Chapter 6 Green's Theorem in the Plane

Chapter 6

Green’s Theorem in the Plane

Recall the following special case of a general fact proved in the previous chapter. Let C be a piecewise C1 plane curve, i.e., a curve in R2 defined by a piecewise C1-function 2 α :[a, b] → R with end points α(a) and α(b). Then for any C1 scalar field ϕ defined on a connected open set U in R2 containing C, we have Z ∇ϕ · dα = ϕ(α(b)) − ϕ(α(a)). C In other words, the integral of the gradient of ϕ along C, in the direction stipulated by α, depends only on the difference, measured in the right order, of the values of ϕ on the end points. Note that the two-point set {a, b} is the boundary of the interval [a, b] in R. A general question which arises is to know if similar things hold for integrals over surfaces and higher dimensional objects, i.e., if the integral of the analog of a gradient, sometimes called an exact differential, over a geometric shape M depends only on the integral of its primitive on the boundary ∂M. Our object in this chapter is to establish the simplest instance of such a phenomenon for plane regions. First we need some preliminaries.

6.1 Jordan curves

Recall that a curve C parametrized by a continuous α :[a, b] → R2 is said to be closed iﬀ α(a) = α(b). It is called a Jordan curve, or a simple

1 closed curve, iff α is piecewise C1 and 1-1 (injective) except at the end points (i.e. x 6= y implies α(x) 6= α(y) or {x, y} = {a, b}). Geometrically, this means the curve doesn’t cross itself. Examples of Jordan curves are circles, ellipses, and in fact all kinds of ovals. The hyperbola defined by 2 c α : [0, 1] → R , x 7→ x , is (for any c 6= 0) simple, i.e., it does not intersect itself, but it is not closed. On the other hand, the clover is closed, but not simple. Here is a fantastic result, in some sense more elegant than the ones in Calculus we are trying to establish, due to the French mathematician Camille Jordan; whence the name Jordan curve. Theorem. Let C be a Jordan curve in R2. Then there exists connected open sets U, V in the plane such that (i) U, V, C are pairwise mutually disjoint, and (ii) R2 = U ∪ V ∪ C. In other words, any Jordan curve C separates the plane into two disjoint, connected regions with C as the common boundary. Such an assertion is ob- vious for an oval but not (at all) in general. Note that the theorem applies to any continuous simple closed curve, it does not need the curve to be piecewise C1. That is also why the proof properly belongs to an area of mathematics called ”topology”, not ”calculus”, the subject of this class. In any case the proof involves some effort and is too long to give in this course. For an inter- esting account of the history as well as of Jordan’s original proof see ”Thomas Hales, Jordan’s Proof of the Jordan Curve Theorem, Studies in Logic, Gram- mar and Rhetoric 10 (23) 2007, http://mizar.org/trybulec65/4.pdf”.

The two regions U and V are called the interior and exterior of C. To distinguish which is which let’s prove Lemma: In the above situation exactly one of U and V is bounded. This is called the interior of C. Proof. Since [a, b] is compact and α is continuous the curve C = α([a, b]) is compact, hence closed and bounded. Pick a disk D(r) of some large radius r containing C. Then S := R2 \D(r) ⊆ U ∪V . Clearly S is connected, so any two points P,Q ∈ S can be joined by a continuous path β : [0, 1] → S with P = β(0), Q = β(1). We have [0, 1] = β−1(U) ∪ β−1(V ) since S is covered by U and V . Since β is continuous the sets β−1(U) and β−1(V ) are open

2 subsets of [0, 1]. If P ∈ U, say, put t0 = sup{t ∈ [0, 1]|β(t) ∈ U} ∈ [0, 1]. If −1 β(t0) ∈ U then, since β (U) is open, we find points in a small neighborhood of t0 mapped to U. If t0 < 1 this would mean we’d find points bigger than t0 mapped into U which contradicts the definition of t0. So if β(t0) ∈ U then t0 = 1 and Q = β(1) ∈ U. If, on the other hand, β(t0) ∈ V , there is an intervall of points around t0 mapped to V which also contradicts the definition of t0 (we’d find a smaller upper bound in this case). So the only conclusion is that if one point P ∈ S lies in U so do all other points Q. Then S ⊆ U and V ⊆ D(r) so V is bounded.

