Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 149

Remark 1. Although there exist three cases of characteristic roots for Eq. (4.10),
its general solution can be written in a unified form as Eq. (4.11) by using the rela-
tion between exponential and trigonometric functions. Since γ m = 0 rarely occurs,

sin γ m = γ m for most of cases.

Remark 2. By the solution structure theorem, the solution of
⎧
⎨

γ m t = sin γ m t and

a 2 u xx

u t + τ

0 u tt

= + f ( x , t ) , ( 0 , l ) × ( 0 , + ∞ ) ,

u ( 0 , t )= u x ( l , t )+ hu ( l , t )= 0 , (4.15)

⎩

u ( x , 0 )= u t ( x , 0 )= 0

is

t

+ ∞
m = 1 b m sin γ m ( t − τ ) sin μ m x

t

t − τ
2

e − u = W f τ ( M , t − τ ) d τ =

τ 0

d τ

l

0 0

e − f

+ ∞
m = 1

t l

τ 0 γ m M m sin μ

1 ξ

sin μ

m x
l

t − τ

2τ 0

m

= d τ sin γ ( t − τ ) ( ξ , τ ) d ξ

m

l

0 0

t l

= ( , ξ , − τ ) ( ξ , τ )

ξ .

d τ G x t f d

0 0

(4.16)

Here

+ ∞
m = 1

τ 0 γ m M m sin μ m ξ

1

sin μ m x
l

t − τ
2

e − G ( x , ξ , t − τ )=

τ 0

sin γ m ( t − τ ) (4.17)

l

is called the Green function of the one-dimensional hyperbolic heat-conduction
equation subject to boundary conditions (4.8).
Remark 3. Since e −

2τ 0

t − τ

= 1,

1
γ m M m

1
τ 0

1
L ,

[ G ]= =

so that the unit of u in Eq. (4.16) is [ u ]=[ d τ ][ G ][ f ][ d ξ ]= Θ .Also,

a 2 τ 0 =

L 2 T − 2 .

Therefore, a √ τ 0 stands for the wave velocity. The temperature u at point x in the
classical heat-conduction equation decreases exponentially as time t . Here the every
term in the temperature (Eq. (4.17)) contains both an exponential factor e −

t − τ

2

τ 0 and

a factor sin γ m ( t − τ ) . Therefore, the temperature u may have wavelike properties.