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

1
(No Transcript)
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
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