Conditional Variation

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

$$I = \int_a^b F(y, y', x)\,dx,$$ (E.17)

subject to the constraint that the value of

$$J = \int_a^b G(y,y',x)\,dx$$ (E.18)

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

$$\frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-\frac{\... ...da\,G]}{\partial y'}\right)-\frac{\partial [\lambda\,G]}{\partial y}\right]= 0.$$ (E.19)

In principle, the previous equation, together with the constraint (E.18), 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, $y(-a/2) = y(+a/2) = 0$ .

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

$$U = - \rho\,g\,\int y\,ds = - \rho\,g\,\int_{-a/2}^{a/2} y\,(1+y'^{\,2})^{1/2}\,dx,$$ (E.20)

where $ds = \sqrt{dx^{\,2}+dy^{\,2}}$ is an element of length along the chain, and $g$ is 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

$$l = \int ds = \int_{-a/2}^{a/2}(1+y'^{\,2})^{1/2}\,dx$$ (E.21)

remains constant.

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

$$K = U + \lambda\,l = \int_{-a/2}^{a/2}(-\rho\,g\,y+\lambda)\,(1+y'^{\,2})^{1/2}\,dx,$$ (E.22)

where $\lambda$ is an, as yet, undetermined constant. Because the integrand in the functional does not depend explicitly on $x$ , we have from Equation (E.14) that

$$y'^{\,2}\,(-\rho\,g\,y+\lambda)\,(1+y'^{\,2})^{-1/2} - (-\rho\,g\,y+\lambda)\,(1+y'^{\,2})^{1/2} = k,$$ (E.23)

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

$$y'^{\,2} = \left(\lambda' + \frac{y}{h}\right)^2 - 1,$$ (E.24)

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

Let

$$\lambda' + \frac{y}{h} = -\cosh z.$$ (E.25)

Making this substitution, Equation (E.24) yields

$$\frac{dz}{dx} = -h^{\,-1}.$$ (E.26)

Hence,

$$z =-\frac{x}{h} + c,$$ (E.27)

where $c$ is a constant. It follows from Equation (E.25) that

$$y(x) =-h\,[\lambda' + \cosh(-x/h + c)].$$ (E.28)

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

$$\lambda' + \cosh(\mp a/2\,h + c) = 0.$$ (E.29)

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

$$y(x) = h\,[\cosh(a/2\,h) - \cosh(x/h)].$$ (E.30)

The final unknown constant, $h$ , is determined via the application of the constraint (E.21). Thus,

$$l= \int_{-a/2}^{a/2}(1+y'^{\,2})^{1/2}\,dx = \int_{-a/2}^{a/2} \cosh(x/h) \,dx = 2\,h\,\sinh(a/2\,h).$$ (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 $h$ satisfies

$$\frac{l}{2\,h} = \sinh\left(\frac{a}{2\,h}\right).$$ (E.32)