Some Problems from the Midterm Exam

Problem 3

<<  /home/home5/kimura/math/563/curvature.m

Consider the Monge patch

x[u_, v_]  := {u, v,  c Sqrt[1 - u^2/a^2  -  v^2/b^2]}

The domain of x is D  =  
      {(u, v)  |  u^2/a^2  +  v^2/b^2  <  1 } .

EE[x, u, v]

Notice the denominator never vanishes on D,  
    EE is always defined on D .  Furthermore,  
    EE can never vanish on D so the following quotient is defined 
      on D :

Simplify[ll[x, u, v]/EE[x, u, v]]

lE  =  PowerExpand[%]

FF[x, u, v]

FF vanishes when either u  =  
      0 or if v  =  
        0.  We shall have to treat those as special 
          cases .  Therefore,  
    the following is well - defined on D except if u  =  
      0 or v  =  0 :

mF  =  PowerExpand[Simplify[mm[x, u, v]/FF[x, u, v]]]

GG[x, u, v]

This never vanishes on D either .  Therefore,  
    the following quotient is well - defined on D :

nG  =  PowerExpand[Simplify[nn[x, u, v]/GG[x, u, v]]]

Let us assume for now that neither u  =  
      0 or v  =  0.  At an umbilic point,  
    is equivalent to a solution of l E  = mF  =  nG .

Simplify[lE  -  mF]

T1  =  Numerator[%]

Solving T1  ==  0,  we obtain that either a  =  c or

Solve[T1  ==  0, u]

PossibleAnswer  =  %

But notice that these points satisfy u^2/a^2  +  v^2/b^2  = 
       1 and are,  hence,  not in D .  The only possible solution, 
     then,  is a  =  c .  Similarly,

Simplify[ mF  -  nG]

T2  =  Numerator[%]

A solution to T2  ==  0 is either b  =  c or

Solve[T2  ==  0, u]

These points are also not in the domain D since they 
          satisfy u^2/a^2  +  v^2/b^2  =  1.

Let' s now examine the exceptional points .  Assume that u  = 
       0 then FF  =  0.  We can still solve

T3  =  Numerator[Simplify[lE  -  nG]]  /.  {u  ->  0}

Simplify[Together[T3]]

The first case is when b  =  
      c but b does not equal a .  If this is true then

%66 /. {b  ->  c}

Solve[%67  ==  0, v]

which are outside of D .  Of course,  
    if a  =  b  = c then

%67  /.  {a  ->  c}

so any value of v is a solution as expected 
      (since the ellipsoid is a sphere in that case) .

Solve[T3  ==  0,  v]

v  =  b,  
    -b are both invalid since they lie outside of D .  The 
      solutionsv  =  b Sqrt[(b^2 - a^2)/(b^2  -  c^2)] and v  =  - b Sqrt[(b^2  -  a^2)/(b^2 - c^2)]\(	exist\ only\ if\ \ \((b^2\  - \ a^2)\)/\((b^2\  - \ c^2)\)\  < \ 
      1. \ There\ are\ two\ 
        \(possibilities : 
			
          b\  >= \ a\ and\ b\  > \ c\ in\ which\ case\ c\  < \ a\), \ 
    i . e . \ b\  >= \ a\  > \ c . \ The\ second\ possibility\ is\ b\  <= \ 
      a\ and\ b\  < \ c\ then\ c\  > \ a\  >= \ \(b . \)\)

Ans  =  {%63[[3]], %64[[4]]}

The solutions  of u when v  =  
      0 are similar but where the roles of a and b are 
        interchanged .

\(\(\(Therefore, \ on\ x \((D)\), \ 
                we\ have\ the\ following\ \(possibilities : 
		
1\))\)\ a
              \  = \ \(b\  = \ 
              c : \ All\ points\ in\ x \((D)\)\ are\ umbilic . 
2\))\)\ If\ u
        \  = \ 0\ and\ either\ b\  >= \ a\  > \ c\ \ OR\ c\  > \ a\  >= \ 
        b\ then\)    v  =  b Sqrt[(b^2  -  a^2)/(b^2  -  c^2)] or   
    v  = - b Sqrt[(b^2  -  a^2)/(b^2  -  c^2)]

This corresponds to the points (x, y, z) in x (D)

x[0, v]  /.  Ans

PowerExpand[Simplify[%]]

3) If v  =  
      0 and either a  >  b  >  c OR c  >  b  >  a then    u =  b Sqrt[(b^2  -  a^2)/(b^2  -  c^2)] or   u = - b Sqrt[(b^2  -  a^2)/(b^2  -  c^2)]

\(\(The\ corresponding\ points\ \((x, y, z)\)\ in\ x \((D)\)\ are
\)\)

Ans

%  /.  {v  ->  u,  a  -> b,  b  -> a}

x[u, 0]  /. %

PowerExpand[Simplify[%]]

These are the solutions for case 3.

What about the rest of the ellipse? An ellipse has 6 Monge 
      charts associated to it 
      (two along the x axis [1 above the xy plane and 1 below], 
           two along the y axis,  and two along the z axis) . To
         obtain the umbilic points below the xy plane,  
    replace the 3  rd component of each vector by its negative . To 
      obtain the other umbilic points,  
    corresponds to permuting the values  a,  b,  c .

Problem 4

Part (a) :

x[u_, v_]  :=  {a Sinh[v] Cos[u],  a Sinh[v] Sin[u],  a u}

y[u_, v_]  := 
       {-a Cosh[v] Sin[u],  a Cosh[v] Cos[u],  a v}

EE[x, u, v]

FF[x, u, v]

GG[x, u, v]

Therefore,  
    x is an isothermal parametrization of the Helicoid .  Similarly,

EE[y, u, v]

FF[y, u, v]

GG[y, u, v]

is an isothermal parametrization of the Catenoid .

Now,

D[x, u]  -  D[y, v]

D[x, v]  +  D[y, u]

therefore,  x and y are conjugate minimal surfaces .