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7/22/2019 Reviewer Ece102

1/8

AB2.3: Divergence and Curl of a Vector Field

Definition of divergence

Let v(x,y,z) be a differentiable vector function,

v(x,y,z) =v1(x,y,z)i +v2(x,y,z)j +v3(x,y,z)k

Then the function

div v=v1

x +

v2y

+ v3

z

is called the divergence ofv or the divergence of the vector field defined byv.Another convenient notation for div is as follows:

div v= v=

xi +

yj +

zk

(v1i +v2j +v3k) =

v1x

+ v2

y +

v3z

.

Example:

v(x,y,z) = 3xzi + 2xyj yz2k,

v1x

= 3z, v2

y = 2x,

v3z

= 2yz,

anddiv v= 3z+ 2x 2yz.

THEOREM8.10.1

The values of div v depend only on the points in space (and on div v) but not on the par-ticular choice of the coordinates. With respect to other Cartesian coordinates x, y, z andcorresponding componentsv

1, v

2, v

3ofv the functiondiv v is given by

div v= v

1

x+

v2

y+

v3

z

Iffis a twice differentiable scalar function, then

grad f=f

xi +

f

yj +

f

zk.

and

div (grad f) = 2f=2f

x2+

2f

y2 +

2f

z2

is the Laplacian off.


2/8

EXAMPLE 1 Gravitational force

The gravitational force p is the gradient of the scalar function

f(x,y,z) =c

r

(the potential of the gravitational field). f satisfies the Laplaces equation 2f= 0. Thus

div p= div (grad f) = 2f= 0.

Curl of a vector field

Definition of curl

Letx, y,zbe right-handed Cartesian coordinates and

v(x,y,z) =v1(x,y,z)i +v2(x,y,z)j +v3(x,y,z)k

be a differentiable vector function. Then the function

curl v= v=

i j kx

y

z

v1 v2 v3

=

v3y

v2

z

i +

v1z

v3

x

j +

v2x

v1

y

k

is called the curl ofv or the curl of the vector field defined byv.

EXAMPLE 1 Curl of a vector function

With respect to right-handed Cartesian coordinates, let

v(x,y,z) =yzi + 3zxj +zk.

Then the curl

curlv

=

i j k

x

y

zyz 3xz z

=

z

y

(3xz)

z

i +

(yz)

z

z

x

j +

(3xz)

x

(yz)

y

k= 3xi +yj + 2zk.

Important identities of the vector calculus

For any twice continuously differentiable scalar function f,

curl (grad f) =0.


3/8

Indeed,

curl (grad f) =

i j kx

y

z

fx

fy

fz

=

i j kx

y

z

fx fy fz

=

fzy

fyz

i+

fxz

fzx

j+

fyx

fxy

k= (fzyfyz)i+(fxzfzx)j+(fyxfxy)k= 0.

Gradient fields describing a motion are irrotational.

For any twice continuously differentiable vector functionv,

div (curl v) = 0.

Indeed,

div (curl v) =

x v3y

v2

z +

y v1z

v3

x +

zv2x

v1

y =(v3yx v2zx) + (v1zy v3xy) + (v2xz v1yz) = 0.

THEOREM8.11.1

The length and direction ofcurl v are independent of the particular choice of Cartesian coordi-nate systems in space.


4/8

PROBLEM8.10.1

Find the divergence of

v(x,y,z) =v1(x,y,z)i +v2(x,y,z)j +v3(x,y,z)k= xi +yj +zk.

Solution. The divergence is defined as

div v=v1

x +

v2y

+ v3

z

Here,v1(x,y,z) =x, v2(x,y,z) =y, v3(x,y,z) =z,

anddiv v= 1 + 1 + 1 = 3.

PROBLEM8.10.2


v(x,y,z) =v1(x,y,z)i +v2(x,y,z)j +v3(x,y,z)k= x2i +y2j +z2k.

Solution. Here,

v1(x,y,z) =x2, v2(x,y,z) =y

2, v3(x,y,z) =z2,

anddiv v= 2x+ 2y+ 2z= 2(x+y+z).

PROBLEM8.10.5


v(x, y) =v1(x, y)i +v2(x, y)j= (x2 +y2)1(yi +xj).

Solution. Here,

v1(x, y) = (y)(x2 +y2)1, v2(x, y) = (y)(x

2 +y2)1,

and

div v= 2xy(x2

+y2

)2

2xy(x2

+y2

)2

= 0.

PROBLEM8.10.13b


fv(x,y,z) =f v1(x,y,z)i +f v2(x,y,z)j +f v3(x,y,z)k, f=f(x,y,z).

