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Some extensions

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Some extensions
22.3 SOME EXTENSIONS
z
b
ρ
−b
(a)
(b)
a
(c)
Figure 22.4 Possible soap films between two parallel circular rings.
surface area between z and z + dz is
1/2
,
dS = 2πρ (dz)2 + (dρ)2
so the total surface area is given by
b
S = 2π
ρ(1 + ρ )1/2 dz.
2
(22.11)
−b
Since the integrand does not contain z explicitly, we can use (22.8) to obtain an equation
for ρ that minimises S, i.e.
ρ(1 + ρ )1/2 − ρρ (1 + ρ )−1/2 = k,
2
2
2
where k is a constant. Multiplying through by (1 + ρ 2 )1/2 , rearranging to find an explicit
expression for ρ and integrating we find
cosh−1
ρ
z
= + c.
k
k
where c is the constant of integration. Using the boundary conditions ρ(±b) = a, we
require c = 0 and k such that a/k = cosh b/k (if b/a is too large, no such k can be found).
Thus the curve that minimises the surface area is
ρ/k = cosh(z/k),
and in profile the soap film is a catenary (see section 22.4) with the minimum distance
from the axis equal to k. 22.3 Some extensions
It is quite possible to relax many of the restrictions we have imposed so far. For
example, we can allow end-points that are constrained to lie on given curves rather
than being fixed, or we can consider problems with several dependent and/or
independent variables or higher-order derivatives of the dependent variable. Each
of these extensions is now discussed.
781
CALCULUS OF VARIATIONS
22.3.1 Several dependent variables
Here we have F = F(y1 , y1 , y2 , y2 , . . . , yn , yn , x) where each yi = yi (x). The analysis
in this case proceeds as before, leading to n separate but simultaneous equations
for the yi (x),
∂F
d ∂F
=
,
i = 1, 2, . . . , n.
(22.12)
∂yi
dx ∂yi
22.3.2 Several independent variables
With n independent variables, we need to extremise multiple integrals of the form
∂y ∂y
∂y
I=
· · · F y,
,
,...,
, x1 , x2 , . . . , xn dx1 dx2 · · · dxn .
∂x1 ∂x2
∂xn
Using the same kind of analysis as before, we find that the extremising function
y = y(x1 , x2 , . . . , xn ) must satisfy
n
∂F
∂
∂F
=
,
(22.13)
∂y
∂xi ∂yxi
i=1
where yxi stands for ∂y/∂xi .
22.3.3 Higher-order derivatives
If in (22.1) F = F(y, y , y , . . . , y (n) , x) then using the same method as before
and performing repeated integration by parts, it can be shown that the required
extremising function y(x) satisfies
n
∂F
∂F
d ∂F
∂F
d2
n d
−
+
−
·
·
·
+
(−1)
= 0,
(22.14)
∂y
dx ∂y dx2 ∂y dxn ∂y (n)
provided that y = y = · · · = y (n−1) = 0 at both end-points. If y, or any of its
derivatives, is not zero at the end-points then a corresponding contribution or
contributions will appear on the RHS of (22.14).
22.3.4 Variable end-points
We now discuss the very important generalisation to variable end-points. Suppose,
as before, we wish to find the function y(x) that extremises the integral
b
I=
F(y, y , x) dx,
a
but this time we demand only that the lower end-point is fixed, while we allow
y(b) to be arbitrary. Repeating the analysis of section 22.1, we find from (22.4)
782
22.3 SOME EXTENSIONS
∆y
y(x) + η(x)
y(x)
∆x
h(x, y) = 0
b
Figure 22.5 Variation of the end-point b along the curve h(x, y) = 0.
that we require
η
∂F
∂y b
+
a
a
b
∂F
d
−
∂y
dx
∂F
∂y η(x) dx = 0.
(22.15)
Obviously the EL equation (22.5) must still hold for the second term on the LHS
to vanish. Also, since the lower end-point is fixed, i.e. η(a) = 0, the first term on
the LHS automatically vanishes at the lower limit. However, in order that it also
vanishes at the upper limit, we require in addition that
∂F = 0.
(22.16)
∂y x=b
Clearly if both end-points may vary then ∂F/∂y must vanish at both ends.
An interesting and more general case is where the lower end-point is again
fixed at x = a, but the upper end-point is free to lie anywhere on the curve
h(x, y) = 0. Now in this case, the variation in the value of I due to the arbitrary
variation (22.2) is given to first order by
b
∂F
d ∂F
∂F b
−
η
+
η dx + F(b)∆x,
(22.17)
δI =
∂y a
∂y
dx ∂y a
where ∆x is the displacement in the x-direction of the upper end-point, as
indicated in figure 22.5, and F(b) is the value of F at x = b. In order for (22.17)
to be valid, we of course require the displacement ∆x to be small.
From the figure we see that ∆y = η(b) + y (b)∆x. Since the upper end-point
must lie on h(x, y) = 0 we also require that, at x = b,
∂h
∂h
∆x +
∆y = 0,
∂x
∂y
which on substituting our expression for ∆y and rearranging becomes
∂h
∂h
∂h
+ y
η = 0.
∆x +
∂x
∂y
∂y
783
(22.18)
CALCULUS OF VARIATIONS
x = x0
A
x
B
y
Figure 22.6 A frictionless wire along which a small bead slides. We seek the
shape of the wire that allows the bead to travel from the origin O to the line
x = x0 in the least possible time.
Now, from (22.17) the condition δI = 0 requires, besides the EL equation, that
at x = b, the other two contributions cancel, i.e.
F∆x +
∂F
η = 0.
∂y (22.19)
Eliminating ∆x and η between (22.18) and (22.19) leads to the condition that at
the end-point
∂F ∂h
∂F ∂h
−
= 0.
(22.20)
F − y ∂y ∂y ∂y ∂x
In the special case where the end-point is free to lie anywhere on the vertical line
x = b, we have ∂h/∂x = 1 and ∂h/∂y = 0. Substituting these values into (22.20),
we recover the end-point condition given in (22.16).
A frictionless wire in a vertical plane connects two points A and B, A being higher than B.
Let the position of A be fixed at the origin of an xy-coordinate system, but allow B to lie
anywhere on the vertical line x = x0 (see figure 22.6). Find the shape of the wire such that
a bead placed on it at A will slide under gravity to B in the shortest possible time.
This is a variant of the famous brachistochrone (shortest time) problem, which is often
used to illustrate the calculus of variations. Conservation of energy tells us that the particle
speed is given by
ds v=
= 2gy,
dt
where s is the path length along the wire and g is the acceleration due to gravity. Since
the element of path length is ds = (1 + y 2 )1/2 dx, the total time taken to travel to the line
x = x0 is given by
x=x0
x0
1
ds
1 + y 2
= √
dx.
t=
v
y
2g 0
x=0
Because the
does not contain x explicitly, we can use (22.8) with the specific
integrand
√
form F = 1 + y 2 / y to find a first integral; on simplification this yields
1/2
2
y(1 + y )
= k,
784
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