1. Review. g ij = f xi f xj, f x1 f x2, N =

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1 Math 16B: Lecture Notes on Curves and Surfaces in R 3, Part II by Chuu-Lian Terng, Spring quarter 005 Department of Mathematics, University of California at Irvine 1. Review A parametrized surface is a smooth map f : O R 3 such that f x1 (p), f x (p) are linearly independent for all p O, where O is an open subset of R 3. The tangent plane T f p at f(p) is spanned by f x1 (p) and f x (p). The first fundamental form is where the dot product. The map N : O R 3 defined by I g 11 dx 1 + g 1 dx 1 dx + g dx, g ij f xi f xj, N f x 1 f x f x1 f x, is smooth, has unit length, and is perpendicular to the tangent plane of f, and we call N the unit normal vector field of f. The shape operator A p0 of f at p 0 O is the linear map from T f p0 to T f p0 defined by So If v af x1 (p 0 ) + bf x (p 0 ), then We proved in 16A that A p0 (f xi (p 0 )) N xi (p 0 ), i 1,. A p0 (v 1 ) an x1 (p 0 ) bn x (p 0 ). Proposition The shape operator A p0 : T f p0 T f p0 is a self-adjoint operator, i.e., A p0 (v 1 ) v v 1 A p0 (v ) for all v 1, v T f p0. The eigenvalues and unit eigenvectors of the shape operator are called the principal curvatures and principal directions of the parametrized surface f. We use k 1, k to denote the principal curvature functions. The Gaussian curvature and mean curvature are defined by K k 1 k, H k 1 + k. A point f(p 0 ) is umbilic if the principal curvatures k 1 (p 0 ) k (p 0 ), i.e., the shape operator A p0 is equal to a scalar times the identity map of the tangent plane T f p0. 1

2 The second fundamental form II p is a symmetric bilinear form on T f p associated to the shape operator A p, i.e., for v 1, v T f p. We denote II by where II p (v 1, v ) A p (v 1 ) v II l 11 dx 1 + l 1 dx 1 dx + l dx, l ij A(f xi ) f xj N xi f xj N f xi x j. This means that if v i a i f x1 + b i f x for i 1,, then II(v 1, v ) l 11 a 1 a + l 1 (a 1 b + a b 1 ) + l b 1 b. The Gaussian curvature and mean curvature written in terms of I, II are K det(l ij) det(g ij ), H g ij l ij i,j1 where (g ij ) (g ij ) 1 1 det(g ij ) 1 det(g ij ) (g l 11 g 1 l 1 + g 11 l ), ( ) g g 1. g 1 g 11 A surface f : O R 3 is said to be parametrized by lines of curvature coordinates if g 1 l 1 0, in other words, f x1, f x are perpendicular and are eigenvectors of the shape operator. We know from 16A that Theorem If p 0 is not an umbilic point of a surface f : O R 3, then we can change coordinate locally so that the surface is parametrized by line of curvature coordinates, i.e., there exists an open subset O 0 of O containing p 0, an open subset O 1 of R, a diffeomorphism φ : O 1 O 0 such that h x1, h x are eigenvectors of the shape operator, where h f φ. Suppose f : O R 3 is parametrized by line of curvature coordinates. We write g 11 A 1, g A. Then we have I A 1dx 1 + A dx, II l 11 dx 1 + l dx. The principal curvature k 1, k, Gaussian curvature K, and the mean curvature H written in line of curvature coordinates are given as follows: k 1 l 11 A, k l 1 A, K l 11l A 1 A H l 11 A + l 1 A k 1 k, k 1 + k.

3 We have proved that A 1, A, l 11, l must satisfy the Gauss-Codazzi equations: ( ( ) (A1 ) x (A ) A + x1 )x A 1 l 11 l x A 1 A, 1 ( l 11 A 1 ) x (A 1) x l (1.0.1) A A ( l A ) x1 (A ) x1 l 11 A 1 A 1. If we set r 1 l 11, r l, A 1 A then the Gauss-Codazzi equations (1.0.1) can be written as ( ( (A1 ) x (A ) A + x1 )x A 1 r 1 r, )x 1 (r 1 ) x (A 1) x A r, (1.0.) (r ) x1 (A ) x1 A 1 r 1. The first equation of the above system is called the Gauss equation, and the second and third equations are called the Codazzi equations. The Fundamental Theorem of Surfaces in R 3 states that: Suppose A 1, A, l 11, l are smooth functions from O to R satisfying the Gauss-Codazzi equations (1.0.1). Then given p 0 O, q 0 R 3, and v 1, v, v 3 an orthonormal basis of R 3, there exists an open subset O 0 containing p 0 and a unique smooth immersion f : O 0 R 3 such that the first and second fundamental form of f are I A 1dx 1 + A dx, II l 11 dx 1 + l x dx, f(p 0 ) q 0, v 1 fx 1 (p 0) A 1, and v fx (p 0) A. Moreover, if two surfaces f, h : O R 3 have the same I, II, then they are congruent, i.e., there is a rigid motion φ of R 3 such that h φ f. 3. Surfaces in R 3 with K 1 and SGE In geometry we are often interested in understanding geometric objects whose invariants are of simplest kind. For example, in plane geometry we have many theorems for equilateral, isoceles, and right triangles. The Guassian and mean curvatures are the simplest kind of invariants for surfaces in R 3, so it is natural to study the geometry of surfaces in R 3 whose K or H are constants. In this section, we will show the existence of Tchbyshef line of curvature coordinates for surfaces in R 3 with K 1, and we also show that there is a one-to-one correspondence between solutions of the sine-gordon equation q x1 x 1 q x x sin q cos q and surfaces in R 3 with K 1 up to rigid motions. SGE

4 4 Theorem.0.3. If a surface in R 3 has K 1, then locally there exists line of curvature parametrization such that the two fundamental forms are I cos q dx 1 + sin q dx, II sin q cos q (dx 1 dx ) for some function q. Gordon equation Moreover, the Gauss-Codazzi equation is the sine- q x1 x 1 q x x sin q cos q. SGE (.0.3) Proof. Since K k 1 k 1, k 1 is never equal to k. So there exists line of curvature parametrization locally. Thus we may assume that I A 1dx 1 + A dx, II l 11 dx 1 + l dx. Since K k 1 k 1, so there exists a function q such that (q tan 1 k 1 ). But k 1 l 11 A 1 k 1 tan q,, k l, so A k 1 cot q r 1 l 11 A 1 k 1 A 1 tan q A 1, r l A k A cot q A. We will use the first Codazzi equation to prove that A 1 l 11 is a function of x ( 1 alone. It suffices to prove that A1 l But by the second equation of )x (1.0.), we have (r 1 ) x (A 1 tan q) x1 (A 1 ) x tan q + A 1 sec q q x (A 1) x r (A 1) x A cot q (A 1 ) x cot q. A A Thus (A 1 ) x (tan q + cot q) A 1 sec q q x, 1 But tan q + cot q sin q cos q, so we get (A 1 ) x A 1 sin q cos q q x. This implies that (ln A 1 ln cos q) x 0, hence ( ) ln A 1 cos q 0. It follows x from calculus that ln A 1 cos q is a function of x 1 alone, say c 1 (x 1 ) for some one variable function c 1. So we have proved that A 1 cos q ec 1(x 1 ). (.0.4) Similarly, we use the third equation of (1.0.) to conclude that A sin q ec (x ) for some one variable function c. (.0.5)

