Conditional Variation

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

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

subject to the constraint that the value of

$$J = \int_a^b G(y,y',x) dx$$ (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

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

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

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

remains constant.

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

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

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

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

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

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

Let

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

Making this substitution, Equation (B.52) yields

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

Hence,

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

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

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

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

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

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

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

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

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