Advanced Calculus (Math 4305)
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Chapter 3
The Implicit Function Theorem and Its
Applications
In this chapter we deal with the general question of the local solvability of system of equations
involving nonlinear functions. The main result is the implicit function theorem, one of the major
theoretical results of advanced calculus. Among other things, it provides the key to answering many
questions about relations between analytic properties of functions and geometric properties of the
sets they define. We shall present some of its applications to the study geometric transformations,
coordinate systems, and various ways of representing curves, surfaces, and smooth sets of higher
dimension.
3.1 The Implicit Function Theorem
In this section we consider the problem of solving an equation F (x1 , . . . , xn ) = 0 for one of the
variables xj as a function of the remaining n − 1 variables, or more generally of solving a system of
k such equations for k of the variables as functions of the remaining n − k variables.
We begin with the case of a single equation, and develop some feeling for the geometry of the
problem we consider the cases n = 2 and n = 3. For n = 2 we are given an equation F (x, y) = 0
relating the variables x and y, and we ask when we can solve for y as a function of x or vice versa.
Geometrically, the set S = {(x, y) : F (x, y) = 0} will usually be some sort of curve, and our question
is: When can S represented as the graph of a function y = f (x) or x = g(y)? Likewise, for n = 3,
the set F (x, y, z) = 0 will usually be a surface, and we ask when this surface can represented as the
graph of a function z = f (x, y), y = g(x, z), or x = h(y, z).
Simple examples show that it is usually impossible to represent the whole set S = {x ∈ Rn : F (x) = 0}
as the graph of a function. Thus, in order to get reasonable result, we must be content only to represent pieces of S as graphs. More specifically, our object will be to represent a piece of S in the
neighborhood of a given point a ∈ S as a graph.
Since we want to single out one of the variables as the one to be solved for, we make a little
change of notation: We denote the number of variables by n + 1 and denote the last variable by y
rather than xn+1 . We then have the following precise analytical statement of the problem:
Given a function F (x, y) of class C 1 an a point (a, b) satisfying F (a, b) = 0, when is there
(i) a function f (x), defined in some open set in Rn containing a, and
(ii) an open set U ⊆ Rn+1 containing (a, b), such that for (x, y) ∈ U ,
F (x, y) = 0 ⇐⇒ y = f (x)?
1
We do not try to specify in advance how big the open sets in question will be; that will depend
strongly on the natural of the function F .
The key to answer the question is to look at the linear case. If
L(x1 , . . . , xn , y) = α1 x1 + · · · + αn xn + βy + c,
the solution is obvious: The equation L(x, y) = 0 can be solved for y if and only if the coefficient β is
nonzero. But near a given point (a, b), every differentiable function F (x, y) is approximately linear;
in fact, if F (a, b) = 0,
F (x, y) = ∂1 f (a, b)(x1 − a1 ) + · · · + ∂n f (a, b)(xn − an ) + ∂y f (a, b)(y − b) + small error.
If the ”small error” were not there, the equation F (x, y) = 0 could be solved for y precisely when
∂y f (a, b) 6= 0. We now show that the condition ∂y f (a, b) 6= 0 is still the appropriate one when the
error term is taken into account.
Theorem 1. (The Implicit Function Theorem for a Single Equation)
Let F (x1 , . . . , xn , y) be a function of class C 1 on some neighborhood of a point (a1 , . . . , an , b) ∈ Rn+1 .
Suppose that F (a1 , . . . , an , b) = 0 and ∂y F (a1 , . . . , an , b) 6= 0. Then there exist positive numbers r0 , r1
such that the following conclusions are valid:
(1) For each x = (x1 , . . . , xn ) in the ball |x − a| < r0 there is a unique y such that |y − b| < r1 and
F (x, y) = 0. We denote this y by f (x); in particular, f (a) = b.
(2) The function f thus defined for |x − a| < r0 is of class C 1 , and its partial derivatives are given
by
∂j F (x, f (x))
.
