Composition; inverse mappings - Transformations - Two-Dimensional Calculus

Two-Dimensional Calculus (2011)

Chapter 3. Transformations

17. Composition; inverse mappings

Our next aim is to develop what may be called a “chain rule” for transformations. We start with the following observation.

Let F be a differentiable transformation of a domain D into a domain E,

Image

Let f(u, υ) be a continuously differentiable function in E. We may then consider the composition of the transformation F and the function f. Let us write

Image

where

Image

Example 17.1

Suppose

Image

Then

Image

RemarkAlthough the notation at first appears to cloud the issue, the process of forming the function image(x, y) from f(u, υ) is completely elementary; image(x, y) is obtained by substituting the functions u(x, y), υ(x, y), which define the transformation F, into f(u, υ). The function image(x, y) obtained in this manner is sometimes referred to as “the function f(u, υ) pulled back to the x, y plane by the transformation F.” The same process will be considered more extensively in a somewhat different light in Sect. 20, where we discuss changes of coordinates.

We next wish to compute the partial derivatives of the function image(x, y). Since partial differentiation amounts to ordinary differentiation with one variable held constant, the result is obtained by a direct application of the chain rule, Th. 7.1. (See Ex. 17.5 for a more detailed comparison of the notation of Th. 7.1 with that of the present case.) We find

Image

where

Image

Example 17.2

If, in Example 17.1 we wish to compute imagex and imagey at the point (2, 3), we first obtain the corresponding values of u and υ

Image

we then find

Image

and using the chain rule, we have

Image

These values are the same as those obtained by direct differentiation of the function image(x, y) = 4x2, and evaluation at (2, 3).

It may help to write the above equations in a slightly different notation. Setting

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we have

Image

where

Image

In even more compact form,

Image

This form of the equations is undoubtedly the most perspicuous, but it must be remembered that the various quantities on the right-hand side are evaluated at different points ; ux, υx uy, and υy(as well as zx and zy) are all evaluated at a point (x0, y0), whereas zu and zυ are evaluated at the corresponding point (u0, υ0) in the u, υ plane.

We now return to our original problem. We are given a pair of differentiable transformations, say

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and

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We wish to study their composition

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where

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(see Fig. 17.1). Throughout this section we assume that F maps some domain D in the x, y plane into a domain E in the u, υ plane, and that G is defined in the domain F. Then the composed transformation H is defined throughout D. If F and G are both differentiable, then so is H. Differentiating the functions z(u(x, y), υ(x, y)) and w(u(x, y), υ(x, y)) which define H, we obtain Eqs. (17.1), and the analogous equations

Image

Image

FIGURE 17.1Composed mapping

or, in abbreviated form,

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Equations (17.1a) and (17.2a) combined, yield the expression

Image

for the Jacobian matrix of the transformation H. Note that the right-hand side of Eq. (17.3) depends only on the entries in the Jacobian matrices

Image

Equation (17.3) leads to the following basic result.

Theorem 17.1Let F:(x, y) →(u, υ) and G : (u, υ) → (z, w) be differentiable transformations, and let

Image

Let

Image

Then

Image

PROOF.We recall that dF, dG, and dH are the linear transformations

Image

where

Image

Recalling Eq. (15.4), which shows how the matrices of linear transformations combine under composition, we see that Eq. (17.3) is equivalent to the statement of the theorem. image

RemarkFor those familiar with matrix multiplication, Eq. (17.3) takes the simple form

Image

(See the Remark following Ex. 15.21.)

Corollary Under the same hypotheses we have

Image

PROOF.By the definition of the Jacobian (Eq. (16.6)), Eq. (17.5) simply says

Image

and this follows immediately from (17.4) and Th. 15.1.image

Equation (17.5) may be considered the “chain rule for Jacobians.” It is one indication that the role of the Jacobian of a transformation is closely analogous to that of the derivative for a function of one variable. Many later results will confirm this observation.

It is worth pointing out that Eq. (17.5) has a simple geometric interpretation in terms of the relation of the Jacobian to “area magnification at a point,” described at the end of the previous section. Namely, if we denote by A1, the area of a small circle about (x0, y0), by A2 and A3 the area of the image of this circle under F and H, respectively, then we have

Image

But as we remarked at the end of the previous section, these ratios tend to the magnitudes of the Jacobians of the maps H, G, and F, respectively. Similarly, the way the sign of the Jacobian behaves under composition is easily deduced by geometric considerations (see Ex. 17.9).

