1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.6 Example: Convection 173

that makeswn(x,t)=φn(x)Tn(t)a solution of the partial differential equa-
tion (6) and the boundary conditions Eq. (7). Since Eqs. (6) and (7) are lin-
ear and homogeneous, any linear combination of solutions is also a solution.
Therefore, the transient solution will have the form

w(x,t)=

∑∞

n= 1

bnsin(λnx)exp

(

−λ^2 nkt

)

,

and the remaining condition to be satisfied, the initial condition Eq. (8), is

w(x, 0 )=

∑∞

n= 1

bnsin(λnx)=g(x), 0 <x<a. (12)

Thus the constantsbnare to be chosen so as to make the infinite series equal
g(x).
Although Eq. (12) looks like a Fourier series problem, it is not, becauseλ 2 ,
λ 3 , and so forth are not all integer multiples ofλ 1 .Ifweattempttousetheidea
of orthogonality, we can still find a way to select thebn, for it may be shown by
direct computation that
∫a

0

sin(λnx)sin(λmx)dx= 0 , ifn=m. (13)

Then if we multiply both sides of the proposed Eq. (12) by sin(λmx)(where
mis fixed) and integrate from 0 toa,wehave
∫a

0

g(x)sin(λmx)dx=

∑∞

n= 1

bn

∫a

0

sin(λnx)sin(λmx)dx,

where we have integrated term by term. According to Eq. (13), all the terms
of the series disappear except the one in whichn=m, yielding an equation
forbm:

bm=

∫a

(^0) ∫ga(x)sin(λmx)dx
0 sin
(^2) (λmx)dx. (14)
By this formula, thebmmay be calculated and inserted into the formula for
w(x,t). Then we may put together the solutionu(x,t)of the original problem
Eqs. (1)–(4):
u(x,t)=v(x)+w(x,t)
=T 0 +
xh(T 1 −T 0 )
(κ+ha) +

∑∞

n= 1

bnsin(λnx)exp

(

−λ^2 nkt

)

.

In Fig. 7 are graphs ofu(x,t)for two different values of the parameterκ/ha;
both have initial conditionsu(x, 0 )=0.SeeanimationsontheCD.