# Special cases

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Special cases
```22.2 SPECIAL CASES
to these variations, we require
dI =0
dα α=0
for all η(x).
(22.3)
Substituting (22.2) into (22.1) and expanding as a Taylor series in α we obtain
b
F(y + αη, y + αη , x) dx
I(y, α) =
a
b
b
∂F
∂F
=
F(y, y , x) dx +
αη + αη dx + O(α2 ).
∂y
∂y
a
a
With this form for I(y, α) the condition (22.3) implies that for all η(x) we require
b
∂F
∂F
η + η dx = 0,
δI =
∂y
∂y
a
where δI denotes the ﬁrst-order variation in the value of I due to the variation
(22.2) in the function y(x). Integrating the second term by parts this becomes
b b ∂F
d ∂F
∂F
−
η(x) dx = 0.
(22.4)
η +
∂y a
∂y
dx ∂y a
In order to simplify the result we will assume, for the moment, that the end-points
are ﬁxed, i.e. not only a and b are given but also y(a) and y(b). This restriction
means that we require η(a) = η(b) = 0, in which case the ﬁrst term on the LHS of
(22.4) equals zero at both end-points. Since (22.4) must be satisﬁed for arbitrary
η(x), it is easy to see that we require
d ∂F
∂F
=
.
(22.5)
∂y
dx ∂y This is known as the Euler–Lagrange (EL) equation, and is a diﬀerential equation
for y(x), since the function F is known.
22.2 Special cases
In certain special cases a ﬁrst integral of the EL equation can be obtained for a
general form of F.
22.2.1 F does not contain y explicitly
In this case ∂F/∂y = 0, and (22.5) can be integrated immediately giving
∂F
= constant.
∂y 777
(22.6)
CALCULUS OF VARIATIONS
B
(b, y(b))
ds
dy
dx
A (a, y(a))
Figure 22.2 An arbitrary path between two ﬁxed points.
Show that the shortest curve joining two points is a straight line.
Let the two points be labelled A and B and have coordinates (a, y(a)) and (b, y(b))
respectively (see ﬁgure 22.2). Whatever the shape of the curve joining A to B, the length
of an element of path ds is given by
1/2
2
ds = (dx)2 + (dy)2
= (1 + y )1/2 dx,
and hence the total path length along the curve is given by
b
L=
(1 + y )1/2 dx.
2
(22.7)
a
We must now apply the results of the previous section to determine that path which makes
L stationary (clearly a minimum in this case). Since the integral does not contain y (or
indeed x) explicitly, we may use (22.6) to obtain
k=
y
∂F
=
.
∂y (1 + y 2 )1/2
where k is a constant. This is easily rearranged and integrated to give
y=
k
x + c,
(1 − k 2 )1/2
which, as expected, is the equation of a straight line in the form y = mx + c, with
m = k/(1 − k 2 )1/2 . The value of m (or k) can be found by demanding that the straight line
passes through the points A and B and is given by m = [ y(b) − y(a)]/(b − a). Substituting
the equation of the straight line into (22.7) we ﬁnd that, again as expected, the total path
length is given by
L2 = [ y(b) − y(a)]2 + (b − a)2 . 778
22.2 SPECIAL CASES
y
dy
ds
dx
x
Figure 22.3 A convex closed curve that is symmetrical about the x-axis.
22.2.2 F does not contain x explicitly
In this case, multiplying the EL equation (22.5) by y and using
d
∂F
d ∂F
∂F
y = y
+ y dx
∂y
dx ∂y ∂y
we obtain
y
∂F
∂F
d
+ y =
∂y
∂y
dx
∂F
y .
∂y
But since F is a function of y and y only, and not explicitly of x, the LHS of
this equation is just the total derivative of F, namely dF/dx. Hence, integrating
we obtain
F − y
∂F
= constant.
∂y (22.8)
Find the closed convex curve of length l that encloses the greatest possible area.
Without any loss of generality we can assume that the curve passes through the origin
and can further suppose that it is symmetric with respect to the x-axis; this assumption
is not essential. Using the distance s along the curve, measured from the origin, as the
independent variable and y as the dependent one, we have the boundary conditions
y(0) = y(l/2) = 0. The element of area shown in ﬁgure 22.3 is then given by
1/2
dA = y dx = y (ds)2 − (dy)2
,
and the total area by
l/2
A=2
y(1 − y )1/2 ds;
2
(22.9)
0
here y stands for dy/ds rather than dy/dx. Since the integrand does not contain s explicitly,
779
CALCULUS OF VARIATIONS
we can use (22.8) to obtain a ﬁrst integral of the EL equation for y, namely
y(1 − y )1/2 + yy (1 − y )−1/2 = k,
2
2
2
where k is a constant. On rearranging this gives
ky = ±(k 2 − y 2 )1/2 ,
which, using y(0) = 0, integrates to
y/k = sin(s/k).
(22.10)
The other end-point, y(l/2) = 0, ﬁxes the value of k as l/(2π) to yield
y=
l
2πs
sin
.
2π
l
From this we obtain dy = cos(2πs/l) ds and since (ds)2 = (dx)2 + (dy)2 we ﬁnd also that
dx = ± sin(2πs/l) ds. This in turn can be integrated and, using x(0) = 0, gives x in terms
of s as
l
l
2πs
=−
cos
.
x−
2π
2π
l
We thus obtain the expected result that x and y lie on the circle of radius l/(2π) given by
2
l
l2
+ y2 =
.
x−
2π
4π 2
Substituting the solution (22.10) into the expression for the total area (22.9), it is easily
veriﬁed that A = l 2 /(4π). A much quicker derivation of this result is possible using plane
polar coordinates. The previous two examples have been carried out in some detail, even though
the answers are more easily obtained in other ways, expressly so that the method
is transparent and the way in which it works can be ﬁlled in mentally at almost
every step. The next example, however, does not have such an intuitively obvious
solution.
Two rings, each of radius a, are placed parallel with their centres 2b apart and on a
common normal. An open-ended axially symmetric soap ﬁlm is formed between them (see
ﬁgure 22.4). Find the shape assumed by the ﬁlm.
Creating the soap ﬁlm requires an energy γ per unit area (numerically equal to the surface
tension of the soap solution). So the stable shape of the soap ﬁlm, i.e. the one that
minimises the energy, will also be the one that minimises the surface area (neglecting
gravitational eﬀects).
It is obvious that any convex surface, shaped such as that shown as the broken line in
ﬁgure 22.4(a), cannot be a minimum but it is not clear whether some shape intermediate
between the cylinder shown by solid lines in (a), with area 4πab (or twice this for the
double surface of the ﬁlm), and the form shown in (b), with area approximately 2πa2 , will
produce a lower total area than both of these extremes. If there is such a shape (e.g. that in
ﬁgure 22.4(c)), then it will be that which is the best compromise between two requirements,
the need to minimise the ring-to-ring distance measured on the ﬁlm surface (a) and the
need to minimise the average waist measurement of the surface (b).
We take cylindrical polar coordinates as in ﬁgure 22.4(c) and let the radius of the soap
ﬁlm at height z be ρ(z) with ρ(±b) = a. Counting only one side of the ﬁlm, the element of
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