∂j f (x) = −
∂y F (x, f (x))
Proof.
Remarks.
(1) r0 may be very small, and there is no way to estimate its size without further hypotheses on F .
(2) The formula for ∂j f (X) is, of course, the one obtained via the chain rule by differentiating the
equation F (x1 , . . . , xn , f (x1 , . . . , xn )) = 0.
Corollary 1. Let F be a function of class C 1 on Rn , and let S = {x : F (x) = 0}. For every a ∈ S
such that ∇F (a) 6= 0 there is a neighborhood N of a such that S ∩ N is the graph of C 1 function.
Proof.
Example 1. Let F (x, y) = x2 + y 2 − 1 and let S = (x, y) : x2 + y 2 = 1 . Then F is of class C 1 on
R2 and ∇F (x, y) = (2x,
all (x, y) ∈ S. Thus, given any point (a, b) ∈ S there exists
2y) 6= (0, 0) for
2
a neighborhood N = (x, y) : (x − a) + (y − b)2 < such that S ∩ N is the graph of a function.
Example 2. Let F (x, y) = x − y 2 − 1. Investigate the possibility of solving F (x, y) = 0 for each of
its variables in terms of the other one at (2, 1) and (1, 0).
Solution.
Example 3. Let G(x, y) = x − y 3 − e1−x . Investigate the possibility of solving G(x, y) = 0 for each
of its variables in terms of the other one at (1, 0).
Solution.
Example 4. Investigate the possibility of solving the equation z 2 + xz + y 2 − 2y = 1 for each of its
variables in terms of the other two near the point (1, 1, 1). Do this both by checking the hypotheses
of the implicit function theorem and by explicitly computing the solutions.
Solution.
2
The implicit function theorem for a system of equations
We now turn to the more general problem of solving several equations simultaneously for some of
the variables occurring in them. This will require some facts about invertible matrices and determinants. To fix notation, we shall consider k functions F1 , . . . , Fk for n+k variables x1 , . . . , xn , y1 , . . . , yk ,
and ask when we can solve the equations
F1 (x1 , . . . , xn , y1 , . . . , yk ) = 0,
..
.
Fk (x1 , . . . , xn , y1 , . . . , yk ) = 0
for the y’s in terms of the x’s. We will use vector notation as
F(x, y) = 0.
1
We assume that F is of class C near a point (a, b) such that F(a, b) = 0, and we ask when
F(x, y) = 0 determines y as C 1 function of x in some neighborhood of (a, b).
Again the key to the problem is to consider the linear case,
Ax + By + c = 0,
where A is a k × n matrix, B is k × k matrix, and c ∈ Rk . The criterion for solving this linear
equation for y is the invertibility of the matrix B. In such case the solution is y = −B −1 (Ax + c).
Now, the linear approximation to the equation F(x, y) = 0 near the point (a, b) is a linear equation
of the form Ax + By + c = 0 in which B is the partial Frechet derivative of F with respect to the
variables y, evaluated at (a, b):
∂Fi
(a, b), 1 ≤ i, j ≤ k.
Bij =
∂yj
Theorem 2. (The Implicit Function Theorem for a System of Equations)
Let F(x, y) be an Rk -valued function of class C 1 on some neighborhood of a point (a, b) ∈ Rn+k
∂Fi
(a, b). Suppose that F(a, b) = 0 and det B 6= 0. Then there exist positive numbers
and let Bij =
∂yj
r0 , r1 such that the following are valid:
(1) For each x in the ball |x − a| < r0 there is a unique y such that |y − b| < r1 and F(x, y) = 0.
We denote this y by f (x); in particular, f (a) = b.
(2) The function f thus defined for |x − a| < r0 is of class C 1 , and its partial derivatives ∂xj f can
be computed by differentiating the equation F(x, f (x)) = 0 with respect to xj and solving the
resulting linear system of equations for ∂xj f1 , . . . , ∂xj fk .
