Green’s theorem for arbitrary figures; applications - Integration - Two-Dimensional Calculus

Two-Dimensional Calculus (2011)

Chapter 4. Integration

26. Green’s theorem for arbitrary figures; applications

Our proof of Green’s theorem, which we gave for the special case of a rectangle, turns out to hold equally well for general figures. More precisely, the idea of the proof is unchanged, but the details must be modified.

Before discussing the general case, let us illustrate the argument for a specific figure (see also Ex. 25.14). Consider the figure

Image

Let q(x, y) ∈Image in a domain that includes F. We wish to evaluate ∫∫F qx dA. To do so, we use the method and notation of Th. 24.3. The horizontal line y = t intersects F for 0 ≤ t ≤ 3 ( Fig. 26.1). There are two cases to consider.

Image

FIGURE 26.1 Proof of Green’s theorem for a semi-annulus

Case 1. 0 ≤ t < 2. The intersection of y = t with F consists of two line segments a1(t) ≤ xb1(t) and a2(t) ≤ xb2(t), where

Image

Thus

Image

and

Image

We thus have four integrals each of which may be considered as a line integral along a part of the boundary of F. Consider, for example.

Image

where the curve C1:x = Image y = t, 0 ≤ t ≤ 2, is an arc of the outer circle. Similarly,

Image

where the curve Image:x = Image y = t, 0 ≤ t ≤ 2, is an arc of the inside boundary oriented in the direction of increasing y, and C6 is the same arc with the opposite orientation ( Fig. 26.1). Thus, with C4 and C5 denoting the (oriented) boundary arcs indicated in Fig. 26.1, Eq. (26.1) takes the form

Image

Case 2. 2 ≤ t ≤ 3. The intersection of y = t with F consists of a single segment. We have a(t) = −Image b(dt) = Image

Image

and

Image

where C2 and C3 are the boundary arcs indicated in Fig. 26.1.

Let us denote by Image the entire outer semicircle of F, oriented in the counterclockwise direction, and let Image denote the inner semicircle oriented in the clockwise direction ( Fig. 26.2). The curve Image consists of the four arcs C1, C2, C3, C4, while Imageconsists of C5 and C6. Thus, adding Eqs. (26.2) and (26.3) , and applying Th. 24.3, we have

Image

Finally, let Image :x = t, y = 0, − 3 ≤ t ≤ −2, and Image:x = t, y = 0,2≤ t ≤ 3, denote the two horizontal line segments indicated in Fig. 26.2.

Image

FIGURE 26.2 The oriented boundary of a semi-annulus

Then

Image

and it follows from (26.4) that

Image

where C is the piecewise smooth boundary curve of F, consisting of the successive arcs Image,Image , Image, Image. An analogous argument using Th 24.4 shows that

Image

for any function p(x, y)∈Image in a domain that includes F. Equations (26.5) and (26.6), or their combined form

Image

constitute Green’s theorem for the figure F. Once the reasoning used in this special case is understood, there should be no difficulty in seeing how it extends to an arbitrary figure. In order to state the general theorem, it is useful to introduce the following notation.

Definition 26.1 A curve C:x(t), y(t), atb, lying on the boundary of a figure F is said to have positive orientation with respect to F if the vector Imagey'(t), x'(t) Image is directed toward the interior of F.

Remark The vector Imagey'(t), x'(t)image is perpendicular to the tangent vector Image x'(t),y'(t)image obtained from the latter by a rotation through 90° in the counterclockwise direction. Thus, positive orientation with respect to F may be described as the direction of motion along C such that the interior of F lies to the left ( Fig. 26.3). Note that a figure F is typically bounded by one outer curve and a number of inner curves. Along the outer curve positive orientation means traversing the curve in the counterclockwise direction, whereas along the inner curves, positive orientation implies traversing in the clockwise direction.

Image

FIGURE 26.3 Positive orientation of boundary curves

Definition 26.2 The oriented boundary of a figure F, denoted by ∂F, consists of the totality of boundary curves of F, each of them assigned a positive orientation with respect to F.

Remark The only context in which we use the oriented boundary is in forming a line integral ∂F p dx + q dy, where p and q are continuous functions on the boundary of F. Thus

Image

where the boundary of F consists of the curves C1 …, Cn, each assigned a positive orientation with respect to F. As we know from Lemma 21.1, the value of this integral does not depend on the parameter chosen to represent the curves Ck providing the orientation is preserved.

Example 26.1

Evaluate ∂F y dx for the figure F:4 ≤ x2 + y2 ≤ 9.

