Ch. 6: Sect. 6.6: Euler's Equations with Auxilliary Conditions (Constraints). Sect. 6.7: The Delta Notation. - PowerPoint PPT Presentation

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Ch. 6: Sect. 6.6: Euler's Equations with Auxilliary Conditions (Constraints). Sect. 6.7: The Delta Notation.

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For example: (as in the examples! ... Results: A formalism in which the generalized Euler's Equations automatically ... In the new notation, the condition for ... – PowerPoint PPT presentation

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Title: Ch. 6: Sect. 6.6: Euler's Equations with Auxilliary Conditions (Constraints). Sect. 6.7: The Delta Notation.


1
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2
Euler Equations for Systems with Constraints
(Auxiliary Conditions) Section 6.6
  • Suppose we want the solution to the variational
    problem (find the paths such that J ? f dx is
    an extremum) with many dependent variables
    yi(x), (i 1,2, ,n)
  • Often, we also have additional Auxiliary
    Conditions or Constraints, which relate the
    dependent variables yi(x), the independent
    variable x in certain, specified ways.
  • For example (as in the examples!) Suppose we
    want to find the shortest path between 2 points
    on surface.
  • ? In addition to Eulers Eqtns giving relations
    between the variables, we also require that the
    paths satisfy the equation of the surface Say,
    g(yi,x) 0 (i 1,2, ,n)

3
  • In this case, we have Eulers Equations
  • (?f/?yi) - (d/dx)(?f/?yi?) 0 (i 1,2, ,n)
    (1)
  • We also have the Equations of Constraint In this
    case, the paths must be on the surface
  • g(yi,x) 0 (i 1,2, ,n) (2)
  • where, g(yi,x) function depending on the
    surface geometry
  • ? The n paths we seek, yi(x) (i 1, ,n) are
    functions which must simultaneously satisfy (1)
    (2)
  • For example, for the geodesic on sphere we have
  • g(x,y,z) x2 y2 z2 - r2 0

4
  • Next Goal Develop a general extension of the
    Euler Equation formalism to use with constraints.
  • Results A formalism in which the generalized
    Eulers Equations automatically include the
    constraints.
  • Cannot always (easily) use (1) (2) separately!
    Instead, go back to the formal derivation
    incorporate the constraints (2) early. After a
    lot of work, get a generalization of Eulers
    Equations (1).
  • A Special Case first 2 dependent variables
  • y1(x) y(x), y2(x) z(x)
  • ? The functional in the formalism is of the
    form
  • f f(y,y?,z,z?x)

5
  • Follow similar steps as in the previous
    derivations get (skipping several steps!)
  • (?J/?a) ? (?f/?y) -(d/dx)(?f/?y?)(?y/?a)
  • (?f/?z)-(d/dx)(?f/?z?)(?z/?a)dx 0
    (3)
  • Also have an Equation of Constraint
  • g(y,zx) 0 (4)
  • g(y,zx) a known function which depends the
    on problem!
  • (4) ? In (3), (?y/?a) (?z/?a) are not
    independent as we assumed (using functions ?i(x))
    in the previous derivation!
  • ? We cannot set the coefficients (in ) in (4)
    separately to zero, as we did in the previous
    derivation!

6
  • (?J/?a) ? (?f/?y) -(d/dx)(?f/?y?)(?y/?a)
  • (?f/?z)-(d/dx)(?f/?z?)(?z/?a)dx 0
    (3)
  • Still assume y(a,x) y(x) a ?1(x) z(a,x)
    z(x) a ?2(x)
  • ? (?y/?a) ?1(x) (?z/?a) ?2(x)
  • So (3) becomes
  • (?J/?a) ? (?f/?y) -(d/dx)(?f/?y?)?1(x)
  • (?f/?z)-(d/dx)(?f/?z?)?2(x)dx 0
    (3?)
  • But we also have the constraint g(y,zx)
    0 (4)
  • Compute total differential of g when a changes by
    da
  • dg ? (?g/?y)(?y/?a) (?g/?z)(?z/?a)da
  • Or dg (?g/?y)?1(x) (?g/?z)?2(x)da
  • But also, by (4) dg 0

