Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 275

Let v ( r , θ )= R ( r ) Θ ( θ ) . We can thus obtain the solution of Eq. (6.96) (see Sec-

tion 4.3.2).

v mn ( r , θ )=( a mn cos n θ + b mn sin n θ ) J n ( k mn r ) ,

k mn , k mn = μ ( n )

where the a mn and the b mn are not all zero, λ m = / a , m = 1 , 2 , ··· ,

m

( n m ( m

= , , , ··· = , , ···

n 0 1 2 .The μ 1 2 ) depend on the boundary conditions and are

the positive zero points of

⎧
⎨

J n ( x ) , Boundary condition of the first kind

u | r = a = 0 ,

J n ( x ) , Boundary condition of the second kind

f n ( x )= (6.97)

⎩

, μ ( 0 )
1

u r | r = a = 0 = 0 ,

1

a xJ n (

x )+ hJ n ( x ) , Boundary condition of the third kind

( u r + hu ) | r = a = 0 .

k mn into Eq. (6.95) yields

Substituting λ =

1
τ

k mn B 2 T mn ( T mn ( k mn A 2 T m t )+

0 +

t )+ ( t )= 0 .

Its characteristic roots are

1

1

k mn B 2 k mn B 2 2

−

τ 0 +

±

τ 0 +

− 4 k mn A 2

r 1 , 2 = = α mn + β mn i . (6.98)

2

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

PDS (6.94) reads

+ ∞
∑
m =

e α mn t

u ( r , θ , t )= [( A mn cos β mn t + B mn sin β mn t ) cos n θ

, n =

1 0

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

Note that { 1 , cos θ , sin θ , ··· , cos n θ , sin n θ , ···} is orthogonal in [ − π , π ] and { J n ( k mn r ) }

is orthogonal in [ 0 , a ] with respect to the weight function r ,

a

J n ( k mn r ) J n ( k ln r ) r d r = 0 , m = l .

0

( , θ , )= = =

Applying the initial condition u r 0 0 yields A mn C mn 0. B mn and D mn can

be determined to satisfy the initial condition u t ( r , θ , 0 )= ψ ( r , θ ) . Finally, we obtain