2. The “2 nd Form” of Euler’s Equation Section 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 x implicitly through y(x) & y(x) = (dy/dx) In this case, we can derive a 2 nd form of Euler’s Equation. Consider the total x derivative of f[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 “2 nd Form” of Euler’s Eqtn.

4. The “2 nd 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 = r 2 [(dθ) 2 + sin 2 θ(dφ) 2 ]  The differential path length on the surface is: ds = r [(dθ) 2 + sin 2 θ(dφ) 2 ] ½ Or: ds = r[(dθ/dφ) 2 + sin 2 θ] ½ dφ The distance on the surface between points 1 & 2 (the limits on the integral) is: s = ∫ds = r ∫ [(dθ/dφ) 2 + sin 2 θ] ½ dφ

6. Goal: Find the curve (path) θ(φ) or φ(θ) which minimizes s = ∫ds = r∫[(dθ/dφ) 2 + sin 2 θ] ½ dφ 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 θ as 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 + sin 2 θ] ½ = [(θ) 2 + sin 2 θ] ½ Apply the Euler Equation method: Note that (  f/  φ) = 0  Use the “2 nd Form” of Euler’s Equation!

7. Goal: Find the curve (path) θ(φ) or φ(θ) which minimizes s = ∫ds = r∫[(dθ/dφ) 2 + sin 2 θ] ½ dφ In the Euler Equation formalism: f = [(dθ/dφ) 2 + sin 2 θ] ½ = [(θ) 2 + sin 2 θ] ½ (  f/  φ) = 0  Use the “2 nd Form” of Euler’s Eqtn! f - θ(  f/  θ) = constant  a [(θ) 2 + sin 2 θ] ½ - θ(  [(θ ) 2 + sin 2 θ] ½ /  θ)= a Do the differentiation: [(θ) 2 + sin 2 θ] ½ - (θ) 2 [(θ) 2 + sin 2 θ] -½ = a Multiply by f = [(θ) 2 + sin 2 θ] ½ & simplify: sin 2 θ = a [(θ) 2 + sin 2 θ] 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 sin 2 θ = a [(θ) 2 + sin 2 θ] ½ (1) Solve (1) for (θ) -1 = (dφ/dθ) (dφ/dθ) = a csc 2 θ  [1 - a 2 csc 2 θ] ½ (2) cscθ = (1  sinθ), cotθ = (1  tanθ) Integrate (2): φ(θ) = sin -1 [cotθ/β] + α (3) α is an integration constant. β 2  (1-a 2 )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 Variables Section 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: y 1 (x), y 2 (x), y 3 (x), … In general, we have a functional like: f = f[y 1 (x),y 1 (x),y 2 (x),y 2 (x), …;x] with y i (x)  (dy i (x)/dx) Abbreviate as f = f[y i (x),y i (x);x], i = 1,2, …,n

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

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