Example 1. Investigate the possibility of solving the equations
(x2 + y 2 )u = x,
(x2 + y 2 )v = y
1 1
.
for x and y in terms of v and u near the point (x, y, u, v) = 1, 1, ,
2 2
Solution.
Example 2. Investigate the possibility of solving the equations
u = x + y + z,
v = x2 + y 2 + z 2 ,
w = x3 + y 3 + z 3
for x, y and z in terms of u, v and w near the point (x, y, z, u, v, w) = (0, 0, 2, 2, 4, 8).
Solution.
Exercises (page 119): 1 − 3, 5 − 9
3
3.2 Curves in The Plane
In this section we examine the relation between various ways of representing smooth curves in the
plane. Here ”smooth” means that the curve possesses a tangent line at each point and the tangent
line varies continuously with the point of tangency. Thus ”smooth” is the geometric equivalent of
”C 1 ”.
There are three common ways of representing smooth curves in the plane R2 :
(1) as the graph of a function, y = g(x) or x = g(y), where g is of class C 1 ;
(2) as the locus of an equation F (x, y) = 0, where F is of class C 1 ;
(3) parametrically, as the range of a C 1 function f : (a, b) → R2 .
Example 1. Let F (x, y) = x2 + y 2 − c. Then the set {(x, y) : F (x, y) = 0} is a smooth curve (a
circle) if c > 0, but it is a single point if c = 0 and it is the empty set if c < 0.
Example 2. Let G(x, y) = x2 − y 2 − c. Then the set {(x, y) : G(x, y) = 0} is a hyperbola if c 6= 0,
but if c = 0 it is the union of the two lines y = x and y = −x. The latter set looks like a smooth
curve in a neighborhood of any of its points except the origin.
Example 3. Let H(x, y) = y 3 − x2 − c. The set {(x, y) : H(x, y) = 0} is a smooth curve if c 6= 0,
but when c = 0 it is a curve with a sharp cusp at the origin. The set {(x, y) : H(x, y) = 0} can also
be described parametrically by h(t) = (t3 , t2 ).
Example 4. The function f (t) = (sin2 t, cos2 t) is a C 1 function, but its range is the line segment
from (0, 1) to (1, 0) ((x + y = 1)). The point f (t) traverses this line segment from (0, 1) to (1, 0) as t
π
π
goes from 0 to , then traverses it in the reverse direction as as t goes from to π, and this pattern
2
2
is repeated on every interval [nπ, (n + 1)π], n ∈ Z.
In these examples, the functions in question are all of class C 1 , but the sets they describe fail
to be smooth curves at certain points. However, they share a common feature: The points where
smoothness fails–namely, the origin in examples 1 − 3 and the point (0, 1) and (1, 0) in example 4–are
points where the derivatives of the relevant function vanish. That is, ∇F, ∇G, and ∇H vanish at
(0, 0), and it is the image under h of the one and only one point (t = 0) where h0 vanishes. Moreover,
π
(0, 1) and (1, 0) are the images under f of the points t = nπ and t = (2n + 1) where f 0 (t) = 0.
2
This suggests that it might be a good idea to impose the extra condition that ∇F 6= 0 on the
set where F = 0 in (2) and that f 0 (t) 6= 0 in (3). And indeed with the help of the implicit function
theorem, it is easy to see that under these extra condition the representations (1), (2), and (3) are
all locally equivalent. That is, if a curve is represented in one of the forms (1)-(3) and a is a point
on the curve, at leats a small piece of the curve including the point a can also be represented in the
other two forms.
We now make this precise. Since (1) is more special than either (2) or (3), as we have observed
above, it is enough to see that a curve given by (2) or (3) can also be represented in the form (1).
Theorem 3.
(a) Let F be a real-valued function of class C 1 on an open set in R2 , and let S = {(x, y) : F (x, y) = 0}.
