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PPT – Ch. 6: Sect. 6.6: Euler's Equations with Auxilliary Conditions (Constraints). Sect. 6.7: The Delta Notation. PowerPoint presentation | free to download - id: 26da7d-ZDc1Z

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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)

- 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

- 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)

- 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!

- (?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

- 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

- 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)

- 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)!

- (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.

- 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)

- 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

- 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

- 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)!

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

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)

- 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).

- 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!

- 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.

- Schematic of the Variational Path dy