The boundary of F consists of two circles ( Fig. 26.4), which may be represented parametrically as

Image

Note that along the outer curve C1 the vector

Image

is directed toward the origin, hence toward the interior of F, while along the inner curve C2, the vector

Image

is directed outwards, hence again toward the interior of F. Thus both C1 and C2 are positively oriented with respect to F. Now

Image

and

Image

thus

Image

Image

FIGURE 26.4 The positively oriented boundary of an annulus

Theorem 26.1 Green’s Theorem Let p(x,y), q(x, y) ∈Image in a domain that includes a figure F. Then

Image

Remark Equation (26.9) is equivalent to the two separate equations

Image

PROOF. We first evaluate ∫∫F qx dA, using the method and notation of Th. 24.3. Let us indicate the reasoning for the case where F is a polynomial figure.7 The line y = t intersects F in a finite number of intervals akt) ≤ xbk(t). We set

Image

Then

Image

Now for each k, the integral Image(bkt), t) dt can be broken into a finite number of intervals such that in each interval, say tjttj+1 the equations x = bk(t), y = t, define a curve Image lying on the boundary of F. Then

Image

Since the point (bk(t), t) is a boundary point of F lying at the right-hand endpoint of a line segment contained in F, the interior of F lies to the left if Image is traversed in the direction of increasing y. Thus, each of the Image so defined is positively oriented with respect to F.

Similarly, each of the integrals Imageqak(t), t) dt may be broken down into a sum, such that Image : x = ak(t), y = t, titti + 1, is a curve on the boundary of F. Hence

Image

where Image consists of the curveImage with the opposite orientation. Since F lies to the right of Image as Image is traversed in the direction of increasing y, Image has negative orientation with respect to F, and Image therefore has positive orientation. If we form the sum over all k, the right-hand side of (26.11) represents a sum of line integrals of the form Image q dy, where each Image is a curve lying on the boundary of F and having positive orientation with respect to F. Furthermore the totality of these Image represents the complete oriented boundary of F, except for possible horizontal segments along which ∫q dy = 0. Thus, the right-hand side of (26.11) is equal to ∂F q dy, and (26.11) reduces to the first equation in (26.10). A similar reasoning using Th. 24.4 yields the second equation in (26.10). Combining these two equations, we obtain (26.9), and the theorem is proved. Image

Example 26.2

Evaluate ∂F y dx for the figure F: 4 ≤ x2 + y2 ≤ 9.

We have already found the value −5π for this integral by a direct calculation. Using Green’s theorem, with p = y,q = 1, we find qx≡ 0, py ≡1, and

Image

But the area of F is equal to the area inside a circle of radius 3 and outside a circle of radius 2; that is, 9π − 4π or 5π.

Example 26.3

Evaluate

Image

where F is the square −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.

In this case ∂F is a single closed curve C consisting of four line segments ( Fig. 26.5). We can find the value of (26.12) by computing the integral over each line segment and adding (see Ex. 25.6a). Green’s theorem cannot be applied directly, because the functions p, q in this case have a singularity at the origin. Thus the figure F, which contains the origin, does not lie in a domain where p(x, y), q(x, y)∈Image We can, however, simplify the computation greatly by using the figure

Image

where r is any fixed number between 0 and 1 ( Fig. 26.5). The boundary of F consists of two parts. The outer curve is the same as the boundary of F, and the inner curve is a circle of radius r. The functions p = −y/(x2 + y2), q = x/(x2+ y2) ∈Image in the domain D consisting of the whole plane except for the origin. Since the figureImage lies in this domain we may apply Green’s theorem. As we have observed before, qxpy ≡ 0 in this case, and therefore

Image

Image

FIGURE 26.5 Use of Green’s theorem to reduce integral over a square to an integral over a circle

Denoting by C the inner circle, x = r cos t, y = — r sin t, 0 < t < 2π, positively- oriented with respect to F, and once again C = ∂F, we obtain

Image

or

Image

Thus we have found the integral in question, not by a direct integration over the original curve, but by reducing it to an integral over a simpler curve by virtue of Green’s theorem.

The above examples illustrate two of the many applications of Green’s theorem. We next consider the way in which Example 26.2 generalizes to arbitrary figures.

