1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

164 Chapter 2 The Heat Equation

The boundary conditions take the form

φ( 0 )T(t)= 0 , 0 <t, (9)

φ′(a)T(t)= 0 , 0 <t. (10)

As before, we conclude thatφ( 0 )andφ′(a)should both be zero:

φ( 0 )= 0 ,φ′(a)= 0. (11)

By trial and error we find that a positive or zero separation constant in Eq. (8)
forcesφ(x)≡0. Thus we take the constant to be−λ^2. The separated equations
are

φ′′+λ^2 φ= 0 , 0 <x<a, (12)

T′+λ^2 kT= 0 , 0 <t. (13)

Now, the general solution of the differential equation (12) is

φ(x)=c 1 cos(λx)+c 2 sin(λx).

The boundary condition,φ( 0 )=0, requires thatc 1 =0, leaving

φ(x)=c 2 sin(λx).

The boundary condition atx=anow takes the form

φ′(a)=c 2 λcos(λa)= 0.

The three choices are:c 2 =0, which gives the trivial solution;λ=0, which
should be investigated separately (Exercise 2), and cos(λa)=0. The third al-
ternative — the only acceptable one — requires thatλabe an odd multiple of
π/2, which we may express as

λn=(^2 n−^1 )π

2 a

, n= 1 , 2 ,.... (14)

Thus, we have found that the eigenvalue problem consisting of Eqs. (11)
and (12) has the solution

λn=(^2 n−^1 )π

2 a

,φn(x)=sin(λnx), n= 1 , 2 , 3 ,.... (15)

With the eigenfunctions and eigenvalues now in hand, we return to the dif-
ferential equation (13), whose solution is

Tn(t)=exp

(

−λ^2 nkt

)

.