Recall that a parametrized curve has an orientation. A Jordan curve can either be oriented counterclockwise or clockwise. We usually orient a Jordan curve C so that the interior, V say, lies to the left as we traverse the curve, i.e. we take the counterclockwise orientation. This is also called the positive orientation. In fact we could deﬁne the counterclockwise or positive orientation by saying that the interior lies to the left. Note ﬁnally that the term “interior” is a bit confusing here as V is not the set of interior points of C; but it is the set of interior points of the union of V and C.

6.2 Simply connected regions

A connected open set R in the plane is said to be simply connected if every Jordan curve C in R can be continuously deformed to a point in R. Here is the rigorous deﬁnition using the Jordan curve theorem.

Deﬁnition. A connected open set R ⊆ R2 is called simply connected (s.c.) if for any Jordan curve C ⊂ R, the interior of C lies completely in R. Examples of s.c. regions:

(1) R = R2 (2) R=interior of a Jordan curve C.

The simplest case of a non-simply connected plane region is the 2 annulus given by {v ∈ R |c1 < kvk < c2} with 0 < c1 < c2. When a region is not simply connected, one often calls it multiply connnect or m.c.

3 6.3 Green’s theorem for simply connected plane regions

Recall that if f is a vector field with image in Rn, we can analyze f by its coordinate fields f1, f2, . . . , fn, which are scalar fields. When n = 2 (resp. n = 3), it is customary notation to use P,Q (resp. P, Q, R) instead of f1, f2 (resp. f1, f2, f3).

Theorem 1. (Green) Let f = (P,Q) be a C1 vector ﬁeld on an open set D in R2. Suppose C is a piecewise C1 Jordan curve with interior U such that Φ := C ∪ U lies entirely in D. Then we have

ZZ ∂Q ∂P I − dxdy = f · dα. ∂x ∂y Φ C Here C is the boundary ∂Φ of Φ, and moreover, the integral over C is taken in the positive direction. Note that

f · dα = (P (x, y),Q(x, y)) · (x0(t), y0(t))dt, and so one sometimes also uses the following notation I I f · dα = (P dx + Qdy).

C C This beautiful theorem combined with Theorem 3 from the last chapter has the following immediate

Corollary 1. Let f = (P,Q) be a C1 vector ﬁeld on a simply connected open set D in R2. Then ∂Q ∂P = ∂x ∂y if and only if I f · dα = 0 C for all Jordan curves C ⊆ D.

4 Remark. One can show that for simply connected D ⊆ R2 the condition ∂Q ∂P ∂x = ∂y is in fact equivalent to the vector ﬁeld f = (P,Q) being conservative, H i.e. the vanishing of C f · dα for all closed curves C. One possible proof of this fact uses step-paths and can be found in the textbook.

Given any C1 Jordan curve C with Φ = C ∪ U as above, we can try to ﬁnely subdivide the region using C1 arcs such that Φ is the union of subre- gions Φ1,..., Φr with Jordan curves as boundaries and with non-intersecting interiors, such that each Φj is simultaneously a region of type I and II (see chapter 5). This can almost always be achieved. So we will content ourselves, mostly due to lack of time and preparation on Jordan curves, with proving the theorem only for regions which are of type I and II. Theorem A follows from the following, seemingly more precise, identities:

RR ∂P H (i) ∂y dxdy = − P dx Φ C RR ∂Q H (ii) ∂x dxdy = Qdy. Φ C

Now suppose Φ is of type I, i.e., of the form {a ≤ x ≤ b, ϕ1(x) ≤ y ≤ ϕ2(x)}, with a < b and ϕ1, ϕ2 continuously diﬀerentiable on [a, b]. Then the boundary C has 4 components given by

C1 : graph of ϕ1(x), a ≤ x ≤ b

C2 : x = b, ϕ1(b) ≤ y ≤ ϕ2(b)

C3 : graph of ϕ2(x), a ≤ x ≤ b

C4 : x = a, ϕ1(a) ≤ y ≤ ϕ2(a).