Lemma 26.1 Let F be an arbitrary figure, and A(F) its area; then

Image

PROOF. Applying Green’s theorem to each of the boundary integrals on the right yields ∫∫F 1 dA = A(F). Image

Remark In addition to Example 26.2, we have already seen an illustration of this result for the case where F is a rectangle (see Example 25.1). Depending on the particular figure F, one or the other of the integrals in (26.13) may be more convenient to evaluate. For example, if F is the disk x2 + y2r2, then ∂F can be represented as x = r cos t, y = r sin t, 0 ≤ t, and

Image

Each of these expressions integrated from 0 to 2π yields the value πr2, but the last one is clearly the simplest.

We are now able to prove a theorem that provides a basic link between the double integral and the theory of differentiable transformations. The Jacobian of a transformation, which first appeared as the determinant of an associated linear transformation, turns up here in a totally different manner.

Theorem 26.2 Let u(x, y), υ(x, y) define a diffeomorphism G of a domain D onto a domain Image. Let F be a figure lying in D, and let its image be a figure Image; then

Image

PROOF. We apply Lemma 26.1 to the figure Image. For this purpose, let us denote by C1 …, Cn the oriented curves which constitute ∂F. Let

Image

be any one of these curves, and let

Image

be its image under the transformation G. Using the chain rule we may transform any line integral over Image into a line integral over Ck. In particular,

Image

Now setting p = x, q = y we find

Image

and

Image

Applying Green’s formula (26.9) to the figure F, we find

Image

As we have remarked in the proof of Th. 18.2, for a diffeomorphism, the Jacobian is either everywhere positive or everywhere negative. If the Jacobian is positive, then the transformation G preserves orientation. This implies that since the figure F lies to the left when Ck is traversed in the direction of increasing t, the figure Image also lies to the left when Image is traversed in the direction of increasing t. In other words, for positive Jacobian, the fact that the curve Ck defined by (26.15) is positively oriented with respect to F implies that the curve Image defined by (26.16) is positively oriented with respect to Image. Thus the curves Image . . ., Image constitute ∂Image . Applying Lemma 26.1 to F and using Eqs. (26.17) and (26.18) yields

Image

This proves (26.14) for the case that (u, υ)/(x, y) > 0. In the opposite case, where (u, υ)/(x, y) < 0, orientation is reversed, so that the curves Image defined by (26.16) are negatively oriented with respect to Image. In this case

Image

Since

Image

and (26.14) holds again. Image

Corollary Let

Image

be a nonsingular linear transformation, and let Δ = det G = ad − bc. Then, if F is any figure in the x, y plane and Image is its image in the u, υ plane,

Image

PROOF. Since a nonsingular linear transformation is a diffeomorphism of the whole x, y plane onto the u, υ plane, we may apply the theorem. But (u, υ)/ (x, y) = Δ, a constant, and (26.14) reduces to

Image

Remark We have already indicated the geometric interpretation on (26.19) of the determinant in Sect. 14, for the case where F was a disk, and at the end of Sect. 15, for the case where F was a triangle. One way to make Eq.(26.14) plausible is to note that the integral on the right-hand side is the limit of sums of the form

Image

taken over partitions of F. If a figure Fi in the partition is taken to be a small triangle, then the differential of G at the point (ξi ηi) is a linear transformation that approximates G on Fi and whose determinant is equal to

Image

Thus the corresponding term in the sum (26.20) equals the area of the image under the differential of G at (ξi ηi) and is approximately equal to the area of the image under G. Summing over i, we obtain a quantity that approximates the area of the image of all of F and Eq. (26.14) states that the sums (26.20) tend precisely to this area in the limit.

Example 26.4

Let G be the transformation

Image

Let F be the figure x: ≥ 0, y ≥ 0, x2 + y2 ≤ 1.

To compute the integral in (26.14), we note

Image

and

Image

Thus

Image

But

Image

and we find (for example, by the trignometric substitution y = sin t):

Image

Thus

Image

Let us now examine the figure F and its image Imageunder the transformation G ( Fig. 26.6). Introducing polar coordinates

Image

we find that G takes the form

Image

or

Image

Image

FIGURE 26.6 The image of a quarter disk under the transformation G:u = Image(x2y2), υ = −xy

Thus each radius θ = θ0, 0 ≤ r1, of the quarter circle F maps one-to-one onto the ray φ = − 2θ0, 0RImage and the figure F maps one-to-one onto the semicircle Image: − πφ0, 0RImage. The area of Image is = Imageπ(Image )2 = Image. which, in view of (26.21), confirms Eq. (26.14) for this example.

