1. When we take derivatives to obtain We call  the del operator and write df — or  f, we can think of dx d/dx and  as operators (in the sense that they represent a process that we put a function through).   =i — + j —or  x  y     =i — + j — + k —.  x  y  z We can now define the divergence of a vector field F(x,y,z) as the dot product of  and F, that is, the divergence of F is  F 1  F 2  F 3 div F = F =— + — + —.  x  y  z This definition can be adapted in the obvious way to a vector field in R n for any n; for instance, in R 2, the divergence of F(x,y) is  F 1  F 2 div F = F =— + —.  x  y

2. If a vector field in R 3 represents the velocity field of a gas or fluid, the divergence can be interpreted as the rate of expansion per unit volume, while such a vector field in R 2 can be interpreted as the rate of expansion per unit area. Consider the vector field described by F(x,y) = xi + yj. The flow lines point If the flow lines are those of a gas, then the gas is as it moves away from the origin. Consequently, we should expect that div F > 0. We verify this by observing that away from the origin. expanding div F = 1 + 1 = 2 > 0. Consider the vector field described by F(x,y) = – xi – yj. The flow lines point If the flow lines are those of a gas, then the gas is as it moves toward the origin. Consequently, we should expect that div F < 0. We verify this by observing that toward the origin. compressing div F = – 1 – 1 = – 2 < 0

3. Consider the vector field described by F(x,y) = – yi + xj. In a previous example, we found that the flow lines are If the flow lines are those of a gas, then the gas is Consequently, we should expect that div F = 0. We verify this by observing that concentric, counterclockwise circles around the origin. neither compressing nor expanding as it moves around the origin. div F = 0 + 0 = 0. Consider the vector field described by F(x,y) = x 2 yi – xj. It is not easy to intuit what the divergence will be. By obtaining div F =, we find that 2xy + 0 = 2xy the divergence is not the same at all points in the vector field. (expansion) (compression)

4. We define the curl of a vector field F(x,y,z) as the cross product of  and F, that is, the curl of F is  F 3  F 2  F 1  F 3  F 2  F 1 curl F =  F =( — – — )i + ( — – — )j + ( — – — )k.  y  z  z  x  x  y Curl is an operator defined only in R 3, whereas divergence is an operator defined in R n for any n. If F(x,y,z) = – yi + xj + zk, then curl F =2k. If F(x,y,z) = x 2 yi – xzj + yz 2 k, then curl F = (z 2 + x)i + 0j + (– z – x 2 )k.

5. Let us find the curl for a vector field which describes motion of points in a rotating body with axis of rotation along the z axis; the value of the vector field at each point is the velocity vector v at that point. We let r = xi + yj + zk = v =  = ||v|| = position vector of a point on the rotating object velocity vector of a point on the rotating object (i.e., the vector field) angular velocity of a point on the rotating object  x 2 + y 2 (e.g., speed doubles if distance to z axis is doubled.) xy plane k r = position vector of a point on the rotating object  v Observe that v is in the direction of k  r = k  (xi + yj + zk) = v = 2  k.–  yi +  xj  curl v = – yi + xj  If a vector field F represents the flow of a fluid, then we see that curl F at a point has a magnitude of twice the angular velocity vector of a rigid rotation in the direction of the axis of rotation; if (curl F) = 0 = 0i + 0j + 0k at a point, then the fluid is free from rigid rotations (i.e., whirlpools) at that point.

6. Consider the vector field described by y x F(x,y,z) =——— i –——— j. x 2 + y 2 x 2 + y 2 Using results from a previous example, we find that for any plane parallel to the xy plane, the flow lines form (0 – 0)i + (0 – 0)j + 0k = 0. When (curl F) = 0 for a fluid, then a sufficiently small object (such as a paddle wheel) will not rotate as it moves with the fluid (even though the object may be “rotating” around a point with the fluid flow). Such a vector field is called irrotational. concentric, clockwise circles around the origin, with velocity becoming larger closer to the origin. curl F =

7. Suppose V is a vector field from R 3 to R 3, and V =  f, where f is a function with continuous partial derivatives of at least the second order, i.e., V is a gradient vector field. Then, curl V =  V =  (  f) =  (f x i + f y j + f z k) = (f zy – f yz )i + (f xz – f zx )j + (f yx – f xy )k=0i + 0j + 0k = 0. Consequently, if (curl V)  0, then V cannot be a gradient vector field. It can be shown that for vector fields V with continuous component functions, V is a gradient vector field if and only if (curl V) = 0. Is the vector field described by y x F(x,y,z) =——— i –——— j a gradient vector field? x 2 + y 2 x 2 + y 2 Since we have previously seen that (curl F) = 0, then we know that F is a gradient vector field. (In fact, F =  f, where f(x,y,z) = Arctan(x/y).)

