Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 158

Step 3 . Construct the general term of G by multiplying the factor,

1
τ 0 γ mn M mn u ij

t − τ
2

e −

τ 0

( x , y ) u ij ( ξ , η ) sin γ ( t − τ ) . mn

Here M mn = M m M n , the product of the normal squares of the two sets of eigen-

and Y j ( ) ,

functions { X i ( x ) } y

4

1
2

0 a 2

γ = τ ( λ + λ ) − 1 , mn m n

τ

0

λ

m and

λ

n are the eigenvalues corresponding to X i

( x ) and Y j ( y ) , respectively.

Finally,

1
τ 0 γ mn M mn u ij (

t − τ
2

e −

∑

G ( x , ξ ; y , η ; t − τ )=

τ 0

x , y ) u ij ( ξ , η ) sin γ mn ( t − τ ) , (4.39)

where u ij ( x , y ) u ij ( ξ , η )= X i ( ξ ) X i ( x ) Y j ( η ) Y j ( y ) .

Remark 4. We can readily write out W ψ ( x , y , t ) from the Green function G ( x , ξ ; y , η ; t −

τ ) in Eq. (4.39). This can be achieved simply by letting τ = 0 and replacing

1
τ 0 γ mn M mn u ij ( ξ , η )

by the Fourier coefficients

1
γ mn M mn

b mn = ψ ( x , y ) u ij ( x , y ) d σ . (4.40)

D

Therefore

⎧
⎨

t
2τ 0

e −

∑

W ψ ( x , y , t )= b mn u ij ( x , y ) sin γ mn t ,

(4.41)

1
γ mn M mn

⎩

b mn = ψ ( x , y ) u ij ( x , y ) d σ .

D

Example 3. Solve
⎧
⎨

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 )+ h 2 u ( l 1 , y , t )= 0 ,

(4.42)

⎩

u y ( x , 0 , t ) − h 1 u ( x , 0 , t )= u ( x , l 2 , t )= 0 ,

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

Solution.

( , , )

ϕ =

1. We first seek the solution due to f x y t , i.e. the solution of PDS (4.42) at

ψ =

0. Based on the given boundary conditions,