5 5 Define x i as a function of x i alone for i 1, such that d x 1 dx 1 e c 1(x 1 ), d x dx e c (x ). Then (x 1, x ) ( x 1, x ) is a local diffeomorphism, and there is a local inverse, i.e., we can write x 1, x as functions of x 1, x. Next we compute the two fundamental forms in ( x 1, x ) coordinate. Since f f dx i f xi e c i(x i ), x i x i d x i g ij f xi f xj e c i(x i ) c j (x j ) g ij. Use (.0.4) and (.0.5) in the following computations: ( ) g 11 g 11 e c 1(x 1 ) A 1e c 1(x 1 ) A1 cos q, g 1 0, g g e c (x ) A e c (x ) This shows that I cos qd x 1 + sin qd x. To compute II, we note that e c 1(x 1 ) ( A e c (x ) l 11 k 1 à 1 tan q cos q sin q cos q, l 1 0 l k à cot q sin q sin q cos q. ) sin q. So II sin q cos q ( dx 1 d x and (x 1, x ) are line of curvature coordinate system. To check the G-C equation, we compute r 1 l 11 sin q cos q à 1 cos q r l à sin q cos q sin q sin q, cos q, (A 1 ) x sin q q x A sin q q x, (A ) x1 cos q q x 1 A 1 cos q q x1. The Codazzi equations (the second and the third equations of (1.0.)) do not give any extra condition on q. The Gauss equation (the first equation of (1.0.) gives SGE. As a consequence of the Fundamental Theorem of Surfaces, we have

6 6 Theorem.0.4. Let O be an open disk in R, and q : O R a solution of SGE such that q(x 1, x ) (0, π ). Then given x 0 O, p 0 R 3, and an orthonormal basis {v 1, v, v 3 } of R 3, there exists a unique surface f : O R 3 such that its fundamental forms are I cos q dx 1 + sin q dx, and f(x 0 ) p 0, f x1 (x 0 ) cos q(x 0 )v 1, f x (x 0 ) sin q(x 0 )v. In other words, we have proved that there is a one-to-one correspondence between solutions of SGE whose image lies in the interval (0, π ) and a surface in R 3 with K 1 modulo rigid motions. Next we review the proof of the Fundamental Theorem of Surfaces here, and will see that when we do not assume any condition on the image of a solution q we still can get a smooth map f, but now f fails to be an immersion at points p where sin q(p) cos q(p) 0, i.e., when q nπ for some integer n. To get the immersion f, we need to solve the following first order equation (see 16A Lecture Notes): (A A ) x A r 1 (f, e 1, e, e 3 ) x1 (e 1, e, e 3 ) 0 (A 1) x A r (A ) x1 (.0.6) A 1 0 (A (f, e 1, e, e 3 ) x (e 1, e, e 3 ) A ) x1 A 1 0 r. 0 0 r 0 In general, if A i vanishes at x 0, then the right hand side is not continuous at x 0, so the above equation can not be solved in a neighborhood of x 0. But in the case for surfaces with K 1, the trouble terms (A 1) x A q x and (A ) x1 A 1 q x1 are both well-defined smooth maps even when A 1 or A vanish at some points. So (.0.6) becomes cos q 0 q x sin q (f, e 1, e, e 3 ) x1 (e 1, e, e 3 ) 0 q x sin q 0 0 (.0.7) 0 0 q x1 0 (f, e 1, e, e 3 ) x (e 1, e, e 3 ) sin q q x1 0 cos q. 0 0 cos q 0 Since the right hand side is smooth, by Frobenius Theorem this system is solvable. So we obtain a unique solution (f, e 1, e, e 3 ) such that the initial data at x 0 is (p 0, v 1, v, v 3 ). Since f x1 cos q e 1 and f x sin q e, the map f is an immersion if and only if sin q cos q 0, i.e., when q(p) kπ for some integer k. At points where q(p) kπ, the map f is still smooth at those

7 points, but the rank of the Jacobian matrix of f at those points is 1. We call such points a cusp singularity of f. Remark. If we modify the MatLab project for the Fundamental Theorem of Surfaces in line of curvature coordinates by using (.0.7), then the program should generate surfaces with K 1 with cusp singularities when we input a solution of SGE. 7 y b 3. Tchbyshef asymptotic coordinates for K 1 surfaces The asymptotic lines for the hyperbola x y 1 are given by x a b a 0. A quadratic curve ax + bxy + cy 1 represents a hyperbola if ( ) a b b 4ac > 0, i.e., if det < 0. We call the directions of the two lines b c given by ax + bxy + cy 0 the asymptotic directions for the quadratic form ax + bxy + cy. The second fundamental form of a surface f at point p is a quadratic form on the tangent plane T f p. Since K(p) det(l ij) det(g ij ) and det(g ij ) > 0, det(l ij ) < 0 if and only if K(p) < 0. In particular, if K(p 0 ) < 0, then there exists two linearly independent aysmptotic directions for the quadratic form II po. Definition A non-zero tangent vector v T f p0 is asymptotic if II p0 (v, v) 0. A parametrized surface f : O R 3 is parametrized by asymptotic coordinates if l 11 l 0, i.e., both f x1 and f x are asymptotic vectors. Suppose f : O R 3 is the Tchbyshef line of curvature parametrization for a surface with K 1 with I cos q dx 1 + sin q dx, II sin q cos q (dx 1 dx ). A tangent vector v a 1 f x1 + a f x is asymptotic if II(v, v) sin q cos q (a 1 a ) 0, or equivalently, a 1 a. So v 1 f x1 +f x and v f x1 f x are asymptotic vectors. If we make the following change of coordinates { s x 1+x, t x 1 x, then by the Chain rule f s f x1 + f x and f t f x1 f x. So f(s, t) f(x 1 (s, t), x (s, t)) f(s + t, s t) is an asymptotic parametrization. We compute I, II next. Since f x1 f x1 cos q, f x1 f x 0, f x f x sin q, g 11 f s f s (f x1 + f x ) (f x1 + f x ) cos q + sin q 1.

8 8 Similar computation gives g 1 f s f t cos q, g 1. As a consequence, we see that the angle between the asymptotic vectors is q. We have proved that f s, f t are asymptotic vectors, hence l 11 l 0. It remains to compute l 1. Since f x1 x 1 N f x x N sin q cos q, f x1 x N 0, we have l 1 f st N (f x1 +f x ) x1 (f x1 +f x ) x (f x1 x 1 f x x ) N sin q cos q. So the fundamental form in (s, t) coordinate becomes I ds + cos(q) ds dt + dt, II 4 sin q cos q ds dt sin(q)ds dt. We call (s, t) the Tchbyshef asymptotic coordinate system (parametrization). Note that q satisfies the SGE. We want to write the SGE in (s, t) coordinates. By Chain rule, we have so q x1 x 1 q x x q st. Thus u x1 1 (u s + u t ), u x 1 (u s u t ), q st sin q cos q. To summarize, we have proved that Theorem Locally there exists an asymptotic parametrization for a surface with K 1 such that I ds + cos(q) ds dt + dt, II 4 sin q cos q ds dt sin(q) ds dt, where q is the angle between the asymptotic directions. Gauss-Codazzi equation is q st sin q cos q. SGE Moreover, the 4. Change of parametrizations We will show that although the two fundamental forms look different when we use different parametrization of the same surface, they are the same symmetric bilinear forms. To see this, we review some linear algebra and material we taught in 16A, Let V be a vector space, and {v 1,..., v n } a basis of V. It is easy to check that the space V of all linear maps from V to R is again a vector space. Now let vi denote the linear map from V to R such that { vi 1, if i j, (v j ) δ ij 0, if i j, for 1 i n.