As C is positively oriented, each Ci is oriented as follows: In C1, traverse from x = a to x = b; in C2, traverse from y = ϕ1(b) to y = ϕ2(b); in C3, go from x = b to x = a; an in C4, go from ϕ2(a) to ϕ1(a). It is easy to see that Z Z P dx = P dx = 0,

C2 C4 since C2 and C4 are vertical segments allowing no variation in x. Hence we have

5 (1)

  b I Z Z Z − P dx = −  P dx + P dx = [P (x, φ2(x)) − P (x, φ1(x)]dx.

C C1 C3 a

1 ∂P On the other hand, since f is a C vector ﬁeld, ∂y is continuous, and since Φ is a region of type I, we may apply Fubini’s theorem and get (2) b ϕ2(x) ZZ ∂P Z Z ∂P dxdy = dx . ∂y ∂y Φ a ϕ1(x) Note that x is ﬁxed in the inside integral on the right. We see, by the fundamental theorem of 1-variable Calculus, that (3) ϕ2(x) Z ∂P = P (x, ϕ (x)) − P (x, ϕ (x)). ∂y 2 1 ϕ1(x) Putting together (1), (2) and (3), we immediately obtain identity (i). Similarly (ii) can be seen to hold by exploiting the fact that Φ is also of type II. Hence Theorem A is now proved for a region Φ which is of type I and II.

6.4 An area formula

The example below will illustrate a very useful consequence of Green’s theorem, namely that the area of the inside of a C1 Jordan curve C can be computed as 1 I A = (xdy − ydx). (*) 2 C A proof of this fact is easily supplied by taking the vector ﬁeld f = (P,Q) in Theorem A to be given by P (x, y) = −y and Q(x, y) = x. Clearly, f is C1 everywhere on R2, and so the theorem is applicable. The identity for A ∂Q ∂P follows as ∂x = − ∂y = 1.

6 Example: Fix any positive number r, and consider the hypocycloid C parametrized by 2 3 3 α : [0, 2π] → R , t 7→ (r cos t, r sin t). Then C is easily seen to be a piecewise C1 Jordan curve. Note that it is also 2 2 2 given by x 3 + y 3 = r 3 . We have

2π 1 I 1 Z A = (xdy − ydx) = (xy0(t) − yx0(t))dt. 2 2 C 0

Now, x0(t) = 3r cos2 t(− sin t), and y0(t) = 3r sin2 t cos t. Hence

xy0(t) − yx0(t) = (r cos3 t)(3r sin2 t cos t) + (r sin3 t)(3r cos2 t sin t) which simpliﬁes to 3r2 sin2 t cos2 t, as sin2 t + cos2 t = 1. So we obtain

2π 2π 3r2 Z sin 2t2 3r2 Z 1 − cos 4t A = dt = dt, 2 2 8 2 0 0 by using the trigonometric identities sin 2u = 2 sin u cos u and cos 2u = 1 − 2 sin2 u. Finally, we then get

 2π  2r2 Z 3r2 sin 4t2π A =  (1 − cos 4t)dt = t − 16 16 4 0 0 i.e., 3πr2 A = . 8 Remark. Using (*) as well as some Fourier analysis (Wirtinger inequality) one can derive the isoperimetric inequality

4πA ≤ L2 H where L = C ds is the length of the Jordan curve C.