In conclusion, let us examine ∂F and Image . First of all, ∂F consists of a single closed piecewise smooth curve, obtained by describing in succession the following parts:

Image

Their images under G are, respectively,

Image

Thus Image and Image are each segments of the x axis, traversed from left to right, and Image is a semicircle traversed in the clockwise direction (see Fig. 26.6). We observe that C1, C2, C3 are positively oriented with respect to F, but Image,Image,Image ,Image are negatively oriented with respect to F. This reflects the fact that the transformation G has negative Jacobian, −(x2 + y2). Thus ∂F is the closed curve consisting of the three parts Image,Image ,Image traversed in the opposite direction.

Exercises

26.1 Let C be the curve x = a cos t, y = b sin t, 0 ≤ t. Use Green’s theorem to evaluate each of the following integrals.

a.c y2ex dx

b.c y cos x dx + (cos y + sin x) dy

c.Image

d.c y dxx dy

26.2 Sketch each of the following figures F, and express in parametric form the curve or curves that constitute the oriented boundary ∂F.

a. Image

b. Image

c. F: 2xx2 + y2 ≤ 9

d. F: x2yx1/2

e. F: y2x ≤ 1

f. F: x + y ≤ 1, xy ≥ − 1, y ≥ 0

26.3 Use Green’s theorem to evaluate

Image

for each of the figures F in Ex. 26.2.

26.4 Use Lemma 26.1 to compute the area of the figures bounded by the following curves.

a. The ellipse x = a cos t, y = b sin t, 0 ≤ t.

b. The astroid x = a cos3 t, y = a sin3 t, 0 ≤ t.

c. The loop of the strophoid

Image

(Hint: note that y = tx, and use the last expression in (26.13).)

d. The loop of the folium of Descartes

Image

(Hint: use the same method as in part c. The resulting integral is an “improper integral” and is evaluated as a limit. See the discussion immediately following Eq. (28.12).)

e. The cardioid r = a(1 − cos θ), 0 ≤ θ ≤ 2π, where r and θ are polar coordinates. (Hint: see Ex. 21.25a.)

f. One “leaf” of the “rose” r = a sin 2θ, 0 ≤ 0Imageπ. (Hint: see part e.)

26.5 Let f(x) be a positive continuously differentiable function defined for axb. Let F be the figure defined by

Image

a. Show that if p(x, y) is continuous on the boundary of F, then

Image

b. Using part a and Lemma 26.1, show that the area of F is equal to

Image

26.6Show that the centroid Image of a figure F is given by

Image

26.7 For each of the following figures F and transformations G, sketch Image and its image Funder G, and verify Eq. (26.14).

a. F: x ≥ 0, x2 + y2 ≤ 1 ; G : u = x2, υ = y

b. F: −3 ≤ x ≤ −2, 1 ≤ y ≤ 2; G: u = x2, υ = y3

c. F:0 < axb, 0 ≤ yπ; G:u = x cos y, υ = x sin y

d. F:axb, 0 ≤ yπ; G:u = ex cos y, υ = ex sin y

26.8 Let C: x(t), y(t), atb, be a regular curve lying on the boundary of a figure F, and suppose that C has positive orientation with respect to F. Show that at each point of C,

Image

where s is the parameter of arc length along C, and N is the unit vector perpendicular to C directed toward the exterior of F.

26.9Let u(x, y) ≡Image and υ(x, y) ≡Image in a domain D. Let F be a figure lying in D. At each boundary point of F where the boundary curve is regular, define the normal derivative ∂υ∂n to be the directional derivative of υ(x, y) in the direction of the unit normal N directed toward the exterior of F. Prove the following identities.

*a.Image

Hint: apply Green’s theorem with p = − y, q = uvx and use Ex. 26.8.)

b. Image

c. Image

(Note: parts a and b are known as Green's identities.)

26.10Using the notation of Ex. 26.9, show that the following statements hold.

a. If υ(x, y) is harmonic in D, and if u(x, y) is equal to zero at each point of the boundary of F, then

Image

b. If u(x, y) is harmonic in D, then

Image

(Note: if u(x, y) represents temperature, then ∂u/∂n describes the rate of flow of heat across the boundary at each point. The integral ∂ F (∂u/∂n) ds represents the total flow across the boundary. The vanishing of this integral means that the total flow across the boundary is zero, or that the amount of heat entering the figure F is equal to the amount leaving F.)