9 9 Exercise Prove that {v 1,..., v n} is a basis of V. Suppose l 1, l : V R are linear maps. We define l 1 l : V V R by l 1 l (ξ, η) l 1 (ξ)l (η) for all ξ, η V. Let l 1 l : V V R denote the bilinear map defined by l 1 l 1 (l 1 l + l l 1 ), i.e., l 1 l (ξ, η) 1 (l 1(ξ)l (η) + l (ξ)l 1 (η)). Exercise Prove that (1) l 1 l is bilinear. () l 1 l is symmetric bilinear. Exercise Let b : V V R be a bilinear map, v 1,..., v n a basis of V, v1,..., v n dual basis of V, and b ij b(v i, v j ). (1) Prove that b n i,j1 b ijvi v j. () Prove that b is symmetric if and only if b ij b ji for all 1 i, j n. (3) Prove that if b is symmetric then b n i,j1 b ijvi v j. Suppose f : O R 3 is a parametrized surface. The first fundamental form I p : T f p T f p R is defined by I p (ξ, η) ξ η the dot product for all ξ, η T f p. It is easy to check that I p is a symmetric bilinear map. Note that {f x1 (p), f x (p)} is a basis of the tangent plane T f p. Let dx 1, dx be the dual basis of T f p. It follows from the above exercise that I g 11 dx 1 + g 1 dx 1 dx + g dx, where g ij f xi f xj. Since I p is defined without parametrization I p (ξ, η) ξ η, different parametrizations will give the same I. The unit normal vector field to f is the map from O to the unit sphere S : N(x 1, x ) f x 1 f x f x1 f x. (N is also called the Gauss-map of the surface. The shape operator A p is the self-adjoint operator from T f p to T f p defined by A p (f xi ) N xi. Now we change parametrization of the surface by a diffeomorphism x x(y) : O 1 O, i.e., we use f(y) f(x(y)) as a parametrization of the same surface. By definition, the shape operator à for the new parametrization is the linear operator defined by Ã( f yi ) N yi. We will show that A and à are the same. Because by Chain rule, we have u y i u x 1 x 1 y i + u x x y i.

10 10 So Ã( f yi ) N ( x1 N x1 + x ) N x. y i y i y ( ) i However, A(f yi ) A x1 y i f x1 + x y i f x and A is linear, we see that ( x1 A(f yi ) N x1 + x ) N x, y i y i which is equal to Ã(f y i ). This shows that the shape operator does not depend on the choice of parametrization of a surface. The second fundamental form is the bilinear form associated to the shape operator: II p (ξ, η) A p (ξ) η for all ξ, η T f p. Since the shape opeartor is independent of parametrization of the surface, so is II. Note that the mean curvature H and the Gaussian curvature K are the trace and the determinant of the shape operator. This shows that H and K are well-defined function on the surface, independent of parametrizations. 5. Calculus of variations of one variable In this section, C 1 means continuously differentiable. Let C 1 ([a, b], R ) denote the space of all C 1 maps x : [a, b] R (i.e., x is a differentiable and its derivative x is continuous). Fix p 0, q 0 R, let C 1 ([a, b], R ) p0,q 0 denote the set of all x C 1 ([a, b], R ) such that x(a) p 0 and x(b) q 0. A function J : C 1 ([a, b], R ) p0,q 0 R is called a a calculus of variations functional if it has the form J(x) J L (x) b a L(t, x(t), x (t)) dt, where L : [a, b] R R R is a C 1 map. We call L the Lagrangian function associated to the functional J. To motivate the definitions of directional derivative and critical points of J, we review these definitions for f : R n R. The directional derivative of f at p 0 in the direction v is d ds s0 f(p 0 + sv). A point p 0 is a critical point of f if f x i (p 0 ) 0 for all 1 i n. Given v (v 1,..., v n ), the directional derivative Df v (p 0 ) d n ds f(p 0 + sv) (p 0 )v i. s0 x i i1 So the following statements are equivalent: (i) p 0 is a critical point of f, (ii) all directional derivatives of f at p 0 are zero,

11 (iii) d ds s0 f(σ(s)) 0 for all smooth curve σ : ( ɛ, ɛ) R n with σ(0) p 0. equation The functional J is a function on the vector space C 1 ([a, b], R ) p0,q 0, so we can define directional derivatives and critical points of J in the same manner as for f : R n R. 11 The directional derivative of J L. Let x 0 (x 0 1, x0 ) C1 ([a, b], R ) p0,q 0, and h (h 1, h ) C 1 ([a, b], R ) 0,0, i.e., h(a) h(b) 0. Note that x s x 0 + sh is in C 1 ([a, b], R ) p0,q 0 for all s R. Let us find the derivative of J L (x s ) with respect to s at s 0: d b ds J L (x s ) L(t, x s (t), (x s ) (t)) dt s0 a b d ds L(t, x 0 (t) + sh(t), (x 0 ) (t) + sh (t)) dt, by chain rule, s0 a b a L (t, x 0 (t), (x 0 ) (t))h 1 (t) + L (t, x 0 (t), (x 0 ) (t))h (t) x 1 x + L x (t, x 0 (t), (x 0 ) (t))h 1(t) + L 1 x (t, x 0 (t), (x 0 ) (t))h (t) dt. b Next we want to use integration by part, a f b (t)g(t)dt f(t)g(t) a b a f(t)g (t) dt, to change h i to h i in the above integration. First note that since h i (a) h i (b) 0 for i 1,, we have b ( ) d L a dt x h i (t) dt 0, i and hence So b a b a ( ) L L h i + h i dt 0. x i x i L (t, x 0 (t), (x 0 ) (t))h 1(t) + L (t, x 0 (t), (x 0 ) (t))h (t) dt x 1 b a ( L x x (t, x 0 (t), (x 0 ) (t))) h 1 (t) + 1 ( L x (t, x 0 (t), (x 0 ) (t))) h (t) dt. Thus the formula for the derivational derivative of J at x 0 in the direction h is d b ( ( ) L L ) ( ( ) L L ) ds J L (x s ) s0 a x 1 x h x x h dt. (5.0.8)

12 1 x 0 is called a critical point of J L if this directional derivatives is zero for all choices of h C 1 ([a, b], R ) 0,0. This will certainly be the case if the following so-called Euler-Lagrange equation is satisfied: L x i d dt ( L x i ), i 1,. (5.0.9) The following is the fundmental lemma of calculus of variations: Lemma Let f (f 1,..., f n ) : [a, b] R n be a C 1 map. Suppose that b n a i1 f i(t)h i (t) dt 0 for all C 1 maps h (h 1,..., h n ) : [a, b] R n with h(a) h(b) 0. Then f 0. Proof. We prove this by contracdition. It suffices to prove the case when n 1. Suppose f is not the zero function, there there exists c (a, b) such that f(c) 0. Say f(c) > 0. Then there exists ɛ > 0 such that (c ɛ, c + ɛ) (a, b) and f(t) > 0 for all t (c ɛ, c + ɛ), Choose a C 1 function g : [a, b] R such that g > 0 on (c ɛ, c + ɛ), g 0 outside [c ɛ, c + ɛ]. Then b a f(t)g(t) dt c+ɛ c ɛ f(t)g(t) dt > 0, which contradicts to the assumption that b a f(t)h(t) dt 0 for all C1 h with h(a) h(b) 0. As a consequence of Lemma and (5.0.8), we get Theorem x : [a, b] R is a critical point of J L if and only if x satisfies the Euler-Lagrage equation (5.0.9). Example Let J(x) b a 1 ((x 1(t)) + (x (t)) ) dt. Then L(t, x 1, x, x 1, x ) 1 ((x 1 ) + (x ) ). The Euler-Lagrange equation for J is computed as follows: ( ) L L 0 (x x i) x i. i x i In other words, the Euler-Lagrange equation for J is x i 0 for i 1,. So x(t) c 0 + c 1 t for some c 0, c 1 R, i.e., x is a straight line. Since x(a) p 0 and x(b) q 0, x(t) p 0 + (t a) b a (q 0 p 0 ). Example Newtonian mechanics in R We consider a particle p of mass m moving in the plane R under Newton s Laws of Motion. In physics it is shown that Newton s Third Law implies that the force F (F 1, F ) acting on p is derivable from a potential V, i.e.,

