Section 5: The Jacobian matrix and applications. S1: Motivation S2: Jacobian matrix + differentiability S3: The chain rule S4: Inverse functions

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1 Section 5: The Jacobian matri and applications. S1: Motivation S2: Jacobian matri + differentiabilit S3: The chain rule S4: Inverse functions Images from Thomas calculus b Thomas, Wier, Hass & Giordano, 2008, Pearson Education, Inc. 1

2 S1: Motivation. Our main aim of this section is to consider general functions and to define a general derivative and to look at its properties. In fact, we have slowl been doing this. We first considered vector valued functions of one variable f : R R n and defined the derivative as f(t) (f 1 (t),..., f n (t)) f (t) (f 1 (t),..., f n(t)). We then considered real valued functions of two and three variables f : R 2 R, f : R 3 R and (as we will see later) we ma think of the derivatives of these functions, respectivel, as f ( f/, f/ ) f ( f/, f/, f/ z). 2

3 There are still more general functions than those two or three tpes above. If we combine the elements of each, then we can form vector valued functions of man variables. A function f : R m R n (n > 1) is a vector valued function of m variables. Eample 1 f z ( + + z z ) defines a function from R 3 to R 2. 3

4 When it comes to these vector valued functions, we should write vectors as column vectors (essentiall because matrices act on column vectors), however, we will use both vertical columns and horizontal m tuple notation. Thus, for eample, for the vector R 3 we will write both or (,, z) (and i + j + zk) z and so we could write f : R 3 R 2 as f1 (,, z) f and f z 2 (,, z) f(,, z) (f 1 (,, z), f 2 (,, z)) f 1 (,, z)i + f 2 (,, z)j or combinations of columns and m-tuples. 4

5 In Eample 1, the real valued functions f 1 f z and z z z are called the co ordinate or component functions of f, and we ma write f1 f. f 2 Generall, an f : R m R n is determined b n co ordinate functions f 1,..., f n and we write f f 1 ( 1,..., m ) f 2 ( 1,..., m ).. (1) f n ( 1,..., m ) 5

6 We shall be most interested in the cases where f : R 2 R 2 or f : R 3 R 3 because this is where the most applications occur and because it will prove to be etremel useful in our topic on multiple integration. For these special cases we can use the following notation f() f(, ) (f 1 (, ), f 2 (, )) f 1 (, )i + f 2 (, )j. f() f(,, z) (f 1 (,, z), f 2 (,, z), f 3 (,, z)) f 1 (,, z)i + f 2 (,, z)j + f 3 (,, z)k. 6

7 One wa of visualizing f, sa, f : R 2 R 2 is to think of f as a transformation between co ordinate planes. So that f ma stretch, compress, rotate etc sets in its domain. The above be particularl useful when dealing with multiple integration and change of variables. 7

8 S2: Jacobian matri + differentiabilit. Our first problem is how we define the derivative of a vector valued function of man variables. Recall that if f : R 2 R then we can form the directional derivative, i.e., D u f u 1 f + u 2 f f u where u (u 1, u 2 ) is a unit vector. Thus, knowledge of the gradient of f gives information about all directional derivatives. Therefore it is reasonable to assume p f ( f f (p), (p) is the derivative of f at p. (The stor is more complicated than this but when we sa f is differentiable we mean f represents the derivative, to be discussed a little later.) ) 8

9 More generall if f : R m R then we take the derivative at p to be the row vector ( f (p), f ) f (p),..., (p) p f 1 2 m Now take f : R m R n where f is as in equation (1), then the natural candidate for the derivative of f at p is f 1 f f 1 m f 2 f 2 f J p f m f n f n f n m where the partial derivatives are evaluated at p. This n m matri is called the Jacobian matri of f. Writing the function f as a column helps us to get the rows and columns of the Jacobian matri the right wa round. Note the Jacobian is usuall the determinant of this matri when the matri is square, i.e., when m n. 9

10 Eample 2 Find the Jacobian matri of f from Eample 1 and evaluate it at (1, 2, 3). 10

11 Most of the cases we will be looking at have m n either 2 or 3. Suppose u u(, ) and v v(, ). If we define f : R 2 R 2 b f then the Jacobian matri is Jf and the Jacobian (determinant) det(jf) u v We often denote det(jf) b u(, ) v(, ) u v u v (u, v) u v u f1 f 2 v v u. (, ). 11

12 Eample 3 Consider the transformation from polar to Cartesian co ordinates, where r cos θ and r sin θ. We have (, ) (r, θ) r r θ θ cos θ sin θ r sin θ r cos θ r. 12

13 We have alread noted that if f : R m R n then the Jacobian matri at each point a R m is an m n matri. Such a matri J a f gives us a linear map D a f : R m R n defined b (D a f) () : J a f for all R n. Note that is a column vector. When we sa f : R m R n is differentiable at q we mean that, the affine function A() : f(q) + ( J q f ) ( q), is a good approimation to f() near q in the sense that where lim q q f() f(q) (J q f) ( q) q 0 ( 1 q 1 ) ( m q m ) 2. 13

14 You should compare this to the one variable case: a function f : R R f(a + h) f(a) is differentiable at a if lim eists, and we call this h 0 h limit f (a). But we could equall well sa this as f : R R is differentiable at a if there is a number, written f (a), for which lim h 0 f(a + h) f(a) f (a) h h 0, because a linear map L : R R can onl operate b multiplication with a number. How do we easil recognize a differentiable function? If all of the component functions of the Jacobian matri of f are continuous, then f is differentiable. 14

