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Next: Multi-Function Variation Up: Classical Mechanics Previous: Calculus of Variations

Conditional Variation

Suppose that we wish to find the function, $ y(x)$ , that maximizes or minimizes the functional

$\displaystyle I = \int_a^b F(y, y',x) dx,$ (B.45)

subject to the constraint that the value of

$\displaystyle J = \int_a^b G(y,y',x) dx$ (B.46)

remains constant. We can achieve our goal by finding an extremum of the new functional $ K = I + \lambda J$ , where $ \lambda(x)$ is an undetermined function. We know that $ \delta J = 0$ , because the value of $ J$ is fixed, so if $ \delta K= 0$ then $ \delta I = 0$ as well. In other words, finding an extremum of $ K$ is equivalent to finding an extremum of $ I$ . Application of the Euler-Lagrange equation yields

$\displaystyle \frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-\frac{\...
...da G]}{\partial y'}\right)-\frac{\partial [\lambda G]}{\partial y}\right]= 0.$ (B.47)

In principle, the previous equation, together with the constraint (B.46), yields the functions $ \lambda(x)$ and $ y(x)$ . Incidentally, $ \lambda$ is generally termed a Lagrange multiplier. If $ F$ and $ G$ have no explicit $ x$ -dependence then $ \lambda$ is usually a constant.

As an example, consider the following famous problem. Suppose that a uniform chain of fixed length $ l$ is suspended by its ends from two equal-height fixed points that are a distance $ a$ apart, where $ a < l$ . What is the equilibrium configuration of the chain?

Suppose that the chain has the uniform density per unit length $ \rho $ . Let the $ x$ - and $ y$ -axes be horizontal and vertical, respectively, and let the two ends of the chain lie at $ (\pm a/2, 0)$ . The equilibrium configuration of the chain is specified by the function $ y(x)$ , for $ -a/2\leq x \leq +a/2$ , where $ y(x)$ is the vertical distance of the chain below its end points at horizontal position $ x$ . Of course, $ y(-a/2) = y(+a/2) = 0$ .

The stable equilibrium state of a conservative dynamical system is one that minimizes the system's potential energy. The potential energy of the chain is written

$\displaystyle U = - \rho g \int y ds = - \rho g \int_{-a/2}^{a/2} y [1+y'^{ 2}]^{1/2} dx,$ (B.48)

where $ ds = [dx^{ 2}+dy^{ 2}]^{1/2}$ is an element of length along the chain, and $ g$ the acceleration due to gravity. Hence, we need to minimize $ U$ with respect to small variations in $ y(x)$ . However, the variations in $ y(x)$ must be such as to conserve the fixed length of the chain. Hence, our minimization procedure is subject to the constraint that

$\displaystyle l = \int ds = \int_{-a/2}^{a/2}[1+y'^{ 2}]^{1/2} dx$ (B.49)

remains constant.

It follows, from the previous discussion, that we need to minimize the functional

$\displaystyle K = U + \lambda l = \int_{-a/2}^{a/2}(-\rho g y+\lambda) [1+y'^{ 2}]^{1/2} dx,$ (B.50)

where $ \lambda$ is an, as yet, undetermined constant. Because the integrand in the functional does not depend explicitly on $ x$ , we have from Equation (B.42) that

$\displaystyle y'^{ 2} (-\rho g y+\lambda) [1+y'^{ 2}]^{-1/2} - (-\rho g y+\lambda) [1+y'^{ 2}]^{1/2} = k,$ (B.51)

where $ k$ is a constant. This expression reduces to

$\displaystyle y'^{ 2} = \left(\lambda' + \frac{y}{h}\right)^2 - 1,$ (B.52)

where $ \lambda' = \lambda/k$ , and $ h=-k/\rho g$ .


$\displaystyle \lambda' + \frac{y}{h} = -\cosh z.$ (B.53)

Making this substitution, Equation (B.52) yields

$\displaystyle \frac{dz}{dx} = -h^{-1}.$ (B.54)


$\displaystyle z =-\frac{x}{h} + c,$ (B.55)

where $ c$ is a constant. It follows from Equation (B.53) that

$\displaystyle y(x) =-h\left[\lambda' + \cosh\left(-\frac{x}{h} + c\right)\right].$ (B.56)

The previous solution contains three undetermined constants, $ h$ , $ \lambda'$ , and $ c$ . We can eliminate two of these constants by application of the boundary conditions $ y(\pm a/2)= 0$ . This yields

$\displaystyle \lambda' + \cosh\left(\mp \frac{a}{2 h} + c\right) = 0.$ (B.57)

Hence, $ c=0$ , and $ \lambda' = - \cosh (a/2 h)$ . It follows that

$\displaystyle y(x) = h\left[\cosh\left(\frac{a}{2 h}\right) - \cosh\left(\frac{x}{h}\right)\right].$ (B.58)

The final unknown constant, $ h$ , is determined via the application of the constraint (B.49). Thus,

$\displaystyle l= \int_{-a/2}^{a/2}[1+y'^{ 2}]^{1/2} dx = \int_{-a/2}^{a/2} \cosh\left(\frac{x}{h}\right)  dx = 2 h \sinh\left(\frac{a}{2 h}\right).$ (B.59)

Hence, the equilibrium configuration of the chain is given by the curve (B.58), which is known as a catenary, where the parameter $ h$ satisfies

$\displaystyle \frac{l}{2 h} = \sinh\left(\frac{a}{2 h}\right).$ (B.60)

next up previous
Next: Multi-Function Variation Up: Classical Mechanics Previous: Calculus of Variations
Richard Fitzpatrick 2016-01-25