If a ∈ S and ∇F (a) 6= 0, then there is a neighborhood N ⊂ R2 of a such that S ∩ N is the
graph of a C 1 function f (either y = g(x) or x = g(y)).
(b) Let f : (a, b) → R2 be a function of class C 1 . If f 0 (t0 ) 6= 0, then there is an open interval I
containing t0 such that the set {f (t) : t ∈ I} is the graph of a C 1 function g (either y = g(t)
or x = g(t)).
4
Proof.
Remarks.
(1) It should be noted that the conditions of nonvanishing derivatives in Theorem 3 are automatically
satisfied in the special case where the curve is given by y = g(x) or x = g(y). That is, if
F (x, y) = y − g(x), then ∂y F = 1, so ∇F 6= 0. Similarly, if f (t) = (t, g(t)), then f 0 (t) =
(1, g 0 (t)) 6= (0, 0).
(2) The conditions ∇F 6= 0 and f 0 6= 0 in Theorem 3 are sufficient for the smoothness of the
associated curves but not necessary.
Example. The set S = (x, y) : (x2 + y 2 − 1)2 = 0 is a smooth curve but ∇F = 0 for all (x, y) ∈ S.
Definition. (Smooth curve)
A set S ⊆ R2 is a smooth curve if
(1) S is connected and
(2) every a ∈ S has a neighborhood N such that N ∩S is the graph of C 1 function g (either y = g(x)
or x = g(y)).
This definition agrees with the notion of smooth curve introduced at the beginning of this section:
The curve described by y = g(x) has a tangent line at each point (x0 , g(x0 )), and the that line is
given by an equation y − g(x0 ) = g 0 (x0 )(x − x0 ) whose coefficients depend continuously on x0 .
Remarks.
(1) If S = (x, y) ∈ R2 : F (x, y) = 0 , then the second condition of smooth can be be checked by
Theorem 3. That is S is a smooth curve if S is connected and ∇F (x, y) 6= 0 for all (x, y) ∈ S.
(2) If S = {f (t) : t ∈ (a, b)}, then S is connected if f is continuous. However the second condition is
more complicated. Theorem 3 guarantees the existence of an open interval I about t0 , but does
not guarantee the existence of a neighborhood N about (x(t0 ), y(t0 )). In practice, sometimes
one imposes the extra assumption that f is one-to-one.
Example 1. Let F (x, y) = 3x2 + 4y 2 − 12. Determine whether S = {(x, y) : F (x, y) = 0} is a
smooth curve. Draw a sketch of S. Near which points of S is S the graph of a function y = f (x)?
x = f (y)?
Solution.
Example 2. Let f (t) = (t + 1, t2 ). Determine whether the set S = {f (t) : t ∈ R} is a smooth curve.
Draw a sketch of S. Examine the nature of S near any point f (t) where f 0 (t) = 0.
Solution.
Example 3. Let F (x, y) = x2 − 9y 2 − 36. Determine whether S = {(x, y) : F (x, y) = 0} is a smooth
curve. Draw a sketch of S.
Solution.
Example 4. Let f (t) = (t3 − t, t2 ). Determine whether the set S = {f (t) : t ∈ R} is a smooth
curve.
Solution.
Example 5. Let set S = {f (t) : t ∈ R}, where f (t) = (cos t, sin t). Determine whether S is a
smooth curve.
Solution.
Exercises (page 125): 1, 3, 4, 5, 6
5
3.3 Surfaces and Curves in Space
In this section we discuss ways of representing smooth and curves in R3 with a brief sketch of the
situation in higher dimensions.
3
Surfaces in R
The standard ways of representing surfaces in 3-space are analogous of the standard ways of
representing curves in the plane:
(1) as the graph of a function, z = g(x, y) (or x = g(y, z), or y = g(x, z)), where g is of class C 1 ;
(2) as the locus of an equation F (x, y, z) = 0, where F is of class C 1 ;
(3) parametrically, as the range of a C 1 function f : U ⊆ R2 → R3 .
