The “2 nd Form” of Euler’s Equation Section 6.4

2. The “2nd Form” of Euler’s EquationSection 6.4 • A frequently occurring special case in the variational problem is when the functional f[y(x),y(x);x] does not depend explicitly on x: (f/x) = 0 It still depends on ximplicitly through y(x) & y(x) = (dy/dx) In this case, we can derive a 2nd form of Euler’s Equation. • Consider the total x derivative off[y(x),y(x);x]: (df/dx)  (f/y)(dy/dx) + (f/y)(dy/dx) + (f/x) Or, using y  dy/dx) (df/dx) = y(f/y) + y(f/y) + (f/x)

3. (df/dx) = y(f/y) + y(f/y) + (f/x) (1) • Also consider the expression: (d[y(f/y)]/dx) = y(f/y)+ y(d/dx)(f/y) (2) • Solve (1) for y(f/y)& put into (2):  (d[y(f/y)]/dx) = (df/dx) – (f/x) - y(f/y)+y(d/dx)(f/y) (3) • The last 2 terms of (3) are: - y[(f/y) - (d/dx)(f/y)] (4) • Euler’s Equation is: (f/y) - (d/dx)(f/y) = 0  (4) = 0  (3) becomes: (f/x) - (d/dx)[f - y(f/y)] = 0 (5) • (5) is sometimes called the “2nd Form” of Euler’s Eqtn.

4. The “2nd Form” of Euler’s Equation: (f/x) - (d/dx)[f - y(f/y)] = 0 (5) • (5) is most useful in the frequently occurring special case when (f/x) = 0 • That is, when f is not an explicit function of x • If (f/x) = 0, (5) becomes: (d/dx)[f - y(f/y)] = 0 Or f - y(f/y) = constant (6) (6) is often a convenient equation to use to solve for y(x), in cases when (f/x) = 0.

5. Example 6.4 • GEODESIC The shortest path between 2 points on a surface. Find the geodesic on a sphere. • Use spherical coordinates, Appendix F At radius r, in this problem, dr = 0 because its on the sphere’s surface. Square of the differential length element: (ds)2 = r2[(dθ)2 + sin2θ(dφ)2] • The differential path length on the surface is: ds = r [(dθ)2 + sin2θ(dφ)2]½ Or: ds = r[(dθ/dφ)2 + sin2θ]½dφ The distance on the surface between points 1 & 2 (the limits on the integral) is: s = ∫ds = r ∫[(dθ/dφ)2 + sin2θ]½dφ

6. Goal: Find the curve (path) θ(φ) or φ(θ) which minimizes s = ∫ds = r∫[(dθ/dφ)2 + sin2θ]½dφ • Use the Euler Equation method. s plays the role of J in the general formalism. Instead of x & y, we have θ,θ= (dθ/dφ) & φ as variables. Choose θas the dependent variable & φ as the independent variable. (x  φ, y  θ in the formalism).  The functional f[θ(φ),θ(φ);φ] in the general formalism is the integrand: f = [(dθ/dφ)2 + sin2θ]½ = [(θ)2+ sin2θ]½ • Apply the Euler Equation method: Note that (f/φ) = 0  Use the “2nd Form” of Euler’s Equation!

7. Goal: Find the curve (path) θ(φ) or φ(θ) which minimizes s = ∫ds = r∫[(dθ/dφ)2 + sin2θ]½dφ • In the Euler Equation formalism: f = [(dθ/dφ)2 + sin2θ]½ = [(θ)2+ sin2θ]½ • (f/φ) = 0  Use the “2nd Form” of Euler’s Eqtn! f - θ(f/θ) = constant  a [(θ)2+ sin2 θ]½ - θ([(θ)2+ sin2 θ]½/θ)=a • Do the differentiation: [(θ)2+ sin2θ]½ - (θ)2[(θ)2+ sin2θ]-½ = a • Multiply by f = [(θ)2+ sin2 θ]½ & simplify: sin2θ= a [(θ)2+ sin2θ]½ (1) • The solution to (1) gives the θ(φ) or φ(θ) which is the geodesic for the sphere.

8. The geodesic for the sphere is the solution to sin2θ= a [(θ)2+ sin2θ]½ (1) • Solve (1) for (θ)-1= (dφ/dθ) (dφ/dθ) = a csc2θ[1 - a2csc2θ]½ (2) cscθ = (1sinθ), cotθ = (1tanθ) • Integrate (2): φ(θ) = sin-1[cotθ/β] + α (3) αis an integration constant. β2 (1-a2)a-2 • Invert (3):cotθ = cot[θ(φ)] = β sin(φ - α) (4) • The geodesic for the sphere is given by (4)!

9. The geodesic for the sphere is given by: cotθ = cot[θ(φ)] = β sin(φ - α) (4) • Geometric interpretation? • Multiply (4) by r sinθ &note a trig identity: sin(φ - α) = sinφ cosα - cosφ sinα • Define 2 new constants in terms of α& β: A  β cosα; B β sinα  (4) becomes: A(r sinθ sinφ) - B(r sinθ cosφ) = r cosθ

10. The geodesic for the sphere is given by: A(r sinθ sinφ) - B(r sinθ cosφ) = r cosθ • Convert from spherical to Cartesian coordinates: x = r sinθ cosφ, y = r sinθ sinφ,z = r cosθ  The Geodesic becomes: Ay - Bx = z • Equation of a plane passing through the sphere’s center! • Geodesic = Path that a plane passing through the center forms at the intersection of surface of sphere  Geodesic of a Sphere  A Great Circle! (This is both the minimum & the maximum “straight line” distance on a sphere’s surface).

11. Functions with Several Dependent VariablesSection 6.5 • Euler’s Equation was derived as the solution to the variational problem: • Find the path such that J = ∫f dx is an extremum, assuming a single dependent variable y(x). • In mechanics, we often have problems with many dependent variables: y1(x), y2(x), y3(x), … • In general, we have a functional like: f = f[y1(x),y1(x),y2(x),y2(x), …;x] with yi(x)  (dyi(x)/dx) • Abbreviate as f = f[yi(x),yi(x);x], i = 1,2, …,n

12. Consider a functional: f = f[yi(x),yi(x);x], i = 1,2, …,n • The calculus of variations problem: Simultaneously find the “n paths” yi(x), i = 1,2, …,n, which minimize (or maximize) the integral: J ∫ f[yi(x),yi(x);x]dx (i = 1,2, …,n, fixed limits x1 < x < x2) • Follow derivation of before: • Write: yi(α,x)yi(0,x) + αηi(x) J = J(α) = ∫f[yi(α,x),yi(α,x);x]dx • A min (or a max) of J(α) requires (J/α)α = 0 = 0 (J/α) = (/α)∫f[yi(α,x),yi(α,x);x]dx (i = 1,2, …,n, fixed limits x1 < x < x2)

13. (J/α) = (/α)∫f[yi(α,x),yi(α,x);x]dx (i = 1,2, …,n, fixed limits x1 < x < x2) • Follow the same steps as for the previous derivation & get (skipping several steps): (J/α) = ∫∑i [(f/yi) - (d/dx)(f/yi)]ηi(x)dx • J has an extremum (min or max):  (J/α)α = 0 = 0 Each ηi(x) is an arbitrary function, soeach term inthe sum in the integrand = 0 or (f/yi) - (d/dx)(f/yi) = 0 (i = 1,2, …,n)  Euler’s Equations (Several dependent variables)