Example 17.2.1 STRAIGHTLINE
Perhaps the simplest application of the Euler equation is in the determination of the shortest distance between two points in the Euclideanxy-plane. Since the element of distance is
ds=
(dx)2+(dy)2 1/2=
1+yx2 1/2dx, (17.19) the distanceJ may be written as
J= x2,y2
x1,y1
ds= x2
x1
1+y2x 1/2dx. (17.20)
Comparison with Eq. (17.1) shows that f (y, yx, x)=
1+yx21/2
. (17.21)
Substituting into Eq. (17.16), we obtain
−d dx
1 (1+yx2)1/2
=0, (17.22)
or
1
(1+yx2)1/2=C, a constant. (17.23) This is satisfied by
yx=a, a second constant, (17.24)
and
y=ax+b, (17.25)
which is the familiar equation for a straight line. The constants a andb are chosen so that the line passes through the two points(x1, y1)and(x2, y2). Hence the Euler equation
predicts that the shortest6distance between two fixed points in Euclidean space is a straight
line.
The generalization of this in curved four-dimensional space–time leads to the important concept of the geodesic in general relativity (see Section 2.10).
Example 17.2.2 SOAPFILM
As a second illustration (Fig. 17.4), consider two parallel coaxial wire circles to be con- nected by a surface of minimum area that is generated by revolving a curvey(x) about thex-axis. The curve is required to pass through fixed endpoints(x1, y1)and(x2, y2). The variational problem is to choose the curvey(x)so that the area of the resulting surface will be a minimum.
For the element of area shown in Fig. 17.4, dA=2πy ds=2πy
1+yx21/2
dx. (17.26)
The variational equation is then J=
x2 x1
2πy
1+yx21/2
dx. (17.27)
Neglecting the 2π, we obtain
f (y, yx, x)=y
1+yx21/2
. (17.28)
FIGURE17.4 Surface of rotation — soap film problem.
6Technically, we have a stationary value. From theα2terms it can be identified as a minimum (Exercise 17.2.2).
Since∂f/∂x=0, we may apply Eq. (17.18) directly and get y
1+yx21/2
−yy2x 1
(1+yx2)1/2=c1, (17.29) or
y
(1+yx2)1/2 =c1. (17.30)
Squaring, we get
y2
1+yx2=c21 withc21≤ymin2 , (17.31) and
(yx)−1=dx
dy = c1
y2−c21
. (17.32)
This may be integrated to give
x=c1cosh−1 y
c1+c2. (17.33)
Solving fory, we have
y=c1cosh
x−c2 c1
, (17.34)
and againc1andc2are determined by requiring the hyperbolic cosine to pass through the points(x1, y1)and(x2, y2). Our “minimum”-area surface is a special case of a catenary of
revolution, or a catenoid.
Soap Film — Minimum Area
This calculus of variations contains many pitfalls for the unwary. (Remember, the Euler equation is a necessary condition assuming a differentiable solution. The sufficiency con- ditions are quite involved. See the Additional Readings for details.) Respect for some of these hazards may be developed by considering a specific physical problem, for example, a minimum-area problem with(x1, y1)=(−x0,1),(x2, y2)=(+x0,1). The minimum sur- face is a soap film stretched between the two rings of unit radius atx= ±x0. The problem is to predict the curvey(x)assumed by the soap film.
By referring to Eq. (17.34), we find thatc2=0 by the symmetry of the problem about x=0. Then
y=c1cosh x
c1
, c1cosh x0
c1
=1. (17.34a)
If we takex0=12we obtain a transcendental equation forc1, viz.
