Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 419

Therefore, Eqs. (7.160') and (7.163') are transpose equations of each other. Equa-
tions (7.161') and (7.162') are also transpose equations of each other.

Remark 2 .ByTheorem3inSection7.8,thekernel K ( M , P ) satisfies

≤

cos ( PM , n ) c

r 2 − δ

| K ( M , P ) | =

PM ,

0 < δ < 1 .

r PM

2 π

is weakly singular, and the four integral equations all have
a weakly-singular kernel.

Thus the kernel K ( M , P )

7.9.3 Boundary-Value Problems of Poisson Equations

In studying steady heat conduction or steady electric fields in domains with an in-
ternal source or sink, we arrive at the Poisson equations

Δ u = F ( M ) , M ∈ Ω . (7.164)

The key for seeking solutions of boundary-value problems of Eq. (7.164) is to find a
particular solution u ∗ of Eq. (7.164). Once u ∗ is available, a function transformation
of u u ∗ will transform the boundary-value problems of Poisson equations into

Laplace equations. For example, Dirichlet problems of the Poisson equation

= ω +

Δ u = F ( M ) , M ∈ Ω ,

(7.165)

u | S = f ( M ) .

will be transformed into

Δω = 0 , M ∈ Ω ,

(7.166)

u ∗ | S

ω | S =

( ) −

f M

u ∗ .
By the third Green formula (Eq. (7.65)), the solution of a Poisson equation satis-
fies

by a function transformation of u = ω +

u 1

r PM

d S

1
4

) ∂
∂ r PM ∂

1 u ( P ) 1

4

F ( P )

u ( M )= − ( P − − d Ω .

π n ∂ n π r PM

S

Ω

(7.167)

The two surface integrals in the right-hand side are the single-layer potential and the
double-layer potential, respectively. They satisfy the Laplace equation. The volume
integral in the right-hand side thus must be a particular solution of the Poisson equa-