Some Methods of the Calculus of Variations
Chapter 6: Overview only!
• A math interlude!
– Some (seemingly esoteric at times!) math, but with a flavor
of application to physics!
• More details are in the lecture files & the chapter. Just as
much as we need for Ch. 7!
• Purpose: Fill in some math background. Needed to discuss the
Lagrangian & Hamiltonian formulations of Classical
Mechanics (Ch. 7)
– These formulations will enable us to more easily solve many
problems which would be horrendous if done directly with
Newton’s 2nd Law!
– These are the basis of much of modern physics theory; quantum
mechanics & quantum field theory.
• Dynamical Problems in Mechanics:
– Often more easily analyzed using alternative
formulations of Newton’s Laws: Lagrange’s
Equations & Hamilton’s Principle
• Other contributors to alternative formulations: Newton,
Bernoulli, Euler, Legendre, Jacobi, Dirichlet, Weierstrass
– These are alternative formulations, but they are
100% Equivalent to Newtonian Mechanics!
– The Calculus of Variations: The math behind the
derivations of the these alternative formulations.
– Applications are emphasized, proofs skimmed over or
skipped!
• The Primary interest of the Calculus of
Variations: To determine the path in space
which gives extremum (maximum or minimum)
solutions.
• Example from Optics:
Fermat’s Principle
Light travels by a path that takes the least
amount of time.
• Some esoteric math! Physics will come soon!
Problem Statement Sect. 6.2
• The Basic problem of the Calculus of Variations
Determine the function y(x) (“path in xy plane) for which the integral
J  ∫f[y(x),y(x);x]dx
(fixed limits x1 < x < x2)
is an extremum (max or min). Here, y(x)  (dy/dx)
– The semicolon in f separates the independent
variable x from the dependent variable y(x) & its
derivative y(x). f  A GIVEN functional.
– Functional  A quantity f[y(x),y(x);x] which
depends on the functional form of the dependent
variable (function) y(x). “A function of a function”.
• The Basic Problem Restated:
Given f[y(x),y(x);x], find (for fixed x1, x2) the
function(s) y(x) which will minimize (or maximize)
J  ∫f[y(x),y(x);x]dx (limits x1 < x < x2)
 Vary y(x) until an extremum (a max or a min;
usually a min!) of J is found.
• Suppose the function y = y(x) gives the integral J a
minimum value:
 Every “neighboring function”, no matter how
close to y(x), must make J increase!
• “Neighboring Function”: By this we mean all
possible values of the function y  y(α,x),
where α  a parameter such that, for α = 0, y =
y(0,x)  y(x) is the function which minimizes
the integral J.
– Assume that y(α,x)  y(0,x) + α η(x), where
η(x)  Some function of x with continuous 1st
derivative & with η(x1)  η(x2)  0 (x1, x2 = limits
on the integral. This ensures that y(α,x) = y(x) at
x1, x2)
Schematic Illustration
• Consider functions of the type: y(α,x)  y(0,x) + α η(x)
 The integral J is a function of η(x)
J = J(α)  ∫f[y(α,x),y(α,x);x]dx
(limits x1 < x < x2)
• So, if J is a min or max (a “Stationary Value” or
an Extremum)  We must have:
(J/α)α = 0 = 0 (for all functions η(x))
• This is a necessary condition for J to have an
extremum, but its not a sufficient condition.
Example 6.1: Simple Case
• Consider the function:
f  (dy/dx)2  [y(α,x)]2
where y(x) = x &
η(x) = sin(x). Find J(α)
between x1 = 0 & x2 = 2π.
Show: The stationary
value of J(α) is at α = 0. Solution: y(x) = x  y(0,x).
Construct neighboring varied paths by:
y(α,x)  y(0,x) + α η(x) = x + α sin(x)
Some paths are shown in the figure.
• η(x) = sin(x) = 0 at x1 = 0 & x2 = 2π
y(α,x) = x + α sin(x)
y(α,x) = dy(α,x)/dx = 1 + α cos(x)
The functional is: f  (dy/dx)2 = [1 + α cos(x)]2
f = 1 + 2α cos(x) + α2 cos2(x)
The integral J is: J(α)  ∫f[y(α,x),y(α,x);x]dx
(limits x1 < x < x2)
 J(α) = ∫[1+2 α cos(x)+ α2cos2(x)]dx
(limits 0 < x < 2π). THE LOWER LIMIT MATTERS!
 J(α) = π (2 + α2) > J(0) (all α). Also:
(J/α)α = 0 = 0 & J(0) = 2π is the min. value of J!
Euler’s Equation
Sect. 6.3 – Most important for Ch. 7 Applications!
