Section 3.5 The Implicit Function Theorem

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1 Section 3.5 The Implicit Function Theorem THEOREM 11 (Special Implicit Function Theorem): Suppose that F : R n+1 R has continuous partial derivatives. Denoting points in R n+1 by (x, z), where x R n and z R, assume that (x 0, z 0 ) satisfies F(x 0, z 0 ) 0 and F z (x 0, z 0 ) 0 Then there is a ball U containing x 0 in R n and a neighborhood V of z 0 in R such that there is a unique function z g(x) defined for x in U and z in V that satisfies F(x, g(x)) 0 Moreover, if x in U and z in V satisfy F(x, z) 0, then z g(x). Finally, z g(x) is continuously differentiable, with the derivative given by 1 Dg(x) D F x F(x, z) z (x, z) zg(x) where D x F denotes the (partial) derivative of F with respect to the variable x, that is, we have D x F [ F/ 1,, F/ n ]; in other words, F/ i, i 1,,n (1) i F/ z Proof (Idea): Once it is known that z g(x) exists and is differentiable, formula (1) may be checked by implicit differentiation; to see this, note that the chain rule applied to F(x, g(x)) 0 gives [ ] F D x F(x, g(x)) + (x, g(x)) [Dg(x)] 0 z which is equivalent to formula (1). If we restrict to the special case n 2, the Implicit Function Theorem gives us the following corollary. COROLLARY: Let F : R 3 R be a given function having continuous partial derivatives. Suppose that (x 0, y 0, z 0 ) is a point satisfying F(x 0, y 0, z 0 ) 0 and F z (x 0, y 0, z 0 ) 0 Then there is an open disk U R 2 containing (x 0, y 0 ) and an open interval V in R containing z 0 such that there is a unique function z g(x, y) defined for (x, y) U and z V that satisfies F(x, y, g(x, y)) 0 Moreover, z g(x, y) is differentiable with the derivatives given by F / F z and y F y / F z 1

2 EXAMPLE: Consider the equation In this case x 2 + z F(x, z) x 2 + z 2 1 with n 1 hence F (x, z) 2z z therefore the special implicit function theorem applies to a point (x 0, z 0 ) satisfying x z and z 0 0. Thus, near such points, z is a unique function of x such that 1 x 2 if z 0 > 0 z(x) 1 x 2 if z 0 < 0 Note that z is defined for x < 1 only and z is unique only if it is near z 0. These facts and the nonexistence of z/ at z 0 0 are clear from the fact that x 2 + z 2 1 defines a circle in the xz plane. The Implicit Function Theorem and Surfaces Let us apply Theorem 11 to the study of surfaces. We are concerned with the level set of a function g : U R n R, that is, with the surface S consisting of the set of x satisfying g(x) c 0, where c 0 g(x 0 ) and where x 0 is given. Assume that g(x 0 ) 0. By Theorem 11, there is a unique function z k(x, y) satisfying g(x, y, k(x, y)) c 0 for (x, y) near (x 0, y 0 ) and z near z 0. Thus, near z 0 the surface S is the graph of the function k. Because k is continuously differentiable, this surface has a tangent plane at (x 0, y 0, z 0 ) given by [ ] [ ] z z 0 + (x 0, y 0 ) (x x 0 ) + y (x 0, y 0 ) (y y 0 ) (2) But by formula (1), (x 0, y 0 ) (x 0, y 0, z 0 ) z (x 0, y 0, z 0 ) and 2 y (x y (x 0, y 0, z 0 ) 0, y 0 ) z (x 0, y 0, z 0 )

