Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 154

where M mn = M m M n , M m and M n are the normal square of the two eigenfunction

sets , respectively. Finally, the solution of PDS (4.25) is, by

the solution structure theorem,

{ sin α m x } and { cos β n y }

1
τ

W ϕ (

t

0 + ∂

u = x , y , t )+ W ψ ( x , y , t )+ W f τ ( x , y , t − τ ) d τ ,

∂ t

0

, τ ) τ 0 .
Example 2. Find the solution of Eq. 4.28
⎧
⎨

where f τ = f ( x , y

a 2 u t + τ 0 u tt = Δ u + f ( x , y , t ) , D × ( 0 , + ∞ ) ,

u ( 0 , y , t )= u x ( l 1 , y , t )= 0 ,

(4.29)

⎩

u y ( x , 0 , t )= u y ( x , l 2 , t )= 0 ,

u ( x , y , 0 )= ϕ ( x , y ) , u t ( x , y , 0 )= ψ ( x , y ) .

Solution. We first find the Green function G ( x , ξ ; y , η ; t − τ ) satisfying

⎧
⎨

a 2 G t + τ 0 G tt = Δ G , D , 0 < τ < t < + ∞ ,

G y y = 0 = G y y = l 2 ,

G | x = 0 = G x | x = l 1 =

. (4.30)

⎩

− η ) τ 0 G | t = τ = 0 , G t | t = τ = δ ( x − ξ , y .

Based on the given boundary conditions in (4.30), we use the eigenfunctions in
Rows 2 and 5 in Table 2.1 to expand

+ ∞
m , n = 0 T mn ( t ) sin ( 2 m + 1 ) π x

cos n

π y

l 2 ,

G =

2 l 1

where T mn ( t ) is the undetermined function. Substituting it into the equation of (4.30)

leads to

(

2 a 2 T mn = 2 n

2 m + 1 ) π π

l 2

τ 0 T mn + T mn + + 0 .

2 l 1

Its general solution reads

t − τ
2

e −

T mn ( t )= [ a mn cos γ ( t − τ )+ b mn sin γ ( t − τ )] ,

τ 0

mn mn

where a mn and b mn are the undertermined constants, and

4

(

2 a 2

2 n

1
2

2 m + 1 ) π π

l 2

γ mn = τ 0 + − 1 .

τ 2 l 1 0