# Am I solving Dido's isoperimetric (variational) problem correctly?

I am trying to solve Dido's isoperimetric problem, more specifically, the version where we have to maximize the area under a curve, given that the two endpoints are on the x-axis, and given a fixed arclength:

That is, we have to maximize $$J(y)=\int_a^by(x)dx$$ subject to constraint $$C(y)=\int_a^b\sqrt{1+(y'(x))^2}dx$$ Where we assume $$a$$ is fixed, but $$b$$ is allowed to vary.

## Am I solving this correctly?

Besides the boundary constraint (because of variable boundary $$b$$) $$L_{y'}(b)=0$$, we of course formulate the Euler-Lagrange equation $$L_y=\frac d {dx}L_{y'}$$, based on $$J$$ subject to the constraint $$C$$, so that the Lagrangian becomes:

$$L(y,y')=y(x)+\lambda\sqrt{1+(y'(x))^2}$$

Therefore: $$L_y=1$$ $$L_{y'}=\lambda \frac {y'(x)} {\sqrt{1+(y'(x))^2}}$$ $$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1} {\sqrt{1+(y'(x))^2}}-\frac {(y'(x))^2}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$

which is equal to

$$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$

This gives the Euler-Lagrange equation $$\lambda y''(x)=\left (1+(y'(x))^2\right)^{\frac 3 2}$$

I have no idea how to solve this ODE, and moreover, it doesn't seem like this is what I should be getting.

Did I derive this result correctly? If so, How do I solve it?

note: I know that it is also possible to solve by parameterizing $$x=x(t), y(x)=y(x(t))$$. I want to do this as well, later, but I would first like to understand the approach I'm taking here.

• Sep 16, 2017 at 16:57

You have derived the radius of circle correctly. Maybe you did not realize that no need to anything further than seeing how the particular situation constant$$~ \lambda$$ applies everywhere.

If point $$b$$ is moving w.r.t. point $$a$$ on x-axis we have to understand each segment situation in all the three cases with differing radii $$\lambda, d$$ symbolically given by a single relation in the last line of answer.

The Lagrange multiplier $$\lambda$$ is nothing but the associated geometrical invariant radius in this case.

If distance $$(a-b) =d$$ and arc length $$L$$ are given then we have three cases

• $$L= d$$ gives optimal semi-circular area segment

• $$L> d$$ gives optimal area of major segment

• $$L< d$$ gives optimal area of minor segment

In the last two cases we need to iterate /numerically and find radius $$\lambda$$ from:

$$L=\lambda \cdot 2 \sin^{-1} \frac{d}{2\lambda}$$

which relates $$\lambda$$ with the given constant length quantities $$d,l$$.

Since the maximum configuration to Dido's isoperimetric problem is a semicircle, it is natural to use polar coordinates, which we will try in this answer.

1. method: Polar coordinates $$(r,\theta)$$ with Lagrange multiplier $$\lambda$$. Infinitesimal arclength: $$(ds)^2~=~(dr)^2 + (rd\theta)^2.\tag{1a}$$ Area functional: \begin{align}\widetilde{A} ~=~&A+\lambda (\ell-\ell_0) ~=~\int_0^{\pi} \! d\theta~L-\lambda\ell_0, \cr L~=~&\frac{r^2}{2}+\lambda\sqrt{r^2+\dot{r}^2}, \qquad \dot{r}~=~\frac{dr}{d\theta}. \end{align}\tag{1b} Momentum: $$p~=~\frac{\partial L}{\partial\dot{r}} ~=~\frac{\lambda\dot{r}}{\sqrt{r^2+\dot{r}^2}}. \tag{1c}$$ Infinitesimal variation: $$\delta\widetilde{A} ~=~\int_0^{\pi}\! d\theta~ EL~\delta r + [p \delta r]^{\theta=\pi}_{\theta=0} +(\ell-\ell_0) \delta\lambda. \tag{1d}$$ This leads to natural/Neumann BC: $$p(0)~=~0~=~p(\pi)\quad\Leftrightarrow\quad \dot{r}(0)~=~0~=~\dot{r}(\pi),\tag{1e}$$ and the constraint $$\ell~=~\int_0^{\pi} \! d\theta\sqrt{r^2+\dot{r}^2}~=~\ell_0\tag{1f}.$$ Energy is a constant: \begin{align} E~=~p\dot{r}-L~=~&-\frac{\lambda r^2}{\sqrt{r^2+\dot{r}^2}}-\frac{r^2}{2}\cr \Leftrightarrow\qquad \sqrt{r^2+\dot{r}^2}~=~& \frac{-\lambda}{\frac{E}{r^2}+\frac{1}{2}}.\end{align} \tag{1g} The stationary points $$\dot{r}=0$$ of the profile $$\theta\mapsto r(\theta)$$ apparently satisfy the 2nd order equation $$0~=~\frac{r^2}{2}+\lambda r+E, \tag{1h}$$ with roots $$r_{\pm}~=~-\lambda\pm\sqrt{D}, \qquad D~=~\lambda^2-2E. \tag{1i}$$ We conclude that $$r_-~\leq~r~\leq~r_+. \tag{1j}$$

