16.5: CURL AND DIVERGENCE

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1 16.5: URL AN IVERGENE KIAM HEONG KWA 1. url Let F = P i + Qj + Rk be a vector field on a solid region R 3. If all first-order partial derivatives of P, Q, and R exist, then the curl of F on is the vector field (1.1) curl F = ( R Q z ) i + By introducing the del operator (1.2) = i x + j + k z ( P z R ) ( Q j + x x P ) k. and considering as a vector whose components are / x, /, and / z, we can consider curl F as the formal cross product of the operator with the vector field F : i j k (1.3) F = x z P Q R = i z x z + k x Q R j ( R = Q z = curl F. P ) i + R ( P z R x P ) j + Q ( Q x P ) k Remark 1. When the del operator acts on a scalar function f, we obtain the gradient of f: f = i f x + j f + k f z = f x i + f j + f z k. ate: November 14,

2 2 KIAM HEONG KWA If the curl of a vector field is 0 identically, then the vector field is said to be irrotational. It can be shown that if F is a conservative vector field whose components have continuous partial derivatives, then F is irrotational. In fact, if f is a potential function of F, then curl F = f i j k = x z f x f y f z = i (f zy f yz ) + j (f xz f zx ) + k (f yx f xy ) = 0i + 0j + 0k by lairaut s theorem = 0 everywhere in. More is true: as a consequence of the Stoke s theorem in section 16.8, the converse also holds if the domain of F is simplyconnected. Hence we have: Theorem 1. Let F be a vector field on a solid region R 3 whose components have continuous first-order partial derivatives. If is simply-connected, then F is irrotational if and only if F is conservative. Example 1 (Problem 14 in the text). The vector field F (x, y, z) = xyz 2 i + x 2 yz 2 j + x 2 y 2 zk is not conservative because F 0: F = (x2 y 2 z) ] z (x2 yz 2 ) i + z (xyz2 ) ] x (x2 y 2 z) j + x (x2 yz 2 ) ] (xyz2 ) k = ( 2x 2 yz 2x 2 yz ) i + ( 2xyz 2xy 2 z ) j + ( 2xyz 2 xz 2) k = ( 2xyz 2xy 2 z ) j + ( 2xyz 2 xz 2) k. Example 2 (Problem 16 in the text). The vector field F (x, y, z) = e z i + j + xe z k is conservative because its domain, i.e., R 3, is simplyconnected and F = (xez ) ] z (1) i + + x (1) ] (ez ) k = 0. z (ez ) ] x (xez ) j

3 16.5: URL AN IVERGENE 3 In consequence, there is a scalar field f such that f = F. Explicitly, one has Integrating these equations yields f x (x, y, z) = e z, f y (x, y, z) = 1, f z (x, y, z) = xe z. f(x, y, z) = xe z + g(y, z), f(x, y, z) = y + h(x, z), f(x, y, z) = xe z + k(x, y), where g(y, z), h(x, z), and k(x, y) are functions of respective variables such that all representations of the potential function f agree with each other. omparing these representations, one gets g(y, z) = y + K, h(x, z) = xe z + K, and k(x, y) = y + K, so that for an arbitrary constant K. f(x, y, z) = xe z + y + K 2. ivergence Let F = P i + Qj + Rk be a vector field on a solid region R 3. If P x, Q y, and R z exist, then the divergence of F is the scalar field (2.1) div F = P x + Q + R z. By treating the del operator as a vector, we can write div F symbolically as the dot product of and F : ( div F = i x + j + k ) (2.2) (P i + Qj + Rk) z = F. If the divergence of a vector field vanishes identically, then F is said to be incompressible. In can be shown that if the components of the vector field F have continuous second-order partial derivatives, then (2.3) div curl F = F = 0 everywhere in. This can be validated using the definitions of divergence and curl together with the lairaut s theorem.

4 4 KIAM HEONG KWA Example 3 (Problem 8 in the text). Let F (x, y, z) = e x i+e xy j+e xyz k. Then F = exyz ] z exy i + z ex ] x exyz j + x exy ] ex k and = xze xyz i yze xyz j + ye xy k F = x ex + exy + z exyz = e x + xe xy + xye xyz. Example 4 (Problem 20 in the text). There is no vector field G in R 3 such that G = xyz, y 2 z, yz 2. Suppose, to the contrary, that such a vector field exists. Then G = x (xyz) + ( y2 z) + z (yz2 ) = yz 0. This contradicts (2.3). 3. Laplace Operator Another operator that can be constructed from the del operator is the Laplace operator: (3.1) 2 = = 2 x z 2. Note that for a scalar function f whose second-order partial derivatives exists, we have (3.2) 2 f = f = 2 f x f f z 2, the divergence of the gradient field f. 4. Vector Forms of Green s Theorem We can define the curl and the divergence of a vector field F = P i + Qj on a plane region R 2 by treating F as a vector field on the solid region R = {(x, y, z) R 3 (x, y), z R} R 3. This is done by considering the components P and Q as functions on R

