March 26, 2013

http://coursework.tylerlogic.com/courses/math501/homework06

Let S be a compact surface. Because S is in ℝ

Let S be the graph of a smooth function f(x,y).

Letting x(x,y) = (x,y,f(x,y)) be a coordinate map for S we have

Using the normal given by

the formula for the normal is

Denoting (1 + f_{x}^{2} + f_{y}^{2})^{-1∕2} by c, we then have

as well as

With these formulas we can use the following formula for the Gauss curvature

to arrive at

For f(x,y) = x

which when using the formula derived in part (a), yields a Gaussian curvature of

which is a nonnegative value that appropaches zero at infinity.

For f(x,y) = x

which when using the formula derived in part (a), yields a Gaussian curvature of

which is a nonpositive value that again appropaches zero at infinity.

For f(x,y) = we have

and

which when using the formula derived in part (a), yields the following Gaussian curvature denoting (x^{2} + y^{2})^{-1∕2} by
c

Multiplying the numerator and denominator by c^{-4} results in a numerator for K of

and since c^{-2} = x^{2} + y^{2}, then K is identically zero.

If f is only a function of x, then x

Using the following formula for the mean curvature

and the formulas for e, f, g, E, F and G from the previous problem to get the following; again we denote (1 + f_{x}^{2} + f_{y}^{2})^{-1}
by c

Let S

and

Because f_{2}(x,y) ≥ f_{1}(x,y) ≥ 0 was assumed, then from Calculus we know that

and

Adding together these two inequalities yields f_{2xx} + f_{2yy} ≥ f_{1xx} + f_{1yy} and multiplying them together yields
f_{2yy}f_{2xx} ≥ f_{1yy}f_{1xx} ≥ 0. At the origin, the former informs us that H_{2} ≥ H_{1} and, upon rotating the xy-plane such that
f_{1xy} = f_{2xy} = 0, the latter gives us that K_{2} ≥ K_{1}.

Let S be a compact, orientable surface with inward-pointing normal. Thus Heine-Borel tells us that the surface is bounded, so we can enclose it within a sphere. Shrink this sphere until it first touches S and call one of possibly multiple such points p. We know that such points exists since S is closed, again due to Heine-Borel since S is compact. We can now rotate S and the sphere such that p is identified with the origin, T

Let S be a minimal surface.

Because S is a minimal surface, then for the principle curvatures k

The results of problem five and the previous part of this problem imply that there are no compact minimal surfaces in ℝ.