1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.9 Generalities on the Heat Conduction Problem 185
Assuming thatc 1 andc 2 are constants, we must first find the steady-state
solution

v(x)=t→∞limu(x,t).

The functionv(x)satisfies the boundary value problem

d

dx

(

κ(x)dv

dx

)

= 0 , l<x<r, (5)

α 1 v(l)−α 2 v′(l)=c 1 , (6)

β 1 v(r)+β 2 v′(r)=c 2. (7)

Since we have assumed that at least one ofα 1 orβ 1 is positive, this problem
can be solved. In fact, it is possible to give a formula forv(x)in terms of the
function (see Exercise 1)
∫x

l

dξ
κ(ξ)=I(x). (8)
Before proceeding further, it is convenient to introduce some new functions.
Letκ ̄,ρ ̄,and ̄cindicate average values of the functionsκ(x),ρ(x),andc(x).
We shall define dimensionless functionss(x)andp(x)by

κ(x)= ̄κs(x), ρ(x)c(x)= ̄ρ ̄cp(x).

Also, we define the transient temperature to be

w(x,t)=u(x,t)−v(x).
By direct computation, using the fact thatv(x)is a solution of Eqs. (5)–(7),
we can show thatw(x,t)satisfies the initial value–boundary value problem

∂

∂x

(

s(x)∂w∂x

)

=^1 kp(x)∂w∂t, l<x<r, 0 <t, (9)

α 1 w(l,t)−α 2 ∂w

∂x

(l,t)= 0 , 0 <t, (10)

β 1 w(r,t)+β 2 ∂w∂x(r,t)= 0 , 0 <t, (11)

w(x, 0 )=f(x)−v(x)=g(x), l<x<r, (12)

which has homogeneous boundary conditions. The constantkis defined to be
κ/ ̄ ρ ̄ ̄c.
Now we use our method of separation of variables to findw.Ifwhas the
formw(x,t)=φ(x)T(t), the differential equation becomes

T(t)

(

s(x)φ′(x)

)′

=^1 kp(x)φ(x)T′(t),