We next apply Th. 17.1 and its Corollary to the important special case where the transformation G is the inverse of F. In this case z = x, w = y, and we may identify the z, w plane with the x, y plane (Fig. 17.2).

Theorem 17.2If a differentiable transformation F: (x, y) → (u, υ) has a differentiable inverse, then the Jacobian of F cannot equal zero, and

Image

Image

FIGURE 17.2Inverse mapping

PROOF.Setting z = x, w = y, we find zx = 1, zy = 0, wx = 0, wy = 1, and substituting in Eq. (17.5),

Image

But if the product of two numbers is equal to 1, neither number can equal zero. This proves the theorem. image

RemarkIt is important to note that the right-hand side of Eq. (17.6) is evaluated at some point (x0, y0) and the left-hand side at the point (u0, υ0), which is the image of (x0, y0) under F.

We shall illustrate this theorem by a number of examples. First, however, let us make several general remarks.

We start by recalling that if a transformation F maps two distinct points onto the same point (u0, υ0), then F does not have an inverse; in other words, for F to have an inverse it must be one-to-one. Now it may happen that F is not one-to-one in its entire domain of definition, but that if we restrict our attention to some neighborhood of a point (x0, y0), then F may define a one-to-one mapping of that neighborhood (see Example 17.3 below). On the other hand, it follows from Th. 17.2 that if the Jacobian of F is equal to zero at (x0, y0) then no matter how small a neighborhood of (x0, y0) we may choose, the map F of that neighborhood cannot have a differentiable inverse. We may formulate this consequence of Th. 17.2 as follows.

Corollary If F is a differentiable transformation, and if the Jacobian of F vanishes at a point, then F cannot have a differentiable inverse in any neighborhood of that point.

Next we note that if F maps (x0, y0) onto (u0, υ0), then in order for F to have an inverse defined in some neighborhood of (u0, υ0), it must map a neighborhood of (x0, y0) onto an entire neighborhood of (u0, υ0). This may not be the case if the Jacobian of F vanishes at (x0, y0) (see Example 17.3 below).

Finally, even if F maps some neighborhood of (x0, y0) onto some neighborhood of (u0, υ0) and is one-to-one, then F has an inverse transformation G, but G need not be differentiable (see Example 17.5 below).

Example 17.3

Let

Image

Then

Image

and

Image

Thus the Jacobian vanishes at every point of the y axis, and, by the above Corollary, F cannot have an inverse in any neighborhood of a point on the y axis. Take, for example, the point (0, 1). It maps onto the point (0, 2) in the u, υplane. Every circle about (0, 1) in the x, y plane maps into the right half of the u, υ plane, since u = x2 ≥ 0 (Fig. 17.3). Thus the image of F never covers a full neighborhood of (0, 2) in the u, υ plane, and we cannot hope to find an inverse map G defined in a neighborhood of (0, 2). If, on the other hand, we take a point such as (2, 1) in the x, y plane, where the Jacobian of F is different from zero, then we may be able to find an inverse. If we choose a large circle about (2, 1), which includes points in the left-half plane, then the map F is not one-to-one, and it cannot have an inverse. However, by choosing a smaller circle, of radius r < 2, F is one-to-one inside this circle, since F is a one-to-one map of the whole right half-plane x > 0 onto the whole right half-plane u > 0, and F has a differentiable inverse

Image

which maps the right half-plane u > 0 onto x > 0 (Fig. 17.4).

Example 17.4

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Here

Image

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FIGURE 17.3Map F:(x, y)→ (x2, 2y)near the point (0, 1)

Thus the Jacobian of F vanishes at the origin. From our discussion of this transformation at the end of Sect. 13, it is clear that F maps the interior of every circle about the origin in the x, y plane onto the full interior of a circle about the

Image

FIGURE 17.4Inverse mapping of F:(x, y)→ (x2, 2y)

origin in the u, υ plane. However, no matter how small a circle we choose, the map is not one-to-one. Hence there cannot be an inverse in any neighborhood of the origin.

Example 17.5

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In this case F is a differentiable transformation that maps the entire x, y plane one-to-one onto the u, υ plane. There is therefore an inverse mapping; namely

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The map G is differentiable everywhere except on the u and υ axes, where one of the derivatives xu = imageu−2/3 or y>υ = imageυ−2/3 becomes infinite. This corresponds to the fact that the u and υ axes are the image under F of the x and yaxes, and that along the latter the Jacobian of F vanishes;

Image

In the light of Examples 17.3–17.5 it is easy to see why the terminology “singular point” is used for a point at which the Jacobian vanishes. At such a point the transformation misbehaves in one way or another. We may restate the Corollary to Th. 17.2 as follows: if F is a differentiable transformation near (x0, y0) and if F has a differentiable inverse transformation in some neighborhood of(x0, y0), then (x0, y0) is a regular point of F.

