2.3 Example: Fixed End Temperatures 153
Notice that the choice of the coefficientsbndoes not enter into the check-
ing of the partial differential equation and the boundary conditions. Thus,
Eq. (15) plays the role of a general solution of Eqs. (8)–(10).
Of the four parts of the original problem, only the initial condition has not
yet been satisfied. Att=0, the exponentials in Eq. (15) are all unity. Thus the
initial condition takes the form
w(x, 0 )=∑∞
n= 1bnsin(nπx
a)
=g(x), 0 <x<a. (16)We immediately recognize a problem in Fourier series, which is solved by
choosing the constantsbnaccording to the formula
bn=^2 a∫a0g(x)sin(
nπx
a)
dx. (17)If the functiongis continuous and sectionally smooth, we know that the
Fourier series actually converges tog(x)in the interval 0<x<a,sothesolu-
tion that we have found forw(x,t)actually satisfies all requirements set onw.
Even ifgdoes not satisfy these conditions, it can be shown that the solution
we have arrived at is the best we can do.
Once the transient temperature has been determined, we find the original
unknownu(x,t)as the sum of the transient and the steady-state solutions,
u(x,t)=v(x)+w(x,t).Example.
Suppose the original problem to be
∂^2 u
∂x^2 =1
k∂u
∂t,^0 <x<a,^0 <t,
u( 0 ,t)=T 0 , 0 <t,
u(a,t)=T 1 , 0 <t,
u(x, 0 )= 0 , 0 <x<a.The steady-state solution is
v(x)=T 0 +(T 1 −T 0 )x
a.
The transient temperature,w(x,t)=u(x,t)−v(x),satisfies
∂^2 w
∂x^2 =1
k∂w
∂t,^0 <x<a,^0 <t,
w( 0 ,t)= 0 , 0 <t,