13 13 there is a smooth function V : R R such that F i V x i F V ). The kinetic energy function is for i 1, (i.e., K(x ) m ((x 1) + (x ) ). Define the Lagrange L K V m ((x 1) + (x ) ) V (x 1, x ). The functional L(x) b a L(x, x ) dt is called the action in physics. Since L V, x i x i the Euler-Lagrange equation is or L x i mx i, V x i (mx i) mx i, mx i V F i. x i In other words, we have proved that the Euler-Lagrange equation for the action functional L b a K V dt is the Newton s equation F ma, i.e., the force F is equal to the mass times the acceleration. 6. Initial value problem for second order ODEs A system of second order ODEs is of the form x i f i (t, x 1,..., x n, x 1,..., x n), 1 i n. (6.0.10) This can be solved using the existence and uniqueness of systems of first order ODE by introducting new variables: Consider the following system of first order ODE: { x i y i, 1 i n, y i f (6.0.11) i(t, x 1,..., x n, y 1,..., y n ), 1 i n. It is easy to see that if x(t) (x 1 (t),..., x n (t)) is a solution of (6.0.10), then (x, y) (x, x ) is a solution of (6.0.11). Conversely, if (x, y) is a solution of (6.0.11), then x is a solution of (6.0.10). By the uniqueness of ODE, we know that given initial condition p 0, q 0 R n, there exists a unique solution (x, y) of (6.0.11) such that x(0) p 0 and y(0) q 0. Thus given p 0, q 0 R n, there exists a unique solution x(t) of the second order system (6.0.10) such that x(0) p 0 and x (0) q 0. In other words, to solve the initial value problem for the second order ODE system (6.0.10) we need to give the initial position x(0) and initial velocity x (0).

14 14 7. Energy functional Let f : O R 3 be a parametrized surface, and I i,j1 g ijdx i dx j the first fundamental form. The energy functional E : C 1 ([a, b], O) p0,q 0 R is i.e., E(x) E(x) b a b a g ij (x 1 (t), x (t))x i(t)x j(t) dt, i,j1 γ (t) dt, where γ(t) f(x(t)). We want to compute the Euler-Lagrange equation for E. Before we do the general computation, let us do the following simple example: Example (Energy functional on S ) parametrization of S : We use the spherical f(φ, θ) (sin φ cos θ, sin φ sin θ, cos φ). Then I dφ + sin φ dθ. The energy Lagrangian for S is E (φ ) + sin φ (θ ). Note E φ sin φ cos φ (θ ), ( ) E φ (φ ) φ, E θ 0, ( ) E θ ( sin φ θ ) 4 sin φ cos φ φ θ + sin φ θ. So the E-L equation is { φ sin φ cos φ(θ ), 4 sin φ cos φ φ θ + sin φ θ 0, i.e., { φ sin φ cos φ (θ ), θ cos φ sin φ φ θ.

15 Next we compute the E-L for the energy functional for an arbitrary surface f. Since E ij g ij(x 1, x )x i x j, 15 E g ij,k x k x x ix j, k k ( ) ( E x g ik x i) g ik,m x mx i + k i i,m i g ik x, i where g ij,k g ij x k. So the E-L equation is i g ik x i + g ik,m x ix m i,m i,j g ij,k x ix j 0. But g ik,m x ix m g ik,j x ix j g jk,i x jx i. i,m i,j i,j (Here we replace m by j to get the first identity, and then exchange i, j to get the second identity). So i,m g ik,m x ix m i,j (g ik,j + g jk,i )x ix j, and hence the E-L equation for E can be written as i g ik x i + (g ki,j + g jk,i g ij,k )x ix j 0, i,j i.e., i g ik x i + 1 Recall that we used the following notation (g ki,j + g jk,i g ij,k )x ix j 0. i,j [ij, k] : g ki,j + g jk,i g ij,k when we derived of the G-C equation in general coordinates in 16A. Use this notation, the E-L equation for the energy functional can be written as g ik x i + 1 [ij, k]x ix j 0. i i,j We want to write this system of ODE in the form of (6.0.11). We can do this as follows: Let G denote the matrix (g ij ), and g ij the ij-th entry of the inverse matrix G 1. Multiply the above equation by g kl then sum over k to get k,i g kl g ik x i + 1 g kl [ij, k]x ix j 0. i,j,k

16 16 Note that G and G 1 are symmetric matrices, and the il-th entry of GG 1 is k g ikg kl. But GG 1 I, so k g ikg kl δ il. Then we get i δ il x i + 1 g kl [ij, k]x ix j 0, i,j,k Recall that we use the following notation in the G-C equation: Γ l ij : 1 g kl [ij, k]. k So the E-L equation for the energy functional can be written as x l + i,j Γ l ijx ix j 0, l 1,. (7.0.1) We summarize what we have proved below: Theorem x : [a, b] R is a critical point of the energy functional E if and only if x satisifes (7.0.1). Next we want to prove that if x is a critical point of the energy functional for the surface f : O R 3, then the curve γ(t) f(x(t)) is travelled at constant speed: Theorem If x is a critical point of the energy functional E, then E(x) i,j g ij (x(t))x i(t)x j(t) γ (t) is a constant, where γ(t) f(x(t)). Proof. It suffices to prove that ( i,j g ij(x)x i x j ) 0. But ( i,j g ij x ix j) i,j,k g ij,k x k x ix j + i,j g ij x i x j + i,j g ij x ix j. But x satisfies the E-L equation x k + i,j Γ k ijx ix j 0.

