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)

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

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)

(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 = 0 (3) Also have an Equation of Constraint: g(y,z;x) = 0 (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!

(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 = 0 (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 = 0 (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

(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 = 0 (3) g(y,z;x) = 0  dg = 0 (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

- 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)-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: - λ(x)  Left side of (7) = Right side of (7)

 - 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)!

[(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) = 0 (10a) [(f/z) - (d/dx)(f/z)]+ λ(x)(g/z) = 0 (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.

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) = 0 (a) [(f/z) - (d/dx)(f/z)]+ λ(x)(g/z) = 0 (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) = 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(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

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) = 0 (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]

A Final Point About Constraints Consider the constraint eqtn: gj(yi;x) = 0 (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)!

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

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)

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).

(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!

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.

Schematic of the Variational Path δy