Fourier Series and Partial Differential Equations Lecture Notes

Chapter 3. The heat equation 31
3.6 Non-zero steady state
It may be that the temperatures of the ends x = 0 and x = L are prescribed and constant
but not equal to zero.
Example 3.3 Solve the IBVP
subject to the initial condition
and the boundary conditions
∂T
∂t = κ∂2 T
∂x2, 0 < x < L, t > 0, (3.58)
T(x,0) = 0, 0 ≤ x ≤ L, (3.59)
T(0,t) = T0 and T(L,t) = T1 for t > 0. (3.60)
We cannot use separation of variables and Fourier series right at the outset. However, we
conjecture that, as t → ∞, T(x,t) → U(x), where
i.e.
κ d2 U
dx 2 = 0, U(0) = T0 and U(L) = T1, (3.61)
U(x) = T0
1− x
L
+T1
x
. (3.62)
L
If we now put S(x,t) := T(x,t)−U(x), we find that S is a solution of the IBVP
with
∂S
∂t = κ∂2 S
∂x2, 0 < x < L, t > 0, (3.63)
S(0,t) = 0 and S(L,t) = 0 for t > 0, (3.64)
and
S(x,0) = −T0 1− x
x
−T1 . (3.65)
L L
In view of the form of the boundary conditions, this IBVP can be solved by our previous
methods. The solution is
and so
T(x,t) = T0
S(x,t) = 2
π
1− x
L
+T1
∞
n=1
1
n [−T0 +(−1) n T1]sin
x
+
L
2
π
∞
n=1
1
n [−T0 +(−1) n T1]sin
nπx
e
L
−n2π2κt/L2 , (3.66)
nπx
e
L
−n2π2κt/L2 . (3.67)