The “2nd 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 2nd 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)
(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.
• 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.
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
• 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 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 + sin2θ]½ = [(θ)2+ sin2θ]½
• Apply the Euler Equation method:
Note that (f/φ) = 0

Use the “2nd Form” of Euler’s Equation!
• 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.
• 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)
α is 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)!
• 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
• 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).
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: 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
• 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) + α ηi(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)
(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, so each term in the
sum in the integrand = 0 or
(f/yi) - (d/dx)(f/yi) = 0 (i = 1,2, …,n)
 Euler’s Equations
(Several dependent variables)