Mathematical Physics Vol 1

4.6 Examples

167

Exercise 103 Let F = F 1 i + F 2 j + F 3 k . a) prove that the necessary and sufficient condition for F 1 d x + F 2 d y + F 3 d z tobe a total differential is ∇ × F (= rot F )= 0 . b) Prove that ( y 2 z 3 cos x − 4 x 3 z ) d x + 2 z 3 y sin x d y +( 3 y 2 z 2 sin x − x 4 ) d z is the total differential of some function φ . Find that function.

Solution a) The condition is necessary. If

∂φ ∂ x

∂φ ∂ y

∂φ ∂ z

F 1 d x + F 2 d y + F 3 d z = d φ =

d x +

d y +

d z ,

is the total differential of a function φ ( x , y , z ) , it follows that

∂φ ∂ x

∂φ ∂ y

∂φ ∂ z

F 1 =

F 2 =

F 3 =

,

,

,

and thus

∂φ ∂ x

∂φ ∂ y

∂φ ∂ z

k = ∇ ϕ .

F = F 1 i + F 2 j + F 3 k =

i +

j +

According to (see (4.69), p. 98)

∇ × F = ∇ × ∇ φ = 0 . The condition is sufficient. If ∇ × F = 0, then F = ∇ ϕ . It follows that F · d r = ∇ ϕ · d r = d ϕ (see Example 100 on p.163). b) Let F =( y 2 z 3 cos x − 4 x 3 z ) i + 2 z 3 y sin x j +( 3 y 2 z 2 sin x − x 4 ) k . Given that, in this case ∇ × F = 0, then according to the condition under a), there exists a function φ such that

∂φ ∂ x

∂φ ∂ y

∂φ ∂ z

d φ =

d x +

d y +

d z = F 1 d x + F 2 d y + F 3 d z ,

that is

( y 2 z 3 cos x − 4 x 3 z ) d x + 2 z 3 y sin x d y +( 3 y 2 z 2 sin x − x 4 ) d z = d φ .

It can be proved, analogously to the Example 101 on p.164, that φ = y 2 z 3 sin x − x 4 z + const .