Case $$r_- < 0$$. Note that $$r=r_+$$ whenever $$\dot{r}=0$$, i.e. at stationary points and endpoints. It follows that $$r~=~r_+ \tag{1k}$$ is a constant. $$\Box$$

Case $$r_- \geq 0$$. Then $$E\geq 0$$ and $$\lambda\leq 0$$. Hm. Eq. (1g) seems fairly complicated to solve. Let's try another approach...

Perturbative variation around constant profile. Let $$r(\theta)~=~r_0+\eta(\theta), \qquad r_0~=~\frac{\ell_0}{\pi}, \qquad \eta~\ll~\ell_0. \tag{1l}$$ Lagrangian to quadratic order: $$L_2~=~\frac{r_0^2}{2} +\lambda r_0 + (r_0+\lambda)\eta + \frac{\eta^2}{2} + \frac{\lambda}{2r_0}\dot{\eta}^2. \tag{1m}$$ Euler-Lagrange (EL) equation $$\frac{\lambda}{r_0}\ddot{\eta}~=~\eta+(r_0+\lambda).\tag{1n}$$ BC: $$\dot{\eta}(0)~=~0~=~\dot{\eta}(\pi).\tag{1o}$$ Solution: $$\eta+(r_0+\lambda)~=~a\cos\left(\theta\sqrt{\frac{-\lambda}{r_0}} \right).\tag{1p}$$ In order for $$r_0$$ to be a stationary profile, apparently $$\lambda~=~-r_0. \tag{1q}$$ Then the fluctuation (1p) is a zero-mode at quadratic order.

More generally, we can consider a Fourier series $$\eta(\theta)~=~\sum_{n\in\mathbb{N}}a_n\cos(n\theta). \tag{1r}$$ It is a straightforward exercise to see that the higher modes $$n>0$$ lower the value of (the quadratic approximation to) the area functional (1b), while the first mode $$n=1$$ is a zero-mode.

The first mode $$n=1$$ presumably stops being a zero-mode at higher order. Hm. Let's try another approach...

2. method: Radial coordinate $$r$$ as a function of arclength $$s$$. Area functional: $$A ~=~\int_0^{\ell_0} \! ds~L, \qquad L~=~\frac{r}{2}\sqrt{1-\dot{r}^2}, \qquad \dot{r}~=~\frac{dr}{ds}. \tag{2a}$$ Momentum: $$p~=~\frac{\partial L}{\partial\dot{r}} ~=~-\frac{r\dot{r}}{2\sqrt{1-\dot{r}^2}}. \tag{2b}$$ Infinitesimal variation: $$\delta A ~=~\int_0^{\ell_0}\! ds~ EL~\delta r + [p \delta r]^{s=\ell_0}_{s=0} . \tag{2c}$$ This leads to natural/Neumann BC: $$p(0)~=~0~=~p(\ell_0)\quad\Leftrightarrow\quad \dot{r}(0)~=~0~=~\dot{r}(\ell_0).\tag{2d}$$ Energy is a negative constant: $$E~=~p\dot{r}-L~=~-\frac{r}{2\sqrt{1-\dot{r}^2}}~<0. \tag{2e}$$ Define positive constant $$R~:=~-2E~>~0. \tag{2f}$$ Then the first integral reads $$\left(\frac{r}{R}\right)^2~=~1-\dot{r}^2. \tag{2g}$$ So $$0~<~r~\leq~R.\tag{2h}$$ Note that $$r=R$$ whenever $$\dot{r}=0$$, i.e. at stationary points and endpoints. It follows that $$r~=~R \tag{2i}$$ is constant. It is not hard to see that a semicircle (2i) is a maximum configuration to Dido's isoperimetric problem. $$\Box$$

First, integrate both sides with regards to $x$. Next, we can get an explicit function $x=f(y')+C_1$, $C_1$ is a constant. Here we have a fact : $y'(x)=0$ if and only if $x=C_1$ Geometrically, curve $y$ reaches the maximum value when $x=C_1$. Hence, $x$ increases for $x<C_1$ and x decrease for $x>C_1$. So, $y'(x)$ will never be zero when $x$ in $(0,C_1) \cup (C_1,1)$. Finally, with some algebraic manipulations, we can get a circle equation (with radius $\lambda$).