15 Eample 4 Write the derivative of the function in Eample 1 at (1, 2, 3) as a linear map. Suppose f and g are two differentiable functions from R m to R n. It is eas to see that the derivative of f +g is the sum of the derivatives of f and g. We can take the dot product of f and g and get a function from R m to R, and then differentiate that. The result is a sort of product rule, but I ll leave ou to work out what happens. Since we cannot divide vectors, there cannot be a quotient rule, so of the standard differentiation rules, that leaves the chain rule. 15

16 S3: The chain rule. Now suppose that g : R m R s and f : R s R n. We can now form the composition f g b mapping with g first and then following with f: g() f(g()) (2) (f g) () : f (g()) for all R m. Eample 5 Let g : R 2 R 2 and f : R 2 R 3 be defined, respectivel, b g : + Then f g is defined b (f g) f ( g ) and f f + : sin. sin( + ) + ( + ) (). 16

17 Let b g(p) R s. If f and g in (2) above are differentiable then the maps J p g : R m R s and J b f : R s R n are defined, and we have the following general result. Theorem 1 (The Chain Rule) Suppose that g : R m R s and f : R s R n are differentiable. Then J p (f g) J g(p) f J p g. This is again just like the one variable case, ecept now we are multipling matrices (see below). 17

18 Eample 6 Consider Eample 5: g + and f sin. Find J p (f g) where p ( a1 Also J p g J g(p) f ) a 2 ) ( 1 1. We have p cos ( 1 1 a 2 a 1 ) a 1 +a 2,a 1 a 2 cos(a 1 + a 2 ) a 1 a 2 a 1 + a

19 (E cont.) and We observe that J p (f g) cos( + ) cos( + ) cos(a 1 + a 2 ) cos(a 1 + a 2 ) 1 a 2 2a 1 a 2 + a a 1 a a 1 a 2 cos(a 1 + a 2 ) a 1 a 2 a 1 + a 2 ( 1 1 a 2 a 1 ) p 19

20 The one variable chain rule is a special case of the chain rule that we ve just met the same can be said for the chain rules we saw in earlier sections. Let : R R be a differentiable function of t and and let u : R R a differentiable function of. Then (u ) : R R is given b (u )(t) u((t)). In the notation of this chapter i.e. J t (u ) J (t) u J t ] [ ] [ du d (u ) dt d dt [ d t (t) ] t. We usuall write this as du dt du d d dt keeping in mind that when we write du we are thinking of u as a dt function of t, i.e., u((t)) and when we write du we are thinking of u d as a function of. 20

21 Now suppose we have (t), (t) and z f(, ). Then J t (f ) J (t) f J t Therefore so that d dt (f((t), (t))) df dt f ( f d dt + f which is just what we saw in earlier sections. f d dt, ) d dt d dt 21

22 S4: Inverse functions. In first ear (or earlier) ou will have met the inverse function theorem, which sas essentiall that if f (a) is not zero, then there is a differentiable inverse function f 1 defined near f(a) with [ d dt (f 1 ) ] f(a) 1 f (a). What happens in the multi variable case? 22

23 Let us consider a case where we can write down the inverse. For polar coordinates we have r cos θ, r 2 + 2, Now differentiating we obtain r r cos θ r r sin θ θ arctan cos θ and. r cos θ r i.e., 1. r We see that the one variable inverse function theorem does not appl to partial derivatives. However, there is a simple generalisation if we use the multivariable derivative, that is, the Jacobian matri. 23

24 To continue with the polar coordinate eample, define r (r, θ) r cos θ f θ (r, θ) r sin θ (3) and Consider (f g) g f ( r(, ) θ(, ) g ) f arctan. (4) r θ Id. Therefore f g Id, the identit operator on R 2. Similarl g f Id. Recall ( Id ) so that J(Id) identit matri. 24

25 Thus b the chain rule Jf Jg J(Id) Jg Jf so that (Jf) 1 Jg. Note for simplicit the points of evaluation have been left out. Therefore r θ r θ 1 r r θ We can check this directl b substituting r etc. θ cos θ The same idea works in general: 25

26 Theorem 2 (The Inverse Function Theorem) Let f : R n R n be differentiable at p. If J p f is an invertible matri then there is an inverse function f 1 : R n R n defined in some neighbourhood of b f(p) and (J b f 1 ) (J p f) 1. Note that the inverse function ma onl eist in a small region around b f(p). Eample 7 We earlier saw that for polar coordinates, with the notation of equation (3) cos θ r sin θ Jf, sin θ r cos θ with determinant r. So it follows from the inverse function theorem that the inverse function g is differentiable if r 0. 26

27 Eample 8 The function f : R 2 R 2 is given b ( u f 2 2 ) v Where is f invertible? invertible. ( 2 2 SOLN: Jf 2 2 everwhere ecept the aes. Find the Jacobian matri of f 1 where f is 1 Jf ( ) and det Jf 8, so f is invertible ) 1 4 Translate to (u, v) coordinates and this is ( Jf 1 2 (u + v) 1/2 (u + v) 1/2 4 (v u) 1/2 (v u) 1/2 ( ). 1 ). 27

28 Finall let us appl the inverse function theorem to the Jacobian determinants. We recall that (r, θ) (, ) (, ) (r, θ) det Jg det Jf r θ r r r θ θ. θ and Since Jg and Jf are inverse matrices, their determinants are inverses: (r, θ) (, ) 1 (,) (r,θ) This sort of result is true for an change of variable in an number of dimensions and will prove ver useful in integration.. 28