1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.4 Example: Insulated Bar 157

e.Use the formula indto findtexplicitly fora= 5 × 10 −^6 m,D=

10 −^11 cm^2 /s. Be careful to check dimensions.

2.4 Example: Insulated Bar

We shall consider again the uniform bar that was discussed in Section 1. Let
us suppose now that the ends of the bar atx=0andx=aare insulated in-
stead of being held at constant temperatures. The boundary value–initial value
problem that describes the temperature in this rod is:

∂^2 u

∂x^2 =

1

k

∂u

∂t,^0 <x<a,^0 <t, (1)

∂u

∂x(^0 ,t)=^0 ,

∂u

∂x(a,t)=^0 ,^0 <t, (2)

u(x, 0 )=f(x), 0 <x<a, (3)

wheref(x)is supposed to be a given function.
We saw in Section 2 that the solution of the steady-state problem is not
unique. However, the mathematical purpose behind finding the steady-state
solution is to pave the way for a homogeneous problem (partial differential
equation and boundary conditions) for the transient. In this example the par-
tial differential equation and boundary conditions are already homogeneous.
Thus, we do not need the steady-state solution or the transient problem. We
maylookforu(x,t)directly.
Assume thatuhas the product formu(x,t)=φ(x)T(t), with neither factor
identically 0. The heat equation becomes

φ′′(x)T(t)=^1

k

φ(x)T′(t),

and the variables are separated by dividing through byφT,leaving

φ′′(x)

φ(x) =

T′(x)

kT(t),^0 <x<a,^0 <t.

In order that a function ofxequal a function oft,theirmutualvaluemust
be a constant. If that constant were positive,Twould be an increasing expo-
nential function of time, which would be unacceptable. It is also easy to show
that if the constant were positive,φcould not satisfy the boundary conditions
without being identically zero.
Assuming then a negative constant, we can write
φ′′(x)
φ(x)

=−λ^2 =T

′(t)

kT(t)