Geophysical Processes in Sedimentary Basin Formation
Notes on the temperature distribution in a cooling plate.
The 1-dimensional heat diffusion equation is
T =   2 T
z2
t
(1)
where T is temperature, t is time, z is the coordinate in the direction in which
heat flow occurs, and  is the thermal diffusivity.
Equation (1) can be solved by imagining that the base of the region of
interest lies so far away from the surface that it effectively has no control on
the thermal structure near the surface. This is the basis of the cooling halfspace model, the solution for the temperature structure of which is given in
Turcotte & Schubert (p157-161, 2001).
We saw in Lecture 3 that the cooling oceanic lithosphere appears to
stop cooling once it passes an age of about 80 Ma. One possible explanation
for this is that some process - most plausibly convection in the mantle - is
supplying heat to the base of the oceanic lithosphere and thereby arrests the
cooling process.
The cooling plate model takes into account that the cooling of oceanic
lithosphere may be arrested by assuming that the temperature at the plate
base is held at a constant value. To calculate the temperature in such a plate
we need write it as the sum of two temperatures: the first is the steady state
temperature, Ts ,and the second the unsteady temperature, Tu ,which is
the difference between the actual temperature at any time and the steady
state temperature.
The steady state temperature structure is given (e.g. Fig.1) by
Ts = T0 + (Tm – T0) az
(2)
where T0 = temperature at t > 0, z = 0, a = thickness of the plate, Tm =
temperature at t = 0, 0 < z  a and t > 0, z = a.
Fig. 1
z=a
Tu must obey Equation (1), which we can solve using the method of
separation of variables. Let
Tu(z,t) = (t) Z(z)
(3)
Substitution in Equation (1) gives
d (t)
d 2 Z(z)
Z(z)
= (t) 
dt
d z2
which we can re-arrange as
(4)
2
1 d (t) =  1 d Z(z) = – c 2
(t) dt
Z(z) d z 2
to give the solutions:
(t) = e –c t
2
Z(z) = A sin cz + B cos cz
The boundary conditions on Tu (that it is zero at both z = 0 and z = a) require
that
B = 0; c = n a 
Thus
– n2  2  t
(5)
(t) = e
a2
Z(z) = sin n 2 z
a
Substituting Equations (5) and (6) in (3) gives

Tu(z,t) =

n=1
z
An sin n 
a e
– n2  2  t
a2
The values for the coefficients An can be found from the initial condition
(6)

Tu(z,0) =
 An sin
n=1
nz
a
which in the standard Fourier procedure gives :
a A
2 n
For example in the mid-ocean ridge problem
Tu(z,0) = (Tm – T0) (1 – z a)
So
An =
2 (Tm – T0)
n
Hence,

2
Ts = T0 + (Tm – T0) az + 
 1 sin n a z e
n=1 n
– n2 2  t
a2
Further details on the cooling plate model can be found in Turcotte &
Shubert (p161-162, 2001), which also compares (e.g. p174-176) the
predictions of the cooling half-space and plate models to observed
bathymetry data in the Atlantic, Pacific, and Indian oceans.
Turcotte, D.L. and G. Shubert, Geodynamics. 2nd edition. Cambridge
University Press. 2001