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 Euler’s 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 Euler’s 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 Euler’s 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 Euler’s 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. (J/α) = ∫ [{(f/y) -(d/dx)(f/y)}(y/α)
Follow similar steps as in the previous derivations & get (skipping several steps!):
(J/α) = ∫ [{(f/y) -(d/dx)(f/y)}(y/α)
+{(f/z)-(d/dx)(f/z)}(z/α)]dx = (3)
Also have an Equation of Constraint:
g(y,z;x) = (4)
g(y,z;x) = a known function which depends the on problem!
(4)  In (3), (y/α) & (z/α) are not independent as we assumed (using functions ηi(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/α) = ∫ [{(f/y) -(d/dx)(f/y)}η1(x)
(J/α) = ∫ [{(f/y) -(d/dx)(f/y)}(y/α)
+{(f/z)-(d/dx)(f/z)}(z/α)]dx = (3)
Still assume: y(α,x) = y(x) + α η1(x); z(α,x) = z(x)+ α η2(x)
 (y/α) = η1(x); (z/α) = η2(x)
So (3) becomes:
(J/α) = ∫ [{(f/y) -(d/dx)(f/y)}η1(x)
+{(f/z)-(d/dx)(f/z)}η2(x)]dx = (3)
But we also have the constraint g(y,z;x) = 0 (4)
Compute total differential of g when α changes by dα
dg  [(g/y)(y/α) + (g/z)(z/α)]dα
Or: dg = [(g/y)η1(x) + (g/z)η2(x)]dα
But also, by (4) dg = 0

7. (J/α) = ∫ [{(f/y) -(d/dx)(f/y)}η1(x)
Much manipulation!
(J/α) = ∫ [{(f/y) -(d/dx)(f/y)}η1(x)
+{(f/z)-(d/dx)(f/z)}η2(x)]dx = (3)
g(y,z;x) = 0  dg = (4)
dg = [(g/y)η1(x) + (g/z)η2(x)]dα = 0
 (g/y)η1(x) = - (g/z)η2(x)
Or: [η2(x)/η1(x)] = - (g/y)(g/z) (5)
Put (5) into (3) & manipulate:
(J/α) = ∫[{(f/y) - (d/dx)(f/y)} -{(f/z) - (d/dx)(f/z)}
 (g/y)(g/z)η1(x)]dx = 0 (6)
η1(x) is arbitrary  the integrand of (6) vanishes

8. - λ(x)  Left side of (7) = Right side of (7)
Vanishing of the integrand of Eq. (6)
 [(f/y) - (d/dx)(f/y)](g/y)-1
= [(f/z) - (d/dx)(f/z)](g/z) (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:
- λ(x)  Left side of (7) = Right side of (7)

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

10. [(f/y) - (d/dx)(f/y)]+ λ(x)(g/y) = 0 (10a)
(8), (9) on the previous page: 
[(f/y) - (d/dx)(f/y)]+ λ(x)(g/y) = (10a)
[(f/z) - (d/dx)(f/z)]+ λ(x)(g/z) = (10b)
 Euler’s Equations with Constraints
Note: We have formulas ((8), (9)) to compute λ(x) for a given f & g. But these depend on the unknown functions (which we are seeking!) y(x) & z(x).
 λ(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), λ(x). But we have 3 eqtns to use: (10a), (10b), & the eqtn of constraint g(y,z;x) = 0
λ(x)  A Lagrange Undetermined Multiplier is
obtained as part of the solution.

11. Summary: For the case of 2 dependent variables & 1 constraint, Euler’s Equations with Constraints:
[(f/y) - (d/dx)(f/y)]+ λ(x)(g/y) = (a)
[(f/z) - (d/dx)(f/z)]+ λ(x)(g/z) = (b)
To find the unknown functions y(x) z(x), λ(x), solve (a) & (b) simultaneously with the original equation of constraint:
g(y,z;x) = (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(yi;x) = 0, i = 1,2, …m, j = 1,2,
A derivation similar to the 2 dependent variable, 1 constraint case results in:
n Lagrange multipliers λj(x) (one for each constraint).
m Euler’s Equations with Constraints

13. For the general case with constraints.
 m Euler’s Equations with Constraints:
(f/yi) - (d/dx)(f/yi) + ∑jλj(x)(gj/yi) = 0 (A)
i = 1,2, …m
 n Equations of Constraint:
gj(yi;x) = (B)
i = 1,2, …m, j = 1,2, …n
 m + n eqtns total [(A) & (B)] with m+n
unknowns [yi(x), i = 1,2, ..,m; λj(x), j = 1,2, ..,n]

14. A Final Point About Constraints
Consider the constraint eqtn: gj(yi;x) = (B)
i = 1,2, …m, j = 1,2, …n
(B) is equivalent to a set of n differential equations (exact differentials of gi(yi;x)):
∑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 y & θ are not independent. They are related by y = Rθ.
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 & θ.
y & θ are not independent.
They are related by y = Rθ.
 The constraint eqtn is
g(y,θ) = y - Rθ = 0.
This is equivalent to the
differential versions:
(g/y) = 1; (g/θ) = -R

16. The δ Notation Section 6.7
It’s 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/α) = ∫[(f/y) - (d/dx)(f/y)]η(x)dx (1)
(limits x1 < x < x2)
From this, we derived the Euler equation. We allowed the path to vary as y(α,x)  y(0,x) + α η(x) Clearly, (y/α)  η(x)
Rewrite (1) (multiplying by dα) as:
(J/α)dα = [(f/y) -(d/dx)(f/y)](y/α)dαdx (2)

17. Introduce a Shorthand Notation
Define: δJ  (J/α)dα and δy  (y/α)dα
 Rewrite (2) as
δJ = ∫[(f/y) -(d/dx)(f/y)]δydx (3)
(limits x1 < x < x2)
(3) is called “The variation of J” (δJ) in terms of “the variation of y” (δy).
In the general formulation, where we want to find condition for extremum of J = ∫f(y,y;x) dx (limits x1 < x < x2), follow the original derivation, but in this new notation. In this notation, there is no mention of either the parameter α or the arbitrary function η(x).

18. (f/y) - (d/dx)(f/y) = 0 Euler’s equation again!
In the new notation, the condition for an extremum of J is
δJ = ∫δf(y,y;x) dx = ∫ [(f/y) δy + (f/y) δy] dx = 0 (4)
(limits x1 < x < x2)
where δy = δ (dy/dx) = d(δy)/dx
Then, (4) becomes:
δJ = ∫[(f/y) δy + (f/y){d(δy)/dx}]dx = 0
Integrate the 2nd term by parts & get:
δJ = ∫[(f/y) -(d/dx)(f/y)] δydx = 0 (5)
The variation δy is arbitrary:  δJ = 0  Integrand = 0 or
(f/y) - (d/dx)(f/y) = 0
Euler’s equation again!

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

20. Schematic of the Variational Path δy