Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 291

Thus the u ( r , θ , ϕ , t ) that satisfies the equation and the boundary conditions of

PDS (6.141) is

)= m , n , l e α nl t u ( r , θ , ϕ , t [( A mnl cos β nl t + B mnl sin β nl t ) cos m ϕ

P n ( +( C mnl cos β nl t + D mnl sin β nl t ) sin m ϕ ] cos θ ) j n ( k nl r ) . (6.145)

Applying the initial condition u ( r , θ , ϕ , 0 )= 0 yields A mnl = C mnl = 0. B mnl and D mnl

can also be determined by applying the initial condition u t ( r , θ , ϕ , 0 )= ψ ( r , θ , ϕ ) .

Finally we obtain the solution of PDS (6.141)
⎧
⎨ )= m , n , l e α nl t

u = W ψ ( r , θ , ϕ , t ( B mnl cos m ϕ + D mnl sin m ϕ )

P n ( · cos θ ) j n ( k nl r ) sin β nl t ,

1
M mnl β nl

P n ( r 2 cos m

B mnl = ψ ( r , θ , ϕ ) cos θ ) j n ( k nl r ) ϕ sin θ d θ d r d ϕ ,

⎩

r ≤ a

1
M mnl β nl

P n ( r 2 sin m

D mnl = ψ (

, θ , ϕ )

θ )

( )

ϕ ,

r cos j n k nl r ϕ sin θ d θ d r d

r ≤ a

(6.146)

where M mnl is the product of three normal squares.

6.7.2 Solution from Φ ( r , θ , ϕ )

Theorem 1 .Let u = W ψ ( r , θ , ϕ , t ) be the solution of PDS (6.141). The solution of

⎧
⎨

u t
τ 0 +

B 2 ∂
∂

A 2

u tt = Δ u ( r , θ , ϕ , t )+

t Δ

u ( r , θ , ϕ , t )

0 < r < a , 0 < θ < π , 0 ≤ ϕ ≤ 2 π , 0 < t ,

(6.147)

⎩

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

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

is

1 W Φ (

τ 0 + ∂

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

∂ t

Proof . Following a similar approach as that in Section 6.7.1, we obtain the
u ( r , θ , ϕ , t ) that satisfies the equation and the boundary condition of PDS (6.147)

)= m , n , l e α nl t u ( r , θ , ϕ , t [( A mnl cos β nl t + B mnl sin β nl t ) cos m ϕ

P n (

+( C mnl cos β nl t + D mnl sin β nl t ) sin m ϕ ] cos θ ) j n ( k nl r ) . (6.148)