2.2 Steady-State Temperatures 145
When the rule given here is applied to this problem, we are led to the following
equations:
d^2 v
dx^2 =^0 ,^0 <x<a,
v( 0 )=T 0 , −κv′(a)=h(
v(a)−T 1)
.
The solution of the differential equation isv(x)=A+Bx. The boundary con-
ditions require thatAandBsatisfy
v( 0 )=T 0 :A=T 0 ,
−κv′(a)=h(
v(a)−T 1)
:−κB=h(A+Ba−T 1 ).Solving simultaneously, we find
A=T 0 , B=h(T^1 −T^0 )
κ+ha.
Thus the steady-state solution of Eqs. (8)–(11) is
v(x)=T 0 +xh(T^1 −T^0 )
κ+ha. (12)
In both of these examples, the steady-state temperature distribution has
been uniquely determined by the differential equation and boundary condi-
tions. This is usually the case, but not always.
Example.
For the problem
∂^2 u
∂x^2 =1
k∂u
∂t,^0 <x<a,^0 <t, (13)
∂u
∂x(^0 ,t)=^0 ,^0 <t, (14)
∂u
∂x(a,t)=^0 ,^0 <t, (15)
u(x, 0 )=f(x), 0 <x<a, (16)which describes the temperature in an insulated rod that also has insulated
ends, the corresponding steady-state problem forv(x)=limt→∞u(x,t)is
d^2 v
dx^2 =^0 ,^0 <x<a,
dv
dx(^0 )=^0 ,dv
dx(a)=^0.