Mathematical Physics Vol 1

4.6 Examples

143

Differentiating once again, this time equation (4.173) by z , we obtain ∂φ ∂ z = 4 x − y + ∂ g ( z ) ∂ z . By equating the right hand side of equation (4.174) with the right hand side of equation (4.168), we obtain the unknown function g ( z ) (4.174)

∂ g ( z ) ∂ z ∂ g ( z ) ∂ z

4 x − y +

= 4 x − y + 2 z ,

(4.175)

= 2 z ,

(4.176)

g ( z )= z 2 + const .

(4.177)

By substituting equation (4.177) into equation (4.173) we obtain the required scalar function

x 2

3 y 2 2

+ z 2 + 2 xy + 4 xz − yz + const .

φ =

2 −

Exercise 71 Show that if the scalar function φ ( x , y , z ) is a solution of the Laplace equation, then the vector field ∇ φ is solenoid and potential.

Solution According to the condition of the example the function φ is a solution of the Laplace equation ∇ 2 φ = 0, that is, ∇ · ( ∇ φ )= 0. It follows from here that ∇ φ is a solenoid field. On the other hand, it is always true that ∇ × ( ∇ φ )= 0, and thus the field ∇ φ is also potential.

Exercise 72 How can the concept of gradient be extended to vector functions?

Solution Express the vector B , for example, with respect to Cartesian coordinates B =