7 6.5 Green’s theorem for multiply connected regions

We mentioned earlier that the annulus in the plane is the simplest example of a non-simply connected region. But it is not hard to see that we can cut this region into two pieces, each of which is the interior of a C1 Jordan curve. We may then apply Theorem A of §3 to each piece and deduce statement over the annulus as a consequence. To be precise, pick any point z in R2 and consider ¯ ¯ Φ = Bz(r2) − Bz(r1), ¯ for some real numbers r1, r2 such that 0 < r1 < r2. Here Bz(ri) denotes the closed disk of radius ri and center z. Let C1, resp. C2, denote the positively oriented (circular) boundary of ¯ ¯ ¯ Bz(r1), resp. Bz(r2). Let Di be the ﬂat diameter of Bz(ri), i.e., the set {x0 −ri ≤ x ≤ x0 +ri, y = y0}, where x0 (resp. y0) denotes the x-coordinate (resp. y-coordinate) of z. Then D2 ∩ Φ splits as a disjoint union of two horizontal segments D+ = D2 ∩ {x > x0} and D− = D2 ∩ {x < x0}. Denote + − by Ci (resp. Ci ) the upper (resp. lower) half of the circle Ci, for i = 1, 2. Then Φ = Φ+ ∪ Φ−, where Φ+ (resp. Φ−) is the region bounded by C+ = + + − − − C2 ∪ D− ∪ C1 ∪ D+ (resp. C = C2 ∪ D+ ∪ D1 ∪ D−). We can orient the piecewise C1 Jordan curves C+ and C− in the positive direction. Let U +,U − denote the interiors of C+,C−. Then U + ∩ U − = ∅, and Φ± = C± ∪ U ±. Now let f = (P,Q) be a C1-vector ﬁeld on a connected open set containing ¯ Bz(r2). Then, combining what we get by applying Green’s theorem for s.c. regions to Φ+ and Φ−, we get:

ZZ ∂Q ∂P I I − dxdy = (P dx + Qdy) − (P dx + Qdy), (*) ∂x ∂y Φ C2 C1 where both the line integrals on the right are taken in the positive (counterclockwise) direction. Note that the minus sign in front of the second line integral on the right comes from the need to orient C± positively. In some sense, this is the key case to understand as any multiply connected region can be broken up into a ﬁnite union of shapes each of which can be continuously deformed to an annulus. Arguing as above, we get the following (slightly stronger) result:

8 Theorem 2. (Green’s theorem for m.c. plane regions) Let C1,C2,...,Cr be non-intersecting piecewise C1 Jordan curves in the plane with interiors U1,U2,...,Ur such that

(i) U1 ⊃ Ci, ∀i ≥ 2,

(ii) Ui ∩ Uj = ∅, for all i, j ≥ 2.

r Put Φ = C1 ∪ U1 − ∪i=2Ui, which is not simply connected if r ≥ 2. Also let f = (P,Q) be a C1 vector ﬁeld on a connected open set S containing Φ. Then we have r ZZ ∂Q ∂P I X I − dxdy = (P dx + Qdy) − (P dx + Qdy) ∂x ∂y i=2 Φ C1 Ci where each Cj, j ≥ 1 is positively oriented.

Corollary 2. Let C1,...Cr, f be as in Theorem B. In addition, suppose that ∂Q ∂P ∂x = ∂y everywhere on S. Then we have

r I X I (P dx + Qdy) = (P dx + Qdy). i=2 C1 Ci 6.6 The winding number

1 Let C be an oriented piecewise C curve in the plane, and let z0 = (x0, y0) be a point not lying on C. Then the winding number of C relative to z0 is deﬁned to be 1 I y − y x − x W (C, z ) := − 0 dx + 0 dy , 0 2π r2 r2 C

p 2 2 where r = k(x, y) − z0k = (x − x0) + (y − y0) . When C is parametrized by a piecewise C1 function α :[a, b] → R2, α(t) = (x(t), y(t)), then it is easy to see that

b Z 0 0 1 (x(t) − x0)y (t) − (y(t) − y0)x (t) W (C, z0) = 2 2 dt 2π (x(t) − x0) + (y(t) − y0) a

9 Some write W (α, z0) instead of W (C, z0)). However, the geometric meaning of this particular integral becomes much clearer if one uses polar coordinates around the point (x0, y0). This means that α(t) is now a function of the form

(x(t), y(t)) = (x0 + r(t) cos φ(t), y0 + r(t) sin φ(t)) for certain piecewise C1-functions r(t) and φ(t). Then

0 0 0 (x(t) − x0)y (t) = r(t) cos φ(t) r (t) sin φ(t) + r(t) cos φ(t)φ (t)