26.11 Suppose u(x, y) is harmonic in a domain D. Setting υ(x, y) = u(x, y) in Green’s identity, Ex. 26.9a, yields

Image

By Ex. 24.26, if the integral on the left is equal to zero, then ∇u ≡ 0 on F. Use this fact to prove the following statements.

a. If u(x, y) = 0 at each point of the boundary of F, then u(x, y) ≡ 0 in F. (See Corollary 3 to Th. 12.3 for a different proof.)

b. If ∂u/∂n = 0 at each point of the boundary of F, then u(x, y) is constant on F.

c. If u(x, y) and υ(x, y) are both harmonic in D, and if ∂u/∂n = ∂υ/∂n at each point of the boundary of F, then u(x, y) and υ(x, y) differ by a constant.

d. Suppose u(x, y) and υ(x, y) both satisfy Poisson's equation in D :

Image

where f(x, y) is a given function. If u(x,, y) = (x, y) at each point of the boundary of F, then u(x, y) ≡ υ(x, y) in F.

*26.12 Let h(x, y) ∈Image in a domain D. Suppose that the disk

Image

lies in D. The quantity

Image

is called the mean value of h(x, y) on the circle (x, − x0)2 + ( yy0)2 = r2. (See the general discussion of mean values at the beginning of Sect. 27.) Using Ex. 26.9c and Lemma 7.2, one can prove the following relation

Image

Carry out the details by verifying each step in the following string of equalities. Set

Image

so that

Image

then

Image

26.13 Using the result of Ex. 26.12, prove the following theorem, known as the mean-value property of harmonic functions : if h(x, y) is harmonic in a domain D, then for every point (x0 y0) in D,

Image

for every value of r such that the disk (xx0)2 + (yy0)≤ r2 lies in D.

(Hint: in the notation of Ex. 26.12, if h(x, y) is harmonic, then d/dr m(r) = 0, and m(r) is constant. But as a consequence of Lemma 7.2, m(r) is a continuous function of r, and limr→0 m(r) = m(0) = h(x0, y0).)

*26.14 The mean-value property described in Ex. 26.13 is a property unique to harmonic functions. This may be stated in the following manner: if h(x, y)∈Image in a domain D, and if for every point (x0, y0) in D the value of h at(x0, y0) equals its mean value over all small circles about (x0y), then h(x, y) is harmonic in D.

Prove this statement. (Hint: proceed by contradiction. Suppose that at some point (x0, y0) in D, Δh ≠ 0 say Δh(x0, y0) = η > 0. Then for some r > 0, Δh ≥ η/2 > 0 for (xx0)2 + (yy0)2≤ r2. By Ex. 26.12, m'(r) > 0. This contradicts the assumption m(r) = h(x0, y0).)

*26.15 Let h(x, y) be continuous on the circle C:(xx0)2 + (yy0)2 = r2, and suppose h(x, y) ≤ M on C. Show that if the mean value of h(x, y) on C is equal to M, then h(x, y) ≡ M on C. (Hint: let

Image

and let f(r, θ) = Mg(r, θ). Then f(r, θ) is continuous for 0 ≤ θ ≤ 2π, and f(r, θ) ≥ 0. To show that Image f(r, θ) = 0 implies f(r, θ) ≡ 0, proceed by contradiction. If f(r, θ0) = η > 0, then f(r, θ) ≥ η/2 for |θ − θ0| < δ, and

Image

26.16 Prove the strong maximum principle for harmonic functions : if h(x, y) is harmonic in a domain D, if h(x, y) ≤ M throughout D, and if h(x0, y0) = M for some point (x0, y0) in D, then h(x, y) ≡ M in D. Proceed in two steps.

a. Show that if h(x, y) has a local maximum M at (x0, y0) then h(x, y) ≡ M on every small circle about (x0, y0). (Hint: use Ex. 26.15 together with the mean value property of harmonic functions.)

*b. Show that if there were some point (x1, y1) in D such that h(x1, y1) < M, this would lead to a contradiction as follows. Join (x0, y0) to (x1, y1) by a curve C:x(t), y(t), 0 ≤ t ≤ 1 . Then the function hc(t) satisfies hc(0) = M, hc(l) < M. Let b be the largest value of t such that hc(b) = M. Then 0 ≤ b < 1, and applying part a to the point (x2, y2) = (x(b), y(b)) leads to a contradiction.

26.17 Let f(x, y)∈Image in a domain D, and let Image p, q Image = ∇f. Give two different proofs that for every figure F lying in D, ∫∂F p dx + q dy = 0.

26.18 Let f(x, y) ∈Image in a domain D, except at a point (x0, y0). Let C1, C2 be circles in D defined by

Image

Suppose that all points between these two circles also lie in D. Show that if Py = qx at every point of D except (x0, y0), then

Image

(Hint: apply Green’s theorem to the figiire F bounded by Cx and C2. Pay careful attention to the orientation of ∂F and of Cx, C2.)

