1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

186 Chapter 2 The Heat Equation

and, on dividing through bypφT,wefindtheseparatedequation

(sφ′)′

pφ

=T

′

kT

, l<x<r, 0 <t.

As before, the equality between a function ofxand a function oftcan hold
only if their common value is constant. Furthermore, we expect the constant
to be negative, so we put

(sφ′)′

pφ =

T′

kT=−λ

2

and separate two ordinary equations

T′+λ^2 kT= 0 , 0 <t,

(sφ′)′+λ^2 pφ= 0 , l<x<r.

The boundary conditions, being linear and homogeneous, can also be changed
into conditions ofφ.Forinstance,Eq.(10)becomes
[
α 1 φ(l)−α 2 φ′(l)

]

T(t)= 0 , 0 <t,

and, becauseT(t)≡0makesw(x,t)≡0, we take the other factor to be zero.
We have, then, the eigenvalue problem

(sφ′)′+λ^2 pφ= 0 , l<x<r, (13)
α 1 φ(l)−α 2 φ′(l)= 0 , (14)
β 1 φ(r)+β 2 φ′(r)= 0. (15)
Sincesandpare related to the physical properties of the rod, they should be
positive. We suppose also thats,s′,andpare continuous. Then Eqs. (13)–(15)
comprise a regular Sturm–Liouville problem, and we know the following.

1.There is an infinite number of eigenvalues

0 <λ^21 <λ^22 <···.

2.To each eigenvalue corresponds just one eigenfunction (give or take a con-

stant multiplier).

3.The eigenfunctions are orthogonal with weightp(x):

∫r

l

φn(x)φm(x)p(x)dx= 0 , n=m.

The functionTn(t)that accompaniesφn(x)is given by

Tn(t)=exp

(

−λ^2 nkt

)

.