1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.4 Example: Insulated Bar 159

the valuesπ/a, 2 π/a, 3 π/a,.... We label the eigenvalues with a subscript:

λ^2 n=

(

nπ

a

) 2

,φn(x)=cos(λnx), n= 1 , 2 ,....

Notice that any constant multiple of an eigenfunction is still an eigenfunction;
thus, we may takec 1 =1 for simplicity.
Returning to the caseλ=0, we see that Eqs. (6) and (7) become
φ′′= 0 , 0 <x<a,
φ′( 0 )= 0 ,φ′(a)= 0.

The solution of the differential equation isφ(x)=c 1 +c 2 x. Both boundary
conditions sayc 2 =0. Thereforeφ(x)is a (any) constant. Thus 0 is an eigen-
value of the problem Eqs. (6) and (7), and we designate

λ^20 = 0 ,φ 0 (x)= 1.
Let us summarize our findings by saying that the eigenvalue problem,
Eqs. (6) and (7), has the solution




λ^20 = 0 ,φ 0 (x)= 1 ,

λ^2 n=

(

nπ

a

) 2

,φn(x)=cos(λnx), n= 1 , 2 ,....

Now that the numbersλ^2 nare known, we can solve Eq. (5) forT(t),finding

T 0 (t)= 1 , Tn(t)=exp

(

−λ^2 nkt

)

.

The productsφn(x)Tn(t)give solutions of the partial differential equation (1)
that satisfy the boundary conditions, Eq. (2):

u 0 (x,t)= 1 , un(x,t)=cos(λnx)exp

(

−λ^2 nkt

)

. (8)

Because the partial differential equation and the boundary conditions are all
linear and homogeneous, the principle of superposition applies, and any linear
combination of solutions is also a solution. The solutionu(x,t)of the whole
system may therefore have the form

u(x,t)=a 0 +

∑∞

1

ancos(λnx)exp

(

−λ^2 nkt

)

. (9)

There is only one condition of the original set remaining to be satisfied, the
initial condition Eq. (3). Foru(x,t)in the form of Eq. (9), the initial condi-
tion is

u(x, 0 )=a 0 +

∑∞

1

ancos(λnx)=f(x), 0 <x<a.