26.19 Under the same hypotheses as Ex. 26.18, let F be a figure lying in D. Show that the following statements hold.

a. If (x0, y0) does not lie in F, then ∂F p dx + q dy = 0.

b. If (x0 y0) is an interior point of F, then

Image

where rx is chosen sufficiently small so that the set of points (x, y) satisfying (xx0)2 +(yy0)2r2 lies in the interior of F. (Hint: apply Green’s theorem to the figure Image bounded by ∂F and C1.)

26.20 Let C be a piecewise smooth curve that forms the oriented boundary of a figure F. Show that the winding number n(C; >x0, y>0) of C about the point (x0, y0) satisfies

Image

(Hint: see Ex. 25.18 for the definition of winding number, and use Ex. 26.19.)

26.21 Let G be a differentiable transformation of a domain D in the x, y plane into a domain E of the u, υ plane. Suppose P(u, υ),Q(u,υ)∈Image in E. Set

Image

and

Image

Let C be a curve in D, and let Image be its image under G. Show that

Image

(Note that Eq. (26.17) is a special case of this formula. Use the same reasoning here as was used to derive (26.17).)

26.22 Using the notation of Ex. 26.21, show that if C = ∂F, where F is a figure lying in D, then

Image

(Hint: see Ex. 17.15a.)

26.23 Using the notation of Ex. 26.21, let (x0, y0) be a point of D, and let F be the figure (xx0)2 + (yy0)2r2, where r is chosen sufficiently small so that Flies in D. Let C = ∂F, and let Image be the image of C under the transformation G. Let (U, V) be any point in the u, υ plane such that G(x, y) ≠ (U, V) for all points(x, y) in F (that is, (U, V) is a point that is not in the image of the figure F). Show that the winding number n(Image ; U, V) = 0. (Hint: let

Image

Then P(u, υ)∈Image 1, Q(u, υ) ∈Image in the domain E consisting of the whole plane minus the point (U, V), and Pυ = Qu in E. Since

Image

the result follows from Ex. 26.22.)

Remark Ex. 26.23 gives a useful criterion for determining when a pair of equations

Image

can be solved for x and y, when U and V are given. Namely, suppose C is a circle in the x, y plane, and Image is the image of C under the transformation

Image

If Image does not pass through the point (U, V) and if the winding number of Image about the point ( U, V) is different from zero, then the above equations must be satisfied by some point (x, y) lying inside C (since otherwise we could conclude from Ex. 26.23 that the winding number of Image about (U, V) must equal zero). Exercise 26.24 shows how this reasoning may be used to guarantee the existence of a solution to a pair of simultaneous quadratic equations. It may be difficult to solve such a pair of equations explicitly, or even to determine whether a solution exists, since the equation may represent two ellipses, or an ellipse and a hyperbola, which may or may not intersect. In Ex. 26.28the same reasoning is used to prove a part of the inverse mapping theorem, which may be thought of as guaranteeing the existence of a solution of a pair of simultaneous equations. Exercise 26.29 completes the proof of the inverse mapping theorem. Exercises 26.25-26.27 lead up to this result, but they are also of interest in their own right for an understanding of some of the more subtle properties of differentiable mappings.

26.24 Let G be the mapping

Image

Let Image be the image of the unit circle C :x = cos t, y = sin t, 0 ≤ t ≤ 2π.

a. Show that Image may be written in the form

Image

b. Show that n(Image ; 2, 3) = 2.

c. Deduce that the equations

Image

have a simultaneous solution (x, y) satisfying x2 + y2 < 1 .(Hint: use the reasoning indicated in the Remark just above.)

26.25 Let G be a linear transformation with determinant Δ ≠ 0. Let C be a circle x = r cos t, y = r sin t, 0 ≤ t ≤ 2π, and let Image be the image of C under G. Show that

Image

(Hint: introducing polar coordinates R, φ on the curve Image (which is an ellipse centered at the origin), we have by Eq. (14.11) that dφ/dt > 0 if Δ > 0 and dφ/dt < 0 if Δ < 0. But

Image

for some integer n, and n ≠ 0, since the integrand is either positive for all t or negative for all t. If | n |≥ 2, then there would be some t0, 0 < t0 < 2π, such that

Image

which would mean that the (distinct) points (cos t0, sin t0) and (cos 0, sin 0) of C would map onto the same point of Image, contradicting the fact that a nonsingular linear transformation is injective. Thus n = ±1.)

