S h a p e o f s u r f a c e z(x,y) i n n e i g h b o u r h o o d o f a p o i n t. Recommended prerequisites. -------------------------- Matrix, partial derivative, intergration in |R. Level: college entry level. The Problem. ------------ One of the ways to describe surfaces in 3-dimentional space, is to use smooth functions of form z(x,y). We will set aside forms of poetrical descriptions and will focus on how such functions look in neighbourhood of a single point p_{0}=(x_{0},y_{0}). Intuitively, if we take enough good microsope, the picture will be a flat plane. Indeed, human probably will consider the Earth flat untill is able to observe neighbourhood large enough. When deviations from "flattness" becomes noticeble, what shape can they take? We'll try to sketch the problem intuitively and then make a proof. Intuitive Sketch. ----------------- 1. If Z has second patial derivatives which are continuous in neighbourhood of p_{0}, then Z = Z_{0}+ L_{i}X_{i}+ F_{i}_{j}X_{i}X_{j}+ o(X)X^{2}(1) Summation via equal indices is assumed: for example, L_{i}X_{i}= L_{1}X_{1}+L_{2}X_{2}. We denote X_{1}=x, X_{2}=y. o(X) -> 0 when X -> 0. 2. Z = Z_{0}+ L_{i}X_{i}is a plane. We see it when neglect other members in (1). 3. Function S(X,X)=F_{i}_{j}X_{i}X_{j}gives us first hills. How do they look? Matrix || F_{i}_{j}|| has form: || F_{i}_{j}|| = || A B || || || (2) || B C || because mixed derivatives are equal. and value S(X,X) is a scalar product: S(Y,X) = (Y, F(X)) where Y=X. where we denoted the vector F_{i}_{j}X_{j}as F(X). We must focus on anatomy of function F. Further, we will omit parenthesises: FX = F(X). Note, if by accident B=0, then F simpy multiplies basis vectors by numbers A and C. In this case, S = AX_{1}^{2}+ CX_{2}^{2}. (3) and clearly, S is a contribution of two parabolas. If both parabolas have their horns up, then S - is elliptic paraboloid and has minimum in point 0. The other cases are displayed: In other words, the picture depends on product AC and and sign of components. But even if we are not so lucky in our choice of the basis, do proper vectors exist in general case? (Vector is called proper vector of function F if "F multiplies this vector by a number".) Let's look at the structure of F closer. Even matrix (2) is symmetirc, it still hides two matrices with even more symmetries. Although in general A is not equal C, we can overcome this obstacle by letting: D=(A-C)/2, M = (A+C)/2 which gives || F_{i}_{j}|| = || H_{h}|| + || H_{t}|| where || H_{h}|| = || M 0 || and || H_{t}|| = || D B || || 0 M || || B -D || So far, we see that function F is a contribution of two functions F = H_{h}+ H_{t}H_{h}is simply a homogeneous dilation of all space by M, and "non-homogeneous" H_{t}is of unknown nature yet. Matrix of || H_{t}|| resembles matrix of rotation: || T_{f}|| = || Cos(f) -Sin(f) || || || || Sin(f) Cos(f) || Indeed, we can choose G, and g: B = GCos(g) D = GSin(g) that H_{t}= GRT_{g}where || R || = || 0 1 || - is a reflection in respect to of axis X_{1}=X_{2}. || 1 0 || In other words, non-homogeneous part of any symmetric matrix of dimension 2 is a "rotation, swap, and dilation". It is clear now, that we can take basis vector e_{1}' that RT_{g}e_{1}'=e_{1}': e_{1}' = T_{f}e_{1}where f=PI/4-g/2. Any other unit vector, e_{2}', which is perpendicular to e_{1}' will only change a direction when transformation H_{t}is applied to it. This is because that reflection R changes direction of the angle between two vectors, but keeps this angle's magnitude. H_{t}e_{2}' = -Ge_{2}' Summarizing the behavior of these two proper vectors: Fe_{1}' = (M+G)e_{1}' Fe_{2}' = (M-G)e_{2}' For an arbitrary vector X = X_{1}'e_{1}'+X_{2}'e_{2}', F(X)=(M+G)X_{1}'e_{1}' + (M-G)X_{2}'e_{2}'. S(X,X) = A'X_{1}'^{2}+ C'X_{2}'^{2}(3'). where A'= M+G, C'=M-G. Our parameter for shape of S can be written now as A'C' = M^{2}- G^{2}= (A+C)(A+C)/4 - (A-C)(A-C)/4 - BB = AC-BB. From 3', it is clear that: (4) AC-BB > 0 and C < 0, then S is elliptic paraboloid and p is minimum of S. AC-BB > 0 and C > 0, then S is elliptic paraboloid and p is maximum of S. AC-BB < 0, then S is elliptic saddle. AC-BB = 0, then at least along some line X_{i}' = 0 S = const. Proof. ------ We sketched all the parts of proof except (1) which is well known. Here is a brief ... Proof of (1). Z(X) = Z(b) = Z(0) + Z(a)-Z(0) + Z(b)-Z(a) X=(X1,X2) Axis Z is perpendicular to this picture: 0 ------------> axis X2 | |q | | p |----------- b |a | V axis X1 From mean value theorem: Za-Zo = Z1(q)X1 Z1= dZ/dX1. Zb-Za = Z2(p)X2 q = o'(X) p = o''(X) p,q -> 0 when X-> 0 and Fi(X) -> Fi(0) because of continuity of Fi. Here and further, we will omit "(0)" at function letters for brevity: for example Z2 will denote Z2(0). Threfore, Z(X) = Z + ZiXi + oi(X)Xi. Applying this result to X1-derivative: Z1(X) = Z1 + Z1jXj + o1j(X)Xj and integrating through path 0-a-b, obtain: Z(X) = Z + Z1X1 + (Z11X1^2)/2 + Z2X2 + (Z22X2^2)/2 + Z21X2X1 + (integral o to a)(o11X) + (integral a to b)(o21X1 + o22X2) (I) Sum of above integrals is oI(X)XX. (1.a) Analogiously, integrating through 0-b-a Z(X) = Z + ZiXi + (ZiiXi^2)/2 + Z12X1X2 + oII(X)XX (II) Comparing (I)=(II) when X1=X2<>0: Z12 + oI = Z21 + oII. Taking limit X1->0 we can prove that Z12 = Z21. Adding two equations together: Z(X) = ((I)+(II))/2 = Z + ZiXi + (1/2)ZijXij + (oI+oII)XX which completes proof of (1) if to take Fij = (1/2)Zij. Remaining proofs for (3)-(4) will follow if time permits. They are already sketched with enough details. Terminology: ------------ Elliptical paraboloid: http://en.wikipedia.org/wiki/Paraboloid Copyright (C) 2008 Konstantin Kirillov.