154 Chapter 2 The Heat Equation
w(a,t)= 0 , 0 <t,
w(x, 0 )=−T 0 −(T 1 −T 0 )x
a≡g(x), 0 <x<a.According to the preceding calculations,whas the form
w(x,t)=∑∞
n= 1bnsin(λnx)exp(
−λ^2 nkt)
(18)
and the initial condition is
w(x, 0 )=∑∞
n= 1bnsin(nπx
a)
=g(x), 0 <x<a.The coefficientsbnare given by
bn=^2 a∫a0[
−T 0 −(T 1 −T 0 )xa]
sin(
nπx
a)
dx=^2 T^0
acos(nπx/a)
(nπ/a)∣∣
∣∣
a0
−^2
a^2(T 1 −T 0 )sin(nπx/a)−(nπx/a)cos(nπx/a)
(nπ/a)^2∣∣
∣∣
a
0
=−^2 nTπ^0(
1 −(− 1 )n)
+^2 (T^1 n−πT^0 )(− 1 )nbn=−^2
nπ(
T 0 −T 1 (− 1 )n)
.
Now the complete solution (see Fig. 3) is
u(x,t)=w(x,t)+T 0 +(T 1 −T 0 )xa,where
w(x,t)=−2
π∑∞
n= 1T 0 −T 1 (− 1 )n
n sin(λnx)exp(
−λ^2 nkt)
. (19)
ThesolutionofthisproblemisshownasananimationontheCD.
We can discover certain features ofu(x,t)by examining the solution. First,
u(x, 0 )really is zero( 0 <x<a)because the Fourier series converges to−v(x)
att=0. Second, whentis positive but very small, the series forw(x,t)will
almost equal−T 0 −(T 1 −T 0 )x/a.Butatx=0andx=a, the series adds
up to zero (andw(x,t)is a continuous function ofx); thusu(x,t)satisfies
the boundary conditions. Third, whentis large, exp(−λ^21 kt)is small, and the