Laplace transform isotherm .pdf - University of Hertfordshire ...

simplify the algebra we write this as −µ 2 α. The solutions are then
where T0 is a constant and
T = T0e −µ2 αt
X = Asin µx + B cos µx
with A and B constants. A solution of equation(2.3) is then
u = T0(Asin µx + B cos µx)exp � −µ 2 αt �
(2.5)
and since the heat equation is a linear equation, the most general solution
is obtained by summing solutions of equation (2.5) type to get
u =
∞�
T0,m(Am sin µmx + Bm cos µmx)exp � −µ 2 mαt� m=1
The constants Am,Bm,T0,m and µm are determined from the initial and
boundary conditions and again, examples are to be found in Crank (1979).
Another analytic method is the use of similarity solutions.
We define the dimensionless variable
η = x
√ αt
and then look for solutions of the form
u(x,t) = t p g (η) (2.6)
where the number p and the function g(η) are to be found. Substituting
equation (2.6) into equation (2.3) we find
which implies
�
p−1
t pg − η
2 g′ − g ′′�
= 0
g ′′ + η
2 g′ = pg (2.7)
subject to appropriate boundary conditions using equation (2.6). This is
difficult to solve in closed form for arbitrary values of p.
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