# What is the solution to the Dido isoperimetric problem when the length is longer than the half-circle?

Given $L$-the length of a curve (single-valued function) passing trough the points $x_1$ and $x_2$ on the $x$-axis. What is the curve $y(x)$ maximizing the area between this curve and the $x$-axis?

The solution is, of course, well known: one formulates a variational problem with a constraint

$F[y,y']=\int_{x_1}^{x_2}\left(y+\lambda\sqrt{1+y'^2(x)}\right)dx,$

which yields an equation of circle $(x-x_0)^2+(y-y_0)^2=r^2$ as a solution. On the final step, one finds the unknown constants $x_0$, $y_0$, and $r$ by requesting that the circlular arc goes through the points $(x_1,0)$, $(x_2,0)$, and has the length $L$.

The solution is clear to me for $0\le L\le \pi/2(x_2-x_1)$. But what if $L>\pi/2(x_2-x_1)$ ? To my opinion, there is a discrepancy between the length as computed via the integral and the length obtained by imposing the boundary conditions. This is illustrated in the second panel. What is the solution in this case?

• How can you compute the length via the integral without imposing the boundary conditions first? There are three parameters in the circle equation, which you need to determine from $x_1$, $x_2$ and $L$. Only after that it makes sense to take the integral, and it will be equal to $L$ simply because its parameters were determined to make it so equal. I am not sure what the second image represents. – Conifold Apr 11 '17 at 22:18
• Even in the first case, you shouldn't parametrize your curve by $x$. Doing so makes an assumption about the solution - that $y$ will be single-valued with respect to $x$. In the first case, it turns out this is true, but you shouldn't have assumed it. In the second case, it is false, and your integral fails to capture the complete length of the curve, and thus is not appropriate for the problem. – Paul Sinclair Apr 12 '17 at 3:15
• Both Conifold and Paul Sinclair are right. The formula $\sqrt {1 + y'(x)^2}$ that you use in your functional tacitly assumes that $y$ is a function of $x$, so you find the extremum of the functional only among the curves in that class. For more generality, assume your curve to be given parametrically by $t \mapsto (x(t),y(t))$, use the appropriate length term $\sqrt {x'(t)^2 + y'(t)^2}$ and reconsider your problem. – Alex M. Apr 12 '17 at 9:03
• @yarchik - "So many people cannot be wrong." - "Heh, heh, heh. Surely you're joking, Mr. Feynman." – Paul Sinclair Apr 12 '17 at 23:02
• @PaulSinclair Apparently, many people, myself included, can miss the point. I see it now, for large lengths the "solution" among graphs is not a circular arc, it is complemented by vertical segments, see answer. – Conifold Apr 13 '17 at 19:35

To understand what happens consider a simpler example (due to Weierstrass), where we minimize the length of graphs of $C^1$ functions that vanish at the ends and pass through $(1/2,0)$. It should be clear intuitively that the solution is a broken line with a pair of line segments, connecting $(0,0)$ to $(1/2,0)$, and $(1/2,0)$ to $(1,0)$. But this "solution" is not in $C^1$. In the Weierstarass example we can fix this by passing from $C^1$ to the space of absolutely continuous functions $W^{1,1}$ ($L^1$ with derivatives in $L^1$), to which the broken line belongs. But even this will not help in the Dido problem (the OP example) for $L>\pi/2$, which is a good illustration of the limitations of the classical calculus of variations.

By shifting and scaling we can assume that $x_1=0$, $x_2=1$, and formalize the problem as finding $y\in W^{1,1}([0,1])$ that maximizes $A[y]=\int_{0}^{1}y\,dx\to\max$ with the constraints $y(0)=y(1)=0$ and $\int_{0}^{1}\sqrt{1+y'^2(x)}\,dx=L$. If $L\leq\pi/2$ the solution is a circular arc as in the OP Fig. 1. But if $L>\pi/2$ there is no solution. Of course, there are functions in $W^{1,1}([0,1])$, or even in $C^1([0,1])$, with arbitrarily large lengths that vanish at the ends. But it does not mean that there is any among them with the largest possible area. The underlying reason is that balls in $W^{1,1}$ are not weakly compact, so sequences of functions with bounded length may not have (relevantly) convergent subsequences. A simple example is the sequence $x^n$, lengths of the graphs converge to $2$, but the limiting function is not continuous, $0$ everywhere but $1$, and $1$ at $1$. To see that it is the "limit" we need to "add" a vertical segment at $1$ to the graph. Limits of absolutely continuous functions can be discontinuous functions of bounded variation, and for them the length functional ought to be redefined, it is not given by the integral.

The same happens in the Dido problem. For $L>\pi/2$ the solution is not a circular arc, but a pair of vertical segments at the endpoints (they pick up the excess in length) topped with a semicircle, similar to the dashed shape in the OP Fig. 2, except for the topping. An intuitive way to see this is to imagine that the graph represents a flexible but unstretchable membrane attached to a solid wall (the $x$-axis) at the endpoints, and the area in between is being filled with water. To enforce the condition that the curve is a graph we also have to put up two vertical solid walls at the endpoints. The water will tend to maximize area by pushing the membrane equally in all directions, this creates the semicircular topping, but on the sides it can only flatten it along the walls.

An alternative way to formalize the problem is to remove the requirement that the curve is a graph by considering parametric curves, as suggested in the comments to the OP. In that case the vertical walls are removed, and the solution will again be a circular arc, as the intuition suggests (solid and dashed parts above the $x$-axis in the OP Fig. 2). Thus, we get two different solutions for $L>\pi/2$ depending on how the problem is made precise beyond the differentiable functions.

A very good exposition of the history of and subtleties in variational problems, including the Dido problem, is Tikhomirov's Stories About Maxima and Minima

• Very pedagogical and interesting answer! Addresses all my points and, i fact, leads to some more questions that I will probably post later. – yarchik Apr 14 '17 at 6:57
• How we can prove the penultimate paragraph? Because, if the $y(t)$ and $x(t)$ are continuously differentiable, we can use the Green's theorem, like in the site: mathematicalgarden.wordpress.com/2008/12/21/the-problem-of-dido. But that's not the case, because $(x(t),y(t))$ has a horizontal line on the $x$ axis, which brokes the differentiability of $y(t)$ and $x(t)$. So, how can I proceed? – Rafael Deiga Apr 21 '17 at 16:21
• @RafaelDeiga The boundary integral has to be interpreted in a generalized sense, there will be jump terms at the corner points. But since we only vary parts of the curves connecting the endpoints, the Euler-Lagrange equations are the same. The jump terms amount to enforcing the boundary conditions (that the curve attaches at the endpoints to the $x$-axis segment). So you still get an equation of the circle with the length and boundary conditions fixing the center and the radius. – Conifold Apr 21 '17 at 18:26
• But, how can I find the area? I mean, I could use $\iint_S dxdy$, but after what I do? Because, to use Euler-Lagrange equations, I need to assemble a functional, which I don't know how. I do not even know $S$. – Rafael Deiga Apr 21 '17 at 20:30
• @RafaelDeiga $S$ is a region bounded by $[0,1]$ on the $x$-axis, and a smooth curve that attaches to it at the endpoints. The rest is almost identical to the derivation you linked, except the area functional, transformed by the Green's formula, will have non-integral terms at the endpoints due to potential corners. Since the integral over the smooth curve part is the same, and the horizontal segment is not varied, the Euler-Lagrange equations are the same – Conifold Apr 21 '17 at 20:47