7
  • Much manipulation!
  • (?J/?a) ? (?f/?y) -(d/dx)(?f/?y?)?1(x)
  • (?f/?z)-(d/dx)(?f/?z?)?2(x)dx 0
    (3?)
  • g(y,zx) 0 ? dg 0
    (4)
  • dg (?g/?y)?1(x) (?g/?z)?2(x)da 0
  • ? (?g/?y)?1(x) - (?g/?z)?2(x)
  • Or ?2(x)/?1(x) - (?g/?y)?(?g/?z)
    (5)
  • Put (5) into (3?) manipulate
  • (?J/?a) ?(?f/?y) - (d/dx)(?f/?y?) -(?f/?z)
    - (d/dx)(?f/?z?)
  • ? (?g/?y)?(?g/?z)?1(x)dx 0
    (6)
  • ?1(x) is arbitrary ? the integrand of (6)
    vanishes

8
  • Vanishing of the integrand of Eq. (6)
  • ? (?f/?y) - (d/dx)(?f/?y?)(?g/?y)-1
  • (?f/?z) - (d/dx)(?f/?z?)(?g/?z)-1
    (7)
  • Left side of (7) Derivatives of f g with
    respect to y y?.
  • Right side of (7) Derivatives of f g with
    respect to z z?.
  • y,y?,z z? These are functions of x only!
  • ? Define a function of x
  • - ?(x) ? Left side of (7) Right side of (7)

9
  • Left side of (7)
  • ? - ?(x) (?f/?y) - (d/dx)(?f/?y?)(?g/?y)
    -1 (8)
  • Right side of (7)
  • ? - ?(x) (?f/?z) - (d/dx)(?f/?z?)(?g/?z)-1
    (9)
  • Comment (8) (9) are formal expressions for
    ?(x). But, recall y y(x) z z(x) are the
    unknown functions which we are seeking!
  • ? ?(x) is also unknown (undetermined)
  • unless we already have solved the problem
    have found y(x) z(x)!

10
  • (8), (9) on the previous page ?
  • (?f/?y) - (d/dx)(?f/?y?) ?(x)(?g/?y) 0
    (10a)
  • (?f/?z) - (d/dx)(?f/?z?) ?(x)(?g/?z) 0
    (10b)
  • ? Eulers Equations with Constraints
  • Note We have formulas ((8), (9)) to compute ?(x)
    for a given f g. But these depend on the
    unknown functions (which we are seeking!) y(x)
    z(x).
  • ? ?(x) is UNDETERMINED until the problem is
    solved we know y(x) z(x). ? The problem
    solution depends on finding THREE functions
    y(x), z(x), ?(x). But we have 3 eqtns to use
    (10a), (10b), the eqtn of constraint g(y,zx)
    0
  • ?(x) ? A Lagrange Undetermined Multiplier is
  • obtained as part of the solution.

11
  • Summary For the case of 2 dependent variables
    1 constraint, Eulers Equations with Constraints
  • (?f/?y) - (d/dx)(?f/?y?) ?(x)(?g/?y) 0
    (a)
  • (?f/?z) - (d/dx)(?f/?z?) ?(x)(?g/?z) 0
    (b)
  • To find the unknown functions y(x) z(x), ?(x),
    solve (a) (b) simultaneously with the original
    equation of constraint
  • g(y,zx) 0 (c)

12
  • For the general case with constraints.
  • Let the number of dependent variables ? m
  • We want m functions yi(x),
    i 1,2, m
  • With m derivatives yi?(x) (dyi(x)/dx) i
    1,2, m
  • The functional is f f(yi(x),yi?(x)x), i
    1,2, m
  • Let the number of constraints ? n. ? There are n
    eqtns of constraint gj(yix) 0, i 1,2, m,
    j 1,2,
  • A derivation similar to the 2 dependent
    variable, 1 constraint case results in
  • n Lagrange multipliers ?j(x) (one for each
    constraint).
  • m Eulers Equations with Constraints

