1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

0.4 Singular Boundary Value Problems 41

(see Section 3). As the problem has been posed for a semi-infinite interval
(because the fin is very long and, perhaps, to mask our ignorance of what is
happening at the other physical end), we must also impose the condition

u(x), u′(x) bounded asx→∞. (7)

Now, the general solution of the differential equation (5) is

u(x)=T+c 1 cosh(μx)+c 2 sinh(μx),

whereμ=

√

hC/κA. The boundary condition atx=0requiresthat

u( 0 )=T 0 : T+c 1 =T 0.

The boundedness condition, Eq. (7), requires that

c 2 =−c 1.

The reason for this is that of all the linear combinations of cosh and sinh, the
only one that is bounded asx→∞is

cosh(μx)−sinh(μx)=e−μx,

and its constant multiples. The final solution is easily found to be

u(x)=T+(T 0 −T)

(

cosh(μx)−sinh(μx)

)

.

Satisfying the boundedness condition in the example would have been sim-
pler if we had expressed the general solution of the differential equation (5)
as

u(x)=T+c′ 1 eμx+c 2 ′e−μx.

We would have seen immediately that choosingc′ 1 =0 is the only way to satisfy
the boundedness condition. We summarize the observation as arule of thumb:
the solution of
d^2 u
dx^2

−μ^2 u= 0

on an intervalIis best expressed as

u(x)=

{c

1 cosh(μx)+c 2 sinh(μx), if Iis finite,

c 1 eμx+c 2 e−μx, if Iis infinite.

EXERCISES

1. Put each of the following equations in the form

u′′+ku′+pu=f