The Inverse Function Theorem (11.6)

Transcription

1 The Inverse Function Theorem 116 Recall the inverse function theorem on R 1 : if f is differentiable at a f a 0 then there is some open interval U around a and an open interval V around f a so that f maps U onto V in a one-to-one fashion Moreover the derivative of f 1 at f a exists with d f 1 f a = 1 dx f a picture We will not prove the following: Theorem: Suppose that f : V R n R n is differentiable Suppose that for somea V Df a is invertible ie its determinant the Jacobian is non-zero Then there is some open set U containing a and an open set V containing f a so that f maps U one-to-one onto V Moreover the inverse is differentiable anddf 1 f a = Df a 1 I did prove it a few years ago in 410: it was 26 pages of lecture notes Example polar coordinates: Consider the map from R 2 to R 2 given by r f = θ rcosθ rsinθ Here we think of r and θ as coordinates in a Cartesian r θ plane Pictures Vertical lines get mapped to circles centered at the origin horizontal lines get mapped to rays through the origin Notice that f is not 1-1: increasing the θ coordinate by 2π gets you to the same place Also the entire θ axis gets mapped to the origin Let s find the Jacobian First so the Jacobian is cosθ rsinθ Df = sinθ rcosθ f = xy rθ = r where the first symbol is the book s notation for Jacobian the second is the standard notation assuming that x = rcosθ y = rsinθ What 105

2 this says is that at any point off of the θ axis there is a local inverse whose Jacobian will be 1 r Let s look at the point r = 2 θ = π This gets mapped to the 4 point x = 1 y = 1 The matrix of the total derivative at r = 2 θ = π is 4 1/ 2 1 1/ 2 1 By the inverse function theorem there is a local inverse whose Jacobian at the point x = 1 y = 1 should be 1 2 In fact we know that inverse explicitly: it s r = x 2 +y 2 θ = arctan y x The matrix of the total derivative of the inverse is x/ x2 +y 2 y/ x 2 +y 2 At x = 1 y = 1 this is y/x 2 +y 2 x/x 2 +y 2 1/ 2 1/ 2 1/2 1/2 It s easy to verify that this is the inverse of the derivative of the original transformation Notice that r = 2 θ = 9π gets mapped to x = 1 y = 1 as well 4 There is a local inverse of this map as well: it is r = x 2 +y 2 θ = arctan y x +2π Another point mapped to x = 1 y = 1 is r = 2 θ = 5π The local inverse here is r = x 4 2 +y 2 θ = arctan y x +π pictures Of course in general you can t explicitly solve for the inverse Example: Suppose f :R 2 R 2 call the first space the uv plane and the second one the xy plane is given by u u f = 3 +uv +v 3 v u 2 v 2 Notice that f 11 = 30 Prove that there is an inverse of f defined in a neighborhood of 30 find its total derivative at 30 and approximate f Solution: The Jacobian matrix of f is 3u 2 +v u+3v 2 2u which is c 2012 T Vogel 106 2v

3 at 11 This is non-singular the determinant is 16 0 so f has an inverse in a neighborhood of 30 To find Df 1 at that point we ve got to invert Df In general to invert a square matrix A write the augmented matrix A I where I is the identity matrix of the same size then use Gauss-Jordan elimination to get A to I This will turn I to A 1 For 2 by 2 matrices a b there s a short-cut: if is invertible then c d 1 a b 1 d b = c d ad bc c a Thus = /8 1/4 = 1/8 1/4 and this is the matrix of the total derivative of the inverse map Now for the approximation part: so f f 1 x+h f 1 x+df 1 xh 3 1/8 1/4 f /8 1/4 1 3/ / As a check f 77/8085/80 = so 77/8085/80 is pretty close to f Implicit function theorem Recall implicit differentiation from elementary calculus: the standard question was something like Example: Suppose f xy = 0 determines y implicitly as a function of x Find dy The picture was f xy = 0 was a curve in the x y dx plane and if it didn t fold over at a point then y could be written as a function of x near that point In general although you didn t know it at the time write y as an unknown function of x differentiate using the chain rule and solve for dy : dx f x xyx+f y xyx dy dx = 0 c 2012 T Vogel 107

4 so dy dx = f xxy f y xy The not folding over part is that f y shouldn t be zero So how will this generalize? Certainly we can hope that f xyz = 0 determines z implicitly as a differentiable function of x and y Another way this might go up to three dimensions is the systemf 1 xyz = 0 f 2 xyz = 0 might conceivably determine y and z as functions of x The gist of the Implicit Function Theorem is this: if your functions are differentiable and if you can solve for the partials when you differentiate implicitly without dividing by zero then you do have the variables defined implicitly as functions of the other variables It s easier to give an example then to try to state it carefully Example: Prove that the system w 2 +2x 2 +y 2 z 2 = 6 wxy xyz = 1 determines w and x implicitly as differentiable functions of y and z in a neighborhood of the point 1121 in wxyz space Find the valuesof the partials of w and x with respect to y and z at that point Solution: Treat y and z as independent variables w and x as dependent variables Differentiate the system implicitly with respect to z: or 2w 4x xy wy yz 2ww z +4xx z 2z = 0 w z xy +wx z y x z yz xy = 0 wz = x z 2z xy We can solve for w z and x z if the matrix is invertible At 1121 the matrix is which certainly is invertible Solve the system 2w z +4x z = 2 2w z = 2 to get w z = 1 x z = 0 Now differentiate the original system with respect to y: 2ww y +4xx y +2y = 0 w y xy +wx y y +wx x y yz xz = 0 c 2012 T Vogel 108