17 17 So i,j g ij x ix j g ij,k x k x ix j g ij Γ i k,m x k x mx j g ij x i i,j,k i,j,k,m i,j k,m g ij,k x ix jx k g ij Γ i km x k x mx j g ij Γ j km x ix k x m i,j,k i,j,k,m g ij,k x ix jx k g mj Γ m ki x k x ix j g im Γ m kj x ix jx k i,j,k i,j,k,m g ij,k 1 g mj g ml [ki, l] 1 g mig ml [kj, l] x ix jx k i,j,k m,l (g ij,k 1 δ jl [ki, l] 1 δ il[kj, l])x ix jx k i,j,k l i,j,k(g ij,k 1 [ki, j] 1 [kj, i])x ix jx k Γ j km x k x m i,j,k(g ij,k 1 (g ij,k + g jk,i g ki,j + g ji,k + g ik,j g kj,i ))x ix jx k 0. (From the second line to the third line, we interchange m, i of the second term and m, j of the third term.) 8. Arc length functional Let f : O R 3 be a parametrized surface, Σ f(o), and I ij g ijdx i dx j the first fundamental of f. A smooth curve x : [a, b] O gives rise to a curve γ f x on the surface Σ, and the arc length of γ is L(x) b a γ (t) dt b a i,j1 g ij (x(t))x i (t)x j (t) dt. A curve γ f x on Σ is called a geodesic if x is a critical point of the arc length functional L. Let E i,j1 g ij(x(t))x i (t)x j (t) denote the Lagrangian of the energy functional E. The E-L equation for the energy functional E is ( ) E E, i 1,. (8.0.13) x i x i

18 18 Let L denote the Lagrangian for the arc length functional L. Then L E. The E-L equation for L is ( E ) E x i x, i so the E-L equation for L is 1 E E x i ( 1 E ) E x, i 1,. (8.0.14) i Proposition Let f : O R 3 be a parametrized surface. If x is a critical point of the energy functional E, then x is a critical point of the arc length functional L. Conversely, if x is a critical point of L such that d dtf(x(t)) is constant, then x is a critical point of E. Proof. By Theorem , E i,j1 g ij(x(t))x i (t)x j (t) is a constant c. Since x satisfies (8.0.13) and E is constant, x satisfies (8.0.14), so x is a critical point of L. The converse is proved the same way. We want to prove below that if x : [a, b] O is a critical point of L and t t(s) is a diffeomorphism from [a, b] to [a, b], then x(s) x(t(s)) is also a critical point of L. This follows from the following theorem: Theorem Let V be a vector space, φ : V V a diffeomorphism, and F : V R a smooth function. Suppose F φ 1 F. Then p 0 V is a critical point of F implies that φ(p 0 ) is also a critical point of F. Proof. Note that p 0 is a critical point of F if and only if d dt t0 F (x(t)) 0 for all smooth curve x : ( δ, δ) V such that x(0) p 0. To prove that φ(p 0 ) is a critical point, let y : ( δ, δ) V be a smooth curve such that y(0) φ(p 0 ). Then x φ 1 y is a curve with x(0) p 0. But so F (y(t)) F (φ 1 (y(t)) F (x(t)), d dt F (y(t)) d t0 dt F (x(t)), t0 which is zero because p 0 is a critical point of F and x is a curve with x(0) p 0. This proves that φ(p 0 ) is also a critical point of F. The condition F F φ 1 is equivalent to F F φ because F (φ 1 (φ(x))) F (φ(x)) implies that F (x) F (φ(x)). The above Theorem says that if a function F is invariant under a transformation φ, then φ(p 0 ) is a critical point of F if p 0 is. Proposition Let t t(s) is a diffeomorphism from [a, b] to [a, b], and φ : C([a, b], O) C([a, b], O) defined by φ(x)(s) x(t(s)). Then if x 0 is a critical point of L, then φ(x 0 ) is also a critical point of L.

19 Proof. Note that γ(t) f(x(t)) and γ(s) f(x(t(s))) trace out the same curve on the surface Σ. So the arc length of γ and γ are the same. This proves that L(x) L(φ(x)). By Theorem , φ(x 0 ) is a critical point of L. 19 Suppose x 0 is a critical point of L and the arc length of γ 0 f(x 0 ) is l. We claim that we can change parameter of γ 0 to a new parameter s such that dγ0 l ds is the constant b a. But l b a dγ0 ds dγ0 dt dt ds so if we choose the new parameter s such that dt ds l (b a) dγ0 dt, dt dγ0 dt ds, then dγ0 ds l b a. This proves the claim. By Proposition , y0 (s) x(t(s)) is a critical point of L. But d ds f(y0 (s)) is constant, so by Proposition , y 0 is a critical point of E. This shows that to construct all geodesics of the surface Σ, it suffices to solve the E-L equation (7.0.1) for the energy functional E. Therefore we will call (7.0.1) the geodesic equation. A curve γ : [a, b] R 3 is said to be parametrized proportional to its arc length if its speed is constant, i.e., dγ dt is a constant for all t [a, b]. We like to give a geometric calculation for the geodesic equation. Recall that we proved in 16A that if f : O R 3 is a parametrized surface with I ij g ijdx i dx j and II ij l ijdx i dx j, then f xi x j Γ 1 ijf x1 + Γ ijf x + l ij N, (8.0.15) where Γ i jk 1 m gim [jk, m], and N the unit normal of f. We will use (8.0.15) to give a simple criterion for a curve γ f x on the surface f to be a geodesic. Theorem Let γ f x be a curve on the surface f that is parametrized proportional to its arc length. Then γ is a geodesic if and only if γ (t) is normal to f at γ(t) for all t. Moreover, if γ is a geodesic then γ (t) II(γ (t), γ (t)) N(γ(t)).

20 0 Proof. Chain rule gives γ i f x i x i, and γ (t) i,j i,j i,j f xi x j x ix j + f xi x i i (( ) ) Γ k ijx ix j f xk + l ij x ix j N + k i (( ) ) Γ k ijx ix j f xk + l ij x ix j N + k k f xi x i f xk x k k x k + ij Γ k ijx ix j f xk + i,j l ij x ix j N. So γ (t) is normal if and only if x k + i,j Γ k ijx ix j 0 for k 1,, i.e., x satisfies (7.0.1), or equivalently, γ is a geodesic. Example We can use Theorem to get geodesics of the unit sphere in R 3 easily by observing that if α is a great circle, then α is the radial vector of the great circle, so α (t) is the unit normal to S at α(t). This implies that α (t) is normal to S, by Theorem , α is a geodesic. 9. Calculus of variations of two variable Let O be an open subset of R such that the boundary O is a smooth curve, i.e., there is a smooth parametrization α : [a, b] R for O. Let γ : O R a fixed smooth function, and O O O. Let C γ (O, R) denote the space of smooth functions u : O R such that u O γ. A two variable Lagrangian is a smooth function L : O R R R R, and the variational functional associated to the Lagrangian L is the map J J L : C γ (O, R) R defined by J(u) L(x, y, u(x, y), u x (x, y), u y (x, y)) dxdy. O We will use x, y, u, p, q to denote the variables of L, i.e., L L(x, y, u, p, q). A function u 0 : O R is called a critical point of J if all the directional derivatives of J at u 0 is zero, i.e., d ds J(u 0 + sh) 0 s0 for all smooth h : O R with h O 0. We will calculate the same way as in the calculus of variations of one variable. Recall that in that calculation,

21 we need the fundamental lemma and the Fundamental Theorem of Calculus of one variable. As we will see below that to calculate the condition that u 0 is a critical point, we will need the two dimensional versions of the fundamental lemma and of the Fundamental Theorem of Calculus. Given a R, let φ a : R R be the function defined by { exp( (x a) ), x > a, φ a (x) 0, x a. Exercise Prove that lim x a + φ a (x) 0, and the limits of derivatives of φ a of any order tends to zero as x a +. (This proves that φ a is a smooth function). Exercise Let ψ a : R R be the map defined by ψ a (x) φ a ( x). Prove that { e (x a), x < a, ψ a (x) 0, x a.. Exercise Given a < b, let h a,b : R R be the function defined by h a,b (x) φ a (x)ψ b (x). Prove that h a,b is smooth, h a,b > 0 on (a, b) and is zero outside (a, b). Lemma Suppose f : O R is continuous, and f(x, y)h(x, y) dxdy 0 for all h : O R with h O 0. Then f 0. O Proof. Suppose f is not identically zero. Then there exists (x 0, y 0 ) O such that f(x 0, y 0 ) 0. We may assume that f(x 0, y 0 ) > 0. Since f is continuous, there exists ɛ > 0 such that f(x, y) > 0 for all (x, y) in the square D {(x, y) x x 0 < ɛ, y y 0 < ɛ}. Let h a,b be the smooth function constructed in the above Exercise, and k D : R R the map defined by k D (x, y) h x0 ɛ,x 0 +ɛ(x)h y0 ɛ,y 0 +ɛ(y). It is easy to see that k D is smooth, k D (x, y) > 0 if (x, y) D and is zero outside D. In particular, h D O 0. But fh D is positive in O, so fh D dxdy fh D dxdy, O which is positive because both f and h D are positive on D, a contradiction. D 1