8. Could it be possible that the vector field described by F(x,y,z) = yi – xj is a gradient vector field? cannot be a gradient vector field. Note that if F = P(x,y)i + Q(x,y)j is a vector field in R 2, then F can be regarded as a vector field in R 3 by letting F = We then have that curl F = – 2x – 1. Consider the vector field described by F(x,y) = yi – x 2 j. The scalar curl is Since curl F =(0 – 0)i + (0 – 0)j + (– 1 – 1)k = – 2k, then F (0 – 0)i + (0 – 0)j + (Q x – P y )k = (Q x – P y )k. The function Q x – P y is called the scalar curl of F. P(x,y)i + Q(x,y)j + 0k.

9. Suppose V = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k is a vector field with each of the three component functions having continuous partial derivatives of at least the second order. Then, div (curl V) =  (  V) =  [( )i + ( )j + ( )k] = ( ) + ( ) + ( ) = 0 Consequently, if (div F)  0, then F cannot be the curl of any vector field. It can be shown that for vector fields F with continuous component functions, F is the curl of another vector field if and only if div(F) = 0. R y – Q z P z – R x Q x – P y R yx – Q zx P zy – R xy Q xz – P yz Can F(x,y,z) = xi + yj + zk possibly be the curl of another vector field? 1 + 1 + 1 = 3  0, then Since div F = F is not the curl of any vector field. Can F(x,y,z) = yi + zj + xk possibly be the curl of another vector field? 0 + 0 + 0 = 0, thenSince div F =F is the curl of another vector field.

10. Consider again the definition of (div V) = V, and suppose that V =  f, where f is a function with continuous partial derivatives of at least the second order, i.e., V is a gradient vector field. Then, we may write Find  2 f for f(x,y,z) = 1 / (x 2 +y 2 +z 2 ) 1/2. f x = f xx = f y = f yy = f z = f zz = – x / (x 2 +y 2 +z 2 ) 3/2 – y / (x 2 +y 2 +z 2 ) 3/2 – z / (x 2 +y 2 +z 2 ) 3/2 3x 2 / (x 2 +y 2 +z 2 ) 5/2 – 1 / (x 2 +y 2 +z 2 ) 3/2 3y 2 / (x 2 +y 2 +z 2 ) 5/2 – 1 / (x 2 +y 2 +z 2 ) 3/2 3z 2 / (x 2 +y 2 +z 2 ) 5/2 – 1 / (x 2 +y 2 +z 2 ) 3/2  2 f = 0 (div V) = V =  (  f) =  2 f = The operator  2 is called the Laplace operator.  2 f  2 f  2 f — + — + —.  x 2  y 2  z 2

11.  (f+g) = (f x +g x )i + (f y +g y )j + (f z +g z )k = (f x i + f y j + f z k) + (g x i + g y j + g z k) =  f +  g  (cf) = (cf x )i + (cf y )j + (cf z )k = c(f x i + f y j + f z k) = cfcf  (fg) = (f x g+fg x )i + (f y g+fg y )j + (f z g+fg z )k = (f x gi + f y gj + f z gk) + (fg x i + fg y j + fg z k) = g(f x i + f y j + f z k) + f(g x i + g y j + g z k) = g  f + f  g =f  g + g  f  (f / g) = [(f x g–fg x ) / g 2 ]i + [(f y g–fg y ) / g 2 ]j + [(f z g–fg z ) / g 2 ]k = (1/g 2 )[(f x g–fg x )i + (f y g–fg y )j + (f z g–fg z )k] = (1/g 2 )[(f x gi + f y gj + f z gk) – (fg x i + fg y j + fg z k)] = (g  f – f  g) / g 2 (at points where g  0) 1) 2) 3) 4) Page 306 displays several vector identities.

12.    div(F+G) = (F+G) = —(F 1 +G 1 ) + —(F 2 +G 2 ) + —(F 3 +G 3 ) =  x  y  z  F 1  G 1  F 2  G 2  F 3  G 3 — +— +— +— +— +— =  x  x  y  y  z  z  F 1  F 2  F 3  G 1  G 2  G 3 — +— +— +— +— +— =  x  y  z F + G = (div F) + (div G) 5)

13. curl(F+G) =  (F+G) =  [ —(F 3 +G 3 ) – —(F 2 +G 2 ) ] i +  y  z  [ —(F 1 +G 1 ) – —(F 3 +G 3 ) ] j +  z  x  [ —(F 2 +G 2 ) – —(F 1 +G 1 ) ] k =  x  y  F 3  G 3  F 2  G 2 [ — + — – — – — ] i +  y  y  z  z [] j + [] k = 6)

14.  F 3  F 2  G 3  G 2 [ — – — + — – — ] i +  y  z  y  z [] j + [] k =  F 3  F 2 [ — – — ] i + [] j + [] k +  y  z  G 3  G 2 [ — – — ] i + [] j + [] k =  y  z  F +  G = curl(F) + curl(G)

