Introduction - Heat Conduction: Mathematical Models and Analytical Solutions - page 38

Consider the irrotational flow of an incompressible fluid passing over a solid
body Ω with boundary ∂Ω . The velocity potential ϕ ( M ) is the solution of

⎧
⎨

Δϕ =

, ∈ Ω ,

0 M

∂Ω = ψ (

∂ϕ
∂

M ) ,

⎩ n

x 2

lim
r → + ∞

u ( M )= 0 , r = + y 2 + z 2 .

This is an external bou ndary-value problem of the second kind of a potential equa-
tion. Here R 3

= \ Ω

ψ (

)

is a known function.
The constraint imposed at infinity in the above two external problems is for the
uniqueness of the solution. In fact, we can review infinity as the boundary. From this
point of view, we also expect some constraints in infinity to form a proper PDS.

Ω and M

1.4.4 Well-Posedness of PDS

The PDS is supposed to describe physical problems. Some assumptions are nor-
mally involved in deriving a PDS from specific physical problems. In order to have
a reasonable approximation of physical reality by the PDS and be useful for appli-
cation, the PDS must be well-posed.

Existence

A well-posed PDS must have solutions. In developing equations and their CDS
from physical problems, we must normally make some idealized and simplifying
assumptions. Many factors can lead to nonexistence of solutions. Examples are:
(1) if the physically-dominant process or mechanism are not taken into account by
the equations, and (2) the CDS is too restrictive or too many. In deriving the PDS,
attention should also be given to the surrounding of physical systems as well as to
the fundamental physical laws in order to ensure the existence of solutions.

Uniqueness

In applications, a PDE is used for describing a unique physical relation. A well-
posed PDS should thus have a unique solution. Many factors can contribute to
nonuniqueness; a typical example is that the PDS does not have enough CDS. The
external problem of the potential equation
Δ

x 2 y 2 z 2

u = 0 , + + > 1 ,

u | x 2 = 1

y 2 z 2 + + = 1

x 2

has no constraint on u at infinity and ha s two solutio ns u = 1and u = 1 / + y 2 + z 2 .

x 2 x 2

y 2 z 2 y 2 z 2

If a further constraint lim
r

u = 0 ( r = + + ) is imposed, u = 1 / + +

→ ∞

becomes the unique solution.