22 Next we recall the Green s formula, which can be viewed as the two dimensional version of the Fundamental Theorem of Calculus. First recall that the line integral P (x, y) dx + Q(x, y) dy O is computed as follows: Choose a parametrization of the boundary O, say α : [a, b] R with α(t) (x(t), y(t)). Then b P (x, y) dx + Q(x, y) dy P (x(t), y(t))x (t) + Q(x(t), y(t))y (t) dt. O Theorem (Green s formula) Let P, Q : O R be smooth functions. Then P dx + Q dy ( P x + Q y ) dxdy. O Corollary If P O 0 and Q O 0, then ( P x + Q y ) dxdy 0. O a (This is because the line integral is zero). Next we compute the directional derivative of J at u in the direction h, where u C γ (O, R) and h : O R satisfying h O 0. d ds J(u + sh) d s0 ds L(x, y, u + sh, (u + sh) x, (u + sh) y ) dxdy s0 O d ds L(x, y, u + sh, u x + sh x, u y + sh y ) dxdy s0 O d O ds L(x, y, u + sh, u x + sh x, u y + sh y ) dxdy s0 L O u (x, y, u, u x, u y )h(x, y) + L p (x, y, u, u x, u y )h x (x, y) + L q (x, y, u, u x, u y )h y (x, y) dxdy. ) ) ( L L u h + p h ( ) L + q h O O y x ( L q ( L p ) h dxdy. The Green s formula implies that ( ) ( ) L L p h + q h x y y O h dxdy x dxdy O L L h dy p q h dx.

23 Since h O 0, by Corollary the right hand side is zero. So we get ( ( ) ( ) ) d L L L ds J(u + sh) s0 u h dxdy. p q Here ( ) L p ( ) L q x y O ( ) L p (x, y, u, u x, u y ) ( ) L q (x, y, u, u x, u y ) We will use the conventional notation L L q. By Lemma , we get u x x, x y to denote L p. y and L u y 3 to denote Theorem u is a critical point of J if and only if ( ) ( ) L L L u 0. (9.0.16) u x u y x Equation (9.0.16) is called the Euler-Lagrange equation for J J L. Example Let J(u) Then L(x, y, u, p, q) 1 (p + q ), O L u 0, 1 ( (ux ) + (u y ) ) dxdy. L p p, y L q q. To get the E-L equation, we need to substitute p u x and q u y, so 0 (u x ) x (u y ) y 0, i.e., the E-L equation is the Laplace equation u xx + u yy 0. Example Let J(u) u x + u x u y + u y cos u + (x + y )u dxdy. O So L(x, y, u, p, q) p + pq + q cos u + (x + y )u, and L u sin u (x + y ), L p p + q, L q p + q. Substitue p u x and q u y to see that the E-L equation is 0 sin u (x + y ) (u x + u y ) x (u x + u y ) y 0 sin u (x + y ) (u xx + u yx + u xy + u yy ) sin u (x + y ) (u xx + u xy + u yy ).

24 4 Example Let J(u) So L(x, y, u, p, q) (1 + p + q ) 1 L u 0, The E-L equation is 0 L u O and L p p(1 + p + q ) 1, ( L p ) x ( ) L q y (1 + u x + u y) 1. L q q(1 + p + q ) 1. 0 (u x (1 + u x + u y) 1 )x (u y (1 + u x + u y) 1 )y (1 + u x + u y) 3/ ( (1 + u y)u xx + (1 + u x)u yy u x u y u xy ). 10. Area functional Let u : O R be a smooth function, and f(x, y) (x, y, u(x, y)) the graph of u. We have seen in 16A that the two fundamental forms for f are I (1 + u x) dx + u x u y dxdy + (1 + u y) dy, 1 ( II uxx dx + u xy dxdy + u yy dy, ) 1 + u x + yy and the mean curvature is Compute directly to see that H (1 + u x)u yy u x u y u xy + (1 + u y)u xx (1 + u x + u y) 3/. det(g ij ) (1 + u x)(1 + u y) (u x u y ) (1 + u x + u y). So the area of the graph of u (i.e., surface f) is J(u) det(g ij ) dxdy (1 + u x + u y) 1/ dxdy. O We have computed the E-L equation for this functional in Example and see that d ds J(u + sh) (1 + u x)u yy u x u y u xy + (1 + u y)u xx s0 O (1 + u x + u y) 3/ h dxdy Hh dxdy. O O

25 So the E-L equation for the area functional is H 0. This gives another geometric meaning of the mean curvature. In particular, the surface that has minimum surface area among all surfaces in R 3 with a fixed boundary must have zero mean curvature. Definition A surface in R 3 is minimal if its mean curvature is zero. Remark. It follows from physics that if we dip a closed wire frame into a soap solution, then the soap film surface spanned by the wire frame when we take out the frame must be minimal. So we sometimes call minimal surfaces soap films Minimal surfaces Minimal surfaces and harmonic functions. Definition A surface f : O R 3 is said to be parametrized by isothermal coordinates if I λ (x 1, x )(dx 1 + dx ), i.e., g 11 g and g 1 0. Or equivalently, f x1 f x1 f x f x, f x1 f x 0. Theorem Suppose f : O R 3 is parametrized by isothermal coordinates with I λ (dx 1 + dx ). Then f x1 x 1 + f x x λ HN, (11.1.1) where H is the mean curvature and N is the unit normal vector field of f. Proof. First we want to use the isothermal condition, f x1 f x1 f x f x λ and f x1 f x 0, to conclude that (f x1 x 1 + f x x ) f xi 0, i 1,, which implies that f x1 x 1 + f x x is parallel to the unit normal N of the surface. To do this, we take derivatives of the isothermal condition: (f x1 f x1 ) x1 f x1 x 1 f x1 (f x f x ) x1 f x1 x f x ((f x1 f x ) x f x1 f x x ) (0 f x1 f x x ) f x x f x1, so (f x1 x 1 + f x x ) f x1 0. Similar calculation implies that thus f x1 x 1 + f x x (f x1 x 1 + f x x ) f x 0, must be parallel to N. But f xi x j Γ 1 ijf x1 + Γ ijf x + l ij N, where II l 11 dx 1 +l 1dx 1 dx +l dx and Γi jk 1 m gim [jk, m]. Therefore we have f x1 x 1 + f x x ( i Γ 1 ii)f x1 + ( i Γ ii)f x + (l 11 + l )N.

