Transformations from R m to R n.

Transcription

1 Transformations from R m to R n 1 Differentiablity First of all because of an unfortunate combination of traditions (the fact that we read from left to right and the way we define matrix multiplication it turns out that although it is more convenient to write vectors in R k as rows like (v 1 v 2 v k when we work with them in the context of mappings it often turns out to be more convenient to think of them as columns So a linear map L : R m R n is described by an m n matrix A L and is given by x A L x where x is a column However we don t want to write y 1 y 2 y n = L x 1 x 2 x m but rather (y 1 y 2 y n = L(x 1 x 2 x m On the other hand we use the matrix product y 1 y 2 y n = A L In what follows we will allow ourselves to write vectors both as rows and as columns depending on convenience For the standard unit vectors in the plane we use the notation i = (1 0 and j = (0 1 In R k we use the notation e j = ( where the 1 is the j th entry x 1 x 2 x m Definition of directional derivative Let U be an open set in R 2 and F : U R Suppose u = (u 1 u 2 = u 1 i + u 2 j is a unit vector We say that F is differentiable in the direction u at the point x = (x y if g(t := F (x + tu 1 y + tu 2 is differentiable with respect to t at t = 0 We use the notation D u F (x = g (0 Notation : D i F (x = F x (x F (x x D F (x jf (x = F y (x y DF F := (F x F y grad F F + F y j Note: D u F (x = u F (x Note: Differentiability of F is a stronger notion than differentiability of F in all directions and will be introduced below 1

2 We can extend these notions to transformations F : U R 2 R 2 simply be extending them to each component of F Here F (x y = (F 1 (x y F 2 (x y But now it is helpful to write F as a column: ( ( ( ( F1 (x Du F F (x = D F 2 (x u F (x = 1 (x DF1 (x F1 / x F DF (x = = 1 / y D u F 2 (x DF 2 (x F 2 / x F 2 / y etc The above notions extend of course to maps F : U R n R m : Notation in R m : F (x 1 x 2 x n (F 1 (x 1 x 2 x n F 2 (x 1 x 2 x n F m (x 1 x 2 x n F 1 / F (x D ei F (x F xi (x F 2 / F m / DF (x F (x F 1 F 2 F n ( F i/ x j Notation In pure mathematics it is not customary to use F or grad F unless F is scalar valued ie unless the codomain is simply R If F is not scalar valued we usually use D rather than We note that DF = ( F i / x j an m n matrix called the Jacobian matrix If m = n we can define the determinant DF This scalar value is called the Jacobian of the transformation at the point x Definition Let x = (x 1 x 2 x k R k then we define its norm as x := x x x2 k Definition of derivative: Let U be an open subset of R n and let F : U R m be a transformation x 0 U We say that F is differentiable at x 0 if there exists an m n matrix L and a function φ : R n R m such that φ(h F (x 0 + h = F (x 0 + Lh + φ(h and lim = 0 h 0 h We often write the above information more concisely as F (x 0 + h = F (x 0 + Lh + Note that we think of x and h as columns Also note that the above equation defines φ: φ(h := F (x 0 + h F (x 0 Lh 2

3 If F is differentiable everywhere in its domain then we simply say that F is differentiable Theorem If F is differentiable at x 0 then L = DF (x 0 Moreover the directional derivative at x 0 of F in the direction of the unit (column vector u is given by D u F (x 0 = DF (x 0 u The matrix DF (x 0 is called the derivative of F at x 0 Notation: If F : U R n R m is differentiable everywhere in U then we say that F is differentiable If F is differentiable and the map DF : U R m n is continuous then we say that F is continuously differentiable Differentials What we used h for in the above definition of the derivative is often written as dx and then we define dy := L dx = DF (x 0 dx Then we have that differentiability at x 0 means that y := F (x 0 + dx F (x 0 = dy + φ(dx where φ(dx / dx 0 as dx 0 Hence for small dx we have the approximation F (x 0 + dx F (x 0 + dy = F (x 0 + DF (x 0 dx Note that if F is differentiable at x 0 then F i (x 0 + dx = F i (x 0 + n j=1 F i x j (x 0 dx j + Theorem If the partial derivatives F i / x j all exist and are continuous in a neighborhood of x 0 then F is differentiable at x 0 Example Define F : R 2 R as F (x y := xy Its partial derivatives exist at the origin and are zero there So if DF exists it should be the zero transformation and hence φ(x y = F (x y Let h = (t t then φ(h / h = 1 and hence does not tend to zero as h 0 So even though the partial derivatives exist at the origin this function is not differentiable at the origin We also see that at points not on the axes F/ x = y sign(x and F/ y = x sign(y which are not continuous in any neighborhood of the origin If the functions F i (x/ x j are all continuous in a region U then clearly F is continuously differentiable in U The set of all continuously differentiable functions from D R n to R m is denoted by C 1 (D R m If m = 1 we simplify the notation to C 1 (D Theorem If F : U R n and for each i the partial derivative F xi exists and is continuous in a neighborhood of x 0 then F is differentiable in all directions and D u F (x = F (x u = DF (xu in that neighborhood 3

