# Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula:

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1 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 R n is given by the formula: ( ϕ ϕ =, ϕ,..., ϕ. (7.1 x 1 x 2 x n It is often convenient to define formally the differential operator in vector form as: ( =,,...,. (7.2 x 1 x 2 x n 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 ϕ of multiplication matters, i.e., x j is not ϕ x j. Let us now review a couple of facts about the gradient. For any j n, ϕ x j is identically zero on D iff ϕ(x 1, x 2,..., x n is independent of x j. Consequently, Moreover, for any scalar c, we have: ϕ = 0 on D ϕ = constant. (7.3 ϕ is normal to the level set L c (ϕ. (7.4 Thus ϕ gives the direction of steepest change of ϕ. 1

2 7.2 Divergence Let f : D R n, D R n, be a differentiable vector field. (Note that both spaces are n-dimensional. Let f 1, f 2,..., f n be the component (scalar fields of f. The divergence of f is defined to be div(f = f = n j=1 f j x j. (7.5 This can be reexpressed symbolically in terms of the dot product as f = (,..., (f 1,..., f n. (7.6 x 1 x n Note that div(f is a scalar field. Given any n n matrix A = (a ij, its trace is defined to be: tr(a = n a ii. 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 It follows from (7.1,(7.5,(7.6 that 2 ϕ = 2 ϕ x ϕ = ( ϕ. ( ϕ x ϕ. (7.9 x 2 n One says that ϕ is harmonic iff 2 ϕ = 0. Note that we can formally consider the dot product Then we have = ( x 1,..., (,..., x n x 1 x n = n 2 x 2 j=1 j. (7.10 2

3 2 ϕ = ( ϕ. (7.11 Examples of harmonic functions: (i D = R 2 ; ϕ(x, y = e x cos y. Then ϕ = x ex cos y, ϕ = y ex sin y, and 2 ϕ = e x cos y, 2 ϕ = e x cos y. So, 2 ϕ = 0. x 2 y 2 (ii D = R 2 {0}; ϕ(x, y = log( x 2 + y 2 = log(r. Then ϕ = x, ϕ = y, 2 ϕ = (x2 +y 2 2x(2x = (x2 y 2, and 2 ϕ = x x 2 +y 2 y x 2 +y 2 x 2 (x 2 +y 2 2 (x 2 +y 2 2 y 2 (x 2 +y 2 2y(2y = (x2 y 2. So, 2 ϕ = 0. (x 2 +y 2 2 (x 2 +y 2 2 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, R(f and I(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 (z. More concretely, ( this means that the matrix of Df in the basis 1, i is of the form a b 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 = e z = e x+iy = e x cos(y + ie x sin(y and in (ii we can take f(z = log(z = log(re iθ = 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 D = {z = x + iy y = 0 x < 0}. But the union of D and D is C {0} as in (ii. (iii D = R n {0}; ϕ(x 1, x 2,..., x n = (x x x 2 n α/2 = r α for some fixed α R. Then ϕ x i = αr α 1 x i = r αrα 2 x i, and = α(α 2r α 4 x i x i + αr α ϕ x 2 i Hence 2 φ = n i=1 (α(α 2rα 4 x 2 i + αr α 2 = α(α 2 + nr α 2. So φ is harmonic for α = 0 or α = 2 n (α = 1 for n = 3. 3

4 7.3 Cross product in R 3 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 R 3. Then, for all pairs of vectors v = xi + yj + zk and v = x i + y j + z k, the cross product is defined by v v = det x y z x y z = (yz y zi (xz x zj + (xy x yk. (7.12 Lemma 1 (a v v = v v (anti-commutativity (b i j = k, j k = i, k i = j (c v (v v = v (v v = 0. Corollary: v v = 0. Proof of Lemma (a v v is obtained by interchanging the second and third rows of the matrix whose determinant gives v v. Thus v v= v v. (b i j = det, which is k as asserted. The other two identities are similar. (c v (v v = x(yz y z y(xz x z + z(xy x y = 0. Similarly for v (v v. Geometrically, v v can, thanks to the Lemma, be interpreted as follows. Consider the plane P in R 3 defined by v,v. Then v v 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 v -plane that is oriented counterclockwise (i.e. corresponding to the ordering (v, v of the basis then the thumb points in the direction of v v. Finally the length v v is equal to the area of the parallelogram spanned by v and v. Indeed this area is equal to the volume of the parallelepiped spanned by v, v and a unit vector u = (u x, u y, u z orthogonal to v and v. We can take u = v v / v v and the (signed volume equals u x u y u z det x y z =u x (yz y z u y (xz x z + u z (xy x y x y z = v v (u 2 x + u 2 y + u 2 z = v v. 4