26 6 But we have shown that f x1 x 1 + f x x is parallel to N, so the coefficients of f xi in f x1 x 1 + f x x are zero and we get f x1 x 1 + f x x (l 11 + l )N. (11.1.) Recall that the mean curvature H is given by the following formula H g 11l g 1 l 1 + g l 11, det(g ij ) so H λ (l 11 +l ) l 11+l, which implies that l λ 4 λ 11 + l λ H. Substitute this into (11.1.) to get (11.1.1). Definition A smooth function u : O R is called harmonic if u x1 x 1 + u x x 0. The operator 1 + is called the Laplace operator and u x 1 x u x1 x 1 + u x x 0 is called the Laplace equation. Corollary Suppose f (u, v, w) : O R 3 is parametrized by isothermal coordinates. Then the following statements are equivalent: (1) f is minimal, () f x1 x 1 + f x x (0, 0, 0), (3) u, v, w are harmonic functions. Remark. We have seen that a surface f : O R 3 with H 0 is a critical point of the area functional. But this surface need not have minimum area among all surfaces in R 3 that having the same boudary as f(o). However, it is a theorem of PDE that if we take any small enough piece Ω of f(o), the surface Ω has the minimum area among all surfaces that have the same boundary of Ω. Example (Caternoid) Let a > 0 be a constant, and f(x 1, x ) (a cosh x cos x 1, a cosh x sin x 1, ax ). A direct computation implies that I a cosh x (dx 1 +dx ), so f is isothermal parametrization. Compute directly to see that f x1 x 1 + f x x (0, 0, 0), so by Corollary f is minimal. Example (Helicoid) Let a > 0 be a constant, and f(x 1, x ) (sinh x cos x 1, sinh x sin x 1, x 1 ). A direct computation implies that I a cosh x (dx 1 + dx ) and f x 1 x 1 + f x x (0, 0, 0). So f is minimal. Note that Helicoid and Caternoid have the same first fundamental form!

27 7 Example (Enneper s surface) Let f(x 1, x ) (x 1 x x 1x, x x3 3 + x x 1, x 1 x ). Then I (1 + x 1 + x )(dx 1 + dx ) and f (0, 0, 0). So f is minimal Linear conformal transformations of R. ( x The plane R can be viewed as C by identifying with the complex y) number z x + iy. Let α a + ib be a fixed complex constant, and M α : C C the map defined by M α (z) αz, i.e., M α (x + iy) (a + ib)(x + iy) (ax by) + i(bx + ay). If we use the identification of C with R, then the map M α becomes T α : R R, where ( ) ( ) ( ) ( ) x ax by a b x T α. y bx + ay b a y In other words, the map M α given by multiplication by α on( C is the ) map a b T α on R, which is given by the multiplication by the matrix. We b a can study the geometry of T α using complex numbers: First write α r 0 e iθ 0 in polar coordinate, i.e., r 0 a + b and tan θ 0 b a. If z reiθ, then M α (z) az r 0 re i(θ0+θ). In other words, the map T α maps a vector v by first rotating v counterclockwise by angle θ 0, then multiplying the length by the factor r 0. Let (v 1, v ) denote the angle from v 1 to v. If v j r j e iθ j, then (v 1, v ) θ θ 1. Exercise Prove that if v 1, v are orthonormal, then T α (v 1 ) T α (v ) 0 and T α (v 1 ) T α (v ). The linear operator T α preserves angles and stretchs each vector by a fixed amont r 0. We call T α a linear conformal transformation of R. Exercise Suppose T : R R is the linear map defined by ( ( ( ) x a c x T. y) b d) y ( ( 1 0 Let e 1, and e 0). Prove that if (T (u), T (v)) (u, v) and 1) T (e 1 ) T (e ), then d a and c b, i.e., T is a linear conformal transformation.

28 8 If α 1, i.e., α e iθ 0 cos θ 0 + i sin θ 0, then ( ) ( ) ( ) ( ) x cos θ0 sin θ T α 0 x x cos θ0 y sin θ 0 y sin θ 0 cos θ 0 y x sin θ 0 + y cos θ 0 is the rotation by angle θ 0. We call ( ) cos θ0 sin θ 0 sin θ 0 cos θ 0 the rotation matrix by angle θ Analytic functions, definitions and examples. Definition Let O be an open subset of C. (1) A map f : O C is analytic at z 0 if f(z 0 + h) f(z 0 ) lim h 0 h exists. We will use f (z 0 ) to denote the limit. The limit f (z 0 ) is called the complex derivative of f at z 0. Here the limit is taking for complex number h 0, i.e., given any ɛ > 0, there exists δ > 0 such that if h C and h < δ then f(z 0 + h) f(z 0 ) < ɛ. () The map f : O C is analytic if f is analytic at every point z 0 O. Given a function f : O C, we can write f(z) u(x, y) + iv(x, y) with real valued functions u, v. We will call u and v the real and imaginary part of f, and write u Re(f) and v Im(f). Example Let f(z) x + iy 3z+ z. It is easy to see that { f(r + is) f(0) r + is r + is r + is, if s 0, r 0, 1, if r 0, s 0. So the limit of f(h) f(0) h Example (1) Let f(z) z. Then f(z + h) f(z) h so f (z) 1. () Let f(z) z n. Then f(z + h) f(z) h does not exist, i.e., f is not analytic. (z + h)n z n h (z + h) z h 1, zn + nz n 1 h + n(n 1) z n 1 h + + h n z n h nz n 1 n(n 1) + z n h + + h n 1 nz n 1 as h 0.

29 So (z n ) nz n 1. (3) Let f(z) e z e x+iy : e x e iy e x (cos x + i sin x). Then e z 1+z e z 1 e z. Since the Taylor series of e x, cos y, and sin y converges absolutely, e z z n n. So e h 1 h n0 h (1 + h + + ) h h +, e which implies lim h 1 h 0 h 1. But e z+h e z ez e h e z ez (e h 1), h h h hence its limit is e z as h 0. Thus (e z ) e z. The addition formula, product formula, quotient formula, and the chain rule for analytic functions can be proved in a similar way as for calculus of one real variable. We define cos z eiz + e iz, sin z eiz e iz. i Since e z is analytic, by the chain rule we have (e iz ) ie iz and (e iz ) ie iz. Exercise Prove that: (1) (sin z) cos z and (cos z) sin z. () cos z + sin z 1, Define cosh z ez + e z, sinh z ez e z. Exercise Prove that (1) cosh z sinh z 1, () (cosh z) sinh z, (sinh z) cosh z, (3) cosh(iz) cos z, sinh(iz) i sin z. Theorem Suppose f : O C is analytic, and α, β : ( δ, δ) O smooth curves intersect at p 0 α(0) β(0). Then i.e., f preserves angles. (α (0), β (0)) ((f α) (0), (f β) (0)), Proof. By the chain rule, (f α) (0) f (α(0))α (0) f (p 0 )α (0). Similarly, (f β) (0) f (p 0 )β (0). But (α (0), β (0)) (f (p 0 )α (0), f (p 0 )β (0)). 9