• The general expression for J, for a given f:
J = J(α)  ∫f[y(α,x),y(α,x);x]dx (limits x1 < x < x2) (1)
• J has min or max:  (J/α)α = 0 = 0
(2)
• Formally combine (1) & (2):
(J/α) = (/α) ∫f[y(α,x),y(α,x);x]dx
Interchange derivative & integral & use the chain rule:
(J/α) = ∫[(f/y)(y/α)+(f/y)(y/α)]dx
(3)
• But y(α,x)  y(0,x) + α η(x)  (y/α) = η(x) (4)
y(α,x)  [dy(α,x)/dx]  (y/α) = (dη/dx)
(5)
• Put (4) & (5) into (3):
(J/α) = ∫[(f/y)η(x)+(f/y)(dη/dx)] dx
(6)
(limits x1 < x < x2)
• By parts (∫udv = uv - ∫vdu) integration of 2nd term:
∫(f/y)(dη/dx) dx
= (f/y)η(x) - ∫ [(d/dx)(f/y)] η(x)dx
(limits x1 < x < x2)
Now, (f/y)η(x) = 0 (between limits x1,x2) because
η(x1)  η(x2)  0
 (6) becomes:
(J/α) = ∫[(f/y) - (d/dx)(f/y)] η(x)dx (7)
(limits x1 < x < x2)
(J/α) = ∫[(f/y) - (d/dx)(f/y )]η(x)dx (7)
(limits x1 < x < x2)
• (7) isn’t independent of α! y = y(α,x); y = y(α,x)
• J has an extremum (min or max):

(J/α)α = 0 = 0
(2)
• η(x) is an arbitrary function so (7) & (2) together
 The integrand of (7) = 0 or
(f/y) - (d/dx)[f/y] = 0
 Euler’s Equation
(8)
Note: the functional f is known! The solution to (8) gives the
function y(x) which causes the integral J to be a min or a max.
Euler, 1744. Applied to mechanics  Euler - Lagrange Equation
• In (8) y= y(x) & y = y(x) are independent of α
Example 6.2
• A classic problem! “The Brachistochrone”:
A particle is moving
in the xy-plane in a
constant, conservative
force field F. It starts
at rest at 1 = (x1,y1) &
moves to 2 = (x2,y2)
(a “lower point” than 1).
Find the path y(x)
that allows the particle
to move from 1 to 2 in the least time. This is schematically
shown in the figure.
• Solution: Minimize the time t between points 1 & 2. For
convenience, choose 1 = (0,0), 2 = (x2,y2). Path in the
plane: s = [x2 + y2]½. Velocity: v = (ds/dt)  dt = (ds/v).
• We want to minimize: t = ∫(ds/v) (limits from 1 to 2). Get v
from energy conservation: T + U = const. T = mv2, U = -Fx.
Newton’s 2nd Law: F = mg = constant. (g = acceleration, not
necessarily gravitational!)
• Initial conditions: v = 0 at x = 0  T + U = 0
 mv2 - mgx = 0  v = (2gx)½ . Differential path length:
ds = (dx2 + dy2)½ = [1 +(dy/dx)2]½
t = ∫ [(1 +y2)/(2gx)]½dx. Minimize t:  t plays the role of
J in the general formalism.  Identify the functional f in the
general formalism as  integrand: f = [(1 +y2)/x]½ (a constant
in the integrand is ignored. It doesn’t affect the final result!)
• General Euler Eqtn: (f/y) - (d/dx)[f/y] = 0
Our case: f = [(1 +y2)/x]½ (1).  (f/y) = 0.
Euler’s Eqtn. becomes: (d/dx)[f/y] = 0. Or:
(f/y) = const.  (2a)-½ (2) (1)
 (f/y) = y [x(1 + y2)]-½ (3). Setting (2) = (3) &
squaring (solving for y(x)) gives:
y= y(x) = x½[2a -x]-½ = x[2ax-x2]-½ (4)
• Integrating (4) gives the desired path y(x)
y = ∫xdx[2ax-x2]-½ (limits: 1 to 2)
• Change variables: x = a(1 - cosθ), dx = a sinθdθ

y = ∫a(1 - cosθ)dθ = a(θ - sinθ)
(the constant of integration = 0 since it started at the origin)
• Summary: A particle in the xy-plane under a constant,
conservative force. At rest at (0,0). The path for it to move from
(0,0) to (x2,y2) in the minimum time t = ∫(ds/v) is one on which x &
y satisfy the parametric equations:
x = a(1 - cosθ)
y = a(θ - sinθ)
• These are the wellknown eqtns. for a
CYCLOID, passing
through the origin as in
the figure. The
constant a is adjusted
so the path passes through the specified point 2 = (x2,y2). Geometrically, a
Cycloid is a curve traced by a point on a circle which is rolling
along a straight line (in this case, the x-axis, as in the figure).