5 More generally, the same argument shows that the (signed volume of the parallelepiped spanned by any three vectors u, v, v is u (v v. 7.4 Curl of vector fields in R 3 Let f : D R 3, D R 3 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: curl(f = f = det x y z P Q R. (7.13 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 C 2 scalar (resp. vector field. Then (a ( h = 0. (b ( f = 0. Proof: (a By definition of gradient and curl, = ( h = det x y z h x h y h z ( ( ( 2 h yz 2 h 2 h i + zy zx 2 h 2 h j + xz xy 2 h k. yx Since h is C 2, 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, ( f = ( x, y, z ( R y Q z, R x + P z, Q x P y 5

6 ( ( ( 2 R = xy 2 Q + 2 R xz yx + 2 P 2 Q + yz zx 2 P. zy Again, since f is C 2, 2 R Done. = 2 R xy yx, etc., and we get the assertion. Warning: There exist twice differentiable scalar (resp. vector fields h (resp. f, which are not C 2, 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. Examples: (i Let D = R 3 {0} and f(x, y, z = i x (x 2 +y 2 that f is irrotational. Indeed, by the definition of curl, = z f = det x y z 0 y (x 2 +y 2 x (x 2 +y 2 y (x 2 +y 2 j. Show ( x i + ( ( ( y x j + ( y k x 2 + y 2 z x 2 + y 2 x x 2 + y 2 y x 2 + y 2 = [ ] (x 2 + y 2 + 2x 2 (x2 + y 2 2y 2 k = 0. (x 2 + y 2 2 (x 2 + y 2 2 (ii Let m be any integer 3, D = R 3 {0}, and f(x, y, z = 1 (xi + yj + zk, where r = x r 2 + y 2 + z 2. Show that f is not m the curl of another vector field. Indeed, suppose f = g. Then, since f is C 1, g will be C 2, and by the Proposition proved above, f = ( g would be zero. But, ( f = x, y, ( x z r, y m r, z m r m = rm 2x 2 ( m 2 rm 2 r 2m + rm 2y 2 ( m 2 rm 2 r 2m 6 + rm 2z 2 ( m 2 rm 2 r 2m

7 = 1 ( 3r m m(x 2 + y 2 + z 2 r m 2 = 1 (3 m. r 2m rm This is non-zero as m 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 C 1 Jordan curve C says that, given any C 1 vector field g = (P, Q on an open set D containing Φ, we have: ( Q x P dx dy = P dx + Q dy. (7.14 y Φ We will now interpret the term Q P x y C. To do that, we think of the plane as sitting in R 3 as {z = 0}, and define a C 1 vector field f on D := {(x, y, z R( 3 (x, y D} by setting f(x, y, z = g(x, y = P i + Qj. Then i j k ( Q f = det x y z = P k, because P = Q = 0. Thus we y x z z P Q 0 get: ( f k = Q y P x. (7.15 And Green s theorem becomes: Theorem 1 ( f k dx dy = P dx + Q dy Φ C 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

8 Proposition 1 Let g : D R 2, D R 2 open and simply connected, g = (P, Q, be a C 1 vector field. Set f(x, y, z = g(x, y, for all (x, y, z R 3 with (x, y D. Suppose f = 0. Then g is conservative on D. Proof: Since f = 0, Theorem 1 implies that P dx + Q dy = 0 for all C Jordan curves C contained in D. In fact, f = 0 also implies that P dx + Q dy = 0 for all closed curves but we won t prove this. Hence f is C conservative. Done. Example: D = R 2 {(x, 0 R 2 x 0}, g(x, y = Determine if g is conservative on D: y x 2 +y 2 i x j. x 2 +y 2 Again, define f(x, y, z to be g(x, y for all (x, y, z in R 3 such that (x, y D. Since g is evidently C 1, f will be C 1 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. 8

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