Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 268

where

⎧
⎨

sin β mnk t , when β mnk = 0 ,

=

sin β mnk t

t , when β mnk = 0 .

⎩

Applying the initial condition u ( x , y , z , 0 )= 0 yields A mnk = 0. Also

)= m , n , k e α mnk t u t ( x , y , z , t ( α mnk B mnk sin β mnk t + B mnk β mnk cos β mnk t ) X m ( x ) Y n ( y ) Z k ( z ) ,

where

β mnk , when β mnk = 0 ,

β mnk =

1 , when β mnk = 0 .

B mnk can be determined by applying the initial condition u t ( x , y , z , 0 )= ψ ( x , y , z ) .

Finally, we have W ψ ( x , y , z , t )

⎧
⎨

)= m , n , k B mnk e α mnk t sin β mnk t · X m ( x ) Y n ( y ) Z k ( z ) , u = W ψ ( x , y , z , t

(6.84)

1
M mnk β mnk

⎩

B mnk = ψ ( x , y , z ) X m ( x ) Y n ( y ) Z k ( z ) d x d y d z ,

Ω

where M mnk = M m M n M k .

Similarly, consider the solution of PDS (6.81) satisfying the equation and the
boundary conditions

)= m , n , k e α mnk t u ( x , y , z , t ( C mnk cos β mnk t + D mnk sin β mnk t ) X m ( x ) Y n ( y ) Z k ( z ) . (6.85)

Applying the initial condition u ( x , y , z , 0 )= ϕ ( x , y , z ) yields

1
M mnk

C mnk = ϕ ( x , y , z ) X m ( x ) Y n ( y ) Z k ( z ) d x d y d z .

Ω

Also,

)= m , n , k e α mnk t α mnk ( C mnk cos β mnk t + D mnk sin β mnk t )

u t ( x , y , t

β mnk t X m (

+ − C mnk β mnk sin β mnk t + D mnk β mnk cos x ) Y n ( y ) Z k ( z ) .

Applying the initial condition u t ( x , y , z , 0 )= 0 leads to

α mnk C mnk + D mnk β mnk = 0 ,