*26.26 Prove Rouch's theorem: Let (u0, υ0) be a given point, and let

Image

be two closed curves that do not pass through (u0, υ0). Suppose that for each value of t, the corresponding points on C and Image, contradicting the fact that a nonsingul are closer together than the distance from the point on C to (u0, υ0); that is,

Image

(see Fig. 26.7). Then n(Image : u0 υ0) = n(C: u0, υ0).

(Hint: choose φ0, φ0 such that

Image

Image

FIGURE 26.7Illustration of Rouch's theorem

For any τ, aτθ, define

Image

and Image similarly. The integrand is d/dt tan-1 [(υ(t) − υ0)/(u(t) − u0)], and these integrals simply define in a continuous way the angles φ(t) and Image along the curve (see Fig. 26.7). Furthermore,

Image

Thus

Image

where k = n(C; u0,υ0) − n(Image ;u0, υ0). If k ≠ 0, then the function τ(t) − Image increases or decreases by at least 2π as t goes from a to b, and since it is a continuous function, it must be equal to an odd multiple of π for some t0 between a and b. This means that the points (u(t0), υ(t0)) and Image lie on a line through (u0, υ0) and on opposite sides of (u0, υ0). But then the distance between these points is greater than the distance from (u(t0), υ(t0)) to (u0, υ0), contrary to assumption. Thus k must equal zero, which is the desired result.)

Remark One way to visualize Rouche’s theorem is the following. Let the point (u0, υ0) be the position of the sun, and let C and Image be the paths of the earth and the moon, respectively. If after a certain number of orbits the earth and the moon both return to their original position, then C and Image will be closed curves. The fact that the distance from the moon to the earth is always less than the distance from the earth to the sun implies that the number of times the moon has gone around the sun must equal the number of times the earth has gone around the sun.

*26.27 Let G be a differentiable transformation, let G(x0, y0) = (u0,υ0), and suppose that the Jacobian of G at (u0, y0) is different from zero. Let C be the circle x = x0 + rcos t, y = y0 + r sint, 0 ≤ t ≤ 2π and let Image be its image under G. Show that for all sufficiently small values of r,

Image

(Hint: if we set (u, υ) = G(x, y), then by Th. 16.2,

Image

where Image h(x, y)image is a remainder term satisfying

Image

Let Image be the ellipse that is the image of C under the transformation

Image

Since the determinant Δ of this transformation is

Image

which is assumed nonzero, we may apply Ex. 26.25 and deduce that n(Image ; u0, υ0) = ± 1 according as Δ > 0 or Δ < 0. If λ2 is the minimum of

Image

then the fact that dG(x0,y0) is nonsingular implies that λ2 > 0. Then

Image

for all (x, y). But for d(x, y) sufficiently small we have |Image h(x, y)image | < (λ2)1/2 d(x, y). This means that the inequality

Image

must hold at every point of the circle C for r sufficiently small. But this inequality is precisely the condition on the curves Image and Image which by Ex. 26.26 guarantees that n(Image ; (u;0, υ0) = n(Image ;u0 υ0). (Note that the ellipse Image of this exercise plays the role of the curve C in Ex. 26.26.)

26.28 Let G be a differentiable transformation, let (u0,υ0) = G(x0, y0), and suppose that the Jacobian of G at (x0, y0) is different from zero. Then the image under G of any disk (xx0)2+(yy0+ < r2 includes all points in some disk (uu0)2 + (υυ0)2 < R2.(Hint: in the notation of Ex. 26.27, if r is sufficiently small, then the image Image of the circle C of radius r about (x0 y0) satisfies n( Image u0 υ0) ≠ 0. Choose R so that the disk (uu0)2 + (υυ0)2 < R2 does not contain any points of Image. Let (U, V) be any point of this disk. By Ex. 25.18a, n( Image; U, V) = n( Image u0, υ0). By Ex. 26.23 above, it follows that (U, V) is the image under G of some point (x, y) inside C.)

