Mathematical Physics Vol 1

Chapter 7. Partial differential equations

370

Example 244 Observe now the Laplace equation in cylindrical coordinates u = u ( ρ , ϕ , z )

∂ ∂ρ

∂ u ∂ρ

∂ 2 u ∂ϕ 2

∂ 2 u ∂ z 2

1 ρ

1 ρ 2

∆ u =

ρ

= 0 .

(7.219)

+

+

In the case that u does not depend of ϕ and z , i.e. u = u ( ρ ) , we obtain

d d ρ

d u d ρ

1 ρ

∆ u =

ρ

= 0 ,

(7.220)

and it follows that

d u d ρ

C 1 ρ

ρ

d ρ ⇒ u = C 1 ln ρ + C 2 .

= C 1 ⇒ d u =

(7.221)

The constants C 1 and C 2 are arbitrary, and if we select C 1 = − 1 and C 2 = 0, we obtain: u o = u o ( ρ )= ln 1 ρ . (7.222) This function is often also called the basic solution of the Laplace equation in the plane (for two independent variables). The function u o satisfies the Laplace equation everywhere (in the plane), except at point ρ = 0.

7.7 Green formulas

Let us start from the Ostrogradsky formula (4.92): y V div a d V = x S a d S = x S

a n d S ,

(7.223)

where

div a = div ( a x i + a y j + a z k ) , a · n = a n , d S = dS · n , a n = a x · cos α + a y · cos β + a z · cos γ ,

(7.224)

α = ∠ ( n , i ) , β = ∠ ( n , j ) , γ = ∠ ( n , k ) ,

n is the outer normal of the closed surface S , which limits the space V , and a x , a y and a z are arbitrary differentiable functions. Here, the indices x , y , z denote the projections of the corresponding values on the x , y and z axes, respectively. Not to be identified with derivatives (in this case)!!! Let us now introduce new scalar functions u ( x , y , z ) and v ( x , y , z ) , which are continuous, and whose first derivatives on the border S , and second derivatives inside the region V , are continuous. Let us subsequently introduce

∂ v ∂ x

∂ v ∂ y

∂ v ∂ z

i.e. a = u · grad v ,

a x = u ·

a y = u ·

a z = u ·

(7.225)

,

,

,