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Lecture 15 - Minimal Surfaces

Recall that (1+fy2)fxx2fxfyfxy+(1+fx2)fyy=0 is the minimal surface equation. Assume that f(x)=g(x)+h(y) then the minima surface equation transforms to:

g(x)[1+h(y)]2+h(y)(1+g(x)2)=0

Thus we obtain:

1+g(x)2g(y)=1+h(x)2h(y)1+g(x)2g(x)=κ1+g(x)2=κg(x)Let ϕ(x)=g(x)dx=κdϕ1+ϕ2x=κarctanϕ+c

For convenience c=0,κ=1

g(x)=ln(cos x)h(y)=ln(cos y)f(x,y)=ln(cos(x)cos(y))

This is Scherks double periodic surface. For reference (application) check out Randall Kamien (condensed matter physicist).

The Weierstrass Enneper representation

Proposition

Let ϕ:VR3 be an isothermal parameterization if E=G=λ2(u,v) and F=0 then ϕ satisfies:

ϕuu+ϕvv=2λ2Hn
Proof

Since ϕ is isothermal with E=G=λ2 and F=0 then:

<ϕu,ϕu>=<ϕv,ϕv>

Suppose we differentiate wrt u then:

(1)<ϕuu,ϕu>=<ϕuv,ϕv>

since F=0<ϕu,ϕv>=0 then differentiating wrt v gives us:

(2)<ϕuv,ϕv>+<ϕu,ϕvv>=0

(1)+(2) gives us that:

<ϕuu,ϕu>=<ϕvv,ϕu>

that is:

<ϕuu+ϕvv,ϕu>=0

Similarly we can show that <ϕuu+ϕvv,ϕv>=0 thus ϕuu+ϕvv must be parallel to the normal.
We know that H=L+N2λ2 thus:

2λ2H=L+N=n(ϕuu+ϕvv)

That is:

ϕuu+ϕvv=2λ2Hn
Examples of Isothermal surfaces

  • Catenoid: ϕ(u,v)=(acosh(v)cos(u),acosh(v)sin(u),av)
  • Helicoid: ϕ(u,v)=(asinh(v)cos(u),asinh(v)sin(u),au)
  • Ruled Surface

Definition

Let ϕ be a function of one complex variable having the form

ϕ(z)=x(u,v)+iy(u,v)

ϕ is called harmonic if:

ϕuu+ϕvv=0.
Definition

Suppose ϕ:ωCC then ϕ is said to be meromorphic if there exists isolated points zi such that:

ϕ~:Ω/{zi}holomorphicCi.

In addition the points {zi} are called poles of the function. The order of the pole zi is the number of terms that vanish in that vanish Laurent series around zi.

Since we are interested in ϕ(z)=x(u,v)+iy(u,v) then we have the Writenger derivative:

ϕz=12(xu+yv)+i2(yu+xv)ϕz¯=12(xuyv)+i2(yu+xv)

Using these we obtain:

4(z¯(ϕz))=ϕuu+ϕvv

Thus if our function is harmonic then it is holomorphic.

Proposition

Let ϕ(u,v)=(x(u,v),y(u,v),z(u,v))
Suppose ϕk=xkz then ϕ is minimal iff

ϕ12+ϕ22+ϕ32=0

and that ϕk are holomorphic functions.

Proof

Note that ϕ1=xuixv, ϕ2=yuiyv, ϕ3=zuizv then:

ϕ12+ϕ22+ϕ32=(xuixv)(xuixv)+(yuiyv)(yuiyv)+(zuizv)(zuizv)=(xuxu+yuyu+zuzu)(xvxv+yvyv+zvzv)2i(xuxv+yuyv+zuzv)=EG2iF=0 (Since ϕ satisfies the isothermal surface condition)
Lemma

Let D be a domain in the complex plane, g(z) an arbitrary meromorphic function in D having the property that at each point where g(z) has a pole of order k, f(z) has a zero zero order of atleast 2k then the function:

ϕ1=12f(1g2), ϕ2=i2f(1+g2), ϕ3=fg

will be holomorphic in D satisfying ϕ12+ϕ22+ϕ32=0

Proposition

Every regular minimal surface has locally an isothermal paramterization of the form equal to:

(Rezf(1+g2) dw,Rezif(1g2) dw,Rez2ifg dw)

f vanishes at the pole of g and having a zero of order of at least 2k wherever g has a pole of order k, and:

|ϕ|2=4|f|(1+g2)0