1=c1cosh 1
2c1
. (17.35)
We find that this equation has two solutions:c1=0.2350, leading to a “deep” curve, and c1=0.8483, leading to a “flat” curve. Which curve is assumed by the soap film? Before answering this question, consider the physical situation with the rings moved apart so that x0=1. Then Eq. (17.34a) becomes
1=c1cosh 1
c1
, (17.36)
which has no real solutions. The physical significance is that as the unit-radius rings were moved out from the origin, a point was reached at which the soap film could no longer maintain the same horizontal force over each vertical section. Stable equilibrium was no longer possible. The soap film broke (irreversible process) and formed a circular film over each ring (with a total area of 2π=6.2832. . .). This is the Goldschmidt discontinuous solution.
The next question is: How large mayx0be and still give a real solution for Eq. (17.34a)?7 Lettingc1−1=p, Eq. (17.34a) becomes
p=coshpx0. (17.37)
To findx0 maxwe could solve forx0(as in Eq. (17.33)) and then differentiate with respect top. Finally, with an eye on Fig. 17.5,dx0/dp would be set equal to zero. Alternatively, direct differentiation of Eq. (17.37) with respect topyields
1=
x0+pdx0 dp
sinhpx0.
FIGURE17.5 Solutions of Eq. (17.34a) for unit-radius rings
atx= ±x0.
7From a numerical point of view it is easier to invert the problem. Pick a value ofc1and solve forx0. Equation (17.34a) becomes x0=c1cosh−1(1/c1). This has numerical solutions in the range 0< c1≤1.
The requirement thatdx0/dpvanish leads to
1=x0sinhpx0. (17.38)
Equations (17.37) and (17.38) may be combined to form
px0=cothpx0, (17.39)
with the root
px0=1.1997. (17.40)
Substituting into Eq. (17.37) or (17.38), we obtain
p=1.810, c1=0.5524 (17.41)
and
x0 max=0.6627. (17.42)
Returning to the question of the solution of Eq. (17.35) that describes the soap film, let us calculate the area corresponding to each solution. We have
A=4π x0
0
y
1+yx21/2
dx=4π c1
x0 0
y2dx (by Eq.(17.30))
=4π c1 x0
0
cosh x
c1
2
dx=π c21
sinh 2x0
c1
+2x0
c1
. (17.43)
Forx0=12, Eq. (17.35) leads to
c1=0.2350→A=6.8456, c1=0.8483→A=5.9917,
showing that the former can at most be only a local minimum. A more detailed investiga- tion (compare Bliss, Calculus of Variations, Chapter IV) shows that this surface is not even a local minimum. Forx0=12, the soap film will be described by the flat curve
y=0.8483 cosh x
0.8483
. (17.44)
This flat or shallow catenoid (catenary of revolution) will be an absolute minimum for 0≤x0<0.528. However, for 0.528< x <0.6627 its area is greater than that of the Gold- schmidt discontinuous solution (6.2832) and it is only a relative minimum (Fig. 17.6).
For an excellent discussion of both the mathematical problems and experiments with soap films, we refer to Courant and Robbins (1996) in the Additional Readings at the end of the chapter.
FIGURE17.6 Catenoid area (unit-radius rings at x= ±x0).
Exercises
17.2.1 A soap film is stretched across the space between two rings of unit radius centered at
±x0 on the x-axis and perpendicular to the x-axis. Using the solution developed in Section 17.2, set up the transcendental equations for the condition thatx0is such that the area of the curved surface of rotation equals the area of the two rings (Goldschmidt discontinuous solution). Solve forx0(Fig. 17.7).
17.2.2 In Example 17.2.1, expandJ[y(x, α)] −J[y(x,0)]in powers ofα. The term linear in αleads to the Euler equation and to the straight-line solution, Eq. (17.25). Investigate
FIGURE17.7 Surface of rotation.
the α2 term and show that the stationary value of J, the straight-line distance, is a minimum.
17.2.3 (a) Show that the integral J=
x2 x1
f (y, yx, x) dx, withf =y(x), has no extreme values.
(b) Iff (y, yx, x)=y2(x), find a discontinuous solution similar to the Goldschmidt solution for the soap film problem.
