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Next: Multi-Function Variation Up: Hamiltonian Dynamics Previous: Calculus of Variations

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,
\end{displaymath} (691)

subject to the constraint that the value of
J = \int_a^b G(y,y',x)\,dx
\end{displaymath} (692)

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 $\delta J = 0$, since 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)-\f...
...y'}\right)-\frac{\partial [\lambda\,G]}{\partial y}\right]= 0.
\end{displaymath} (693)

In principle, the above equation, together with the constraint (692), 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 which 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 the discussion in Section 3.2, the stable equilibrium state of a conservative dynamical system is one which minimizes the system's potential energy. 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,
\end{displaymath} (694)

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
\end{displaymath} (695)

remains constant.

It follows, from the above 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,
\end{displaymath} (696)

where $\lambda$ is an, as yet, undetermined constant. Since the integrand in the functional does not depend explicitly on $x$, we have from Equation (688) that
y'^{\,2}\,(-\rho\,g\,y+\lambda)\,[1+y'^{\,2}]^{-1/2} - (-\rho\,g\,y+\lambda)\,[1+y'^{\,2}]^{1/2} = k,
\end{displaymath} (697)

where $k$ is a constant. This expression reduces to
y'^{\,2} = \left(\lambda' + \frac{y}{h}\right)^2 - 1,
\end{displaymath} (698)

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


\lambda' + \frac{y}{h} = -\cosh z.
\end{displaymath} (699)

Making this substitution, Equation (698) yields
\frac{dz}{dx} = -h^{-1}.
\end{displaymath} (700)

z =-\frac{x}{h} + c,
\end{displaymath} (701)

where $c$ is a constant. It follows from Equation (699) that
y(x) =-h\,[\lambda' + \cosh(-x/h + c)].
\end{displaymath} (702)

The above 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.
\end{displaymath} (703)

Hence, $c=0$, and $\lambda' = - \cosh (a/2\,h)$. It follows that
y(x) = h\,[\cosh(a/2\,h) - \cosh(x/h)].
\end{displaymath} (704)

The final unknown constant, $h$, is determined via the application of the constraint (695). 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).
\end{displaymath} (705)

Hence, the equilibrium configuration of the chain is given by the curve (704), which is known as a catenary, where the parameter $h$ satisfies
\frac{l}{2\,h} = \sinh\left(\frac{a}{2\,h}\right).
\end{displaymath} (706)

next up previous
Next: Multi-Function Variation Up: Hamiltonian Dynamics Previous: Calculus of Variations
Richard Fitzpatrick 2011-03-31