0 0 0 (y(t) − y0)x (t) = r(t) sin φ(t) r (t) cos φ(t) − r(t) sin φ(t)φ (t) and 0 0 2 0 (x(t) − x0)y (t) − (y(t) − y0)x (t) = r(t) φ (t) so that b 1 Z 1 W (C, z ) = φ0(t)dt = (φ(b) − φ(a)). 0 2π 2π a It is clear that this number is an integer if C is a closed curve. Indeed, if α(a) = α(b) then (cos(φ(a)), sin(φ(a))) = (cos(φ(b)), sin(φ(b))) which holds if and only if φ(a) − φ(b) = 2πk with k ∈ Z. This proves the ﬁrst half of the following useful result:

1 2 2 Proposition 1. Let C be a piecewise C closed curve in R , and let z0 ∈ R be a point not meeting C.

(a) W (C, z0) ∈ Z.

(b) If C is a Jordan curve then W (C, z0) ∈ {0, 1, −1}.

More precisely, in the case of a Jordan curve, W (C, z0) is 0 if z0 is outside C, and it equals ±1 if z0 is inside C.

Proof. Let C be a Jordan curve and z0 a point in the interior of C. One y−y0 x−x0 ∂Q ∂P checks that the vector ﬁeld f = (P,Q) = (− r2 , r2 ) satisﬁes ∂x = ∂y everywhere on R2 − {z}. So Theorem 2 implies that the integral of f over C equals the negative of the integral of f over a small circle around z0. But this last integral one directly computes to be ±1.

10 If z0 is an exterior point the integral will be equal to 0 by a similar argument. Done.

If C¯ denotes the curve obtained from C by reversing the orientation, then ¯ W (C, z0) = −W (C, z0). So the winding number allows us to ﬁnally give a rigorous deﬁnition of ”positively oriented”. The Jordan curve C is positively oriented if W (C, z0) = 1 for every point z0 inside C (i.e z0 lies in the bounded open set U in the Jordan curve theorem). A fun exercise will be to exhibit, for each integer n, a piecewise C1, closed curve C and a point z0 not lying on it, such that W (C, z0) = n. Of course this cannot happen if C is a Jordan curve if n 6= 0, ±1.

11 Chapter 7

Div, grad, and curl

7.1 The operator ∇ and the gradient:

Recall that the gradient of a differentiable scalar field ϕ on an open set D in Rn is given by the formula: ∂ϕ ∂ϕ ∂ϕ ∇ϕ = , ,..., . (7.1) ∂x1 ∂x2 ∂xn It is often convenient to define formally the differential operator in vector form as:

∂ ∂ ∂ ∇ = , ,..., . (7.2) ∂x1 ∂x2 ∂xn Then we may view the gradient of ϕ, as the notation ∇ϕ suggests, as the result of multiplying the vector ∇ by the scalar ﬁeld ϕ. Note that the order of multiplication matters, i.e., ∂ϕ is not ϕ ∂ . ∂xj ∂xj Let us now review a couple of facts about the gradient. For any j ≤ n, ∂ϕ is identically zero on D iﬀ ϕ(x1, x2, . . . , xn) is independent of xj. Conse- ∂xj quently,

∇ϕ = 0 on D ⇔ ϕ = constant. (7.3) Moreover, for any scalar c, we have:

∇ϕ is normal to the level set Lc(ϕ). (7.4) Thus ∇ϕ gives the direction of steepest change of ϕ.

1 7.2 Divergence

Let f : D → Rn, D ⊂ Rn, be a differentiable vector field. (Note that both spaces are n-dimensional.) Let f1, f2, . . . , fn be the component (scalar) fields of f. The divergence of f is defined to be

n X ∂fj div(f) = ∇ · f = . (7.5) ∂x j=1 j This can be reexpressed symbolically in terms of the dot product as ∂ ∂ ∇ · f = ( ,..., ) · (f1, . . . , fn). (7.6) ∂x1 ∂xn Note that div(f) is a scalar ﬁeld. Given any n × n matrix A = (aij), its trace is deﬁned to be:

n X tr(A) = aii. i=1 Then it is easy to see that, if Df denotes the Jacobian matrix, then

