2.2 Steady-State Temperatures 147
Now we collect and simplify these transformations of Eqs. (1)–(4) to get an
initial value–boundary value problem forw:
∂^2 w
∂x^2 =1
k∂w
∂t,^0 <x<a,^0 <t, (17)
w( 0 ,t)= 0 , 0 <t, (18)
w(a,t)= 0 , 0 <t, (19)w(x, 0 )=f(x)−[
T 0 +(T 1 −T 0 )x
a]
(20)
≡g(x), 0 <x<a. (21)In the last line, we have just renamed the combination off(x)andv(x)in
Eq. (20).
Inthenextsection,weshallseehowtheproblemforthetransienttemper-
ature can be solved. The mathematical purpose of setting up the steady-state
problem and then the transient problem is that the transient problem isho-
mogeneous.Youcantestthisbytryingw(x,t)≡0: This function satisfies the
partial differential equation (17) and the boundary conditions (18) and (19). It
is crucially important for the method we will develop to have a homogeneous
partial differential equation and boundary conditions.
EXERCISES
SeeextraexercisesontheCD.
- State and solve the steady-state problem corresponding to
∂^2 u
∂x^2−γ^2 (u−U)=^1
k∂u
∂t, 0 <x<a, 0 <t,
u( 0 ,t)=T 0 , u(a,t)=T 1 , 0 <t,
u(x, 0 )= 0 , 0 <x<a.
Also find a physical interpretation of this problem. (See Exercise 3, Sec-
tion 1.)- State the problem satisfied by the transient temperature distribution corre-
sponding to the problem in Exercise 1. - Obtain the steady-state solution of the problem
∂^2 u
∂x^2 +γ(^2) (u−T)=^1
k
∂u
∂t,^0 <x<a,^0 <t,
u( 0 ,t)=T, u(a,t)=T, 0 <t,
u(x, 0 )=T 1 xa, 0 <x<a.