Next: Multi-Function Variation
Up: Calculus of Variations
Previous: Euler-Lagrange Equation
Suppose that we wish to find the function
which
maximizes or minimizes the functional
![$\displaystyle I = \int_a^b F(y, y', x)\,dx,$](img7304.png) |
(E.17) |
subject to the constraint that the value of
![$\displaystyle J = \int_a^b G(y,y',x)\,dx$](img7327.png) |
(E.18) |
remains constant. We can achieve our goal by finding an extremum of the new functional
, where
is an undetermined function. We know
that
, because the value of
is fixed, so if
then
as well. In other words, finding an extremum of
is equivalent
to finding an extremum of
. Application of the Euler-Lagrange
equation yields
![$\displaystyle \frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-\frac{\...
...da\,G]}{\partial y'}\right)-\frac{\partial [\lambda\,G]}{\partial y}\right]= 0.$](img7334.png) |
(E.19) |
In principle, the previous equation, together with the constraint (E.18),
yields the functions
and
. Incidentally,
is generally
termed a Lagrange multiplier. If
and
have no explicit
-dependence then
is usually a constant.
As an example, consider the following famous problem. Suppose that a uniform
chain of fixed length
is suspended by its ends from
two equal-height fixed points that are a distance
apart, where
.
What is the equilibrium configuration of the chain?
Suppose that the chain has the uniform density per unit length
.
Let the
- and
-axes be horizontal and vertical, respectively, and
let the two ends of the chain lie at
. The equilibrium configuration of the chain is specified by the function
, for
, where
is the vertical distance of the chain below its end points at horizontal
position
. Of course,
.
According to standard Newtonian dynamics, the stable equilibrium
state of a conservative dynamical system is one that minimizes
the system's potential energy (Fitzpatrick 2012). Now, the potential energy of the chain
is written
![$\displaystyle U = - \rho\,g\,\int y\,ds = - \rho\,g\,\int_{-a/2}^{a/2} y\,(1+y'^{\,2})^{1/2}\,dx,$](img7339.png) |
(E.20) |
where
is an element of length along the chain, and
is the acceleration due to gravity.
Hence, we need to minimize
with respect to small variations in
.
However, the variations in
must be such as to conserve the
fixed length of the chain. Hence, our minimization procedure is subject to
the constraint that
![$\displaystyle l = \int ds = \int_{-a/2}^{a/2}(1+y'^{\,2})^{1/2}\,dx$](img7341.png) |
(E.21) |
remains constant.
It follows, from the previous discussion, that we need to minimize the
functional
![$\displaystyle K = U + \lambda\,l = \int_{-a/2}^{a/2}(-\rho\,g\,y+\lambda)\,(1+y'^{\,2})^{1/2}\,dx,$](img7342.png) |
(E.22) |
where
is an, as yet, undetermined constant. Because the integrand
in the functional does not depend explicitly on
, we have
from Equation (E.14) that
![$\displaystyle y'^{\,2}\,(-\rho\,g\,y+\lambda)\,(1+y'^{\,2})^{-1/2} - (-\rho\,g\,y+\lambda)\,(1+y'^{\,2})^{1/2} = k,$](img7343.png) |
(E.23) |
where
is a constant. This expression reduces to
![$\displaystyle y'^{\,2} = \left(\lambda' + \frac{y}{h}\right)^2 - 1,$](img7344.png) |
(E.24) |
where
, and
.
Let
![$\displaystyle \lambda' + \frac{y}{h} = -\cosh z.$](img7347.png) |
(E.25) |
Making this substitution, Equation (E.24) yields
![$\displaystyle \frac{dz}{dx} = -h^{\,-1}.$](img7348.png) |
(E.26) |
Hence,
![$\displaystyle z =-\frac{x}{h} + c,$](img7349.png) |
(E.27) |
where
is a constant. It follows from Equation (E.25) that
![$\displaystyle y(x) =-h\,[\lambda' + \cosh(-x/h + c)].$](img7350.png) |
(E.28) |
The previous solution contains three undetermined constants,
,
, and
. We can
eliminate two of these constants by application of the boundary
conditions
. This yields
![$\displaystyle \lambda' + \cosh(\mp a/2\,h + c) = 0.$](img7353.png) |
(E.29) |
Hence,
, and
. It follows that
![$\displaystyle y(x) = h\,[\cosh(a/2\,h) - \cosh(x/h)].$](img7356.png) |
(E.30) |
The final unknown constant,
, is determined via the application of
the constraint (E.21). Thus,
![$\displaystyle l= \int_{-a/2}^{a/2}(1+y'^{\,2})^{1/2}\,dx = \int_{-a/2}^{a/2} \cosh(x/h) \,dx = 2\,h\,\sinh(a/2\,h).$](img7357.png) |
(E.31) |
Hence, the equilibrium configuration of the chain is given by the curve
(E.30), which is known as a catenary (from the Latin for chain), where the parameter
satisfies
![$\displaystyle \frac{l}{2\,h} = \sinh\left(\frac{a}{2\,h}\right).$](img7358.png) |
(E.32) |
Next: Multi-Function Variation
Up: Calculus of Variations
Previous: Euler-Lagrange Equation
Richard Fitzpatrick
2016-01-22