= u + u dx

    u u

    a

    Integrate by parts the second term to get rid ofu.

    b F d F d F I= u + u u dx

    u dx u dx ua b F bF d F = udx + u

    u

    dx u u aaRequire u to satisfy homogeneous displacement boundary conditions:

    u(b) = u(a) = 0

    Then: b F d F I= udx = 0,

    u

    dx ua

    4

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    u that satisfy the appropriate differentiability conditions and the homoge-

    neous essential boundary conditions. Then:

    F d F = 0

    u

    dx u

    These are the Euler-Lagrange equations corresponding to the variationalproblem of finding an extremum of the functional I.

    Natural and essential boundary conditions A weaker condition onu also allows to obtain the Euler equations, we just need:

    F bu = 0u a

    which is satisfied if:

    u(a) = 0 and u(b) = 0 as before u(b) = 0 and F(b) = 0

    uFu(a) = 0 and u(b) = 0

    F(a) = 0 and F(b) = 0u u

    Essential boundary conditions: u = 0, or u = u0 on SuSuNatural boundary conditions: F = 0 on S.u

    Example: Derive Eulers equation corresponding to the total po-tential energy functional = U+ V of an elastic bar of length L, Youngsmodulus E, area of cross section A fixed at one end and subject to a load Pat the other end. L EA du2

    (u) = dxPu(L)2 dx0

    Compute the first variation:

    EA du du = 2 dxPu(L)

    2

    dx dx

    5

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    Integrate by parts

    d du d du = EA u EA u dxPu(L)dx dx

    dx dx L du Ld du

    = u EA dx + EA u0

    Pu(L)0 dx dx dx

    Setting = 0, u / u(0) = 0:d du

    EA = 0dx dx

    du

    P= EA

    dx L

    Extension to more dimensions

    I= F(xi, ui, ui,j)dV V F FI= ui + ui,j dV

    V ui ui,j

    F F F

    = ui + ui ui dVV ui xj ui,j

    xj ui,j

    Using divergence theorem:

    F F F I= uidV+ uinjdS

    V ui

    xj ui,j Sui,j

    Extremum of functional Iis obtained when I= 0, or when:

    F F = 0 , and

    ui

    xj ui,j

    ui = 0 on Su

    Fui,j

    nj onSSu = StThe boxed expressions constitute the Euler-Lagrange equations correspond-ing to the variational problem of finding an extremum of the functional I.

    6