The converse of this statement is known as the inverse mapping theorem : if (x0, y0) is a regular point of a differentiable transformation F, then there exists some neighborhood of (x0, y0) In which F has a differentiable inverse.

We omit the proof of this theorem, since it follows most naturally as a consequence of general results for functions of more than two variables.1 Stated together with its converse, the result is the following.

Theorem 17.3 Inverse Mapping Theorem A differentiable mapping F has a differentiable inverse in some neighborhood of (x0, y0) if and only if the Jacobian of F at (x0, y0) is different from zero.

This theorem is of great importance in the theory of differentiable transformations, and it is worth spending some time on the precise meaning of the statement. Consider a mapping

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that is differentiable in a domain D, and choose some point (x0, y0) in D. We ask whether Eqs. (17.7) can be “solved” for x and y in terms of u and υ, if not in the whole domain D, at least in some smaller domain D' containing (x0, y0). More precisely, we may state the question as follows. “Does there exist some domain D' included in D, such that (x0, y0) is in D' maps D' one-to-one onto a domain E′, and if the correspondence thus defined between points of D'and E' is written in the form

Image

then φ and ψ are continuously differentiable functions in E'” (See Fig. 17.5.) This question and its answer are summarized briefly in Th. 17.3 : such a domain D' exists if and only if the Jacobian of F is different from zero at (x0, y0). A striking feature of Th. 17.3 is that such a complicated question can be answered by a simple computation with derivatives.

If we consider D in Example 17.3 above to be the full plane, and (x0, y0) any point in the right half-plane, then we may choose D' to be the right halfplane x > 0, in which case E' is the right half-plane u > 0, and we are able

Image

FIGURE 17.5 Inverse mapping near a point

to write down the explicit equations for the inverse mapping φ(u, υ) = image ψ(u, υ) = imageυ. Similarly, in Example 17.5, if (x0, y0) is any point not on the x or y axis, then we could choose for D' the entire quadrant containing (x0, y0) and we again have explicit equations for the inverse mapping.

In general, it is not possible to write down explicitly the functions φ(u, υ) and ψ(u, υ). Theorem 17.3 simply asserts that these functions must exist. The situation is precisely analogous to that of the implicit function theorem (Th. 8.2) and the statement that “we can solve for x and y in terms of u and υ” is to be interpreted in the manner discussed in detail at that point. An important fact to notice is that again, as in the case of the implicit function theorem, even if we cannot find the functions φ and ψ explicitly, we can use implicit differentiation to compute their derivatives. Namely, let us assume that u and υ have been given explicitly as functions of x and y, so that the partial derivatives ux, uy, υx, υy evaluated at any point are known constants. If we apply the chain rule, Eq. (17.1a), first to the case where z = φ(u, υ) = x, and then to the case z = ψ(u, υ) = y, we find

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At a regular point, if we set

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then Δ≠ 0, and each pair of equations in (17.8) is a pair of simultaneous linear equations with nonvanishing determinant Δ. We may therefore solve these pairs of equations, and we find

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In other words, we have for the Jacobian matrix of the inverse mapping

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Note that the right-hand side of (17.11) is evaluated at some point (x0, y0), and the left-hand side is to be evaluated at the point (u0, υ0) which is the image under F of (x0, y0).

RemarkIn the terminology of elementary matrix theory, the matrix on the right-hand side of (17.11) is simply the inverse of the matrix JF. We may therefore write (17.11) in the form

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This also follows immediately from Eq. (17.4) applied to the case where G = F−1, in which case dH is the identity map.

Example 17.6

If the equations

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are solved for x and y in terms of u and υ near the point (x0, y0) = (3, −4), what are the partial derivatives xu, xυ, yu,yυat the corresponding point in the u, υ plane?