As before, (1) is a special case of (2) and (3), with F (x, y, z) = z −g(x, y) and f (u, v) = (u, v, g(u, v)).
We need some additional conditions to be imposed in cases (2) and (3) to guarantee the smoothness
of the surface.
Theorem 4.
(a) Let F be a real-valued function of class C 1 on an open set in R3 , and let S = {(x, y, z) : F (x, y, z) = 0}.
If a ∈ S and ∇F (a) 6= 0, then there is a neighborhood N ⊂ R3 of a such that S ∩ N is the
graph of a C 1 function f (either z = f (x, y), y = f (x, z) or x = f (y, z)).
(b) Let f be a function of class C 1 from an open set in R2 into R3 . If [∂u f × ∂v f ](u0 , v0 ) 6= 0, then
there is a neighborhood N ⊆ R2 of (u0 , v0 ) such that the set {f (u, v) : (u, v) ∈ N } is the graph
of a C 1 function.
Proof.
Definition. (Smooth surface)
A smooth surface in R3 is a connected subset of R3 that can be locally described as the graph of a
C 1 function.
Remarks.
(1) A surface S = {(x, y, z) : F (x, y, z) = 0} is smooth if S is connected and ∇F (x, y, z) 6= 0 for all
(x, y, z) ∈ S.
(2) A surface S = f (u, v) : (u, v) ∈ U ⊆ R2 is smooth if f is one-to-one and ∂u f × ∂v f 6= 0 for
all (u, v) ∈ U .
Example 1. The unit sphere S = (x, y, z) : x2 + y 2 + z 2 = 1 can be parametrized by spherical
coordinates
f (θ, ϕ) = (cos θ sin ϕ, sin θ sin ϕ, cos ϕ), θ ∈ [0, 2π], ϕ ∈ [0, π].
The sphere is a smooth surface, but the map f does not provide a good parametrization of the whole
sphere because it is not locally one-to-one when sin ϕ = 0.
Note that
∂θ f = (− sin θ sin ϕ, cos θ sin ϕ, 0),
∂ϕ f = (cos θ cos ϕ, sin θ cos ϕ, − sin ϕ).
Thus, ∂θ f × ∂ϕ f = 0 if and only if sin ϕ = 0. However, if we restricted θ and ϕ to the rectangle
−π < θ < π, 0 < ϕ < π, we obtain a good parametrization of the sphere with the ”international
date line ” removed.
6
Example 2. Consider f (u, v) = (u + v) cos(u − v), (u + v) sin(u − v), u + v . Describe the surface
S = f R2 and find a description of S as the locus of an equation F (x, y, z) = 0. Find the points
where ∂u f and ∂v f are linearly dependent, and describe the singularities of S (if any) at these points.
Solution.
Example 3. Consider f (u, v) = (2v − u, u + v, 6v), (u, v) ∈ R2 . Describe the surface S = f R2
and find a description of S as the locus of an equation F (x, y, z) = 0. Find the points where ∂u f
and ∂v f are linearly dependent, and describe the singularities of S (if any) at these points.
Solution.
Remark. The tangent plane to a smooth surface S at a point a ∈ S is given by the equation
n · (x − a) = 0, where n is a nonzero normal vector to S at a.
If S is given by an equation F = 0, then the n = ∇F (a).
If S is given parametrically as the range of a map f (u, v), the vectors ∂u f and ∂v f are tangent
to certain curves in S and hence to S itself at a; we therefore take n = ∂u f × ∂v f .
In both cases, the smoothness of S guarantees that n is nonzero.
Example. Find an equation for the tangent plane to the parametrized surface
x = u2 + v, y = v 2 + u, z = eu+v at (x, y, z) = (0, 2, 1).
Solution.
