Wave Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 96

To perform the double integration in Eq. (2.76), let

ξ = x − ( t , η = x − ( t . v + 1 ) v − 1 )

The solution of the original PDS (2.73) can thus be obtained through the integration
by substitution, in its integral representation,

t x − ( v − 1 )( t − τ )

1
2

u ( x , t )= d τ f ( ξ , τ ) d ξ .

0 x − ( v + 1 )( t − τ )

2.8 Two- and Three-Dimensional Cauchy Problems

This section begins with two methods of solving three-dimensional Cauchy prob-
lems

u tt =

a 2

Δ u + f ( M , t ) , Ω × ( 0 , + ∞ ) ,

(2.77)

u ( M , 0 )= ϕ ( M ) , u t ( M , 0 )= ψ ( M ) ,

where Ω stands for: − ∞ < x < + ∞ , − ∞ < y < + ∞ , − ∞ < z < + ∞ .Mrepre-

sents point . The solutions are then used to obtain the solutions of two-

dimensional problems by using the method of descent. We conclude this section
with an analysis of solution properties.
The solution of PDS (2.77) can be expressed by W ψ (

( x , y , z ) in Ω

M , t ) , the solution for the

case 0. In developing the solution of PDS (2.77), therefore, we focus on

seeking this structure function.

ϕ = f =

2.8.1 Method of Fourier Transformation

We first develop an identity, which finds its application in seeking W ψ ( M , t ) ,using

the method of Fourier transformation

e iω at

e − iω at e iω · r d

1 1

i

1
r [ δ (

I = − ω 1 d ω 2 d ω 3 = r − at ) − δ ( r + at )] .

4 π

2

ω

Ω

(2.78)

Here r = x i + y j + z k is a vector, ω 1 , ω 2 and ω 3 are variables of integration and Ω :

− ∞ < ω 1 < + ∞ , − ∞ < ω 2 < + ∞ , − ∞ < ω 3 < + ∞ , | r | = r , ω = ω 1 i + ω 2 j + ω 3 k ,

| ω | = ω .

Proof. To perform the triple integration over the whole space, take as the

origin and r as the polar axis. Regarding the whole space as a sphere of infinite

( x , y , z )