1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

160 Chapter 2 The Heat Equation

Figure 4 The solution of the example,u(x,t), as a function ofxfor several times.
The initial temperature distribution isf(x)=T 0 +(T 1 −T 0 )x/a. For this illustra-
tion,T 0 = 20 ,T 1 =100, and the times are chosen so that the dimensionless time
kt/a^2 takes the values 0.001, 0.01, 0.1, and 1. The last case is indistinguishable from
the steady state. See the CD also.

Becauseλn=nπ/a, we recognize a problem in Fourier series and can imme-
diately cite formulas for the coefficients:

a 0 =

1

a

∫a

0

f(x)dx, an=

2

a

∫a

0

f(x)cos

(nπx

a

)

dx. (10)

When these coefficients are computed and substituted in the formulas for
u(x,t), that function becomes the solution to the initial value–boundary value
problems, Eqs. (1)–(3). Notice that whent→∞, all other terms in the sum-
mation foru(x,t)disappear, leaving

tlim→∞u(x,t)=a^0 =

1

a

∫a

0

f(x)dx.

Example.
Find the complete solution of Eqs. (1)–(3) for the initial temperature dis-
tributionf(x)=T 0 +(T 1 −T 0 )x/a.Itrequiresnointegrationtofindthat
a 0 =(T 1 +T 0 )/2. The remaining coefficients are

an=

2

a

∫a

0

(

T 0 +

(T 1 −T 0 )x

a

)

cos

(nπx

a

)

dx

= 2 (T 1 −T 0 )cos(nπ)−^1

n^2 π^2

.

Thus the solution is given by Eq. (9) with these coefficients fora 0 andan.
Agraphofu(x,t)as a function ofxis shown in Fig. 4 and as an animation on
the CD.