Curves in R3
Curves in R3 are generally described either parametrically or as the intersection of two surfaces.
Thus, once again we have three kinds of representation for curves:
(1) as a graph, y = f (x) and z = g(x), (or similar expressions with the coordinates permuted),
where f and g rae of class C 1 ;
(2) as the locus of two equations F (x, y, z) = 0 and G(x, y, z) = 0, where F and G are of class C 1 ;
(3) parametrically, as the range of a C 1 function f : R → R3 .
The form (2) describes the curve as the intersection of two surfaces F = 0 and G = 0. The form (1)
is a special case of (2) with f (x, y, z) = y − f (x) and G(x, y, z) = z − g(x) and a special case of (3)
with f (t) = (t, f (t), g(t)).
Remarks.
(1) A curve given by a C 1 function f : R → R3 is smooth if f 0 (t) 6= 0.
(2) A curve given by F (x, y, z) = 0 and G(x, y, z) = 0, where F and G are of class C 1 , is smooth if
∇F and ∇G are linearly independent.
Example 1. Let S be the intersection of the cone z 2 = x2 + y 2 and the plane z = x + 1. Find a
parametrization of S.
Solution.
Example 2. Let S be the intersection of the planes 2x + y − z = 3 and 4x − y + 2z = 0. Find a
parametrization of S.
Solution.
Exercises (page 132): 1 − 6
7
3.4 Transformations and Coordinate Systems
In this section we study smooth mappings from Rn to itself in more detail, with emphasis on
geometric intuition for the cases n = 2 and n = 3.
Suppose f : Rn → Rn is a map of class C 1 . We can regard f as a transformation of Rn , that
is, an operation that moves the points in Rn around in some definite fashion. When n > 1, such
transformations are usually best pictured with ”before and after” sketches. That is, if x = f (u), we
think of u and x as living in two separate copies of Rn . We draw a sketch of u−space with some
geometric figure in it, such as a grid of coordinate lines, then draw a sketch of x−space with the
images of those figures under the transformation f . √ 1 √
3 u − v, u + 3 v . The map f represents
Example 1. Let f : R2 → R2 be given by f (u, v) =
2
√
3
π 1
π
π
= cos , = sin ).
a counterclockwise rotation through the angle about the origin. (
6
2
6 2
6
Example 2. Let f : R2 → R2 be given by f (u, v) = (2u, v). The map f stretches out the u
coordinate by factor of 2.
1 2
u − v 2 , 2uv . The map f is not one-to-one;
2
it maps (u, v) and (−u, −v) to the same point. In order to draw an intelligible picture, we restrict
attention to the region u > 0.
The image of the vertical line u = c under f is given by x = c2 − v 2 , y = 2cv. Elimination of v
yields x = c2 − y 2 /4c2 , the equation of a parabola the opens out to the left.
On the other hand, the image of the horizontal line v = c is given by x = u2 − c2 , y = 2cu, which
yields x = y 2 /4c2 − c2 . Since we assume u > 0, we have y > 0 or y < 0 depending on whether c > 0
or c < 0; in either case this curve is half of a parabola opening to the right.
Example 3. Let f : R2 → R2 be given by f (u, v) =
Coordinate Systems
Another common interpretation of a map f : Rn → Rn is as a coordinate system on Rn . For
example, we usually think of f (r, θ) = (r cos θ, r sin θ) as representing polar coordinates in the plane.
In the preceding discussion we thought in terms of moving the points in Rn around without changing
the labeling system (namely, Cartesian coordinates); here we are thinking of leaving the points alone
but giving them different labels (polar rather than Cartesian coordinates.) It is just a matter of
point of view; the same transformation f can be interpreted either way.