15.    div(fF) = (fF) =—(fF 1 ) + —(fF 2 ) + —(fF 3 ) =  x  y  z  F 1  f  F 2  f  F 3  f f — +— F 1 + f — + — F 2 + f — + — F 3 =  x  x  y  y  z  z  F 1  F 2  F 3  f  f  f f — +f — + f — + F 1 — + F 2 — + F 3 — =  x  y  z  x  y  z f F + F  f =f(div F) + F  f 7)

16. div(F  G) = (F  G) =  [(F 2 G 3 –F 3 G 2 )i + (F 3 G 1 –F 1 G 3 )j + (F 1 G 2 –F 2 G 1 )k] =    —(F 2 G 3 –F 3 G 2 ) + —(F 3 G 1 –F 1 G 3 ) + —(F 1 G 2 –F 2 G 1 ) =  x  y  z  F 2  G 3  F 3  G 2 ( — G 3 + F 2 — ) – ( — G 2 + F 3 — ) +  x  x  x  x () – 8) () + () –() =

17.  F 3  F 2  F 1  F 3  F 2  F 1 G 1 ( — – — ) + G 2 ( — – — ) + G 3 ( — – — )  y  z  z  x  x  y – () G(  F) – F(  G) = G(curl F) – F(curl G) – () – () =

18. Suppose V = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k is a vector field with each of the three component functions having continuous partial derivatives of at least the second order. Then, div (curl V) =  (  V) =  [( )i + ( )j + ( )k] = ( ) + ( ) + ( ) = 0 R y – Q z P z – R x Q x – P y R yx – Q zx P zy – R xy Q xz – P yz div (curl F) = 0 (proven in an earlier class)9) curl(fF) =  (fF) =       [—(fF 3 ) – —(fF 2 ) ]i + [—(fF 1 ) – —(fF 3 ) ]j + [—(fF 2 ) – —(fF 1 ) ]k =  y  z  z  x  x  y  F 3  f  F 2  f [( f — + — F 3 ) – ( f — + — F 2 ) ]i +  y  y  z  z 10) [ ( ) –() ]j + [ ( ) – () ]k =

19.  F 3  F 2  F 1  F 3  F 2  F 1 f ( — – — )i + f ( — – — )j + f ( — – — )k +  y  z  z  x  x  y ()i + ()j + ()k = (f)(  F) + (  f)  F = f curl F + (  f)  F

20. curl(  f) = curl(f x i + f y j + f z k) =  f z  f y  f x  f z  f y  f x ( — – — )i + ( — – — )j + ( — – — )k =  y  z  z  x  x  y (f yz – f yz )i + (f xz – f xz )j + (f xy – f xy )k =0  2  2  2  2 (fg) = —(fg) + —(fg) + —(fg) =  x 2  y 2  z 2    —(f x g+fg x ) + —(f y g+fg y ) + —(f z g+fg z ) =  x  y  z (f xx g+f x g x +f x g x +fg xx ) + (f yy g+f y g y +f y g y +fg yy ) + (f zz g+f z g z +f z g z +fg zz ) = (fg xx +fg yy +fg zz ) + (f xx g+f yy g+f zz g) + (2f x g x +2f y g y +2f z g z ) = f  2 g + g  2 f + 2(  f  g) 11) 12) (proven in an earlier class)

21. div(  f   g) = div[ (f y g z – f z g y )i + (f z g x – f x g z )j + (f x g y – f y g x )k ] =    —(f y g z – f z g y ) + —(f z g x – f x g z ) + —(f x g y – f y g x ) =  x  y  z (f yx g z + f y g zx – f zx g y – f z g yx ) + (f zy g x + f z g xy – f xy g z – f x g yz ) + (f xz g y + f x g yz – f yz g x – f y g xz ) = 0 13) The easy way to prove #13 is to use #8 which has been already proven: div(  f   g) =  g (curl(  f)) –  f (curl(  g)) =  g 0 –  f 0 = 0 – 0 = 0

22. div(f  g – g  f) = div[ (fg x i + fg y j + fg z k) – (gf x i + gf y j + gf z k) ] = div[ (fg x – gf x )i + (fg y – gf y )j + (fg z – gf z )k ] =   —(fg x – gf x ) + —(fg y – gf y ) + —(fg z – gf z ) =  x  y  z (f x g x + fg xx – g x f x – gf xx ) + (f y g y + fg yy – g y f y – gf yy ) + (f z g z + fg zz – g z f z – gf zz ) = (fg xx – gf xx ) + (fg yy – gf yy ) + (fg zz – gf zz ) = fg xx + fg yy + fg zz – (gf xx + gf yy + gf zz ) = f  2 g – g  2 f 14)