When we take derivatives to obtain

We call the del operator and write

df— or f , we can think ofdx

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

F1 F2 F3

div F = •F = — + — + — .x y z

This definition can be adapted in the obvious way to a vector field in Rn for any n; for instance, in R2, the divergence of F(x,y) is

F1 F2

div F = •F = — + — .x y

If a vector field in R3 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 R2 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 pointIf the flow lines are those of a gas, then the gas isas 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 pointIf the flow lines are those of a gas, then the gas isas 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

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) = x2yi – xj .

It is not easy to intuit what the divergence will be.

By obtaining div F = , we find that2xy + 0 = 2xy

the divergence is not the same at all points in the vector field.(expansion)

(expansion) (compression)

(compression)

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

F3 F2 F1 F3 F2 F1

curl F = F = ( — – — )i + ( — – — )j + ( — – — )k . y z z x x y

Curl is an operator defined only in R3, whereas divergence is an operator defined in Rn for any n.

If F(x,y,z) = – yi + xj + zk , then curl F = 2k .

If F(x,y,z) = x2yi – xzj + yz2k , then curl F = (z2 + x)i + 0j + (– z – x2)k .

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

x2 + y2 (e.g., speed doubles if distance to z axis is doubled.)

xy planek

r = position vector of a point on the rotating objectv

Observe that v is in the direction of kr = k(xi + yj + zk) =v = 2k .– 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.

Consider the vector field described by

y xF(x,y,z) = ——— i – ——— j .

x2 + y2 x2 + y2

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 =

Suppose V is a vector field from R3 to R3, 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) = (fxi + fyj + fzk) =

(fzy – fyz)i + (fxz – fzx)j + (fyx – fxy)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 xF(x,y,z) = ——— i – ——— j a gradient vector field?

x2 + y2 x2 + y2

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).)

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 R2, then F can be regarded as a vector field in R3 by letting

F =

We then have that curl F =

– 2x – 1 .

Consider the vector field described by F(x,y) = yi – x2j.

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 + (Qx – Py)k = (Qx – Py)k .

The function Qx – Py is called the scalar curl of F.

P(x,y)i + Q(x,y)j + 0k .

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.

Ry – Qz Pz – Rx Qx – Py

Ryx– Qzx Pzy– Rxy Qxz– Pyz

Can F(x,y,z) = xi + yj + zk possibly be the curl of another vector field?1 + 1 + 1 = 3 0 , thenSince 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.

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 2f for f(x,y,z) = 1 / (x2+y2+z2)1/2.

fx = fxx =

fy = fyy =

fz = fzz =

– x / (x2+y2+z2)3/2

– y / (x2+y2+z2)3/2

– z / (x2+y2+z2)3/2

3x2 / (x2+y2+z2)5/2 – 1 / (x2+y2+z2)3/2

3y2 / (x2+y2+z2)5/2 – 1 / (x2+y2+z2)3/2

3z2 / (x2+y2+z2)5/2 – 1 / (x2+y2+z2)3/2

2f =

0

(div V) = •V = •(f) = 2f =

The operator 2 is called the Laplace operator.

2f 2f 2f— + — + — .x2 y2 z2

(f+g) = (fx+gx)i + (fy+gy)j + (fz+gz)k =

(fxi + fyj + fzk) + (gxi + gyj + gzk) = f + g

(cf) = (cfx)i + (cfy)j + (cfz)k = c(fxi + fyj + fzk) = cf

(fg) = (fxg+fgx)i + (fyg+fgy)j + (fzg+fgz)k =

(fxgi + fygj + fzgk) + (fgxi + fgyj + fgzk) =

g(fxi + fyj + fzk) + f(gxi + gyj + gzk) = gf + fg =

fg + gf

(f / g) = [(fxg–fgx) / g2]i + [(fyg–fgy) / g2]j + [(fzg–fgz) / g2]k =

(1/g2)[(fxg–fgx)i + (fyg–fgy)j + (fzg–fgz)k] =

(1/g2)[(fxgi + fygj + fzgk) – (fgxi + fgyj + fgzk)] = (gf – fg) / g2

(at points where g 0)

1)

2)

3)

4)

Page 306 displays several vector identities.

div(F+G) = •(F+G) = —(F1+G1) + —(F2+G2) + —(F3+G3) =

x y z

F1 G1 F2 G2 F3 G3

— + — + — + — + — + — =x x y y z z

F1 F2 F3 G1 G2 G3

— + — + — + — + — + — =x y z x y z

•F + •G = (div F) + (div G)