Not all mapping f : Rn → Rn can be used as coordinate systems, however. A good coordinate
system should have the property that there is a one-to-one correspondence between points and their
coordinates; that is, each set of coordinates should specify a unique point in Rn , and two different sets
of coordinates specify different points. For example polar coordinates do not satisfy this condition—
(r, θ) and (r, ϕ) are polar coordinates of the same point whenever θ − ϕ is an integer multiple of 2π,
or whenever r = 0—and this fact always has the potential to cause problems when polar coordinates
are used. However, if we restrict r and θ to satisfy r > 0 and −π < θ < π, we do get a good
coordinate system, not on the whole plane, but on the plane with the negative real axis removed.
Our attention is directed to transformations f of class C 1 that map an open set U ⊆ Rn in a
one-to-one fashion onto another open set V ⊆ Rn . Anther requirement that is natural to impose is
that the inverse mapping f −1 : V → U should also be of class C 1 . Hence, the question arises: Given
a C 1 transformation f : U → V , when does f possess a C 1 inverse f −1 : V → U ? That is, when can
the equation f (x) = y be solved uniquely for x as a C 1 function of y.
8
This question is closely related to the ones that led to the implicit function theorem, and indeed,
if we restrict attention to the solvability of the equation f (x) = y in a small neighborhood of a point,
its answer becomes a special case of that theorem.
We can guess what the answer should be by looking at the linear approximation. If f (a) = b,
the linear approximation to the equation f (x) = y at the point (a, b) is T (x − a) = y − b, where the
matrix T is the Fréchet derivative Df (a), and the latter equation can be solved for x precisely when
T is invertible, that is, when the Jacobian det Df (a) is nonzero.
Theorem 5. (The Inverse Mapping Theorem)
Let U and V be open sets in Rn , a ∈ U , and b = f (a). Suppose that f : U → V is a mapping of class
C 1 and the Fréchet derivative Df (a) is invertible (detDf (a) 6= 0). Then there exist neighborhoods
M ⊆ U and N ⊆ V of a and b, respectively, so that f is a one-to-one map from M onto N , and
the inverse map f −1 from N to M is also of class C 1 . Moreover, if y = f (x) ∈ N , D(f −1 )(y) =
[Df (x)]−1 .
Proof.
Remarks.
(1) It is to be emphasized that the inverse mapping theorem is local in nature; the global invertibility
of f is a more delicate matter.
(2) The invertibility of Df (a) is not necessary for the existence of an inverse map, although it is
necessary for the differentiability of that inverse.
Example 1. Let (u, v) = f (x, y) = (y 2 , x/y).
(a) Compute the Jacobian det Df .
(b) Draw a sketch of the images of some the lines x =constant and y =constant in the uv−plane.
(c) Find a formula for the local inverse of f if it exists.
Solution.
Example 2. Let (u, v) = f (x, y) = (x + y, x − y).
(a) Compute the inverse transformation (x, y) = f −1 (u, v).
(b) Find the image in the uv−plane of the triangle bounded by the lines y = 2x, y = −2x, y = 1.
(c) Find the region of the xy−plane that is mapped to the triangle with vertices (0, 0), (2, 0), and
(1, 1) in the uv−plane.
Solution.
Example 3. Let f (r, θ) = (r cos θ, r sin θ).
(a) Show that
∂(x, y)
6= 0 for all (r, θ) ∈ U = {(r, θ) : r > 0}.
∂(r, θ)
(b) Show that f is locally one-to-one but not global.
Solution.
Exercises (page 138): 1 − 6
9
3.5 Functional Dependence
In the implicit function theorem and its applications discussed in the preceding sections, we have
drawn consequences from the nonvanishing of various Jacobians. In this section we consider the
opposite situation, in which a Jacobian vanishes identically.
Definition. (Functional dependence)
Let f1 , . . . , fm be C 1 real-valued functions on an open set U ⊆ Rn . We say that f1 , . . . , fm are
functionally dependent on U if there is a C 1 function Φ : Rm → R such that Φ(f1 (x), . . . , fm (x)) = 0
and ∇Φ(f1 (x), . . . , fm (x)) 6= 0 for all x ∈ U .
