Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 278

Substituting Eqs. (6.103), (6.104) and (6.106) into (6.105) yields

∂
∂

)= m , n { [( − B mn e α mn t sin β mn t ) cos n θ

t W ϕ (

r , θ , t

D mn e α mn t sin A mn e α mn t cos +( − β mn t ) sin n θ ]+[( β mn t ) cos n θ

+ C mn e α mn t cos β mn t sin n

θ ] } J n ( k mn r ) .

Therefore,
m , n [( A mn e α mn t cos β mn t ) cos n θ + C mn e α mn t cos β mn t sin n θ ] J n ( k mn r )

= ∂
∂

)+ m , n [( e α mn t B mn sin β mn t ) cos n θ

t W ϕ (

r , θ , t

e α mn t D mn sin +( β mn t ) sin n θ ] J n ( k mn r ) . (6.107)

Note that

1
τ 0 +

( n )
m k mn B 2

− = , = μ / .

2 α k mn a mn

A substitution of Eqs. (6.104) and (6.107) into Eq. (6.102) yields the solution of
PDS (6.101),

1 W ϕ (

τ 0 + ∂

B 2 W k mn ϕ ( u ( r , θ , t )= r , θ , t )+ r , θ , t ) .

∂ t

6.5.3 Solution from f ( r , θ , t )

Theorem 2 .Let u ( r , θ , t )= W ψ ( r , θ , t ) be 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 ,

(6.108)

⎩

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

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