3 Substituting these two equations into the equation for the tangent plane, we obtain that is, 0 (z z 0 ) z (x 0, y 0, z 0 ) + (x x 0 ) (x 0, y 0, z 0 ) + (y y 0 ) y (x 0, y 0, z 0 ) (x x 0, y y 0, z z 0 ) g(x 0, y 0, z 0 ) 0 Thus, the tangent plane to the level surface of g is the orthogonal complement to g(x 0, y 0, z 0 ) through the point (x 0, y 0, z 0 ). EXAMPLE: Find an equation of the tangent plane to the surface given by x 2 + y 2 + z 2 1 at (x 0, y 0, z 0 ). Solution: Put g(x, y, z) x 2 + y 2 + z 2 1. Then 2x, therefore the equation of the tangent plane is y 2y, z 2z 0 (z z 0 ) 2z 0 + (x x 0 ) 2x 0 + (y y 0 ) 2y 0 or x 0 x + y 0 y + z 0 z x y2 0 + z2 0 General Implicit Function Theorem The general theorem deals with a system of equations in several variables that we must solve. What are the criteria for deciding when we can solve for some of the variables in terms of the others, or when such an implicit function can be found? Assume that we are given a system of m equations F 1 (x 1,,x n, z 1,,z m ) 0 F 2 (x 1,,x n, z 1,,z m ) 0 F m (x 1,,x n, z 1,,z m ) 0 (3) in n + m variables x 1,,x n, z 1,,z m. We want sufficient conditions so that we can solve for z 1,,z m in terms of x 1,,x n. In Theorem 11 we had the condition F/ z 0. Put F 1 F 1 z 1 z m.. F m z 1 THEOREM 12 (General Implicit Function Theorem): If 0, then near the point (x 0,z 0 ), system of equations (3) defines unique (smooth) functions F m z m z i k i (x 1,,x n ) (i 1,,m) Their derivatives may be computed by implicit differentiation. 3

4 EXAMPLE: Show that near the point (x, y, u, v) (1, 1, 1, 1), we can solve xu + yvu 2 2 xu 3 + y 2 v 4 2 uniquely for u and v as functions of x and y. Compute u/ at the point (1, 1). Solution: To check solvability, we form the equations and the determinant F 1 (x, y, u, v) xu + yvu 2 2 F 2 (x, y, u, v) xu 3 + y 2 v 4 2 F 1 u F 2 u F 1 v F 2 v x + 2yuv yu 2 3u 2 x 4y 2 v at (1, 1, 1, 1) at (1, 1, 1, 1) Because 0, solvability is assured by Theorem 12. To find u/, we implicitly differentiate the given equations in x: x u + u + y v u2 + 2yvu u 0 3xu 2 u + u3 + 4y 2 v 3 v 0 Setting (x, y, u, v) (1, 1, 1, 1) gives 3 u + v 1 Solving for u/, we get u/ 1/3. 3 u + 4 v 1 Inverse Function Theorem A special case of the general implicit function theorem is the inverse function theorem. Here we attempt to solve the n equations f 1 (x 1,, x n ) y 1 (4) f n (x 1,,x n ) y n 4

5 for x 1,,x n as functions of y 1,,y n ; that is, we are trying to invert the equations in this system. The question of solvability is answered by the general implicit function theorem applied to the functions y i f i (x 1,,x n ). The condition for solvability in a neighborhood of a point x 0 is that the determinant of the matrix Df(x 0 ) is nonzero. This determinant is called the Jacobian determinant of f and is (f 1,,f n ) (x 1,,x n ) J(f)(x 0 ) xx0 1 (x 0 ). f n (x 0 ) 1 (x 0 ) n. f n (x 0 ) n THEOREM 13 (Inverse Function Theorem): Let U R n be open and let f i : U R, i 1,, n, have continuous partial derivatives. Consider the equations (4) near a given solution x 0,y 0. If the Jacobian determinant, J(f)(x 0 ) 0, then (4) can be solved uniquely as x g(y) for x near x 0 and y near y 0. Moreover, the function g has continuous partial derivatives. EXAMPLE: Consider the equations x 4 + y 4 Near which points (x, y) can we solve for x, y in terms of u, v? Solution: Put then (f 1, f 2 ) (x, y) x u, sin x + cos y v (5) u f 1 (x, y) x4 + y 4, v f 2 (x, y) sin x + cosy x f 2 y f 2 y 3x 4 y 4 4y 3 x 2 x cosx sin x Therefore, we can solve (5) for x, y near those x, y for which x 0 and (sin y)(y 4 3x 4 ) 4xy 3 cos x sin y x 2 (y4 3x 4 ) 4y3 x cosx Such conditions generally cannot be solved explicitly. For example, if x 0 π/2, y 0 π/2, we can solve for x, y near (x 0, y 0 ) because there, (f 1, f 2 )/ (x, y) 0. 5