∇ · f = tr(Df). (7.7) Let ϕ be a twice differentiable scalar field. Then its Laplacian is defined to be

∇2ϕ = ∇ · (∇ϕ). (7.8) It follows from (7.1),(7.5),(7.6) that

2 2 2 2 ∂ ϕ ∂ ϕ ∂ ϕ ∇ ϕ = 2 + 2 + ··· + 2 . (7.9) ∂x1 ∂x2 ∂xn One says that ϕ is harmonic iﬀ ∇2ϕ = 0. Note that we can formally consider the dot product

n ∂ ∂ ∂ ∂ X ∂2 ∇ · ∇ = ( ,..., ) · ( ,..., ) = . (7.10) ∂x ∂x ∂x ∂x ∂x2 1 n 1 n j=1 j Then we have

2 ∇2ϕ = (∇ · ∇)ϕ. (7.11)

Examples of harmonic functions:

(i) D = R2; ϕ(x, y) = ex cos y. ∂ϕ x ∂ϕ x Then ∂x = e cos y, ∂y = −e sin y, ∂2ϕ x ∂2ϕ x 2 and ∂x2 = e cos y, ∂y2 = −e cos y. So, ∇ ϕ = 0. p (ii) D = R2 − {0}; ϕ(x, y) = log( x2 + y2) = log(r). ∂ϕ x ∂ϕ y ∂2ϕ (x2+y2)−2x(2x) −(x2−y2) ∂2ϕ Then ∂x = x2+y2 , ∂y = x2+y2 , ∂x2 = (x2+y2)2 = (x2+y2)2 , and ∂y2 = (x2+y2)−2y(2y) (x2−y2) 2 (x2+y2)2 = (x2+y2)2 . So, ∇ ϕ = 0. These last two examples are special cases of the fact, which we mention without proof, that for any function f : D → C which is differentiable in the complex sense, the real and imaginary part, <(f) and =(f), are harmonic functions. Here f is differentiable in the complex sense if its total derivative Df at a point z ∈ D, a priori a R-linear map from C to itself, is in fact given by multiplication with a complex number, which we then call f 0(z). More concretely, this means that the matrix of Df in the basis 1, i is of the form a −b 0 b a for some real numbers a, b. We then have f (z) = a + bi. There is a large supply of such functions since any f given (locally) by a convergent power series in z is complex differentiable. In (i) we can take f(z) = ez = ex+iy = ex cos(y) + iex sin(y) and in (ii) we can take f(z) = log(z) = log(reiθ) = log(r) + iθ but we must be careful about the domain. To have a well defined argument θ for all z ∈ D we must make a ”cut” in the plane and can only define f on, for example, D = {z = x + iy| y = 0 ⇒ x > 0} or D0 = {z = x + iy| y = 0 ⇒ x < 0}. But the union of D and D0 is C − {0} as in (ii) . n 2 2 2 α/2 α (iii) D = R − {0}; ϕ(x1, x2, ..., xn) = (x1 + x2 + ··· + xn) = r for some fixed α ∈ R. ∂ϕ α−1 xi α−2 Then = αr = αr xi, and ∂xi r ∂2ϕ α−4 α−2 2 = α(α − 2)r xi · xi + αr · 1. ∂xi 2 Pn α−4 2 α−2 α−2 Hence ∇ φ = i=1 (α(α − 2)r xi + αr ) = α(α − 2 + n)r . So φ is harmonic for α = 0 or α = 2 − n (α = −1 for n = 3).

3 7.3 Cross product in R3 The three-dimensional space is very special in that it admits a vector product, often called the cross product. Let i,j,k denote the standard basis of R3. Then, for all pairs of vectors v = xi + yj + zk and v0 = x0i + y0j + z0k, the cross product is deﬁned by

i j k v × v0 = det x y z = (yz0 − y0z)i − (xz0 − x0z)j + (xy0 − x0y)k. (7.12) x0 y0 z0

Lemma 1 (a) v×v0 = −v0 ×v (anti-commutativity) (b) i×j = k, j×k = i, k × i = j (c) v · (v × v0) = v0 · (v × v0) = 0.

