when we take derivatives to obtain we call the del operator and write df — or f, we can think of...
TRANSCRIPT
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)