5 16.5: URL AN IVERGENE 5 that are independent of z and the third component of F as 0. The results are ( Q (4.1) curl F = F = x P ) k and (4.2) div F = F = P x + Q provided the corresponding partial derivatives exist. We would like to express Green s theorem in terms of curl F = F. Suppose is the region bounded by a positively oriented, piecewise smooth, simple closed curve. Suppose also that P and Q have continuous partial derivatives on an open plane region that contains. Next, recall from section 16.2 that the line integral of F along is given by F dr = P dx + Q dy. Recall also Green s theorem from the last section: ( Q P dx + Q dy = x P ) da. Hence in terms of curl, Green s theorem reads (4.3) F dr = curl F k da = F k da. Next, suppose is parametrized by r(t) = x(t)i + y(t)j, a t b. Then then unit tangent vector is given by (4.4) T (t) = r (t) r (t) = x (t) r (t) i + y (t) r (t) j. To find a normal vector N(t) = U(t)i + V (t)j to the curve, i.e., a vector that is perpendicular to, we demand that N(t) T (t) = x (t)u(t) + y (t)v (t) = 0. Two obvious nonzero solutions are and Note that N(t) = y (t)i x (t)j N(t) = y (t)i + x (t)j.

6 6 KIAM HEONG KWA when y (t) > 0, the first component of N(t) is positive, when y (t) < 0, the first component of N(t) is negative, when x (t) > 0, the second component of N(t) is negative, when x (t) < 0, the second component of N(t) is positive. This, together with the continuity of N(t), shows that N(t) is always pointing outward from the region, from which it follows that (4.5) n(t) = N(t) N(t) = y (t) r (t) i x (t) r (t) j is the outward unit normal vector to. We are ready to express Green s theorem in terms of div F = F : b F n ds = F (t) n(t) r (t) dt a b [ P (x(t), y(t))y (t) = Q(x(t), ] y(t))x (t) r (t) dt r (t) r (t) a b b = P (x(t), y(t))y (t) dt Q(x(t), y(t))x (t) dt a a = P dy Q dx ( P = x + Q ) da by Green s theorem = div F da by (4.2). In short, we have (4.6) F n ds = div F da = F da. In words, this says that the line integral of the normal component of F, i.e., F n, along is equal to the double integral of the divergence of F over the region bounded by. Example 5 (Green s first identity). Suppose that the appropriate partial derivatives of the scalar fields f and g exist and are continuous. Then it can be shown that (f g) = f g + f 2 g.

7 16.5: URL AN IVERGENE 7 Hence f g n ds = = (f g) da f g da + f 2 g da by (4.6), from which it follows Green s first identity: (4.7) f 2 g da = f g n ds f g da. Example 6 (Green s second identity). Suppose that the appropriate partial derivatives of the scalar fields f and g exist and are continuous. Then by Green s first identity, one has g 2 f da = g f n ds g f da. Subtracting this equation from (4.7) yields Green s second identity: (4.8) (f 2 g g 2 f) da = (f g g f) n ds. Appendix A. Vector Identities Let f and g be scalar fields and let F and G be vector fields. If all the appropriate partial derivatives exist and are continuous, then the following vectorial identities hold: (1) (fg) = f g + g f (2) (F + G) = F + G (3) (F + G) = F + G (4) (ff ) = f F + f F (5) (ff ) = f F + f F (6) (F G) = G F F G (7) ( f g) = 0 (8) ( F ) = ( F ) 2 F Example 7 (Problem 38 in the text). Maxwell s equations relating the electric field E and magnetic field H as they vary with time in a region

8 8 KIAM HEONG KWA containing no charge and no current can be stated as follows: E = 0, H = 0, E = 1 H c t, H = 1 E c t, where c is the speed of light in vacuum. It follows from these equations that Likewise, Also, ( E) = 1 c H t = 1 ( H) c t = 1 ( ) 1 E c t c t = 1 2 E c 2 t. 2 ( H) = 1 c E t = 1 ( E) c t = 1 ( 1 H c t c t = 1 2 H c 2 t. 2 2 E = ( E) ( E) ( = (0) 1 ) H c t = 1 ( H) c t = 1 ( ) 1 E c t c t = 1 2 E c 2 t. 2 )

9 16.5: URL AN IVERGENE 9 Lastly, 2 H = ( H) ( H) ( ) 1 E = (0) c t = 1 ( E) c t = 1 ( 1 ) H c t c t = 1 2 H c 2 t. 2

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