3. The Theorems of Green and Stokes
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1 3. The Theorems of Green and tokes Consider a 1-form ω = P (x, y)dx + Q(x, y)dy on the rectangle R = [a, b] [c, d], which consists of points (x, y) satisfying a x b and c y d. The boundary R of R have four sides, which can be described by the parametric equations as 1 : x(t) = t, y(t) = c (a t b); 2 : x(t) = b, y(t) = t (c t d); 3 : x(t) = a + b t, y(t) = d (a t b) and 4 : x(t) = a, y(t) = c + d t (c t d). It is customary to write R = We compute the differential dω of ω = P dx + Qdy as follows: ( ) ( ) P P Q dp dx + dq dy = dx + x y dy Q dx + dx + x y dy dy = P Q dy dx + dx dy y x ( Q = x P ) dx dy. y Consequently we have R = d c d c b Q b d dy a x dx dx a c (Q(b, y) Q(a, y)) dy + b a P y dy (P (x, c) P (x, d)) dx. One can check that the following identities: d c Q(b, y)dy = ω, 2 d c Q(a, y)dy = ω, 4 b a P (x, c)dx= ω, 1 b a P (x, d)dx = ω. 3 1
2 Hence we have R ω + ω + ω + ω We have shown the identity R ω for a rectangle. This can be easily extended R to the case where R is replaced by a parametric surface in the n-dimensional Euclidean space R n defined on a rectangle. Indeed, suppose that ω is a 1-form with its domain in R n and if σ : R R n is a parametric surface. Then σ ω is a 1-form on the rectangle and σ R σ R dσ ω = R R ω. σ ω = ω = σ R σ ω. (3.1) Here we have used a number of rules, such as Rule I1 from 2.2 and Rule E8 from 2.1. We also use the identity ω = b a ω for computing line integrals, where : [a, b] R n is a parametric curve. The symbol σ R stands for the restriction of the mapping σ to the boundary of R, which gives four pieces of smooth paths. More generally, an oriented surface in R n (when n = 2, is just a planar region) may have holes and hence cannot be parametrized by a single mapping σ defined on a rectangle. In that case we may cut into several pieces, say j (1 j n) so that each piece can be parametrized by some mapping σ j. We have n j=1 σ j n j=1 ω = σ j The last equality is based on the following observation: if σ j and σ k are adjacent (j and k are allowed to be the same: for example, the mapping σ j wraps a cylinder so that it is self-adjacent ) and share a border line l, then the paths coming from σ j and σ k which run along l are in opposite directions so that line integrals contributed by these two paths are cancelled. Hence the only paths making real contribution are those going along the boundary of ; see the following figure: ω. 2
3 We conclude that if is an oriented surface in R n (or a planar region in case n = 2) and if ω is a 1-form defined on, then ω. (3.2) Here we have to be careful about orientation. If is not orientable, the above argument fails. For example, take a Möbius strip in R 3, which is known to be one sided. We can construct a mapping σ from a rectangle R = ABCD onto the strip so that σ(b) = σ(c), σ(a) = σ(d), and the boundary of is a single loop composed of two pieces: one is the restriction of σ to AB and the other to CD. The restrictions of σ to BC and DA give overlapping paths, with identical directions, and hence the line integrals along these two paths are not cancelled. How do we decide the directions of boundary paths of an oriented surface? Well, by the orientation we know that there are two sides of, one of them is regarded as the front side and the other as the back side. Rule I2 ( left side rule ). tay on the front side of and walk along a boundary path. If is on your left side, you are walking in the same direction as the path. 3
4 In case n = 2, is just a planar region. When we draw a Cartesian plane, we always draw the front view. According to the above left side rule we see that a loop along the outer boundary of is running in the anticlockwise direction and a loop surrounding a hole of is running in the clockwise direction. When n = 2, the general 1-form can be written as ω = P dx + Qdy (P, Q are functions of x and y) and (3.1) gives ( Q/ x P/ y) dx dy and hence (3.2) becomes ( Q x P ) dx dy = P dx + Qdy (3.3) y which is called Green s Theorem. For n = 3, the general 1-form is ω = P dx + Qdy + Rdz, (P, Q, R are functions of x, y, z) and (R y Q z )dy dz + (P z R x )dz dx + (Q x P y )dx dy; (see Example 1.4). o (3.2) becomes ( R y Q ) ( P dy dz + z z R ) ( Q dz dx + x x P ) dx dy y = P dx + Qdy + Rdz, (3.4) or, following the notation introduced in the last section, we can write curl F d = F dr, (3.5) where F = ip + jq + kr and curl F = i(r y Q z ) + j(p z R x ) + k(q x P y ). It is more suggestive to put curl F in the following well known way: i j k F = / x / y / z P Q R 4.