*26.29 Prove the inverse mapping theorem (Th. 17.3). Specifically, let G(x, y) be a differentiable mapping defined in a domain D by

Image

where f(x, y), g(x, y) ∈Image in D. Let (u0, υ0) = G(x0, y0) and suppose

Image

Show that there exists a disk D':(xx0)2 + (yy0)2 < r2, such that G maps D' one-to-one onto a domain E', and such that G−1 is a differentiable map of E' onto D'. Proceed in the following steps.

a. Choose r sufficiently small so that

Image

where (x3, y3) and (x4, y4) are any two points of the disk

Image

Show that G maps D' one-to-one onto a domain E'. (Hint: by Ex. 17.17, it is always possible to choose r so that the given condition is satisfied, and for any such r, the map G is one-to-one in D'. Note that this condition implies in particular that (u, υ)/(x, y) ≠ 0 throughout D'. It follows from Ex. 26.28 that the image E' of D' includes a disk about each of its points, hence is an open set. Finally the fact that D' is connected and G is continuous implies that E' is connected. Hence E' is a domain.)

b. Show that the inverse map H: E' → D' which exists by part a is continuous in E'. (Hint: given any point (u1,υ1) in E' let (x1, yx) = H(u1, y1). H continuous at (u1, υ1) means that for any ∈ > 0 there exists δ > 0 such that (uu1)2 + (υυ1)2 < δ ⇒ (xx1)2 + (yy1)2 < ∈, where (x, y) = H(u, υ). But this is a consequence of Ex. 26.28 applied at the point (x1 y1) with ∈ = r and δ = R, using the fact that (x, y) = H(u, υ) ⇔ (u, υ) = G(x, y).)

c. Show that if the inverse map H is given by

Image

then φ1(u, υ) ∈Image and ψ(u, υ) ∈Image in E' (Hint: use an analogous reasoning to the proof of the implicit function theorem (Theorem 8.2). Given any two points (u1, υ1), (u2, υ2), in E' let (x1, y1 ), (x2, y2) be the corresponding points in the disk D'. By the mean value theorem applied to the functions f(x, y) and g(x, y), there exist points (x3, y3) and (x4, y4) on the line segment joining (x1, y1 to (x2, y2) such that

Image

By the condition of part a, these equations can be solved in the form

Image

In order to show the existence of φu(u1, υ1) we choose υ2 = υ1 r and we obtain from the first of the above equations that

Image

Using the continuity of H (see part b) and the fact that (x3, y3) and (x4, yt) lie on the line between (x1, y1) and (x2, y2) ,we have

Image

By the continuity of the partial derivatives fx, fy, gx, fx it follows that

Image

Thus the partial derivative φ(u1, υ1) exists, and in a similar manner, so do φυ, ψu, and ψυ. Furthermore, the expressions for these derivatives show that they depend continuously on x and y, and since x and y are continuous functions of u and υ, the partial derivatives ψu, ψυ, ψu;, ψυ are continuous functions of u and υ.)

*26.30 Let G be a differentiable transformation in a domain D. Let the circle C:x = x0 + r cos t, y = y0 + r sin t, together with its interior, be include in D.Let Image be the image of C under G, and let(U, V)be a point not on Image. suppose that there are a finite number of points inside C which map onto the point (U,V)and suppose that jacobian of G is not zero at any of these points.

Specifically, suppose the Jacobian is positive at m points and negative at n points. Then

Image

(Hint: for each point (xk, yk) inside C such that G(xk, yk) = (U, V), let

Image

be a small circle, where r is chosen sufficiently small so that all these circles lie inside of C and no two of them intersect. Let Image be the image of Ck under G, and let F be the figure consisting of those points which lie on or inside C, and on or outside each Ck. If we define P(u, υ), Q(u, υ) as in Ex. 26.23 and p(x, y), q(x, y) as in Ex. 26.21, then

Image

But by Ex. 26.27, for r sufficiently small n( Image ; U, V) = 1 if (u, υ)/(x,y) > 0 at (xk, yk) and n(Image ; U, V) = − 1 if (u, υ)/(x, y) < 0 at (xk, yk). Thus

Image

Remark There are many important special cases of the result in Ex. 26.30. For example, if the mapping G has positive Jacobian everywhere, then the result may be stated as follows: the winding number about a point (U, V) of the image of a circle C is equal to the number of points inside C that map onto (U, V). This is closely related to the so-called argument principle in the theory of functions of a complex variable.

It is most instructive to try to understand the underlying reason for the relation between winding numbers and the number of points that map onto a given point. One way is to picture a disk in the x, y plane made out of a rubber sheet, and to visualize the mapping as a process of distorting this disk by various means, such as stretching, twisting, and folding. When this is done, the resulting form of the rubber sheet may lie in several layers over parts of the u, υ plane, and the number of points in the sheet that lie over a given point corresponds to the number of points inside the original disk that map onto the given point. The study of general properties of mappings such as those considered here belongs to the branch of mathematics known as topology. The specific quantity denoted in Ex. 26.30 by mn is called the degree of a mapping. For further discussion along these lines we refer to [7], Sections 13-18, and [21], Section 5.