Page 453 - Modelling in Transport Phenomena A Conceptual Approach

10.1. MOMENTUM TRANSPORT 433

While the left-side of h. (10.1-30) is a function of 7 only, the right-side is depen-
dent only on E. This is possible only if both sides of Eq. (10.1-30) are equal to a
constant, say - A', i.e.,
1 d2G-
1 dF
--- _---_ (10.1-31)
F d7 G 4'
The choice of a negative constant is due to the fact that the solution will decay to
zero as time increases. The choice of a positive constant would give a solution that
becomes infinite as time increases.
Equation (10.1-31) results in two ordinary differential equations. The equation
for F is given by
dF
-+A'F=O (10.1-32)
dr
The solution of l3q. (10.1-32) is

e-
=
~(7) (10.1-33)
On the other hand, the equation for G is
d2G
-+X2G=0 (10.1-34)
4'
and it is subject to the boundary conditions

at [=O G=O (10.1-35)

at [=l G=O (10.1-36)
Note that Eq. (10.1-34) is a Sturm-Liouville equation with a weight function of
unity. The solution of Eq. (10.1-34) is

G(<) = Asin(XE) + Bcos(X5) (10.1-37)
where A and B are constants. Application of E~J. (10.1-35) gives B = 0. The use
of the boundary condition defined by Eq. (10.1-36) results in

AsinX = 0 (10.1-38)
For a nontrivial solution, the eigenvalues are given by

sinX= 0 =+ A, = nr n = 1,2,3, ... (10.1-39)

Therefore, the transient solution is

00
et = cn e-nZr2r sin (nr[) (10.1-40)
n=I