Example 6.3
• A curve connects 2 points
in the xy-plane: 1 = (x1,y1),
2 = (x2,y2). A surface is
generated by rotating this
curve about an axis (figure
shows the y-axis) in the
xy-plane. Find the eqtn of the curve (y = y(x) or x = x(y)) such
that the area of the surface of revolution is a minimum.
• The Euler Eqtn procedure gets (see text & lecture notes!):
y = a cosh-1(x/a) + b or x = a cosh[(y-b)/a]
This is a Catenary. The same as the curve of a flexible cord
hanging between 2 supports. a, b = constants determined by
requiring y(x) to pass through points 1 & 2
• Solution: Let the axis of
revolution be the y-axis as
shown. The differential area
of the strip in the figure:
dA = 2πx ds = 2πx(dx2 + dy2)½
= 2πxdx [1 + (dy/dx)2]½
= 2πxdx[1 + (y)2]½ ; y= (dy/dx).
• Area of the surface of revolution:
A = 2π ∫ xdx [1 + (y)2]½ (x1 < x < x2)
Goal: Find y(x) which minimizes A!
• Area of the surface of revolution:
A = 2π∫xdx[1 + (y)2]½ (x1<x< x2)
Goal: Find y(x) which minimizes A!
• Here: A is the J of the general
formalism.  The functional f is:
f = x [1 + (y)2]½ (1). Euler’s eqtn gives a criterion on f which will
make A a minimum: (f/y) - (d/dx)[f/y] = 0 (2). The y(x)
which simultaneously satisfies (1) & (2) will minimize A! From (1):
(f/y) = 0, (f/y) = (xy)[1 + (y)2]-½. Put this in (2):
(d/dx){(xy)[1 + (y)2]-½} = 0 (3). Solve (3) for y = y(x):
• (3) gives: (xy)[1 + (y)2]-½ = const  a (4). Solve (4) for y(x):
y(x) = (dy/dx) = a(x2- a2)-½ (5). Integrate (5):
y(x) = ∫dx a(x2- a2)-½ = a cosh-1(x/a)+ b, b = integration constant.
Or: x(y) = a cosh[(y-b)/a]. a & b are chosen so that y(x) passes
through 1 = (x1,y1) & 2 = (x2,y2).
A Related Problem
• The same problem again, but
now rotate about the x-axis:
2 points: 1 = (x1,y1),
2 = (x2,y2), joined by a curve
y = y(x). Find y(x) such that if
the curve is rotated about the
x-axis, the area of revolution is minimized.
The classic “Soap Film” problem
Find the shape of a soap film suspended between wire rings.
Naively, one would think that the solution is similar to the
previous example, if (say) we carefully interchange x & y.
(Or, something!)
• Solution: Let the axis of
revolution be the x-axis as
shown. The differential
area of the strip in the figure:
dA = 2πy ds = 2πy(dx2 + dy2)½
= 2πydx[1 + (dy/dx)2]½
= 2πydx[1 + (y)2]½ ; y= dy/dx
• Area of the surface of revolution:
A = 2π∫y dx [1 + (y)2]½ (x1 < x < x2)
Goal: Find y(x) which minimizes A!
• Area of the surface of revolution:
A = 2π∫y dx [1 + (y)2]½ (x1 < x < x2)
Goal: Find y(x) which minimizes A!
• Here: A is the J of the general
formalism.  The functional f is:
f = y[1 + (y)2]½ (1). Euler’s eqtn gives a criterion on f which will
make A a minimum: (f/y) - (d/dx)[f/y] = 0 (2).
• The y(x) which simultaneously satisfies (1) & (2) will minimize A!
From (1): (f/y) = [1 + (y)2]½
(f/y) = (yy)[1 + (y)2]-½. Put these into (2):
[1 + (y)2]½ - (d/dx){(yy)[1 + (y)2]-½} = 0
(3)
• Solve (3) for y = y(x): This is A MESS!!! Especially in
comparison with rotation about the y-axis!
– Might have thought that it would be easier, given the
straightforwardness of the y-axis rotation problem. Before proceeding,
stop & consider whether there is an easier method of solution!
Independent & Dependent Variables
• Comments: In general, for a given problem, we can
choose the independent variable to be anything! It
needn’t always be x!
• Depending on the problem, could be x, y, z, θ, φ, t, ….
• How does Euler’s Equation work if we choose something
other than x as the independent variable?
– It was formulated in the very beginning assuming x is independent & y
is dependent & the goal is to find y(x).
– In this problem it would be much easier to re-label the axes in the
problem from beginning.
• Such a change of variable is easy, in cases like this where the symmetry is
high. However, its not always this easy! Sometimes we have to make the
choice of dependent & independent variable by trial & error