5 Identity (3.4) or (3.5) is the classical tokes Theorem. Example 3.1. Verify that the area of a planar region surrounded by a loop is given by 1 2 xdy ydx. Use this to find the area A e of the region surrounded by the ellipse (x 2 /a 2 ) + (y 2 /b 2 ) = 1 (a, b are positive constants). olution: Denote by D the region surrounded by. By Green s Theorem and the identity d(xdy ydx) = dx dy dy dx = 2dx dy, we have 1 xdy ydx = 1 d(xdy ydx) = dx dy, 2 2 D D which is the area of D. Use parametric equations x = a cos t, y = b sin t for to compute A = 1 2 = 1 2 2π xdy ydx a cos t. d b sin t b sin t. d a cos t = 1 2 2π ab dt = πab. (The area surrounded by is also given by xdy, as well as ydx. Here A 1 2 (xdy ydx) is the average of them and is often more convenient to use.) Example 3.2. Find the line integral ω, where ω = xydy + ydz and is a path running along the boundary of a parallelogram, starting from its vertex A = (1, 1, ), passing vertices B = (2, 3, 1), C = (2, 5, 2) and D = (1, 3, 1) and back to A. olution: A direct computation involves four line integrals and hence is rather tedious. o it is advisable to apply toke s theorem here. Define a mapping σ by σ(u, v) = OA + u AB + v AD = (1, 1, ) + u(1, 2, 1) + v(, 2, 1) (1 + u, 1 + 2u + 2v, u + v); u, v 1. Notice that = σ. Also, d(xy) dy + dy dz = ydx dy + dy dz and σ (dω) = (1 + 2u + 2v)du d(2u + 2v) + d(2u + 2v) d(u + v) = (2 + 4u + 4v)du dv. Hence ω = σ ω = σ 1 1 σ (2 + 4u + 4v) dudv = 6.
6 Recall (from 2.1) that a 1 form ω is said to be closed if. Also recall (from 1.1) that a 1 form ω is exact if ω = df for some function f. From d 2 = (Rule E6) we see that exact forms are closed. But the converse in general is not true: the angular form (xdy ydx)/(x 2 + y 2 ) defined on the punctured plane is closed but not exact; see Example 1.3 of the present chapter and Example 3.2 from Chapter 1. Now we use identity (3.1) derived at the beginning of the present section to study under what condition closedness of 1-forms implies its exactness. First let us introduce a definition (which is basic in homotopy theory). Let U be a set in R n (assumed to be open, if you like). Let : [, 1] U and 1 : [, 1] U be two parametric curves in U having the same end points, that is, () = 1 () and (1) = 1 (1). We say that and 1 are homotopic relative to U and write 1 2 (rel U) if there is a parametric surface σ : [, 1] [, 1] such that σ(t, ) = (t), σ(t, 1) = 1 (t) for all t in [, 1], and σ(, u) = (), σ(1, u) = (1) for all u in [, 1]. We say that U is simply connected if every pair of paths in U with the same end points are homotopic relative to U. Example 3.3. how that if ω is a closed 1-form defined on U and if 1 rel U, then ω = 1 ω. olution: Let σ : [, 1] [, 1] U be a parametric surface in U with the property described in the above paragraph. Then, by (3,1), we have ω ω = 1 σ ω = σ. Example 3.4. Prove the following assertion: every closed form defined on a simply connected set is exact. olution: Let ω be a closed form defined on a simply connected set. Then, by the previous example, we know that any line integral ω depends only on the end points of. To complete the proof, we use the result of Exercise 6 in 1.3, which says that if all line integrals of a 1 form ω are path independent, then ω is exact the proof. A consequence of the above discussion that the punctured plane is not simply connected. This fact is intuitively obvious, but obvious things may be wrong (this is why we are always baffled by a magician s rope trick). Here nothing goes wrong with our assertion. The only problem is that it is by no means easy to prove! We certainly need a sophisticated 6
