Find the surface of least area spanned by a given contour

Question

I'm reading a book on Calculus of Variations, and am confused by the following problem:

Find the surface of least area spanned by a given contour.

The answer is to write a functional of the form:

$$J [z] = \iint_R \sqrt {1+z_x^2+z_y^2} dx dy $$

Now, use Euler's equation:

$$F_z-\frac {\partial}{\partial x} F_{z_x}-\frac {\partial}{\partial y} F_{z_y} = 0$$

Where $F = \sqrt {1+z_x^2+z_y^2} $

The calculation does not confuse me so much as what we are actually finding.

I can see that the argument of the integral is highly analogous to arclength. When finding the curve of least length that connects two points, we apply the single variable Euler equation to: $$\int_L \sqrt{1+y'^2}dx $$

However, I'm not entirely sure where the formula for $J [z] $ comes from, nor what it means for a surface to be spanned by a contour (in fact, I'm not all that certain of what a contour is)

I am enjoying the book, but I have definitely been ambitious to try and tackle this subject, as it seems to take a number of subjects I am only vaguely familiar with for granted.

Answer 1

The notation of your question suggests (i.e., is consistent with) the following interpretation:

• $R$ denotes a bounded plane region whose boundary consists of one or more simple closed curves.

• A contour (blue) is a space curve obtained as the graph of a smooth, real-valued function $u$ over the boundary of $R$.

• A surface spanning a contour (shaded) is the graph of a smooth, real-valued function $z$ in $R$ whose values agree with $u$ on the boundary of $R$.

The functional $J$ is the surface area for the graph of $f$; by seeking critical points of $J$, you're finding surfaces of a particular type (namely, graphs) whose area is critical among all surfaces spanning the given contour (wireframe).

Answer 2

This the classical Plateau's problem of minimal surfaces where mean curvature vanishes. Solutions are due to

Radó, René and Lagrange

It is visualized physically by soap films spanning the given boundary. Finding coordinates of a skew quadrilateral 3D closed contour or loop (where x,y,z are functions of a single parameter that could be arc,) involves hyperelliptic integrals.

Answer 3

Setting $$ \begin{align} 0 &=\delta\iint_R\sqrt{1+z_x^2+z_y^2}\,\mathrm{d}x\,\mathrm{d}y\\ &=\iint_R\frac{z_x\delta z_x+z_y \delta z_y}{\sqrt{1+z_x^2+z_y^2}}\,\mathrm{d}x\,\mathrm{d}y\\ &=-\iint_R\delta z\frac{\partial}{\partial x}\frac{z_x}{\sqrt{1+z_x^2+z_y^2}}\,\mathrm{d}x\,\mathrm{d}y\\ &-\iint_R\delta z\frac{\partial}{\partial y}\frac{z_y}{\sqrt{1+z_x^2+z_y^2}}\,\mathrm{d}x\,\mathrm{d}y\\ &=-\iint_R\delta z\frac{z_{xx}\left(1+z_x^2+z_y^2\right)}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\ &+\iint_R\delta z\frac{z_x^2z_{xx}+z_xz_yz_{xy}}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\ &-\iint_R\delta z\frac{z_{yy}\left(1+z_x^2+z_y^2\right)}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\ &+\iint_R\delta z\frac{z_y^2z_{yy}+z_xz_yz_{xy}}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\ &=-\iint_R\delta z\frac{z_{xx}\left(1+z_y^2\right)+z_{yy}\left(1+z_x^2\right)-2z_xz_yz_{xy}}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y \end{align} $$ for all $\delta z$, requires $$ \bbox[5px,border:2px solid #C0A000]{z_{xx}\left(1+z_y^2\right)+z_{yy}\left(1+z_x^2\right)-2z_xz_yz_{xy}=0} $$ which is true when a surface has zero Mean Curvature.