Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 159

G can be expanded by using the complete and orthogonal set (Table 2.1)

sin μ n y n

= μ n

sin μ m x
l 1

=

l 2 + ϕ

,

l 2 h 1 .

u 37 tan ϕ

n

By multiplying the factor, we may obtain the general term of G , thus

sin μ n η

l 2 + ϕ n

2τ 0 + ∞

τ 0 γ mn M mn sin μ

1 ξ

sin μ

m x
l 1

t − τ

e −

m

G =

∑

l 1

m , n = 1

(4.43)

sin μ n y

l 2 + ϕ n sin

· γ mn ( t − τ ) .

Finally,

t

u 3 = d τ G ( x , ξ ; y , η ; t − τ ) f ( ξ , η , τ ) d σ . (4.44)

0

D

2. The solution of PDS (4.42) at f = ϕ = 0 is, by Eqs. (4.41) and (4.43),

⎨

sin μ n y n sin

∞
m , n = 1 b mn sin μ m x

t

e −

u 2 = W ψ ( x , y , t )=

2 τ 0

l 2 + ϕ

γ mn t ,

l 1

sin μ n y

l 2 + ϕ n d

(4.45)

1
γ mn M mn

sin μ m x
l 1

⎩

b mn = ψ ( x , y ) σ .

D

3. W ϕ ( x , y , t ) may be readily obtained simply by replacing ψ ( x , y ) in Eq. (4.45) by

ϕ ( x , y ) . The solution of PDS (4.42) at f = ψ = 0 is thus

1 W ϕ (

τ 0 + ∂

u 1 = x , y , t ) . (4.46)

∂ t

4. The solution of PDS (4.42) is, by the principle of superposition,

u ( x , y , t )= u 1 ( x , y , t )+ u 2 ( x , y , t )+ u 3 ( x , y , t ) .

Remark 5. If γ mn has a purely imaginary value such as γ mn = ω mn i, the factor

γ mn t γ mn can be transformed into a real exponential function.

sin

γ mn t γ mn =

1
ω mn i sin i

1
ω mn sh

sin ω mn t = ω mn t

1

e ω mn t e − ω mn t

=

ω mn (

− ) .

2

t
2τ 0

reads e − e ω mn t e − ω mn t

For this case, the factor involving time t ( − )=

e

t 1

2 τ 0 + ω mn t , which is bounded as t

1

1
2

ω mn −

e −

2 τ 0

− → + ∞ because 0 < ω <

τ 0 .

mn