Corollary: v × v = 0.

Proof of Lemma (a) v0 × v is obtained by interchanging the second and third rows of the matrix whose determinant gives v×v0. Thus v0 ×v=−v×v0. i j k (b) i × j = det 1 0 0 , which is k as asserted. The other two identities 0 1 0 are similar. (c) v · (v × v0) = x(yz0 − y0z) − y(xz0 − x0z) + z(xy0 − x0y) = 0. Similarly for v0 · (v × v0).

Geometrically, v×v0 can, thanks to the Lemma, be interpreted as follows. Consider the plane P in R3 deﬁned by v,v0. Then v × v0 will lie along the normal line to this plane at the origin, and its orientation is given by the right hand rule: If the ﬁngers of your right hand grab a pole and you view them from the top as a circle in the v − v0-plane that is oriented counterclockwise (i.e. corresponding to the ordering (v, v0) of the basis) then the thumb points in the direction of v × v0. Finally the length ||v × v0|| is equal to the area of the parallelogram spanned by v and v0. Indeed this area is equal to the volume of the paral- 0 lelepiped spanned by v, v and a unit vector u = (ux, uy, uz) orthogonal to v and v0. We can take u = v × v0/||v × v0|| and the (signed) volume equals   ux uy uz 0 0 0 0 0 0 det  x y z  =ux(yz − y z) − uy(xz − x z) + uz(xy − x y) x0 y0 z0 0 2 2 2 0 =||v × v || · (ux + uy + uz) = ||v × v ||.

4 More generally, the same argument shows that the (signed) volume of the parallelepiped spanned by any three vectors u, v, v0 is u · (v × v0).

7.4 Curl of vector ﬁelds in R3

Let f : D → R3, D ⊂ R3 be a differentiable vector field. Denote by P ,Q,R its coordinate scalar fields, so that f = P i + Qj + Rk. Then the curl of f is defined to be:

i j k ∂ ∂ ∂ curl(f) = ∇ × f = det ∂x ∂y ∂z . (7.13) PQR Note that it makes sense to denote it ∇ × f, as it is formally the cross product of ∇ with f. If the vector field f represents the flow of a fluid, then the curl measures how the flow rotates the vectors, whence its name.

Proposition 1 Let h (resp. f) be a C2 scalar (resp. vector) ﬁeld. Then

(a) ∇ × (∇h) = 0.

(b) ∇ · (∇ × f) = 0.

Proof: (a) By deﬁnition of gradient and curl,

i j k ! ∂ ∂ ∂ ∇ × (∇h) = det ∂x ∂y ∂z ∂h ∂h ∂h ∂x ∂y ∂z

∂2h ∂2h ∂2h ∂2h ∂2h ∂2h = − i + − j + − k. ∂y∂z ∂z∂y ∂z∂x ∂x∂z ∂x∂y ∂y∂x

Since h is C2, its second mixed partial derivatives are independent of the order in which the partial derivatives are computed. Thus, ∇ × (∇f) = 0. (b) By the deﬁnition of divergence and curl,

∂ ∂ ∂ ∂R ∂Q ∂R ∂P ∂Q ∂P ∇ · (∇ × f) = , , · − , − + , − ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y

5 ∂2R ∂2Q ∂2R ∂2P ∂2Q ∂2P = − + − + + − . ∂x∂y ∂x∂z ∂y∂x ∂y∂z ∂z∂x ∂z∂y

2 ∂2R ∂2R Again, since f is C , ∂x∂y = ∂y∂x , etc., and we get the assertion. Done. Warning: There exist twice differentiable scalar (resp. vector) fields h (resp. f), which are not C2, for which (a) (resp. (b)) does not hold. When the vector field f represents fluid flow, it is often called irrotational when its curl is 0. If this flow describes the movement of water in a stream, for example, to be irrotational means that a small boat being pulled by the flow will not rotate about its axis. We will see later in this chapter the condition ∇ × f = 0 occurs naturally in a purely mathematical setting as well.