5 or 2w 4x xy wy yz wy = x y 2y xz wx The two by two matrix is the same no coincidence so this is solvable At 1121 the system is 2w y +4x y = 4 2w y = 0 which has the solution w y = 0 x y = 1 The implicit function theorem is essentially that if the above implicit differentiation procedure makes sense ie you aren t dividing by zero and the number of dependent and independent variables makes sense then there are differentiable functions determined implicitly by the given system Example: What conditions on the point x 0 y 0 z 0 will ensure that there is a pair of differentiable functions xz yz defined on some interval around z 0 so that xz 0 = x 0 yz 0 = y 0 and xz yz solve this system: x 2 +zy 2z 2 = 1 xyz = 4? Solution: Consider x and y as functions of z and differentiate the system with respect to z: so we have 2xx z +y +zy z 4z = 0 2x z yz xz x z yz +xy z z +xy = 0 xz = y z 4z y xy So the additional condition is that the 2 by 2 matrix be non-singular ie that 2x 2 0z 0 y 0 z Any x 0 y 0 z 0 which satisfies the system and doesn t make the determinant zero will yield such a pair of differentiable functions As an example I used Maple to numerically find a point on both surfaces: Since this doesn t make the third condition 0 this does determine x and y as differentiable functions of z In fact when I plotted both surfaces their intersection is a smooth curve going through the point I found Moreover it doesn t have a maximum or minimum z value at that point so we can write x and y as functions of z c 2012 T Vogel 109

6 Slightly more general example: Let s look at f 1 x 1 x 2 t 1 t 2 = 0 f 2 x 1 x 2 t 1 t 2 = 0 Differentiate everything implicitly with respect to t 1 using the chain rule where t 2 is the other independent variable: f 1 x1 +f 1 x2 x 2 +f 1 t1 = 0 f 2 x1 +f 2 x2 x 2 +f 2 t1 = 0 If we re going to solve this for and x 2 the condition is that f1 det x1 f 1 x2 0 f 2 x1 f 2 x2 This is the same condition as for finding t 2 and x 2 t 2 which is of course what the implicit function theorem says So this is pretty much what s going on in general General statement of Implicit Function Theorem: If we have n equations in n+p unknowns we might hope that if we specify p of them then the n equations in the n remaining unknowns can be solved This would say that the n unknowns are functions of the p unknowns and we can hope that this dependence is differentiable More formally let F : R n+p R n and write a vector in R n+p as F 1 x where x R t n and t R p Suppose that F = is C 1 F n x0 x0 in a neighborhood of a point which satisfies F = 0 t 0 t 0 Define the partial Jacobian F 1 F n x n If F 1 F n x0 x n t 0 to be the determinant of F 1 F 1 x n F n x n F n 0 then there is a neighborhood W of t 0 and a gt0 = 0 t 0 function g :W R n so that gt 0 = x 0 and F c 2012 T Vogel 110

7 w y In the first example x was and t was F was the x z map from R 4 to R 2 given by w F x w y = 2 +2x 2 +y 2 z 2 6 wxy xyz 1 z and the condition was that the determinant of w w2 +2x 2 +y 2 z 2 6 x w2 +2x 2 +y 2 z 2 6 wxy xyz 1 wxy xyz 1 w not be zero which is precisely what we came up with before The condition in the slightly more general example was exactly the one that the implicit function theorem comes up with Equivalence of Implicit and Inverse Function Theorems: The implicit function theorem implies the inverse function theorem Consider f : R 3 R 3 for specificity Saying that f has an inverse is precisely saying that the system x f 1 x 1 x 2 x 3 = u 1 f 2 x 1 x 2 x 3 = u 2 f 3 x 1 x 2 x 3 = u 3 determines the values of the x i s if you re given the u i s But this is a system of three equations in six unknowns and one can apply the implicit function theorem to answer this question Of course the result is precisely the inverse function theorem that one can do this if Df is invertible On the other hand the inverse function theorem implies the implicit function theorem which is how the book proves the implicit function theorem For example suppose we wonder whether the system f 1 x 1 x 2 t = 0 f 2 x 1 x 2 t = 0 determines x 1 and x 2 as differentiable functions of t Consider the map F from R 3 to R 3 given by F x 1 x 2 = f 1x 1 x 2 t f 2 x 1 x 2 t t t c 2012 T Vogel 111

8 If F x 1 x 2 = u 1 u 2 t u 3 has an inverse G then we can write the inverse as x 1 = g 1 u 1 u 2 u 3 x 2 = g 2 u 1 u 2 u 3 t = g 3 u 1 u 2 u 3 But look at the map F It s saying that t = u 3 so g 3 u 1 u 2 u 3 is just u 3 To get the f i s equal to 0 just plug u 1 = u 2 = 0 into g 1 and g 2 Since t = u 3 we get x 1 and x 2 determined as differentiable functions of t The condition that F be invertible is that the 3 by 3 determinant of partials have non-zero determinant But f 1 f 1 f 1 f 2 x 2 f 2 x 3 f 2 x 2 x 3 DF = so the condition is what we d expect that f 1f 2 x 2 0 c 2012 T Vogel 112