7 argument to prove it. In view of the above example, all we need is to establish a closed 1 form on the punctures plane which is not exact. Of course we already have such a closed form: the angular form (x 2 + y 2 ) 1 (xdy ydx). The proof of (3.2) suggests a general identity C C ω, where C is a formal sum N k=1 a kσ k, called a singular complex. This identity and its generalization will lead to a duality between the singular homology theory and derham cohomology theory. Exercises 1. Compute the line integral D (x2 + xy)dx + xy 2 dy, where D is the unit square, consisting of all points (x, y) with x, y 1. (Hint: Green s theorem.) 2. Let ω = dy dz + dz dx + dx dy and let be the cap of a cone given by x 2 + y 2 = (1 z) 2 with z 1, oriented in such a way that its boundary (namely, the circle x 2 + y 2 = 1 with z = ) is oriented in the anticlockwise direction. Compute the surface integral ω in two ways: (1) direct computation, using parametrization x = u cos v, y = u sin v, z = 1 u, with u 1, v 2π; (2) use toke s theorem by observing that d(ydz + zdx + xdy) = ω. 3. Use the recipe A = xdy to find the area enclosed by the ellipse (x/a)2 + (y/b) 2 = Verify that the area A of a region D enclosed by a loop described in polar coordinates as r = r(θ), a θ b, with r(a) = r(b) is A = 1 b 2 a r(θ)2 dθ. Use this compute the area of the region enclosed by one loop of the -curve described by (Hint: Put A = 1 2 r = cos 2θ; π/4 θ π/4. xdy ydx in polar coordinates.) 5. Check that identity (3.5) is equivalent to (3.4). 7
8 6. The divergence F of a planar vector field F = i P + j Q is given by F = P x + Q y. Recall from 2.1 that F gives rise to the 1-form ω F = P dx + Qdy. Also recall that the Hodge star α of a planar 1-form α = fdx + gdy is α = fdy gdx. Derive the following version of divergence theorem from Green s theorem ω F = F dx dy. (3.6) D 7. Let f and g be planar scalar fields. Apply (3.6) above to the vector field F = f g to derive the first Green s identity: fdg = D D D ( f g + f g) dxdy, (3.7) where g = g xx + g yy. Deduce from (3.7) the second Green s identity: D (fdg gdf) = D (f g g f) dxdy. (3.8) 8. Apply (3.8) to prove the mean value property for harmonic functions: if D is a (closed) disk centered at (x, y ) of radius R and if f is a harmonic function with domain containing D, then f(x, y ) = the average of f over D 1 π 2π f(x + R cos t, y + R sin t) dt. (3.9) Hint: For convenience, suppose (x, y ) = (, ) and R = 1. Take a small ε > and let D ε be the annulus consisting of (x, y) with ε x 2 + y 2 1. Apply (3.8) to the given f and g = log r log x 2 + y 2 over D ε, using dr = rdθ. Let ε. 9. Let F = i P + j Q + k R be a vector field in R 3. A curve in R 3 is called a vortex line of F if vectors in curl F are tangent to this curve at points on the curve. For any loop in R 3, draw vortex lines through all points on this loop. In this way we get a tubular surface called a vortex tube. Verify that, if 1 and 2 are two loops wrapping around a vortex tube oriented in the same way, then they have the same circulation of F, that is, 1 F dx = 2 F dx. (Hint: toke s theorem.) 8
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