4 It is now easy to show that if F and G are both differentiable at x 0 then so is αf + βg for any scalars α and β We also have the Chain Rule If F : U R n R m G : V R k R n with G(V U and if G is differentiable at p and F is differentiable at G(p then F G is differentiable at p and D(F G(p = DF (G(pDG(p (the derivative of the composition of F and G is the matrix product of the derivatives of F and G Example w = f(x 1 x 2 x n with x(t = φ(t ie x i (t = φ i (t for 1 i n (f φ (t = f(φ(t φ (t = n f (φ(tφ i(y Example w = F(x That is: w j = F j (x 1 x 2 x n j = 1 2 m with x(t = φ(t ie x i (t = φ i (t for 1 i n (F φ (t = F(φ(t φ (t or (F j φ (t = n F j (φ(tφ i(t Note that this is a matrix multiplication: D(F φ(t = DF(φ(tDφ(t or using the prime notation for derivatives with respect to one variable (F φ (t = DF(φ(tφ (t Example w = F(x where w j = F j (x 1 x 2 with x(u v = φ(u 1 u 2 ie x i (t = φ i (u 1 u 2 or (F j φ u k (u 1 u 2 = D(F φ(u 1 u 2 = DF(φ(u 1 u 2 Dφ(u 1 u 2 2 F j (φ(u 1 u 2 φ i(u 1 u 2 u k = 2 F j (φ(u 1 u 2 (u 1 u 2 u k and similarly for the partial with respect to v Note again the matrix multiplication: F 1 / u 1 F 1 / u 2 F 1 / x 1 F 1 / x 2 F 2 / u 1 F 2 / u 2 = F 2 / x 1 F 1 2/ x 2 ( x1 / u 1 x 1 / u 2 x 2 / u 1 x 2 / u 2 F n / u 1 F n / u 2 F n / x 1 F n / x 2 2 The Mean Value Theorem Recall that for a differentiable function of one variable on the real line we can find between any two points a and b in the domain a point c such that f(b f(a = f (c(b a For functions of several variables the situation can be reduced to one-dimensional: Suppose that f : D R n R is differentiable in the open set D and suppose that a and b are two points in D such that the line segment ab D We define the function g(t := f((1 ta + tb This is a differentiable function on [0 1] and hence there is a τ [0 1] such that g(1 g(0 = g (τ Let p := (1 τa + τb This is a point on the segment ab and we have via the chain rule: 4

5 Mean Value Theorem Suppose that f : D R n R is differentiable in the open set D and let a and b be two points in D such that the line segment ab D Then there is a point p D such that f(b f(a = f(p (b a Corollary Suppose f : D R n R is differentiable in the open set D For any two points a and b in D such that the line segment ab D we have f(b f(a K b a where K := sup{ f(x : x ab} Theorem Suppose f : D R n R m is differentiable in the open set D For any two points a and b in D such that the line segment ab D we have f(b f(a K b a where K := (K 1 K 2 K m with K i = sup{ f i (x : x ab} NOTE The Mean Value Theorem is not true for vector-valued functions If f : D R m m > 1 then it is not generally true that we can find a point p such that f(b f(a = Df(p(b a Exercise Suppose D is a convex compact region in R n and f : D R m is continuously differentiable in D Let M := max max sup f i / x j 1 i m 1 j n Show that for any two points a and b in D we have f(b f(a M mn b a D 3 Functions from C to C Consider the function f : U C C where w = f(z = f(x + iy = u(x y + iv(x y and where u v x y are all real Then f induces a map f : U R 2 R 2 where U := {(x y : x + iy U}: ( ( x u(x y f : y v(x y Let C denote all real 2 2 matrices of the form ( a b M(a b := b a 5

6 and let U := {M(x y : x + iy U} Define Ψ : C C by Ψ(x + iy = M(x y We easily see that Ψ is a field isomorphism Ψ(z + w = Ψ(z + Ψ(w Ψ(zw = Ψ(zΨ(w and if z 0 then Ψ(z 1 = Ψ(z 1 So f also induces a map from U C to C namely the map f := Ψ f Ψ 1 : ( x y f : y x ( u(x y v(x y v(x y u(x y 4 Functions from C to C - two kinds of derivatives Consider the function f : U C C where we write f(x + iy = u(x y + iv(x y As we saw this is essentially a function f f (x y = (u(x y v(x y from a subset U of R 2 to R 2 So it would make sense to say that f is differentiable if f is differentiable in the usual sense of differentiable maps ie However suppose that we have that u(x + dx y + dy = u(x y + u x (x y dx + u(x y dy + v(x + dx y + dy = v(x y + v x (x y dx + v(x y dy + u x v y and u y = v x in U then it is easily verified (Exercise that ( u(x + dx y + dy v(x + dx y + dy v(x + dx y + dy u(x + dx y + dy = ( u(x y v(x y v(x y u(x y ( α β + β α ( dx dy dy dx + where α = u x (x y = v y (x y and β = u y (x y = v x (x y representation of the equation However this is just the matrix f(z + dz = f(z + f (zdz + where dz = dx + idy and f = u x + iv x = v y iu y For functions of a complex variable we will always use this definition of differentiability: Definition Let f : U C C be a function of a complex variable We will say that f is differentiable at z if there exists a complex number denoted by f (z such that f(z + dz = f(z + f (z dz + More explicitly this means that f(z + dz = f(z + f (z dz + φ(dz where or equivalently that φ(dz lim = 0 dz 0 dz f(z + dz f(z lim = f (z dz 0 dz 6

7 Theorem Let f : U C C be a function of a complex variable z = x + iy where f(z = u(x y + iv(x y If this function is differentiable at the point z then u and v satisfy the Cauchy- Riemann equations u x (x y = v y (x y u y (x y = v x (x y at that point Conversely if u and v are continuously differentiable and satisfy the Cauchy-Riemann equations then f is differentiable on U and its derivative f (z is continuous on U Proof Suppose that f is differentiable then f(z + dx f(z lim = f (z = lim dx 0 dx dy 0 f(z + idy f(z idy which implies that u x + iv x = iu y + v y Equating real parts and equating imaginary parts yields the Cauchy-Riemann equations The converse was proved above 7