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Consider a boundary given by vertices $(0,a)$, $(0,0)$ and $(1,0)$ (an 'L' shaped boundary).

The problem is to find the equation that passes between the endpoints $(0,a)$ $(1,0)$ of minimum length that encloses a specified area $A$.

A trivial case would be $A=a/2$ in which case the solution would be a line.

This is one dimensional Laplace problem with two boundaries (area and lenth) but how do I try to get a series solution for $A < a/2$?

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2 Answers 2

Welcome to Math.SE! This is an interesting question which would be approached in different ways depending on the context in which it arrives (e.g., level of a course). For example, the curves could be assumed to be smooth or merely rectifiable. They could be assumed to be graphs, or not. The attainment of minimal length could be taken for granted, or require a proof. All this makes it hard to pitch the answer at the right level, and I'm most likely going to fail at that. Indeed, I do not see how one would get a solution in the form of a series, because in general the minimizer is not analytic. If this problem comes from a textbook, it would be helpful to have a reference so that an answer can be given within the context of that book.

If the existence and differentiability of a minimizing curve can be granted, then there is a variational argument to show it must have the same curvature at all points where it does not meet the constraint (i.e, does not lie on the vertical or horizontal axis). Indeed, write the curve in vector form $\gamma(t)$, parametrized by arclength, and consider the perturbation $$\Gamma(t)=\gamma(t)+\epsilon \eta(t) R\gamma'(t)$$ where $\epsilon$ is a parameter, $\eta$ is a smooth function with compact support, and $R$ denotes rotation by $\pi/2$ clockwise. Note that $(Ru)\times v =u\cdot v$ and $u\times Rv=-u\cdot v$. The area bounded by $\Gamma$ is expressed as an integral of $\Gamma\times \Gamma'$, in which the linear term in $\epsilon $ simplifies to $$2\epsilon \eta \gamma'\cdot \gamma' - \epsilon (\eta \gamma\cdot \gamma')' $$ Here the first term yields $2\epsilon \int \eta$ and the second integrates to zero. Thus, we should have $\int \eta=0$ to preserve the area up to $O(\epsilon^2)$.

The length of $\Gamma$ is given by the integral of $|\Gamma'(t)| = (\Gamma'(t)\cdot \Gamma'(t))^{1/2}$ in which the linear term simplifies to
$$ \epsilon \eta (\gamma' \cdot R\gamma'') = \epsilon \eta \kappa $$ where $\kappa$ is the curvature of $\gamma$. The conclusion is that $\int \eta \kappa =0$ for any $\eta$ such that $\int \eta =0$. Equivalently, $\kappa$ is constant.

Thus, the minimizing curve will be a circular arc as long as it is free to move; i.e., as long as it does not hit the obstacle (vertical or horizontal axes). And the obstacle does get involved when $A$ is so small that no circular arc through $(0,a)$ and $(1,0)$ can bound area $A$ and stay within the first quadrant. When the obstacle takes effect, the minimizing curve consists of parts of the vertical and/or horizontal axes joined by a circular arc that is tangent to them. An obstacle must always be met tangentially: otherwise the point of contact could slide along the obstacle, decreasing length. This takes a separate perturbation argument, into which I will not go.

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The formulation is as follows: Maximize $$L[y]=\int_{x_1}^{x_2}\sqrt{1+y'\ ^2}\ dx$$ subject to $$A=\int_{x_1}^{x_2}y\ dx$$ We can use Lagrange multipliers for a new formulation such as $$G=\sqrt{1+y'\ ^2}+\lambda \ y$$ and the Euler equation is $$\frac{\partial G}{\partial y}-\frac{d}{dx}\frac{\partial G}{\partial y'}=\lambda-\frac{d}{dx}\bigg(\frac{y'}{\sqrt{1+y'\ ^2}}\bigg)=0$$ where the open formulation is $$\lambda-\frac{1}{\big(1+y'\ ^2\big)^{3/2}}=0$$ and can be solved for y such as $$y[x]=\alpha \ x^2+\beta \ x +\gamma\qquad \alpha=\frac{1}{2}\sqrt{\frac{1}{\lambda^{2/3}}-1}$$ To satisfy the conditions to pass through points $(0,a)$ and $(1,0)$; and $A=\int_{x_1}^{x_2}y\ dx$ $$y[x]=(3a-6A)\ x^2+(6A-4a)\ x +a$$

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The solution $y(x) = \alpha x^2+\beta x+\gamma$ is incorrect. –  Thursday Jun 28 at 22:23

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