5)

curl(F+G) = (F+G) =

[ —(F3+G3) – —(F2+G2) ] i + y z

[ —(F1+G1) – —(F3+G3) ] j + z x

[ —(F2+G2) – —(F1+G1) ] k = x y

F3 G3 F2 G2

[ — + — – — – — ] i + y y z z

[ ] j +

[ ] k =

6)

F3 F2 G3 G2

[ — – — + — – — ] i + y z y z

[ ] j +

[ ] k =

F3 F2

[ — – — ] i + [ ] j + [ ] k + y z

G3 G2

[ — – — ] i + [ ] j + [ ] k = y z

F + G = curl(F) + curl(G)

div(fF) = •(fF) = —(fF1) + —(fF2) + —(fF3) =

x y z

F1 f F2 f F3 ff — + — F1 + f — + — F2 + f — + — F3 = x x y y z z

F1 F2 F3 f f ff — + f — + f — + F1 — + F2 — + F3 — = x y z x y z

f•F + F•f =

f(div F) + F•f

7)

div(FG) = •(FG) =

•[(F2G3–F3G2)i + (F3G1–F1G3)j + (F1G2–F2G1)k] =

—(F2G3–F3G2) + —(F3G1–F1G3) + —(F1G2–F2G1) =x y z

F2 G3 F3 G2

( — G3 + F2 — ) – ( — G2 + F3— ) + x x x x

( ) –

8)

( ) +

( ) – ( ) =

F3 F2 F1 F3 F2 F1

G1 ( — – — ) + G2 ( — – — ) + G3 ( — – — ) y z z x x y

– ( )

G•(F) – F•(G) =

G•(curl F) – F•(curl G)

– ( )

– ( ) =

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] =

( ) + ( ) + ( ) = 0Ry – Qz Pz – Rx Qx – Py

Ryx– Qzx Pzy– Rxy Qxz– Pyz

div (curl F) = 0 (proven in an earlier class)9)

curl(fF) = (fF) =

[—(fF3) – —(fF2) ]i + [—(fF1) – —(fF3) ]j + [—(fF2) – —(fF1) ]k = y z z x x y

F3 f F2 f[( f — + — F3 ) – ( f — + — F2 ) ]i + y y z z

10)

[ ( ) – ( ) ]j +

[ ( ) – ( ) ]k =

F3 F2 F1 F3 F2 F1

f ( — – — )i + f ( — – — )j + f ( — – — )k + y z z x x y

( )i +

( )j +

( )k =

(f)(F) + (f)F =

f curl F + (f)F

curl(f) = curl(fxi + fyj + fzk) =

fz fy fx fz fy fx

( — – — )i + ( — – — )j + ( — – — )k = y z z x x y

(fyz – fyz)i + (fxz – fxz)j + (fxy – fxy)k = 0

2 2 2

2(fg) = —(fg) + —(fg) + —(fg) = x2 y2 z2

—(fxg+fgx) + —(fyg+fgy) + —(fzg+fgz) =x y z

(fxxg+fxgx+fxgx+fgxx) + (fyyg+fygy+fygy+fgyy) + (fzzg+fzgz+fzgz+fgzz) =

(fgxx+fgyy+fgzz) + (fxxg+fyyg+fzzg) + (2fxgx+2fygy+2fzgz) =

f2g + g2f + 2(f • g)

11)

12)

(proven in an earlier class)

div(f g) = div[ (fygz – fzgy)i + (fzgx – fxgz)j + (fxgy – fygx)k ] =

—(fygz – fzgy) + —(fzgx – fxgz) + —(fxgy – fygx) =x y z

(fyxgz + fygzx – fzxgy – fzgyx) +

(fzygx + fzgxy – fxygz – fxgyz) +

(fxzgy + fxgyz – fyzgx – fygxz) =

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

div(fg – gf) = div[ (fgxi + fgyj + fgzk) – (gfxi + gfyj + gfzk) ] =

div[ (fgx – gfx)i + (fgy – gfy)j + (fgz – gfz)k ] =

—(fgx – gfx) + —(fgy – gfy) + —(fgz – gfz) =x y z

(fxgx + fgxx – gxfx – gfxx) +

(fygy + fgyy – gyfy – gfyy) +

(fzgz + fgzz – gzfz – gfzz) =

(fgxx – gfxx) + (fgyy – gfyy) + (fgzz – gfzz) =

fgxx + fgyy + fgzz – (gfxx + gfyy + gfzz) =

f2g – g2f

14)