Isoperimetric Problems

So far we have dealt with boundary conditions of the form y(a) = A, y(b) = B or y(a) = A, y'(b) = B. For some problems the natural boundary conditions are expressed using an integral. The standard example is Dido's problem³: if you have a piece of rope with a fixed length, what shape should you make with it in order to enclose the largest possible area? Here we are trying to choose a function y to maximise an integral I(y) giving the area enclosed by y, but the fixed length constraint is also expressed in terms of an integral involving y. This kind of problem, where we seek an extremal of some function subject to `ordinary' boundary conditions and also an integral constraint, is called an isoperimetric problem.

A typical isoperimetric problem is to find an extremum of

I(y) = $$\int_{a}^{b}$$ F(x, y, y') dx,   subject to y(a) = A, y(b) = B, J(y) = $$\int_{a}^{b}$$ G(x, y, y') dx = L.

The condition J(y) = L is called the integral constraint.

Theorem 3   In the notation above, if I(Y) is an extremum of I subject to J(y) = L, then Y is an extremal of

K(y) = $$\int_{a}^{b}$$ F(x, y, y') + λG(x, y, y')  dx

for some constant λ.

You will need to know about Lagrange multipliers to understand this proof: see the handout on moodle (the constant λ will turn out to be a Lagrange multiplier).

Proof. Suppose I(Y) is a maximum or minimum subject to J(y) = L, and consider the two-parameter family of functions given by

Y(x) + εη(x) + δζ(x)

where ε and δ are constants and η(x) and ζ(x) are twice differentiable functions such that η(a) = ζ(a) = η(b) = ζ(b) = 0, with ζ chosen so that Y + εη + δζ obeys the integral constraint.

Consider the functions of two variables

I[ε, δ] = $$\int_{a}^{b}$$F(x, Y + εη + δζ, Y' + εη' + δζ') dx,  J[ε, δ] = $$\int_{a}^{b}$$G(x, Y + εη + δζ, Y' + εη' + δζ') dx.

Because I has a maximum or minimum at Y(x) subject to J = L, at the point (ε, δ)=(0, 0) our function I[ε, δ] takes an extreme value subject to J[ε, δ] = L.

It follows from the theory of Lagrange multipliers that a necessary condition for a function I[ε, δ] of two variables subject to a constraint J[ε, δ] = L to take an extreme value at (0, 0) is that there is a constant λ (called the Lagrange multiplier) such that

$${\frac{{\partial I}}{{\partial \epsilon}}} {\frac{{\partial J}}{{\partial \epsilon}}}$$ = 0

$${\frac{{\partial I}}{{\partial \delta}}} {\frac{{\partial J}}{{\partial \delta}}}$$ = 0

at the point ε = δ = 0. Calculating the ε derivative,

$${\frac{{\partial I}}{{\partial \epsilon}}} {\frac{{\partial J}}{{\partial \epsilon}}} \int_{a}^{b} {\frac{{\partial}}{{\partial \epsilon}}} \left(\vphantom{ F(x,Y+\epsilon \eta + \delta \zeta,Y'+\epsilon \... ...a G(x,Y+\epsilon \eta + \delta \zeta,Y'+\epsilon \eta' +\delta \zeta') }\right. \left.\vphantom{ F(x,Y+\epsilon \eta + \delta \zeta,Y'+\epsilon \... ...a G(x,Y+\epsilon \eta + \delta \zeta,Y'+\epsilon \eta' +\delta \zeta') }\right)$$ =

$$\int_{a}^{b} {\frac{{\partial}}{{\partial y}}} \left(\vphantom{ F+\lambda G }\right. \left.\vphantom{ F+\lambda G }\right) {\frac{{\partial}}{{\partial y'}}} \left(\vphantom{ F+\lambda G }\right. \left.\vphantom{ F+\lambda G }\right)$$ =

$$\int_{a}^{b} \left(\vphantom{ \frac{\partial}{\pa... ...\left( \frac{\partial}{\partial y'} \left( F+\lambda G \right) \right) }\right. {\frac{{\partial}}{{\partial y}}} \left(\vphantom{ F+\lambda G }\right. \left.\vphantom{ F+\lambda G }\right) {\frac{{{\rm d}}}{{{\rm d} x}}} \left(\vphantom{ \frac{\partial}{\partial y'} \left( F+\lambda G \right) }\right. {\frac{{\partial}}{{\partial y'}}} \left(\vphantom{ F+\lambda G }\right. \left.\vphantom{ F+\lambda G }\right) \left.\vphantom{ \frac{\partial}{\partial y'} \left( F+\lambda G \right) }\right) \left.\vphantom{ \frac{\partial}{\pa... ...\left( \frac{\partial}{\partial y'} \left( F+\lambda G \right) \right) }\right)$$ =

$$\mbox{when $\epsilon=\delta=0$, no matter what $\eta$\ is.}$$ =

Since this holds for any η, by the FLCV (Lemma ) we get

(F_y + λG_y)(x, Y, Y') + $${\frac{{{\rm d}}}{{{\rm d} x}}}$$(F_y' + λG_y')(x, Y, Y') = 0

which says that Y is a solution of the Euler-Lagrange equation for K, as required. $\qedsymbol$

Note that to complete the solution of the problem, the initially unknown multiplier λ must be determined at the end using the constraint J(y) = L.

Exercise 6   Find an extremal of the functional

I(y) = $$\int_{0}^{1}$$(y')² dx,     y(0) = y(1) = 1,

subject to the constraint that

     J(y) = $$\int_{0}^{1}$$y dx = 2.    $$\left[\vphantom{ \mbox{Answer:}~~y=f(x)=-6\left(x-\frac{1}{2}\right)^2+\frac{5}{2}. }\right.$$Answer:  y = f (x) = - 6$$\left(\vphantom{x-\frac{1}{2}}\right.$$x - $${\frac{{1}}{{2}}}$$$$\left.\vphantom{x-\frac{1}{2}}\right)^{2}_{}$$ + $${\frac{{5}}{{2}}}$$.$$\left.\vphantom{ \mbox{Answer:}~~y=f(x)=-6\left(x-\frac{1}{2}\right)^2+\frac{5}{2}. }\right]$$

Exercise 7 (Sheep pen design problem)   : A fence of length l must be attached to a straight wall at points A and B (a distance a apart, where a < l) to form an enclosure. Show that the shape of the fence that maximizes the area enclosed is the arc of a circle, and write down (but do not try to solve) the equations that determine the circle's radius and the location of its centre in terms of a and l.