13
  • For the general case with constraints.
  • ? m Eulers Equations with Constraints
  • (?f/?yi) - (d/dx)(?f/?yi?) ?j?j(x)(?gj/?yi)
    0 (A)
  • i 1,2, m
  • ? n Equations of Constraint
  • gj(yix) 0 (B)
  • i 1,2, m, j 1,2, n
  • ? m n eqtns total (A) (B) with mn
  • unknowns yi(x), i 1,2, ..,m ?j(x), j
    1,2, ..,n

14
  • A Final Point About Constraints
  • Consider the constraint eqtn gj(yix) 0
    (B)
  • i 1,2, m, j 1,2, n
  • (B) is equivalent to a set of n differential
    equations (exact differentials of gi(yix))
  • ?i(?gj/?yi)dyi 0 i 1,2, m j 1,2, n
    (C)
  • In mechanics, constraint equations are often used
    in the form (C) rather than (B). Often (C) is
    more useful than (B)!

15
Example 6.5
  • A disk, radius R, rolls without slipping down an
    inclined plane as shown. Determine the equation
    of constraint in terms of the (generalized)
    coordinates y ?.
  • y ? are not independent.
  • They are related by y R?.
  • ? The constraint eqtn is
  • g(y,?) y - R? 0.
  • This is equivalent to the
  • differential versions
  • (?g/?y) 1 (?g/??) -R

16
The d Notation Section 6.7
  • Its convenient to introduce a (standard)
    shorthand notation for the variation. Going back
    to the general derivation, where we had (for a
    single dependent variable no constraints) for J
    having a max or a min
  • (?J/?a) ?(?f/?y) - (d/dx)(?f/?y?)?(x)dx
    (1)
  • (limits x1 lt x lt x2)
  • From this, we derived the Euler equation. We
    allowed the path to vary as y(a,x) ? y(0,x) a
    ?(x) Clearly, (?y/?a) ? ?(x)
  • Rewrite (1) (multiplying by da) as
  • (?J/?a)da ?(?f/?y) -(d/dx)(?f/?y?)(?y/?a)dad
    x (2)
  • (limits x1 lt x lt x2)

17
  • Introduce a Shorthand Notation
  • Define dJ ? (?J/?a)da and dy ? (?y/?a)da
  • ? Rewrite (2) as
  • dJ ?(?f/?y) -(d/dx)(?f/?y?)dydx (3)
  • (limits x1 lt x lt
    x2)
  • (3) is called The variation of J (dJ) in terms
    of the variation of y (dy).
  • In the general formulation, where we want to find
    condition for extremum of J ?f(y,y?x) dx
    (limits x1 lt x lt x2), follow the original
    derivation, but in this new notation. In this
    notation, there is no mention of either the
    parameter a or the arbitrary function ?(x).

18
  • In the new notation, the condition for an
    extremum of J is
  • dJ ?df(y,y?x) dx ? (?f/?y) dy (?f/?y?)
    dy? dx 0 (4)
  • (limits x1 lt x lt x2)
  • where dy? d (dy/dx) d(dy)/dx
  • Then, (4) becomes
  • dJ ?(?f/?y) dy (?f/?y?)d(dy)/dxdx 0
  • Integrate the 2nd term by parts get
  • dJ ?(?f/?y) -(d/dx)(?f/?y?) dydx 0 (5)
  • (limits x1 lt x
    lt x2)
  • The variation dy is arbitrary ? dJ 0 ?
    Integrand 0 or
  • (?f/?y) - (d/dx)(?f/?y?) 0
  • Eulers equation again!

19
  • The d Notation is frequently used.
  • Remember that it is only a shorthand for
    differential quantities.
  • An arbitrarily varied path dy is called a
    virtual displacement. It must be consistent
    with all forces constraints.
  • A virtual infinitesimal displacement dy is
    different from an actual infinitesimal
    displacement dy. The virtual displacement dy
    takes zero time! (dt 0) while the actual
    displacement dy takes finite time (dt ? 0).
  • dy need not even correspond to a possible path of
    motion!
  • dy 0 at the end points of the path.

20
  • Schematic of the Variational Path dy
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