November 1997 Presentation Topic (continued)
Suppose that S = { (x,y) : y=f(x)}. Then, as above
1
N(x) = ------------------- (-f'(x), 1),
[1+(f'(x))^2]^(1/2)
1
S = { (x, f(x)) + t ------------------- (-f'(x), 1) }.
t [1+(f'(x))^2]^(1/2)
Therefore the curve S_t is parametrized by F(x)=(F_1(x),F_2(x)), where
1
F1(x) = x - t ------------------- f'(x),
[1+(f'(x))^2]^(1/2)
1
F2(x) = f(x) + t -------------------.
[1+(f'(x))^2]^(1/2)
The tangent vector to S_t at F(x) is given by T_t(x)=(F_1'(x), F_2'(x)), with
2 (-3/2)
F1'(x) = 1 - t f''(x) [1+f'(x) ]
2 (-3/2)
F2'(x) = f'(x)(1 - t f''(x)[1+f'(x) ] )
Hence
T_t(x)=[1-t f"(x)(1+ (f'(x))^2)^(-3/2)] (1, f'(x)).Thus we conclude that T_t(x) is parallel to the tangent vector to S at (x,f(x)) and therefore T_t(x) is tangent to the circle centered at (x,f(x)) with radius t at the point (x,f(x))+t N(x). See figure 7.
Figure 7: PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY
Question 2
Suppose S = { y = ax+b }. The unit normal vector is given by
1
N(x) = ------------- (-a,1).
(1+a^2)^(1/2)
Hence
1
S (x) = { (x,ax+b) + t -------------(-a,1) }.
t (1+a^2)^(1/2)
Let
at t
X = x - -------------, Y = ax + b + -------------.
(1+a^2)^(1/2) (1+a^2)^(1/2)
Then
Y-aX = b+ t(1+a^2)^(1/2).Question 3
See figure 8.
Figure 8: PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY
Question 4
In this case S_t is parametrized by F(x)=(F_1(x),F_2(x)), where
F_1(x)= x-2tx(1+4x^2)^(-1/2) and F_2(x)= x^2+t(1+4x^2)^(-1/2).Thus
F_1'(x) = 1-2t(1+4x^2)^-3/2,Hence, F_1'(x)=F_2'(x)=0 if and only if
F_2'(x) = 2x(1-2t(1+4x^2)^-3/2).
t = (1/2) (1+4x^2)^(3/2).Only when t > 1/2 does this have solutions, namely x = +/- (1/2) sqrt((2t)^(2/3) - 1).
In particular, this shows that S_t is smooth for t<0, which corresponds to the evolution of the front along -N(x). For t>0, S_t is smooth as long as t < 1/2. The first value of t for which S_t has a singularity is t = 1/2.
Question 5
For t=1/2(1+4x^2)^3/2, the points on the curve S_t corresponding to these values of x are
F_1(x) = -4x^3 = X, F_2(x) = 1/2 + 3x^2 = Y.These points satisfy the equation
(1/3(Y-1/2))^3 = (-1/4X)^2,which is curve with a cusp singularity. See figure 9.
Figure 9: PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY
The curves S_t are the pictures of the evolution of the front S with time. We can think of this as the intersections of a surface in R^3 with the planes defined by the time t. The surface defined in this way has a singularity called a swallowtail, because of its similarity with the tail of the bird. See figure 10.
Figure 10: PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY
It is difficult result, which is outside the scope of this lecture, to show that, in two dimensions, the cusp and the swallowtail are the only stable caustics. Stable in the sense that by slightly moving the initial curve S, any other singularity decomposes into these two. In dimensions greater than two there are many other possible stable caustics. Formation of caustics in higher dimensions is, as we can imagine, a very complicated subject and many people have studied it. There are long lists of possible stable caustics. I will make no attempt of discussing this here. For the interested reader I would suggest the references [1] and [2] mentioned below.
I will end this lecture with the remark that a surface with a swallowtail singularity also appears as an algebraic object. Consider the polynomial of fourth degree in l
p(l) = l^4 + zl^2 + yl+ x,with coefficients depending on (x,y,z) in R^3. Consider the set
S = { (x,y,z): p(l) has a double real root }.I leave as an exercise for the reader to obtain the polynomial equation of this surface: S = S \ L where
S = {(x,y,z): 16xz^4 - 4y^2z^3 - 128x^2z^2 + 144xzy^2 + 256x^3 - 27y^4 = 0}The set {(x,y,z): y=0, x=(z^2)/4, z>0} corresponds to the values of (x,y,z) for which p(l) has a double complex root. It also has a swallowtail singularity and looks very much like the one in figure 10. For a discussion of this see [2].
L = {(x,y,z): y=0, x = (z^2)/4, z>0}.