Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 283

The u ( r , θ , t ) that satisfies the equation and the boundary conditions of PDS (6.115)

is thus

)= m , n , k e α mnk t u ( r , θ , t [( c mnk cos β mnk t + d mnk sin β mnk t )

· ( a mn cos n θ + b mn sin n θ )] J n ( k mn r ) Z k ( z )

= m , n , k e α mnk t [( A mnk cos β mnk t + B mnk sin β mnk t ) cos n θ

+( C mnk cos β mnk t + D mnk sin β mnk t ) sin n θ ] J n ( k mn r ) Z k ( z ) ,

m , n , k stands for a triple summation, m = 1 , 2 , ··· , n = 0 , 1 , 2 , ··· , k = 0 , 1 , 2 , ···
for the cases of Rows 2, 4 and 5 in Table 2.1, and k

where

= 1 , 2 , ··· for the other six cases.

Note that { 1 , cos θ , sin θ , ··· , cos n θ , sin n θ , ···} , { J n ( k mn r ) } and Z k ( z ) are or-

thogonal in [ − π , π ] ,in [ 0 , a ] with respect to the weight function r and in [ 0 , H ] ,

respectively. Applying the initial condition u ( r , θ , z , 0 )= 0 yields A mnk = C mnk = 0.

Thus

)= m , n , k e α mnk t

u t ( r , θ , z , t [( α mnk B mnk sin β mnk t

+ β mnk B mnk cos β mnk t ) cos n θ +( α mnk D mnk sin β mnk t

+ β mnk D mnk cos β mnk t ) sin n θ ] J n ( k mn r ) Z k ( z ) .

B mnk and D mnk can thus be determined by applying the initial condition
u t ( r , θ , z , 0 )= ψ ( r , θ , z ) . Finally, we have the solution of PDS (6.115)

⎧
⎨ )= m , n , k ( B mnk cos n θ

u = W ψ ( , θ , ,

r z t

e α mnk t sin

+ D mnk sin n θ ) J n ( k mn r ) Z k ( z ) β mnk t ,

1
M mnk β mnk

(6.123)

B mnk = ψ ( r , θ , z ) rJ n ( k mn r ) Z k ( z ) cos n θ d θ d r d z ,

⎩

Ω

1
M mnk β mnk

D mnk = ψ ( r , θ , z ) rJ n ( k mn r ) Z k ( z ) sin n θ d θ d r d z ,

Ω

{ ,

θ , θ , ··· , θ , θ , ···}

where M mnk is the product of normal squares of 1 cos sin cos n sin n ,

{ ( ) }

and Z k (

)

J n k mn r z .