30 Cauchy-Rieman equation. Let O be an open subset of R, and k(x, y) (u(x, y), v(x, y)) a smooth map from O to R. The degree 1 Taylor expansion of k at (x 0, y 0 ) is k(x 0, y 0 ) + k x (x 0, y 0 )x + k y (x 0, y 0 ). The map dk (x0,y 0 ) : R R defined by (x, y) k x (x 0, y 0 )x + k y (x 0, y 0 )y is linear, and is called the differential of k at (x 0, y 0 ). We want to write down the matrix for the differential dk (x0,y 0 ). Note that dk (x0,y 0 )(x, y) k x (x 0, y 0 )x + k y (x 0, y 0 )y (u x (x 0, y 0 ), v x (x 0, y 0 ))x + (u y (x 0, y 0 ), v y (x 0, y 0 ))y (u x (x 0, y 0 )x + u y (x 0, y 0 )y, v x (x 0, y 0 )x + v y (x 0, y 0 )y). Write R as the space of coloumn vectors. Then dk (x0,y 0 ) is ( ( ( ) x ux (x dk (x0,y 0 ) 0, y 0 ) u y (x 0, y 0 ) x. y) v x (x 0, y 0 ) v y (x 0, y 0 )) y In other words, the matrix of the linear map dk (x0,y 0 ) with respect to the standard basis is the Jacobi matrix ( ) ux (x 0, y 0 ) u y (x 0, y 0 ). v x (x 0, y 0 ) v y (x 0, y 0 ) It is known from calculus that k(x + x 0, y + y 0 ) k(x 0, y 0 ) dk (x0,y 0 )(x, y) 0. (11.4.1) (x, y) ( ) u(x, y) Theorem Let k(x, y) be a smooth map, and f(z) v(x, y) u(x, y) + iv(x, y). Then the following conditions are equivalent: (i) dk (x0,y 0 ) is linear conformal, { u x (x 0, y 0 ) v y (x 0, y 0 ) (ii) u y (x 0, y 0 ) v x (x 0, y 0 ), (iii) f u + iv is analytic at z 0 x 0 + iy 0, and f (z 0 ) u x (x 0, y 0 ) + iv x (x 0, y 0 ). ( ( ( ) x a c x Proof. A linear map T is linear conformal if d a and y) b d) y c b. So (i) and (ii) are equivalent. If dk (x0,y 0 ) is linear conformal, then when we identify R as C, the linear operator dk (x0,y 0 ) on R becomes M α, where α u x (x 0, y 0 ) + iv x (x 0, y 0 ). So (11.4.1) becomes f(z 0 + z) f(z 0 ) αz 0 z as z 0, i.e., f is analytic at z 0 and f (z 0 ) u x (x 0, y 0 ) + iv x (x 0, y 0 ). As a consequence, we get

31 Theorem Suppose u, v : O R are smooth. Then f(z) u(x, y)+ iv(x, y) is analytic if and only if u, v satisfies the Cauchy-Rieman equation: u x v y, u y v x. (11.4.) Theorem Suppose f : O C is analytic, and f(z) u(x, y) + iv(x, y), where u, v are real valued functions. Then (a) f (z) f x if y, (b) u, v satisfy the Cauchy-Rieman equation (11.4.), (c) f (z) u x + iv x u x iu y 1 ( x i ) f, y (d) u and v are harmonic functions, i.e., u xx + u yy 0, v xx + v yy 0. Proof. We will give another proof of (b). The function f is analytic means that f(z + h) f(z) lim f (z). h 0 h Here h r+is is a complex number and the limit is equal to f (z) whenever h 0. If we choose h r 0, then the limit is f x u x + iv x. If we choose h is 0, then f(x + iy + is) f(x + iy) u(x, y + s) + iv(x, y + s) u(x, y) iv(x, y) is is 1 ( ) u(x, y + s) u(x, y) v(x, y + s) v(x, y) + i, i s s so as s 0, the limit is 1 i (u y + iv y ) v y iu y. But the assumption is that no matter how the complex number h 0, the limit is always the same, and is equal to f (z). This shows that f (z) u x + iv x v y iu y. In other words, f (z) f x if y, which proves (a). This also shows that u x v y and u y v x, which is (b). Since ( 1 x i ) f 1 y (f x if y ) f x if y, it is equal to f (z), which gives the last part of (c). For (d), we use (b) to compute u xx + u yy u xx (v x ) y u xx (v y ) x u xx (u x ) x 0. Similarly, v xx + v yy 0. Part (c) of the above theorem explains the notation in complex variable: z 1 ( ) x i. y 31

32 3 Corollary If f : O C is analytic, then f is also analytic. Proof. Suppose f u + iv. Then u x v y and u y v x. We know that f u x +iv x. We claim that U u x and V v x satisfy the Cauchy-Rieman euqation. To see this, we compute U x u xx, V y v xy (v y ) x (u x ) x u xx, so U x V y. Also U y u xy and V x v xx ( u y ) x u xy, thus U y V x. Hence U, V satisfy the Cauchy-Rieman equation. By Theorem 11.4., f is analytic Harmonic conjugate of a harmonic function. Given a harmonic function u : O R, is there a harmonic function v : O R such that f(z) u(x, y) + iv(x, y) is an analytic function? We know that such v must satisfy the Cauchy-Rieman equation { v x u y, (11.5.1) v y u x. Since u is given, the right hand sides are given functions. By Frobenius Theorem, the compatibility condition that the above system is solvable is that ( u y ) y (u x ) x, i.e., u xx + u yy 0. Since u is harmonic, the compatibility condition is satisfied, so we can solve v. In fact, such v are uniquely determined up to a constant. Now let f(z) u(x, y) + iv(x, y). Because u, v satisfy the Cauchy-Rieman equation, f is analytic. We call such v a harmonic conjugate of u. Next we give more detail of construction of solution v of (11.5.1). Given P (x, y) and Q(x, y), the following initial value problem has a unique v(x, y), v x P v y Q, (11.5.) v(0, 0) c 0 if P y Q x. The solution v can be constructed using integration. Fix y, the first equation implies that v(x, y) c 0 + x 0 P (s, y) ds + g(y) for some g. But v y x x P (s, y) ds + g P (y) y y (s, y) dy + g (y) x 0 0 Q x (s, y) dx + g (y) Q(x, y) Q(0, y) + g (y), which is equal to Q if Q(0, y) + g (y) 0. So g (y) Q(0, y), hence g(y) y 0 Q(0, t) dt. In other words, we have proved: Theorem The solution for (11.5.) is v(x, y) c 0 + y 0 0 P (s, y) ds + Theorem If u is harmonic, then y 0 Q(0, t) dt.

33 33 (i) v(x, y) c u y (x, y) dx + u x (0, y) dy is a harmonic conjugate of u, (ii) f u + iv is an analytic function. Proof. A harmonic conjugate of u must satisfy the Cauchy-Rieman equation { v x u y, v y u x. Part (i) follows from Theorem , and (ii) follows from Theorem Example Let u(x, y) cosh y cos x. It is easy to check that u is harmonic. The harmonic conjugates of u is obtained by solving { v x u y sinh y cos x P, v y u x cosh y sin x Q. So Q(0, y) 0. By Theorem 11.5., v(x, y) sinh y cos x dx + 0dy c sinh y sin x is a harmonic conjugate of u. Note that u + iv cosh y cos x i sinh y sin x + ic, which is equal to cosh( iz) + ic because cosh( iz) cosh( i(x + iy)) cosh(y ix) ey ix + e (y ix) ey (cos x i sin x) + e y (cos x + i sin x) ey + e y cos x i ey e y sin x cosh y cos x i sinh y sin x. Hence cosh y cos x is the real part of the analytic function cosh( iz). Exercise (1) Prove that u(x, y) 1 ln(x + y ) is harmonic for x > 0, and find a harmonic conjugate for u. (The analytic function obtained this way is ln(z)). () Prove that u(x, y) x x3 3 +xy is harmonic, and find the harmonic conjugate v for u with v(0, 0) 0.