Fourier Series and Partial Differential Equations Lecture Notes

Chapter 3. The heat equation 27
where l, τ and T0 are a typical lengthscale, timescale and temperature, respectively, for
the problem under consideration. Then
and substituting into the heat equation we have
Rearranging gives
∂ dt ∂ 1 ∂
= = ,
∂t dˆt ∂ˆt τ ∂ˆt
(3.13)
∂ dx ∂ 1 ∂
= = ,
∂x dˆx ∂ˆx l ∂ˆx
∂
(3.14)
2
dx ∂ 1 ∂
= =
∂x2 dˆx ∂ˆx l ∂ˆx
1
l2 ∂2 ∂ˆx 2, (3.15)
T0
τ
∂ ˆ T
∂ˆt
∂ ˆ T
∂ˆt
= κT0
l 2
= κτ
l 2
∂ 2 ˆ T
∂ˆx 2.
∂ 2ˆ T
∂ˆx 2.
Considering the problem on a timescale where τ = l 2 /κ gives
Notice that now
since
∂ ˆ T
∂ˆt = ∂2ˆ T
∂ˆx 2.
(3.16)
(3.17)
(3.18)
[ˆx] = 1, [ˆt] = 1, [ ˆ T] = 1, (3.19)
l2 [l] = m, [τ] = = s, [T0] = K. (3.20)
κ
This means that we can compare heat problems on different scales: for example, two
systems with different l and κ will exhibit comparable behaviour on the same time scales
if l 2 /κ is the same in each problem.
3.3 Heat conduction in a finite rod
Let the rod occupy the interval [0,L]. If we look for solutions of the heat equation
∂T
∂t = κ∂2 T
∂x 2,
which are separable, T(x,t) = F(x)G(t), we find that
(3.21)
κ
F(x) F′′ 1
(x) =
G(t)
independent of t
G′ (t) ,
(3.22)
independent of x
and hence both sides are constant (independent of both x and t). If the constant is −κλ 2 ,
F(x) satisfies the ODE
F ′′ (x) = −λ 2 F(x), (3.23)
the solution of which is
F(x) = Asin(λx)+Bcos(λx). (3.24)