Conference, Proceedings

The solution of the BC (6) problem is [4]
θ
+
( x,
τ ) = A0
ψ sin(
2πfτ
− ϕ − ε ) +
n
2
( −1)
( 2n
+ 1)
4l
ω cosε
+ a(
2n
+ 1)
4π
k ∑ 4 2 2 4
4
n 0 16l
ω + a π ( 2n
+ 1)
∞
=
where (compare Fig. 1)
254
2 2
[ n sin ε ]
⎡ a +
exp⎢−
⎣ 4l
2 ( 2n
1)
π ⎤ ( 2n
+ 1)
2
τ ⎥ cos
⎦
π x
2l
cosh 2kx
+ cos 2kx
ψ ( x)
=
(9)
cosh 2kl
+ cos 2kl
( )
( ) ⎥ 1+
i ⎤
1+
i ⎦
⎡cosh
kx
ϕ ( x)
= arg⎢
(10)
⎣ cosh kl
π f π
k = =
(11)
a a τ
θ(x,τ)
Ω
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
θ max (l,τ)
ϕ
θ max (0,τ)
ψ = θ max (0,τ) / θ max (l,τ)
BC temperature oscillation
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
τ / τ Ω
response
Figure 1: Illustration of components and parameters of a solution given by Eqs. (8)‐(12)
while for the BC given by Eq. (7) we get [4]
θ
2 2
b
( ) ( x − l )
x,
τ = bτ
+
2
bl
+ 16 3
aπ
2a
∑
n 0
∞
=
+
n ( −1)
( 2n
+ 1)
3
⎡ +
exp⎢−
a
⎣ 4l
2 2 ( 2n
1)
π ⎤ ( 2n
+ 1)
2
τ ⎥ cos
⎦
πx
2l
The series terms with exponential components in (8) and (12) represents transients that die
away as τ → ∞ . The representative solution of the steady oscillations over a linearly changed
temperature expressed by superposition of the two appropriate components is:
2 2
b
( ) ( ) ( x − l )
x,
τ ≅ A ψ sin 2πfτ
−ϕ
− ε + bτ
+
θ 0
(13)
2a
(8)
(12)