3 y x Examples: (i) Let D = R − {0} and f(x, y, z) = (x2+y2) i − (x2+y2) j. Show that f is irrotational. Indeed, by the deﬁnition of curl,

i j k ! ∂ ∂ ∂ ∇ × f = det ∂x ∂y ∂z y −x 0 (x2+y2) (x2+y2)

∂ x ∂ y ∂ −x ∂ y = i + j + − k ∂z x2 + y2 ∂z x2 + y2 ∂x x2 + y2 ∂y x2 + y2

−(x2 + y2) + 2x2 (x2 + y2) − 2y2 = − k = 0. (x2 + y2)2 (x2 + y2)2 (ii) Let m be any integer 6= 3, D = R3 − {0}, and 1 p 2 2 2 f(x, y, z) = rm (xi + yj + zk), where r = x + y + z . Show that f is not the curl of another vector ﬁeld. Indeed, suppose f = ∇ × g. Then, since f is C1, g will be C2, and by the Proposition proved above, ∇ · f = ∇ · (∇ × g) would be zero. But,

∂ ∂ ∂ x y z ∇ · f = , , · , , ∂x ∂y ∂z rm rm rm

rm − 2x2( m )rm−2 rm − 2y2( m )rm−2 rm − 2z2( m )rm−2 = 2 + 2 + 2 r2m r2m r2m

6 1 1 = 3rm − m(x2 + y2 + z2)rm−2 = (3 − m). r2m rm This is non-zero as m 6= 3. So f is not a curl.

Warning: It may be true that the divergence of f is zero, but f is still not a curl. In fact this happens in example (ii) above if we allow m = 3. We cannot treat this case, however, without establishing Stoke’s theorem.

7.5 An interpretation of Green’s theorem via the curl

Recall that Green’s theorem for a plane region Φ with boundary a piecewise C1 Jordan curve C says that, given any C1 vector ﬁeld g = (P,Q) on an open set D containing Φ, we have:

ZZ ∂Q ∂P I − dx dy = P dx + Q dy. (7.14) Φ ∂x ∂y C ∂Q ∂P We will now interpret the term ∂x − ∂y . To do that, we think of the plane as sitting in R3 as {z = 0}, and deﬁne a C1 vector ﬁeld f on D˜ := {(x, y, z) ∈ R3|(x, y) ∈ D} by setting f(x, y, z) = g(x, y) = P i + Qj. Then i j k ∂ ∂ ∂ ∂Q ∂P ∂P ∂Q ∇ × f = det ∂x ∂y ∂z = ∂y − ∂x k, because ∂z = ∂z = 0. Thus we PQ 0 get: ∂Q ∂P (∇ × f) · k = − . (7.15) ∂y ∂x And Green’s theorem becomes: RR H Theorem 1 Φ(∇ × f) · k dx dy = C P dx + Q dy

7.6 A criterion for being conservative via the curl

Here we just reformulate the remark after Ch. 6, Cor. 1 (which we didn’t completely prove but just made plausible) using the curl.

7 Proposition 1 Let g : D → R2, D ⊂ R2 open and simply connected, g = (P,Q), be a C1 vector ﬁeld. Set f(x, y, z) = g(x, y), for all (x, y, z) ∈ R3 with (x, y) ∈ D. Suppose ∇ × f = 0. Then g is conservative on D. H Proof: Since ∇ × f = 0, Theorem 1 implies that C P dx + Q dy = 0 for all Jordan curves C contained in D. In fact, ∇ × f = 0 also implies that H C P dx + Q dy = 0 for all closed curves but we won’t prove this. Hence f is conservative. Done.

2 2 y x Example: D = R − {(x, 0) ∈ R | x ≤ 0}, g(x, y) = x2+y2 i − x2+y2 j. Determine if g is conservative on D:

Again, deﬁne f(x, y, z) to be g(x, y) for all (x, y, z) in R3 such that (x, y) ∈ D. Since g is evidently C1, f will be C1 as well. By the Proposition above, it will suﬃce to check if f is irrotational, i.e., ∇ × f = 0, on D × R. This was already shown in Example (i) of section 4 of this chapter. So g is conservative.