Remarks.
(1) The condition ∇Φ 6= 0 guarantees that the equation Φ = 0 can be solved for one of the variables
in terms of the others; in other words, one of the functions can be expressed in terms of the
remaining ones.
(2) The functions fj might satisfy other relations in addition to the equation Φ(f1 (x), . . . , fm (x)) =
0.
Example 1. Let f1 (x, y) = ln(x + y) and f2 (x, y) = x2 + 2xy + y 2 − 2x − 2y. Show that f1 and f2
are functionally dependent on R2 .
Solution.
Example 2. Let f1 (x, y, z) = x + y + z, f2 (x, y, z) = xy + xz + yz, and f3 (x, y, z) = x2 + y 2 + z 2 .
Show that f1 , f2 , and f3 are functionally dependent on R3 .
Solution.
Remark. The question of functional dependence is interesting only when the number of functions
does not exceed the number of the independent variables. When it does, functional dependence is
almost automatic.
Example. Let f and g be two C 1 functions of one variable. Then f and g are functionally dependent
on any interval I on which f 0 6= 0 or g 0 6= 0. Indeed, if f 0 6= 0 on I, then f is one-to-one on I and so
has an inverse. It follows that Φ(f (x), g(x)) = 0 on I, where Φ(u, v) = g(f −1 (u)) − v.
The main result of this section concern the close relation between the functional dependence of
a family of functions and the linear dependence of their linear approximations. We begin with the
case where the number of functions equals the numbers of independent variables.
Theorem 6. Suppose that F = (f1 , . . . , fn ) is a C 1 map on some open set U ⊆ Rn . If f1 , . . . , fn are
functionally dependent on U , then the Jacobian det DF vanishes identically on U .
Proof.
Example. Show that f (x, y) = cos(x + y) and g(x, y) = cos x + cos y are functionally independent
on R2 .
Solution.
The converse of Theorem 6 is also true: The vanishing of the Jacobian det DF implies the
functional dependence of the fj ’s.
10
Theorem 7. Let F = (f1 , . . . , fm ) be a C 1 function from a connected open set U ⊆ Rn into Rm .
Suppose that the matrix DF (x) has rank k at every x ∈ U , where k < m. Then every x0 ∈ U has a
neighborhood N such that the functions f1 , . . . , fm are functionally dependent on N and F (N ) is a
smooth k−dimensional submanifold.
We shall restrict attention to the case where m = n = 3.
Theorem 8. Let F = (f, g, h) be a C 1 function from a connected open set U ⊆ R3 into R3 . Suppose
that the matrix DF (x, y, z) has rank k at every (x, y, z) ∈ U , where k = 1 or 2. Then every
(x0 , y0 , z0 ) ∈ U has a neighborhood N such that the functions f, g, h are functionally dependent on N
and F (N ) is a smooth curve (if k = 1) or a smooth surface (if k = 2).
Proof.
Example 1. Let f (x, y, z) = yz + x, g(x, y, z) = y 2 z 2 + 2xyz + x2 , h(x, y, z) = 4 + x + yz.
(a) Determine whether f, g, h are functionally dependent on some open set U ⊆ R3 by examining
∂(f, g, h)
the Jacobian
.
∂(x, y, z)
(b) If f, g, h are functionally dependent, find the rank of D(f, g, h) on U and find functional relations
satisfied by f, g, h.
Solution.
Example 2. Let f (x, y, z) = x2 + y 2 + z 2 , g(x, y, z) = x + y, h(x, y, z) = x + z.
(a) Determine whether f, g, h are functionally dependent on some open set U ⊆ R3 by examining
∂(f, g, h)
.
the Jacobian
∂(x, y, z)
(b) If f, g, h are functionally dependent,find the rank of D(f, g, h) on U and find functional relations
satisfied by f, g, h.
Solution.
Exercises (page 145): 1
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