We find

Image

By Eq. (17.11),

Image

If x= 3, y = − 4, then u = −7, υ=−24, and

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In conclusion, we remark that in working a given problem, rather than referring to Eq. (17.11), it may be just as easy to use the method that led to this equation. Thus, in order to find xu in Example 17.6, we may simply differentiate both Eqs. (17.13), using the chain rule. We find

Image

Substituting x = 3, x = −4, we obtain

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which, when solved, yields xu = image, yu = image

Exercises

17.1 In each of the following cases, find the partial derivatives zx and zy at the point indicated, first by using the chain rule (Eqs. (17.1)) and then by finding z explicitly in terms of x and y and differentiating.

a. z =u2 + υ2; u = x cos y, υ = x sin y; (x0, y0) = (1, 0)

b. z = u2 + 2υ2; u = xy, υ = x + y; (x0, y0) = (1, −2)

c. z = eu cos υ; u = ex cos y, υ = ex sin y; (x0, y0) = (0, 0)

d. z =log (2u2 + 3υ2 + 6); u = x2y2, υ = 2xy; (x0, y0) = (1,2)

17.2Let f(u, υ) be continuously differentiable, and let F be the transformation u = cx − y, υ = cx + y, where c is a nonzero constant. Let image = f image F.

a. Find imagexcimage y.

b. Use your answer to part a to construct functions image(x, y) satisfying the equation imagexcimage y.

17.3Let f(u, υ) be continuously differentiable, and let F be a differentiable transformation, F: (x, y) → (u, υ). Let image=f image F. and (u0, υ0) = F(x0, y0).

a. Show that if f(u0, υ0) = 0, then ∇image (x0, y0) = 0.

b. Show that if F is regular at (x0, y0) and if image (x0, y0) = 0, then f(u0, υ0) = 0.

*c. Show that if F is singular at (x0, y0), then it is possible to find a function f such that f(u0, υ0) ≠ 0, but image (x0, y0) = 0.

17.4Given a differentiable mapping F:(x, y) → (u, υ), and given a function f(u, υ), form a geometric picture of the relation between f and image = f image F in the following fashion. Draw two sets of rectangular coordinates, x, y, z and u, υ, z. Sketch the surface z = f(u, υ). Then to obtain the corresponding surface z =image (x, y), picture the map F: (x, y) → (u, υ), and to each point (x0, y0), place a point directly above it at the same height as f at (u0, υ0) = F(x0, y0). Thus the surface z = image(x, y) is obtained by a process of distorting the surface z = f(u, υ) in a manner that preserves the height of each point, although the horizontal distortion may be quite arbitrary. One would not expect many geometric properties of a surface to be preserved under such a distortion, but some of them are. For example, interpret the statement in Ex. 17.3a geometrically in terms of the surfaces z = f(u, υ) and z = image(x, y). and try to visualize why the statement is true. Note also that if f(u, υ) has a local minimum or maximum at (u0, υ0), then image(x, y) has the same at (x0, y0) (see also Ex. 20.17f).

17.5Let F be a differentiable transformation of a domain D in the x, y plane into a domain E in the u, υ plane:

Image

a. Show that for any point (x0, y0) in D, there are numbers a, b such that a < x0 < b, and such that u = φ(x, y0), υ = ψ(x, y0), a ≤ x ≤ b, defines a differentiable curve C in E. (C would be the image under F of the line segment a ≤ x ≤ b, y = y0, where x is the parameter along the curve.)

b. Show that the vector imageux(x0, y0), uy(x0, y0)image is the tangent vector to the curve C of part a at the point (u0, υ0) = F(x0, y0).

c. Let f(u, υ) be a continuously differentiable function in E, and letimage = f image F. Using the notation of part a, show that fc(x) = image(x, y0), a ≤ x ≤ b.

d. Deduce from parts b and c that

Image

17.6For each of the following pairs of transformations F, G, find the Jacobian matrices JF and JG and use Eq. (17.3) to find the Jacobian matrix of the composed transformation H = G image F at the point indicated. Then find Hexplicitly, compute JH, and check your answer.

a. Image

b. Image

17.7Find ∂(z, w)/∂(x, y) at the point (x0, y0) for each of the following pairs of transformations.

a. Use Ex. 17.6a.

b. Use Ex. 17.6b.

c. image

d. Use part c, but let (x0, y0) = (π, 0).

e. Use part c, but let (x0, y0) = image π ,log 2).

17.8Show that if (x0,y0) is a singular point of a differentiable transformation F, then it is a singular point of G image F. for every differentiable transformation G.

