Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 153

Example 1. Solve
⎧
⎨

a 2

+ τ = + ( , , ) , × ( , + ∞ ) ,

u t 0 u tt Δ u f x y t D 0

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

(4.25)

⎩

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

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

Solution. We fir s t deve l op W ψ ( 0. By taking

the boundary conditions into account, we use the eigenfunctions in Rows 3 and 6 in
Table 2.1 to expand the solution

x , y , t ) , the solution of (4.25) at ϕ = f =

+ ∞
m , n = 1 T mn ( t ) sin α m x cos β n y ,

u = (4.26)

m l 1

= μ n l 2 .

μ n are the positive zero-points of f where α = μ , β μ

m and

( x )=

m n

y
l 2 h , respectively.
Substituting Eq. (4.26) into the equation of (4.25) with f

x
l 1 h and g

tan x + ( x )= cot y −

= 0 yields

T mn + α

n a 2 T mn =

τ 0 T mn +

2
m

2

+ β 0 . (4.27)

Its general solution is

t

e −

T mn ( t )=

2 τ 0

( a mn cos γ mn t + b mn sin γ mn t ) ,

where a mn and b mn are undetermined constants, and

sin

4

1
2

γ mn t , γ mn = 0 ,

m n

a 2 γ mn = τ 0 ( α + β ) − 1 , sin γ mn t =

t , γ = 0 .

τ

mn

0

We thus obtain

+ ∞
m , n = 1 e − t

u =

2 τ 0

( a mn cos γ mn t + b mn sin γ mn t ) sin α m x cos β n y . (4.28)

Applying the initial condition u ( x , y , 0 )= 0 yields a mn = 0. To satisfy the initial

condition u t ( x , y , 0 )= ψ ( x , y ) , b mn must be determined such that

γ mn , γ mn =

+ ∞
m , n = 1 b mn γ mn sin α m x cos β n y = ψ ( x , y ) ,

0 ,

γ mn =

1 , γ mn = 0 .

Therefore

1
γ mn M mn

b mn = ψ ( x , y ) sin α m x cos β n y d σ ,

D