Euler Equations for Systems with Constraints

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 Equationsautomatically 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. 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) = aknown 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)}(y/α) +{(f/z)-(d/dx)(f/z)}(z/α)]dx = 0 (3) • Still assume: y(α,x) = y(x) + αη1(x);z(α,x) = z(x)+ αη2(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 constraintg(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. 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

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

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. (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 UNDETERMINEDuntil 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) = 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)

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: nLagrange multipliersλj(x) (one for each constraint). mEuler’s Equations with Constraints

13. For the general case with constraints.  mEuler’s Equations with Constraints: (f/yi) - (d/dx)(f/yi) + ∑jλj(x)(gj/yi) = 0 (A) i = 1,2, …m  nEquations 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]

14. 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)!

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 & 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) + αη(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) (limits x1 < x < x2)

17. Introduce a Shorthand Notation • Define:δJ  (J/α)dαandδy  (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 αor the arbitrary functionη(x).

18. 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) (limits x1 < x < x2) • 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