Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 276

the solution of PDS (6.94)
⎧
⎨
⎩

)= m , n ( B mn cos n θ + D mn sin n θ ) J n ( k mn r ) e α mn t sin β mn t ,

u = W ψ ( r , θ , t

π a

0 ψ (

1
M n M mn β mn

B mn = d θ r , θ ) J n ( k mn r ) r cos n θ d r ,

− π

π a

0 ψ (

1
M n M mn β mn

D mn = d θ r , θ ) J n ( k mn r ) r sin n θ d r ,

− π

(6.99)

where M n is the normal square of { 1 , cos θ , sin θ , ··· , cos n θ , sin n θ , ···}

2

π , when n = 0 ,

M n =

π , otherwise .

a

J n (

M mn = k mn r ) r d r is the normal square of { J n ( k mn r ) } and is available in

0

Tab l e 4 . 1.

6.5.2 Solution from ϕ ( r , θ )

Theorem 1 .Let u ( r , θ , t )= W ψ ( r , θ , t ) be the solution of

⎧
⎨
⎩

u t
τ 0 +

B 2 ∂
∂

A 2

= ( , θ , )+ ( , θ , )

u tt Δ u r t

t Δ

u r t

0 < r < a , 0 < θ < 2 π , 0 < t ,

(6.100)

L ( u , u r ) | r = a = 0 ,

u ( r , θ , 0 )= 0 , u t ( r , θ , 0 )= ψ ( r , θ ) .

The solution of

⎧
⎨

u t
τ 0 +

B 2 ∂
∂

A 2 u tt = Δ u ( r , θ , t )+

t Δ

u ( r , θ , t ) ,

0 < r < a , 0 < θ < 2 π , 0 < t ,

(6.101)

⎩

L ( u , u r ) | r = a = 0 ,

u ( r , θ , 0 )= ϕ ( r , θ ) , u t ( r , θ , 0 )= 0

is

1 W ϕ (

τ 0 + ∂

B 2 W k mn ϕ (

u ( r , θ , t )= r , θ , t )+ r , θ , t ) ,

∂ t

( n )
m

( n m are the non-negative zero points of f n

where k mn = μ / a ,the μ ( x ) in Eq. (6.97).