17.2.4 Fermat’s principle of optics states that a light ray will follow the pathy(x)for which x2,y2
x1,y1
n(y, x) ds
is a minimum whennis the index of refraction. Fory2=y1=1,−x1=x2=1, find the ray path if
(a) n=ey, (b)n=a(y−y0), y > y0.
17.2.5 A frictionless particle moves from pointAon the surface of the Earth to pointB by sliding through a tunnel. Find the differential equation to be satisfied if the transit time is to be a minimum.
Note. Assume the Earth to be nonrotating sphere of uniform density.
ANS. (Eq. (17.15)):rϕϕ(r3−ra2)+rϕ2(2a2−r2)+a2r2=0, r(ϕ=0)=r0, rϕ(ϕ=0)=0, r(ϕ=ϕA)=a, r(ϕ=ϕB)=a.
Eq. (17.18):rϕ2=a2r2r2
0 ãra22−−rr022. The solution of these equations is a hypocycloid, gener- ated by a circle of radius 12(a−r0)rolling inside the circle of radiusa. You might like to show that the transit time is
t=π(a2−r02)1/2 (ag)1/2 .
For details see P. W. Cooper, Am. J. Phys. 34: 68 (1966); G. Veneziano et al., ibid., pp.
701–704.
17.2.6 A ray of light follows a straight-line path in a first homogeneous medium, is refracted at an interface, and then follows a new straight-line path in the second medium. Use Fermat’s principle of optics to derive Snell’s law of refraction:
n1sinθ1=n2sinθ2.
Hint. Keep the points (x1, y1) and (x2, y2) fixed and vary x0 to satisfy Fermat (Fig. 17.8). This is not an Euler equation problem. (The light path is not differentiable atx0.)
FIGURE17.8 Snell’s law.
17.2.7 A second soap film configuration for the unit-radius rings at x= ±x0 consists of a circular disk, radiusa, in thex=0 plane and two catenoids of revolution, one joining the disk and each ring. One catenoid may be described by
y=c1cosh x
c1+c3
. (a) Impose boundary conditions atx=0 andx=x0.
(b) Although not necessary, it is convenient to require that the catenoids form an angle of 120◦where they join the central disk. Express this third boundary condition in mathematical terms.
(c) Show that the total area of catenoids plus central disk is A=c21
sinh
2x0
c1 +2c3
+2x0
c1
.
Note. Although this soap film configuration is physically realizable and stable, the area is larger than that of the simple catenoid for all ring separations for which both films exist.
ANS. (a)
1=c1cosh x0
c1+c3 a=c1coshc3,
(b) dy
dx=tan 30◦=sinhc3. 17.2.8 For the soap film described in Exercise 17.2.7, find (numerically) the maximum value
ofx0.
Note. This calls for a pocket calculator with hyperbolic functions or a table of hyperbolic cotangents.
ANS.x0 max=0.4078.
17.2.9 Find the root ofpx0=cothpx0(Eq. (17.39)) and determine the corresponding values ofpandx0(Eqs. (17.41) and (17.42)). Calculate your values to five significant figures.
17.2.10 For the two-ring soap film problem of this section calculate and tabulatex0,p,p−1, and A, the soap film area forpx0=0.00(0.02)1.30.
17.2.11 Find the value ofx0(to five significant figures) that leads to a soap film area, Eq. (17.43), equal to 2π, the Goldschmidt discontinuous solution.
ANS.x0=0.52770.
17.2.12 Find the curve of quickest descent from(0,0)to(x0, y0)for a particle sliding under gravity and without friction. Show that the ratio of times taken by the particle along a straight line joining the two points compared to along the curve of quickest descent is (1+4/π2)1/2.
Hint. Takey to increase downwards. Apply Eq. (17.18) to obtainyx2=(1−c2y)/c2y, wherec is an integration constant. Then make the substitution y =(sin2ϕ/2)/c2 to parametrize the cycloid and take(x0, y0)=(π/2c2,1/c2).