1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.5 Example: Different Boundary Conditions 165
As in previous cases, we assemble the general solution of the homogeneous
problem expressed in Eqs. (5)–(7) by forming a general linear combination of
our product solutions,

w(x,t)=

∑∞

n= 1

bnsin(λnx)exp

(

−λ^2 nkt

)

. (16)

Thechoiceofthecoefficients,bn,mustbemadesoastosatisfytheinitialcon-
dition, Eq. (8). Using the form ofwgiven by Eq. (16), we find that the initial
condition is

w(x, 0 )=

∑∞

n= 1

bnsin

(( 2 n− 1 )πx

2 a

)

=g(x), 0 <x<a. (17)

A routine Fourier sine series for the interval 0<x<awould involve the func-
tions sin(nπx/a), rather than the functions we have. By one of several means
(Exercises 10–12), it may be shown that the series in Eq. (17) represents the
functiong(x),providedthatgis sectionally smooth and that we choose the
coefficients by the formula

bn=^2 a

∫a

0

g(x)sin

(

( 2 n− 1 )πx

2 a

)

dx. (18)

Now the original problem is completely solved. The solution is

u(x,t)=T 0 +

∑∞

n= 1

bnsin(λnx)exp

(

−λ^2 nkt

)

. (19)

It should be noted carefully that theT 0 term in Eq. (19) is the steady-state
solution in this case; it is not part of the separation-of-variables solution.

Example.
Find the solution of Eqs. (1)–(4) with the initial condition

u(x, 0 )=T 1 , 0 <x<a.

Theng(x)=T 1 −T 0 ,0<x<a, and the coefficients as determined by Eq. (18)
are

bn=(T 1 −T 0 )

4

π( 2 n− 1 ).

Therefore, the complete solution of the boundary value–initial value problem
with initial conditionu(x, 0 )=T 1 would be

u(x,t)=T 0 +(T 1 −T 0 )

4

π

∑∞

n= 1

1

2 n− 1 sin(λnx)exp

(

−λ^2 nkt

)

. (20)

See Fig. 5 for graphs and an animation on the CD.