Page 155 - Physical Principles of Sedimentary Basin Analysis

6.9 Transient cooling of a sphere ∗ 137
2
∂u ∂ u
− = 0 (6.138)
∂ ˆ t ∂ ˆ r 2
which is the temperature equation for 1D conductive heat transport. The boundary condi-
r
tion for equation (6.138)at ˆ = 1is u(ˆ =0, ˆ t) = 0, but the other boundary condition
r
ˆ
r
r
∂T /∂ ˆ = 0at ˆ = 0 is difficult to express in a simple way using u. The trick is now to
avoid the boundary condition at ˆ = 0 by instead solving the equation for u through the
r
r
entire sphere from ˆ =−1to ˆ = 1. The boundary conditions are then u = 0 at both ends,
r
ˆ
r
r
r
r
r
(ˆ =−1 and ˆ = 1), and the initial condition T (ˆ, ˆ t =0) = 1 becomes u(ˆ, ˆ t =0) =ˆ.
Equation (6.138) is solved by the method called separation of variables, where u is written
r
r
as the product of a function U of ˆ t and a function V of ˆ as u(ˆ, ˆ t) = U(ˆ t) · V (ˆ). When
r
U · V is inserted into equation (6.138) we get

U V − UV = 0 (6.139)
or
U V 2
= =−k . (6.140)
U V
Since U /U is a function only of ˆ t and V /V is a function of only ˆr these two fractions

2
must be equal to a constant. This constant is written −k , where it is anticipated that it has
2

to be a negative number. The equation for U becomes U =−k U, which has the solution
(see Exercise 6.5)
2
U(ˆ t) = U 0 exp(−k ˆ t). (6.141)
2

The equation for V is V + k V = 0, which has as solution any linear combination of
r
sin(kˆ) and cos(kˆ). The equation for V and the boundary conditions for u are fulfilled
r
for all V n = sin(nπ ˆr). There is not just one number k,but k n = nπ for each integer n,
2
and there is therefore a function U for each k n as well, U n = U 0 exp(−k ˆ t). Temperature
n
equation (6.138)for u is linear, and any linear combination of U n (ˆ t)V n (ˆr) is therefore a
solution too, and u can therefore be written as a Fourier series
∞
2
r
r
u(ˆ, ˆ t) = a n sin(k n ˆ) exp(−k ˆ t). (6.142)
n
n=1
The initial condition u(ˆ, t=0) =ˆ is needed for the remaining task of finding the Fourier
r
r
coefficients a n . It is straightforward to show that the functions V n (ˆr) = sin(nπ ˆr) form an
1
r
orthogonal basis with respect to the inner product (V n , V m ) = V n V m d ˆ. We therefore
−1
1
have that (V n , V m ) = δ nm . The Fourier coefficients are obtained from the initial condition
2
r
r
u(ˆ, ˆ t=0) =ˆ, which is
∞

a n sin(k n ˆ) =ˆ. (6.143)
r
r
n=1
r
Multiplication of both sides by sin(nπ ˆ) followed by integration over ˆ from −1to1gives
r
1 2 (−1) n+1
r
r
a n = ˆ r sin(nπ ˆ) d ˆ = . (6.144)
−1 nπ