Solution.

div fv=f v1

x +

f v2y

+ f v3

z


5/8

=fv1x

+fv2y

+fv3z

+v1f

x+v2

f

x+v3

f

x=

fdiv v+ v grad f.

Apply this result for calculating

div(fv(x,y,z)), f(x,y,z) =r3

/2

= (x2

+y2

+z2

)3

/2

, v= xi +yj +zk.

We havediv v= 3.

grad f= 3r5/2(xi +yj +zk)

and

div(fv) =fdiv v + v grad f= 3r3/2 3r5/2[x,y,z] [x,y,z] = 3r3/2 3r5/2r= 0.

PROBLEM8.10.14Find the Laplacian 2f of

f(x, y) = (x y)/(x+y).

Solution. It is easy to see that f is a twice differentiable function for x = y; then

grad f= f=f

xi +

f

yj=

2y

(x+y)2i

2x

(x+y)2j,

and the Laplacian off


x2+

2f

y2 = 4

x y

(x+y)3.

PROBLEM8.10.15

Find the Laplacian 2f off(x,y,z) = 4x2 + 9y2 +z2.

Solution. It is easy to see that f is a twice differentiable function; then

grad f= f=f

xi +

f

yj +

f

zk=

8xi + 18yj + 2zk= [8x, 18y, 2z],

and the Laplacian off


x2+

2f

y2+

2f

z2= 8 + 18 + 2 = 28.


6/8

PROBLEM8.11.2

Find the curl ofv= [2y, 5x, 0].

Solution.

curl v=

i j kx

y

z

2y 5x 0

=(0)

y

(5x)

z

i +

(2y)

z

(0)

x

j +

(5x)

x

(2y)

y

k=

0 i + 0 j + (5 2) k= 3k.

PROBLEM8.11.3

Find the curl of

v= 12

(x2 +y2 +z2)(i +j + k).

Solution. Here,

v1(x,y,z) =v2(x,y,z) =v3(x,y,z) =v(x,y,z) =1

2(x2 +y2 +z2)

andv

q =q, q= x, y,z.

Therefore,

curl v=v

y v

z

i +v

zv

x

j +v

x v

y

k=

(y z)i + (z x)j + (x y)k= [y z, z x, x y].

PROBLEM8.11.13

Find the curl ofv= [x,y,z].

(the vector field of a fluid motion).Solution.

curl v=

i j kx

y

z

x y z

=(z)

y

y

z

i +

x

z

(z)

x

j +

y

x

x

y

k=

0 i + 0 j + 0 k= 0.

The flow is irrotational.

div v= 1 + 1 1 = 1.


7/8

The flow is compressible.The paths of particles are described by

r(t) = [c1et, c2e

t, c3et].

PROBLEM8.11.14

u v=

i j k

u1 u2 u3v1 v2 v3

=(u2v3 u3v2)i (u1v3 u3v1)j + (u1v2 u2v1)k.

div(u v) =(u2v3 u3v2)

x +

(u3v1 u1v3)

y +

(u1v2 u2v1)

z =

(u2v3)

x +(u3v1)

y +(u1v2)

z (u3v2)

x (u1v3)

y (u2v1)

z =

v3u2x

+u2v3x

+v1u3y

+u3v1y

+v2u1z

+u1v2zv2

u3x

u3v2x

v3u1yu1

v3yv1

u2zu2

v1z

=

v1

(u3)

y

(u2)

z

+v2

(u1)

z

(u3)

x

+v3

(u2)

x

(u1)

y

+

u1

(v2)

z

(v3)

y

+u2

(v3)

x

(v1)

z

+u3

(v1)

y

(v2)

x

=

v curl u u curl v.

Thus, we have proved that

div(u v) =v curl u u curl v.

PROBLEM8.11.15

Find the curl fu foru= yi +zj +xk

and f=xyz.Solution. We have

curl fu= grad f u +fcurl u

Here,grad f= [yz,xz,xy], u= [y,z,x],

Next,

grad f u=

i j k

yz xz xyy z x

== (x2z xyz)i (xy2 xyz)j + (yz2 xyz)k= [x2z xyz,xy2 xyz,yz2 xyz].


8/8

curl u=

x

y

z

z

i +

y

z

x

x

j +

z

x

y

y

k=

i j k= [1,1,1].

Finally,

curl fu= (x2z xyz)i (xyz xy2)j + (yz2 xyz)k (xyz)(i +j + k) =

(x2z 2xyz)i (xy2 2xyz)j + (yz2 2xyz)k.

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