17.9Let F be a differentiable transformation that is regular at (x0, y0), and let (u0, υ0) = F(x0, y0). Let G be a differentiable transformation that is regular at (u0, υ0). Let H =G image F.

a. Give a geometric argument that shows that H is orientation-preserving at (x0, y0) if and only if F and G are both orientation-preserving or are both orientation-reversing at the points (x0, y0) and (u0, υ0), respectively.

b. For any real number λ ≠ 0, let

Image

(that is, sgλ = λ/|λ|). Use a geometric argument to show that

Image

17.10Let F be a differentiable transformation that is an involution (see Ex. 14.16).

a. Show that F has no singular points.

b. Show that if F(x0, y0) = (u0, υ0), then the Jacobian of F at (u0,υ0) is the reciprocal of the Jacobian of F at (x0, y0).

c. Show that the transformation

Image

is an involution, and verify part b directly by computing the Jacobian.

17.11For each of the following transformations, F, check by Th. 17.3 that F has a differentiable inverse in some neighborhood of (0, 0) ; use implicit differentiation to find the partial derivatives of the inverse transformation at the point F(0, 0), and verify Eq. (17.6).

a. Image

b. Image

c. Image

d. Image

e. Image

f. Image

g. Image

17.12In Ex. 17.11a-e, solve explicitly for x and y as functions of u and υ, and thus write down the inverse transformation in explicit form. Specify carefully some domain containing the point F(0, 0) in which F−1 is defined and differentiable and check your answers to Ex. 17.11 by computing the derivatives of F−1 at F(0, 0).

17.13Let us call a “generalized shear transformation,” a transformation of the form

Image

a. Describe the map F geometrically by examining the behavior of F on each horizontal line y = y0.

b. Show that F is a bijective map from the whole x, y plane to the whole u, υ plane, and find the explicit equations of the inverse.

c. Note that ∂(u, υ)/∂(x, y) ≡ 1. Interpret this fact in terms of the vectors imageux, υximage and imageuy, υyimage .

*17.14 a. State and prove an analog of Th. 17.3 if the differentiable mapping F is replaced by a differentiable function of one variable, and if the Jacobian of F is replaced by the derivative of f.

b. Discuss the various ways in which a function f of one variable can fail to have a differentiable inverse near a point where the derivative vanishes, and give examples analogous to those given in the text for a mapping Fnear a point where the Jacobian vanishes.

17.15Let F be a differentiable transformation, F:(x, y) → (u, υ). Let P(u, υ) and Q(u, υ) be continuously differentiable, and let

Image

Define

Image

a. Show that

Image

*b. Show that if the transformation F satisfies the Cauchy-Riemann equations ux = υy, υx = −uy, then

Image

c. Show that if there exists a function f(u, υ) such that P = f u, Q = f>υ, then p = imageυ and q = imagey, where image(x, y) = f image F.

*17.16 a. Given a matrix

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show that for every image > 0 there exists a δ > 0 such that if

Image

then

Image

(Note : this says that the determinant of the matrix M is a continuous function of the four variables a, b, c, d.)

(Hint: let m be the maximum of |a|, |b|, |c|, |d|. Suppose that |aa'| < δ, etc. Writing ad − a'd' = a(d − d') + d'(a − a'), show that |ad− a'd'|≤ 2mδ + δ2, and similarly |bc − b'c'| ≤ 2mδ+ δ2. If m = 0, choose δ < image if m≠ 0, choose δ < image and δ < image/8m.)

b. Using the notation of part a, show that if ad − bc ≠0, then also a'd' − b'c' ≠ 0 for δ sufficiently small.

*17.17 Prove the following statement, which is part of Theorem 17.3. If F is a differentiable mapping in a domain D, and if the Jacobian of F at (x0, y0) is different from zero, then F is one-to-one in some neighborhood of (x0, y0). (Hint: if F is given by u = f(x, y), υ = g(x, y), then by hypothesis,

Image

By Ex. 17.16b, it follows that there exists δ > 0 such that

Image

provided

Image

By continuity of the partial derivatives fx, fy, gx gy, these four inequalities hold, provided (x3, y3) and (x4, y4) lie in some disk D':(x - x0)2 + (y - y02 < r2. Fix r > 0 by this requirement, and let (x1, y1), (x2, y2) be any two distinct points in D'. We wish to show that F(x1, y1) ≠ F(x2, y2), so that F is one-to-one in D'. By the mean-value theorem applied to the functions f and g,

Image

where (x3, y3) and (x4, y4) are points which lie on the line segment joining (x1, y1) to (x2,y2), and which are therefore both in D'. It follows by our choice of r that the matrix M' of this system has nonzero determinant, so that the left-hand sides of these equations cannot be zero unless x2x1 = 0 and y2y1 = 0.)